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ELECTRIC CIRCUITS EIGHTH EDITION

W.Nilsson James Prcfessor Emeritus Iowa StateUnirelsitJ

Susan A. Riedet Marquette Unive$ity

Upper SaddleRiver,New Je$ey 07458

Dataot File Libtott of CongresCataloging-in-Publicatiafl

Vice President and EditoriatDilectot,Ecs: Marcia J Hotton Editorial Assistanr Carck Sntder Acquisitions Editor: Mt.rdel McDonald Executive Managing Editor. Vince O'Blien ManagingBditor: David A. Georye hoduction Editor Rore reldli/t Director of Creative ServicestPaul Belfanti Cte3tile Dilector: han Lopez Aft Djnector. Ionethan B oyldn Managing Editor, AV Management and hoduction: Patdctd au /ru AfiEditor Xidohong Zhu Photo Researcher:,tcon -Drsarro Manu{acturing Manager:/4leris-Heydt Long ManuJacturing Buyer Zbd McDo'", Marketing Managei ?m Grrigan

O 2008,2m5,2001,2000.1996PearsonEducation,Inc. PeaftonPrenticeHall PeaNonEducation.Inc. Upper Saddle River, NJ 07458 All rights resemed.No parl of this book rnay be reproduced, in any form or by any means,without permission in witing from the publisher. Peanon Prentice Hall6 is a trademark of PearsonEducation,Inc The aurhor and publisher ofrhis book have used their best efforts in preparing this book.These efforts include the development, research,and testing of the theoiies and prograns to determine their effectiveness The author and publisher mate no warranty of any kind, exFessed or inplied. with regard to these programs or the documentation contained in this book. The aulhor and pubtisher shal not be liable in my event for incidental or consequential danages in connection with, or adsiDg out of, the fumishhg, performance,or use of these prcgams PSpi@ is a regislered trademark of Caden@ Design Systems. hinted in the United States of America 1 0 9 8 7 6 5 4 3 2 r

r s B N0 - l , l - l , t s 1 ? 5 - r Pedson Education Ltd., ro"dd Pedson Education Australia Pty. Ltd., S)dne] PearsonEducation Singapoie, fte. Ltd. PearsonEducation North Asia Ltd., Hong KonC PearsonEducation Canada lnc., ?r/o,to PearsonEducaci6n de Mexico, S.A. de C.V PearsonEducation Japan, k , PearsonEducation Malaysia, Pte. Ltd. Peanon Education, lnc., Upper Sadt e Rire,. NewJersey

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PhotoCredits coyer Image Courtesy of Coibis/RF, Royalty Free. Chapter 2 Practical Pe$pecaiy€r Courtesy of Corbis/NY, Ron Chapple. Figue 2-9: Couitesy of Corbis/NY, Thom Lang. Chapt€r 3 Practical Per6pective: Coufiesy of Getty Images/Creative Express. Chapler 4 Practical Perspective: Couitesy of Getty Images/Ciealive Express. Chapt€r 5 Practical PeNpectiver Courtesy o{ Getty lmag€s/PhotodiscRed, Akira Kaede. Chapter 6 Practical P€rspective: Courtesy of Getty Images,iPhotonica,Ron Rovtar. Chapter 7 Practicel Perspective: Courtesy ol Getty Images/PhotodiscGreen. Chapter 8 Pnctical P€Bpective: Courtesy of Getty Images/Image Source Royalty Free,Image Source Pink. Chspter 9 Prsctical P€rspectivs Courtesy of Getty Images/PhotodiscGreen, Steve Cole. Chapter 10 Practical Penp€crive: Courtesy of Alamy Images Royalty Free. Chapter 11 Practical Persp€ctive: Cou esy of Coibis/NY, Rolf Vennenbernd. Chapter 13 Practicsl Persp€ctive: Courtesy of Getty Images/PhotodiscBlue. Chapter 14 Praclical Persp€ctive: Courtesy of Corbis/RE Tom Grill. Chapter 15 Pncticrl P€rspective: Courtesy of Corbis^lY Dana Hoff.

Preface The eighth edition of Electlrc Cilclrr is a carefully planned revision to the most widely used introductory circuits text of the past 25 years.As this book has evolvedover the yearsto meet the changinglearningstylesof students,importantly,the underlying teachirg approachesand philoso phiesremainunchanged.Thegoalsare: . To build an understandingof conceptsand ideasexplicitlyin termsof previousleardng. . To emphasizethe relationshipbetween conceptualunderstanding and problem-solvingapproaches. . To provide studentswith a strong foundation of engineeringpractices

WHYTHISEDITION? When planr rg for the eighth editjon ievision of Electic Citcuits, carcfnl thought was given to how \ve should best update this classictext to improve upon the success o{ precedingeditionsandmakethe eightl edition ascom, pelliry as the first.Througha thoroughreviewprocessthat includedboth inshuctors and students who currendy use tlectli. C;rc&itsand those who use other texts, our revision plan was formed. wllat emerged from this exercisewas a clear pictue of what matters most to instructors and stu derts. With this feedback in mind, we made the following changes: . Problem solving is fundamental to the study of circuit analysis.The authorsput their pdmary eftbrt into updatingand addingnew endof-chapter problems. The result is a ftesh text with approximately 80o/"new or revised problems comparedto the previous edition. Having a wealth ofnew problemsto assignand work is a key to suc cessin any circuits course. . The eighth edition iepresentsa major redesignto the text. Careful attentionwaspaid toward how to presentthe material tex1,figures, and aitwork-in a clean,clear mannerthat would facilitatelearning and encouragereading.The seventhedition was the fi$t introductory circuits text to recognize the changing needs of today's students with a modern,four-colordesign.The eightheditionrefinesthis color treatmentwith a morc pedagogjcallycoherentpresentation. ' Navigation was imprcved by the addition of page numbers to the chapterobjectives,less rcliance on icons whete nameswere more effective, ard the updated organization of end-of-chapter problems by sectionlevels Additionally,the layout was enhancedto limit the instarceswhereExamplesspill over onto multiple pages. . All artwork, pholos, and imageshave been modernized and enhanced to present a crisper illustration of the key elementsand application of circuit analysis.

. Recognizing ttrat moie classpreparation and studying is happening online wilh t}re use of additional resources, ttre development of online resources for tle eighth edition represents a sigdfica[t improvement from the seventh edition. From online, automatically graded homework, to study aids and an e-book, all of this and more is now available on an easy{o-navigate website for students and College textbooks excel at presenting complicaled mateiial in a clear, straigltforward mannei, Authon and publishers spend countless hous developingthe bestpossibleleamingaid for studentsand teachhg aid for instructors. Prcntice Hall is committed to working with authors to create textbooksandsupporthg resourcesthat enablebetterteachingandbetter student leamhg. The eigh0l edition of tlectric Cir.cr.itr is one such example,It setthe standardfoi circuitseducation25 yean ago and il continues that trend today.

HALLMARK FEATURES Chapter Problems Userc of Electic C cuits have consistently rated the Chapter Problems as one ofthe book'smost attmctivefeatures.Inthe eighth edition,theie are ovei 1000problems with approximately 80% that are new or revised from the previous edition. Problems are organized at the end of each chapter by

PracticalPerspectives The eighth edition continues the use of Practical Perspective inftoduced with the chapter openers.They offer examplesof real-world circuits, taken from real-world devices.Most chapters begin wittr a bdef descdption of a practical application of the material that follows. Once the chapter mateda1 is presented,the chapter concludes witl a quantitative analysis of the application along with a Practical Perspective problem. This enables you to unde$tand how to apply the chapter contents to the solution of a realworld problem.

Assessment Problems Each chapter begins with a set of chapter objectives At key points in the chapter,you are askedto stop and assessyoul masteryof a particular problems.Ifyou are able to objectiveby solvingore or more assessment problems givel solve the assessment for a objective, you have mastered that objective.

Examples Every chapter includes many examples that illustrate the concepts presented in the text in the folm of a numeric example. There are over 130 examples in this text. The examples are intended to illustrate the applicationoI a particularconcept,and alsoto encouragegood problemsolving skilts.

Equations andConcepts Fundamental Throughout the text, you will seefundamental equatioN and concepts set apart from the main text.This is done to help you focus on some of the key piinciples in electric circuits and to help you navigate tbrough the important topics.

Tools Integrationof Computer Computer tools can assiststudents in the learning processby pioviding a visual representation oI a circuit's behavior, validating a calculated solu_ tion, rcducing the computational burden of morc complex circuits, and iterating toward a desired solution using parameter variation. This computational suppofi is often invaluable in the design process The eighth edition ircludes the support of PSpice, a popular computer tool Chapter problems suited for explomtion with Pspice are so marked.

Emphasis Design The eighth edition continues to support the emphasison the design of circuits in many ways First, several of the Practical Perspective discussions focus on the design aspects of the circuits. The accompanying Chapter Problems continue the discussion of the design issuesin these pnctical examples.Second,desigl-odentedChapterPrcblemshave been labeled explicitly, enabling students and instructols to identify those problems with a design focus.Third, the identification of problems suited to explo_ ration wit}I Pspice suggestsdesign opporturdties using this soflware

Accuracy All text and problems in the eig.hthedition have undergone our strict hallmark triple accuracychecking process,to ensure the most error-ftee book possible.

ANDINSTRUCTORS FORSTUDENTS RESOURCES ha[[.con/ni lsson www.pren The eighth edition of Elsctlic Circuits comeswith Prentice Hall's poweful new suite of student and instructol online resources

ForStudentsi Online homework and Dractice with immediate feedback and inte grarede-bookuslDgPH UradeAssl\l . Online Study Guide that highlights the key conceptsof electric circuits . Additional book and coursespecificresouces

,

ForInstructors: . Assignable, automatically graded online homework with PH cradeAssist . Digital version of all figur€s from the book . Interactive Learning classroomPowerPoint slides . Sample Chapter Tests . Additional book and coursespecificresources

ADDITIONAL OFFLINE RESOURCES ForStudents: . Student Study Pack-This new resoutce teachesstudents techniques for solvingproblemspresentedin the text. Organizedby concepts, this is avaluableproblem-solvingresourcefor all levelsof students. . Introduction to Pspice Manual-Updated for the eighth edition, the manual comeswith the latest available releaseof the software on CD.

ForInstructors: . Instructor Solutions Manual-Fully worked out solutions to end-ofchapter problems. ' Instructor Problem Bank A tremendous new resoruce with many additional problems and conespo ding solutions to problems not found in the text.This is a great tool for qeating homework and exams.

0rdering0ptionsl Ek!:tric Citcuits wilh PH Grade Assist Online Homework Access: ISBN 0-13-514291-1 Elecftic Citcuits wft SfitdentStudt Pa&:ISBN 0,13 51,4290-3 Eled c Circuits \NithInrrcduction to Pspice Ma utl..ISBN 0-13,514292-X

PREREQUISITES In writing tlte fint 12 chapters of the text, we have assumed tlat the reader has taken a course in elementary differential and integral calculus. We have also assumedthat the rcader has had an introductory physics coune, at either the high school or university level, that iritroduces the concepts ol energy,power, electric charge,electric current, electdc poten, tial, and electromagnetic fields. In writing the final six chapters,we have assumedthe studenthas had,or is enrolledin, an introductorycoune in differential equations.

c0uRsE 0m0Ns The lext hasbeen designedfor usein a one-semestei, two-semester, or a thiee-quartersequence. . Single-semestetcourserAlter coveringChapters1-4 and (hapten 6 10 (omitting Sections?.7 and 8.5) the instructor can choosefrom Chaprer5 (operational amplifiels), Chapter 11 (tbree-phasecircuirs).Chapters13

and 14 (hplac€ methods), and Chapter 18 (Tko-Port Circuits) to develop the desiredemphasis . Ttro-semestersequencerAssuming three lectuies per week, tle fiIst nine chapters can be covered during the first semester, leaving ChaptersI0- l8 for thesecond.emesler. ' Academicquartu schedrlerThe book can be subdividedinto three parts:Chapten 1-6, Chapters7-12,and Chapters13-18. The introduction to operational amplifier circuits can be omitted vithout interfercnce by the rcader going to the subsequentchapten. For examplq if Chapter 5 is omitted, the instructor can simply skip Section7.7,Sectron8-5, Chapter 15, and those problerns and assessingobjective problems in the chaptercfolowing Chapter 5 that pertain to operational amplifiers. There are seveml appendixesat the end of the book to help readers make effective useof t]left matlematical background.Appendix A review$ Cramer's method of solving simultaneous lirear equations and simple matrix algebra;complex numbers are reviewed ir Appendix B;Appendix C contains additional mateiial on magnetically coupled coils and ideal transfomels; Appendi{ D contabs a biiel discussior of the decibeliAppendixE is dedicated to Bode diagEmsiAppendix F is devoted to an abbreviated table of trigonometdc identities that are useful in circuit analysis;and an abbreviated table of useful integrals is given in Appendix G Appendix H prcvides answersto selectedsuggestedprcblems

ACKNOWLEDGMENTS We continue to expressour appreciation for the contributions of Norman Wittels of Woicester Polytechnic Institute. His contributions to the Practical Perspectives geatly enhanced both this and the previous two editions. Jacob Chacko, a transmission and distribution engineer at the Ames Municipal Electric System, also contributed to t}le Practical Perspecrives. Special tlanks also to Robert Yahn (USAF), Stephen O'Conner(USAF), andwilliam Oliver (BostonUniversity)for their con tinued intercst in and suggestionsfor this book. There were many hard-working people behind the scenesat our publisher who deserve our thanks and gmtitude for their efforts on behalf of the eigl[h editiorl At Pretrtice Hall, we would like to thank Michael McDonald, Rose Kernan, Xiaohong Zhu, Lisa McDowell, Jonathan Boylan, David A. Georye, Tim Galigan, and Scott Disanno for their continued suppo and a ton of really hard work. The authon would also like to acknowledge the stafl at GEX Publishing Servicesfor their dedication and hard woik in typesetting this text. The many revisions of the text were guided by carelul and thorough reviews tom professors.Our heanJelt thanks to: . Paul PanayotatoER tgerc University . Evan Goldsrein, University ofwathington . Kalpathy B. Sundaram, Uniretsity of Central Flotidtl . Andrew K. Chan, feta, A&M Unitersity . A. S^taai-Jazi,Viryinia PolytechnicInstitute and State University . Clifford H. Grigg, Rose-Hulman Institute of Technology

. Karl Bithdnger, Univetsit! ofwashington . Carl Wells, Iryr8fttnglonState Unirercity ' Aydir I.Karsilayan,kasA&M UnireNity . Ramakant Sdvastava,University of Florida . Michel M. Mala$iz, Univetsity of Mithigan, Ann Arbor . Christopher Hoople, Rocheskr Institute ofTechnolosy . SannasiRamanan, RochesterItlstitute ofTechnologj . Gary A. Hallock, U/.ivenitt) ol Texasat Austin The authom would also like to thanl Ramakant S vastava of the Udversityol !loflda.aDdlhe AccuraryRe!iewTeamal CEX Publisbing Senices, for their help in checking ttre text and all the problems in the eighthedition. We are deeply indebted to the many instructors and students who have offered positive feedback and suggestionsfor improvement. We use as many of these suggestionsas possible to continue to improve the content, the pedagogy, and the presentation. We are honorcd to have the opportunity to impacr the educational expeiience ol the many thousands of future engheers who will tum the pagesof this text. JaMEsW. NlssoN A, REDEL SUSAN

BriefContents Chapter 1 Chapter 2 Chapter 3 Chapter4 Chapter5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Appendix A AppendixB Appendix C Appendix D Appendix E Appendix F Appendix G Appendix H

List of Examplesxiii Preface xvii CircuitVariables 2 CircuitEtements 22 SimpteResistiveCircuits 56 Techniques of CircuitAnalysis 92 The0perational Amplifier 154 Inductance, Capacitance, andMutualInductance186 Response of First-orderRLandRCCircuits 228 NaturalandStepResponses of RtCCircuits 284 Sinusoidal Steady-State Analysis 330 SinusoidalSteady-State PowerCalculations390 Balanced Three-Phase Circuits 432 Introductionto the LaplaceTransform 466 TheLaplaceTransform in CircuitAnalysis 506 Introductionto Frequency SelectiveCircuits 566 ActiveFitterCircuits 606 FourierSeries 656 TheFourierTransform 698 Two-Port Circuits 730 TheSotutionof LinearSimultaneous Equations759 Complex Numbers781 Moreon Magneticatly Coupted CoilsandldealTransformers 787 TheDecibel 797 BodeDiagrams799 An Abbreviated Tableof Trigonometric Identities 817 An Abbreviated Tableof Integrals 819 Answers to Selected Problems821 Index 839

Contents Listof Examptesxiii Prefacexvii Chapter1 CircuitVariables 2 1.1 1.2 1.3 1.4 1.5 1.6

Etectri.atEngineering: An 0verview 3 TheInterrationatSystemof Units 8 CircuitAnalysis! An overview t0 VoltageandCurrent 11 TheldeatEasicCircuitEtement 12 PowerandEnergy 14 Sunnoty 16 Prcblehs 17

Chapter 2 CircuitElements22 2.1 2,2 2,3 2.4 2.5

PrdcticqlPerspective: E[ectri.slSoIety 23 VoltageandCuffentSources24 Etectrical Resistan(e (0hm'sLaw) 28 Constrlction of a CircuitModet 32 Kirchhoff's Laws 36 Anatysis of a CircuitContaining Dependent Souraes42 ProcticolPetspective: ElectficolSolety 46 Sumndry 47 Prcblens 48

Chapter 3 SimpleResistive Circuits 56 ProcticolPerspedive,A neor Window DeIrcster57 nesistors in Series 58 Resistors in Paraltel 59 TheVoltage-Divider andCurrent-Divider LITCUIIS

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VottageDivisionandCurrentDivision 65 Measuring VoltageandCurrent 68 Measuring Resistance-The Wheatstone Bridge 71 3 . 7 Delta-to-Wye (Pi-to-Tee) EquivatentCircuits 73 PructicalPe5pedive:A RearWindow Defrcster 76 Sumnory 79 Problems 80

Chapter 4 Techniques of CircuitAnalysis 92 Pnctical Pe6pedive:Circuitswith Reolistic Resistors 93 4.7 Terminology94 4.2 Introduction to the Node-Vol.tage Method 97 4.3 TheNode-Vottage Methodand0ependent Sourcest00 4.4 TheNode-Voltage Metbod:SomeSpecial Cases 101 4.5 Introduction to the Mesh-Current MethodJ05 4.6 TheMesh-Current MethodandDependent SourcesJoZ 4.7 TheMesh-Current Method:SorneSpecial Cases 109 the Node-Vottage MethodVersus the Mesh-CurrentMethod 112 4.9 Source Transfoimations116 4.10 Th6v€nin andNortonEquivalents 119 4.11 Moreon Deriyinga ThAvenin Equivatent12J 4.12 Ma)dmum Power-fransfer126 4.13 Superposition 129 Pncticol Pe6pectivetCitcuitswith Realistic Resistors 133 Sunnary 137 Problens 138

Chapter 5 TheOperational Amplifier 154 5.1 5.2 5.3 5.4 5,5 5.6 5.7

PrdcEcqlPe6pective:StruinGqges 155 operational AmptifierTerninals 156 lerminalVottages andCurrents 156 TheInverting-Amptifier Circuit 161 TheSumming-Amptifier Circuit 163 TheNoninverting-Amptifier Circuit 164 TheDifference-A$plifier Cir.uit J65 A MoreReatistic Modelfor the 0perational Ampliliet 170 ProcticalPerspective: Sttuin 6ages 173 Sunnary 175 Problens 176

Chapter6 Inductance. Capacitance, andMutualInductance186 Pnctical Perspective: ProximitySwitches 187 TheInductor 188 TheCapacitor195 Series-Paraltel Cornbinations of Inductance andCapacitance200 6.4 MutuatInductance203 6.5 A CloserLookat MutualIndudance 207 PructicalPerspective: Prcxinity Sdtches 214 Sunmary 217 Prcblens 218

Chapter9 SinusoidaI Steady-State Analysis 330

6.1

9.1 9.2 9.3 9.4 9.5 9.6 9.7

Chapter7 Response of First-0rderfl andflf, Circuits 228 7.1 7.2 7.3 7,4 7,5 7.6 7.7

PructicolPerspedive:A FloshingLight Citcuit 229 TheNaturalResDonse of an fl Circuit 230 TheNaturalResponse of an 8CCircuit 2j6 TheStepResponse ot fl. andRCCircuits 240 A Generat Solutionfor SteDandNatural Responses248 Sequentiat Switching 254 Unbounded Response258 TheIntegratingAmplifiet 260 ProdicdlPe6pective:A FlqshingLight Circuit 263 Sumnory 265 froDens

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Chapter8 NaturalandStepResponses of RLCCircuits 284 8.1 8,2 8.3 8.4 8.5

PtocticalPetspedive:Anlgnition Cir.uit 285 Introduction to the NaturalResDonse of a Palattet nIC Circuit 286 TheFormsof the NaturalResDonse of a Paratl.et RICCircuit 291 TheStepResponse of a Parallel RICCircuit 301 TheNaturatandStepResponse of a Series ilc Circuit 308 A Circuitwith TwoIntegratingAmplifiers 312 Ptodicol Percpective: An lgnition Circuit 317 Summam Szu

Probte;321

9.8 9.9 9.10 9.11 9.12

PracticdlPetspective: A Househotd Distribution Circuit 331 TheSinusoidal. Sowce 332 TheSinusoidal. ResDonse335 ThePhasor 337 ThePassive CircuitElements in the Freouencv Domain 342 Kirchhoff!Lawsin the Frequency Donain 346 Series.Parallel. andDelta-to-Wye SimDlifi(ations348 Source Transformations andTh6venin-Norton Eouivalent Circuits 355 lhe Node-Vottage Method 359 TheMesh-Current Method 360 TheTransformer361 TheldealTransformer365 PhasorDiagrams-t72 Pnctical Percoective: A HouseholdDistribution Circait 375 Sunnary 375 rnoDtems J/b

Chapter10 Sinusoidal Steady-State PowerCalculations390 10.1 10.2 10.3 10.4 10.5 10.6

PructicolPerspedive:HeotingAppliancs 391 Instantaneous Povrer392 Average andReactive Power 394 ThermsValueandPowerCatculations399 Comptex Power 401 PowerCatcllations 403 Maximum Power Transfer 410 ProcticdlPerspedive:HeqtingApptisnces 417 Sumnoty 419 Prcblent 420

11 Balanced Chapter Three-Phase Circuits 432 11.1 11.2 11.3 11.4 11.5

ProcticolPe6pective:Trunsnission ondDlstibution oI Electic Powet 433 Balanced Three-Phase Voltages4j. Three-Phase VoltageSources435 Anatysis of the Wye-Wye Circuit 436 Anatysis of the Wye-Delta Circuit 442 PowerCa[.ulations in Balanced Three-Phase Circsits 445

11.6 Measuring Average Powerin Three-Phase Llrcut$

452

Ptsdicol Perspedive:Transnission on.t DisttibutionoJElectic Powet 455 Su nary 456 ftootens

45/

Chapter 12 Introduction to the Laptace Transform 466 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9

Definition of the LaplaceTransform 46l TheStep Function 468 TheImputseFunction 470 functionaITransfofids 474 0perationatTransfortf.s 475 Apptyingthe LaplaceTransform 481 InverseTransforms 482 Potesand Zerosof F(s) 494 Initial- and Finat-Value Theorems 495 Sumnaty 498 Ptoblens 499

Chapter 13 TheLaplace Transform in CircuitAnalysis 506 Pncticol Pe$pective:SurgeSuppressors507 1 3 . 1 CircuitElernents ifl the s Domain 508 13.2 f,ircuitAnatysis in the s Domain 5Jl 13.3 Applicatiohs512 13.4 TheTransfer function 526 TheTransfer Function in Partiatfraction Expansions528 13.6 TheTransfer Functionandthe Convotution Integrat 537 13,7 TheTransfer Fundionandthe Steady-state SinusoidalResponse 537 13.8 TheImpulsefunctionin Ciiclit Analysis 540 Procticd[Pe6pective:SurgeSuppressors548 Summary549 Probtems 550

Chapter 14 Introduction to Frequency SelectiveCircuits 566 PrccticalPe6pective:PushbuttonTelephone LtrcutEs 5b/

14.1 SomePretiminaries 568 14.2 Low-Pass Fillets 570 14.3 High-Pass Filtets 577

14.4 Bandpass Fitters 582 14.5 Bandreject Fitters 593 ProcticqlPerspe.tive:PushbuttonTe[ephone Circuits 598 Sunnary 599 Problens 599

Chapter15 ActiveFilterCircuits 606 Prdcticdl PeBpedive: BqssVolume Confiol 607 15.1 First-order Low-Pass and High-Pass 75.2 15.3 75,4 15.5

iitters 608 ScaLing612 0p AmpSandpass andBandreje€t Fitters 615 fligherorder0p AmpFitters 622 Narrowband Bandpass andBandreje€t ntters

DJD

PructicolPerspective. EdssVolune Control 642 Sunndry 644 Prcblens 646

Chapter 16 Fourier Series 656 16.1 FourierSeriesAnalysis: An overview 658 16.2 ThefourierCoefficients659 16.3 TheEffectof Symmetry on the Fourier Coefficients662 16.4 An Alternative Trigonometric Formof the FourierSeries 668 16.5 AnApptiration670 16.6 Average-Power Calculations with Periodic funcDonso/5 16.7 ThermsValueof a PeriodicFunclion 678 16.8 TheExponentiat Formof the Fourier sede. 679 16.9 AmptitudeandPhaseSpe.tra 682 Sunnory 685 Prcblens 686

Chapter 17 TheFourier Transform698 17.1 TheDerivation of the FourierTransform699 17.2 TheConvergence of the FourierIntegral 70J 17.3 usingLaplace Transforms to FindFourier franstofifis 703 17.4 FourierTransforrns in the Limit 706 17.5 SomeMathematical Properties7r8 17.6 0perational Transfo(ms710

17.7 CircuitApplications714 17.8 Parseval! Theorem 717 Sunnary 724 Problens 725

Chapter18 Two-Port Circuits 730 gquations 731 18.1 TheTerminat 18.2 TheTwo-PortPa'amelers 732 18.3 Anatysis of the Terminated Two,Port Cit.uit 741 18.4 Interconneded Two-Poft Ctr.uits 747 Sunnaty 751 Prcblens 752

AppendixA TheSolutionof Linear SimultaneousEquations 759 A.1 Pretiminary Steps /59 4.2 Crame/sMethod 760 4.3 TheCharacteristic Determinant760 4.4 TheNurnerator Determinant760 4.5 TheEvaluation of a Determinant761 4,6 Mati.es 764 A.7 MatrirAtgebra 765 A.8 Identity,Adjoint,andInverseMatri(es ZZ0 4.9 Partitioned Matrices 772 4.10 Apptications776

Appendix B Complex Numbers781 8.1 Notation 781 8.2 TheGraphirat Representation ot a Complex Nurnber 782 8.3 Arithmeti.operations /83 8.4 UsefulIdentities 785 8,5 TheIntegerPowerof a Complex Number 785 8.6 TheRootsof a Complex Number 286

AppendixC Moreon Magnetically Coupled CoilsandIdeal Transformers787 C,1 Equivatent Circuitsfor Magnetically Coupted Coils 787 C.2 TheNeedfor ldealTransformers in the Equivalent Circuits 792

Appendix D TheDecibet 797 Appendix E BodeDiagrams799 E.1 E.2 E,3 E.4 E.5 8.6 E.7 E.8

Reat,First-0rder PolesandZeros 299 Straight-lineAmptitudePlots 800 MoreAccurate AmplitudePtots 8ra Straight-Line Phase AngtePlots 805 BodeDiagrarns: Complex PolesandZeros 807 AmotitudePtots 809 Correcting Straight-Line AmplitudePtots 810 Phase AnglePtots 813

AppendixF An Abbreviated Table of Trigonometric Identities 817 AppendixG An Abbreviated Tableof Integrals 819 AppendixH Answers to Selected Problems821 Index 839

Listof Examptes 2 Chapter of IdeatSources26 2.1 lestingInterconne.tions of IdeatIndependent 2.2 Testinglnterconnedions Sources27 andDependent for a vottage, Current.andPower 2.3 Catcutating Circuit 31 SimpleResistive 2.4 Construding a CircuitModelof a Ftashtight33 a Circuit[{ode[Basedon Terminal 2.5 Constructing Measurements 35 CurrentLaw 39 2.6 UsingKirchhoff's 2.7 UsingKirchhofftVottageLaw 39 Laws 2.8 Applyingohm'sLawandKirchhoff's Current 40 to Findah Unknown 2.9 Constructing a Cir€uitl4odelBasedonTerninal Measurements 41 2.10 Applyingohm'sLawand(i(hhoff's Laws Vottage 44 to Findan Unknown Law 2.11 ApFlyingohm'sLawandKirchhoff's in an Arnplifiertircuit 45

3 Chapter 61 3.1 ApptyingSeries-ParattetSirnptification Circuit 63 3.2 Anatyzing the vottage-Divider Circuit 64 3.3 Analyzing a Current-Divider 3.4 UsingVoltageDivisionandCurrentDivision to Solvea Circuit 67 Ammeter 69 3.5 Usinga d Arsonval Vottmeter Z0 3.6 Usinga dArsonvat Transform75 3.7 Applyinga Delta-to-Wye

4.6 4.7 4.8 4,9 4.10 4.11 4,12 4.13

Chapter5 an 0p AmpCircuit J60 5.1 AnatyzinE

6 Chapter 6.1 6.2 6.3 6.4 6.5 6.6

4 Chapter 4.1

4.4 4.5

ldentifying Node.Eranch,Mesh,and Loop in a Circuit 95 Method 99 Usingthe Node-Vottage Methodwith Usingthe Node-Vol.tage Dependent Sources 100 Meihod 106 Usingthe Mesh-Cuffent Methodwith Usingthe Mesh-C{rrrent DependentSources 108

Method tlnderstanding the Node-Voltage Method 113 Versus Mesh-Curent andMesh-Current [omparingthe Node-Vottage Methods115 to Sotve Transformations UsingSource a Cit.uit 117 Transformation UsingSpecialSource 118 Techniques Equivalent of a Circuit Findingthe Th6venin Source 122 with a Dependent Equivatent Usinga Test Findingthe Thdvenin SoUwrce 124 for MadmumPower the Condition Catcutating Transfer 128 to Solvea Circuit 132 UsingSuperposition

Giventhe Current, Determining the Vottage, of an Indlctor 189 at the Terminats the Current,Giventhe Vottage. Determining at the Terminats of antndu.tor 191 Powet Vottage. Determining the Current, andEnergy for an Inductor 193 Power, and Detefmining Current,Vottage. 197 for a capacitor Energy Finding?).p, and,lr)Inducedby a Triangutar CurrentPutsefor a Capac.tlor198 for a Circuit Equations FindingMesh-Current coits 206 with Magneticatty Coupled

7 Chapter of an the NaturalResponse Determining RLCircuit 234 of an the Natual Response 7.2 oetermining Inductors 235 iI Circuitwith Parallel of an the NaturalResponse 7,3 Determining nfCircuit 238 7.1

xiv

Lkt of Exanptes

7.4

Determining the NaturalResponse of an RCCircuitwith SeriesCapacitors2-39 7.5 Determining the StepResponse of an RLcir.uit 244 7.6 Determining the StepResponse of an RCCitcuit247 7.7 Usingthe General SolutionMethodto Findan RCCircuit'sStepResponse250 7.8 Usingthe Generat SotutionMethodwith Zero lnitiat Conditions251 7.9 Usingthe General SolutionMethodto Findan ,(LLrrcurl5 >Iepxesponsez5l 7.10 Determining the StepResponse of a Circuit with Magneticatl.y Coupted Coils 253 7.11 Anatyzing an Rl Circuitthat hasSequentiat Switching 255 7.72 Analyzing an RCCirclit that hasSequential Switching 257 7,13 Finding the Unbounded Response in an RCCircuit 259 7.14 Anatyzing an IntegratingAnptifier 261 7.15 Anatyzing an IntegratingAmptifierthat has Seqoentiat Switching 262

Chapter 8 8.1

Findingthe Rootsof the Characteristic Equation of a ParallelRLCCircuit 290 8.2 Findingthe overdamped NaturalResponse of a Paraltet RaCCircuit 293 8.3 Catrulating Bran(hCurrents in the Natural Response of a Panllel RLCCir.uit 294 8.4 Findingthe Underdamped NaturatResponse of a Parattet nlC Circuit 297 8.5 Findingthe CriticattyDamped Naturat ResDonse of a ParallelRICCircuit 300 8.6 Findingthe overdamped StepResponse of a Parattet f,lCCircuit 304 8.7 Findingthe Underdamped step Response of a Parattet RICCircuit 305 8.8 Findingthe CriticaltyDamped StepResponse of a Parattet faf Circuit 305 8.9 Comparing theThree-Step Response Forms 306 8.10 FindingStepResponse of a Paratl.et ilc Cir.uit with Initiat Storedlnergy 306 8.11 Finding the Underdamped NaturaI Response of a SeriesRaf,Circuit 310

8.12 Finding the Underdarnped StepResponse of a SeriesRtCCircuit -?11 8,13 Anatyzing TwoCascaded lntegrating Amptifiers 3l.t 8.14 Anatyzing TwoCascaded IntegratingAmplifiers with Feedback Resistors316

Chapter 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16

Findingthe Characteristi€s of a Sinusoidal Cufient 334 findingthe Characteristics of a Sinusoidal Vottage 334 Translating a SineExpression to a Cosine Expression-t.t4 Catcutating the rmsVatueof a Triangutar Waveforrn3J5 Adding Cosines UsingPhasors341 Impedances in Series 349 Combining Combining Impedances i'n Seriesandin Paraltet 351 Usinga Delta-to-Wye Transform in the F equency Domain 353 Perforrnihg lransformations Source ir| the Frequency Domain 356 Findinga Th6venin Equivalent in the Frequency Domain 35l Usingthe Node-Vottage Methodin the Frequenqr oomain 359 Usingthe Mesh-Current l4ethodin the Frequenqf Donain 360 Anatyzing a Linearlransformer in the Frequenqr Domain 364 Anatyzing an IdealTransformer Circuitin the trequency Domain 370 UsingPhasorDiagrams to Analyzea Citcuit 372 UsingPhasorDiagrams to AnalyzeCapacitive LoadihgEffects 373

Chapter 10 10.1 Cattulating Average andReadivePower 396 10.2 MakingPowerCalculations Involving HousehotdAppliances 398 10,3 oetermining Average PowerDetivered to a Resistor by a Sinusoidat Voltage 400

Listof Examphs xv

10.4 10.5 10.6 10,7

Power 402 Catcutating Complex Average Power 406 Calcutating andReactive Power in Pardllet Loads 40l Calculating Balancing PowerDelivered with Power Absorbed in an acCircuit 408 10,8 DeteminingMa$mumPower Transfer without LoadRestrictions41-t 10.9 DeteminingMaximum Power Transfer with LoadImpedance Restri.llon 414 10.10FindingMaximum Power Transfer with I pedance AogleRestidions 414 10.11FindingMaximunPower Transfer in a Circuit with an IdealTransformer4J5

11 Chapter 11,1 Anatyzing a Wye-Wye Cit.uiI 440 11.2 Anatyzing a Wye-Detta Circuit 444 11,3 Catcul.ating Powerin a Three-Phase Wye-Wye Circuit 449 11.4 Catculating Por,rer in a Three-Phase Wye-Delta

14.4 14.5 14.6 14.7

Loading the Seriesna fiigh-Pass Fitter 580 Designing a Bandpass Fitter 587 Designing a Parattet RaCBandpass Fitter 588 Determining Effed of a Nonideal Voltage Source on a fLC Bahdpass Fitter 589 14.8 Designing a SeriesRtCBandreject Fitter 596

15 Chapter 15.1 0esigning a Low-Pass 0p AmpFitter 609 15.2 Designing a High-Pass 0p AmpFitter 611 15,3 Scal.ing a SeriesRLCCir.uit 613 15.4 Scating a Prototype Low-Pass 0p Arnp FiLlet 614 15.5 Designing a Broadband Bandpass 0p Anp Filtet 618 15.6 Designing Bandreject a Sroadband 0p Amp Filtet 621 15.7 Designing a Fourth-order Low-Pass 0p Amp fitrer

bz5

15.8 CaLcutating Butterworth Transfer Functions628 11.5 Catcllating Three-Phase Powerwith 15.9 Designing a Fourth-order Low-Pass an Unspecified Load 450 ButterworthFilter 631 11.6 Computing Wattmeter Readings in Three-Phase 15.10Determining the orderof a Eutterworth Citcuits 454 Filtet 634 15.11AnAlternate Approach to Determining 12 Chapter the orderof a Eutterworth fitter 634 12.1 UsingSteptunctionsto Represent a Function 15.12Designing a High-4Bandpass Fitter 638 oT ilntre uuralon 4/u 15.13Designing a High-oEandreject Fitter 64, tltcull

45u

13 Chapter 13.1 Deriving the Transfer tundionof a Circuit 527 13.2 Anatyzing the Transfer Fundion of a Circuit 529 13.3 Usingthe Convol|ltior Integralto Find an oljtp|lt Signat 535 13.4 Usingthe Transfer Function to Find the Steady-State Sinusoidal Response539

Chaptor 14 14.1 Designing Filtet 574 a Low-Pass 14.2 Designing a SeriesfC Low-Pass titter 575 14.3 0esigning a SeriesRl High-Pass tillet 579

Chapter 16 16.1 Findingthe FourierSeriesof a Triangutar Waveform with NoSymmetry660 16.2 Findingthe FourierSeriesof an odd Function with Symmetry667 16.3 Catcutating Formsof the Trigonometric Folrier Sedesfor Periodic Voltage 669 16.4 Calcutating Average Powerfor a Circlit with a Periodic VoltageSo[.e 677 16.5 Estirnating the rmsValueof a Periodic tun(tion 679 16.6 Findingthe Exponential Formof the Fourier Series 681

Slnusoidal Steady-state Response716 17.3 ApplyingParevaltIheoren 719 17.4 ApptyingParsevalt Theorem to an Ideal Eandpass tilter 720 17.5 AppMngParseval's fheoremto a Low-pass Fiatet 721

Measurements 235 18.3 Findingt parameters fromMeasurements ard Table18.1 738 18,4 Anatyzing a Terminated Two-port Circuit 246 18.5 Analyzing Cascaded Two-port Circuits Z5O

Variables Circuit 1.1 EtectricatEngineering: An overviewp. 3 1.2 TheInternationatSystemof Units p. 8 1.3 CircuitAnatysisrAn overviewp. 10 1.4 Voltageand (urrent p. 1? 1.5 TheIdeal BasicCircuitEtementp. 12 1.6 Powerand Energyp. 14

1 Unde6tand andbe abteto use5I unitsandthe standadprefi:esfor powersof 10. 2 Knowandbe abteto useihe dennitionsof 3 Knowafd be abt€to us€ihe defifitionsof 4 Beabteto usethe passive siqnconvention to caLcutaie the powerfor af idealbasiccjrcujt €tementgivenits vottageandcunent.

ElectricaI engineering is an excitingand challcngingprofcssion for anvone who has a genuine interest in, and aptitude lor, applied scienceand mathematics.Over the past century and a hall, electrical engineershave played a dominant rolc in thc developnrentof svstemsthat have cbangedthe v',aypeople live and work. Satellitecommu cation links,telephones,digital com puters, tcicvisions,diagnosticand surgical medical equipment, assembiylinerobots,and eleclricalpower tools are representative componentsof systemsthat define a modern technological socicty.Asan clcctricalengincer.you canparticipatein this ongoing technologicalrevolution by improvilg and refining these existingsystemsand by discovcdngand dcvclopingnew systems to meet the needsot our ever-changing socjety. As ]'ou embark on the study of circuit analysis,you nccd to gain a feel for where tlris study fits into the hierarchyof topics that conlprisean introductionto electricalengineering.Hencewe begin by presentingan oveNiew of electricalelElineering,some ideasabout an engineeringpoint of view as jt relatesto circuit analysis.and a reviervof the internationalsystemof units. Wc then describegenerallywhat circuit analysisentails.Next, we intloducethe conceplsolvollage and current.Welbllow these conceptswith discussiolof an ideal basicelementand the necd for a polarity reference system.We conclude the chapter by describinghow current and voltagerclatc to powcr and cncrgy.

An0verview 1.1 EtectncatEngin€€nng:

An Overview Engineering: 1.1 Etectrical Electrical engineeringis the profession concetned with systemsthat oroduce.transmit, analmeasureelectric signals.Elect cal engineeing ;ombines the pbysicist'smodels of natural phenomena ith the mathematician'stools lor manipulatingthose modelsto producesystemsthat meet practicalneeds.Electiical systemspervadeour lives;theyare found ir homes, schools,workplaces, and transportation vehicles everywhere We beginby presentinga few examplesfrom eachofthe five major classificationsof electricalsystems: . commu catlonsyslems ' computel systems t control systems . power sysEms . signal-processing systems Then we describehow electrical engineersanallze and designsuchsystems Conrmunication systems are electrical systems that generate' transmit. and distribute information. Well-known examples include television equipment, such as cameras,transmitters, receivers,and VCRS; radio telescopes,used to explore the univeise; satellite systems,which return images of other planets and our own: radar systems,used to coordinateplane fl ights;ard telephonesystems. Figure 1.1depictsthe major componentsof a modern t€lephonesystem. Sta ing at the left of the figure,inside a telephone,a microphone turns sou d waves into electric signals.These signalsare caded to a s$ritchingcenter where they are combined with the sigmls from tens,hundreds, or thousands of othei telephones.The combined signals leave the switchingcenter;tleir form dependson the distancethey must travel.In our example,tley are sent througl wires in underground coaxial cablesto a microwavetransmissionstation.Here, the signalsare transformedinto microwavefrequenciesand broadcastfrom a tansmission antena through air and space,via a communicationssatellite,to a rcceivingantenna.The microwave receiving station translates the microwave signals into a folm suitablefor furtler transmission,perhapsas pulsesof light to be sent through fiber-opticcable.On arival at the secondswitchingcenter,the combinealsignalsare separated,and each is routed to the approprrate telephone,where an earphoneacts as a speakerto convert the leceived electricsignalsback hto soundwaves.At eachstageof the process,electric ci.cuits operate on tlle signals. Imagine the challenge involved in designing, building, and operating each circuit in a way that guarantees that all of the hundredsof thousandsof simultaneouscallshave high quality Computer syslemsuse electric signalsto processinformation mrging from word processingto mathematicalcomputations.Systemsrange in size analpower from pocket calculato$ to personal computers to supercomputersthal perform suchcomplextasksas plocessingweather data and modelingchemicalinteractionsof complexorganicmolecules. Thesesystemsincludenetworksof microcircuits,or iniegratedcircuitssYstem. postage-stampsized assembliesof hundreds,thousands,or millions of tigure1,1 A Atetephone

Circuit Vanabbs electricalcomponentsthat often operateatspeedsandpowerlevelsclose to fundamenlal physical limits, including th.3speedol light and the thermo dynamic laws, Control systemsuse electricsignalsto regulateprocesses, Examples include the control of tempemtures, pressures,and flow mtes in an oil refinery;the fuel-airmixlure in a fuel-injectedautomobileengine;mechanismssuchasthe molors,doors,andlightsin elevators;andthe locksin the Panama Canal.The autopilot and autolanding systemsthat help to fly and land airyIanes aie also familiar control systems Powersyst€msgenerateand distributeelectricpower.Electricpower, which is the foundatior of our technology-based society,usuallyis generated in large quartities by nuclear, hydroelectric, and thermal (coal-, oil-, or gas-fired)generato$.Poweris distributedby a grid of conductorsthat crissooss tlle country. A major challenge in designing and operating such a systemis to provide sufficient redurdancy and control so that failure of any piece of equipment does not leave a city, state, or region completely

figure1.2 a A tT scanofan adLrtt head.

Figure1.3 A Anajrplane-

Signal-processingsystens act on electric signals that represent information. They translorm the signalsand the informarion contained in tlem into a more suitable form. There are many different ways to process the signalsand their information. For example,image-processing systems gathermassivequantitiesof data from orbiting weathersatellites,reduce the amount of data to a manageablelevel,and transform the remaining data into a video imagefor the eveningnewsbroadcast.A computerized tomography(CT) scanis another€xampleofan image-processing system. It takessignalsgenemtedby a specialX-ray machineand transformsthem into an imagesuchas the one in Fig. 1.2.Although the original X-ray signals are of little useto a physician,once they are processedinto a recognizableimagethe informationthey containcanbe usedin the diagnosisof djseaseand injury. Considerable interaction takes place among the engineering disci plines involved in designhg and operating these five classesol systems. Thus conmunications engineersuse digital computeis to control the flow of information,computers contain control systems!arrd control systems contail computers,Po\ver systemsrequire extensive communications sys temsto coordinatesafelyand reliably the operationof components, which may be spread acrossa continent. A signal-processingsystemmay involve a conmunicationslink, a computer.and a control system. A good exampleof the interaction among systemsis a commercial airplane,such as the one showl in Fig. 1.3.A sophisticatedcommunicatiom systemenablesthe pilot and the air traffic controllerto monitor the plane'slocation.permittingthe air traffic controllerto designa safeflight path for a of the nearbyaircraftand enablingthe pilot to keep the plane on its designatedpath. On the newestcommercialairplanes,an onboard compulersystemis usedfor managingenginetunctions,implementingthe navigation and flight control systems,and gererating video information scrcensin the cockpit.A complexcontrol systemusescockpit commands to adjustthe positionand speedofthe airplane,producingthe appropriate signals to the engines and the control surfac€s (such as the wing flaps, ailerons, and rudder) to ensure the plane rcmains safely airbome and on the desiredflight path.Theplane must have its own power systemto stay aloft and to provide and distributethe electricpower neededto keep the

1.1

cabin lights on, make the coffee, and show the movie. Signal-processing systems reduce the noise in air taffic communications and tlansform information about the plane's location into the more meaningful form of a video display in the cockpit. Engineering challengesabound in t}le design of each of these systemsand their integration into a coherent whole. For example,these systemsmust operate il widely varying and unpredictable environmental conditions. Perhaps the most importalt engineering challenge is to guarantee that sufficient redundanry is incorporated in the designsto ensurethat passenge$arive safelyand on time a! their desired destinations. Although electrical engineersmay be interestedprimarily in one area, they must also be knowledgeable il other areas that interact with this area of interest. This intemction is part of what makes electrical engi neering a challenging and exciting profession.The emphasisin engineer ing is on makingthingswork,soan engineelis free to acquireand useany technique,ftom anyfield, tlat helpsto get t}lejob done

CircuitTheory In a field asdiverseas electricalengineering,you might well askwhether all of its brancheshave anythingin common.The answeris yes rlectric circuits.An elect'riccircuit is a mathematicalmodel that approximates the behaviorof an actualelectricalsystem.Assuch,itprovidesan important foun{:lationfor leaming-in your later coursesand as a practicing engineer-the details of how to design and opefate systemssuch as tnose iust described.The models,the mathematical techniques,and the language of circuit theory will form the intellectual framework for your future engi neering endeavors. Note that the term electtic circuit is commonly used to reter to an actual electrical system as well as to the model that represents it. In this text. when we talk about an electdc circuit, we always mean a model, unlessotherwisestated.Itis the modelingaspectof circuit theorythat has broad applications acrossengineering disciplhes. Circuit theory is a specialcaseof elestromagneticfield theory: the study of static and moving electdc charges.Altlough generalized Iield theorl might seemto be an approp ate starting point for investigating electric signals, its application is not only cumbersomebut also requires the use of advanced mathematics. Consequendy,a course in electromagnetic field theory is not a prerequisite to urderstanding the material in this book. We do, ho ever, assumethat you have had an introductory physics course ilr which electdcal and magneticphenomenawere discussed. Three basicassumptionspermit us to use circuit theory,rather than electromagnetic field theory to study a physical systemrepresented by an electric circuit. These assumptionsare as follows: 7. Electrical effects happen instantaneously throughout a system.We can make this assumption because we know that electric signals travel at or near the speed of light. Thus, if the system is physica y small, elecfic signals move through it so quickly that we can consider them to affect every point in the systemsimultaneously A system that is small enoughso that we call make this assumptionis system. calleda lumped-paramet€r

An0verview Engjneenng: EL€ciricat

2. The net charge on ewty component in the system is always zero. Thus no component can collect a net excess of charge, although some components,as you will learn latei, can hold equal but oppositeseparatedcharges. 3. Therc is no magnetic co pling betteeenthe componentsin a system. As we demonstrate later, magnetic coupling can occtrr within a componenL Thafs it; there are no other assumptions.Using circuit theory provides simple solutions (of sufficient accuracy) to problems that would become hopelesslycomplicatedif we were to use electromagneticfield theory These belefits are so great that engineeN sometimes specifically design electical systemsto ensurethat theseassumptioNare met. The importance of assumptions 2 and 3 becomes apparent after we inlroduce the basic circrlit elementsaltd the rules for analyzing interconnected elements. However,we needto take a closerlook at assumption1.The question r\ "How smalldoesa pb).icalsyslemhare ro be ro quatifyas a lumped parameter system?"We can get a quantitative handle on the question by noting that electdc signals propagate by wave phenomena. If the wavelength of the signal is large compared to the physical dimensions of the system, we have a lumped-parameter system.The wavelength I is the velocity dividedby the repetitionrate,or frequency,of the signal;thatis,,\ = c/f. The frequency/ is measuredin hertz (Hz). For example,power systems in the United States operate at 60 Hz. If we use t]le speed oI light 103m s) as lhe velocir) ot propagation. 1r - 3 lhe walelenglhis 5 l0'm. If rbe po\rer(ystemot inrere\ric pbysically smatterlhan $is wavelength,we can representit as a lumped-parametersystemand use circuit theory to analyzeits behavior.How do we definemarel? A good N\e 1sthe rule of 1/10ri: if rhe dimensionoI the systemis 1/10th (or smaller.)of the dimension of the wavelength,you have a lumped-parameler system.fhus,aslong asthe physicaldimensionof the power systemis less than 5 x 10)m,we cantreat it as a lump€d-parameter system. On the other hand, the propagation frequency of mdio signalsis or tlle order of 10vIIz. Thus the wavelength is 0.3 m. Using the rule of 1/10ttr, rtre relevant dimerNionsof a connunication syslemthat sendsor receivesmdio signalsmust be less tian 3 cm to qualify as a lumped-parameter system. menever any of the pertinent physical dimensionsof a systemurder study approachesthe wavelength of its signals,we must use electromagneticfield theory to aralyze that system. Thioughout this book we study circuits de ved ftom lumped-parameter systems.

ProblemSotving As a practicingengineer,you will not be askedto solve problemsthat have already been solved. Wlether you are trying to improve the perfomance of an existing systemor qeating a rew system,you will be working on unsolved problems.As a student, however, you witl devote much ol your attention to the discussionof problemsalready solved.By reading about and discussinghow theseproblemswere solvedin the past,and by solving related homework and eram problems on your own, you w.ill begin to develop the skills to successfully attack the unsolved problems you'll faceas a practicingengineei.

AnoveMew Engineering: 1.1 Etectrical Some general problem-solving prccedures are presented here Many perlainlo lhinkingaboutand organLing)our solulionslr0legy rhem ol proceedingwith calculations. 6elor€ 1.. Identify what'sSiven and what's to be found h problem solving,you need to know your destilation before you can selecta route tbr getting tlere. What is the problem asking you 10 solve or find? Sometimesthe goal of tle prcblem is obvious; other times you may need to paraphrase or make lists or tables of known and unknowr inJormation to seeyour objective. The problem statement may contain extraneous information tlat you need to weed out betore proceeding.On the other hand' it may offer incomplete information or more complexities thar can be handled given the solution methods at your disposal. In that case, you'll need to make assumptionsto fill in the missinginformation or simpliJy the problem context. Be prepared to circle back and reconsider supposedlyextraneousinformatior and/or your assumptionsif youl calculationsget boggeddown or produce an answerthat doesn't seemto make sense. 2. Sketcha cicuil diagram or other risual model. Translathg a verbal problem description into a visual model is often a useful step in the solution process.If a circuit diagam is aheady provided' you may need to add inlormation to it, such as labels, values' or leference directions.You may also want to redraw the circuit in a simpler, but equivalent, form. Later il this text you vrill Ieam the methods for developing such simplified equivalent clrcuits 3. Think of several solution methoh and decide on a w6Jrof choosing among theru'f\is colnse will help you build a collection of analytical tool$ several of which may work on a given pioblem. But one method may pioduce fewer equations to be solved than another, or it may require only algebra bstead of calculus to reach a solulion. Suchefficiencies,if you can anticipate them, can sfeamline your cal_ culations consideiably. Having an alternative method in mind also givesyou a path to pursue iJ your first solution attempt bogs down 4. Calculate a solution. Your planning up to tltis point should have helped you identify a good analytical method and the correct equationstor lhe probtem.Not\ comesrhe\olulionof lhoseequalions. Paper-and-pencil, calculator, and computer metlods are all available for performing t]le actual calculations of circuit analysis Efficiency and your insxuctor's prefeiences will dictate which tools you shoulduse. 5. UseWw creatirit!.ltyou suspectthat yolu answeris off baseor if the calculationsseemto go on and on without moving you toward a solu_ tion, you shotlld pause and consider altematives.You may need to revisit your assmlptions or selecta diffeient solution method. Or, you may need to take a less-conventionalproblem_solvingapproach'such asworking backwad from a solution.This text provides answen to all of the AssessmentPrcblems and many of the Chapter Problems so that you may wo* backward when you get stuck. ln the real world, you wonl be given answersin advance,but you may have a desired prcblem outcomein mind from which you can work backward.Other seative approachesinclude allowing yourself to see parallels witl

CircuiiVariabtes other t!?es of probleris you've successfullysolved, following your intuition or hunches about how to proceed, and simply setting the prcblem asidetemporadly and coming back to it later. 6. Test your sol tion. Ask yourself whether the solution you've obtained makes serlse.Does the magnitlde of the answer seemiea sonable?Is the solution physically realizable? You may want to go furtler and rework tle problem via an alternative method. Doing so will not only test the validity of your odghal answer,but will also help you develop your intuition about the most efficient solution methodsfor various kinds of problems.In the real \qorld, safetycdtical designsare always checked by several iDdependent means. Getting into the habit ol checking your answeis will benefit you as a studentand as a practicingengheer. Theseproblem-solving stepscannot be used asa recipe to solve every prob lem in this or any other cource.You may need to skip, changethe order of, oI elaborate on certail stepsto solve a particular problem, Use thesesleps as a guideline to develop a problem-solving style that works for you.

1.2 TheInternational Systemof Units Engineen comparc theoretical results to experimental results and compare competingengine€rhgdesignsusingquantitativemeasures. Modern engineering is a multidisciplinary profession in which teams of engineers work together on projects, and they can connnunicate their results in a meaningful way only if they all use the same units of measure. The Intemational System of Units (abbieviated SI) is used by aI the major engheeringsocietiesand most engifleersthroughoutthe world;hencewe useit in this book. Th€ SI units are basedon sevend€lired quantities: . length ' fime . elect c current . thermod]'namic tempemture . amount of substance . luminousintensity

:rI4*l:lrlillrri Qumtily

BasicUnit

Length Mass

kg

Tine Thermodynamic temperature AmouDt of substane

K

1.2

System ofUnits TheIntenationat

Thesequantities,along rvith the basicunit and s]'mbo1for each,are lisled in Tablc 1.1.Although not sldclly SI units,the familiar time unils of minute (60 s), hour (3600s), and so on arc often usedin engineeringcal culalions.In addirion.defined quanlitiesarc combinedto form deriv€d units.Some,suchas force,energl',power,and elcctriccharge.you already krow through previousphysicscou$es.Table 1.2lists the derived units usedin this book. Ir manycases, the SI unit is either ioo smallor too largeto useconveprefixcs niently. Standard correspondingto polve$ of 10, as ljsted in Table 1.3,are then appliedto the basicunit. All o{ lhescprefixesare correct.but engineenoftcn useonly the onesfor powersdivisibleb]' 3;thus centi,deci,deka,and hectoare usedrare\,.Also, engineersoflen selectthe prefix lhat places the base number in the range between I and 1000. Supposethat a time calculationyieldsa result of 10 ' s,that js,0.00001s. Most engineers would dcscribe this quantity as 10 /is. that is. j ps. 10 = 10 x 10j s,mlher than as0.01ms or 10,000,000

prefiresfor powersof 10 objective1-lJnderstandand be abteto useSI units and the standard 1.1

1.2

How many dollarsper millisecondwould the federalgovemmenthaveto collectto retire a delicit of$100 billion in one year?

If a signalcantmvelin a cableat 80%of the speedof light,whatlengthof cable,ir inches, rcpreserts1 ns?

Answer:9.45'.

Answ€r: $3.17/ms. NOTE: Abo try ChaptetPrcblems1.1,1.3,and L6.

Pr€fixesto signify TABLE1.3 Standardized Prefu

Symbol

Pol€r

t0 r3 TABLE 1.2 DerivedUnits in SI Qufllity

UnitNrne (Slnbol)

pico

p

hetv (Hz)

106

newtoD(N) joulc (J)

N m

nilli

1 0 3 t0 l

J coulonb (c) Electricpotential

vo1l(V)

d

h C/V

Magnetic llux lrenry (H)

1o' t0l

kilo

sienens(s)

l0' l|)

I/C

ohm (o) Electricconduclance

10 r' 10v

M

106

G

10e

T

l0''

10

1.3 CircuitAnalysis:An 0verview

modet forelechcatenqitiguret.4 A A conceptuaL needng design.

Before becoming involved in the details of circuit analysis, we need to take a broad look at engineeing design,specilicallythe designof electric circuits. The purpose of this oveNiew is to prcvide you with a perspective on where circuit analysis fits within the whole of circuit design. Even though this book focuseson circuil analysis,w€ try to provide opportudties for circuit design where appropriate. A11engineeringdesignsbegin with a need,as shownin Fig. 1.4.This need may come from the desire to improve on an existing desig , or it may be something brand-new A careful assessmentof the need results in design specifications,which are measumble characteristics of a proposed desigl. Once a designis proposed,the designsp€cificationsallow us to assesswhether or not the design actually meets the need. A concept for the designcomesnext. fhe concept derives ftom a complete understandingof the designspeciJicationscoupled with ar insight into t]!e need,which comesfrom education and experience.The concept may be realized as a sketch, as a written description, or ir some other fom. Often t]le next step is to translate tle concept into a mathematical model. A commoily usedmathematicalmodel for electrical systemsis a circuit model. lhe elements fhat comprise the circuit model are called ideal circuit components. An ideal circuit component is a mathematical model of an actualelecldcalcomponent,like a battery or a light bulb. It is important for the ideal circuit component used in a circuit model to iepresent the behavior of the actual electrical component to an acceptable degree of accuracy.The tools of circuit snalysis, the focus of tlis book, are then applied to ttre circuit. Circuit analysisis basedon mathematical techniques and is used to predict the behavior of the circuit model and its ideal circuit components.A comparison between the desired behavior, from t}le design specifications,and the predicted behavior, from circuit analysis,may lead to rcfinements ir the c cuit model and its ideal circuit elemerts. Once the desiredandpredictedbehavioraie in agreement,a physicalprototypecan be constructed. The physical proroqpe is an actual electrical system,constnrcted ftom actualelectricalcomponentsMeasuiementtechniquesare usedto determine the actual,quantitativebehaviorof the physjcalsystem.This actual behavior is compared vrith the desired behavior from the designspecifications and the predicted behavior from circuit analysis.The comparisons may result ir rcfhements to the physicalprototype,the circuit model,or botb Eventually, this iterative process,ir which models, components,and systems are coflthually rcfined, may produce a design that accurately matches the designspecifications and ttrus meets the need. From this desciption, it is clear that circuit analysis plays a very importantrole in the designprocess.Becauseciicuit analysisis appliedto circuit models,practicingengineerstry to use mature circuit models so that the resultingdesignswill meet the designspecificationsin the filst iteration.In this book. we use modelsthat have been lestedfor between 20 and 100 years; you can assume that they are matuie. The ability to model actual electrical systemsvrith ideal circuit elements makes circuit theory extremely useful to engineers. Sayingthat the interconnectionof ideal circuit elementscan be used to quantitativety predict th€ behavior of a system implies that we can

1.4

Voltage andCLrent

describethe interconnectionwith mathematicalequation$For the mathematicalequationsto be useful,wemustwrite them in termsof measurable quantities.Inthe caseof circuits.thesequantitiesare voltageand current, whjch we discussin Section 1.4.The study of circuit analysisinvolves understandingthe behavior of eachideal circuit elementin terms of its voltage and current and understandingthe constraintsimposedon the voltageand currentas a resultof interconncctingthe ideal elcmcnts-

1.4 VottageandCurrent The conceptof electricchargeis the basisfor describingall electricalpheof electriccharge. nomena.Let's review someimportant characteristics . The charge is bipolar, meaDingthat electrical effects are described in termsof positiveand negativecharges. . The electric chargeexistsin discretequanrities,which are integral multiplesof the electroniccharge,1.6022x 10-1eC. . Electical effectsare attributcdto both the sepafationof chargeand chargesin motion. ln circuit theory, the separation of charge creates an electric force (vol! age),andthe motion of chargecreatesan electricfluid (cunenr). The conceptsof voltageand cu ent are usetul from an enginee.ing point ol view becausethey can be expressedquantitatively.Whenever positiveand negativechargesare separated, energyis expended.Voltage is the energyper unit chargecreatedby the separation.We expressthis ratio in differential form as

(r.1)

4 Definitionof voltage

0 = the voltagein volts, 1, = the energyinjoules, 4 = the charge ir coulombs. The elect cal effectscausedby chargesin motion dependon the rate of charge flow The rate of charge flow is klown as the electric currcnl, which is expressedas

,=#,

(1.2) < Definitionof current

j : the currentin amperes, q = the chargein coulombs, I : the time in seconds. Equations1.1and1.2are definitionsfor the magnitudeofvoltage and cu ent,rcspectively.The bipolarnatureof electdcchargerequircsthat we assignpolarity referencesto thesevariables.we will do so in Section1.5.

11

t2 Although current is made up of discrete,moving elechons, w€ do not need to considerthem hdividualy becauseof the enormousnumber of them. Rather, we can think of electrons and thei corresponding charge as one smoothly flowing entity. Thus, i is heated as a continuous variable. One advantageof usingcircuit modelsis that we canmodel a compo, rent stricUy in terms of tie voltage and current at its terminals. Thus two physically different components could have the same relationship betweenthe teminal voltage and terminal current. lf they do, for purposesof circuit analysis,they are identical.Once we know how a component behaves at its terminals, we can anal'ze its behavior in a circuit. However,when developingcircuit models,we are interestedin a component's intemal behavior.We might want to know, for example,whether charge conduction is taking place because of free electrons moving through the dystal lattice structureof a metal or whetherit is becauseof electronsmovingwithin the covalentbondsof a semiconductormaterial. However, these concerns are beyond the rcalm of circuit theory In this book we usec cuit modelsthat have alreadvbeen develoDed: we do not di.cu\\ho$ compoDent modelsarede\eloped.

1.5 TheIdealBasicCircuitElement

Figure 1.5A Anid€albask circuit ehment.

An ideal basiccircuit el€menthasthree attributes:(1) it hasonly two terminals,which are points of connection to other circuit components;(2) it is describedmathematicallyin terms of current and/or voltage;and (3) it cannotbe subdividedinto other elements. We usethe word deal to imply that a basiccircuit elementdoesnot exist as a realizablephysicalcomponent. However,aswe discussed in Seclion1.3.ideal elementscan be connected in order to model actual devices and systems We use the word Dartcto imply that the circuit elementcannotbe further reducedor subdivided into other elements.Thus the basic circuit elemelts form the bui]ding blocks for constructing circuit models, but they themselvescannot be modeledwith any other type of element. Figur€ 1.5is a rcpresentation of an ideal basic circuit element.The box is blank becausewe are making IIo commitment at this time asto the type of circuitelementit is.In Fig.1.5,the voltageacrossthe teminals of the box is denoted by o, and t}le cunent in tlle circuit element is denoted by i. The polaiity reference for the voltage is indicated by the plus and minus signs, and the reference direction for the curent is shown by the arrow placed alongsidethe current. The interprctation of these rcIercnces given positive or negativenumerical valuesof o and i is sunmarized in Table 1.4.Note that

.u voltagedrop from terminal1 to terminal2

i

voltage rise from terminal 1 ro terminal 2

vollageise ftomterminal2to terminall

voltagedrop from terminal2 to terminal1

positivechargeflowing from terminal1 ro terminal2

positive charge tlowing from terminal 2 to terminal I

negativechargeilowing ftom terminal2 io tenninal1

negative charge flowi.g fiom terminal 1 to termilal2

1.5

TheldeatEasicCiftuitEtement

the notion ofpositivechargeflowingin one directionis equiv algebraically aleni lo the notion ofnegativechargeflowingin the oppositedircction Thc assignmcntsof the rcferencepolarily for voliage and the reftr encedircction for curreni irc entirely arbitrary.Howevel.olrccyou have assignedthe Iefercnces,you must wrilc all subsequentequations to agr;e \vith the choscnreferencesThe most widely Lrscdsign convention applied lo theseretcrencesis called the passivesign conv€niion,which we use throughoul thjs book The passivesign conventioncan bc staled

Whcneverthe referencedirectionIor the currentin an elementis in the directionof the re{ercncevoltagcdrop acrossthe elemcnt(asin Fig.1.5).usea positivesignjn any exprcssionthat relateslhc voltage ro Lhccuff
-l Passivesign convention

We applythis signconvertion in all the analysesthat follo$r Ourpurpose for inlroducirg it even before we havc introduccd the dilTereni lypcsofbasic circuit elenentsis !o impresson )routhe Iact that the selection of polaity referenccsalong with the adoption ot the passivcsign conventior is rol a function of the basicelementsnor thc type oI interconncctionsnadc with thc basic elements.We present thc applicatlon ard interpretalionof the passivesignconventionin power calculationsin Seclion1.6.

Objective2-Know and be abteto usethe definitionsot voltageandcu ent 1.3

The currentat the terminalsof thc clement Fig.1.5is

The expressionfor the chargeeDtedngthe upper t€rminalofFjg.1.5 is

1 i :20e 5ooo'A, r > 0.

Calculatethe total charge(in microcoulombs) enteringthe elementat its upper terninal-

Answer: 4000/..C. NOTE: Also try Chapter Ptoblem 1.9

- ( . r . , le'" c.

Find lhe maximumvalueofthe curlenl entering the terminalifa = 0.03679sr.

Answer: l0 A.

14

1.6 PowerandEnergy Power and energy calculations also are important in circuit analysis.One reason is that although voltage and current are useful variables in the analysis and design of electrically based systems,the useful output of the systemoften is nonelectrical, and this output is conveniently expressedin terms of power or energy.Another reason is that all practical deviceshave Iimitations on the amount of power that they can handle. In the design process,therefore, voltage and current calculations by tlemselves are not sufficient. We now relate power and ene.gy to voltage and curent and at the sametime use the power calculation to illustrate ttre passivesEn convention. Recall from basicphysicsthat power is the time rate of expending or absorbing energy. (A water pump rated 75 kW can deliver more lite$ per secord than ore rated 7.5 kW.) Mathematically, energy per unit time is er-Dressedin the form of a derivative. or

,:'H,.,

Definitionof power>

(1.3)

the power rn watts, the energyin joules, the time in s€conds.

Thus1W is equivalent to 1 J/s. The powei associatedwith the flow of chargefollows directly from thedefinitionof voltageandcurrentin Eqs.1.1and 1.2,or

aw

/ aw\( aq\

dt

\ dq l\dt J'

Power equation>

(1.4)

p : t}le power in watts, 1)= the voltage in volts, i : the curent in amperes,

1.6

Equation 1.4 shows that the pow€r associatedwith a basic ckcuit element is simply the pmduct of the current in the element and the voltage adoss the element. Therefore, power is a quaDtity associatedwith a pair of tei_ minals, and we have to be able to tell from our calculation whether power is being delivered to the pair of teminals or extracted from it. This information comes ftom the corect application and interpretation of tle passive sign convention. If we usethe passivesign convention, Eq. 1.4 is correct if the rcference alirection for the current is in the direction of the rcference voltage drop acrossthe terminals. Otherwise, Eq. 1.4 must be written with a minus sign. In other words, if the current reference is il the direction of a refererce voltage rise acrossthe teminals, the expressionfor the power is

Power andEneqy

+ (b)p = -?)i

(1.5)

'The algebraic sign of power is based on charge movement tbiough voltage drcps and rises.As positive chargesmove ttrrough a drop in voltage,they lose energy,and as they move through a rise in voltage, they gain energy.Figure 1.6 summarizesthe relationship between the polarity rcferencesfor voltage and current and the expressiol for powerWe can now state the rule for interprcting the algebraic sign of power: If the power is positive (that is, if p > 0), power is being deiivered to the circuit iNide the box. If the power is negative (that is, iJ p < 0), power is being extracted from the circuit inside the box.

For example, suppose that we have selected the polarity refeiences shown h Fig. 1.6(b). Assume further that our calculations for the curent and voltage yield ttre following numeical results: j=4A

and D:

1 0V .

Then the power associatedwith the terminal pair 1,2 is

? = _(_10x4): 40w. Thus t}Ie circuit inside the box is absoibing 40 W. To take this analysis one step furthel, assumethat a colleague is solv_ ing the same problem but has chosen the reference polarities shown in Fig.l.6tcl.Theresuldngnumerical\aluesdre i:-4A,

0:10V,

and

p:40w.

Note that interprcting these results in terms of this reference system gives the same conclusions that we proviously obtained-namely, that the circuit inside the box is absorbing 40 W In fact, any of the reference systems il Fig. 1.6 yields tlis sameresult.

,

I;:]

l - - : l

*-113

(c)p=-d

41.-_stl}-l .

L - - - I *----f:--] 1 (d)p = ,i

andth€expr€sion rcferences figure1.6a Potadty

signof power atgebraic { Interpreting

76

objective3-Know andusethe definitionsofpowerandenergyt Objective 4-Be abteto usethe passive sign 1.5

Assumethat a 20 V voltagedrop occursacross an clcmcnt from tcrminal2 to lcrninal l and that a curcnt of4 A entcrstcminal2 a) SpeciJythe valuesof d and t for the polarity ret
Calculatcthc total cncrgy (injoulet dclivered lo the circuit clement. Answer: 20 J.

1.7

Answen (a) Circuit1.6(a):r,= 20V,t: 4A; circuii1.6(b):1]: 20V,i = 4A; c 20V,t = 4A: c i r c u j1t . 6 ( c ) : = circuit1.6(d):0: 20 V. t : 4 A; (b) absorbing; (c) 80w: 1,6

A higi-voltagedirecl curenl (dc) lransmissioll line betweenCelilo,Oregonand Sylmar, C ,litufDiI i. opcrrtirgJt 800kV andcarr\ i19 1800A, asshowr. Calculalelhe power (in megawalts)at the Oregonend ofthe line and statethe directionofpower flow

Assumethat the voltageat the terminalsofthe elementin Fig.1.5correspondingto the currcnt in AssessnentProblem1.3is

kv, 1]: 10e-s00e

I > 0

Answ€r: 1,140MW Celilo to Sylmar.

NOTE: Also try ChapterProblems1.12,1.17,1.24, and 1.26.

Srimnrary 'lhe

InternationalS]'stemof tjnits (SI) enablesengjneers to communicatc in a mcaningful way about quanlilalive rcsults.Tablc 1.1summarizcsthc bascSl unils;Table1.2 prcsentssomcuscill dcrivcdSI units,(Seepages8 and9.) Circuit analysisis basedon the variablesofvoltage and current.(Seepaser1.) Vollageis the erergy per unit chargecreatedbt' charge sepantion and has the SI unit of \olt Q) = dlNldll). (Seepage11.) Cunenl is the rate ofchargetlow and hasthc SI unit of ampere(, : d4ldr). (Sccpagc I l.) The id€albasiccircrit elemenlis a two-terminalcompoDert that cannot be subdivided;ii car be described malheDralically iD tenns ofits terminalvoltageard currenl. (Seepage12.)

The passivesign conventionusesa positivesign in the expressionthal relales lhe voltagc and cu cnt at thc le niials oI an elemenl when thc relcrcncc dircction Ior the cu enl lhroughthe elemenlisin thc dircclionof the relerence vollage drop across the element. (Scc page13.) Pofler is energy per unit of iime and is equal to the productofthe terminalvoltageand curren! it hasthe SI unit of$'att (t : dflrlll = ot). (Seepage1a-) The algebraicsignof power is interpretedasfollows: . If p > 0, power is being deliveredto the circuit or cfcurt component, . If r, < 0, power is beingextracledtuomthe circuil or circuii component.(Seepage15.)

17

Problems Seclion 12 L1 There are apprcximately 250 million passenger vehicles rcgistered ir the United States. Assume that the batteiy in the average vehicle stores 440 watt-hours (Wh) of energy. Estimate (in gigawatt-houis) tlle total energy stored in U.S.passengei vehicles, 1.2 The line desuibed in AssessmentProblem 1.7 is 845 mi in lengtl. The line contains lour conductors, eachweighing2526lb per 1000ft. How many kilo $ams of conductorare in the line? L3 The 4 giga-blte (GB : 10ebytes) flash memory chip for an MP3 player is 32 rnm by 24 rnn by 2.1mm.This memory chip holds 1000thee-minute songs a) How many seconds oI music fit into a cube whosesidesarc 1 mm? b) How many bytes of memory are stored in a cube whosesidesaie 100pm? 1,4 A hand-held video player displays 320 x 240 picture elements(pixels)in eachframe of the video.Each pixel requires 2 bytes of memory. Videos are dis' played at a mte of 30 frames per second.How many minutes of video will fit in a 30 gigabyte memory? 1,5 Some species of bamboo can grow 250 nn/day. Assume individual cells in the plant are 10 pm long. a) How long, on average,does it take a bamboo stalkto $ow 1 cell length? b) How many cells are added in one week, on average? 1.6 One liter (L) oI paint covers approximately 10 m2 of wall. How thick is the layer before it dries? (/{tnr: 1L=1x106mm3.)

35 /rA curent, due to the flow of electrons.What is the averagenumber of elechonsper secondthat flow past a fixed reference qoss section that is perpendicular to the direction of flow? 1,9 The current enteriry tlle upper terminal of Fig. 1.5is i : 24 cos4000tA. Assume t}le charge at t}le upper terminal is zerc at the instant the current is passingthrough its maximum value.Find the expressionfor q(t). L10 How much eneryy is extracted ftom an electron as it flows through a 6 V battery from tlle positive to tbe negarivererminal: F\pfess )ouf answef iD attojoules. Sections1.F1.6 1,11 Ore 9 V battery supplies 100 rA to a camping flashlight. How much energy does the battery supply in 5 h? 1,12 Two electiic circuits, repiesented by boxes A and B, are connectedas shown in Fig. P1.12.Therefer ence dhection foi ttre current i h the inteiconnection and the reference poladty for the voltage ?.r across the inter connectiotr are as shown in the figure.For eachof the following setsof nume cal values,calculate the power in ttre interconnection and state whether the power is flowing ftom A to B or a) i:5A, b) i: 8A, c) i:16A, d) t:

u: l20Y r:25OY t,: -1s0v

10A, 0:

480V

figuruP1.12 Section 1.4 A curent of 1200A exists in a copper wfue,wrth a circular cioss-section (iadius = 1.5 nrm). The current is due to free electrons moving through the wfue at aIt average velocity of o meten/second If the concentntion of free electrons is 1de elecrrons per cubic meter and iJ they are uniformly dispersed tbroughout the wire, then what is the averagevelocity of an electon? 1.8 In electronic circuits it is not unusual to encounter currents in the micioampere iange. Assume a

1.13 The references for the voltage and cu ent at the terminal oI a circuit element are as shown in Fig.1.6(d).ThenumedcalvaluesIor o andt are 40V and 10A a) Calculate the power at the teminals and state whether the power is being absorbedor deliveredby the elementin rhe box.

18

va'iabtes Circuit b) Given that the current is due to electronflow, statewhetherthe electronsare enteringor leaving terminal2. c) Do the electronsgain or loseenergyasthey pass thmugh the elementin the box?

1.14 RepeatProblem1.13with a voltageof

60V.

1.15 Wlen a car hasa dead battery it can often be started by connectingthe battery from another car acrossits terminals. The posilive terminals are connected together as are the negative terminals.The connection is illustrated in Fig. P1.l5. Assume the crment i in Fig.P1.15is measuedandfound to be 30A. a) Which car hasthe deadbattery? b) If this connectionis maintainedfor 1 min, how much energyis transferredto the deadbattery'?

a) Find tlre power absorbedby the element at 1=10ms. b) Find the total energyabsorbedby the element. 1,19 The voltage and current at the terminals of the circuit elementin Fig.1.5are shownin Fig.Pl.19. a) Sketchthe power versuslplot for 0 < t < 50 s. b) Calculalethe energydeliveredto the circuit ele ment at t : 4, 12, 36,ard 50 s" fiqureP1.19

1](v) t0

Figure P1.15 2 0 2

8 1

8 10

1,16 The manufacturer of a 9V diy-cell flashlight battery saystlat the battery will deliver 20 mA for 8u conrinuoushours.During thal time lhe vollageqill drop tromq v lo o V Assumethe drop in !ollageis linear witl time.How much energydoesthe battery deliver in tlis 80 h interval? 1,17 The voltage and current at th€ t€rminals of the cir cuit elementin Fig.1.5are zerolor , < 0. For , > 0 they are ?) = e

s00,

i : 30

?-1500! v,

]0 Il.4 0

2024'2832364044

s2 s6 r(s)

-1.0

mA 40e56'+ 10e-1s00i

a) Find the power at r - 1 ms. b) How much energyis deliveredto the circuit element between0 and 1 ms? c) Find the total energydeliveredto the element 1.18 The voltage and curent at the terminals of fte cir 6trG cuit elementin Fig.1.5aie zerofor I < 0.Fort > 0 they are

t' = 400eroo'sin 200t V, j = 5e-1m'sin200tA.

1.20 lhe voltage and cu ent at the terminals of t}le cir 6pre cuit elementin Fig-1.5are zerofor 1 < 0. For I > 0 r):75 - 75a1oauv, j = 50e rm/mA. a) Find the maximumvalue ofthe power delivered to the circuit. b) Find the total energydeliveredto the element.

Probt€ms19 1:1 The voltage and current at the teminals of the eleP*r€ ment inFig.1.5 are 36 sin 200rt V,

r':

1,23 The voltage and currcnt at t}Ie teminals of the ciitsi'c crit element in Fig. 1.5 are zero for t < 0. For t > 0 they are

i = 25 cos 2007t A. 1)- (16,000r+ 20)e-eo'V,

a) Find the maximum value of the power being delivered to the element. b) Find t]le maximum value of the power beirg erlracted frcm the element. c) Find the aveiage value of p iJl the interval 0
L22 The voltage and cu:rent at the teminals of an auto6PIcrmobile battery du ng a charge cycle are shown in fi$.P1.22. a) Calculate tne total charge tmnsfered to the battery b) Calculate the total energy transferred to the battery.

Figure P1.22 t,(v) 12 10

t = (128r + 0.16)?{0' A. a) At what instant of time is ma-\imum powel delivered to the element? b) Find the maximum power in watts. c) Flnd the total energy delivered to tlle element in millijoules.

1.24 The voltage and clment at the terminals of tlte cir6flc cuit elementin Fig.1.5are zerofor t < 0andt > 3s are In rhejniervalbelween0 and 3 s lbe er?ressjons b:t(3

t)V, 0
i=6-4tr,4.,0
6 4

z 12

lo

20

l(ks)

(A)

16 14 12 10 8 6 4 2 1,2

t6

20 r(kt

1.25 The voltage and curent at the terminals of the cirE Ic cuit element ir Fig. 1.5 are zero for t < 0. For t > 0 mey are + s)eroo'V, ?)= (10,000r

1> 0;

t = (40r + 0.05)e4orA,

r > 0.

a) Fird the time (in milliseconds) when the power deli\eredlo lhe ciJcuitelemeDlis maximum. b) Find the maximumvalue olp in milliwatts. c) Find the total energy delivered to the circuit element in millijoules.

20 '.26 The numerical values for the currents and voltages in the circuit in Fig.P1.26are givenin Table P1.26. Find the total power developed in the circuit P1.26 Figure

5.0 2.0 3.0 -5.0

150 250 200 400

1.0

50 350 400

I

-2.0

g

350

6-0

P1.26 TABI.E Etemenl

Currenr (A)

Yoltage (n\7)

150 150

0.6

100 250 300

-0_8 -0.8

rigureP1,28

2.0 7.2

300

f

1.28 The numerical values of the voltages and currents in the interconn€ction seenin Fig. P1 28 are grven in Table P1.28. Does tlle interconnection satisfy t]le Powercheck?

4 fl3

r27

Assumeyou are an engineerin chargeof a project and one of your subordinate engineen reports that the interconnection in Fig. P1.27 does noi pass the power check. The data for the interconnection are givenin TableP1.27. a) Is the subordinate corect? Explain your answer' b) If the subordinate is corect, can you find the error in the data? P1.27 Figure

,

+

Elem€nt

.

L

l

+

Volt.ge (\,)

36

250 -250

2a

t

108 32

+ f g

48

h j

80 80

100 150 350 200 - 150 -300

129 One method of checkirg calculationsinvolving interconnectedcircuit elementsis to see that the total power deiivered equals the total power absorbed (conservation-of-energy principle). Witl this thought in mind, check the interconnection in Fig. P1.29and statewhether it satisfiesthis power check.The current and voltagevaluesfor eachelement arc given in Table P1.29.

1.30 a) In the circuit shown in Fig. P1.30, identity which elemenls are'absoibing power and which are detilefiDgpower. usi0g the pa$ive sign conventron. b) The numerical values of tle currents ard voltagesfor each element are given in Table P1.30. How much total power is absorbed and how much is delivered in this circuit?

Figure P1.29 figureP1.30

!

,t" "l

1.6 2.6 1,.2 1.8 f g j

80 60 50 2t)

1.8 3-6 3.2

30 40 30 -20

2.4

30

L

+

Elemenl

volraee(mv)

b

300 -100 200 2M 350

I g h

200 250 50

!

c

+

Curenl (irA)

25 10 15 35 25 10 35 10

Elements Circu'it Thereare five ideal basic circuit elements:voltascsourccs. 2.1 VoLtage and CunentSourcesp. 24 2.2 ELectric;L Resistance (Ohmt Law) p. 28 2.3 Con5trudionof a Circuitl4odetp. 32 2.4 Kir€hhoff'slaws p. .t6 2.5 Anatysisof a CircuitContainingDependent

Undertandth€ symbots for afd the behaviorof the fottowifgideatbasiccircujtel€ments; jndependent voltag€andcuir€ntsources, d€pefdertvotiageaid cuiientsourc€s, and Beabteto stat€0hmt taw,Kirchhoffscurrent ta& and (irchhoffsvottagetaw andbe abteto usetheselawsto anatyze simpt€cjrcujtr. Knowhowto caLculate th€ powerfor €ach elementjf a simpl€circujtandbe abteto detefmjiewhetheror not the powerbatarces

22

cunent sources,r'esistors, inductols.and capacito$.In this chaptcr wc discussthc chaaactcristicsof voltage sources,current Altlrough this ma,yseemlike a small num sources,and r-esistors. pracber of elementswith which to beginanalyzingcircuits,mal1y tical systemscan be modclcdwith iust sourcesand resistors.They are alsoa LlseftLl startingpoint becauseo[ their relativesimplicity; thc mathcmaticalrclationshipsbctwcen voltage and current ill sourcesand resistorsare algebraic.Thusyou will bc ablc to bcE:in leaning the basic techniquesoI cjrcuit analysiswith only algebraic manipulations. We will postponeintroducinginductols and capacitorsuntil Chapter6, becausetheir userequiresthal you solveintegral and However. the basicanalyticaltechniqucs differential ecluations. for solvingcircuitswith induclols and capacitorsare the sameas those introduced in this chapter.So, by the time you need tcr begirlmanipulatingmore difficult equations,you shouldbe vcry familiar with the methodsof writirg them.

PracticaI Perspective Etectrical. Safety

"Danger-High VoLtage." Thiscommonty seenwarning is misteading. Attformsof energy, including etectricaL energy, can gut ifs not ontythe vottagethat harms. be hazardous. The staticetectriciry shockyou receive whenyou watkacross a carp€t andtoucha doorknob is annoying butdoesnotinjure. Yetthat sparkis caused by a vottagehundr€ds or thousands oftimeslargerthanthevoltages that cancause harm. Theetectricat energy that canactuatty cause injuryis due to etectricai currentandhowit ftowsthroughthe body.Why, then,doesthesignwarnof highvottage? Eecause ofthe way poweris produced eLectrical anddjstributed, it is easierto determjrevoLtages than cuffents.Atso, most etectdcaL produce sources constant, specified voltages. So the signs warnaboutwhatis easyto measure. Determining whether andunderwhatcondjtjons a sourcecansupptypotentiaLty dangerous currents is moredifficutt,asthisrequjres anunderstanding of etectrical engineering. Before wecanexamine thisaspect of eLectricaL safety, we haveto learnhowvoltages andcurrents areproduced andthe reLationship between them.Theetectrical behavior of objects,

suchas the humanbody,is quite comptexand often beyond comptetecomprehension. To atLowus to predjctand controt phenomena, eLectfical we usesimpLiryjng modelsjn whjchsimple mathematjcaL retationships betweenvottageand current jn reaLobjects.Suchmodapproximate the actuaL retationships elsandanalyticaL methods formthe coreofthe electrjcaL engineering technjques that wjLlattow usto understand attetectrjcaL oheromena. incldding oseretdLirgto electdcaI safety. At the end of this chapter,we witt usea sjmpteetectric circuitmodelto describehow andwhy peopleareinjuredby etectriccurrents,Eventhoughwe rnayneverdeveLop a compteteand accurateexptanation of the etectricaL behaviorof the humanbody,we can obtaina ctoseapproximation usjng simptecjrcuit modeLs to assessand improvethe safetyof etectrjcalsystemsand devices.DeveLoping modetsthat pro, videan understanding that h jmperfectbut adequate for solvinq practicalprobtems lies at the heartof engineering. I4uch of the at of etectricat engineering, whichyou wil[ Learnwith js in knowingwhen and how to sotvedifficutt experience, probtems by usingsimptiryingmodets.

23

24

2.1 VoltageandCurrentSources ideal voltageand curent sources,we need to consider Befoie aliscussing the general nature of electncal sources.An elechical sourc€ is a devjce that is capable oI converting nonelectric en€rgy to electric energy ard vice versa.A discharging battery converts chemical energy to electric energy, whereas a battery being charged converts electric energy to chemical energy.A dynamo is a machhe that conveits mechanical energy to electdc mode,it is energyandvice versa.If operalingin the mechanical-to-electic energy, it is to mechanical ftom electric called a generator. If transforming about these to remember thing referred to as a motor. The impotant sourcesis tlat theY can eitler deliver or absorb elect c power, generally maintaining either voltage or current. This behavior is of particular interest for circuit analysisand led to the creation of the ideal voltage sourceand the ideal current sourceas basiccircuit elements.Thechallenge is to model practical sourcesin telms of the ideal basic circuii An ideal voltage source is a circuit element that maintains a prescibed voltage across its terminals regardless of tle curent flowing ilt those terminals. Similarly, an idesl cuEent source is a circuil element that maintains a presoibed current thmugh its terminals regardlessof the volf age aooss those teminals. These circuit elements do not exist as practical devices-they are idealizedmodelsof actualvoltageand currentsources. Usingan idealmodelior currenlandvollagerourcesplacesan important restriction on how we may descdbe them mathematically. Becausean ideal voltagesourceprovidesa steadyvoltage,even if the current in the it is impossibleto speciry'the current in an ideal voltage elementchanges, source as a function of its voltage. Likewise, iJ t}le only inlormation you have about an ideal ctment sourceis the value of curent supplied,it is impossibleto determinethe voltageaooss that current sourceWe have sacrificed our ability to relate voltage and current in a practical source for the simplicity of using ideal sourcesin circuit analysis. Ideal voltageand curenl sourcescan be further describedas either independent sourcesor dependent sources An independent sowc€ estab lishes a voltage or current in a circuit without relying on voltages or cur rents elsewherein the circuil. The value of the voltage or current supplied is specifiedby the value of the independentsourcealore. In contrast'a avoltageor curent whosevaluedependson dep€ndenlsouce establishes tie value of a voltage or current elsewherein the circuit. You cannot spec,A ( - ) ify the value of a dependent source unlessyou know the value of the volt ageor currenron whichil depends. The circuit symbolsfor the ideal independentsourcesale shown rn Fig.2.1.Note that a circle is usedto rcpresentan independentsourceTo completely specify an ideal independent voltage source in a circuit, you (b) G) must include the value of the supplied voltage and the reference polarity, ideal independforG)anideatinde- as shownin Fig.2.1(a).Similarly,to completelyspecifyan Figure2.1^ Ihecircuiisymbob pendent vothge source and(b)anjdeatindependent ent cuirent source,youmust includeth€ valueofthe suppliedcurrentand its rcferencedirection,asshownfuFig.2.1(b).

Y L

(b Y

2-1 Vottag€ andCunent Sourc€s 25 The circuit symbolsIor the ideal dependentsourcesare shown in Fig. 2.2.A diamond is used to representa dependert source.Both the depeDdentcurrent sourceand the dependentvoltagesourcemay be controlled by either a voltage ot a current elsewherein the circuit, so there are a total of four variations,as indicated by the symbolsin Fig. 2.2. Dependent sourcesare sometimescalled controlled sources. To completely specify an ideal dependent voltage-controlled voltage source!you must identify the controlling voltage,the equationthat permits you to compute the suppliedvoltage ftom the controlling voltage, and the referencepoladty for the suppliedvoltage.ln Fig.2.2(a),tlrecontrollirg voltage is named o,, the equation that determinesthe supplied

I, = Itrx.

? t G)

( c)

?

and the referencepolarity for ?r is as indicated.Note that ri is a multiplying constantthat is dimensionless. Similar requircments exist for completely specifying the other ideal (b) (d) dependentsources.InFig.2.2(b),the controllingcurrcnt is i' the equation Figure2.2 A Thecircujtsymbols for G)an ideat for the supplied voltage ," is Ds = p1\,

the relerencepolarity is as show4 and the multiplyingconstantp hasthe dimensionvolts per ampere.In Fig. 2.2(c),the conrroling volrageis o-, the equationfor the suppliedcurrentrsis rs:

dl)r,

the referencedirectior is asshown,and themultiplying constantd hasthe dimensionamperesper volr. In Fig.2.2(d),the controling currentis j,, the equation for the supplied current i, is i, = Bi,, the referencedirection is as shown,and the multipllng constantp is dimensionless. Finally, in our discussionoI ideal souices,we note that they are examplesof activecircuit elements.Anactiveelementis one that models a device capableof generatingelectdc energy.Passiveetementsmodel physicaldevicesthat cannot generateelectric energy Resistors,inductorE and capacitors are examples of passivecircuit elements.Examples 2.1 and 2.2 illustrate how the characteristicsof ideal indepefldent and dependentsourceslimit the t'?es ofpermissibleinterconnectionsofthe

dependent voltage-controthd vottas€ (b) an source, jd€atdependent current-controtted vothgesoufte,(c)an jdeatdependent vottage-conhoLled currcnisource, and (d) anjdeatd€pendent cunent{ontrolLed cunentsource.

26

C cuitEtements

TestingInterconnections of ldealSources

Using t}le defidtions ofthe ideal independentvoltage and current sources,state which interconnections in Fig. 2.3 are pemissible and which violate tlle constraints imposed by t}le ideal sources.

10v

10v

Solution Conneclion(a) is valid. Each sourcesuppliesvolt age aooss the samepair of terminals,marked a,b. This requires that each sourcesupply the samevoltagewiti the samepolarity,which they do. Connection (b) is valid. Each source supplies current tbrough the same pair of terminals, marked a,b.This requiresthat eachsourcesupplythe same currentin the samedirection,which they do. Conrection (c) is not permissible. Each source supplies voltage across the same pair of terminals, marked a,b.This requires that each sourcesupply the samevoltage with the samepolarity, which they

(b)

+\

LOV

/+

5V

G)

Connection(d) is not permissible.Each source suppliescurrenr througl the samepair oI teminals, rnarked a,b.This requiresthat each sourcesupply the same current in the same direction, which they Connection(e) is valid.The voltagesourcesupplies voltage acrossthe pair of terminalsmarked a,b.The current sourcesupplies curent through the same pair of terminals.Becausean ideal voltage sourcesuppliesthe samevoltageregardlessof the current, and an ideal curent sourcesuppliesthe sane currentregardlessof the voltage,thisis a permissibleconnection.

fortuampte 2.1. Figur€2.3a Thec'rcuits

(d)

2.1

Voltage andCurentSources

TegtingInterconnections of IdeatIndependent andDependent Sources

Using the definitions of the ideal independent and depende0r sourcer. sralewhichiDterconneclrons in Fig. 2.4 are valid and which violate the consuamN imposedby the ideal sources.

T\\

|),=3r,,

Solution b

Connection (a) is invalid. Both the independenr souce and t])e dependent souice supply voltage across the same pair of terminals, Iabeled a,b.This requircs that each source supply the same voltage with the samepolaity.The independent sourcesupplies 5 Y but the dependentsouce supplies 15V Connectiotr (b) is valid. The independent volragesource suppliesvoltage acioss the pair of terminals maiked a,b. The dependent current source suppliescurent through the sarnepair of terminals. Becausean ideal voltagesourcesuppliesthe same voltage regardless of cuirent, and an ideal currert source supplies the samecurent regardlessof voltage,this is an allowable contrection. Connection(c) is valid. The independenrcurrent source supplies crment tlrough the pait of termhals marked a,b.The dependentvoltage source supplies voltage aqoss the same pafu of terminals. Because an ideal cufient source supplies the same current regardless of voltage, and an ideal voltage source supplies the same voltage regaidless of currcnt, this is an allowable connection. Connection(d) is invalid. Both the irdependent sourceand the dependent sourcesupply cunent through ttre samepair of terminals, labeled a,b.This requkes that each source supply the same curent in the same reference dircction. The independent souce supplies 2 A, but the dependent source suppfies 6 A iII the opposite direction.

(.)

r ) /", ="3", , ( t +\

b (b)

t)

'.=4r.

G)

I

T i, b (d)

Figure2.4lr Thecircuitstor Examph 2.2.

28

Objective1-understand ideal basiccircuit etements 2.7

2.2

For the circuit shown, a) what vdl e ofzr is rcquired in order for the interconnectionto bc valid? b' for lhi. valu

For lhc circtlil shown, a) What valueof d is requiled in order for the interconncctionto be valid? b) For the valueof a calcularedin part (a), fhd with the 25 V sonrcethe power associated

Answer; (a) 0.6 A/Vl (b) 375 w (37sw absolbed)

Answer: (a) -2 V; (b) 16 w (16 w delivered).

NOTE: Abo try ChuptetProblens 2 2 dnd2 3.

fiesista$ee{$hm's!-aw} 2.2 El.cctfiea{ +\\.-

R

a resktorhavinga figur€ 2.5 :" Thecjrcuitsynrbotfor

Resistrnc€is lhe capacityof mateials to impede the llow ot curcnt or' thc flou of clectdc charge.ll]c circuil clement usc'l to more specifically, model ihis behavioris the resistor-Figure2 5 showsthe circuit symboltor Lhcresistor,with R denotingthc resislancevalueof the resistor' Conceptually,we can undcrstandresistarceif we think aboul the moving electronsthat nake up clectriccurrentiltcracting with and being resistedby the atonic structurcof the rnaterialthrough which they are noving. In lhe cou6c of thesc interactions,somc amount of electric eDergyis conr,ertedto thcrmal encrgyand dissipaledin the [om1 of heat' Tlis cfTectmay be undeslrable.However,many uselu] electdcaldevices take advantageot resislancehealing.includingstoves.toasters,rrons.and Most marerialsexhibit measumbleresislanceto current The amouDl of resistancedcpendson the matcrial Metals such as copper and alu minlrm have smal1values of resistance,making then good choiceslbr wiriig usedto conductelcctriccurcnl ln fact' when rcpresenledin a cir cuit diagram.copperor aluminumwiring isn't usuallymodeledas a resN tor; thc resistanccof the wire is so small compared1()the resistanceoI other elementsilr the circuit that we can neglcctthe lYiringresistanceto simplify 'For the diagran. purposesof circuil analysis.we musl rcference the currcnt in the resistorio the tcrminal voltage.We can do so in two waysieilher in

2.2

ELectricat Resktance (ohmtLaw)

29

the direction of the voltage drop acrossthe resistor or in the direction of the voltage rise acrossthe resistor,as shown in Fig. 2.6.If we choose the former, the rclationship betweenthe voltage and current is

?r- iR,

(2.1) .{ 0hm'staw

il

r : the voltagein volts,

^ 'l

i = the curent in amperes, r=iR

R : the resistancein ohms,

r=

,R

Figure2.6.q Twopossibte r€ference choices forthe drrentandvottage atthe ternrinals of a resistor, and

If we choosethe secondmethod,we must wnte u = -iR,

:1

\2.2)

where r,, j, and R are, as before,measuredin volts,amperes,and ohms, respectively. The algebmicsignsusediII Eqs.2.1 and2.2are a directconsequerceof the passivesignconvention,which we introducedin Chapter1. Equations 2.1 and 2.2 are known as Ohm's law after Georg Simon Ohm, a German physicistwho establishedits validity early in the nineteenth century Ohm's law is the algebraic relationship between voltage 8r} and current for a resistor. In SI units, resistanceis measured in ohms.The Grcek letter omega (O) is the standard symbol for an ohm. The circuit Flgure2.7d. Thecircuit symboLfor an8 O rcsjstor. diagramsymbolfor an B O resistoris shownin Fig.2.7. Ohm's law expressesthe voltage as a function of the current. However, expressingthe curcnt as a function of the voltage also is convenient.Thus, frcm Eo.2.1. (2.3) or,fromEq.2.2, 12.4)

Thereciprocal of theresistance is rcferredto asconductance, is sym(S).Thus in siemens bolizedby theletterG, andis measured " = * "

\2.5)

An 8 O resjstor has a conductancevalue of 0.125S In much of tie profes sionaliiterature, the ur t usedfor conductanceis the mho (ohm spelledbackward), which is symbolized by an inverted omega ((J). Therefore we may also descdbean 8 O resistor ashaving a conductanceof 0.125nho, ((').

30

Circuit Etements We use ideal resistors in circuit analysis to model the behavior of physical devices.Using the qualifier ideal rem'inds us that the resistor model makes several simplifying assumptions about the behavior of actualresistivedevices.lhe most impo ant ofthese simplifyingassumptions is that the iesistanceof the ideal resistoris constantand its value aloesnot vary over time. Most actual resistive devicesdo rot have constant resistance. and their resistance does vary over time. The ideal rcsistor modelcanbe usedto representa physicaldevicewhoseresistancedoesn't vary much from some constant value over the time period of interest in the circuit analysis.Inthis book we assumethat the simplifyhg assumptions about resistanc€devicesare valid, ard we thus use ideal resistors in circuit analysis. We may calculate the powel at the termhals of a resistor in several ways.The first approach is to use tle defining equation and simply calculate the product of the terminal voltage and curent. For the refeience systemsshoM in Fis.2.6.we x'dte \2.6)

w h e n o= i R a n d (2.7) when1' = -iR. A secondmethodof expressingthe power at the terminalsol a resistor expresses power in terms of the currcnt arld the resistance. SubstitutirgEq. 2.1into Eq. 2.6,weobtain p=ui=(iR)i

Power in a resistorin termsof currentb'

P=i2R

(2.8)

Likewise,substitutingEq.2.2 into Eq.2.7,we have p = -1)i : -(-i

R)i = i2R.

(2.e)

Equations2.8 and 2.9 are identicaland demonstrateclearly that, regard_ lessof voltagepolarity and currentdirection,the powel at the terminalsof a resistor is positive. Therefore, a resistor absorbspower from the circuit. A tlird methodof expressingthe power at the teminals of a iesistor is in terms of the voltage and resistance The expressionis independent of the Dolaritv references.so

Power in a resistorin tennsof vottageF

p

R

(2.10)

?.2

Etectricat (0hm'sLaw) Resistance

31

Sometimesa resistor'svaluewlll be er?rcssedasa conductance rather than as a resistance.Usirg tlle relationship betweer resistance and conductancegivenin Eq.2.5,we may also wrire Eqs.2.9 and 2.10in lerms of the conductance, or i,

(2.11) l?.12)

Equations2.G2.12providea vadety ofmethodsfor calculatingthe power absorbedby a resistor. Each yields the sameanswer.In analyzing a circuit, look at the informationprovidedand choosethe power equationthatuses lbal inlormaliondirecll). Example2.3 illustratesthe applicationof Ohm,s law in coniunction with an ideal sourceand a resistor.Powercalculationsat the terminalsofa iesistor also aie illustrated.

Cal.cutating Vottage, Current,andPou/er for a SimpteResistive Circuit Ir eachcircuitin Fig.2.8,eittrerthe valueof z or, ts

I ) 1 A 1 , , ,8 o

50v

rb= (50x0.2)- 10A. The voltager). in Fig.2.8(c)is a dse in rhe direction of the current in the rcsistor. Hence

(a)

f ) r a " .: o o

0.2s

The currenti, in the resistorwith a conductance of 0.2 S in Fig. 2.8(b) is ir the direction of rhe voltagedrop acrossthe resistor.Thus

(b)

(

)50v

(c)

0c= -(1)(20)= 20v.

25n

'.1

(d)

tigure2.8A Thecircuih forExampL€ ?.3. a) Calculatethe valuesof 0 and i. b) Determineth€ power dissipatedin eachresistor.

The curent id in the 25 O resisrorin Fig.2.8(d) is in the direction of the voltagedse affoss the resistor.Therefoie

. 5 0 ' 2 5 b) The power dissipatedin eachof the four resistorsis

r8l ( 1 ) 1 8=) 8 w . r s o= ; = 500w, pors: (s0),(0.2)

Solution a) The voltageoain Fig.2.8(a)is a drcp in the direction of the curent in the resistor.Therefore,

,o : (1X8)

I 7n\) = 20w, r:on= _fr = t1t(20)

{50)2 = ( 2)12s)- 100w. PL'a= i-

of ideal sourcesand resisHaving introducedthe generalcharacteristics lors,wc next showhow to setheseelementsto build the circuit model of a practicalsystem-

objective2-Be abteto state and useohm's Law. . . 2.3

For the circuit shown.

For the circuit shown,

a) Ifos = 1 kV and ls = 5 mA, find the vahre of R and lhe power absorbed by tlre resistor b) Ifis : 75 mA and thc powcr dcliveredby the voltagesourceis 3W,find os,n, and th€ power absorbedby the rcsistor, c) If R - 300 O and the power absorbedby R is 480 mW. find ts and z,g.

a) ff is = 0.5A and G : 50 mS,find ?s and t}repower deliveredby the curent source b\ lr ds - l5 V rnd lhe poqcr Lleli\efed lo rhe conductoris 9 W.find the conductanc(G and the sourcecurrentis. c) If G : 200pS and thc power deliveredto the conductanceis 8 W find is ard z's.

r)

G

Answer: (a) 10V5 W; (b) 40mS,0.6A; (c) 40mA,200v

Answer: (a) 200kO,5 W: (b) 40v s33.330,3 W; (c) 40mA,i2V 2.6and2.8. NOTE: Alsotty ChapterProblems

2.3 eonstruction cf a eireuitiv!*dei We have alreadyslatedlhat onc reasonfor an interestin the basiccircuit elementsis thal they can be uscd 1o constructcircuit modelsof practical systens.Thcskill rcqujred1odevelopa circuitmodel ofa deviceorsysten is as conplex as thc skill rcquired to solvethe derived circuit.Although thc skills requiredto solvecircuits,I'oualsowill need this text emphasizes other skills in the practiceof clcctricalengineering,and one of the most impoilanl is modeling. We deveiopcircuit modelsin the next two cxamples.ln Exampie2.1 we construcla circuit modei basedon a knowledgeofthc bchaviorof the system'scomponentsand how the componenlsare intcrconncctcd.ln Example2.5we createa circuitmodelbymeasuringthc tcrminalbehavior

2.3

Consiiuction ofa Cncuitllodet

Constructing a CircuitModetof a Ftashtight Consrructa circuitmodel oia flashlight.

SoLution We chosethe flashlighl to illustrate a practical system becauseits components are so familar. Figure 2.9 shows a photograph of a widely available flashlight. When a flashlight is regarded as an electrical system,the componentsof primary interestare the batteries, the lamp, the connector, the case,and the switch. We now consider the circuit model for each A dry-cell battery maintahs a reasonably constart teminal voltage if the current demand is rot excessive.Thus if the dry-cell battery is opemting witlin its intended limitq we can model it wrth an ideal voltage source.The prescribed voltage ther is constant and equal to the sum of two dry-cell value$ The ultimate output of the lamp is light energy, which is achieved by heating the filament in the lamp to a tempeiaturehigh enoughto causeradia tion in the visible range. We can model tlte lamp with an ideal resistoi. Note in this casethat although the resistor accounts for the amount of electric enelgy converted 10 thefitlal energy,it does not predict how much of the thermal energy is converted to liglt eneigy.The rcsistor used to represent the lamp does predict the steady current dmin on the batteiieE a charactedsticof the systemthat also is of inteiest.In thismodel,Rr symbolizestle lamp resistance. The connector used in the flashlight se es a dual role. First, it provides an electdcal conductive patll between the dry cells and the case.Second,it is formed into a springy coil so that it also can apply mechanical pressure to the contact between the batteries and the lamp.The puipose of this mechanical pressureis to maintain contact between the two dry ce s and between the dry cels and the lamp. Hence, in choosing the $rirc lor the connector, we may find that its mechanicalprope ies are more important than its electrical properties for the flashlight design. Electdcally, we can model the €onnector with an ideal resistoi, labeled Rr. The case also se es both a mecharical and an electdcal purpose Mechanicaly, it crntains all the other componentsand provides a $ip for the peNon usingit. El€ctrically,it provides a connectionbetween

Flglre 2.9 A A flashtight canbeviewed asanetectricat system.

other elementsin the flashlight. If the caseis metal,it conductscurent betweenthe batteries and the lamp. If it is plastiq a metal strip inside the case connects the coiled connector to the switch. Erther way, an ideal resjstor,which we denote &, models the electri, cal connectionprovided by the case. The final component is the switch. Electically, the switch is a two-statedevice.It is either oN or oFF.An ideal switch offen no resistanceto the current when it is in the oN state, but it offers iDfinite resistance to curent when it is in the oFF state. Thesetwo statesrepresentthe limiting valuesof a resislor: rharic,rheO\ slalecoffespoDds ro a resi, tor with a numedcal value of zero, and the oF! state correspolds to a resistor with a numerica] value of infinity. The two extreme values have the descriptive names sho circuit (R = 0) and open circuit (R : m).Figure 2.10(a)and (b) showttregraphical representation of a short circuit and an open circuit, respectively. The symbolshown in Fig.2.10(c)represents the fact that a switch can be either a shofi circuitoran open circoit,dependingon the position oI its contacts,

34

Etements Circuit

(b) OFF

*/-

ON

G) (b)openci(uit. cncuit. Figure2.10A Circuit synbols. G)Short G)swiich.

We now construct the circuit model of the flashlight.Staiting with the dry-cell balteries,the positive terminal of the first cell is connectedto the negativeterminalof the secondcell,asshown1n Fig.2.11.Thepositiveterminal of the secondcell is connected to one terminal of the lamp. The oth€r teminal ofthe lamp makescontactwith one sideof lhe .sirch,and lhe other.ide of lhe s\\ilchis connect€dto the metalcase.Themetal caseis then con nectedto the negativeteminal of the fust dry cell by mcans of the metal spring.Note that the ele ments form a closed path or circuit. You can seefte closed path formed by the connected elements in Fig.2.11.F$re 2.12showsa circuit model for the flashlight.

offtashtight components. Figure2,11 ,s Thearangement

a flashtight. Figure2.12 A A ciftuitmodettor

We can make some generalobservationsabout modeling hom our flashlight example:First. in developing a circuit model, the eli?cr'icdlbehavior of each physical component is of primary interest. In the flashlight model, three very diJferent physical components a lamp, a coiled wire, and a metal case-are all representedby the samecircuit element (a resistor), becausethe electrical phenomenon taking place in each is the same Each js presenting resistanc€10the current flowing through the circuit Second.circuit modelsmay need to accountfor undesiredas well as desiredelectricaleffects.For example,the heat resultingftom the resist ancein the lamp producesthe light, a desiredeffect.However,the heat resultingfrom the resistancein the caseand coil representsan unwanted or parasiticeffect.It drainsthe dry cellsand producesno usefuloutput Suchparasiticeffectsmust be consideredor the resultingmodel may not adequalelyrepresentthe system. And finally,modelingrequiresapproximation.Even for the basicsystem representedby the flashlight,w.3made simplifying assumptionsir developjngthe circuit model.For example,we assumedan ideal switch,

2.3

Construction ofa Ciftuit!1odet

35

but in practicalswitches,contactresistancemay be high enoughto interfere with proper operationofthe system.Our model doesnot predicttlis behavior. We also assumedthat the coiled connector exerts enough pressureto eliminateany contactiesistancebetweenthe dry cels. Our model does not predict the eflect of hadequate pressure.Our use of an ideal voltage source ignores any internal dissipation of energy in the dry cells, which might be due to the parasiticheatingjust mentioned.We could account foi this by adding an ideal resistor between the source and the lamp resistor.Our model assumesthe internal lossto be negligible. In modelhg the flashlightasa c cuit,we had a basicunderstandingof and accessto the internal components of the system.However, sometimes we know only the terminal behavior of a device and must use this infor mation in constructingthe model.Example2.5 exploressucha modeling problem.

Construding a Cir.uitModelBasedon Terminat Measurements The voltage and current arc measued at the tenninals of the device illushated in Flg. 2.13(a),and the values oI o! and ir are tabulated in Frg. 2.13(b). Constructa circuit model of the deviceiiside t}le box.

,, (v) t (A) -40

-10

20

-5

0 20 40 (")

5

10

Solution Plotting the voltage as a function of the cunent yieldsthegraphsho*nir Fig.2.14(a).The equation of tle lille in this figue illustratesthat the termhal voltageis directlyproportionallo the terminalcurrcnt, t'! = 4i. In tems of Ohm's law, the device insidethebox behaveslile a 4 ( I resistor.Therefore, lhe cfcuir modelfor lhe de\iceiniide rhe bo-\is a 4 O resislor, asseenh Fig.2.14(b). Wecomebackto this techniqueof usingterminal characteristics to constmcta circuit modelafter i n t f o d u c i n g K i r c b h o f fs l a $ s a D d c i r c u i r a n a l y s i s .

(b)

Figure2.13 A lhe (a)device and(b)datator Exampte 2.5.

r,,(v) ,10 20 20 40

G.)

(b)

Flgure2.14 ^ (a)Thevatues ofor versus t, forthedevicejnFig.2.13.(b)Ihe cncLrit modet forthedevjce in Fi9.2.13.

NOTE:

Assessyow undetstunAing ofthk

efumpk by tyine Chapter Prcbkms 2.10 and2.11.

36

Cncuit ttements

2.4 Kirchhoff'sLaws

a R 1

b

R "

Flgure2.15l Circuitmodetofthe flashtight with assigned voltage andcurcntvariabtes.

A circuit is said to be solved when tle voltage across and the current in everyelementhavebeendetemhed. Ohm\ law is an important equation for deriving such solutions. However, Ohm's law may no1 be enough to provide a complote solution. As we shall see in trying to solve th€ flash light circuit from Example2.4,we need to usetwo more important alge braic relationships, knom as Kirchhoffs laws,to solve most circuits. We begh by redrawing the circuit as shown in Fig. 2.15, with the switch in the oN state.Nole that we have also labeled the current and volt agevadablesassociated with eachresistorand the currentassociated with the voltage source. Labeling includes reference polarities, as always.For convenience,we attach the same subscripl to the voltage and current labelsaswe do to the resistorlabels.InFig.2.15,wealsoremovedsomeof the terminal dots of Fig. 2.12and have insertednodes.Terminaldots ate the start and end pointsofan individualcircuit element.A node is a point wheretwo or more circuit elementsmeet.It is necessary to identify nodes in order to use K chhoff's currert law. as we will see in a moment. In Fig.2.15,the nodesare labeleda, b, c, and d. Node d connectsthe battery and the lamp and in essencestrelches all the way acrossthe top of t]le diagrum, though we label a single point for convenience.The dots on eithet side of the switch irdicate its terminals, but only one is needed to representa node,so only one is labelednode c. For the circuit shown in Fig.2.15,we can idenlify sevenurftnowns: t,, t1,i., i, 1J1, z'.,and ?,r.Recall that o" is a known voltage,as it represents the sum of the terminal voltagesof the two dry cells,a constantvoltage of 3 V The problem is to find the sevenunknown va ables.From algebra,you know that to find r unkno$,nquantitiesyou must solve,?simultaneousindependentequations.From cur discussionof Ohm's law in Section2.2,you know that three of the necessaryequationsare

or : irRr,

(2.13) \2.14) 12.15)

What about the otler four equations? The interconnection of circuit elements imposes consttaints on the relationshipbetweenthe teminal voltagesand current$Theseconstraints are rcferred to as Kirchhoff's laws, after Gustav Kirchhoff, after Gustav Kirchhoff, who first stated them in a paper published in 1848.The two laws that state the constiaints in mathematical form are known as Kirchloff's current law and Kirchhoff's voltage law. We can now state Kfuchhoffs cunent law

Kirchhoff's currenttaw(KCL)>

The algebraic sum of all the cufients at any node in a circuit equalszero,

To use Kirchhoff's curent law, an algebraic sign co[esponding to a reference direction must be assigned to every current at the node. Assigning a positive sign to a current leaving a node requires assigninga negative sign to a cu[ent entering a node. Convenely, giving a negative sign to a current leaving a node requires giving a positive sign to a current enteringa node. Applying Kirchhoff's curent law to the four nodes in the circuit shownin Fig. 2.15,usingthe conventionthat currentsleavinga node are considered positive, yields four equations: j"-i1 :0,

(2.16) (?.17)

-ic-r:t), q

\=t)'

(2.18) (2.1e)

Note that Eqs.2.1G2.19are not an independentset,becauseany one of the four can be dedved hom the ottrer tbree. In any cilcuit witl r nodes. l' 1 independent crment equations cail be dedved from Kirchhoffs current law.r Let's disiegardEq. 2.19 so that we have six independent equations, namely,Eqs.2.13-2.18. We needone more,which we cande ve from K cbloff's voltage law. Before we can state K chhoffs voltage law, we must define a dosed psth or loop. Starting at an arbitra ly selected node, we trace a closed path in a circuit througl selected basic circuit elements ard returr to the original node without passingtlrough any intermediare node more tlan once.The circuit shown in Fig. 2.15 has only one closed pat}l or loop. For example, choosing node a as the starting point and tracing tlte circuit clockwise,we form the closed path by moving through nodes d, c, b, and back to node a.We can now state Kirchhoff's voltase law: The algebraicsum of all the voltagesaround ary closedpath in a circuit equalszero. To use Kirchhoffs voltage law,we must assignan algebnic sign (ieferencedirection) to eachvoltage ir the loop. As we tiace a closedpath, a voltagewill appeai either asa rise or a drcp in the tracing direction. Assigning a positive sign to a voltage rise requiies assignirg a negative sign to a voltage drop. Conversely,giving a negative sign to a voltage rise rcquires givhg a positive sign to a voltage drop. We now applyKirchhoffs voltagelaw to the circuitshowllin Fig.2.15. We elect to trace the closed path clockwise, assignhg a posilive algebraic signto voltagedropr Startingat node d leadsto the expression 11-1)c+1)1

I We say more about lhis obseNalion in Chapier ,1,

.=U.

(2.20)

{ Kirchhoff'svottageLaw(RVL)

38

CircuitEtemenis

which represents the severth independent equation needed to find the sevenunknown circuit variables mentioned earlier' The thought of having to solve seven simultaneous equations to find the curent delivered by a pair oI dry cells to a flashlight lamp is not very appealing. Thus in the coming chapte$ we introduce you to analytical techniques that will enable you to solve a simple one-loop circuit by writing a single equation. However, before moving on to a discussionof these circuit techniques, we need to make several observations about the aletailedanalysisof the flashlight circuit ln general,these observations are true and therefore are important to the discussionsin subsequentchapters. They also suppo the contention that the flashlight circuit can be solvedby defininga singleunknown Firsi, note that if you know tlte curent ill a resistor, you also know the voltage across the resistor, because current and voltage are directly relatod through Ohm's law. Thus you can associateone unknown variable with eachresistor.eithel the current or the voltage Choose,say,the current as the unknovn vadable. Then, once you solve for ttre unknown current in the resistor, you cal! find the voltage acrossthe resistor. In general, if you know the current in a passive element, you can find the voltage across it. greatly reducing the number of simultaneous equatrons to be solved. For example,in ttre flashlight circuit, we eliminate the voltages ?., ot, and ,l as unknowns.Thus at the outset we leduce the analytical task to solving four simultaneous equations rather than seven of conThe secondgeneralobser\alion felaleslo rhe coDsequences curto Kiichhoff's nectingonly two elementsto form a node.According you know the if to a node, rent law, when only two elementsconnect element. the second you know it in also current in one of the elements, In other words,you need define only one unlnown current for the two elements.Whenjust two elementsconnectat a singlenode,the elements are said to be in series.The importance of ttris second observationis obviouswhen you note that each node in the circuit shown in Fig 2.15 Thus) ou needto delineonl) one unknown invol!eronly rqo elemenls. current.The reasonis that Eqs.2.16-2.18lead directly to

\2.2r) which states that if you know any one of th€ element currents,you know them all. For example,choosingto use ir as the unknown elimi nates4, i., and ir.The problem is reducedto determiningone unknown, Examples 2.6 and 2.7 illustrate how to write circuit equations based on Kirchhoffs laws. Example 2.8 illustrates how to use Kirchhoff's laws and Ohm's law to find an unknown current. Example 2 9 er?ands on the technique prcsented in Example 2 5 for constructing a circuit mod€l ibr a device whose terminal characteristics are known.

39

UsingKirchhoff's CurrentLaw Sum the curents at each node in the circuit shown in Fig.2.16.Note that there is no connectiondot (.) in the center of the diagram, where the 4 {) branch crossesthe branch containing the ideal current

Sotution In wdting the equations,we use a positive sign for a crment leavirlg a node.The four equations are i\ + ia n o d eb

i2+i3

i2

is: 0,

Figure2.16A Thecjrcuit forExanrpte 2.6.

i 1- i 6 - t , = 0 , i3 - i4 - i.:0,

ib

node d

4+ta+ic:0.

UsingKirchhoff's VottageLaw Sum t}le voltages around each designated path ln the circuit shown in Fig- 2.17.

Solution In writing the equations,we use a positive sign for a voltage drop. The four equations are -r'1 + ti2 + ?r4- 2rb

path a

60

parh b

rJa+?]3+zrs:0,

path c palh d

z'3: 0,

ab-14

-r.

- ,r

ir.

uc

?,6

q , rr7

?)5:0,

,J - 0.

Figure2.u A lhecircuit forExampte 2.7.

40

CncuitEtements

Current Lawsto Findan Unknown Applyingohm'sLawandKirchhoff's a) Use Kirchhoff'slawsand Ohm\ law to find i, in the circuit shownin Fig.2.18. 10()

-

5oa

)tzov

We obtain the secondequation from Kirchhoff's vollage law in combination with Ohm's law. Nolinglrom Ohm c law lhrt ,- ic Int. and d' is 504,we sum the voltagesaroundthe closedpath

(t

forExanrpte ?.8. Figur€2.18A lhecircuit

b) Tcst the solution for i, by verirying that the total power generatedequalsthe total power dissipated.

Solution

120+10t,+50ir=0. In writing this equation,we assigneda positive sign to voltage drops in the clockwise direc tion. Solving these two equations for i, and it yields

t.:

3A

and

ir=3A.

in rh( 50 Q re.i.lorrs b) fte po$ef dirsipaled

a) We begh by redrawingthe circuit and assigning an unknown current to the 50 O resistor and unknown voltages across the 10 O and 50 O resisto$.Figure2.19showsthe circuit.The nodes arelabeleda, h, andc lo rid the discussion.

p5oo= (3)'(50) : 4s0 w. The power dissipatedin the l0 O resistoris 21oo=( 3),(10)=90W.

.

10() ,,, The power deliveredto the 120V sourceis

- )rzov

= -120i" = p121,y

jn Fig.2.18,withth€ shown Figure2,19A Thecircuit i1,0., andrr d€fined. uiknowns Becausei, also is the current in the 120V source, we have two unknown currents and lherefore must deive two simultaneousequa tions involving l, and 4. We obtain one of the equatioN by applying Kirchhoff's current law to eithernodeb or c. Summingthe currentsatnode b and assigning a positive sign to the cunents leavingthe node gives il

io-6:0.

120(-3) : 360 W

The power deliveredto the 6 A sourceis

pra =

u(6),

P6a =

but

tr = 50t1: 150 V

-1s0(6) : -900 w'

The 6 A source is delivering 900 W, and the 120 V source is absorbing 360 W. The 1otal power absorbed is 360 + 450 + 90: 900 W Therefore,the solution verifies that the power deliveredequalsthe power absorbed.

41

a CircuitModelBasedon TerminalMeasurements Construding The terminal voltage and terminal current were measuredon the deviceshown in Fig. 2.20(a),and the valuesof tr and i ar€ tabulatedin Fig.2.20(b).

:r -

,

,, (A)

+

t Device ' l I

I |

:

-

I

30v

10o

Figlre2.20A (a)Devjce and(b)datatorExampLe 2.9.

a) Construct a circuit model of the device inside b) Using this circuit model, predict the power tlis devicewill deliverto a 10 O resistor.

(b) jn t/ torihe devic€ Figurc2,21 i (a)Thegftphof r, versus r i s .2 . / 0 ( a )(.b i I 5 p' e \ Lf , n gc i ' c i' L - o d e t t r ! \ e d e r c e ' . to a 10 O r€sjstor. Fig.2.20(a), connected

Solution a) Plotting the voltage as a function of the current yieldsthe giaph shownin Fig.2.21(a).Theequation of the line plotted is u':30

5'!

Now we treedto identiJy the componentsof a circuit model t}Iat will produce the same rclationship between voltage and current. Kirchhoffs voltage law tells us tlat the voltage drcps aqoss two components in sedes add. From the equa tion, one of those components produc€s a 30 V drop regardless oI t}le current. This component can be modeled as an ideal independent voltage source.The other component producesa positive voltage drop in the direction of t}le current ir. Becausethe voltage drop is propo ional to the current, Ohm's law tells us that this component canbe modeledas an ideal resistorwith a value of 5 (r. The resulting circuit model is depicted in the dashedbox in Fig.2.21(b).

b) Now we attach a 10 O rcsistor to the deuce m Fig. 2.21(b) to complete the circuit. Kirchhofs current law tells us that t}le current in the 10 f} resistor is the sameast]Ie current in the 5 {) resistor. Using Kirchhoffs voltage law and Ohm's law, we can write the equation for the voltage drops around the circuit, stafiing at the voltage source and proceeding clockwise: -30+5i+10i:0. Solving for t, we get i=24.. Because this is the value of currcnt flowmg in the 10 O resistor,we canusethe powerequation p : j2R to computethe power d€liveredto this rcsistor: Proo - (2)'(10) = 40 w

42

0bjective2-Be ableto stateanduseohm'slawandKirchhoff's currentandvoltagelaws For the circuit shown,calculate(a) i5;(b) "r; (c) ?,r;(d) z,\;and (e) thc power deliveredby the 24V sorrce.

2.5

2.7

Answ€r: (a) 2 A; (b) 4Y (c) 6 V;

(d) 14v; (e)a8w 30

21V

a) The terminalvoltageand terminalcurrcnt weremeasuredon the deviceshown,The valuesof 2,,and j, are providedin the table. Using thesevalues,createthe st(aightlil1e ptot of 4 versusir-Computethe equalionof the lire and usethe equarionto constructa cfcuit model for the deviceusjngan ideal voltagesourceand a resistor. b) Use the model constructedh (a) to predicr the power that the devicewill deliverto a 25 O resistor.

Answer: (a) A 25 V sourcein serics$iith a 100O resistor;

( b )1 w

1A 21)

2.6

Use Ohm's law and Kirchhoffs iawsto find the valueoIn in the circuit shown-

Answer:R = 4().

200v

(lt)

2.8

120\:24!)

8r)

(b)

RepeatAssessment Problem2.7 bul usethe equationofrhe graphedline to constructa circuit model containingan ideal currentsource and a resistor,

Answer: (a) A 0.25A cunert sourceconnected betweenthe terminalsof a 100O rcsistor;

(b) 1w. \OTf:

A L , ' ! r j ( h d p l e r P t u b | , n 5 2 . 1 1 . 2 . 1 ' , 2 . 1 8 . d n J) . 1 u .

2.5 A*nlysisqf e eircuitCcmt*iming Sependent 5o$rees We concludethis introductionto clenenlary circuit analysiswith a discus sion of a circuitthat containsa depeident source.asdepictedin Fig.2.22. 5i,r We want to use Kirchholls laws and Ohm\ la$'to find o, in this cir cuit. Before writing equations,il is good practiceto examinethe circuit diagramclosel_v. This rvill help us identify the information that is known aDdthe informationrje musl calculate.It may also help us dcvisca slrat Figure2.22Z A circuitwitha dependent source. egyfor solvingthe circuitusingonly a few calculations.

2.5 Anatysis of a Cjrcujt Contajning Depend€nt Sources A look at the ckcuit in Fig.2.22revealsthat . Oncewe know to,we can calculateoousingOhm's law . Once we know i^, we also know the cu enl supplied by the dependent . The curr€ntin the 500V sourceis i^. Thereare thus two unknowncurents, iA and io.Welleed to constructand solvetwo independentequationsinvolving thesetwo currerts to produce From the circuit,noticethe closedpath containingthe voltagesource, the 5 O resistor, and the 20 O resistor. We can apply Kircbloff's voltage law aro nd this closed path. The resulting equation containsthe two

500: 5ia + 20t,.

\?.?2)

Nowweneed to generatea secondequationcontainingthesetwo cur, rents. Consider the closed path formed by the 20 O resistor and the dependentcurent souice.If we attemptto apply Kirchhoff'svoltagelaw to thjs loop,we fail to developa usefulequation,becausewe don't know the value of the voltageaooss the dependentcurrent source.In faclj the voltageaooss the dependentsourceis oo,whichis the volragewe ate trying to compute.Writing an equationfor this loop does not advanceus towarda solution.For tlis samereason,wedo not usetie closedpath conlainhg the voltagesource,the 5 O resistor,and the dependentsource. Thereare three nodesin the circuit.so we turn to Kirchhoff'scurrent law to generatethe secondequation.Node a connectsthe voltagesource andthe 5 O resistor;aswe havealreadyobserved.the currentin thesetwo elementsis the same.Eithernode b or node ccan be usedto constructthe secofldequatior from Kirchhoff'scurrentlaw We selectnode b and producethe following equation: lo=iA+5tA=6ta

\2.23)

SolvingEq$2.22 and 2.23for the currents,we get j^:4A' in:A

4,.

12.24)

Using Eq.2.24 and Ohm's law for the 20 O resistot,we can solvefor the voltage tro: ?)o= 20to : 480 V' Think about a circuit analysisslrategy before beginning to write equation$As we havedemonstrated, no1everyclosedpath providesan oppor tunity to write a useful equation based on Kirchhoff's voltage law. Not every node provides for a useful applicatior of Kirchhoff's current law Somepreliminary thinking about the problem can help in selectingthe most fruitful approachand the most usefulanalysistools for a particular

Ciruit Ehments

problem. Choosing a good approach and the appropriate tools will usually reduce the numbei and complexity of equations to be solved.Example 2.10 illustrates another application of Ohm\ law and Kirchhoff's laws to a circuit with a dependentsource.Example2.11involvesa muchmore complicated circuit, but with a careful choice of analysistools, the analysjsis relatively uncomplicated.

Vottage Lawsto Findan Unknown Apptyingohm'sLawandKirchhoff's a) Use Kirchhoff's laws and Ohm's law to find the voltage,r, asshowninFig.2.23. b) Show that your solution is consistent with the constraintthat the total power developedjn the circuit equalsthe total power dissipated. 2A

Applying Ohm's law to the 3 O resistor g.Ives the desiredvoltage: ?)o: 3to - 3V b) To computethe power deliveredto the voltage sources,we use the power equationin the form p : ol. The powei deliveredto the independent volragesourcels

: -16.7W. p : (10x 1.67) Eramph 2.10. Figure 2.23A Thecircuittor

The power delivered to the dependent voltage

p = (3r"x-i"): (5)( 1) = -5 w.

Solution a) Acloselook at the circuitin Fig.2.23revealsthat: ' There are two closedpaths,the one on the left with the curent ir and the one on the right with the curent i,. . Once i, is known, we can compute 2),. We need two equations for the two currents. Becausethere are two closedpaths and both have voltage sources,we can apply Kirchhoft's voltage law to eachto give the following equations: 10 _ 6i' 3i"=2i.+3i.. Solvingfor the curents yields

i" : 1.67A,

Both sourcesare developing power, and th€ total developedpower is 21-7W lo compurelhe powerdeli\eredlo LheIe.i< ton, we use the po er equation in the form p - lTR.Thepoqer deliveredro rheb Q resisloris

p = (1.61)2n: M.1w. The power deliveredto the 2 O resistoris

P : (1)'(2:)= 2w ' The power d€liveredto the 3 O iesjstoris

P = (1F(3):3w. The resiston ail dissipatepower, and the total power dissipatedis 21.7 W equal to lhe total power developed in the sources.

2.5

Anatysjs ofa CncujtContaining DependentSources

Apptying0hm'sLawandKirchhoff's Lawin an AmptifierCircuit The circuitin Fig.2.24representsa commonconfiguration encounteredin the analysisand designof transistor amplifien. Assume that the values of all the circuit elements'_R1,R2,Rc. RE,Ucc,and U$a) Developthe equationsneededto determinethe current in each element of this circuit. b) From these equations, devise a formula for compufing is in terms of the circuit element values.

(3) iE- iE- ic-0. A fourth equation iesults ftom imposing the constraintpresentedby the sedesconnectionof Rc and the dependentsource: (4) ic = BiB We tum to Kirchhoff's voltage law in de ving the remaining two equalions We need to selecttwo closedpaths in order to useKirchhoffs voltage law. Note that the voltage across the dependentcurrent souice is unknown, and that it cannot be detemined from the source current Bis. Thereforc, we must select two closed patbs that do not contain tlis dependentcurrent source. We choose the patls bcdb and badb and specifyvoltagedrops aspositiveto yield

(s) Uo+ (6) Figure2.24A Thecircuit forExampte 2.11.

Solution A careful examination of the cfcuit reveals a total of six unknown currents,designatedir, rj, is, ic, ir, and jcc. In definingthesesix unlnowr curents, we usedrheobservalion thal rheresisror R( i\ in series with the dependentcurent source Bir. We now must derive six independentequationsinvolving thesesix unknowrs. a) We can derive three equations by applying Kirchhoff's curent law to any thiree of the nodes a, b,c, and d. Let's usenodesa, b, arld c and label thecurrenlsaqay from lhe 0odesasposili!e: (1) i1 + tc - icc: 0, (2) iB+ i2

it=0,

iERE

i2R2: 0,

- 4& + Ucc

i2R2: 0.

b) To ger a single equation for i, in terms oI t}le known circuit variables,you can follow thesesteps: . Solve Eq. (6) for t1,and substitutethis solurion for jl into Eq. (2). . Solvethe transfomed Eq. (2) for tr, and sub stitute tiis solution for i, into Eq. (5). ' Solvethe transformedEq. (5) for ir, and sub stitute this solution for iE into Eq. (3). Use Eq. (4) to eliminateic in Eq. (3). . Solve the transformed Eq. (3) for is, and rearange the termsto yield

(uccR)/(&+ R2\-uo (nlRt/(Rr + n,) + (1 + p)RE

(2.25)

Problem2.27asksyou to verify thesesteps.Note that once we know ir, we can easilyobtain the remanlng currents.

CncuitElements

pow6rfor eachetement objective3-Know howto calcutate in a simplecircuit 2.9

For the circuit shownfind (a) the currentir in (b) thc voltageo in volls,(c) the microamperes, iotal power gcncratcd,and (d) the total power absorbed.

c) ihe power deliveredby the independentcurrenl source! d) the power deliveredby the contlolledcurrent source, €) the lolal power dissipaledin the lrvo rcsislors.

Answer: (a) 25 /..A;

(b) 2 v: (c) 61s0/,w; (d)61s0/..w.

Answer: (a) 70 V;

(b)210w (c) 300w; (d)40w; (e)130w

2,10 The currentt{ in the circuitsho*n is 2A. Calculate b) Lhepower absoibedby the independent vollagesource, NOTE: Also tt| ChapterPrcblems2.21und 2.28.

PracticalPerspective ttectrical Safety At thebeginning of thischaptetwesaidthat current throughthebodycan caus€ injury.Lefsexamine thisaspect of etectricaI safety. jnjuryis dueto burrs.However, Youmightthjnkthat etectrjcaI thatis jnjury js notthe case,Themostcommon etectricat to the n€rvous system. Nerves useetectrochernical signats, andetectrjc currents candisruptthose signats. Whenthe currentpathjnctudes onLysketetaI musctes, the effects paratysis (cessation caninctude temporary of neruous signats) or involuntarymuscle contractions, whicharegeneralty notfifethreatening. Howevet jncLudes whenthe currentpath nerves andmuscLes that controL th€ supply js muchmorese ous.Temporary paral of oxygen to thebrain,the problem ysisof thesemusctes muscanstopa pe6onfrombreathing, anda sudden cLecontraction candisruDt thesiona[s that reoutate heartb€at. Theresuttis

47

deathin a few btoodto the brain,causing a hattin theflowof oxyqenated js gjven a range immediatety. Tabte 2.1shows emergency aid minutes untess in this reactlons to variouscurlentlevets.Thenumbers of physiotogicat froman anatysis of accidents they are obtajned tableare approxinate; experiments on it is not ethicatto perform eLectricaL obviousty, because, orless people. witllimitcurrenttoa fewmilLiamperes Good etectrjcal design conditions. underal[ possible

PhylioLoglcal Reaction p€rcephbLe BarcLy

Heat stoppage

3 5 5 0m A 5 0 - 7 0m A 500mA

rvot* Datatakenfrcm w F.coop€t,Ehcttico!SaJetyEngtnging, 2d ed. (London:ButteNodh, N.Y.:l,4arcel Dekket1988). P@cttditledr?dlsdrtry(Monti@tlo, 1936);andc. D,Winburn,

eLectrjcal modelof the humanbody.The Nowwe devetop a sinpLified so a reasonable starhngpointis to of current, bodyactsas a conductor potentiatly dangerous Figurc2.25showsa modelthe body0singresistors. onearmandoneleg of a A voLtage difference existsbetween situahon. model ofthehunanbodyjn 2.25(b) shows aneLectricaL human being.Figure eachhave Thearms,tegs,neck,andtrunk(chestandabdomen) Fig.2.25(a). path is through the Notethatthe of thecurrent a characterjstic resistance. deadLy arnngement. trunk,whichcontains the heart,a potentjatty bodywjtha vottase Figure2.25 A (a)A human oneam andoneteg.(b)A simdjfference between

bysoling Chapter ptifiedmodetofihe humanbodywitha vottasedifyourunderstonding of thePradicalPerspective NO|E: Assess oneatmandoneteg. ference bebveen 2.34-2.38. Prcblens

Summary The circuit elementsintroduced in this chapter are voltagesources, curent sources,and rcsisto$: . An ideal voltage sourc€ maintains a prescribed voltage regardlessof the current in the device.An ideal q|rletrt sourcemaintains a prescdbed cuffent regardless of the voltage acrossthe device.Voltage and current sources are eitler indepetrd€nt, that is, Dot influenced by any other curent or voltage in the ckcuit;or d€petrdent,that js! determined by some other current or voltage ir the circuit. (Seepages24 and 25.)

. A resistor constrains its voltage and current to be proportional to each other. The value of the proportional constant relating voltage and current in a resistor is called its resislance and is measured in ohms (Seepage28.) Ohm's law establishes the proportionality of voltage and current in a resistor. Specifically, \2.?6)

4S

Circuit Etements

if the current flow in the resistor is in the directiol of the voltagedrcp acrossit, or n=-iR

12.27)

if the curent flow in the resistoris in the direction of the voltagedse aooss it. (Seepage29.) By combidng the equation for power, p : ot, with Ohm's law,we can determinetle power absorbedby a

p=izR-uzlR.

(2.28)

through connecting elementq starting and ending at the samenode and encounteringintermediatenodeso y onceeach.(Seepages36 38.) fhe voltagesand cullenfs of interconnectedcircuit elementsobey Kirchhoffs laws: . I(irchhoffs current law states that the algebraic sum of all the currentsat anynode in a circuitequalszero. (Seepage36.) . KirchhofPs voltage law statesthat the algebraic sum of all the voltages aiound any closed path in a circuit equalszero.(Seepage37.)

(Seepage30.) . Circuits are described by nodes and closed paths. A node is a point wheretwo or more circuit elementsjoin. When just two elements connect to form a node, t]!ey are said to be in seri€s.A clos€dpath is a loop traced

A circuit is solvedwhen the voltageacrossand the cur rent in every elementhave been detemined. By combining an undentandingof independentand dependent sources,Ohm's laq and Kirchhoff's laws, we can solve many simple circuits.

Problems Section2.1

figlre P2.2

2.1 a) Is the interconnection of ideal sourcesin the circuil in Fig.P2.1valid?Explain. b) Idenrify which sourcesaie developing and which sourcesare absorbingpower. c) Verify that the total power developedin thecircuit equalsthe total power absorbed. d) Repeat (a)-(c), reversing the polarity of the 10V source. figureP2.1

10v

2.3 If the interconnectionir Fig. P2.3is valid, find rhe total power developedbythe voltagesources.Ifthe interconnectionis not valid,explainwhy. FigureP2,3

30v 10v 2.2 Tf rhe inlerconneclion in Fig.P2.2is valid.find rhe power developed by the current sources.If the interconnectionis not valid,explainwhy.

Probtems49

2.4 If the interconnection il Fig. P2.4 is valid, find ttre total power developed in the circuit. If the inteiconnection is not valid, explain why.

a) Is the interconnection in Fig. P2.7valid? Er?lain. b) Can you find the total energy developed in the circuit? Explain.

Flgurc P2.4

Figurc P2.7

10v 20v 5A

The inteiconnection of ideal sourcescan lead 10 an indeterminate solution. With this thought in mind, explail why the solutions foi ot and t'2 in the circuif in Fig. P2.5 are not unique. FlgueP2.5

100v

l)

, o A (I

i)

troov( cnv{

t

3ir

If the interconnection in Fig- Y2.8 is valid, find the total powerdevelopedin the circuit.Il the irterconnection is not valid, explain why.

FlgureP2.8

50v t0A

MA

10A

25 If the interconnection in Fig. P2.6 is valid, find the total power developed ir the circuit. ff the interconnectionis not valid,explainwhy. Figure P2.6 6V

2.9 Find the total pov/er developedir the circuit m Fig.P2.9it a. = 100V andis : 12 A. tigureP2.9

50

Etenenb Circuit FlgurcP2.12

Seclions2.2-2.3

r (A) p 0v)

2,10 The teminal voltage and terminal current were measuredon the deviceshownin Fig.P2.10(a).The values of ? and i are given in the table of Fig. P2.10(b).Use tlle valuesin the table to construct a circuit model for the device consisting of a singlercsistor.

z

8 10 12

Figure P?.10 t (mA) ?,(v) -20 -160 -10 80 10 80 20 160 u0 30


- ? --E

100 400 900 1600 2500 3600 (b)

(b)

2.13 A pair of automotive headlamps is connected to a

2.tl A variety of voltage source values were applied to

l2 V barrer) via tbe arfangementsshown in Fig. P2.13.In the figure, the tdangular symbol V is used to hdicate that the terminal is coturected directly to the metal ftame of the car-

the device shown in Fig. P2.11(a).The power absorbed by the device for each value of voltage is recordedin the table givenill Fig.P2.11(b).Usethe values in the table to construct a circuit model for ttre device consisting of a single resistor.

a) Construct a circuit model using resisto$ and an independent voltage source.

Flgure P2.11

b) Identify the corespondencebetweenthe ideal circuit element and the s]'mbol component that it rcpresents.

--E

! (v) 10

---__l /A

(

lDevie I

l- f - rl l

Y G)

),,

' (mwl 25.0 6.25

6.Zs 15

t0

25.0 56.25

20

100 (b)

2.12 A variety of curent source values were applied to tlle device shown in Fig. P2.12(a). The power absorbed by tle device for each value of curent is recorded in the table given in Fig. P2.12(b).Use t})e values in the table to construct a circuit model for the device consisting of a single resistor.

tigurcP2.r3

51 2.14 Tho voltage atrd cunent werc measured at the terminals of ttre device shown in Fig. P2.14(a). The resultsare tabulatedin Fig.P2.14(b). a) Construct a circuit model for this device using an ideal current source and a resistor, b) Use ttre model to predict the amount of power the device will deliver to a 5 O resistor. Figwe P2.74 i, (A)

100 180 260 340 420

0 8 12 16

mhals of the device shown in Fig. P2.15(a). The resultsare tabulatedin Flg.P2.15(b). a) Construcl a circuil model for lhis device usiDg an ideal voltage source and a resistor, b) Use the model to predict the value of ii when or

Figure P2.15 ?,',(V)

r ----: :L

l

t

; G)

40 35 30 25 18 8 0

10 20 30 40 50 55 (b)

L15 The voltage and curent were measured at the ter-

l r

figureP2.16

(b)

(")

fr--l -

r)

Use your circuit model to predict the open-circuit voltagg of the current souice. Wlat is the actual open-circuit voltage? E)plain why the answersto (d) and (e) are not

,, (A)

0

50 58 66 '74

z 6 8

82 90

IO

(b) (b)

LM The table in Fie. P2.16(a) gives the rclationship betweenthe teminal current afid voltage ot the Factical constantcunent sourceshoM in Fig.P2.16(b). a) Plot i, versus ,'". b) Construct a circuit model of this current source that is valid for 0 < z'" < 30 V, based on the equationof tle line plotted in (a). c) Use your circuit model to predict the currcnt delivered to a 3 kO resistor.

2.17 The table in Fig. P2.17(a) gives the ielationship betweenthe terminal voltage and current of the practical coDstantvoltage sourcesho$n in Fig.P2.17(b). a) Plot ,J versus ir. b) Construct a circuit model of ttre piactical source that is valid for 0 < i" < 225 mA, based on the equation of the line plotted in (a). (Use an ideal voltage source in serieswitl an ideal rcsistor.) c) Use your circuit model to predict the curent delivered to a 400 O resistor connected to the teminals of tle practical souce. d) Use your circuit model to predict the orrrent delivered to a short circuit connected to the terminals of t}Ie practical source. e) What is the actoal short-circuit current? f) Explain why the a$wers to (d) and (e) are not Figurc P2.r7 '75

0

60 45 30 20 10

75 150

225 300 400 500 (a)

(b)

52

Circuit ELements

Section 2.4

FigureP2.21

2,18 Given the circuit shownin Fig.P2.18,tind """ a) the valueofi., b) the value of 4, c) t]le value of o,, d) the power dissipated in each resistor, e) the power deliveiedby rhe 50 V source.

250 10()

5() 180V

Flgure P2.18 2.22 For rhe circuit shown in Fig. P2.22,find (a) R and 6trc (b) the power supplied by tle 125V source.

50v

r,ilzo o

800

2,19 a) Find the currents il and t, in t}le circuir in Fig.P2.19. b) Fitrd tlle voltage !'o. c) Verify that t}le total power developed equalsrhe total power dissipated.

ligurcP2.22

R

125V

30f,r

2.23 The variable rcsistor R in rhe circuir in Fig. p2.23 is EEc adjusted until z'dequals 60 V Find the value of R.

Flgure P2.19 300

Fig(reP2.23 450

80()

90f,)

180r) 18()

2.20 The current ia in the cftcuir shown h Fig. p2.20 is Errd 2 mA. Find (a) io (b) is; and (c) rhe power deljvered by the independent current source. FlgueP2,20 1k()

'.Jizro

4k()

3ko

221 The cunenl i, in the circuit in Fig. p2.21 is 4 A. """ a) Find i1. b) Find the power dissipatedin eachrcsisror. c) Vedfy thar the rotal power dissipated in rhe circuit equalsthe power developedby the 180V

2.24 The voltage across ttre 15 kO resistor in the circuit 6nd in Fig. P2.24is 500V positive at rhe upper terminal. a) Find the power dissipated in each resistor. b) Find the power suppliedby the 100 rLA ideal cufient source, c) Verify tiat the power supplied equals t}le total power dissipated. li$rc ?2-24

2r5 The cuffents ia and ib in the circuil in Fig. P2.25are 6fl4 4 A and 2 A, respectively. a) Fitrd is. b) Find the power dissipated in each resistor. c) Find 0s. d) Show that the power delivered by the current source is equal to the power absorbed by all the other elements, Figu€P2.25

2.28 a) Find tie voltage o, in the circuit in Fig. P2.28. """ b) Show that t]le total power genemted in t}le circuit equalsthe total power absorbed.

tigureP2,28

15.2 V

72o' 4(}

24o"

2r9 Find (a) t,, (b) ir, and (c) i, in the circuitin Eg. P2.29.

zzn The curents

i1 and i2 in the circuit in Fig. P2.26are 10 A and 25 A, respectively. a) Find the power supplied by each voltage source. b) Show that the total power supplied equals the total power dissipated in the resistors

Flgurc P2.29 1ko

5000

Figure P2.26

100r} 2.30 Find rJ1and os in the circuit shown in Fig. P2.30 6xa when oo equals250 mV (Hirtr start at tle dght end of the circuit and work back toward os.) S€dion 2.5 227 Derfte Eq.2.25. Hint: Use Eqs (3) and (4) fiorD Example 2.11 to expressiE as a function of is. Solve Eq. (2) for i, and substitute the result into both Eqs. (5) and (6). Solve the "new" Eq. (6) for tr and subshtute this rcsult into tle "new" Eq. (5). Replace tr in the "new" Eq. (5) and solvefor is. Note that becausercc appearsonly in Eq. (1), the solution for is involves the manipulation of only five equatiofls.

FiguleP2.30 10()

12.5I)

54

cncuitEtements

2,31 Foi thecircuitshownin Fig.P2.31,calculate(a) ia and Ede o" andO) showdratthe powerdevelopedequalsthe powerabsorbed, rigure P2.31

in excessof two, Show how one four-way switch plus two three-way switches can be connected between a and b in Iig. P2.33(c) ro control the lamp from three locations. (Hirr The four-way switch is placed between the three-way switches.) Flgure P2.33

tat]l

t| \\ l l 232 For the circult shoxn n Fig.2.24,Rr = 20kO, Emr R, : 80ko, Rc : 5000, RE = 100(), vcc = 15v, y0 : 200mY and B : 39.Calculateis, ic, ;8, 2,3d, ,bd,,2,4,oab, icc,ando13.(NoterIn the doublesubscript notation on voltagevariables,the first subscript is positive with respect to tlle second subscript. SeeFig.P2.32.) figureP2.32 3

Il +

R,i ,,,,

tt z. r3 l1 Position 2

11---21

f1

|1 l l 1

\ V / lI

n l

l

?

-t]

9

1 3 4 l Position2

Position 1 (b)

1 d

Sectiono2.1-2.5 2.33 It is often desirable h designing an electric wiring pffiTil| systemto be able to contrcl a single appliance from two or more locationE for example, to control a Iighting fixtue from both the top and bottom of a stairwell. In home wfting systems,this type of control is implemented \irith three-way and four-way switches.A tbree-way switch is a three-terminal, two-positior switch, and a four-way svdlch is a fourteminal, two-position switch.The switchesare shown schematica[y in Fig. P2.33(a), which illustrates a three-way switch, and P2.33(b),which illustmtes a four-way switch. a) Show how two three-way switches can be connected between a ard b in the circuit ill Fig. P2.33(c) so that the lamp I can be tumed oN or orT lrom two locations b) If the lamp (appliance) is to be controlled hom more than two locationE four-way switches are used in conjunction with two thrce'way switches. one four-way switch is required for eachlocation

tc/

234 Suppose t]le power company installs some equiptigl;ii! ment that could provide a 250 V shock to a human being.Is the cuffent that results dalrgerous enough to warmnt posting a warning sign ard taking otlei piecautions to prevent such a shock? Assume that if the souce is 250 Y the resistance of the arm is 400 O, the resistance of the trunk is 50 O, and the rcsistance of the leg is 200 O. Use ttre model given inFig.2.25(b).

2.35 Basedon the model and circuit sho$n in Fig.2.25, pll|;l#; draw a circuit model ol the path of current tlrougl tlle humar body for a person touching a voltage source with both hands who has both feet at the samepotential as the negative teminal of the volt-

55 2.36 a) Using the values of resistance for arm, leg, and ""'"'''"' lrunl Drovidedin Pfoblem).14,calculatelhe po.", di.ripuredin the arm.leg.andlr unJ. b) The specificheat of water is 4.18 x 10rJ/kg'C, so a massof water M (in kilograms)heatedby a power P (in watts)undergoesa rise in temperalure ar a mre glven oy dT

i=

2.39 . 10-aP '^^' 1 "

M

Assuming that the mass of an arm is 4 kg, the massof a leg is 10 kg, and t]le massof a trunl is 25 kg, and that the humanbody is mostlywater, how many secondsdoesit take the alm,leg, and living tissue? trunk to dsethe 5'C that endangers c) How do the valuesyou computedin (b) compare vrith the few minutes it takes for oxygen sta ation to injure the brain? 2.37 A person accidently $abs conducton connected to each hand ?lltfff*Jreach end of a dc voltage source,one in a) UsiDg the resistancevalues for the human body provided in Problem 2.34, what is the mhimum

source voltage that can produce electdcal shock sufficient to causepamlysis, preventing the person ftom letting go of the conducton? b) Is there a significant dsk of this type ol accident occurring while servicing a pe$otal computer, which typically has 5 V and 12 V souces l 2.38 To unde$tand why the voltage level is not the sole nI oI polentialiniur] due to elecrrical ,lli,l*l, '' _' delelmina shock sbock,con.iderlhe caseol a .laticeleclricily mentioned in the Practjcal Perspective at the start of this chapter.W}IeIl you shuffle youi feet acrossa carpet, your body becomes charged The effect of this charye is that your entire body iepresents a vollage potential.w]len you touch a metal doorknob, a voltage difference is created between you and the doorknob, and curent flows-but the conductionmaterialis air, not your body! Supposethe model of the spacebetweenyour hand and the doorknob is a 1 MO resistancewhat voltage potential exists between your hand and the doorknob ifthe current causingthe mild shock is 3 mA?

Circuits SimpleResistive 3.1 Resistorsin Seri€sp. 58 3.2 Resistorsin Parattetp. 59 and Current-Divider 3.3 TheVoltage-Divider Circlits p. 62 3.4 Vottag€Divrsionand CurrentDivisionp. 65 3.5 li'leasuring Voltageand Cunent p. 68 3.6 t'leasuringResistance-TheWheatstone Bndge p. 71 (Pi-to-Tee)Equivatent 3.7 DeLta-to-Wye Circuitsp. 7.t

in connected I 8e ableto rccogfjzeresistors and useth€ rutesfor s€riesandin paratLet serjesconiected resjstolsand combining paraiteL-connected resistors to yietdequivatent and 2 Knowhowto designsjmptevottage-divjd€r cunent-divid€r circuits. 3 Beabt€to usevoltagedivjsionandcur€it to sotvesjmpi€cjrcuits. djvisionapprcpriatety the readingof an ammeter 4 Beabteto determine curfenLbe whenaddedto a circuitto measure abteto deteminethe readjngof a voLtmeter vottage. whenadd€dto a cjrcujtto measure bfidgeis usedto howa Wheatstone 5 Understand 6 Knowwhenandhowio used€Lta-to-wye equival€itcitcuitsto sotvesimpb cncuits.

0ur analytica[ toolbox now containsOhm's law and Kirchhofls laws.In Chapter2 we usedthesetools in solvingsimple circuits In this chapterwe continue applying these tools,but on more_ complexcircuits.The greatercomplexitylies in a greaternumbelThis chapoI elementswith more complicatedinterconnections. ter focuseson reducingsuchcircuitsinto simpler,equivalentcircuits.We continLreto focus on telatively simple circuits for two reasons:(1) It givesus a chanccto acquaintourselvesthoroughly with the laws underlyirg more sophisticatedmethods,and (2) it allowsus to be introducedto some circuits that have important application.. engineering The sourcesin the circuits discussedin this chapterare limited to voltageand curent sourcesthat generateeither constant voltagesor curents; that is,voltagesand cunents that are invatant with time. Constantsourcesale often called dc sources.The dc standsfordirectcurleat,a dcscriptionthat hasa historicalbasis but can seemmislcadingnow. Historically,a direct current was definedas a currentproducedby a constantvoltage.Therefore,a coNtant voltagebecameknown asa direct cunent, or dc,voltage The use of dc for conrldtf stuck,aDdthe terms dc cuftent arr'ddc volldg?arc now universallyacceptedin sciencealld engineering to meanconstantcurrentand constantvoltage.

PracticaI Perspective A RearWindow Defroster js an examTherearwindowdefrostergrid on ar automobite pteof a resjstivecircuitthat pedormsa usefulfunction.one suchgridstructureis shownon the teft of the figurehere.The gid conductors can be modeledwjth resistors,as shownon the ight of the figure.Thenumberof hoizontaIconductors varieswith the makeandmodeL of the carbut typicattyranges from9 to 16. Howdoesthis gridworkto defrostthe rearwindow?How are the propertiesof the g d determined? We witt answer thesequestions in the Practical Perspective at the endofthis chapter.The circuitanatysisrequiredto answ€rthesequestiors adsesflom the goaLof havinguniformdefrostingin bothth€ horizontalandverticaL directions.

57

58

Simpte Resistjve Circuih

3.1 Resistors in Series 'ii

In Chapter 2, we said that when just two elementsconnectat a single node.lheyare said to be in series.Seri€\-connected circuit eternentscanr lhe samecuffenr.Theresi.rorsin rhe circuilshowniD frg.3.t are connecledin series. \ e canshowlhal lhe.e fesistorc cafry lh; .amecurrerr b\ appllingKirchhoffscurrenllaw to cacbnodein rh; circuir.The scrics interconnectionin Fig.3.1requiresthat

i

\t'i

) "

R,

RO

Rr

n ; c - i r - i " jn sedes. figurc3.1 A Resjstors connected

R,

RO

h

g

is=4=

RS

f

e

i2=ir=ia:

is= -i6= i. ,

(3.1)

which statesthat if we know any one ofthe sevencurrents,we know them all.Thus we can redrawFig.3.1as shownin Fig.3.2,retainhg rhe idenrity of the singlecurent i". To find i., we apply Kirchhoff's voltage law around the single closed toop.Delinirg thevoltageacrosseachresistorasa drop jn the directionof rr grves

Figure3.2 A Series resistors witha sjngteunkno\i/n

-ol + i"Rr + irn2 + irR3 + j.Rj + lrR5 + trn6 + irRT:0,

(3.2)

z ' .= t " ( R+1R 2 + & + & + R s + R 6 + R ? ) .

(3.3)

The significanceof Eq.3.3 for calculatingir is that the sevenresistorscan be replacedby a singleresistorwhosenumeiical value is the sum of the individualresistors. that is. R e q= R 1+ R 2 + R 3 + R 4 + R 5 + R 6 + R 7

\3.4)

and (3.5) h Figure3.3 A A simplified version of the cjrcujtshown in Fig.3.2.

Combining resistocin series>

Thus we can redrawFig.3.2assho$n in Fig.3.3. In general,if tr resistorsa.reconnectedin series,the equivalenlsircle re.islorhasa resrstance equallo lhe sumot lhe & re,i.rancir.or

(3.6)

Note that the resistanc€of the equivalentresistoris alwayslarset than !harot lhe largcslresi.torin rheseriesconneclion.

3.?

in PaEttet Resisto6

59

Another way to think aboutthis conceptof an equivalentresistanccis ro visualizethe string of resistorsas beinginsidea black box. (An elcctri + cal engineerusesthe term blackbor to imply an opaquecontainer;that is, the conterts are hidden from view The engineeris theD challengedto modei ihe contentsof the box by studyingthe relationshipbetweenthe h R7 R6 .It5 voltageand current at ih termillals.)Determiningwhether the box concircuit tains k resistorsor a single equivalentresistoris impossible.Figure 3.4 Figure3.4 A Theblackboxequjvahniofthe in fi9. 3.2. shown illustratesthis methodofsttdying the circuit shownin Fig.3.2.

in Faratlel 3.2 Resistors When two elemerts connectat a singlenode pair, they are said to be in parallel.Parall€l-cotrnecled circuit elementshavethe samevoltageacross their terminals.Thecircuitshom in Fig.3.5illust.aleslesistorsconnected in parallel.Don't make the mistake of assumingthal two elementsare parallelconnectedmerelybecausethey are lined up in parallelin a circuit elementsis that diagran.The definingcharactedsticof parallel-connected jn parattel. they have the samevoltageacrosstheir terminals.In Fig.3.6,you can see Figure3.5 A Resistors belwecntheir respecthat Rr and R3 are not parallelconnectedbecause, R2 rive termhals.anotherresistordissipatessomeofthe vollag.3. Resislorsin parallel can be reducedto a single equivalentresistor usingKirchhoff'scurrent law and Ohm's law. as we now demonstmte.In the circuit shownin Fig.3.5,we lel the curenis ir, tr. ii. and ia b.3the curWe also let the positive rentsin the fesistorsRr througl Ra,respectively. referencedirectionfor eachresislorcurrentbe down firough th.3resistor, Figrre3.6 { Nonparalhtresistors. that is,from node a to node b.From Kirchhoff'scunent la\ i r : t 1+ t r + i r + i 4 .

(3.7)

The parallelconnectionot the resistorsmeansthat the voltageacrosseach resistormust be the same.Hence,fromOhm'slaw,

rrR,: irRz= 4Rr = raRr- r,

(3.8)

Therelore,

Rr' :t

l, :I R1

and (3.e)

60

simpte Resktjve Circuits

Substitutirg Eq.3.9into Eq.3.7yields

,. " : " \[^r, '

1

1

1 \

&, k. R,l

(3.10)

from which i

r

1

1

1

1

1

,,=R*:4-&-&-&

(3.11)

Equation3.11is whatwe setout to show:that the four resistorsin the circuit shownin Fig.3.5canbe replacedby a singleequivalentresistor.The r€sisto$shown circuit shownin Fig. 3.? illustratesthe substitution.For k resistorsconFlgorc3.7 A RepLacing th€ foff palaLtet nectedir parallel,Eq.3.11becomes in Fiq.3.5witha singleequivat€nt resistor.

combining resistors in patallel>

13.\2)

Note that the resistanceof the equivalent resistor is alwayssmaller than the rcsistance of the smallest rcsistor in the parallel comection. Sometimes, usirg conductancewhen dealing witl resistorsconnectedin parallel is more cotrvenient,In that case.Eo. 3.12b€comes

'-Fl' jn palaLteL. Figure3.8,| Tworcsistols connecied

+G2+

G " q => G i : G l

+Gk

(3.13)

Many times only two resistors are connected in pamllel Figure 3.8 illustrates this special case We calculate tle equivalent resistance ftom Eq.3.121

1 1 1 R*=&-&:

n2+ R1 ' Rt&

R.R"

(3.14)

(3.15)

Thus for just two resistorc h parallel the equivalent resistarce equals the product of the resistaDcesdivided by the sum of the resistance$ Remember that you can only use this result in the special caseofjust two rcsistors in Darallel. Examole 3.1 illustrates the usefulnessof these results.

3.2

Relisto6inPanttet

61

ApplyingSeries-Parall.et simpLifi cation Find i,. rr . aod i2 in lhe ci-rcuilshownin Fig.3.s.

40

t20v

+

ttl:18o ',{ 6 0

Solution We begin by noting that the 3 f,) resistor is in senes wirh rbe6 Q resistor.weLherelorereplacelhis series combination with a 9 O resistor, reducing the circuit to the one shown in Fig. 3.10(a).We now can replace the parallel combhation of the 9 O and 18 O resistors with a singleresistanceof (18 x 9)/(18 + 9), or 6 O. Egure 3.i0(b) shows t]Iis turtler reduction of the circuit.The nodesx and y marked on all diagmms facilitate tracing through t]le reduction of the circuit. From Fig. 3.10(b) you can veiify that iJ equals 120/10,or 12 A. Figure3.11showsthe iesult at this point in the analysis.We added the voltage ?)1to Using Ohm's helpcladfy the subsequentdiscussion. law we compute the value of ,1:

'72v. 1\ = (12)(6)-

1)1 :72

9

v Figur€3.9 a Th€crrcujtfor E,dmph3.1.

4f)

r,l$rso r,,f9 f )

l20v y

G) 40

120V

x

+\ '"

60

(3.16)

But t 1is the voltage drop fmm node x to node y, so we can return to the circuit shown in Fig. 3.10(a) andagainuseOhm'slaw to calculatetr and ir. Thus,

- 72 : 4 A ' 18 1 8

3()

v (b) jn Fjg.3.9. Figue3,10a A sinpLification ofthecjrcuitshown

(3.17)

4f)

+\ 12A 9

:8A.

120V

(3.18)

We have found t}le thrce specified currents by using series-parallel reductions in combination with Ohm'slaw'

6()

v figuru3.r1 A lhecircuit of Fig.3.10(b) showins thenumedcaL

Belore leaving Example 3.1, we suggest that you take the tlme to show that the solution satisfies Kirchloffs curent law at every node and Kirchloff's voltage law around every closed path. (Note that there are thee closed paths that can be tested.) Showing t]Iat the power delivered by the voltage souce equalsthe total power dissipatedin the rcsistors also is informative. fsee Prcblems 3.3 and 3.4.)

62

SimphRsistiveCircuits

objective1-Be abteto recognize resistors conneded in seriesandin parattel 3.1

Fortbecircuitshown,find (a) thevollageo, (b) thepowerdeliveredto thecircuilby the cufent source, and(c) thepowerdissipated in the 10() resistor.

7.211

101)

Answer: (a) 60Vr (b) 300w; (c) s7.6w NOTE: Alsotry Chapter Problems 3.1,3.2,3.5, and3.6.

3.3 The\foltage-Sivider

andeurrent-Sivider eireuits

(")

(b)

Figure3.12a (a)A votiase-divider circuitand(b)the vottagsdivids circuitwithcrrent ; indicaied.

Al timss especiallyin eleclronic circuils developingmore than one voltagelevel from a singlevollagesupplyis necessary. One way of doing this is by usinga voltsge-divid€rcircuil,suchasthe one in Fig.3.l2. We analyze this circuit by directly applying Ohm's la$' and Kirchhoff'slaws.Toaid the analysis, we introducethe curent i asshownin Fig.3.12(b).From Kilchhoff's current la\ Rr and R, carry the samecurrel1l.Applyilg Kilchhoff's voltagelaw aroundthe closedloop yields r.-lRr+iRz,

Rr+R,

(3.20)

Nowwc canuseOhm'slawto calculale ul and,2: R r 01 =rr(1 : ?rnr + R2,

(3.21)

R, .0 2= u < 2 : x s R t + R,

13.22)

Equations3.21and 3.22show that ,r and o, are fractionsof o,- Each ftaction is the ratio of the resistanceacrosswhich rhe divided voltageis definedto the sum ofthe two resistances. Becausethis ratio is alwayslcss than 1.0.the divided voltagesor and ?J2are alwaysiess than the source If you desirea particularvaluc of ?)2,and o, is specified,an infinile numberofcombinarionsof Rr and R, yicld thc propcr ratio.For example, supposethat !. equals15 V and 1J2 is to bc 5 V Thon r2/o, = i and,liom

3.3

TheVottage'Divider andCurrcntDjvider Circuiis

Eq.3.22,wefind that this mtio is satisfiedwhereverR, : ;R1.Other factors that may enter into the selection of R1, and hence R2, include the powei lossesthat occur in dividing the source voltage and the effects of connecting the voltage-divider circuit to other circuit components. Consider connecting a resistor Rz fu parallel witl R2, as shown in Fig. 3.13.The resistorR, acts as a load on the voltage-djvidercircuit.A load on any circuit consistsof one or more circuit elementsthat draw power from the circuit. With the load Rr connected,the expressionfor the output voltagebecomes -'

'" Rr + R.q

\3.23) Figure3.13A A voLtage djvjder connected to a toadR..

RtRr R2+ RL

(3.21)

Substituting Eq.3.24into Eq.3.23yields R2

Rlt1 + (,R/R')l + R"' Note that Eq. 3.25 reduces to F'q.3./2 as Rz+c!, as it should. Equation 3.25showsthat, as long as R, >> R2,the voltageratio r,/?i" is essentially undisturb€d by t})e addition of the load on the divider. Atrother chaructedstic of the voltage-divider circuit of interest is the sensitivity of the divider to tie tolerances of the resistors.By roler.ar?ce we mean a range of possible values.The resistancesof commercially available resistorsalways vary within some percentageof theh stated value. Example3.2illustratesthe effectof resistortolerancesin a voltage-divider circuit.

Anatyzing the Voltage-Divider Circuit The resistorsused in the voltage-dividercircuit shownin Fig.3.14have a toleranceof t10%. Find the maximumand minimum valueofo,.

25kO

t00v

Sotution From Eq. 3.22,the maximum value of r)ooccurswhen R is l0oohighandRr is l0'o lo\\.and(beminimum value of t', occurswhen R2 is 10% low and Rr is 10% high.Thereforc

,lma\) = ,,(Illrn) =

100kI)

Figur€3,14 A Thecircujtfor Exampi€ 3.2.

(100v110) = 83.02 v. 1101-(100x90) e0 +--

: 76.60v.

Thus,in makingthe decisionto use10% resistorsin this voltage divider, we recognize that the noload orltput voltage will lie between 76.60and 83.02V

64

Simpte R€sistive Circuits

TheCurrent-Divider Circuit

t

o , * 1,, '*

rigure3.15A Ihedrnent-divider circuit.

The curent-divider cfucuit showr in Fig- 3.15consistsof two tesistors connectedin paralleladoss a current source.The current divider is designed to divide the current i. between R1 and R2. We find t]le relationship between ttre curent i, a.ndthe current in each resistor (that is,i1 and ir) by directly applying Ohm's law and Kirchhoffs current law. The voltage aoossthe parallelresistomis (3.26) Ffom Eq.3.26, ,"'. =

R, , R, + R7-"'

13.27)

,"'. =

R' , R, + Rz-''

(3.28)

Equations3.27and 3.28showthat the current dividesbetweentwo resis tors in parallel such that the curent in one resistor equals the curent entering the parallel pail multiplied by the other resistanceand divided by the sum of the rcsistors.Example 3.3 i ustratesthe use of the cuflentdivider equation.

Analyzing a Cufient-Divider Circuit Find the power dissipatedin the 6 f) resistor shown inFig.3.16.

and the power dissipated in the 6 O resistor is p=(3.2)'(6)=61.44w.

Solution

1.6r}

First,we mustfind the curent in the resistorby simplifying the circuit with series-parallelrcductions ltus, the circuit shownir Fig. 3.16rcducesto the one shownin Fig. 3.17.We find the qrllent i, by usingthe formulafor currentdivision: 16

j" = _--(10) tb+ 4

frro 3on

figure 3.16 A Theci,cuitforErampte l.l.

:8 A.

Note that i, is the cunert in the 1.6() iesistor in Fig. 3.16.We now car turther divide i, between the 6 O and 4 O resistols.The current in the 6 O resistoris

rd: _(8)

t)

: 3.2A,

10A

I )

160

4{)}

Figure3.17 A A sjmptjfication oftheciftuitshown in Fig.3.16.

3.4

Vottage Division andCrrentDivision

0bjective2*(now how to designsimplevottage-dividerand cuff€nt-divid€rcircuits 3.2

^) Find the no-loadvalueof?J,in the Find n, when nr is 150kO. c) How muchpower is dissipatedin thc 25 kO resistorifthe load terminalsare accidentally shon-circuited? d) What is the maximumpower dissipatedjn the ?5 kO resistor?

3,3

a) Find the valueof R that will cause4 A of currentto flow throughthe 80Q resistorin the circuit shown. b) How muchpower will lhe resistorR from part (a) nced to dissipale? c) How much power will the currentsource genemtelor the valueof R froln pafi (a)?

60() 100 20A

It 800

(a)150v; (b)133.33 V (c)1.6W (d)0.3w

Answ€r: (a) 30O; (b) 7680w; (c) 33,600 W.

NOTE: Also tty ChaptetPnblems 3.13,3.15,antl3.21.

3.4 tfqlltaqefiivisisn

a$deurrentllivision We canrow generalizethe resrlts from analyzingthc voltagedivider circuit in Fig.3.12and tlre currentdivider circuil in Fig.3.15.Thegeneralizations will yield two addiiional and very usctul circuit analysislechniques known asvoltagedivisionand currentdivision.Considerthe circuil shown inFig.3.18. The box on tlre left can cortain a singlevoltagesourceor any other combinationof basiccircuitelementsthat resultsin thc vollag€?)shownin the tigure.To the right of the box are r resistorsconncctcdin series.We division. are interestedin finding the voltagedrop ol acrossan arbitraryresistorRl figure3.18t Cjrcuitusd to ittustEievottage in terms of the voltager. We start by usingOhm's law !o calculatei, fte currentthrough all of the resistorsin serics,in lerms of the current o and

' n l + R 2 + - + R , , R.q

(3.2e)

SjmphResjstive Cjrcuitr

The equivalentresistance, Rcq,is the sum ol the r resistorvaluesbecause the resistorsare in series,as shownin Eq. 3.6.We apply Ohm's law a second time to calculatethe voltagedrop o/ acrossthe resistorRj. usingthe cureni t calculatedin Eq.3.29:

Vottaqe-division equation>

ur:

iRt :

RI =-1)

(3.30)

Note that we used Eq.3.29 to obtain the right-hand side oI Eq.3.30. Equation 3.30 is the voltage division equation.It saysthat the voltage drop ri acrossa singleresistorRi ftom a collectionof se es-connected resistorsis pioportionalto the total voltagedrop , acrossthe setofseriesconnected resislors.The coDstantof propor'tionality is the mtio of the single resistanceto the equivalert resistanceof the seriescomected set of resistors, or R'/Red. Now considerthe c cuit showr in Fig. 3.19.The box on the left can contain a single curent source or any other combination of basic circuit elemenls that results in the current i shown in the figure. To tlre right of the box are r resistorsconnectedin para el. We are inrerestedin finding the curent ii throughan arbitraryresistorRr in terms of the current i. We stafi by usingOhm'slaw to calculatez),thevoltagedrop acrosseachof the resistorsin parallel.in termsof the current; and the r resistors: 0 = (R1 R2 ... iln,) : rReq.

(3.31)

Theequivalent resistance ofn resistors in parallel,Rcq,canbe calculated usirgEq.3.12.We applyOhm'slawa second timeto calculate thecurrent l, throughtheresistorR;,usingthevoltageo calculated in Eq.3.31:

Current-division equationt

n

R,

Note that we used Eq.3.31 to obtain the right-hand side of Eq- 3.32. Equation 3.32is the current division equation.It saysthat the current i through a singleresistorRr {rom a collectior of parallel-connected resisto$ is proportionalto the total current i srpplied to the set of parallelconnectedresistors.The constantof proportionality is the ratio of the

tigure3.19A Circuii used to ittustrate cunentdivjsion.

3.4

Vottage Divisjon andCurrent Division

6l

equivalentrcsislanceof the parallel-connecled serofresistorsto the siDgle resrstance. or Rcq/&. Note that the constanlof proportionaliryin thc cul' rent divisionequationis rhe inverseof rhc constantof proportionalilyil1 the voltagedivisionequationl Examplc 3.4 usesvoltage divislon and current division to solve for voltagesand culrenlsin a circuit.

UsingVottageDivisionandCurrentDivisionto Solvea Circuit [Jsecurrenr divisionto find thc cullett i, and use voltagedivisionto find the voltager,, for the circuit ]n Fig.3.20.

Solution We san use Eq. 3.32if we can find the equivalcnl tcsislanceof the four parallel branchescontaining rcsislors.Symbolically,

Figur€3.20 ,",ThecircuitforExamph 3.4.

Req= (36 + 44) l0 (40 + 10 + 30) 24

= 8 01 0 8 0 2 4 -

I * 1 * 1 lJ0 r o s 0

t

=60.

-t

Applyhg Eq.3.32.

i, =

6

t(8A)

= 2A.

We can usc Ohm's law to find thc \,ottagedrop acrossthe 24 () resistor:

This is also tlle voltage drop acrossthe branch con tainingthe 40 Q, thc 10 O, and the 30 0 resistorsin .
,,=l!,**u,: ,av.

1]: (24)(2) = 48 V.

obiective3-ge abteto usevottageandcurrentdivisionto sotvesimplecircuits 3.4

a) Use voltage division to determine the voltagc ?),acrossthe 40 O resistorin lhe circuit shown. b) Use o,,from part (a) to determincrhe current throughthe 40 J) resistor,and usethis currcnt and currelt divisionto calcu]atethe currentin the 30 O resistor. c) How much poweris absorbedby lhe 50 O

400

',o,,1 60v

rrl

+ 70o

Answer: (a) 20V; (b) 166.67I1lA;

(c) 347.22mw. NOTE: Alxo ttr Chapter Problens 3.22 and 3.23.

50(:}

,0,,

68

citcujts Simpte Resistive

Vottageand€urrent 3.5 Measuring When working with actual circuits, you will oJten need to measue voltagesand cullents.we will spendsometime discussingseveralmeasuring deviceshere and il the next section,becausethey are relativelysimpleto analyze and offer practical examples of the curent- and voltage-divider configurations we have just studied. An .nmet€r is an hsuument designedto measue curent;it is placed in seies with the circuit element whose cuuent is being measured A the connected to measure Figurc3.21A Anammeter voltrnet€ris an instument d€signedto measurevoltage;itis placedin par' the connected to measure curentin Rr. anda voLtmeter allel with the elementwhosevoltageis beingmeasured.Anideal ammeter or voltmeter has no effect on the circuit variable it is designedto measure That is, an ideal ammeler has an equivalent resistanc€ of 0 o and functions as a short circuit in sedes with the element whose current is being measured.An ideal voltmeter has an i inite equivalent resistance and thus functions as an open circuit in parallel with ttre element whose voll age is being measu.ed.The configurations fol an ammeter used to measure the current in R1and for a voltmeter used to measurethe voltage irl 1R2 are depictedin Fig.3.21.The ideal modelsfor thesemetersin the samecilmodet forthejdeatamme cuit aie shownin Fi9.3.22. Figue3.22A A short-circujt jdeatvoLtmeter. modeLtotthe ier,andanopen'circuit There arc two broad categoriesof meters used to measurecontiluous voltages and currents: digital mete$ and analog meters. Digital meters measde the continuous voltage or current siglal at disclete points in time, called the sampling times.The signal is thus conve ed from an analog sigml, which is continuous ir time, to a digital signal, which exists only at discrete hstants in time. A more detailed e4lanation of th€ workings of digital meters is beyond the scopeof this text ard coufte. However, you are lilely to seeand use digital meten in lab settingsbecausethey olfer several advantagesover analog meters.They introduc€ less resistanceinto the circuit to which they are connected,they are easier to co rect, and the precisior of the measurement is greater due to the nature of the rcadout

diagram 0f a dl\rsonvat Figue3.23A A sch€matjc

tigue 3.24A A dcammeter crrujt.

Figue3.25a A dcvottmeter cncuit.

Analog m€t€rs are based on the d'Arsonval meter movement which implements the readout mechanism.A d'A$onval meter movement consistsof a movable coil placedin the field of a pemanent magnet wlen current flows in the coit, it creafesa torque on the coil, causingit to rotate and move a pointer across a calibrated scale.By design, the deflection of the pointer is directly proportional to the current in t]le movable coil- The coil is chaiactefzed by both a voltage rating ard a curent rating. For example, one commercially available meter movement is rated at 50 mV and 1 rnA. This meansthat when tlle coil is carrying 1 ntA, the voltage drop affoss the coil is 50 mV and the pointer is deflected to its full-scale position. A schematrcillustration of a d'Arsonval meter movement is shown in Fig.3.23 An analog ammeter consists of a d'A$onval movement in parallel with a rcsistor, as shown in Fig. 3.24.The purpose of the pamllel resistor is to limit the amount of curent in the movement's coil by shunting some of it ttrrough R,4.An analog voltmeter consistsof a d'Arsonval movement in series with a resistor, as showtr in Fig. 3.25. Here, the iesistor is used to limit ttre voltage drop across tlle meter's coil. In both meteis, the added rhefull-scale feadingof rhemelermo\emenl. resistorderermines From thesedescriptionswe seethat an actualmeter is nonideal;bottr the added resistor and tlle meter movement i troduce resistance in the

3.5 Measudns vottage andCureni

69

circuit to which the meter is attached.In fact, any instrument used to make physical measu{ements extracts energy from the system while making measurements.The more energy extracted by the instnments, the more severcIy the measurement is disturbed, A real ammeter has an equivalent resistanceUrat is not zero, and it thus effectively adds resistanceto the circuit itr series with the element whose current the ammeter is readins. A real\ohmelerbasan equivalenr resistance lhal is not inLinite, so it eFectively adds resistance to the circuit in pamllel with the element whose voltage is being read. How much these meters disturb the circuit beinA measured deDends on LheeltecliveresistaDce of the meter.compared;lh the resista;ce in t]le circuit. For example,using the rule of 1/10th, the eflective resistanceof an ammetershould be no moie than 1/10th of the value of the smallest iesistance in the circuit to be sure that the current being measured is nearly t]le samewitl or without the ammeter. But in an analoe meter. the vaiueoI resistanceis determinedby lhe desiredtull-scalefead;g \re wisb to maLe, and it cannot be arbitarily selected. The following examples illustrate the calculations involved in detemining the rcsistanceneeded it an analog ammeter or voltmeler. The examplesalso consider the resulting effeclive resistanceof the meter when it is insefied in a circuit.

lJsinga d'Arsonval Amrreter a) A 50 nV 1 mA d'Arsonval movement is |o o€ used in ar ammeter with a full-scale reading of 10 mA.Determine R,a. b) Repeat(a) for a full-scalereadingof 1A.

b) Wlen the full-scale deflection of the ammeter is 1A, R,amust cary 999mA when the movement carries1mA.In tlis case,then, 999x10-3R,{=50x10-3,

c) How much resistance is added to the or€urt when the 10 mA ammeter is inserted to mcasurc R,{ : 50/999 ! 50.05mO. d) Repeat(c) for the 1A arnmeter.

c) Let R- rcpresert the equivalelt resistanceof the ammetei. For the 10 mA alnmeter,

50mV 10mA

Solution or, alternatively, a) From the statement of the problem, we know that when the current at the terminals of the ammeter is 10 mA, 1 mA is flowing rlrough the meter coil, which means that 9 mA must be dive ed through R4. We also know that when the movement caries 1 mA, the drop acrossits terminalsis 50mV Ohm'slaw requiresthat

9x103Rn-50x103 R1 : 50/9 : 5.555O.

r50)r50l9 | xm:sn+(50/g=so. d) For the 1 A arnmeter

50mV or, alternatively,

(5oxso/eee) :

().

0.050 50+ (s0/eee)

70

SimpLe Resistjve Cncuits

Usinqa d Arsonval Vottmeter a) A 50 mV, 1 nA d'Arsonval movementls to be usedin a voltmeter in which the full-scalercad ing is 150V Determinenr. b) Repeat(a) for a lUll scalereadingoI5V c) How much resistancedoes the 150 V meter insertinto the circuit? d) Repeat(c) for the s V neter.

b) Fora tul] scalereadingof5 Y

50x103:nJso(s), R, : 4950O. c) If we let R", representthe equivaleniresistance

R- =

Solution

t0

or, altematively,

a) Full-scaledeflectionrequjrcs 50 mV acrossthe meier movement.and the movementhasa resisl 'lterefore anceof 50 O. we apply Eq. 3.22with = 50nV. Rr - RtR, : 50,D.= 150.and1)2 -

150.v= ,l,. r5o.ooo 'A

n,,, - 149.950+ 50 - 150,000O. d) Then,

5rl

= ,ooo ^-' : l]L ,,. t0'A

s0. 10':=-(1s0). K/ + )ll

Solvingfor Rr gives

or, alternatively.

R- : 4e50+ 5U= 50U0A.

R. = 149,950O.

objective4-Be abteto determine the readingof ammet€rs andvottmeterc 3.5

a) Find the currentin the circuitshown. b) lfthe ammeterin Example3.5(a)is usedto measurethe cullent, what will it read?

3.6

a) Find the voltage1,acrossthe 75 k0 resistor in the circuit shown. b) lfthe 150V voltmeterof Example3.6(a)is usedto measue l]re vollage,whatwill be ttre reading?

75kO

Answer: (a) 10mA; (b) 9.524mA. NO'.E: Abotry Chapter Probtems 3.30a

Answer: (a) 50Y (b) 46.15v 3.33.

3.6

I'leasuingResistance-TheWheatstoneBddqe 71

3.6 Measuring ResistanceTheWheatstone Bridge Many difrerent circuit conJigurationsare used to measue resistance,Heie we will locus on just one, the mealstone bddge. The Wheatstone brialge circuit is used to precisely measureresistancesof medium values,that is.itr the range of 1 O to 1MO. In commercial models of tie Wheatstone bridge, accuracieson the order of +0.1oloare possible.The bridee circuit consistsof four resislorra dc voltagesource.anl a delector.The;sisraoce of one of the four rcsiston can be varied, wbich is indicated in Fis_3.26 bv lhe arrow tlu-oughRr. Tbe dc \ohage sourceis usuallya bartery,whicb is indicated by the battery symbol for the voltage source z, in Fig. 3.26.The detector is generally a d'Arsonval movement in the micioamD ranqe and is bidqecircuit. caueda galvanometef.Hgure 3.2bsho\rsthe circuil aflansemenlot lhe Figure3.25^ TheWheatstone resislaoces balteryanddetectofq bereRt. R.. andR3areL;owo resislors and R, is the unknown resistoi. To find the value of R,, we adjust the variable resistoi R3until theie is Iro cunetrt in thg galvanometer, We then calculate the unknown rcsistor from the simole exDiession

&:

-R.- R r .

The derivalion of Eq. 3.33 follows directly from ths apptication of Kirchhoffs laws to the bddge circuit. We redmw the bridsi circuit as Eg. 1.27ro sho\r (be curenrsappropriateto lhe derilationof Eq. .J.l3. W}len is is zero, that is, when the bddge is balanced, Kirchhoff\ curent law requires that Q3q tz = tt.

(3.35)

Now, becauseis is zerc, there is no voltage drop acrossthe detector, and thercfore points a and b are at the samepotential. Thus wher the bddge is balanced,Kirchhoff's voltage law requires that i:Rs : i,R ,

(3.36)

iRI : i2R2.

(3.37)

CombiningEqs.3.34and 3.35with Eq.3.36gives i1R3 : 12R".

(3.38)

WeobtainEq.3.33by firsrdividing8q.3.38by Eq.3.37andr}tensolving the resultingexpressionfor Rr: R: Rr

R, Rz'

(3.39)

a R, -'/

Figure 3.27l A batanced wheatstone bfidse (ls = 0).

Simpte Resistive Circuitr

R.

ft,: *n:.

(3.40)

Now that we have voriticd rhe validity of Eq.3.33,severalconments aboul lhe r'esullare in ordcr. First.note that ifihe ratio R /R1is unily.the unknown resistorR, cquals Rr. ln this case,the bridge resislorn3 must vary over a rangcthat includesthe valueR.. For example,iI lhe unknown lvere1000O and Rl couldbevariedfrom0 to 100(), the bridge resistance couldneverbe balanced.Thus to covera wide rangeof unknownresislors. we must be able to vary the ralio R2/Rr. In a commercialWheatstone bridge,Rr aDd 112consisloI dccimal values of resistancesthat can be switched into the bridge circuil. Normall,v,the decimal values are 1, 10,100,and 1000O so |hal the ratio R2/Rrcan be varied fron 0.001to 1000in decimalsteps.ThevariablcrcsistorRr is usuall]'adjustableiD iDle gral valuesofresislancefrom :l to 11,000O. Altlough Eq.3.33 inplies that R, can vary from zero to infinit),.the practicalrarge ofR, is approxinalely I () to 1 MO. Lower resistances are whealslone because of thermodifficrlt to measureor a standard bridgc eleciric voltages generatedat the juncliorls of dissimilar metals and bccauseof thermalheatingeffects lhat is,i'R eIiecls. Highcr rcsjstances arc ditTicultto measureaccuratelybecauseof leakagecurrcnts.In other words.if R, is large,the currentieakagein lhe eleclricalinsulationmay be conparableto the curfent in the branchesoI lhe b dgc circuit.

. obiectives-lJnderstandhowa Wheatstone bridoeis usedto measure resistance The bridgecircuit shownis balancedwher R1: 100O../1,- 1000O,and Rj - 150O. Thc bridgc is energizedfrom a 5V dc source. a) What is the valueofR,? b) Supposceachbridge resistoris capableof dissipating250mW Can the bridgebe left in thc balancedstatewithout exceedingthe power dissipatingcapacityofthe resistors, therebydamagingthc brjdgcl NOTE: AIso tr! Chupter Problem 3.4ti.

Answer: (a) 1500{); (b) yes.

3.7

DeLta to'Wye(Pi-to-lee) EquivaL€ni Cncuiis

3.7 Detta-to-Wye (Pi-to-Tee) Equivalent

Circuits + The bridge configuratior in Fig. 3.26 introduces afl interconnectiofl of resistancesthat warrantsfurther discussion.If we reDlacethe salvanomeler$1lhrtsequi\alenrresisrance R,. qe candraq rle circuir.houn in Fig. 3.28.We cannot reduce the interconrected rcsiston of this circuit to a singleequival€ntresistance acrossthe teminals ofthe battervif reslricted to t]le simple seriesor parallel equivalent circuits introduced earlier in this chapter. The hterconnected resistors can be reduced to a single equivalent resistorby meansof a delta-to-r,ye(A{o-Y) or pi-to tee (,7-ro-T) equivalentcircuit.l The rcsiston Rl, R2,and R- (or R3,n- and R,) in the circuit shown in Fig.3.28 are referred to as a delta (d) intercorn€ction becausethe interconnectionlooks like the creek letter A. It also is refened to as a pi int€rconnectionbecausethe A can be shapedinto a ?' without disturbing the electricalequivalenceof the two conJigurations. The electrical equivalencebetween the A and z interconnectionsis aDDarentin Fig.J.2,r. The resistorsRl, R-, and R3 (or Rr, R- and R,) in the circuit shown in Fig. 3.28are referred to as a rrye (Y) interconnectionbecausethe interconnectior can be shapedto look ljke the letter Y It is easierto seethe y shapewhen the interconnection is drawn asin Fig. 3.30.The y configuration also is rcferred to as a tee (T) int€rconnection becausethe Y structure can be shapedinto a T structure without disturbing the electdcal equivalenceof the two structures.The electrical equivalenceof the y and t}le T configurationsis apparentfrom Fig.3.30. Figure3.31illustratesthe A-to-Y (or r {o-T) equivalentcircuit tiansformation. Note that we cannot transfom t}le A interconrection into the Y interconnection simply by changing the shape of the interconnections. Sayingthe A-connectedcircuit is equivalentto the Y-connectedcircuit meansthat the A configuration can be replaced witi a Y configuration to mal(e the termhal behavior of the two configurationsidentical.Thus if each circuit is placedin a black box, we can't tell by extemal measuremelts whetherthe box containsa set of A-coinectedresistorsor a set of Y-comectedresistors.This conditionis true only if the resistaflce between correspondingterminal pails is the samefor each box. For elample,the re'istance belweenterminal.a and b mu.l be lhe ."me $herher$e use the A-connectedset or the Y connecteds€t.For eachpair of terminalsin

r A andY sLuctnrcsare presenrin a variely of usefulcircuits,rcr jusr resislivc nelworks Hencethe Aro-Y transfomationis a helptulioot in circuit analysis.

Figure3.28A A resistjve netwo*generated bya Wheaistone bridsecircuit.

F i g u r e 3 . 2 9 L configuration AA vjewed asa r

Figlre3.304 A Y niuctureviewed asa T stiuciure.

Figure3.31A TheA to-Y transformation.

'14

Simple Resktive Cncuits

the A-connected circuit, the equivalent rcsistance can be computed using seriesand pamllel simplifications to yield

Rat :

R.(R" + nb) - R 1 +R2, &+R6+R.

(3.11)

R.(R, + &) = R : + R : , R,+nb+R.

(3.42)

Rr(R. + R,) = R 1 +R}

(3.43)

Ra+Rb+R.

StaightJorward algebnic manipulation of Eqs. 3.41 3.43 gives values for the Y-connected resistors in terms of the A-connected rcsistors reouired for the A-to-Y eouivaletrt circuit:

RaR.

-Rd+ R, + &'

n.& R,+R6+R.' &Ra &+Rr+R.

11.11)

(3.45)

13.46)

Revening tle A-to-Y trarsfomation also is possible.fhat iq we can start with the Y shucture aDd replace it with an equivalent A structure. The expressionsfor the three A-connected resistors as functions of the three Y-connected rcsistors are

n1R2+R2R3+R3R1 R1 n1R2+R2R3+R3Rr R2 n1R2+n2R3+R3R1

\3.47)

(3.48)

(3.1e)

Example3.7illustiatesthe useof a A{o-Y hansfomation to simplify tlle analysisof a circuit.

3.7

(ncrirs (Pi-ro-TeF) Detta-to-Wye Equivdlert

Applyinga Delta-to-Wye Transform Find the current and powei supplied by the 40 V souce in the circuit shownin Fig.3.32.

resistance acrossthe terminalsof the 40V sourceby se es-parallel simplifications:

(501450) : 55 + -_= 80 (). R," -' I ll(,

The final step is to note that the circuit reduces to an 80 f) resistor across a 40 V souice, as shown ln Fig. 3.35, from which it is apparent t]lat the 40 V source delivers 0-5A and 20 W to the circuit.

Figure3.32A lhecircuit forExampte 3.7.

Solution We are interestedoDly in the current and power drain on the 40 V source,so the problem hasbeen solved once we obtain the equivalent rcsistance acrossttre terminals of the source.We can find this equivalenl resistanceeasily after replacingeilher the upper A (100,125,25 O) or the lower A (40, 25, 37.5O) with its equivalent Y We choose to replacethe upper A- We then computethe threeY resistances, defined in Fig. 3.33,from Eqs.3.44 to 3.46.Thus,

Figure3.33A TheequivalentY rcsistor.

37.50

100x 125= 5 0 r ) , 250 125 25 250

-^_^

100 25 250

.^ ^

Substituting the Y-rcsistoN into t]le circuit shown iD Fig. l.J2 produce( the ciJcuit sholn in Fig.3.34-From Fig.3.34,wecan easilycalculatethe

Flgure3.34t Atrandom€d version ofthecncuitshown in Fjg.3.32.

,J' jn thesimptification Flgure3.35 ^ Thefinatstep of thecjrdrit s h o wjnn F i g . 3 . 3 2 .

75

76

Simph Resistive Cncuits

0bjective6-Xnow wh€nandhowto usedetta-to-wye equivaterit circuits 3.8

Use aY-to-A tansforinaticjnto find the voltage , in Lhecircuit.horrn.

105{}

Answer:J5\. NOTE: Abo try Chapter Prcbletns 3.52,3.53, and3.51.

PracticalPerspective A ReaIWindow Defroster A nodel of a defrostergridis shownin Fig.3.36,wherer and], denotethe horizontalandverticaL spacingof the grid etemerts.Giventhe dimensions of the grid, we needto find expressions fof eachresistorjn the gfid such that the powerdissipatedper unit tengthjs the saniejn eachconductor. ThiswjlLensureuniformheatjrg of the rearwindowin both the .v and l directjons. Thuswe needto find vatuesfor the gridresistors that satis! the foUowinqretationshiDs:

:,,(+) =(+):-(+) "(+) -(+)

(3.50)

'(+),(+) (+)='(+):,'(+):, (3.51)

tigur€3,36e Modetof a deftustagrid.

(3.52)

,,(?)=',(*)

(3.53)

Webegjnthe anatysis of the grid by takingadvantage of its structure. portion (i.e., Notethat if we disconnect the lowef of the circuit the resjstor i1, iz, il, andib areunaffected. Thus,instead &, R,/,Ra,andR5),the curr€nts of analyzjng the circuitin Fig.3.36, we can analyzethe simplercircujtin

PmcticaLP€rspectjve 77

Fig.3.37.NotefurtherthatafterfindingR1,R2,R3,Ra,andR, in the circuit in Fig.3.37,we havealio foundthe vaLues for the remaining rcsjstors, since Rt -

R1

Rz, (3.54)

R" : RA,

,

R

.

Ra - R,.

Beginanalysisof the simpLified grid circuit in Fig. 3.37 by writing expressions for the currentsi1, i2, i?, andib. Tofind ir, descrjbe the equivaFigure 3.37a A sjmptified modeL ofthe lent resistance in paratlelwith R3:

-

^^

x r ( R 1+ 2 & )

_ (& + 2R.)(R2+ 2R) + 2R2Rb (Rt I R, 2RJ Forconvenience, definethe numerator of Eq.3.55as D = (\

+ 2R)(R2 + 2R) + 2R2Rb,

(3.56)

andtherefore

(Rr+R,+2R")

(3.57)

It foLtows directtythat

'.=&

-'

R.

Ydc(Rr+Rr+2&) D

(1 58)

Expressjons for i1 and 12can be found djrectlyfrom ib usjngcurrent division.Hence '

iiR1

%,R'

R--R, r'R.,

D

and .

ib(R] + 2R4) (Rt' R, 2R

vd.(Rr + 2&) D

Theexpression fori3 js simpLy

.

''

v,t R3

(3.61)

Sinrpte Resistive Circuits

in Eqs.3.50-3.52to deriveexpressions for Nowwe usethe constraints R., nb, R2,andR3 asfunctionsof R1.FromEq.3.51, R,

RI

R^=LR,=,rR,.

13.62) Thenfrom Eq.3.50we have

., =(1)'^,.

(3.63)

Theratio(ir/ir) is obtained directty fromEqs.3.59and3.60: Rz tJRz : + + i2 Rr 2R" Rr zoRl

(3.64)

WhenEq.3.64js substitutedinto Eq.3.63,we obtain,aftersomeaLgebraic (seeProbiem maniputation 3.69),

Rr=(t+2o)' Rr.

(3.65)

Theexpression for R6 asa fundjon of R1js derivedfromthe constraint imposedby Eq.3.52,nametythat /: \2

n o :l r l n .

(3.66)

\tb/

Theratio (i/i/,) js derivedfrom Eqs.3.58and3.59.Thus,

n

\_

(.\ + R2 + 2R")

ib

(3.67)

WhenEq.3.67is substituted into Eq.3.66,weobtain,aftersomeatgebraic manipuLation 3.69), GeeProbtem ^ ^'-

(L + 2c)2oR1

4(1j "r,

(3.68)

FinaLly, the expression for R3canbe obtainedfromthe constnintgiven jn Eq.3.50,or

o.: i1)'",,

(3.6e)

sLrmmary 79

RzR: D

i3

onceaqain,aftersomeatgebraic (seeProbLem maniputation 3.70),the expression for R1canbereduced to

(t + 2o)zoR1 4(I + o)'

(1 + 2o\1 (I+o)'

R 'l : - R , .

(3.70)

Theresutts of ouranaLysis aresumnarized in Tabte 3.1.

Rz

(1 + 2o\!R1 I

+ 2d\a

a *'F"'

R3

yourundeRtanding NqTE:Assess of thePrddicalPetspective bytryingChopter Problens 3.71-3.73.

5ummary . Sedes resistorccan be combined to obtain a single equivalentresistanceaccordingto the equation R"q=)Rr:Rr+R2+.

. When currelt is divided betweenpara]lel resistors,as shown in the figure, the curent Urrougheach resistor canbe found accordingto the equations

+R&. .

(Seepage58.)

''

. Pamll€l resistors can be combined to obtain a single equivalentresistanceaccordingto the equation

| R.q

: r e

Ri

I

I

r

Rr

R2

Rfr

-'

Whenjust two resistorsare in parallel,the equationlbr equi\dlcnlresisrance canhe simplifiedro gi!e ,."q

.R R, Rr + R2.

(Seepages59-60.) . When voltage is divided between sedes resistors,as shownin the figure,the voltageacrosseachresistorcan be found accordingto the equations

", ,'-

Rl

,,

Rr

Rr+ Rr'' R:

*="r*&-'" (Seepage62.)

ir"

R

l

2

Rr + tRr'" R1

l)

' r l n, nl

Rr + Rr'"

(Seepage64.)

Voltagedivisionis a ciicdt analysistool thar is usedto find the voltagedrop acrossa singleresjstancefrom a collectionof series-coDnected resistances wheDthe vol! asedroDacrossthe collectionis known: Rj t '] ' = - ? ) ' Kcq

where oi is the voltage drop acrossthe resistanceRl and o is the voltage drop acrossthe se es-connected resistanceswhose equivalent resistanceis R.q. (See page66.)

80

R€sistjve Circuits Simpt€

Curr€trt division is a circuit analysis tool that is used to find the cunent through a single resistanceftom a collection of paiallefconnected resistanceswhen t]Ie curient into ttre collection is kno$n: R"" ,j_

Digital met€rs and analog meterc have intemal resisG ance,which influences ttre value of the circuit variable beins measured.Meters based on the d'Arsonval meter mov;ment detbemtely include internal rcsistance as a way to limit t]le current in the movemenf's coil. (See page68.)

p.,

where ir is the curent thmugh the iesistance Rj and i is the current into the parallel-connected resistances whose equivalent resistanceis Req.(See page 66 )

A voltmeter measuresvoltage and must be placed in parallel with the voltage being measuled.An ideal vollmeter hasinfinite inlemal resistanceand tlus doesnot alter tlle voltage being measured.(Seepage 68.)

. An ammeter measurcs current and must be placed in serieswith the curent beingmeasured.Anideal ammeter has zero internal rcsistance and thus does not alt€r the currentbeingmeasured.(Seepage68.)

The Wh€atstone bddg€ circuit is used to male precise measuementsof a rcsistoi's value usingfour resistors,a dc voltage source,and a galvanometer'A Wheatstonebiidge js balancedwhen the resjstorsobey Eq. 3 33, resulting in a galvanometerreading of 0 A. (Seepage 71.)

. A circuit with three resistors connected in a A configuration (or a ?]'configuiation) can be transformed into an equivalent circuit in which the ttrree resistors are Y connected(or T connected).The A-to-Y transformationis given by Eqs.3.44-3.46;the Y-to-A hansformationis givenby Eqs.3.47-3.49.(Seepage74.)

Problems Sections3.1-32

b) simplify the circuit by replacing ttre seriesconnectedresistonwith equivalentresistors.

3,1 For each of the circuits shown, a) identify the resistors connected in serieq

P3.1 Figure 240{l

5ko

3ko 8ko

300()

1800

1400

10v 200()

400

soo :ov( ) 60f)

450

3oo

81 3.2 For eachof tle cftcuitsshowr in Eg. P3.2, a) identify the rcsistorsconnectedin parallel, b) simplify tle circuit by replacing the parallelconnectedrcsisto$ with equivalentrcsistorc, 3,3 a) Find the powerdissipatedir eachresistoriII the cLcuit showr)in Fig.3.9. b) Fhd tle powerdeiiveredby the 120V source. c) Showthat the powerdelivetedequalsthe power dissipated. 3,4 a) Showthat the solution of the circuit in Fig. 3.9 (seeExample3.1) satisfiesKirchhoff's current law at junctionsx andy.

b) Show that the solution of the circuit in Fig.3.9 satisfies Kirchhoff's voltage law around every closed loop.

3,5 Find lhe equivale0rresisrance seenby the sourcein each of the circuits of Problem 3-1.

3.6 Find the equivalentresistanceseenb' lhe sourcein each of the circuits ol Problem 3.2.

3J Eind the equivalent resistance R"b for each of the circuitsin Fi9.I,3.7.

3.8 Find the equivalent resistance Rabfor each of the circuits in Fig. P3.8.

Figure P3.2

Flgure P3.7 2A 10()

7ko 14()

Figure P3.8

300

140

12[r

30

500

20a ' a

4{O 300

\1

2{l

G)

z7a 24o" 120

82

Simple Resisiive Cncuitr

3.9 a) In the circuitsin Fig.P3.9(a)-(c),find the equivalentresistance-Rab, b) For eachcircuit find the powerdeliveredby the

S€€{ioE3.}3.4

c) Using tle results of (b), destn a resistive network with an equilaieDlresislaDce o[ using 1 kO resistois. d) Using tle rcsufts of (b), design a resistive network with atr equivalent resistanceof 5.5kO using 2 kO resistors.

3.13 a) Calculate the noload voltage t, foi the voltage-

3.10 Find the powerdissipatedin the 30 O resislorin the EnG c cuir ir Fig.P3 10

!!99!!1

FigueP3.10

3,11 For the circuit in Fig. P3.11calculate

divider circuit shoM in Fig. P3.13. b) Calculate the power dissipated in & and R2. c) Assume tlat only 0.5 W resisto$ are available. The noload voltageis to be the sameas in (a). Specify the smallest ohmic values of R1 and Rr. tigur€P3.13

t60v

b) the power dissipatedin the 12 O resistor. c) the power developed by the curent souice. Figure P3.U 10()

3,14 In the voltage-divider circuit shown in Fig. P3.14,

12A

the no-Ioad value of oo is 4 V Wlen the load resisf ance Rr is attached acrossthe teminals a and b, ?r, drops to 3 V Find Rr. Figure P3,14 40()

1800

-t*l-?

3.12 a) Find an expressionfor the equivalent resistance oI two resistois of value R in parallel. b) Find an expressionfor the equivalent resistance of ,?resistomof valueR in parallel.

-r

(:)

tigurcP3.9 14() 144v

16O

18fI

100 14o

10()

.l]]]

3.,,'

lR.

l_ *___l

Probtems83

3.1S The no-load voltage in the voltage-divider circuit shown ir Fig. P3.15is 20 V The smallestload resistor !ryq!!11 that is ever connecled 1ot]le divider is 48 kO. When the divider is loaded, r', is not to drop below 16V a) Design the divider circuit to meet the speciJications just mentioned. Specify the numerical value of R1and R2. b) Assume the powei ratings of commercially availableresistorsare 1/16, 118,114, 1, aad2 W. Wlat power rating would you specify? Figure P3.15 n1

3.18 There is often a need to produce more than one voltage using a voltage divider. For example, the memory components of many penonal computen require voltagesof -12 V, 5 V and +12 V, all with respect to a common reference terminal. s€lect the valuesof R1,R2,ard R3in the circuitin Fig.P3.18to meet the followitrgdesignrequirements: a) The total power supplied to the divider circuit by the 24 V source is 80 W when the divider is unloaded. b) The three voltages,all measuied with respect to the commonreferenceterminal,are ,r = 12 V, ?)2-5V,andt'3= 12V.

+ 100v Fig(reP3.18

3.16 Assume tie voltage divider in Fig. P3.15 has been constructed from 0.15 W rcsistors, How small can R, be before ore of tlle resistorsin the divider is operating at its dissipation limit? 3.17 a) The voltage divider in Fig. P3.17(a)is loaded with the voltagedivider shownin Fig.P3.17(b); that iq ais connectedto a', and b is connectedto b'. Find !',. b) Now assumethe voltagedivider in Fig.P3.17(b) is connected to the voltage divider in Fig. P3.17(a)by meansof a current-controlled voltagesourceasshownin Fig.P3.17(c).Findr.. c) What effect does adding the dependent-voltage source have on the opention of the voltage divider that is connectedto the 480V source?

3,11)A voltage divider like that in Fig. 3.13 is to be designedso that o. : /,.r'"at no load (R, : oo) and ?, = aor at full load (Rr = R,). Note that by defia) Showthat

figuleP3.17

20ko

L

R, = -R"

60ko

o

AR

and k - d ^ ^ ^'qt _ rf-

20ko 80,000 i 180ko

b) Specify the numedcal values of R1 and R2 iI & : 0.85,a = 0.80,andRo : 34 kO. c) If x'" = 60 V, specify the maximum power that will be dissipatedin R1and R2. d) Assume the load resistoi is accidentally short circuited. How much power is dissipated in R1 and Rr?

84

Simpte Resjrtiv€ Cjrcuits

3.20 a) Show that the current in the ,tth branch of the 6BG circuit in Fig. P3.20(a)is equal to tle source current is times the conductance of the *th branch divided by the sum of the conductanceqthat is,

iFr b) Use the result derived in (a) to calculate the current in the 6.25 f,) resistor in t]le circuit in Fig.P3.20(b). Figuru P3.20

b) Using your result from (a), find tle current flowiog i0 the 240 O resisrorlrom lei lo right. c) Using your result form (b), use curent division to find the current in the 140 O resistor. 324 a) Find the voltage ?J,in the circuit in Fig. P3.24. """ b) Replacethe 30V sourcewith a generalvoltage source equal to 14.Assume % is positive at the upper teminal Find o, as a function of %. hg,rrcP3.24

5ko

60ko

1ko

15 k{)

30v

3.2S Find ?1 and ?J2in the circuit itr Fig. P3.25EPr'r Figure P3,25 120

50()

321 Specify tle iesistorc ir the circuit in Fig. P3.21 meet the following design criteda: is=5mA;0s:lvti=4i21 i2 : 84; and i3 : 5la. FigueP3.21

3.26 Find o, in the cfcuit in Fig. P3.26. '5"!' tigureP3.26

R. i.

Look at the circuit in Fig.P3.1(a). a) Use current division to fhd the current flowing from top to bottom ir the 10 kO resistor. b) Using your result from (a), find the voltage drop affoss the 10 kO resistor, positive at the top. c) Using your result ftom (b), use voltage diusron to find the voltage drop acrossthe 6 kf,) iesistor, positive at t])e top. d) Using 'our resuit lrom (c). use !oltage division to find the voltage drop acrossthe 5 kO resistor, positive at the left. Look at the circuit in Fig. P3.1(b). a) Use voltage division to find the voltage drop acrossthe 240 O resistor, positive at tle left.

10ko 3ko 2kO +ir,- 4kO 15kO

12kO

327 Find i. and is in tlle circuit in Fig. P3.27. """ Figure tl,zz 10()

6'75 V

3.28 For the cftcuit in Fig. P3.28, calculate (a) t, and tsdc (b) the power dissipated in ttre 15 O resistor.

Figure P3.31 60f)

figuruP3.28 4()

2ll

329 The curent in the 12 O resistor in the circuit in Fig.P3.29is 1A, asshoM. a) Find 0s. b) Find tlle power dissipated in the 20 O resistoi.

The ammetei described in Problem 3.31 is used to measruethe current jo in the circuit in Fig.P3.32.What is the percetrtageof elror ir the measuredvalue? figureP3.32

tigureP3.29 20() 1.20

5f)

20mA

2tl

t)

3,n

l1A

I-/

t2f} 3()

6f)

Section3.5

A d'Arsonvalvoltmoteris shownin Fig.P3.33. Find t}le value of Ro lor eachof the follovrhg full-scale (a) 100v, (b) 5V,and(c) 100mV readinss: figureP3.33

3.30 a) Showfor tle ammetercircuit in Fig. P3.30that the current in the d'Arsonval movement is always1/25thof thecurent beingmeasured. b) Whatwould tle fractionbe if the 100mV 2 mA movementwereusedin a 5 A ammeter? c) Would you expect a unifom scale on a dc . d'Arsonvalannneter? FigunP3.30 100mV.2 mA

(25/n)a 3.31 The ammeter in the circuit in Fig. P3.31has a resistanceof 0.5 O. What is the percentageof error in the reading of this ammeter if ^. ,/oerror:

measwedvalue "\ / ___________ _^^.r | \ IUU/ I varue rrue \ /

a 3.34 Suppose tlle d'Arsonval voltmeter described in Problem3.33is usedto measurethe voltageaqoss the 24 O resistorin Fig.P3.32. a) What wil tle voltmeter read? b) Usinglhe definilion of the percentageoferror in a meter reading found ir Problem 3.31, what is the percentageof eror in tle voltmeter rcading? A shunt resistor and a 50 mV, 1 mA d'Arsonval movement are used to build a 10 A ammeter.A resistanceof 0.015O is placedacrossthe terminals of tle ammeter. Wlat is the new full-scale range of the ammeter?

86

5impte Resjstive Circuits

3.36 A d'Arsonval movement is rated at 1 mA and 50 nV Assume0.5 W precisionresistorsare available to use as shunt$ What is the largest full-scalereadingammelerlhal canbe designedlEyplain. A d'Anonval ammeter is shown in Fig. P3.37. Design a set of d'Arsonval ammeten to rcad the following tull-scale cunent readings: (a) 5 A, (b) 2 A, (c) 1 A, and (d) 50 mA. Specify the shunt resisror R,afor each ammeter-

c) What will the voltmeter read if it is placed aooss the 20 kO resistor? d) Will the voltmeterreadingsobtainedin parts(b) and (c) add to the readingrecordedin pal1 (a)? Explain why or why not.

Figure P3.39

Figurc P3.37

3.40 You have been told that tlle dc voltage of a power

3.3E 1!e elementsin the circuit h Frg.2.Z havethe folowadc ing values:R1 = 20 kO, R, - 80 kO, rRc= 0.82kO, Rr: 0.2kO, Vcc= 7.5V,% : 0.6V,andB = 39 a) Calculatethe valueof/s inmicroamperes. b) Assume that a digital multimeter, when used asa dc ammeter,has a resistanceof 1kO. If the meter is inseted between terminals b and 2 to measurethe current is, what will the meter read? c) Using the calculated value of iB in (a) as the correct value, what is the percentage of elror in the measurement? l.Jg The vohage-dividercircufl shown in Fig. PJ.Jq is designedso that the noload output vollage is 7/9ths of the irput voltage.A d'Arsonval voltmeter having a sensitivity of 100 O/V and a full-scale rating of 200 V is used to check the operation of tlte a) What will the voltmeter read if it is placed aqoss t]le 180V source? b) What will the voltmeler rcad if it is placed acioss the 70 kO iesistor?

supply is about 500V Wren you go to the instrument Ioom to get a dc voltmeter to measue t]Ie power supply voltage, you find that there are only two dc voltmeters available.The voltmeters are rated 400 V full scaleand have a sensitivity of 1000 O/V. a) How can you use the two voltmeters to check the power supplyvoltage? b) What is the maximum voltage that can be measured? c) If ttre power supply voltage is 504 Y what will each voltmeter read?

3.41 Assume that in addition to the two voltmeters describedin Problem3.40.a 50 kQ precisjonresistor is alsoavailable.The 50 kO rcsistoris connected in sedeswith the sedes-colnectedvoltmeters,fhis circuit is then connected acrcssthe terminals of the power supply. The reading on the voltmete$ is 328V Wlat is the voltage of the power supply?

3.42 Th€ voltmeter shown in Fig. P3.42(a) has a fu[scale rcading of 800 V The meter movement is rated 10OmV and 1 mA. Wlat is the percentage oI error in the meter readingif it is used to measure the voltageo in t}le circuit of Fig.P3.42(b)?

Prcbl€ms87 tigurcP3.42 -----800V

3.44 Assume in designing the multLange voltmetet shown in Fig. P3.44that you ignore the rcsistanceof the meter movement,

r----.

R-: I 100mv/ /\ t 'A\-/

T

a) Specify the values of R1,R2,and R]].

3.5mA

b) For eachof the three ranges,calculatethe percent ageof er:rorthat this desig[ strategyprcduces.

co*'on

tiglre P3.44 (b)

50v 3.43 A 600 kO resistor is connected from the 200 V ter' minal to the common termhal of a dual-scile volt' meter, as shown iD Fi& P3.43(a). This modified voltrneter is then used to measurethe voltage across the 360 k(! resistor in the circuit ir Fig. P3.43(b).

20v 2V

a) Wltat is the reading on lhe 500 V scale of the meter? b) What is the percentage of ellor in the measued voltage?

3.,15Design a d'Alsonval volfieter

that will have the three voltage mnges shown in Fig. P3.45. a) Specify the values of R1,R2,and R3.

FigureP3.43

| 500v

b) Assume that a 500kO resistor is connected between the 100 V terminal and the common terminal. The voltmeter is then connected to an unknown voltage using the common teminal and the 200 V teminal. The voltmeter reads 188V What is the unknown voltage? c) W}lat js the maximum voltage the voltmeter h (b) tigureP3.45

r-- -- -l

500v| I I

Modified

360ko

(b)

88

5jmple Resistiv€ Circuits

3.46 The circuit model of a dc voltage souce is shown in Flg. P3.46.The following voltage measurementsare made at 1Ie terminals of the source: (1) With the terminals of the source open, the voltage is measured at 80 mY ard (2) with a 10 MO resistorconnected to the terminals, the voltage is measured at 72 mV All measurements are made with a digital voltmeterthat hasa meter resistanceof 10 MO, a) What is the internal voltageof the source(or) in millivolts? b) Wlat is the internal resistanceof the source (Rr) ir kilo-ohms? figureP3.46 R"

3,50 Find tle power dissipatedin the 18 O rcsistorir lhe 6r.r circuit in Fig P3.50. figureP3.50

3.51 In the Wleatstone bddge circuit shown in Fig. 3.26, the ratio R2/Rr catr be set to the following values: 0.001,0.01,0.1,1, 10, 100,and 1000.The resistorR3 can be varied from 1to 11,110O, in inffements ol 1 (1. An unknown resistor is knoim to lie between 4 and 5 O. Wlat should be the setting of the R/Rl ratio so that the unknown resistor can be measured to foui signiJicantfigures?

352 Use a A-to-Y transformation to find the voltages ,1 and t,2in the circuit in Fig. P3.52. Sectiotrs3.6-3.7 3.47 Assume the ideal voltage source in Fig. 3.26 is replaced by an ideal cuirent source. Show tlat Eq.3.33is still valid. 3.48 The bddge ctucuitshown in Fig.3.26 is energized 6@ ftom a 21 V dc source.The bddge is balanced when Rl = 800o, R' = 1200o,and R3 : 600o. a) What is the value of &? b) How much curent (in miltamperes)doesthe dc source supply? c) Wlich resistor ir t]le circuit absorbs the most power? How much power does it absorb? d) Which resistor absorbs the least power? How much power does it absorb?

Figure P3.52 28f} 16|l

3.53 a) Find the equivalent rcsistance Rabin the chcuit in Fig. P3.53by using a d-to-Y transfomation invol!iDgrberesislo^Rr. Rr.and Ra. b) Repeat (a) ushg a Y{o-A transfonnatior involving resiston -R2,Ra,and R5. c) Give two additional A-to-Y or Y-to-A transformationsthat could be usedto find Rab. figuruP3.53

3.49 Find the detector curent i,i in the unbalanced Edc bridge in Fig. P3.49 if the vollage drop across the detector is negligible. Figure P3,49

60f}

20o"

Prcblems89 3,54 Find the equivalent resistance R.b in the circuit in rigureP3.54 15()

100 fia

3.58 a) Find the resistance seen by the ideal voltage source in the circuit in Fig. P3.58. b) If oab equals 600 Y how much power is dissipatedin the 15 O resistor? Figure P3.58 a

20

60J) 80()

140f)

38()

15(}

60J) 3.55 In tlle circuit in Fig. P3.55(a) the device labeled D P5E.Erepresentsa component that has the equivalelt circuit shown in Fig. P3.55(b).The labels on the terminals of D show how the device is connected to the circuit. Find ?,xand the power absorbedby the device. Figure P3.55

3.59 Use a Y-to-A transformation to find (a) i,; (b) irt 6dc (c) 4; and (d) the power delivered by the ideal current sourceir the circuit in Fig. P3.59. tigureP3.59 320{))

3.56 Find id and the power dissipated in t}le 140 O resisEtrG tor in the circuit ir Fig. P3 56.

3.60 For the circuit showrl in Fig. P3.60,find (a) ,1, 0) o, k) t , and (d) the power supplied by tle voltage

Flgure P3.55 224

20 tL

Figure P3.60 300

240Y

) "n-'..-. Itogrruon 10()

182()

80

120

3.57 Find nabin the circuit in Fig. P3.57. Figure P3.57

1.8kO

3.61 Derive Eqs.3.44-3.49from Eqs.3.41 3.43.Thefollovrhg lwo hints should help you get stated in the dglt dhectioni a) To find Rl as a tunction of &, R6, and Ra,filst subtractEq. 3.42ftom Eq.3.43and then add this result to Eq. 3.39.Use similar manipulationsto find R2 and R3 as functions of R,, R6,and R..

90

Simpte Resistjve Cncuits b) To fitrd Rb as a function of n1, R2, and R3,take advantage of the derivations obtained by hint (1),namely,Eqs 3.44-3.46. Note that theseequations can be divided to obtait R2

R^ = ;, ^b

R. R, = ;Rb,

or

b

and xr

Ro

&:&.

!--Rr-

t--..1 d

Attenuator

R2^

. ' ^^ , : x ^ '

3.64 a) The fixed-attenuator pad showl in Fig. P3.64 is

Now usetheseratios in Eq.3.43to eliminate& and &. Use similarmanipulationsto find Raand & as functions ol R1,R2,and R3. Show that the expressions for A conductances as lunctions of the three Y conductancesare

called a brillgel tee. Use a Y{o-A transformation to show that Rab : RL if R = RL. b) Showthat when R : RL,the voltageratio o,/o, equals0.50. FlgueP3.54

GzGz Gr+G2+G3' GGz Gt+GziGt' GtGz G\+G2+G3'

"

i________,_

pad Fixed-attenuator

^

r, :

1

^

o.

r. :

1

&.

erc.

I '

The designequationsfor the bddged-teeattenuator circuit in Fig. P3.65are

2RRi

Seclions3.1-3.7 Resistor networks are sometimes used as volume control c cuits. In this application, they are refened to as resr.rtdrceatknu&tors or pads. A typrcal fixed-attenuator pad is shown in Fig. P3.63. In designiDg an drlenuado0pad.lbe circuildesigner will select the values of R1 and R2 so that the iatio of r,/oi and the resistanceseenby the input voltage souice R"6 both have a speciJiedvalue. a) ShowthatifRab: RL,then R i : 4 R r ( R 1+ R ) , R2

2R1 -r R2 + RL b) Select the values of R1 and R2 so that Rab- RL = 600O and 0,/?,r : 0.6.

3n-RL 3R + RL' when R2hasthe valuejust given. a) Designa fixed attenuatorso tlat oi : 3oowhen RL:600o b) Assume the voltage applied to the input of the pad desigtred in (a) is 180 V Which iesistor in the pad dissipates t}le most power? c) How much power is dissipatedin the resistorin part (b)? d) Which rcsistor in the pad dissipatesthe least e) How much power is dissipated ill the resistor in part (d)?

Probtems91 Figure P3.65

show that the percent elror in the approximation of o, in koblem3.66 is /A P\P.

o/oenor = -::il:i:. 100. (fi, + &r)ll4 b) Calculate the perce error in od,using the values h Problem 3.66(b).

3.68 Assume the eror in t'd in the bridge circuit in 3.66 a) For the circuit shown in Fig. P3.66 rhe bridge is balancedwhen AR = 0. Showthat if A-R << R, the bridge output voltage is approximately

-An& U"^ = - r ) , -

( R , + R 4 r-

Fig.P3.66is not to exceed0.5%.Whatis tlle largest percent change in Ro that can be tolerated? 3.69 a) Derive Eq.3.65.

b) DedveEq.3.68. ,:lNfl[ii, 3.70 DeriveEq.3.70.

b) Given R, : 1 kO, R3 : 500O, n4 = 5 kO, and oin - 6 V, what is the approximate bridge output voltageif AR is 3oloof Ro? c) Find the actual value of 1'" in part (b). Figure P3.66

3,fl Suppose the grid structurein Fig.3.36is 1 m wide and the veitical displacementof fie five hodzontal gdd linesis 0.025m. Specifythe numericalvaluesof R1 R5and R, - Rd to achievea unilorm power dissipationof 120rym, usinga 12V powersupply. (Hrrri Calculate (' filsr,thenR3,R1,Ra,Rr, andR2 in thatorder.) 3,72 Checkt}le solutionro Problem3.71by sho$ringthat J#;lHlEthe total power dissipatedequalsthe powerdevele;ai- opedby the 12V sowce. 3.73 a) Design a defroster grid in Fig. 3.36 having five horizonral conductors lo meel

the following T!ryla! specifications:The gdd is to be 1.25m wide, the ry,ryL. ve ical sepamtion between conductors is to be

3.67 a) llperce Ifpercent eror is definedas approxrmale value

r] , roo

0.05 m, and the power dissipation is to be 150Wm when the supplyvoltageis 12V b) Check your solution and make sure it meets the design specifications.

Techniques of Circuit Analysis Sofar, we haveanalyzedretativetysimpleresistivecircuits 4.1 T€rminotogyp. 94 4.2 Introductionto the Node-Vottage 4.3 The Node-VoLtigc Methodand Dependent 4.4 The Noile-Vottage Method:SomeSpeciaL Casesp. ?01 4.5 tntroductionto the l4esh-[urrent l4ethodp. 105 4.6 Thel'lesh-Current l4ethodand Depend€nt 4.7 Tte li,lesh-Current l,lethod:Som€Spe.iat t.ses p. 109

4.8 Th€Node-Voltage Merhod V€rsus the M€sh-Current ethodp. 112 4.9 Source lransformations p. 116 4.10Thevenin andNortonEguivalents p. 119 4.11 ltlor€on Derivinga Th6venin EquivaGnt p. 1?3 4.12 i\,laximunPowerTransferp. 126 4.13 Superpositionp. ?29

1 Understaid andbe abteto us; ihe iode-voltage nrethod to soLve a circuit2 Understand andbe abiet0 us€the mash-cunent methodio solvea circuit. 3 Beableto decidewhetherth€node-vottage methodor the m€sh-cunent methodis the preturred approach to soLvjng a partjcutar circuit. 4 Understand sourcetransforflatiorandbe abLe to us€it io sotvea circuit. 5 Und€Etand the conceptofthe Th6venin ard Noton equjvaLent circujtsafd be abteto construct a Theverjnor Nortonequivatent for a 6 Knowtlr€ conditionfor maximum powertransfer to a resistivetoadandbe ableto calcuLate the vatueofthe Loadresistorthatsatisfiesthis

g2

by applyingKirchhoff's laws in combinationwith Ohm's law We can use this approachfor all circuits,but as they becomestructurally more complicatedand involve more and more elements, this direct methodsoonbecomescumbersome.Inthis chapterwe introduce two powerful techniquesof circuit analysisthat aid in the analysis of complex circuit structures: the node-voltage method and the mesh-curent method.Thesetechniquesgive us two systematicmethodsof describingcitcuitswith the minimum Ilumberof simultaneousequations. In additioll to these two generalanalyticalmethods,in this chapterwe also discussother techniquesfor simplifyingcircuits. We have alreadydemonstratedhow to use sedes-parallelreduc tions aIId A-to-Y transformatjonsto simplify a circuit'sstructure. We now add sourceffansfor'rnationsand Thdvenin and Norton equivalentcircuitsto thosetechniques. We also considertwo other topics that play a role in circuit analysis.One,maximumpower transfer,considersthe conditions necessary to ensurethat the power deliveredto a resistiveload by a sourceis maximized.Th6venin equivalentcircuits are usedin establishingthe ma,\imum power transfer conditions.The final topic in this chapter,superposition,looks at the anatysisol circuitswith more than one independentsource.

Perspective Practicat Resistors Circuits withReatistic the effectofimprecise In the tastchapterwebeganto expLore on of a circuit;specificaLty, resistorvatueson the peformance are manufac the perfomanceof a voltagedivider Resistors vaLues, andanygiven turedfor ontya smatlnumberof discrete witl varyfiom its statedvalue resistorfroma batchof resistors wjth tighter toterance,say within sometoteBnce.Resistors than resjstors wrth greatertolerance, 1ol0, are moreexpensive it in a circuitthat usesmanyresjstors, say 100/".Therefore, whichresisto/svatuehas woutdbe importantto understand perfornrance ofthe circljt. on the expected the grcatestimpact

In otherwords,we wouldtiketo predjctthe effectof varying vatueon the outputof the circuit.If we know eachresistor's resistormustbe veryctoseto its statedvatue that a palticul"ar for the crrcuitto functioncofiectly,we can then decldeto to achieve a tr'ghtertoLerance spendth€ €xtramoneynecessary on that resisto/svatue. valueon the ExpLorirg the effectof a circuitcomponenfs 0ncewe have cifcujfsoutputisknownassensitivityanaLysis. the topic of presented technjques, additionatcircuitanatysis wL[be examjned. sensitivjtyanatysis

93

94

ofCncuitAnatysis Techniques

4.1 Ternninology To discussthe more involvedmethodsof circuit analysis,we must define a few basic terms. So far. all the circuits presentedhave been planar circuits-that is, those circuits that can be drawn on a plane with no oossing branches.A circuit that is drawn witb crossingbranchesstill is consideredplanar if it can be redrawn with no crossoverbranches.For example,the circuit shown in Fig. 4.1(a) can be redrawn as Fig 41(b); the circuits are equivalentbecauseall the node connectionshave been maintained.Therefore,Fig.4.1(a) is a plarar circuit becauseit can be redmwn as one. Figure 4.2 shows a nonplanar circuit-it cannot be redra$n in sucha way that all the node connectionsare maintainedand no branchesoverlap.The node-voltagemethodis applicableto both pla_ nai and nonplanarcircuits,whereasthe mesh-currentmethod is limited to planar circuits. (b)Ihesame cncuii rigurc4.1A (a)A planar cncuit. rednwn to venry thatit is ptanar RI

R2

RS Fiqlre 4.2 d A nonplanarcircuit.

Describing a Circuit-Thevocabulary In Section1.5we defined an ideal basiccircuit element.Whenbasiccircuit elementsare interconnectedto form a circuit, ihe resulting interconnectionis describedin terms of nodes,paths,branches,loops,and meshes.We defined both a node and a closed path, or loop, jn Section2.4.Here we restalethosedefinitions and then define the terms ail of thesedefinitionsare path,blanch, ardmesh.Foi your convenience, presentedin Table 4.1.Table 4.1 also includesexamplesof eachdefinition taken fromthe circuit in Fig.4.3(seepage95.).whichare developed in Example4.1.

Exanple Fron Fig. (3 A point wheretwo or more circuit elenentsjoin A nodeehere three or more circuitelemenrsjoin path

;

A traceof adjoiningbasicelementsvilh no elementsincludedmore than once A path that connectst\vo nodes A path {hich connectstwo esential nodeswithout pasiDg throughan essentialnode Apath whoselastnodeis the sameasthestartingnode Aloop that doesnotencloseany olher loops A circuir that can be drawn on a plane lvith Do

,r Rr 2r-Rr

R5 R6- Ra ,? Rj R, ?r-Rr-nj Fis.4.3is aplanarcncuit circuit Fig.4.2is a nonplanar

4.1 Teminotogy 95

IdentifyingNode,Eranch,MeshandLoopin a circuit For the circuit in Fig 4.3,identify a) a[ nodes.

b) all essentialnodes. c) all brarches. d) all ess€trtialbranches. e) all mesh€s. 0 two paths that are not loop6 or essentialbranches. g) two loops that are trot meshes.

Sotution a) The nodes are a, b, c, d, e, f, and g. b) The essentialnodesare b, c, e,and g. c) The branchesare ?J1,tb, Rt, R2,R3, R4,R5, R6, R?,and 1. d) The essential branches are Rz - Rl 2,2 R4,R5,R6,R7,and I e) The meshes are \ - R1- R5 - R3 - R2, u 2 R 2 R 3- R 6 - R 4 , R 5 - R 7 - R 6 , a n d

& r .

2 )

R, R z d R 3

",(

\ )

jttustnting nodes, bmnches, nreshes, Figure4.3 r A circujt

D R1 ns R6 is a path, but it is not a loop (becauseit does not have the same starting and ending nodes), nor is it an essentialbmnch (becauseit doesnot connect two essentialnodes), o, - R2 is also a path but is neither a loop nor an essentialbranch,for the samereasons. g) ?J1 Rl R5 - Re - Rr - ?'2 is a loop but is not a mesh,becausethere are two loops wilhin it. / R5 Rois alsoa loop but ool a mesh.

NOTE: Assessyour understandingof this motelial bt ttting Chopter Prcblettus4.2 snd 4.3

Equations-HowMany? Simultaneous The number of unktrown currents in a circuit equals the number of blaltrche$ ,, where the current is not known. For exadple, the circuit showtr itr Fig. 4.3 has nitre branches in which the cu[ent is uoknown. Recal that we must have D itrdependent equations to solve a circuit with b unknown currenls If we bt ,r rcpresent the number of nodesin the circuit, we can derive ,l 1 independent equations by applying Kirchhoffs current law to any set of'' - 1 nodes.(Application of the current law to the ,rth node doesnot generatear iDdependentequation,becausethis equation can be derivedfrom the previousrl - 1 equation$SeeProblem4,5.) Becausewe need b equationsto describea giveDcircuit and becausewe can obtain n - 1 of these equations from Kirchhoff's current law, we must apply Kirchhoff'svoltagelaw to loops or meshesto obtain the remaining b (/r l) equations. Thus by counting nodeE meshes,and brancheswhere the cuuent is unknoM, we have establisheda systematicmethod for writing the necessarynumber of equationsto solve a circuit. Specifically,we apply 1 nod€s and Kirchhoff's voltase law to Kirchhoff's curent law to n

96

Techniques of CncuitAnatysh b - (n - 1) loops (or neshes). These obseryations also are valid in terms of essential nodes and essential branches.Thus iI we lel ne represent the number of essentialnodes atrd 4 the number of essential brancheswhere the current js unlno*n, we car apply Kirchhoffs current law at n. - I oodes and Kirchhoffs voltage law around 4 - (n - 1) loops or meshes. In circuits, the Dumber oI essential nodes is less than or eoual to the number of nodeq and the number of essential bmnches is less than or eoual to the number ot branches. Th us it is ofletr convenient to useessentialnodes and essential branches when anallzing a circuit, because they produce fewer independent equations to solve. A circuit may consist of disconnectedparts.An example of such a circuit is examinedin Problem4.3.The statementspe ainingto the number of equationsthat can be derivedfrom Kirchhoffs curent law,r - 1, and (n - l), apply to connectedcircuits.If a circuit has voftage laqwib ,?nodesand b branchesand is made up of r parts,the current law can be appliedn s times,and the voltagelaw 6 - /, + r times.Anytwo separate partscanbe connectedby a singleconductor.This connectionalways causestwo nodesto fofin one node,Moreover.nocu[ent existsin the single conductor,so any circuit made up of s disconnectedparts can always be reducedto a connectedcircuit.

TheSystematic Approach-AnIllustration /-a R5

) R z a R : r

i

R,c

:

' r ! R6

\

i

J n4 f

We now illustmte this systematic approach by using the circuit shown in Fig. 4.4. We write the equations on the basis of essential nodes and branches. Th€ circuit hasfour essentialnodesand six essentialbranches, denoted ir - ,;, for which the current is unhown. We derive thrce of the six simultaneous equations needed by applying Kirchloffs current law to any three of the four essentialnodes.We use the nodesb,c,and e to get

-

Fi$re 4.4a ne circuitshown in Fig.4.3withrix unknou,n bEnchcunents defined.

it+i2+i6-I:0, ir-ir

i5

4 + i4

i2 = 0.

0. (4.1)

We derive the remaining ttrree equations by applying Kirchhoff's voltage law around three meshes.Becausethe circuit has four meshes,we need to dismiss_one mesh.Wechooseri? 1, becausewe don't know the voltage across1,, Using the otler threemeshesgives R1il + R5i, + 6(R, + Rt - 1,r : 0, -i3(n, + Rl) + iaR6+ i5na

t2 = 0,

-irR5+i6R7-loR6=Q.

' W. saymole aboutthisdcchionin Section4.7.

(4.2)

a.2

hiroduction to theNod*VoLtage Method

91

RearrangiDgEqs.4.l and4.2ro tacilitare lheir.olulionyieldslheser

-ir + i + 0t3+ 0i4+ 0i5+ t6 = 1, i1 + Oi2 i) + ola i5 + 0t6:0, qir- i2+ h + i4 + 0i5 + 0i6 = 0,

R1i1+R5i2+(Rr+R3)t1 + ota+ ots+ 0t6= r,1, 0i1+ 0i' - (R' + R3)i3+ R6i4+ Rai5+ 0i6= or, 0i1

R5i2 + 0i3

R6ia + Ois + R?i6 = 0.

\4.3)

Note that summing the curent at t}le nth node (g in this example) gives is

ia

i6+I:0.

(4.4)

Equation 4.4 is not independent, because we can deiive it by summhg Eqs.4.1and tlen multiplyingtle sum by -1. Thus Eq.4.4 is a linear combination of Eqs. 4.1 and theiefore is not itrdependent of them. We now cally the procedureone stepfurther.By introducingnew vadables,we can des€ribea circuit with just,1 - 1 equationsor jusr b - (n - 1) equations. Therefore these new variables allow us to obtain a solurion by manipula! ing fewer equationq a desirable goal even if a computer is to be used to obtain a nume cal solution. The new variables are known as rode voltages and mesh curents.The rode-voltage method enables us to describe a circuit in tems of r? 1 equations;tle mesh-currentmethod enablesus to desffibe a circuit in tems of be (na 1) equations.We begin in Section 4.2 $dth the nodevoltagemethod. NOTE: Assessyow unde$tunlling of this mateflil by trying Chaptel Prcblens 1.1 an.l 1.4

4.2 Introductionto the Node-Vottage Method We introducethe node-voltagemetlod by usingthe essentialnodesof the circuit. The first step is to make a neat layout of the circuit so that ro branchesclossover and to mark clearlythe essentialnodeson the circuit dia$am, as in Fig.4.5. This circuit has three essentialnodes (/,": 3); therefore, we need two (r?" - 1) node-voltage equations to descdbe the circuit. fhe next step is to select one of the three essential nodes as a rcferencenode. Although theoretically the choice is arbihary, pmctically the choice for the reference node often is obvious.For examDle.the Dodewith

10v

Figlre 4.5 A A ciruit usedto ittustntethe node-vottage method of circujtanatysk.

98

Techniques of CncLrit AnaLysis

1f,) ^ lq

jn Fig.4.5witha Figure4.6 A Thecncuitshown rcference nodeandthe nodevoltages.

10v

Figure 4.7A Computation ofthebranch currcnt t.

the most branchesis usually a good choice.The optimum choice of the reference node (if one exists) wi become apparent after you have gained someexpefience usinglhi\ method.In rhe circuirsboqDiD fig.4.5.rbe lo\rernodeconnect5 rhemoslbranchelso $e useir asLhereference node. We flag the chosenreferencenode with the symbolV,as in Fig.4.6. Atter selectingthe referencenode,we definethe nodevoltageson the circuit diagram. A nod€ volfage is defined as the voltage rise from the reference node to a nonreference node, For tlis circuit, we must define two node voltages,which are denotedol and 2'2in Fig-4.6. We are now ready to generate the node-voltage equations.We do so by first writing the cufient leaving each branch connected to a nonreference node as a function of the node voltagesand then summinb thesecunents to zem in accordancewith Kirchhoff's current law. For the circuit in Fis. 4.6. lhe currenl awayfrom node I lhroughthe I O res;\rori( lhe rohag;rop acrosst]le resistor divided by the resistance(Obm's law). The voltage drop acrossthe resistor, in the direction of the curent away hom the node, is r)r - 10.Thereforet]le cunent in the 1 J) resistoris (o1 10)/1.Figure4.7 depictstheseobsenations.Itshowsthe 10 V-1 O branch,with the appropriate voltages and culrent. This samereaso ng yields the curent in every branch where the current is unknown.Thus the current away ftom node 1 thtough the 5 O resistoriso/5, and the cunent awayfrom node l thrcughthe 2 O resistor is (or ?)r)/2.The sum of the three curents leavingnode 1 must equal zero;thereforethe rode-voltageequationderivedat node 1is r'r - 10 1

r'1 5

2

'

14.5)

The node-voltageequationderivedat node 2 is

2

1

0

\4.6)

Note that the first terminEq.4.6 is tie current awayfrom node 2 through the 2 O rcsistor, the second term is the current away ftom node 2 tlrough the 10 O resistor,ard the third term is the cu[ent away from node 2 through the current source, Equations 4.5 and 4.6 are t]te two simultaneous equations that descdbethe circuit shownin Fig.4.6in tems of the node voltagesor and ?)2. Solving for r'1 and ?,l2yields .l-

r_ r,-

10n -ll

1 t-o tl

= 9.09V

- 1 0 . 9 v1.

Oncethe node voltagesare knowi\ all the branchcurents canbe calculated.Once theseare knoM, the branchvoltagesand powen can be calculated.Example4.2illustratesthe useofthe node-voltagemethod.

4.2

llethod to the NodeVottage Introduction

99

Method Usingthe Node-Voltage a) Use the node-vollagemethod of circuit analysis to fhd the branchcurrentsi., ib,and ic in ihe cir cuit shownin Fi9.4.8. with eachsource,and b) Find the powcr associated statewhetherthe sourceis deliveringor absorbing power.

5o 50v

1 ) ' | , n n , , $ r o( ot

42 Figur€4.8 .i, ThecircuitforEranrph

SoIution a) We beginby notingthatthe circuithastwo essential nodes;thirs we need to write a singlenodeWe selectthe lower node as voltageexpression. the rcference node and define the unknown node voltage as r,1.Figure 4.9 illustratesthese deci' sions.Surming the currenls away fiom node I generatestlle node-vohageequation 50

0,-

s

,1 +i]+i

1

0

0l

4

-3:0 0

l-./Tl'<po\re' a*uciatcdt\irh Ih.
gives Solvingfor ?J1

= ) A

; : ' '

'"

in Fig.48 witha rcference shown figure4.9 r. Thecircuit or. node voliage nodeandtheunknown

10

40 ' 4 0

p3A=

3?,r:

3(40)=

120w (deh'ering).

We check thesecalculationsby noiing that the total deliveredpower is 220W.The total power absorbedby the three resisrorsis 4(5) + 16(10) + 1(40), or 220W, as we calculalcd and as it

method andb€ableto usethe node-voltag€ 1-Understand Obiective l)

For the circuit sho\ln, use the node-voltage methodto lind r,1,r,2,ancli1. How much power is deliveredto the cfcuit by the 15A source'l

c) Repeat(b) for the5 A source. Answ€r:(a) 60V 10Y 10A;

(b)eoow; (c) s0w.

5r} FL

--

)rro, |uon *"n

I

+

}rn , (l

100

Iechniqu€5 ofCncLrit Anay5is

4.3 TheNode-Voltage Method andDependent Sources Ifthe circuitcontahs dependentsources, the node-voltageequationsmust be supplementedwith the constraintequationsimposedby the presence of the dependentsources.Example ,1.3illustratesthe applicationof the node-voltage method to a circuit containing a dependent source.

Usingthe Node-Voltage Methodwith Dependent Sources Use the node-voltage metiod to find the power dis(ipatedin lhe 5 Q resisrorin lhe circuilsho$n in Fig.4.10.

5f)

+

20o-

8b

10(l

Summing the curents away from node 2 yields 01- x1

u2

Ir2- 8lp

^

As \raritten,these two node-voltage equations con, tain tluee unknowns, namely, 01,?r2,and id. To elim inate id we must expressthis controlling current in terms of the node voltages,or

Figure4.10A Thecircuit forExampte 4.3.

Substitutingthis relationshipinto the node2 equation simplifies thetwonode-voltage equations to

SoLution We begin by noting t]Iat the circuit has three essential nodes,Hence we need two node-voltage equations to describethe circuit. Four branchestemtnare on the lower node, so we select it as the rcfcrcrce node. The two unknown node voltages are defined on t}te circuit sho$,nin Fig. 4.11. Summing the currents away ftom node 1 generatesthe equation o-, ' 2 0 o, + j +

2

2

s

"

br+16u2=0. Solving for ?J1atd zJ2gives

q:16v and

u-, - t )' , = n

0

0.75u1-0.2u2- 10,

'

t'2 = 10V.

4.4

l,lethod: Some speciatcases 101 TheNodeVotiage

2{)

16

51)

1

2{t

10

lsr) : (1.14)(5): 7.2w. A good exercise10build your problen solving intuition is to reconsiderthis example,usingnodc 2 as the rcferencenode. Does it make the analysis

tigure 4.11 ;1lThecircuitshownin Fig.4.10,\i/itha refelen.e nodeandthe nodevoLtaqes.

method objedive l-ljnderstand and be able to usethe node-vottage 4.3

a) Usc the node-voltagemcthod to find the power associated with eachsourcein ihe

3it

6f)

b) Statewhetherthe sourceis delive ng power

2A

to the circuitor extractingpower from the

Answen(a)p5ov: 150W,11= -144W. zre = -80 w;

50v

80

4 ' , (

(b) all sourcesare deliveing power to the clrcurt.

NOTE: Also try ChaptetProbkms 1.19and 424.

f4ctltsd: 4.4 TheFlodc-tfo{tage

5*rn*Special{ias*s I i0{] 2 When a voltagesourccis the onll' elemenlbctweentwo essentialnodcs, the node-voltagemethod is simplified.As an example,look at the circuit in Fig. 4-12.There are lhree cssentialnodesin this circuit. which means 100v that two simultaneousequationsare needed.From thesethree essential nodes,a rcferencenode hasbccn chosenard two olher nodeshave been latreled.But the 100V sourceconstrainsthe voltagebetwcennode 1 aDd the retercncenode to 100V This meansrhat there is only one unknown nodc voltage(rr). Solutionoflhis circuitthus involvesonl]' a singlenode figurc4.123 A circuit\vitha knownnodevottaqe. vollageequationat node2i ,

10

2

.

,

50

,

14.7)

But rr = 100V. so Eq.4.7 canbe solvedlbr tr. 1)2= 125V.

(4.8)

Knowing riz,we cancalculatethe curent in everybranch.Youshouldverify that the current into node 1 in lbc branchcontainingthe independent voltagesourceis 1.5A.

102

Iechniques of CircuitAnatysis

Figufe4.13A A cjrcuitwitha dependent votiage source connected betw€en nodes.

In generul,when you use the node-voltagemethod to solvecircuits that havevoltagesourcesconnecteddirccrly betweenessentialnodeqthe number of unknown node vollages is reduced. The reason is that, whenever a voltage sourceconnectstwo essentialnodes.it constraiN tle differencebetweenthe rode voltagesat thesenodesto equalthe voltag€of the source.Taking the time to seeif you can ieduce the number of unknowns in this way will simplily circuit analysis. Supposethat the circuit shownit Fig.4.13is to be analyzedusingthe node-voltagemethod.The circuit containsfour essertial nodes.so we anticipatewriting three flode-voltageequations.However,two essential nocles arc connected by an independent voltage source, and two other essentialnodesare connectedby a current-controlleddependentvoltage source,Hence,there actuallyis only one unknownnode voltage, Chooring$hich node lo useas rhe reterencenode in\olvesseveral possibililies.Either node on each side of the deDendentvoltase source look. atbacrivebecause. iJ cho<en. oDe of lhe ;ode volldger;ould be known to be eitler +10id oeft node is rhe rcference) or 10td(righr node is the reference)The lower node looksevenbetter becauseone nodevoltage is irnmediately known (50 V) aIId five branches terminate there. We tierefore opt for the lower node as the reference. Figure 4.14showsthe redrawn circuit, with tie reference node flagged and the rode voltages defined. Also, we intrcduce the current i becausewe cannot express the cufent in the dependent voltage souce branch as a functior of the node voltageso2and 2)3.Thus,at node 2

5 jn fig.4.13.wiihthe rigure4.14a Thecncuit shown setected node voliages defined.

5

0

'

"

14.e)

'

aIId at node 3

fr-; +:o

(4.10)

We eliminatei simplyby addingEqs.4.9and 4.10ro get I)

Ur

5

i

'I)1

5(J too

-

"'

(4.1L)

TheConcept of a Supernode

4A

Figure4.15A Consideing nodes 2 and3 to bea

Equation 4.11 may be written directly, without resorting to the intermediaie steprepresented by Eqs.4.9and4.10.Todo so,we considernodes2 and 3 10 be a single node and simply sum the currents away from the node in terms of the node voltages02 and ?J3. Figure 4.15illustrates this approach. When a voltage sourceis betweentwo essentialnodes.we can combine those nodes to form a supenode. Obviously, Kirchhoffs current law must hold for the supemode.Ir Fig.4.15,srartingwirh rhe 5 O bmnchand moving counterclockwise aroundthe supernode,wegeneratethe equation b) -

5

Di

'01

50

u1

100

-

"'

(4.r2)

SpecjaL Cases 103 l'4ethod: Some 4.4 TheNode_Vottage which is identical to Eq. 4.11. Creating a supemode at nodes 2 and 3 has madethe task of analyzingthis circuit easier'It is tlerefore alwaysworth taking the time to look for this tlpe of shortcut before writing any equations. After Eq.4.12hasbeendedved,thenext stepis to reducethe expression to a single unknown node voltage. FiISt we eliminate ,1 from the equation becausewe know that ,r = 50 V. Next we expressD3as a func_ tion of 02: ,3=12+10i0.

(4.13)

We now expressthe current controlling the dependent voltage source as a functionof the node voltases:

u2 50

14.14)

UsingEq$4.13and4.14andor = 50VreducesEq4.12to ( 1

, , \ s o I- 5 -# . # )

:10+4+1,

r2(025)= Is,

FromEqs.4.13 and4.14: 60

50 5

24, in circuitshown Figsre4,16A Thetnnsistoramptifier 69.2.?4.

?r3:60+20=80V.

Analysisof the AmplifierCircuit Node-Voltage Let's use tlle node-voltagemethod to analyzethe circuit ftst introduced in Section2.5and shownagah in Fig.4 16 wlten \le used the branch-current metlod of analysis in Section 2.5, we faced the task oI writing and solving six simultaneous equations Herc we will show how nodal analysiscan simplify our task. The ciicuit hasfour essentialnodes:Nodes a and d are connectedby an independentvoltagesourceas are nodesb and c Thereforethe prcblem reduces to finding a single unknown node voltage, because (n" 1) - 2 = 1 Using d as the referencenode,combinenodesb and c the voltagedrop acrossR2astb, andlabel the volf hto a supernode,label agedrop acrossRE asoc,asshowllinFig.4.17 Then,

* ,,fn

uo =u" r*-Bi":0.

,'-, '-' ,-\

f i g u r e4 . 1 7A l h ei r c u ' t+ o w r : r F 9 . 4 . 1 6w. i r h rdentified vottages andthe lupernode

104

Techniques of Cncuit Analysis We now eliminateboth tc and iB from Eq.4.15by noting rhar

.,.

(in

Brr)Rr.

\4.16)

'Dc:n6-Ur'

14.17)

Substituting Eqs.4.16 and4.17inroEq.4.15yields - l R

' r R

.

rl

_

B ) R ,I

l

R

- n o rI I P,R/

'-'

SolvingEq.4.18for ob yields

yc.n (1 + B)Rr + %R1R, RrR2 + ( 1+ B ) R E ( R I + R )

(4.1e)

Using the node-voltagemethod to analyzethis c cuit reducesthe prob lem bom manipulatingsix simultaneousequationsGee Problem2.27)ro manipulatingthree simultaneousequations.You shouldverify rhat,when Eq.4.19is combinedwith Eqs.4.16and 4.17,the solutionfor is is identical to Eq.2.25.(SeeProblem4.30.)

Objective l-t

ndectandandbeabt€to usethe node-voltag€ method

4.4

U s e t b e n o d e - v o l l a g em e t h o d l o l i n d , . i n r h e cifcuil .bo\ n.

4,5

U(e (henocle-voltage merhodlo find , in lhe circuitsbo$D.

Answer:8V Use lhe node{ollaqe melhod to find L, in lh,

NOTE: Also try ChapterProbkms 4.21,426,and 1.27.

30r)

4.5

llethod Introduction to the Mesh-Curent

105

4.5 Introductionto the Method Mesh-Current As stated in Section 4.1, t}]e mesh-currentmethod of circuit analysis (/,e 1) equations.Recall enablesus to describe a circuit in terms of r" tlat a mesh is a loop with no otler loops inside it. The circuit in Fig.4.1(b) t+ is shown agai in Fig. 4.18,with cu ert arrows inside each loop to distinguish it. Recall also that the mesh-curert method is applicable only to planar circuits. The circuit in Fig. 4.18 contahs seven essential branches where the current is unknown and four essentialnodes,Therefore, to solve (4 1)] mesh- Figue4.18A Thecjrcujtshown in Fjg.1.1(b), withthe it via the mesh-curent method, we must *.dte four [7 A mesh cunent is the current that exists only il the penmeter of a mesh. On a circuit diagram it appears as eitier a closed solid line or an almost-closed solid line that follows tlte pedmeter of the apprcpriate mesh.An arrowhead on the solid line indicates the reference direction for the mesh current. Figuie 4.18 shows the four mesh ctments that describe tle circuir in Fig. 4.1(b). Note that by definition, mesh currents automatically satis{y Kirchhoff's curent law. That is, at any node in the circuit, a given mesh current both enten and leaves the node. Figure 4.18 also shows that identifying a mesh current il terms of a bmnch current is not always possible. For example,the mesh curent t2 is not equal to any branch cutent, whereasmesh currents i1, 4, and i4 can be identified with branch currenls. Thus measudng a mesh curent is not always possible; note that tlere is no place where an ammeter can be insefted to measure t]]e mesh current i2. The fact that a mesh current can be a fictitious quantity doesn't mean that it is a uselessconcept. On the contrary, the mesh-curent mefiod of circuit analysis evolves quite naturally ftom the bmnch-current equations. We can usethe circuit in Fig.4.19to showthe evolutionof the mesh_ current technique.We begin by using the branch currents (ir, i2, and i3) to formulate the set of independent equations. For this circuit, 4 = 3 and }?, = 2, We can write only one independentcurent equation,so we need two independent voltage equations Applying Kirchhoff's curent law to the upper node and Kirchhoff's voltage law around the two meshesgeneidevetopnrent to itLusirate tiglre4,19A A circuiiused ates the followins set of equations: analysis. method of circuit 0fthemesh-cunent (1.20) iI: i2 + i3, ?)r = i1R1 + i3R3, -D2 = izR2

(4.21) \4.22)

iR}

We rcduce this set of three equations to a set of two equations by solving Eq.4.20for i3 and then substitutingthis expressioninto Eqs.4.21and4.22: ?,r = ir(R1 + R3) -r2:

4nj,

i7R3+ tr(R, + R3).

\4.23) (4.24)

we can solve Eqs 4.23 and 4.24 for i1 and t, to replace the solution of three simultaneous equations \.ith the solution of two simultaneous equa1 curent tions.We derived Eqs.4.23 and 4.24 by substitutingthe 'l"

106

Techniques of CncuitAnaLysis equations into the 6d (,?" - 1) voltage equations.Thevalue of t}le meshcurrent method is that, by defining mesh curents, we automatically eliminate the ,1e- 1 curent equations. Thus the mesh-curent method is equivalent to a systematicsubstitution of the /,a 1 curent equations into the b" - (n" 1) voltageequations.The meshclrllentsir Fig.4.19that are equivalent to eliminating the branch current i3 frolxl Eqs.4.21 al:td4.22 arc shownin Fig.4.20.We now apply Kirchhoff'svoltagelaw aroundthe two mesheq expresshg all voltages acrossrcsistois ilr tems of the mesh cuirents,to get the equations ?r1: ianl + (ia tb)R3, -?,, - (rb tJn3 + i6R .

tigure4.20A Me5h cuirents ia andtb.

\4.25) (4.26)

Collecting tlle coefficienls of ir and ib in Eqs.4.25 and 4.26 gives u1 = ia(n1 + R3) ibR3, -1tz: ia& + tb(n, + Rr).

\4.?7) (4.28)

Note ttrat Eqs.4-27and 4.28and Eqs.4.23and 4.24are identicalin form, wilh the mesh currentsia and ib replacingthe branch curents 4 and i2. Note alsothat the branchcurrentsshownin Fig.4.19can be expressedin termsofthe meshcurents shownin Fig.4.20,or \4.29) (1.30) (4.31) The ability to write Eqs.4.294.31 by inspectionis crucial to the meshcurent methodof circuit analysis.Onceyou know the meshcurrents,you also know the branchculrenls.And once you know the bmnch currents, you cancomputeanyvoltagesor powersolinterest. Example4.4 illustrateshow the mesh-currentmethod is usedto find souce Dowersand a branchvoltape.

Usingthe Mesh-Cunent Method a) Use the mesh-currentmethod to determinethe power associated with eachvollagesourcein the circuitshownin Fig.4.21. b) Calculatethe voltageo, acrossthe 8 f) rcsistor.

SoLution a) To calculate the power associatedwitl each so[rcej we need to know the culent in each source. The circuit indicates that these source curents will be identicalto meshcurrents.Also, note that the circuit has sevenbrancheswhere

Figure4.21A Thecjrcujtfor Lyampt€ 4.4. the current is unknown and five nodes.Therefore (,? 1) : 7 - (5 i)] we need three [b mesh current equations to descdbe the circuit. Figure 4.22showsthe three mesh currentsused to descdbethe ctucuitin Fig. 4.21.If we assume

4.6

llre l,4esh-Cutrent l4ethod andDependent sources 107

that the voltage drops are positive,the three mesh +\

,! , 8 a ? ', 7' I

4 0 + 2 t a+ 8 ( i . t b ) : 0 , +6ir+

ll('r,

lb) + 4i. + 20 = 0.

6(,.

(4.32)

Your calculatorcan probably solvetheseequations, or you can use a computer tool. Cramer's method is a useful tool when solvirrg tlree or more simultaneous equatioDsby hald. You cdr review this important tool in Appendi\ A. ReofgJntingtqs.4.32 in anriciparion of using your calculator,a computer program, or Cramer's methodgives 10i, - Stb+ 0i. = 8i. + 20ib oir

)

/+

? 6 ! 'r \ I

figure4.22,\ Thethr€€mesh to anatyze the cu enisused circuit shown in Fi9.4.21. Thc mcshcnrrent i" is identicalwith the branch 'n crrrrcnr rl'c 4rr\ source..orheposer ds\ociatcd with this sourceis prov = -40i, : -224 W. The minussignmcansthat this sourceis delivering power to the nelwork. Thc currcnt in the 20 V sourceis identical 10 thc mcsh current ic:

6;c =

6lb + 101.=

.io

60

2ll

prov - 20i" = -16 W.

(4.rr)

The 20 V soulcealso is delivcringpower to the

The threemeshcurrentsare b) The branch curfent in the 8 () resistor in the direction of the voltage drop 0, is i" - tb. ib:20A,

r. :

0.80A.

,,:

8(1"

lb)

objective2-llnderstand and be able to usethe mesh-current method 4.7

Use lhe mesh-currentmethodto find (a) the power deliveredby the 80V sourceto the circuit shownand (b) the power dissipaledin the 8 O resistor.

Answer:(a)400W; (b)50w

30(:}

80v

NOTE: Abo try Chaptet Problens 4.31 snd 1.32.

Stethed 4.6 TiieMesh-eurrent amdmependent So|'srees II lhe circuilconlainsdependenlsources.the nesh-currenlequalionslnusl be supplenenledby lhe approprialeconslraiil equalions.Exanple 4.5 illustralesthe applicationof the mesh curenl method when the circuit includesa dependenlsource,

ti - 28.8V.

108

lechniques of CircuitAnatysis

Usingthe Mesh-Current Methodwith Dependent Sources Use lhe mesh-curentmethod of c cuit analysisto determinethe power dissipatedin the 4 O resistor in the circuitshowninFig.4.23.

15 t6

Figure4.23 A Thecjtcujtfor Exampte 4.5.

Solution This circuit has six branches where the current is unknown and four nodes.Thereforewe need three mesh curents to describe the circuit. They are delined or the circuit shownin Fig.4.24.Thethree meshcurent equationsare

so

l')

qo

''';),, 15 i,L

50: 5(i1- r') + 20(rr ir), 0 = 5(i, 0 = 20(b

Figure4.24 a, Thecjrcuitshown in Fig.4.23 withthethrce

11)+ 1i2+ 4(iz - h), Q + 4(i

i) + 15i/".\4.34)

We now er?ress the branch currcnt controlling the dependentvoltage source in terms of the mesh

Becausewe are calcuiatingthe power dissipatedin the 4 O resistor,we computethe mesh currentsi2 and i3: i2=26A,

iO: iI

i3,

which is the supplementaleq ation imposedby the presenceof the dependent source. Substituting Eq. 4.35 into Eqs. 4.34 and collectingthe coefficientsofi1, i, and i3 in eachequationgenerates

50:25ir-5i2-20h, 0:

5i1+ 10i2 4i3,

0:

54

4iz+9h.

j3:28A

11.35)

The curent in the 4 f) resistor o ented from left to right is i3 - i2 , or 2 A. Therefore the power dissipatedis p4n= (4

i2)2(4:): (2)'(4) = 16 w.

What if you had not been told to use the meshcurrent method? Would you have chosentlre nodevoltage method? It reduces tlre problem to finding one u*nown nodevoltagebecauseof the presence oftwo voltagesourcesbetweenessentialnodes.We presentmore about makingsuchchoiceslater.

4 . 7 T h eI l e s h - C L r rMr eent ht oS do : mS e p e c i a t i a s e s109

objective2-lJnderstand andb€ableto usethe mesh-current method 4,8

a) Determinethe numberofmesh-cunent equationsneededto soh'ethe circult shown.

Answert (a) 3; (b) 36]V.

b) Use the mesh-currentmethodto find how much power is being delivered to t}le vollJgesource. dependenl

4,9

Usethemesh-current methodto lind ?r,in lhe circuit sho$n.

Answer: 16V NOTE: Also try ChapterPnblems 4.37dnd 4.38.

4.7 TheMesh-eurrent Method: Son"re SpeeiaI eases l0 f) W})ena branchincludesa currentsource,thcmcsh currcntmethodrcquircs Thc circuit shownin Fig. 4.25depictsthe somc additionalmanjpulations. natureof the problem. We havedefinedthe meshcurrenlsia,ib, and ic.as well as the voltage acrossthe 5 A curent sowce,to aid the discussion. Nole thal the circuil conlainslive essentialbrancheslvhere the cur'renlis uDknownand four l00v essenlialnodes.Hence we need 10write lwo 15 (4 1)] nesh'current equationsto soivethe circuit.Thepresenceof the curreni sourcereduces the three unknown mesh cufents to tlvo such currents.becauseit corjttustrating 4.25a A circujt nresh anatysis when strainsrhe differencebetweeni" and t. to equal5 A. Hence,if we know 1.. Figure a branch contains an independent cLrrrent source. we krow i., and vice versa. However,whenwe attemptto sum the voltagesaroundeitler mesha or rneshc- we must inlr'oduceinlo the equationsthe unknown voltage acrossthe 5 A currentsoulce,fhui for rnesha:

r00=3(i"-lb)+t,+6ia,

50=4ic

1]+2(i.

ib)

(4.36)

\4.37)

110

Techniquesof Ci'cuitAnaLysis We now add Eqs.4.36and 4.37to eliminate? and obtaitr

50=9i"-5ib+6r..

(4.38)

Summingvoltagesaroundmeshb gves 0 = 3(ib i") + 10rb+ 2(ib - iJ.

(4.3e)

We rcduceEqs.4.38and4.39to two equationsand two unknownsby using the constraint that (4.40)

We leave to you tlle verification that, when Eq. 4.210is combined witl Eqs.4.38 and 4.39,the solutions for the tbree mesh currents a-re h = 1.75A,

ib : 1.254.,

and

t" = 6.75A

TheConcept of a Supermesh 10(}

l:t)

100v 60

50v

4lL

shown in Fiq.4.25, iltustrattigure4.25,| Thecircuit ingtheconc€pt of a supemesh.

We can derive Eq. 4.38 without introducing tle unknown voltage o by using the concept of a supemesh. To qeate a supermesh, we mentally rsmove the current sourceftom the circuit by simply avoiding tlis branch when lTiting the mesh-crrrent equations.We expressthe voltages around the supermesh in terms of ttre original mesh currents. Figure 4.26 illusfiates tho supermesh concept. When we sum the voltages around t]le supemesh (denoted by the dashedline), we obtair the equation -100 + 3(i. - ib) + 2(tc

,b) + 50+4i"+

6ia=0,

(4.4\)

which reduces to 50=9ia-sib+66.

o, i,

a'f,,)

\4.42)

Note that Eqs.4.42 and 4.38 are identical. Thus the supemesh has eliminated the need for introducing the unknown voltage aooss the curent sourc€.once agaio, taking time to look carefully at a circuit to identify a shortcut such as this provides a big payoff in sirnplifyhg t]le analysis.

Mesh-Cunent Analysisof the AmptifierCircuit in Fig.2.24with the tigure4,27a Thecircujtshown mesh cun€nts i", ib, andt".

We canusethe circuitfirst introducedin Section2.5 (Fig.2.24)to illustnte how the mesh-cu ent method works when a bmnch contains a dependent current source.Figure 4.27showsthat crrcuit, with the three mesh currents denoted t,, ib, and i". This circuit has four essentialnodes and five essential

a.7

Tle !1esh-(unent l4erhod: Some Special Cases 111

bmnches wherc the current is unlmown. Therefore we know that the citcuit can be analyzed in terms of two t5 - (4 - 1)] mesh-curent equations.Although we defined three mesh cunefis in Fig. 4.27,the dependent crurent source forces a constaint between mesh currents ia and ic, so we have only two unlnown mesh currents, Using tle concept of the supermesh,we iedmw the circuit as show[ in Fig. 4.28. We now sum the voltages arormd the supermeshin terms of the mesh cunents i!, ib, and L to obtait Rri^ + 'Dcc + RE(tc

ib) - Vo = 0.

\4.43)

The mesh b equation is

R2tb+Yo+nr(ib-i")=0.

Figure4.28A Th€circuitshownjn Fig.4.27, depicting (4.44) the sup€nnesh crcated bythepresence ofthe dependent cuftentsourc€.

Theconstaint imposedby the dependentcunent sourceis piB=i,-i..

14.45)

The bmnch current contiollhg tle dependent curent source, expressed as a function of the mesh currents! is (4.46)

FromBqs.4.45 and4.46, jc=(1 r B)'a- Ir,b.

14,47)

WenowuseEq.4.47to eliminateicfrcm Eqs.4.43and4.44: fn1 + (1 + p)RElia- (1 + B)REib=V)-Ucc, (t

B\Rd" - [R/ + fr - B)Rr]t"- -Yo.

(4.48)

14.4eJ

You should veriiy that the solution of Eqs. 4.48 and 4.49 for ia and ib gives

. h:

uoR2- uccR2- ucc(1 + B)RE RrRtr (1 -r B)R CR,+ R, '

-uo& - (1+ B)REUcc

R1R'.r (1 + B)R,(R1+ R )

\4.50)

(4.51)

We also leave you to verify that, when Eqs. 4.50 and 4.51 are used to fhd iB, the resultis the sameas that givenby Eq.2.25.

112

Iechniques of Circuit Analysis

objective2-Understand and be abteto usethe nesh-currentmethod 4.10

Use the mesh-curent method to find the power dissipatedin the 2 O resistor in the circuit shown.

30

Answer 72 W: 4.11 Use the mesh-curlentmethodto find the mesh currert ia in the circuit shown.

i0A

Answer:15A. 4.12 Use the mesh-currentmethodto find the power dissipatedin the 1 O resistorin tlle ci!cuit shown.

10v

NOTE: AIso try ChsptetPrcblems1.41,4.12,4.17,and 4.5A.

4.8 TheF{ode-VoltaEe MethodVersus the Mesh-eecrremt Methsd The greatestadvantageofboth the node-voltagcand mesh-currentmethodsis that they reducethe numberofsimultancousequationsthatmust be manipulated.Thcyalsorequirethe analystto be quite systematicin terms oforganizingand wdling theseequations.Itis naturalto ask,lhen,"When is the node-voltagemethod preferred to the mesh-currentmelhod and vice versa?"As you might suspect.there is no clear-cutanswer.Askinga number of questions,however,may help you identify the more efficient methodbelbre plunginginto the solutionprocess: . Does one of the melhods result in fewer simultaneousequations . Does the circuil contain supernodes?If so, using the node-voltage method will permil you to reduce the number of equalions to

4.8 Th€Node Voltase Method Versus th€l,lesh Current Method . Does the circuit contain supermeshes?If so, using the mesh-current method will p€rmit you to reduce the number of equations to be solved, . Will solving some portion of the c cuit give the requested solution? If so, which method is most efficient for solving just the pertinent portion of the circuit? Perhaps the most important obse ation is that, for any situation, some time spentthinking aboutthe problemin relationto the variousanalytical apprcachesavailable is time well spent.Examples 4.6 and 4.7 ilustrate the processol decidingbetweenthe node-voltageand mesh-curentmethods

Understanding the Node-Vottage MethodVersus Mesh-Current l{ethod Find the powei dissipaled in the 300 O resistor in the circuit showl in Fig.4.29.

3000 l! 150{} \

1000

500

5000

\50t{ t256v ,000 (

)

rooo "uu(

Figure 4.29A Thecircuit forBGnrpte 4.6.

dependentvoltage source between two essentral nodes,we have to sum the currents at only two nodes.Hencethe prcblem is reducedto writing two node-voltageequationsand a constraintequation. Becausethe node-voltagemethod r€quires only three simullaneousequations,it is the more altractive approach. Once the decision to use the node-voltage method has been made,the next step is to selecta refercnce node,Tlvo essentialnodes ir the circuit in Fig. 4.29 merit consideration. The fiist is the referencenode in Fig.4.31.Ifthis nodeis selected,one of the unknown node voltagesis the voltage acrossthe 300() resistor,namely,?2 in Fig. 4.31. Once we know this voltage, we calculate the power in the 300O resistorby usingthe expression prmo : oji 100.

Solution To find tlre power dissipated in the 300 O resistor, we need to find either the curent in the resistor or tlle voltage across it. The mesh-current method yields tlle curent in ttre resistor; this approach iequires solving five simultaneous mesh equations, as depicted in Fig. 4.30. In writhg the five equations,we must includethe comtraint iA : r,. Before going furtler, let\ also look at the cir cuit in termsof the node-voltagemethod.Note that, once we know the node voltages, we can calculate eitler the current in the 300 O rcsistor or the voll age acrossit. The circuit has four essentialnodes, and therefore only tlree node-voltage equations are requiredto descibe the circuit.Becauseof the

1.28V

jn Fig.4.29,withthenve Figure4.30A Thecircuit shown

114

Techniques ofCircuii Analysis we compare thesetwo possiblercference nodes by mea]ls of the lollovring setsof equations.The first set pertains to the sircuit shown in Fig. 4.31,and the secondsetis basedon the circuitshownin Fig.4.32.

300{) 100_O I ir'r250_O

set 1 (Fig4.31) At thesupernode, Llr_

r)

jn Fjg.4.29,wjtha Figur€4.31 a Th€circujtshown

r00

Note that, in addition to selecting the reterence node,we definedthe threenode voltages"1, ?,r,and ?.,3and indicated that nodes 1 and 3 form a supernode, becausethey are cormected by a dependent voltage source.It is understood that a node voltage is a risefrom the referencenode;tlerefore,in Fig.4.31, we have not placed tlle node voltage polarity refererces on the circuit diagram. The second Dode that merits considemtion as the reference node is the lower node in the circuit, as shoM h Fig. 4.32.It is attractivebecausert has the most branchesconnectedto it, and the nodevoltage equations are thus easierto $'rite. However, to find eitlei the current in the 300 f,, resistor or the voltage acioss it requies an additional calculation oncewe know the node voltagesoaand oc.For example, the curent in the 300 O iesistor is (.,. - r,,)/300.$hereactbe voltageacfo.
250

200

,r\ + 256 150

^

\

rol]o

"b

250o ,.:q!o

'*o n(

"

l2s6v moo(

)'''

\

128)

- + 1r2+ 128

u3

D l

500

From the supemode,the constraint equatron$ t)t :

)UrI :

Ut

Ltl

-

. Set2 (Fig4.32)

0" - 256

1)"

2c0-

,c

1500

-

500

.100

250

' 4oo

-ij

3000

400

i jb : : : +X -) - D - + D-) 300

o -(D

D

ot

150

u. + 128

5oo

-

-

,a - 0b

1oo

1).- rb

250

-

-

oa

0c

3oo

3oo

=''

-'

From the suDemode.the constraint equation is 128V

Figure4.32A Thecircujtshown in Fis.4.29 withan attemative r€ference node.

50(r" ub: 50ia: -ff

,") 6

You should veify that the solution of either set leadsto a power calculationof 16.57W dissipated in the 300 O rcsistor.

4.8

TheNode-Vottage Meihod Versus thel4esh-Current l,4ethod ,15

Conparing the Node-Voltage andMesh-Current Methods fiDd rhevoltage,, in rhecjrcuitsho$nin Fig.4.33.

Sotution At firsl glance,the node-voltagemethod tooks appealing,becausewe may define the unknown voltage as a node voltage by choosing the lov/er terminal of the dependent curent source as the reference node, The circuit has four essential nodes and two voltage-contmlleddependentsources,so the node-voltage method requires manipulation of three node-voltage equations and t\ro constraint equatiorN. Let's now turn to the mesh-currentmethodfor finding oo. The circuit contains three meshes,and we call use the leftmost one to calculatezr.,If we let i. denote the leftmost mesh curent, then oo = 193 10ia.The presenceof the two curent sourcesreducesthe problem to manipulatinga single supermeshequation and two constmint equations.Hence the mesh-curent method is the more attractive technique here.

Figure4.35A Thecircujtshown in Fig.4.33 withnode vottages. To help you compare the two approaches,we summaize both methods.The mesh-curent equations arebasedonthe circuitshownin Fig.4.34,and the node-voltageequationsare basedon the circuit shownin Fig.4.35.ThesupermeshequationIS 193 : 10i, + 10tb+ 10tc+ 0.80d, alld t]re constraint equations are ta : 0.4?,r:0.8tc;

lb

1)d=

7.5id and

ic - ib = 05 We usethe coNtraint equationsto write the supermeshequationirl termsofla:

4f)

2:4

160 : 80i,, or ia : 2 A, 1). = 193 - 20 : 173 Y.

2A

The node-voltageequationsare D. 6J)

7.5()

193

10

80

o.+,tr+\f

: o,

o,-(z'b+0.80,

tigure4.33A Ihecncuit forExampte 1.7.

10 0b+0.80d-r,"

++o.s+

10

The constraintequationsare

- (16+ 0.8r'd) J^

U r = , 0 ,, O = l & 60

7.5{)

8r)

jn Fig.4.33wjthth€threemeshcurrents. Figun4.34A Thecncujtshown

10

l-

We usethe constraintequationsto reducethe nodevoltage equations to three simultaneous equations involving o,, oa, and ob. You should vedfy that the node-vollage approach alsogi\esd. 173V.

116

Techniques of CncujtAnatysis

objective3-Decidingbetw€en the node-vottage andmesh-current methods 4.13 Find the power deliveredby the 2A current sourcein the circuit showr.

4.14 Find the power deliveredby the 4A current sourcein the circuitshown.

15()

2A

25V

Answer: 40W: NOTE: AIso try ChapterProblens 4.54and 1.56.

4.9 S*ure* Tyaslsf*r$fi a"iioris

G)

rj

Even though t1renode-voltageand mesh-currentmethodsare powcrful tcchniqueslor solvingcilcuits,weare still irterestedin methodsthat canbe usedto simplifycircuits.Sedes-parallel reductionsand I{o-Y transfbrma tions are alreadyon our list of simplifyingtechniques. We begincxpandnrg this lisl with source transformations. A solrlc€ transformation. shown in Fig.4.36,allows avoltagesourcein serieswitha resistorto be replacedbya curent sourcein parallelwith the sameresistoror vicc vcrsa.Thedouble headedarrow emphasizes tiai a sourcetransformationis bilateral that is, lve canstartwith eitherconfigurationand dedvethe other. We needto find the relationshipbctwcorlo, and ir thal guamnteesthe two configurationsin Fig. 4.36are equivalentwith respect1()iodes a,b. Equivalenceis achievedif an_vresistorR, experiencesthe samecurrent flo\l:,and thus the samevoltagedrop,whetherconnectedbetweennodes a,bin Fig.4.36(a) or Fig.4.36(b). SupposeR, is connecledbetlveennodes a,b in Fig. 4.36(a).Using Ohm\ law.thc currentin n, is

Figure 4.36r. Source transtormatioN. 14.52) Now supposethc sameresistorR, is connectedbetweennodesa,b in Fig.4.36(b).Usingcurrent division,lhe cunent in R, is

If the lrvo circuits in Fig. 4.36are equivalent.these resistor currents must be the same.Equatingihe righlhand sidesofEqs.4.52and:1.53 andsirnplitying, ,'=R

14.54)

4.9

SourceT6nsformations 717

When Eq.4.54 is satisfied for the circuits in Flg.4.36, tle curenl itr Rr is the same for both circuits in tle figure for all values of Rr. If the current through R. is the samein botl circuits5then the voltage drop acrossR, is the sameir both circuitq and tle circuits are equivalent at nodes a.b. lf lhe polarily of d. is reversed.rhe orie0rarionolir mu
UsingSource Transformations to Solvea Circuit a) For the circuil showr in Fig.4.37,find the power associated wifi the 6V souce. b) State whether the 6 V sourceis absorbitrgor delivedngthe powercalculatedin (a).

4{}

+'

60

30o

5{}

t20o

40v

10{} Flgure4.37A Thecircujtfor E\ampte 4.8.

Solution a) ff we study the circuit show! ill Fig. 4.37,knowitrg rhat lhe power asqociated witb fte o V sourceis of interest, several approachescome to mind, The circuit has four essential nodes and six essential brancheswhere the curent is unlnown, Tnus we can find the current in the brarch containitrg the 6 V source by solving either thrce 16 (4 - 1)l mesh-currert equations ot three [4 1] nodevoltage equations, Choosing the mesh-curent approach involves solving for the mesh current that cor:respondsto the branch current in the 6 V source. Choosing the trode-voltage approach involves solving for the voltage adoss t}le 30 O

resistor, from which the branch current in the 6 V source can be calculated.But by focusing on just one branch culrent, we can first simptily tlle circuit by using sourcetransformations, We must reduce the circuit in a way tlDt preserves $e idenriryof lhe braDchconlatning the 6 V source.We have no leasol to preserve the identity of the branch cotrtaining the 40 V source.Beginnhg witl tlis btanch, we can transform the 40 V sourcein serieswith the 5 f,) reslstor into an 8 A curent sourc€ in parallel witl a 5 O resistor,asshownin Fig.4.38(a).

(a) First step

(b) secondstep

(c) Thnd step

(d) Fourth step

Flgur€4.38 Siep-by-step sjmplification of ih€ circujtshown in Fig.4.37.

118

Techniques of Circuit Anatysis Next, we can rcplace the pamllel combination of the 20O and 5O resistorswith a 4O resistor. This 4 O resistor is in parallel with the 8 A source and tlerefore can be replaced with a 32 V source in series with d 4Q resisror, as sho\rn in Fig. 4.38(b).The 32 v sourceis in seiies with 20 O of resistanceand,hence,can be rcplaced by a currcnt sourceof 1.6A in parallel with 20 l), asshown in Fig.4.18(c).Tbe 20 A and l0 0 parauelle\i. tors can be reduced to a single 12 O resistoi. The parallel combination of the 1.6A cu:rent souice

t t

,zi\

(

Y

-

),, tR"

I

L -,,----a---

b

/n

(

P6v= (0 82s)(6)= 4 9s w' b) The voltagesourceis absorbingpower.

A question that arisesftom use of the souce transformation depicted in Fig.4.38is,"Wlat happensif thereis a resistanceR, ir parallelwith the voltage source or a rcsistance Rr in se es with the current source?" In both cases,the resistancehas no effect on the equivalent circuit that prc_ dicts behavior with rcspect to terminals a,b. Figure 4.39 summarizes this

l

Y

and the 12O resistortiansfoms hto a voltage sourceof 19.2V in serieswidt 12o. Figure4.38(d) showsthe result of this last transformation.The curent in the directionof the voltagedrop across the 6 V sourceis (19.2- 6)/16, or 0.825A. Theiefore the power associatedwith the 6 V

)u,

+b

-r--*T1."- (r-] f),;fn

'"

The two circuitsdepictedin Fig.4.39(a)are equivalentwith iespectto terminals a,b becausethey produce the same voltage and cuirent in any resistorRr insertedbetweennodesa,b.The samecan be saidfor $e circuits in Fig.4.39(b).Example4.9 illustratesan applicationof tlle equivalent circuitsdepictedin Fig.4.39.

-r__L....-, r_____l_,

(t)i

fn

(D)

a Figure circuits containing 4.39 Equivatent otin series in paratteL witha voltage source rcsistance

@

Techniques sour€e T]ansformation Usingspecial

I

I a) Use sourcetransfomations to find the voltage ooin the circuit shownin Fig.4.40. I by rhe25uv lolrage ut nnd rnepo*eraeleloped I source. I c) Find the power developedby the 8 A current I source. I I

1250

250V

{ J8A

1000

150

100 forErampL€ 4.9. tigure4.40A Thecjrcuit

I Solution

25o"

I

I d) Webeeinbv removinstbe I25 Q and l0 Q fesisror". l-.*ut. the l2'5O re\islor is connecled I the 250V \oltagesourc€and the l0 Q across I resisror is conDected i0 rerieswjlh lhe R A curI rent source.we also combinethe .eries-conI nectedresistorsinto a singleresistanceof 20 O. I Figure 4.41 showsthe simplified circuit. I

5f)

25A

250V

1)

(f)ro '.|,oo

20o"

F i g u e4 . 4 1A A q i_ p t ' F evde l s ' oonf l h e , c u t s ' r o r .i l Fig.4.40.

4.10 Th-;venin andNofton Equivatents119 We now use a source transformation to replacethe 250V sourceand25 O resistorwith a 10A sourcein parallelwith the 25 O rcsistor,as shoqnin Fig.4.4).Wecdnnoq simpliJylhe circuit showr in Fig. 4.42 by using Kircbhoff's current law to combinethe parallelcurent sources into a singlesource.The pamllel resistorscombine into a singleresistor Figure4.43showsthe Iesull.Hencer), = 20 V. b) The currentsuppliedby the 250V sourceequals the currentin the 125O resistorplus the current in the 25 O resistor-Thus ''

250 125

250 20 25

.-^ "'

-

souce, posilive at the upper terminal of the D J+ 8 ( 1 0 ) = r o = 2 0 . o t

.N.:

and the polver developedby the 8 A sourcers 480W Nole that the 125O and 10 O resistors do not alfect the value of ,". but do affect the power calculations.

f ) r o aJ z : o( { ) r , r , } r , , r , o Figure4.42A Thecircuiishownin Fig.4-41aftera source

Thereforethe power developedby the vottage : (250)01.2) = 2800W. /25ov(developed) c) To find &e power developed by the 8 A current source,we fi$t find the voltage acrossthe source. It we let o" represent the voltage aqoss the

t)

lUO

Figurc4.43d Thecircuitshown in Fig.4.42aftercombjning

objective4-llndersiandsouicetransformation 4.15 a) Usea seriesdf sdurceti:;sfonilalions to find thevoltageo in thecircoifshown. b) How mtch powerdoesthe 120V source deliverto ttre circuit?

(a)asV (b)374.4w NOTE:

AIso tty Chaptel Problens 1.59 and 4.62.

4.10 Th6venin andNortonfrquival,ents At timesin circuit analysis,we wanl 10concentrateon what happensat a speciJicpair ofterminals.For example,whenwe plug a toasterinto an out let, we are interestedprimarily in the voltageand currentat the tcrminals of the toaster.Wehavelittle or no interestin the effectthat connectingthe toasterhason vollagesor curents elsewherein the circujt supplyingthe outlet.We can expandthis irterest in terminal behaviorto a set of appli, ances,eachrequiringa different amountof powcr.We then are interested in how the voltage and current delivered at thc oullel changeas we changeappliances.Irotherwords,wewant to focuson thc bchaviorol the circuitsupplyingthe outlet,but only at the ortlet terminals.

1.6r} 8()

120

Techniques of Cncuit Anatysk

Th6venin and Norton equivalents aie circuit simplification techniques that focus or termiml behaviorand thus are extremelvvaluableaids in "ndllsi..Althoughbere$e drscuss them a. theypenainro re.isri\eciF nelqork @ntaining cuitE Th6venin and Norton equivalent circuits may be used to represent independcnrand any circuitmadeup of lineat elements. We can best describe a Th6venin equivalent circuit by relercnce to Fig. 4.44,which representsany circuit made up of sources(both hdepend, (a) (b) ent and dependent) and rcsisto$. The letters a and b denote the pair of ter, minals of interest.Egure 4.44(b)showsthe Th6venjnequivalent.Thus,a Flgure4.44a (a)A generatcircuit. (b)TheThdvenjn Th€ve n €quivalent circuit is an independent voltage souce yrh in series witl a resistor Rll, , which replacesan interconnection of sourcesand rcsis, tors.Tfus seriescombination of yrh and Rr, is equivalent to the orighal circuit in the sensethat,if we connectthe sameload acrossthe tetminalsa,b of each circuit, we get the samevoltage and current at the terminals of the load-l]!is equivalenceholds for all possiblevalues of load resistanc€. To represent the original circuit by its Thivenin equivalent, we must be able to delermine the Th6venin voltage /rl1 and the Thevenin resistance Rr,. First, we note that if the load resistanceis infinitely large,we have an open-circuil condition. The open-circuit voltag€ al the terminals a,b in the circuit shownin Fig.4.44(b)is V16.By hlpothesis,this must be the sameasthe open-circuitvoltageat the terminalsa,bin the o ginal circuit.Therefore,lo calculatethe Th6veninvoltageyrh, we simplycalculate the open circuit voltageir the originalcircuit. rigure4.45.4A circuit used io ittunrate aThevenjn Reducingt}le load resistance to zerogivesus a short-circuitcondition. If we placea short circuit acrossthe terminalsa,b of the Th6veninequiva lcnt circuit, the shortcircuit current directed ftom a to b is

By h!?othesis,this shorlcircuit current must be identical to the short-circuit currentthat existsin a short circuit placedacrossthe terminalsa,b of the originalnetwork.From Eq.4.55,

Rrr, =

figure4.46A Thecjrcuitshown in Fig.4.45 with ierminaba andb shot cncuit€d.

8r)

\4.56)

Thusthe Th6veninresistanceis the ratio ofthe open-circuitvoltageto the sho -circuit curent.

Finding a Thdvenin Equivalent To find the Th6venin equivalent of the circuit shown in Fig. 4.,15,we first calculatethe open ciicuit voltageof o,b.Note that when the teminals a,b are open,thereis no currentin the 4 O resistor.Therefore the open,circuit voltageoabis idcnticalto the voltageacrossthe 3 A curent source,Iabeled We find the voltage by solvinga singlenode-voltageequation.Choosing the lower node as the referencenode,weget

-[* 3 2 V[ )

zrr - 25

I-

Figue4,47A TheTh6venin equivatenr ofrhecircuit shown in Fig.4.45.

vn

s

- 6D1-

(4.51)

1 1 : 3 2v .

(4.58)

solvingfor or vields

4.10 Th6v€nin andNorton Equivatentr121 Hence ttre Th6venin voltage for the circuit is 32 V The trext step is to place a sholt circuit affoss the terminals and calculate the rcsulting short-cftcuit culrent. Figue 4.46 shows the circuit vrith the short in place. Note that the shot-circuit curent is in tle direction of the opetr-circuit voltage drop auoss the terminals a,b. If the shon-circuit curent is in tlle direction of the open-circuit voltage rise aqoss the temioals.a miDussign must be insenedin Eq.4.5o. The short-circuit curent (isc) is foutd easily once ,2 is known. Therefore tle problem reducesto finding 1,2with the short in place.Agaiq iJ we usethe lower node asthe reforence node. the equation for ,, becomes

,,-25 5 2

0

-

\4.59)

SolvingEq.4.59for ?]2gives u2:16Y'

(4.60)

Step1: Source transformation

HeDce,the shofi-circuitcurrentis .

1

6

| s _ / =

(1.61)

Wenow find theTh6veninresistanceby substitutingthe numericalresults fromEqs.4.58 and4.61inroEo.4.56:

^*=*:?='n

Step2:

14.62)

Figure 4-47showsthe Th6venin equivalent for the circuit showtrin Fig. 4.45. You should verify that, if a 24 O resistor is connected across the terminals a,b in Fig. 4.45, the voltage acrossthe resistor will be 24 V and the current in t}le resistor will be 1 A, as would be the casewith the Thdvenit circuit iD Fig. 4.47.This same equivalence between the circuit in Figs.4.45 alrld4.47holds for any resistor value cornected between rodes a,b.

TheNortonEquivalent

Step3: Souce kansformationj series resistorscombined,producing the Thevenin equivalent circuit

A Norton equirelent chcuit consists of an independent current source in parallel with the Norton equivalent rcsistance.We call dedve it from a Thevenin equivalent circuit simply by making a source tansfomation. Thus the Norton curont equals the shorlcircuit current at tlle terminals of interest, and the Norton resistanceis identical to the Th6venin resistanc€.

UsingSourceTransformations Step1:

Sometimeswe can make effective use of source transformations to derive Souce traDsfornation, producing the Norton equivalent circuii a Th6venin or Norton equivalent circuit. For example,we can derive the Th6venin and Nortor equivalents of tle circuit shown in Fig.4.45 by making tle series of source tansformations shov,n in Fig. 4.48.This technique is most useful when the network cotrtains only independent sources,The presenceof dependent sourcestequires retainhg the identity of the controllhg voltages and/or curents, and this constraint usually prohibits conthued reduction of the circuit by source tmnsformations We discuss tlle problem of finding the Th6venin equivalent when a circuit contains Figure4.48A Step-by-step denvation of theIh€venin dependentsouces in ExamDle4,10. jn Fjg.1.45. andNoironequivatenis ofthecjrcujtshown

122

Anatysis Techniques of Circuit

Source of a Circuitwith a Dependent Equivalent Findingthe Thdvenin Find the Th6venin equivalent for the circuit con taining dependentsourcesshownin Fig-'1.49.

2ko

a in Fig.4.49wiihtermjnaLs Figure4.50 s Thecircuitshown

eqLrilatent a Th6venin Figure4.49 a A circuitus€dto itLustraie sources. dependent whenthecircuitcontajns

As the voltagecontrollingthe dependentvoltage sourcehas been reduced to zero, the current controlling the dependent current source is

Sotution The first slep i]l analyzingthe circuit in Fig 4'19 js to recognizethat the current labcled i. must be zero. (Note the abscnceof a retum path for ir to enter the leffhand portion of the cjrcuit.) The open circuit, or Th6venin.voltage will be the voll ageacrossthe 25 O rcsistor.Withi" : {), Yrh=oob:(

2 0 t ( 2 s )=

500i'

5 2000

Combiningthesetwo equationsyieldsa shortcircuit i," =

20(2s) =

Fromi,. and Yrhwe gel

The current i is

-

':

5-30

: 2ooo

5

50 mA.

x 10e= 100O.

3yft

,(no '

In writing t}le equation for i, wc recognize that the Th6veninvoltageis identicalto the control voltage wllen we combinethesetwo equalions,weobtarl

Figure 4.51illustratesthe Th6veninequivalent Note that the ref for the circuit shownin Fig.4.,19. voltage the Th6venin erence polarity marks on ,{.51 preceding equathe agreewitb sourcein Fig. tion for Yft.

Yrh=-5v To calculatethe short-circuitcurent, we place a short circuil acrossa,b.Whenthe terminalsa,b are shortedtogelher,thecontrol voltage? is reduccdto zero.Therelore,with the short in place,the circuit shor\n in Fig. 4.4,rbecomerlhe one \ho$n in Fig. 4.50.WitI the sho.t circuit shuntingthe 25 O resistor.all the curent from the dependentcunent sourceappearsin the short,so

100()

('

in cncuitshown equjvahntforihe Figure4.51',. TheTh6venin Fil.4.49.

4.11 r,4ore on 0eriving a Th6venin EquivaLent123

objective5-Understand Th6venin andNortonequival€nts 4,16 FindtheTh6veninequivalentcircuitwith respect to the termimls a,b Ior the circuil shown.

7 2l l

Answer: Yab= Yrh = 64-8V, Rft : 6 O. 4,17 Find thc No on equivalentcircuirwith respeci to the teminals a,b tbr dre circuit sho!v!. 15A

Answe.: 1N = 6 A (directedtoward a), RN : 7.5 O. 4.18 A voltmeterwirh an inlernalresistance of 100kO is usedto measurethe voltage?ABin the circuil shown.Whalis the volrmeterreading?

12kO

Answer 120V NOTE: Aho try ChaptetProblems4.63,4.66,and4.67.

4.11 More*n $erivi*g* Th$v*n-in Hquivx[ent The lechniquefor determiningnrh rhat wc discussedand iltustraredin Scclion4.10is not ahvaysthc easiestmerhodavailable.Two other meth odsgenerallyare simplerto use.Thefirst is usefulif thc netlvorkcontains 2 5 V only indepcndentsourccs.To calculatonft for such a nerwork,wc first deactivatcall independcnlsourcesand theDcalculatethe resistanceseen looking inlo lhe network allhe designaredrerminatpair.A voltagcsource is deactivaledby replacingit $'ith a shorl c cuit.A curent sourceis deac- Figure4.52 .r A cjrcuitusedto itlustctea Th€venin tivatedby replacingit wirh an open circuil.Fof example,considerrhe cir, cuit shown in Fig. 4.52.Deactivatingthe independcnlsourcessimplifies the circuit to the one shownin Fig-4.53.The resisranceseenlooking into 5(' 4(:} the terminalsa.bis denotcdR"b,which consistsofthc 4 O resistorin se es with ihe parallelcombinalioDsof thc 5 and 20 Q rcsislors.Thus.

R

R

,

. r2o_8,,.

(4.63)

Nole that thc derivation of nTn with Eq.,1.63is much simpler than the sane derivationwith Eqs.,l-57+.62.

figure4.534 Thecircuitshown in Fig.4.52afterdeac tjvatjonof thejndependent soLrrces.

124

lechniques0fti,cuitAnatysis

If the circuit or network contains dependent sources,an alternative procedure for finding the Th6venin resistance Rrr is as follows. We fi$t deactivate all independent souces, and we then apply either a test voltage sourceoi a test culrent sourceto the Th6venin terminals a,b.The Th6venin resistanceequals tle ratio of the voltage acmss the test source to t}Ie current delivered by the test source.Example 4.11 illustrates this alternative procedure for fhding Rrh, using the sameciicuit as Example 4.10.

Equivalent Usinga TestSource Findingthe ThEvenin Find the Thdvenin resistance Rrh for the circuit in Fig.4.49,usingthe alternativemethoddescdbed.

5olution We filst deactivate the independent voltage source ftom t}Ie circuit and then excite the circuit ftom the teminals a,b with either a test voltage souice or a test current soruc€.If we apply a test voltage source, we will know t}Ie voltage of the dependent voltage sourceard hencethe controlling curlent i. Therefoie we opt for t]le test voltage soluce.Figure 4.54shows the circuit for mmputirg the Th6venin resista[ce.

The externally applied test voltage source N denoted o?, and the current that i[ delivers to the ciicuit is labeledir. To find the Thdveninresistance, we simplysolvethe circuitshownin Fig.4.54for the ratio of the voltage to the current at the test source; that is,Rrh = Dr/ir. From Fig.4.54,

i,=fr+zoi, -30" i=;'nA.

(1.64)

\4.65)

We then substituteEq.4.65 into Eq.4.64 and solve the resulthg equation for the mtio t r/i?l b.r 25

60t)r 2000'

i.f,7 _ 6 ur 25 200

(4.66)

50 5000

1 100

(4.61)

From Eqs.4.66and 4.67, Figure4.54

Anatternatjve method for compuiing the

nrn=3=tooo.

(4.68)

In general,thesecomputationsare easierthan thoseinvolvedin computing the shofcircuit current. Moreover, in a netwoik containing only resiston and dependent sources,you must use the alternative method, because the ratio of the Th6venin voltage to the shorFcircuit current is indeteminate.That is.it is the ratio 0/0.

4.11 Moreon Deriving a Th€venin Equivatent 125

objective5-Understand Th6veninand Nortonequivatents 4.19 Find the Th6veninequivalenrcircuitwilh respect to the lerminals a,b lor the circuit shown. - ?rab = 8V,Rrh: Answeri Y1.h

1O.

4.2O Find thc Th6venin equivalent circuit with respectto the terminalsa.b for the circuit showr. (girtr Define the voltage at the left most node as?),and wrile two nodal equations with ynr asthe righr node voltage.) Answer: Y1]r: oab= 30 V. Rft : 10 O.

160;r

NOTE: Ako trt Chaptet Ploblems 1.71 dnd 4.77.

Usingthe Th€venin Equivalent in the AmplifierCircuit At trmeswe canuseaThivenin equivalcnlto reduceone portion of a cir cuir ro greatll'simplify analysisof the larger network.Let,s reiurn to the circuil first introduced in Section 2.5 and subsequentlyanalyzed in Sections4.4and 4.7.To aid our discussion. we redrcw the c cuit and identified the branchcurrentsofinterest,asshor\ltlin Fig.4.55. A. L'urprerrou\analvsi\hds\no$n./s i, Inc ke) ro tindingrhcorlrer ' r branchcurrents.We redrawrhe circuil as shownin Fig.4.56to prcpareto replacethe subcircuilto rhe left of % wirh its Th6veninequivatcnt.you shouldbe able to determinethat lhis modificalion has no etlect on the branchcurrentsi1,ir, iB, and iE. h Now we replace the circuit made up oI y.t., R1, and R2 with a Thdveninequivalent,with respectto the terminalsb,d.TheThdvenir volr ir , ageand reslslanceare

VccRz Rr+R2

-\€ -\p.€.' 4.55, apolica io- o d eq na a li9u.,e r4.6e, ' r no r ( u ra r narvsE.

Rrn: R1 + Rr'

14.10)

of CircuitAnaty5is 126 Te.hniqLres

with the Th6veninequivalent,the circuit in Fig. 4.56 becomesthe one shownhFig.4.57. We now derive an equation for iB simply by surnming t]le voltages around the left mesh. ln writing this mesh equation, we recognize that i, : (1 + B)t,. Thus,

vn,: R*i, +va+ RE(1+ B)iB,

\1.71)

irom which

i,,

Vm

d sho$,n Figure4.56a A modified veEionofthe circLrit in Fig.4.t5.

V,t

Rrh + (1 + B)R,

\4.72)

Wiren we substituteEqs. 4.69 and 4.70 into Eq. 4.72,we get the same er?ressionoblainedin Eq.2.25.Notelhatwhen we haveincoiporatedthe Th6veninequivalentinto the original circuit,we can obtain the solution for is by wdting a singleequation.

,r,

jn Pig.4.56modified tigurc4.57A Theciftuitshown equivatent. bya Th€venin

power maximum figure4.58A A circuitdescribing

PowerTransfer 4,12 Maxircrum Circuit analysisplaysan importantrole in the analysisof systemsdesigned to transferpower lrom a sourceto a load,We discusspower transferin terms of two basic types of systems.The first emphasizesthe efficiency of the power transfer. Power utility systemsare a good example of this type becausethey are concernedwith the generation,tmnsmission,and distribution of large quantitiesof electricpower. If a power utility systemis inefficient,a large peicentageof the power generatedis lost in the trans and thus wasted. missionand distributionpiocesses, The secondbasictype of systememphasizesthe amount of power transfeired. Communication and instrumentation systemsare good examples becausein the transmissionof information,or data,via electdcsignals,the power availableat the transmitteror detector is limited. Thus, transmittingas much of this power as possibleto the receiver,or load, is desilrable. In such applicationsthe amount ol power being lransferredis small,so the efficienryof transferis not a primary concen. We now considermaximumpowertransfeiin systemsthai canbe modeledby a purely resistivecircuit. Ma-\imum power transfer can best be described wilh t}le aid of the circuit shown in Eg. 4.58.We assumea resistivenetwork containingirdependentand dependentsourcesand a designatedpair of terminals,a,b,to problemis to determinethe value which a load,Rr, is to be connected.The

4.12 l,4aximum Power Transfer 127 ol Rr lhat permiLsmaximum power deliver) ro Rr. The first slep in thjs processis to fecogniTelhat a fesisrivenerworkcan aiwaysbe feplacedby its Thdvenin equivalent.Therefore, we redraw the circuit sho$n ir Fig. 4.58 as the one shown.in Fig. 4.59. Replacing the original network by its Thdvenin equivalent geatly simpiifies the task of finding Rr. Derivation of Rr rcquires expressing the power dissipated in R, as a function of the three circuit parameters yTh,Rft, and Rr, Thus rigur64,59 A circuitus€d to d€t€rmine thevalue of nr for maximum power transfer.

p=PRL:(#E)""

\4.73)

Next, we rc€ognize that for a given circuiq yft and Rrn will be fixed. Therefore the power dissipated is a function of the single variable Rr. To find the value of R, that mMi;izes the power, we use elementary calculus. We begin by writing an equation for the dedvative ofp with rcspect to Rr:

^:'^l

{^Trlr + ^r)_ - KL. Z\Krh + ltL)

(Rft+Rf

l

\4.74)

The dedvative is zerc and p is maximized when (^Tb -

r(L.,-,

1l(r(Kft

-

Rrl

\4.75)

SolvingEq.4.75ields

(476) < Condition for m.ximumponertransfer Thusmaximumpower,hansferoccurswhenihe load resistanceR, equals lheTheveDi0 resisraoce Rrb.To findthema"\ilnum povr'er de[!eredto Rr. we simplysubstituteEq. 4;76]JJtoEq.4.731

^ =v'*a" "* enD'

4R.

\4,77)

The analysisof a circuit wheii the Ioad resistotis adiustedfor maximum powerhansferis illustratedin Example4.12.

128

Techniqu€s ofCircuit Anatysh

Cal.cutating the Condition for MaxinumPower Transfer a) For the circuit shownin Fig. 4.60,find the value of Rr_ that results in maximum power being transferredto Rt.

25o" 300v

R.

300 360V

-

jn Fig.4.60by Figure4.61 3. Reduction ofthe circuitshown means of a Thdvenin equivateni.

) rsoo b

Figure4.50a Thecjrcuitfor ExampLe 4.12.

b) Calculate the maximum power tlat can be deliveredto Rr, c) When R._is adjustedfor maximumpower tiansfer, what percentageof the power deliveredby the 360V sourcereachesRr?

b) The maximumpowerthat can be deliveredto Rr is

,*"=(S)i,a: 'o*. c) When Rr equals25 O, the voltage?,b is

/too\ j(2s) : 150v. ,"h : l= \rtr/

Solution a) The Th6veninvoltagefor the circuitto the left of t]reterminalsa,b is

v* :

= :oo,r. '*141:eo;

The Th6venin resistanceis (150V30)

Replacingthe circuit to the left of the terminals d.bwilh ilsThevenin givesusrhecirequivalenr cuit shown ir Fig. 4.61,which indicatesthat Rr_ must equal25 J) for maximumpower transfer.

From Fig. ,1.60,when o"b equals150V the current in the voltagesourcein the dnecdon ofthe voltagerise acrossthe sourceis

(=

360 150- 3210 0 = / _A :ro

Therefore, thesource is delivering2520W to the p, = t,(360)= 2s20w. The percentageofthe sourcepowerdeliveredto the load is

900 ^ 100:35.71%. 25?]J

a.1l

Supe,positio1 n29

obiective6-Know the conditionfor andcatcutate powertransferto resistiveload maximum 4.21 a) Find the valueofI that enablesthe circuit shownto delivermaximumpower to the terminalsa.b, b) Find the maximumpower deliveredto R.

4.22 A..u.nelhdl rhccircuuin Assessmenr Problenr4.21isdeliveringmaximumpower to the load resistorR. a) How much powcr is the 100V sourc€delivering to the network? b) Repeat(a) for the dependentvoltage c) Wlul percentageof the total power gcner atedby theseiwo sourcesis deliveredto the load resistor-R?

Answen(a)3 O; (b) 1.2kw

Answer: (a) 3000W: (b) 800w; (c) 31.s8%.

NOTE: AIsott! ChaptetPtublems 1.79and4.80.

4.13 5up*rg:**iti*n A linear systcn obeys the prirciple of superposilion.which statesthat whenevcra linear systemis excited,or driven,by more than one indc pendentsoulceoI energy,the total responscis the sum of the individual responsos. An individual responseis thc resull of an independentsource acting alone. Becausewe are dcaling lrilh circuits made up of inter connccledliDear-circuitelements.wc canapplythe prirciple ofsupcrposi tion direclly to the analysisof suchcjrcuilswhen they are driven by lnore thanore independentenerg,\'sourcc.Atpresent,werestrictthediscussion 10 simple resistivenetworks:however.the principle is applicable1()any Superposition is appliedin both the analysisand thc designoI circuiis In ana\zing a complcxcircuil with multiple independcntvoltageand current sources.therc are oflen fewer,simpler equations1{)solvewhen the effectsof the indepcndcnlsourcesare consideredone at a 1ime.Applying superyositioncan thus simplify circuit analysis.Be aware.though,that somedmes applyingsrpcrpositionacluallycomplicates thc analysis,producrl1glnore equationsto soh'cthanwith an alternativemethod.Superposition is requiredonly if thc jndepeDdentsourcesin a circuit are lriDdamental]y different.In thcsccarly chapters, all independentsourcesare dc sources, so superposition is nol required.We introducesuperposilionhere ir anticipaiion ot latcr chaple$ir whichcircuiiswill rcquie i!.

130 Iechniques of cncuiiAnatysk Superpositionis applied in design to synthesizea desired circuit responsetlat could not be achievedin a circuit with a singlesource.Ifthe desiredcircuitresponsecan be written as a sum of two or more terms,the response can be realized by including one hdependent source for each term ol the rcsponse.This approach to the desigt of circuits with complex responsesallows a designei to consider several simple designsinstead of one complexdesign, We demonstrate t}le superposition pdnciple by using it to find the bmnch curents in the circuit shoMr in Fig. 4.62.We begin by finding the bmnch curents resulting from the 120V voltage source.We derote those 120V 12A currents with a pdme. Replacing the ideal current sourcewith an oper circuit deactivatesiq Fig. 4.63 shows this. The branch currents ir this circuit tigure4.62A A circuitusedto iLLustrate supeposition. are the resultof only the voltagesource. We can easily find the bmnch curents in the circuit in Fig. 4.63 once we know the node voltageacrosst}le 3 O resistor.Denoting this voltage

q

120V

120 6

r . r -

11.t8)

from which 01 :30 v.

jn Fjg.4.62u/jththe Figue4.53 Th€circuitshown curentsource deactivat€d.

(4.79)

Now we can wite the expressionsfor the branch currents ti - ti directly:

ii = 14-lq

= 15A,

=;=10A, .

724

jn Fig.4.6?wjththe Figure4.64A Thecircuiishown voltage source deactivaied. 6{)

2tl

3

(4.81)

0

-

(4.82)

To find the componentol the bnnch cu:rents resulting from the curent source,we deactivatetle ideal vollage sourceand solve the circuit shown in Fig. 4.64.The double-prime flotation for the currents hdicates they are the componentsof the total cunent resulting from the ideal cuffent source. We determinethe branchcurrentsin the c cuit showr h Fig.4.64by first solving for the node voltages across the 3 and 4 O resisto$, respectively. Figurc 4.65 shows tie two node voltages. The two rode-voltage equationsthat describethe circuit are

t2 A.

3 Figlre4.55-r Thecircujtshown in Fig.4.64showjng the nodevottag€s rr and?a.

(4.80)

U4

6 0l

2 ,4

.^

0,

14.83)

\4.84)

1.13 tupeeosition131 Sol\iDgEqs.4.83and 4.84for ,r aDdd.. we get a3 : -72 V,

(4.85)

ot=-24V

(1.86)

Now we can write the branch currents ii tlrough ii directly in tems of the node voltages?r3and 04:

ii: =:::2

A,

\4.87)

il=l=

aA.

(4.88)

3

l= 3

-12 + 24

(4.8e)

ii=i:i=-sa.

(4.e0)

To furd the branch currents in the original citcuit, that is, the curents iI, i2, 4, and ia n FIE 4.62, we simpty add rhe cunents given by Eqs.4.87-4.90 to the currentsgivenby Eqs.4.804.821

ir:i\+ii:15+2:17A,

\4.91)

i2=ii+i5=10-4=6A,

14.e2)

it= i\+ ii:5 + 6:11 A,

(1.93)

i + = i L + i i =5 - 6 :

(4.e4)

1A.

You shouldveriry that tle curents givenby Eqs,4.914.94 aretlle corrcct valuesfor the bmnchcurrentsir the circuit shownin Fig.4.62. Whenapplyingsuperpositionto linear circuitscontainingboth independentand dependentsources,you must recognizethat the dependent sourcesare neverdeactivated.Example4.13illustratesthe applicationof superpositionwhen a circuit containsboth dependert and indeDendent

132

]echniqles of CjrcujtAnatysjs

UsingSuperposition to Solvea Circuit Use the principleof superpositionto find o, m the circuit shownin Fig.4.66.

when the 10 V souce is deactivated, the circuit reducesto the one shownin Fig. 4.68.We have addeda reference nodeandthe nodedesignations a, b, andc to aid the discussion. Summingthe curientsawayfromnodea yields _2

2

-! 0

5

O.du';- 0. or rr:

B.i

u.

Summingthe currentsawayfrom nodeb gives

Figure4.66 A ThecjtcujttofExamph 4.13.

0 . 4_0+( -

4D'\+ Db

Sotution

5=0.or

l0

2iL:

50.

I)b=21L+ll' L

We begin by linding the component of oo resulting from the 10 V source.Figure4.67showsthe circuit. With the 5 A source deaclivated,r,l, nust equal ( 0.ari!X10).Hence,oi must be zero,tlre branch containingthe two dependentsourcesis open,and

to find the valuefor 0i. Thus, 5ui = 50, or l)l = 10 V. From the node a equation,

4:2*@) : Bv.

soii = 80, or 06 = 16 V. The value of o, is the sum of z'i,and o;, or 24 V

10v

Figure4.57A Thecircuit shown in Fig.4.66withthes A

tigure 4.68 A Thecjrcuitshown in Fig.4.66wjththe 10V

NOTE: Assessyour understanding of this matetial by trting Chapter Problems 1.91 and 4.92.

PEctjcatPersp€ctive 133

PracticalPerspective Circuits with Realistic Resistors It is notpossibte to fabricate identicaI etectricaL components. Forexampte, produced resis(ors fromihe )amemanu%ctuing process car varyin vatue byasmuchas20%.Therefore, in creatjng anelectrjcal systern thedesigner mustconsider the impactthat component variation wilt haveon the per fornance of the system. onewayto evaluate thisjmpactis by performing sensitiviry permitsthe designer anatysis. Sensitiviw anatysis to calcutate theimpactof variations in the component vatues onthe outputofthe system.WewitLseehowthis information enabtes a designer to speci! an acceptable component vatuetoterance foreachofthesystem's conponents, Considerthe circuitshown in Fig.4.69. Toittustrate sensitiviw anatysjs, wewjt[investigate thesensitivity ofthenodevottages rl and?2to changes jn theresistor Rt. Usingnodalanatysis wecanderivetheexpressjons fof r,1 and22asfunctions of the circuitresistors andsource currents. Theresults aregivenin Eqi.4.95and4.96: - lRz(Rr+ R4)+ n.Ral/rr} Rr{RrRy's, (Rl+nr(R3+R4)+R3n4 -'

\4.95)

R3R4[(R1+ R)Ir2-

RJ 91] ( R 1+ R ) ( R 3 + R 4 ) + R 1 & '

(1.e6)

Ihe sensitivjtyof ?,1with respectto Rr is foundby differentiating Eq.4.95 with respectto R1, andsimitarty the sensitivityof zJ2with respectto R1 js foundby differentiating Eq.4.96with respect to R1.Weget

drr dRr

[R3& + Rr(R3+ n4)]{R3R4rs, lR3R4+ R (R3+ Ri)11s1} I(R1 + R )(R3+ R4) + R3R4l2 \4.97)

d?,r, R3R4{R3R418, lRr(R3 + R4) + R3R4U81} d& l(Rr + R)(R3+ R4) + R3R4l'

Figure4.69A Cjrcuitusedto introduce sensiiivity ana\6is.

(4.e8)

134

ofCircuitAnalysis Techniques

to itLustrate vatues withactualcomponent anexampLe Wenowconsider and 4.98. theuseof Eqs.4.97

EXAMPLT in Fig.4.69are: in thecircuit vatues ofthecomponents Assume thenominat R r = 2 5 0 o ; R z = 5 O ; R 3 = 5 0 O ; R a = 7 5 O ; I s 1= 1 2 A a n d to predictthe vatuesof 01 and t'2 if 1, = 16A. ljse sensitivityanatysis from its nominalvatue. 10% the vatueof Rr is differentby

Solution of 01andtr2.Thus vaLues FromEqs.4.95and4.96wefindthe nominal -'

) 5 { 2 7 5 0 t l o r 5 ( 1 2 5-, J 7 5 0l 2 l 30(125)+ 37s0

and - 25(12)l 37s0130(16)

1-',=

3odt+3dj=eov

(4.100)

to NowfromEqs.4.97and4.98we canfind the sensitivityof t)1and?J2 in Rr. Hence changes

- 137s0 + s(12s)112} + 5(125)l- {37s0O6) d?-'1 _ 137s0 dRr l(30x125)+ 37sol' -

(4.101)

ivla,

and

duz 37s0{3750(16) [5(12s)f 37s0]12]l LtR (?soo/ : 0.5v/o.

(4.102)

Assume that givenby Eqs.4.101and4.102? Howdoweusethe resutts : 22.5 O. Then is, Rr value, that its noninat Rl is 10% lessthan A?)1wittbe ARr = -2.5 11un, ,0. ,.101prcdicts t 7 \ aa - l';Jf-25)

14581v'

predicts vatue,ouranatysis if Rr is 107olessthanits norfiinal Therefore, that01witLbe v. or : 25 - 1.4583: 23.5417

(4.103)

P'acticatPe,spe.iive135

Simjtarty tor Eq.4.102wehave A?],: 0.5(-2.5): 1.25V, ?,,: 90 1.25= 88.75V.

\4.104)

jn Eqs.4.103and4.104by substituting Weattemptto confirmthe resuLts thevatueR1 : 22.5O intoEqs.4.95and4.96.When wedo,theresutts are r"r = 234780V, 02= 88.6960 V.

(4_105) (4.106)

Whyis therea difference betweenthe valuespredjctedfromthe sensitivity anaLysis andthe exactvatuescomputedby substitutingfor Rr in the equationsfor or and?'2?Wecar seefromEqs.4.97and4.98that the s€rsjtiviry of 01 andz)2with respectto Rr is a functjonof Rr, because R1 appears in the denominatorof both Eqs.4.97 and 4.9a. Thjs meansthat as Rl changes, the sensitivities changeandhencewe cannotexpecttqs,4,97and 4.98 to give exactresuttsfor largechangesin R1. Notethat for a 10% change1n11, the percenterrorbetweenthe predjctedandexactvatuesof js smatl.Specjficatty, 0?)1 and0o2 the percenterrorin Dt = 0-2713"/0 aft the errorin ,2 : 0.06'764/". P€rcent Frornthis exampte, we can seethat a tremendous amountof workis jn the invotved ifwe areto determjne the sensitivjtyof?r and02to changes remajning component vaiues,nametyR2,R3,-R4,/sr, and1s2.FortunateLy, Pspicehasa sensitivityfonctionthat witl pedormsensitivityanatysjs for us. Thesensitivityfunctionin Pspicecatculates two typesofsensitjvjty. Thefirst is knownas the one-unjtsensitiviry,and the secondis knownas the 1ol. sensitivity.In the examptecircuit,a one-unitchangeir a resistorwould changeits vatueby I O and a one-unitchangein a currentsourcewoutd changeits vaLue by 1A. In contEst,1% sensitivityanatysis determines the effectof changjng resjsto$oI sourcesby I o/oof th€ir nominaI vaLues. Theresultof PspicesensjtivityanaLysis of the circujtin Fig.4.69 js shownin Table4.2. Because we are analyzinga Unearcircuit,we can use superpositjon to predictvaluesof zJland1)2if morethan onecomponent's vaLuechanges.Forexanpte,L€tus assumeR1 d€creases to 24O and R2 decreases to 4 O. FromTabte4.2 we cancombinethe unit sensitjvityof or to changes in Rr andR2 to get A1)'

ln,

+

Az,,

*

:oss::

- 541'/:

4.8337Y/a.

Simjtarty, Ar) an'

Ar, 05- 65 tln

-ov

o

Thusif bothRr andR, decreased predict by 1 O wewouLd Dt=25 + 4.8227= 29.8331v, 12: 90 '7 = 83Y.

136

Technjques of cncuitAnatysis

TABLE 4.2 Etement

Pspice Sensitivity Analysis Res ts Elenent

.!:r......_ - rillJ

ElenentSensttlvlty

orinaliz€dSensitivity

09tt"/,u1q,.,.,...,...,....(-v-9!!:1l9TJl!l

(a) DCSensitivities oJNodeVoLtase V1 R1 R R3 R4 IG1 !G2

2

25 5 50 75 72 16

0.5833 -5.477 0.45 0.2 ,74.58

o.7458 -o.27Q8 0.225 0.15 -\,15 2

(b) Sensitiities of 1utput Vz R1 R R3 R1 IG1 IG2

25 2

5 50 75 72 76

0.5 6,5 0.51 4.24 -72.5 15

0.125 0.325 Q,27 0.18 ?.4

If wesubstitute Rr = 24O andn, : 4 O jnto Eqs.4.95and4.96weget ut = 29'793Y' 1tz= 82759Y' In bothcases ourprcdictions arewithjna fractjon of a vottoftheactualnode vottage vatues. Circujtdesigners usethe resuttsof sensitivity anaLysis to determine whi€hcomponent vatuevarjatjon hasthe greatest impactonthe outputof the circuit.Aswecanseefromthe Pspice sensitivity anaLysis in Tabte 4.2, q and,2 aremuchmoresensitive the nodevoLtages to changes in R2than to changes in Rr. Specificatty, r\ is (5.417 /0.5833)or approximatety 9 timesmoresensitive to changes in R2thanto changes in R1 andtt2js (6.5/0.5)or 13timesmoresensitive to changes in R2thanto changes in Rl. Hence in theexampte circuit,thetolerance on R2mustbemorestringentthanthetolerance on R1if it js important to keepo1ando2closeto theirnominal vatues. yau understanding ll0TE: Assess af thisPradicolPe5pective byttyingChapter Problems 4,1054.107.

Summary 137

5ummary For the topics in tlis chapter,mastery of somebasictems, and the conceptsthey represent,is necessary. Thoseterms are nod€, ess€ntisl node, path, brsnch, essentialbmnch, m€sh, and planor circuit. Table 4.1 piovides definitions and examplesof theseterm$ (Seepage 94.) Two new circuit anaiysjstechniqueswere introducedin tlis chapter: . The node-voltage method works with both planai and nonplanar circuits. A reference node is chosen ftom among the essentialnodes.Voltage variables are assignedat the remaining essentialnodes,and Kirchhoff'scurrentlaw is usedto write one equation per voltage variable.The number of equations is ,?"- 1, where r, is the number of essentialnodes. (see pagee7.) . The mesh-current method works or y with planar circuits.Mesh cunents are assignedto each mesh, and Kirchhoff's voltage law is used to write one equation per mesh. The rumber of equations is 6 - (, - 1), where b is the number of branchesin which the curent is unknown,andn is the numberof nodes.The mesh currents are used to find the bralch curents. (Seepage105.)

. Th6v€nin equiElenrs and Norton equivalents allow us to simplifya circuitcompris€dof sourcesand resistors into an equivalent circuit consisting oI a voltage source and a series resistor (Th€venin) or a current souce and a parallel resistor (Norton). The simplified circuit and the original circuit must be equivalent in terms of their terminal voltage and curent. Thus keep in mind that (1) the Thdvenin voltage (yrJ is the open-cjrcuit voltage aooss lhe teminals of the odginal cjrcuit, (2) the Th6venin resistaflce (Rn) is the ratio of the Th6venin voltage to the shortcircuit curent across the terminals of t]re oiginal circuit; and (3) the Nonon equivalent is obtained by performing a source transformation on a Th6venin equivalent.(Seepage119.)

Maximum power transfer is a technique for calculating the maximumvalueofp that canbe deliveredto a load, Rr. Maximum power transferoccu$ when Rl = Rrh, thc Th6venin resistanceas seen from the resistor Rr, The equationfor the maximumpower tmnsferredis

v+d '

Several rew circuit simplification techniques were introducedin this chapter: . SourcetraNformationsallow us to exchangea voltage source (o") and a sedesresisror (R) for a current qouce (1.) and u paralielfesisror(R) and \ icever.a. The combirations must be equivalent in terms oI tleir terminal voltage and current- Tefininal equiva, lenceholdsprovidedthat

(Seepage116.)

4Rt

(Seepase 126.)

In a circuit witl m ltiple independent sources, superyositionallowsus to activateone sourceat a timc and sum the resultingvoltagesand curreDtsto det€r' mine the voltagesand currentsthat existwh€n all inde pendent sources are active, Dependent sources are never deactivatedwhen applying superposition.(See pase 129.)

138

Techniques of Cjrcujt Anatysis

Problems Sectiotr4.1 41 Assune ttre current is in the circuit in Fig. P4.1 is known.TheresistorsR1 Rs are alsoknown, a) How many unknoM currents are there? b) How many independent equations can be witten usingKirchhoff'scurent law (KCL)? c) Wdte an indeperdent set of KCL equations. d) How many independent equations can be derived from Kirchhoff's voltage law (KVL)?

c) How many branches are theie? d) Assume tlat tle lower node in each part of the circuit is joined by a single conductor. Repeat fie calculationsin (a)-(c). Flgufe P4.3

nr

l)

jn, r'J

e) Write a set of hdependent K\aL equations. Figur€ P4.1

{) pr,

R5

4.4 a) If only the essential nodes and bnnches are

42 For the circuit shown in Fig. P4.2,state the trumerical value of the number of (a) hanches, (b) brarches wherc the current is urkno\rn, (c) essentialbmnches, (d) essentialbrancheswherc the cunent is unknown, (e) nodes,(f) essentialnodes,and (g) meshes. FigureP4.2

identified in t}le circuit in Fig. P4.2, how many simultaneousequationsare neodedto descdbe ttre circuit? How many of these equationscan be derived using Kirchhoff's current law? c) How many must be derived using Kirchhoff's voltagelaw? d) Wbarrso meshes shouldbe dvoidedin dpplying the voltage law? 4.5 A current leaving a node is defined as positive. a) Sum the currents at each node in the circuit showr in Fig. P4.5. b) Show ttrat any one of the equations in (a) can be derived from tlle remaining two equation$

43 a) How many separate parts does the circuit in Fig. P4.3 have? b) How many nodes?

i)

,,ll^, ,,llo, ili 3

PrcbL€ms 139 Section 42

FiglreP4.10 18{))

46 Use t}le node-voltage method to find 2,. in the cir6rc cuit in Fig.P4.6. 128V

FigureP4.6

a7 ^) Find

t]le power developed by tle 3A current source in the circuit in Fig. P4.6. Find the power developed by the 60 V voltage source in the circuit in Fig. P4.6. c) Verify that tle total power developed equalsthe total power dissipated.

411 The circuit shoM in Fig. P4.11is a dc model of a Atrft rcsidential power distdbution circuit. a) Use the node-voltage method to find rhe branch currentsi1 i6, b) Test your solutior for the bianch currents by showingthat the total powei dissipatedequals the total power developed. Figure P4.11 1{) 125V

4.8 A 10 O resistoris connectedin serieswith the 3A 6flft current source in the circuit in Fig. P4.6. a) Find ?,,. b) Find the power developed by the 3,{ currenr

125V

c) Find the power developedby the 60 V voltage d) Ved$' thatthe totalpower developedequalsthe total power dissipated. e) What effect wil any finite resistance connected in serieswith the 3,{ current source have on the valueof o.?

t3 4.12 Use the node-voltage method to find ?,1and 02 in am the circuit in Fig P4.12. tigureP4.12 4f}

80|l

4.9 Use the node-voltage method to find ?J1atd z)2in the circuit sho$n in Fig. P4.9. FlglreP4,9

25tl

4.13 Use the node-voltagemethod to find how much E!'c power the 2A sourceextracts ftom the circuit in Fig.P4.13. Figure P4.13

410 a) Use the node-voltage method to find the bmnch curreirts ir ;€ in the circuit shom in Fig.P4.10. b) Find the total power developed in the circuit.

140

Iechniques ofCjrcuit Anatysis

4.14 a) Use the node-voltagemethodto find t r, 02,and t'3 in ttre circuit in Fig. P4.14. b) How much power does the 640V voltage source deliver to t}le circuit?

Figure P4.17

20o'

figureP4.14 3A

2.54

+ 640V

+

, , J ' 0 o "j,' o ( 1

l) 2A

12.8A 4,18 a) Find the node voltagestt, ,2, and r'3in the circuit in Fig.P4.18.

2.5A

4.15 Use the node-voltagemethod to find the total power tstrG dissipatedin the circuit in Fig. P4.15.

b) Find the totalpower dissipatedin the circuit. tigur€P4.18

Figure P4.15 + l

30v

)

25ll 31.25()

50()

5oo

(

500

4,16 a) Use the node-vollage method to show that the output voltage o, in the circuit in Fig. P4.16 is equal to the averagevalue of the sourcevoltages. b) Fhd o, if 2,1-150V, r'2 = 200V, and 03 = -50 V-

20()

5r}

1 ) 2 , "" i r o o or | : o o o ^ {

\_-,/ 15()

25o" +

(:

38.5V

5 i,,

4.19 Use the node-voltagemethod to calculate the Fr't po$ef delivered vollagesourcein by rhedependenl the circuit in Fig.P4-19. figurcP4.19 5()

10J)

'jlsoo

80v tigureP4.16

150

420 a) Use the node-voltagemethod to find the total power developed in the circuit in Fig. P4.20. b) Check your answerby finding the total power absorbedin the circuit. Seclion 4.3 4.17 a) Use the node-voltage method to find o, in the circuitin Fig.P4.17. b) Find the power absorbedby the dependentsource. c) Find the total power developed by the independent sources.

tisul€P4.20 5()

300

-

t)

Jtso J:oo Jruo

5ir

Prcbtems14X S€ction 4.4

FigureP4.24

500()

421 Use the node-voltage method to find the value of o, Atrc ir the circuit in Fig. P4.21. Figure P4.21 800J)

10v

4r5 Use the node'voltage method to find the value of o, EnG in the circuit in Fig.P4.25. Figure P4.25 4.22 Use the trode-voltage method to find i, in rhe cir6nc cuit in Fig. P4.22. Figure P4.22 10A

423 a) Use the node-voltagemethod to find the power

+

426 Use the node-voltage method to find r', and the poser deliveredb) lhe 40 V voltagesoufcein the circuitin Fig.P4.26.

dissipated in the 5 O resistor in the circuit in Fig. P4.23. b) Find the power supplied by the 500 V source.

rigureP4.25

40v

figureP4.23

417 Use the node-voltage method to find o, in the cir6trc cuit in Fig.P4.27. Figure P4.27

424 a) Usethe node-voltagemethodto find the bmnch currentsil, j2,and i3in the circuit in Fig.P4.24. b) Checkyoursolutionfor tr, tr, andi3by showing that the power dissipatedin the circuit equals thepowerdeveloped.

100

50v

rals

| , .

l ' v l

10o

: r oo

+

i3oo,,

142

Technique5 of Cncuit AnatJ6k

428 Assume you are a project engineer and one of your Bfl(! staff is assigned to analyze the circuit shown in Eg. P4.28.The referencenode and node numbers given on the figurc were assigned by the analyst. Her solution gives the values of D3and oa as 235 V and 222V, respectively. Test these values by checking the total power developed in the circuit against t}le total power dissipated. Do you agee vrith the solution submitted by the analyst? Figure P4.28

(30)ro

FigureP4.31 3 r)

40v

+ 2A

4A

r' ,,1+*'

+ 64V

1.5o

4.32 a) Use t]le mesh-currcnt method to find tlte total powei developed in the circuit itr Fig. P4.32. b) Check your answer by shovring that the total power developed equals the total power dissipated. Figurc P4.32

2tl I

1()

3

I+ ti ') 25ov( zoa

4r}

F-

10{}

40r}

4

q

10r)

3.2ra

Da

110V

429 Use the node-voltagemethod to find the power develBnc oped by the 20 V sourceh the circuit in Fig. P4.29.

2A

4.33 SolveProblem4.10usingthemesh-current method.

Figur€ P4.29 35ia

434 SolveProblem4.11usingttremesh-currentmetlod. 4.35 SolvePrcblem4.22usingthe mesh'curent metlod.

3.125rA

436 SolveProblem4.23usingthe mesh-currentmethod.

Sectior 4.6 430 Showthat when Eqs.4.16,4.17, and 4.19are solved for is, the rcsult is identicalto Eq.2.25

Section45 431 a) Use the mesh-current method to find the branch currentsi, lr, and i. in the circuitin Fig.P4.31. b) Repeat (a) if the polarity of the 64 V sourceis revelsed.

4.37 Use the mesh-current medod to fitrd the power dis" tsrG sipated in the 8 O resistor in the circuit ir Fig. P4.37. Figure P4,37

16f}

4A

Probt€ms 143 438 Use the mesh-current method to find the power 6dc delivered by the dependent voltage source in the chcuit seenin Fig.P4.38.

FigureP4.41

tigureP4.38

4.42 a) Use t]le mesh-curent method to solve foi iA in the circuit in Fig. P4.42. b) Find the power deliveied by the independent 439 Use the mesh-curent metlod to fhd the power si,G developedin 1hedeperdent voltage sourcein the circuit ir Fig. P4.39.

c) Fhd the power deliveredby the dependentvol! age soulce, FiglreP4.42

Flgure P4.39

9800

1.8kO

2.65u!

' , 1 * ,o, n 15()

200ir

4.7kA

250

4,43 Use tle mesh-currentmethod to find the total power 6dc developedin the circuit in Fig. P4.43.

125V

FigneP4.43

4.40 a) Use the mesh-curentmetlod to find ,l, in the circuit in Fig.P4.40. b) Findthepowerdeliveiedby thedependent souce FigureP4.40 2A

0.5 ,I

12o"

165V

4.,|4 Use the mesh-cuffent method to fhd the total power 6dc developedin tlle circuit in Fig. P4.44. S€ction 4,7 441 a) Use the mesh-curent method to find how much power the 12 A current souce delivers to the circuir in Fig. P4.41. b) Find the total power delivered to tle circuit. c) Check your calculations by shovring that the total power developedir the circuit equalsthe total power dissipated.

FigureP4.44

25{}

20()

144

Technjques of Cncuit Anatysis

445 a) Use the mesh-currentmethodto find the power defiveredto the 2 O resistor in the ckcuit in Fig. P4.45. b) What percentage of the total power developed in the circuit is delivered to the 2 {) resistor?

b) Repeat(a) if the 3 A curent sourceis replaced by a short circuit. c) Explah why the answersto (a) and (b) are the same, 4.50 a) Use the mesh-clrlrent method to find the branch curents in i. i" il the circuit in Fig. P4.50. b) Check your solution by showingthat the total power developedin t}le circuit equalsthe total power dissipated.

tigur€P4.45 L Zr ^

Figure P4,50

10v

100f}

4.46 a) Use the mesh-curent method to determine which sourcesil the circuit in Fig. P4-46are genemting power. b) Find the total power dissipated it the circuit. figureP4.46 2A

4.51 a) Find the branch currents ia - ie for the circurt

5 ()

shownin Fig.P4.51. Check your answersby showingthat the total power generated equals the total power dissipated. FigureP4.51 1 5d

*tc

4.47 Use the mesh-currcnt method to find the total xnc power dissipatedin the circuit in Fig.P4.47_ FigureP4.47 3{)

9{))

21,48Assume the 18 V sourcein the circuit in Fig. P4.47is ""c increasedto 100V Find the total power dissipated in tlle circuit. 4.49 a) Assume t}le 18V sourcein the circuil inFig. P4.47 is changedto -10 V. Find the total power dissipated in the circuit.

S€clion 4.8 452 The circuit in Fig. P4.52is a directcurrent ve$ion 6rt! of a typical tbree-wire distdbution system. fhe resistoisR", Rb,and R" representthe rcsistances of the three conductorsthat connectthe three loads Rr, R2,and R3to the 110/220Vvoltage supply.The resistoft Rl and R2 representloads connectedto

Prcblens145 the 110V circdts, and R3 tepresentsa load con, nectedto the 220V circuit. a) What circuit analysis method will you use and why? b) Calculateq, 2,2,and q. c) Calculatethe power deliveredto Rr, R2,and R3. d) What percentageof the total power developed by the sourcesis deliveredto the loads? e) The Rb branch representsthe neutral conductor in the distribution circuit. What adverse effect occurs if the neutml conductor is opened? (Hir?r: Calculateot arld r'2and note that appliancesor loads designedfor use in this circuit would have a nominalvoltagerating of 110V)

Rr=18o Rr=54625O

1t0v

A 4k O resjstor is placed in parallel with the 10 nA curent source in the circuit in Fig. P4.54.Assume you have been askedto calculatethe power developedby the curent source, a) Which methodof circuit analysiswould you rccommend?Explain why b) Find tlre power developedby the currentsource. 4.56 a) Would you usethe node-voltage or mesh-current method to lind the power absorbed by the 10 V source in the circuit in Fig. P4.56?Explain your choice. b) Use the method you selectedin (a) to find the power.

Figure P4.52 R. = 0.1O 110V

Flgure P4.54

Figure P4.56

r,E.

R, = 110.5 o

1qv

4.53 Show that whenever R1 : R, in the circuit in Fig. P4.52,the curent in the neutral conductoris zero. (,r1/rti Solve for t}le neutral conductor curent as atunction of Rr and -R).

d54 Assume you have been askedto filld the power dissiFc, pdledin lhe I kO resi.rorin rhecircuirin Fig.P4.54. a) Which method of circuit aflalysis would you rcc onrmend? Explain why. b) Use your rccommendedmethod of analysisto find tle power dissipated in the 1 kO resistor. c) Would you changeyour recommendationif the problem had been to find the powei developed by lhe l0 mA currenrsource? Lxplain. d) Find the power delivered by the 10 mA current source.

10A

2i,

4.57 The vaiiable dc current source in the ciicuit in Fig. P4.57is adjustedso that the power developed by the 15A currentsourceis 3750WFind the value

Figure P4.57 15A \-./ '7.2 {l

420y

) ,/

15[) 200 40()

' n (

146

Iechniques of CjrcuitAnatysis

4.58 The variable dc voltage source in the circuit in Anc Fig.P4.58is adjustedso that i, is zero. a) Find the value of Ydc. b) Check youi solution by showing the power developedequalsthe power dissipated-

4.61 a) Use source transfomations to find o, in the cir, cuit in Fig. P4.61. b) Find the power developed by the 340V source. c) Find the power developedby the 5 A current d) Vedfy that the total power developed equals the total power dissipated.

figureP4.58

figureP4.61 5{)

340V

15()

+ 20() 10r)

Section 4.9 459 a) Use a sedes of source transformations to find the cunent i, in tle circuit in Fig. P4.59. b) Verify your solution by using the node-volrage method to find i,.

4.62 a) Use a seriesof sourcetransformationsto find i, in tlre circuit in Fig. P4.62. b) Verify your solution by using the mesh-cufient m€thodto find i,. Figure P4.62 l0A

Figure P4,59 2.7kO

l)

$ 2 . : r oJ r r . o( f

4.60 a) Find t})e current in the 10 kO resistor in the circuit in Fig. P4.60 by making a succession of appropriate source tmnsformations. b) Using the rcsult obtained in (a), work back through the circuit to lind the power developed by the 100V source. figurcP4,60 20ko

Seclion4.10 4.63 Find the Th6venin equivalent with respect to rhe 5r.r telminals a,b for the circuit in Fig. P4.63. FigureP4.63

rko

5ko

* ) r o o v$ s o r (ot ) , : . o f u o o o 1ko

10(}

10ko

80

Probtems 147 4.64 Fnd the Thdvenin equivalentwith respectto the 5rft terminalsa,b for the circuit in Fig.P4.64.

Figure P4.67

4ko

3ko

FigueP4.54

't2{r

2a 468 a) Find the Th6venin equivalent with respect to the terminals a,b for the circuit in Fig. P4.68by find' ing the open-circuit voltage and the shortc cuit

1,2v

4.65 Find the Th6venin equivalent with iespect \c - !erminals a.bfor lhe ciJcuitin Fig.P4.o5.

b) Solle for rheThdleninre.isrance b! removing the independent sources.Compare your result to the Th6veninresistancefound in (a). FiglreP4.68

Figure P4,65

20o"

4.66 Find the Norton equivalentwith rcspectto the terErft minalsa,bin thecircuith Fig.P4.66. FiglreP4.65

t5 ko

4.69 An automobilebatterywhenconnected to a car radio,provides 12.5Vto theradio.When connected provides11.7Vto theheadto a setofheadlights,it lighfs.Assumethe mdio canbe modeledasa 6.25O iesistoi and the headlightscan be modeledas a WhataretheTh6veninandNorton 0.65{) resistor. equivalents for thebatlery? 470 DetermineL ando, in the circuitshownin Fig.P4.70 tsr.r whenR, is 0,2,4,10,15,20,30,50,60, and70O.

6() 4.67 A voltmeterwith a resistanceof 100kO is used to amE measrre the voltage ,"b in the circuit in Fig. P4.67. a) What is tlte voltmeterreading? b) mat is the percentage of eiror in the voltmeter r€ading if the percentage of error is defined as l(measured actual)/actuallx 100?

60()

148 4fl *rft

Techniques ofcncuitAnatysis Determinethe Theveninequivalentwith respectto the terminals a,b for the circuit shown in Fig. P4.71.

liglureP4.74

Figure P4.71 9800

16(l 5 x l0 5.u,

1140

4.72 Find the Th6venin equivalent with respect to the EEe teninals a,b for the circuirseenin Fig.p4.72. ligtrrcP4.72

30ra -\.;.-

zko

5ko

) ,,120ko

4,75 A Thdvenh equivalent can also be determined trom measurements made at the pair of terminals of interest,Assume the following measurements were made at the terminals a,b in the circuit ill Fi+.P4.75. When a 15 kO resistoris connectedto the terminals a,b,the voltage o,b is measuredand found ro be 45V

10ko

:50ko

96{)

40ko

When a 5 kO resistoris conrected to the terminals a,b, the voltage is measuredand found to be25v' Find the Th6venin equivalent of the network with respectto the teminals a,b.

4.73 Wlten a voltmeter is used to measure the voltage zre En.r in Fig.P4.73,itreads7.5V a) Wha[ is the resistanceofthe voltmeter? b) What is t}le percentageof erroi in the voltage measurement?

tigureP4.75

Figure P4.73 _^

0.4v

4.74 When an ammeteris usedto measurethe currenttd 6tr( in the circuit shown in Fig. P4.74,ir reads 10 A. a.) Wlat is the rcsistance of the ammete( b) What is the percentage of error in the current measurement?

4.76 The Wheatstone bddge in the circuit shown in tsda Fig.P4.76is balancedwhenR3equals1200O.If tlle galvanometerhas a rcsistanceof 30 O how much cuuent will tlte galvanometer detect when the bridge is unbalancedby setting R3 to 1204O? (ftrf Find the Th€veninequivalentwilh respecrto the galvanometerterminals when R3 = 1204O. Note that once we have found rhis Th€venin equivalent, it is easy to find the amount of unbalanced curent in the galvanometer branch for different garvanometer movements.)

Probt€ms 149 Figure P4.76

Figure P4.79 50()

----,,

1200C) 120V

60f,l

€qoo

200(:}

100v

+)60i 4.80 The variable resistor (R, in tle circuit in Fig. P4.80 6dc is adjusted for maximum power hansfer to RL. a) Find the numericalvalueof RL. b) Find the maximumpouer tmnsferredto RL.

S€ctiotr4.11 4.77 Find the Th6venin equivalent with respect to the 6trc terminalsa,b in the circuitin Fig.P4.77.

Figure P4,80

tigureP4.77

10()

120 4.81 The vaiable resistor in lie circuit in Fig. P4.81 is 6flc adjustedfor maximumpower traNfer to R,. a) Find the valueof R,. b) Find the maimum power that can be delivered to R,.

4.78 Fird the Th6venin equivalent with respect to the ade terminalsa,b for the circuit seenin Fig.P4.78.

FigureP4.81 4k{)

1.25 ko

Figure P4.78 10{)

120 4.82 What percentageof th€ total power developedin Er( the circuitin Fig P4.81is deliveredto R. when R"is set tbr maximumpower transfer?

Section 4.12 4.79 The variable resistor (&) in the circuit in Fig. P4.79 Eflc is adjusted until the power dissipatedin the resistor is 1.5W. Find the values of R" that satisrythis condition,

4.83 A variable resistor R, is coinected across the teram minals a,b in the circuit in Fig. P4.?2.The vadable resistoris adjusteduntil maximum power is translerred to R,. a) Find the value of Ro. b) Find the maxjmumpower deliveredto Ro. c) Find the percentage of the total power devel oped in the circuit that is delivered to R,.

150

lechniquesof ftcojtAnatysjs

4.84 a) Calculate the power delivered for each value of n, usedin Problem4.70. b) Plot the power delivercdto R, ve$us the resistc) Ar what value of R, is the power delivered to Ro a maximum?

b) Find the maximum power. c) Find the percentage of the total power developed in the circuit that is delivered to R,. tiguleP4.88 2A

40

4,85 The variable resistor (Ro) in tle circuit in Fig. p4.85 6trc! is adjusted for maximum powet transfer to R,. What percentageof the total power developedin the ciicuit is delivered to R,?

5fI

Figure P4.85

4.89 The vadable rcsistor in the circuit in Fig. P4.89 is 6flc adjusted for maximum power transfer to _R,. a) Find the numedcalvalueof &. b) Find the maximum power delivered to R,. c) How much power does ttre 280 V source detiver to the circuit when R, is adjustedto the value found in (a)? FlgureP4.89 486 The vadable resistor (R") in the circuit in Fig. p4.86 6trc is adjusted for maximum power transfer to R . a) Find the value of &. b) Find t}le maximum power that can be delivered to R,.

50 i4

10f)

so

20a 0.5725 a^

FigureP4.86 14ia 4.90 a) Find the value of the variable resistor R, in the circuilin Fig.P4.S0lhal wijl fesullin ma\imum power dissipation in the 6 O resistor. (I*nr: Hasty conclusions could be hazardous to your career.) b) Wlat is the maximum power rhar can be delivered to the 6 O resistor? 4.87 w}lat percentageof the total power developedin Atrc the circuit in Fig. P4.86is delivered to R ? 4.88 The va able rcsisror (R,) in t})e circuit in Fig. p4.88 tsdd is adjusted ulltil it absorbs maximum power ftom the circuit. a) Find rhe value of Rd.

figureP4.90

Probtems 151

Section4,13

Figure P4.94

491 a) Use the principle of superpositionto find the voltageo in tle circuitof Fig.P4.91. b) Findthepo\aerdissipaled in lhe20n resistoFiguleP4.91

4.95 Use superposition to solve for i, and z), in the cir6nrr cuit in Fig.P4.95. FigureP4.95 400 30()

7 . 5A

20tr

492 Use the principle of superpositionto find the vottagex' in the circuit of Fig. P4.92.

I

Figore P4,92

4()

180V

:60 O

80f}

250

496 Use the principle of superposition to find the curBflc rent L in the circuit shown in Fig. P4.96.

r-

Figure P4.95 1.8()

10r)

50v

A 12o" I ,)4.s

4.93 Use the principle of superposition to find Erra rent L in the circuit in Fig. P4.93. Flgurc P4.93 5()

10J)

f):oetr+o

) s ov

497 a) In the ctucuit in Fig. P4.97,before the 10 mA current source is attached to ttre terminals a,b, the current io is c?lculated and found to be 1.5 mA. Use superyosition to find the value of i, after the current source is attached. b) Verily your solution by finding i, when all three sourcesare acting simultaneously, figuruP4.97

4.94 Use the principle of superposition to fhd r,. in the 6da ckcuit in Fig. P4.94.

15{)

mv

l)

r,J$ roro r,Jfrsr,o rsro (t Jroro

152

Techniques ofCjrcuitAnatysis

Sections4.1-4.13 4.98 Laboratory measurements on a dc voltage source yield a teminal voltageof 75 V with no load connectedto the soutceand 60 V when load€dwiti a 20 O,rcsistor. a) What is the Th6veninequivalentwirh respectto the terminals of the dc voltage source? b) Show tlat the Th6ve n resistanceof the source is given by the expression / t:-"

Rrh=[-1 \ r,,

\

1JRr. /

4.100 Assume your supervisorhas askedyou to detemine the power developedby the 16V sourcein the circuit in Fig.P4.100.Before calculatingthe power developed by the 16V source,the supervisorasksyou to submit a proposal describirg how you plan to attack the problem. Futhemore, he asks you to explain why you have chosenyour prcposedmethod of solution. a) Describe your plan of attack, explainingyour reasoning. b) Use rhe merhod you have outlined in (a) to Iind t}le power developed by the 16 V source_ Figure P4.100 16V

q'i = the Th6venin voltage, ?, = the terminal voltage correspondiflg to the load resistanceRL. 4,99 Tko ideal dc voltage souces are connected by electdcal conductors that have a resistanceof / O/m, as shown in Eg. P4.99.A Ioad having a resistance of R O moves between the two voltage soucen Let .r equal the distancebetween the load and the source ?)1,and let .L equal t}re distancebetween the souices a) Show that

8r,

4.101 Find the power absorbedby the 2 A current source Edc in the circuit in Fig.P4.101 Figure P4.101 30

8{)

UrR.+R(U, O)J u=--. RL + 2rLx 2rr' b) Show that the voltage ?,will be minimum when "Y:-l

-0r +

c) Find i when a = 16km,?)1: 1000V, u2 : 1200V, R : 3.9 O, and r' : 5 x 10 s O/m. d) What is the minimum value of o for the circuit of part (c)? Figure P4.99

4102 Find 01, r)2,and r,3in the circuit in Fig. P4.102. figureP4,102 0.20

0.2A

110V

180 110V | < - l - _

Probtems 153 4.103 Find i in the circuit in Fig. P4.103.

4,105 Assume the nominal values for the components in t!61!q,v!

tigureP4.103

the circuirin Fig.4.60are: R. = 25 Q: R2 5 O: R3 - 50 O;Ra : 75 O;1s1= 12A;and1s2= 16A. Predict the values of q and D2if 1s1deqeases to 11A and all other componerts stay at thefu nominal value$ Check your predictions using a tool like Pspice or MATLAB.

4.106 Repeat Problem 4.105 if 1s2increasesto 17 A, and Jrliflfii! all otler components stay a1 t]reir nominal values. 5;;i Check your predictions using a tool like Pspice or MATLAB-

4.1o7 Repeathoblem 4.105if Is1 deqeasesto 11A ard 1()

t! !t!!!v!

/d2increase. ro l7 A. Checkyour prediclions using a tool lile Pspice or MATLAB.

4108 Use the results given in Table 4.2 to predict tle val-

4.t04 For the circuit ir Fig.4.69 derive the expressionslor the sensitivity of ?r and 1,2to changesin the source currents1"j and1"r.

ues of 1)1and q if Rt and n3 increaseto 10oloabove their norDinal values and R2 and Ra deqease to l0o. belovr' lbeiJnomiDai!alues./st and /s2remajn at theh nominal values.Compare your predicted values of or and ?2 with their actual values.

The0perationaI Amplifier Theelectroniccircuitknownas an operationaI amptifierhas 5.1 operationalAmptifierT€rminalsp. 156 5.2 TerminatVottagesand Currentsp. 156 5.3 TheInverting-AmptifietCitctit p. 161 5.4 TheSumming-AnpLifiet Citcuit p. 163 5.5 TheNoninverting-AnptifletCircu:tp- 164 5.6 TheDifference-Amplifier Circuit p. 165 5.7 A MoreReatisticl4odetforthe operational

Beabteto nam€th€ five op ampteminalsand describe and usethe vottageandcurreft constrairtsandthe r€sultingsimptjficatjons t h e yL e a tdo i n a r i d e a t o pa m p . Beableto anaLyze simptecircujt' coitajning ideatop anrps,and'ecosniz€the fottowjngop ampcjrcujc:irveftingampfifier,summing amptifier,nonjnvertirsampLjfi er, anddiffererce Understand the morereaListjc modetfor ar op ampandbe abteto usethis modetto anatyze simpt€cjrcujtscontainingop amps.

becomeincreasinglyimpofiant- However, a detailed analysisof this circuit requircs an understandingof electronicdcvicessuch as diodcs and transisloN.You may wondel, then, why we are introducirlgthe circuit bcforc discussingthe circuit's electronic componcnts.There are severalreasons,First,you can developan appreciationfor how the opcrationalanlplifiel cat1be usedas a circuit buildirg block by focusingon its terminal bchavior.At an introductory level,you need not fully understandthe operation of the electronic componentsthat govem termiDal behavior. Second,the circuit model of lbe operationalamplificr requires thc useof a dependelt source.Thusyou havea cbanceto usethis type oI sourceill a practicalcircuit rather than as an abstractcircuit component.Third, you can combinc thc operationalamplificr with resistorsto pedorm son,cvery usefuLlhDctions,suchas scaling,summing,sign changing,and subtracting.Finally, after iltroducing inductorsand capacitorsin Chapter 6, we cal1show you how to use the opcrational ampliiier 1()designintegrating and dillercntiatingcircuits. Our iocuson the termiral bchaviol of the operationalampli fiel irnpliestaking a black box approachto its operation;thar is, we arc not interestedin the internal structureof the ampliliernor in the cuuents and voltagesthat existill this structure.Thc important thing to remernberis that the intemal belraviorof the anplifier accourtsfor the voltageand cu ent constraintsimposedat the terminals.(For now,we ask that you acccpttheseconstrairlts on liith.)

PracticalPerspective StrainGages jn a metaL Howcou[dyoumeasure theamount of bending bar jn physjcaLty suchasthe oneshown the fiqurewjthout contactjngthebar?onemethod woutdbeto usea straingage.A straingagejs a typeof transducer. A transducer is a devjce that neasures a quantityby convefingjt into a moreconvenientform.Thequantiwwe wishto measure in the metal baris the bending angte,but measuring the angtedjrectty is quitedifficuttand coutdevenbe dangerous. lnstead,we attacha strajngage(shown in the Ljredrawing here)tothe netatbar.A straingageis a gridof thjn wjreswhoseresjstancechanges whenthewiresaretengthened or shortened: \t

AR = 2R=

L

whereR is the resistance of the gageat rest, AI/a js the fractionatlengthening of the gage(whichis the definitionof "strajn"),the constant2 is typicalofthe manufacture/s gage factor,andAR is the changejn reshtance dueto the bending pairs of stEin gagesare attachedto of the bal, TypicaLty, oppos'tesidesof a bar.Whenthe ba-is bent,the wiresin one pajr of qagesget longerand thinner,increasjngthe resistance,whjlethe wjresin the otherpajf of gagesget shorter andthjcker,decreasing the resistance. But howcanthe changein resistance be measured? one way woutdbe to usean ohmmeteLHowever, the changejn resistance experienced by the strain gaqejs typjcattynruch smatterthan couldbe accuratety measured by an ohmmeter, pairs gages Usuatlythe of strain are connectedto form a Wheatstone bridge,and the vottagedjfferencebetweentwo legsof the bridgeis measured. In orderto fiake an accurate

fieasurement ofthe voltagedifference, we usean operationat ampufiercircujt to amptit/, or increas€,the voltagedifference.After we intfoducethe operationalampLifier and some of the impotant circujtsthat emptoythesed€vjces,we witL presentthe circujt usedtogetherwith the strain qaqesfor rreasudng rheamounr of beldingin a metalb:,,. Theoperationat amptifiercircuitfirst cameinto existence asa basicbujLding btockin analogcomputers. It wasreferred to asopetutional beca$eit wa5usedto impLement the mathematicalopentionsof integation, differentiation,addition, sign changing,and scaLing.In recentyears,the rangeof hasbroad€ned appLication beyondimptem€rtingmathenratjcaIoperations; however, the orjginaInamefor the cjrcuitper sists,Engineers andtechnjcians havea penchantfor creating technicaljafgon; hencethe operationalampLifieris wideLy knownasthe op amp.

155

156

TheoperationatAmptifier

5.1 Sperationa[ .{rnptifierTerrnina[s Becausewc are stressingthe temhal behaviorof the opemtionalamplifier (op amp), we begin by discussingthe terminalson a commercialty availabledevice.In 1968,FairchildSemiconductorinlroducedan op amp that hasfound widespreadacceptance:the /.A741. (The /rA prefix is used by Fairchiidto irdicate a microcircuitfabricationof the amplifier.)This amplifieris availablejn severaldi{ferentpackages. For our discussion, we assumean eighllead DIP.I Figure 5.1 showsa top view of thc package, with the terminal designalionsgiven alongsidethe terminals.The rerminalsof primary inierestare . invertinginput ' noninvertingi put . output . positjvepower supply(r+) . negativepower supply(Y ) The remahing thrce teminals are of little or no concern.The ofliet null terminals may be used in an auxiliary circuit to compensatefor a degadation Figure5.1 .,Ihe eishtteadDIPpackase (topview). in performance becauseof aging and imperfectiors. Howcver, the degradalion in most casesis negligible.so the offset terminals often are unused and P o s i r n Lp o $ e rs u p p l ) Nonir\LLIrns play a secondaryrole in circuit analysis.Terminal B is of no interest simply F\.t Inpur l.+ t\\ becauseit is an unusedterminal;NCstandsfor no connection, whichmeans I >-outDur nrverrrng-F J/ that the terminal is not connectedto the amplifier circuit. i n'D u t l - - 4 . Figure5.2 showsa widelyusedcircuit symbolfbr an op amp thar con rcgrrre nowersupprv tainsthe five teminals ofprimary interest.Using word labelsfor thc terFigure5.2 n Thecircuitsymbottor anopeEtional minals is jnconvenientin circuit diagams, so we simplify the rerminal designationsin the following way. The roninve ing input terminal is labeledplus (+), and thc inverting input terminal is labeled minus (-). The power supplyterminals,which are alwaysdralvnoutsidethc triangle, are marked y- and y_. The terminal at the apex of the triangularbox is alwaysunderstoodto be the output terminal.Figure5.3 summarizesthese sirnplifieddesignations. Figure 5.3rr A simptified circuit symbotforan opanrp.

5.2 TerminaI Voltages andeurrents We are now readyto introducethe terminal voltagesand curtentsusedto descdbethe behaviorof the op amp.The voltagevadablesare measured from a commonrefercncenode.2 Figure 5.4 showsthe voltagcvariables with lheir referencepolarities.

Conmonnode tigure5.4 ,g lemjnatvottage larjabtes.

t DIP is an abbeviation lor .fual d lte pd.nda..'Ihis neans thar rhe temrinals on eachside of tlre packageare in linc, and that lhe temnrah on opFsite sidesof the packagelho line up. I l]1ecotunor nodc is extemai to the op anp It h the reterene temnral ol the circuir in whicl the op anp is embcdded.

5.2 TeminalVottages andturrents 157 All voltages arc considered as voltage sesfrom the common node. This conventionis the sameas that used in tle node-voltagemettrodof analysis.A positive supply voltago (ycc) is connectedbetween l^ and the common node. A negative supply voltage (-ycc) is connected between y- and the common node. fhe voltage betweetr the irverting input terminal and the commonnode is denotedo,. The voltagebetween the noninvertinginput termhal and the common node is designatedas o,. The voltage betweenthe output lermiltal and the common node is Figue 5.5 shows the curent variables with their reference dircctions, Note that all the current rcference d[ections are into t]le terminals of the operational amptJier: i, is the curent into the inverting input terminal;i? is the curent into the noninveting input terminal; io is the current into the output terminal;i.* is the curent irto the positive power supply terminal and i" is the curent itrto the negative power supply terminal. The teminal behavior of the op amp as a linear circuit element is characterized by constaints on the input voltages and the input currents. The voltage constraint is dedved from the voltage transfer characteristic of the op ampintegratedcircuit and is picturedin Fig.5-6. The voltage transfer characteristic descdbes how the output voltage varies as a function of the input voltages;that is, how voltage is tmnsferred ftom the input to the ouQut. Note that for the op amp, the output voltage is a function of the difference between the input voltages,o, - o,,. The eouationfor the voltasetransfercharactedstic is

\:$

- "^)

A(Dt,-D")<-Ucc. - Vcc = A(up rn) < +Ucc, A(np n") > +Ucc.

Figure5.5 A Ie,minatcu,reni vadabe5

(5.1)

tigure5.6A Thevoliage transfer chancteristic ofan We seeftom Fig. 5.6 atrd Eq. 5.1 that the op amp has three distinct iegions of opemtiotr. When the magnitude of the input voltage difference ( o, - o, ) is small, t]Ie op amp behaves as a linear device, as the output voltage is a Linearfunction of tlle input voltages.Outside this linear region, t]Ie output of the op amp satumtes,and the op amp behavesas a nonlinear device, becausetlle output voltage is no longer a Lhear functiol of the input voltages.When it is opemting linearly, the op amp's output voltage is equal to the difference in its input voltages times the multiplying constant, or gaitr.,4. wllen we conline the op amp to its linear operaling region, a corstraint is imposed on the input voltages,o? and r'" . The constraint is based on tpical numerical values for ycc and ,4 in Eq. 5.1.For most op amps,t]le recommendeddc power supply voltages seldom exceed20 V and t]Ie gain, ,4, is mrely lessthan 10,000,or 104.We seefrom bottr Fig.5.6 and Eq.5.1 that in the linear region, the magnitude of the input voltage differerce ( l r ? , , I m u s l b el e s st b a o2 0 l 0 " . o r 2 m V

The0p€rabonat Amplifi er

Typically, node voltages in th€ circuits w€ study are much larger than 2 mV,so a voltagedifferenceoflessthan 2 mV meansthe two voltagesare essentiallyequal.Thus,when an op amp is constrainedto its linear operal ingregion andthe nodevoltagesarernuchlargerthan2 mV the constraint on the input voltagesof the op amp is

Inputvottageconstraint for idealop ampts

(5.2)

Note tlat Eq.5.2 characterizes the relationshipbetweenthe input voltages for an ideal op amp;thatiq an op ampwhosevalueof.4 is infinite. The input voltage constrajnt jn Eq.5.2 is called the yiltual short condition at the input of the op amp.It is natural to ask how the virtual short is maintained at the input of the op amp when the op amp is embeddedin a circuit, thus ensudnglinear operation.Theansweris that a signalis fed back from the output terminal to the inverting inpul ter minal. This configuration is known as negative feedback becausethe signal fed back ftom the output subtiacts from the input signal.The negative feedback causes the input voltage difference to declease. Becausethe oulput voltage is proportional to the input voltage difference,the outpul voltage is also decreased,and the op amp operatesin its linear region, If a circuit containing ar op amp does not provide a negative leedback path from lhe op amp output to the invertinginput, then the op amp will nonnally saturate.The difference in the input signals must be extremely smallto preventsaturationwith no negativefeedback.B t evenif the circuit providesa negativefeedbackpath for t}le op amp,linearoperationis not ensured. So how do we know whetler the op amp is operating in its linear region? The aNwer is,we don't!We deal with this djlemmaby assuminglin ear operation, peforming the circuit analysis,and then checking our results for coltradictions. For example,supposewe assumethat an op amp in a circuit is operating in its linear region, and we compute the output voltage of the op amp to be 10 V On examiningthe circuit, we discover that ycc is 6 Y resulting in a contradiction, becausethe op amp's output voltage can be no larger than Vcc, Thus oul assumption of linear operation was invalid, and the op amp output must be saturated at 6 V We have identified a constrainton the input voltagesthat is based on the voltage transfer characteristicof the op amp integrated circuit, the assumptionthat the op amp is restricted to its linear operating rcgion and to typical valuesfor ycc and.4. Equation 5.2 representsthe voltage constmint for an ideal op amp, that is, wittr a value of,4 that is infinite. We now tum oul attention to the constraint on the input currents. Analysis of the op amp integrated circuit reveals that the equivalent resistanceseenby the input termimls of tlre op amp is very large,tpically 1 M O

5.2 Te,minatVottages andcu'renis 159 or more. Ideally, the equivalent input resistanceis infinite, resulting in the

(5.3) < Inputcurrentconstraint for ideatop amp

Note ttrat the curent constraint is nor based on assuming the op amp is confined to its linear operating region as was t}te voltage constraht. Together, Eqs. 5.2 and 5.3 folm the constaints on terminal behavior that define our ideal op amp model. From Kirchhoffs current law we know that the sum of the currents entering the operational amplfier is zero, or

ip+ir-

ia

i,

r. =u.

(5.4)

Substituting the constraint given by Eq.5.3 into Eq.5.4 gives

io:

Q , .+ i . ) .

The significance of Eq. 5.5 is that, even though the curent at the input terminals is negligible, there may still be appreciable curent at the output terminal. Beforc we starl analyzingcircuils containing op ampElet's further simplify tlle circuit symbol.w1len we know that the ampliEer is operaringwithin its linear rcgion, the dc voltages+Vcc do not enter into t}Ie circuit equations. In this case,we can remove the power supply teminals from tle s]'rnbol and the dc power suppliesfrom the circuit, as showt ir Fig. 5.7.A word of caution: Becausethe power supply teminals have been omitted, therc is a dangerof inJening from the symbol that i, + in + i, = 0. We have already noted that suchis not the case;that is,i, + i, + i, + i., + i.- : 0. In other words! the ideal op amp model conshaint that i, : i, : 0 does no1 imply that i" : 0. tigure5.7 A The0pampsyrbotwiththe powersuppty Note that tle positive and negative power supply voltages do not have to be equal ir magnitude. In the linear operating region, z, must lie between the two supply voltages.For example, if V+=15V and y = -10V,then-10V< D. < 15V.Be awarealsothatthevalueof.4 is not constant under all operating conditions. For now, however, we assumet]tat it is. A discussion of how and why the value of .4 can change must be delayed until after you have studied the electronic devices and componentsused to fab cate an amplifier. Example 5.1 ilustrates t}le judicious applicationof Eqs.5.2 ard 5.3. When we use these equations to predict the behavior of a circuit containing an op amp, in effect we are using an ideal model of the device.

160

lhe operationaL Amptifier

Anatyzing an 0p AmpCircuit Theop ampin thecircuitshownin Fig.5.8is ideal. a) Calculate 1,, if ?J!- 1 V and ,b = 0 Vb) Repeat(a) for 0o : 1 V and ob : 2 V. c) Ifo, = 1.5V, specifythe rangeof rb that avoids amplifiersaturation.

t ! ! 1 0 0k O

The current constraint requires i, = 0. Substiluting the values for the three currerts into the node-voltageequation,weobtain 7 25

1). 100

''

Hence, t, is 4V, Notc that because,, lies between j 10V, the op amp is in its linear region of opemtion. b) Using the sameprocessas in (a),we get ap:ah:un=)y,

'"

25

1-2 : 25

100

ff-"'

Figure5.8 ..' Theci,cuitforEMmpte 5.1.

,25 :

1 25.^,

,1tl(]

Sotution a) Becausea legative feedback path exists from the op amp's output to its inverting irput througl the 100 kO resistor, let's assumethe op amp is confined to its linear operating region. We can write a rode-voltage equation at the inverting input terminal.The voltage at tlle irverting input terminal is 0, as r'r, : x'b = 0 from the connectedvoltage source, and uh:1)p from the voltage constraintEq. 5.2.The node'voltageequationat

Therefore, ?,o= 6 V. Again, ?, lies within + 10 V. c) As beforqo, = op = ob,andi25= itoo.Because ?]a:15V, 1 , 5- 0 b

25

0o-0t

100

Solving for ob as a function of o, gives I.'b=i(o+"i

From Ohm'slaw,

,,r = ru. ,,,)/25:

l

-

mA.

= (?Jd 0,)/100:0,/100 mA. 4o1r

Now, if the amplfier is to be within the linear regior of operation, -10 V < 0, < 10V. Substituting these limits on o, into the expression lor ?b,we seethat ob is limited to -0.8V<'Db<3.2V.

r ceu, i t 1 6 1 5 . 3 T h eI n v e d i n g - A m p (t ji F

objective1-Use voitageand curent constraintsin an ideat op amp 5.1

AqsumerbarIhe op amp i0 rhecircuir,hown is ideil.

80ko

a) ( alcuialedolbr rhe louowingvaluc\oi, : -0.6. 1.6,and -2.,1V. 0.4,2.0,3.5, h) SpeciJ) lhe rang.-ol d, requiredlo a\ord amplifier saturation. A n s w e r :( a ) 7 . 1 0 . - 1 5 . 3 . 8 . a nld0 \ ; rbr 2\ ,.1l\. ''IOl L: Also lry Charpr Ptublen\ 5.1-5.3.

5.3 TheXnvefri*lg-Amp['!fier efrcsit We are now readyto discussthe operationof someimportant op amp cir, cuils,usirg Eqs.5.2 and 5.3 to model the behavior of the device itseli Figure5.9showsan inverting-amplifiercircuit.We assumethat thc op amp is operatingin jts linear regior Note that, in addition to thc op amp,the circuit consistsof lwo resistors(Rl and R,), a voltagesignalsource(1,,), anda shortcircuitconnectedbetweenthe noninvefiinginput terninal and We row analyze this circuit, assumingan ideal op amp.The goal is ro obtainan expressionfor the output voltage,.ro,as a functionof the source voltage,or.Wc employa sirgle node voltageequationar the invertingterminal ofthe op amp,givenas Figurc5.9 ). AnjnvedjngimpLifier circuit. is + if: in. (5.6) The vollageconstraintof Eq. 5.2 setsihe voltageat un = 0, becausethe voltageat tlp = 0. Therefore, (5.7)

'

.

J

1)a D

..t

(5.8)

Now we invoke the constraintstatedin Eq.5.3,namely.

(5.e) SubstitutingEqs.5.7-5.9into Eq.5.6 yieldsthe soughrafter resuit: -R, (5.10) { Inverting-amptifierequation

162

TheoperationatAmptifier Note that the ouQut voltage is an inverted, scaledreplica of the input. The sign revercalfrom input to ouQut is!of course,the reasonfor refening to tlle circuit asan invertnS amplifier.The scalingfactor, or gain,is the ratio -Rf/Xs. The result given by Eq. 5.10is valid only if the op amp shownin the circuit h Fig.5.9is ideal;that is,iL4 is infinite and the input resistanceis infinite. For a practicalop amp, Eq.5.10 is an approximation,usually a good one. (We say more about this later.) Equation 5.10 is important becauseit tellsus thatifthe op amp gain,4 is large,we canspecifythegain ofthc invertingamplifierwith the externalresistorsRf and Rr.The upper limflon rhegain.R/ R..is derermined b) rhepot\er supplyvoltagesand the value of the signalvoltageo,. Ifwe assumeequal y- : fcc, we get power supplyvoltages,thatis,y" :

ro < Vcc'

lnr I

= l4*l u-'

R. p

- lY-l

t,"l

(5.11)

For example,if ycc : 15 V and o" : 10 mV, the ralio Rl /R, must be less than 1500. In the inverting amplifier circuit sho*'n in Fig. 5.9,the resistor Rf pro videsthe negativefeedbackcornection.That is,it connectsthe output ter minal to the inveting input termiml. If Rf is removed, the feedback pattr is openedand the amplifier is saidto be operatingopen loop.Fi9].lle5.10 showsthe open-loopoperation. Opedng the feedbackpath drasticallychangestlle behavior of the circuit.First,the output voltageis now Figure5.10.d AninleidngampLifier operating

(5.12)

assuming asbetorethat V* : V : ycc; th€nlo,l < VccfA lor llnear operatjon. Because theinve ing inputcurent is almostzero,thevoltage drcpaoossR"is alnostzero,andtheinve inginputvoltagenearlyequals thesignalvoltage, o"; thatis,on - os, Hence,theop ampcanoperateopen loopin thelinearmodeonlyif ltr, < ycc/A.If ?)"> ycc/A, theop amp simplysatuates.In particular, if tr, < ycc/A, the op ampsatumtesat +ycc, and if tr, > ycc/A, the op a1npsaturates at -ycc. Becausethe path,the relationship shownin Eq.5.12occu$whenthereis no feedback gainof the op amp. valueofA is oftencalledtheopen-loop

0bjedive2-Be abteto analyze sinpl; circuitscontaining id€al.dpamps 5.2

\ollcgeu. in thecircuitin lhesource Assessment Problem.s,lis-640 mV.The 6UkIt IeeoDacx resrslor rsreplaceo b) a \anableresislorR. . Whar.angeot R, alloqslhe

NOTE: Abo try Chapter Problems 5.8 and 5.9.

,

inverting amilifier to opeiate in its lheaf region? Answer: 0 < Rl < 250ko

5.4 lhesumnins Anrptjfier cjrcujt 163

5.4 TheSumming-Amplitier Circuit The output voltageof a summingamplifier is an inveited,scaledsum of the voltagesappliedto the input of the amplifier.Figure5.11showsa sum ming amplifierwith three input voltages. We obtain the relationshipbetweentlte output voltage o. and the threeinput voltages,zra,?h,and o., by summingthecurrentsawayfrom the invertinginput termiMll

&

*--*

'r1o

tn - uc

un

uo

1-+--+t":o

(5.13) Fisure 5.11.{ A summins amptifier.

Assumingan ideal op amp,we cannsethe voltageand curenl constraints together with thc ground imposed at z), by the circuit to see that rn= DD: o zndin = 0.ThisreducesEq.5.13ro

"=

(\

\e*

Rt

+4'b+

Rr\

&"(/'

(5.14) r€ Inverting-sumningamplifier equation

Equation5.14statesthal the output voltageis an inverted,scaledsum of the thrce input voltages. If R" : Rb = Rc : &, then Eq.5.14reducesto

(5.15)

Finaily,if we make Rt = Rr, the output voltageisjust the irverted sum of the input voltages. That is, 1,, =

(oa+ 4+

0c).

Although we illustratedthe summingamplifier wjth just three input signals,the number oI input voltagescan be inffeased asneeded.For example, you might wish to sum 16 individually recorded audio signalsto form a single audio signal.The summing amplifier conJigurarionin Fig. 5.11could irclude 16 djfferent input resistorvaluesso that cach of the input audio tracks appears in the output signal with a differeni amplification factor. The summhg amplifier thus plays the role of an audio mixer. As with invefiing amplifier circuits, the scaling factors in summirg-amplifier circuitsare deternired by the exlemalresistorsRr, na, Rb,R., . . . , R,.

164

TheopeationatAmptifier

ideatop amps 0biective2*Be abteto anatyze simptecircuitscontaining 5.3

a) Find r, jr the circuit shownifo. : 0.1V and tb = 0.25V. b) If tl6 = 0.25V, how large canoabe before the op amp saturates? c) lf x'" - 0.10V, how largecan 1]l)be before I he op ampsarurares? d) Repeat(a), (b), and (c.)wjth the polaity of

(c) 0.5v: (d) -2.s,0.2s,and2v

'!_bIeVCned.

Answer: (a) -?.5 v; (b) 0.15v; NOTE: Alsotty ChaptetProblems 5.12,5.1i,dnd5.15.

5.5 Thef{oninverting-Amptitr'er Cireuit Figure 5.12depiclsa nonirveriing amplifier cilcuit. The signalsourceis representedby us in serieswith the resistorRg.In derivhg the expression Ior the oulput voltageas a function of the souce voltage,we assumeaD ideal op amp operatingwilhin its linear region.Thus,as belore, lve use Eqs.5.2 and 5.3 as t}le basisfor the dedvation.Becausethe op amp iDput curenl is zero,we can write a, = 1" o"4,1to- tq. 5.2,2,, = u! as well Now,becausethe input currert is zero (i, : ,/, : 0), the resistorsRf and R, form an unloadedvoltagedivider acrossD..Therefore,

u,R,

Figurc5.12A A noninverling amplifier.

xJ+R/

(5.17)

SolvingEq.5.17for o,,givcsusthesoughtafterexprcssion:

Noninverting-amptifi er equationv

,,,:

R"+Rl R,

0s.

(5.18)

Opcrationin the linear regionrequireslhat

o,*or.ly.. & lr" Note again that, becauseof the ideal op amp assumption,we can expressthe output voltageas a functionoftheinput voltageandthe external resistors in this case,R" and Rt.

5.6

TheDiffeence-Amptifie, tncuit

objective2-Be abteto analyze simptecircdtscontaining ideatop anps 5.4

Assumethat the op ampin the circuirshown is ideal. a) Find the output voltage when tlle vadable resistoris set to 60 kO. b) How large can n" be before the amplifier saturates?

Answer (a) 4.8V;

(b)7sko. NOTE: Abo try ChapterPnblems 5.17and 5.18.

5.6 The$ifference-"4r'nptifier efreuit Thc output voltageof a differenceamplifier is proportionalto rhe dificr enccbetweentlle two inpul voltages.Todemonstrate, we analyzethc diJfcrence-amplifiercircuit showDin Fig.5.13, assumiDgan ideal op amp operatingin its linear region.We derive thc relationshipbetweeno, and lhe two input voltagesDaand ?rbby summingthe currentsawayfrom the inve ing input nodc: Dn -

Xa

'nr

R.

1)a

(5.1e)

fir

Becausethl3op amp is ideal,weusethe voltageand currentconstraintsro

\5.20) ",r = 0e =

Rd RJ

Rd

0Lr.

\5,2r)

Conbining Eqs.5.19,5.20,and 5.21 gives the desired relarionship:

'

1id(R"+ Rb)

R,(R.+ t?,r)"

Rr,

n"

"'

\5.??)

Equation5.22showsthar the output voltageisproportionalto the diflerencebetweenascaledreplicaofob and a scaledreplicaoI?i".Ingeneral the scalingfactor applied to ob is not the samc as that applied to oa. Howcver,Lhescaiingfactor applied to each input voltagecan be made cqualby setting

R.

n"

Rr

Ra

\5.?3)

Figurc 5.13i,\ A differcnce amptjner.

165

166

The0pe6tjonat Amplifier

When 8q.5.23 is satisfied, the expiession for the output voltage reducesto

simptifieddifference-amp[iIier equation>

15.24)

Equation 5.24 indicatesthat the output voltage can be made a scaled replica of the difference between the input voltages ob and oa.As in the previous ideal amplifier circuits, t]le scaling is controlled by the external resisto$, Furthermore, tlte relationship between tle output voltage and the input voltages is not affected by cornecting a nonzeio load resistance acrossthe outDut of the amDlfier.

TheDifference Amplifier-Another Perspective We can examine t]te behavior of a difference amplifier more closely if we redefine its inputs in tems of two othei voltages. The fbst is the difiercnlial mod€ input, which is the difference between the two input voltagesin Fig.5.13: (5.25)

The second is tlre commotr mode input, which is the average of the two input voltagesin Fig.5.13:

r.

= (o^+ Di/2.

\5.?6)

5.6

TheDjfference-Amptjfi€r Circujt 167

UsingEqs.5.25and5.26,we cannowrepresentthe originalinput voltages, oaand 0b,in tems of the differential modeatrdcommonmodevoltases.

lta:

D6

I \5.27)

t|)dn,

I rb:1)d+l:dn

(5.28)

SubstitutirgEqs.5.27and5.28into Eq.5.22givesrhe outpurof the differenceamplifier in terms of the differertial mode atrd commonmode voltaees:

, t t"n"Rd - RbRc

R"(n. + Rd)

.I

Rd(Ra+R6)+Rb(R"+Rd)l

= ?4!nod

,RJRJ RJ + ,4dm0dm,

j'd-'

15.?9) (5.30)

where Ad is t}le common mode gain and .4dn is tlle differential mode gain. Now, substitute Rc = Ra and Rd : Rb, which are possible values for Rcand Rd rharsatisfyEq.5.23,into Eq.5.29:

/nt\

\4/** Thuq an ideal dilference amptifier has /d = 0, amplifies only rhe differential mode portion of the input voltagg and eliminates the cormon mode po ion of the input voltage.Figure 5.14 shows a difference-amplifier circuit lvith differential mode and common mode input voltagesin place of

1!4 2

3ss Equation 5.30 provides an important pe$pective on t}le function of 2 the difference amplfier, since in many applicatioDs it is the differential mode signal that contains tle infomation of hterest, whereas the common mode signal is the noise found in all electric signals.For example,an electrocardiograph electrode measures the voltages produced by your body to regulateyour heartbeat.Thesevoltageshave very small magni- rigure5.14A A differcnce ampLifierwith common anddjff€r€ntiat mode inputvottages. tudes compared with the electdcal noise that the electrode picks up from mode sourcessuch as lights and electrical equipment. The noise appears as the common mode poftion of the measured voltagg whereas tle hea rate voltages compdse the differential mode portion. Thus an ideal difference amplifier would amplify only the voltage of interest and would suppress the noise.

168

TheoperationatAmptjfi€r

Measuring Difference-Amplifi er PerformanceTheCommon ModeRejectionRatio Ar ideal difference amplifier has zero conrmon mode gain and nonzerc (and usually large) differe tial mode gain. Two factors have an influence on tle ideal common mode gain-resistance mismatches(that is,Eq. [5.23] is not satisfied)or a nonidealop amp (that is,Eq. [5.20]is not sarisfied). We locus here on the effect of resistancemismatcheson the performance of a difference amplifier. Supposethat resistorvaluesare chosenthat do not preciselysatisfy Eq. 5.23.Instead,the relationshipamongthe resistorsR,, Rb,R", and Rdis R"

R^, , ' -€)R

i;:(1

X" = (1 - €)R.

and

Rr = Ra,

(1 - €)Rb

and

R, = R",

Rd:

(5.32)

where € is a very small number.We car seethe effect of this resistarce mismatchon the commonmode gain ofthe differenceamplifierby substituting Eq.5.33into Eq.5.29and simplifyingthe expressionfor ,4..,: R,(1 -€)Rb-RaRb

n,[R"+(1 - €)Rb]

15.34)

-exr Ra+(i-€)Rb -eRr

-- R , . &

(5.36)

€ is verysmall, Wecanmakethe approximation to giveEq.5.36because andthercforc(1 €) is approximately 1 in the denominator of Eq.5.35. Note that, when the resiston in the differenceamplifier satisfyEq. 5.23, € : 0 andEq.5.36givesAcn = 0. Now calculatethe effectof the resistancemismatchon the differential mode gain by substitutingEq. 5.33into Eq. 5.29and simplityingthe expression for ,4dm: (r - e)n(n" + Rb) + R6[R"+ (1 - €)nb] 2RalRa+ (1 €)Rbl 'l (e/2)Ra n.I R"L R , . ( 1 " ) & ]

(5.37)

(5.38)

5.6 lhe Differcnce-Amptifier Circuir 169

= R,[4L'

(€/2)xiI

(5.re)

R "+ R ' ]

We usethe samerationalefor the approximationin Eq.5.39asin the com_ putation of r cm.Wlen the resistors in the difference amDtifier salisfv Fq.5.)J.c = 0 and Eq 5.lq gi!ec?4o,- Ro RI The conrmon mode r€jection ratio (CMRR) can be used to measure how nearly ideal a difference amplifier is. It is defined as rhe rario of rhe diflerenrial modcgajnlo rhecommonmodesrin: a. ) (5.10)

The higherthe CMRR,the more nearlyideal rhe differenceamplifier.We can seethe effect of resistancemismatchon the CMRR bv substitutins Eqs.5.36and 5.39into Eq.5.40:

* *",1 lltt ,o,+r,,,^,

C M R R- l - '

-l

-€fb/(Rr

+ Rb)

\5.41)

R"(1 €/2)+ Rbl

-&

1 1 +R b / R] " I

\5.4?)

15.13)

From Eq. 5.43,if the resistorsin the differenceamDlifier are matched. . 0 rnd CVRR - rc. L\en it rhe iesislorsare mi.marched. ue can minimize the impact of the mismatch by making the differential mode gain(Rb/Ra)very large,therebymaking the CMRRlarge. We saidat the outsetthat anotherreasonfor nonzerocornrnonmode gain is a nonideal op amp. Note that the op amp is itself a difference amplifier,becausein the linear operatingregion,its output is proporrional to tlle difference of its irputs; that is, r,, : ,4(rp , o,,). The outpur of a nonidealop amp is not strictlypropofiional to the differencebetweenthe inputs (the differentialmode input) but also is compdsealof a common modesignal.Internalmismatchesin the componentsofthe intesratealcircuil make(hebeha\ior ot rheop amp Donideal. in rhe.ameua."y rharrhe resistor mismatchesin the difference-amplifier circuit make its behavior nonideal.Even though a discussionof nonideal op amDsis bevond the .copeol lhiste)\r.you ma! notelha(lhe CM RR i, orrenu:edin i.se*ing how neaily ideal an op amp'sbehavioris.In fact,it is one ofthe main wavs of mting op amps in pnctice. NOTE: Assesstour undentanding of thb materiat by trying Chapter Prcblems 5.32and5.33.

170

lheopentionat tunptifier

5.7 A MoreReatisticModetfor the

0perational Amptifier We now consider a more realistic model tlat piedicts the perfomance of an op amp in its linear region of operation. Such a model includes three modifications to the ideal op amp: (1) a finite inpul iesistance, Rj; (2) a finite oper-loop gain,,4;and (3) a noMero output iesistance,Ro.The circuit shown in Fig. 5.15illustr4tes the more realistic model. A \rp - 1,), Wheneverwe use the equivalentcircuit shownin Fig.5.15,we disregard the assumprionsthat t^ = Dp (Eq. 5.2) and ti = ,e = 0 (Eq. 5.3). Furthermore,Eq. 5.1 is no longer valid becauseof the presenceof the nonzero output resistance, Ro. Arother way to undentand the circuit That is,we canseethat shownin Fig.5.15is to reverseour thoughtprocess. the circuitreducesto the idealmodelwhen R; + co, 14- m , and Rd+ 0. For th€ /rA741 op amp,the tlpical valuesof &, ,4, and Ro are 2 MO, 10,, and 75 O, respectively. Figure5.15A Anequivalent cncuitforanopeEtional Although the pres€nceof Ri and Ro makes the analysisof circuits containing op amps more cumbeNome,such analysisremains straightforward. To illustmte, we anal'ze both an invefiing and a noninverting amplifier, ushg the equivalent circuit shown in Fig- 5.15.We begin with tle inverting amplifier.

Analysisof an Inverting-Amplifier €ircuit using the MoreReatistic0p AmpModet If we use the op amp circuit showl in Fig. 5.15,the circuit for the inverting amplifier is the one depicted in Fig. 5.16.As before, oul goal is to express ttre output voltage, o,, as a function of the souice voltage, r'r, We obtain the desiredexpressionby writing the two node-voltageequationsthat describe the circuit and tlen solving the rcsulting set of equations foi tro. In Fig.5.16,the rwo nodesare labeleda and b. Also note that 2,, : 0 by vi ue of the extemal short-c cuit connection at the noninverthg input terminal. The two node-voltaee eouatioN are as follows: node a:

Dn-bs

un , bn

R,

nooe o:

-

+

uo

15.11)

xr

D" - A(-D,\

(5.15)

(ncuit. tigure5.16A Aninveiring-amptifier We rearrangeEqs.5.44and 5.45so that the solution for o, by Cramer's methodbecomesaDoarent:

lt

r

r\

1

1

\a*^,*4/" 4%-A.,"'

(5.40)

(*-+)... (+.;)":"

15.47)

5.7 A li4orc ReaListic l4odeL torihe 0perationat Amptifrer 171 Solving for 0" yields

A + (R"/R)

".*). (t.').*" +('.

(5.48)

Note that Eq.5.48reducesto Eq.5.10asn, '0, &+ co, and "4 + oo. I{ the inve ing amplifier shown in Fig.5.16 wero loaded at its output term;nal(wi$ a loadre.istaoce ot R. ohms.rhe relaliooship belwee;,. and o" would become

A + (R./Rf)

D.

t + A

--

.t.*).("r"X' -4)-4 R,\

1,..(5.49)

&

Analysisof a Noninverting-Amplifier €ircuitUsing the MoreRealistic 0p AmpModel w}Ien we use the equivalent circuit shown in Fig.5.15 to analyze a roninve ing amplifier, we obtain the circuir depicred in Fig. 5.17.Helq the voliage source os, in se es with tle resistatce Rs, represents tlle signal source.The resistor Rr denotes the Ioad on the amplifiet. Our analvsis consislsof derivingaDexpre<(ioDtor ,, as a flrncrionol 0". We do so by writingrhenode-\ollage equadoDs ar nodesa andb. Ar nodea,

un

on-xg

Dn-Do

^,

&-Rr+R,'

= "^

(5.501

and at node b, Figure 5.17A A noninverting amptifrer cncuit.

Becausethe currentin Rs is the sameasin Rr, we have

rp-Dg

I"-

Dg

RI+R"

15.52)

172

Theop€ntjonal Amptifi€r We use Eq. 5.52 to eliminate o/ from Eq. 5.51, giving a pair oi equations itrvolving the unknown voltages on ard 2,. fhis algebraic manipulation leadsto

lt

r

r\

/r\

a\n.' n"-n,' \)

/

t

\

"\E)-%\R".R,/

(55r)

(t en, tl I r\ " lf n , r n- ," , - 4 l ' ".\ & - & ' & /

f en. l - ',Ln rn,* n,rl

(5'51i

Soivjngfor lJ, yields

[(R/ + x") + (R,R"lARt\]Ds

4 * f1, * *.; +

^,,' :

Rc

&

Rr

&

RtR"+ (Rrl f"x^r + Rs)

R,

{5.55)

R1R. r R1R,- RrR.

RB,

&

Note that Eq.5.55 reducesto Eq.5.18 when R,+0, ,4+oo, and Ri+ co. For the unloaded (Rz = c.) noninverting amplifier,Eq. 5.55 simplifies to

'

(nr+&)+

- . |n . ( "I . R. ;\L

R"R"/ARID|

^ | n , +" n , \r .. ^ R" R,

)'

A:,R-,IR,

(R,

4XRi

R!)l (5.56)

Note that, in the derivationof Eq.5.56 from Eq.5.55,1(, reducesto (& + Rs)/n,.

Pra.ti.atPe6pe.tive 173

objective3-Understandthe morerealisticmodelfor anop amD 5.6

The invertingamplifierin the circuit shownhas an input resislanceof500 kO, an output resisf ance()15kO, and an openloop gain of300,000. Assumcthat the amplifier is operatingin its linear region.

Answer: (a)

19.9985: (b) 69.995pV;

(c) 5000.3s o; (d) 20,0pV,5ko

a) Calcnlatethe volragegain (r,,/us) of the anlpnner. b) Calculatethe valueof z',,ir rticrovoltswhcll D c= 1 V ' c) Calculatethe resistanceseenby thc signal source(0s). d) Rcpeat(a)-(c) usingthe ideal modetfor the op amp. NOTE: AIso tly Chaptcr PtobLems5.42dnd 5.43.

PracticalPerspective StrainGages Changes in the shapeof etasticsotidsareof greatimpotanceto enqjr€ers whodesjgrstructures that twjst, stretch,or bendwhensubjected to exter, nalforces.An ajrcraftframeis a primeexampte of a structurein whichengi neersmusttaker'ntoconsideration etasticstrain.ThejnteLtigent apptjcatjon of strain gagesrequiresinformationabout the physicatstructureof the gage,methodsof bondingthe gageto the surfaceofthe structure,andthe orjentationof the gageretativeto the forcesexertedon the structure.our purposehereis to pointout that straingagemeasurements areimportantjn engjneering apptications, anda knowledge of eLectric cjrcujtsis germane to their oroDeruse. Thecircuitshownin Fjg.5.18provid€s onewayto measure the change in resistanceexperienced by strain gagesjr appLicatjons Ike the one

F+ARa

fR-tR

Figure5.18n Anopampcircujtusedfor measuing thechange in nraingage

r00ko

174

Th€0pe,aiionaLAmplifier

in thebeginning ofthischapterAswewit[see,thiscitcuitis the described thetwo familiardifference amptifier, wrththe strainqagebridgeproviding Thepairof strajngagesthat are whosedjfference is amptified. vottages R + An in the bridge lengthened oncethe baris benthavethe vatues that are amp$fier, whereasthe pairof straingages feedingthe difference thiscircuitto discover havethevaL0es -R- AR. Wewittanatyze shortened jn resistthe outputvottage, o, andthe change the retationship between bythestrainqages. ance,AR experienced at thattheopampis ideal.wdtingtheXcLequations Tobeqin,assume inputterminat5 ofthe opampwesee theinvedingandnoninverting

R+AR

R A R

:R

L)n

,

AR'

Dn

Ua

&

=R+AR-&

'

(5.58)

for the vottageat the roninEq.5.58to get an expression Nowrearrange vertingtermjnaIof the op amp:

(5.5e)

* *,(-!-al*.a) in its Linear regjon, so Asusuat, wewitlassume thattheopampis operating bethe expression for ?, in Eq.5.59mustaLso r', = o, andtheexpression theright-hand sideof Eq.5.59in ptaceof zJ, for r',. Wecanthussubstitute maniputation, in Eq.5.57andsotveforr,. AftersomeaLgebraic

r' =

Rr(2AR)

F'

r1;1't'"t

(5.60)

js verysmalL, by straingages in resistance experienced Because thechange (AR)'<< R2,soR2 (AR)' - R' and Eq.5.60becomes Rf

(5.61)

where6 : AR/R. NOTE: Assets,ou anderstandinq of this Practical Perspectiveby tlying Chapter Prcblem 5.48.

summary175

Surnnrary The equatior that definesthe voltage transfercharac tedsticofan ideal op amp is

A(t;o - r") < V66, l-Ur., ro = \A\rp D), -Ucc = A(up - u") < + Vcc, A(ur 1),1> + ucc, I + t.., where .{ is a propofiionaUty constantknoM as the open-ioopgain, and ycc representsthe power supply voltages.(Seepage157.)

An inverting amplifier is an op amp circuit producing an output voltagethat is an inverted,scaledreplica of the input. (Seepage161.)

A sunming amplifier is an op amp circuit producing an outp tvoltage thatis ascaledsum olthe input voltages. (Seepage163..)

A feedback path between an op amp's output and its nrverting inpnt can constrain the op amp to its linear operatingregion where ,', = ,4(?,r o,.). (See page157.)

AnoDinvertingamplifieris an op amp circuitproducing an output voltage that is a scaledrcplica of the input voltage.(Seepage164.)

A voltageconstmintexistswhen the op ampis confined lo its linear operatingregion due to typical valuesof Ucc and A. If the ideal modeling assumprionsare made-meaning A is assumedto be infinite the ideal op amp modelis characterizedbythe voltageconstmint

A difference amplifier is ar op amp circuil producing an oulput voltagethat is a scaledreplicaof the input voitagedifference.(Seepage165.)

rP:

0"

(Seepage158.) . A currenl constraini further characterizes the ideal op amp model,becausethe ideal input resistanceof the op ampintegratedcircuitis infinite.This curent constraint is givenby iP: i" = 0'

The two voltageinputs to a differenceamplifier can be used to calculate the common mode and difference mode voltageinputs,om and ud.. The oulput from the difference amplitier can be written in the form r,-/m0cm+r

dnrid,i,

where ,4cnis the common mode gain, and ,4dmis the differentialmode gain.(Seepage167.)

(Seepage159.) . We consideredboth a simple,ideal op amp model and a more reaiistic model in this chapter.The differences betweenthe two nodels are asfollows: Simpli6cdModel

More Reslislic Model

Infinite input resislance

Finite lnput resistrnce

Infinite openloop gain

Piniteopenloop gain

Zefo output resistance

NoMero oueur resislancc

(Seepage170.)

Tn an idealdiflerencedmplifierA.n,= 0. To mca(ure how nearly ideal a differenceamplifier is. we usc rhe commonmode rejectionratio:

C M R :R l + 1 . l^.ml

An ideal difierence amplifier has an infinite CMRR. (Seepage169.)

176

lheopentionat Amptifier

Probtems S€ctions5.1-5.2

5.3 Find i, in t}te circuit in Fig. P5.3if the op amp is ideal.

5,1 The op amp in the circuit in Fig. P5.1 is ideal. tsPItta) Label the five op amp term.inalswitl their names.

Figurc P5.3

9ko

b) W}tat ideal op amp constraht determines the value of i,? What is this value? c) W}lat ideal op amp constrairt detemines the valueof (rr - 0,)?What is this value? d) Calculate t'o. figureP5.1

8ko

5,4 A voltmeterwith a tull-scalereadingof 10V is used tsP!.rto measurc the output voltage in the circuit in Fig. P5.4.What is tlle reading of the voltmeter? Assurnethe op amp is ideal. Fig!reP5.4 3.3MO

The op ampin the circuit in Fig.P5.2is ideal. a) Calculate2,.if o" : 1.5V and?rb= 0 V b) Calculate 1,oif z,! : 3 V and ?,b : 0 V.

c) Calculate ,, if o. : 1 V and z'o : 2 Y. d) Calculate0, if ou : 4 V and ,b = 2V. e) Calculatet, if o, = 6 V and ,b - 8 V. f) If ob = 4.5V, specifythe range of o, such that the amplifierdoesnot saturate. Figuru P5.2

55 The op amp in tlle circuit Ed.r Calculate the folowinC:

in Fig. P5.5 is ideal.

b) 0, ")h d) i, Fig!reP5.5

160ko

20ko 9V

z0 ko u;60kO

Prcblems177

5.6 Find ir (in microamperes)in the circuirin Fig.P5.6.

5ko

5.9 a) Design an inverting amplifier using an ideal op amp t]tat hasa gain of 4. Use only 10 kO resisto$. b) Il,ou wirb lo amplill a 2.5V inputsignalu.ing the circuil you designedin part (a), what are the smallest power supply signals you car use?

5.10 a) The op amp in the circuit sho\an in Fig. P5.10 is ideal. The adjustable rcsistor R^ has a ma-ximum valueof 120kO, and a is rcstrictedto the range of 0.25 = d < 0.8. Calculatethe range of o, if 5.7 A circuit desigrer claims t}le circuit in Fig. P5.7 will an output voltage that will vary berween pl$flg;Eproduce _j9 as os variesbetween0 and 6 V Assumethe op 6mi amp is ideal. a) DIaw a graph of the output voltage od as a functionof theinputvoltage?s foro < zs < 6V b) Do you agree with the designer'sclaim?

b) If d is not restricted, at what value of a will tle op amp satumte? figureP5.10

Figur€P5,7 15 k()

5.11 The op amp in the circuit in Eg. P5.11is ideal. a) Find t]le rarge of values for (' in which the op amp does not saturate. b) Find i, (in microamperes)when d :

Section 5J 5"8 The op amp in the circuit oI Fig. P5.8 is ideal. """ a) What op amp circuit conliguration is tlis? b) Calculate0d. tiguleP5.8

100ko

figureP5.11 30k{}

178

Theoperationat Amptjfi€r

5.15 Designan in\ erlrngsummingamplilier50thal

Section5.4 5,12 The op ampin Fig.P5.12is ideal. """ a) What circuit configuration is sholl,n in this figure? b) Find r, if ou : 0.5V, ?]b = 1.5V, and ?''=-25V c) The voltages0aand ,'bremainat 0.5V and 1.5V, respectively.w}lat are the limits on ?c if the op amp operateswithin its Linearregion? Figure P5,12

180ko

1!9!€ll

u":

-(3r, + sxb+ 4Dc+ 2Di.

If the feedbackresistor(RJ) is chosento be 60 k0, draw a circuit dia$am of the amplifier and speciJy the valuesofRa, Rb,&, and Rd. 5.16 Refer to t}le circuit in Fig. 5.11, where the op amp Efl( is assumedto be ideal. Given that R. = 4kO, Rb : 5 kO, & = 20 kO, ?,, - 200mV, ,b : 150mV, zrc= 400mV, and ycc - a6 V, specify the range of Rl for which tlle op amp operates within its linear region. SecfionSj 5.17 The op amp in the circuit of Fig.P5.17is ideal. a) What op amp circuit configumtion is this? b) Find ?. in termsof ?". c) Find the range of values for os such that z,odoes not saturateand the op amp remainsinits linear regionof opention. Figure P5.17 32kO

5.13 a) The op amp in Fig. P5.13 is ideal. Find 1,oif ?]a= 16V, ub: 12V, 1J.= -6V,and,Ud= 10V. b) Assume th ,'c, and ?rdretain their values as given in (a). Specifythe range of ?,bsuchthat the op amp operateswithin its linear region. Figure P5.13

330ko

33kO

5.14 The 330kO feedback resistor in the circuit in Afln Fig.P5.13is ieplacedby a variableresistorRJ. The voltagesoa- od have the samevalues as given in Problem5.13(a). a) Whal value of .lRfwill causethe op amp to saturate?Note that 0 < nf < co. b) Wlen Rt hasthe valuefourd in (a), what is the current (in microamperes)into the output ter minal of the op amp?

5.18 The op amp jn the circuit shown in Fig. P5.18is ideal. """ a) Calculate o, when z'sequals 3 V b) Specifythe range of valuesof ts so that the op amp operatesin a linear mode, c) Assumethat z)sequals5 V and that the 48 kO resistor is replac€d with a variable resistor.What value of the variable resistor will cause the op amp to saturate? Figure P5.18 .18kO

hobtens 179 5.19 The op amp in the circuit of Fig. P5.19is ideal. """ a) What op amp circuit configurarion is this? b) Find r]" in terms of 0r. c) Fird the range of values for Dr such that o, does not saturateand the op ampremainsin its linear regionof opemtion.

Flgure P5,21

Figure P5,19

40ko

5.20 The op amp in the circuit shown in Fig. P5.20 is Erft ideal.Thesignalvoltageso" aIId ob are 500mV and 1200mY respectively. a) Wlat circuit conJigurationis showl in tlle figure? b) Calculate r', in volts. c) Find i, and 4 in microamperes. d) wllat are the weighting factors associatedwith ?aand ,b? Figure P5.20

The circuit in Fig. P5.22is a noninverting summing amplifier.Assume the op amp is ideal. Design the lr9r!!1 circuit so that ra=4ud+1)b+2u., a) Specifythe numericalvaluesofRb, R., and Rr. b) Using the values found in part (a) for R/, RD,and R., calculate (in microamperes)i,, ib, and tc when ?)a= 0.75V, 0b : 1.0V, and r,. - 1 5 V-

Figure P5.22 72kO

5,1kO

4.7kO

5.21 The op amp in tbe noninverting summing amplifier Erie of Fig.P5 21is ideal. a) Specifythe valuesof &, Rb,and Rcso thar 0o:60a+3rb+41)". b) Usingthe valuesfound in part (a) for R/, Rb,and n., find (in microamperes)ia, ib, i., ie, and is when ?,'a= 0.5V, ?b : 2.5 V, and ?. : 1 V.

Section 5.6 5.23 a) Use the principle of superpositionfo clerive Eq.5.22. b) Derive Eqs.5.23and 5.24.

180

Ihe0perationat AmpLjfier

524 The op amp in the circuit of Fig. P5.24is ideal. What value of Rf will give the equation uo: 15 zDd, for this circuit.

Design the difference-amplfier circuit in Fig. P5.27 so that ?,, : 7.5(r,b 0,), and the voltagesourceob !rg|!r!. seesan jnput resistanceof 170kO. Specifythe values ot Ra,Rb,and Rf. Use the ideal model for tle op amp.

Figure P5.24 Figure P5.27

15kO

5.25 The op amp in the adder-subtracter c cuit shown in 6trc Fig. P5.25is ideal. a) Find oDwhen oa = 0.4V, ?b : 0.8V, ?rc: 0.2V, and?')d:06V' b) ff o,, r'., and 2d are held constant,what values of 0b will not saturate the op amp?

5.28 Select the values of Rb and Rr in the circuit in *fJfljr Fis. Ps.28 so t}lat ,r, : 4000(ib

r").

The op amp is ideal.

figureP5.25 375kO

Flgufe P5.28

5.29 Designa differenceamplifier(Fig.5.13)to meet tne 5.26 The resistors in the difference amplifier shown in Ma Fig.5.13are R" = 20 kO, Rr : 80 kO, R" = 47 kO and Rd = 33 kO. The signal voltages ?raand ,6 are 0.45and 0.9 V respectively,and ycc : 19 V. a) Find 1)". b) W[at is the resistance seen by the signal c) What is the resistance seen by the signal

following cdteda: I)a :zDb - 5a^.The resistance seen by the signal source rJbis 600kO, and tlle resistanceseen by the signal source o, is 18kO when the output voltage o, is zero. Specilf the valuesof Ra, Rb, nc, and Rd.

to,r_!ll1

The op amp in the circuit of Fig. P5.30is ideal. a) Plot Daversusa when RJ : 4Rr and DE= 2V. Use inuements of 0.1 and note by hypothesis that0 < a < 1.0.

pmbtems181 b) Wdte atr equation for the staight lire you plotted in (a). How are the slope and intercept of the line relatedto os and the ratio Rr/R1? c) Using the results from (b), choose values for ?is and tllreratio Rr/R1 suchthata" = -6a + 4.

5.33 In tle differerce amplifier shom in Fig. P5.33,what rangeofvalues of Rr yieldsa CMRR > 750?

Figure P5.33 47kA

Figure P5.30 Rr

-10v

531 The rcsistor Rr in the circuit in Fig. P5.31is adjusted EtrG until the ideal op arnp saturates.Specill' lRrin kilohms.

Sectiotrs5.1-5.6 5.34 a) Show that when the ideal op amp in Fig. P5.34is operating in its Linearregion,

Flgure P5.31

.

3r,

, a : ^ .

b) Show that the ideal op amp will saturate when

R(tUcc - zr,E)

T0 the dif{erence ampUfier sbosD in fig. P5.J). compure (a) the differential mode gain, (b) the commonmode gain,and (c) rhe CMRR. Flgufe P5,32

100ko

;

tigur€P5.34

182

lhe 0peBtionat Anptifier

5.35 The c cuit insidethe shadedareain Fig.P5.35is a arc constant current source for a limited range of val uesof Rl, a) Find tle valueofi, for R, : 2.5kO. b) Find the maximum value for Rr for which i_ wiil have the valuein (a). c) Assumethat RL = 6.5kO. Explainthe operation of the circuit. You can assumethat in : i? - 0 under all operatingconditions. d) Skeichir versusR' for 0 < R, < 6.5kO.

5,37 Assume that the ideal op amp in the circuit in Ers Fig.P5.37is operatingin its linear region. a) Calculate the power delivered to the 16kO b) Repeat (a) witi the op amp removed from the circuit,that is,with the 16 kO resistorconnected in the serieswith t}le voltage source and the 48 kO resistor. c) Find the ratio of the power found in (a) to that found in (b). d) Does the insertionof the op amp betweenthe source and the load serve a useful purpose? Explain.

FigurcP5.35

Figure P5,37

48kO 320mV

itit s.38 Assumethat the ideal op amp in the circuit seenir Fig.P5.38is opemtingin its lineai region. 5.36 The op ampsin the circuitin Fig.P5.36aie ideal. """ a) Find 1". b) Find the value of the right sourcevoltage for which i, = 0.

Figure P5.36

500mV

a) Showthatr',: [(Rr + Rtlnr]?r. b) What happensif Rl + oo and R2+ 0? c) Explain why this circuit is referred to as a vol! agefollower when -R1: co and R, = 0.

Probt€ms 183 figureP5,38

Figur€ P5.40 2ko

18kO

3.9kO

5.39 The two op ampsin the circuit in Fig. P5.39are Edc ideal. Calculate 2,1and 1,,2.

5.41 The voltageos shownin Fig. P5.41(a)is appliedto 6ntr the inverting amplifier shown in Fig. P5.41(b). Sketchrroversust, assumingfie op ampis ideal.

Figure P5.39 FigureP5.41 2Y

' \ / \ / 0.s l q 1.5 1.0 2.s 3.q 3.s y'.o /G

4.7kO 2

(a) 75tO

5.40 Th€ signalvoltage o! in the circuit shownin Fig.P5.4.0 Fa., is describedby the followingequations:

r<0, ?,s= 4 sin(srl3)rV,

8.2kO

0 < t < oo.

Sketch?r,ve$us ,, assumingtlte op ampis ideal.

(b)

184

The0peGtionat AmpLifier

Section 5.7

Figure P5.44 240ko

5.42 Repeat Assessment Problem 5.6, given tlat the Efld irverting amplifier is loaded with a 500 O resistor.

5.,(} The op amp in t}le noninverting amplifier circuit of xrc Fig. P5.43has an itrput resistanceof 440 kO, an output resistance of 5 kO, and an openloop gain of 100,000.Assume that t}le op amp is operating in its linear region, a) Calculatethe voltage gain(1,"/D). b) Find the inve ing and noninverting input voltageson and o/ (in millivolts) if lls = 1 V. c) Calculate the difference (?i? on) in microvolts when t,s = 1V. d) Fhd the current drain in picoamperes on rhe signal source rrswhen l)s = 1 V. e) Repeat(a)-(d) assumhgan ideal op amp.

2lmko

J 5.45 Repeat Problem 5.44 assumingan ideal op amp.

5.216Assume rbe inpur resislanceof the op amp in qr+ Fig.P5.46is infinireandilc outputresi.rance i. /ero. a) Find o, as a function of ?s and the open-loop galn,{. b) What is the value of t,o if os : 0.5V and / : 150? c) What is the value of 1), if o" = 65Y ,tt. d) How largedoes.4haveto be so that o,is 98% of its valuein (c)?

FigureP5.46

150ko

s.44 a) Find the Th6venin equivalent circuit vdth rcspect lo lbe ourpul lermjnals a.b lor the inverrjng ampliJier of Fig. P5.44.The dc signal souice has a value of 150 mV The op amp has an itrput resist ance of 500 kO, aJI output resistance of 750 O and an open-loop gain of 50,000. $4lat is rbe ourpurresistance of the in\efling amplifier? c) lvhat is the re.isrance fin ohms)seenb) thesig, nal source ?rswhen the load at the teminals a,b is 150O?

5.47 DedveEq.5.60.

P'obtems185 S€ctions5.1-5.7

5.50 a) IfperceDterroris deliredas

5.48 Supposethe stmin gagesin the bridge in Fig. 5.18 pPRi.Hthave the value 120 O I 1%. The power supplies to the op amp are I 15 V, and the reference voltage, OEr,is taken from the positive power supply. a) Calculate tie value of Rt so that when the strain gage that is lengtlening reaches its maximum Iengtl, the output voltage is 5 V b) Supposethat we can accurately measure 50 mV changes in the output voltage. What change in strain gage resistance can be detected in milliohms?

r!:r!!lv! approximate value

-'l"'*,

show that the percent eroi in the approximation of I'o in Problem 5.49is AR (R + Rf) . %eror= = -:vlot,. ^ t^-z^t b) Calculatethe percent erloi in ?r,for Problem 5.49.

5.49 a) For the circuit showr ir Fig. P5.49,show rhar if AR << R, the output voltage of the op amp is ,ll|;iH;, approximately iiiiii

do !;;

-R,rR + n.l , -

.(-^R)2,,0.

b) Find ?,,if Rr = 470kO, R = 10 kO, AR : 95 O, and !'in = 15 V. c) Find tie actual value of oo in (b). Figure P5.49

5.51 Assume the percent error in the approximation of pRflH;lod in the cncuit in Fig. P5.49 is not to exceed 1ol". Fna Wlat is the largest percent change in 1Rthat can be toleiated?

5.52 Assume the resistor in the variable branch of tlle p'#flI#hbridge circuit in Fig. P5.49is n - AR. Ma; a) What is the expressionfor o, if AR << R? b) What is the expression for the percent erroi in 0o as a function of R, Rr, and AR? c) Assume the resistance in the variable arm of the bridge circuit in Fig.P5.49is 9810O and the vallresof R. &. and ," are the same as in Problem5.49(b).Whatis the approximatevalue What is the percenteror in the approximation ot ro \rhen the variable arm resi.ranccj( 9810()?

Inductance, Capacitance andMutual Inductance 6.1 TheInductor p 188 6.2IheCapaeitot p 195 6.3 Series-Parattet Combinations of Inductance p ?00 and Capacitance 6.4 Mutuatlnductan€ep 203 6.5 A CtoserLookat llutuallndtdan.e p 2A7

Knowandbe abteto usethe equations for voLtage, curcnt, pow€r,anden€rgyjr an jnductor;understafdho\aan inductorbehaves jn th€ presence of cofstait cLrrrent, andthe requnement that the cunentbe continuous in Knowandb€ abt€to usethe equatjorsfor voltage,cunent.power,andeneryyin a capacitor; undestandhowa capacjtorbehaves in the presence of consiantvottag€,andthe rcquirement tlrat th€ vottagebe continuous in a Beabteto combineinductoBwith iiitiat condjtions in sen€sandin panltetto forma singt€equivatent inductorwith an jfiiiat coidjtion;be abteto combir€capacjtoBwjth jfitial coidjtionsjn senesandjf pamttetlo forma singLe equivatent capacitorwith an Urde6tafdthe basiccofceptof mutual inductance andbe abteto write m€sh-cufleit equatjoisfor a citruit containingmaqneticatly couptedcoitsusjfs the dot conveniion

186

We begin this chapter by introducingthe last two ideal circuit elementsmentioneclin Chaptet 2, namely.inductorsand capacjtors.Be assuredthat the circuit analysistechniquesintroducedin Clrapters3 and 4 applyto circuitscontaininginductorsand capac itols. Thercfolc, once you Lrlde$fand the tcrminal bebavior oI these elementsin terms of current and vollage, vou can usc KirchholT'slaws to desc be anv intetconnectionswith the other basicclcments.Like other compoflents.inductorsand capacitols are easier 1()describeirl terms of circuil variablesrather than electromaSnetic ficld variables.T-Iowever, before we focuson the circuit desc ptions,a blief reviervof thc field conceptsul1der-lying thesebasicelcmcntsis in order'. An incluctor is ar1electdcal component that opposesany changein electrical current. It is composedoI a coil of wire wound ar-ounda supportingcore wbosematerial may bc lnagnetic or nonmagnetic.'l'he behaviorof inducto$ is baseclon phenomena associatedwith magnetic fields. The source of thc magnetictleld is chargein motioD,or current. If thc culrent is varvingwith tjrne.the magneticfield is varyingwith time.A timevaryiig ma[iletic licld inducesa voltagein any conduciorlinked by ihe ticld. The cir-cuitparameter of ind[ctance relates the inducedvoltage1()the cufrent.We discussthis quanlilativerela tiorNhipin Section6.1A capacitoris an electricalcomponenl that colsists of two corrductorsseparatcdby an insulator or dielectricmatcrial.The capacitoris the only device other than a battery ihat can store electricalcbarge.Thebehaviorofcapacitorsis basedon phel1omena associated with electricficlds.Thesourceof the electricficld is separationof charge,or voltage.lf the voltagc is varying with time,the electliclield is varyingwith time.A time-varyingelec ic field producesa displacementcurrel1lin the spaccoccupiedby thc field. The circuit parameterof capacitancelelates the dis placementcurrentto thc voltage,wherethe displacementcurcnt is equal to the conductioncurrent at the terminalsof thc capacitor. Wc discusslhis quantitative rclationship in Section 6.2.

PracticaI Perspective ProxinitySwitches Theetectricaldeviceswe usein our dailytjvescontainmany switches. l4ostswitchesaremechanicat, sucha5the oneused in the ftashlightintroduced in Chapter2. l4echanical switches usean actuatorthat is pushed,putLed, stid,or rotated,causjng two piecesof conductingfiretalto touch and createa shot circuit. Sometjmesdesignerspr€fef to use switches withoutmovingparts,to r'ncrease the safety,retjabitity,convenience,or novettyof their products.Suchswitchesare caLtedproxjmityswitches.Proxjmityswitchescan empioya varietyof sensortechnotogies. For exarnpl€, someetevator doorsstayopenwhenever a Ljghtb€amis obstructed. Anothersensortechnotogyused in proximityswitches detectspeopleby responding to the djsruptionthey causein etectrjcfields.Thistype of proximityswjtchis usedin some desklampsthat turn on andofFwhentouchedandin etevator buttonswith no movingparts(as shownin the figure).The switchjs basedon a capacitorAsyouareaboutto dtscover in this chapter,a capacitoris a cifcuit €Lement whoseterminaL characterjstjcs are deteminedby etectricfietds.!!hen you proximjtyswitch,you producea changein toucha capacitjve

the vatueofa capacitor, causinga voltagechange, whichacti vatesthe switch.Thedesignof a capacjtjvetouch-sensjtjve swjtchis the topic ofth€ Practicat Perspective exanrple at the endof this chapter.

187

188

and14utuat lnductance Inductance, Capacjtance,

Section6.3describestechniquesusedto simplifycircuitswith seriesor parallelcombinationsof capacitorsor inducto6. Energy can be stored in both magnetic and electic field$ Hence you should not be too surprisedto learn that inducto$ and capacitorsare capableof storingenergyFor example,energycan be storedin an inductor and then releasedtofirea sparkplug.Energycanbe storedin acapac_ itor andthenrcleasedto fire a flashbulb.Inidealinductorsand capacitors, or y as much energycan be extractedas hasbeen stored.Becausehductors and capacitoiscannotgenerateenergy,they arc classifiedas passive In Sections6.4and 6.5we consideithe situationin which two circuits are Linked by a magretic field and thus are said to be magnetically cou_ pled.In this case,the vollageinducedin the secondcircuit canbe related to the time-varyilg cment in the first circuit by a parameterknofir as mutual inductance. The practical significance of magnetic coupling and unfoldsaswe study1ie relationshipsbetweencurrent,voltage,power, We introducethese severalnew parametersspecificto mutual inductancerelationshipshereand then describetheir ulility in a devicecalleda transformer in Chapters 9 and 10-

6.1 TheInductor Inductanceis the circuit parameter usedto describean inductor. Inductance is synbolized by tlle letter l,, is measuredin henrys (H), and is represented graphicatly as a coiled wire a reminder that inductance is a consequence of a conductorlinking a magneticfield. Figure 6.1(a)showsan inductor. Assignhg the rcfererce direction of the current in lhe direction of the volf agedrop acmsstheteminals ofthe inductor,asshownh Fig.6.1(b),ields

- t equation Theinductorr ts

,di

(6.1)

where o is measuredin volts,L in henrys,i in amperes,and I in second$ Equation6.1reflectst}le passivesignconventionshownin Fig.6 1(b);that (a) is, the current reference is in the diroction of the voltage drop acrossthe inductor.If the current referenceis in the direction of the voltagerise, L F-,ffi< Eq.6.1is written witl a minussign. Note fromEq.6.1 tlat the voltageaoossthe termjnalsofan inductor i is proportional to ttre time rute of change of the current in the inductor. (Dl We canmake two important obsenationshere.First,if the currentis con' stant, the voltage acoss the ideal inductor is zeio. Thus the inductor Figure6.1 A (a)Thegraphic symbotfor aninductor behaves as a short circuit ir the presence of a constant, or dc, current. (b)Assigning of /, henrys. refer€nc€ withaninductance in an inductor;that is,the Second, cu en[ cannotchangeinstantaneously fottowjng the pasvottage andcmenttotheinductor, curent cannotchangeby a finite amountin zero time. Equation 6.1 tells us that this changewould require an irfinite voltage, and infinite voltages are noi possjble.For example,when someoneopens the switch on an L

. f f i -

6.1

Thelnductor 189

inductive circuit in an actual system,the current irutially contlnues to flow in the ah across the switch, a phenomenon called a/cr'r& The arc across the switch prevents the current ftom drcpping to zero instantaneously. Switching inductive circuits is an important engineering prcblem, because arcing and voltage surgesmust be controlled to prevent equipment damage.The first step to urderstandbg the natue of this problem is to master tle hfoductory material presented in this and tle following two chapteG Example 6.1 illusirates the application of Eq. 6.1 to a simple circuit.

Determining the Voltage, Giventhe Current.at the Terninalsof an Inductor The independent current soruce in the circuit shownin Fig. 6,2-generates zerc curent for t < 0 anda pulse10te"A, for t > 0.

t=0,

r<0

100mH i=10te 5lA, ,>0

c:)u = Ldild.t = (0.1)10€5!(1-5, =e "(1 5 l ) v , l > 0 ; ? ,: 0 , t < 0 . d) Figure 6.4 shows the voltage waveform. e) No;the voltageis proportionalto dt/dt, not i. f) At 0.2 s,which coresponds to the moment when /t/d/ is pa$ing rhrougbzero and changingsign. g) YeE at t : 0. Note that the voltage can change instantaneouslyacross the teminals of an inductor,

Fisure5.2r Th€cjrcujt forb(ampl€ 6.1.

a) Sketch the curent waveform. b) At what instant of time is the current maximum? c) Er?ress the voltage adoss the lemrinals of the 100mH irductor as a function of time. d) Sketch tlle voltage waveform. e) Are the voltage and the clrlrsnt at a maximum at t])e same time? f) At what instant of time does the voltage change polarity? g) Is therc ever an instantaneouschange in voltage aqossthe inductor?If so,at what time?

r (A)

Flgure5.31Thecunentwaveform forExample 6.1.

?r(v)

Sotution a) Figue 6.3showsthe curent waveform. b) dildt = lo(-ste-st + s 1= 10e 5'(1 - sr) Als;dildt = 0 whent = s. (SeeFis.6.3.) l

Figure6.4

ThevoLtage waveform fo' E\ample 6.L

190

Inductance,Capacitance,and Mutuatlnductance

Currentin an Inductorin Termsof the Voltaae Acrossthe Inductor Equation6.1expresses the voltageaoossthe terminalsofan inductorasa function oI the curent in the inductor. Also desirable is the ability to expressthe curent as a function of the voltage.To find i as a function of o, we startby multiplyingboth sidesofEq.6-l by a differentialtime dt:

=,(*),,

(6.2)

Multiplying the rate at which i vadeswith tby a differenrialchangein rime generatesa differentialchangein t, so we $,riteEq.6.2 as adt = Ldi.

{6.3)

We next integrateboth sidesof Eq. 6.3.For convenience, we interchange the two sidesofthe equationand write

"l*,'o'=1,"' ' o'

(6.4)

Nole that we useI and i as the variablesof irtearation. whereasI and I becomelimirsonlhe integral..Thcn. trom Lq.0.4.

Theinductori - z' equationb

1 r 4.t) : i I 1'!1r + j(ro),

(6.5)

where i(r) is the current coresponding to 1,and i(r0) is the value of the inductor curent when we initiate the integration,namely,t0. In many practicalapplications,to is zero and Eq- 6.5becomes

iI'.*.

(0).

(6.6)

Equations6.1 and 6.5 both give the relationshipbetweenrhe volrage and current at the terminalsof an inductor. Equation 6_1exDresses the \olrageasa funcrionot currenr.u hereasLq.o.5i\pfesse\lhe curfcnlasa tunction of voltage.In bolh equationsthe referencedirectionfor the current is in the directionof the voltagedrop acrossthe teminals. Note that i(lo) cardesits own algebraicsign.If the initial cu enr is in the samedircc tion as the reference direction for i, it is a positive quantity. If the initial current is in the oppositedirection,it is a negativequantity.Example 6.2 illustratesthe applicationof Eq. 6.5.

6.1 TheInductor 191

Determining the Current,Giventhe Voltage. at the Terminals of an Inductor The voltage pulse applied to the 100 mI{ inductor shown in Fig. 6.5 is 0 for t < 0 and is given by the expfession

r=0, 100mH

1)= z\te-|tuV,t>0

a(t) = 20tu ntV Figure6.5

foit > 0. Also assumei : 0 for I < 0.

lhe circuitfor Exanpte 6.2.

a) Sketch the voltage as a function of time. b) Find the inductor currenaas a function of time. c) Sketch the current as a functiotr of time.

Solution a) The voltage as a lunction of time is shorm ln Fig.6.6. b) Thecunentin thehductor is 0 a = 0.Therefore, thecurrentfor t > 0is 1= t

2 I

I t0

10te10'- s 10)A,

Flgure6.6 A ThevoLtage wavefo,n fortuample 6.2

r (A)

| 2 r . x et c d t + o

t ",o' ll, = 2 0 0 11 m ( 1 0 + r 1 )l l . =2(1-

r<0

r> 0

c) Figure 6.7 showsthe cment asa function of time.

Figure5.7 Thecun€ntwavetorm forExampte 6.2.

Note ir Example 6.2 that i approaches a constant value of 2 A as t increases.We say more about this rcsult after discussingth€ energy stored in an hductor,

PowerandEnergyin the Inductor The power and energy relafionships for an inductor can be derived directly ftom the current al1d voltage relationships. ff the curent reference is in rhe dircction of t]le voltage drop across the terminals of the inductor.the Doweris

(6.7)

192

Inductance, CaDacitance, andlluiuatInductanc€

Reqgmbe-Ilhlt pqwg.r.istu wa!t!, voltageis in voltq atrd aurrent is in amperes.If we expressthe inductorvoltageas a function of tlte hductor currcnt,Eq. 6.7becomes

'i:lt:',;1.. :: ': :'i'i:Li:i!.. ':.

Power in aninductor >

..

\t dt .:i.

(6.8)

We can also expressthe curent it terms of the voltage:

o:,1!1,"',*. r,"t)

(6.e)

Equation 6.8 is useful in expressing the energy stored in the inductor. Power is the time mte of expending energy,so d.w

,.di

(6.10)

Multiplying both sidesof Eq.6.10 by a differential time gives the differential relationship dw : Li .1i.

(6.11)

Both sidesof Eq.6.11 are integiatedwith the understandingthat the refeietrce for zerc energy correspotrdsto zero curent h the inductor. Thus

I

Energy in anindudor>

dx=Llyd)J,

$u,iffi

(6.12)

As before,we use different s]'mbolsof integration to avoid confusion with the limits placedon the integrals.In Ed: 6.12,the energyis itr joules, inductanceis in henrys,and curent is in amperes. To illustrate the applicatiotrof Eqs.6.7and 6.12,we return to Examples6.1and6.2by meansof Example6.3.

6.1 lheInductor 193

Determining the Current, Voltage, Power. andEnergy for an Inductor a) Por Example6.1,plot i, o,p, and x, versustime. Line up the plots veftically to allow easy assess ment of each variable's behavior. b) In what time interval is eneryy being stoied in the inductor? c) In what time interval is energy being extmcted from the inductoi? d) What is the maximum energy stored in the hduclor? e) Evaluate tl]e integrals

f" J" oo'

and

e) Fiom Example6.1, i : 10t?-5'A

and

? , : e 5 ' ( 1- 5 4 v .

Therefore, p = 1)i = Tote 1or- 5or2e 1orw.

/.l",oot, 0.6

and comment on thet significance. Repeat (a) (c) for Example 6.2.

0.8

1'(v)

In Example6.2,why is there a sustainedcurent ir the inductor asthe voltageapproaches zero?

Solution a) The plots of i, o, p, and ?.'follow direcdy from the er?ressions for i and t, obtained in Example 6.1 and are .hown^in I ig. 6.8.ln particulaf.p - ,t. andl/, - (;)Lt. b) An increasing energy cun'e indicates that energy is being stored.Thus erergy is being stored in the time intenal 0 to 0.2 $ Note that this correspondsto the irterval when p > 0. c) A decreasingenergy curve indicates that energy is being extracted.Thus energy is being extracted in rherimeinrerval0.2s lo oo.Nole thatrhiscorrespondsto t}le inte al when p < 0. d) From Eq.6.12 we seethat eneigy is at a naxlmum when current is at a maximum; glarcing at the gmphscontirms this.From Example 6.1,ma)omum curent : 0.736A. Therefore,unax : 2'7.07mJ.

u (nJ.)

figure 6.8 A Thevarjabtes t, r, p, andu venusr for Example 6.1.

194

Induchnce, Capacitance, aid l,lutuat Inductance

Thus

ta.2

f,tu lo.z rat : tul*t-to' - tt l"

/

'{t#. *[#r-,,-

g) The application of tle voltage pulse stores ene4y in the inductor. Because the inductor is ideal,this energycannotdissipateafter the vottage subsidesto zero, Therefore, a sustained current circulates in the circuit. A losslessinductor obviously is an ideal circuit element. Practical inductors rcquire a rcsistor in the circuit model. (More about this laler.)

',]):'

= 0.2e) : 27.0'7tJ,l,

, (v) /-

f "1at lm = IoLlmt to, t, Jate',t ]0.,

'{3#..e[#, *'-

"]);

i (A)

: ^0.2e 2 = -21.M mJ.

Based on the definition of p, the area undei the plot of p ve^us r represents the energy erpeDded over the intenal of inregration. Hence tlte integmtion of t}le power between 0 and 0.2 s representsthe energystored in the inductor during tiis time interval. The inteFal of p ovei the interval 0.2 s co is the energy extracted. Note that in this time interval, all the energyodginallystoredis removed;thatis,after the cment peak haspassed,no energyis stoied in the inductor. f) The plots of o, 4p, and 1, follow dircctly from the er?ressions for ?,and i given in Example 6.2 and are shown in Fig. 6.9. Note that in this case the po$er is alwals posilire.and henceenergyis alwaysbeing stored during the volrage pulse.

? (-w)

u (mr)

0.4 0.5 0.6 Figure5.9 l. Theva abtes 1',t, p, and?ove6usr for Exampte 6.2.

'

6.2

TlreCapacjtor 1 9 5

for vottage, current,power,andenergyin aninductor objective1-Know andbeab[€to usethe equations 6.1

The curent sourcein the circuit showngeneratesthe curent pulse lr(t)=O,

t<0,

is(l) = 8€ 3m'- 8d 12oo'A,

r > o.

Fnrd (a) u(0); (b) the iDstantof time,greater than zero,when the voltageo passesthrough zerc;(c) the expressionfor the power deiivered to the inductor:(d) the instantwhen the poNer deliveredto the inductoris maximum;(e) the naximum power;(t the instantof time when the storederergy is maximum;and(g) the maximum energystoredin the inductor.

An5wer (a) 28.8V (b) 1.s4ms; (c) 76.8e-6ooi + 384e15oo' - 307.2e-2{otw. r > 0; (d) 411.0s i!s; (e) 32./2w; (f) i.54rns; (g) 28.s7mJ.

NOTE: Also try Chaptet Pnbkns 6.1 and 6.3.

6.2 Thefapacitor The circuit parameteroI capacitanceis representedby the leller C. is measuredin farads(F), and is synbolizedgraphicallyby tlvo short paral -lel conductiveplates,as shown in Fig.6.10(a).Becausethe farad is an (a) extrernelylargequartity ofcapacitance.practicalcapacitorvaluesusually lic in the picofarad(pF) to microtarad(,.rF)range. C Thc graphic symbol for a capacitoris a reminder that capacitance occurswheneverelectricalcoDduclorsarc scparatedby a dielectric,or insulating,material. This condition implics that electric charge is nol lransportedthrough the capacitor.Although applying a voltage to the (b) terminalsofthe capacitorcannotmove a chargethroughthe dielectric.it a capacitor. can displacetI chargc within the dielectric.As lhe voltage varies with Figure6.10;]: (a)Ihecircuiisymbotfor (b)Assignjnq vottage and cu entto the rcfercnce time, the displacementot chargealso varies with lillle, causingwhat is thep$sivesignconvention. capacitor, totLowing knowr as the displecem€ntcunent. fiom a At the terminals.Lhedisplacementcurrentis indistinguishablc conductioncurcnt. The current is propolioDal 1()thc ralc at which the voltageacrossthc capacitorrarieswith time, or, malhematicall-v,

t =r # ,

(6.13) n; capacitori

whcre i is measuredin amperes,C in larads.?rin volts,and tin seconds.

t, equation

196

Inductance, capacitance, andtlutuat Inductance Equation 6.13reflects the passivesign convention shown in Fig. 6.10(b); that is,the current referenceis in the direction of the voltaqe drop acrossthe capacilor.It tbe currenl referencei\ in fie direclion oirhe volraserise. Eq.o.lJ i\ wrirrenwitha mjDu.srgn. Two importantobsenationsfollow from Eq. 6.13.Iirst.voltagecarmot change instantaneouslyacrossthe terminals of a capacitor. Equation 6.13 indicares tbrl sucha changewouldproduceintinirecurrenr,a phrsical impossjbiiir).second. it the vollageacrosslhe lermtnal.is con.ranr,rhe capacrtorcurrent is zero.The reason is that a conduction current cannot be establishedin the dielectdc male al of the capacitor. Only a time_varying voltage can produce a displac€mentcurrent, Thus a capacitor behavesas an open circuit in the presenceof a constant voltage. Equation 6.13givestle capacitorcurent as a function of the capacitor voltage.Expressingthe voltageas a lunction of the cu ent is alsouseful. To do so,we multiply both sidesof Eq. 6.13by a differcnriattime dr and then integrate the resulting differentials:

id! = Cdo

l))'*=il,"''*

Carrying out the integralion of the left-hand side of the secondequa, tlon grves

Capacitor?)- i equationb

,r,t=ll',0,*,<^t.

{6.14)

In many practicalapplicarionsof Eq. 6.14,rhe initial rime is zero; thar is, lo = 0. ThusEq.6.14becomes ,(t) :

Il,',0,,^o,

(6.15)

We can easily derive the power and energy relationships lor the capacitor. From the definitionof power,

powerequationts Capacitor

(6.16)

,='11f,""*. ^,",] Combiningthe definitionof energywith Eq.6.16lelds du : C1)d't),

16.r7)

6.2 TheCapacitor197 from which

I d.x: C I jdy,

(6.18) .{ Capacitorenergyequation

In the derivationof Eq.6.18,the referencefor zeroenergycorresponds to Examples 6.4 and 6.5 ilustrate the application of the current, voltage, power, and energy relationships for a capacitor.

Determining CurrentVottage, Power, andEnergy for a Capacitor The voltage pulse described by the following equations is impressed auoss the terminals of a 0.5 /rF

Solution a) FromEq.6.13,

{li,,"

1)0):

t < 0s; 0s<1<1s;

( (0.sx 10 6)( 4a(i r)) :

a) Derive the expressionsfor the capacitor current, powei, and energy, b) Sketch the voltage, current, power, and eneigy as functions of time. Ljne up the plots ve ically. c) SpeciJytle interval of time when energy is being stored in the capacitor. d) Specify the interval of time when energy is being deliveredby the capacitor. e) Evaluate tlle integrals

I'

pdt

and

* 10-9(0)= 0, [ { o s , : 1 ( o . s1x0 6 \ ( :=) 2 p " A ,

l* oo,

and comment on their significance.

1< 0s; 0s rs.

Theexprcssion for the poweris derivedfrom Eq.6.16:

: ,,,*, 1,oiuo, 1) = -8e 2( 1) (4e (' 1)(-2rt [

/rW,

t < 0s; 0s=1<1s; , > 1s.

The energy expressionfollows directly fmm Eq.6.18:

t < 0s; 0s 1 s ,

: o,',r, { l,to.a,u,'

198

Inductance, Capaciiancq andMutuat Inductance

b) Figure 6.11 shows the voltage, clurentj power, and eneryy as functions of time.

z'(v)

c) Energy is being stored in the capacitor whenevei the power is positive. Hence energy is being storedin the illteival0-1s. d) Energy is being delivered by the capacitor whenever the power is negative.Thus energy is being deliveredfor all t geater than 1 s. e) The integral of p d/ is the energy associatedvrith the time inte al cofiesponding to the limits on the integral.Thus the first integralrepresentsthe eneryy stored in the capacitor between 0 and 1 s, whereas the second integral represents the energy returned, or delivered, by the capacilor in the interval 1 s to co:

2 I 0 1

rG)

z p @w) 8

tr tt ,lr t a r = , 8 t d t: 4 i l : 4 p I . I ,,0 J0

rG)

l0

u (pJ) /N

I

J

tF

" e a, = , r-8c

')dt = (-8)

- 2r,-lJlN

,It

il

z

= -a /,J

ll

The voltage applied to the capacitor rctums to zero astime inoeases without limit, so the energy returned by this ideal capacitor must equal the eneigy storcd.

!@!|

| || I

Figure5.11 .A Thevariabt€s x,,i, p, and?ove6us,for Exanple 6.4.

Finding,tl,p and ?, Inducedby a Triangularcurrent Pulsefor a capacitor

I An uncharged0.2/,F capacitoris ddvenby a tlianI gularcurrentpulse.Thecurrentpulseis desciibedby

I

rG)

tn

,
s000/A. 0. r-20ss: ",". -. - J s o o o , a 2 , 0 < r< 4 0 p s ; l|o . . z r - 4ops to

I a) Deri!e lhe er"ression.for the capacilorvollage. Wwer, ana enereVfor eachof the four time rnterI I valsneededto descdbethe current.

b) .P]glt: y lersusI AligotheplotsasspecI Ineornilflr"l rneprevlous exampres, I c) Why does a voltage rcmain onthecapacitor aJter I I thecurrentrcturnsto zero?

Solution

4

Flli

== arr are zero' ?':r;iii"D

o=5x106 l1sornlar+0=12.5x10er' v, Jo' p .,i 025 lorrl'w' w:1Ctl

= 15.625 < 101'14J.

For20ps< , < 40ps, a: S x tOu/ (0.2- S000r)dr+ 5. Jzo^

6.2

(Notcthal 5 V is lhe voliageon thecapacitor at theendof thc prccedirgiDlerval.) Then,

Thetapa.to

199

t (mA)

| = (106r- 12.5x 10er2 10)V. t (i.t

- (62.5x 10tr13 7.5)<10et,+ 2.5x 105r- 2)w,

,,(v)

. =lc,t, = 05.625x 1or'ia 2.5 x 10qrr+ 0-125x 106t2

I (,..s)

? + lo 5)J. p ('nw) 500 400 300 200 100

a = 10V,

1r =

1

^

tclr

= 10pJ.

0

b) The excitationcurrcnt and the resullingvoltage, po\\,er.and cncrgy are ploned in Fig.6.12. c) Note thal thc power is alwayspositive for the duralion of the cullenl pulse,which meansthat encrgyis conlinuouslybehg storedin the capacitor. When the curreit returDsto zero,the stored energy is lrapped becausethe ideal capacitor oflers no meansfor dissipatingenergy lhus a vollageremainson the capacitorafter i rclurns

r0

20

30

40

50

60

10

20

30 ,10 50

60

r (pt

" 0,1) l0 8 6 2 0

I (!t

Figure6.126.Thevariabhs i, r, p, andu versus r for txamph 6.5.

obl'edive2-Know andbeableto usethe eqqations for voltage,current,power,andenergyin a capacitor 6,2

Thc vollageal t11etermimls ofthe 0.6pF capacilorshowDin the figure is 0 for I < 0 and 40e rs'0mrsin 30.000tV tor t > 0. Find (a) i(0); power (b) the deliveredto the capacitorat r = 7/80 ms; and (c) the energystoredin lhe capacitorafi : d80 ms. 0.6pF

NOTE: Also rty ChapterProblems6.14aru16.t5.

Answer: (a) 0.72A; (b) -649.2 mw; (c) 126.13pt. 6.3

Thc curcnt in the capacitorofAssessment Problenr6.2is 0lbr r < 0 and 3 cos50.0001A for r > 0. Find (a) ?)0)i (b) the ma"{imum po}'er delivered to the capacitor at any one ilNtant of time;and (c) the maximumenergystoredin the capacitor at any one instant of time. Answer: (a) 100sin 50,000tV, / > 0; (b) 150w;(c) 3 mJ.

200

inductance,Capacitance,andl4utuatlnductance

6.3 Series-Paratlel Combinations of Inductance andCapacitance Lt + ? L

Lz + L 2

L3 + r 3

Just as serieepaiallel combinalions of resistois can be reduced to a sinAle equivalenl reristor. series-parallel combinarion5 ot inducrors or capacir6n car be reduced to a single inductor or capacitor. Figure 6.13 shows inductors in series,Here, the inducto$ are forced to cat:rythe samecurent;thus we define only one curent for the s€ries combination. The voltage drops acrossthe hdividual inductors are

=i'.11<.1T*J i Figure 6.13A Inductors in sedes.

, Lt

Lz

d i, . - t t. a

0z-Li

di

and

dt u -t.d,.

L1

@

The voltageacrossthe seriesconnectionis i(to) 1l i Leq: L1+ L2+ L3

01-tLt+ut

L,l;t.

i'g'e (,0)

Flgure 6.14 Anequivatent circujt forinductors in series canying aninitialcuri€nt t(d.

from which it should be apparent that the equivalent inductance ofseriesconnected inductors is the sum of the individual irductances. For ', induc-

Combining inductors in series>

f,.t,;

(6.19)

If the original inductors carry an initial curtent of i(to), the equivalent inductor carries the same initial curent. Figure 6.14 shows the equivalent circuit for se es inductors carrlng an initial culrent. Inductorsin parallelhavethe sameterminalvoltage.Inthe equivalent circuit, the current in each inductor is a function of tle terminal voltage and the initial curent in the inductor. For the tlree inductors in pamllel shown in Fig. 6.15,the currents for the individual inducton arc

jnductors PiguE6.15A Three in parattet.

;

1 . f I r a I + ir(ro),

1 r 1 r

+ i2Gd,

+ /3(to).

(6.20)

6.3

Series-ParattetCombinatjonsof Inductance andCapaciiance201

The current at the terminals of the tlree parallel inductors is the sumot the inductor currents:

i=ir+i2+h.

\6.?1)

Subsrituring Eq.6.20into Eq.6.21ields I

\.r

l\

f'

- + ; - l l u d r + ir(ro)+ iz(ro)+ 4(ro).

(6.22)

Now we can interpret Eq. 6.22in telms of a single inductor; that is,

. 1 r

t:

I

\6.23)

lltar+

CompadngEq.6.23with (6.22)yields

- 1: - + 1 -+ L"o

L,

1 Lt

1 Lt

;(ro)= 400)+,,G0)+ 4(ro).

(6.24)

L=1+ Lcq

\6.25)

L1

L+! L2

L3

j(ro.)= ir(,o)+ t (rJ + i(ro) Figue 6.16shows the equivalent circuit for the three parallel inductoE in Fig.6.15. The results er:pressed in Eqs. 6.24 and 6.25 can be exterded to , hductors in pamllel:

Leq

figure 6.16 A An equjvatent cjrcujtfor ihreeinductoE in parattet.

(6.26) < Combining inductors in paratlel

(6.27) a Equivateni inductance initiaI current

Capacitois connected in sedes can be reduced to a single equivaletrt capacitor.The reciprocal of tle equivalent capacitanceis equal to the sum of t]le rcciprocals of the individual capacitances.If each capacitor cades its own initial voltage, the initial voltage on the equivalent capacitor is tle

202 Inductancq Capacitance, andltlutuat Inductance algebmic sum of the initial voltageson tlle individual capacirors,Figure 6.17 and t]Ie following equations summadzetheseobservations:

Cornbining capacitors in seri€s>

(6.28)

fquivatentcapacitance iniiial voltage>

(6.2e)

We leave tlle dedvation of the equivalent circuit for sedes-connected capacitors as an exercise.(SeePrcblem 6.30.) The equivalent capacitanceof capacitors connected in parallel is simply the sum of the capacitancesof the individual capacitors, as Fig. 6.18 and the follouring equation show:

Co|nbining capacitors in paraltet>

(6.30)

Capacito$ connected in parallel must carry the samevoltage.Therefore, if fiere is an initial voltage acrossthe origbal pamllel capacitors,this same idtial voltage appeaN acrossthe equivalent capacitance Ceo.The derivation of the equivalent circuit for pamllel capacitors is left ai an exercise. (Seehoblem 6.31.) We saymore about se es-parallel equivalent circuits of inductors and capaciton in Chapter 7, where we interpret results based on their use.

;--r-r''T

+

cl

c1

q (ro)

IT

l + "^i"t^s (a)

.

- _),

;

1

_)"- T

1

;1 l

(a)

_('.q L=!+ L+...*! cr c c"

cea,

1,(d = ?,(d + !(h) + " + 1'"(ro) (b)

C e 1 , =C r + C 2 + - + C n

Figure 5,17A Anequivatent cjrcuittor capacitors connected in (d)Ihesedes series. caDacirors. (b)Tteeqriva.ent crcur.

(b) figure6.18"| Anequjvatent cjrcuitfor capacitors connected in palattet. (a) CapacjtoB (b)Ihe equivatent in paratteL. cncuii.

6.4

l4utuatlnductance203

objective3-Be ableto combine inductors or capacitors in seriesandin paral(elto forma singLe equivaLent indudor 6.4

The initial valuesoflr and l2 in the circldt showl are + 3 A and 5 A, respective]y. The voltageat the terminalsofthe paralielinductorsfort > 0is 30e"mV. a) Il lhc paralleliDductorsare replaccdby a singicinductor',what is its irduclancc? b) Whal is Lheinitial currentand ils leference directior in the equivalcn1inductor? c) Use the equivalcntinduclor to find l(r). d ) I n d ; { , 1 a n Lr lt ) . \ e r i r } r r J . r r e . o l u r ' o n . fbr i1G),ir(,), and ,0) satisfyKirchhofl's

An5wer: (a) ,18mH; (b) 2 A,|p; (c) 0.125e5, 2.125A,1 > 0; ( d )r r 0 ) = o l e s r + 2 . 9 4 . r > o , jy'r) = o.o25e5' 5.025A, r > o. 6.5

Thc currenl ai the terminalsof the two capaci lors showDis 240? "'l.A for I > 0. Thc inilial valuesofur andq are 10Vand 5V. respectivellCalcuiatethe total cncrgy trapped in the capacitoKasr + N. (H,irij Don'lcombine the capacitorsin series lind thc energy lrappedin each,and then add.)

240mH

Answer: 20AJ. NOTE: ALto try ChapterProbler8 6.21,6.22,6.26, urul6.27.

6.4 t4util;linducta*se The magneticijcld we consideredin orn studyof iiduolors in Secrion6.1 wasrestrictedto a singlecircuit.We saidthat induclancei! the parameter that relatcsa voilage 1()a time-\,aryingcuflcnt in the samecircuit;thus, jnductanccis nore preciselyreferredto as scll inductance. We now considerthe situationin which lwo circuitsare linkcd by a - + v V - t , _ magnclic[e]d.In this case.the voltageinducedin rhe secondcircuil can bc rclaled lo the time-varyingcurrcnl in the first circuir bv a paramcter kno\n asnutual inductance. Thc circuil showr in Fig.6.i9 reprcscnlstlvo magnclicallycoupled coils.The scll nrductancesof the tr,o coils are labeled/,1 ard lr, and the mutual inductanccis labeledM. The doublc headedarrow adjacentto M indicatesthc pair oI coitswith this valuc of Fig!ie6.19 i. Twomagneticatty coupted coiLs. mutualinducQnce.Tlisrotation is necdcdparlicularlyin circ its conrrin ing morc than oDepair ofrnagneticallycouPledcoils. The easiestway to analvzecircuitsconlainhg murualinducranceis 1{) uscmeshcu ents.Theproblemis 1owdle thecircuitequationsthat describe lhe circuitin termsof thc coil currents.First,choosethc rclcrencedirection lbr eachcoil current.Figurc6-20lholvs arbitrarilysclcctcdrelcrencecurrerls.After choosingthc rcfcrcncedirectionsfor i and ;2.sum the voltages aroundeachclosedpath.Becauscol lhe mutualinductanceM. thcrc will be Figure6.20i Coitcurients tLandl2 usedto describe two vollagesacrosseachcoil.namcll a self-hducedvoltageand a nutually tlrecncuitshownin Fiq.6.19.

.,o ',1[,', i^'

204

Inductance,Capacitance,andllutualInduchnc€

r------.d -r. .l'rra |

I

*Qrl.,l ["? *" t . l Figurc5.21A Thecjrcuitof Fig.6.20withdotradded toth€ coitsindicating the potarity ofthe muiuatly

Dotconvention for mutuattycoupl€dcoits>

induced voltage.The self induced voltage is the product of the self, inductance of the coil and the 6rs1derivative of th€ cunent in tftat coil. The mutually induced voltage is the pioduct of tlrc mutual inductanceof the coils and the fust derivative of the curent in the other coil. Consider the coil on the left in Flg. 6.20whose selJ-itrductarcehasthe value la. The selJ-induced voltage adoss this coil is ar(dildt) and the mutualy hduced voltage aooss this coit is M(di/dr). But what about the polariries of rheserwo voltages? Using the passivesign convention, the self-induced voltage is a voltage drop in the direction of the curent producing the voltage. But the poladty of the mutually induced voltage dependson the way t]le coils are wound in relation to tte reference directior of coil clrllents. In general, showing ihe details of mutually coupled windings is very cumbersome.Instead,we keep track of the polarities by a method known as t}Ie ilot convention,in which a dot is placed on one lerminal of each whding, as shown in Fig.6.21.These dots carry the sign informatior and allow us to draw the coils schematically rather than showing how they wrap around a core structure. The rule for using the dot convention to determine the polarity of mutually induced voltage can be summaized as follows: When the reference direction for a curent enters the dotted teminal ol a coil, t]le reference polarity of the voltage that it induces in the othei coil is positive at its dotted terminal. Or, stated alternatively,

Dotconvention for mutuattycoupted coits (alternate)>

W}len tie reference dircction for a current leaves the dotled terminal of a coil, the reference pola ty of the voltage that it induces in the other coil is negative at its dotted terminal. For the most part, dot markings will be provided for you in the circuit diagrams in this text. The important skill is to be able to wdte the appropriate circuit equations given your underctanding of mutual inductance and the dot convention. Figu ng out where to place the polarity dots if they are not given may be possible by examining tie physical configura, tion of an actual circuil or by testing it in fhe laboratory. We will discuss these procedures afrer \i'e discussthe use of dot markings. In Fig. 6.21,the dot convention rule indicates that the reference polarity for the voltage induced in coil 1 by the current i2 is negative at tie do! ted terminal of coil 1.This loltaEe (Mdi2/dt) is a voltage dse with respecr to 4. The voltage i duced in coil 2 by the curent ilis M dh/.Lt, ^nd its rcferencepolarity is positive at the dotted terminal of coil 2. This voltage is a voltage ise in the direction of 4. Figue 6.22 sho*s t}le self- and mulually inducedvoltagesaoosscoils 1 and 2 alongwith their polarity maiks.

M

+

Ll

tiguru6.22a Thes€lfandmutuattyinduced vottases appearjng across ihecoits shown in Fig.6.21.

6.4 Muiuatlnductance 205 Now let\ look at the sum of the voltagesaroundeachclosedloop.In Eqs.6.31and6.32,voltageises in the referencedircctioll ofa currentare negative:

-u" - i,R, t t,d" " d t

t,*, * t,*

d

, o ' t. - - o .

(6.31)

= o.

\6.3?)

,*

TheProcedure for Determining DotMarkings We shift now to two methods of determining dot markings. The first assumesthat we know the physicalarangement of the two coils and the mode of eachwindingin a magneticallycoupledcircuir.The following six steps,applied here to Fig.6.23,determinea set ofdot markirgs:

,l

a) Arbitradly select one terminal say,the D terminal- of one coil and mark it with a dot. b) Assigna currentinto the dotted termiml and labelit rD. ?) c) Use the ight-hand iule1 to determinethe direction of the magneric lsrel field establishedby iD iruide the coupl€dcoilsand label this field dD. rA.rbitrarily d) Arbitrarily pick one terminal of the secondcoil-say, terminal A and assigna curent into this terminal, showirg the current as iA. (Step1) e) Use the ght-handrule to detemine t}le direction of the flux established by ta ir ide the coupled coils and label this flux dA. Figure6.23A A setof coilsshowing a method for d€t€rmining a set ofdot nrarkings. f) Compare the direcrions of the two fluxes dD and dA. If the flrlxes have the samercference direction, place a dot on the terminal of the second coil wherethe test current (iA) enters.(In Fig.6.23,the fluxesdD and dA havethe samereferencedirection,and thereforea dot goeson terminal A.) If the fluxes have different reference directioni place a dot on the teminal of the secordcoil wheretie testcurrentleaves. The relativepolaritiesof magneticallycoupledcoilscan alsobe determhed experimentally. This capability is important becausein some situations,determininghow the coilsare wound on the core is impossible.One experimentalm€thodis to connecta dcvoltagesouice,a resistor,a switch, and a dc voltmetei to the pair oI coiiE as shownin Fig.6.24-The shaded box coverjng the coils implies tlat physical inspection of the coils is nol possible.The resistorRLimitsthe magnitudeof tie currentsuppu€dby the dc voltagesource, Flgure 6.24A AnexperimentaL setup tordeteririning The coil terminal connected to the positive terminal of tle dc source via the switch and limiting resistor receives a polarity mark, as show, in Fig.6.24.Wlen the switchis closed,the voltmeterdeflectionis observed.If the momentarydeflectionis upscale,the coil teminal connectedto the positiveterminalof the voltmeterreceivesthe poladty mark.ffthe deflection is dowNcale, the coil terminal connected to the negative telmhal of ure voltmeter ieceives the polarity mark. Example6.6 showshow to usethe dot markingsto fomulate a set of circuitequationsin a circuit containingmagneticalycoupl€dcoils. -' s".ti*ion

otru.uauyttawonpage207.

206

Inductance,Capacitance,andMutualInductance

FindingMesh-Current Equations for a Circuitwith Magenticatty Coupted Coits a) Write a set of mesh-crment equations that descdbe the circuit in Eg. 6.25 in tems of the currents i1 and ,2. b) Verify that if there is no energy stored in tlle clrcuit at r - 0 and if ic = 16 - 16e-5'A, tie solutions lbr i1 and i2 are i 1= 4 + 6 4 a 5 t - 6 8 e 4 t A ,

i1(0)=4+64-68=0, i,(0)=1

i2= 7 - 52a5t+ 51e4tA..

t.) 8H

b) To check the validity of i1 and i2, we begin by testhg 1}leinitial and final values of i1 and i2.We know by h}?ot}lesis that t1(0) = ir(0) = 0. From the giver solutions we have

20f,l

reu f i'-i aoo

5 2 + s 1= 0 .

Now we observe that as l approachesinlinity the source current (is) approachesa constatrt va.lue of 16 A, and ttrerefore the magnetically coupled coils behave as shon circuils.Hence al I - m lhe circuil feducesro rhat (howo in Fg. b.26. From Fig. 6.26 we see that at I : co the thrce rcsistors are in pamllel aqoss the 16 A source. The equivalent resistanceis 3.75 O and thus the voltageacrossthe 16 A curent sourceis 60V It tbllows that 4(co)

Figure6.25 Thecircuit forExample 6.6.

60 20

60 60

t(,:o): r @ = t o

Solution a) Summingtle voltagesaroundthe il meshyields 4d

- I3dtlis t2) .20(i -i)

l5(;

-r")=0.

Thesevaluesagreewith the final valuespredicted by t}Ie solutions for 4 and i2. Finally we check the solutions by seeing if they satisfy the differential equatiom dedved in (a). We will leave this fhal check to the readei via Problem6.37.

The i mesh equation is 20lir

) l,) | 60i, + lb:{/, dt-

i"r '

si, 8"1' - ". dt

Note that the voltage aqoss the 4 H coil due to the current (is - i), that is,gd(is - i)/dt,1s a voltage drop in the d ection of tt. The voltage induced in the 16 H coil by tle curent i1, that is, 8dtldl, is a voltage rise in tle direction of i2.

50

20a

Figure5.26A lhecircuiiof Lumpte 6.6when r = co.

6.5

A Ctoser Lookat l4utual Inductance 207

0bjective4-Use the dot convention to writemesh-current equations for mutuallycoupted coils 6.6

a) Wite a set of mesh-curtentequationslor the circuit in Example6.6if the dot on the 4 H inductor is at the righl-hand telminal, the referelce djrectionof is is reversed,alrd the 60 O resistoris increasedro 780 (). b) Verify that ifthere is no energysroredin rhe circuitat r : 0, andif is - 1.96 1.96e4tA, the solutionsto the differentialequations derived in (a) of this AssessmentProDremare

Ans$,er:(a) 4(dirldr) + 25ir + a(rlizlrh) 20i2 : sis s(disldt) 8(di/dt) - 20ir + 16(rli2/ dt) + 800i2 - 16(di:r/dt); (b) vedfication.

ir = -0.,1 11.6e{'+ 12€-5'A, /2 -

0.Ul

0'1o,""

,' 5 A

NOTE: Also try ChaptetPrcblen 6.34.

6.5 A fl[oserLookat Mutual3*duetaalee In order to fully explainthe circuit paramercrmutual inductance,and to examinethc limilationsandassunptionsmadein the qualitativediscussion presentedin Section6.4,wc beginwith a more quanritativcdescdptionof self-inductance thanwasprcliousiy providcd

A Reviewof Setf-Inductance :lhe conceptofinductancecan be tracedto Michael Faraday. who did pio neeringwork in this area in t]re early 1800s.Faradaypostulatedthat a magnelicfield consistsof lines of force surroundingthe current-carrying conductor.Visualizetheselines of lbrce as energystoring elastjcbands that closeon themselves.As the currentincreasesand decreases. the clastic bands(that is.thc liDesof force) spreadand collapseaboutrhe conductor.The voltagejnducedin the conductorisproportionalto rhe numberoI lines that collapseinto, or cut. thc conductor.This imageof inducedvolf ageis expresscdby what is callcdFaradays law;lnar N,

,:n'

(6.33)

where,\ is rcferredto asthc flux linkageand is neasuredin weber{urns. Howdowegetfrom Faraday'slaw to the definitionofinduclaDcepresentedin Scction6-1?Wecanbeginto draw thisconnectionusingFig.6.27 Figure6.27A Represeniation of a magnetjc fieldtink

208

Inductanc€, CaDaciiance, andlllutuat Inductance

The linestbreadingthe N tums and labeledd representthe magnetic lin€s of foice that mal(e up the magnetic field. The shength of th€ mag' netic field dependson the strengttrof the current,and the spatialorientation of ihe field depelds on the diection of the current. The right-hand rule rclates the odentation of the field to the direction of the cment: w}len the fingers of tlle right hard are rrrapped around the coil so that the fingers point in the directior of tlle curent, tle thumb points in the direction of that potion of the magneticfieldinsidethe coil.Theflux linkageis the product of the magneticfield (0), measuredin webers(Wb), and the numberof turns linted bv the field fN\: 16.34) The magnitude of the flu\, d, is rclated to the magniiude of the coil currentby the relationship

O: sNt,

(6.35)

whereN is the numberof tums on the coil, and O is ttrepermeanceofthe space occupied by the flux. Permeance is a quantity that describes the magneticpropertiesof this space,and assuch,a detaileddiscussionofpermeancejs outsidethe scopeof this text. Here,we need only obsene that, when the spacecontainhg the flux is made up of magnetic materials (such as iron, nickel, and cobatt),the permeancevarieswith the flux, giving a nonlinearrelationshipbetween4 and t. But when the spacecontaining the flux is comprised of noimagnetic materialsr the permeance is constant, giving a linear relationship between d and i. Note from Eq. 6.35 that the flux is alsopropoitional to the numberof tums on tle coil. Here, we assumethat the core material-the spacecontaining the flux is notunagnetic.Then, substituting Eqs.6.34 and 6.35into Eq. 6.33yields

d^

= N;:

d(No)

N;tENrl

= Nza!! _ t!!

(6.36)

which showsthat self-inductance is propoitional 10 tlle square of the number of tums on the coil.We make useof this observationlater. The polarity of the induced voltage in tie circuit in Fig. 6.27reflects the reaction of the field to the cwent creating the field. For example,when i is increasing, dt/dl is positive and o is positive. Thus energy is required to establishthe magnetic field. The product ?i givestlre rate at which energy is stored in the field. Wlen the field collapses,di/dt is negative,and again the polarity of the induced voltage is in opposition to the change.As the field collapsesabout the coil, energy is returned to the circuit. Keeping in mind this furtler insight into the concept of sef-inductance, we now turn back to mutual inductance,

6.5 A ctoser Look at l{utuat Inductance209

TheConcept of MutualInductance Figure 6.28 shows two magnetically coupled coils.You should verily that the dot markings on the two coils agreewith the direction of the coil windings and curents shown.The number of turns on each coil are N, and N". respeclively.Coii I is energizedb) a trme+arling cunenr tour.. rlai establishesthe current ir itr the Nl turns. Coil 2 is not etrergized and is open. The coils a-rewound on a ronmagnetic core. The flux produced by the cment il can be divided into two components,labeled du and d21. The flu\ component d11is tlle flux produced by ir that links only the N1 Figure 6.28A Twonragneticatly coupted coib. turns.The component 421is t}le flrlx produced by i1 that links the N2 turns and the N1 tums. The tint digit in the subscript to the flux gives the coil number, and the second digit refers to the coil current. Thus dj r is a flux linking coil 1 and produced by a current in coil 1,whereasd2r is a flux lhking coil2 and producedby a currentin coil 1. The total flux linling coil 1 is dl, tle sum of d11and d\: .lt + 6*

et:

(6.37)

The flux dr and its components dx and d2r are related to the coil curent i1 asfolows:

6t = 94rir 0i

(6.38)

: 9l ,Ni1,

(6.3e) (6.40)

where 9r is the permeance of the spaceoccupied by the flux d1, 9rr is the permeanceof the spaceoccupied by the flux d1r, and 921is the permeance of t}tespaceoccupiedby the flux d21.SubsrirutingEqs.6.38,6.39, and 6.40 into Eq. 6.37 yields the telatioNhip between the permealce of the space occupied by the total llux dr and tle pemeances of rhe spacesoccupied by its componentsd11andd2r: 91=gr+@21.

(6.41)

We use Faraday's1awto derive erpressions for r)r and o2:

: N,*@u, ,' =# =o'ry;fu o,') : N]g.11+s^\*

= *p,*

= ",*

(6.42)

and d(Nzdzt) dt

= N,w,g,,\

=u.frg,,N,i,1 (6.43)

210

Induchnce,Capacitance,andl,4utuallnduciance

The coefficient of dt/dt i]) Eq. 6.42 is the self-inductance of coil 1. The coefficientof dtldl in Eq. 6.43is the mutual inductancebetweencoils 1 and 2.Thus \6.44)

The subscript on M specifiesan irductance that relates the voltage induced in coil2 to the curent in coil 1. The coefficient of mutual inductance sives D2= M2r

di. dt.

(6.45)

Note that tle dot convention is used to assignthe polarity rcference to ?)2 in Fig.6.28. For the coupled coils in Fig. 6.28, exciting coil 2 from a time-varying current source (i2) and Ieaving coil 1 open produces the circuit arrangement shown h Fig. 6.29. Again, the polarity reference assignedto or is basedon the dot convention. The total flux linking coil2 is Figure5.29A Themagnetkatly coupted coil50f Fig.6.28,withcoit2 exciied andcoill open.

6z:6zz+4o.

(6.46)

The flux d, and its components d22afil dD are related to the coil curent i2 as follows:

O2= g2N2i2,

\6.47)

4n : qrNzrz,

(6.48)

0D: grzN2i2.

\6.49)

The voltagesu2and ur afe

u=ff=uZo,ft=1,ff,

(6.50)

- Nfi2{to*. ,,:ff =fiw,q-l

(6.51)

The coefficient of mutual inductancethat relates the voltage induced in coil 1 to the time-varyingcurrent in coil 2 is the coefficient of dizldt in Eq.6.51: Mp = N1N20p.

(6.52)

For nonmagnetic materials, the pemeances O12and @r are equal, and rherefore MD:

M2r = M.

(6.53)

Hence for linear circuits with just two magnetically coupled coils, attaching subscriptsto the coefficientofmutual inductanceis not necessary.

6.5 A Ctos€r Loolat l,iutuat Inductance211

MutualInductance in Termsof Self-Inductance The value of mutual inductance is a furction of the seiJ-inductances.We derivetlis relationshipasfollows.From Eqs.6.42atrd6.50,

Lr = Nlgr,

(6.54)

L2 = N1!t2,

(6.55)

respectively. FromEqs.6.54 and6.55, LrL2- N?N1!trs2.

(6.56)

We now use Eq.6.41 and the corresponding expressionfor @2to write

L[2 = N1Ni(gr + @2)(922+ gn).

(6.57)

But for a linearsystem, O21: 9p, soEq.6.57becomes

Lr4= (N1N2s1)2(1. #)('.

#)

=*(,H)('.H)

(6.58)

Replacing the two terms involving permeances by a single consIant expresses Eq.6-58in a more meaningfulform:

i=("H)(.H)

(6.5e)

Substituting8q.6.59 into Eq.6.58yields

tf

= kzLrLz

(6.60) < Retating setf-indudances andmutuar inductance usingcouptingcoeffiaient wherc the constantt is called t}le coefficientof coupling.Ac€ordingto Eq.6.59,1/fr'musrbe$eater than1,whichmeansthat ,t musrbetessthan1. In fact,the coefficientof couplingmustlie between0 and 1,or 0
(6.61)

The coefficient of coupling is 0 when the two coils have no conmon flux; that is, when dD = drl = 0. This condition implies that 912 : 0, and Eq. 6.59 indicates that Yle - cr', or tr : 0. If there is ro flux linkaqe betweenthe coils,obviouslyM is 0.

212

IndLrctance, Capacitance, andl4uiuat Inductance

The coefficient of coupling is equal to 1 when 411and 422are 0. This condition implies that aI the flux that finks coil 1 also links coil 2. In terms of Eq. 6.59,gr1 = @z : 0, which obviouslyrepresentsan ideal state;in ieality, winding two coils so that tley share precisely the sameflux is physica y impossible. Magnetrc matedals (such as alloys of iron, cobalt, and nickel) createa spacewith high permeanceand are usedto establishcoefficients of coupling that approach unity. (We say more aboul this important quality ofmagneticmaterialsin Chapter9.) NOTE: Assessyou undeqtanding of this material b! tying Chapter Problems6.12and6.43.

EnergyCalculations

Figur.5.30 A Thecircuit used io denve thebasic

We concludeour first look ai mutual inductancewith a discussionof the total energy stoied in magnetically coupled coils. In doing so, we will confirm two observationsmade earlier: For lirear magnericcoupling, (1\ MD: M^ = M,^nd(2) M: krZE, where0 < t < 1. We use the circuit shown ir Fig. 6.30 to dedve the expression for the total energy stored in the magtetic fields associaled$rith a pair of iinearly coupled coils. We begin by assumirg tiat the currents ir and t2 are zero and that this zero-current state corresponds to zero energy stored in the coils.Then we let i1 increasefrom zero to some arbitrary value 1r and compute the eneigy stored when ir = 1r. Becausei:0, t]le total power input into the pair of coilsis ?,r4,and t}le energystoredis

I.

d1t,= Lt.la idir 1

W:;L1Ii.

^

(6.62)

Now we hold ir constant at 11 ard increasei, from zero to some arbitrary value 12. During this time interval, ttre voltage induced in coil 2 by il is zero because1r is constant.The voltage induced h coil1by i2 is M Driizldt. powefinpullo thepairof coilsis Therefore.lhe di. p=IrMn;+i2o2. The total energystoredin the pair of coilswhen i2 : 12is tw

J*

o*:

tt,

J"

ItMldi,+

th

J)

L,idi,,

w =w+ rJ,MD+ tk4. :)r,4 +lkt1+ t,r,u,,.

(6.63)

6.5 A Ctoser Look at l4utuat Inductance If we reverse the procedure-that is, if we fkst increase 4 from zero to 12 and then inuease i1 from zero to 11-the total energy stored is

w =l,r?,1b4*r,t,u,,.

16.64)

Equations 6.63 and 6.64 express the total energy stored in a pair oI lin, early coupled coils as a function of t}le coil curents, the selJ-inductances, and t]le mutual inductance. Note that the only diflerence between rhese equations is the coefficient of the cu ent product 1rI2. We use Eq. 6.63 if i1 is establishedfirst and Eq.6.64ifi, is estabtished first. When the coupling medium is linear, the total energy stored is the sameregardlessofthe order usedto establishlr and 12.The reasonis that in a linear coupling, tlle resultant magnetic flux depends only on the final valuesof 4 and i2, not on how the clllrents reached their final values.If the resultant flux is the same,the stored energy is t}le same.Therefore, for linear coupling, M12 : Mz1.Also, because11arrd 12are arbitrary values of i1 and i2, respectively,we rcpresert the coil currents by their instantaneous values 4 and i2. Thus, at any instant of time, the total energy stored it the couDledcoils is

-Ol=7,& +lr,i1+ui,i,.

(6.65)

We de ved Eq. 6.65 by assuming that both coil curents entered polarity-marked terminals. We leave it 10 you to vedfy that, if one current enters a polarity-marked terminal while the other leaves such a teminal, the algebraic sign of the telm Mi1i2 reve6es Thus, in general,

*1t7: ,\,i? + f,r,i} r ui,i,

(6.66)

We use Eq. 6.66 to show that M cartrtot exceed \/LJ;. The magnetically coupled coils are passive elements, so the total energy stored can never be negative.If ?r(r) can never be negative,Eq. 6.66indicates that the quanlrty t

,Li'

-

l

)Lri)

- tt|.t,

must be greatertlan or equalto zero when it and i2 are eilher both positive or both negative.The Limitiry value of M corresponds to setting the quantityequalto zero:

I,t*)"a-

Mi1i2:0.

(6.67)

To find the Limiting value of M we ^dd and subtract the telm ilirvall2 to tle left-handsideofEq.6.67.Doing so genetatesa term that

I E;. \!7'

/ _ \ + iiz\.t/LrL2 - M ):0.

(6.68)

214

I n d u c t a nC . ea, p a . j t a n . € ,M a nudt u a t h d u ( t a n ( e

The squaredlerm in Eq. 6.68can never be negative,but it can be zero. Thereforcu(t) > 0 only if

^/Lrt2 > M,

(6.69)

which is anofter way ofsayingihat

M = k^/L;L; (o=&<1). We derivedEq.6.69 by assumingthat 4 and i2 arc either both positiyeor both negative.However,we get the sameresult if il and i2 are of opposile sigD,becausein this casewe obtainthe limiting valucof Mby selectingthe plus signin Eq.6.66. NOTE: Assessfoar unde^tanding of thi,Jmaterial by tJing Chapter Problems6.17and6.18.

PracticaI Perspective ProxinitySwitches At the b€qinningof thjs chapterwe jntroducedthe capacitiveproxjnity swjtch.Therearetwo fonis of this switch.Theoneusedjn tableLarnps js basedon a sinqle-electrode switch.It is Leftto your jnvestigationin Probtem6.50. In the exampleher€,we considefthe two-etectrode swjtch usedin elevatorcatlbuttons.

EXA14PLE Th€eLevatof call buttonis a snrallcupinto whichthe fingeris inserted,as shownin Fig.6.31.Thecupis nadeof a metatring eLectrode anda circutar ptateetedrodethat areinsutated fromeachother Sometimes two concentric ptasticare usedinstead.The eiectrodes rings embedd€d in insuLating are coveredwith an insutatingtayerto preventdirectcontactwith the metat. Theresuttingdevicecanbe modeted asa capacitor, asshownin Fig.6.32,

cl

. Figlre 6,31"'. Anetevator cattbutton. G) Fronivieu (b)Sideview.

-i--.

Figure5.32 ^4.A capacitor modeL ofthetwo-eteclrode prcximity switchusedjn etevator caLlbuttons.

PractjcatPe6p€ctive 215

Untikemostcapacitors, proximjty thecapacitive switchpermits youto cl insertanobject,suchasa finger,between yourfintheetectrodes. Because geris muchmoreconductive thanthe insutating covering surrounding the etectrodes, the circuitresponds asthoughanother electrodg connected to ground,hasbeenadded. Theresuttis a three,terminal circuitcontainjng threecapacitors, asshownin Fig.6.33. Theactualvatues of the capacitors in Figs.6.32and6.33arejn the rangeof 10to 50pP,depending ontheexactgeometry of theswitch,how Figure6.33 A A circujtmodeLofa capacitive the fingeris jnserted, switchactjvated gtoves, whether the person byfingert0uch. is weadng andso forth. proxjmiry Forthefotlowing problems, assume that all capacjtors havethesamevalue of 25pF. Atsoassume the elevator catlbuttonis ptaced in the capacjtive €quivatent of a vottage-divider circuit,asshown in Fjg.6.34. a) Catcutate theoutputvoltage wjthnofingerpresent. b) Catcutate theoutputvottage whena fingertouches thebutton.

tt:r -.,Tr.,-

Solution a) Beginby redrawing thecircujtin Fig.6.34withthecatlbuttonreplaced by its capacitive modet catlbuttoncncuit. fromFig.6.32.TheresuLting circuitis shownjn Figure6.34 A Anelevator Fig.6.35.Writethecurrent equation at thesingLe node:

".

d(u +) * c'. dI4 : o . t1l

(6.70)

Rearrange this equationto produce a differentialequationforthe output vottageo(t): dn _ CI dt" dt Cr+C2 rlt'

(6.71)

Finatty, integrateEq.6.71to find the outputvoLtagei

,{ti:-

C.

u , ( r )+ u ( 0 ) .

Figure6.35 A A modet ofthe etevaior caltbutton cjrtujtwithnofingerpreseni. (6.12)

Theresuttin tq. 6.72shows thatthesedes capacitor circuitin Fig.6.15 formsa voLtage dividerjust asthes€ries resistor cjrcujtdjdin Chapter 3. In bothvottage-divider circuits, theoutputvoltage doesnotdepend on thecomponent vatues butontyonthejrratio.Her€,C1= C2 = 25 pF, sothecapacjtor ratjois Cr/C, : 1. Thustheoutputvoltage is 0(t)=0.5r,(r)+"(0).

(6.73)

Theconstant termin Eq.6.73is dueto theinitiaL charge onthecapacitor, Wecanassume thato(0) : 0 y, because thecircuitthatsenses theoutputvoltage etiminates theeffectoftheinitiatcapacjtor charge. Therefore, thes€nsed outoutvoltaoe is 0(i) = 0.5,,(1).

16.74)

216

Inductance. CaDacitance. andMuiuatlnductance

b) NowwerepLace the catLbltton of Fig.6.34withthe modet of theactivatedswitchin Fig.6.33.Theresultisshown in Fig.6.36.Again,wecal cutatethecurrents leaving theoutplt node: d ( u - u- s- c) )^ddr D - e t d^ -d-u0 . c,1,

(6.15)

Reafianging for?,(r)resutts to writea differential equation in dD

du"

Ct

A: cn c. + qn'

(6.76)

FinatLy, sotving the differentiai equation in tq. 6.76,wesee rlt):

?,"(r)+ o(0).

c 1+ c 2 + c r

\6.77)

If Cr = C2 = Cz = 25pF. ?'(t) = 03331'r(t)+ t'(0)'

(6.78)

As before,the sensing circujtetiminates ?r(0),so the sensed output voltage is o(t) = 0.3330"(r).

(6.7e)

Comparing Eqs.6.74and6.79,weseethat whenthe buttonis pushed, the outputis onethird of the inputvoltage.Whenthe b0ttonis not pushed, theoutputvoLtage is onehaLfoftheinputvoLtage. Anydropin outputvottage is detected by the etevato/s controlcomputef anduttimateLy resuLts in theelevator a ivingat theappropriate floor. yourundestanding NoTE:Assess of thisProctical Perspective by tryingChaptel Problens 6.49ond6.51,

Figure6.36d A modet oftheei€vator cattbutton cncuitwhen activated byfinger touch.

Sumnrary2 7 7

Surnmary Inductanc€is a linear circuit parameterthat relalesthe voltageinducedby a time-varyingmagrericfield to the currentproducingthe field. (Seepage188_) Capacifance is a linear circuit pammeterthat relatesthe cment induced by a time-varying electric lield to the voltageproducingthe field. (Seepage195.) Inductorsand capacitorsare passiveelemenls;theycan store and releaseenergy,but they cannot generateor dissipateenergy.(Seepage188.) The instantaDeous poweral the terminalsof an inductor or capacitorcan be posilive or negative,dependingon whetherenergyis beirg deliveredto or extractedfrom

.

and Capacitors

(v)

,=L*

i=iJ,,,a"+4t,,1 p:Iii-Li#

(w) (r)

.o=ilia,+x1to;

(v) (A)

. doesnot permit an instantaneous changein its termi. does permit an inslantaneouschangein fusteminal voltage . behavesasa shortcircuitin the presenceofaconsrant terminalcurrent(Seepages188-189.)

p:ri=CD*

(w)

w:icf

(r)

TABLE 6.2 Equationsfor Series-and Pahllet-Connected Inducto15.nd Capacitors

. doesnot permit an instantaneous changein its teminal voltage . cloespemit an instantaneous changein ils terminal . behavesas aII open circuil in t]re presenceof a con stantterminalvoltage(Seepage196.) Equationsfor voltage,curent, power, and encrgy in ideal inductorsand capacitorsare givenin Table6.1. Inductorsin seriesor in parallel can be repiacedby an equivalenlinductor.Capacitorsin seriesor in parallel can be replacedby an equivalentcapacitor-The equations are summaizedin Table6.2.SeeSection6.3 for a discussionon how to handle the initial conditionsfor seriesand parallel equivaientcircuitsinvolving inductols ano capacrton.

r..:zr+l+. +f ceq: c|+ c?+ ,,, + c, Mulual inductance, M, is the circuit parameter relating r h c v o l l a g ei n d u c c d i n o n e c i r c u i t t o a l i m e - v a r y i n gc u r . rent in another circuil. Specifically, dri r,t=Lri+Mi;

Jt.

_. ,tiL - dit + L )u) = tulr. dt ,lt

218

Inductance Capacitance,andl4utuatlnductance

where ?)1and i1 are the voltage and cment in circuit 1, and ?)2and i2 are the voltage and current in circuit 2. For coils wound on notunagneticcore<.Mp - M1] - V (Seepage210.) . The dot conv€ntion establishesthe polarity of mutually inducedvoltages: When ths referencedir€ction for a curent enters the dotted teminal of a coil, tle referencepolarity of the voltage that it inducesh the other coil is positive at its dotted terminal. Or, altematively, When the rcferencedirection for a curent leaves t]Ie dotted teminal of a coil, the refererce polarity of the voltage that it inducesin the other coil is negativeat its dotled teminal.

the ielationship between the self-hducfance of each winding and the mutual inductance between windhgs is

M : k^/LJ4. The coefffcient of coupling, &,is a measureof the degiee of magnetic coupling. By definition, 0 < I < 1. (See page 211.)

The energy stored in magnetically coupled coils in a linear medium is related to the coil currents and inductancesby tlre relationship

*:LS,ii *l,i?+ uri,. (seepaee213.)

(Seepaee20a.)

Probtems Figure P6.2

Secrion6,1 6.1 The triangular current pulse shoM in Fig. P6.1 is """ appliedto a 375mH inductor. a) Write the expressionsthat desc be i(t) ill the foul intelvalsl < 0,0 = t = 25ms,25ms< l < 50ms,andt > 50 ms. De ve the expressionsfor the inductor voltb) age,power, and energ]. Use the passivesign convention.

fr

0

(a)

1 2 (b)

6.3 The current in the 4 mH inductor in Fig. P6.3 is

Figure P6,1 i (nA) 400

known to be 2.5 A for I < 0. The inductor voltage for t > 0 is giver by the expression Ir(t) : 30e3rmv, 0*
0

t(ns)

25

50

,(m0

6.2 The voltage at the terminals of the 300 /-.H hductor """ in Fig.P6.2(a)is shownin Fig.P6.2(b).The hductor cu[ent i is known to be zerofor I < 0, a) Derive tlre expressionsfor i for t > 0. b) Sketchtversuslloro< I < @.

figureP6.3 iL(.4

u,@

,1mH

prcbrems219 The curent in a 100/rH inductor is known to be iL = zotelt A Ior t>0. a) Find the voltage aqoss the inductor for t > 0. (Assume the passivesign convention.) b) Find the power (in microwarts) at the rerminals of the inductor when r : 100ms. c) Is the inductor absorbing or delivering powei at 100ms? d) Find the energy (in microjoules) srored in the inductor at 100ms. e) Find t}le maximum energy (in midojoules) slorediD rhe inducloraDdthe time (in micro seconds)when it occurs.

The current in and $e voltage acrossa 2.5 H inductor are known to be zero for t = 0. The voltage acrosslhe inductofis gjren bv the graph in Fig.po.5

lort > 0.

a) Derive the expression for the curent as a function of tim; in the intenals 0 = t < 2s, 2 s < r < 6 s , 6 s < r = 1 0 s ,1 0 s< , < 1 2 s , a n d 1 2 s= t < c o . b) For r > 0, what is the current in the ilductor when the voltage is zero? c) Sketchiversus,foro< t < c.o. figureP6.5

, (v)

6.7 a) Find the inductor curent in tle circuit in Fig. P6.7 if t) = 250 sin 10001V,l, = 50 mH, and (0) : -5 A. b) Sketch r', i, p, and tl) veNus r. In makhg rhese sketches,usethe fomat usedin Fig. 6.8.plot over one complete cycle of the voltage waveform. c) Describe tle subintervals in the lime inte al between 0 ard 27 ms when power is being abso$ed by the inductor.Repeat for the subintervals when power is being delivered by the inductor. Figure P6,7

fl 6.8 The current ir a 15 mH inductor is known to be t<0; i : Ale-zm, .+ A2e-3,m,A,

t > 0.

The voltage affoss the inductor (passive sign convention) is 60 V at I = 0. a) Find the e4)ression for the voltage across the inductor for r > 0. b) Fitrd the time, greater than zero, when the power at t]le terminals of the inductor is zero. 6.9 Assume in Problem 6.8 that the value of the voltage asoss the inductor aff = 0is 300 V insteadof 60V a) Find the nume cal expressions for j and o for

r>0.

b) Specify the time itrtervals when the inductor js storing eneryy and the time intenals when the inductor is delivedng eneryy. .) Show that the total energy extracted from the inductor is equal to t}le total energy stored.

6.10 The current in a 2 H inductor is 6,6 The curent in a 20 mH inductor is known ro oe 7 + (15 sin 140r- 35 cos1zl0r)€20'mA for I > 0. Assume the passivesign convention. a) At what instant of time is tle voltage acrossthe ilductor maximum? b) What is the maximum voltage?

i =254,

t=0;

i = (81 cos 51 + 82 sir sl)e-,A,

I > 0.

The voltage adoss the inductor (passive sign qjnvention) is 100V a : 0. Crlculate the power at the terminals of the hductor at t = 0.5 s. State whether the inductor is absorbingor deliverhg power.

220

Inductance,capacjtance,andMutuatlnductance

6.11 Initially there was no energy stored in the 20 H inductor in the circuit in Fig. P6-11when it was placed acioss the terminals of the voltmeter. At t = 0 the inductor was switched instantuneously to positionb whereitiemainedlor 1.2s beforereturning instantaneously10 position a. The d'Arsonval voltmeterhasa full-scalereadingof25V and a sensitivity of 1000 O/V. W}lal will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of 1ie d'Arsonval movement is negligible?

Section6, 6.14 A 0.5pF capacitoris subjeciedto a voltagepulse having a duration of 2 s.The pulseis describedby the following equations: ( qot'v.

o=1<1s;

, . ( t ) = 1 a 0 ( 24 ' v , 1 s < r < 2 s i elsewher(. t0 Sketchthe cuuent pulsethat existsin the capacitor duringthe2sinterval.

Figure P5.11

6.12 Evaluate the integral

1,"oo, for Example 6.2. Cornment on the significance oI the result. 6.13 The expressionsfor voltage, power, and energy derived in Example 6.5 involved both integntion and manipulation of algebraic expressions.As an engineer, you cannot accept such results on faith alone.That is, you should develop the habit of asking yourself,"Do theseresultsmake sensein tems of the known behavior of the c cuit ttrey porport 1I) describe?" With tlese tloughts in mind, test the expressionsof Example6.5 by performingthe following checks: a) Check the expressionsto see whether the voltageis continuous ir passhg from one time interval to the next. b) Check the power expression in each interval by selecting a time within the inteflal and seehg whether it gives the same result as the corresponding product of o and i. For example,test at 10 and 30 ps. c) Check the eneryy expressionwithin eachinterval by selectinga time within the interval and seeing whether,the-eneigy equation gives the same rcsultas;Co'. Use 10 and30 ps astestpoints.

6.15 The rectangular-shapedcurrent pulse shown in End Fig-P6-15is applied to a 0.2IrF capacifor.The initial voltage on the capacitoris a 40 V drop in the relerenco direction of the currert. Derive the expression for the capacitor voltage for the time intervalsin (a)-(c). a) 0 < t < 100ps; b) 100ps < t < 300l..s; c) 300ps < r < oo; d) Sketch r(l) over the interval -1001.s < t< 500ps. Flgure P6.15

r(/d

6.16 The voltage at the terminals of the capacitorin Eflc Fig.6.10is known to be

"

r - rr: [ --rov. 1m,(4 + sin3000r) cos3000r v I > 0. lto ro" AssumeC:0.5/rF. for I < 0. a) Findthecurent in thecapacitor b) Fnrdtlre curent in the capacitortbr r > 0.

Prcbtems 221 c) Is there an instantanoous change in the vottage aooss tle capacitor at I : 0? d) Is tlere an instantaneous change in the current in the capacitor at t = 0? e) How much energy (in mioojoules) is stored in the capacitorat, = co?

6,17 The curent shown in Fig. P6.17 is applied to a 6flft 0.25pF capacitot. The idtial voltage on the capacitor is zero. a) Fird the charge on the capacitor at t = 30 ps. b) Find the voltage on the capacitor a1r = 50 ps. c) How much energy is stored in the capacitor by this culrent? figureP6.17

6.19 The voltage across the terminals of a 0.4/r,F

t

fzsv,

r5m'+,42€ ]soev. I > 0 . [erre The initial qrllent ir the capacitor is 90 mA. Assume t}Ie passivesign convention. a) What is the initial eneigy srored ir the capacitor? b) Evaluatethe coefficients,4land.42. c) What is the expiessionfor the capacitor current? Section 6.3 6.20 Assume tlat the initial energy stored in the inductors of Fig. P6.20is zero. Find the equivalent inductance with respect to the terminals a,b. figureP6.20 21H

6.21 Assume that the initial energy stored in the inducton of Fig. P6.21is zero Find rhe equivalent iDductancewith respectto the terminalsa,b. 6.18 The initial voltage on the 0.2 rlF capacitor shown in 6tre Hg. P6.18(a)is -60.6 V. The capacitorcurrent has the wavefom shownin Fig.P6.18(b). a) How much energy, in miffojoules, is stored in the capacitorat t = 250ps? b) Repeat(a) for t = oo. Flgure P6.18

O.2pF --

-60.6V

(a)

Flgure P5,21 12H

5H

14H

222

Inductance, Capacitance andl4utuat Inductance Figurc P6.23

622 The two parallel inductors in Fig. P6.22 are con-

necledacrosslhe terminaL of a black bo),al / - 0, The resulting voltage u for t > 0 is known to be 1800€ m' V. It is also known that ir(o) = 4 A and ir(0) = -16 4. a) Replacerhe originalinducror.with an equivalent inductor and find i(r) for I > 0. b) Find ir(t) for t > 0. c) Find i2(t) foi t > 0. d) How much energy is delivered to the black box in the time intenal0 < 1 < oo? e) How much energy was initially stored in the par' allel inductors? f) How much energy is tiapped in ttre ideal inductoft? g) Show that your solutions for tr ard t, a$ee with the answerobtainedin (0.

32H

624 For the circuit shown in Fig. P6.23,how many milliseconds after the switch is opened is the energy delivercdto the black box 80o/oof the total energy delivercd?

625 Find t}le equivalent capacitancewit}l respect to the terminals a,b for tlle circuit shown in Fig. P6.25. Figure P6.25 12p.F 8V 20 pF

FigureP6.22

--rts?1PF - L 28 pF

24 p.F

5V

;,oi10H,?0)l 2Y 6.:26 Find the equivalent capacitancewith respect to the terminals a,b for the circuit shown in Fig. P6.26. figu|eP6.26 8nF

6.23 The thee inducton in tlle circuit in Fig. P6.23 are

30v +

conrected across the termimls of a black box at t = 0. The resulting voltage for I > 0 is knorn to be u. : 7250e25'Y If tr(0) = 10 A and i(0) : -s A, ftrd

a),,(o); b) ',(r),r > 0;

18 DF

1 5 V+ 5.61F, //

40 Dr b.-

+ 5V

6.n

l0v +

c) il(r),r > 0; d)ir(r),r>0;

The two series connected capacitors in Fig. P6.27 are connected to the terminals of a black box at I = 0. The resultingcurrent l(t) for t > 0 is known pA. to be 900e 15oot

e) the initial energy stored in the three inducto$; f) the total energydeliveredto the blackbox;and g) the energy tmpped in tlre ideal inductors

a) Replace'he origjnalcapacirors wirh an equi!dlent capacitor and find z"(r) for / > 0. b) Find 01(t)for t > 0.

ProbLems 223 c) Find r'r(r) for r > 0. d) How much energyis delivercdro the black box in the time interval0 < r < oo? e) How much energy was initially stored in the seriescapacitors? f) How much energy is trapped in the ideat capacitors? g) Showt]tat the solutionsfor or and o, agreewith the answerobtainedin (1). Figure P6.27

45V

15V +

6.?2 The four capacitorsin the circuit jn Fig.P6.28are connected acrossthe tenninals of a black box at 1 = 0. The resulting current i, for a > 0 is kllown to be

d) the percentageof the initial energystoredthat is deliveredto the black box| and e) the time, in miliiseconds, it takes to deliver 5 /..J to the blackbox. 6.30 Derive the equivalert circuit Ior a seriesconnection of ideal capaciton.Assumethat eachcapacitorhas its o$,ninitial voltage.Denote theseinitial voltages as rr(r0), o2(r0),and so on. (Ht rr Sum rhe voltages acrossthe stdng of capacitors,recognjzingthat the seriesconnectionforcesthe current in cachcapacitor to be the same.) 6.31 Derive the equivalentcircuit for a parallel connection of ideal capacitors. Assumethat the initial volf age acrossthe paralleledcapacitorsis u(lo). (dir?rr Sum the currentsinto th€ string of capaciton,recognizing that the parallel connection forces the voltageacrosseachcapacitorto be the same.)

Sections6.1-{,3 6.32 The cment ir the circuit in Fig. P6.32is knowl to be i, : 5043o0'(cos60001+ 2sin60001.1mA for / > 0+.Find ?,r(0+)and z,l0+). Figure P6.32

it = 50e-^a'tt A. If o"(0) : is v, t.(0) = 45 v, and z'd(o): 40 V, tind the following for / > 0: (a) olr, (b) ?,,0),

320Q

(c)o.(1),(d)trlr, (e)n(0, and(f) ,,(,). Figure P6.28

6.33 At t = 0, a seriesconnectedcapacitotand inductoi are placedacrossthe terminalsof a black box, as shownin Fig.P6.33.For r > 0, ir is known that j, : e-30'sin6ot A If0c(0) = tigureP6.33

6.29 For the circuitin Fig.P6.28,calculate a) the initial eneigystoredin the capacitors; b) the final energystoredin the capacitors; c) the total energydeliveredto the black box;

300V lind ", for I > 0.

224

Inductance. Calacitance. andlllutuat Inductance

Section 6.4 6.34 There is no energy srored in the circuit in Fig. P6.34 at the time the switchis opened. a) Derive the dilferential equation that govems the behaviorot 12 if Lr - l0 H. L, - 40 H. M:5H,andRd:90O. b) show that when ie = 10;' - 10 A, t > 0, tlle differential equation derived in (a) is satisfied when i2 = at - 5e 225tA, t > lJ. c) Find the expressionfor the voltage ?1 acrossthe current source. d) what is the initial value of x'r? Does tlis make sensein terms of known circuit behavior?

Figure P6.34

l),

LI

6.35 Let o. represent the voltage acrossthe 16 H inductoi in the circuit in Fig. 6.25.Assume D, is positive at thedot.AsinExample6.6,i"= 16 - t'n5' O. a) Can you find 2o without having to differentiate the expressionsfor the currents? Explain.

6.37 a) Showthat the differential equationsdedvedin (a) of Example6.6canberearangedasfollows: di.

4;

di"

+ 2sit - 8i

di, 8dt

- 2Oi2= si. - 8

di"

dt:

di, di. 20r, I t6-i + 80i,- t6dt

dt

b) Show that the solutions for it, and 12given in (b) of Example6.6satisfythe differentialequations given in pa (a) of this problem. 6,38 The physical construction of four pairs of magneticaly coupled coils is shown in Fig. P6.38. (See page 225.) Assume that the magnetic flux is confined to the core material ir each structure. Show two possible locations for the dot markings on each pair of coils. 6.39 The polarity markings on two coils are to be determhed expeimentally. The experimental setup is shoM in Fig. P6.39.Assumethat tho terminal connected to the negative termhal of tle battery has been given a poladty mark as shown. When the switch is close4 the dc voltmeter kicks upscale. Where should the polarity mark be placed on the coil connected to the voltmeter? rigureP6.39

b) Derive tle expressionfor r',. c) Check your answer in (b) using the approp ate current derivatives and inductances,

6.36 I-et DsreFesentthevoltageacrossthe currentsource in the circuitin Fig.6.25.Thereferencefor ?,sis positive at theupperteminal of the currentsource. a) Find-ts as a Iunctionof time whents = 16b) wlat is the initial valueof os? c) Fitrd t])e expressionfor the powerdevelopedby the currentsource, d) How muchpoweris the currentsourcedeveloping whenI is infinite? e) Calculatethe power dissipatedin eachresistor wherlt is infinite.

6.40 a) Show that the two coupled coils in Fig. P6.40can be replaced by a single coil having an inductance of aab : a1 + L2 + 2M. (Hint: Express o^bas a tunction of tab.) b) Show that if the connectionsto the terminals of the coil labeled a2 are reveNed, Ldb= Lr + Ia 2M. Figure P6.40 M

a5:r._._ri'^_'b

Pmblems225 rigureP6.38

0)

G)

6.41 a) Show that the two magneticaly coupled coits in Fig.P6.41canbe feplaced by a singlecoithaving a.ninductance of

(d)

b) Show that if the magnetic polarity of coil 2 is reversed,then Lrlo - M'

M2 hlo Lr+ L2 2M

Lr+ L2+2M Fi$re P6.41

(Airi Let it and i2 be clockwise mesh curents in the left anddght "whdows" of Fig.P6.41,respectively. Sum t]te voltages around the two meshes. In mesh 1 let ?,abbe t})e unspecified applied voltage.Solve for d4/dt as a function of oab.)

226

Inductanc€, Capacitance, andl{otuat Inductance

S€.tion 6.5 6,42 TWo magnetically coupled coils are wound on a nonmagneticcore.The se[-inductanceof coil 1 is 250mH,the mutualinductanceis 100mH, the coef ficient of coupling is 0.5, ard tlle physical structure ofthe coilsis suchthat 91r = 922. a) Find 12 and the turns ratio N/N2. b) If Nr = 1000,what is the valueofgr and 0r? 6.43 The self-inductances of two magnetically coupled coils are l,r = 400 riH and 12 : 900 &H. The coupling medium is nonmagnetic. If coil t has 250 tums and coil 2 has 500 tums, find 9x and 921 (in nanowebersper ampere) when lie coefficientof couplingis 0.75. 6.,14 Two magneticallycoupled coils have self-inductances of 52 mH and 13 mH, respectiveryThe mutual induc tance betweenthe coils is 19.5mlL a) What is the coefficient of coupling? b) For these two coils,what is the largest value that ,14can have? c) Assume that the physical structure of these coupled coils is such that Or : 92. What is the turns ratio N/N, if Nl is the number of tums on the 52 mH coil? 6.45 The self-inductances of two magtetically coupled coils are 288 mH and 162 InH, respectively. The 2B8 mH coil has 1000 tums, and the coefficient of coupling between the coils is Z. The coupling medium is nonmaglelic. When coil 1 is excited with coil2 open,thefluxlinking only coil 1 is 0.5aslarge as the flux linkhg coil 2. a) How manyturff doescoil2 have? b) What is the value of 02 in nanowebersper c) What is the value of 9n in nanowebersper ampere? d) What is the ratio (drxldrr?

6.47 The self-inductances of thl3 coils in Fig. 6.30 are L1 : 25 mH andL2 : 100mH. If the coefficientof couplingis 0.8, calculatethe energystored in the systemin millijouleswhen (a) t1 : 10 A, 4 : 15 A: (b) i1: 10A, i= 15A; (c) tr = -10A, '2 = 15 A; and (d) t1 = 10 A, i2 : -15 A. 6.48 The coefficient of coupling in Prcblem 6.47 is increasedto 1.0. a) If ,r equals10 A, what value of i resultsir zero storedenergy? b) Is there any physicallyrealizablevalue oft2 that canmake the storedenergynegative? Sections6.1-6.5 6.49 Rework the Practical Penpective example, except pll$lf?i! that this time, put the button on the hottom of tie divider circuit, as shown in Fig. P6.49.Calculate tlre output voltage o(t) when a finger is present. Figuru P6.49

t"(t)

6.50 Somelamps are made to tum on or off when the ,l$llJjhbaseis touched.Theseusea one-tenDinalvariation of the capacitive s$itch circuit discussed in the PracticalPerspective. Figue P6.50showsa circuit model ol such a lamp. Calculate the change in the voltage t'(t) when a person touches the lamp. A(.umeall capacirorr are i0iliallyJischarged. Figure P6,50 10pF

6,46 a) Startingwith Eq.6.59,show Oratthe coefficient of couplingcanalsobe expressedas

' J(#X#) b) On the basis of the fracnons 62\/6r and.OdQ, explainwhy k is lessthan 1.0.

..{-

=ti

Lamp Person10pF

Probtems 227 6.51 In the Practical Perspectiveexample,we calculated the output voltagewhen the elevarorbufton pPEIifiIHLE is the upper capacitor in a voltage divider. In Problem 6.49,we calculatedthe voltage when the button is the bottom capacitorin the divider, and we got the same tesult! You may wonder if this will be true for all suchvoltagedividers.Calculate the voltagedifference(fingerversustro finger) Ior the circuits in Figs. P6.51(a) and (b), which use two identicalvoltage sources.

FigucP6.51 Fixed

_p5 pF + )(t)

,"(r)

; \ ""(,)

-f

No finger

?5pF ri^"a

capacitor

E5 pF + ?,(r) Finger

Response of First-0rder Rl andRCCircuits 7.1 TheNaturatResponse of an RL[itcuirp.230 p.236 7.? TheNaturatResponse of anRCaircuit 7.3 TheSt€pResponse of f,l and RCCitcuits p. 24A 7.4 A GeneratSoLutionlor Stepand NaturaL p. 248 Responses 7.5 SequentiatSwitchingp- 254 p. 258 7.6 UnboundedResponse 7.7 TheIntegnting Amplifier p. 26,

1 Beableto deiermireihe faturatfesponse of botlrRLaid RCcncuits. 2 Beabteto determine the steprcsponse of both 3 Knowhowto anatyze circuitswith sequentjat swiiching. 4 B€abt€to analyze op ampcncujtscontaining rcsistors anda sjngl€capacitof.

228

In Chapter6, we notedthatanimportanlalt buteo[ inductois and capacitors is their ability to storeenergy.We are now in a positionto determinethc currcntsandvoltagesthat alisewhen cnergyis either releasedor acquiredby an inductor or capacitor ln responseto all abrupt changein a dc voltaE:cor curent source, In this chaptcr.we will focus on circuits that consistonly of sources,resiston,and either (but nol both) inducton or capaci' tors. For brevity, such configurationsare called lla (.esistorinductor) andRC (resistor-capacitor) circuits. Our analysisof R, and RC circuitswill be dividedinto three phases.II1 the first phase,wc considcrthe curents and voltages that arisewhen storederergy in an inductor or capacitoris suddenly releasedto a resistivenetwork. This happcnswhcn thc inductor or capacitoris abruptlydisconnectedfronr its dc source. Thuswe canreducethe circrlit to one ofthe two equivalenttbrms shownin Fig.7.1on page230.Thecurents and voltagesthat arise in this configurationare leferred to asthe natuml tesponseof the circuit,to emphasizethat the natureof the circuit itsclf,not cxtcr nal sourcesof excitation.doterminesits behavior. Tn tbe secondphaseof our analysis,we considerthe cunents and voltagesthat arisewhenenergyis beingacquircdby an inductor or capacitordue to the suddenapplicationof a dc voltageor cunent source.This responseis referred to as the step rcsponse. The proccssfor finding both the natural and stepresponses is the same;thus.jnthe third phaseof our analysis, we developa general method that can be usedto find thc responseof RL and nC cjrcuitsto any abruptchangein a dc voltageor cuflent soulce. Figure7.2on page230showsthe lour possibilitiesfo. the gen eral configurationol RZ and RC circuits.Note that when there are no indcpendentsourcesin the circuit.the Th6veninvoltageor Norlon cunent is zero, and the circuit reducesto one of those shownin Fig.7.1;thatis,we havea natural-response problem. na and RC circuits are also known as first-order circuits, becausetheir voltagesalrd culrents are describedby first ordcr ditferential cquations.No matter how complex a cjrcuit may

PracticaI Perspective A Fl.ashing LightCircuit Youcan probablythink of rnanydjfferentapplications that requirea flashingtight.A stitLcamerausedto takepicturesjn low light conditionsemptoysa brightftashof Lightto ittuminatethe sceneforjust tongenoughto recordthe inrageon fitm. GereralLy, the cameracannottakeanotherpjctureuntjl the circuitthat createsthe ftashof Lighthas'te-charged." otherapplicatjons usefLashing tightsas warnjfgfor hazards,such as tat[ antennatowers,constructionsites, and secureareas.In designingcircuitsto producea flashof tight the engineermustkrow the requir€nr€nts of the apptication. Forexampte, the designengineerhas to knowwheth€rthe flashis controtLed manuatLy by operatinga switch(asin the caseof a camela)of if the flashis to repeatitsetfautomaticattyat a predetermined rate.Theengineeratsohasto knowif the flashinglight js a permanent fixture(ason ar antenna)or

a temporary instatlation(as at a construction sjte). Another qre\fionrhathas[o beanswFred'. power wherLrer a sorrcei. r€adiLy avaitabte. Manyofthe cjrcuitsthat areusedtodayto controlflashjng Lightsare basedon €tectroniccircujtsthat are beyondthe scopeof this text. Neverthetess we can get a feel for the jnvotvedin designing thoughtprocess a fLashing tight circuit by anatyzjng a circuitconsisting of a dc voLtage source, a reststor, a capacr'tor, and a lampthat is designed to discharge a jn flashoftight at a criticatvottage. Sucha cjrcuitisshown the figure.Weshatldjscuss this cjrcuitat the endofthe chapter.

229

230

Response of Fi6t-0rder fi andRrCircuits

c.,r (a)

(b)

FiguE7.t A Thetu/oforms ofthecncuits fornatuGt (a)Rrcncujt. (b)nrcircuit. rcsponse.

appear,if it can be reduced to a Th6venir or Norton equivalent connected to the terminals of an equivalent inductor or capacitor, it is a first-order circuit. (Note that if multiple inductors or capacitors exist in the original circuit, they must be intercoinected so that they can be replaced by a single equivalentelement.) Aftei introducing tlle techniques for analyzing the natural and step rcsponsesof fi$t-order circuits, we discusssome special casesof interest. The first is that of sequentialswitching,involving circuits in which switching can take place at two or more hstants in time. Next is the unbounded rcsponse.Finally, we anallze a useful circuit called the integrating amplifier.

Rrh

I

l

-

fl"* Y

7.1 TheNaturat Response of an RLCircuit

.j, I

G) vrh

t

L

(b)

Yrh Rft

The natural responseof an Ra circuit can best be desc bed in terms of the circuit shown in Fig. 7.3.We assumethat the independent current source generatesa constantcurent of 1"A, and that t]re switch has been in a closed position for a long time. We define the phrase a long time morc accumtely later in this section.For now it meansttrat all currents and voltageshave rcached a constant value.Thus oDly constant, or dc, currents can exist in the circuit just prior to t]le switch's being opened, and therefore the inductor appears as a shot c cuit (adi/dl : 0) prior to the release of the stored energy. Becausethe inductor appears as a short circuit, the voltage acrossthe inductive branch is zero, and there €an be no current in either Ro or R. Therefore, all the source current 1, appea-rsin the inductive branch. Fhding the natuial response requires finding the voltage and curent at the terminals of the resistor after tlle switch has been opened, that is, after the source has been disconrected and the inductor begins releasing energy.If we let t = 0 denotethe instantwhen the switchis opened,the problem becomesone offinding ?(r) and t(t) for I > 0. For t > 0, the cir, cuit shownin Fig.7.3reducesto tlle one showr in fig.7.4 on the next page.

Derivingthe Expression for the Current (d)

tigure7,2 A Fourpossible fi'st-order circuits. (a)Aninductor connected t0 a Th6venin equjvateni. (b)Aninductor connected to a Norionequjvatent. connected to a Th6venin equjvatent. k) A capacitor (d)A capacitor connect€d to a Norton equivaLent.

Figure 7.3l' AnRtcircuit.

To find i(t), we use Kirchhoff's voltage Iaw to obtain an expressioninvolving i,R, and a. Summingthe voltagesaroundthe closedloop gives

t4n^r=o,

(7.1)

wherewe uselhepassivesigncorvention.Equation7.1is known asa firs! oider ordinaly differential equation, becauseit contains terms involvhg the ordinary derivative of the unknown, that is, dy'dt. The highest order derivativeappeaing in the equationis 1;hencethe telm fitst-otdet. We can go one step fu her in descdbing this equation. The coeffi cients in the equation, R and a, are constants; that is, ttrey are not ftrnctionsof either the dependentvariableior the independentvariablet.Thus ttre equation can also be described as an ordinary differcntial equation witl constant coefficients.

7.1

To solve Eq. 7.1,we divide by a, transposerhe rerm involving i ro the right-hand side, and then multiply both sidesby a differenrial time dr. The resuttrs di,

-4tot

TheNatumt RespoNe ofanil Circuit 231

-,=.'8,

1 7 . 2 ) F i g u r7e, 4A t t p , , . 1 ' ts o w -i - t g . / . ? .f o ., - 0 .

Next,we recognizethe lelrhard sideofEq.7.2 as a differertial changein the current i, that is, di. We now divide through by j, gettitg di

i

R, = -70,

(7.3)

We obtair an explicit expressionfor i as a function of r by integrating both sidesofEq.7.3. Usingr and I asvariablesof integiationyields

-:1,"'*, li,,'+=

17.4)

in which i(lo) is the cu[ent corespondingto time r0,and j(1) is the current correspondingto time t. Here, r0 = 0. Therefore,carrying out the indicaledintegrationgives

' ( 0 (0)

R L"

(7.5)

Basedon the definitionofthe naturallogarithm, i\t) = i\o)e tR Ltt

(7.6)

Recall from Chapter 6 that an instantaneous change of current cannot occur in an inductor. Therefore, in the first instant after the switch has beenopened,the curent in the inductor remainsunchanged.Ifwe use0 to denotethe timejustprior to switching,and 0+for the time immediately Figure 7.5A Thecurrent response forthecncuitshown Iollowing switching, then jn Fjg.7.4. ,(0-):(0+)=10,

d Initial inductorcurrent

where,asin Fig.7.1,10denotesthe initial curent in the inductor.The initial current ir the inductor is oriented in the same directiorl as t]re reference directionof i. HenceEq.7.6 becomes itJ\:

r^" lR/L)t

(7.7) { Naturalresponse of annt circuit

which showsthat the currentstartsfrom an initial value10and decreases exponentiallytowardzerc as r increases. Figure7.5showsthis response. We derivethe vollageacrossthe rcsistorin Fig.7.4ftom a direct appli cationof Ohm'slaw: r:

iR=

IoRe-\RtL)t, t >0+.

(7.8)

232

Response of Fnst-0rd€r RtandRrCjrcuits Note that in contrust to the expressiotrfor the curent shown in Eq. 7.7, the voltage is defined only for I > 0, not at t = 0. The reason is that a step change occurs in the voltage at zero. Note that for r < 0, the derivative of the current is zerq so the voltage is also zero. (This result follows hom 1)=Ldi/dt-\.Jmus

"(0):0,

(7.e)

t (0-) = 1oR,

(7.10)

wherer{0+) is obtainedfrom Eq.7.8 withr = 0+.1Withthis stepchangeat an hstant in time, the value of the voltage at 1 = 0 is unknowr. Thus we use I > 0- in defining the rcgion of validity for these solutions. We dedve the power dissipated in the resistor ftom any of the follow-

p = izR,

(7.11)

Whicheverform is used,theresultingexpressioncan be reducedto p

lzRe-2(RtL)t,

t > 0+,

17.12)

The energy delivered to the rcsistor during any intenal of time after the swilchhasbeenopenedis

-= | ,,": fo'na",,'',,a,

- "u'*'", :4fu13*1t = L*tZrt - e-2tR/Lr),t >0.

(7.13)

Note ftom Eq. 7.13 tlat as t becomesiniinite, the energy dissipated in the iesistor approachesthe initial energy stored in t]te inductor.

TheSignificance of the TimeConstant The e\pr-essions lof j(1)(tq 7.7iand d tEq. 7.8)includea termot rhe "'. " plorm The coefLicienrot l-namel]. n L determinesrbe rate al which the cwent or voltage approacheszero. The reciprocal of this ratio is the time constsntof the c cuit. denoted

Timeconstant for Bl circuit>

\7.11)

t we can de6ne the er?ressio.s 0 and 0+ more fomaly. The exprcsion r(0 refeB to the ) linit or the variable x as t +0 fron tne left. oi lron negalive tine. The expresion r(0+) relers to the limil of ihe rariable x as r + 0 fron the right, or fton posilive tine

7.1 TheNatumt Response ofanfit Circuii 233 Using the time-constantconcept,we write the expressionsfor curtent, voltage, power, and energy as ilt)= Ioea',

t>0,

1 ) ( t )= I o R e - ' / 1 , p = I1Re2th,

*=

t

-

rut^r,

t>0+, t>0+,

e-o,,t. t>0.

(7.15) (7.16) \7.17) (7.18)

The time constant is an important parametet fot firsForder circuits, so mentioning several of its charactedstics is worthwhile. First, it is convenient to fhink of the time elapsedafter switching in terms of integral multiples of r. thus one time constant after the inductor has begun to release its storedeneigy to the resistor,the curent has been reducedto tl, or apprcximately 0.37of its initial value. Table 7.1 gives the value of ?-t" for integal multiples of r from 1 to 10. Note that when the elapsed time exceedsfive time constants,the current is less than 1ol" of its initial value. Thus we sometimes say that five time constants after switching has occuned, the currents and voltages have, for most practical purposes,rcached tlle final values. For single time-constant circuits (first-order circuits) with 1% accuacy, the phnse a /org dme implies that tive or more time constants have elapsed.Thus the existenceof current in the R-L circuit shown in Fig.7.1(a) is a momentary event and is refered to as the tramient r€sponse of the circuit. The response that exists a long time after the switching has taken place is called the steady-stateresponse.The phrase d long ,tne then also means the time it takes the circuit to reach its steady-statevalue. Any first-oder circuit is characterized,in part, by the value of its time constant.If we have no method for calculating the time constant of such a circuit (perhaps becausewe don't know the values of its components), we candetermineits valuefmm a plot of the circuit'snaturalresponse. That's becauseanother important characteristic of the time constant is that it givesthe time required for the current to reach its final value if the curent continues to change at its initial rate. To illustmLe, we evalrratedi/ dt at 0* and assumethat the cunent continuesto changeat this mte:

firot=-lr,: lt

2r 3r 5r

3.6788x 1.3534x 4.9'787x x 1.8316

10_1 10 1

10n 10-, 6.7379\ 10 3

2.4i88x 10 3 9.1188x 10+ 8r 3.3546x 1or 9r 1.2341x 1or 10t 4.5400x 10 5 6r '7r

(7.19')

Now, if i starts as 10 and decreasesat a constant rate of 1o/r amperesper second.the exDressionfor i becomes Io

11.zoJ

Equation 7.20 indicates that i would reach its fhal value of zero in r seconds.Figue 7.6 showshow this graphic interpretation is useful in esti. mating tle time conslant of a circuit from a plot of its natural rcsponse. Such a plot could be generated on an oscilloscopemeasuring output currcnt. Drawing the tangent to the nalural responseplot at t : 0 and readhg Figure7,6 A A g€phicjntenretatjon of thetimecor the value at which the tangent intersects the time axis givesthe value of r. stantoftheRt cjrcuitshown in Fig.7.1.

234

Response of Fnst-oder Rlandfc Cncuits Calculating the natural responseof an RL circuit can be summarized

Calculating the naturalresponse of Rl circuitts

1. Find the initial current,10,throughthe inductor. 2. Find the time constant o{ the circuit, r : a/F 3. Use Eq. 7.15,Ioe '/', to genemtet(l) from 10and i A other calculations of interest follow ftom knowing i(1). Examples7.1 and 7.2 illustratethe numericalcalculatioNassociated with the natuml responseofan Rt circuit.

Determining the NaturalResponse of anil Circuit The switch in th€ circuit shown in Fig. 7.7 has been closedfor a long time before it is opened at t : 0. Find a) iro) for t > 0, b) i,(t) for t > 0*, c) ?,o(,)for r > 0+, d) the percentage of the total energy slored tn the jn the 10 O resistor. 2 H inductorthat is dissipated 2Q

0 . 1 o, i , l 2 H

)'"

b) We find the current in the 40 O resistor most easilyby usingcurrent division;lhatis, '"

'"10

10 + 40

Note that this er?ression is valid for I > 0* becausejo = 0att:0 . The inductorbehavesas a shor circuit prior to the s$ritch behg opened, producing ar instantaleous changein the cufent i". Then, t,(r)= -4d5rA,

100

40f)

Figure7.7 4 ThecircuiiforExampte 7.1.

r>0*.

c) We find the voltage r, by direct applicationof Ohm's law: D.(t) = 40i": -160e 5'v,

r > 0'.

d) The power dissipatedin the 10 O resistorls

Solution a) The switchhasbeenclosedfor a long time p or to r :0, so we know the voltage across the inductor must be zero at r : 0 . Therefore the initial cuirent in the inductor is 20 A at r = 0 . Hence,iz(O-) also is 20 A. becausean instantaneou.changein rbe cuffenrcannotoccufiD an inductor. We replace the resisrive circuit connectedto the terminalsof the inductor wu a singleresistorof 10 (): R.q=2+(40 10)=10O. The time constantof the c cuit is I/Rcq, or 0.2s, giving the expressionfor the inductor current as ,r(r) - 20e-5,A, r>0.

,? D N ' u r = ? - 2 5 b 0 lf0 W . t(l

/'0'

The total energydissipatedin the 10 O resistoris u )^ " ( t \ :

/I 2 5 o 0 P r u ' d: l - 2 5 b 1 . .lo

The initial energystoredin the 2 H inductorrs l _ 1 ru(O)= -Zi'(0)::(2)(4n0)

= 40uJ.

Thereforethe percentageoI energydissipatedin the 10 O resistoris

=o.to.. ffirroot

7.1 TheNatumt ResDons€ ofanNlCircuit 235

Determining the NaturalResponse of anBl.Ci]cuitwith ParallelInducto]s In the circuit shown in Fig. 7.8,the initial currents in inducto$ 1,1 and Z2 have been established by sourcesnot shown.Theswitchis openedat I = 0.

:

7.6

t > 0,

= 5 . 7 o zP A . / > 0 + .

i,: --

a) Find i1,i2,and ir for I > 0. b) Calculate the inilial erergy stored in the paiallel

2.4e2' A,

Nole that the expressionsfor the inductor cunents it and ,2 are valid lor r > 0, whereasthe expression for the resistorcurent i3 is valid for t > 0*.

c) Determine how nuch energy is stored in tl]e inductorsast+co. d) Sbowthar$e roralenergydelivefedro lhe re.istive network equals the dilference between the resultsobtainedin (b) and (c).

,o1

8()

Solution

jn Fig.7.8. ofthecircuit tisure7.9 A simptification shown

a) The key to finding curents ir, i2, and i3 lies in knowing the voltage o(t). We can easily find o(r) if we reduce the circuit shown in Fig. 7.8 to the equivalent folm shown in Fig. 7.9. The parallel inductors simplify to an equivalent inductance of 4 H. carryingan i0ilialcurrentof l2 A.The resi* tive network reduces to a single rcsistance of 8 O. Hence the initial value of i0) is 12 A and rhelimeconslanr is 4/8.or n.5s. Iherelore

b) The initialenerg)sroredin rheinduclor(h 1 1 d: -(5X6a) + -(20)(16):320J.

i(t)=12e)1A,

j1'1.6A c)As t+co, and i21.6A. Thereforc, a long time after the switch has been opened, the energy stored in the two inductors is t _ t d J2J. ^(5Xl.o) -:(20X l.ol

t>0.

d) we obtainlbe toralenergydeliveredlo the resisrive network by integrating the expression for powerfromTerolo infinir): lhe ioslcnraneous

Now z'(l) is simply tlle prcduct 81,so .,(t) = 96e2'Y,

t>o+.

u::

The circuit showsthat o(l) : 0 at t = 0 , so the expressionfor D(r) is valid for r > 0*. After obtaining o(,), we can calculatei1,i, and 4:

"-4t )"

ln

i,' =: I 96e2'dr - 8 :Jo =1.6-9.6e2A, t>0,

i,:j l'sa"^a',t 4{}

(5H)

,ln

= 1152:,l -+

1 I l

L1

I pdt:

lo I r = 0 r2QoH) 1

Figurc 7.8A Thecncuit forExamph 7.2.

40{))

,rrn

"l

/.m

I 1rs2e4,dt

.la

: 288J.

This result is the dillerence between the initially stored eneryy (320 J) and the energy trapped in the paftllel induclors (32 J). The equivalent inductor for the parallel inductorc (which predicts the terminal behavior of the parallel combination) has an initial energy of 288 J; that is, the energy stored in the equivalent inductor representsthe amount of energythat will be delivered to the rcsistive netwoik at the terminals of the original inductors. 10r)

236

Response of First-order 8l andRrCircuits

objective1-Be ableto determine the naturaL response of both [l andRCcircuits 7.1

The swirchin thc circuil shorvnhasbeenclosed for a lo g time and is openedat t - 0. a) Calcular.3 the initial valueof;. b) Calcularethe inirial cncrgv sloredin the

7,2

c) w}lal is the time corstant of the circuit lbr t>0'l d) Wlat is thc nunericat expressionfor ;0) for e) What percentageof the initial energystored hasbeendissipatedin the 2 r) resislor5 ms afler the switchhasbeenopcned?

120V

Answer: (a) -12.5 Ar (b) 62snJ; (c) a nis: (d) -12.5a1sorA.r > 0r (e) e1.8o/.. At r : 0, the s*itch in the circuit shownnoves instantaneously from posiliona to posilionb. a) Calculate,. for t > 0'. b) What percentageof the initial cncrgyslored in the inclucroris eventuallydissipaledin the ,1O resistor'?

8c "'V. t > 0: (b) 80%. (a)

NOTE: Also try ChapterPrabkn$7.1 7.3.

ResBrnne 7 . 2 l-heN*tsrraI *'i'e* R{ eirer"rit

F i g r r e7 . 1 0 , 1 A nf C c i r c u i i .

rigurc7,11;! Thecir.uitshownin

As mcntionedin Section7.1,the naturalresponscot an RC circuil is aral, ogousto thal ofan Rl, circuit.Consequentl],. wc don't treal lhe RC circuit in thc sane detail aswe did the nl- circuit. The Daturalresponseof an RC circuil is dcvclopedfroD the circuit shownin Fig.7.10.Webeginb]' assumingthal lhc ss,ilchhasbeenin position a lbr a long time,allowilrgthe loop nade up of thc dc vollagesource ys.the resistorn1. aDdthe capacitorC to reacha steadystalc condilion. Recall fionr Chapler6 that a capacitorbehavesas an opcn circuil il1 the presenccof a constantvoltage.'lhusdre voltage sourcccaniot suslaiDa currcnt,and so lhe sourcevoltageappearsacrossthe capacitortenninals. In Scction7.3,we will discusshow thc capacitorvollageactuallybuildsto the sleady-state value ofthc dc voltagesource,but for now the important pojnl is that when the switch is moved from posiiioDa to position b (at i = 0), the voltageon the capaciloris y-. Becausethere can be no instantancouschangein the voltageat the tcrminalsoI a oapacitor.the problem rcduces1lrsolvingthe circujtshownin Fig.7.l1.

7.2 TheNaturaL Response of annCCircuit 237

Derivingthe Expression for the Voltage We can easily find tlle voltage ?(t) by thinldng in tems of node voltages. Using the lower junction between R and C as the reference node and summing tlle curents away from the upper junction between R and C gives

c4 * {:0.

(7.21)

Comparing Eq. 7.21 with Eq.7.1 shows that tlte same mathematical techniques can be used to obtair the solutior for z(t). We leave it to you to showthat D(i) = u(o\e1tRc, t-0.

\7.22)

As we have already roted, t]le initial voltage on tle capacitor equals the voltage sourcevoltage %, or

(7.23) < Initial capacitor vottage where yo denotesthe ioitial voltage on the capacitor.The time constant for the nC circuit equals the product of the resistarce and capacitance, namely,

i:rli,;#!l

(7.24) < Tine constant for BCcircuit

SubstitutingEq$ 7.23arrd'7.24n\b Eq.1.22yields (7.25) < NalulalresDonse of annc circuii which indicates that tle natural rcsponse of an RC circuit is an exponential decay of the initial voltage. The time constant RC governs tle mte of decay.Figure 7.12 shows the plot of Eq. 7.25 and the $aphic interpretation of the time constant. After determiningD(l), we can easilydedve the expressionsfor i,p, and 1,:

44 = !!

=\-t1,

, - ()*,

vo

1,(t,= ve r' a(t) :vo

17.26J Flgure 7.12 ThenaiuraLrcsponse ofanif circuit.

p-Di

= 2 R e - 2 ' , , t. > O + .

:?,a,,

17.27)

17.28)

238

Response of Fi6t-0rder Rrandfr Cjrcujts Calculating the natural responseol an RC circuit car be summarized as followsi

Catcutating the naturalresponse of an fC circuit>

l . Find the initial voltage, y0, acrossthe capacitor. 2. Find the time constant of the circuit, r = RC. 3. Use Eq.7.25,?(t) : U& 't', to Eerleifre1)(1)from yo and r. Al1 other calculations of interest follow from knowing t)0). Examples7.3 and 7.4 illustratethe numericalcalculationsassociated rvith the naturalresponseofan RCcircuit.

Determining the NaturalResponse of an fC Circuit The switchin the circuit shownin Fig.7.13hasbeen in position r for a long time. At I : 0, the swltch moves instantaneously to position y. Find a) oc(t) for l > 0, b) 0,(t) for r > 0*, c) to(4 for r > 0*, and d) the total energydissipatedin the 60 kO resistor.

b) The easiestway to find o,(r) is to note that the resistive circuit forms a voltage divider across the teminals of the capacitor. Thus

, r')

48 - b0. -_ z-'v. s;rcfl)

r> u .

This expression for ,"(t) is valid for / > 0+ becausetro(0-)is zero.Thuswe havean instantaneouschang€in the voltageacrossthe 240kO

c) We find the current i,(t) ftom Ohm's law: Figure7.13A Thecircujt forExampte 7.3.

i"(t\ :

r,"(t) : e 2 5 r mA, 60 x 103

t>0+.

Solution a) Becausethe switch has been in positionr for a long time, the 0.5/rF capacitor will charge to 100V and be positive at the upper terminal. We can replace ttre resistive network conlected to the capacitorat a : 0' with an equivalentresist, ance of 80 kO. Hence the time constantof rhe circuiris (0.5 x 10-1(80 x 1d) or 40 ms.Then, "c(t) = 100e' 5'V, / > 0.

d) Thepowerdissipatedin the 60kf,l tesistoris p60ko0): t3(r)(60x 103)- 60e-50,mw, r > 0-. The total energydissipatedis

7.2 ThelaructRespolse o'ar nr('(uit

239

Determining the NaturalResponse of an 8CCircuitwith SeriesCapacitors The initial voltages on capacitors C1 and C2 io the circuit shown in Fig. 7.14 have been established by sourcesnot shown.The switch is closed at t : 0. a) Find o(t, or(t), and o(t) for / > 0 and t(t) for r>0-. b) Calculare rbe inirialenergysroredin lhe capacitors C1and Cr. c) Determine how much energy is stored in tlle capaqlonast+oo, d) Show tlat the total energy delivered to the 250kO resistor is ttre difference between the resultsobtainedir (b) and (c).

r(, Cl(5pF) "(r)

+

1'(r_)250ko

+ 24V

c2Q0 t/F) nzl)

Figure7.14A Thecjrcuit forE\ampte 7.4.

Solution

l(4

a) Once we know o(t), we can obtain the cunent t(t) from Ohm's law. After determining i(l), we can calculateq(1) and ?rr(l) becausethe voltage aooss a capacitor is a function of the capacitor cunent. To find ?0), we roplace the series-connected capacitoN with an equrvalent capacitor. It has a capacitanceof 4 pF and is chargedto a voltage of 20 V Therefore, the circuit shoM ir Fig. 7.14 reduces to the one shown in Fig. 7.15, which revealsthat the initial value of o(t is 20V and that the time constantof the circuit is (4)(250) x 10 r, or 1 s.Thus the exprcssionfor o(r) is r(t) = 20e-'V,

250ko Figure7.15 A A simphfication of theckcuitshown in Fjg.7.14.

b) Theinitial energystoredir C1is 1 t,j - :(5 \ 10n)(16) = a0 sJ. The initial energy stored in C, is

t>0. w2:

The current i(t) is ,/,1 i ( r l= : * : 8 o e i I)U,WU

!A.

/>o+.

- (16e-' i.'6

, r ( r )= - =

20) V,

4

t > 0,

tt

I 80 . to6e'dr + U

&'o : 40 + 5760 : 5800pJ.

rr+

20V

and u2+ +20V.

Ttere[orelhe energysloredin rbe two capaci fols is

.\t.ta

= ( 4 e . + 2 0 ) V ,r > 0 .

x 10 )(s76) = 5760pJ.

The loralenergysroredin rherwocapacilors is

Knowing i(t), we calculate the expressionsfor 0(t) and or0): r0b /' r,(t) = --| 80 . l0-6e "dx 5 ./o

;Q0

rr& =

1 , + 20) x 10-"(400): 5000pJ. ;(5

240

Response of First-order RLandRCCircujts

d) The totalenergydelivcredto the250kO resistoris f- 4onu 2' r l t- R t . ' r u J lpd! l^^^ / ./ r Z)lr.(rlru Comparingthe results oblained in (b) and (c) 800riJ : (5800

Thc enelgy storedin the equivalcntcapacitorin Fig.7.l5 is +(,1x 10 6)(400),or 800/..J.Because this capacilorpredictsthe terminal bchavior of the original series-connectedcapacilorsjthe encrgyslored in the equivalentcapacitolis the encrgydeliveredto the 250kO rcsislor.

5000)/1J.

objective1-Be abteto determin€ the naturalresponse of bothnt andnCcircuits 7.3

The switchin the circuit shownhasbeen closed for a long rime and is openedat r : 0. Find a) the initial valueof?r(/), b) the time constantlol r > 0c) the numericalexptessionfor ?r(l)after the switchbasbeenopened. d) thc inilial energystoredirl the capacitor,and e) the lengthoI timc requiredto dissipate757o of ihe initially storedenergy

7,4

The switchin the circuit shownhasbee! croseo for a long time beforebejngopenedat r = 0. a) Find 1r,,(l)for r > 0. b) Whal percentageofthe initjal encrgystored in thc circuit hasbeendissipatedafter lhe svilch hasbeen open for 60 rns?

50kJ)

rsv )r, Answen (a) 200Y (b) 20 ns; (c) 200€ 5o'V, r > 0j (d) 8 mJ;

''uloo*,

Answer: (a) 8e '5' + 4e Iu v. L > 0; (b) 81.05o/o.

(c) 13.86ms. NOTE: ALto try ChapterPrcbleft 7.21und7.21.

7.3 TheStenll{esponse*f ffA

ai'dfff e!rcu'its We arc now readyto discussthc problemof firding the currentsand vol! agcsgeneratedin first-ordcrna or RC circuitswhen either dc vollageor currenl soulcesare suddcnlyapplied.Theresponseof a circuit to the sudden applicationofaconstantvollageorcurrent sourccis referredto asthe

7.3

lhe SteDResDome offt andRCCncuits 241

steprcsponseof ttre circuit.In presentingthe stepresponse, we showhow the circuit rcspotrdswhenenergyis beingstorediI) the inductoror capacitor. Webeginwith the stepresponseof an RL circuit.

TheStepResponse of an nl Circuit To begin, we modiry ttre firsrorder circuit shown in Fig. 7.2(a) by adding a switch.We usethe resultingcircuit, shownin Fig.7.16,in developingthe stepiesponseof an Ra circuit.Energy storcdin the inductor at the time the switch is closed is given in terms of a nonzero initial current i(0). The task is 10find t]le eeressions for the current in the c cuit and for the voltage acrossthe inductor after the s$ritch has been closed.The procedure is the sameas that usedin Section7.1;we usecircuit analysisto derive the differential equation that descdbes the cftcuit in terms of the variable of intercst,and then we useelementarycalculusto solvethe equation. AJter the switchin Fig.7.16has been closed,Kirchhoffs voltagelaw rcquires that

v":Ri+L+,

11.2e)

which can be solved for the current by separating the variables t and r ard then integrating.The first stepin this approachis to solveEq.7.29for the deivative di/ dtl

Ri+u" dt

, * )

L

(7.30)

Next, we multiply both sides of Eq. 7.30by a differential time dl. This step reduc€s the leffhand side ot the equation to a differential change in the ctrllent, Thus

.h*

a \

,

R/.

dr-

L\i-

- ; )v"\ 0,,

v.\ R)/lt

We now separatethe variablesin Eq.7.31to get di

-R,

t-tu/R)

L

"''

Figure 7.15A A circujt used to ittustrate ihestep tesponse ofantst-order ru cncuit.

\7.31)

242

Response of Fiui-order Rl andRrCircuiis and then integate both sidesof Eq. 7.32.Using .r and y asvariables for th€ intesration. we obtain [i(t)

dx

-R

f .

1 , " , - , v 1 4 =7 1 , ' t '

(7,33)

where 10 is the crment at I = O and j(t) is the curent at any t > 0. Performing tlle integration called for in Eq.7.33 generatesthe exprcssion . n

i(t) - (4/R)

hlvr$

=

-R

Lr'

\7.34)

from which

i(t) (ulR) I o - ( WR )

Stepresponse of & circuit>

(7.35)

When the initial energy in the inductor is zero, 10 is zeio. Thus Eq. 7.35

i0 =f; \^"t*ro'

(7.36)

Equation 7.36 indicates that after tlle switch has been closed,the currcnt hcreases exponentially from zero to a final value of y,/R. The time constalt of the circuit. Z/R. detemines the mte of hclease. One time constant after the switch has been closed. the cunent will have reached approximately 6370of its final value, or

u,':\-\^"'=o.o:zr*.

17.37)

If ihe currentwereto continueto increaseat its initial rate.it wouldreach its final valueat t = 7: that is.because di clt

-

4( '\.,,,

u -,,.

R \ , , /

L

(7.38)

7.3

TheStepResponse ofRt andffCCjrcuits

the initial rate at which i(1) inoeases is

_u L' firot

\7.39)

If the current were to continue to increaseat this rate, the erpression for i would be

'=l''

(7.40)

from which, at t = 7,

. uL

u

I, R

R.

\1.41)

Equations7.36and ?.40are plottedin Fig.7.1?.The valuesgivenby Eqs.7.37 and7.41arealsoshownin thisfigure. anitrductoris ady'dt,sofromEq.?.35,forI > 0-, Thevoltageacross

TX.E), r"r,r, R)'

= (v" - ry!.\RtL,.

Q.4z)

The voltage across the inductor is zero before the switch is closed. Equation 7.42 indicates fiat the inductor voltage jumps to I{ - 1oRat the instant the switch is closed and then decaysexponentially to zeio. Doesthe valueof o at t = 0* make sense?Becausethe initial current is 10 and the inductor prevents an instantaneous change i[ curent, the Figure7.17.. Thest€presponse 0fthenLcircujt current is 10 in the instant after the switch has been closed.The voltage shownin Fiq.7.16when10 = 0. drop aooss the resistor is 1oR,and t]le voltage impressedacrossthe inductor is the souce voltage minus t]le voltage drop, that is, ys 10R. When the initial inductorcurent is zero,Eq.7.42sjmplifiesto r : Vse IRILY

17.13)

If the initial cuirent is zerq the voltage acrossthe inductor jumps to 14.We also expect the inductor voltage to approachzero as1inoeases,becausethe curent in the circuit is approachhg the constantvalue of VJ-R.Figure 7.18 showsthe plot of Eq.7.43 and the relationship between the time constant and the initial rate at which the hductor voltage is deueasing. If there is an initial current in the inductor, Eq. 7.35 gives the solution for it. The algebraic sign of 10is positive if the initial current is in ttre same 0 ? 2 t 3 r 4 r directionasi;otherwise,lo carriesa negativesign.Example7.5 illustrates ve66 tjme. the application of Eq.7.35 to a specfic circuit. figure7.18A Inductorvottag€

244 Response of Fjrst-OrderRL andfr Circujts

Determining the StepResponse of an Rl Circuit The switchin the c cuit shownin Fig.7.19hasbeen in position a for a long time. At I = 0, the swltcn movesfrom positiona to positionb.The switchis a make-before-break type; that is, the connectionat position b is establishedbefore the conneclionat position a is broken,so there is no interruption of currentthroughthe inductor.

b) The voltageacrossthe induclor is

a) Find the expressionfor t0) for, > 0. b) What is the initial volrageacrossthe irducrorjust afterthe switchhasbeenmovedto positionb? c) How many milliseconds after the switch has been moveddoesthe inductorvoltageequat24V? d) Does this initial voltagemake sensein terms or circuitbehavior? e) Plot both t(r) and ,l(r) versusr.

The initial inductorvoltageis

dt 7 ' = L ,

: 0.2(200e-10,) :40e lo'V, t>0+. 2(0-.) = 40 V. c) Yes; in the instant after the switch has beefl movedto positionb, the inductor sustainsa current of 8 A counterclockwisearound the newly formed closedpath. This current causesa 16 V drop acrossthe 2 O resistor.This voltage drop rddc ro lhe dfop acros.lhe.ource.producinga 40 V drop acrossthe inductor. d) We find the time at which the inductor voltage €quals24V by solvirg the expression 24 : 4De tol tor t:

':

1 , 4 0

1o-t

= 51.08x 10 l - 51.08ms. figure 7.19 A ThecircuitforExamph 7.5.

Solution a) The switchhasbeenin positiona for a long time, so the 200 mH inductor is a short circuit across the 8 A current source.Therefore,the irductoi carriesan initial curent of 8 A. This currenr is orientedoppositeto the referencedirectionfor i; Jhus10is -8 A. w}len the switchis in posiuonD, the final valueof i will be 24l2,or 12A.T\e time constaDtof the circuit is 200/2, or 100 ms. Substitutingthesevaluesinto Eq.7.35gives

r = 12 + (-8 =12

e) Figure7.20showsthe graphsoIt0) and r,(r) versus/. Note that the instantof time when the cur rert equalszero correspondsto the instant of time when the inductorvoltageequalsthesource voltageof24V aspredicredby Kirchhoff,svolrage law 1J(v)t(A) 40 32 21 1(r 8

, (nt

12)€4'rr

2 0 et n A , t > 0 .

figure7.20.d Thecurrent andvottaqe wavetorns for

7.3

TheSiepRespoNe of{L andRfCircujts 245

of bothnl' andnCcircuits obj€ctive2-Be abteto determine ihe stepresponse 7-5

\O7 f:

Assumethal the switchin the circuit showDro Fig.?.19hasbeenin positionb for a long timeand at r = 0 ir moves10positiona.Find (a);(o*); (tr) z(o-); (c) r, t > 0: (d) i(r), r > 0; and (e) ,l0), t > 0+. Al,o rt Chaplrt Prohlrn'

Answer: (a) l2A: (b) 200v: (c) 20ms; (d) -8 + 20?-' A, I > (e) -200eso'V,I> 0 .

-..1.1-.13.

We can also describethc voltage?J(r)acrossthe inductor in Fig.7.16 directly,Dotjusl in lemrsol lhe circuit current.Webeginby noling thal thc voltage acrossthe resistoris the ditferencebetweenthe sourcevollage and the inductorvollage.Wc$.rite 11.44) $ h c r cv i . a c o n \ t d n r . D i r t e f e n l i rbtoj nt h! r i d c \ u i r hc \ t e c lr o r i ' n e ) i e l d \

dt

Rdt

(7.45)

Theq iI we mulliply eachside of Eq. 7.15by the induclalce a, wc gct an expressionfor thc voltageacrossthe inductor on the lefl hand sidc,or (7.46)

PuttingEq.7.46into standardIoIm yiclds d x R

(7.47)

You shouidvedly (in Problen17.40)that the solutionto Eq.7.47i! idcnti cal io that givenin Eq.7.42. At this poinl, a generalobservationabout the step responseoI an Rl, circuil is pertincnt.(This observationwill prove helplul latcr.) When we derivedlhe dificrcntial equatior for the inductorcurrenl,we obtained Eq.7.29.Wenow rcwritc Eq-7.29as (li A

R. t

L

V" L

\1.48)

Obscrvc that Eqs.7.47 and 7.48 have Lhesame form. Specifically,each equatcsthc sum ofthe firsi derivativeoI lhe variablcand a constanttimes the variableto a constantvalue.In Eq.7.47,thcconstanton the right-haDd sidehappensto be zero;hencethisequationtakcson the sameform asthe

246

Response of First-order Rl andfftCi(uits

natuml rcsponseequationsin Section7.1.Ir bolh Eq.7.47 a$d,Es.7.48. rbe consla0lmultiplyinglhe dependent !ariableis the rec;procal of tbe time constant, that is, R/a = 1/r. We encounter a similar situation iII the de vationsfor the step responseof an RC circuit.In Section7.4.we will use these observations to develop a general approach to finding the llatural and step responsesof Ra and RC circuits.

TheStepResponse of an nCCircuit We can find the step responseof a firsforder RC c cuit by analyziry the circuit shownin Fig.7.21.For mathematicalconvenience, we choosethe Norton equivalent of the retwork comected to the equivalett capacitor. SummingrhecuJrenFa\rayfiom tbe lop nodein Fig.r.:t generiresrhe differcntial equation Figure 7.21AA circuit used to iltustnterhe step response ofa first-oder Rrcjrcuit.

,!; ,3 =,"

(7.49)

Division of Eq.7.49by C gives

dt

-

RC=V

(7.50)

CompadngEq. 7.50vrith Eq.7.48rcvealsthat the form of the solutionfor oc is the same as that for the curent in the inductive circuit. namelv. Fq. 7.15.Tberelore. b) simpl)substitlrling rhe appropriare variabtes anl coefficients, we can wdte the solutiotr for ?c directly. The translation requircs that 1, replace 14,C replace a, 1/R reptace R, and y0 replace 10. We get

Stepresponse of anfC circuit>

(7.51)

A similar derivation for the curent in the capacitor yields the differential equation d dt

i

l RC'

^ ''

\7.52)

Equation ?.52has the sameform as Eq. ?-47,hencethe solution for j is obtained by usjng the same translations used for the solution of Eq.7.50.Thus

i-(r.-\\t,'^",,-6.,

\7.53)

where y0 is the initial value of oc, the voltage acrossthe capacitor. We obtainedEqs.7.51and 7.53by usinga mathematicalamlogy to the solution for the step rcsponse of the inductive circuit. Let's see whether these solutions for the RC circuit make sensein tems of known circuit behavior.From Eq.7.51, note tlat the initial voltage affoss the capacitoris yo,the final voltageacrossthe capacitoris IrR, and t]te time

ofil andRCCncuits 247 7.3 TheStepR€sponse

constantof the circuit is RC.Also note that the solutionfor oc is valid for t > 0. fhese obse ationsare consistentwith the behaviorof a capacitor in parallelwith a resistorwhendrivenby a constantcurrentsource. Equation7.53predictsthat the curent in the capacitorat t - 0- is 1" - yo/R.Thispredictionmakessense because thecapacitor voltagecannot changehstantaneously, andthereforethe initial curent in the rcsistor is yo/R,The capacitorbranchcurent changeshstantaneouslyftom zero att = 0 tols - yo/Rat I : 0+.Thecapacitorcuirentis zeroat I : l)o. Also rote that the final valueof 1, = 1"R. Example?.6illustrates howto useEqs.7.51and7.53to fhd the step resDonse of a first-orderRC circuit.

Determining the StepResponse of annCCircuit The switch in t}Ie circuit shown in Fig.7.22 has been in position 1 for a long time. At t = 0, t})e switch moves to position 2. Find a) 0.(t) for / > 0and b) i"(r) for t > 0*.

The value of the Norton current sourceis the ratio of the open-circuit voltage to the Th6venin resistance,or 60/(40 x 10) : -1.5 mA. The resullrng Norronequivalenlcifcuit is sboqn in Fig. 7.23. From Fig.'7.23, I,R = -60V and RC:10ms. We have already noted tiat o,(0) : 30 V, so the solutionfor z'. is .t). = =

Figule 7.22A lhecircuiitor Exampte 7.6.

vo,

io :

160x 103

(-7s) = -60 v (40+ 160)< 103

( 60)]rlm,

60 + 90e-1mrv, ,>0.

b) We write the solution for i, directly from and Eq. 7.53 by noting -that 1, = -15mA : (30/40) x 0.7s mA: 10-r, or U/R

Solution a) The switch has been in position 1 for a lorg time, so the initial value of ?, is 40(60/80),or 30 V To take advantageof Eqs.7.51and 7.53,wefind the Norton equivalent with respect to the terminals of the capacitorfoi I > 0.To do this,we beginby computing the open'circuit voltage, which is given by the 75 V source divided across the 40kO and 160kO resistors:

60+ [30

2.25ermt mA, t > o+.

We check the consistencyof the solutions foi o, and L by noting that t,,

C?

(0.25

t11

=

l0 6)(-o000er'"',

2.25elau mA.

Becausedp.,(D-)/dt:0, the expressionfor i, clearly is valid only for t > 0-.

Ner1, we calculate the fh6venin iesistance, as seento the right of the capacitor, by shoning t}le 75V source ard making sedes ard pamllel combinatioN of tlle resistors: RTl' : 8000 + 40,000lL160,000= 40 kO

Figlre 7.23 a Theeqoivatent circuiifort > 0 forthecircuit s h o wjnn F j g . 7 . 2 2 .

248

Response of Fnsi,order Rl andffCircuih

objective2-8e abteto determine the stepresponse of bothnl andfC circuits 7.6

a) Find the expressionfor the voltagcacross the 160kO resistorin the circuirshownm Fig.7.22.Let this volragebe denoleuui, anu assumethat the reference polarity for the voltageis posjtivea1the upper termillalof rhe 160kf) resistor.

b) Specifythe interval of time for which the expressionobtainedin (a) is valid. Answer: (a) -60 + 72e 1t)rtyl ( b )I > 0 ' .

NOTE: Also try ChaptetPtublems7.50und 7.51.

7.4 A SeneralSolutiog'l for $tep ftndNsturelResponses arr , - l )

V* .J /G\

G)

The generalapproachLo finding cither the natural responseor the step responseof the first order Rl, and RC circuitssbownin Fig.7.24is based on their differential equationshaving the sanrclorm (compareEq.7.48 and Eq.7.50).Togeneralizethe solutionofthesc lour possiblecircuits,we let r(r) representthe unknown quanriry,givingr(,) four possiblevatues.Ir canrepresentthe currentor voltageat the lcrminalsof an inductoror rhe current or voltage at the terminalsof a capacitor.From Eqs.7.47.7.4tj. 7.50.aDd7.52.wc knolv that thc differentialcquariondescribingany one of rhe four circuilsin Fig.7.24rakesthe fornr

i/p L

\7.51)

dt-; (b)

wherethe value ofthe constant1( canbe zero.Becauscthe sourcesin rhe circuit are conslantvoltagcsaDd/orcurrcnls,the final value of .r will be conslantlthal js,Lhefinal valuemust satisfyEq. 7.54,and.rvhenx reaches its firralvaluc.lhe derivativedr/dl must be zero.Hencc r, = Kr. where.r/ represeDts the final valueofthe variable. We soivcEq.7.54by separatingthevariabtes,beginnil1gby solvingtor thc first derivative:

frr

(x

..

Figure7.24a Fourpo$jbhfirt,ordercircuits. connected to a Th6venin equivatent. {a)Anjnductor (b)Aninductorconnected to a Nodonequivateni. connected to a Th6venjn equivateni. G)A capacjtor (d)A capacitor connected to a No,tonequjvahnt.

Kr)

-1x

-r1) (1.56)

In w ling Eq.7.56,we usedEq. 7.55to subsriruter, for Kr. We now mul tiply both sidesofEq.7.56 by dr and divideby.r .r/ to olrtain

o'

:1at.

17.51)

7.4

A GenentSotuiion forSteDandNaturat ResDon5es249

Next,we integrateEq. 7.57.To obtair asgenerala solutionaspossible,we usetime lo asthe lorqerlimit and I asthe upperlimit. Time t0 corresponds to the time of the switchingor other change.Previouslywe assumedthat r0 : 0, but t}Iis changeallows the switchingto take place af any time. Using& and o asslanbolsof integration,we get

f't') a"

t J'1';u-t7

1' f

r du. 1J"

(7.58)

Canying out the integiation called for in Eq.7.58 gives

(7.59) < ceneralsolutionfor naturalandsteo resDonses of f[ andfC cirauits The importanc€of Eq.7.59becomesappaientif we write it out in words: the unknown the final variable as a - value of the variable function of time tlre idtial thefinaLl value of the - vaiueof rhe x I variable variablel

€

rlne$tulrd)

(7.60)

In many cases,the time of switching that is, t0 is zero, W}len computing the step and natural rcsponses of circuits, it may help to follow thesesteps: 1. Identify the va able of interest for t}le circuit.For RCcircuits,it is most convenient to choose the capacitive voltage;for -Ra circuits, it is best to chooset}le inductive cuffent 2. Detemine tlle initial value of the variable, which is its value at 10. Note that if you choosecapacitive voltage or inductive current as your variable of interest, it is not recessary to distinguish berween t : l; and I : td.2This is becausethey both are continuous variables.If you choose another variable, you need to remember that its initial value is defined at , = 16. 3. Calculatetle fhal value of the variable,which is its value asI - m. 4. Calculate the time constant for the circuit. With these quantitieE you can use Eq. 7.60 to produce an equation describing the variable of interest as a function of time. You can then find equations lor other circuit variables using the circuit analysis techniques introduced in Chapters 3 alld 4 or by rcpeating the preceding stepsfor the other vadables. Examples 7.7-7.9 illustrate how to use Eq. 7.60 to find the step respoff e of an -RCor Rl, circuit. 'Tne expressions r; andrf areanalogous to (' and0'. ttus r(rt) is the linit of r(r) asr + ,0 lron the left, andrG is the linit of r(t) ast - rofron the rieh!.

< Calcutating the naturalor steprcsponse of Rl or nCcircuits

250

Response of Fi6t order8r andfC (i(uits

Usingthe General SolutionMethodto Findannf CircuifsStepResponse The switchin the circuit showr in Fig.7.25hasbeen in position a for a long time. At r = 0 the switch is moved to position b. a) What is the initial valueof 0c? b) What is the final valueof 0c? c) What is the time constantofthe circuit when the switchis in positionb? d) Wlat is the expressionfor z,c(r) when I > 0? e) Wrat is the expressionfor i(r) when r > 0+? t) How long after the switch is in position b does the capacitorvoltageequalzero? g) Plot "c(1) and i(r) venus r.

Solution a) The switch has been in position a for a long time, so the capacitor looks lite an open circuit. Thercfore the voltage acioss the capacitor is the voltage acmsstle 60 (} resistor.From tie voltage, divider rule, the voltageacrossthe 60 O resistor is a0 x [60/(60 + 20)], or 30 v As rhe refer ence for oc is positiveat the upper lerminal of the capacitor,wehavez,c(o): 30 V. b) After the switchhasbeenin positionb for a long time,the capacitorwill look like an open circuit in terms of the 90 V source.Thus the final value ofthe capacitoivoltageis + 90V. c) The time constantis T:RC

= (400x 103x0.5 x 10{) = 0.2s. d) Substitutingthe appropriatevaluesfor z/, t'(0), and r irto Eq.7.60yields 0c(t)=90+( 30- 90)e{' = 90 120ts'V, I > 0. e) Here the value for r doesn't chaflge.Thus we need to find only the initial and final values for the curent in the capacitor.When obtainingthe initial value, we must get t}le value of t(0+), becausethe currett in the capacitorcan change instantaneously. This current is equal to the current in the iesistor, which from Ohm,s law is [90 ( 30)]/(400 x 103)= 300/,A. Nore that whenapplyingOhm'slawwe recognizedrharthe

Figure7.25A Thecircuit forExample 7.7. capacitorvoltagecannotchangeinstantaneously. The fiMl value of i(t) - 0, so i0)-0+(300-0)e-5. = 300e-srpA, r>0+. We could have obtahed this solution by diffbrentiating tlle solution in (d) and mulriplying by the capacitance.You may want to do so for yourseli Note that this altemative approachto finding r(r)alsopfedrclc lhedisconlinu] dl I = 0. f) To find how long the switch musr be in position b before the capacitorvoltage becomeszero, we solve the equation derived in (d) for rhe rime when 0c(t) : 0: 120t5r = 90 or

-

,.

720 90'

, =1r,-/\43\ / = 57.54ms. Note that when ,c = 0, i = 225pA and the voltagedropacross the400kO resistoris 90V g) Figure7.26showsthe graphsof oco) and i0) , (i.A) t)cry) 300 250 200 150 100 50

Figure7.26a Thedrr€ntandvottage waveforms tor Exanrpte 7.7.

7.4 A CeneratSotutjon torStepandNaturalResponses 251

Usingthe General SolutionMethodwith ZeroInitiaLConditions The switch in t]le circuit shown in Fig.7.27 has been open for a long time. The initial charge on the capacitor is zero. At 1 = 0, the swirch is closed.Find the expressior for

Substitutingtlese valuesirto Eq.7.60genemtes

(, =o+(3 -o\ett5tn: a) i(r) for t > 0- and

: 3 e 2 o o ' m Ai,> 0 + . b) ,0) when t > 0-.

30ko

b) To find ?,(1),we note ftom the circuit that it equalsthe sum of the voltage acrossthe capacitor and the voltageacrossthe 30 kO resistor.To find the capacilor volrage (which is a drop in the direcrion of the current), we note that its inirial value is zero and its final value is (7.5)(20),or 150VThe time constantis the same as before, or 5 ms. Therefore we use Eq. 7.60 to wnte

Figure7.27A Thecircujttor ExampLe 7.8.

z'c(l)- 150+ (0

SoLution

150)e'mr

= (150- 150e,oo,)v,

a) Becausethe initial voltage on the capacitoris zero,at the instantwhen the switchis closedthe currentin the 30 kO branchwill be

(7.5X20) (0) =

I > 0.

Hencethe expression for thevoltagez'(,)is z(t) = 153- lsOeioo'+ (30x3)eioo'

50

: 050

6oa'rir) v,

I > 0*.

= 3 mA.

The final value of the capacitor current will be zero because the capacitor eventually will appear as an open circuit in terms of dc cwent. Thus if : 0. The time constant of the circuit will equal the product of the Th6venin resistance(as seen ftom the capacitor) and the capacitance. Thercforer = (20 + 30)10'(0.1)x 10 " : 5ms.

As one checkon this expression, note that rt predictsthe hitial valueof the voltageacross the 20O resistoras 150 60, or 90 V The instant t}re switch is closed,the curent in the 20ko resisroris (7.5)(30/50), or 4.5 nA. This curent producesa 90V drop acrossthe 20kl) resistor,confirming the value predictedby the solutiofl,

252

andRCCircuits Response 0f First-0rderft

SolutionMethodto Findan RLCircuifsStepResponse Usingthe General The switchin the circuit shownin Fig.7.28hasbeen open for a long time.At I : 0 the switchis closed. Find the expressionfor

constantis 80/1,or 80 ms.we useEq. 7.60to foi 0(t): wite theexpression oO=0+(15

o)e'l3orrr'

a) "0) whenI > 0- and : 15t12stV,

I > 0+.

b) i(t) when r > 0. b) We have alreadynoted that tlle initial value of the inductor cuirent is 5 A. After the switchhas been closedfor a loDgtime, the inductor current reaches20/1,or 20A. The circuittime constantis 80 mq so the expressionlor i(t) is

i(t) : 20 + (s - 20)e r25t

= (20 - 15?-125,) A,

r > 0.

7.9. Figure7.28 A ThecjrcujtforExampLe

We determine t}lal the solutions for t'(l) and i(l) agree by noting that

Solution a) The switch has been open for a long time, so the initial current i]l th.3 inductor is 5 A, onented lrom top to bottom. Immediately after the swrtch closes,the current still is 5 A, and tierefore the initial voltage across the inductor bemmes 20 5(1),or 15V The fhal valueofthe inductor voltageis 0 V WitI the switch closed,the time

NOTE:

1' 5i] :80 x 10 3[15(12.5)e : 15e l'r'v,

r > ot.

Assessyour ande$anding of the seneral solution method b) trrng Chapter Problems 7 53 a d 7-54-

Example7.10showsthat Eq.7.60 can even be used lo find th€ step responseof somecircuitscontainingmagneticallycoupledcoils

7.4

A GenentSotution forStepandNaturaL Responses253

Determining StepResponse of a Circuitwith Magnetical.ty Coupted Coils There is no energystoredin the circuit in Fig.7.29 at the time the switchis closed. a) Find the solutions for i., u., ir and i2. b) Show that the solutions obtained in (a) make sensein termsoI known circuit behavior,

'.3 H

120V

Figure7.29a Thecircuit forExampte 7.10.

Solution a) For the circuit in Fig.7.29,the magneticallycoupled coils can be replacedby a singleinductor having an inductance of LrL2 - Ml Lr+ L2- 2M (SeeProblem6.41.)It follows thal the circuit in Fig.7.29canbe simplifiedasshownin Fig.7.30. By hypothesisOleinitial valueof i, is zero. From Fig. 7.30we seethat the final value of i. will be 120/7.5or 16A. The time constantof the circuit is 1.5/7.5or 0.2 s.It follows directly from Eq.7.60that io=16

1 6 e - s t A ,t > 0 .

fhe voltage r,, follows from Kirchloff's voltage law.Thus, D.: 120

7 5i"

- 12(le-5tV.t>O+. To find i1 and i2 we first note from Fig.7.29that ^air

.diz

. dir

dt

at

dt

di

i :

.1t

IT

j j

"-diz /11

d;

n,tt-n Therefore

Figure7.30,d ThecjrcuitinFjg.7.29wjththe magneiicaLty coupLed coitsreptaced byanequivatent cojt.

Because i2(0) is zero we have

a=

.ltj

-

because

ale " dx

8+8€s'A,

r>0.

Using Kircbioffs current law we get iI:24 2 4 es 'A , t > 0. b) First we observethat i,(0), ir(0), and ir(0) are all zero, which is consistent with the statement that no energyis stored in the circuit at the instant the switch is closed. Next we obsene r',(0-) = 120V, which is consistentwith the fact that i,(0) = 0. Now we observethe solutions for ir and 4 are consistentwith the solution for o, by observing dt dt '

.lt

follows from Fig. ?.29 tlat + iz,

1.5H

t1l

At

= 36oe st - 24oe sl : 720e-sty, t>O*, di.

u-:6:+15::

di"

: '720e 5t - 600e 5l : 120e-5tv,

t > 0'

254

Response of Fnst-order Rl andnr Cncuitr The final values of 4 and i2 can be checked using flux linlages. The flux linking the 3 H coil (11) must be equal to the flux linldng the 15 H coil (12),because

The tinal valueoflr is

n(co) andthetioalvalueofi is

', : dn'

ir(oo) = The consistencybetween these final values for i1 and i2 and the final value of the flux Linkage can be seenfrom the expressions: 3i1 + 6i2 Wb-turns

^r(co) = 3i,(@) + 6i(,.o)

= 3(A) + 6( 8) : z wb{urm,

6i1 + 15i, Wb-tums. Regardless of which expression we use, we obtain

= 6Qa) + E(

,\1 = i2 : 24 - 24ts' Wb-tums. Note the solution for It or 12 is consistent with the solution for rro. The final value of the flux linking either coil 1 or coil 2 is 24 Wb-tums, that iE I(co) = r,(oo) : 24 wb-tums.

^,(o.) : 6i1(oo)+ 15ir(,tro) 8) : 24 wb-tums.

It is worth noling that the final values of 11 ard i2 can only be checked via flu\ linlage becauseat t : oo the two coils are ideal short circuits. The division of curent between ideal short circuits cannot be found from Ohm's law.

NOTE: Assessyour understanding of this material bt using the genersl solution method to sotve Chaptet froDlems /.ol and /.o /.

7.5 Sequentiat Switching Whenever switching occuN more that once in a circuit, we have sequential switching.For example,a single,two-position switch may be switched back and fo ]\ or multiple switchesmay be opened or closed in sequerce.The time reference for all switchings cannot be I : 0. We detemine tlle voltages and cu ents generated by a switching sequenceby using tle techniques described prcviously in this chapter. We derive the er:pressionsfor o(t) and i(, for a given position of the switch or switches and rhen use these solutions to determine the initial conditions for the next position of the switah or switches. With sequential switching problems,a premium is placed on obtaining the inilial value r0o). Recall that an)'thing but inductive curents and capacitive voltages can change instantaneously at the time of switching. Thus solving fi$t for itductive currents and capacitive voltages is even more pertinent in seque ial switching problems. Drawing the circuit that pefiains to each time irterval in such a problem is often helpful in the Effmples 7.11 and 7.12 illustrate the analysis techniques for circuits with sequential svritching.The fircl is a natual responseproblem with two switching times,and the secondis a step responseproblem.

SequentjatSwjtchjng 255

7.5

Analyzing an ff Circuitthat hasSequential. Switching The two switches in the circuit shown in Fig. 7.31 havebeenclosedfor a long time.At r - 0, switch1 is opened.Then,35ms later,switch2 is opened. a) Find ir(l) for 0 < , < 35 ms. b) Find iz for t > 35 ms. c) Wlat percentage of the initial energy stored in the 150 mH inductor is dissipatedin the 18 O r€sistor? d) Repeat(c) for the 3 O resistor. e) Repeat(c) lor the 6 O resistor.

F i g u r € 7 . 3 2 l^t e L i ' L i . s h o w l i n - i g . 7 . l l . r o '0r .'

150mH

,:35ms

40 \ )60v

-1i-

18()

Figure 7.33A Thecncujtshown in Fig.7.31, for0 < r =35 ms.

3r} 2

: 1 , 2 A 1 6ll

tt. 150inH

18()

b) When t:35ms,

the value of the inductor

ir = 6e rL = 1.48A Figure7.31 A Thecircujtfor Exampte 7.11.

Thus, when switch 2 is opened, the circuit rcducesto the one shown in Fig. 7.34,and the time constant changesto (150/9) X 10-3, or 16.67ms The expressionfor i. becomes

Solution

oo3s) iL : 1.48e4411 A,

a) For t < 0 both switches are closed, causing the 150nH induclor ro shorr-flrcurlthe l8 O resistoi. The equivalent circuit is shown in Eg. 7.32.We determine the initial current in the inductor by solving for ir(o-) ir the circuit showr in Fig.7.32. AJter making several soutc€ transformations,we find 4_(0-) to be 6 A. For 0 < t < 35 ms, switch 1 is open (switch 2 is closed),which disconnectsthe 60 V voltage souice and the 4 O and 12 O resis' torc from the circuit. The inductor is no longer behaving as a short circuit (becausethe dc source is no longer in the circuit), so t}Ie 18 O rcsistor is no lorger shortcircuited. The equivalent circuit is shown in Fig.7.33.Note tlat t}le equivalent rcsistance across the teminals of the inductor is the parallel combhation of 9 O and 18 O, or 6 O. The time constantof the circuit is (150/6) x 10 3, or 25 ms,Theieforc the expressionfor ir is

r > 35 ms.

Note that the exponentialfunction is shiftedin time by 35 ms.

150mH l, (0.035)= 1.48A

tigure7.34A Thecjrcuitshown in Fig.7.31, forr > 35ms. c) The 18 O resistor is in the circuit only during the first 35 ms of the switching sequence.During tlfs intelval, tle voltage auoss the rcsistor is ) u, = o.-ts:(6Pro') tll

iL = 6e Nt A,

0 < , < 35 ms.

=-36eNtY,

0
256

Response of Fi'st-o'der8land8r CkcLriis

fhe powerdissipated in the 18I) r€sistoris 'D

.'? -2-j2?

3 0\ \ .

Iti

o

Hence the energydissipatedin the 3 O resistorfoi I > 35msis

r. ismr. [:,

Henc€ the energy dissipated is

fi,

r0,035 '72e!o'dt I .lo

w:

""

3(36)e2 3e 12o\t o o35)dt

lo.or:

,o'

i4o/, 3,tt

N --120(r-0035) 1^ rzu l00rs

: 1 0 8 P' 3

lo

: 0.9(1 e{)

t0a __

- 845.27nL Theinitial energystoredh the 150mH inductoris

The total energydissipatedin the 3 O resistoris : 563.51+ 54.73 ?{)3o(otal)

' r', : -){ 0 . 1 5 ) { 3=6 )2 . 7J : 2 7 0 0 m J .

: 618.24 mJ. Theretore (845.2'7 /2'700)x 100, or 31.31ol"of the initial eneryystoredin the 150mH inductor is dissipated in the 18o resistor. d) For 0 < t < 35ms, the voltageacrossthe 3 O

The percentageof the initial energystoredis 6 tJl )/

== tA t

leslstof1s I uL\.^.

,r!!=l;l(r)

\ r I 1 =

u3tt -

e) Becausethe 6 O resistoris in serieswith the 3 O resistor,the energydissipatedand the percentageof the initial energystored$rillbe twice that ofthe3Orcsistor u6o(rotal) = 1236.48mJ,

12e"Y.

Thefefore lheenergy dissipated in lheI O resis tor in the fiNt 35msis *, /o.n"t++" -----dr _ | r ,n\

:0.6(1

? ' 3)

- 563.51 mJ.

lou : 22.90",".

and the p€rcentageof the initial energystoredis 45.80%.Wecheckthesecalculationsby observing that 1236.48+ 618.24 + a45.27 : 2699.99ml and 31.31+ 22.90+ 45.80:100.010/..

For t > 35ms,the cunent iII the 3 O resistoris 1a)e{o('oo3s) A. ha= iL= (6e

The small disqepanci€s in the summations are the result of ioundoff errors,

7.5 Sequ€ntiat Switching 257

Anal.yzing anRf Circuitthat hasSequentiat Switching The uncharyed capacitor in the circuit showt in Fig. 7.35 is iritialy switched to terminal a of the throe-positionswitch.At t = 0, the switchis moved to position b, where it rcmains lor 15 ms.After the 15 ms delay,the switch is moved to position c, where it remains indefinitely. a) Derive tle numerical expression for the voltage acrossthe capacitor, b) Plot the capacilor voltage versus time.

100ko b

Figure7.35a lhecircujt fortxampte 7.12.

c) When will the voltage on the capacitor equal

200v? Solution a) At the instant the switch is moved 10 position b, the initial voltage on the capacitor is zero. If tlle switch were to remain in position b, the capacitor would eventually charge to 400 V The time constant of the circuit when the switch is in position b is 10 m$ Therefore we can use Eq. 7.59 with t0 : 0 to write the expression for the capacitor vorEge: u:400

+ (0

: (400

In wiiting the expression for o, we recognized that lo : 15 ms and tlat this expression is valid only for I > 15 ms. b) Figure 7.36 showsthe plot of o versus r. c) The plot in Fig. 7.36 rcveals that the capacitor voltage will equal 200 V at two different times: once in the inte al between 0 and 15 ms and once after 15 ms.We find the fiISl time by solving

200=400-400tlmr',

400)ermr

400erm) v,

0 < , < 15 ms.

Note thal. becauserhe swirch remainsin position b for only 15 ms,this expressionis valid or y for t]le time interval from 0 to 15 ms.Aftcr ure switch has beer in this position for 15 ms, the voltage on the capacitor will be ?r(15mt = 400 - 400e-rs: 310.75V. Therefore,when the switch is moved to position c, the initial voltage on the capacitor is 310.75V Wit}l the switch in position c, the final value of the capacitor volaageis zero, and the time constantis 5 m$Again, we useEq.7.59to write the er?ression for the capacitor voltage:

which yields 11: 6.93ms. We find the second time by solving the expression -oors) 200 : 31O;75etno(t In this case,t2: 17.20ms.

r,(v) 300 200

/:310.75.

ao(/ od:J

100

?, = 0 + (310.75- o)€ 20r(!ools) - 310.75?-200(r-0015) V, 15ms < r.

Flgure 7.35^ Thecapacitor votiage fortxampte 7.12.

258

Response of Fifst-Order nl andRCCjrcujis

objective3-Know howto anatyze circuitswith sequential switching 7.7

In the circuit sho$'n.switchI hasbeenclosed r n d . $ i r c h) h ,\ . . c n o p c l o r : rl o ' r gl i r n eA .l t - 0. s$,itch1 ls opcncd.Ther l0 ms later, switch2 is closcd.Find

S$ir(hJ iI rheci-cuir.h^qn hl. bccnop 1s.

7,8

a) r.(r) for 0 < r < 0.01s. b) ,tJ.(/)for r = 0.01s. c) dre total cncrgydissipatedin the 25 kQ d) rhe lolal energydissipatedin thc 100ko

20 AFFrI

b

- 1

0.8o

9ll

l f ok ( l

\

i3o

r:(l 2H

ii 3o

60

Answer: (a) 80.rrrrlv; oorrVl (b) 53.63e5o1r Answer: (a) (3 3c t'5')A,0 < , < 1s; (b) (-4.8 + 5.98r r'25(/r) A. r > I s.

(c) 2.e1nJ; (d) 0.2enrJ. NOTE: Aha iry ChapterProbl ts7.72and7.76.

7.6 i,inbcu*ti{:d RrsFs*se A circuil responsemay grox'.rather lhan decay.exponentiallywith timc. This ttpe of responsc. calledan unbomded response.is possiblcif rhe cir cuit containsdcpcndcn! soulces.Ir that case,the Th6vcnin cquivalenl resistance with rcspccl10the lerminalsofeither an inductoror acapacilor nav be negaiivc.This negadveresistancegeneratcsa ncgativctine con stant.and thc rcsullingcurerts and voltagesincreascwithoul limi1.Tn aD actual circuit. thc rcsponseeventuallyreachesa limiting valuc whcn a componentbreaksdo\\n or. goesinto a saturationstate,prohibiling rur ther increasesin voltagcor cult'enl. wlren we considcrunboundedresponses, the conceptof a tinal laluc is confusirg.Hcncc,ralher lhan usingthe step responsosolutiongi!en in Eq.7.59.wedcrivc lhe differentialequationthat describcsthc circuil con, taining thc negalile resistanceand then solvc it using the separariorof variablcs lcchnique.Example 7.13 prcscnts an exponenlialll,growing responscin termsoI lhe vollageacrossa capacitor-

Response 259 Unbounded

7.6

Response in an Rccircuit Findingthe Unbounded a) When the switch is closed in the circuit shown m Fig.7.37.lhe \ollage on the capaciloris l0 V Find the expressionfor ?),for t > 0. b) Assume that the capacitor short-cilcuits when its terminal voltage reaches 150 V How many millisecondselapsebefore the capacitorshorG circuits?

metfod r e d t of r d R n F i g u r €7 . 3 8^ T f er e s r - l o i ( e

-5 ko 7.13. Figure7.37 l Th€crrcujtfor Exampt€

jn ofthecjrcujt shown Fig!rc7.39l A simptjfication fig.7.31.

Sotution a) To find the Th6venin equivalent resistance with respectto the capacitor teminals, we use the tes! souce method describedin Chapter 4-Figure 7 38 shows the {esulting circuit, where ?,r is the test voltage ald ir is the test current. For th expressed in volts,we obtail .

rr:

UT

10

- -/l )rL2 0 t +

1)T

,

For , > 0, the ditferential equation describing the circuit shownin Fig.7.39is

_3 15 1s_oyy_:g 111

,a

r0_r: o.

Dividing by the coefficient of the first derivative yields

2-mA.

4

Sol\ingfof tbe rario ,r"/ir yieldsthe fie\enin

Rrr, :

5

= -5 kO

d

t

,'r,u^=o. "

We now usethe separationof variablestechnique to find 0"(r): q(t):10e4utY,

with this Th6veniil- resistance,we car simplify lo lheone\hownin thecircuirsho\rnin Fig.7.37 Fig.7.39.

t > 0.

b) o, : 150V when?au'- 15.Therefore,4ot= ln 15, and/ = 67.70ms.

NOTE: AssessyoLr understandingof thk material by trying Chapter Probkms 7 86 and 7.87.

The fact that hterconnected circuit elements may lead to everhcreasing currents and voltages is impofant to engineels. If such inter: connections are unhtended, the resulting ckcuit may experienc€ unexpected,and potentially dangerous,component failures'

260

Response of Fi6t-0rder fi andRrCircuiis

7.7 TheIntegratingAmplifier

tigurc7.40e Animeg,aLing anplifier.

Recall from the introduction to Chapter 5 tlat one reason for our interest in t})e operational ampMier is its use as an integrating amplifier. We are now ready to analyze an integrating-ampiifier circuit, which is shown in Fig. 7.40. The purpose of such a circuit is to generate an output voltage proportionalto the integal of the input voltage.InFig.7.40,we addedthe branch curents it and i,, along with the node voltages o, and o?, to aid our analysis. We assumetlat the operational amplifier is ideal- Thus we take advantageof the constraints \7.61) \7.62)

Becauseu? : 0. (7.63)

dD^

(7.64)

Hence,from Eqs.7.61,7.63, and'7.64,

du.

n:

1

R{r,,"

\7.65)

Multipling both sidesof Eq.7.65 by a differentialtime dl and thetrintegrating from lo to I generatesthe equation 1 R.

t"'

,,r d) + ?,o(10).

(7.66)

In Eq. 7.66,t0representstlle instantin time whenii'ebeginthe inte$ation. Thus ?r,(to)is the value of the output voltage at that time. Also, because bn: tp: 0, t"00) is identical to the idtial volrage on tlre feedback caDacitorC,, Equation 7.66 states that the ouQut voltage of an inregrating amplifier equalsthe initial value of the voltage on the capacitor plus an inverted (minus sign), scaled (1/&Cl) replica of tle inregral of the input voltage. ff no energyis storedin the capacitorwhenintegation commences, Eq.7.66

ol0 -

1 l nq J,"o'ot

\7.67)

7.7

TheIntegBiingAmptifier 261

If 0Jis a step changeifl a dc voltage level, the output voltage will vary linearly with time. For example, assumetiat the input voltage is the rectangular voltagepulseshownin Fig.7.41.Assume alsothat the initial valueof ,"(1) is zero at the hstant or steps from 0 to y,. A direct application of Eo.7.66vields

."=

l-,v^t

+ 0, 0
(7.6s) tigure7.41,i, Aninputvothgesjgnal.

W})en t Liesbetween 11ar'd2t1,

-#,'*' #1"'"-'* u^ fi"cr

'

2uR,Cr'"

t1
(7.6e)

Figuie 7.42 shows a sketch of o,(t) versus r. Clearly, the output voltage is an inverted, scaledreplica oI the integal of the input voltageThe output voltageis propo ional to the integralof the input voltage only ifthe op amp operateswithin its linear range,that iE i{ it doesn'tsa! urate.Examples7.14 and 7.15 fuitler i ustratethe analysisof the inte- Figure7.42A Theoutputvottage of anjniegcting gratn€ amplfier.

Analyzing an IntegratingAnplifier Assume that the numerical values for the signal voltage shown in Fig. 7.41 are y-:50mV and 11: 1s. This signal voltage is applied to the integrating-amplifier circuit sho\\.n in Fig. 7.40.The circuit parametersof the amplifier are R" = 100kO, Cf = 0.1 pF, and Vcc : 6 V. The initial voltage on the capacitor is zerc.

Forl
a) CalculatezJ,(t). b) Plot r',0) venus ,.

Solution a)For0
(1oox 10)(0.1x 1oi) =

5l V,

0
50x103r+0 Figure7.43 A Theoutputvottage for Exampte 7.14.

262

Response of Fnst'0rder il andRCCircuits

Analyzing an IntegratingAmptifierthat hasSequentialSwitching At the instant th€ switch makes cofltact with terminal a in th€ circuitshownin Fi9.7.44,the voltageon the 0.1/rF capacitoris 5 V The switch remainsat terminala for 9 ms and then movesinstantaneously to terminal b. How many millisecondsafter making contactwith terminalb doesthe opemtionalampli fier saturate?

Thus,9 msaftertheswitchmakescontactwith termiDala,theoutputvoltageis 5 + 9. or 4V Theexpression for theoutputvoltageafterthe switchmovesto terminalb is

D' ^ : 4

.

+ 5V

-4

0.1pF

I

t, ^.

nd\J I t0 . ../,r ^ o.

-

800(1 9x10-3)

: (11.2- 800' V. Duringthistimeinterval,thevoltageis deoeas ing,and the operational amplifiereventuallysatu mtesat 6 V. Thereforewe setthe expression for oo equalro 6 \ lo obldinrheraruradon rimer_

tigure 7.44 A Thecjrcuitfor Examph 7.15.

11.2

Solution

800r, : -6,

The expression foi the output voltage during time the switchis at terminala is

,": -s- ft;l'etota:, = (-s + 1000tv.

t' = 215 ms Thus the integrating amplifier saturates21.5 ms after makirg contact with terminal b.

From the examples,we seethat t}le integrating amplifier can perform the integration function very well, but oniy within specified limits that avoid salurating the op amp. The op amp saturates due to the accumulation of chargeon the feedbackcapacitor.We canpreventit from satuat ing by placing a rcsistor in parallel with the feedback capacitor.We examine such a circuit in Chapter 8. Note that we can convert the integrating amplifier to a differentiating amplifierby interchangingthe input resistanceR" and the feedbackcapacitor Cr.Then tln" u. = -R'Ct clt

\7.70)

We leavethe derivationof Eq.7.70asan exercisefor you.The differentiating amplifieris seldomusedbecausein practiceit is a sourceofunwanted or noisysignals, Finally, we can desigr botl integrating- and differentiating-amplifier circuitsby usingan inductor insteadof a capacitor.However,fabricating capaciton for inlegmted-circuitdevicesis much easier,so inductorsare rarely used in integrating amplifiers.

Pa.trct Pespe.tive 263

obiective4-Be abteto analyze op ampcircuitscontaining resistors anda singtecapacitor 7.9

Thereis no energystoredin rhe capacitorat the lime the switchin the circuit nakes contact wilh termitul a.The switchrcnrainsaLposition a lor 32 ms and then movcsjnstantaneously to positionb. How nany millisecondsafter making contactwith terminal a doesthe op amp

0-2r.F

7.10 a) Whcn the switchslosesin the circuir shown.there is no energystoredin thc capacitor.How long doesit rakc to saturale lhe op amp? b) Repeat(a) with an initialvollage on the capacitorof l Ypositivc at the upper lermlnal. t0 ko

.10ko

0.8to

Answer:262ms-

Answer:(a) 1.l1ms; (b) 1.76ns.

NOTE: Aho rry Cha?@ Ptublems7.92and 7.93.

PracticaI Perspective A Ftashing LightCircuit Wearenowreadyto anatyze the ftashjnglight cjrcuitintroduced at the stat of thjs chapt€rand shownin Fig.7.45. Thetampjn this circuitstats to conductwhenever the lampvoltagereaches a valuey-i,. Durjngthe time the larnpconducts, it canbe modeledas a resistorwhoseresistance is X,. TheLampwiL[contjnueto corductuntiIthe Lanrp voLtaqe dropsto the value y, i. Whenthe [ampis not conducting, it behaves asan opencjrcuit. Beforewe devetop the analyticaL expressions that descrjbe the behavior Figurc7.45 A fla!hingtight. ,cuit. of the circuit,let us developa feelfor howthe circuitworksby notingthe fotlowing.Fi6t, whenthe lampb€haves as an opencjrcuit,the dc voLtage sourcewiL[chargethe capacitor viathe resistor,l?towarda valueof14voLts. However, oncethe tampvottagereachesy,,a,,it stats condLrctjng andthe capacitorwittstartto discharge towardthe Th6veninvottageseenfromth€ t€rrnjnats of the capacitor.But oncethe capacjtorvottagereaches the cutoff vottageof the lanp (y",i,,),th€ Lampwjlt act as an opencircujtandthe capacitorwit[ stat to recharg€. Thiscycleof chargjrganddischarqing the capacitor is summarized in the sketchshownin Fig.7.46. In drawingFig.7.46we havechoserr = 0 at th€ instantthe capacitor startsto charge.Thetjrner, represents the instantthe Lampstartsto con- Figlre7.46; LampvoLtage versus tine for the duct,andr. is the endof a comptetecycte.WeshouldaLsomentionthat jn c i f c u iitn f i g . 7 . 4 5 .

Respome of Fjrsi-Orderfl andRf Circuih

constructing Fig.7.46we haveassumed the circuithasreachedthe repeti tive stageof its operation.0ur desjgnof the ftashingtjght circuitrequires we deveLop the equationfor r'r(t) asa functjonof V-*, y.,i., %, R, C, and R, for the intervats0 to t, andt, to t". Tobeginthe anaLysis, we assume that the circuithasbeenin operation for a long time. Let t : 0 at the jnstantwhenthe Lampstopsconducting. Thus,at / = 0, the tampjs modeled asan opencircuit,andthe voLtage drop acrossthe tampjs yni,,,as shownin Fig.7.47. Frofirthe circuit,wefind Dr(c') = %' r,.(0) : ynh,

Egure7.47A Theftashing tjghtcircuit att = 0, when ihetamp is notconducting.

r _ RC, Thus,whenthe Lamp is not conductrng, "r(t) : 14 + (Y-r"

he-iac.

Howlongdoesit takebeforethe Lampis readyto conduct? Wecanfind this tjme by settingthe expression for or(r) equaL to y."" andsoLvjng for t. If we callthis vatuet., then ^^-

''

RL

'-

U-it

\

t -""

{'

Whenthe [ampbeginsconductjng, it canbe modeLed asa resistance Rz, as seenin Fig.7.48. In orderto find the expression for the vottagedrop acrossthe capacitorjn this circuit,we needto find the Th€venin equivatent as seenby the capacitor. Weleaveto you to show in Problem7.106,that whenthe Lampis conducting,

Figure7.48A Ihe ftashjng tjghtcjrcuiiai, = ro, whenthelampis conductjng.

DL(.t)= vrh+ (y.*

yrh)e ('-,t"

and

'

RRLC

n+R/'

Wecandetermine howlongth€ lampconductsby settjngthe aboveexpres'o' s on lor r/ (/) to l-_. andso-ving tr. r,J,giving RR1C . %'^ - v,. 'n y.n yrh R Rr your understanding NqTE: Assess ofthis PradicdLPerspediveby ttying Chdpter Prcblems 7.103-/,105.

265

5ummary A iirst'order circuit may be reducedto a Th6venin(or Norton) equivale.rtconrectedto either a singleequivalent inductoror capacitor.(Seepage230.) The natursl r€sponse is the currents and voltages that exist when stored energy is releasedto a circuit that containsno independentsources.(Seepage228.) The time conslant of an Xa circuit equals the equivalert inductance divided by the Thevenin resistalc€ as viewed from t1leterminalsof the equivalentinductor. (Seepage232.) The tim€ constentof an RC circuit equalsthe equivalent capacitancetimes the Thdvenin resistance as viewedfrom the terminalsof the equivalentcapacitor. (Seepase237.) The step r€spons€ is the cuuents and voltages that resultfiom abrupt changesin dc sourcesconnectedto a circuit.Storedenergymay or may not be presentat the time the abrupt changestake place.(Seepage240.)

The solutionfor either the natural or step responseof boti R-Land RC circuitsinvolvesfinding the initial and final value of the curent or voltage of interest and the time constantof the circuit. Equations 7.59 and 7.60 summarizetlis approach.(Seepage249.) Sequential switching in first-order circuits is analyzed by dividing the analysisinto time intervalscorrespon ding to specificswitchpositions.Initial valuesfor a particular interval are determined ftom the solution correspondingto the immediatelyprecedjnginte al. (Seepage25,1.) An unbound€d r€sponseoccurs when the Th6venir resistanceis negative, which is possible when the first order circuit contains dependent sources.(See page258.) An integratingamplifier consistsof an ideal op amp,a capacitorin the negativefeedbackbianch,and a resis, tor in serieswith the signalsource.It outputs the inte gral of the signal source,within specifiedlimits that avoidsaturatingthe op anp. (Seepage260.)

Prob[erns Sectiotr7,1 7.1 In the circuit shownin Fig. P7-1,the switchmakes contactwith positionb just beforebreakingcontact with positiona.As alreadymentioned,thisis known as a make'belbrebreak switch and is dcsignedso that the switchdoesnot interrupt the cment in an inductive circuit. The inte al of lime between "making" and "breaking"is assumedto be negligible.Theswilchhasbeenin the a positionfor a long time.A1 t = 0 the switchis thrown from positiona to posilioDb. a) Determincthe initial currert in the inductor. b) Determine thl9time constantof the circuit for r>0. c) Find i. 1)r,and 02for 1 > 0. d) What percentag€of the initial energystoredin the irductor is dissipatedir the 20 O resistor 12 ms after the switchis tbrown fiom positiona to positionb?

figurcP7.1

5o

7.2 The switchin the circuit ir Fig.P7.2hasbeenclosed EDGfor a long lime before openingat 1 = 0. a.) Find t1(0 ) and i,(r). b) Find 4(0+) and,,(0+). c) Find tio) for r > 0. d) Find tr(1) for r > 0*. e) Explain why t (0 ) * tr(o-). figureP7.2

266

RlandRfCircuits Response of Fnst-oder The switch shown in Fig. P7.3 has been open a long time before clositrgat, = 0. a) Find i,(0 ).

Figure P7.5

b) Findtr(o-). 10A

c) Find i"(0*). d) Find i,(0-). e) Find t,(c.). f) Find tr(,ro). g) Wdte tlle expressionfor ir(,) for t > 0. h) Find z'r(0 ). i) Find r'r(0*). j) Fird ?,.(co). k) Wite tlle expressionfor !'r(t) for t > 0'. l) Wdte the expressionfor i"(l) for I > 0*.

rent source, o represent tlre fracrion of initial energy srored ir the inductor that is dissipated in to seconds,ard a represent the inductance. a) Show that

L1LI1|0 - o)l 2t. b) Test the expression derived in (a) by using it to find the value of R in Problem 7.5.

Figure P7.3 50(l

7.6 In the €ircuit in Fig. P7.5,Iet 1s represent tlle dc cur-

100()

2000

7,4 ln the circuit in Fig. P7.4, the voltage and current expressionsare o = 100t30!v,

r >0+;

i = 4e4'A,,

t>0.

Find a) R. b) r (in miltiseconds). c) L. d) tle idtial energy stoied in the inductor. e) the time (in milliseconds)it takes to dissipate 80% of the initial stored energy. Figure P7.4

7,7 The switch ir ttre circuit in Fig. P7.7 has been open 6fl( for a long time. At , = 0 t]le switch is closed. a) Determinei,(0") and i,(m). b) Determinei,(t) for t > 0*. c) How many millisecondsafter the switch has been closedwill t]le curent in the switch equal 3.8A? figureP7.7

12o"

160

80v

'(r=0

+\

20nH 40

|

8()

7,8 The switch in the circuit in Fig. P7.8hasbeen closeda 6dc long time At I : 0 it is opened.Find "o(r) for l > 0. FigueP7.8 15O \ )80v

50O -r

i s oo

lf} 0.2H o.a:60r)

20o"

2A The switch in tlle circuit seen in Fig. P7.5 has been in posirion 1 for a long time. At r : 0, the svritch moves instantaneously to position 2. Find the value of R so tlar 50ol. of the initial energy stored in the 20 In}I inductoris dissipatedin R in 10lrs.

7.9 Assume that the switch in the circuit in Fie. P7.8has been open for one time constant. At this instant, what percentage of t}le total energy stored in the 0.2 H inductor has been dissipated in ttre 20 O

Prcbtems ?67

7.10 In the circuit shown in Fig. P7.10,the switch has 6fl( beenin positiona for a longtime.AtI :0,itmoves instantaneously from a to b. a) Find r"(t) for I > 0. b) W}lat is the total energydeliveredto the 1 kO rcsistor? c) How manytime constantsdoesit taketo deliver 95%of the energyfound in (b)?

7.14 The switch in the circuit in Fig. P7.14 has been 6tr4 closed for a long time before openirg at t = 0. Find ?o(t) for r > 0'. tigur€P7,14

50nH

tigureP7.10

7.11 The switch in the circuit in Frg. P7.11 has been in 6nc position 1 for a lory time. At t = 0, the switch moves instantareously to position 2. Find ?o(t) for t > 0*. figlre P7.11 7A 1

7.15 The switchin Fig.P7.15hasbeenclosedfor a long time beforeopeningat t : 0. Find a)iL(t),t>o. b)rL$),t>o-. c)iA(r),r>0-. Figure P7.15 3r)

50

96mH

')s.rv

2 15()

5(}

50 ia

4.5f,)

T<{2--l*!l i,. 100o. :9o 2oomH:

:'

20()

l;

200():

7,16 Wlat percentage of the initial energy stored in tlle inductor in the circuit in Fig. P7.15is dissipated by t}Ie cunetrt-contiolled voltage soulce? 7.12 For the circuit of Fig. P7.11,what percenlage of the initial efergy stored in the inductor is eventually dissipated in t]te 20 O resistor? . 7.13 In the circuit in Fig. P7.13, the switch has been closed for a long time before opening at I = 0. a) Find the value of a so that r'"(t) equals0.25 ?,,(0-) whenr:5ms. b) Find the percentage of the stored energy tlat has been dissipatedin the 50O re stor when t - 5ms.

7.17 The two switchesin tle circuit seenin Fig. P7.17are synchronized.The switches have been closed for a long time before opening at t = 0. a) How many microsecondsafter the switches are open is the eneigy dissipatedin the 60 kO resistor 257oof the initial energy stored in tho 200mH hduclor? b) At the time calculated in (a), what percentageof the total energy stored itr the inductor has been dissipated? Fig!reP7.17

Figure P7.r3 200nH

)',."

40ko

20ko

:60ko

30ko

268

R€sponse of First'order Rl andRrCircuits

7.18 The 220 V 1 O source in the circuit in Fig. P7.18is 6nc inadvertendy short-circuited at its terminals a,b.At the time the fault occurq the circuit has beer in operation for a long time. a) Wlat is the initial value of the current iabin the short-circuit connectionbetween terminals a,b? b) Wlat is the final value of the current iab? c) How many microseconds after the short circuit has occurred is the current in the short equal ro 210A?

S€cfion 7.2 721 The svritch in the circuit in Fig. P7.21 has been in position a for a long time and 1)2= 0V. At t : 0, the switchis thmwn to positionb. Calculate a) i,?r, and ?r,for, > 0*. b) the energy storcd in the capacitor aft = 0. c) rhe energy happed in the circuit and the total energy dissipated h the 5 ko resistor if the switch remains in position b indefinitely. Figure P7.21

FigueP7.18

4.'7kA

1()

a

b

5kO +tj.

12{r

60(])

2nH

15nrH

220V 7.22 In the circuit in Fig. P7.22 the voltage and current exPressionsare ?r:100e-1m0lv, r > 0;

b

7.19 The two switches shom in the circuit in Fig. P7.19 6tG operate simultaneously,Prior to t : 0 each switch has been in its indicated position for a long time. A1 I : 0 the two switches move instantaneously to their new positions. Find a) 0"0), t > 0-. b) ,o(r),I> 0. FigueP7.r9

j - 5e-1000! mA,

I > 0+-

Find a) R.

b) c. c) ' (in minisecondt. d) the initial €nergystoredin the capacitor. e) how many microsecondsit takes to dissipate 80o/, of the initial eneryy stored in the capacitor. Figj'Je ?7.22

80 mH "o

i^

7.23 The switch in the circuit in Fig- P7.23 is closed at ED.r I = 0 aft€r beingopen for a long time. 7.m For the circuit seenin Fig. P7.19,find a) the total energydissipatedin the2.5 kO resistor. b) the energy trapped in t]le ideal inductors

a) Find tr(r) and i(0 ).

u) nna 4(o+)ana;z(o+). c) Explahwhyil(o) = t1(0*).

Probtems 269

d) Explain why i(0 ) + i(0+). e) Findil(t) for t > 0. l) End t (t) for t > 0*. tigure P7.23

726 In tllle circuit sho*n in Fig. W.26, both switches operate together; tlat is, they either open or close at the same time. The switches are closed a Iong time before opening a1I - 0. a) How many microjoules of energy have been dissipatedin the 12 kO resistor 2 ms after the switchesopeD? b) How long does it take to dissipate95% of rhe initialy stored erergy? Figur€ P7,26

7.24 The switch in the circuit in Fig. P7.24 has been in position a for a long time. At r : 0, the swifch is thown to positionb. a) Find t,(t for t > 0+. b) What percentage of the initial energy stored in the capacitoris dissipatedin the 4 kO resistor 250 /,csafter the switch has been tfuown?

7.25 Both switches in the ckcuil in Fig. P7.25have been sflc closed for a long time.At t = 0, both switchesopen simultaneously. a) Find roc) lor t > 0+. b) Find ,"(1) for , > 0. c) Calculate tle energy (in microjoules) trapped in the circuit.

7.27 The switch in the circuit in Fig. P7.27has beenin En@posirion 1 for a long time before moving to position 2 a - 0. Findl,(t) foi t > 0t. figtr'e P7.21

7.28 The switch in the circuit seen in Fig. P7.28has been in position x for a long time. At r : 0, the switch moves instantaneously to posirion y. a) Find d so that the time constantfor I > 0is1ms. b) For the a found in (a), find t,a. Flgure P7.28

20ko Figure P7.25

10ko

270

Response of Fnst-0rder Rtandrr Cinuits

7.29 a) In Problem 7.28, how many microjoules of energy !!re generated by the dependent curent source during the time the capacitor discharyes to 0V? b) Show that for t > 0 the total energy stored and genemted in the capacitive circuit equals t]Ie total energy dissipated.

730 After the circuit in Fig. P7.30has beer in operation tsd.r for a long time! a screwdriver is inadvertently connected aqoss the terminals a,b. Assume the resis! ance of ttle screwdriver is negligible. a) Find the current in the screwdriver at I : 0+ and b) Derive the expression foi the curent ir the screwdriver for t > 0+.

7.32 At the time the switch is closed in the circuit in Fig. P7.32,the voltage acrosstlle paralleled capacitors is 30 V and the voltage on the 200 nF capacitor is 10V a) Wlat percentage of the initial energy stored in the three capacitors is dissipated in the 25 kO resistor? b) Repeat(a) for the 625O and 15 kO resistois. c) What porcentage of the initial energy is tapped in the capacitors? tigureP7,32

l0 nF

tigureP7.30 7.33 The switch in the circuit seen in Fig. P7.33has beer arft closed for a long time. The switch opens at 1 = 0. Fhd the numerical expressions tor i,(t) and oo(t) when t > 0*. Figure P7.33

7,31 At the time the switch is closed in the circuit shown in Fig. P7.31,the capacitors are charged as shown. a) Find ?J,(t)for t > 0'.

60ko

,J4

b) Wlat percentage of the total ene4y initially stored ir the three capacitors is dissipated in the

25kO rcsistor? 0 Find 2,1(tfor t > 0. d ) Find ?r20)for t > 0. e) End the energy (in milijoules) trapped itr the idealcapacitois. Figure P7,31

+ + 0.6pF "l

" 25kO

Section 7.3 7,34 The switch ir the circuit sho$n in Fig. P7.34 has Eflft been closed foi a long time before opening at I : 0. a) Find tlle numerical expressions for i.(t) and ,"(l)fort > 0. b) Find tlre numedcal values of 0r(0+) and o,(0+). FlgureP7.34

Probtems 271 7.35 The switch in tle ckcuit shown in Fig. P7.35 has Btrc been in position a for a long time. At r:0, the switch moves instantaneously to position b. a) Find the numerical e4ression for io(t) when

Figur€ P7.38

b) Find the numerical expression for o,(r) for l>0*. FiqueP7,35

7.36 After the switch in the circuit of Fig. P7.36has been open for a long time, it is closed at I : 0. Calculate (a) the initial value of i; (b) the final value of i; (c) the time constant for t > 0; and (d) the nunedcal expressionfor i(t) when I > 0. Flgur€ P7.36 5kO

4kO

75mH

7.39 The switch in the circuit iII Fig. P7.39 has been closed for a long time. A student abruptly opens the switch and repo s to her instructor that when the switch opened, an electdc arc with noticeable per, sistence was established across the switch, and at the same time the voltmeter placed aqoss the coil was damaged. On the basis of your aralysis of the circuit in koblem 7.38,can you explainto the student why this happened? Pigure P7.39

20kO

737 The currert and voltage at the teminals of tlle inductorin the circuit in Fig.7.16are

(D : (10- 10r5m') A, r>0; u(t\ : 20oesm'v, l>0*. a) Specify the numerical values of %, R, 1,, and I. b) How many miliseconds after the switch has beencloseddoesthe energystoredin tlle inducior reach 25% of its final value? ?.38 The svritch in the circuit shown in Fig. P7.38 has been closed for a long time. The switch opens at l:0.Foit>0*i a) Find I'o0) as a function of 1s, Rr, R2,and a. b) Explain what happers to ?,(t) as R2 gets larger and larger. c) Find osw as a function of 1s, Rt, R2,ard L. d) Explain what happens to osw as R2 gets larger and larger.

7.4O a) Derive Eq.'7.4'7byf st conveting the Th6venin equivalent in Fig. 7.16 to a Norton equivalent and then summing the curents away from the upper [ode, using the inductor voltage o as the variable of interest. b) Use the separation of variables technique to find the solution to Eq. 7.47.Verily that your solution agees with the solution given in Eq- 7.42. 7.41. The swilch in tle circuit in Fig. P7.41has been open 6ntr a long time before closing ar r = 0. Find ,"G) for t>0. Figure P7.41 87.2mH

20()

10O

09?]i

40O

272

Rtandlr Circujts Response of Fitst-od€r

7.42 The swilch in the circuit in Fig. P7.42has been open 6xa a long time before closhg at I = 0. Filld o,(1) for r>0-. P7.42 Figure 100

),,o

150 5 0 " i2mH

:

l'*(

,n ot^(

7.43 There is no energy stored in the inductors a1 and az at the time the switch is opened in the circuit shown in Fig. P7.43. a) Derive t]le exp.essionsfor the curents i1(t) and i20)forl > 0. b) Use the expressionsderived in (a) to find i1(x') and i2(oo).

7,46 The make-before-break switch in the circuit of 6r,t! Fig.P7.46hasbeen in positiol a for a long time At t = 0, tle switch moves instantaneously to position b. Find a)?,,(,),r>0*. b)t10),t>0. c)4(t),t>0. Figure ?7.46

15o r,+

?.47 Theswitchir thecircuitin Fig P747hasbeenopen 6ire a long time before closingat t = 0. Find o,(t) for

FiglreP7.43

u

iz,.l)l

figureP7.47

t=0 40 mH

5rI 10A

7.44 The switch in the circuit in Fig. P7.44 has been in 6xd position 1 for a long time. At I = 0 it moves instantaleously to position 2. How many milliseconds after the switch opentes does ?t"equal 80 V?

60 nH

ligve P7.44 7.48 The switch iII the circuit in Fig. P7.48 has been in tsdG position i for a long time. The initial charge on the 15 nF capacitor is zero At t : 0, the switch moves iNfantaneouslyto positiony. a) Find 0.0) for t > 0". b) Fhd 1)(, for t > 0. tigur€P7.48 ?,45 For the circuit in Fig. P7.44,find (in joules): a) the total energy dissipated in [he 80 O resisfor; b) the energy trapped in the inductors; c) tle idtial energy stored in tlle inductors.

7.49 For the circuit in Fig. P7.48,find (ia microjoules) a) the energydeliveredto the 200kO resistor; b) the energytmpped in the capacito$; c) the initial energy stored in the capacitors. 7.50 The switch in tle circuit shown in Fig. P7.50 has antr been closed a long time before opening at r = 0. For, > 0+,find a) ro\t). b) i,(r).

c) 4(r). d) i2O. e),r(0-). figueP7,50

19

4ko

Figure P7.53

1g.ko" 120V

4;

6125tO ,:0 150ko 40nF

7.54 The switch in the circuit of Fig. P7.54 has been in position a for a Iong time. At t - 0 the switch is movedto positionb. Calculate(a) the initial voltage on tlre capacito! (b) the fhal voltage on the capacitor; (c) tlre time constant (in micmseconds) for t > 0; and (d) tle length of time (in microseconds) required for the capacitor voltage to reach zero after the switch is moved to position b. Figure P7.54

20ko 7.51 Thecircuit in Fig.P7.51hasbeenin operationfor a 6trc Iong time.At t = 0, the voltagesouce dropsfrom 100V to 25V andthe currentsourcereversesd ection. Find r,,(r) for , > 0. Figrre P7.51 5ko

15kO 75V 10k()

2ko

100v

7.52 The curent and voltage at le terminals of the capacitor in the ckcuit in Fig. 7.21 are i(r) : 50e-'sm'mA' v. ,{r} - (80 - 80P25m')

t > 0t; r > r.

a) Specify t}le numerical values of 1", U, R, C, and r, b) How many midoseconds after the switch has been closed does the energy stored in the capacitor reach64% of its final value? 7.53 Assume that the switch in the circuit of Fig. P7.53 has been in position a for a Iong time and that at 1 = 0 it is moved to position b. Find (a) oc(o+); (b) ?,c(oo);(c) r for I > 0; (d) (0*); (e) oc, r > 0; and (f) ,, t > 0+.

755 The switch in the circuit shownin Fig. P7.55has 6rc beencloseda long time beforeopeningat I : 0. a) Whatis the initial valueof i,(l)? b) Whatis thefinalvalueof i,(tx c) Whatis tlle time constantof the circuitfor t > 0? d) What is the numericalexpressionfor i.(t) when r>0-? e) What is the numericalexpressionforu,,(t) when

figureP7.55 6.8kO 75V

274

Response of First-ord€r nt andfr Circujts

7.56 The switch in the circuit seenin Eg. P7.56hasbeen in Edc position a for a long time. At t : 0, the swrtchmoves instantaneouslyto position h For t > 0*, find

b),,(r). c) r.lt).

7.59 The switch in the circuit shown in Fig. P7.59 has EdG been in the oFF position for a long time. At 1 = 0, the svdtch moves instantaneously to the oN position. Find o,(l) fo. t > 0. Figure P7.59 30 x ldia

d) ,g(0-).

7,60 Assume that the switch in the circuit of Fig. P7.59 6na has been in ttre oN position for a long time before switching instantaneously to the oFF position at t = 0. Find ?r,(t) for 1 > 0.

7.57 The switch in the circuit seenin Fig. P7.5?has been 6tra in position a lor a long time. At t = 0, the switch moves instantaneously to position b. Find oo(r) and i,(t)forr>0'. Figure P7.57

.i,r,r-! ,,,,"

7.61 a) Derive Eq.7.52 by tust converting the Norton equivalent circuil shown in Eg. 7.21to a Th6venin equivalent and then summingthe voltagesaround the closedlooA usilg the capacitorcurrent i asthe relevant vadable, b) Use the separalion of variables technique to find the solution to Eq. 7.52.Verify that your solution agees with that of Eq.7.53. 7.62 There is no energy stored in tlle capacitors C1 and C2at the time the switch is closed in the circuit seen in Fig.P7.62. a) Dedve the expressions for o(t and zJr(t) for t>0. b) Use t})e expressionsderived in (a) to find o(oo) and 02(').). Figure P7.62

,r(t) 758 The switch in the circuit shown in Fig. P7.58opensat mPi.r , : 0 alter behg closed for a long time. How many milliseconds after the switch opens is the energy stored in the capacitor 90% of its final value? rig'lle P7.58

?,n, 7.63 The switch in the circuit of Fig. P7.63 has been in sP,m position a for a long time. At t : 0, it moves instantaneously to position b. For t > 0*, find a) polt).

b) c) d) e)

i,(r). ?,(0. ,10. theenergyfappedin thecapacitonast + oo

Figure P7.63

7.67 There is no energy stored in the circuit in Fig. P7.67 EPj.Eat the time the switch is closed. a) Find i,0) for t > 0. b) Ftnd u"0) for r > 0*. c) Fird ir(,) for r > 0. d) Find i(D for I > 0. e) Do your answen make sensein terms of knowlr circuit behavior?

100v

I

Figure P7.57 -+i"

7.64 The switch in the circuit in Fig. P7.64 has been in 6Ee position a for a Iong time. At I = 0, the switch moves instantaneously to position b. At the instant the switch makes contact $rith terminal b, switch 2 opens.Find o,(r) for t > 0.

t+

.

10v

7.68 There is no energy stored in the circuit in Fig. P7.68 Edc at the time the switch is closed. a) Find (i) for I > 0. b) Fhd r,(t for t > 0*. c) Find r,r(r) for 1 > 0. d) Do your answe$ make sensein terms of known circuit behavior? Figure P7.68

Section 7,4 7.65 Therc is no energy stored in t}le circuit of Fig. P7.65 at the time the switch is closed. a) Fird io(t) for I > 0. b) Find 0o(t)for t > 0*. c) Find t1(t for I > 0. d) Find irQ) for t > 0. e) Do your answersmake sensein terms of known circuit behavior? FiWleP7.55 ,/10riH\.

200v

8H

d'

7.69 Repeat Problem 7.68 if the dot on the 8 H coil is at 6nc the top of the coil. Section 7.5 7.70 In the circdt in Fig. P7.70,switch A has been open E ic and switch B has been closed for a long time. At t = 0, switch A closes One second after switch A closes,switchB opens.Find tr(t) for t > 0. Figure P7.70

7,66 Repeat (a) and (b) in Example 7.10if the mutual inductance is reduced to zero,

l0 nH

276

Response ofFirst-0rderfl andRCCncujts

7.71 There is no energy stored in the capacitor in the cir 6trr! cuit ir Fig. P7.71when switch 1 closesat r :0. Three microsecondslater, switch 2 closes.Find od(t) fort > 0.

ligurc P7.74

tigureP7,71 r:0 + 50/rs

+

7.72 The action of the two switchesin the circdt seenin *rI( Fig. It.72 is asfollows. For t < 0, switch 1 is in position a and switch 2 is open.This state has existed for a long time. At t = 0, switch 1 moves instantaneously frcm position a to position b, while switch 2 remains open. Tivo hundred fifty microseconds after switcb I operales.switch 2 closes.remain( closed for 400 ps, and then opens.Find o,(t) 1 ms after switch 1 moves to position b.

7.75 The switch in the circuit shom in Fig. P7.75has EPicbeenin positiona for a long time.At I :0, the switchis movedto positionb, whereit remainsfor 100ps.The switchis then movedto positionc, whereit remainsindefinitely.Find a) (0). b) t(25ps). c) i(200ps). d) 0(100 i.is). e) r(100+ps). FigureP7.75

IrgurcP7.72

500v

7.73 For the circuit in Fig.P7.72,how manymilliseconds after switch 1 movesto position b is the energy sroredin the inductor4% of its initial value?

+ J

96()

480f)

20nH

276 The suritch in the cftcuit in Fig. P7.76 has been in Edc position a for a long time. At I = 0, it moves instantaneouslyto positionb, whereit remainsfor 250ms before moving instantaneouslyto position c. Find I,otbrr > 0. tigureP7.76

7.74 The capacitor in the circuir seen in Fig. P7.74 has Efld been charged to 494.6mV Aft = 0, switch 1 closes, causingthe capacitor to discharye hto the resistive network. Switch 2 closes50 &s after switch 1 closes. Find t])e magnitude and direction of the current in the second switch 100 ps after switch 1 closes.

120()

4{}

50-A

7.77 In the ckcuit itr Fig. P7.77,switch t has been in posisPI.r tion a and switch 2 has been closed for a long time. At t - 0, switch 1 moves instantareously to position b. Four hundred microseconds later, switcb 2 opens,rcmahs open for 1ms, and then recloses. Find ,, 1.6 ms after switch 1 makes contact with teminal b.

0+40Ops

35kO

Figure P7.80

I J sko

7.81 The voltage waveform shown in Fig. P7.81(a) is Erla applied to the circuit of Fig. P7.81(b). The initial voltage on the capacitor is zero. a) Calculate0o(.). b) Make a sk€tchof r',(t) ve$us t Figure P7.81

7.78 For the circuit in Fig. P7.77,what percentage of the initial energy stored in the 50 nF capacitor is dissipatedin the 10kO resistor?

7.79 The voltage wavefonn shown in Fig. P7.79(a) is am applied to the circuil of Fig. P7.79(b). The initial curent in the inductor is zero. a) Calculate0"(l). b) Make a sketch of ?,,(t) venus L c) Find io at I : 4 &s. FlgueP7.79 ?,"(v)

20

.---.1

1t)

6 (a)

t(nd (b)

7.82 The voltage signalsourceh dre cncuit in Fig.P7.82(a) is geneiatingthe signalsholvl in Fig.P7.82(b).Thereis no stored energyat t = 0. a) Derive tlre expressionsfor o,(t) that apply in the i n r e r l a lrs. 0 : 0 = t < l 0 m s ; 1 0 m s < r < 20ms;and20ms < t
20ko

tigureP7.82 R=50kO

G)

(b)

7.80 The cunent sourcein the circuit in Fig. P7.80(a) 6r,c generales the currentpulseshownh Fig.P7.80(b). Thereis no energystoredat t : 0. a) Derive the numedcalexpressionsfor o,(t) for the time inteivalst < 0, 0 < 1< 50ps, and 5 0 p s< , < o o . b) Calculale ,, (50 Fs)aod,, (50 p./ c) Calculatei" (50- /rs) andi. (50' l..s).

278

ResDonse of First-0rder RlandRCCncuiis

7.83 The cunent souce in the circuit h Fig. P7.83(a) 6r,( generatesthe current pulse shown in Fig. P7.83(b). There is no energy stored at I - 0. a) Derive the erpressions for i,(r) and oo(t for the time inte als t<0; 0
?5x ldra

25V

i"(0+); io(0.0005-); and

,,(0.000s).

c) Calculate o,(0 ); o,(0+); 1),(0.0005 ); and o,(0.0005*). d) Sketch t"0) veNus I for rhe inrerval -2ns <, < 2ms. e) Sketch Oo(t) veisus r for tle inteival -2ms < t < 2ms, Flgure P7.83 k (nd)

2ko

tigureP7.85

7,86 The gap in the cfcuit seenin Fig. P7.86will arc over Btr( whenever the voltage aqoss the gap reaches36 kV The initial curent in the inductor is zero. The value of P is adjusted so tle Th6venir resistance with respect to tle terminals ol tle inductor is -3 kO. a) WIat is the value of B? b) How many microseconds after the switch has been closed will the gap arc over? tigureP7.86

20 0.1rrF

0.sr(mt (a)

(b)

Section 7.6 7.84 The inductor curent in the circuit in Fig. P7.84 is AnG 2; mA at the instant the suritch is opened. The inductor will malfunctior whenever the magnitude of the inductor current equals or exceeds 12 A. How long after the switch is opened does the inductor malfunction?

7,87 The svritch in the circuit in Fig. P7.87 has been tsflft closed for a long time. The maximum voltage rating of the 16 nF capacitoi is 930 V How long after the switch is opened does the voltage acrosstle capacitor reach the maximum voltage mtirg? Figure P7.87

4ko

^ . ^ r _ ,_ .

Figurc P7.84

2ko

tjo

rxn n,r

7.85 The capacitor in the cLcuit shol*il in Fig. P7.85 is Eqc charged to 25 V at tle time tle switch is closed. If the capacitor ruptures when its terminal voltage equalsor exceeds50 kV how long does it take to rupture the capacitor?

7,88 The circuit sho$,n in Fig. P7.88 is used to close the switch between a and b for a predetermined Iength of time. The electdc relay holds its contact arms down as long as the voltage aooss the relay coil exceeds5 V When tle coil voltage equals 5 V the relay contacts return to their initial position by a mechanical spring action. The switch between a and b is initially closed by momentarily pressing the push button. Assume that the capacitor is fully charged when the push button is first pushed dov.n.

,robtems279 The rcsistance of t]te relay coil is 25 kO, and the inductance of the coil is negligible. a) How long wi[ the switch between a and b remaitrclosed? b) Write the numerical expression for i ftom the time tlle relay contacts first open to the time the capacitor is completely charged. c) How many milliseconds (after the circuit between a and b is interrupted) does it take t])e capacitor to reach 85o/oof its final value?

7.91 Therc is no energy stoied in the capacitofi in the Aflc circuit shown in Fig. P7.91 at the instant the two switchesclose. a) Find x', as a funclion of ?r,,,b, R, and C. b) On the basis of the result obtained in (a), describerhe operaLionof lhe circuit. c) How long will it take to saturatethe amplifier if t,a = 10mv; ?)6 = 60mv; R = 40ko; C - 25 n\ and Ucc: 72Y?

Figure P7.88 Flgure P7.91

S€ction 7,7 7,89 The eneryy stored in the capacitor in the circuit 6trc shown in Fig. It.89 is zero at the instant the switch is closed. The ideal operational amplifier ieaches saturation in 3 ms Wtat is the numerical value of R in kilo-ohms? tiguruP7.89

7.92 At the instant the switch of Fig. P7.92is closed, the 6r,a voltage on the capacitor is 16 v. Assume an ideal operational amplifier. How many milliseconds after the switchis closedwill t}le output voltageoo equalzero?

60 nF Figurc P7.92 + 16V 10k[]

7.90 At the instant the switch is closed iI} the circuit of Fig. P7.89,t}te capacitor is charyed to 5 Y positive at the ight-hand terminal. If the ideal operational amplifier satumtes in 8 ms.what is the value of n?

280

Response of First-order Rl andnrCircuits

7.93 At the time the double-pole switch in the circuit tsdG shown in Fig. It.93 is closed,the initial voltages on the capacitors are 45 mV and 15 mV as shown.Find the numedcal expressionsfor r'"(t), olt, ard ,10) that are applicableaslong asthe idealop ampoper atesin ils linear range. Rgure P7.93

0

()ko

-

12.5nF !r(t) +

') -6V

4OkO +

7.95 Repeat Problem 7.94 with a 4 MO resistor placed EpI( acrossthe 50 nF feedback capacitor. 7.96 fhe voltagesourcein the circuit in Fig. P7.96(a)is Erc generating the triangular waveform shown it Fie. P7.96(b).Assume the energy stored in the capacitoris zerc at t = 0. a) Derive the numerical expressions for r',(r) for t h e f o l l o $ i n g L i m e i n l e r v a l ' :0 r r 5/rs: 5ps < t < 15ps;and15/..s= t < 20Fs. b) Sketchthe output wavelorm between0 and 20,.s. c) ff the triangular input voltage continuesto repeat itsef for r > 20 /-.s,what would you expect the ouFut voltage to be? Er?lain. Figure P7.95

10mV e0) 15

25nF

7.94 The voltage pulse shown in Fig. P?.94(a) is applied Edc to the ideal integrating amplifier shown in Fig.P7.94(b).Derive the nume cal expressions for o,(t) when t,(0) : 0 for the time intenals a)t<0. b) 0 < r < 50 rns. c)50ms
50 nIF

Sections7.1-7,7 7.97 The circuit showr in Fig. P7.97 is known as an astable multivibrutor ar].dfinds wide application in pulse circuits The purpose of this problem is to relate tle charging and discharying of the capacitors to the operation of the circuit. The key to analyzing the circuit is to undentand the behavior of the ideal transistor switch€sT1 and T2.The circuit is designedso that the switchesautomaticallyalternate between oN and oFF.When Tt is oFF,T2 is oN

Probtems 281 and vice versa.Thus in the analysisof tltis circuit, we asstmlea switch is either ONor OFF,We also assumc ttrat the ideal transistor switch can change its state instantaneously. In other words! it can snap from oFFto ONand vice velsa,When a transistor switch is oN, (1) the base curent lb is geater than zcl{r, (2) the teminal voltage ob€is zero, and (3) the terminal voltage o@ is zeio. Thus, when a transistor switch is oN, it presents a sho circuit between the terminals b,e and c,e. When a transistor switch is oFF,(1) the terminal voltage obeis negative, (2) the basecurent is zero, and (3) there is an open circuit between the termhals c,e, Thus when a hansistor switch is oFF, it presents an open circuit between tle teminals b,e and c,e.Assume that T2 has been on and hasjust snapped oFFrwhile T1 has been oFF and has just snapped oN. You may assumethat at this instance, C2 is charged to tle supply voltage ycc, and the charge on Cr is zero. Also assume C1 = C2and R1 : R2 = 10Rr. a) Derive t]le expression for ob!2during t]le interval that T2 is OFF. b) Derive tlle expression for o@2during the interval that T2 is oFF. c) Find the length of time T, is oFF. d) End the value of r'""2 at the end of the interval that T2 is oFF. e) Derive the expressionfor ib1duritrg the interval that T2 is oFF. I) Find the value of ib1 at the end of the interval that T2 is oFF. g) Sketch ?r."2versus t during the interval that T2 is oFF, b) Stelch tbt versusr during tbe inter!al lhal T2

figureP7.97

7.98 The component values in the circuit of Fig. P7.97 are Ucc : 9V; Rr : 3 kO; Cl - C2 : 2 nF; and R 1: R 2 : 1 8 k O . a) How Iong is T2 in the oFFstate during one cycle of opemtion? b) How long is T2 in the oN state du ng one cycle of operation? Repeat(a) forT1. a) Repeat(b) forT1. At the first instant after T1 turns oN, what is the value of ib1? At the instant just before Tr tums oFF,what is tle value of ib1? g) What is the value of ?ce2at the instant just before T2 tums oN?

7.99 Repeat Problem 7.98 with C1 :3 nF and C2:2.8nF. All other component values are unchanged.

7.100 The astable multivihator cftcuit iII Fig. P7.97 is to satisly tlte lollowing criteda: (1) One traNistor switch is to be oN for 48/rs and oFF for 36ps for (3) ucc: 5v; each cycle; (2) RL:2k0t (4) R1 : Rr; and (5) 6R, < R1 < 50-Rr.wllat are the limjting values for the capacitors C1 and C2?

7.101 The circuit shom in Flg. P7.101 is ktrown as a m onostabIe multfuib r ator. -lhe adlectlve monostlib k is used to descdbe the fact that t]le circuit has one stable stale. That is, if left alone, the electronic switch T2 will be oN, and Tr $rill be oFF.(The operation of the ideal hansistor switch is descibed in Problem 7.97.) T2 can be turned oFF by momentarily closing the s$ritch S. After ,t returns to its open position, T2 will rctum 1oits oN state. a) Show that if T, is oN, Tr is oFF and will stay oFF. b) Explain why T, is turned oFFwhen S is momentadly closed. c) Show that T2 will stay oFFfor RC ln 2 s.

242

Response of Fjrst-order flLandRf Circuits

FigureP7.101

continue to conduct until the lamp voltage drops to 5 V Wlen the Iamp is not conducting, it appears as an open circuit. 14 = 40 V; R : 800kO; and

+

+

a) How many tim€s per minute will the lamp turn onl b) The 800 kO rcsistor is replaced with a variable rcsistor R. The resistance is adjusted until the lamp flashes12 times per minute. What is the valueof n?

7.105 In t]le flashinglight circuit shown in Fig.7.45,the lanp can be modeledas a 1.3kO resjstorwhen it is pillfffii, _ ;itrii conducting.The lamp triggers at 900 V and cuts off at 300V 7.102 The parametervaluesin the circuit in Fig. P7.101 a) If % : 1000V, R : 3.7kO, and C : 250pF, are Ucc = 6 V; R1 : 5.0kO; Rr = 20 kO; how many timesper minute wil the light flash? C = 250pF; and R : 23,083O. b) What is the averagecurrent in milliampsdeUv a) Sketch 0@2versus t, assuming that after .t is ered by t]le source? momentarilyclosed,it remains open until the c) Assume the flashinglight is operated24 houn circuit has rcachedits stablestate.AssumeS is per day If the cost of power is 5 cenKper kilowatt closedat I : 0. Make your sketchtor the interhour, how much doesit cost to operatethe light val 5
hoblerTrs 283 7.107 The relay shol*n in Fig. P7.107connects rhe 30 V dc p:Sfi*iiEgenentor to t}le dc bus as long as the relay cment is greater tlan 0.4 A. If the relay current drops to 0.4 A or less,the sprhg-loadedrelay immediarely connectsthe dc busto the 30V standbybatteryThe resistanceof the rclay winding is 60 O. The inductanceof the relay windingis to be detemined. a) Assume t]le pdme motor ddving the 30 V dc genemtor abruptly slows down, causingthe generated voltage to drop suddenly to 21 V What value of a will assure that the standby battery will be coinected to the dc bus in 0.5 seconds? b) Using the value oI L determined in (a), state how long it will take the relay to ope{ateif the generatedvoltagesuddenlydropsto zero.

FigureP7.107

30v gen

DC loads

Natural andStep Responses of RICCircu'its In this chapter,discussion of the ratural rcsponseand step 8.1 Introduction to th€ NaturatResponse of a P.ratLeL flc Circuitp. 286 8-2 TheForms of th€ NaturatResponse of a PaElbtnlc Circuitp. ?91 8.3 TheStepResponse of a ParalLeL RLCCitcuit p. 3a1 8.4 TheNaturalandStepResponse of a Series RLCffrctljt p- 3A8 8.5 A CircuitwithTvoIntegrating p. 31? Amptifi€rs

1 B€abteto determine the naturalrespoiseand the stepresponse of paratlet RtCcncuits. 2 Beabteto determine the natuat response and the st€prcsponse of seriesRLfcjrcuits.

284

responseof circuits containingboth inductors and capaciton is limited to two simplesffucturesrthe parallelRZC circuit and the seriesnLC circuit.Findingthe naturalresponseof a parallelRaC circuit consistsof finding the voltagecreatedacrossthe parallel branchesby the releaseof elergy storedin the inductoror capacitor or both. The task is defined in terms of the circuit shown in Fig.8.1on pagc286.Theinitial voltageon the capacitor,V6,representsthe initial energystoredin the capacitor.Theinitial current throughthe inductor,10,representsthe initial energystoredin the illductor.If the individualbranchcuuents are of interest,you can find them after detemining the terminalvoltage. We derivethe stcpresponseofa parallelRl,Ccircuit by using Fig.8.2on page286.Weare interestedin the voltagethat appears acrossthe parallelbranchesas a result of the suddenapplication of a dc current source.Energy may or may not be stored in the circuit when the currellt sourceis applied. Finding the natural responseof a seriesRLC circuit consists of finding the current generatedin the se esconnected elements by the releaseof initially storedenergyin the inductor,capacitor, or both.lhe task is defined by the circuit shown in Fig.8.3 on page286.As before,the initial inductorcurrent,lo, and the initial capacitorvoltage,yo,representthe initially stored energy.If any of the individual element voltagesare of interest,you can find them after determi ng the current. We describethe stepresponseof a seriesRZCcircuit in terms of the circuit shown in Fig. 8.4.We are interestedin the curent resultingfrom the suddenapplicationof the dc voltage source. Energy may or may not be stored in the circuit when the switch is closed. If yolr havenot studiedordinarydifferentialequations,de vation of the natural and step responsesof parallel and se.iesRaC circuitsmay be a bit difficult to follow. However,the resultsare important elough to warant presentationat tbis time.We begin with thc natural responseof a parallel RZC circuit and cover

Perspective Practical AnIgnitionCiJCuit weinkoduce the stepresponse of anRlf cirIn thjschapter ignitioncircuitis based onthetransient cuit.Anautomobile operresponse of anRlCcircuit.In sucha circuit,a switching in the currentin an jnductive ationcauses a rapidchange windingknownasanignitioncoit.Theignjtjoncoilconsists in series. This coupted coitsconnected of two magneticatly The is alsoknownas an autotransfomer. seriesconnection to the batteryis referred to as the primary coil connected to the sparkptugis referrcd wjndingandthe coilconnected in winding.TherapidLy chanqing current to asthesecondary (mutuai prjmary via magnetic coup$ng the windinginduces in the secondary winding. inductance) a veryhighvoLtage Thisvottage, whichpeaksat frorn20 to 40 kV is usedto ignitea sparkacross the gapof the spafkp[ug.Thespark ignitesthefueL-air mixture in thecytinder.

A schematic diagramshowingthe basiccomponents of an jgnition systemis shownin the accompanying figure. In (as opposedto mechanicat) todays automobite,eLectronic switchingis usedto causethe rapidchangein the primary ofthe electronicswitching windingcurrent.An undefstanding that is circuitrequiresa knowtedge of etectroniccomponents beyondthe scopeof thjs text. However, an anatysisof the ignjtion circujtwjt[ seweas an introdlcoLder,ionventionaL in the designof a encountered tion to the typesof probtems usefulcircuit.

Capacitor .

245

286

NatuEtand StepResponses of RrcCircuits

C

Figure 8.1A A cjrcuitused to itlunrate ihenatumt response of a paraLtet Rr circuit.

tlis material over two sections: ofle to discuss the solution of the differertial equation that describesthe cicuit and one to prcsent the three distinct Iorms that the solution can take. After introducine these lhreelorms.$ e shos lhat rhe(dmetormsapplyro the srepre.pon.eof a parallel RLCciicuit as well as to the natural and step responsesof sedes RIC circuits.

8.1 Introductionto the Natural Response of a Paraltel RLCCircuit Figure 8.2A A circujt used t0 jltustmte thestep rcspon* ofa panLlet tu circuit-

va

The fint stepin findingthe naturalresponseof t}le circuitshownirl Fig.8.1 is to derive the dilferential equationthat the voltage ?i must satisfy.We chooseto find the voltagef st.becauseit is the samefor eachcomponent. After that, a branch current car be tound by using the currenfvoltage relationshipfor the branch compoDenr. We easilyobtain the differential equation for the voltage by summing the curents away from the top node, whereeachcurrentis expressedas a function ofthe unknown voltage?:

u Figure 8.3A A circuit used io jltustrate thenaturat r€sponse of a sedes rur cjrcujt.

1f '

n'i.lnoo'*

4*cfi=0.

We eliminate the htegral in Eq.8.1 by ditrerertiating once with respect to I, and,because10is a constant,weget

na-i Figure 8.4A A circuit used i0 jltustctethe step response of a s€ries SLfcjrcuit.

(e.r)

(8.2)

We now divide through Eq. 8.2 by the capacitanceC and arrange the deivatives in descendingorder: d2r

1 du

't)

^

(8.3)

ComparingEq. 8.3 vith the differential equarionsderived h Chaprer? reveals that tley differ by the presenceof the term involving the second deivative. Equation8.3is an ordinary,second-orderdiffeietrtialequation with constantcoefficients Circuits in this chaptercontain botll inducto$ and capacitors,so the diJferential equation describingthesecircuits is of the second order.Therefore, we sometimescall suchcircuitssecond-ord€r circuits.

TheGeneral Sotutionof the Second-0rder Differential Equation We can't solveEq. 8.3 by separatingthe variablesand integmting as we were able to do with the firstorder equationsin Chapter7. The classical approachto solvingEq.8.3 is to assumethat the solutionis of exponential folm, that is. to assumethat the voltase is of the form r, = Ae',. whereA ands aie unknownconstants,

(8.4)

8.1 Intrcduction totheNaturat Resoonse ofaParaltetRlC Circuit 297 Before showhg how this assumptionleadsto tle solutionof Eq. 8.3, we needto showthat it is mtional.Thestrcngestargumentwe canmakein favorof Eq.8.4is to noteIrcm Eq.8.3that the seconddeivativeof the plusa constant plusa constant solution, timesthefint derivative, timesthe solutionitself,mustsum to zero for all valuesof L This can occur only if higher order derivativesof the solutionhavethe sameform as the solu tion.The exponentialfunction satisfiesthis cdterion.A secondargument in favor of Eq. 8.4is that the solutionsof all the firsl-order equationswe derivedin Chapter7 wereexponential.It seems reasonable to assume that the solution of the second-orderequationalso involvesthe exponential function. If Eq.8.4is a solutionof Eq.8.3,it mustsatisfyEq.8.3for all valuesof t. Substituting Eq.8.4into Eq.8.3generates theexpression

er'e'rfi*rff=0,

-

, 1 \ RC-i)

(8.5)

which can be satisfied for aI values of t oDly iJ ,4 is zero or the parenthetical term is zero,becauseer' f 0 for ary finite valuesofrt. We cannotuse ,4. = 0 as a geneml solution becauseto do so implies tlat the voltage is zero for all time-a physical impossibility if energy is stored in either the inductor or capacitor. Therefore, in order for Eq. 8.4 to be a solution of Eq.8.3,theparentheticalterm in Eq.8.5 must be zero,or

(8.6) < Characteristic equation,parattel ruCcircuit Equation 8.6 is called the characteristic equation of the differential equation becausethe roots of this quadmtic equationdetemine the mathematicalcharacterof t'O). The two roots of Eq.8.6 are

'

I 2RC

2RC

(8.7)

(8.8)

288

Natunt andStepRespons€s of fr Cjrcuits If either root is substitutedinto Eq.8.4, the assumedsolutionsatisfiesthe giver differential equation,that is, Eq.8.3. Nore fiom Eq.8.5 thaf this resultholdsregardlessof the valueof,4. Therefore,both u = Ard,' ar.d D = A2ett satisfyEq.8.3.Denotingthesetwo solutionsz,iand o2,respectively, we can show that their sum also is a solulion. Specifically,if we le1 'D = z)1+ D2 - Aret

(8.e)

+ A2erI,

du=

(8.10)

Arfie\' + A2sae,'

(8.11)

SubstitutingEqs.8.9-8.11into Eq.8.3 gives /

4/"('i

t

- - rn )\ o n R;,

I

*

I

\': ^.,

' . ,rJ -\ o

re.rzr

But each parenthetical term is zero because by definition 11 and 12 are roots of the characteristicequation.Hence the natural iesponseof the parallel RaC circuit shown in Fig. 8.1 is of the form 0:

Ares\t + A2e5,r.

(8.13)

Equatior 8.13 is a repeat of the assumptionmade in Eq.8.9. We have shown that z'r is a solution, ?)2is a solution, and tJ1+ ,2 is a solution. Therefore,the geneml solutionof Eq. 8.3 has the form givenin Eq. 8.13. The roots ofthe characteristic equatjon(st and sr) are determiDedby the circuitparametersR, a, and C The initial conditionsdeterminethe values ofthe comtants/1and,42. Note that the form of Eq.8.13must be modified if the two rcots sr and r, are equal. We discussthis modification when we turn to the criticallydampedvoltageresponsein Section8.2. The behaviorofo(t) dependson the valuesof sl and 12.Thereforethe first step in finding the natural response is to determine the roots of the characteristicequation.We retun to Eqs.8.7 and 8.8 ard rewrite them t u s i D gnao l a r i o\n i d e l yu s e d i nl h e k l e r a l u r e :

\=-d+l&4, "r= -'

\/&

(8.14)

"1,

(8.15)

8 . t - r r ' o d r r i o l . o h el a r u r a l R e 5 D oo,tsaeP a n l t eRrl t ( i l ( r i r

(8.16) < Nep€rfrequenqr, paralletRICcircuit

(8.17) < Resonant parattel radianfrequency, Rlf circuit These results are summarized in Table 8.1. The eeonent of e must be dimensionless, so both sr and r, (and hence a and o0) must have the dimension of the reciprocal of time, or frequency.To distinguishamongthe ftequeflciess1,12,a, and oo, we usethe fotlowingterminology:sr and s2are refeired to asmmplexftequencies,a is called the neper frequenct', and oo is the resonant m.li&n ftequenq.'I]Je full significance of this teminology unfolds as we move t}lrough the remainingchaptersof this book. All theseftequencieshave the dimensior of angular frequency per time. For complex frequencies, the neper frequency, and the rcsonant radian frequency, we specify values using the nni,tndilns per second(rad/s).The nature of the roots 11and s2depends on the ^valuesof c and o0. There are three possible outcomes. First, if oii < a'. both roots will be real and distirct. For reasonsto be discussed late^r,the ^voltageresponseis said to be overdamped in this case.Second, if 06 > l, both s1and s2will be complexand,in addition,will be conjugatesof each other. In this situation,ttre voltage rcsponseis said to be underdamped.The third possibleoutcomeis that oB = a2.In this case,rl and 12will be real and equal.Here the voltage rcsponseis said to be critically damped. As we shall see, damping affects the way the voltage responseleachesits final (or steady-state)value.We discusseach case separatelyin Section8.2. Example 8.1 illustmtes how the numerical values of 11 and 12 are detemhed by the valuesofR, a,and c-

Terminology rj-

a+Vd'

s1:-d1

Resonant radian frequency

"'

2RC I

.vzr

va-

06 onl

289

290

Naturat andStepRespons€s ofnr Cjrcuits

Findingthe Rootsof the Characteristic Equation of a PaEttetfl"CCircuit a) Find the roors of the chamcteristic equation that govems the transient behavior of the voltage shownin Fig.8.5if R : 200 O, L = 50 mH, and C = 0.2 p.F. b) Will tle responsebe overdamped,underdamped, or critically damped?

C

Figure8.5l A cjrcujt used to illustrat€ th€natuntr€sponse of a palattet Rrrcircuit.

c) Repeat(a) ard (b) for R : 312.5O. d) What value of R causesthe responseto be criticaly damped?

c) ForR = 312.5O, 106 = 8000rad/s, d = ..:-.^ (oz)){u.rl

SoIution

& : 64 x 106: 0.64 x 103rad'ls'.

a) For the given values of -R,a, and C,

As ofr remains at 103rad2/s2, -

2RC 1

," _r,{.,na_Cs. (400)(0.2)

r, = -8000 - ./6000rad/s.

/ rnl ! 1n6r

(In electrical engineering,the imaginary number V-1 is representedby the letter i, becausethe letter j representscu[ert.) In this case,t}le voltage responseis underdamped since ofr > L d) For critical damping, a' : (,3, so

FrcmEqs.8.14 and8.15, \ - -1.25^ td + r"ftsox = -12,500+ 7500:

to3

to3

5000rad/s,

\2= -r.25^ ro'- lffirsox ro' rd :

12,500 7500:

rr = -8000 + i6000 radls,

20,000 rad/s.

b) The voltageresponseis overdampedbecause oZ < a).

(#)'=#='.'' #='*' 106

(2 x 10)(0.2)

: 250().

Circuit 291 ofa PaEttetRlC s.2 TheForms ofthe NatuBtResDonse

of parattelruCcircuits obj€dive 1-Be abteto determinethe naturalresponseandth€ st€presponse 8.1

The resistanceand inductanceofthe circuit in Fig.8.5are 100O and 20 mH, rcspectively. a) Find the valueof Clhat makesthe voltage responsecriticallydamPed. b) If Cis adjustedto give a neperfrequencyof 5 krad/s,find the valueof C and the roots of the charactedsticequation. c) If C is adjustedto give a resonantlrequcncy of20 krad/s,find the value of C and the equation, roots of the characteristic

Ansu,er:(a) 500nF; (b)c:1pF, 11: 5000+ j5000rad/s, r, = 5000- i5000rad/s; (c) C - r2snF, '!r - -5359rad/s, s2: -14,647radls

NOTE: Also by ChapterPtublem8.1.

of th* !$aturai$tesB*nse 8.2 TlteForsns 8l{ Ciieuit ef a Far*tte!. Sofar we have seenthat the behavior of a second-orderR LC circuit depends on the valuesof rt and 12,which in turn dependon ihe circuit paramete$ R, l-. and C.Thercfore, the first step ill finding the natural responseis to calcu late thesevalucsand,relatedly,dctcrmine whetherthe rcsponseis over , under-,or criticallydamped. Completing the description of the natural resPonserequires finding two untnown coetficientqsuchas A1and,4, in Eq.8.13.Themcthodusedto do this is basedon rnatchingthe solulion fbr the natural rcsponscto the initial conditionsimposedby the circuil, which are ihe idtial value of the cufent (or voltage) and the initial value of lhe Iirsl derivative of the omenl (or voltage). Note that thesesamejnitial conditions,plus thc final value of the variable,wil of a second-order circuit alsobe neededwhenfindingthe stepresponsc In this section, wc analyze the natuml responseform {or each of the three types of damping,beginningwith the overdampedresponse.As we will see, the response equations, as well as lhe equations for evaluating the unknown coefficients,are slightly different for each of the three dampjng configurations.This is why we want to determine al thc outsetof the Problcm whether the responseis over . under-.or critically damped.

VoltageResponse Theoverdamped equationarereal anddistinct,thevoltWhen lhe rootsof the charactenstic solu ageresponseof a parallelRLC circuitis saidto be overdamped.The tion for the voltage is of the lorm r-A1e\r+A2et1t,

(8.18)'j! voLtage naturaIresponse-ovetdarnped parattetnlf circuit

NatuntandSiepResponses 0f flr Circuitjs where J1and s2are the roots of the characteristic equation, The constants ,41 and ,42 are detemined by the initial conditiorq speciiically from the vaiuesof r(0 ) andd0\0 ) dr, wbich in rurn are determinedtr;m |he initial voltage on the capacitor, y0, and the initial current in tle inductor. 1n. \ext. we showhow to uselhe inilial\okageon lhe capacilordodthe initial current in the inductor to find ,41and ,42.First we note from Eq.8.18 rharA and,42.FirslwenoleftomEq.8.l8rhal

z'(0'):Ar+,4r, dr(o+)

: srAr + szA2.

(8.1e) (8.20)

wilh rr and12known.lhe taskot tinding/4| and,42reduces to finding u(0 ) and /rfo ) dI. The valueot r{0-) is rhe inirial vokageon lbe capacitor y0.We get the initiat value of dr/dt by first finding the currenr i; the capacitor branch at I - 0+.Then,

dr'(0*)

,.(0.) C

(8.21)

We use Kirchloffs current law to fhd the initial current in the capac_ itor bmnch. We ktrow that the sum of the three branch currents at r : 0+ musl be zerc. The currcnt in the resistive branch at r : 0+ is the initial voltage % divided by the resistance, and the curent in the inductive branch is 10.Using the reference system depicted in Fig. 8.5,we obtain -%

ic(o)

\8.2?)

After finding the numericalvalueoft6(0+),we useEq.8.21to fhd rhe jniIial \ante of da/dt . We can summarize the processfor finding the overdamped response, o(l), asfollows: 1. Find the ioots of the characteristic equation, st and s2,using the valuesof R, Z, and C 2. Find r'(0+) and do(O+)/dl using circuit analysis. 3. Find the values of "{r and ,,t2 by sotving Eqs. 8.23 and 8.24 simultaneously:

u(0-):A1+4, db(o+)

,c(o-) C

(8.23) s1A7+ s2Az.

\8.24)

4. Substitutethe valuesfor sl, 12,,41,and,42 into Eq.8.18 to determinerheexpression for ,(r) tor r > 0. Examples 8.2 and 8.3 illustrate how to find the overdamped responseof a pamllel RIC circuit.

Rr Cncuit 293 Reslonse of a Parattet 8.2 TheForms oftheNatunt

NaturalResponse of a Parattet 8tCCircuit Findingthe overdamped For the circuit in Fig. 8.6, o(0"):12V. ir(0') : 30 mA.

and

a) Find the initial current in each branch of the b) Find the initial value ofdt'ldr. c) Find the expressionfor 0(t). r d) skelch,(r) in rheinrerval0

Examph 8.2. Figure8.6 a Thecircuittor 250ss.

Solution a) The inductor preventsan instantaneouschange in its current,so the initial value oI the inductor currentis 30 mA: ir(0 ) : tr(0.) - i-(0-) : 30mA. The capacitorholdsthe initial voltageacrossthe parallel elementsto 12V Thus t}le initial current in tbe resistive branch, ir(O-), is 12/200, or 60 mA. Kirchhof|s currentlaw requircsthe sum of the currents leaving the top node to equal zero at everYinstant-Hence

rc(O) = rr(0) iR(o) - 90mA. Note that if we assumedthe inductor current and capacitor voltage had reached their dc values at the instant that energy begins to be released, tc(o-) = 0. In other words,there is an instantaneouschangein the capacitor current at I : 0. b) Becauseic : C(.tu/dt),

du('') dt

0.2pF

the form ofEq.8.i8.We find the co-efficientsA1 and ,4, from Eqs.8.23 and 8.24.We've already determinedrr, r,, r'(0+),and dt'(0*)/dr, so 12: Ar+ A2, 450 x 1d: 5000/r - 20,000/r. We solve two equations for ,4r and ,4, to obtajn Ar : 1.4y and A2 - 26 V. substitutingthese valuesinto Eq.8.18ields the overdampedvolt ageresponse: '"' t \. r . .. ( t p-smolt- 2bp) ,(/) As a check on these calculations,we note that the solution yields o(0) : 12 v and da(o*)/dt

:

4s0,000v/s.

d) Figure 8.7 showsa plot of o(l) versus1 over the interval0 < r < 250ps.

,(, (v) 12 10

9 0 . r 0 r: 4 5 0 k v / s .

0.2 x 10-6

c) The roots of the characteristicequation come from the values of R, Z, and C. For the values specifiedand from Eqs.8.14and 8.15alongwith 8.16and It.17,

. t'T tF c, - -t.25 . t0" \,{5= -12,500+ 7500= 5000rad/s, r.25' l0'- \,/i5b25 r0' u' 12 : -12,500- 7500: 20,000 rad/s. Because therootsarerealanddistinct,weknow is overdamped andhencehas thattheresponse

rcsponse for Example 8.2. Figure8.7 A Thevottage

294

Naturaland StepRespomes ofR|rCircuits

Catcutating BranchCurrents in the NaturalResponse of a parattet flc Circuit Derivc the expressionsthat describe the rnrce branch currents tx, ir, and r. in Examplc 8.2 (Fig-Il.6) during thc lime the storcdenergyis bcing

A second approach is to find rhe current in the capacitivcbranch first and then use the tacl that iR + it. + ic = 0. Let's usc this approach.The current in thc capacitivebranchis

,.(r)= Solution

5r{r0' 520,000ar0,000/ : 0.2 x 101j(70.000e )

We know the voltage acrossthe thrcc branches frollr the solLrtionin Example8.2,narncly, solro' + 26e 2oooo') v, 44 = ( L1e

= (14e tooo/ 104e'oooo/)mA,

. > 0.

The currcntin the resistivebranchis theD i"tt\:

:I , 70. s000, + I tn, :0000r \, ,m, ,r A . ' : 200

u.

'I}lcre

rre two waysLofind the curent in thc inductive branch.Onc wav is to usc the integralrelationship lhat existsbclweenthe currentand rhc voltage allhe terminalsoI an inductor: it(t) =

c+

lI',uo*

t1o

Notc thal tc(0*) = 90 mA, which agreeswirh th.rcsull in Examplei1.2. Now we obtain the iDductivebranch curreni &om the relationship iL(t) = -iRQ)

ic(t)

= (56d 5ooor 26d 2o,ooo,) mA,

Use the integralrelationshipbelweeni, and ? to find rhce\pfes.i.nfor r, in fig.U.o.

Answer: tr(1) = (56e taht

8.3

26e 20m,1 m,r, r - U.

The elcmentvaluesin the circuit showtrare R - 2kO, Z = 250mH, and C : 10 nF. The iDitialcurrent10in the inductoris 4 A, and the initjal voltageon the capacitoris 0V The output signalis the voltage2,.Find (a) t,r(o+); (.h)ic(1'); (c) dr(c')/.Lt, (d) ,41:(e),4r;ard (f) ?J(1) when I > 0.

NOTE: AIso try ChapterPrcblens 8.2,8.5,an(t8.19.

r > rr.

We leave it to you, in AssessmentProbtem8.2,10 show that the integral relation alluded to leads to the saDe result. Note that thc expressionfor iz agreeswith thc initial inductor curreDt,as it musl.

obiedive 1-Be abteto determinethe naturalresponseand the stepresponse of parattet f,lf circuits 8.2

I > 0+

(a) o; (b),rA; (c) ,l x 103v/s; (d) 13.333 V; (e) -13,333V; - e rn,mo (0 13,333(e-10.0''0' ) v.

nrcircuit ofih€Naturat Response ofa Paiatkt 8.2 rheFonns

295

TheUnderdamped VoltageResponse Whetr06 > l. $e roots of lhe characrerislicequadoDare complex.and the response is underdamped, For convenience, we express the rcots sl and J2as

:

a + j\/;{-i (8.25)

(8.26)

(8.27) < Damped radianfrequency

(8.28)

(8.?9)

+ i(/r

- A)snt adt).

< vottagenatoralresponse-undedamped DarattelruCcircuits

296

N a t u € t a nSdt e pR e s p o 6 e sRoLf t c i ( u i t s

At this point in the transitionfrom Eq. 8.18to 8.28,replacet}le arbitiary constants,4r + ,4, and t(A1 - At with new arbitary constantsde oted 81 and B, to get o = € "'(Br cos,rdt + 82 sin.dd, : B1e "t cosadt + B2e-"tsin.odt.

The constants-Brand 82 are real,not complex,becausethe voltageis a real tunction.Don't be misledby the fact that B, : j(,4r - ,4r). In this and thus81 and 82 underdampedcase,,4rand ,42are complexconjugates, are real.(SeeProblems8.13and 8.14.)Thereasonfor definingthe underdampedresponsein terms of the coefficients81 and 82 is that it yieldsa simpler expression for the voltage, ,. We determine B1 and B, by the initial energystoredin the circuit,in the sameway that we found,4r and ,42 for the overdamped response:by evaluating ?Jat t : 0' and its derivative at I : 0*. As with sl and sr, a and ro2are fixed by the circuit pammeters R, L,ancl C. For the underdamped response,the two simultaneous equations that determifle 81 and B2 are

(8.30)

u(0'):uo-81,

at'(o) _ i.(o) : d

t

C

dBr + odBz.

(8.31)

Fint, Let's look at the generalnatue of the underdampedresponse. the trigonometric functioff indicate tlat this responseis oscillatory; that is, the voltage alternates between positive and negative values.The rate at which the voltage oscillates is fixed by .dd.Second,the amplitude of the oscillation decreaseser?onentially. The iate at which the amplitude falls off is detemined by a. Becaused detemines how quickly the oscillations subside,it is also referred to as the damping factor or dsmping coeflicient. That explains why (1),? is called the damped radian frequency. If there is no damping, d = 0 and th€ frequency of oscillation is o0. Whenever there is a dissipativeelemenqR, in the circuit, a is not zero and the ftequencyof oscillation, ar, is less than o0. Thus when .Yis not zeio, the frequency of oscillationis saidto be damped. The oscillatory behavior is possiblebecauseof the two Opes of energystomge elementsin the circuil the inductor and the capacitor. (A mechanical analogy of this electric circuit is that of a masssuspendedon a spring, where oscillation is possible because erergy can be stored in both the spring and the moving mass.)We saymore about the characteistics of the underdamped responsefollowing Example 8.4, which examines a circuit whose response is underdamped. In summary, note that the overall processfor finding the underdamped responseis the same as that tbr the overdampedresponse,althoughthe responseequalionsand the simultaneous equalions used to find the constants are slighdy different.

Response ofa ParattetRlr Cncuit 297 8.2 TheFomsoftheNatunL

NaturalResponse of a Parattet flc Circuit Findingthe Underdamped In the circuit shown in Fig. 8.8, Vo:o, Io: 12.25mA.

and

a) Calculate the ioots of the characteristic equation. b) CalculateD and do/dt at t : 0'. c) Calculate the voltage responsefor t > 0. d) Plot o(t) versus I for the time interval 0
explicitly. However, this example emphasizes $b! r' a0d\? arekno$.nascomple\frequencies. b) Becauseo is the voltageaqossthe terminalsof a capaclror,we nave 0(0)=u(0-)=Yo:0' Because?(0+) = 0, the current in the rcsistive branch is zero at t = 0+. Hence the current in the capacitor at I = 0- is the negative of the hductor current:

tc(0') = 0.125pF

( 12.25)= 12.2sr]l,4.

Thereforethe initial valueof the derivativer d u { o + ) { 1 2 . 2 5 ) (r1) 0 - - - - - , ,

d/

Flgure8.8 a Theci(uitfo' E{anrpte 8.4.

:

-:9Xli(n)V/s.

(0.r2s)(10 )

c) From Eqs.8.30and 8.31,Br = 0 and B, =

oR non -'''''

c 100V.

Solution Substitutingthe numerical valuesof a, od, 81, and 82 into the exprcssion for ?r(t) gives

a) Because 1

106

2RC 2(2q1d9.12s\

',77

: 200rad/s,

- "f-.'ff = 103md/s, (8X0.125) \

.3 > "'.

0(t) = looa'misin 97980r v,

t > 0.

d) Figure8.9 showsthe plot oI t (t) ve$us I for the fkst 11 ms after the stored eneryy is released.It clearly indicatesthe damped oscillatorynature of the underdampedresponse.The voltage 2{t) approachesits final value,alternating between values that are greater than and less than the fiml value.Fu hemore, theseswingsabout the final value deqease exponentialy wittr time.

Therefore, the responseis underdamped. Now,

a6: ^,/S ]:

r,,iFJxld:

roor,66

: 979.80radls, st:

a + jod:

s2 = -a - jad =

-200 + j97980rad/s, 200

j979.80rad/s.

For t}le underdamped case,we do not ordinadly solvefor J1 and 12becausewe do not usethem

, (v) 80 60 40 20 0 -20 40

7 8 9

tesponse for Exampte 8.4. Figure8.9 A ThevoLtage

,(*)

298

Naturatand StepResponses ofntaCircuih

Characteristics of the Underdamped Response The underdampedresponsehasseveralimporranl characterisrics. Firsr.as the dissipatiyelossesin the circuil decrease. rhe pcrsistenceof thc oscillations incrcaser and the frcquencv of the oscillalionsapproachcso0. I'r other words.asR+,6. thcdissipationin thecircuir in Fig.8.ltapproaches zero bccausep : i)7R. As R-- co. a+ 0. which tells us lhat o,j + rr0. When a : 0. the maximum amplirudeof rhc voltage rentainsconslant: thus the oscillationat .r0is sustained.In Example8.4,if /t wcrc increascd to infinily, the solutionfor o(r) \\'ouldbccorne 1Jt) : 98 sin 1000rV.

r > 0.

'lhus,

iD this casethe oscillarior is sustaincd,the maximum amplitudeof thc voltageis 98 V and the frequencyof oscillationis 1000rad/s. Wc Draynow describequalitarivelythc differencebetwccnan undcrdiDped and an ovcrdampedresponsc.In an underdampcdsystem.thc lcspoNe oscillates,or "bounces,"aboul ils fiDal value.This oscilLalionis also rcferred to as rinqing. In an ovcrdamped system,the responsc approachesits final value without ringing or in what is sometjmcs dcscibed asa "sluggish"manner.Whcn specifyingthe dcsircdresponscof a sccondordersystcmryou mav want to reachthe final value in the sho(csl tiDe possiblc.and you may not be concernedwith smaLloscillations about that final value.If so,you would designthe systcmcomponentsto achievean underdampedresponse.On the other hand.you nlay bc concornedthat thc responsenot exc€edits final value.pcrhapsto ensurcthat oomponentsarc not damaged.Insucha case,youwould dcsignthe systcm componentsto achievean overdampedresponse,and you would havc to :rc(€pri relalivcl)\lu\ ri.e ro rhclinJlvalu(.

obj€ctivel-Be ableto detemine the naturaland the st€p responseof paraltelilc circuits 8.4

A 10 trI}I inductor,a 1pF capacitor,and a variable resistorare connectedin parallelin the circuit shown.The resistoris adjustedso thal the roots ofthe characteristicequationarc -8000 ,r i6000 rad/s.The initial volrageon the capacitoris 10Y and the initial cuffent in thc inductor is 80 mA. Find a) Ri b) dr(o+)ldt; c) Br and 82 in the solutionfor ?);and

a);r(r). NOTE: Aho tty ChdpterProblems8.3 antl8.20.

Answ€r: (a) 62.5O; (b) -240,000v/s; (c) -Br: 10v, B, = -80/3 V; (d) rr.(r)= 10e3mo,[8cos6000t + (82/3)sin600011 mA whenI > 0.

Response of a PanLtel RrcCncLrit 299 8.2 TheForms ofthe NaturaL

TheCriticattyDamped VoltageResponse The second-ordercircuit in Fig.8.8is critically dampedwhen ofr : o', or o0 : a. When a circuit is critically damped,the responseis on the verge of oscillating.In addition, the two roots of the characteristicequation are real and equalt that is,

1 2RC

(8.32)

When this occun, the solution for the voltage no longer takesthe form of Eq. 8.18.This equation breaksdown becauseif rr = 12 = a, it predictsthat D = \ A 1+ 4 ) e d :

AGd,

(8.33)

where ,40 is an arbitrary constant. Equation 8.33 cannot satisfy two inde' pendent initial conditions (y0, 10) with only one arbitrary constant, A0. Recallthat the circuitparametenR and Cfix d. We can trace this dilemmaback to the assumptionthat the solution takesthe form of Eq.8.18.wllen the roots of the characteristicequation are equal, the solution for the differential equation takes a different folm, namely

/8 14) < Vottage naturalresponse-criticall, parattelf,lCcircuit damped Thus in the caseof a repeated root, the solution involves a simple exponenlial telm plus the product of a linear and an exponential term. The justification of Eq. 8.34 is left for an intrcductory course in differential equations.Finding the solution involves obtaining ,1 and D2 by follovring the same pattem set in tle overdamped and underdamped cases:We use the initial values of ttre voltage and the derivative of the voltage with respect to time to write two equations containing Dr andlot D2. From Eq. 8.34,the two simultaneousequationsneededto detemine D, ard D, are

a(.r)=uo:D2, do(o) _ ic(o*)_ Dr

(8.35)

aD2.

(8.36)

As we can see,in the case of a critically damped response,both the equation for o(l) and the simultaneous equations for the constants,r and D2 diJfer ftom tlos€ lor over- and underdamped responses,but the general approach is the same.You will rarcly encounter critically damped systems in practicg largely becauseoo must equal a €xactly.Both of these quartities depend on circuit parameten, and in a real circuit it is very difficult to choosecomponent values that satisfy an exact equality relationship. Example8.5illustratesthe approachfor finding the criticallydamped resoonseof a oarallelRIC circuit,

300 NatuDtandStepResponses ofiltOrcurts

Findingthe CriticattyDamped NaturaIResponse of a Paratlel ntc Circuit a) For thc circuit in Example8.4 (Fig.8.8),find the valucoIn thairesultsin a criticallydampedvolt

Eqs.8.35and 8.36,r, : 0 and ,i = 98,000V/s. Substjtutinglhesevaluesfor a. ,1, and ,2 inlo Eq.8.34gives

b) Calculate!(r) for r > 0. c) Irlot 1r(l)vcrsusrfor 0 < t = 7 ms.

? ) ( i )= 9 8 , 0 0 0 t i ? r o m ' rv>, 0 . c) Figure 8.10 showsa plot of 2,0) versus1in thc intervall)= I < 7ms.

Sotution - r f f o m I \ J n r n , c8 . 4 .$ < k n o w r h " r u ; Thereforefor crilical damping.

l0'.

1 )RC.

"(v) -12 21

106 : 4000(). (2000)(0.12s)

8 0

b) Fron the solutionof Example8.4.we know that u(0*) - 0 and dr(0+\ldt = 98,000v/s. From

1

, (mt

Figure8.10,i llre vottage rcsponse for Exanrph 8.5.

obiedive 1-86 abteto determin€the nai|jraland the st€p r€sponse of paralletf,lCcircuits 8,5

The resistorin ihe circuit in Assessment PfoblemR.4i. idtu.led tof crilicaldamping The inductanccand capacitance valuesare 0.4H and 10 /..F,respectively. Ihe inirial energy storedin thc circuit is 25 mJ ard is distributed equallybetweenthe ind ctor and capacitor. Find (a) R:(b) yor(c) 10;(d) Dr and ,, in the soiution Ior t';and (e) in. I > 0*.

Answer: (a) 100O:

(b)s0v; (c) 2s0mA; (d) -50.000v/s,50 V; (e) tn(r) = (-500re500.+0.50rs00)A, t=t)'

NOTE: AIso *j ChapterProblems8.4dnrl8.21.

A Summary of the Results We concludeour discussionofthe parallcl R/,Ccircuir\ naturalresponse with a bricf sunmary of the results.The first slep iD finding the natural responscis to calculatethe roots of the characteristicequation.You thcn knowimmediatelywhetherthe rcsponseis overdamped.underdampcd,or criticallydamped. If the roots are real and dislincl (o3 < a2), the responseis over dampcdand the voltageis

n(.t): A'd."'+ Aze''.

8.1 TheStepRespo6e ofa Pa,altetRtC Circuit

\/7-', 1 2RC'

" -

l

"u

Lc

The values of -41and ,42 are determined by solving the following simultaneousequatlons: D(0*)= A1 +A2, dr'(0+r l.{0+l d

t

L

If the ioots are complex ..rA> d' t}le response is underdamped ard the voltageis D(t) - B1e "'cosadt + B2e-"'sinadt,

-,

r-a

"

The values of Br and B, are found by solving t}le following simultaneous 1)(0-):uo:Br, dDt\+\ - ' i.tq+ | ( =-aB -u"B dr li the roots of the characteristicequation are real and equal (dA : a'), the volfage responseis a(t) = D1rc-a'+ Dp d', where a is as in the other solution forms, To determine values for the constants,1 and D2,solvethe folowing simultaneousequations: a(O-):Uo=D2,

/u(01

ic(u+)

8.3 TheStepResponse of a Paraltel

RICCircuit Finding the step responseof a parallel RaC circuit involvesfinding rhe voltage acrossthe parallel branchesor the current in the individual branchesas a result of the sudden application of a dc current source.Therc may or may not be energy stored in the circuit when the current source is applied.The task is representedby the circuit shown in Fig. 8.11.To Figlre8.111A circuitLrsed to de(fibethestep of a paGlletfr circuit. develop a general approach to finding the step responseof a second-order rcsponse

301

102

Natunt andSiepResponses of,?rrCircujts circuit, we focus on finding the current in the inductive branch (ir). This current is of particular interest becauseit does not approachzero as I increases.Rather, after the switch has been open for a long time, the inductoi current equals Uredc sourcecurrent I Becausewe want to focus on tlle lechdque for finding rhe step response,we assumethat tie initial energy stored in tlle circuit is zero.This assumptionsimplifies t]le calculations and doesn't alter the basic processinvolved. In Example 8.10 we will see how the presenceof idtially stored energy enters into the general procedure. To find the inductor current ir, we must solve a second-order differ, ential equation equated to the forcing lunction 1, which we derive as fol lows.From Kirchhoff's curent law. we have iL+iR+ic:1,

i,r!*c!:r.

(8.37)

Because - cllL

(8.38)

we get (Lu dt

- r12i, dt2'

(8.3e)

SubstitutingEqs.8.38and 8.39into Eq.8.37gives L (li, iL+=-:1

i LC.:-L dr

(8.40)

For convenienc€,we divide througl by -LC and rearrange terms: d2i, dt!

I

dit

RC d!

i,

t

LC

LC

(8.41)

Compadng Eq. 8.41 with Eq. 8.3 reveals that the presence of a nonzero term on the dght-hand side of the equation alters the task. Before showing how to solve Eq. 8.41 directly, we obtain the solution indirectly. When we know the solution of Eq.8.41, explai ng the direct approach will be easier.

TheIndirectApproach We can solve foi ir indhectly by first finding the voltage o. We do this with the techniquesintroducedin Section8.2,because the differentialequation that ?Jmust satisfyis identicalto Eq. 8.3.To seetlis, we simply return to Eq. 8.37and expressi, as a functioDof r; thus

l[',0,*i*rolt='

18.42)

8.3 lheSteD ResDonse of a Panttet flrCircuit 303 Differentiating Eq. 8.42 once with respecl to t reduces the righfhand side to zero because1 is a constant.Thus o L

d2u

ldu t< dt

^dz\ ar

1 cla RC dt

^

u LC

(8.43)

As discussed in Section8.2,the solutionfor t, dependson the roots ofthe characteristic equation. Thus the three Dossiblesolutions are b=Aretn+A2es|t,

(8.44)

b = Bpn'

(8.45)

a=

Dle-"t

cos tJdt + B2e-"'sin adt, + Dze-"t.

\8.46)

A word of caution: Because there is a souce in the circuit for l > 0, you must tate into account the value of the sourcecurrent at t = 0* when you evaluate the coefficients h Eqs.8.4ll-8.46. To fhd the three possiblesolutions for ir, we substitute Eqs.8.44-8.46 into Eq. 8.37.You should be able to vedry, when this has been done, that the three solutions for i, vrill be iL:I+Aid't+Aie"1t, iL:

I + B'Ie-'cosaat + Bie "Isind,1t.

iL=I+Dita"t+Die"t,

\8.47) (8.18) (8.4e)

where Ai, Ai, Bi, Bi, Di, and Di, arc arbitrary constants. Ir each case,the primed constantscan be found indiiectly in tems of the arbitrary constantsassociatedwith the voltage solution. However, this approach is cumbe$ome.

TheDired Approach It is much easier to find tle primed constants directly in tems of the initial valuesof the responsefunction.Foi the circuit being discussed,we would find the pimed constants from iz(o) and dir(0)/dt. The solution for a second-order differential equation witl a constant forcing function equals the foiced responseplus a responsefunction identical in form to the natural response.Thuswe can alwayswrite the solution for the steo resoonsein the form I lunctionof lhe samefolm I I = r/ + asthenat"*i *"p.*.J' |

(8.50)

304 Naturat andStepResponses of Rtf Circujts

ol thesameformI ' = -. vr + lfunctron I astheDot-"r*"p.'*J'

(8.51)

where 1t and yi represent tbe final value of the response function. The fillal value may be zero, as was,for example,the casewith the voltage x' in tlle c cuit in Fig.8.8. Examples 8.G8.10 illustrate ttre technique of finding the step responseofa parallelRaCcircuit usingthe direct approach.

Findingthe overdamped StepResponse of a ParalteI flc Circuit The initial energystoredin the circuit in Fig.8.12is zero. At t : 0, a dc current source of 24 mA is applied to the circuit. The value oI the resistor is 400 (). a) Wlat is the initial value of ir? b) What is the initial value of diLldt? c) wllat are the roots of the characteristicequation? d) W}lat is t}le numerical expression for iz(r) when t>0?

c) From tlle circuirelements.weobtain

.a-== o

r h. r 0 " ,

a' =25r103. Becauseofr < l, fie roots oI the chamcteristic equationarercal anddistinct.Thus ,1 : 5 x 104+3 x 104= -20,000 rad/s, .r2:

Figure8.12.AThecircuiitorExampte 8.6.

^:=

\z))\z)) 1 loe 'RC ')(4uu,(25, t

5x10*

3 x 10'= -80,000 rad/s.

d) Because the rootsof the characteristic equation arerealanddistinct,theinductorcurent response will be overdamped. Thusir(l) takest}le form of Eq.8.47,namely, iL= If + A\es)t+ Aietxt

Solution a) No erergy is stored in the circuit prior to the applicationofthe dc curent source,so the initial current in the inductor is zero. The inductor prohibits ar instantaneouschangein inductor current; therefore il(o) = 0 immediately after the switchhasbeenopened. b) The initial voltage on the capacitor is zero before the switchhas been opened;ther€foreil will be zero immediately after. Now, because u: LdiL/dt,

!:tot : o.

> Indudorcurrentin overdamped parattet RICcircuitstepresponse Hence,ftom thissolution,the two simultaneous equations thatdetermine A\ andAi are i,(0)=./,+A'.+A\=0. ;(0)

= sL,ai+ 'rAi:0

Solvingfor ,4i and Ai gives , 4 j: 3 2 m A a n d A i = 8 m A . Thenumerical solutionfor ilo) is iLo : Qa , 32e20,m& + 8e-s0,1\J0' ) mA, I > 0.

8.1 lhesteoResoonse ola Para.teL flt LtrLut

305

Findingthe Underdamped StepResponse of a Parattet RICCircuit The resistorin the circuit in Example8.6 (Fig.8.12) is increasedto 625O. Find ,z(1)for I > 0.

Here, d is 32,000 rad/s, rrr is 24,000 rad/s, and is 24 mA. 4 As in Example8.6,Bi andBi aredetermined from the initial conditioN Thusthe two simultare-

Solution rr(0)=1f+Bi=0,

BecauseI, and C remain fixed, oA has the same value as in Example 8.6; that is, o; = 16 x 103. Increasing n to 625O decreases a to 3.2 x 10amd/s. With ofr > a', the roots of the charactedsticequationare complex.Hence

*Q)

3.2 x 10"+ j2.4 x 10"rad/s,

= dBt - "B't: o.

B't : -21m4

3.2 x 10- j2.4 x 10"rad/s. The cu{ent responseis now underdamped and givenby 8q.8.,18:

Bi:

-32 mA.

Thenume cal solutionfor ir(t) is itul = It + Btc"'cosd.t. Be't:inuut.

paratler > Inductorcurrentin underdamped ruCcircuitstepr€sponse

iLO = (24

24e 32ac4t cos24,0}0t

- 32e'2ooo'sin 24.ooor) mA. r > o.

Findingthe CriticatlyDamped StepResponse of a ParallelRlc circuit The resistorin the circuitin Example8.6 (Fig.8.12) is setat 500O. Find ir for I > 0.

Again, Di and Di are computed from initial i L ( O )= I r + D ! 2 = 0 ,

Sotution We know tlat oAremainsat 16 x l0s.WithRsetat 500O, d becomes4 x 104s 1, which corresponds to c tical damping. Therefore tlre solution for i_(1) takesthe form of Eq.8.49:

!!,or=o',-oo,,:0. d

'

Thus ,i =

itltl:

l

mA/s and Di: 960,000

24m1..

It + D\te "' + Diad.

Thenumericalexpression for iz(t) is > Inductorcunentin criticalLydampedpanlleL nlc circuit step response

iLO : Q1 - 960,000rerq0m'24t10m0') mA, r>0.

306

NatuEtand SiepResponses 0fRlfCncuits

Comparing the Three-Step Response Forms a) Plot on a single graph, over a range lrom 0 to 220!,s, the overdamped, underdamped, and cdtically damped responses derived in Examples8.6-8.8. b) Use the plots of (a) to find the time requiredfor tr to reach90o/.ofits final value. c) On the basisolthe rcsultsobtainedin (b), which responsewould you specityh a designt}lat puts a premiumon reaching90% ofthe final valueof tlte output in the shortest time? d) W}lich responsewould you specify in a design that must ensure that the final value of the current is neverexceeded?

Sotution a) SeeFig.8.13. b) The final value of ir. is 24 mA, so we can read the timesoffthe plots correspondiq to ir = 21.6mA. Thusrd,l= 130/.rs,r.r = 91 ps,andt"d : 74 ps. c) The underdampedresponseieaches90o/oof the final value in the fastesttime,so it is the desired responsetype when speedis the most importart designspecification.

(/l = 625O) Underdanped

26 22 18 l4 10

1n=+oooy fi- licalltlo'!'a-'p.a = (1l daDpcd 500O) tC

2 i)

140

180

I(it

Figure8.13.rr Thecun€niptotsfor Example 8.9.

d)From the plot, you can see that the underdamped responseove$hoots the final value of current, whereasneither the critically damped nor the overdampedrespoNe producescurents in excessof 24 mA. Although specifying either of the latter two responseswould meet the design specificatior, it is best to use the overdamped fesponse. Ir $ould be impraclicalto requirca design to achievethe exact component values tlat ensurea criticallydampedresponse.

FindingStepResponse of a ParallelRLCCircuitwith Initial StoredEnergy Energy is stored in the circuit in Example 8.8 (Fig.8.12,withR = 500 O) at the instantthe dc cur, rent source is applied.The initial current in the inductoris 29 mA, and the initial voltageacrossthe capacitor is s0 v Find (a) iL(o); (b) diL(o)/dt: (c) tr(r) for I > 0; (d) u(D for t > 0.

b) The capacitor holds the initial voltage acrossthe inductor to 50V Therefore

r;(o-)

: 50,

lo1l:fl

" 103:2oooA/s.

So[ution a) Therecannotbe an instartaneouschangeof cu. rent in an inductor, so the initial value of ir in the first instant after the dc curent source has been appliedmust be 29 mA.

") From the solution of Example 8.8,we kno\q that the currentresponseis citically damped.Thus iL(t) = Ir + D\te "' + Die ""

8.3 TheStepResponse ofa Panttetilrcncujt 307 Thus the nume cal expressionfor ir(l) is d

^:^

4 0 . 0 0r0a d / s a n d / , - ' 4 m A .

Notice that the effect of the nonzero initial stoied energy is on the calculations for the constants Di and ,i, which we obtain tom the nntial conditions.First we use the initial value of the inductorcurent:

iLC) = (.24+ 2.2 x l06reroooor + se-aoooo) mA,

t > 0.

d.)We can get the expressionfor ?J(t),t > 0 by using the relationshipbetweenthe voltage and cment in an inductor:

tr(0) = 1r + Di - 29mA., T(, : L;

from which we get

Di:29

24

' 1o r) 2.2

Thesolutionfor ,i is

106)(-4o.ooo)reroooo'

+ 2,2 X I06e-4aIUt

=(0-)

+ (5)( 40,000)?ro!m/l x 10 3 :

+ 50? {,000,V, I > 0_ 2.2 x 106tua1,auJt

D'r=2000+aDtz : 2000+ (40,000)(5 x 10 ) : 2200Als : 2.2 x 106nA/s.

To check this result, let's verify that the initial voltage acrossthe inductor is 50 V:

D()t)-

2.2x 106(ox1) + 500):50v.

StepResponses of RtCCircuits 308 NatuGtand

8.4 TheNaturalandStepResponse of a SeriesRICCircuit +

Figure 8.14A A circuiiused to itLustrate thenaturat response of a series Rlrcircuit.

The procedures for finding the natural or step respoNes of a seriesRZC circuit are the same asthose used to find the natural or step responsesof a parallel RIC circuit, becauseboth circuits are describedby differertial equationsthat have t}le sameform. We begin by sumnine the voltages aroundthe closedpath in the circuit shownin Fig.8.14.Thus ,ii

Ri

i

L",'

fl

^lidttva-j.

dt

(8.52)

cJo

We now differentiateEq.8.52oncewitl respectto I to get _ d' i - i - o .tr c

_di ar

(8.53)

which we can reanange as dzi

Rdi

i

,t,'-iA-i

"'

18.54J

ComparingEq. 8.54with Eq. 8.3 revealsthat tiey have the sameform. Therefore,lofind the solutionof Eq.8.54,we follow the sameprocessthat led us to the solutionofEq.8.3. From Eo.8.54. the characteristic eouation for the sedesRIC circuit is

Characteristic equation-series f,aCcircuit >

(8.55)

R 2t.

/RY \i)

(8.56)

(8.57)

The neperfrequency(d) for the sedesRIC circuit is Neperfrequenq'-series Rlf circuit>

. = ,| ,"0u.

(8.58)

andtheexpression for theresonant radianfrequency is Resonant radianfrequency-series nlc circuit>

(8.5e)

Note tlat tle neper frequency of the seriesRIC circuit differs ftom that of the parallelRaC circuit,but the resonantmdian frequenciesare the same.

8.4 TheNatuaLand StepResponse of a5e esffrCjrcujt 309 The currentresponsewill be overdamped,underdamped,or critically damped accordjng to whether rofr < c', rofr > c', or tafr = a', respectively. Thusthe threepossiblesolutionsfor the curent are asfollows: i(t\ = nt""'

+ "\rer,t (overdamped), i(t, : Bf-"' cos.Dat+ B2e "'sinadt (underdamped), i(t) = 1;rt" "r + Dr€-"! (criticallydamped).

(8.60)

f8.ol) < CurrentnaturaL response formsin series ,- _- l(ta orcults

When you have obtainedthe natural current response,you can find the natural voltage responseacrossany circuit element. To verifl that the procedure for finding the step responseof a series RaCcircuit isthe sameasthatfor a parallelRIC circuit,we showthat ihe differentialequationthat descibes the capacitorvoltagein Fig. 8.15has the sameform as the differential equation l]rat describesthe inductor current in Fig.8.11.For convenience, we assumethaazero energyis storedin the circuii at the instantthe switchis closed. Applying Kirchhoff'svoltagelawto the circuitshowr in Fig.8.15gives Figure8.15A A circuitusedto ittushate thesiep rcsponse of a se.ies fircncuit.

V - R i + L ! + D . '.

(e.63)

/11

Th€ current(i) is relatedto the capacitorvoltage(oc) by the expression (8.64)

from which (8.65)

Substitute Eqs.8.64 and 8.65 into Eq.8.63 and write the resulting

d?t4

R dua

* - i a -

:!_ LC

V I.C

(8.66)

Equation8.66hasthe sameform as Eq. 8.41;thereforethe procedurefor finding ,c parallelsthat for finding ir. The three possiblesolutionsfor z'c are asfollows:

D . - V + A\e"lt+ A?\r (oveidamped),

(8.67)

r c = V f + Bie "' cosodt + Bie-"' sln @,1 (undeidamped), (8.68) { Capacitor vottagestepresponse formsin series nlC circuits "t D c : u f + Dtte + Diad (criticallydamped), (s.6e) whereyl is the final valueof rc. Hence,from the circuil shownin Fig.8.15, the final value of z'c is t}le dc source voltage y. Example8.11and 8.12illustratethe mechanicsof finding the natural and steDresDonses ofa seriesRIC circuit.

310

NaturaLand StepResponses ofnlf Circuits

Findingthe Underdamped NaturalResponse of a SeriesRLCCircuit The 0.1pF capacitorin the circuit shown rn Fig.8.16is chargedto 100V At l : 0 the capacitor is dischargedthrougha seriescombinationof a 100mH inductoranda 560O resistor. a) Findj(r) for r > 0. b) Findoc(r)for r > 0.

the initial conditions.The inductor current is zero before the switch has been closed, and henceit is zeroimmediatelyafter.Therefore (0)=0=81. Io find 82.we evalualeJir0 l Jr. Fromlhe cil cuit, we note that,becausei(0) : 0 immediately alter the switch has been closed,there vall be no voltagedrop acrcssthe resistor.Thus the initial voltageon the capacitorappearsacrossthe ter minals of the inductor, which leads to the

di(0-]' Figur€8.16 A lhe circuiifor E)Gnpte 8.11.

di(o-)

Solution a) The filst step to finding i0) is to calculatethe root\of rhecharacleri(lic equadon, For rbegi\en

@ 6 =_

(ld)(r01 (100x0.1)

= 1000A/s. Because B1 : 0, co,so00r- Tsinab0o/j. : - a00?,cuu'(24

Ll1

Thus d(0)

= 960082,

R " 2 L 8, :

= -!!L ^ |n' 2(100)

: 2800rad/s. Next,wecornparearfrto a2andnotethat oA > 02, l=7.84x106 = 0.0784x 103. At thispoinl.$e know lbal lbe re\poDse is underdamped andthat the solutionfor i(t) is of the folm i(t) : B( "t cosa,tt+ B2a"t sinridt, where o = 2800rad/s and od: 9600md/s. Thenumericalvaluesof B1 and ,, comefrom

=T=rr*'t"

t000 s 0.1M2 A. 9600

The solution foi i(t) is t(/)

go00/A. r j 0. 0.104)e-2m'\in

b) To find oc(t), we can use either of the followitrg relationships:

"= -tl''o'*'ooo' ,1,

.- = t R + L 1 . llt

Wichever expressionis used (the secondis reconmended), the result is oc(r) = (100cos9600r+ 29.17sin 9600t)e'm'V,

r > 0.

8.4 TheNatunL andStepResponse of a sedesSLrCjrcuii 311

Findingthe Underdamped Step Response of a Seriesffc Circuit No energyis storedin thc 100mI-I inductor or the 0.4r.F capacitor when the switch in the circuit shownin Fig.B.l7 is closed.Find z,c(t)for t = 0.

The roots are complex.so the voltagc respoise is underdamped.'nrus 0c0):

'18 + Bid lroo'cos4800r + rle rao(r'sin4800r,r > 0.

No energyis stored in thc circuil initially. so both ,c(0) and d?rc(o-)/lt arc zcro.Ther, 0c(0):0=,18+Bi, F i g u r e8 . 1 7h T h e. i r . u i tf o rE a n r p l8e. i 2 .

dt).(0'\ __:t)

Soh'ingfor ai and Bi yields

5olution

ai : -18v,

The roots ofthe characteisticequationare

:

=4'rn0Ri 1400R

280

//zso\. -

+ 1 / l

|

0.2 v\0.2/

Bi:

106

tn.r)(rl4)

- 14V.

Therefore,the solutionfor ,(.(r) is

: ( 1400+ i1800)rad/s.

Dc(t): (48 - 48eraoo'cos4s00r

r, = (-1100 - j1800)rad/s.

- 14r r{orsin 4800r)V.

objective2-Be ableto determine the naturalresponse andthe stepresponse of s€riesftf circuits 8.7

The switchin the circuitshownhasbeenin positiona tor a long 1ime.At t = 0- it movesto positionb. Find (a) l(0+);(b) "c(0+); (c) dl(0')/dr; (d) .rl, rr:and (e) t(D for I > 0.

Answer: (a) 0;

(b)s0Y (c) 10,000 A/s; (d) ( 8000+ 16000)radls, ( 8000 j6000)rad/s; (e) (1.67€Mo'sin60001) A for r > 0. NOTE: ALto try Chteter Problems8.15-8.17

Findoc(l) for I > 0 for thecircnitin Assessment Problem8.7. Answer: [100 e 3mi(50cos 6000t + 66.67sin60m01V for r > 0. 8,8

r > 0.

312

Natunt andStepR€sponses offtr Circuits

8.5 A Circuitwith TwoIntegrating Amplifiers A chcuit containirg two inregraring amplifiers coDnectedin cascadel is also a second-order circuit; that is, the output voltage of the second integrator is related to t}le input voltage of the first by a second-order ditrerential equation.We begin our analysisof a circuit containing two cascaded amplifierswith tle circuit shownin Fig.8.18. We assumethat the op amps are ideal. The task is to derive tlle differential equation that eslablishes the relationship between ?, and Dp.We beeh tle derivation by summirg the currents at the inverting input terminal of the first inte$ator. Becausethe op amp is ideal,

0-u" n =K tj + c , d; tt o

u . , ,:)0 .

(8.70)

FromEq.8.70, dDot

dt

I

(8.71)

&Cl"g'

Now we sum the cullents away ftom the inverting input teminal of t}le second integrating amplifierl

0-z)-, R, dbo

rJt Differcntiating Eq.8.73gives Euo dt'

c,frtoo"t: o. 1

l.8.t2)

(8.73)

&C)"" 7 doo1 R{2 dL

(8.74)

We find the differetrtial equation that govems the relationship betweetr o, and ?rsby substitutingEq.8.71into Eq.8.741

d2u" : 1 1 d,' &(\-R C:*

(8.75)

CI

+

Figure8.18A Twointegratjng ampLifieu connected in Gscade. ' In a cascademnnection, the output siglal oI the ii6r mplifrd signal tor the secondanplifier.

(ror in Fig. 8.18) is the inpul

8.5 A CiftuitwiihTwoInt€grating AmpLjfiels 313

Example 8.13 illustrates tlle step responseof a circuit containing two cascaded integrating amplifierc.

Analyzing TwoCascaded IntegratingAmplifiers No energyis storedin the circuit shownin Fig.8.19 when the input voltage ?s jumps instantareously from 0 to 25 mV

0 . 1F F

1/,F

9V

a) Derive the expressionfor r'"0) for 0 < t< r""t. b) How long is it before the circuit satuiates?

Sotution

Figure 8.r9A Thecjrcuit forExampLe 8.13.

a) Figure 8.19irdicates that the amplifier scaling

1

1000

n1c1 (2s0x0.1) 1 RrC,

1000 (500)0)

-

No\ becauset's : 25mV for I > 0, Eq. 8.75

-

= (40X2)(25. 10 )) : 2.

To solvefor z',,we let da^ dt' then, dp(r\ ::2,

a n d d R 1 )= 2 d t .

Hence

becausethe energy stored in the circuit ini tially is zero, and the op amps are ideal. (See Problem 8.53.)Then, dr^ i=2t,

and Do: ( + 1).(0).

But ,"(0) - 0, so the experssionfor rn becomes p.:t2, o
1.

Solvingfor 0oryields tc\4

|

ldx, I d.y-2 .rE(o) Jo frcm which

80)-8(0):2r. However,

Be)=4#:o,

0ot:

t

Thus, at r = 3 s, Dor= 3 V, and, becausethe power supply voltage on the first integrating ampli{ieris +5 V, the circuit teachessaturation when the secondamplifier saturates. When one of the op ampssaturate$we no longer can use the linear model to predict the behavior of the

NOTE: AsselsJout understandingof this materiul b), tying Chaptet Prcblem 8.58.

314

Natuct andSteD R€sDonses ofRlrCncuits

TwoIntegratingAmptifierswith Feedback Resistors Figue 8.20depicts a variation of the circuit shown in Fig.8.18. Recall from Section 7.7 that the reason the op amp in the integrating amplifier saturates is the feedback capacitor's accumulation of charge.Here, a resistor is placed in parallel wittr each feedback capacitor (C1 and C2) to overcome this problem. We rederive the equation for the output voltage,oo,and detemine the impact of these feedback resistors on the integrating amplifiels ftom Example 8.13. We begin the derivation of the second-orderdifferential equation that relates o,1 to Dsby summing ttre currents at the inverting input node of the firct irtepmtor:

u

us 0 uo\ ^d.^ =R:+c;(0R"

u,r= ) 0.

(8.76)

WesimplifyEq.8.76to read th:or dr

1. R cr"''

rc RJ r'

18.77)

For convenience,we let zt = R1C1and write Eq. 8.77 as

rlu"r dt

u.\ rr

-rs R,C,

(8.78)

The next stepis to sumthe currentsat the invertinginput terminalof the secondintegrator:

?f.?

+c,*@_1,.)=0.

Figure8.20l Cascaded inhgratinganpLifien withfeedback resistorr.

(8.7e)

8.5 A Circuit MthTwolntegnting Amplifier 315

Werew te Eq.8.79as -Do1

dD.

uo

i

,r= nc,

(8.80)

where 12 = R2C2.DifJerentiating Eq.8.80 yields &uo,lduo 1. d o . 1 : -RC -; dt dt' dP

(8.81)

From 8q.8.78, door

-por

ilt

rr

_

oe RuCt'

{8.82)

andftom Eq. 8.80,

o"r=-Rcr*-ryr..

(8.83)

We use Eqs 8.82 aDd 8.83 to elimiuate du.l/dt from Eq.8.81 and obtain the desired relstionship:

*

(:,

',)o'"(i),. - a.,',1r,,, (8.81)

From Eq. 8.84,the chamctedstic equation is

.

/1

1\

\ar

r2,/

r'+i---ls

|

r -0. 7172

(8.85)

The roots of the characteridtic equation ate real, mmely, -1 (8.86)

-1 (8.87)

Example8.14illustratesthe analysisof the stepresponseof two cascaded iDtegating amplifien when tlle feedbackcapacitorsare shuntedwith feedbackrcsistors.

316 Naturat andStepR€sponses 0f RarCircuitj

Resistors Analyzing TwoCascaded IntegratingAnplifierswith Feedback The parametersfor the circuit shown in Fig. 8.20 are R. - 100kO, Xr = 500kO, Cr = 0.1pF, 'Ihe Rb : 25 kO, R2 : 100kO, and C, = 1 r!F. power supply voltage for each op amp is t6 V. The signal voltage (os) for the cascadedintegrating amplifiers jumps from 0 to 250 mV at t : 0. No energyis stored in the feedbackcapacitomat the instantthe signalis applied. a) Find tlle numerical expressionof the differential b) find r'.0) for I > 0. c) Fhd the numedcal expression of the differential equation for t,1. d) Find o,(4 for t > 0.

Solution a) Fromthe numeical valuesof the circuit paramete$, we have z1 : R1C1- 0.05s;r: - RuCz : 0.10s, and or/RuCrRrC,: 1000v/s'. Substituthg tlesevaluesinto Eq.8.84gives

d'u^ du^ ---+ + 301: + 200r- = 1000. d(

The solutionfor uotbus takestbe forml

u"=5+A\erat+Ate-t't. With tr,(0) : 0 ard dr.,"(o)/dr= 0, the numedcal valuesof ,4i and Ai arc A\ : 10V and ,4i - 5 V. Therefore,the solulionfor z', is

.,,()=(5-lOe

o

5 ? - 2 0v?. r I - - .

The solution assumesthat neither op amp saturates.We have already noted tlat the final value of ?,,is 5 V which is lessthan 6 Y hencethe secondop amp does not satuiate,The final value of r,,1 is (250 x 1{r)(-500/100), or -1.25v. Therefore, the first op amp doesnot saturate,and our assumptionand solution are co ect, c) Substituting the numerical values of the parameters hto Eq. 8.78 generatestle desired differcntial equation:

++2oD^.= 25.

dt

b)The roots of the charactedstic equation are sr = -20md/s and s2 = -1orad/s. The final value of o, is the input voltage times the gain of ea€h stage,becausethe capacito$ behave as open circuits as t + oo. Thus,

d) we have aheady noted tie hitial and final values of o,r, along wit]I the lime constant rt. Thus we wdte the solution in accordancewith the technique developed in Section 7.4:

aar= -1.2s+lo-

( ' I 0 )^r 500) 100)= 5 v . , , ( m )= 1 2 5 0 t00 15

(-1.25)Ienat

: -1.25 + 1.25e1vV, t > 0.

NOTE: Assessyow andetstandinq of this matetial by tjing Chapter Prcblem 8.59.

P,acticalPerspective 317

PracticaIPerspective AnlgnitionCircuit NowLetus returnto the conventionat ignitionsystem introduced at the beginning of the chapter. A cjrcujtdiagram of the systemis shownjr Fig.8.21.Consider the circujtcharacterjstics that provide the energyto jgnitethefuet-airmixture in thecytinder. First,the maxjmum vottage avai.L ableat thesparkptug,tr"p,mustbehjqhenough to ignitethefuet.Second, the voltageacross the capacitor mustbe limitedto prevent arcingacross jn theprimary points.Thjrd,thecurrent theswitchor distributor winding of theautotransformer mustcause sufficient energy to bestoredin thesystem to ignjtethe fuet-airmixturein the cyfinderRemember that the energy storedjn the circuitat the instantof swjtching is proportionaL to the primarycurrent squared, thatjs, 4,0= laf(o).

I

EXAMPLE a) Findthe maximum vottageat the sparkptug,assuming the fottowingval uesin the circuitof Fig.8.21: Ud.= 12Ir/,R= 4A, L=3mH, C:0.4pF,anda:100. o ) Whatdistancemustseparatethe swjtchcontactsto preventarcingat the time the vottageat the sparkplugis maximum?

Solution a) Weanatyze the circuitin Fig.8.21to find an expression for the spafk pLug voLtage r'"r.Wetimitouranatysis to a studyof thevoLtages in the circuitpriorto the firingof the sparkptug.Weassume that the current in the pdnarywindingat thetjmeof switching possihasits maximum blevalueydJR,whereR is the total resistance primary in the cjrcuit. Weatsoassume that the ratjoof the secondary voltage(oJ to the pri(r, is the sameastheturnsratjoN2/N1.Wecanjustjry maryvottage thisassumption asfotlows. Withthesecondary circuitopen,thevottaqe induced in thesecondary windjng is _- d i 1v,.

a- .

llt

(8.88)

ard ll-evoltage indJced in tl'e pr'mary w noingis ,di

xt - L--,:. .lt

(8.8e)

It fottows fronrEqs.8.88and8.89that (8.e0) 1

L

'

Figure8,21A ThecircLrit djagnmofthe conveF tionatauiomobite ignitionsystem.

318

Naturatand St€pRespons€s offLr Circuits

It is reasonable is thesamefortheftuxes to assume thatthepermeance in iron-core hence autotansforner; Eq.8.90reduces to d11anddz1 the u2

NlN2g

N2

n = - n T =N ' = ^

(8.e1)

Wearenowreadyto analyze the voLtages in the ignitioncircuit. Thevatues ofR, a, andC aresuchthat whentheswitchis opened, the primarycoiLcurrentresponse is underdamped. lJsingthe technjques devetoped in Section 8.4andassuming f = 0 at theinstanttheswitch pdmary is opened, theexpression forthe coiLcurrent is foundto be

, =*" -[-,-n . (;).'"*,],

(8.e2)

(seeProbtem in the primary 8.62(a).) Thevottage induced winding is of theautotransformer Va, ^, .

..li

(8.e3)

(SeeProbtem 8.62(b).)It foLtows from Eq.8.91that bz:

-av* -, . St[@dl. M?

(8.e4)

ThevoLtage across thecapacitor canbederived ejtherby usingthe relationship

".:lf''a,,,"101

(8.e5)

or by summing the vottages aroundthe meshcontaining the pfimary winding: o':U'r'-iR-Li'

/11

(8.e6)

In eithercase,wefind D": ud"[I

a"'cosadt + Ke-"'sinadtl,

(8.e7)

PracticatP€rspectjve 319

rl "- .= A \

r

"\ ).

RC

(SeeProblem8.62(c).)As canbe seenfrcm Fjg.8.21,the vottage acrossthe sparkptugis t'sp=Yd.+02

= v* - *"-'

""n'ot

=r".lt- -l=n* "^,",1,.

(8.e8)

Tofindthe maxirnum positive vatueof 0sp,wefindthesmattest vatueof timewheredosp/dtis zeroandthenevaluate Dspat this instant.The expression for tnaxis

r . * = I , u n' ( " 1 a,t \d

(8.ee)

/

(SeeProbLen 8.63.)Forthecomponentvatues in theprobLem statement,

-'

R 2L

4 x 1 d:

666.67ftd/s.

and

$

= 28.8sa.81rad/s. rooo.arr'

Substit0tjng thesevalues into Eq.8.99gives /",{ - 53.63 ss. NowuseEq.8.98to findthe maximum sparkptugvottage, o,p(r--): tse(r-*) = -25,975.69 V. D)

Thevottage across thecapacitor at lnaxis obtained fromEq.8.97as ac?n,J : 262.15Y. ThedieLectric strengthof air is approximatety 3 X 106V/m,so this resuLttelts us that the switch contactsmust be separatedby x 106,ot 87.38, 262.1513 arcingat thepointsat tffi. /,Lmto prevent

320

offr Cncujts Naturatand stepResponses

must consideration andtestjngof ignitionsystems, In the design ofthespa* pluq thewidening fuet-airmixtures; begivento nonuniform theretatjonship of theptugetectrodes; gapovertimedueto theerosion speed; thetimeit takes andengine sparkptugvottage between avaitabte the prinarycurrentto buitdup to its initiaLvalueafterthe switch to ensureretjabie required of maintenance is ctosed; andthe amoLlnt oPeration. jgnitionsysten of a conventional anatysis wecanusethepreceding ngin mechanical switchi hasreptaced switching whyetectronic to exptain and on fuel econorny First,the currentemphasis today'sautomobites, a sparkptlg witha widergap.'[his,jn turn, requires exhaust emissions (up highervottages sparkptugvoltageThese requires a higheravajtabLe Etectronic mechanjcaL switching. with be achieved 40 kV) cannot to jn the primary windjngof atsopermjtshigherinitjalcurrents switching is in thesystem energy theinitialstored Thismeans theautotransformer. condiand running mixtures range of fuelair a wider and hence larger, crrcuite[mswitching Finalty, theeLectronic tionscanbeaccommodated. effects thedeteterious Thismeans inatestheneedforthepointcontacts. from the system. be removed point arcjng can contact of bytryingchapter Perspective yourunde5tanding of theProctical N?TE:Assess 8.64 and 8,65. Problens

Summary The characf€risticequation for both the parallel and seriesRLC circuitshasthe form s2+20s+o3:0' where d : 1/2RC for the parallel circuit, d : R/24 for the seriescircuit, and (,6 = 1/aC for bot}l the paralel and seriescircuits.(Seepages287and 308.) The roots of the characteristic equation are "'z =

o + \/"'-

(Seepage288.) ofseriesard The form of the naturalard stepresponses parallel RtC circuitsdependson the valuesof I and rofi;suchresponsescan be oterdamp€d,underdamped, or critically rlamped.These terms describe the impact of the dissipative element (R) on the respons€ The nePer ftequency,a, reflectsthe effectofR. (Seepage289.) The responseof a second-ordercircuit is overdamped, underdamped, or critically damped as shown in Table8.2. In determinhg the nafuml respoNe of a second-order clrcuit, we fust detemine whether it is over-, uncler', or

critically damped,and then we solve the appropiate equationsas shownin Tirble8.3. In determining the st€p respons€of a second-order cir' cuit, we apply the apFopriate equations depending on the damphg, as shown in Table 8.4 For each of the three forms of response,the unknown coefficients(i.e.,the,4s, B s, and ,s) are obtained by evaluating the circuit to find the initial value of the response,r(0), and the initial value of the first derivadi(0)/lr. tive of the response, When two integrating amplifien v,/ith ideal op amps are connected in cascade,the output voltage of the second integrator is related to the input voltage of the fiISt by an ordimry second-order differential equation Therefore' the techniquesdeveloped in this chapter may be used to analyze the behavior of a cascadedintegrator' (See page312.) We can overcome the limitation of a simple integrating amplifier-the saturation of the op amp due to charge accumulatingin the feedbackcapacitor-by placing a resistor in parallel with the capacitorin the Ieedback path. (Seepage314.)

Prcbtenrs321

TABLE 8.2

TheResponse ofa S€lond-Order Clrcultis overdanped. Und€rdanped, or Critic.ttyDamped

TXe Circuit is

When

Qualitrlive Nrture of the ResDorse

"'>

Ttre voltageorcurrent approaches its final valuewithout oscillation

,3

Underdamped

d <.4

The voltaSe or cunent oscillales about its final value

Criticallydanped

i =;t,

The voltage or current is on the verge of oscillaring about its firal value

TABLE 8,3 In Determining the NattllatRespons€.of a Second-Order Circuit,WeFirstDetermine Wherher it is Over-,Und€r, or Criticatty oamped, andlhenWeSotvethe Appropriate Equations Danping

Natural R€sponseEquations

CoefiicientEqurlions

x(t)=ACt'+Aze,t

r(0):Ar+A,: dr/tlt(o) - 4\+

x(r) : (Br cos@dr+ B, sino,rre-"r

Azsz

rlo) = Bi dx/.tt(ot:-aBt+@a8,.

where o, = \,ttCrilicaly darnped

TABLE 8.4

x(r) : (Dn + D)e "t

xlo) : Dr, .1r/dt(0)= Dt

dD2

In D€termining the StepResponse of a Second-order Circult,WeApplythe Appropriat€ tquationsDepending

on the Danping Danping

Step Respome Equationsr

Coefricient EquatioIs

r(t)=xr+A\e\t+Aferl

x(0)=xt+A\+Ail dx/ddo): A\ sr + Ai s2

x(t) = xt + (B'rcos@dt+ Bi sitioat)e-"t

x(0)-Xr+B\: dx/.11(D): -dB\ + oraBi

Giticallydamped

x(t) - Xt + D\te-"' + Die "'

r(0\=xt+Dil dx/dt(o)=D\-dDi

'wheie xl is thefinalralue ofr(r).

Frob[ervls Sections8.1-8.2 8.1 The resistance,inductance,and capacitancein a parallelnlc c cuit are 5000O, 1.25H, and 8 nF, respeclively. a) Calculate the roots of the characteristic equation tllat describethe voltageresponseofthe circuit. b) Will the responsebe over-,under-,or critically damped?

.) What value of R will yield a dampedfrequency of6 krad/s? What are the roots of the characteristicequation for tlle valueoIR found in (c)? e) WtratvalueofR will resuitin a criticallydamped response?

322

NatuEtand StepResponses ofRICCiruits

8.2 Suppose the capacitor in the circuit shown in Fig. 8.1 has a value of 0.05 pF aDd an initial voltage of 15 V. The initial curent in the inductor is zero. The rcsulting voltage responsefor I > 0 is 'noor + 20P 2oolro'V ,(l, 5e

8.7 The resistance in Problem 8.6 is increased to m'j.r 2.5kO. Find the expressionfor ?,(/)for t > 0. 8.8 The resistance in Pioblem 8.6 is increased to EFlc 12500/3 O. Find the expressionfor o(t) for r > 0.

a) Determine the numerical values of R, a, a, b) CalculatetRo), ir(t), and ic(t) for t > 0*. 8.3 The natural voltage response of the circuit in Fig.8.1is o(t) : 125ar0mr(cos 30001 2sin3000Dv, r>0, when the capacitor is 50nF. Fird (a) t; (b) n; (c) Yo;(d) 10;and (e) ir0). 8.4 The voltage responsefor the circuit in Fig. 8.1 is known to be D(t) = Dfe4aoot+ D2e-aatu, t > o. The initial current in the inductor (It is 5 mA, and the initial voltage on the capacitor (yd is 25 V The inductor has an hductance of 5 H. a) Fitrd the value of R, C, D\, and.D2. b) Find ic(t) for / > 0+. 8.5 The idtial value of the voltage ?Jin the circuil in Fig. 8.1 is zero, and the initial value of the capacilor cunent, ic(O-),is 15 mA. The expressionfot the capacitor currert is known to be i"(t) = AF16t+

A2eat' t>o+,

when R is 200 O. Find a) the valueofa, (,o,a, C,Aband A)

( ,,,,.0''!o)=

.liL(0)

r?ip(o) u(0) d t :

1 rc(o)\

L - R

c

)

b) the expressionfor r(t), t > 0, c) the expressionfor in(t) > 0, d) the expressionfor ir(l) > 0. 8.6 The ciicuit elementsin the circuit in Fig.8.1 aie 3 ' c R : 2 k O , C = 1 0 n F ,a n d a : 2 5 0 m H . T h e i n i tial inductor curert is 30mA, and the initial capacitorvoltageis 90V a) Calculatethe initial current in each branch of the circuit. b) Find 0(t) for t > 0. c) Find iz(r) for r > 0.

8.9 The natuml responsefor the circuitshownin Fig.8.1 is known to be -o'-.'300)v. />-. u ( / )- - 1 2 ( r If C : 18 pF, find i1(0') in milliamperes. 8,10 In the circuil shown in Fig. 8.1, a 5 H inductor is 6tr4 shunted by a 8 nF capacitor, the resistor R is adjusled for c tical damping, Vo25V, and 1o = -1 mA. a) Calculate tle numedcal value of R. b) Calculate r'(t) Ior t > 0. c) Find 0(t) when tc(t) = 0. d) What percertage of the initially stored energy remains stored in the circuit at the instant ic(r) is 0? 8.11 In the circuit in Fig. 8.1, R = 2r}, a = 0.4H, Bdo c:0.25F,U.J:0V,and10 = -3A. a) Find r,(t) for t > 0. b) Find the filst three values of I for which do/dt is zero. Let these values of I be denoted t1, t2, and 13. c) Showthat 13- 11= Zt. d) Show thaft2 - 11: Zd/2. e) Carcuhte?,(11), ?,(tr),and 0(t3). f) Sketch0(l) versustforO < t < t2. 8,12 a) Find 2(l) for I > 0 in the circuit in koblem 8.11 if the 2 O resistor is removed from the circuit. b) Calculate the ftequency of t (t) in hertz. c) Calculatethe ma,\imum amplitude of r,(t) in volts 8.13 Assume tlle underdamped voltage response of the circuitin Fig.8.1is *ritten as r(t) = (Ar + A)a"'cos a,i + j(At - Az\e-tsinadt The initial value of the inductor current is 10, and t}le initial value of the capacitor voltage is y0. Show tlat ,4, is the conjugate of,41. (lllrt Use the same processas oudinedin the text to find,41 and ,42.)

Prob.€ms 323 8.14 Show that the results obtained from Problem 8.13rharis.rhe expression\ tor,4r atrd.4r-are consistent with Eqs.8.30and 8.31in the text

820 The resistor in the circuit of Fig. P8.19is inffeased tstrc from 1.6 kO to 2 kO and t]le inductor is deffeased from 1 H to 640 mH. Find zJ.(t) for t > 0.

8.15 The resistor in the circuit in Example 8.4 is changed strr! to 4000/\2 o.

8.21 The resistorir the c cuit of Fig.P8.19is deqeased 6dc from 1.6kO to 800O, andtle inductoris deqeased from 1 H to 160 mH. Find oo(t) for I > 0.

a) Find the numerical expressionfor o(1) when t>0. b) Plot rr(1) ve$us t for the time intenal 0 < t < 7 ms. Compare this response with the one in Example 8.4 (R=20kO) and Example8.5 (R - 4 kO). In particular,compare peak values oI o(t) and the times when these peak values occur. 8,16 The switch in the circuit of Fig. P8.16 has beetr in 6plft position a for a long time. At t = 0 the switch moves instantaneously to position b. Find ?r,(t) for t>0. Figffe P8.16

t=0

Section 8.3 8.22 For the circuit in Example 8.6, find, for 1> 0, 6*E (a) ,0); (b) iR(t); and (c) tc(t). 8.23 For the circuit in Example 8.7, find, for , > 0, Em (a) ,,0) and (b) ic(1). 8.24 For the circuitin Example8.8,find o(t) for t > 0. 8.25 Assume that at fhe instant the 15 mA dc current Er!4 souce is applied to the circuit in Fig. P8.25,the initial curent in t}le 20 H inductor is -30 mA, and the initial voltage otr the capacitor is 60 V (positive at the upper terminal). Find the expressionfor ir(r) for t > 0 ifR equals800O. Figure P8.25

8.17 The capacitor in the circuit of Fig. P8.16 is decreasedto 1 rF and the itrductoris mueasedto 10H. Find o,(t) for 1 > 0. E.18 The capacitor in the circuit of Fig. P8.16 is decreasedto 800pF and t]le inductor is increasedto 12.5H. Find t"(t) for r > 0. 8.19 The two switchesin the circuit seen in Fig. P8.19 6flc operate synchronously.When switch 1 is in position a, switch 2 is in position d. When switch 1 moves to position b, switch 2 moves to position c. Svritch t has beenin positiona lor a long rime.Ar r 0. rhe switchesmove to their altenate positions.Find !,aG)forr>0.

l) '.,ullr" *,t 826 The rcsistancein the circuit in Fig. P8.25is changed Atrc ro 1250O. Fird ir(, for r > 0. 827 The resistancein the circuit in Fig. P8.25is changed tsdi! to 1000O Find ir(r) Ior, > 0. 8.28 The switch in the circuit in Fig. P8.28has been open E .r for a long time before closing at t = 0. Find o"(t) for r > 0. figureP8.28

Figure P8.19

1ko 60v 62.5nF

1.6kO

6.25pF

324

Natuntand SieoResDonses of{r Cncuits

8.29 a) For the circuit in Fig.P8.28,find i, for I > 0. """ b) Showthat your solutionfor i, is consistentwith the solutionfor r,. in Pioblem8.28.

Figure P8.33

1kO

830 Thereis no eneryystoredin the circuit in Fig.P8.30 6pr.rwhenthe switchis closedat r = 0. Find r"(t) for t>0.

,-r- 10/,r'

1.6H

FigoreP8.30

r:0'

|

834 Use the circuit in Fig. P8.33 'strcr a) Find the total energy deliveied to the incluctoi. b) Find tlle total energy deiivered to the equivalent

+

6.25pF,1c) Find the total energy delivered to the capacitor. d) Find the total energydelivered by the equivalenl cturenl source, e) Check the results of parts (a) through (d) against the conseraation of energy pdnciple.

8.31 a) For tle circuit in Fig. P8.30,find i, for I > 0. """ b) Show that your solution for i. is consistent with the solution for 0, in Problem 8.30. 8.32 The switch in the circuit in Fig. P8.32 has been 6trc open a long time before closing at t = 0. At the time the switch closes,the capacitorhas no stored energy.Find tto for t > 0.

8.35 The switch in the circuir in Fig. P8.35 has been EPI.[ open a long time before closingat t - 0. Find iL(l) fort > 0. tigureP8.35

Figure P8.32

3000tr+ 31.25 rnH=500 72V

1 25rlF

8.33 The switch in the circuit in Fig. P8.33has been open ^"" a long time before closing at t : 0. Fhd a) ?o(t)for, > 0+, b) iro) foi t > 0.

8.36 Switches 1 and 2 in the circuit in Fig. P8.36are syn6trt! chronized. Wlen switch 1 is opened, switch 2 closes and vice versa. Switch t has been open a long tirne beforeclosirg at. : 0. Find ir0) for r > 0.

rigureP8.35 5kO

\ )240v

J w r c o1r v Switch

-1-

\,=0 Switch 2

-5 sF

trl 8 0 H

1ko

(

60mA

PlobLems325

Section 8,4 8.37 The initial energy stoied in the 50 nF capacitor in the circuit in Fig. P8.37 is 90 /iJ. The initial energy stored in the inductoi is zero. fhe roots of the characteristic equation that describesthe natuml behavior of the current i are -1000 ai and 4000 s-1. a) Find the numerical values of R and a. b) Find the numerical values of i(0) and di(0)/d1 lrnmediately after the switch has been closed. c) Find t(t) for t > 0. d) How many miqoseconds after the switch closes does the curent reach its maximum value? e) What is the maximum value of i h milliamperes? f) Fbd 0r(t) for r > 0. FlgureP8,37

FigueP8.40

{rvJ 72Ov

g.41 In the circuit in Fig. P8.41,the resistoris adjusted Mc for critical damping. The initial capacitor voltage is 90 V and the initial inductor current is 24 mA. a) Find the numerical value of R. b) Find the numerical values of i and dt/d/ immediately after the switch is closed. c) Find 0c0) for t > 0.

Flgure P8,4r

8,38 The current in the circnit in Fig. 8.3 is knom to be = Bre-3m!cos600t + B2?{e sin 600t, I > 0. The capacitor has a value of 500 /rF; ttre initial value of the curent is zero; and the initial voltage on the capacitor is 12 V Fhd the values of R, a, 81, ard ,2.

8,39 Fhd the voltage aqoss the 500 ,.F capacitor for the circuit describedin Prcblem 8.38.Assume tle refererce polariry for the capacitor voltage is positive at t]le upper termiftl.

8.42 fhe switch in the circuit in Fig. P8.42has been In position a for a long time. At I = 0, the swrtch moves iNtartaneously to position b. a) What is the initial valueof?)d? b) what is the initial valu.eof dDJ dt? c) What is the nume cal expressionfor 2,,(t)for t>0?

FlgoreP8.42 8.40 The switch in the circuit shown in Fig. P8.40 has 6rft been closed for a long time. The switch opells at I : 0. Find a) t"(l) for t > 0, b) 0"(r) for t > 0.

326

Natu6tand StepResponses ofRlCCircuiis

8.43 The make-before-break switch in the circuit shown 6PIcr in Fig. P8.43has been in position a for a long time. At , = 0, the switch is moved instantaneouslyto position b. Find i(t) for t > 0. FigureP8.43

100mH

8.48 The switchin the circuit shownin Fig.P8.48has beenclosedfor a long time beforeit is openedat t - 0. Assumethat the circuit parametersare such that the rcsponseis underdamped. a) Derive tle expressionfor o,(t) as a function of us, ct,od, C, andR fot t > O. b) Derive the expression for the valueof I when the magnitudeof r', is ma-\imum. FlglreP8.48

8.44 The switch in the circuit shown in Fig. P8.44has B{c been closed for a long time. The switch opens at t : 0. Find ?a(t) for t > 0. tigun P8.44

30rI 10()

8()

10f}

100v

8.45 The initial energy stored in tle circuit in Fig. P8.45 6n(r

i. ?er^

Fin,l

r /,\ +^r,

>

n

8.49 fhe cftcuitparametenin the circuitof Fig.P8.48 a r e R = 1 2 0 O , a : 5 m H , C : 5 0 0 n F F ,a n d 0s = -600 V a) Expressr'"(t) numericallyfor t > 0. b) How manymiqosecondsafter the switchopens is t}le inductorvoltagemaximum? c) Wlat is the maximumvalue of the inductor voltage? d) Repeat(a)-(c)withR reducedto 12O. 8.50 The circuit showtr in Fig. P8.50 has been in opera6'IG tion for a long time. At t = 0, the sourcevoltage suddenly drops to 100V Find o,(r) for r > 0.

tigureP8.45 Figure P8.50 4()

40 mH

8.,16the capacitor in the circuit shown in Fig. P8.45 is changed to 100 nF. The idtial energy stored is still zero Find od(t) for r > 0.

8.47 The capacitor in the circuit shown in Fig. P8.45 is changedto 156.25nF. The initial energy stored is still zerc. Find 0o(l) for t > 0.

8.s1 The swilch in the circuil of Fig. P8.51 has been in position a for a long time. At I = 0 the switch moves instantaneously to position b. Find

a) r'10') b) du.(o')ldt c) ?,'o(t)for t > 0.

tigureP8.51

the switch has been closed.At the time when the switch is closed, tlere is no energy stored in either the capacitoror the inductor. Find the rume cal values of -Rand L. (H,nL Work Problem 8.53firsr.) 12kO

,.,,, mv

28v **+

8.52 The two switchesin ttre circuit seen in Fig. P8.52 sflft operate s]'nchronously.men switch 1 is in position a, switch 2 is closed.wlten srvitch1 is in position b, switch 2 is open. Svritch t has been in position a for a Iong time. At t : 0, it movesinstantaneouslyto position b. Find ?.G) for t > 0.

8.55 Show that, if no energy is stored in the circuit showr in Fig. 8.19 at the instant rJsjumps in value, then db./dt eqlralszeroatt =0. 8.56 a) find the equarion foruo(r) for 0 < r < r.urin the circuit sho$n in Fig.8.19if o,1(0) : 5 V and ?,,(0) = 8 V. b) How long does the circuit take to reach satumtion? 8.5? a) Rework Example 8.14with feedbackresistors R1 and R2removeil. b) Rework Example 8.14with r',(0) : -2v ^nd

0,(0)= 4v.

Figule P8.52

200r} 400O 100mH b\

|

+

160v1600o

8.53 Assume that te capacitor voltage in tle circuit of Fig. 8.15 is underdamped.Also assumethat no energyis stored in the circuit elementswhen the switch is closed. a) Showthat doc/dr = (SolaiVe-t sn(l.at. b) Show that doc/dt :0 when t = nr/r,,r, where n : O,7,2,.... c) Let t"= nnld,j, and show that !'c(t,)

:v - v (-1)"e-*"1"

d) Showthat "

1. oc1i-V Td'- uc(t) - V

whereZd = t3 - tr,

854 The voltage acrossa 200 nF capacitor in the circuit of Fig. 8.15 is described as follows: After the switch has been closedfor severalseconds,the voltage is constant at 50 V The fbst time the voltage exceeds 50 Y it reaches a peak of 63.505V This occurs ?r/12 ms after the switch has been closed.The second time the voltage exceeds50 V it reachesa peak of 50.985V This secondpeak occursd4ms aftei

Section 8,5 E.5E The voltage signal of Fig. P8.58(a) is applied to tsr.r the cascaded integrating amplifiers shown in Fig. P8.58(b). There is no energy stored in the capacitorsat the instant the signalis applied. a) Derive the numerical expressionsfor to(t) and r',l(f) for the time intervals 0 < t < 0.2s and 0.2s
328

Naturatand StepResponses off|r Cjrcuits

8,59 The circuit in Fig. P8.58(b) is modified by adding a *fl( 250 kO resistor in parallel with the 2 pF capacitor and a 250kO resistor in parallel with the 4pF capaciror. As in Problem8.58.thereis no energ! stored in the capacitorsat the time the signal is applied. Derive the numerical expressionsfor ,o(l) and o,1(t) for the time intervals0 < I < 0.2 s and r > 0.2s.

8.60 a) Derive the differential equation tlat rclates the output voltage to the input voltage for the cir, cuit shownin Fig.P8.60. b) Compare the result with Eq. 8.75 when RrCr: R2C2= RC in Fig.8-18. c) What is the advantageof the circuit shown in Fig.P8.60?

is availableand by successive integrationsgenerates d.r/dt ard r. We can synthesizethe coeffi, cients in the equations by scaling amplifien, and w^ecan^combine the terms requircd to genemte d:x/dt" by rlsir]'g a summing amplifiei. With theseideasin mind, analyzethe inlerconnectron shown in Fig. P8.61(b).In pa iculai, describe the purpose of each shadedarea in the circuit and describethe signalat the points labeledB, C, D, E, and R assumingthe signalat A representsdr4df.Also discussthe parametersR;Rl, C1; R2,C2; R3, Ra; R5,.lR6; and R7, Rs in terms of the coefficients in the differential equation.

Seclions8.1-8.5 8.62 a) Deive Eq.8.92. ,lli;L!{"',b) Derive Eq. 8.93. c) DeriveEq.8.9.

FlgureP8.50

8.63 Derive Eq.8.99.

8.64 a) Using the same numedcal values used in the pl$fi*ii€ Practical Perspective example ir the text, find the instant of time when the voltageacrossthe capacitor is ma-{imum, b) Find the maximumvalue of 2c. c) Comparethe valuesobtainedin (a) and (b) witl t-". and 2,.0-aJ. 8.61 We now wish to illustmte how several op amp circuits can be interconnectedto solve a differential equation. a) Derive the differential equation for the springmass system shown in Fig. P8.61(a). (See page 329.)Assume tlat the force exerted by ttre springis directly proporional to the springdisplacement,that the massis constant,and that the ftictional force is diectly prcportional to the velocity of the moving mass. b) Rewrite the differential equation derived in (a) so that the highest order derivative is expressed as a ftrnction of all the other terms in the equa tion. Now assumethat a \olt^ge equalto C xI dt2

8.65 The values of the parameters in the circuit in pP#fl*i;lFig.8.21 are R = 3o; Z, = 5mH; C = 0.2spF; Vd"= 72Y; ard a = 50. Assume the switch opens whell the primary windingcurrentis 4A. a) How much energy is stored in the circuit at 1=0-? b) Assumethe sparkplug doesnot fire.Wlat is the ma-ximumvoltage available at the spark plug? c) What is the voltage acrossthe capacitorwhen the voltageacrossthe spark plug is at its maximum value?

tigureP8.61

SinusoidaI Steady-State AnaLysis Thusfar, we havefocusedon

9.1 Th€Sinusoidat Sour.ep. 332 p. 335 9.2 Th€Sinusoidat Response p- 337 9.3 ThePhasor 9.4 ThePassive CncuitEtements in the Frequencf Donainp. 34? 9.5 Kirchhofi's Lawsin th€ FreguencJ Doln'ainp. 346 9.6 Series, Parattet, andDetta-to-Wye p. 348 SimpLifications 9.7 Source Transfonnations and Th6venin-Norton Equivat€nt Circuitsp. 155 9.8 TheNode-VoLtage l,lethodp. 359 9.9 Theliesh-Cuffent l4ethodp. 360 p. 361 9.10TheTransformer 9.11ThetdealTransform€r p. 365 9.12 PhasorDiagramsp. 372

phasorconcepts 1 Understand andbe ableto perfofma phasortransformard an inveEe 2 8e abteto transform a circuitwith a sinusoidat sourcejnto the frcquency domainusjngphasor 3 Knowhowto usethe fottowjnqcircuitanatysjs techniques to sotvea circuitjn the ftequency . . .

Knchhoffstaws; Series,parattet, anddelta-to-wye

.

Vottageandcunentdivision; Th6venin and Nortonequivatents;

.

Nod€-voitage method,and

.

iqesh-cuffent method.

4 8e abLeto anatyze circuitscoitainingtjfear transformers usingphasormethods. 5 UndeEtand the ideattraisformerconstraints andbe abteto anatyze circujtscontainingideat tnnsformeuusingphasormethods.

330

with constantsources;in this chapterwe are now ready to considercircuits energizedby time-varyingvoltageorcurent sources.lnparticular,weare interestedin sourcesin whichthe valueofthe voltageor currentvades sinusoidally. Sinusoidalsourcesand their eflecton circuit behavior fbrm an impo ant areaof studyfor severalreasons. First,the generation, transmission,distribution.and consumptionof electic energyoccur under essentiallysinusoidalsteady-state conditions. Second,an understandingofsinusoidalbehaviormakesit possible to predict the behavior of circuits with nonsinusoidalsources. Third, steady-state sinusoidalbehavioroften simplifiesthe design of electricalsystems.Thus a designercanspellout specifications in termsof a desiredsteadystatesinusoidalresponseand designthe circuit or systemto meet thosecharacteristics. If the dcvicesatisfies the specifications, the designerknows that the circuit will respondsatisfactoly to nonsinusoidalinputs. The subsequentchaptersol this book are largely basedon a thoroughunderstandingof the techniquesneededto analyzecircuitsddven by sinusoidalsources. Fortunately,the circuit anatysis and simplificationtechniquesfirst introduced in Chapters 1-4 work for circuitswith sinusoidaias well as dc sources,so someof the materialin this chapterwill be vety familiar to you.The challengesin first approachingsinusoidalanalysisincludedevelopilg the appropdatemodeling equationsand working in the mathematicalrealm of complexnumbers.

PracticaI Perspective A Househotd Distribution Circuit Power systems thatgenerate, transnrit, anddjstribute eLectricatpoweraredesjgr€d to operate in the sjnusoidaL steady state.Thestandard househotd distribution circuitusedin the ljnitedStatesis the lhrce-wlte, 240/72A V circuitshownin theaccompanying figure. Thetransformer is usedto reduce the utititydjstrjbution voltage frofi 13.2kVto 240V.Th€center tap onthesecondarywjndingprovides the 120V service. Theoperating frequency of powersystems in the ljnitedStates is 60 Hz.Soth 50 and60 Hzsystems arefoundoutside the ljnitedStates.

Thevottage ratingsatluded to abovearermsvatu€s. Thefeason for definingan rmsvalueof a time-varying sjgnaljs in Chapter 10. explained

331

332

sjnusoidaL Steady State Anatysis

9.1 TheSinusoidat Source A sinwoidal voltag€ source (independent or dependent) produces a volt age that varies sinusoidally with time. A sinusoidal curr€nt source (independent or dependent) produces a curent that varies sinusoidally with time. In rcviewing t]te sinusoidal function, we use a voltage source,but our obseFations also apply to current sotuces, We can express a sinusoidally varying function with either the sine function or the cosine function. Although either works equally well, we cannot use both functional forms simultaneously.We will use the cosine functionthroughoutour discussion. Hence,we write a sinusoidallyvarying voltageas

0=Y-cos(ot+d,).

(e.1)

To aid discussionof the parametefi in Eq.9.1, we show the voltage venus time plot in Fig.9.1. Note that the sinusoidalfunctior repeatsat regularinte als.Sucha functionis calledperiodic.One panmeter ofinterest is the lengthof time requiredfor the shusoidalfunctionto passthroughall its possiblevalues. This time is rcfered to as the pedod of the function and is denoted Z. It is measued in secondsThe reciprocalof 7 givesthe number of cyclesper second,or the ftequency,of the sinefunction and is denotedi or figur€9,1A A sinusoidaL vottage.

r=|

\9.2)

A cycle per second is referred to as a hertz, abbreviated Hz. (The term ctcla per se@nd ftrely is used in contemporary technical literature.) The coefficient of t in Eq. 9.1 contains the numerical value of f or /. Omega (id) representsthe angular frequency of the sinusoidal function, or a = 2Tf :21tlT (tadians/seco ).

(e.3)

Equation 9-3 is based on the fact that t}le cosine (or sine) function passes through a complete set of values each time its argument,o,, passes though 2' rad (360'). From Eq. 9.3,note that, whenevert is ar integral multiple of f, the aryumentot increasesby an integmlmultiple of 2r md. The coefficient y, gives the maximum amplitude of the sinusoidal voltage.Because+1 boundsthe coshe function,ty- boundsthe amplitude. Figure 9.1 shows these chancte stics. The angle4 in Eq.9.1 is known as the phaseangle of the sinusoidal voltage.It detemines the value of the sinusoidalfunction at I : 0; thereforc, it fixes the point on the periodicwave at which we start measudng time. Changing the phase angle d shifis the sinusoidal function along the

9.1 TheSinusoidatSourc€ 333 time a-xisbut has no effect on either the amplitude (y.) or the angular fre' quency ((r). Note, for example,that reducing 4 to zero shifts the sinusoidal functionshoM il Fig.9.14/@ time units to the riglt, asshoxn in Fig.9.2. Note also that if d is positive, tle siousoidal function shifts to the left, whereasif d is negative,the function shifts to the right. (See Problem 9.4.) A conrment vrilh iegard to the phase angle is in ordei o, and d must carry the same units, becauset}Iey are added together in the argument of tlle sinusoidal furction. With (dt expressedin iadians, you would expect d to be also. However, d nomally is given in degrees,and .rt is converted frcmFis.9.1 vottase from radians to degreesbefore t]le two quantities are added,We continue tigure9.2A Thesinusoidat to thenshtwhen this bias toward degees by expiessing the phase angle in degrees.Recall shjfted d : 0. from your studies of tdgonometry tlat the convenion from radians to degreesis givenby

''//N,/A i[", '"qf".v

\.v

(numberof degrees)= 1q{lnumber of radians).

(e.4)

Arother important characteristic of the sinusoidal voltage (or currenr) is its rms value.The rms value of a periodic function is defined as the square root of the mean value of the squared funclion, Hence, if D = V^cos(ot + d), tlle rms valueof lJis

=rEl'*\,; ,,^" *",,.* *,

(e.5)

Note ftom Eq. 9.5 that we obtain ttre mean value of the squared voltage by integrating I over one period (tlat is,ftom t0 to t0 + I) and then dividing by the range of integration, r. Note furtller that the starting point for the inregratjon roi. arbirrar'. The quanril! underlhe fadicai.igDiD Eq.4.5reduces to vt 2. (See Problem 9.6.) Hence the rms value of o is

(9.6) < rns vatueof a sinusoidal voltagesolrce The rms value of the sinusoidal voltage depelds only on t]re maximum amplitude of ?, namely, y-. The rms value is not a function of either the frequency or the phaseangle.We stressthe importance of the Ims value as it relatesto power calculationsin Chapter10 (seeSection10.3). Thus,we can completely describea specificsinusoidalsignaliJ we know its frequency,phaseargle, and amplitude (either the ma-'rimumor tlte rrns value).Examples9.1,9.2,and 9.3 ilustrate thesebasicpropediesof the sinusoidalfunction. In Example 9.4,we calculate the rms value of a periodic function, and in so doing we clariry the meaning oI root mean squarc.

334

SinusoidatSteady-StateAnatysis

Findingthe Characteristics of a SinusoidaI Current A sinusoidal current has a ma-\imum amplitude of 20 A. The current passesthrough one complete cycle ir 1 ms.The magnitude of the cu ent at zerc time is 10A. a) What is the ftequency of the current in hertz? b) What is the frequencyitrradiansper second? c) Wrire lhe expressioD lor itr' u(ing lhe cosine function. Express d in degrees. d) Wlat is the rms valueof the current?

Solution a) From the statementof ttre problem,Z : 1 ms; hence/:1/Z:1000H2. b) ti : 2nf = 2mut radls. c) Wehavet(r) : I,cos (ot + d) : 20cos(20002r + d), bur (0) : 10A. Therefore10 - 20cosd andd = 60".'Ihus the expiessionfor i(l) becomes t(r) = 20cos(2000't+ 60'). d) Fromtle derivationof Eq.9.6,thermsvalueof a sinusoidalcurrent is 1-/\4. lherefore the rms valueis 20/\4, or 14.14A.

Findingthe Characteristics of a Sinusoidat Vottage A sinusoidal voltage is given by the expresston 1' : 300cos(1202, + 30'). a) What is the period of the voltage in milliseconds? b) What is the frequencyin hertz? c) What is the magnitude of a at t = 2.778ms? d) Whatis the rms valueof ?J?

Sotution a) From tllle erFession for t, a : 120r :f2'dls. Bec?l$ea=2n/T,T -2nla =e s, or 16.667ms. b) The frequency is 1/2, or 60 Hz. c) Frcm (a), o = 2n1L6.667:th]uf.,nt t = 2.'7'78 m\ ol is nearly 1.047 rad, or 60'. Therefore, 0(2.778ms) = 300cos(60" + 30") = 0v.

:21213v. d)u^, - 300l.va

Transtating a SineExpression to a CosineExpression We can translatethe sine function to the cosine tunclionby subtracting90' (r/2 rad) from the aryumentoflhe sinefunction.

Sotution a) Verification involves direct application of the trigonometric identity c o s ( a- B ) : c o s a c o s B + s i n d s i n B . Weleta: (dt + d andB : 90".As cos90' = 0 and sin90" : 1, we have

a) Verify this tmnslationby showingthat sin(.,,t+ d) : cos(rrt+ d - 90'). co(a

b) Usethercsultin (a) to expresssin((or+ 30")as a cosinefunction.

B ) = s i n a - s i n ( ' r t + d ) = c o s ( o l+ P

b) From (a) we have sin(.rt + 30") - cos(or + 30' - 90') - cos((,r

90").

60").

9.2

lhe Sjnusoidat Response 335

Catculating the rmsVatueof a Triangutar Waveform Calculate tle Ims value of the peiiodic tdangllar cureDrsho\ n in Fig.c.3.Express)our a0s$efiD terms of the peak cunent 1/.

rP. r/4 0 Figure9.4 A r' versurr.

The anal)'ticalexpressionfor i in the interval 0 to T/4is 4T

Fiqurc9.3 l Pefiodic tnangular current.

i=1r.0
Solution FromEq.9.5,thermsvalueofi is

The area under the squaredfunction for one periodis

+1,"^" Interpretingthe integralunderthe radicalsignas the areaunderthesquaredfunctionfor aninterval of one period is helpful in finding the rms value. The squaredfunction with the areabetween0 and f shadedis sho\r.nin fg. 9.4,which also indicates that for this particular function,the areaunderthe squarcdcurrent for an inteflal of one period is equalto four timesthe areaunderthesquared currentfor theintenal0 to f/4 seconds;that is,

p+r ^ t2,T tr416t',^ i'dt:4 | I -fit='. ) tJh Ja Themean,or avemge,valueof the functionis simply the areafor oneperioddividedby the period.Thus " t2T

The Ims value of the crment is the square root of this mean value. Hence I.

NOTE: Assesslow undelstanding of thit mate/ial b! tying Chapter Prcblems 9.1,9.5,9.8.

Response 9.2 TheSinusoidaI Before focusing on the steady-state response to sinusoidal sourceq let's consider the prcblem in broader termq that is, in terms of the total rcsponse.Such an ove iew will help you ke€p the steady-statesolution in perspective.fhe circuit shown in Fig. 9.5 describes the general naturc of the prcblem.There,o" is a sinusoidalvoltage,or 0,:

Y-cos(@l+ d).

(e.7)

bya sinusojdat Figure9,5 r AnRl cncuitexcited voltase source.

336

Sjnusoidatst€ady-StateAnatysjs For convenience,we assumethe initial current in the circuit to be zero and measurctime from the moment the switchis closed.The task is to derive the expressionfoi i(l) when t > 0. It is similar to finding the step response of an Rl, circuit, as in Chapter 7. The only difference is that the voltage sourceis now a time-varyingsinusoidalvoltagerather than a constant,or dc, voltage. Direct application of Kircbhoff's voltage law to the circuit shownin Fig.9.5leadsto the ordinarydifferentialequation

tfi , ai = y- cos(.,r+ d),

(e.8)

t]le formal solution of which is discussedin an introductory course in differential equations.We ask those of you who have not yet studied differentialequationsto acceDtthat the solutionfor r N

w,

\,F;;t:

cos(<,- 0)e 'R tt '

, ,u- . vR' ul

,cos(dr

d

d). (9.9)

where d is defined as the angle whose tangent is da/R. Thus we can easily detemine d for a circuit driven by a sinusoidal source of knor,n frequency. We can check the validity oI Eq. 9.9 by dererminingthat ir sarisfies Eq. 9-8for all valuesof t > 0; tlis exerciseis left for your explorationin Problem9.10. The fiIst teim on the dght-hand side of Eq. 9.9 is refered to as the tmnsi€nt component of the current because it becomes infinitesimal as time elapses.The second term on the dght-hand side is known as the steady"state component of the solution. It exists as long as the switch remainsclosedand the sourcecontinuesto supplythe sinusoidalvoltage. In this chapter,we develop a techniquefor calculatingthe steady-state response directly, thus avoiding the problem of solving the differential equation. However, in using this tecbnique we fodeit obtaining either the tnnsient component or the toral response,which is the sum of the transientand steady-state components, We now focuson the steadystateportion of Eq. 9.9.It is impo ant to remember the following chamcteristics of the steady slate solution: 1. The steady-state solutionis a sinusoidalfuncrion. 2. The fiequercy of the responsesignal is identical to the frequency of t]le source signal. This condition is always true in a linear circuit when the circuit parameters,R, Z, and C, are constant. (ff frcquencies in the responsesignalsare not presentin the sourcesignals, thereis a nonlinearelementin the circuit.) 3. The maximum amplitude of the steady-stateresponse,in general, diJfers ftom the maximum amplitude of the source.For the circuit beingdi.scussqd,-lhe maximumampliludeof rhe fesponse signalis UI V R' d L andthe maKimum amplitudeol $e .ignalsource is U-. 4. Th€ phase angle of the responsesignal, in general, differs from the phase angle of the source.For the circuit being discussed,the phase angle of the current is d I and that of the voltage source is 4.

These characteristics are worth remembering because they help you unde$tand tlle motivatior for the phasor method, which we intoduce in Section9.3.In particular,note that once the decisionhas been made to find only the steady-stateresponse,the task is reduced to finding tlle mlximun amplitude and phase angle of the response signal. The waveform and frequencyof tlle responseare alreadyknown. NOTE: Assessyow underctandingof this matefial by ttying Chaptel Prcbkm9.9.

9.3 ThePhasor The phasoris a complexnumberthat caries the amplitudeand phase angleinformationof a sinusoidal function.rThephasorconceptis moted in Euler'sidentity,whichrelatesthe exponential functionto the trigonometricfunction; e ' r "= c o s d l i s i n d .

(e.10)

Equation9.10isimpo antherebecause it givesusanotherwayof exprcsshg thecosheandsinefunctions.Wecanthinl( of the cosinefunctionasthe realpart of the exponentialfunctionandthe sinefunctionasthe imaghary partof theexponential function;that is, cosd: n{d'},

(e.11)

sin0 = S{dr},

(9.12)

ancl

where $t means"tle real part of" and S means"theimaginarypafi oi" Becausewe have aheady chosen to use the cosine function in analyzing t}le sinusoidalsteadystate (see Section9.1),we can apply Eq.9.11 directly. In particular, we write the sinusoidal voltage function given by Eq.9.1in the form suggested by Eq.9.11:

0=Y-cos(ol+d) = v,,$?idG'+d)] - vnn Ietatetaj

(9.13)

Wecanmovethe coefficienty. insidethe aryumentof the rcal part of the function without alteringthe rcsult.we can alsoreversethe order of the twoexponential functionshsidetheaigumentandwriteEq.9.13as t = n{vneEd@'}. I II you feel a bit uneasyabour cotupler nhbe4

peruse Apperdix B

\9.14)

338

SinusordaLsteddy-SiateAnalysis

In Eq.9.14,notethat the quantity y,er't is a complexnumber that carries the amplitude and phase angle of the given sinusoidalfunction. This complex number is by definition the phssor reprcsentation,or phasor lransform. of the qiver sinusoidal function. Thus

Phasor transform>

(9.15)

where t})e notation g{yncos (a,t + 4)} is read "the phasortransfom of y- cos (ot + d)." Thus the phasor tmnsform transfers the sinusoidal function from the time domain to the complex-number domain, which is also called the fr€queng, domain, since the responsedepends,in general, on .r. As in Eq. 9.15,throughoutthis book we representa phasorquantity by using a boldface letier. Equation9.15is the polar form of a phasor,butwe alsocan expressa phasor in rectangular form. Thus we rewrite Eq. 9.15 as

Y:U^cosf+jU-si\6

(e.16)

BotI polar and rectangular folms are useful in circuit applications of the phasorconcept. One additionalconrmentregardingEq. 9.15is in order.The frequent occu ence ot the exponential function ete has led to an abbreviation that lendsitsell to text malerial This abbreviationis the anslenotation

11i!: = letE' Weusethis notationextensivelyin the materialthat follows.

InversePhasorTransform So far we have emphasizedmoving frcm ttre sinusoidal function to its phasor tmnsform,However,we may also revene the process,That is, for a phasor we may write the expression for the sinusoidal function. Thus for V = 1.001_4:, the expressionfor o is 100cos(or - 26") becausewe havedecidedto usethe cosinefunction for all sinusoids. Obsene that we cannot deduce the value of d from the phasor. The phasor ca ies only amplitude and phase information. The step of going from the phasor tmnsform to the time-domain expression is rcferrcd to as frnding the inversephasor tnnsform and is fomalized by the equation

g- rlv^ei4\ - n {v^erQ etd},

\9.t7)

9.3 ThePhasor 339 l{y-ere} is read as"the inversephasortransformof wherethe notation@ " yfter{ Equation 9.1? indicates that to find tlle inve$e phasor ffansform, we multiply the phasor by e,oiand then extract the real part of the product. The phasor transform is useful in circuit analysis becauseit reduces the task of finding the maximum amplitude and phaseangle of the steadystate sinusoidal responseto the algebra of complex numbe$. The following observations verify t}lis conclusion: 1. The transient component vanishes as time elapses,so the steadystate component of the solution must also satisfy the differential equation.(SeeProblem9.10[b].) 2. ln a linear circuit diven by sinusoidalsources!the steady-state response also is sinusoidal, and the frequency of t}le sinusoidal responseis the sane as the ftequency of the sinusoidal source. 3. Usirg the notationintroducedinEq.9.11,we canpostulatethat the steady-statesolution is of the form n{Aejpetd}, where A is the maximumamplitudeof the responseand B is the phaseangleof the response. 4. When we substitute the postulated steady-srat€solution into the diJferential equation, the exponential term d'' cancelsout, leaving the solution for,4 and B in the domai[ of complex numbers. We illustrate these obse ations with the circuit shown in Fig.9.5 (see D.33t. We know that the steadv-statesolution for the current i is of the folm

i""(t) : n{I^eiqetd},

(9.18)

where the subscdpt "ss" emphasizesthat we are dealing with t}le steady statesolution.When we substituteEq. 9.18into Eq. 9.8,we generatethe expressron

n{jaLl^eiqeid}

+ n{RI^etpeid} = nlune!'tejdt}.

(e.1e)

In dedvingEq.9.19we recognizedthat both differcntiationand multiplicatioflby a constantcan be taken insidethe real part of an operation.We alsorewrotethe right-handside of Eq.9.8,usingthe notation of Eq.9.11. From t]le algebra of complex numbers, we know that the sum of the real parts is the same as the real part of the sum.Thereforc we may reduce the left-handsideoI Eq.9.19to a singletelm:

n{(joL

+ R)Ineiqejd} : nlv-eiadd}.

(e.20)

Recall that our decision to use t}le cosine Junction in analyzing the responseof a circuit in the sinusoidalsteadystateresultsir the useoI the

340

SjnusojdalSteady-StaieAnatysis

n operatorin de ving Eq. 9.20.If insteadwe had chosento usethe sine function in our sinusoidal steady-state analysis!we would have applied Eq.9.12directly,in placeofEq.9.11, and the resulrwould be Eq.9.21: 3 {(jaL + R)I -eipetd} : 3 {v.etdei.,}.

(e.21)

Note that the complexquantitieson either sideofEq.9.21 are identicalto thoseon either sideofEq.9.20.Whenbolh the real and imaginarypartsof two complex quantities aie equal, then the complex quantities are themselvesequal.Therefore,tom Eqs.9.20and 9.21, (j@L + R)I^etu = Unda,

R+ joL

\9.22)

Note that db' has been elimimted from the detemination of the amDlilude r/.r and phaseangle{6r ot lhe respon.e.t}us. tor lhis cLcujl,ihe taskof finding/. and p in\olve.rheatgebraic manipulation of rhecomy-C'and R - idl. Nole thal \ e encounrered ple\ qudntities bolh polaf andreclanguiar iormc. An important warning is in order: The phasor transform, along with the inversephasortransform,allowsyouto go back andforlh betweenthe time domain and the frequency domain. Therefore, when you obtain a solution,you are eithei in the time domain or the ftequencydomain.you cannotbe in both domaim simultaneously. Any solution that containsa mlxture of time domain and phasordomainnometrclatureis nonsensical. The phasor transfom is also useful in circuit analysisbecauseit applies direcUy to the sum of shusoidal functions. Circuit analysis involves summmg curents and voltages,so the importanceof this observationis obvi_ ous,We can formalize this DtoDertv as follows: If

1J:IJr+

D2+

- +Dn

1s.23)

whereall the voltagesor tle right-handsideare sinusoidalvoltagesofthe samefiequency, ther

V=V+Vr+..+Vn.

19.?4)

Thus the phasor representation is the sum of the phasors of the individual terms.We discussthe developmentofEq.9.24 in Section9.5.

9.3 ThePhasor 341 Before applying the phasor tramfom to circuit analysis,we illusfate its usefulnessin solving a problem witl which you are already familiar: adding sinusoidal functions via trigonometdc identities. Example 9.5 showshow the phasor transform greatly simpiifies this type of pioblem.

AddingCosines UsingPhasors ff)r = 20cos((rl 30') and yr: 40cos(al + 60'), express) = )rr + ),2as a single sinusoidal function. a) Solveby usingtrigonometricidentities. b) Solveby ushg the phasorconcept.

3732 in thesoLutjon torl. Figure9.6 r A rjghttriangte used

Solution a) Filst we expand both y1 and 12, using the cosine of the sum of two angles,to get

b) we can solve the problem by using phasorsas follows:Because

)r : 20cos,)tcos30" + 20 sin.dt sin30';

Y=YltY2.

)2 = 40cos.rtcos60' 40sin(l)t sin 60". Adding )r and )r, we obtain

Y=Y1 +Y,

y - (20cos30+ 40cos60)cosrrt + (20 sin 30 = 3'7.32cos at

/'\1 1) \tv+ tz

cosdt

: 201_!: + 401!L

40 siII 60) sin {r,

: (17.32

U.64sir]'at.

: 37.32 + i24.64

To combhe these two terms we teat the co-efficients of the cosine and sine assidesof a dg:ht triangle (Fig. 9.6) and then multiply and divide the righFbandsideby lhe hypotenuse. Our expres'ion

y : 44.721-:

then,from Eq.9.24,

)4 64

]

\

sin ot J

j10) + QO + j34.64)

= 44;72//33.43'. Once we know the phasor Y, we can write the corresponding trigonometdc tunction for y by taking the invene phasortmnsform: : n{44;72ei33a3et"t} y = g 1I44;tzei3343\

/

- 44.?2(cos33.43'cosdt - sin33.43"sinot). Again, we invoke the identily involvhg the cosine of the sum of two argles and wdte y : 44.12cos(@t+ 33.43").

= 44;72cos(at+ 33.43"). The superiority of the phasor approach foi addirg sinusoidal fulctions should be apparent. Note that it requ es the ability to move back and forth between the polar and rectargular forms of complex numbers.

342 SinusoidaL Steady-State Anatysjs

objective1-Undetstandphasorconcepts andbeableto performa phasortnnsformandanlnversephasortransform 9,1

(c)r1.r8/ 26.s7'A; (d)33e.e0M!L mv.

Find the phasortransformof eachtigonometric function: a) lJ - 170cos(377r 40') V. b) t : tOsin (1000r+ 20') A. c) t: [scos(or + 36.87')+ 10cos(a,t

- s3.13)lA. d) u : l300cos(20,000,r +,15') 100sin(20,000rr + 30')l rnv.

Answer: (a) 1701 lp: Vi

(b)10l zq A;

9.2

Find the time-domainexpressioncorespon, ding to eachphasor:

a)v=18.61:1fv. b) | : (201!!: - 501:lq) InA. c) v : (20+ J80 301J!) v. Answer: (a) 18.6cos(ol - 54') V; (b) 18.81cos(.dr + 126.68') mA; (c) '/2."/9cos(at + 97.08') v.

NOTE: Also tt)rChapterPrcblen 9.12

9.4 TheFassive CircuitF[en'lents in the Freqr:ency Ssrmailr Thc systematicapplicatior of thc phasor transform in circuit analysis rcquirestwo steps.First, we must establishthe relationshipbetweenthe phasorcurrent and the phasorvoltageat the terminalsof the passjvccir cuil elements.Second,we must develop the phasor-domainversion of Kirchhoff'slawq which we discussin Section9.5.In this section.wc establish the relationshipbetweenthe phasorcurrent and voltageat rhc lerminalsof the rcsislor,inductor,and capacitor.We bcgin with the resjstorand usethe passivesignconventionin a1lthe derivations.

TheV-I Relationship for a Resistor R

figure9.? .:hA rcsistjve eLement carryjnq a sinusoidat

From Ohm'slaw,if the currcnt in a resistorvaricssinusoidallywirh time thatis,ifl = l cos(@t+ di) the voltageat the rerminalsofthe rcsislor, as shorvnin Fig.9.7,is

t] = RU_ cos((,r + drl = R/_[cos(or+ r)].

(e.25)

where1u is the maximumamplitudeof the currenrin amperes and di is thcphaseangleof thecurrent.

Domarn 9.4 ThePassive Circuit Etements in theFrcquercy The phasor transfom of this voltage is

V:RI^dd,:RI.ll.

|s.26)

But 1-14 is the phasor representation of the sinusoidal cunetrq so we can write Eq.9.26as

,v-;$

\9.27)

which statestlat the phasor voltage at the terminals of a resistor is simply the resistance times the phasor current. Figure 9.8 shows the circuit diagram for a resistor io the ftequency domain. Equations 9.25 and 9.27 both contain another important piece of information-namely, that at the teminals of a resistor, there is no phase shift between the cuirent and voltage. Figure 9.9 depicts this phase relationship, wheie the phase angle of both the voltage and the currert waveforms is 60'. The signals arc said to be in phos€ becausethey both reach corresponding values on their rcspective curves at the same time (for example,they are at thek positive maxima at the sameinstant).

phasorvoltageand < Relationship between phasorcurrentfor a resisior H^-a

R

*Jy_ I equivaLent cncuitof Figure9.8 AThe frequency-domajn

for an Inductor TheV-I Retationship We derive the relationship between the phasor cu ent and phasor voltage at the lerminals of an inductor by assuminga sinusoidal cuffert and using Ldildt to est,blish the coresponding voltage. Thus, for i = 1- cos (,)t + di), the expressionfor the voltage is

-aLl#n(at

+ q.

andcurrigure9.9A A ptotshowins thatthevoLtage arcin phase. rentatthetemjnabofa rcsktor

(9.28)

WenowrewriteEq.9.28usingttrecoshefunction: u = -aLl^cos(at

+ 0i

90").

The phasor representation of the voltage given by Eq. 9.29is

phasorvottageand (9.30) < Retationship between Dhasor curent for anindudor

344

Sinusojd at Steady-State AnaLysis Note tlat in deriving Eq. 9.30we used the identity

e F' : cos90. _ i siD90. = _j. Equation 9.30statesthat the phasorvoltage at the terminalsof an inductor equalsja,Z times the phasor cuffenL Figure 9.10 shows the frequercy-domainequivalentcircuit for the inductor. lt is important to I Irotethat the relationshipbetweenphasorvoltageand phasorcurrent for figuE 9.10 Theftequency-domain equivaLent circuit an inductor appliesas well for the mutual hductancein one coil due to currentflowingin anothermutuallycoupledcoil.That is,the phasorvoltageat the terminalsof one coil in a mutuallycoupledpair of coilsequals i&rM timesthe phasorcurrentin the otler coil. Wecanre*.riteEq.9.30as

v = (aL/p:)r ^19 = aLIn /@L + 90)'.

rigure9.11 A pLotshowing thepharepLationrhip betl,een thecunentandvoLtage at th€tenninals of an inductor (tr = 60").

(e.31)

which indicatesthat the voltageand current are out of phaseby exactly 90'. In paiticular, thevoltageleadsthecurrentby 90', or,equivalently, the currentlagsbehindthevoltageby 90".Figure9.11illustrates thisconcept of voltageleadingcurrentot cunentlaggingvohage.For example,the voltage ieachesits negativepeak exactly90" before the cufient reachesits negativepeak.The sameobservationcan be madewith respectto the zero-going-positive crossingor the positivepeak. We can alsoexprcssthe phaseshift in secondsA phaseshift of 90' corespondsto one-founhof a period;hencethe voltageleadsthe current by T/4, or a7second.

TheV-I Relationship for a Capacitor We obtain the relatiodship between the phasor current and phasor voltage at the terminals of a capacitor ftom the derivation of Eq. 9.30. In other words. if we note that for a caDacitor that

andassume that D=U-cos(at+0),

then

| : j.,,cv.

{e.32}

9.4

Donain 345 Th€Palsive CircuitEt€mentsin th€ Ftquency

Now if we solve Eq. 9.32 for the voltage as a function of the current, we get

(e.33) < Relationshipbetweenphasorvoltageand phasorcurrentfor a capacitot Equation 9.33demonstratesthat t]le equivalent circuit for the capacitor in the phasordomainis asshownin Fig.9.12. The voltage acrossthe terminals of a capacitor Iagsbehind t]Ie current by exactly 90". We can easily show this relationship by reniting Eq. 9.33as

1/jac -(.---.

iFigure domain equivalent circuit 9,12a Thefreqoency

v : \

7''oo't^707

:4rp,

not.

(e.34)

lte altemative way to exprcssthe phaserelationship contained in Eq.9.34 is to say that the current leads the voltage by 90". Figure 9.13 shows the phase relationship between the curent and voltage at the terminals of a

Impedance andReactance Flgule9.13l A pLotshowing thephas€r€tationship

We conclude t}lis discussion of passivec cuit elements in the frequency between the cunentandvottage atth€t€rminals ofa domain with an impo ant observation.Wlen we compare Eqs. 9.27,9.30, capacitor (0i = 60'). and 9.33,we note that tley a-reall of the folm

(9.35) < Deflnitionof impedance

where Z represenfsthe impedence of the circuit element. Solving for Z in Eq.9.35,youcanseethat impedanceis the ratio of a circuitelement'svoltagephasor to its cunent phasor.Thus the impedance of a rcsistor is R, the impedance of an inductor is io-L, the impedance of mutual inductance is jdM, afld the impedance of a capacitor is 1/loc. In all cases,impedance is measuredin ohrns.Note that, altlough impedance is a complex number, it is not a phasor,Remembei, a phasor is a complex number that showsup as the coefficient of d''. Thus, althougl all phasors are complex numbers, not all complex numbers are phasols. Impedance in t]Ie frequency domain is the quantity analogous to resistance,inductance,and capacitanceir the time domain. The imaginary part of the impedance is called r€actance.The values of impedance and readance for each of the component values are sunrmarizedin Table 9,1. And finally, a reminder. If the refererce direction for the curent in a passivecircuit element is in tlle d ection of the voltage dse acrossthe ele' ment, you must insert a minus sign illto the equation that rclates the volt' age to t]te cullenl.

CiIcuia Element

Impedence

Resistor

R

j( 1/"'c\

-r/@C

346 5inusojdat Steady-StateAnatysis

l

9.5 Kirchhoff'sLaws in the Frequency Domain Wepointedoutin Section9.3,withreference to Eqs.9.23 ard 9.24,rharrhe phasortranslormis usefulin circuit analysisbecauseit appliesto the sum of sinusoidalfunctions.We illustratedthis usefulnessin Example9.5.We no$ lofmalizelhisobser\drion by de!elopingKifchbotts taw;in lhe trequencydomain.

Kirchhoff'sVoltageLawin the Frequency Domain We begin by assumingthat 01 - o, representvoltagesarourd a closed path in a circuit.We alsoassumethat the c cuit is oDeratinein a sinusoidal slead)srate.TbusKirchhoftr\oltage lawfequiresihdr Dt+x2+

+un=0,

(e.36)

which in the sinusoidalsteadystatebecomescomplex y,1cos(a,t + 0) + V,1cos(at + 0, + .

+ V^.cos(d + Ah)=0. \9.37)

We now useEuler's identity to write Eq. 9.3?as n{V.eia,ej^} + n{vh,eta,ei't} +

. + nIv-,eta"et,t}

(e.38)

9.5 Krchhoffs tawsin theFr€quency Domain which we rewrite as Yi{v^,4,ejd + v,.4'ejd +

. + v^,ejs"etd}=t).

(e.3e)

Factoring the term d't lrom each term lelds

nltv-ce -v,,.ce--.

n{(V +v,+ .

v - c a t e t " t-l. .

r v , ) e , o ' }: 0 .

(e.40)

Bur d'i + o, so (e.41) < KVLin the frequency donain which is the statement of Kirchhoff's voltage law as it applies to phasor voltages.Inotler words,Eq.9.36appliesto a set ofsinusoidalvoltagesin the time domain, and Eq. 9.41 is the equivalent statement in the frequency domain,

Kirchhoff'sCurrentLawin the Freguency Domain A similar derivation aDDliesto a set of sinusoidal currents Thus if \9.4?)

then

(9.43) < KCLin the frequency donain where11,I2,-. , I, are the phasorrepresentations of the individualcui Equations 9.35,9.41, and9.43form thebasisfor circuitanalysis in t}le frequencydomain.Note that Eq. 9.35has the samealgebmicfolm as Ohm'slaw,andthat Eqs.9.41and9.43stateKirchhoff'slawsfor phasor quantities.Therefore you mayuseall the techniquesdevelopedfor analyzing resistivecircuitsto find phasorcurents and voltage$You needlearn no new analytictechniques;the basiccircuit analysisand simplification toolscoveredin ChapteN2+ canall be usedto analyzecircuitsin the frcquencydomain.Phasorcircuit analysisconsistsof two fundamentaltasks: (1) You must be able to col1struclthe frequetrcy-domain model of a circuit; and (2) you must be able to manipulatecomplexnumbeE and/or quantitiesalgebmically.We illustiate theseaspectsof phasoranalysisin th€ discussionthat follows,begiffing with seiies,parallel,and delta{olinr'esimDlitications.

349

SinusojdalSteady-SiaieAnalyris

9.6 Series,Para[[e[, andDelta-to-Wye Simplifications The rules for combining impedances in sedes ot parallel and for making deltato-wye tnnsformations are the same as those for resistofi, The oDly difference is that combinirg impedancesinvolves tle algebraic manipulation of complex numbers

Combining Impedances in Series andPara[[el

tiguE 9.14

Impedancesin sedes can be combined into a single impedance by simply adding the individual impedances,The circuit shown in Fig. 9.14defines the problem in general terms.The impedancesZr, 22, . . . , Z, arc connectedin seriesbetween terminals a,b.Wlen impedancesare in series,they carry the samephasor cunent I. From Eq. 9.35,t]Ie voltage drop aqoss each impedan\t is ZJ,221, .. , Z"1, and ftom Kircbloff's voltagelaw, impedances in sedes.

Yab:Zl+

Z2l+ .. + Z"I

=(21+22+,,. + z")1.

(e.44)

The equivalent impedance between terminals a,b is

: zr+ 22+

. + Zn-

Example 9.6 ilustrates a numerical application of Eq. 9.45.

Ie.15)

Parattet, 9.6 Senes, andDetia to Wye Simptificalions 349

Impedances in Series Combining A 90 O resistor,a 32 mH inductor, and a 5 pF capacilorare connectedin se es acrossthe terminils uf r .inu.oiJcl \olrrg< suurce.r\ shoqn in Fig.9.15.Thesteady-state expressionfor the source voltage?."is 750cos(5000t + 30") V. a) Construct thc ficqucncy domain equivalenl circuit. b) Calculatethe sleadystalecurrenlibl the phasor melhod.

The phasor iransform of ?,"is

v = 7s0llqv. Figure 9.16 illustrates the ftequency-doman equivalentcircuitof the circuii shownin Fig.9.15. h) we computethe phasorcurrentsimplyby dividirg rherolr"geoi rhe\ olrage\oufceb) rheeqli\ alentjmpedancebetweenrhe ierminalsa,b.From Eq.9.,15, z^b = 9lt + j16o

90O

32nH

j40

- e0+ i120: 1sol!l!a J). Thus '750 /30"

Figure9.15 J. ThecircuitforExamph 9.6. W e m a ) n o ! \ ! \ J r e t l ' e . t e a d ) - s t a r ee \ p r e . s i o n lbr i directly:

Solution

i : scos(s0oor23.13') A.

a) From thc cxprcssion lor 11, we have rr = 5000 md/s. Thereforethe inpedanceoI lhe 32 mH inductor is x 10 ) : 1160-,, zL = jaL = j(5000:)(.32

140{}

and the impedanceof the capacitoris

"'

-t

' aC

1n6

' (sooo)(s)

Figlre 9.16 ,t Thefrequency-domain equivatent cncuitofthe circuiishownin Fig.9.15.

objective3-Know howto usecircuitanatysis techniques to sotvea circuitin the frequency domain 9.6

Usingthevaluesofresistance andinductance in the circuil of Fig.9.15,let V" = 125l:q V

b) the magnitudcofthc steady-state output curenf ,,

a n d @ - 5 0 0 0 r a Lcl F i n d

a) the value of capacjaance that yields a steady'stateoutput current I with a phase angleof 105'. NOTE: Also try Chaptet Prcbleng.2l.

Answet (a) 2.86pF; (b) 0.982A.

350

SinusojdaLSteady-StateAnalysis Impedancesconnectedin parallelmay be reducedto a singleequivalent impedance by the reciprccal relationship 1

1

1

1

\9.46)

Figue 9.17depicts the parallel connection of impedances.Note that when impedancesare in parallel, they have ttre same voltage acrosstheir terminals. We de ve Eq. 9-46 directly from Fig. 9.1? by simply combidng Kirchhoff's current law with the phasor-domainvenior of Ohm's law, that is,8q.9.35.From Fig.9.17,

I = 1 1+ 1 2 + . . . + I , ,

figlre9.17 Impedancesin panLLeL.

\9.47)

Canceling the common voltage term out of Eq. 9.47rcveals Eq. 9.46. From Eq.9.46,for the specialcaseofjust two impedancesin pamllel, Z.Z,

(e.18)

We can also expressEq. 9.46 in terms of admiltatrce, defined as the reciprocal of impedance and denoted Y. Thus

1 ^ 7 , -

lB (siemens).

\9.49)

Admittanceis, of couise,a complexnumber,whosercal part, G, is called cotrdu€itarce and whose imaginary part, B, is called susceptar€e. Lite admittance, conductanc€ and susceptanceare measured in siemens (S). Using Eq.9.49in Eq.9.46,weget Yoo:Yt+Yz+.

+Y".

(9.50)

The admittance of each of the ideal passive circuit elements also is worth noting and is summarized in Table 9.2. Example 9.7 illustrates the application of Eqs. 9.49 and 9.50 ro a spe, cilic circuit.

Circuil El€ment

Adnfttrnce (Y) j(-r/aL)

Capacitor

1/.,1

9.6

Sedes. ParatLel. andDeLta-to'Wve SimDtifications351

Combining Impedances in Seriesandin Parallel The sinusoidalcurrent souce in the circuit shown in Fig. 9.18produces the cunent ir : 8 cos200,000tA. a) Construct the frequency-domain equivalent crcull, b) Find the .leady-srare expre<sioos tor u. ir. tr. ancli3, Figure9.18l ThecircuiiforExampte 9.7.

Solution a) The phasor transfom of the current souce is 8 A; tle resistors transform directly to tle ftequency domain as 10 a.nd 6 O; the 40 /rH inductor has an impedance of j8 O at the given frcquency of 200,000 rad/s; and at this ftequency the 1 pF capacitor has an impedance of q.lq sho$srbelrequenc) -domain i5 O. Figure equivalent circuit atrd symbols representing the phasor tiatrsforms of the unknowns. b) The circuit shownin Fig. 9.19indicatesthat we can easily obtain the voltage across the current source once we know the equivalent impedance of the tlree pamlel branches. Moreover, once we know V, we can calculatethe three phasor currents11,12,and 13by usingEq. 9.35.To find the equivalent impedance of the three branches, we first find the equivalent admiltance simply by adding the admittances of each bnnch. The admittanceof the first branchis 1

v,:_:0.1S.

the admittance of the secondbmnch is Y"

1 6+18

6-i8 = 0.06 - j0.08 S, 100

Ir

Figure9.19 A Thefrcquency-donain equjvat€nt cjrcujt.

TheVoltagev is

v: zr:40/ 36.87"V. Hence 40/ -36.87' r4 /-36.87' - 3.2 j2.4 \. ,^ 40/ -36.87' b+/a and 40/ -36.8'7" = I /s3.r3" : 4.8+ j6.4A. 13= 5 ,/ 90' Wecheckthe computationsat this point by vedfying that

and the admittance of the third bmnch is

Y . : -l/ = ,o.zs. J The admittarceof the tbree branchesis Y:Yr+Y2+Y3 = 0.16+ ./0.12

- 0.2149:s. The impedance at the current souce is

z=!=s/-36.87"a.

1 1+ 1 2 + 1 3 = I Specificay, 3 . 2 j 2 . 4 j 4 + 4 . 8+ j 6 . 4 = 8 + i 0 . steady-slare Tbe corresponding Iime-domain - 36.87')V, t, = 40cos(200,000t t1 - 4 cos(200,000t 36.87')A, - 90")A, t2 : 4cos(200,000t j3 = 8cos(200,000t + 53.13')A.

352

SinusoidatSteady'StateAnatysjs

donain techniques to sotvea circuitin the frequency 3-Know howto usecircuitanatysis Objective 9,7

A 20 O resistoris connectedin parallelwith a pardllelcombinalion is 5 mH Inducror.Thir connectedin serieswith a5 O resistorand a 25tF cdpacilor. ol lhi\ Inlera) Calculate the impedaDce the frequency is 2 krad/s. conneclion if br Repear(a' for a frequenc)of8 krad s. c) At what finiie frequency does tlle impedance of the interconnection become purely resistive? at tbe trcquency d) Whari. the impedance founJin (c'l

Answer:(a)9 - j12O; . ( b ) 2 1+ i 3 O ; (c) 4 krad/s; (d) 1so. 9.8

fte interconnection described in Assessment Problem9.?is connectedacrossthe teminals of a voltagesourcethat is generating , = 150cos4000 \. Wlal i! lhe ma\imum 'amplituale olthe curent in the 5 mH inductor?

Answ€r: 7.07A.

\OTE: Al5u try ch!ryn ProbLcn,a 24-o 23

Transformations Detta-to-Wye The A to-Y transformalionlhat we discussedin Section3.7 with regard to resistivecircuits also appliesto impedances.Figure 9.20 delines the A connected impedances along with the Y-equivalent circuit. The Y impedancesas functionsofthe A impedancesare

Z&. za+zb+2.'

z2 tnnsfomatjon. Figure9.20, Thedetta-to-wye

z)

z.z^ z^+ 26+ zc Z,Zr z\+zb+2.

(e.51)

(e.52)

(e.53)

The A to-Y transfonnationalsomay be reversed;thatis,we can start with the Y structureand replaceit with an equivalentA structure.The .\ impedancesasfuncdonsoftheY impedalcesare

z t z ) + z z z 3 +z 1 z 1 zl z1z2 + z2z3 + z1z1

z2 ztz2+ z2z1+ z3zr z3

\9.54)

9.6 Sedes, Pa6ltet, andDetta-to-Wve SimDtifications The processusedto deriveEq$ 9.51-9.53or Eqs.9.54-9.56 is the same as that used to derive the corresponditrg equations for pure resistive circuit$ In fact, comparing Eqs. 3.44-3.46 'r th Eqs. 9.51-9.53, and Eq$ 3.47-3.49witl Eqs.9.5,19.56,revealsthat the symbolZ hasreplaced t}le symbol R. You may want to review Problem 3.61 concerning the deri vation of the Alo-Y transformation. Example9.8illustratest]le usefulness ofthe A-to-Y transfomation in Dhasorcircuit analvsis.

Usinga Delta-to-Wye Transform in the Frequency Domain Use a A-to-Y impedance transformation to find Io, Ir,12,13,Ia,Is,V, and V2in the cicuit inFig.9.21.

the Y imDedanceconnectins to teminal c is

10( j20) 30 + j40

3.2 j2.4 A,

and the Y impedance connecting to termillal d is 2.4{l

120&lv+b j20a

_

(20+ j60x j20)

^

JU + 14|J, Inse ing the Y-equivalent impedarces into the circuit, we get ttre circuit showr in Fig 9.22, which we can now simplify by series-pamlel reductions. The impedence of the abn branch is Z^h"= 12 + j4 - j4 = DA,

Figur€9.21 l Theciicuitfor Exampte 9.8.

and the impedance of the acn branch is z^_:

Solution First note that the circuit is not amenable to series or parallel simplification as it now stands,A A-to-Y impedatrcetansformation allows us to solve for all the branch currents without rcsofiing to eithei the node-voltage or the mesh-curent metlod. If we rcplace either the upper delta (abc) or the lower delta (bcd) with its Y equivalent, we can further simplify the resulting circuit by series-parullel combhations. In decidn€ which delta to replace, the sum of the impedances arourd each delta is worth checkingbecause this quanrirllormsthe deoomi nator for the equivalent Y inpedance$ The sum around tle Iower delta is 30 + j40, so we chooseto eliminate it from the circuit. The Y impedance connecting to terminal b is

(20 + j60X10) : 12 + j4A,

30 + 140

63.2+ j2.4

j2.4

3.2 : 60 A.

j2.4A

jn Fjg.9.21,wiihthe tow* Figure9.22 a Thecircuitshown dethr€ptaced byjis €qujvatent wve.

354

sjnusoidatsteady-stateAnaLysjs

Note that the abn brarch js in palalel with the acn bmnch,Thereforc we may replacethesetwo branches with a singlebranch having an impedanceof

I!-

" 18()

160)r12)

-juo

Combiningthis 10O resistorwith the impedance betweenn and d reduces the circuitsho$nin Fig.9.22to tle oneshownin Fig-9.23.Fromthelat

I1r =

1201! 18 - j24

= 4 /s3.r3' = 2.4+ j3.2A.

Once we know I1],we can work back through the equivalentcircuits to find the branch currents in the original circuit. We begin by noting that Io is the curent in lhe branchnd of Fig.9.22.Therefore v"d: (8

j24)t - 96

j32V.

F i g u r e9 . 2 3a A r m p l i 5 erde ' s i o ol f l \ . , i l r i h o a n 1 lig.9.22.

To find the branch cunents 13,I1, and 15,we must fint calculate the voltages V and Vr. Refering to Fig.9.21,we note that

1)R

v.:1)0/0'

( i 4 ) L= = + I

Y2:12019:

(63.2+ j2.4)r2- 96

t8v. 104 jiv. _ ,

We may now calculate the voltage Vai because

Wenowcalculate thebranchcurrents13,Ia,and15:

V=V",+Y,a and both V and Yndare known.Thus vai : 120

96 + j32 :21 + j32v.

We now computethe branchcu[ents I"b. and laci: 21 + i32 I =2+iJA. I,bn: 12 ,

)4+ j32

,l

8

In termsof the branchcurrentsdefinedin Fig.9.21,

r,=!"I=J-'if

e,

r,=ffi=] ;r.oe, v^ rs = _1

=;

)6

+ i4.rJA.

We checkthe calculationsby noting that

1 1: I " b . : 2 + j : A ,

2 2 6j1.6+ j4.8 = 2.4 + j3.2 : t{,, 14+15= + 3 t5

I,=I..,=fi*;f,e

' ', l, I t, - 1 , i - , u - / |.o= 2 - I 8 - | .

We check the calculations of Ir and l, by noting that 4

l1+ 12: 2.4 + j3.2 = l1'.

A

l,r12 .1 ,n t

t -' Ri -

R

t6

n/;-;;+/a.8-1,.

9.7

5orceTratrformation5 andTh6venin-Norton Equivahnt Circuits

objective3-Know howto usecircuit analysistechniquesto solvea circuit in the frequencydomain 9.9

Use a A-to Y transfoimationto find thc current I in ihe circuit showD.

14(])

AnswenI= 4/28.01'A. NOTE: Also trf ChaptetPrcbkng34.

9.7 $oureeTrnnsfqarmatiens ar:ri Thdweilf cr-hlorton iquivatentCircuits The souce lralsformationsintroducedin Section4.9 and the Thdvenin, Norton equivalentcircuitsdiscussedin Scction 4.10are analyticalrechniqucs that also can be applied to frequencydomain circuits.We prove ihe validily of these techniquesby follorving the sameprocessused in Scctions4.9 and,l.10,exceptthat $'e substituteinpedance (Z) for resistance (R). Figure 9.24 shows a sourcctrunslormationequivalentcircuit Figure9.24A A souretransfomation in the qilh the Ionr
Frequency-domain maycontain

->

Figrre9,26Ir ThefrcqLrenry-domain version of a Norton

SinusoidaL St€ady.State Anatysis

PerfornlngSource Tnnsformations in the Frequency oonain Use the concept ofsource transformation to find tle phasor voltage Vo in the circuit shown in Fi& 9.27.

j3a

lo

o2o io6o

O2o i0.6o

Figure9.28 a Thefilstsiepjn reducing th€ circujtshov/n in Fig.9.27.

tigure9,27,t Theci(uitforExanple 9.9,

1.8O j2.4O

O.2A j0.64

Solution Wecanreplacethe sedescombination of the voltagesource(404) andtheimpedance of I - i3 O with the parallel combinationof a current source andthe 1 + j3 O impedance.The sourcecurrentis zt{l

40 = r ! _____::r + J r ;(l lu

_ /3) = a _it2A.

Thuswe canmodily the circuit shownin Fi9.9.27to the one shownitr Fig. 9.28.Note that the polarity referenceof the ,10V sourcedeterminesthe referencedirectionfor I. Next, we combirc the two parallel branches into a sitrgleimpedalce,

( 1+ i 3 x e - j 3 ) = 10

Also note that we have reduced the circuit to a simple series circuit. We calculate the current Io by dividitrg the voltage of the source by the total series

1.8+ j2.4 A,

which is in parallel with the cu[ent source of 4 i12 A. Another source tmnsformation converts this parallel combination to a se es combination consisting of a voltage source in serieswith the impedanceof 1.8 + t2.4 O.The voltageofthe voltage source is

v = (4

Figure9,29 ^ Thesecond st€pin reducing thecircuitshown in Fig.9.27.

36 jr2 12 jr6

. -

=

r2(3 jD 4(3 - j4)

-'1 0 + i ) 7 -:- = 1.56+ ,1.08A. It

j12X1.8+ j2.4)- 36 - j12V.

Using this sourcetransfomation, we rcdraw the circuitasFig.9.29.Notethe polarityof the voltage source.We addedthe current Io to the circuit to expeditethe solutionfor Yo.

We now obtainthe valueof V0by multiplying lo by theimpedance 10 - i19: V0= (1.56+ i1.08)(10- j19) :36.12 - jr8.84v.

9.i

Source Tmnsfomations andThavenin-Norton Equivatent Circuiis 357

Findinga Th6venin Equivatent in the Frequency Donain Find the Th6veninequivalentcircuitwith respectto terminalsa.b for the circuit shownin Fig.9-3w.

j44{r

-j40 (l

risure9.31A A sjmptified veBion ofthecircuit shown jn Fjg.9.30.

We relateth€ controllingvoltageY to the currentI by noting ftom Fig.9.31thal

tigure 9.30 A lhe circuitfo, ENample 9.10.

v,=100-10L Then,

Solution We first determinethe Thdvenin equivalenl volf age.This voltageis the open-circuitvoltageappearingat terminaisa,b.We choosethe referencefor the Th6veninvoltageas positiveat terminal a.We can make two source transformations relative to the 120V 12 O, and 60 O circuit elementsto simplify this pofiion of the circuit.At the sametime, tiese transformationsmust prese e the identity of the controlling voltage \ becauseof the dependert voltagesource. We determinethe two sourcetransformations by first replacing t}le se es combination of the 120 V sourceand 12 O resistorwith a 10A currenl sourceill parallel with 12 O. Next, we replacethe parallel combinatior of the 12 and 60 O resistors $rith a single 10 O resistor. Fina y, we replace the 10 A souce in parallel with 10 O with a 100 V souce in serieswith 10 O. Figure 9.31 showsthe resultingcircuit. We addedthe currert I to Fig.9.31to aidturther discussion, Note that once we know the current I, we cancomputethe Th6veninvoltage.Wefind I by summingthe voltagesaroundthe closedpath in the circuitshownin Fig.9.31.Hence 100 : 10I

j40l + 120I + 10v":

(130 - j10)I + 10v..

I :

30

900 : 18/ -126.8'7'A. J40

we now calculate va: Y,:

100

180/-126.87' :20A + jt44V.

Finally,we note from Fig.9.31that Yrh=10V,+120I = 2080 + 11440 + 120(18)/ - 784 j)qQ, 815).

126.87"

)0.t7'V.

To obtainthe Th6veninimpedance,wemay use any of the techniquespreviouslyused to find the Thdvenin resistance.We ilustrate the testsource method in this example.Recall that in using this mcthod, we deactivate all independent sources Irom the circuit and then apply either a test voltage sourceoI a test current sourceto the terminaisof interest.The ratio of the voltage to the current at the sourceis the Th6veninimpedance.Figure 9.32 showsthe result of applyjngthis techniqueto the cncuit shownin Fig.9.30.Note that wc chosea lest voltage sourceVr. Also note that we deactivated the hdependentvoltagesourcewithan appropriate \horr-circuir andpresenedlhe idenrir)of V.

358

Steady StateAnatysis Sinusoidal

_ -v.(e + i4)

-i4o o

120(1 j4)

Ir=I.+It

9+i4\ 1 2 )

v? (. 10 140\ v?(3 12(10 YZrh:7:

i4) j40)'

0 12

/384O

equivateni forcatcutating thelh6venin Figure9.32a Acjrcuit Figure9.33depictstle Theveninequivalentcircuii. The branch curents I, and Ib have been added 10 the circuil to simplify the calculation of l? By straightforwardapplicationsof Kirchhoff's circuit laws,you should be able to verify the following relationships: T

=

10

-138.4O

v- - 10L, cncuitshown equjvahntforthe Figure9.33 a TheTh€venin Fig. in 9.30.

lr' =

to sotvea circuitin the frequeicydomain techniques objective3-Know howto useorcuit anatysis in expre.sionfor L'o(r) 9.10 | ind lhe steady-slale lhe circuitshowoby u.inglhe lechniqueol the .inusoidalvollage sourcerranslormalions. solucesare

wilh respecr lo 9.11 FindrheThe!eni0equivalenl

. t6rmiuls a,bin thd circuit shown.

z,1= z0cos(4ooor+ 5:li3")v, zb : 96 si! 40001V. 15mH

-

,!.r-l

Answer: 48 "''" cos(4000t *-: + 36.87')v . ^'-*:' ) 'IOTL: Al'o try Chapw ProblPnta 40.att. ando 1-

Vir,= v"h: r0/4s' v;

9.8

TheNode-Voltage Method 359

9.8 TheNode-Voltage Method In Sectio s 4.2-4.4,we intioduced ttre basicconceptsof the node-voltage method of circuit analysis.The same concepts apply when we use the node-voltage method to analyze frequency-domain circuits. Example 9.11 illustrates the solution of such a cicuit by the node-voltage technique. AssessmentProblem 9.12 and many of the Chapter Problems give you an opportunityto usethe node-voltagemethodto solvefor steady-state sinu soidalresponses,

Usingthe Node-Vottage Methodin the Frequency Domain Use the node-voltage method to lind the branch currents Ia, Ib, and t in the circuit shown in Fig.9.34.

1() 10.6/E

iza

t l$ { 1r0ooo

v1(1.1+ 10.2) y2:10.6 + j21.2. Summingthe curents awaylrom node 2 gives

t ,

i)

5f,}

Multiplying by 1 +/2 and collecting the coefficientsof Y1and V2generatesthe expression

l t b

r.,+-i5r) 20t,

v,

v2.v2

!

-

r+ tl

20I, ^

----=

f

)

/)

The controlling current I* is Figure9.34 A Thecjrcujtfor Erampt€ 9.11.

I,:

Solution We can describe the circuit in terrns of two node voltagesbecauseit containstbree essentialnodes. Four branchesterminateat the essentialnode that stretchesaqoss the bottom of Fig.9.34,so we useit as the relerencenode.The remainingtwo essential nodesare labeledl and 2, and the appropriatenode voltages are designatedV and V2. Figuie 9.35 reflectsthe choiceof referencenode and the terminal labels.

1

j2A 2

J- ; + , , . to

|

s0

20r,

Summingthe curents awayfrom node 1 yields

v

v-v,

10.6+-+-r_^:=0 ru r+ tz

1+j2'

Substituting tlis expression for I, into the node 2 equation, muitipling by 1 + j2, and collecting coefficientsof V1and V2producesthe equatron

-sY+(4.8+J0.6)Yr:0. Thesolutionsfor V andV2are v = 68.40 j16.80V, v, = 68 j26 V. Hencetie bmnchcurrentsare I. :

t0

= 6.84 y'.68A,

v -v

I, = ,I a Ib =

v,

v

Figur€9.35L lhecircuit shown in Pig.9.34,withthenode vottages defined.

Y-V

I,.: =

lz

20r-=

.-.....

J]

: 3.76+ lr.bSA, -1 44

j1192 A,

= 5 . 2+ 1 1 3 . o A .

To checkour work, we note that Ia + I. = = Ir = =

6.84 - j1.68 + 3.76 + j1.68 10.6A, Ib + !c: 1.44 j11.92+ 5.2 + j13.6 3;76+ j1.68A.

360

Sieady StateAnatysis Sinusojdal

objective3-Know howto lse circuitanatysis techniqlesto sotvea circuitin the frequency domaih 9.12 Use lhe nodc voltagemethodto lind the sleady fbr o0) in the circuitshown.The slateexpression sirusoidalsourcesare i = l0cos(,tArDd L, - 100si d/ V. $ lrer
204

,t, Irr"

Answer: o(r) = 31.62cos(50,000t 71.57") V. NOTE: ALto ttr ChapterProblens 9.51tud9.56.

StretXr*d 9.9 The$4esh-{urrcnt Wc can also usc thc ncsh curcnt mcthod to analyzcficqucncy-domain circuits.The proceduresused in frequency-domainapplicationsare the sameas thoseused in anal]'zingresistivecircuits.In Sections4.5 .1.7,we introduccdthc basiclcchniqucsof thc mcsh currcnt mcthodl we demonstratc thc cxtcnsion of this mcthod to frcqucncydomain circuits in Example9.12.

Usingthe Mesh-Current Method in the Frequency Domain Use the mesh-currentmethod io find the voltages V r .\ ' . " n d \ n r h e c i f c . r.i hr . $ I i r l : 9 . o . J h .

tir

t2O1

lO

/l!)

72Q

Sotution

150i0:

The circuit has two meshesand a dependentvoitage source.so we rnusl wrrle two mesh-current equationsand a constraintequation.Thereference direction for the mesh currentsIr and I, is clockwise.as shownin Fig.9.37.Oncewe know Ir and Ir, q e c . n e d . l ) f n d r h - u n k n ^ \ n\ o l r a C cS. .u n m , n ! the voltagesaroundmesh I gives

rer,

t ,t,

i,'

Figure9,36...!.Thecjrcuitfor Exampte 9.12.

1s0= (1 + l2)rr+ (r2- i16)(rr r,). j2tl 1s0 : (13 - r14)Ir - (12 - 116)I,.

1 2O : r I .

Sunming the voltagesaroundmesh2 generaleslhe 0 = (12

j16)(1,

/3{}

11)+ (1 + j3)r' + 39r..

Figure9.37revealsthat the controllingcurrentI, is ttre differencebetweenIr and lrr that is. the conl- : Ir

12.

V

i--

/160

Figure9.37r| l4esh cuftents used to solve thecncuitshown jn Fiq.9.36.

9.10 TheIransformer 361

Substitutingthis consrraintlnto lhe mesh 2 equa rionandsimplihing rhefesulring e\pre*ion grve.

- ( 2 6+ 1 1 3 ) r r . 0=(27+j16)r1 Solvingfor Ir ard I, yields Ir = -26 j52 A. Iz = -21

t 5 0+ v 1 + v :

12

jt04 + 150 j130

78+j234=0,

jr04v,

v,: (12 ji6)r, = 12 + j104y.

1 5 0+ 7 8 - j 1 0 1+ 1 2 + j101: 0,

v2 + vr + 39I. =

Thcthreevoltages are

7iJ+ i234V.

We chocklhesecalcularions bv sumningthe voliagesaroundclosedpaths:

j5lt A,

l,=-2+j6A. vl =(1 + j2)Ir

39I, =

1 5 0 + Y 1+ v 3 + 3 9 I a = 1 5 0 + 7 8 j 1 0 4 + 1 5 0

v3: (r + j3)r, = 1s0 J130v.

78 + i234 - 0.

jr30

objective3-Know howlo usecircuitanatysis technques to solvea circuitin the freguency domain 9.13 Use the mesh-currentmethodto find the phasor curent I in the circuit shown.

1{l

i2a

0.75V,

\'+ Answer:I:29+

j2 -29.U 13!t:A.

NOTE: Also try ChapterPrcblens9.58and9.61.

9.10 TheTransforrmer A transformeris a devicethat is basedon nragneticcoupling.Tiansformers areusedin both communicatioD and powercjrcuils.Incommunicaiioncir cuits.the transformeris usedto matchimpedances and eliminatedc signals tiol1] portions of the system.In power circuits,UaNformers are used to establishac voltagelevelsthat facilitatethe transnission,distribution,and consumptionof electricalpolver.A knowlcdgeof the sinusoidalsteady, statebehaviorof lhe lransformeris requiredi]l the anaiysisof both communication and power systems.In this section.we will discussihe sinusoidalstcadystatebehaviorof the linear transform€r,which is found primarily in comnuication circuits.In Section9 11,we will deal with the id€al transformer,which is usedto modcl the ferromagnetictransformer Iound in powersystems.

l5o

362

Sinusoidat Steady StateAnalysis

Before shning we make a usetul observation.When anal'zing circuits cortaining mutual inductanceusethe meshorloop-currert method for writing circuit equations The node-voltagemetlod is cumbersometo usewhen mutual inductarce in involved. This is becausethe currents in the vadous coils c-annotbe *ritten by inspection as lurctions of tlle node voltages.

TheAnatysis of a LinearTransformer Circuit A simple iraNform€r is formed when two coils are wound on a single core to ensure magnetic couplirg. Figure 9.38 showstle ftequency-domain circuit model of a system that uses a transformer to connect a load to a source.In discussingthis circuit, we refer to the transformer wirding connected to the source as the primary winding and the winding connected to the load as the secordary winding. Based on this terminology, the transfomer circuit parameten are Fisure 9.38l lhefr€quency domain cncuiinrodetfora trandormer used to connect a toad to a source.

R1 = the resistanceof the p mary winding, R2 : the resistanc€of the secondary\ dnding, a1 = the self-inductance of the primary witrding, 12 = the self-inductance of the secondarywinding, M = the mutual inductance. The internal voltage of the sinusoidal source is v', ard the intemal impedance of the source is 2,. The impedance ZL representsthe load connected to the secondary winding of t}re transfolmei, The phasor currents I1 and 12represent the piimary and secondarycurents of the transfonner, iespectively. Analysis of the circuit in Fig. 9.38 consistsof finding 11and 12as functions of the circuit paramergrcY", 2,, Rb Lr, L2, R2,M, ZL, ald a.We arc also interested in finding the impedance seenlookirg into the transformer ftom the terminals a,b.To find 11and 12,we fiIst write t}le two mesh-crlrrent equations that descdbe the circuit:

- jaMr2,

(e.57)

0 = -joMr1 + (R2 + jal2 + Z)r2.

(e.58)

vs= (zs+ Rr+ jaL)\

To facilitatethe algebraicmanipulationof Eqs.9.57and 9.58,we let Zr=Z"+Rr+j'tLl,

(e.5e)

Zn= Rz+ jtiL2+ ZL,

(e.60)

where Z1r is the total self-impedance of the meshcontainingthe pdmary winding of the transfomer, and Zn is the total self-impedance of tle mesh containing the secondarywinding. Based on tlle notation intoduced inEqs.9.59and 9.60,thesolutionsfor lr and 12lrom Eq$ 9.57and 9.58are (e.61)

z\zn + ZnZ22 !

v

=',

lL

(e.62)

9.10 IheTGnsformer353 To the intemal source voltage \, tle impedanco appea$ as V"/I1, or

v" = zi"' i

znz22 + ozMz

(e.63)

The impedanceat the teminals olthe sourceis Zinl

2 " " = 2 . .+ + -

z " : R .+ i u L , +

-Z.,so 0jM'

(R2+ jaL2+ Z)'

(e.61)

Note tllat the impedance Z.b is independent of tle magnetic polarity of the tiansformer. The reason is that the mutual hductance aoDearsin fq. q.64a. a \quaredquanriry. This impedance i( ot paruicuta; inreresr becauseit showshow the tmrsformer affects the impedance of tle load as seen hom the source. Without the transformet, the load would be connected directly to the source, and the source would seea load impedance of ZL; with tle transformer, the load is connected to the source through the transfoner, and the sourceseesa load impedancethat is a modilied versionofZL, asseenin the third term ofEq.9.64.

Reflected Impedance The third term ir Eq. 9.64is caled the r€flected impedance (2.), because it is t]le equivalent impedance of the secondary coil and Ioad impedance tlansmitted, or reflected, to the pdmary side of the transformer. Note that the reflected impedance is due solely to the existence of mutual induc, tance;that iq if t]te two coils ate decoupled, M becomeszero, Z, becomes zero, and Zubreduces to the self-impedanceof the primary coil. To consider reflected impedance in more detail, we first exprcss the load imDedancein rcctansular form: ZL:

RL + jXL,

(e.65)

wherc the load reactance-YL cardes its own algebmic sign. In other words, -YL is a positive number if the load is inductive and a negative number if tlle load is capacitive.We now use Eq.9.65 to *rite the reflectedimpedancein rectansularform:

R2+ RL+ j@L2 + XL) a'M2l(R, + RL) - j(oL2+ xL)l (R2 + RL)? + (alz + XL)'

=

ffin*

- j(aL,+ xL)1 + RL)

(e.66)

The derivation of Eq. 9.66 takes advantage of the lact that, when ZL is written iD rectangular form, the self impedance of the mesh contai ng the secondarywinding is Zn=

R2+ RL+ j@L2+ XL\.

(e.67)

SinusoidaL Steady StateAnaLysis Now observefrom Eq.9-66that the self-impedance of the secondary circuit is reflected into the primary circuit by a scaling factor of (oMl Zrr\', atfi, that the sign ol the reactive component (ol2 + XL) is revelsed.Thus the linear transformerreflectsthe conjugateof the selfimpedanceof the secondarycircuit (ZiJ into the primary winding by a scalarm tiplier. Example9.13illustratesmeshcurrent analysisfor a cir c t containinga linear transfbrmer.

Analyzing a LinearTransformer in the Frequency Domain The parametersof a certainlinear transformerare R1 : 200 O, R2 : 100O,1-r = 9 H, L2 = 4H, and ,t = 0.5. The transformer couples an impedanc€ consistingof an 800{} resistorin seies with a 1 /rF capacitorto a sinusoidalvoltagesource.The 300V (rms) sou(ce has an internal jmpedance of 500 + j100 O and a frequencyo1400rad/s. a) Construct a ftequency-domain equivalent circuit of the system. b) Caiculate the self impedance of the primary

transformer. In constructing the circuit in Fig.9.39,we madethefollowingcalculations: jo.1 = i(400xe) = j3600o, jo., = j(a00)(a): j1600o, M=0.5{e)(,0:3H. j{,M = i(400x3) : 11200 o, 1

d) Calculatetlle impedarce reflectedinto the primary winding. e) Calculate the scaling factor for the reflected impedance. f) Calculale the impedance seen looking into t}le pdmary terminals of the transformer, g) Calculatethe Th6veninequivalert witl respect to the terminalsc,d.

a) Figure9.39shows thefrequency-domain equiva lentcircuit.Notethattheintemalvoltageof the phasor.andthat sourcese es asthe reference the terminalvoltages of thc V andU represent

^=

ldL

l4uu

zr = 500+ j100+ 200+ j3600: 700+ i3700o. c) The self-irnpedance of the secondarycircuit is ZL = 100 + j1600+ 800 - j2500 - 900

vL i3600o

figure9.39& Ihefrcquency-domajn equivatent circuittor Exampte 9.13.

j900 O.

d\ The impedance re0ecr(Ll iDro rhe primary winding is

=

/

"t "r n "n

\ l9oo j9o0 ;(900

\2 l/on^!;onn'

+ i900) = 800+ i800o.

e) Thescalingfactorby whichZi, is reflectedis 8/9.

Ir

300!t v

l2s00o.

h) t}e ,elf-impedance of rhepflmdrycircuilis

7 =l "'

Sotution

t06

_=

c) Calculatethe self-impedanceof the secondary

j2s00o

s.11 TheIdeatTnnsformer 365 f) The irnpedanceseen looking inlo the primary terminalsofthe transformeris the impedaDceof l h e n r i m a r yu i n L l n gp l u . r n e r e l l c ( l e di m p e d -

TheTh6veninimpcdancewil be equalto thc impedance of the secondarywindilg plus the impedance reflectedfrom the prinary when ihe voltagesource is replacedby a shot circuit.flus

zdh :200 + j3640 + 800 + i800 : 1000+ j4400 ().

Zrh: 100 F 11600 + (

g) The Thdvcninvoltagewill equalthc open circuit valuc oI Vcd.The open circuil v.1lueof V.d will equal j1200 limes the open circuil value of lr. The opcn circuit valueof lr is

Ir:

^#$-)','nn ,.'0,

= 1'71.09 + j1224.26 A. 'IheTh6vcnin equivalent is shownin Fig.9.40.

3001!: 700+ j3700

1 7 1 . 0{9)

O /t22.1.26

19.67/ 79.29'mA. Therefore

= i120aQ9.6'7 / 19.29')t 10 3 : 95.60 /10.71'V.

Figure9.40;i TheThevenin equivalent circuittortxampLe s.13.

objective4-Be abteto anatyze circuitscontaining lineartransform€rs usingphasornethods 9.14 A linear transformercouplcsa load consisting ofa 360 O resistorin serieswirh a 0.25H induclor to a sinusoidalvollagesource,as shown.The voltagesourcehasan internal impedanceof 184 + JOO and a mlximum volt, ageof245.20Y and jt is operatingat 800rad/s. The transformerparamelersare Xr : 100O, l,1 = 0.5 H, R' : ,10o, lz = 0.125 H, arrd k : 0.4.Calculale(a) t}le reflectedjmpedarce; (b) .he primary currerq ard (c) thc secondary

Answer: (a) 10.24 j7.68Q, (b) 0.5cos(800t- 53.13")A; (c) 0.08cos8001 A.

NOTE: Also ttf Cha4er Problems 9.72and 9.73.

9.11 TheXdeal Tramsfornrer An ideal tmnsformerconsistsoftwo magneticallycoupledcoilshavingN1 and N2 turns.rcspcctively, and exhibitingthcsethreeproperries: 1. The coefficientof couplingis unity (ft = 1). 2. The self-inductance ofcach coil is infinite (L1 = l,, : ,ro). 3. The coil losscs,due !o parasiticresistance, are negligible.

366

Anatysis Sinusoidat Steady State Understanding the behavior of ideal transformen begins with Eq. 9.64 which describesthe impedance at the terminals ofa sourcecolnected to a Iinear transformer. We reDeatthis eouation below and examine it further.

ExploringLimitingValues A useful relationship between the input impedance and load impedance, as given by Z.h in Eq. 9.68, emergesas -L1and a2 each become infinitely large afld, at the sametime, the coefficient of couplhg approachesunity:

zzb:

-

zt1 a --

=\+jaLj+

zs

(e.68)

(R2 + jaL2+ ZL)

Tiansformers wound on ferromagnetic cores can approach tl s condition. Even though such transformers are nonlinear, we can obtain some useful informationby constructingan idealmodelthatignoresthe nonlinearitiet To show how Zub changeswhen & : 1 and lt and a? approach infinity, we first irtroduce the notation

zn= R2+ RL+j@L2+ x)

= R2+ jxL

andtlen rearrange Eq.9.68: 7," " - R,

d2M2R,, t ---------+ t-iloL '\

Rt

xz,

-2M2x,, \ - --_= |

Riz xiz'

= R,b + jx,b.

(e.6e)

At this point, we must be careful with the coefficient of j in Eq. 9.69 because,as -L1and -L2approachinfinity, this coetficientis the dillerence betweentwo large quantities.Thus,before letting Z1 and -L2increase,we write the coefficientas Xdb = 'tLI

(d'L)(dL)

X22

RL + x,?2

: d L , \/ t

aL.X-

4;;rL

\

(,.70)

wherewe iecognizetha1,when k = 1, M' = tr1l2. Puttingthe telm multiplying oar over a common denominator gives

X^6= rLt(

R4.+ -L,x, + xi \ # l

Riz + xi

,/

(e.71)

9.11 TheldealTEnsformer

Factoring(dl, out ofthe numeratorand denominatorofEq.9.71yields

xL + (4,, + x?)lak

Lr

Lz (Rzl.L)2 + [1 + (xJdh)]2'

\9.72)

As & approaches1.0,the ratio lr/lz approachesthe constantvalue of (&/Nr)', which followsfrom Eqs.6.54and 6.55.Thereasonis rhar,as the couplingbecom€sextremelytight,the two permeances 91 and g2 become equal.Equation9.72then reducesto

(**)'

XL,

(9.73)

asal + oo,l4 +,ro, andt + 1.0. The samereasoning leads to simpliJication of the reflected resistance in Eq.9.69: a2M2R22

L.

R,+ x4,

L2

--

(t)'*,

\9.14)

Applying the resultsgivenby Eqs.9.73and 9.74ro Eq.9.69yields

Z*=

R t . (**)'

Rz+

(+)'

(Rr + jxr.

le.75)

Comparethis resultwith t]le result in Eq.9.68.Herewe seethat when the coefficient of coupling approaches unity and the selJ-inductancesof the coupled coils approachinfinity, the tmnsformer reflects the secondary windingresistanceand the load impedanceto the primary sideby a scalhg factor equal to the tums ratio (N/Nr) squared. Hence we may describe the terminal behavior of the ideal translormer in terms of two characteristics. First, the magnitudeof the volts per tuln is the samefor eachcoil.or

l"l=i-'l

-i

(e.76)

Second,themagnitudeofthe ampere-turNis the samefor eachcoil,or

llrNr : llzNz.

(e.11)

!r

We are forced to use magnitudesignsin Eqs.9.76and 9.77,becausewe (b) havenot yet establishedieferencepolaritiesfor the currentsand voltages; we discussthe rcmoval of the magnitude signsshortly Figure9.41A Theci(ujis us€dto veriirth€ vottsFigure 9.41showstwo lossless(Rl = R2 : 0) magneticallycoupled pertum andampere-turn rcLabonships for anjdeal coils.We useFig.9.41to validateEqs 9.76and 9.77.InFig.9.41(a),coil2 is

368

SinusojdalSteady"StateAnaL!^h

open; in Fig. 9.41(b), coil 2 is shorted. Althougi we carry out the following analysis in tems of shusoidal steady-state operation, the results also apply to instantaneousvalues of ?,and i,

Determining the VottageandCurrentRatios Note in Fig. 9.41(a) that tle voltage at t}Ie terminals of the open-circuit coil is etrtirely the result of the curent in coil 1;therefore \, = jaMlr

(9.78)

The current in coil 1 is

\ :

19.19)

i_L,

From Eqs.9.78and 9.79,

,.-to,. For unity coupling,the mutual inductanceequals lEf,

(e.80)

so Eq. 9.80

E;

(e.81)

For unity coupling, the flux linking coil 1 is the same as the flux linking coil2, so we need only one permeatrce to describe the self-inductance of eachcoil.ThusEq. 9.81becomes

n:ffin=ffn R',- i,

for anideal Vottage retatlonship > transfonner

r

lv

(e.82)

(981)

Summing tie voltages aroundtheshortedcoilof Fie.9.a1(b)lelds 0=-jaM\+juL2r1,

(9.81)

fiom which,for & = 1, \ =h = L, lz M \/Lrh

(e.85)

s.11 TheIdeatTmnstornrer 369 Equation9.85is equivalentto

I1N1 : I2N2.

(9.86) { Currentretationship for anideal iransformer

Figure 9.42 shows the graphic symbol for an ideal transfomer. The verticallinesin the symbolrepresentthe layersof magneticmaterialfrom which terromagneticcoresare often made-Thus,the symbol remirds us that coilswound on a ferromagneticcore behavevery much like an ideal

-;.lt.

^,1I l', 2 r l I IdealI

symbotfor anjdeat There are seveml reasonsfor tiis. The ferromagnetic material creates Figure9.42A Thegraphic a spacewith high permeance.Thus most of th€ magnetic flux is trapped inside the core material, establishingtjght magneticcoupling between coils that share the same core. High permeancealso meaff high selfinductance,becausea : N'S. Finally, feromagnetically coupled coils efficiently transfer power from one coil to the other. Efficiencies in excess of 9570are common,so neglectinglossesis not a cripplingapproximation lbr manyapplications.

Determining the Potarityof the Vottage andCurrent Ratios We now tuln to the removal oI the magnitude signs from Eqs. 9.76 and 9.77.Note that magnitudesignsdid not show up in the derivationsof Eqs-9.83and 9.86.Wedid not needthem therebecausewe had established reference polarities for voltages and relerence directions for currents. In addition,we knew the magneticpolarity dotsof the two coupledcoils. The rulesfor assignirgthe proper algebraicsigr ro Eqs.9.76and 9.77 arc asfollows: If Olecoil voltagesVland V2are bothpositiveor negativeat the dot markedtermiMl, use a plus sign in Eq. 9.76.Otherwise,usea negatrvesrgn, Ifthe coil currents11and I, are both dircctedinto or out of the do! marked teminal, use a minus sign in Eq.9.77. Otherwise,use a plus sign.

"{ Dotconvention for idealtransformers

The four circuitsshownin Fig-9.43illustmte theserules.

.-ll"i I

u,,,ll I

i ldeal

vt=vt. Nrll = (a)

NrI,

[:E'

&r,-&t: (b)

vr -Yu, NrIr = NuIu G)

l/'= &' Nrlr = NrI, (d)

figure9.43A Circujts thatshowthepropsatgebran signsfor retating thetermjnat voLtag€s andcunent5 ofan ideattranstormer.

370

Sjnusojdalsieady-StaieAnatysis

Nr = 500

.;l

N, = 2500

F. +

+;l 1r5[. +

" lll v: ",llf v: lIdealL (a)

The ratio ofthe turns on lhe two windingsis an important parameter of the ideal tnnsformer. The turns ratio is defined as either Nr/& or N/Nt; both mtios appearin various writings.In this text, we use d to denotethe mtio Ny'N1,or

lldeal (b)

N,

(e.87)

+ ; l 1 A : 1 F .+

" lli v. ) t ( lra*rL k) Figule9,44A Thrcewaysto showthatthetumsratio of anideattransfornrer is 5.

Figure 9.44showsthree waysto representth€ turns ratio of an ideal transformer.Figure9.44(a)showsthe numberof turns in eachcoil explicitly. Figure 9.44(b)showsthat the rario NxlNr is 5 ro 1, and Fig. 9.44(c) showsthat the ratio Nx/Nr is 1to i. Example 9.14illustmtestle analysisof a circuit containingan ideal transformer,

Anatyzing an ldealTransformer Circuitin the Frequehcy Domain The load impedanceconnectedto the secondary windingofthe ideal transformerin Fi9.9.45consists of a 237.5mO resjstor in series with a 125pH inductor. If the sinusoidalvoltagesource(r,s)is generat ing the voltage 2500cos400t V, find the steady stateexpressions fof:(a) t1;O) t1; (c) tr;and (d) or.

0.25O 5 mH

237.5m{}

tigurc9.45A The(ircuitforExanipte 9.14.

Solution a) We begin by constructing the phasor domain equivalentcircuit.The voltage sourcebecomes 25004 V; the 5 mH inductor convertsto an impedanceof t2 O; and the 125/rH inductor conveitsto an impedanceof j0.05 O. The phasor domainequivalentcircuit is showninFig.9.46. It followsdirectlytom Fig.9.46that

0.25(} _rl

l2soou ) v

./2|l

0..2375 o 10r1

_---Il

v.

i0.05o

Figure 9.46A Phasor domain circuit forExampte 9.14.

2s001y= (0.2s+ j2)Ir + v1, and

+ jo.0s)rrl. v : 10v,: 10(0.2375 Because 12: 10Ir

v1 = 10(0.237s + 10.05.)1011 : (23.7s+ js)rr Therefore

p: 2s00

\ = 100/ -16.26'A. Thusthesteady-state er?ression for i1is tr : 100cos(400t 16.26')A. b) V = 2s00lp: $00 // 16.26)(0.2s+ j2) = 2500- 80 - jl85 = 2120 j185: 427.06/-4.3'7' y. Hence

(2a+ 17)ry

xr : 21n.06cos(400t 4.37')Y.

Iransformer 371 9.11 TheldeaL

c) I? = 10Ir : 1Un /-16.26' A. Therefore t, = i000cos(400r- 16.26)A.

d) V, = 0.lV :242.11/-4.37'V, glvlDg

Dz= 242;7Icos(400t- 4.37')V.

TheUseof an IdealTransformer for Impedance Matching Ideal transformerccan alsobe usedto raiseor lower the impedancelevel of a load.The circuitshownin Fig.9.47illustratesthis.The impedanceseen by the practicalvoltagesource(Vsin sedeswith Zr) is Vr/Ir. The voltage and currentat the terminalsof the load impedance(V2and Ir) are related to H and Ir by the transforrncrlurns ratio;thus v'=

(e.8e)

Thereforethe impedanceseenby the praclicalsourceis

Yr

L

-

u ' [ ) Y /

._, -

J

l : +-

.r +

.

i

q

.

+

l

\ L ] | ' l z , ) T -I .

(9 88) fitut" 9.47g Ur;rganideat tnnsform€r to coupte a toadt0 a source.

lt = alz'

-.

l

L+.---_lldeall

v"I

and

ztN=it-

- I r .), Z,t

1V aj t,

(e.s0)

but the ratio Vz/I, is the load impedanceZL, so Eq.9.90becomes

z, =iz, Thus,the ideal transfo(mer'ssecondarycoil reflectsthe load impedancc back to the primary coil, with the scalingfactor 1/ar. Note that the idealtransformerchang€sthe magnitudeof ZL but does not alfect its phaseangle.wlether ZrN is grealeror lessthan ZL depends on the turns ratio d, The ideal transformer or ils practical counterpart, the ferromagnetic core transformer can be usedto matchthe magnitudeof Zr, to the magnitude of 2.. We will discusswhy this may be desirablein Chaptcr10.

obtectlire5-Be ableto anatyzealrcuitsreithideal transformers 9.15 The sourcevoltagein lhe phasordomaincircuit in the accoEparyingfigureis 25 A kV. Fird the amplitudeand phaseangleof Vzalld 12. Answer V2= 1868.15 //142,39'Yi h = 125 '/216.87' A. NOTE: Also try ChapterProblem9.n.

372

SinuroidatSteady-StaieAnatysis As we shall see,ideal transformersarc used to increaseor deoease voltagesftom a sourceto a load.Thus,ideal transformersare usedwidely in t}le elect c utility industry,where it is desirableto decrease.or step down,the voltagelevel at the powerline to saferresidentialvoltagelevels.

9.12 Phasor Diagrams When we are Ning the phasormethod to analyzethe steady-state sinusoidaloperationofa circuit,a djagramofthe phasorcwents and voltages may give further insight into the behavior of the circuit. A phasor diagram shows the magnitude and phase angle of each phasoi quantity in the complex number plane,Phaseanglesare measuredcounterclockwiseftom the positive real a\is, and magnitudes are measuredfrom t}re origin of the axes.For example,Fig. 9.48showsthe phasor quafiines n 1q, D 1lE:, Figure9.48A A guphicrcpGsentation of phasols.

Figlre9.49A Thecomplex numb€r -7 - j3 :' 7.62 /-156.80'.

51_15:,an.dB / 170".

Constructingphasordiagramsof circuit quantitiesgenerallyinvolves both currents and voltages,As a rcsult, two diflerent magnitude scalesar:e necessary,one for currents and one for voltages,The ability to visualize a phasorquantityon the complex-numberplare canbe usefulwhenyou are checkhg pocket calculator calculations. The twical pocket calculator doesn'toffer aprinloul ofthe dataentered.Butwhenthe calculatedangle is displayed,you can compare it to your mertal image as a check or whether you keyed in the appropriate values.For example, suppose that you are to computethe polar form of -7 - i3. Witlout making any calculations,you should anticipatea magnitudegreaterthan 7 and an angle in the third quadrant that is more negativeihan 135' or lesspositive than 225",asillustratedin Fig.9.49. Examplesf.i5 and 9.16ilustrate the constructionand use ol phasor diagrams.We use such diagramsin subsequentchapten wheneverthey giveadditionalinsightinto the steady-state sinusoidaloperationof the circuit under irvestigation.Problem 9.76showshow a phasordiagramcan help explainthe operationof a phase-shifting circuit.

usingPhasorDiagrams to Analyzea Circuit For the circuit in Fig. 9.50,use a phasordiagmm to find tlre value of R that will cause the curent rbroughthrr fesislor. t/?.ro lagrhesourcecunenl.l,. by 45" when o = skrad/s.

.ll,. t)

,.,]o.z-n

f *nopr

Figure9.50 A Theci,cuiiforEranple 9.15.

R

Solution By Kirchhoff's curent laq the sum of the currents IR, lr, and Ic must equal the source current Ir. If we assumethat the phaseangie of the voltage V- is zero,we can draw the current phason for each of tlle components.The cur.ent phasor for the induc tor is givenby

lL=

v- 1!: i(s0n0x0.2r0 )

- v^ / -90' ,

9.12 Phasor Diagmms373 the current phasor for the capacitor is given by

u^

1!: L = j7(s000x800 106) : 4U,,1y:,

see.summingthe phaso$ makesan isoscel€striangle..orhelengrhof lhe curfenrphasorfor there(i\ tor must equal 3y-. Therefore,the value of the resistoris i o.

and the currentphasorfor the resistoris givenby

rn=

v. 1l:

t)4, = :!!!

/^.

R

Thesephasorsare shownin Fig. 9.51.The pha, sor diagramalso showsthe sourcecurrent phasor, sketchedasa dotted line,whichmust be t}le sum of the curent phason of the threecircuit components and must be at an angle that is 45' more posjtive thar the current phasorfor the resistor.As you can

r

I l

I a = V,'lR

Fiqure 9.514 Thephasordiag€m forthecurr€nts in tig.9.50.

UsingPhasorDiagrams to AnalyzeCapacitive LoadingEffects The circuit in Fig. 9.52has a load consistingof the parallel combhation of the rcsistor and inductor. Use phasor diagramsto explore the effect of adding a capacitor acrosstbe terminals of the load on the amplitude of Y if w€ adjust Vr so that the amplitude of Vr remains constant. Utility companies use this teclmique to control the voltage drop on their lines.

I

v, n' J r,,1i,r,

L

Figure9.53 d, Thefrequ€ncy-domain equivatent ofthecncuit in Fig.9.52. Ll

.

I ),"

]

-

+

".

R.J

L,

Figure9.52 A Th€circuitfor Examph 9.16.

Solution We begin by assumingzero capacitanceacrossthe load.AJter constructinglie phasordiagramfor the zero-capacitance case,wecanadd the capacitorand study its eflect on the amplitude of Y, holding the amplitudeof VL constant.Figure9.53showsthe fre, quency-dom"in equi\alenro[ rhe circul shownin Fig.9.52.Weaddedthe phasorbranchcurrentsI,Ia, and Ib to Fig.9.53to aid discussion.

Figure9.54showsthe stepwiseevolutionof the phasordiagram.Keepin mind that we are not interestedin specificphasorvaluesand positionsin this example,but ratherin the generaleffectof addinga capaciloracrossthe terminaisof the load.Thus,we want to developthe relative positionsof the phasorsbefore and after the capacitorhasbeenadded. Relaling the phasor diagram to the circuit shownin Fi9.9.53revealsthe following poinls: a) Becausewe arc holding the amplitude of the load voltageconstant,wechooseV! asour referencc-For convenience, we place this phasor on the posilivereal a-\is. b) We know that I" is in phasewith Vr and that ils magnitudeis Vrlnz. (On the phasor diagam,

374

sinLrsoidatSt€ady-Stat€Aratysjs

R,I\

\I

(5)

phasor diagnm shown ir Fig. 9.57 depicts tlese obse ations.Thedotted phasomrepresentthe pertinent currents and vollages before tlle addition of the capacitor. Thus,compadngthe dotted phason of I, RrI, j.darl, and V" witl tlren sold counteryarts clearly showsthe effect of adding C to the circuit. In particular, note that this reduces t]te amplitude of the source volrage and still maintains tle amplitude of the load voltage. Practically, this rcsuli means that, as the Ioad inffeases(i.e.,as Ia and Ib increase),we can addcapacitorsto the system(i.e.,inqeaseL) so tlat under heavy load conditions we cao maintain VL withoul ircreasing the amplitude of th€ source voltage.

Fiqure9.54A Thestep-by-step evoLuiion ofthephasor jn Fiq.9.53. djagram torthecircuit the magnitude scale foi t]le cunent phason is independent of the magnitude scalefor the voltagephasois.) c) We know that Ib lags behind VL by 90" and that its magdtude is IVL/o42. d) The line curent I is equal to the sum of Ia and Ib. e) The voltage drcp across -R1is in phase with the line curent, and the voltage drop across ioal leadsthe line currentby 90". f) The source vollage is the sum of tte load voltage and the drop along the Line; that jsr Vs : Vt

Flgure9.55,| Theaddition 0f a capacitorto the circujtshown in Fi9.9.53.

+ {& + jdL)r.

Nore rhatrhecompleledphasordiagfamshownin step 6 of Fig. 9.54 clearly shows the amplitude and phase angle relationships among all the currerts and voltages in Fig. 9.53. Now add tbe capacitor branch sho\rn in Fig. 9.55.We are holding VL constant,so we conslrucl the pbasordiagfamtof lhe circuil in fig. r).55 followingthe samestepsastlose in Fig.9.54,except that, in step 4, we add the capacitor currert I to the diagram. In so doing, I. leads VL by 90', witl its magnitude being lVL(,Cl. Figure 9.56 shows tlle elfect of Ic on the line clmenr Both the magnitude and phase angle of the litre cunent I change with changesir the magnitude of lc. As I changes,so do the magnitude and phase angle of the voltage drop along the line. As the drop along the line changes, the magnitude and phase angle of Vs change.The

Figure9.55 A The€ff€ctof the capacitor cunentId onthe tjne

I Figure9.57 Theeffectofaddinq a load-shuntjng capacitorto jn Pig.9.53ifV, is hetdconsiant. thecircuitshown

NOTE: Assessyow ndentanding of this material by ttying Chaptet Ptoblens 9.81 and 9.82.

sunrmary375

Practical Perspective A Househotd Distribution Circuit Letusreturnto the househotd djstribution circuitinhoduced at the beginningof thechapter. Wewittmodjfu the circuitsLightty byaddingresistance to eachconductor onthesecondary sideofthehansformerto simutate more accuratety the residentiat wiringconductors. Themodified cicujt is shown in Fig.9,58.In Probtem 9.85youwitl catcuLate thesix bEnchcurrents on the secondary sideof the distributjon transforner andthenshowhowto , dlcuLate in theprimary thecurrcnt winding. yourunderstanding NoTE:Assess ofthis PrddicolPerspective byttyingChdpter Prablems 9.85and9.86. 1()

-r1 12O/X

_ -v, - 2{ , r- } 1201!Y

-v

1r)

| | {20 r)

L

roo

{11

l:{o

t6

+T.

Figure9.58,{ Distribution cjrcuit.

5ummary . The general equation for a sinusoidsl source ls x = U^cos(at + d) (voltag€source),

i = I -cos(dt + 4) (currentsource), where y- (or 1-) is the maximum amplitude,o is the tequency,and 4 is tie phaseangle.(Seepage332.) . The frequency,(o,of a shusoidal responseis the sameas the frequency of the sinusoidal soorcedrivirg the circuit. The amplitude and phaseangle of the respome are usually differentfrom thoseof the source.(Seepage335.) . The best way to find the steady-statevoltages and currents ir a circuit d{iven by sinusoidal sourc€sis to perform the analysisin the frequency domain.The following

mathematical transfoms allow us to move between tie time and frequency domains. . The phasor transform (from the time domain to the frequency domain): Y = V-eto = g{Yu cos(d, + d)}. . lhe inve$e phasor transform (ftom the frequency domainto the time domain):

s) {u^eto= ft {v^daet""\. (Seepages 337-338.) Wlcn working with sinusoidally varying signals, rememberthat voitageleadscurrent by 90' at the terminalsof an inducloi, and curent leadsvoltageby 90' at the terminalsof a capacitor.(Seepages342-345.)

376

SinusoidatSteady-StateAnatysis

hnpedance(Z) plays the same role in the frequency domair as resistance, induclance,and capacitanceplay in the time domain. Specifically,the relationship betweenphasor curent and phasor voltagc for resistors,inductorEand capacitorsis

v:zt, where the referencedirectior for I obeys the passive sign convention. The reciprocal of impedance is admittance(y), so anotier rvayto er?ressthe cullentvoltagerelationshipfor resistors,inductors,and capacitors in the frequencydomainis

V = UY, (Seepages 345and350.) All of the circuit analysis techniquesdeveloped in Chapters2-4 for resistivecircuits also apply to sinusoidal steady-statecircuits in the ftequency domain. ThesetechniquesincludeKVL, KCL, se.ies.and parallel combinationsof impedances,voltage and current division, node voltage and mesh curent methods, source transformations and Th6venin and Norton

The two-winding linear traNfonner is a coupling device made up of two coilswound on the samenonmagnetic core.R€ll€ctedimpedanceis the impedanceof the sec, ondarycircuit asseenfrom the terminalsofthe prjmary circuit or viceversa.Thereflectedimpedanceof a linear tansformer scenlrom the primary sideis the conjugate ofthe self impedanceof the secordarycircuit scal€dby thelactor ('rMl Z22l)2.(.Seepages361and363.) Thc two winding id€al transforn€r is a linear transformer with the fo owing special properties:perfect coupling (k : 1), infirite self-inductancein each coil (Lt = L2 = .r.), and losslesscoils (R1 : R, = 0). The circuit behavior is govemedby the tuns ratio d = Nr,/Nl. In particular,the volts per turn is the same fot each winding,or

and the anpere tums are the samelor eacbwinding,or NrIl = + NrIr. (Seepase36s and 366.)

TABLE 9,3

Impedan(eand ReLated Valu€s Inpedanre (Z)

j(

t/@c)

Adnittance (Y)

Rer(tance

tl@c

of

j(

I/aL)

-1/aL

Froblems Section9.1 9.1 A sinusoidalvollageis givenby the exprcssion lj = 100cos(240rr + 45") mV. Find (a) / in hertz; (b) r in mitliseconds;(c) y-: (d) o(0);(e) din degreesandradiansr(f) the smaltcst positivevalueof/ at whicht) : 0: and (g) the small est positive value of r at which .lz,/dl = 09.2 Ina singlegraph,sketchc : 60cos(or + d) versus ot for d = -60', 30",0",30',and60'. a) Statewhetherthe voltagefunction is shifting !o the right or left asd becomesmore positive. b) What is the direclior of shift if d changesfrom 0ro 30'l

9.3 Att :

250/6 ps. a sinusoidalvoltageis known to be zero and goingpositive.The voltageis next zero aLt = 1250/6/.s. Il is also known that lhe voltage is75Vatl = 0. a) What is the frequcncyot Din hefiz? b) What is the expressionfor ??

9-4 A sinusoidalcu ert is zero at l:150!,s

and rncrea.ing dt x rrte or 2 l0az { q. fhc ma\imum amplitudeof the voltageis 10A. a) w}Iatis the ftequencyof ? in radianspersecond? b) What is the expressionfor u?

probtems377 Considerthe shusoidalvoltage o(r) = :170cos(1202t

60') V.

a) mat is the maximum amplitude of the voltage? b) What is the frequency i]) heitz? c) Wlat is the tiequency in mdians per second? d) mat is t}le phase angle in mdians? e) What is the phaseanglein degees? f) What is the period h miliseconds? g) What is the fi^t time after l : 0 that zr : 170V? h) The sinusoidal tunction is shifted 125/18 ms to the right along the time axis.What is the expression for 0(t)? i) What is the minimum number of milliseconds that the functioD must be shifted to the dghl if the expressionfor t'(t) is 170sin 1202t V? j) What is the mhimum number o{ milliseconds tlat the functiotr must be shifted to the left if the expressionfor o(l) is 170cosl2htt y? 9.6 Showthat th+T

/

Jr'

c) Find the nume cal value of i after the switch has been closed for 1.875rns. d) Wlat arc the maximum amplitude, frequercy (in radians per second), and phase angle of the steady-state current? By how many degreesare the voltage and the steady-state cment out of phase?

o

9.10 a) Ve Jy that Eq.9.9 is the solutionofEq.g.s.This canbe done by substitutingEq.9.9 into the lefthard side of Eq. 9.8 and tlen noting that it equalsthe dght-hand side for all values of I > 0. At l - 0, Eq. 9.9 should reduce to the initial value of the current, b) Because the tiansient component vanishes as time elapsesand becauseour solution must satisly the differential equation for all values of t, the steady-statecomponent, by itself, must also satisfy the differential equation. Verify this observation by showing that the steady-state componentof Eq.9.9 satisfiesEq.9.8.

v2 T

v'. cosz(dr+ 6S,tt= :ir L

The Ims value of the sinusoidal voltage supplied to the convenience outlet of a US. home is 120 V What is the maximum value of the voltage at the outlet? 9.8 Find the rms value of the half-wave rectified sinu soidalvoltageshown. Figu|e P9.8

r=v^sin+t,0
Sectiotr9.2 9.9 The voltageappliedto the circuit shownin Fig.9.5 atl = 0is 100cos(400t + 60') V. The circuitrcsistance is 40 O and the initial current in the 75 mH irductor is zero. a) Find t(r) for I > 0. b) Write the expressions for the transient and steady'statecomponentsof i(t).

Sections9.3-9.4 9.11 Use the conceptof the phasorto combinetlle following sinusoidalfunctionsinto a singl€ trigono metricexpression: a) -y: 100cos(3001 + 45') + 500cos(300r 60'), : b) y 250cos(3771 + 30") 150sin(377,+ 140"), y c) 60cos(100t+ 60') - 120sin(100r- 125") + 100cos(100t+ 90"),and d) ) : 100cos((,,t+ 40') + 100cos(ot+ 160") + 100cos(o,- 80").

9.12 A 50 Hz sinusoidal voltage with a maximurh amplitude ot 340 V at t : 0 is applied acrcss the teminals of an inductor. The maximum amplitude of the steady-statecurrent in the inductor is 8.5 A. a) wlat is the frequency of the inductor culrent? b) If the phaseangleof the voltageis zero,what is the phaseangleofthe current? c) w}lat is t}le inductive reactanceof the inductor? d) W}Iat is the inductance of the inductor in millihenrys? e) What is the impedance of the inductor?

378

SjnusojdalSteady'StateAnatysis

9.13 A zl0kHz sinusoidalvoltagehas zero phaseangle and a maximum amplitude of 2.5 mV When this voltageis applied acrossthe teminals of a capacitor, the resultingsteadystate cunent has a maxrmum amplitudeof 125.67/..A. a) What is the hequency of the currcnt in radiafls per second? b) What is the phaseangleof the current? c) What is the capacitivercactanceof the capacitor? d) What is the capacitanceof the capacitor in microfarads? e) W}lat is tie impedance of the capacitor?

9.14 The expressionsfor the steady-statevoltage and

9.16 A 400 O resistor, a 87.5 mH inductor, and a adc 312.5nF capacitor are conrected in series.The series-connectedelements are energized by a sinusoidal voltage sourcewhosevoltage is 500cos (8000t

+ 60")v. a) Draw the frequency domain equivalent crcurt. b) Referencethe curent in the direction of the voltageiise adoss the source,and find the pha sor current, c.) Find the steady-state er?ressionfor i(r). 9.17 a) Show thar at a given frequency o, the circuits in Fig.P9.17(a)and (b) will have the sameimpedancebetweenthe terminalsa,b if

current at the temhals of the circuit seen rn Fig.P9.14are

p-= -''

R2

1 + azR34'

z's : 150cos(800011+ 20") V,

1 + l/lR74

ts = 30sin (8000rr + 38") A a) Wlat is the impedanceseenby the source? b) By how many micmsecondsis the currert out of phasewith the voltage?

.'4c, b) Find fte values of resistance and capacitance thal whenconnectedirseries will havethe same impedance at 80 Lrad/. as rharot a 5n0O re(i. tor connectedin parallelwith a 25 nF capacitor.

tigureP9.14 Figure P9.l 7

^ ,1" 1

-l "l G) Sections9.5 and 9.6 9.15 A 20 O rcsistorand a 1 pF capacitorare connected En( in parallel.This parallelcombinationis also in par allel with the sedescombinationof a 1 O resistor and a 40],.Hinductoi.Thesetlueeparallelbranches are diven by a sinusoidalcurrent source whose - 20') A. currentis 20 cos(50,0001 a) Draw t}le frequency-domain equivalent circuit. b) Reference the voltage acrossthe currenr source as a rise in the direction of the sourcecurrent, and find the phasorvoltage. c) Find the steadystateexpressiorfor ir(r).

9.18 a) Show that at a given frequency rr, the circuits in Fig 9.17(a)and (b) will have the sameimped, ancebetweer the terminalsa,b if p^:

"

-

'

f + d'R?C?

.rn,cl cl

r + d,Rici

(Hirt: The two circuits will have the same impedanceif they havethe sameadmittance.)

Problens379 Find the valuesof resistanceand capacitancethat when connected in paralel will give the same impedanceat 20 krad/s as that of a 2 kO rcsistor connecred in qerierrilh a capdcirance ot50 r 9:19 a) Showthat, at a givenftequency{r, the circuitsln Fig.P9.19(a)and (b) will have the sameimped ancebetweenthe terminalsa,b if

9.22 a) For the circuit shown in Fig. P9.22,find the frequency (in radians per second) at which tlle impedanceZ.b is purely resistive. b) Find the valueof Z.b at the frequencyof (a).

FigureP9.22

.,LiR2

ff + &rl'

R) + -' L!'

Find the valuesof resistanceandinductancethat when connecled in series will have the sane impedance al 20 krad/sas rharol a 50 kO resistor connectedirl parallelwith a 2-5H inductor.

^+" "?.

9.23 Find the impedance Z,b in the circuit seen in

FigureP9.19

Fig.P9.23.ExpressZ"b in both polar and rectangular form.

' \

figureP9.23

-110o

(a)

10f,)

j10o

9.20 a) Show that at a given frequency d, the circuits in Fig.P9.19(a)and (b) will have the sameimpedancebetweenthe terminalsa,b if

R1+ a'Ll

_

j20a

R +.?L1

(alirt The two circuits will have the same impedance if lheyha!e rhesameadmilrance.) b) Find the valuesof resistanceand inductancetnar when connectedin parallel will have the same impedanceat 10 krad/s as a 5 kO resistorconnectedin serieswith a 500mH inductor.

9.21 Three brancheshaving impedancesof 4

j3 O, rc+ jnA, and -i100 O, rcspectivelyj are connected in parallel. What are the equivalent (a) admittance,(b) corductance,and (c) susceptance of the pamllel connection in millisiemens? (d) lf theparallelbranche.are e\citedtroma sinusoidalcurent sourcewhere i = 50 cos.rt A, what is the maximum amplitude of the current in the purely capacitivebrarch?

9.24 Find the admittance yab in the circuit seen in Fig.P9.24.Er?ress yd, in both polar and rectangular form. Give the value of yabin millisiemens.

Pigure P9.24 y'2.8O

/10O

13.60

380

SinusoidatSteady-StateAnatysjs

9.25 Find the steady-stateexpiession for io(t in the cirBla cuit in Fig. P9.25if o" : 750 cos50001mV. FiglreP9.25

0.4/,F

9.29 The phasor current L in tlle circuit shown in sPrc Fig. P9.29is 40 A rllA. a) Find Ib, \, and Vs. b) Ifra = 800 rad/s, wdte the expressions for ib(l), t"(r), and ?,s0). Figur€ P9.29 25'|"

trzoo r{l

r

l40 + t80mA

9.26 The circuit sholin in Fig. P9.26 is operating m the sinusoidalsteadystate.Find the valueof., if L = 100sin(ot + 81.87")mA, t's = 50 cos(ot - 45") V FiglreP9.26 4000

930 a) For the circuit shownin Flg.P9.30,find the steadystate exprcssionfor oo iJ is : 5 cos (8 x 105t)A. b) By how many nanosecondsdoes 2l. lag is? tjqureP9.30

40mH

12()

9.n Find the steady-stateexpressionfor o, in the circuit ofFi9 P9.27it iR : 200cos5000tmA.

9.31 Thecircuitir Fig.P9.31is opelatingin tie sinusoidal 6*c steadystate.Fird rr,0) iJ i"(, : 15cos8000rn . Figore P9.31

rbrreP9.27

5ko

+

240A7u,

r ) I 2.5pF

800 48rnH

9.32 Find Ib and Z in the circuit shown in Fig. P9.32 iJ

928 The circuit h Fig. P9.28is openting ir tle sinu-

vs : 60av andr, : 5l:2q:A.

Eilc .oidalsleady(late.Findlbeslead)-srare e\pression tigureP9.32 for od(t) if ?Js: 64 cos 8000t V. Figure P9,28 31.25nF

j5o+

r,J -18o

Probtems 381 9.33 End the value of Z in the circuit seenin Fig. P9.33if

vs = 100- i50v,Is:20 + i30A,andvr = 40 + i30v.

Flgure P9.35

figureP9.33

10ko

j5o

F -i10o 9,37 The frequency of the sinusoidal current souce in E,IG the circuit in Fie. P9.37 is adjusted until o. is in phasewith ls. a) Wlat is the value of o in radians per second? b) If i! = 2.5 cosl.)t rnA (where o is tle frequency found in [a]). whal is lhe slead]-stareer?ression lor 1).'l

9.34 Find Z,b for tlle circuit shown in Fig P9.34. Flgule P9.34

Figure P9.37

9,35 The frequency of the sinusoidal voltage source in 6plft the circuit in Fig. P9.35is adjusted until ttre current i, in phasewith I's. a) Find the frequency ir hertz. b) Find tne steady-state expression for i. (at the frequency found in [a]) if rs - 10 cos(,t V. FlglreP9.35 1500

10() 4mF

2H

9.38 The circuit shom in Fig. P9.38 is operating in the sirusoidal steady state. The capacitor is adjusted until tlte current is is in phase $rith the sinusoidal voltage I's. a) Specify the capacitarce in miffofarads if ?Js: 250 cos 1000t V. b) Give t}le steady-state expression for is when C hasthe valuefound in (a).

Figore P9.38

9.36 a) The ftequency of the sourcevoltage in the circuit in Fig. P9.36 is adjusted until is is in phase with r,s.W}Iat is the value of o in radians per second? b) If o, = 45"o".t V (where (l) is the frequency foundin [al).whatis thestead]-srate expression

382

Sinusoida l Stea dy-state Anatysis

9.39 a) The sourcevoltagein the circuit in Fig.P9.39is os : 96 cos10,00OtV. Find thevaluesofl, such that is is in phase with os when the circuit is opemting in the steady state. b) For the valuesof a found in (a), find rhe sreadystate er?ressions foi is. Figure P9.39 62.5nF

9.43 Find the Th6venin equivalent circuit with respect to the terminalsa,b for the circuit shownin Fig.P9.43. figureP9,43 j12 tL 12f,l

Section 9,7 9.40 Use souce transfomations to lind ttre Thdvenin equivalentcircuit with respectto the terminalsa,b for.the circuitshownin Fig.P9.40.

DA

Figure P9.40 /40 ()

9.44 The device h Fig. P9.44 is represented it the fre-

9.41 Use souice tmnsformationsto find the Norton equivalent circuit with respect 10 the terminals a,b lbr the circuitshownin Fig.P9.41.

quency domain by a Norton equivalent. When an inductor having an impedance of j100 O is connectedacross the device, the value of Vo is 1007_l![ mV. W]en a capacitor havhg an impedance of j100 O is cornected acrossthe device,the value of I0 is -31?lq mA. Find the Norton current IN and the Norton impedance ZN. rigureP9.44

Figure P9.41 J30O 16[A

9.42 The sinusoidal voltage source in the c cuit in Fig. P9.42 is developirg a voltage equal to 22.36cos(5000r+ 26.565')v. a) Find the Th6veninvoltage with respecrto rhe terminalsa,b. b) Fhd the Th6venin impedancewith respectro the terminalsa,b. c) Draw the Th6veninequivalent.

9.45 Find Zabin the circuit shownin Fig.P9.45whenthe circuitis operating at a frequency of 1.6Mracvs. FiguleP9.45

15ra

Probtems 383 946 Find the Th6venin impedance seen looking into the telmhals a,b of the circuit in Fig. P9.46 if the frcquency of operation is 25 krad/s. Figrrc P9.46 5

t 0 _

".'l

1ko

9.50 The circuit shown in Fig. P9.50is operating at a frcquency of 10 krad/s. Assume a is rcal and lies between 50 and + 50,that is, 50 < a < 50. a) Find the value of a so that the Th6venin impedance looking into the terminals a,b is purely

_

19i,\ a nr

What is tlle value oftheTh6ve n impedance for the a found in (a)? c) Can a be ddjusredso lhal lhe fierenjn impedanceequals5 + /5 O? If so,what is the d) For what values of a vrill the Th6venin imped-

ancebe inductive?

9.47 Find the Norton equivalentwith respectto terminalsa,bin the circuit of Fig.P9.47.

Figure P9.50

FigueP9.47

Section 9.8 9.51 Use the node-voltage method to find Va in the circuit in Fig.P9.51. 9,48 Find the Thdvenin equivalent circuit with respect to t}le terminals a,b of the circuit shown in Fig. P9.48.

Figure P9.5r

j40a

rigureP9.48 600O j150O

1004 v

+

60{)

40r)

v,

j20a

9.52 Use the node-voltagemethod to find the steadyAflft stateexpression for o,(t) in the circuit in Fig.P9.52if 9.49 Find the Norton equivalentcircuit with respectto the terminalsa,bfor the cLcuit shownin Fig.P9.49 whenVr : 25 A V. Figure P9.49

?Jsr= 10cos(5000t+ 53.13')V, 0s2= 8 sin50001V. figuru P9.52

./250O

384

Sinusoidat Steady-state Anatysjs

9.53 Use the node voltage method to fhd t}Ie steady-state expressionsfor t}le branch curents ia and ib in tle circuit seen in Fig. P9.53if za : 100sin10,0001V axd ?b : 500cos10,000t V.

Sectior 9.9 9.57 Use the mesh-curent method to find the st€ady slateexpressionfori.(r) in the cncuit in Fig.P9.57if t)x = 60 cos40,0001V,

Figure P9.53 tb = 90sin(40,000r + 180')v.

5 r.F

Figure P9,57 1.25tlF

20tl

9.54 Use the node-voltagemethod to find the phasor voltageVs in the circuit shownin Fig.P9.54.

9,58 Use the mesh-currentmethod to find the sleadystateer?ressionfor oo(r)in the circuit in Fig.P9.52.

tigureP9,54

9,59 Use the mesh-currentmethod to find the phasor curent Is in the circuit in Fig.P9.54.

/3()

9.60 Use the mesh-currenrmethod to find the bmnch currenlsla. Ib. lc. and ld in lhe circuil sho$n in Fig.P9.60.

510"A

Figure P9,60

5-9!I v

2CA

9.55 Use the rode-voltagemethod to find V, and I, in the circuitseenin Fig.P9.55.

r.,i50

FiguEP9.55

5o i:o

100r1ry v

500 v

j50o

9.56 Use the node-voltage method to find tle phasor voltage V. in the circuit showrl in Fig. P9.56.Express the voltagein both polar and rectangularform_ figureP9,55

lr

9.61 Use the mesh-currentmethod to find the steadysrft state expressionfbr o, in the cfucuit seen in Fig.P9.61.ifrr equals72cos5000rV. FigrrcP9.61 500nF

15-CA

Secrions9.5-9.9 9.62 Use the concept of voltage division to find the 6nc steady-state expression for ?.(1) in the circuit in Fie.P9.62it ts = 75cos50001V. tigureP9.62 300O 400mH

9.63 Use the €oncept of curent divisior to find the ErI( steady-stateer?ression for ,, in tlre circuit in Fig.P9.63if is - 125cos500t

9.67 The op amp in the circuit seenir Fig. P9.67is ideal. tsdc Find the steadystate expressior for o,(t) when ?Jr: 20 cos10bl V Figure P9.57

Figure P9.63

40 ko

250c)

50r} 20 pF

1H r0k0 |

r-\?6 v

9.64 The sinusoidal voltage source in the circuit shown tsPld in Fig. P9.64 is generaring tie volrage os = 1.2cos1001V.If the op amp is ideal,what is the steady-state expressionfor o.(l)? Fig(reP9.54

200ko

9.65 The 1 /,F capacitor in the c cuit seenin Fig. P9.64is 6PIG replaced witl a variabl€ capacitor. The capacitor is adjusteduntil the output voltage leads the input voltageby 120". a) Find the values of C in microfarads. b) Write the steady-state er?ressionfor uo(r)when Chas the valuefound in (a)9.66 The op amp in the circuit in Fig.P9.66is ideal. """ a) Find the steady-state expressionfor o,trl. b) How largecanthe amplitudeofos be beforethe amplifiersaturates?

9.68 The opeiational amplfier in the circuit shown in sPI.r Fig. P9.68is ideal. The voltage of the ideal sinu\oidalsuurce is ,s = I0co.2 l0sr \ . a) How small can C, be before the steadystate output voltage no longer has a purc sinusoidal wavelbrm? b) For the value of C, found in (a), write the steady-state expressionfor r),. Figurc P9.58 12.5nF

386

SinusoidatSteady-stat€Anatysi,

9.69 a) Fhd tle hput impedance Zab for the circuit in Fig. P9.69.Express Zab as a function of Z and I<

wheref : (R,/nr.

b) If Z is a pure capacitive elemenr, what is the capacitanceseen looking into the teminals a,b? tigureP9.69

Seclion 9,10 9.72 a) Flnd the steady-state expressions for the currents is and iL in the circuit in Fig. P9.72 when ?s : 200 cos 10,000tV. b) Find the coefficient of coupling. c) Find t}le energy srored ir the magnetically coupled coils at, = 50zps and t : 1002,.s. F,gweP9.72 5()

0.5mH 1nH

9.70 For the circuit in Fig.P9.57,suppose D, = 5 cos80,000rV ,,, = -2.5 cos320,000t V.

15r}

lmH

9.73 The sinusoidal voltage source in the circuit seen in 6prft Fig. P9.?3 is operating at a frequency of 50 kmd/s. The coefficient of couplhg is adjusted until the peak amplitude of ir is maximum. a) What is the valueof ,t? b) Wlat is the peak amplitude of i1 if .'s : 369cos(5 x 104r)v ? Figur€ P9.73

a) What circuit a.nalysis techniquemustbe usedto fhd the steady-state expressionfoi i.(t)? b) Findthe steady'state expression for t,(r)?

20o"

75[)

1000

3000

5 rnH

9.fl For the circuit in Fig.P9.71suppose rr1= 20cos(2000r 36.87")V + 16.26)V r', = 10cos(5000r a) Wlat circuit analysistechniquemustbe usedto find thesteady-state expression for 2,(1)? b) Findthesteady-state expression for o.(t). Figure P9.71 100pF

9.74 For t}le circuit in Fig. P9.74, find the Th6venin equivalent with respect to t}le terminals cd. figuleP9.74

15r} 225U + v Gns)

jsoo

300

j20o

9.75 The value of I itr the c cuit in Fig. P9.75is adjusted so that Z,b is purely resistive when zo- 25 krad/s. F]]trdZ^b.

PobLers 387

40()

b) Show that if the polatity terminal of either one of the coils is reveNed that -

Z

L /

N \,

t1--:t 9.?6 A series combhation of a 150 O resistor and a 20 nF capacitor is comected to a sinusoidal voltage sourceby a Uneartransformer.Tbesourceis oper ating at a iiequercy of 500 krad/s.At this frequency, the intemal impedanceof the sourceis 5 + i16 O. The rms voltage at the terminals of the souce is 125V when it is not loaded. The parameters of the Lhear transformer are n1 : 12 O, 11 : 80 pH, R2 : 50 O, L2 : 500 pH, and M : 100 PH a) Wlat is the value of the impedance rellected into the primary? b) What is the value of the impedance seen ftom the terminals of the pmctical source?

Section9.11 9.7? Find ttre impedance Zabin the circuit in Fig. P9.77if

zL=2

tigureP9.79

9.80 a) Show that the impedance seen lookhg into t]le termimls a,b in th€ ci(cuit in Fig. P9.80is given by the expression

+ i1'50 A '^, = ('

FigureP9.77

.t)'"

Show that iI the polarity terminals of either one of the coilsis revened,

9.78 At first glance,it may appear trom Eq. 9.69 that an inductive load could make the reactanceseenlooking irto the pdmary terminals (i.e., Xab) look capacitive. Intuitively, we know tlis is impossible. Show that -l.ab can never be negative if ,YL is an inductive rcactance. 9.79 a) Show tiat the impedance seen looking into the terminals a,b in the circuit in Fig. P9.79is given by the expression

(.,

r^": ('-ft)'"'.

388

sinusoidatSteady-StareAnatysjs

Section9.12

Sections9.1-9.12

9.81 The pammeters in the circuit showl in Fig. 9.53 are Rr = 0.1 O,arl1 = 0.8 A. R2 : 24 A,oL2 : 32 A. andvl=240+J0V. a) Calculatethe phasorvoltagey. b) Connect a capacitor in parallel with rhe inducror, hold VL constant, and adjust the capacitor until the magnitude of I is a minimum. What is the capacitive reactance?What is the value of V,? c) Fird the value of the capacitivereactancethat keeps the magnitudeof I as small as possible and thal at the sametime makes

lYl : Yr =240v 9.82 Show by using a phasordiagramwhat happensro lhe magnirude andphasedngleo[ the vollage.,, in the circuit in Fig.P9.82as R, is vaded from ze{o to infinity. The amplitude and phase angle of the sourcevoltageare held constantasRr varies, Figure P9.82

9,83 a) For the circuit shown in Fig. p9.83,compurey and V. b) Constructa phasor diagram showingthe relationshipbetweenV", V, and the load voltageof

4401!: v.

c) Repeat parts (a) and (b), given rhar rhe load voltageremainsconstantat 440 A V. when a capaqtrve reactance of -22 O is connected acrossthe load terminals, FiguEP9.83

+ 0.2rl

-

_

l

j1.6O + 14011!: V

j22n j22A-;:

9.84 You may have the opportunity as an engineering gracluateto sene as an expe witnessin lawsuits involving either personalinjuy or property damage.As an example of the type of problem on which,ou may be askedlo gj!e an opinion.consider the following event.At the end of a day of fieldwork,a famer returnsto his famstead,checks his hog confinemert building,and finds to his dismay that the hogsare dead.Theproblem is rraced to a blown fuse that causeda 240V lan motor to stop. The loss of vertilation led to the suffocation of the livestock.The interupted fuse is locatedin the main switchthat connectsthe farmsteadto the electrical service. Before the insurance company settlesthe claim,it wantsto know if the electriccircuit suppling the farmsread functioned properly. The lawyersfoi the insurancecompanyatepuzzled becausethe farmer'swife,who wasin the houseon the day of the accidentconvalescingfiom minor surgery,was able to watch TV dudng the aftei noon. Furthermore,when she went to the kitchen to start ptepaing the evening meal, the electric clock indicatedthe conect time. The lawyershave hired you to explain (1) why rhe etectricctock in the kitchen and the televisionsetin the living room continued to operate after the fuse in the main switch blew and (2) why rhe second fuse in rhe main switchdidn't blow after the fan motor stalled. After ascertainingthe loads on the tlnee-wire distribution circuit prior to the interruption of fuseA, you ate able to constructthe circuit model shown m Fig. P9.84on page 389.The impedancesof rhe line conductors and the neutral conductot are assumednegligible. a) Calculatethe branchcurents 11,I2j13!14,15, and 16priorto rbe inlerruplionot fuseA. b) Calculate the bmnch currents after |nc ulerruption of fuse A. Assume the stalled fan motor behavesas a shoit circuit. c) Explain why the clock and televisionset were not affectedby the momentaryshort circuit that intefupted fuseA. d) Assume the fan motor is equipped with a rhermal cutout designedto interrupt the motor circuil if lhe molor currenrbecomeserce.si\e. Would you er?ect the thermal cutout to operate?Explain. e) Explain why fuseB is not interrupredwhen the fan motor stalls.

ProbL€ms 389 FisunP9.84 FuseA (100A) lro-irta.y 1201U'

circuil nrernPrs fusc A

| 1,-

X I t2 ; :-

24l)

t.l 15A 1 2{ }

r20[t

brunchcarryingI, is modelingthe neutral conductor. Our purposein analyzingthe circuit is to show the importance of the neutral conductor in the satisfactory operation of the circuit. You are to choose the method lor analyzing the circuit. a) Showthat I, is zero if -R1:.R2.

15A

b) Showthat V : Vz if R' = Pr. c) Open the neutral branch and calculate Y and V2 if R1 : 60 o, R2 = 600o, and n.r = 10 o. d) Closethe neutralbranchand repeat(c). e) On the basis of your calculations, explain why the neutral conductoris never fused in such a manner that it could open while the hot conducrors are energveo.

o 16.r0 F;;;"t..

FuseB (100A)

935 a) Calculatethe branchcurrentsll 16inthecircuit

J,liil'jfr.. in Fig.P9.s8. b) Fhd tie primary curent Ip.

tigureP9.87 _IL

9.86 Supposethe 40 O resistancein the distributioncirRndL cuit in Fig.P9.58is replacedby a 20 O resistAnce

. +

a) Recalculate the branch curent in the 2 O resistor,12. 14tffkv b) Recalculate lhe pfima4 cufrenr.I! c) On the basis of your answers,is it desirable to have the rcsistanceof the two 120 V loads be equal? 9.87 A residentialwidng circuitis shownin Fig.P9.87.In ,lH;lLliEthis model, the resistor R3 is used to model a 240V appliance (such as an electric range), and the resistors Rr and R2 are usedto model 120V appliances (such as a lamp, toaster,and iron). The branches carrying 11 and I, are modeling what electricians refer to as the hot conductors in the circuit, and the

0.02O 0.020

j0.02o" t0.02tt

rzsffv

v,Jn, o 10.03

' + o.o3 o

:f;'

l2stffv _

O 0.02

-1

Rr

v.€R, jDUA +tz

9.88 a) Find the pdmary current Ip foi (c) and (d) in Problem9.87. b) Do your answersmake sensein termsofknown circuitbehavior?

SinusoidaL Steady-State Power CaLculations Powerengineeringhas evolvedinto oneof the importantsub10.1 Instantaneous Powerp. J92 10.2 Averageand R€activePowerp. J94 10.3 Therms Vatueand Pow€r p. 398 CaLculations 1a.4 Compter ?owet p. 401 10.5 Power(aLculationsp. 40J 10.6 l4arimumPowerTransf€rp. 410

Understand the fotLowinq ac powerconcepts, thejr rctationships lo oneanother,aid howto catcuLate themin a circuit: . Instant€feous powet . Averag-a (reat)power; .

power;ard Comptex

Und€rtardthe conditioifor maximum reaL powerd€tjvered to a toadjn an ac circuitandbe abteto calcutate the toadimpedance reqLired to d€Ljver maximum reatpowerto the load. Beabtelo catcutate att foms ofac powerin accircuitswirh tireartransfornrers andii accircuitswith ideattnnsformers.

390

disciplineswithin electricalengineering.The range of problcms dealingwith the delivery of energyto do work is considerable, from determining the power rating within which an appliance operatessafclyand efficiently.to designingthe vast array of gcn crators,tlansfbrme$, and wires that provide electric energy to householdand industrialconsumers. Nearly all electlicenergyis suppliedin the tbrm of sinusoidal voltagesand curents. Thus, atter oul Chapter g discussionof sinusoidalcircuits,this is thc logical place to considetsinusoidal steadystatc power calculations.We are pimarily intcrcsled in the averagepower deliveredto or suppliedfron a pair of terminals as a result of sinusoidalvoltagesand currents.Other meas ures. such as reactive power, complcx power, and apparent power,will also be presentcd.The conceptof the rnrs value of a sinusoid,briefly introducedin Chapter9, is particularlypcrtincnt to power calculations. We begin and cnd this chapterwith two coDceptsthat should be very familiar to you hom previouschapters:thc basicequation for power (Section 10.1) and maximum power transler (Section10.6).In bctween,we discussthe generalprocessesfor analyzingpo\\,er,which will be familiar from your studies in Chapters1 and 4, althoughsome additional malhematicaltech niqucs are required here to deal with sillusoida],rather than dc, signals.

Perspective PracticaI App[iances Heating voLtages andcur the steady-state In Chapter9 we caLcutated In thjs sources. rentsin eLectrjc circuitsdfivenby sinusoidal we power The techiiques in such circuits. chapterweconsjder rnanyof the etect caLde\rces developareusefuIfor anatyzing sourcesare the prewe encounterdaity,becausesinusoidat doninant meansof providingeLectricpowerin our homes, schoots, andbusinesses. is heaters,which one cornmonclassof electricaldevices jnctude energy. ExampLes transforfir etectricenergyinto thermaL electric atovesand ovens,toasters,irons, etedric water dryers,andhairdryers eLectric cLothes heaters, spaceheaters, in " hedle is poie-'oI onFol rherrilicatde.iqrroncerrs js Themorepower sumption,Power impoffantfor two reasonsl

andthe moreheat a heateruses,the moreit coststo operate, it canproduce. I4anyelectricheatershavedifferentpowersettingscorre spondingto the amountof heatthe devjcesupptiesYoumay wonderjusthowthesesettjngsresuttin differentanrountsof exampleat the end of heatoutput.ThePracticatPerspechve hair dryer the designof a handheLd this chapt€rexanrines figure). with threeoperatiigsettingsGeethe accompanying provides for three different YouwjLtseehow the design powerlevets,which correspordto three djfierentL€vetsof heatoutPut.

391

392

SinusoidaL Steady State Power catcutations

10.1 Instantaneous Fower We begin our investigationof sinusoidalpower calculationswith the familiar circuitin Fig.10.1.Heie, o and i arc steady-state sinusoidalsignals. Using the passivesign convention, the power at any instant of time is p - ui. figur€10.1A Theblackboxreprcsentatjon of a circujt usedforcalcutating power.

(10.1)

This is inslatrfaneouspower. Remember that iJ the reference direction of tlle current is in the direclion of the voltage rise, Eq. 10.1must be w tter with a minus sign.Instantaneouspower is measuredin watts \qhen the voltage is ir volts and the current is in amperes.First, we write expressions lbr ?)and i: r,: Vhcos(at + 0.), i:

(10.2)

I -cos (at + 0),

(10.3)

wherc d. is the voltage phaseangle,and 0; is the currelt phaseangle. We are operating in the sinusojdal steady state,so we may choose any convenient relerence for zero time. Engineers designing systems that transfer large blocks of power have found it convenient to use a zero time coirespondingto the instantthe curent is passingthrougha positivemaximum. This reference system requires a shift of both the voltage and cur ient by pr.ThusEqs.10.2and 10.3become r-U-cos(at+0,-0),

(10.4) (10.5)

Whenwesubstitute Eqs.10.4 and10.5intoEq.10.1,rheexpression Ior rhe instantaneous powerbecomes p = V^I ^cos (at + d0

0i) cosor.

(10.6)

We could useEq. 10.6directlyto find the averagepower;however,by simply applyirg a coupleof t gonometricidentities,we canpur Eq. 10.6irto a much more inlormative form. We beginwith the t gonometricidentityr

co.ocosB

]..,,"

to e4and Eq. 10.6;lettinga : at + p!

o :T*"r,

L Secenlry 8 in Appendir F,

ei)+ W

u'

]*.,"

- B,

,i and B = al gives

cos(2at + o, - oj).

(10.7)

Powet 393 10.1 Instantaneous

Now use the trigonomeidc identity c o s ( d+ B ) :

c o s a c o s P- s i n d s i n B

to expandthe secondterm on the right-handsideofEq.10-7'which gives

e =).ns(0,-

+Lf o1)

cos @,

cos2at 01)

!t!o"in1o,- e,1"a2,,

(10.8)

Figure 10.2 depicts a representative relationship among t', ', and p, basedtn t}le assumptionsd, = 60' and d = 0" You can seethat the freouencv of the instantaneouspower is twice the frequency of the voltage or c'urrent.Ttris observation ah; follows directly from the second two terms on the right-hand side of Eq. 10 8. Therefore, the iNtantaneous power goes through two complete cyclesfor every cycle of either the voltage or ihe curt.ot. Also ttote that the instantan€ouspower may be negative for a po(ion of each cycle,even if the network between the terminals is passive in a completely passive network, negative power implies that energy storealin the inductors or capacitors is now being extracted.The facf that the instantaneous power varies with time in the sinusoidal steady_state oDeration of a circuit explains why some motor-driven appliances(such as refrigerators) experienci vibration and require resilient motor mountings vibration. to Dreventexcessive \ e a r en o wr e a d \t o u s eE q l U . 8 l o l i n d l h e a \ e r a g e p o w e r a l l h e l e r minals of ttre circuit ;epresented by Fig. 10 1 and, at the same time, introduce the concept of reactive power'

3V-r^

v^I2

0

4,

ilM

Gadiant

power, andcun€mvdsusor for vottage, figure10.2l. Instantaneous opention sinusoidal steady-state

394

Sinusoidat Steady StatePower CatcLrLations

10.2 Average andReactive Forser We beginby noting that Eq.10.8hasthree terms,which we can rewrite as follows: p:P+Pcos2ot-Qsin2at,

(reat)powerF Average

Reactivepowerts

(10.e)

r =h!cos(e,- e,:t,

(10.10)

Q:2;Lslr (o"-a).

(10.11)

P is called the averagepower, and O is called the r€active pow€r. Average power is sometimescalledreal power,becauseit descdbesthe power in a circuit thal is transformedfrom electricto nonelectricenergy.Altho gh the two terms are inierchangeable, we prinarily use the term dvenge It is easyto seewhy P is calledthe averagepower.The averagepower associatedwith sinusoidalsignalsis the averageof the instantaneous power over one period,or,in cquationlorm,

+1,.^"

(10.12)

where 7 is the period of the sinusoidalfunction.The limits on Eq. 10.12 imply that we caninitiate the integmtionprocessat any convenienttime ro but that we must terminatethe integrationexactlyone period later. (We could integrateovcr nZperiods,wherer is ar integer,providcdwe multi ply the integralby 1/rI.) We couldfind the avemgepolverby substitutingEq.10-9directlyinto Eq. 10.12and then performingthe integration.But note that the average value ofp is given by the fi6t term on the right-handside oI Eq. 10.9, becausethc integral of both cos2ol and sin2ot ovcr one pe od is zero. Thusthe avemgepower is givenin Eq. 10.10. We candevelopa better understandingofall the temsinEq.10.9 and the relationshipsamongthem by examiningthc power in circuitsthat are purely resistive.purely hductive, or purely capacitive.

Powerfor PurelyResistive Circuits If the circuit betweenthe terminalsis purely resislive,thevoltageand current are in phase,whichmeansthat d! : dr.Equalion 10.9then reducesto 0

0.005 0.01 0.015 0.02 0.025 rine G)

Figlre10.3A Instantaneous reatpowerandaverage powerfora purelyresistilecjrcujt.

p = P + Pcos2ot.

(10.13)

power cxpressedin Eq. 10.13is referred to as the The instantaneous instatrlatr€ous real power. Figure 10.3sholvsa graph of Eq. 10.13for a

Power 395 andReactive 10.? Averaqe

purely resistivecircuit,assumingo = 377 rud/s.By defini representative tion, the averagepower,P, is the averageof p over one period Thus it is easyto seejust by looking at the graph that P = I for this circuit Note from Eq. 10.13that the instantaneousreal power can never be negative, whichis alsoshownin Fig.10.3.Inother words,power cannotbeextracted from a purely resistivenetwork. Rather, all the electiic energyis dissi_ patedin the form ofthermal energy.

Powerfor PurelyInductivecircuits If the circuit betweenthe terminalsis purely inductive,the voltage and currentare out ofphaseby precisely90'.In particular,thecurrentlagsthe voltage by 90" (that is, dr: d! 90'); therefore 0u - 0t: 199' 11. power then reduccsto expressionfor the instantaneous p : -Qsir,2dt.

(10.14)

In a purely inductive circuit, the average power is zero. Therelbre no transfomation of energy ftom electdc to nonelectric form takes place The inslantaneouspower at the terminalsil a purely inductivecircuit is continually exchangedbetween the circuit and the source driving the circuit, at a ftequencyof 20. In other words,when 2 is positive,cnergy is with the inductiveelements, beingstoredh the magneticfieldsassociated and when p is negative,energy is being extracled from the magnetic fields A measureof the power associatedwith purely inductivecircuitsis the reactive power O. The narne reactivepower comesfrom the characterization of an inductor as a reactiveelement;itsimpedanceis purely reactive. Note that averagepower P and reactivepower O cary the same dimension.To distjnguishbetweenaverageand reactivepower,we usethe units na, (W) for avemge power and var (vrlt-amp ledctive, ot yAR) tor power for a represenreactivepower.Figure 10.4plots the instantaneous : 377 rad/s and O : 1 VAR lative purely inductive circuit, assumingo

1.1)

c (\ AR)

-0.5 1.0 0

0.005 0.01 0.0i5 0.02 0.025 rine G)

rcatpower,average Fiqlre10.4A Instantaneous inductive circuit powerfor a purcLy power, andreactiv€

Circuits for PurelyCapacitive Power If the circuit betweenthe terminalsis purely capacitive,the voltageard current are precisely90' out of phase.In this case,the current leadsthe voltageby 90' (that is,0' = do + 90'); thus,dN 0i : -90'. Thc er?respower then becomes sionfor tie instantaneous p =

Q sin2ar.

(10.15)

1.{J 6 E

0.5 0

-0.5 Again, the averagepower is zero,so there is no transformationof energy from electric to roflelectnc form. ln a purely capacitive circuit, the power -1.0 is conthually exchanged betweer the source driving the circuit and the a with the capacitiveelements.Figure10.5plots the electic field associated 1.5 power for a represenaativepurcly capacitive circuit, assuminstafltaneous 0 0.005 0-01 0.015 0.02 0.025 = = 1VAR. it.rgu 377 radlsandQ rine G) Note that the decisionto use the current as the referenceleadsto Q reatpowerandaveraqe beingpositivelor inductors(that is,0? di = 90" ard negativefor capac tigrre 10.5s Instantan€ous pur€ly capacitive cjrcuit. powerfor a in recognize tris difference (that Power engineets is,d,, 9'= 90'. itors

396

Srnusoidal Steady StatePower taLcutatjons

the algebraic sign of O by saying that inductors demand (or absorb) magnetizingvars,and capacitoNfumish (or deliver) magnetizingvars.We say more about tlfs convention later.

ThePowerFactor The angle d, P, plays a role in the computation of both avemge and power reactive and is referred to as the power factor angle.The cosine of this angle is called the pow€r faclor, abbreviated pf, and the sine of this angleis calledthe reactivefactor,abbreviatedrl Thus

Powerfactor>

(10.16)

rr : sin(d,

0,).

(10.17)

Knowing the value of tlle power factor does not tell you the value of t}le power factor angle,because €os(9! d,) : cos (di - d!). To completety describethis angle,we use the descriptive phraseslagging power factor and leading power f|clor. Lagging power factor implies that current lags vol! age-hence an inductive load. Leading power factor implies that current leadsvoltage hencea capacitiveload. Both t}te power factor and the ieactive factor are convenient quantities to use in descdbingelectrical loads. Examplel0.l illustratesthe interptetationofP and O on the basisof a numedcalcalcurauol

Calculating Averdge andReadivePower a) Calculatethe averagepower and the rcactive po\reral the tefminalsol lhe ner$orksho$n in Fig.10.6if

100cos(or + 15') V, I : 4sin ((,)t- 15') A. b) State whether the network inside the box is absorbingor delivedngaveragepower. c) Slale whelher rhe net$ork inside lhe bor i. absorbingor supplyingmagnetizingvars-

Figure10.6 ^ A pairo'teTinaL used_0rca.cutaring poqe..

Solution a) Because i is expressedin terms of the sine function, the filst step ir the calculation for P and O is to rewrite i as a cosine function: t - 4 cos(ot

105') A.

We now calcnlate P and C directly from Eqs.10.10and 10.11.Thus

P-

I

( = -100w. ;(100)ra)cosIls Iosij

= 173.21 vAR. 1rs- ( 105)l Jrool+;.in b) Note from Fig. 10.6the use of the passivesign convention. Because of this, the negative value of -100W meansthat the network inside the box is delivering avemge power to t]te terminals. c) The passivesignconventionmeansthat,because O is positivg t]le network inside the box is absorbingmagnetizingvaISat its terminals.

Power 397 10.2 Avenqe andReactjve

to oneanother, andhowto catruateth€min a circuit their retationships Obiective 1-Unde6tandac powerconcept!, 10.1 for (dctol lhelollo$ing.etsolvolr"gernJ currenl,calculatethe real and rcactivepower in the line betweennetworksA and B in the clrcuit shown.In eachcase,statewhetherthe powerflow is from A to B or vice versa.Also statewhethermagnetizingvars arc beingtrallsferred from A to B or vice versa.

a) 1)= 100cos(or 45")V; + 15")A. i = 2ocos(a,r

Answer: (a) P = 500 W (A 10B), B66.03VAR (B to A);

a=

( b ) P : -866.03w (B toA),

a =500vAR

( . ) P = 500w (A to B),

866.03vAR (A to B); (d)P:

500 W (B to A),

o - -866.03VAR (B ro A).

b) 0:100cos(dt 4s') t = 20cos(@t+ 165')

t:

(A to B)r

100cos(.,t - 45") V ; 20 cos(or - 105') 100cosol V; 20 cos((,r + 120")

10.2 Computothe power Iactor and the reactivefac_ tor lor the network insidethc box in Fig.10.6, whosevoitageand curent are describedin Example10.1.

Hirt:Use -i to calculatethe power and reac_

Answer pf - 0.5leading:f - -0.866. NOTE: Also tJ Chaptet Prcblen 14.2.

Ratings Appliance Avcragepower is usedto quantity lhe powcr needsof householdappliances.Theaverageporvcrrating and eslimatcdannualkilowatt hour consumptionof some comn1onappliancesarc presentedin Tablc l01.lhe cnergy consumplionvalues are obtained by estimatingthe number of hoursannuallythat the appliancesare in use.For example,a coilccmaker hasan cstimatedannualconsumptionof 140kwh and an averagcpower consumptionduring operation of 1.2 kW. Thcrcfore a cofleenaker is assumed10be in operation1,10/l-2,or 116.67.hourspcr year.or approxi narely l9 ninutes per da),. Exampie 10.2usesTablc 10.1 to determinc whether tbur common appliancescan all be in opcrationwithoul excccdingthe currenl carrying capacjtyof the household.

398

Sinusojda t Stea dy'State Power catcuLaiions

MakingPowerCalculations InvolvingHousehotd Appliances The branch circuit supplyhg the outlets in a typical home kitchen is wired with #12 conducror and is protectedby either a 20 A fuse or a 20 A circuit breaker.Assume that the following 120V appliances are in operation at the same time: a coffeemaker, egg cooker, hying pan, and toaster. Will the crcurt be interruptedby the protectivedevice?

Avenge Wattag€

Est kwh Consumed AnnurlJl

Solution From Table10.1,thetotal averagepower demanded by 0re four appliancesis P : 1200+ 516 + 1196+ 1146 = 4058W. The total cunent in the protectivedeviceis 40s8 -]]:1". _ : 12{l e 33.82A. Yes,the protectivedevicewill interrupt the circuit.

trsr.kwh Appliance

Average

Consumed

Wrttrge

Annurryr

Health and beauly

1,200 Dishwashel Egg cooker Fryins pan Mixer oven. microsave (only)

1.201 516 1,196 127 1,450 12,20t) 1,146

140 165 14 100 2 190 596 39

LrDdry Clothes dryer Washingmachi.e. automatic

103 4,219 4,811

Fan (circulaiins) Heater (portable)

860 25'7 88 7,322

860b 37'1 43 1'76

600 15 279

25 0.5

Home entertrinmenl Radio Tclevision,color.tube type Sotid-staterype

993

Comforl conditioning Dehumidifier

Sunlamp

Clock

4.856 572

Quick recovery t}?e Air onditioner (roon)

Han dryer

'71

86

240 145

528 320

2 630

1

7 46

a) Basedon nomal usage.When usingtheseligureslorprojectionn such factors as thc size ol the specilic .ppliane, the geographiml area ot use,and nrdividual usageshould be laken into consideration, NoIe Lhat the wartagesarc not additive, since all units are .ormauy noi in oPeralionai the samctime b) B.sed oD 1000hous of operatiod per year.This figure will vary Nidelt depending on the area and the speciiic sire of the unit. See EEI-lub *76 2, Air Condidoring Usage Srudy." for an estinare JdL/.z: Edisor Elecrric Insiituie.

NOTE: Assessyur uru)erytandingof this materiljl4i trying Chapter Ptoblen n3.

catcutations399 10.3 ThennsVatue andPowe,

10.3 ThermsValueandPower

Catculations In introducing the rms value of a shusoidal voltage (or €urrent) in votiase apptjed to ihe Section9.1,we mentionedtlat it would play an important role ir power Figure10.7A A sinusojdat calculations.We can now discussttris role. Assume that a sinusoidal voltage is applied to the terminals of a resistor, as shown in Fig. 10.7,and that we want to delermine the average power deiiveredto the resistor.From Eq.10.12,

'^= i .1l

['r+rv)-cos'(dt + O) ,

"'

R

,,

:*l j/'v;-.l.t+6,1 ^ L t J n

I

1.

(10.18)

ComparingEq. 10.18with Eq. 9.5 revealsthat the avemgepower delivered to R is simply the rms value of the voltage squared divided by R, or (10.19)

R If the resistoris carrying a sinusoidalcurent, say.I-cos (i'))t+ dr), tlle averagepower delivered to the iesistor is

P : rl-"R.

(10.20)

The rms value is also referred to as the effective value of the sinusoidal voltage (or cullent). The rms value has an interestingproperty: Given an equivalent resistive load, R, and an equivalent time period, Z, the rms value of a sinusoidal source delivers the sameenergy to R as does a dc sourceof the samevalue.For example,a dc souce of 100V delivers the sameenergyin ? secondsthat a sinusoidalsourceof 100V!,." delivers, assumingequivalentload resistarces(see Problem 10.11).Figure 10.8 demonstratesthis equivalence.Energywise,the effect of the two sources is idertical. This has led to the term efrectire value being used interchangeablywith rmr ral,ra The average power given by Eq. 10.10 and the reactive power given bv Eo.10.11canbe written in tems of effectivevalues:

r:)cos@,

e1)

:!,$"""t'"-'t : %ffI.ncos(d, - 0J;

+

?, = 100V (rnt R

+

Y. : 100v (dc) R

vatueof!" (100V ms) detiversthe Figure10,8A lhe effective v" (100V dc). samepowertoR asthedcvottage

(10.21)

400

SinuroidatSteadjr 5tatePor,/er Caru aiions

and,by sirrilar maripulaLion, O = I/.i}I.irsin(d"

d).

170.22)

The effectivevalue ofthe sirusoidalsignalin powcr calculationsis so widcly used that voltage and current ralings oI circuils and equipment involvcd in porer utilizationare given ir terms o[ rms valucs.For examplc. thc voltagerating of residentialelectdcwiring is oltcn 240V/120 V voltagcs seNice.Thesc|'oliagclcvcls are the rms valuesof the siDrrsoidal supplicdby thc utilit)' companfwhich prorides power allwo voltagclevcls !o accommodatelow-voliage appliances(such as lclcvisions) and highcr voltagc appiiances(such as electric raDges).Appliancessuch as eleclriclamps.irons.and toastersall carry rms ratingson their namcplates. For examplc,a 120V 100w lamp hasa resistanoeof 120'/100,or 144O. ard drawsan rnis currcnt ot 120/144.of 0.833A. The peak valuc of thc lamp currentis 0.833\4, or 1.18A. The phasorlransfbrmof a sinusoidalfunction may also be expressed in lerms of the nns valuc.Thcmagnitudeof the rns phasoris equalto thc rms valLreol lhc sinusoidalfunction.If a phasoris baled on thc rms valuc. we indicalethisbycithcr an explicitstatement,aparenlhelicai"rms'adja cenl1|)lhe phasorquantit]'.or the subscript"efl" asin Eq.10.21. ln Exanplc l0.3.wcillustratetheuseofrms values{or calculalingpowcr.

bysinusoidaI vottage Determining Average PowerDelivered to a Resistor a) A sinusoidalvoltagehaving a n dmun anplitude of 625 V is applied to the terninals of a 50 O resistor.Find the averagepower delivered

Eq- 10.19.the averagepower delivered1()$c 50 f) rcsistoris r44lrJ4rl

|::] -=lqnb.2sw. b) Rcpcat (a) by first finding the c rrent in the

SoIution a) Thc rms value of the sinLrsoidaivoltage is 625iy2. ot approxi ately 4,11.94V From

1J

b) Thc maximum amplitude of the culrenl in thc resislor is 625/50.or 12.5A. The rms value of lhe currcnt is 12.5/rD. or approximaleiy 8.84 A. Hcnce the averagepower delivered lo P = (8.84)'50: 3906.25w.

obiective1-ljnderstand ac powerconcepts,their relationshipsto one another,and howto calculatethem in a orcuit 10.3 The periodictrlangularcurrentin Example9.4, repcatedhere.hasa peak valueof 180mA. Find thc avcragcpo\r'crthat this currentdelive$ to a 5 k() rcsistor.

NO'[E: Aln nj ChapterPrcblen 14.13.

Power 401 10.4 CompLex

Power 10.4 Complex Before proceedingto the vadousmethodsof calculatingreal and reactive power ir circuits operating in t]le sinusoidal steady state,we need to introduce and define complexpower.Complex power is the complexsum of realpower and reactivepower,or ('10.23) .{ ComptexPower As you will see,we can compute the complex power directly from the voltageand curent phasorsfor a circuit.Equation 10.23car then be usedto computethe averagepower and the reactivepower.becauseP : !t{S}

andO = s{s}.

Dimensionally, complex power is the same as average ol ieactlve power. However, to distinguish complex power ftom either average or reactive power, we use the units volt-amps (VA).Thus we usevolt-amps for complex power, watts for average power, and vars for reactive power, as summarizedin Table 10.2. Another advantageof usingcomplexpoweris the geometricinterpretation it provides. When working with Eq. 10.23,think of P, Q, and S as the sidesofa right tdangle,asshownin Fig.10.9.Itis easyto showthat the angle I in the power tdangle is the power factor angle d. 0i. For the right triangleshownin Fig.10.9,

_o

Unils

(r0.24)

of P andI (bqs'll0.l0l and[10.I l]. recpecri\ely). But ftomlhe delinirioD\

P : ateragc poler

Figure10,9 a A powettiangte.

a

(ubl^/2)sin(er 0) (v^I ^12) cos(0. - 0) = tan(d, - 0J.

(10.25)

Therefore, d : 0, Pi.The geometric rclations for a right triangle mean power triangle dimensions(tie three sidesand the also that the foul power factor angle) can be determined if any two of the foui are kno{'n. The magnitude of complex power is referred to as apparent pow€r. Soecificallv.

power (i0.26) d Apparent Apparent power, like complex power, is measuied jn volt-amps The apparentpower,or vollamp, requhementofa devicedesignedto convert electric energy to a nonelectric form is more important than the average power requirement.Although the averagepower representsthe useful output of the energy-converting device,the apparenl power representsure vollamp capacityrequired to supplylhe averagepower.As you can see

402 SinusoidaL Steady-state Power Catcuhtjons from the power triargle in Fig. 10.9,unlessthe poiver factor angleis 0' (that is, the deviceis purely resistive,pf: 1, and O = 0), the volt amp capacity required by the device is larger than the average powet used by the device.As we will see in Example 10-6,it Dakes senseto operale devicesat a power lactor closeto 1. Many useful appliances(suchas refrigerators,fans,air conditioners, fluorescentlighting fixtures,and washingmachines)and most industrial loadsoperateat a laggingpower factor.The power factor of theseloads sometimesis correctedeither by addinga capacitorto the deviceitself or by connectingcapacitorsacross the line leeding the load; the latter method is often used for large industdal loads.Many of the Chapter Problemsgive you a chanceto make somecalculalionsthat conect a laggingpower factor load and improve the operationof a circuit. Example 10.,1usesa power triangle to calculateseveralquaniities associated with an electricalload.

Catcutating Power Complex An electricalload operatesat 240V rms.The toad absorbsan dveragenower of 8 kw ar d lagging power factor of 0.8. a) Calculatethe complexpower ofthe load. b) Calculatethe impedanceofthe load.

Solvingfor /cn, I cn = 416'7A' We alreadyknow the angle of the load impedance,because it is the power factor angle: 0:

Solution a) The power lactor is describedas lagging,so we know that the load is inductive and that the algebraicsign of the reactivepower is positive. From the power triargle shownin Fig.10.10, P:

l s l c o sd ,

Q : l s l s i0n. Now,because cosd = 0.8, sind - 0.6.Therefore P

cos '(0.8): 36.87".

We also know that , is positive becausethe power lactor is lagging,indicatirg an inductive load. We compute the magnitude of the load impedarcefrom its definition as the ratio of the magnitudeof the voltage to the magnitudeof

_,

v,n)

24n

IL\l/

= 10kvA. sl' = r]] =:t:_j: (l-ll cosft O = iosin0:6kVAR,

z : 5.'76/36.8'7"A = 4.608+ j3.456().

s:8 + i6kvA. b) From the computation of the complex power of thc load,weseethat P = 8 kW. UsingEq.10.2l, P :

feflI,ffcos (0,

: (240)1.r(0.8) : 8000w.

A)

Figure10.10A A powe,irianqle.

10.5 Power Catcuiations 403

10.5 PowerCatculations We are now ready to develop additional equations that can be used to calculate rcal, reactive,and complex power.We begin by combining Eqs.10.10, 10.11.and 10.23to set

s =bfcos(e,

\/l

0,t+ ja;sinrc,

- dj) + jsin (r, =k!r*"re.

:\t!r,lu, ) -

0,1

dr)l

u,t: !vJ-/ lo, - a,t.

(to.?7)

If we use the elfective values oI the sinusoidal voltage and curent, Eq.10.27becomes

s : v"ttr"11_!3"_4.

(10.28)

Equations10.2?and 10.28are importantrelationshipsin power calculations becausethey show that if the phasor current and voltage are knowr at a pair of terminals, the complex power associatedwith that pair of terminalsis either one half the productof the voltageand the conjugate ofthe current,or the productofthe rms phasorvoltageand the conjugate of the rms phasor culrent. We can show this lor the rms phasor voltage Flgure10.11AThe phasorvottage andcunentassocjand curent inFig.10.11asfollows: atedwjrha panofhnninals.

s:V

.t11!ui)

: V"rl "rreilo' o) : Ve[eio,Iexe ja,

: v"d:r.

power (10.29)
Note that {fi : Iefier', follows from Euler's identity and the trigonomef ric identities cos( d) : cos(d)and sin(-0) = -sin(9) lene-ra = /effcos( 0,) + jlensin( - 1effcos(9r)

= I3r'

jlcffsin (d,

di)

404

SinusoidatSteady-StatePowerCatcuLations [he \amedefivalionlechnrque couldbe dppliedro Fq. 10.27lo yield

s: fvl

(10.30)

Both Eqs.10.29and 10.30are basedon the passivesignconvention.lf the current refeience is ilt the direction of the voltage dse aqoss the terminals,we inse a minus sign on the right-hand side of each equation. To illustratethe use of Eq. 10.30in a power calculation,let's use the samecircuit that we usedin Example10.1.Expressedin termsofthe phasor representationofthe teminal voltageand current,

v : 100lll

v,

|= 41 !!lA. Therefore

I s = (100Z!114 ,, 105'

200Llq

= -100 + j173.21VA. Oncewe calculatethe complexpower,we canread off both the real and rcaclivepowen,because S : P + jO. Thus P:

100w,

Q : 113.21vAR Theinterpretations signsonP andO areidenticalto tlose ofthe algebraic givenin thesolutionofExample10.1.

AlternateFormsfor Complex Power

figure10.12 Thegenentcjrcujtof Fjg.10.11 reptaced withanequivatent impedance.

Equations10.29and 10.30haveseveralusefulvadations.Here,we usethe Ims valueform of the equations,becauserms valuesare the most commoD type of represertation for voltages ard curents in power computations. The first variarion of Eq. 10.29is to replace t}le voltage with the product of the currenttimesthe impedance. In other words, e can alwaysrepresentthe circuit insidethe box ofFig.10.11by an equivalentimpedance, asshownin Fie.10.12.Then. Ys = Zlat

(10.31)

10.5 Power GLculrtions405

Substituting Eq.10.31into Eq.10.29ields

S=ztd*t : r"nl'Z : LdlR + jx) = l.nl'R + jll.n2x = P + jQ,

(10.32)

frcm which

P-lr,tr2R:r2^R, =:P.x. o = lr"$l,x

(10.33)

(10.34)

In Eq. 10.34,-Y is the reactatrce of either the equivalent inductance or equivalent capacitanceof the circuit. Recal from our earlier discussionof reactancetlat it is positive for hductive circuits atrd negative for capacitive cfucuits. A second useful variation of Eq. 10.29comes ftom replacing the current with t]le voltage divided by the impedance:

/ v-"\,

s-Y,rl - l= \ z

/

lv-",= P - i O . :;

(10.35)

Notethat if Z is a Dureresistiveelement.

-

lv"nl' R '

(10.36)

andif Z is a pule reactiveelement, ^

V"al2

(10.37)

In Eq. 10.37,X is positive for aDinductor and negative for a capacitor. The folo$.ing examples demoDstrate various power calculations in circuits opeiating in the sinusoidal steady state.

406 Sinusoidal Steady-state PowoCatcuLaiiom

Calculating Average andReactive Power In the circuit shownin Fig. 10.13,a load having an impedanceof 39 + j26 O is fed ftom a voltage source through a Line having ar impedance of 1 + j4 (). The effective,or rms,value of the source voltageis 250V a) Calculatethe load cunent IL and voltageVL. b) Calculatethe aveiageand reaclivepower delivered to the load. c) Calculatethe averageand reactivepower deliveied to the line. d) Calculaiethe averageand reactivepower sup plied by the source.

lQ

i4a

'

)Ji9*, Source -F

39f,l

1,1

P6p" Line

*l+Load

Figure10.13,AThecircuitfor Examph 10.5. c) The average and reactive power delivered to the line are most easily calculated from Eqs. 10.33 and 10.34becausethe line currentis known.Thus

P = (51(1)= 2sW

Solution

vAR. o = 6),e) : 1oo

a) The Lineard loadimpedancesare in seriesacross the voltagesource,sothe ioad currentequalsthe voltdgedi! idedb) lhe loralimpedance. or

lr

250'/0' ,c u^ + . , ^

a

/ru

1 3- 5

- 3 o8 - ' A 0 m ,

Becausethe voltage is given in terms oI its rms value,the curent also is rms.The load volt ageis the product of the load current and load impedance: vL = (39 + /26)lL = 234

= T4.36/

j13

Note ihat the reactivepower associatedwiththe line is po.iri\e becduserhe line reaclancei( inductive. d) One way to calculatethe averageand reactive power deliveredby the souce is to add the complex powerdeliveredto the line 1()that delivered to the load,or S:25+i100+975+./650 : 1000 + j750 vA. The complexpowerat the sourcecanalsobe calculatedfrom Eq.10.29:

s" = -250ri.

3.r8'v (nnt.

b) The averageand reactivepower delivered10the load can be computedusingEq. l0.29.Theiefore

Theminussignis insertedin Eq.10.29whenever thecurrentreference is in thedirectionofa volt agerise.Thus s": -250(4+ j3) :

S : vlli = (234-./13)(4+ 13) - 975+ j650vA. Thus the load is absorbingan averagepower of 975W and a reactivepower of650VAR.

(1000+ j7s0)vA.

power Th€minussignimpliesthal both average andmagnetizing reactivepowerarebeingdeliveredby the source. Note that this resullagrees wilh the previouscalculationof S, as it must, because the sourcemustfunish all the average andreactivepowerabsorbedby the line andload.

10.5 Pow€r Catcutations 407

Calcutating Powerin ParallelLoads The two loadsin the circuit showniD Fig.10.14can be desoibed as follows: l-oad 1 absorbs an average powerof BkW at a leadingpowerIactorof 0.8.Load 2 absorbs20 kVA at a laggingpo\eerfactor of0.6. 0 05O

s kw t36 87'

ervln *

,'ffi

{) /O.50

16KVAR 12kW (b)

(a)

10KVAR

20kw

G) Figure10.14A Thecjrcuit forExampte 10.6. a) Determine the power factor of the two loadsin parallel. b) DetelmiIle the apparent power required to supply t})e loads,tie magnitude of tlre current,I", and the aveiage power loss in tlre transmissionline. c) Given that the frequencyof the sourceis 60 Hz, compute the value of the capacitorthat would correct the power factor to 1 if placed in parallel with the two loads.Recomputethe valuesin (b) for the load with the correctedpower factor.

tisur€ 10.15 a (a)ThepoweftdangL€ for toad1. (b)The powertrianste for Load 2. (c)Thesun ofthe powertrianstes.

It follows that

s = 20.000+ j10.000vA, and

j10.0n0

+ r ; =20,n00 250

= 80 + 140A.

Therefore

Solution

I":

a) All voltageand curent phasorsin this problem are assumedto represent effective values, Note ftom the circuit diagram in Fig. 10.14 that I, : 11+ 12.The total complexpower absorbed by the two loadsis

s = (250)rl = (250)01 + rr)' = (2s0)ri+ (2s0)ri = .t1+ s2. We cansumthe complexpowersgeometricallyl usingthe powertrianglesfor eachload,asshown in Fig.10.15. By hlpothesis,

roru '1499 (.li) 8000- j6000vA, + j20,000(.8) 20,000(.6) 12,000+ i16,000vA.

80

J40: 89.44/

26.57'A.

Thus the power factor of the combined load is pf - cosro

26.57') - 0.8q44iaggints.

The power factor of the two loadsjn parallel is laggingbecausethe net reactivepowerispositiv... b) The apparentpower which must be suppliedto theseloadsis

kvA. ls : 20 + i10l = 22.36 The magnitudeof the cunent that suppliesthis apparentpower is rJ :

80

j40l : 89.44A.

The averagepower lost in the line resultsirom the currentflowing throughthe line resistance:

: 4oor.v 4r^": l\l'n : (B9.44)'(o.os) Note thatthe powersuppliedtotals20,000+ 400 - 20.400w.even rhougbrhe load. requne a total oI only 20,000W.

408

SinusoidatSt€ady-Stat€ PowerCatcutatjons

c) As we can see from tlle power triangle in Fig.10.15(c).we cancorrectthe powerlactor to 1 if we place a capacitor in parallel with the existing loads such that the capacitor supplies 10 kVAR of magnetizingreactivepower.The value of the capacitoris calculatedas follows.First,find the capacitivereactancefrom Eq. 10.37:

apparentpower once the power factor hasbeen

l s l: P : 2 0 k v A . The magnitudeof the current that suppliesthis apparentpower is

lr,l:

lv->

o

The average power lost in the line is tlus

(2s0),

: 320w. PI"" : r"l,R: (80),(0.0s)

10.000 :

Now, the power supplied totals 20,000 + 320 = 20,320W. Note that t}le addition of tie capacrtor hasreduced the line Iossfrom 400W to 320W

6.25A.

Recallthatthereactiveimpedance o{ a capacitor is l/oc, and @= h(60) : 376.99rad/s, if the sourceftequencyis 60IIz. Thus, -1 ax

-l

20.00080 A. 250

= 44.4u.F.

(3'76.99)(. 6.2s)

The addition of the capacitor as the third load is representedin geometricform as the sum of the two power tiangles sho$n in Fig. 10.16.When the power factor is 1, the apparentpower and the averagepowerare the same,asseenftom the power triangle in Fig. 10.16(c).Theiefore, the

1OKVAR +

20kw (")

10KVAR (b)

20kw (c) figure 10.16d (a)Thesumofthe powertiangtes for Loads 1 and2. (b) Thepower inangtefot a 421.1ttFcapacitot at 60 Hz. ( c ) I h es u mo ft h ep o w e r i n a n si hn sG ) a n d( b ) .

Batancing PowerDetivered with PowerAbsorbed in an acCircuit a) Calculatethe total averageand reactivepower deliveredto eachimpedancein the circuitshown in Fig.10.17.

jO

l2O+l

v-

tO

v, lI,

/lO

3qr"

b) Calculatethe averageand reactivepowen associatedwith €achsourcein the circuit.

\=1s0l{rv c) Veril, that th€ average power deliveied equals tlte averagepower absorbed,and that the magnet izing reactive power delivercd equalsthe magnet izing reactive power absorbed.

V vl = (78 1104) = v2 (72+i104)v v vi = (150 1130)

Ir = ( 26 ./s2)A 11: ( -2 +./6) A

Ir=( 24 r58)A

Figure10.17A Thecircuilwithsotutjon, for Example 10.7.

10.5 Power Catculations 409

Solution

b) The complex power associated with the independentvoltagesourceis

a) The complexpower deliveredto tle (1 +J2)O impedanceis

l

s' " = - : v " r = P " + r o " 2 "

51 =

=

l

;vJi

= PI+ jQ1

l : -:(150X 26 + i52)

:1950

l

26+ ls2) t(78 t104X

I = :(3380 + 16760)

j3900vA.

vollagesourceis Nore lhdr tbe independent absorbingan averagepower of 1950W and deliverhg 3900VAR. The complexpower associated with the current-controlled voltage

= 1690+ j3380VA. Thus this impedanceis absorbhg an average power of 1690 W and a ieactive power of 3380VAR. The complexpowerdeliveredto the (12 j16) O impedance is

I s. = -(]gr.)(li): P. + io. 2 -

=:( -

1

s2:

;v2r,

:

;(72

: P2 + jQ2

+ tI04\(-2 - j6\

Thereforethe impedance in the veiticalbmnch is absorbing240W anddelivedng320VAR. The complexpower deliveredto the (1 + j3) O impedance is l

sr=ivrr,=Pr+lol I

s8s0 i5070vA.

Both av€ragepow€r and magnetizirgreactive power arc being deliveredby the dependent c) fhe total powerabsorbedby the passiveimped!ollagesourceis aDces aDdthe iDdependeDl

: 240 - j320VA.

:;fi50

78 + /234)(-24+ 158)

i13u)( 24 + i58)

: 1970+ j5910VA. This impedanceis absorbing1970 W and 5910VAR.

Por,wra : Pr + P2 + P3 + Ps : 5850W. The dependentvoltagesourceis the only circuit elementdeliverhg averagepower,Thus Pd"ri,.,.d: 5850W. Magnetizingreactive power is being absorbed by the two horizontalbranches.Thus Qou,o,r"a: Or + 03 : 9290 VAR. Magnetizing reactive power is being delivered by the independentvoltagesource,tlrecapacitor ir the vertical impedance branch, and the dependentvoltagesource.Therefore Qaai,-.a : 990 VAR

410

Power Catcutations SinsoidatSteady-state

acpowerconcepts, theirretationships to oneanother. andhowto catcutate themin a circuit objective1-Understand (c) 23.52W. 94.09vAR: (d) 1152.62w'-376.36VARi

10.4 The load impedancein the circuit shownrs shuntedbt' a capacirorhavirg a capacitivereactanceof 52 O. Calculaie: a) the rms phasorsVL and IL, b) the averagepower aDdnagnetizingreactive power absorbedby rhe (39 + j26) o load c) the averagepower and magnetizingreactive power absorbedby rhe (1 + 14) o line d) lhe averagepower and nagnetizingreactive power deliveredby thc source,and e) the magnetiziDg reacdvepower deliveredby the shuntingcapacilor. 10

jl tl

\

+ \ 25010.

t)ii,:;

sourcc -

Lnro

-t-

Answer: (a) 2s2.20/ 4.s4'V(ms), s.38/ 38.23'A(rms); (b) 1129.09w,752.73 VAR;

''

39{)

(e) 1223.18VAR. 10.5 The rms voltageat the lermirlalsof a load is 250VThe load is absorbjngan avcragcpower of40 kW and deliveringa magrclizingrcactive power of30 kVAR. Derive hvo equivalcnl impedancemodelsof the load. Answer: l (t rn\e e. uilh tr.-SQ ol .nnocrii\e {) ii parallclwith 2.083O reactance:1.5625 of capacitivereactance. 10,6 Find tho phasorvoltageV, (rms) in the circuit shownif loadsLr and l, are absorbing15 kVA 1 l 0 . hf l l r g g r r r g a n d 6 k V A0dR p l e a d : n g . respectively. ExpressY in polar form.

tlo

t26lI

l

load

Answer: 251.61 /15.91'Y.

and10.22. NOTE: ALtotry ChaptetPlohlent 10.17,10.21,

10.6 $'iaxin*r$Psw*r'nr*n$r*er

power Figlte10.18,1 A circuitde$rjbingmaxjmum

Recallfrom Chapter4 that certaiDsysrems for exanplc.th osethat transmit information via electricsignals deperd on bcjng ablc to transfera maximumamountofpower from the sourceto lhe load.Wc now rccxamine maximum powef transferin the contert oI a sinusoidalsteadystatc network. beginning$'ith fig. 10.18.We musr determinethe load impcd anccZr that rcsultsin thc dcliveryof maximum averagepower 1lrlermi nals a and b. An]' lincar nctwork n1aybe viewedfrom the terninals oI lie load in tcrms of a Thivcnin equivalentcircuit.Thus the task reduces1() finding the valueof Zr_that resultsin maximun averageporverdelivcrcd io Zr. in thc circuit sho\\,nin Fig.10.19.

10.6 Maximum Power I€nsfer 411 For maximum averagepower tmnsfer, ZL must equal tie conjugate of the Thdvenin imoedance:that is.

(10.38) < Condition for rnaximum avengepower transfer WededveEq. 10.38by a straightforwardapplicationof elementarycalculus.Webeginby expressingZrh and ZL in rectangularform: Rrh+ jxrn,

(10.3e)

RL + jXL.

(10.40)

Zrh:

ZL:

In botl Eqs. 10.39and 10.40,the reactance telm caries its own algebraic sign positive for inductance and negative for capacitance.Because we are making an average-power calculation, we assumethat tie amplitude of the fh6venin voltage is er?ressed in terms of its ims value.We also use the Th€venin voltage as tlle reference phasor. Then, from Fig. 10.19,the Ims value of the Ioad current I is

-t =

v.. " (nr, + R, + j(xrh + x)'

110.4t)

The averagepower delivered to the load is

P : lrl'Rl.

\rc.42)

yields Substituting Eq.10.41 inroEq.10.42

lvrr'l'Rr (Rrh+Rrf+lxrn+lf;2

(10.43)

When workiq with Eq. 10.43,always remember that Vrh, Rrx, and -Yrh are fixed quantities, whereas R1 and -YL aie independent variables. Therefore, to maximize P, we must lind the values of RL and -YL where aPlARL andAPIAX L arc both zero.From Eq. 10.43,

aP _

dxr

-lV11'2RL().r +

[(Rr + Rr,)'+ (xL + xrif]''

(10.44)

ap _ lvl.hl'l(Rl+ nrh)'+ (xL+ x]x), - 2RL(RL+ Rft)] ' dfir [(Rr + Rr + 6L+ Xr,.),], (10.45)

412

SjnusoidatSteady-StatePowerkLcutations

FrcmEq.10.44,aP laXL is zerowhen Xt:

-Xn.

(10.46)

From Eq. 10.45,aP/aRLis zero when

R': \/6 + 6';t#

(10.41)

Note that when we combineEq.10.46with Eq.10.47,bothdeivatives are zerc when ZL - Zit.

PowerAbsorbed TheMaximum Average The maximum average power tiat can be delivered to Zr when it is set equal to the conjugate of Zrh is calculated directly ftom the circuit in Fi.9.1O.19.When ZL = Zih, the Ims load cu(ent is V11,/2R1,and the maximum averaseDower delivered to the load is

vrr,l'Rr 1 v*l' 4ElL

4 Rr-

(10.48)

If the Thdvenin voltage is er?ressed in terms of its maximumamplitude ratherthan its rms amDlitude.Eq. 10.48becones

-

1V;

(10.49)

MaximumPowerTransfer WhenZ is Restricted Maximum averagepower can be delivered to ZL only if ZL can be set equal to the conjugateof Zrh. Th€re are situationsin which this is not possible.First, RL and -l.L may be restricted to a limited range of values.In this sitlration,the optimum condition for RL and -YLis to adjust xt-:q !9C! !9 4I0 as possible and then adjust RL as close to VRih + (xL + xr 'as possible(seeExample10 9). A second type of restdction occuN when the magnitude of ZL can be vaded but its phase angle cannot. Under this restriction, the greatest amount of powei is lransferred to the load when the magdtude of ZL is setequalto the magnitudeof Zrh;that is,when

lz'l = zrnl The pioof ol Eq.10.50is left to you asProblem10.40.

(10.50)

10.6 l4aximum Powerlmnsfer 413

For purely resistivenetworks,maximum power transferoccurswhcn the load resistanceequals the Th6venin resistarce.Note that we first dedved this result in the introduction to maximum power transfer in Chapter4. Examples10.8-i0.11 illustrate the problem of obtaidng maimum power traNfer in the situationsjustdiscussed.

Determining Maximum Power Transfer withoutLoadRestrictions a) For lhe circuit shownin Fig.10.20,determinetl1e impedanceZL that resultsin maximum average Powertransferredto ZL. b) What is the maxirnumaveragepowertranslered to the load impedancedelerminedin (a)?

Solution

5{}

20t111+

jlo

20f)3 ioo

ZL

Figure10.20A ThecircuitfofExampte 10.8.

a) We beginb! dererminints tbeThd!eninequi\alent with respectto t1leload teminals a,b.After two source lransformations involving the 20 V source,the 5l) resistor,andthe 20 O resistor,we simplifythe circuitshownin Fig.10-20to the one sbownin Fig.10.21.Then, 16 /0' v * = ;q + l F. r,-(l o

i30

lo)

53.13' - 11.52- 115.36V.

= 19.2 /

4r)

figure 10.21n A sinptification of Fjg.10.?0bysource

we lind the Th6veninimpedanceby deacrivar ing the independentsourceand calculatingthe impedanceseen looking into the terminals a and b. Thus.

i1.68o

Zn

{- i6Jr4+ r3l +i/r

/o

For maximum averagepower transfer,the toacl impedancemust be thc conjugateofZrh, so

+j1.68o

zL = 5.76 + j\.68 0. b) We calc ate the maximun averagepower deliveredto Zr from the circuil shownin Fig.10.22,in which we replacedthe orighal network with ils Th6veninequivalent.From Fig. 10.22.the rms magdtudeofthe load currertI is

: 1.1785 A.

2(s;76)

F i E n e7 0 . 2 2t l l F r ' L | . \ o ^ i l t ' 9 . 1 0 .0 . r i r ' l l . originat network reptaced byjts Thdvenin equivaLent.

The averagcpoNer deliveredto the load is P = 1:ft(5.76)- 8W.

414

sinusoidatSt€ady-StatePowercaLcutatjons

Maximum Power Transfer with LoadImDedance Restriction Determining a) For t]Ie circuil shown in Fig. 10.23,what value of ZL results in maximum averagepower fiansfer [o ZL? What is t]Ie maximum power in milliwatts? b) Assume tlat the Ioad resistance can be varied between 0 and 40000 and that the capacitive rcactance of the load can be varied between 0 and 2000 O. Wlat settings of -RL and XL hansfer the most average power to the load? What is the maximum averagepower that can be hansferredundortheserestrictions?

Solution a) If therc are no restrictionson RL and XL, the load impedanceis set equal to t})e conjugateof the outputor theTh6veninimpedance.Therefore RL : 3000O and -Yr -

4000O,

ZL : 3000 J4000o. Becausethe sour€evoltagejs givenin terns of its rmsvalue,the averagepowerdeliveredto ZL is I

Inl

'\

: inw P: ii: 4 Jt't J{) .t

= 8 . 3m 3 w.

b) BecauseRL and -trL are restdcted,we fi^t set -trr as close to -4000 O as possible: thus

X, - :?@Q!. !qt. \ e set R, as clo.e to Ril. + (xL + xft)'as possible.Thus

o. Rt = \r6000t+ F2000+ 4000t= 360s.55

30000 l4O00Oa 101!l

b figure 10.23 a Tle c'(Lit fo tvarple5 l0.a dld t0.t0.

Now,becauseRL can be varied ftom 0 ro 4000 O, we can set RL to 3605.55O. Therefore,the Ioad impedanceis adjustedto a valueof Zt = 3605.55 - j2000 A. With ZL set at this value, the value of the load cufent is

lar =

101!

: 1.M89 /-16.85' 'lj{.

+ j2000 6605.55

The averagepower delivered to the load is p : (1.4489x 10-3f(3605.s5): 7.57mW. This quantity is the maidmum power that we can deliver to a load, given the restictions on RL and XL. Note that tlis is less than the power that canbe deliveredif there are no restrictions; in (a) ne toundrhar$e candetiver8.31mW

Power Transfer AngteRestridions FindingMaximum with Impedance A load impedance having a constart phaseangle of 36.87"is connectedacrossthe load terminals a and b in the circuit shown in Fig. 10.23.The mag tude of ZL is varied until the averagepower delivercd is the most possibleunder the given restriction. a) Specify ZL in rectangular form. b) Calculate the averagepower deliveied to ZL.

Solution a) From Eq. 10.50,we kl)ow that the magnitude of ZL must equal the magnitude of Z.rh.Therefore

lz"l = lztl:

= s000 o. 3000+ J40001

10.6 Maxinun Power Transfer415 No\ as wc know that the phaseanglc of Zr is 36.87',wc have zL = 5000/ -36.8'7' :4000

j3000l).

b) With ZL sel equal to ,1000 13000O, the load

lar =

Iu : 1.4U2/-8.13'mA, 7000+ j1000

and thc averagepower deliveredto the load is P : (1.4142x 10 )' (4000) :8mw. This quantilyis the maximumpower that can be delivered by this circuit to a load impedance whoseangleis constantat 36.87'-Again, this quantilyis lessthanthe maximumpowcrthal can bc deliveredif thereareno restrictionson ZL.

0bjediv€ 2-Understand the conditionfor maximunreal powerdetiveredto a Loadin an ac circuit 10.7 The sourcecurrentin the circuit shownis 3 cos5000r A.

3.6mH

a) Wlat impedanceshouldbe connecred acfussrenninals d.blof narimum rrcrage power lransfer? What is the averagepower transferredto the impedancein (a)? c) Assumethat the load is resrrictedto pure

resistance, What sizeresistorconnccted acrossa.b will result in the maximum aver, agepower transferred? What is the averagepower transferredto the resistorin (c)?

j10 o; (b) 18W: (c) 2n6 a; (d) r7.00w.

Answen (a) 20

NOTE: Also tt! Chopkt Ptoblens 10.11,10.19,and 10.50.

FindingMaximum Power Transfer in a Circuitwith an IdealTransformer The variable rcsistor in lhe circuit in Fig. 10.24is adjusteduntilmaximum averagepoweris delivcrcd ro RL. 8.10 /0' o) Whatis rhcvclu( oi Rr jn ohmsl

b) What is the maximum averagepower (n wattsl deliveredto Rz?

v t..)

b Figure10,24.t Thecircuitfortxamph10.11.

416

Sjnusoid aLsteady-state Power Calcutations

Sotution a) We first find the Th6venin equivalent wlth respectto the terminals of Rr. The circuit for delermrning lhe opencrrcuir!ollagein sho\rnin Fig. 10.25.The variablesVr, Y2, 11,and 12have beenaddedto expediatethe discussionFirst we note the ideal transformer imposes the following constraints on the vadables H, V2, 11,and 12:

Ir+

840!ry v (nnt

F i g u r e1 0 . 2 5d T 5 pL i r r i . u . " d. o f i i d l f e l h e v e - i 1 v o t r a 9 e .

v,- = lv. 4

t, :

lr, 4 -

The open circuit value of 12is zero, hence 11is zero. It follows that lr+

qV= 8a0Av,

\z = 2101!Y.

From Fig. 10.25we note that Vrh is the negative of V2,hence

84otll

vrh= -210av. The circuit shown in Fig. 10.26is usedto determine the short circuit current. Viewing 11and I, asme.hcurrenrs.lhe lwo mesbequalions afe

840A:80Ir 0 :2012

20I, + V, 35()

2011+!2.

Wlen thesetwo meshcurent equationsare combined with theconstraint equations we get 840l!|=

Figure10.26 A Thecircujtusedto catcuLat€ theshodcircuii

210d1

35{)

40Ir+ L

v

b Loaded tot nraxjmum tigure 10.27 A TheThcvenin equjvatent

0 = 2 5-L + 4- . Solving for the short circuit value of I, yields

r, = 6A. Thereforethe Theveninresistanceis )10 Rrh=-:=35O Maximum power will be deliveredto Rr when nr equals35 O.

b) The mlximum power delivered to -Rr is most easily determined using the Th6venin equivalent. From the circuitshownin Fig.10.27we have /

",^\2

/--*=l-:^

l(r)r=rr)w.

\ / u . /

P,adkaL Pe,spedr@ 417

0bjective3-ge abLeto catcutate all formsof ac powerin accircuitswith tineartransformers andideal transformers 10.8 Find thc averagepower deliveredto the 100O resistorin the circuit shownif rs : 660cos5000tV. 340

4O-trH -

::

^ 100 LnH

375{}

l^u-qg.

,.,1-\,-"

--j

100ll

Answer:612.5w' 10.9 a) Find the averagepower deliveredto thc 400O resislorin the circuitshownif 0s = 248cos10,000rV. b) Fird the averagepower deiiveredto the 375 O resistor. c) Find the powerdeveiopedby the idealvol| agesource.Checkyoul resultby showjngthe powerabsorbedequalsthe powerdeveloped

400f)

Answer: (a) 50W; (b)49.2w; (c) 99.2W 50 + 49.2= 99.2W. 10.10SolveExample10.11ifthe poladtydol on the coilconnected to terminala is at thelop. Answer: (a) 15 (); (b) 73sw. 1 0 . 1 15 o l \ cF \ a m p l e l 0 . l l i f I l < \ u l r a g e . o L ri c. e reduced10146lq V rmsandtheturnsratiois reversed to 1:4. Answer: (a) 1460O; (b) s8.4w.

NOTE: Also trt ChaptetPrcbkns 10.41,10.45,10.58, and 10.59.

PracticaI Perspective Heating Apptiances A handheldhair dryercontainsa heatjngelement,whichis just a resjstor heatedby the sinusoidalcurrentpassingthroughit, and a fan that btows the warmair surrounding the resistorout the front of the unit. This js shownschematicaLV in Fig.10.28.Theheatertub€in this figureis a resis tor madeof coiLednichromewire.Nichrome is an attoyof iron, chromium, andnjckel.Twopropetiesmakeit idealfor usein heaters.Fj6t, it is more resistivethan rnostotherrnetals,so lessiraterialjs required to achjevethe neededresistance. This atLowsthe heaterto be very cornpact.Second, untik-"manyotherrnetats, nichrofire doesnot oxidiz€whenheatedredhot in ajr Thusthe heateretementtastsa longtime. js shownin Fiq.10.29.This A circuitdiag|am for the hairdryercontroLs is the only partof the hairdryercjrcuitthat givesyou controloverthe heat setting.Therestofthe cjrcuitprovides powerto the fan motorandis not of inter€sthere.Thecoitedwirethat comprises the heatertube hasa connec, tion partwayatongthe coit,dividingthe coitinto two pieces.Wemodelthis Figur€10.28!. Schenratic Eprerntationofa jn Fig.10.29with two seriesresjstors, Rr and R2.Thecontrolsto turn the

418

SinusojdatSteady-StatePowerCatcutations

switchin whjchtwo dryeronandsetect theheatsettingusea four-position pairsoftermina15 in thecircuitwiLtbeshorted together by a pairof diding whichpairsof terminaLs metalbars.Thepositionof the switchdetermines Themetalbarsareconnected areshorted together. by aninsutator, sothere is noconduction Dathbetween the Dairs of shotedtermina15. Thecjrcujtjn Fig.10.29contains fuse,Thisis a protective a thermaL 1 : 1 1 Butif the temperature neaf device that normalty acts like a short cjrcuit. OFF L M H the heaterbecomes dangerousty high,the thermalfusebecomes an open Figure forthehairdryer circujt,djscontinuing 10.29.4A circuiidiagnm the flow of curfentandreducjng the rjskof fire or protection injury.Thethermalfuseprovides in casethe motorfajtsor the is not airftowbecornes btocked. Whjtethe designof the protection system Thermalfuse partof thG exampte, it is important to pointoutthat safetyanai.ysjs is an pat of anetectrjcaI essentjaI enginee/s work. Nowthat we havemodeted the controts for the hajrdryer,lets deter mjnethecl'rcuitcomponents that arepresent forthethreeswjtchsetbngs, Tobegin,the circujtin Fig.10.29js redrawn in Fjg.10.30(a) for the Lolf switch setting. The open-circujted wires have been removed for ctariry.A 1 1 1 1 OFF L M H simptified equivatent circuitis shownin Fiq.10.30(b). A sjmilarpairof fig(a) uresis shownfor the ilEorur'4 setting(Fig.10.31)and the rrcr'rsethng (Fig.10.32).Notefromthesefiguresthat at the rowsettjng,the vottage source seestheresistors Rl andR2in series; at ther,4E0rui1 settinq,thevottagesource seesonLythe R2 resistor; andat the HrGH settinq,the vottage jn paralLel. source seestheresistors

Figure10.30a (a)ThecircujtjnFis.10.29red€wn fortheL switchsetiing.(b) A simptjfied equjva-

Thernalfuse

i l 1 1 OFF L

M H

1

1

f

OFF L M H

i

G)

rigurc10.31,a(a)IhecncLrit in tiq.10.29 edrawn (b)AsimpLjfied forthe,roruswitch settinq. equjv-

Figure10,32a (a)Th€cjrcujtjnFjs.10.29 redrawn equivatorih€ HI6H switch5ettins.(b)A simptjfied lentcircuiifor (a).

your undestonding NoTE:Assess af this PrddicalPerspective by ttyingChapter ProbLems 10.66-10.68.

Summary419

Summary . tNtantaneous power is the producl of the instartaneousteminal voltage and current,or ? = trt. The positive sign is used when the referenced ection for the current is from the positive to the negativereference polarity of the voltage.The frequency of the instantaneouspower is twice the freqoency of the voltage(or current).(Seepage392.)

. The power faclor is the cosire of the phase angle betweenthe voltageand the currenl:

. Av€rage power is the averagevalue of the instantaneouspower over one period.It is the power converted Irom electric to nonelectric form and vice versa. This conversionis the reason that averagepowet is also refeired to as real power.Averagepower.witl the passivesignconventior,isexpressedas

The reactiye lactoi is the sine of the phase angle betweenthe voltageand the current:

- e,:t r = lv; ^"o"1e,

pf=cos(d,,9r. The terms laSginSand leadtnSadded to tle description of the power factor indicate whether t}le current is lag' gingor leadingthe voltageand thus whetherthe load is inductiveor capacitive.(Seepage396.)

rf : sin(or

d,).

(Seepage396.) Complex pow€r is the complex sum of the real and reac-

S-P+jQ

- 4n/encos(o, 0,).

:

(Seepag€394.)

ftt'= *t'u v?n z'

Reactiv€ power is the electdc power exchanged between the magnetic field of an inductor and the sourcethat drives it or betweenthe electriclield of a capacitorand the sourcethat drivesit. Reactivepower is never converted to noneiectric power. Reactive power,with the passivesignconvention,is expressedas

(Seepage401.) ' Apparentpoweris the magnitudeof the complexpower:

sl=\/P\e' (Seepage401.)

1

Q = ;u^I -sin(.o,

0)

= %dd sin(d, - d'). Both average power and reactive power can be eeressed in terms of ejther peak (y-, 1-) or effective (%fi, 1ed current and voltage. Effective values are widely used in both householdand industrial applica tior].s.Effectire ,alue and /ms value are interchangeable termsfor the samevalue.(Seepage394.)

The watt is usedas the unit for both instantaneous and real power.The mr (volt amp reactive,or VAR) is used as the unit for reactive power. The volt-arnp (VA) is usedas the unit for complexand apparentpower.(See page401.) Maximum pow€r transfer occurs in circuits operating in the shusoidal steady state when the load impedance is the conjugate of the Th6venin impedanceasviewed from the terminalsof t})eload inpedance.(Seepage410.)

420

SinusoidatSt€ady'StatePowerCatcutatjons

Problems Sections10.1-10,2

figureP10.5

80nF

10.1 Showthat the maximumvalueof the instantan€ous po\rergivenbyEq.Io.qi. P VP) ' STandthal theminimumvalueis P - t/ e' 9 . 102 The following setsof valuesfor o and i pe ain to the circuit seenin Hg. 10.1.For eachset of valueE calculateP and I and state whether the circuif insidethe box is absorbingor delivering(1) average powerand (2) magnetizingvals. a) o = 340cos(ot+ 60')V, j = 20cos(.dr+ 15') A. b) ?,:75cos(ot- 15') V, j = 16cos(.dt+ 60') A. c) ?J- 625cos(ot+ 40")V, i - sln(at + 240")A. d) 2 = 180sin(ol+ 220')V, t = 10cos(ar+ 20")A.

10,6 Find the averagepower dissipatedin the 20 O Btrc resistor in the chcuit seel1 in Fig. P10.6 if is = 15cos10,0001A FiglreP10.5

10.3 a) A universitystudentis makhg coffeewhile listeningto a Buck'sgameon themdio At the same time,her roommateis watchingthe Packenol1a solid-stateTV All theseappliancesaie supplied ftom a 120V branchcircuit protectedby a 20A circuit breaker, If the roommate tuns on a portableheater,is the circuitbreakertdpped? b) Will the roommatebe able to use the heaterif t}le coffemakeris unplugged? 10.4 Find tlle averagepower delivered by the ideal curent source in the circuit in Eg. P10.4 if mA. is = 30cos25,0001 Figure P10,4

10ta

lmH

1) r"Jfz.si,r

20o'

10.7 The load impedarce in Fig. P10.7 absorbs 40 kw and 30 magnetizing kvars. The sinusoidal voltage source develops 50 kW. a) Find the values of capacitive line reactance that will satisfy these constraitrls. b) For each value of line reactarce found in (a), show that the magretizing vars developed equalsthe magnetizirgvars absorbed.

2A

t)

f +od

figue P10.7

40r.H

10.5 The op amp in the circuit shown in Fig. P10.5 is ideal. Calculate t}le average power deiivered to the 1 kO resistor when t's = 4 cos50001V.

n" ^

2500rc| v(nnt

-/Xo

Problems 421 10.8 A load consistingof a 1350 O resistorin parallel with a 405 mH inductor is connected aooss the terminals of a sinusoidalvoltage source og, where oc = 90 cos25001V a) What is the peak value of the instantaneous pow€r deliveredby the source? b) Wlat is the peak value of the instantaneous power absorbedby the source? c) What is t]le averagepower delivered to the load? d) What is the rcactivepowerdeliveredtotheload? e) Does the load absorb or generate magnetiz ing vars? f) Whar is the power factor of the load? g) What is the reactivefactor of the load? 10.9 a) C-alculaiethe real ard reactive power associatecl with each circuit element in the cifcuil in Fig.P9.56. b) Verify that the avemge power generated equals the avemgepower absorbed. c) Verify that the magnetizing vars generated equalthe magnetizingvarsabsorbed. 10.10 Repeat Problem 10.9 for the circuil shown in Fig.P9.57.

Figure P10.13

10,14 Fhd the rms value of the periodic current shor,n in Fig.P10.14. figureP10.14 i (A)

Section103 10.11 A dc voltage equal to fdcV is applied to a resistor of R O. A sinusoidal voltage equal to z)sV is also appliedto a resistoroI R O. Show that the dc voltagewill deliver th€ sameamount of energy in 7 seconds (where r is the period of the sinusoidal voltage) as the sinusoidal voltage provided yd" equals the rms value of o,. (1tir?t Equate the two er?ressionslbr the energy delivered to tle resistor.) 10.12 a) A personalcomputer with a monitor and keyboard requires 60 w at 110V (rms). Calculate the rms value of the cuirent carried by its power cord. b) A laser printer for the peisonal computer in (a) is ratedatB0W at 110V (ms).If this printer is pluggedinto the samewall outlet as the computer, what is the rms value of the current drawn from the outlet? 10.13 a) Find tne rms valueof the peiodic voltageshown in Fig.P10.13. b) If this voltage is applied to the terminalsof a 12 O resistor,what is the averagepower dissipatedin the resistor?

40

s0

.(mt

10.15 fhe pedodiccurrentshownin Fig.P10.14dissipates an averagepower of 24 kW in a resistor.wlrat is the valueof the resistor?

Sections10.4-10.5 10.16 Find the averagepower,the reactivepower,ard the 6'c apparentpower absorbedby the load in t}le circuit in Fig.P10.16if isequals30cos 100tmA. Figure Pr0.16

422

Power 5inusoid aLSteady-State Calcutations

10.17 a) Find the average power, the reactive power, and ttre apparent powei supplied by t]le vollage source in the circuit in Fig. P10.17 if 0g = 50 cos10sl V. b) Check your answer in (a) by showins

c) Check your answer in 0a- = )8"t"

(a)

by

1()

i8r)

2a0.llV (rmt

showing

10.20Two 660V (rms) loads are connectedin parallel. The

tigu|ePl0.u

two loads draw a total averagepower of 52,800W at a power factor of 0.8 leading. One of t]Ie loads dmws 40 kVA at a power factor of 0.96lagging.Wlat is the power factor of the other load?

7.50

10.18 The voltage Vs in the frequency-domain circuit shownio Fig.P10.18is 340 A v (rms). a) Find the average and reactive power for the voltage source, b) Is the voltage source absorbingor deliverirg averagepower? c) Is the voltage source absorbing or delivedng magnetizingvars? d) Find the average and reactive powers associated with each impedance branch in the circuil. e) Check fie balan€e between delivered and absorbed averagepower. f) Check the balance between delivered and absorbedmagletiang vars. tigureP10.18 50()

- )

rigureP10.19

80()

1

-lloo o

j60()

10.19 a) Find VL (rms) atrd d for the circuit in Fig. P10.19 iJ the load absorbs 250 VA at a lagging power factor of 0.6. b) Consrucl a phasordiagramof each solulion obtainedin (a).

ln.2I The thrce loads in lhe circuit in Fig. P10.21can be described as follows: Inad 1 is a 12 o resistor in series with a 15 mH inductor; load 2 is a 16 rrF capacitor in serieswith a 80 O resistor; and Ioad 3 is a 400 Q resistor in sedeswith tle pamlel combination of a 20 H inductor and a 5 pF capacitor. The ftequency of the voltage source is 60 Hz. a) Give the power factor and reactive factor ot each load. b) Give the power factor and reactive factor of the composite load seenby the voltage source. Figure P10.21

1022 Three loads are connectedin parallel acrossa 2400V (ms) line, as sho$n in Eg. P10.22.Load 1 abso$s 18 kw and 24 kvAR. Load 2 absorbs60 kvA at 0.6pf lead.I-oad 3 absorbs18 kW at uDrrypower tactor. a) Find the impedance that is equivalent to the tbreeparallelloads. b) Find t}le power factor of the equivalent load as seenfrom tbe line'sioput terminals, Flgure P10.22

Probtems 423 1023 The tlree loads in Problem 10.22are fed frcm a line having a sedesimpedance0.2 + j1.6 O, as shown inFig.P10.23. a) Calculate the ms value of the voltage (Y) at the sendhg end of the line. Calculate the average and reactive poweN associated with the line impedance. ") Calculate the averageand reactive powen at the sending end of the line. d) Calculate the efficiency (T) of tie line if the efficiency is defhed as 4 = (P6"6/P.."ai"*"") X 100. tigureP10.23 0.2o j1.6o

Figure P10.25 0 . 0 1 O1 0 . 1 O

+

0.01oj0.1o ll.2i

The three loads in the circuit shown in Fig. P10.26 are 51 : 5 + j2 kVA, S, : 3.75 + j1.5 kVA, and s3=8+j0kvA. a) Calculate the complex power associated wit]I each voltage source,\1 and Vs2. b) Vedfy that the total real and reactive power delivered by the sources equals the total rcal and reactive power abso$ed by the network. figuruP10.26

2400i0"

0.050

12s48V (mt

t0.u

The three parallel loads in the circuit shown in Fig. P10.2 can be described as follows: Load 1 is absorbingan avenge power of 24 kW and 18 kVAR of magnetizing varsi load 2 is absorbing an average power of 48 kW and generating 30 kVAR of magnetiziiig reactive powert load 3 coNists of a 60 O rcsistor in paiallel with an inductive reactance of 480 O. Fhd t})e rms magnitude and tlre phase angle of Vs iJ V. = 2400 lq V(Ims). tigurcP10.24

j10o +

10.25 The two loads showr in Fig. P10.25can be described as follows: Load 1 abso$s an average power ol 24.96 kW and 47.04 kVAR magnetizing reactive power; load 2 has an impedance of 5 - j5 O. fhe voltage at the teminals of the loads is 48O^vAcos12ont Y . a) Find the Ims value of the souce voltage. b) By how many microseconds is the load voltage out of phasewith the source voltage? c) Does t]Ie load voltage lead or lag the source voltage?

12s{r V (rmo

to27 The three loads in the circuit seen in Fig. ?10.27 are desffibed as follows: Load 1 is absorbing1.8 kW and 600VAR; load 2 is 1.5kVA at a 0.8pf lead;load 3 is a 12 O rcsistorin parallelwith an inductor that hasa reactanceof 48,(!. a) C-alculatethe average power and t])e magnetizing reactive powei delivered by each source if Y"t = Ysz:120 lq V (rms). b) Check your calculations by showing your results are consistentwith the requirements SP.

= Sp.

Sn.

: Sn.

Figure P10,27

424

sinusoidatsieady-StatePow€rCalculatjons

10.28 Supposethe circuit shown in Fig. P10.27represents a residential distribution circuit in which the impedances of the service conducton are lregligible and Vrl = Vgz= 120 7!|| V(rms). The three loads in the circuit are L1 (a coffeemaker,a frying pan and an egg cooker); L, (a hairdryer, a sunlamp, a fan, and an automatic washing machine);and q (a quick recovery water heater and a range with an oven).Assumethat all tlese appliancesare in operation at the sametime.The senice conductorsare protected with 100A circuit breakerr Will the serv ice to this residencebe interrupted?Explain.

10.29 The steady-statevoltage drop between the load and the sendingend of the iine seenin Fig. P10.29is excessive.A capacitor is placed in paralel with the 250 kVA load and is adjusted until the steady-state voltageat t}le sendingend of the line hasthe same magnitude as the voltage at the load end, that is. 2500V (ms). The 250 kVA load is operatingat a power factor of 0.96 lag. Calculatethe size of the capacitoi in microfarads if the circuit is operating at 60 Hz. In selectingthe capacitor,keep in mind the need to keep the power lossin the line at a reasonablelevel. tigureP10.29 1f}

figurcP10.30 20

j20o I 1380

7200[f

v G-t

+

Souce !'--Line

i460o 'F

Load

10.31 A group of small appliances on a 60 IIz system r€quires25 kVA at 0.96pf laggingwhenoperatedat 125V (rms).The impedanceof the feedersupplying the appliancesis 0.006+ j0.048O. The voltage at the load end ofthe feederis 125v (rms). a) What is the rms magnitude of the voltage at the source end of the feeder? b) What is the averagepower Iossin the feeder? c) What sizecapacitor(jn microfarads)at the load end of the feederis neededto improve the load power tactor to unity? d) Aftef the capacitor is installed, what is t}le Ims magnitude of the voltage at the source end of the feeder if t}le Ioad voltage is maintained at 125V (rms)? e) What is the averagepower loss in the feeder for (d)?

isI) 25ixl& V (rms)

10.30 a) Find the averagepower dissipatedin the lire m Fig.P10.30. b) Find the capacitive reactance that when connected in parallel with the load will make the load look purcly rcsistive. What c) is the equivalent impedanceof the load in (b)? d) Find lhe averageponer dissipaled i0 lhe line when the capacitive reactance is connected acrossthe load. e) Expressthe power lossin (d) as a percentageof the power lossfound in (a).

10.32 A factory has an electrical load of 1800kW at a lagging power factor of 0.6. An additional variable powei factor load is to be added 10 the factory. The new load will add 600 kW to the real power load of the factory. The power faclor of the added load is to be adjustedso ttrat the overall power factor of the factory is 0.96lagging. a) Specifythe reactivepower associated$rith the addedload. b) Does the addedload absorbor deliver magnetizing vars? c) w}lat is the power factor of t]le additional load? d) Assumethat the voltageat the input to the factory is 4800V (rms).What is the rms magntude of the currert into the factory before the variablepower factor load is added? e) Wlat is the Ims magnitude of the current into the factory after the variable power factor load hasbeeDadded?

Probtems 425

10.33 Assume the factory describedin Problem 10.32is fed from a line having an impedance of 0.02 + j0.16 O. The voltage at the factory is maintained at 4800V (mN). a) Fird the avemge power loss io the Lhe before and after the load is added. b) Find the magnitude of the voltage at the sending end of the line before and after t]le Ioad is added.

tlgurcP10.36

o

1oo a .is!-q 3qc v (r'nt

+

;-

i30 ^t29-9.

j70.o

I L$a

j100o b

10.34 Considerthe circuit describedin Problem9.76. a) What is the Ims magnitude of the voltage adoss the load impedance? b) What percentage of the average power devel oped by the practical source is delivered to the load impedance?

10.37 a) Find the average power delivered by the smusoidalcurrentsourcein the circuitof Fig.P10.37. b) Find the average power delivered to the 1 kO rcsistor, Flgure P10.37

10,35 a) Find t}le six branch currents la Ir in tle circuit in Fig.P10.35. b) Find the complex power ir each branch of the crcult. c) Check your calculations by veriflng that the average power developed equals the average power dissipated. d) Check your calculations by vedfing that the magnetizing vars genemted equal the magnetizing vaIS absorbed.

101UmA

Flgure P10.35 10,38 a) Find the average power dissipated in each resistor in the ctucuit ir Fig. P10.38. b) Check your answer by showing that the total power developed equals the total power absorbed.

1() j2 o /jlo\

" tl+ )lt" ff'"

20E | . ,

It'

I

jl o Id

iio

rr

10 Pig(eP10.38 40

10.36 a) Find the averagepower deliveredto ttre 40 O resistorin thecircuitin Eig.P10.36. b) Find the averagepower developedby the ideal sinusoidalvoltagesource.

o Fjrrd Z ab. d ) Show tlat the average power developed equals

the averagepowei dissipated.

-j8 0

426

SinusojdatSteady StatePowerCatculations

10.39 The sinusoidal voltage source in the circuit in Fig. P10.39is developingan rms voltage of 680V The 80 O load in the circuit is absorbing 16 times as much averagepower as the 320 O load. The two loadsare matchedto the sinusoidalsourcethat has an interml impedanceof 136 A kO. a) Speci$'the numedcalvaluesof dr and dr. b) Calculatethe power deliveredto the 80 O load. c) Calculatethe rms valueof the voltageaqossthe 320 O resistor.

10.42 The phasor voltage Vrb in the cjrcuit shown m Fig. P10.42is a80 A VtmO when no external load is connectedto the terminalsa,b.When a load having an impedanceoI 100 + j0 O is connected acrossa,b.thevalueof\b is 240 j80 V (rms). a) Find the impedancethat should be connected acrossa,b for ma-\imum avemge power transfer. b) Find the maximumaveragepowertransferredto the load of (a).

Figu|e P10.39

680U

10.43 The load impedanceZL for the circuit shown in Fig. P10.43 is adjusted until ma-\imum average poweris deliveredto ZL. a) Find the maximum average power delivered to ZL. b) what percentageof the total power developed in the circuit is deliv€redto ZL?

Seclion 10.6 10.40 Prove that if only the magnitude of the load impedance cdn be vafied.mosl a\erdgepouer i\ transferred to the load when ZL - Z\. (Hint: In deriving the expressionfor the averageload power,write the load impedance(ZL) in the fom ZL: ZLlcos9 + j ZL s;\0, and note that only lZLl is variable.) 10.41 a) Determine the load impedancelor the circuit shownir Fig.P10.41that will resultin maximum averagepower being transferredto the load if o = 10 krad/s. b) Determine the ma-\imum average power delivered to the load ftom part (a) if r)s: 90 cos10,000tV. Figure P10.41

figue P10.43

l10o 2500'V lrms)

lD.U For the frequency-domaincircuit in Fig. P10.4,1, calculate: a) the Ims magflitude of V,. b) the averagepower dissipatedin the 70 O resistorc) the percentageol the averagepower gen€rated by the ideal voltage sourcethat is deliveredto the 70 O toadresistor. Figure P10.44

2.5nF

j40 {)

10o

' 36011I

v (rno

l

l

+

r 3 o n - ] ; + o ov , 70o"

PrcbLenis 427

10.45 The 70 O rosistor in the circuit in Fig. P10.44 is replaced with a variable impedance-Z,. Assume Z, is adjustedfor maximum avemgepower transfet to za. a) What is the maximum average power that can be delivered to Z,? b) What is the average power developed by the ideal voltage source when maximum average poweris deliveredto Zo?

10,46 Suppose an impedance equal to the conjugate of the Th6venin impedance is connected to the terminals c,d of the circuit sho*lr in Fig. P9.74. a) Find the average power developed by the sinusoidalvoltagesource. b) What percentageofthe power developedby the source is lost in the linear transformer?

10.47 The variable resistor in the circuit shown in Fig. P10.47is adjusteduntil the averagepower it absorbsis maximlrm. a) Find R. b) Find the maximum averagepower. tigureP10.47 18()

\ 630xr

i6r}

8o

llso

F;

Figure P10.48

100

j.29a

600Cv Gmt

10.49 The peak amplitude of the sinusoidal voltage souce in the circuit shown in Fig. P10.49 is 15014 V, and its period is 2002],.s.The load resistor can be vaded ftom 0 to 20O, and the load inductoi canbe variedftom 1 to 8 mH. a) Calculate the average power delivered to the load when Ro : 10 O and L, : 6 mH. b) Detemine the settingsof R, and Z, that will rcsult in the most avemgepower being translelred to Ro. c) w}lat is the most avenge power in (b)? Is it gieater than tlle power in (a)? d) If tlere are no constaints ofl & and L,, what is the maximum average power that can be delivered to a load? e) wllat are the values of R, and -L. for the condition of (d)? f) Is the avemgepower calculatedin (d) larger than thal calculatedin (c)? figureP10.49

10.44 The variable resistor Ro in the circuit shown m Fig. P10.48 is adjusted until maximum avetage poweris deliveredto Ro. a) What is the value of Ro in ohms? b) Calculatethe aveiagepower deliveredto R.. c) If R" is replaced with a variable impedance .Z,, what is the mlximum avemge power that can be deliveredto Zo? d) In (c), what percentageof the circuit's developedpower is deliveredto the load Z.?

10.50 a) Assume tlat R, in Fig. P10.49 can be varied between 0 and 50O. Repeat (b) and (c) of Problem10.49. Is the new avenge power calculatedin (a) geater than that found in Problem10.49(a)? c) Is the new avemgepower calculatedin (a) less than ttrat found in 10.49(d)?

Power Catcutations 425 SjnusoidaL Steady-State

10s1 The

sending-end voltage in the c cuit seen in Fig. P10.51is adjusted so that the rms value of the load vokage i. dLra)s 4800V The variablecapacitor is adjusted until the averagepower dissipatedin the lire resistanceis minimum. a) If the frequency of tle sinusoidal source is i0 o0 Hz. whal is rhe \alue ol lhe capacilance miuofarads? b) If the capacitor is removed ftom the circuit, what percentageinqease in the magnitude of V" is necessaryto maintain 4800V at the load? c) If ttre capacitor is removed from t]le circuit, what is the percentage inuease in line loss?

10,54 The polarity dot on a1 itr the circuit in Fig. P10.52is a) Find the value of k tlat makes t , equal to zero b) Fird the power developed by the souce when k has the value found in (a). 10.55 The impedance ZL il the circuit in Fig. P10.55 is adjusted for maximum average power transfer to ZL. The intemal impedance of the sinusoidal voltagesourceis 8 + J56O. a) What is the maximum average power deiivered to ZL? b) What percentageof tle averagepower delivercd to the lirlear transformer is delivered to ZL?

Figure P10.51 1()

Figure P10.55

i8fl

- - l

I

a800 10:v (rmo 7600a

j40o

v (rmt

10.52 The values of the paiameten in the circuit shown i n F i g . P 1 0 . 5 2a r e L r 4 0 m H : / ) - l 0 m H : k-0;75|. Rs= 20O; and Rr-140o. ff : find ?,s 240\4 cos 4000t V, a) the rms magnitude of ?), b) the averagepower delivered to RL c) the percentage of the avenge power generated by the ideal voltage sourcethat is delivered to RL. P10.52 Figure Rs

+\

L1

k ,l L,1 -'

source+L

+

in the circuit in Fig. P10.52is adjustable. a) wlat value ot RL will resull iD the ma"\imum averagepo\{er being transfe ed to RL? b) What is the value of the maximum power transferred?

j*Load*

10.s6 a) Find the steady-stateexpressionfor the currents is and iL in the circuit in Fig. P10.56 when t's = 70 cos50001V. b) Find the coefficient of coupling. c) Find the energy stored in the magnetically coupled coils a : 100,7 /rs and I : 200r' ps.

e)

1) g)

h)

10,53 Assume the load resistor (R,

_ _ _Tralslo!!9r

Find the power deliveredto the 30I) resistor. If tle 30 O resistor is replaced by a variable rcsistor RL, what value of RL will yield maxl_ mum averagepower tmnsfer to RL? What is the maximum averagepower in (e)? Assumerhe 30 O resisloris replacedb' a variable impedance ZL. What value of ZL will result in maximum averagepower transler to ZL? What is t}le ma-{imum average power in (g)?

tigue P10.56 10O

zr,Il 30[}

Probtems 429

10.57 Find the impedanceseenby the ideal voltage source in the circuit in Fig. P10.57when Z, is adjusted for maximum averagepower transfei to Zo, tigureP10.57 i{Qr

I0o 801!8

10.58 Findthe averagepowerdeliveredto the 4 kO resistor in lhe circuitofFlg.P10.58.

10.61 a) ff & equals2520turnq how many tums should be plac€d on the N2 winding of the ideal transfom)er in the circuit in Fig P10.61so that maximum averagepower is delivered to the 50 O load. b) Find the average power delivered to the 50O load. Find the voltage V. d) What percentageof the power developed by the ideal current source is delivered to the 50O resistor?

tigureP10,61

Figure P10.58 10(}

5ko

-e;l1,2.sft

1:4f.lfl

loo1fvA I llt ll I l {nas,Y// |

llt \

lt\

|Ideal[1 ]Ideall

160

lsko '-ll

t l

10.59 The ideal transformer connected to the 4kO load in Problem 10.58 is replaced with ar ideal transfomer t])at has a tums mtio of 1:l?. a) W}lat value of d results in maximum average power being delivered to the 4 kf,l resistor? b) Wlat is the maximum averagepower? 10.60 a) If Nl equals 1500 tums, how many tums should be placod on the N2 winding of the ideal tmnsfomer in the circuit seen in Fig. P10.60 so that maximum average power is delivered to the 3600O load? b) Find t}te average power delivered to the 3600O rcsistor. c) wltat percentage of the average power delivered by the ideal voltage source is dissipated in the linear tra.nsformer?

10.62 The variable load resistor RL in the circuit shown ttr Fie. P10.62is adjusted for maximum average power transfer to RL. a) Fhd the maximum averagepower. b) What percentageof the averagepower developed by the ideal voltage source is delivered to RL when RL is absorbingmaximum avenge powei? c) Test your solution by showing that the power developed by the ideal voltage source€quals the power dissipated in the circuit.

Figure P10.62

Figurc Pl0.60 t-200 JO;--\

it#,9 ,"1 I

JzO

t24 A

4f}

'l [ra.,r

&1 |

3ooo o

2sE

v G.t

0.25f}

430

SinusoidatSteady-StatePowercaLcutatjons

10.63 Repeat Problem 10.62 foi the circuit shown in 4trc Fig.P10.63.

tigureP10.65 30ko j10 kO

Figure PI0.63

20ko lko

10ko

1001ry

10.64 The sinusoidal voltage source h the circuit in Fig. P10.64is operating at a frequency of 50 kmd/s. The variable capacitive reactance in the circuit is adjusted until the average power delivered to the 160 O resistor is as large as possible. a) Find the value of C in micofarads b) When C has the value found in (a), what is r}le averagepower delivered to tlle 160O resistor? c) Replace the 160O resistor with a va able resistor R,. Specify t}le value of R. so that mlximum averagepower is delivered to Rd. d) What is the maximum average power that can be deliveredto R,?

Sections10.1-10.6 10,66 The hair dryer in the Practical Perspective uses a tl$ll&rr 60 Hz sinusoidal voltage of 120V (rms). The heatei element must dissipale 250 W at the Low setting, 500 W at the MEDTUMsetting, and 1000 W at tie HrcHsetting. a) Fhd the value for resistor R2 using the specification for the MEDrI,.Msetting, using Fig. 10.31. b) Find the value for tesistor using t]te specification for the Low setting, using the results frcm pal1 (a) and Fig.10.30. c) Is the specification for the HrGHsetting satisfied?

Figure P10.54

25r} 2s0Lqf

20o./50O

160r)

ill'' i

(

*,

IdealI

10,65 The load impedance ZL in the circuit in Fig. P10.65 is adjusted until maximum average power is tansIened to ZL. a) Specifythe valueof Zr if Nr = 15,000turnsand N2:5000turns. b) Specily the values of IL and VL when ZL is absorbing ma-\imum avemge power,

10.67 As seenin Problem 10.66,only two independent plllflI#hpower speciflcationscan be madewhen two rcsis;;iA-__tors makeup t}le heatingelementof the hair diyer. a) Show tiat the er:prcssionfor the rxcH power iating(PIr)is

-"

P4. Pv-Pr'

where Pff = the MEDIUMpower rating and Pt = the Low power mthg. b) ff Pr - 250 W and PM = 750W, what must the r{rGHpower rating be?

431 10.68 Speci$'the valuesof Rr and R2in the hair dryer cirplgljil - cuit in Fig. 10.29if the low power rating is 240W 6e and ttre high power tating is 1000W Assumethe supply voltage is 120 V (rms). (flird: Work Problem10.67fint.)

tigureP10.59 R3

Rr

R3

It"l"3"

10,69 If a third resistoris addedto the hair dryer circuit in Fig.10.29,it is possibleto designto ttrreeindependp?Jiflfii€ LOW MEDIUM HIGH -;;if- ent power specificationsIf the resistorR3is added in serieswith the Thermalfuse,then the correspondingro\\. MEDIUM . and FncHpowercircuitdja- 10.70 You havebeengiventhe job of redesignhgthe hair grairs are as shown in Fig. P10.69.ff the three dryer described in Problem 10.66 for use itr J€l|;:,#;E power settirgs are 600 W, 900 W and 1200W, The\randardsupplyvoltaeejn Eoglandis o6c, England. \,/r'balresisrorlalueswill you usein respectively, (rms). whenconnectedto a 120V (rms) sup220 V ;i ply,whatresistorvaluesshouldbeused? your designto meetthe samepowerspecifications?

Ba[anced Three-Phase C'ircuits Thre€-Phase 1 1 . 1BaLanced Vottagesp. 434 17.2 Three-Phas€ VottageSourcesp. 4J5 1 1 . 3Analysisofthe Wye-WyeCircuit p. 4i6

r\.4 Anatysisofthe Wye-DeltaCircuit p. 44? in Batan(edThree-Phase 1 1 . 5Po$erCaLcuLations 11.6 i\,leasuring AveragePowerinThfee-Phas€

(now howto anaLyze a baLanced, thrce,phase wY€wY€cornectedcircujt. Knowhowto aratyzea batanced, thlee-phase wye-detta coinectedcircuit. power(averag€, B€abteto catcuLat€ reactive, jn anyihrc€ phasecircuit. andcomptex)

432

Generating,transmitting, distributing, and using largeblocks of electricpower is accomplishedwith three-phasecircuits.The complehensivcanalysisof such sysfemsis a field of study in its own right; \ve cannot hope to covcr it in a singLechapter. Fortunatel),,an uldcrstanding of only the steady-statesinusoidal behavior of balancedthree-phasecircuits is sufficientfor enginccrswho do not specializein powcr systems. We deline what .rve mean by a balancedcircuit later in the discussion. Thc samecircuit analysistechniquesdiscussedin carlier chapter-scan be applied to either ulbalanced or balancedthree-phasecircuits. FIeLewe use thesc familiar techniquesto developsevcralshortcutsto the analysisof balancedthrcc phasecircuits. Fol economic rcasons, three-phase systems are usually designedto operatein the balalced statc.Thus.in this irllroductory treatrnent,we canjustify consideing only balancedcircuits. The analysisol unbalancedthree-phasecircuits,which you will encounterifvou studyelectricpowe. in later coulses,reliesheavi l ) o n r n u n d ( | . t a n J i nogl b J l c n c ecdi r c u i r . . ffle basicstructureof a three-phasesystemconsistsof voltage sourcesconnectedto loads by meansof translonnersand tlansmissionlil1es. To analyzesucba circuit,we can reduceit to a voltagesou.ceconnectedto a load via a linc.The on1ission oi the transformersimplitiesthe discussionwithout jeopardizinga basic understalding of the calculations involved. Figure i].1 on pagc434 showsa basiccircuit.A dcfining cbalacteristicof a bal anced three-phasccircuit is that it contains a set of balanced three-phasevoltagesat ils source.We begin by consideringthese voltagcs.and then we move to the voltage and currenl relationshipsfor the Y-Y and Y A circuits.After consideringvoltageand current in suchcircuits,we concludewith sectionson power and power rneasutement,

PracticaI Perspective Transnission andDistribution of tlectricPower In this chapterwe introducecircuitsthat are designedto handtelargebtocksof etectrjcpower.Theseare the cjrcujts that areusedto transportelectricpowerfromthe generating ptantsto bothindustrialand residential customers. Weintroducedthe typicalresidentiat customercircuitas usedjn the linitedStatesasthe designperspective in Chapter9. Nowwe introducethe type of circuitusedto deiiveretectdcpowerto an entireresidentiai subdivision. imposedon the desjgnandoper 0ne of the constraints is ation of an etect c utitjty the requirement that the utitjty maintainthe rms voitage[eve[at the custome/sprcmjses. Whethertighttytoaded,as at 3:00 am, or heavityloaded,as at midafternoon on a hot, humidday,the utitity is obtjgated to supptythe samermsvottage.RecatL frofi Chapter10 that a capacjtorcan be thought of as a sourceof magnetizing vottageLeveLs vars.Therefore, onetechniquefor maintaining on a utitity systemis to placecapacitorsat strategic[ocationsin the distributionnetwork.Theideabehindthis techniqueis to use the capacitorsto supptymagfetizjngvars

ctoseto the loadsrequirjng then, as opposed to sending themoverthe linesfrornthe generator. WeshalLitLustrate this conceptafterwe haveintroduced the anatysis of bal:n.a/ rhra.-nh:(o.ir."it(

434

Balanced Thrce-Phde Circuitr

figure11.1a A basicthrce-phase c'rcuit.

Voltages 11.1 Balanced Three-Phase A set of balancedthiee-phasevoltagesconsistsof three sinusoidalvolt agesthat have idenlical amplitudes and frequencies but are out of phase with eachotler by exacdy120'. Standardpmcticeis to refer to the three phasesas a, b, and c, and to use the a-phaseas the referencephase.The tlnee voltages are refened to as the a-phasevoltage, the b-phas€ voltag€, and the c-phasevoltage. Only two possiblephaserelationshipscan exist betweenthe a-phase voltage and the b- and c-phasevoltages.One possibility is for the b-phase voltageto lag the a-phasevoltageby 120', in which casethe c-phasevoltage must lead the a-phasevoltage by 120'. This phaserelationshipis known as the abc (or posifiye) phase s€quence.The only other possibility is lor the b-phasevoltageto lead the a-phasevoltageby 120",in which casethe c-phasevoltagemust lag the a-phasevoltageby 120'.This phase relationship is knowr as the acb (or negative) phase s€quenc€.In phasor nolation,thetwo possiblesetsof balancedphasevoltagesare

Y"- v^1g:, vb = v- / -r20"' vc = u^ / +'t20',

v^= v- 1!1. Yh = u^ / +720",

v" = u^// -120".

FiguE1r.2 A, Phasor diagBms ofa batanced s€tof thee-phase vottases. (a)Iheabc(positivdfquence. (b)rheacb(nesatjve) seqLrence.

\11.?)

EquatioDs 11.1 arc for the abc, or positive, sequence.Equations 11.2 Figure 11.2showsthe phasordiaare for the acb,or negative,sequence. gramsof the voltagesetsin Eqs.11.1and 11.2.Thephasesequenceis the clockwise order of the subscripts around the diagram ftom V.. The fact tlat a thee-phase cftcuit can have one of two phase sequencesmust be taken into account whenever two such circuits operate in panllel. The circuitscan operatein parallelonly il they havethe samephasesequence.

Voliag€ Sou(es 11.2 Three-Phase

Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero. Thus, from either Eqs. 11.1 or Eqs.11.2,

\+vb+Y=0. Becausethe sum of the phasorvoltagesis zero,the sum of the instanta neousvoltagesalsois zero;that is, (11.4)

wi!ding

Now that we know t}le natue of a balancedset of three-phasevolf ageq we can state the filsl of the analyticalshortcutsalluded to in the iniroductionto this chapter:Ifwe know the phasesequenceand one voltagein the set,we know the entire set.Thusfor a balancedthree-phasesystem,we can focuson determiningthe voltage (or curent) in one phase, becauseonce we know one phase quantity, we know the others NOTE: Assessyow understandingof three-phasewltages by trying ChapterProbkms 11.1and 11.2.

Vottage5ources 11.2 Three-Fhase A three-phasevoltage source is a genemtor with tbree separate windings winding distributed around th€ periphery of the stator. Each winding comprises voLtage soLrrce. onephaseof the generator.The rotor oI the generatoris an electomagnet Figure11.3A A skekhofa thrce-phase diven at synchronousspeedby a prime mover,suchasasteamor gasturbine. Rotation of the electrcmagnet induces a sinusoidal voltage in each winding.The phasewindingsare designedso that the sinusoidalvoltages induced in them are equal in amplitude and out of phase witl each other by 120'. The phasewindingsare stationarywith respectto the rctating so the frequencyof the voltageinducedin eachwindingis electromagnet, the same.Figure11.3showsa sketchof a two_polethree-phasesource. Theie arc two ways of interconnecting the separatephasewindings to form a three'phasesource:in either a wl,e (Y) or a delta (A) conliguratior. Figure11.4showsboth,with ideal voltagesourcesusedto model the phasewindingsof the three-phasegenerator.The commonterminalin the /l in Fig.11.4(a),iscalledthe neuhal terminal Y-conrectedsource,labeled of the source.The neutral teminal may or may not be available for exterSometimes,tle impedanceof each phasevrindingis so small (compared with other impedancesin the circuit) that we need not account tbr it in modeling the generator;the model consistssolely of ideal voltage sources,asin Fig. 11.4-However, if the impedance of eachphasewinding is not negligible, we place the whding impedance iII series with an ideal sinusoidal voltage source. All windings on the machhe arc of the same (b) construction, so we assumethe whding impedances to be identical. The ofanideat wjnding impedanceof a tluee-phasegeneratoris hductive Figure 115 rigure11.4A Thetwo basicconn€ctions (a)A Y connected source and thrce-phas€ soutce. showsa model of sucha machine,in which is the windingresistance, (b) A A-connected sourc€. of thewinding ) , i. rheinductivereaclance

436

Batan.ed ThrcePhase Ci(uits

Flglre11.5A A modeLofa thrce-phase soLrfte wiih winding inrpedance: source; and G)a y-connected (b) a A-connectedsource.

Becausethree-phasesourcesand loadscanbe eitherY-connected or A-connected,the basic circuit in Fig. 11.1represents four different configurations: Load

Y

A A A

We begin by analyzing the Y-Y circuit. The remaining tluee arrangements can be rcduced to a Y-Y equivalent circuit, so anal]sis of the Y Y chcuit is the key to solving all balanced three-phase arangements. We then illustrate tie reduction of tle Y-A arrangement atrd leave the analysis of the A-Y and A-A arrangements to you in the Problems.

11.3 Anatysisof the Wye-Wye Circuit Figure 11.6illustrates a geneml Y-Y circuit, in which we included a fourth conductorthat connectsthe sourceneutml to the load neutial,A fourth conductor is possible only in the Y-Y arangement. (More about this later.) For convenience,we tmnsformed the Y connections into ..tipDed o!ertees. ln f-,g.ILo.7sa.7sa.afidZc. represenl lhe inlerDalimped;nce associatedwith each phasewinding of the voltag e genento\ Zh, Z h, nnd Zr" represent the impedance of the Linesconnecting a phase of the source to a phaseofthe load;Zo is the impedanceol the neutral conductorconnectingthe sourceneutral to the load neutral;and.Za, ZB, ard Zc rcpreseDlthe impedance of eachphaseo[ the load.

Circuit 437 11.3 Anatysis oftheWye-Wye

YYsystem figure11.5A Aihte+phase

We can describ€this circuit with a single node-voltageequation. Ushg the source neutral as the refercnce node and letting VN denote the noalevoltagebetweenthe nodesN and n, we fird that the node-voltage equationis

vN za

\ za+ zb+ ze

zR+z1b+Zgr

Zcl

v", Zb* Zs"

This is the general equation for any circuit of the YY configuntion depictedin Fig. 11.6.But we can simplify Eq. 11.5significantlyif we now coNider the formal definition of a balancedthree-phasecircuit. Such a circuit satisfiesthe following criteria: 1. The voltage sourcesform a set of balanced three-phase voltages. In Fig. 11.6,tlis meansthat V, ., Vb., and Vcn are a set of balanced tbree phasevoltages. 2. The impedanceof each phaseof the voltage sourceis the same-In Fig.11.6,this meansthatZsa: Zcb = Zcc. 3. The impedance oI each line (or phase) conductor is the same.In Fig.11.6, thismeansthatZ1u: Zyo: 26 4. The impedanceof eachphaseof the load is the sameln Fig 116, this meansthat ZA = Zs : Zc. There is no restriction on the impedance of a neutal conductor; its value has no effect on whethei the systemis balanced. If the circuit in Fig.11.6is balanced,wemay rewrite Eq.115 as "^(;.;)

va,i+vb.+v."

Z a = Z ^ + Z r ^ + Z E a =Z B + 2 6 l

zb

216: Zc+ Zk+ Zc.'

three-phase for a balanced 4 conditions circuit

438 Batanced Ihrce-Pha5€ Circuitr The dghFhandsideofEq.11.6 is zero,becauseby hypothesisthe numerator is a set of balanced three-phase voltages and Zd is not zero. The only valueof VN that satisfiesEq.11.6 is zero.Therefore,for a balancedthreephasecircuit, Y.r = 0.

(11.j)

Equation 11.7is exhemely importart. If VN is zero, there is no differencein potential betweenthe sourceneuhal, n, and the load neutral,N; consequendy,tlre curent in the neutal conductor is zerc. Hence we may either remove the neutral conductor from a balarced Y-Y configuration (lo = 0l or replaceit with a perfedsbortcircujtberreenrhenodisn and N (\ - 0). Both equivalentsare convenientro usewhen modelingbalanced three-phasecircuits. We now turn to the effect that balanced conditions have on the three line currents.With reference to Fig. 11.6,when the system is balanced, the threeline currentsare V"', - VN

z a + z r a + z c ^ z4 , vuo Yrq

\s:

Zsl

I"c =

Figurc11.7A A singte-phase equjvaLent cjrcuit.

Zlot

zt'

Zs6

Y'" - VN

v"'"

z . + 2 . " + 2 , . " zb

(11.8)

(11.9)

(11.10)

We seethat the three line curents form a balanced set of three-phasecurrents;that is, the current in each line is equal in amplitude and frequency ard is 120' out of phasevrilh the other two line currents.Thus. if we calculate the curent I,A and we know the phase sequence,we have a shortcut for Iinding IbB and lcc. This procedure paraltels the shorcur used to find the b- and c-phasesource voltages from the a-phasesource voltage. We can useEq.11.8 to constructan equivalentcircuil for the a-phase of the balancedY-Y circuit. From this equation, the curent in the a-phase conductor line is simply tl)e voltage generated in the a-phasewinding of the generator divided by t]le total impedance in the a-phaseof the circuit. Thus Eq. 11.8describesthe simplecircuit shownin Fig. 11.7,in which rhe neutral conductor has been replaced by a pedect short circuit. The circuit in Fig. 11.7 is referred to as the singl€-phase equivalent circuit of a balanced three-phasecircuit. Because of the establishedrelationships between phases,otrce we solve this circuit, we can easily write down the voltages and cu(ents in the other two phases.Thus, drawing a singlephase equivalent circuit is an important first step itr analyzing a t]lieephasecircuit. A word of caution here. The current in the neutral conductor in Fig.11.7is IaA,which is not the sameasthe curent in the neutml conduclor o[ tbe bdlanced rbree-pha(e ciJcuit. whichi. Io: I.a + IbB + lcc.

(11.11)

Thus the circuit shown in Fig- 11.7gives tlle correct value of the line current but only the a-phasecomponent of the neutral current. Whenever this

oftheWye-Wye Circuii 439 11.3 Anaiysis single-phaseequivalent circuit is applicable, the line currents form a balancedtluee-phas€set,and the ight-hand sideofEq.11.11 sumsto zero. Oncewe know the line currentin Fig.11.7,calculatioganyvoltagesof interest is relatively simple. Of particular interest is the relatlonship between the line{oljne voltages and the line{o-neutral voltages. We establish this relationship at the load termhals, but our obsenations also apply at the source terminals. The Line{o-line voltages at fie load termi_ nalscan be seenin Fig.11.8.They are VaB,VBc,and VcA,where the double subscript notation indicates a voltage drcp from the first-named node to the second.(Becausewe are interesledin the balancedstate,we have C omitted the neutralconductorfromFig.11.8.) vottages The line{o-neutral voltages are VAN, VBN, and VcN. We can now Figure11.8A Linetoljne andtjne-to-neutmt descibe the line+oline voltages in terms of the line-to-neutral voltages, using Kirchhoff's voltage law: VAB = VAN

VBN,

(lt.r?)

VBc = VBN

vcN,

(11.13)

VcA:VcN-VAN.

(11.14)

To show the relationship between the iine+oline voltages and t]le Ushg the line{o-neutralvoltages,we assumea positive,or abc,sequence. reference, line{o-neutralvoltageofthe a-phaseasthe

var : voA,

(11.15)

YsN: Va/ -720' ,

(11.16)

YcN = u6 / +120',

111.17)

where /d, representsthe magnitude of the line{o-neutral voltage. yields SubstitutingEqs11.15 11.17into Eqs.11.1211.14,respectively,

\;/ts- vo 191- vb/ 120'= \/av6/30",

(11.18)

Ysc= va/ 120'- u,t/120" = '\a3va / -9o',

(11.19)

Yce=Vq, /1s0". /120" V$1!:= \a3vb

\\1.24)

revealthat Equations11.18-11.20 1. The magnitude of the Line+o-line voltage is a5 dmes the magnitude of the line-to-neutralvoltage. 2. The line-toline voltagesform a balancedthree-phaseset of voltages. volt3. The set oI line-to-linevoltagesleadsthe set of Line-to-neutral agesby 30". We leave to you the d€monstration that for a negative sequence,the only changeis that the set ollinetoline voltageslagsthe set of line to-neutral

440

BaLanced lhrce'Phase Cncuitr

voltagesby 30".The phasordiagramsshownin Fig. 11.9summarizetlese observations. Here, again,is a shortcutin the analysisof a balancedsystem: If you know the lirle-to-neurralvoltage at somepoint in the circuit, you can easily determinethe line-toline voltage at the samepoint and

rigurc11.94 Phasordjaqnms showjng the r€taijonshipbetwe€n tineio-tineandtine-to{eutrat voltages in a batanced (a)Theabcsequence. system. (b)Theacb

We now pauseto elaborate on teiminology. Line voltage refers to the voltageacrcssany pair oflines;phssevoltag€refersto the voltageacross a single phase.Line current refers to the curent in a sillgle line; phase cullent refers 10 current in a singlephase.Obse e that in a A connec tion,line voltage and phasevoltage are identical,and in aY connection. line cunent and phasecurrent are identical. Becausethree-phasesystemsare designedto handlelarge blocks of electric power, all voltage and current specilications are given as rms vai ues.Wlen voltage ratings are given, they refer specifically to the rating of the line voltage.Thus when a three-phasetransmissionline is rated at 345kV, the nominal value of the rms line-to-lhe volraseis 345.000v' In fiis chapterwe er?re..all vokagesandcurrenlsar r*t "ufr... Finally,the Greek letter phi (d) is widely used in the literature ro denotea per-phasequantity.Thus Vp,I!r, Zd, Pd, and Od are interpreted as voltage/phase,current/phase,impedance/phase,power/phase,and reactivepower/phase, respectively. Example11.1showshow to usethe observationsmadeso far to solve a balancedthree-phase Y-Y circuit.

Anatyzing a Wye-Wye Circuit A balanced three-phaseY-connected generator wit}l positive sequence has an impedance of 0.2 + j0.5 O/d and an internal voltageof 120y/d. The genemtoi feeds a balanced tbiree-phase Y-connected load having an impedance of 39 + J28O/d. The impedanceof lhe lire connecring the generatorto the load is 0.8 + i1.5 O/d.The a-pha.eintemal\ollageol the generalori, specified asthe referencephasor.

Sotution a) Figure 11.10showstle single-phaseequivalent circuit. b) The a-phaseline cunent is

12019: (0.2+ 0.8+ 39)+ j(0.5+ 1.s+ 28)

= a) Construct the a-phase equivalent circuit of Ine system,

1201y 40+ i30

= 2.4/

36.8'7' A.

b) Calculatethe threeline curents IaA,IbB,and Icc. c) Calculatethe three phasevoltagesat rhe load, VaN, Vs1, and VcN. d) Calculate the line voltages VAB,VBc, and VcA at the termhals of the load.

a, 0.2o /0Jo

a 0.8o j1.5o A 39f)

-

)tzoruvv",

e) Calculatethe phasevoltagesat the terminalsof the generator, V-, Vb,, and Vm. l) Calculate the line voltages Vab,Vbc,and Vcaat the terminalsof the generator. g) Repeat(a) (0 for a negativephasesequence.

i28O N

rigure11.10A ThesjngLe-phas€ €qujvatenr cjrcujr for Example 11.1.

Circuit 441 11.3 AnaLysis of theWye-Wye

For a positivephasesequence, lbB = 2.4/ -'t56.87"A, r.c = 2.4/83.13"A. c) Thephasevoltageat the A terminalof the loadis v^tr = G9 + j28)(2.4/ -36.87' ) = 115./2/ -1.19" V. For a positivephasesequence, vBN= 115.22/ -121.19'v , Vcr = 11522 ,/11881' V tl1eline voltages d) For a positivephasesequence, leadthe phasevoltagosby 30"; ttrus vAB: (\6 /30')vA}r = 199.58/28.81" V,

g) Changing the phase sequence has no ellect on the single-phaseequivalent c cuit.Th€ three lme

l"A:2.4 /-36.87' A,

r6s= 2.4191jl:A, rcc= 2.4l:E5g:4. Thephasevoltagesat the load are v^N : 1L5.22 / 1.19"v, vBN: 115.22 /118.81"V, YcN = 115.22/

the line voltages For a negativephasesequence, by 30': lagthephasevoltages vAB : (\61:lq)vAN

Vnc = 199.58 / 9119'V,

: 199.58 /-31.19'V,

Vcr = 19958,/14881' V' vollage lhesource is al thea lermjnalo[ el Thepha(e van= 120- (0.2+ j0.5)(2.4 / -36.87") = 120 1.29/31.33" : 118.90- j0.67 = 118.90 / 0.32"V. Fora positivephasesequence, / 12032'Y, \." = 118.90 V. v* : 118.90,/119.68" 0 The line voltagesat the sourceterminalsaie v"b = (1,6 /30')v-

121.19"Y

vBc = 199.58 /88.81"V, vcA - 199.58 /-151.19'V. at theteminalsof thegenerThephasevoltages atof are V,. = 118.90 / 0.32"V, Vb. = 118.90 /119'68"V, !6 - 118.90/

12032' V.

The line voltagesat the terminalsof the genera_

v"b: (\4/

3o')v-

- 20s.94 V, / 30.32'

= 205.94 /29.68' V, ybc: 2n5.94/ 90.32'V,

\.

vq - 205.94/ 149.68'V.

va:

= 2fr5.94 /89 68"Y ' 205.94/ 150.32'v.

442 BatanedThree'Phas€ Circuits

objective1-Know how to analyzea balanced,thrce-phasewye-wyecircuit 11.1

'l'he

voltagefromA to N in a balancedtbreephasccircui!is 2,10/ 30' V. 11lhephase sequcnceis positive,what is the valueof VRcl

Answ€r: 4,56el 120." 11,2 The c-phasevoltageof a balancedthree-phase Y conlectedsystemis 450/ 25'V.If the phasesequelceis negatjve.whatis the va]uc ofv^ts? Answet '779.421j! V. 11,3 The phasevoltageal the terminalsot a balancedthree'plras()Y-connectedload is 2400V The load hasan impedanceof 16 + j12 O/., and is fed from a line havingan impcdanceof 0.10 + i0.13{) O/.r. The Y-connectedsourceat the sendingcnd oI lhe line hasa phase

sequenceofacb and an internal impedanceoI 0.02 + /0.16 O/.r. Use rhe a-phasevolrageat the load as thc relerenceand calculatc(a) the

[:,""Tlill."t*"1'+;Jt].:;!r],:i:iTi" internal phasclo'neutral voltagesa! the source.

Va'n.x',,. and Vcn. Answer: (a) I"A = 120/-3687' A, IDB= 120/83.13" A. and rcc : 120/ 156.87' A; (b) V,b = 4275.02/ 21J.38"V, Ir''".= 1n5.02 /9162' y, ar'd Vca= 1275.02/ 14838' Vi (c) Y,,. - 2182.05L):g V, Vb.,,- 2482.05/121.93' V, and V.,. - 2182.05/ 118.07' V_

NOT E: ALU)tn ChaptetPnhLens Il.8 11.10.

11.4 A*eNy*ris clf:igt*';tliy*-ileLi:a Cireui.i U the load in a thrcc phasecircuit is conncctedit a detra,ir can bc rrans iormed into a wyc by usingthe delta-tow]'e translormationdiscusscdin Seclion9.6.Whcn the load is balanced.ihc impedanceof eachlcg ol lhe wl'e is one third thc iilpedance of eachlcg ol the delta.of

Retationship between three-phase detta-connected andwye-connected r,' imPedance

rigurc11.11., A single?hdeequivatent cncuit.

-

z! , : ]

(11.21)

which follows direclly fiom Eqs 9-51 9.53.After the I ioad has been replacedb,r'its Y equivalent,the a-plrasccan be modeled bl' rhe single, phaseequivalcntcilcuit shoiJnin Fig-11.11. We usc this circuit to calculatcthe line currents,and wc then usethe tine c flcnts lo find the currcntsin eac]rle.qof the orignralI load.The relationshipbelweenthe line currcntsaDdthe currentsin cachleg of the delta cai be derivedusingthc circuit shownin Fig.11.12. Whcn a load (orsourcc)is conrectedin a delta.thc currerrineach leg of ihc dclla is tire phasecurrcnl, and rhe voltage acrosseach leg is rhe phasevoltage.Figure 11.12showsthal, in rhe I configuralion.fie phase voltagcis identicalto the linc vollage.

Cncuit 443 11.4 Anatysis ofih€Wye-Detta To demonstrate the relationship between t]le phase currents and Line curents, we assume a positive phase sequence and let 1d represent ttre masnitudeof the Dhasecurrent.Then

les : I o19:,

111.22)

rec= Io 1-lT,

(11.23)

rcl'= Ia1lT.

\11.24)

In wriling these equations, we arbitrarily selected IAB as the reference pllasor, we can write the line currents in terms of the phase currents by direct applicationof Khchhoff'scunent law:

used to estabtish the Figure 11.t2a A circuit Ljne cunents andphase curr€ntsin rctationship between

Iua-Ias-Io\

: Io 1!l - Io1)4) - ',Bto 1 !,

(11.25)

IbB:IBc-IAB

: lo / 120'- 1,r19: = \nI6 / -15o", tc:

111.26)

Ica Inc = I+ 1&l

lao/ 12o"

= .\trto12Y.

\11.21)

revealsthat the magniCornparingEqs.11.2511.27with Eqs.11.22-11.24 the magnitude of t]le phase currents tude of the line curents is \4 times and that the set of line currentslagsthe set ofphasecurrentsby 30". We leaveto you to verify that,for a negativephasesequence,theline curents are 1/3 times larger than the phase currents and lead t}le phase (b) cunenls by 30'. Thus,we have a shoitcut for calculating line curents from (or A-connected phasecuirents vice versa) for a balancedthree-phase diagrams shov/ing the Figure11.13A Phasor load. Figure 11.13summarizesthis shotcut graphicaly.Examp]e 11.2 relationship tinecurrents andphase curcntsin b€tween (b)The illustiates the calculations involved in analyzing a balanced three phase a A-connected sequence. toad.G)Thepositjve load. circuithavhg a Y-connectedsourceand a A-connected

444

Batanced Thr€€-Phase Circuits

Anatyzing a Wye-Detta Circuit The Y'connectedsourcein Example 11.1 feeds a A-connected load through a distribution line having an impedance of 0.3 + i0.9 O/d. The load impedanceis 118.5+ i85.8 O/d. Use the a-phase htemal voltage of the generator as the reference.

39.50

a) Consfuct a single-phaseequivalenr circuit of the three-phase system. b) Calculalethe line currentsIaA,IbB,and Icc. c) Calculatethe phasevoltagesat the load terminals. d) Calculate the phase currents of the load. e) Calculatetle line voltagesat the soruceterminals.

Solution a) Figurc 11.14showsthe sitrgle-phase equivalent circuit.Theloadimpedanceof theY equivalentis 118.5+ 185.8

= 39.s+ j28.60/0.

b) The a-phaseLinecurent is

1201l: r,_r: (0.2+ 0.3+ 39.s)+ i(0.5+ 0.9+ 28.6) 120 = -; .:/0' = 2.4/ 4u + lJu

36.87"A

Hence lB = 2.4/-156.87' A, t c:2.4 /83.13"A. c) Becausethe load is A connected,the phasevoltages are the same as the line voltages To calcu, late the line voltages,we first calculate VAN. vAr,{= (39.5+ p8.6)(2.4 / 36.87') = 117.04/ 0.96" y. Becausethe phasesequenceis positivg the line voltage VABis

P8i A

N

N

Figsr€11.14l. Thesingle-phase equivalent circujtfor Example 11.2. d) The phasecurrents of the load may be calculated directly from the line currents: / 1 \

r ^ a = l - , -//3 0 ' l t .,' " \vJ = 1.39/ 6.87'A.

Oncewe know IAB,we alsoknow tlle other load pnaseculrenls: lec = 1.39/ 126.8'7' A, lc^ - 1.39/113.13'A. Note that we cancheckthe calculationof IaB by using the previously calculated VAB and the impedanceof the A-comectedload;that is, 202.72/29.04" \aa _ 26 118.5+ /35.8 = 1.39/ 6.n' A. e) To calculatethe litre voltageat tte terminalsof the souce, we first calculateVm. Figure 11.14 showstlat Vr is the voltagedrop acrossthe line impedanceplustlle load impedance,so Yn : G9.8+ j2e.s)(2.4 / -36.87') : 118.90 /-032'y. Theline voltageVabis

v"b: (15 /30")v"",

vAB : (\6 /30') vAN

= 202;72 /29.04"v.

v,b:205.94 V. /29.68"

\Bc:202.72 / 90.96"y, \tcA : 202;72/149.04"V.

Therefore \r" = 205.94 /-90.32' Y, v.": 205.94/149.68"V.

Thereforc

l1.r

. B.,, "d \ "e +.,e t' . i

P o t r e( .r. . t . c .o

445

wye-delta conne
11.6 The line voltagcVAr at the terminalsoI a balancedthree-phaseA-connectedload is 4160A V. Thc line current I"A is

6e.28 1 )!: A.

Answer: 13.86/

a) Calculatethe per-phaseimpedanceof the load if ihe phasesequenceis positive. b) Rep€at(a) for a negativephasescquence.

45- A-

11.5 A balancedthree-phase,\-connecledload js fed from a balancedthree-phasecircoil.The referenceibr thc b'phaseline cullenl is tot'arcl the load.Thevaluc of the currentir thE b-phas< i. ' '_lq A. Tfthe pltu'..
Answer: (a) L041 lLttl

(b) 104lllq

o.

11.7 The line voltageal lhc tcrmjnalsofa balanced A-connected load is ll0V Eachphaseof the load consistsofa3.667 O resistorin pamllel\rilh is the l1]ag a 2.75() inductiveimpcdance.What nit de of the ounenl in thc lioe feedingthe load?

85' A.

NOTE: AIso tty ChapterProblems11.11-l1.16.

in S*la*e*d 11.5 F*rvsr{t*i*uiatisnr {'ireuits Thrse-Phase So lar. we have limited our analvsisof balancedthree phasccitcuits to dclcrmining curfents aDd voltages.we no\\, discussthree phasc po$'er We begin by considcringthe averagepower delivcrcd to a calculations. balancedY connectedload-

AveragePowerin a BatancedWyeLoad Figurc11.15showsaY conncctcdload,alongwith its perlinentcurrentsand with aDyone phascbi' vollagcs.Wc calculatethe averagcpoNcr associated ulirg thc tcchniquesintroducedin Chapter10.WithEq. 10.21as a starung \lith the a phaseas poinl,wc cxpressthe averagepowcr associated PA :

v,\\ llaAlcos('?,A

diJ,

r,:

A

(11.28)

whcrc 0rA and PiAdenole the phascanglesof VANaDdI,,^, rcspcctively Using llrc notation introducedin Eq. 11.28,we can find the powcr associatedwilh thc b- and c-phases: (d"B ,iB); Pn = XrNllbulcos P. = LVCN ll.c cos(d"c

,ic).

(tt.?9)

Y toadusedto introduce Fig!re11.15,i A batanced power in thre+plrase circuits. avsage caLcutatioB (11.30)

are \rrillenin icrms currcntsandvoltages In Eqs.11.2811.30,a1lphasor oI lhe Ims vahe of the siDusoidalfunctionthey represent.

446 Eatanced Three-Phase Circuits In a balancedth ree-phasesystem,the magnitude of eachline-to-neutral voltage is the same,as is the magnitude of each phase current. The argument of the cosine functions is also the same fot all three phases-We emphasizetheseobsenationsbyintroducing the following notation: : VrNl = YcNl, Yp= V,111l

(11.31)

1{= IoAl- IbBl:

\11.32)

Iccl,

and

0a: AA

qiA = 0B

eiB = 0,c - qic

(11.33)

Mor€over,for a balancedsystem,the power deliveredto eachphaseof th€ load is the same,so PA: PB = Pc = Pi - Uil6cosei,

(17.34)

whereP4 representsthe aveiagepower per phase. The total averagepower deliveredto the balancedY-connectedload is simplythreetimesthe power per phase,or ga. Pr : 3P,b: 3U6I,bcos

(11.35)

Expressingthe total power in termsof the rms magnitudesof the line volt, age and cment is also desirable.If we let yL and 1L representthe rms magnitudesof the lire voltage and current,iespectively,we can modify Eq. 11.35asfollows:

Tota[reatpowerin a baLanc€d three-phase loadF

r, =z(ft)t*"e, (11.36)

In deriving Eq. 11.36,we recognizedthat, for a balancedY-connected load,the magnitudeof the phasevoltageis the magnitudeofthe line voltage divided by \6, and that the magnitude of the line current is equal ro the magnitudeofthe phasecurrenl.WhenusingEq.11.36to calculatethe total power deliveredto the load, rememberthat 9d is the phaseangle betweenthe phasevoltageand current.

ComplexPowerin a BalancedWyeLoad We can also calculatethe reactivepower and complexpower associated with any one phaseof a Y-connectedload by usingthe techniquesintroducedin Chapter10.For a balancedload,the expressions for the reactive

TotaLreactivepowerin a batanced three-phaseload ts

Qt: Uil$lr'sd,

\11.37)

Qr=3Qo=\f3VLILsin,i.

(11.38)

11.5 Power CatcuLatiom in Batanced Three-Phase Cncuits 447 Equation 10.29is the basisfor expressingthe complexpower associated with anyphase.For a balancedload, Sd = VaN{A : VBN4B = Vo,fic : VdU,

(11.39)

where Yd and I4 repr€sent a phase voltage and curent taken from the samephase.Thus, in general, \11.40)

(11.41) < Totatcomptex powerin a batanced threephaseload

PowerCatculations in a Balanced DeltaLoad If the load is A -connected,the calculation of power-reactive or complex is basicallythe sameas that for aY-connectedload.Figure 11.16showsa A-connected load, along with its pertinent currents and voltages"The Dower associatedwith each Dhaseis Pa = lvABlIABlcos(d,AB- daB),

111.12)

PB = lvBcllBc cos(d,s6 - d;6d,

\r1.43)

Pc : lvcAlllcAlcos(dvcA oicr.

(1144) rigur"rl.ro r I a-connected toadused to discuss power catcuLatjons.

Foi a balancedload,

lVag:lYsc:lVcr:V6

(11.45)

lle-e = llecl = llcc =10,

(11.16)

0,as-oies:d,sc-diBc

= 0&^

9ic^ - 00,

PA = PB = Pc = P6=Uol6cos0o

B r,,^1

(17.47)

(11.48)

Note that Eq. 11.48is the same as Eq. 11.34.Thus,itr a balalced load, regardlessof whether it is Y- oi A -connected,the aveiage power per phase is equalto the productof the rms magdtude of the phasevoltagg the rms magnitudeoI the phasecwent, and the cosineof the anglebetweent]le Dhasevoltaseand current,

448

Balanced Ihree-Phde Ci,cuits

The lotalpoqerdelivered ro a balanced A-connecred loadrs Pr = 3Pq = 3UoIacos06

=-(*)*"*

: \6yLlLcosd+.

(11.49)

Nore that Eq. 11.49is the sameas Eq.11.36.Theexpressions for reactive power and complex power also have the samefolm asthose developed for the Y load:

sitt0q; Q4,= V,pI,p

(11.50)

Qr : 3Qo: 3U,bI$ine6.

(11.51)

Sd=Pa+jO{=Va4;

\11.52)

sr=3sc=\a3vLILA4.

(11.53)

Instantaneous Powerin Three-Phase Circuits Although we are primarily interestedin average,rcactive,and complex power calculationq the computation of the total instantaneous power is alsoimportant.In a balancedthree-phasecircuit,this power hasar interesting propefiy: It is invariant vrith time!Thus the torque developed at t]le shaft of a three-phasemotor is constant,which in turn meanslessvibration in machinerypoweredby three phasemotors. Let the instantaneousline to-neutral voltage ?rANbe the reference, and,as before,dd is the phaseangle0,A - 9jA.Then,for a positivephase sequence,the instantaneouspower in each phaseis ?A = oANiuA= U-I ^cos ar cos@t

ed),

pB = oBNibB: U^I-cos(ot - 120') cos(ot

d{

pc = ocNi"c= U^I^cos(at + 120')cos(@t

dd + 120"),

120"),

where y- and 1- representthe maximum amplitudeof the phasevoltage and line currert, respectively.The total instantaneous powei is the sum of phasepowerqwhich rcducesto 1.5y-1-cosdd; that is, the instantaneous pr:

pA+ pB+ pc:7.5VhI-cos06.

11.5 Power CaLcutations in Batanced Three-Phase Cncuits 449 Note this result is consistent with Eq. 11.35 s:f;lce V^= \/=2Vo and 1- = \41d (seeProblem11.29). Examples11.3 11.5illustmte power calculationsin balancedthreephasecircuits.

Calculating Power in a Three-Phase Wye-Wye Circuit a) Calculatethe averagepower pei phasedelivered to the Y'connectedload of Example11.1. b) Calculate the total average power delivered to the load. c) Calculatetle totalaveragepowerlostin the line. d) Calculate the total averagepower lost in the generalor, e) Calculate the total number of magnetizing vars abso$ed by the load. f) Calculatethe total complexpower deliveredby

c) Thetotalpowerlostin thelineis = 13.824 Prne- 3(2.4F(0.8) w. d) Thetotalinternalpowerlostin thegenerator is = 3.4s6w. Ps- = 3(2.4)r(0.2) varsabsorbed e) Thetotal numberof magnetizing by the load is sin3s.68" o? = \6(199.s8X2.4) = 483.84 VAR.

50lution

f) The total complex power associatedwith the

a) FromExample11.1,y6 = 115.22y, 16 = 2.4 A, andgd = 1.19 (-36.87): 35.68'.Therefoie Pa = (11s.22)(2.4) cos35.68" :224.64W. The powerper phasemay alsobe calculated from I'aRd,or Po: Q.q' Ge):224.64w. b) The total aveiagepowerdeliveredto the load is Pr = 3Pa = 673.92W. We calculatedttre line voltagein Example11.1,so we may alsouse Eq.i1.36: Pr : v3(199.58)(2.4) cos35.68' :673.92W.

sr = 3s{ = -3(120)(21)/3687' = -691.20- j518.40VA. Theminussignindicates thattheintemalpower andmagnetizingreactivepowerare beingdeliveredto t}le circuit.Wecheckthis resultby calculatingthe total ard reactivepowerabsorbed by the circuit: P : 673.92+ 13.824+ 3.456 : 691.20w(check), I : 483.84+ 3(2.4f(1.s)+ 3(2.4f(0.5) = 483.84+ 25.92+ 8.64 : 518.40 VAR(check).

450

Three-Phase Circuits Baknced

Wye-Delta Circuit Power in a Three-Phase Calculating a) Calculate the total complex power delivered to the A-connectedload of Example11.2. b) Wlat percentage of the avemge power at the sending end of tle Lineis delivered to the load?

Solution

b) The total power at the sendhg end of t]le distribution Iine equals the total power deliveied to the load plus the total power lost in the line; therefore p",,*:

a) Using the a-phase values from the solution of Example11.2,weobtain

682.56+ 3(L4\,(0.3) = 687.74W.

Yo = \as:20272

/29O4'Y' -6.8'7" : : A. t4 IAB 1.39/

Using Eqs.11.52and 11.53,wehave

sr : 3(N2;72 /2e.M")(1..3e 159:) = 682.56+ j494.21Y4.

The percentageof the avemgepower reaching or 99.25ol".Nearly the load is 682.561687.74, power the average at the input is 100'/. of impedanceof to the Ioad because the delivered quite to the load line is small compared the impedance.

Load Three-Phase Power with an Unspecified Catculating A balanced tlree-phase load requfues480 kW at a lagging power factoi of 0.8. The load is fed fmm a lhe having an impedance of 0.005 + j0.025 A/0. The line voltage at the tem)inals of the load is 600V a) Consrruct a single-phase equivalent chcuit of the system. b) Calculate tle magnitude of the line current. c) Calculate the magnitude of the Line voltage at t]le sending end of the line. d) Calculate the power factor at the sending end of the lille.

Solution

/6on\ (=lt;=(160+j120)1f.

t:^ = 5Tt.35/36.87' A. Therefore,I.{= 577.35,z 36.87"A. Themagnitudeof the Linecurent is the magnitudeof IaA: A. IL- 577.35 We obtain an altemative solution for 1L from the expression Pr = \f3vLlLcos 0p

a) Figure 11.17 shows tle single-phase equivalent circuit. We arbitmrily selectedthe line-fo'neutral voltage at the load as the reference,

0.005 o

b) Theline currentIh is givenby

p.m5(l

: \€(600)lL(0.8) : 480,000 w; 480,000

160kW at 0.8lag

16(600x0.8) 1000 .J3

circuiitor Hgure11.17A Thesingt€-phase equivatent Exampte 11.5.

: 5'17 .35A.

11.5 Power Cahutations in Batanced IhreePhase Cncuits 451 c) Ib calculatcthc magniludeof the line voltagcat the sending cnd. we firsl calculate Vxf Fron Fig.11.17,

s : (160 + jl20)10r + (577.35t(0.005+ j0.025)

V""=Va1;i21I.,1 600

10.rrn5 ,0 02<){s-7r\ /

An alternatjvemethodlor calculatingthe power factor is to first calculatethe complexpower at the sendingend of the line:

16r- )

= 16r.67+ j128.33kVA = 206.41 /38.41'kVA.

= 35'7.st 1lA:v. Thus

The po$'crfactor'is pf = cos38.44'

tJ_: v3I," = 619.23 V. d) Thepowcrlaclorat lhe sendingendof the line is ihc cosineoI the phaseanglebetr,cenVrn andI.^: pf: cos[1.s7' ( 36.87')l

= 0.783lagghg. Fina y, if we calculatethe total complexpower al the sending end, after first calculatingthe magniludeof the lire current,we may use this valueLocalculateYL.'fhatis. \a3YL1L= 3(206.4r)x 10r,

= cos311.44'

3(206.4r ) x 10r \6(577.35)

= 0.783lagging.

V. 619.23

0bjective3-Be abteto catculate power(average. reactive, andcomptex) in .ny three-phas€ circuit 11.8 The lhree-phase averagepowerratingof the central processirg udt (CPU) on a mainframe digiialcomputeris 22,659WIhe three-phase line supplying the con1puterhasa line voltage rating of 20BV (ms). The line curent is 73.BA (rms). The computer absorbsmagnetizirg VARS. a) Calculatethe total magnetizingreaclive power absorbedby the CPU b) Calculatethe powerfactor.

11,9 The complexpower associated with eachphase ofa balancedload is 384 + l28B kVA. The line voltageat the terminalsofthe load is 4160V a) w}Iat is the magnitudeof the line currenf feedingthe load? b) The load is delta connected,and the impedance of each phase consistsof a resistancein pa.allel with a reactance.Calculate R and -L. c) The load is wye connected.and the irl1pedanceof eachphaseconsistsofa resistancein serieswith a reactance.CalculateR and-Y. Answerr (a) 199.95A:

Answer: (a) 13,909.50 VAR: (b) 0.852lagging. NOTE: Also tryCh.tptetPrcbtems I1.24and 11.25.

(b)R - 4s.07o,x = 60.09o; ( c )R : 9 . 6 O 1 ,x : 7 . 2 1 O .

Ihrce-Phase circuits 452 Batanced

Power 11.6 Measuring Average in Three-Phase Circuits

rerminals

The basjcinstrumentusedto measurepower in three-phasecircuitsis the el€ctrod].namometer wattmeter.It containstwo coils-One coil, calledthe currentcoil,is stationaryand is designedto carry a cunent proportionalto the load current.The secondcoil, calledthe potentialcoil, is movableand carriesa currentprcportional to the load voltage.The impo ant features ofthe wattmeterare sho\r'I)in Fig.11.18. The average deflection of the pointer attached to the movable coil is proportional to tle product of the effective value of the current in the currcnt coil, the effective value of the voltage impressedon the potentral coil, and the coshe of the phase angle between the voltage and current. The direction ilr which t}re pohter deflects dependson the instantaneouspolarity of the currentcoil current and the potential-coil voltage.Therefore each coil has one terminal with a polarity mark usually a plus sign but sometimes t]le double polarily maik t is used.The wattneter deflects upscale when (1) the polaity-marked termhal of the current coil is toward the source,and (2) the poladty-marked teminal of tle potential coil is con nected to the sameline in which the current coil has been inserted,

Pointer

figure11.18a. Thekeyfeatures ofthe ehctrcdynamometer wattmeter.

TheTwo-Wattmeter Method

poweris Figure 11.19A A genentcircuii whose supptjed by, conductor. I""

z , t =z 8

\\r

ia Fisure r1.20A A circuit used to anatyze the power method two-wattmeter of measuring average d€tivercd to a batanced Load.

Considei a generalnetwork hside a box to which power is suppliedby r conductinglines.Sucha systemis shownin Fig.11.19. If we wish to measure the total power at the terminals of the box, we need to know n - 1 currents and voltages. This follows because if we choose one teminal as a reference, ttrere are only n - 1 independent voltages.Lilewise, only n - 1 independent curents can exist in the n conducto$ entering the box-Thus the total power is the sum of n - 1 product terms;thatis,p : 1)i1 + azi2+ .. + a" i" 1. Applying this general observation, that for a threeconductorcircuit,whetherbalancedor not, we need only two wattmete$ to measurethe total power.For a four-conductorcircuit, we need three wattmeters if the three-phase circuit is unbalanced, but only two wattmetersifit is balanced,becausein the latter casethereis no currentin the neutml lhe. Thus, only two wattmeters are needed to measure the total averagepower ill any balanced thrce phase system. The two-wattmeter method reduces to detemining the magnitude and algebraic sign of the average powor indicated by each wattmeter. we can describethe basicproblem in terms of the circuit shownin Fig.11.20, wh€re the two wattmetersare indicatedby the shadedboxesand labeled Iry1and W2.The coil notarions cc and pc stand for curent coil and potential coil, rcspectively. We have elected to insert the curent coils of the wattmeters in lines aA and cC Thus, line bB is the rcference line for the two potential coils.The load is connected as a wye, and the per-phaseload impedanceis designated^s26= Zl4!.Tlis s ^ generalrcpresentation, as any A-connected load can be repiesented by its Y equivalenl fu hermoie, for the balanced case,the impedance angle 0 is unafrected by the A-to-Y transformation.

11.6 l4easudng Average Powerin Three-Phase Circuits We now develop general equations for the rcadings of the two wattmetef. We assumethat the curent drawn by the potential coil of the wattmeter is neglgible compaied with the line current measuredby the current coil. We further assumethat tlle loads canbe modeled by passivecircuit elementsso that the phaseangle of the load impedance(d in Fig. 11.20)lies between 90" (pure capacitance) and +90' (pure inductance). Finally,we assumea Positive phasesequence. From our introductory discussion ol the avetage deflection of the wattmeter, we can see that wattmeter 1 will resDond to the Droduct of lvasl. I a. andrhecosineot theanglebetweenVABaDdl,A. ll we denole this wattneter reading as Wl, we can wite W1 :

VABIIa\ cos 01

- I/111cosd1,

(11.54)

It followsthat

IV, : VcBltcl cosd, : VLILcosq2.

(11.55)

In Eq. 11.54,91is tlre phaseanglebetweenVABard l"A, ard in Eq- 11.55, 02is tlte phaseangle between VcB and lcc. To calculate Wr and Wr, we express 0t and d2in terms oI tle impedance angle 0, which is also tlle same as the phase angle between the phase voltageand cuffent.For a positivephasesequerce,

e r - 0 + 3 0 "= 0 0 + 3 0 ' ,

(11.56)

02=0 30":00-30'.

(11.57)

The derivationof Eqs. 11.56and 11.57is left as an exercise(see hoblem 11.32). wllen we substitute Eqs.11.56and 11.57into Eqs.11.54 and11.55, respectively, we get w\=uLILcos(0,b+30'),

(11.58)

try2= VLILcos(0,, - 30').

(11.59)

To find the total power, we add W1and W2;thus Pr : Wr + W2 = 2YLILcosd{ cos30" : \a3YL1Lcosd!r,

(11.60)

which is the expressionfoi the total power in a three-phasecircuit. Therefore we have confimed that the sum of the two wattmeter readings yieldsthe lotal averagepowei.

454

Batanced lh€e'Phase Cncuits A closerlook at Eqs.11.58and 11.59revealsthe following about the readingsof the two wattmeteN: 1. ff the powerfactoris greaterthan0.5,both wattmetersreadpositive. 2. Ifthe powerfactor equals0.5,onewattmeterreadszero. 3. If the power factor is lessthan 0.5,one wattmeterreadsnegative. 1. Revening the phasesequencewill interchangethe readingson the two wattmeters. These obse ations are illustrated in the following example and in Problems11.,13F11.51.

Wattmeter in Three-Phase Computing Readings tircuits Calculatethe readingof eachwattmeterin the circuit in Fig. 11.20if the phasevoltageat the load is 120v and (a) zd) = B + j6 o; (b) zd : 8 j6 O; (c) Zd : 5 + j5V3 o; ar,d (.d)Z$: 10 1_ll:4. (c) VerjJy for (a) (d) thar th.. sum of the wattmeter rcadingsequalsthe total power delivercd to t}le load.

c) zo = s(I + j\5) - 101!!

o, vL : 120\6v,

and IL : 12 A.

wr = (i20vr(12) cos(60' + 30') : 0, w, = (120\aO(12) cos(60' 30') : 2160W.

d) zb = 101 lt:a'vL IL=124.

Solution a) z4 : 10/36.8'7'A,VL = 120\6V, and IL - 120/10- 12A. tv1: (120\4)(12)cos(36.87'+ 30") :979;75W, r, : (120\6)(12)cos(36.87'- 30') - 2416.25 W. b t l , ! - l 0 - 3 0 . 8 70 . l r - l 2 0 v l v . x n d IL : 120/10: 12A. u/1: (120\6)(12)cos(-36.87"+ 30') = 24'76.25 W,

= 120\4v, and

lll

= (120\6X12)cos( 75'

w2 = (120\6X12)cos( 75"

w, 30'): 1763.63 30'):

W. 645.53

e) Pr(a) = 3(12)'(8)= 34s6w. w + t4r2= 979.75+ 24'76.25 = 3456W, Pr(b) = Pr(a) : 3456W w1 + w2 = 24'76.25+ 979.75 = 3456W Pr(c) : 3(12),(s)= 2160w, wr+w2-0+2160 : 2160W, : 1118.10w, Pr(d): 3(12f(2.5882)

w, = (120\4)(12)cos(36.B7' 30') = 979.75 W.

wt + w2 = 1163.63 645.53 = 1118.10W.

your anderstading of the tuo-trattmetermethodby ttying ChapterPrcblems11.41and11.42. NOTE: Assess

PracticatPe6pectiv€ 455

PracticaI Perspective Transmission andDistribution of Etectric Power At thestartof this chapter we pointedout the obLigation utiLities haveto maintajn premises. the rmsvoLtage levelat theircustome/s Atthough the acceptabte deviation froma nominal [eve[mayvaryamong different utititjes we witl assume for purposes of discussion that an attowabte toterance is 15.8%.Thusa nornlnal rmsvottage of 120V couLd range frcm113to pointed 127V. WeaLso outthatcapacjtors strategjcalLy located onthesystemcoutdbeusedto support vottage levels. Thecjrcuitshown in Fiq.11.21represents a substation ona Midwestern municipal systen.Wewitl assume the system is batanced, the [ine-to{ine vottage at thesubstation is 13.8kV, the phase impedance ofthe djstribution tineis 0.6 + J4.8O,andthe toadat the substation at 3 PMon a hot, humiddayin Jutyis 3.6MW and3.6maqnetizing MVAR. lJsjng the Line-to-neutral vottage at the substation asa reference, the sjngte-phase equivalent circuitfor the system in Fiq.11.21js shownin figurc11.21.{ A substation connecied to a power Fig.11.22.TheUnecurr€ntcanbe calcutated fromthe expression for the ptantviaa thr€€-phase tine. powerat thesubstation. compLex Thus,

(r.2+ j1.2)106.

#$rtIt fotiows that

{-,, = 1s0.61+ j150.61A IaA= 150.61- 1150.61 A.

n

rigure11.22A A singte phase equivaLent circuit for thesyst€m in Fig.11.21.

ptantis The[ine-to-neutraL voLtage at thegenerating V"-

t -l R . -O: n VJ

-1 0

r 0 . 6- 1 4 . 8 x l 5 0 . o L il50.ol)

= 8780.74 + j632.58 = 8803.50 lllq

v.

Therefore plantis the magnitude ofthe linevoLtage at thegenerating v,bl = \6(8803.50)= 15,248.11v. W€areassuming theutiLityis required to keepthevoLtage Levet within + 5.8o/o of thenominal vaLue. Thismeans themagnitude ofthe Line-toiine voltageat the powerptantshoutdnot exceed 14.6kV nor be lessthan 13kV. Therefore, themagnitude ptant ofthe linevoLtage at thegenerating probLems coutdcause for customers. When the magnetizing varsaresuppfied bya capacitor bankconnected to thesubstation bus,thelinecurrent IiA becomes I"a = 150.61+ J0A.

N

456

Eatanced Ihree-Phase cncLrits

ptantnecessary Therefore the vottageat the generating to maintaina [ine-to-tine vottageof 13,800 V at the substatioris 1J-800 v, 1 4 . 8 ) f l 5 0 -. /60t ) \3 '/0"-(0.6= 805'7 .8O+ j'722.94

: 80eo.r'7 1:13:v. Hence v. lv"bl : \6(8090.17) : 14,012.58 lhis voLtage tevethLtswithin the aLLowabLe langeof 13 kv to 14.6kV. your undetstonding Njft: Assess of this PtocticalPerspective by tryingAoptel Prcblems 11.52(a)-(b) dnd11.53-11.55.

Sunrmary Wlen analyzing balarced thee-phase circuits, the fust stepis to transform any A connectionshto Y connections, so that the overall cicuit is of t}te Y-Y configuration. (See page436.) A single-phaseequivalent circuit is used to calculate the line curent and the phasevoltagein one phaseof the Y Y structure. The a-phase is normally chosen for this purpose.(Seepage438.) Once we know the line curent and phasevoltage in the a-phaseequivalentcircuit,we cantake anallaicalshorf cutsto find any current or volragein a balancedthreephasecircuit,basedon the followingfacts: . The b- and c-phasecurents and voltagesare identical to tle a-phasecurrent and voltage exceptfor a 120'shift h phase.Ina positive-sequence circuit.the b-phasequartity lags tlle a-phasequantity by 120", and the c-phasequantity leadsthe a-phasequantity by 120".For a negativesequencecircuit,phasesb and c are interchangedwith respectto phasea. . The set of line voltagesis out of phasc with the set of Fha\e\ollagesh) LJ0".Theplucor mtnu.,igncorre spondsto positive and negativesequencgrespectively. . In a Y-Y circuit the magnitudeof a line vollage is \6 timesthe magnitudeof a phasevoltage.

. The set of line currentsis out of phasewith the set of phasecurents in A-connectedsourcesand loadsby +30".The minusor plus sign correspondsto positive and negative sequence,respectively. . The magnitudeof a line currentis \4 timesthe mag nitude of a phasecurent in a A connectedsource or load, (Seepages439 and 443.) The techniques for calculating per-phase average power, rcactive power, and complex power are identical to thoseintroducedin Chapter10.(Seepage445.) The total real, reactivg and complex powei can be determined either by multiplying the conesponding per phase quantity by 3 or by using tlle expressionsbased on line currentandline voltage,asgivenby Eqs 11.36,11.38, and 11.41.(Seepages146and447.) powerin a balancedthree-phase The total instantaneous circuit is constant and equals 1.5 times the average powerper phase.(Seepage448.) A wattmetermeasuresthe avemgepower deliveredtoa load by using a current coil connectedin serieswith the load and a potential coil connectedin parallel wirh rhe load. (see page,l52.) The total averagepower in a balancedthree-phasecircuit can be measuredby surnmingthe readingsof two wattmetersconnectedin two different phasesoI the circuit.(Seepage452.)

Probhms457

Problems All phasor voltagesin th€ folowi|rg Probl€ms ar€ stated in t€rmsofthe rms value,

e) ?,, : 339cos(dt + 30") V, ?)b: 339cos(4,1- 90') V,

Section11.1

0c = 393cos(ot + 240') V.

11.1 What is the phasesequenceof eachof the following setsof voltages? a) 1]a= 120cos(o, + 54') V, t)b : 120cos(or

66') V,

I) r'" = 3394sin((,t + 70') V, t,b : 3394cos((rr - 140")V, lJ" : 3394cos(.rt + 180")V.

?]": 120cos(.,,1+ 174")V. b) 1]i : 3240cos(ot

26') V,

ll,3 Vedfy that Eq. 11.3is true for eitler Eq. 11.1 or Eq.11.2.

,b : 3240cos(ol + 94") V, tr. = 3240cos(rt - 146') V. Section11.2 11.2 For eachset of voltageEstatewhetler or not t]te voltagesform a balancedthrce-phaseset.ff the setis balanced,statewhether the phasesequenc€is positive or negative.If the set is not balanc€d,erylain why.

11.4 Reler to the circuitin Fig.11.5(b).Assumethat there are no erlemal comections to the terminals a,b,c. Assume furtler that t}le three windings are from a balancedthree,phase generator. How much currenr will circulate in the A-connected genemtor?

a) ?a : 339cos37?l V, ?)b= 339cos(37?r

120') V,

t'. = 339cos(377r + 120")V. b) ," : 622sin377rV, Db: 622sn(3'7'7t- 240')v, D.: Ozsln(311t + 240')Y. c) ?]a= 933sin377rV, 0b = 933sin(377r + 240') V, 0" = 933cos(377t + 30') V. d) ?,! : 170sin(,))t+ 60') V, 0b = 170sin (or + 180') V, t'c = 170cos(.rr - 150")V.

Section11.3 11.5 a) Is the circuit in Fig. P11.5a balancedor unbalancedthree-phasesystem?Explain. b) Find I,. Figure P11.5

458

Eatanced Three-Phase Circuits

11.6 a) Find I, in the circuit in Fig.P11.6. 6a€ b) Find vAN. c) Find VAB. d) Is the circuit a balancedor unbalancedthreephasesystem? 11.7 Find the Ims value of L in the unbalanced three 6dc phasecircuit seenin Fig.P11.7. 1L8 The time-domain er?ressiois for three line{o-neutral voltagesat the teminals of a Y-connectedload are oAN: 7967cosol V, t'BN: 7967cos((,t + 120') V, ocN:7967cos(ot - 120")V. What are the time-domain expressionsfor the three lhe-to'line voltages?.,A8, ,Bc, and ?cA? 11.9 The magnitude of the line voltage at the terminals of a balancedY-connected load is 12,800V The load impedanceis216 + j63 O/d. The load isfed from a line tiat hasanimpedanceof 0.25 + j2 Ab. a) What is tlre magnitude of the line cu[ent? b) What is the magdrude of tlre line voltage at the source?

11.10 The magnitude of t]le phase voltage of an ideal balanced tlree-phase Y-coinected source is 4800 V The souce is contrected to a balancedY-connected load by a distribution line that has an impedance of 2 +j16 O/d. The load impedanceis 190+ /40 O/d. The phase sequence of ttre source is acb. Use the a-phase voltage of the source as the referenc€. Specify the maglitude and phase angle of the following quantitiesr (a) the three line currents, (b) the threeline voltagesat the source,(c) the tbreephase voltagesat the load, and (d) tle thee line voltages at the load. 11.11 A balanced tlree-phase circuit has t}le folowing chamctenstics: . Y-Y connected; . The line voltage at tlte source, Vab, is

120^v51yv; . . .

FigurcP11.6

0.2O ll.6O

a

0.8O jJ.4O A

03a

b

0.7o

j5.6o

B

u0/-t20" v 0.4O .i2.8O c

L6A

j1.24

C 78O j50O

0.8o

j4o

B loo

i4O

C 1590 j114.4O

t2.4

7aO

/ 5 to 5O

Ta

j52a

Figure P11.7

0 . 2o

jt.6o

b

480t128V 480!2q" V o2A

jL6A

i 2 4 .o 4

2OA c

0.8O

The phase sequenceis positive; is2 . i3ab: The lineimpedance fhe load impedanceis28 + j3'7 AlO. a) Draw the single phase equivalent cfucuit for the a-phase, b) Calculatedthe line currentin the a-phase. c) Calculated the line voltage at the load in the a-phase.

probtems459 Section11,4 11,1j, A balancedA-connectedload hasan impedanceof 360 + i105 O/d. The load is fed througha line havhg an impedanceof0.1 + j1 O/d. The phasevottage at the teminals of the load is 33 kV The phase sequenceis positive. Use VABas the reference. a) Calculate the three phase currents of tle load. b) Calcutate the three line currents. c) Calculate the thiee line voltages at the sending end of the line.

11,13 A balalced Y-connected Ioad having an impedance of 96 - j28 O/4 is coinected in parallelwith a balanced A-connectedload having an impedanceof 144 + j42 A/ 6.me paralleled loads are fed frcm a line having an inpedance of i1.5 O/4. The magni, tude of the Line-to-neutral voltage of the Yload is

7500 v

a) Calculate tie magnitude of the current ir t}le line feedingthe loads. b) Calculate the magnitude of the phasecurent it the A-connectedload. c) Calculate the magnitude of the phase current in tie Y-connected load. d) Calculate the magnitude of rhe line volrage ar t}le sendhg end of the line.

11.14 A balanced, three-phase circuit is characterized as follows: . Y-A contrected; . Souice voltagein the b-phaseis 201:9q V; . Sourcephasesequenceis acb; . Lhe impedanceis 1 + j3 O/4; . Load impedanceis 117 - j99A/0. a) Draw the singlephaseequivalent for the a-phase. b) ( alculared lhe a-phase linecurreDl. c) Calculated the a-phase line voltage for the three-phaseload.

11,15 In a balanced ttree,phase system, the source $ a balanced Y with art abc phase sequenceand a line voltage Vab= 208 q V. The load is a balanced1' in pamllel with a balancedA. The phaseimpedance of the Y is 4 + i3 O/4 and the phaseimpedance of

the A is J iq O d. The tine impedance is 1.4 + j0.8A/0. Draw the single phaseequivalenr circuit and useit to calculated tle line voltage at t}Ie load in the a-phase.

11.16 An abc sequencebalancedthree-phaseY-connected sourcesuppliespower to a balanced,three-phase A-connected load with an impedance of 12 + j9 O/d. The sourcevolragein the a-phaseis 1204qV. The line impedance is 1 + j1Ab. Draw the single phase equivalent circuit for the a-phaseand useitto find the currentin the a-phase of the load.

11.f

A balanced three-phase A-connected source is showninFig.P11.17. a) Find the Y-connected equivalent circuit. b) Show that the Y-connected equivalent circuit delivers the same open-circuit voltage as the original A-connectedsource. c) Apply an external short cftcuit to the terminals A, B, and C. Use the A,connected source to find the thee line curents I"A, IbB, and lcc. d) Repeat (c) but use the Y-equivalent source ro tind ttre thrce line currents. rigurcP11.17

'72U,!tLl2w

460

Three-Phase cncLrits Salanc€d

1L18 The A-connectedsourceof Problem 11.17is con' nected to a Y-connectedload by meansof a balanced three-phase distribution line. The load impedance is 957 + j259 O/d, and t}le line impedanceis1.2 + j12 A/6. a) Construct a single-phaseequivalent c cuit of me sysrem, b) Determinetlre nagnitude of the line voltageat the teminals of the load. c) Determine the magnitude of the phase current in the A-source. d) Detemine the magnitudeof the line voltageat the terminals of th€ source. 11.19 A three-phaseA connectedgeneratorhasan internal impedanceoI0.6 + j4.B O/d. Wlen the load is removed from the generator, the magdtude of the terminal voltageis 34,500V The generatorfeedsa A-connectedload through a transmissionline with an impedanceof 0.8 + j6.4o"/6. The per-phase impedanceof the load is287'7 j864A. a) Construct a single-phaseequivalent circuit. b) Calculatethe magnitudeofthe line current. c) Calculate the magnitude of the line voltage at the terminalsof the load. d) Calculate the magnitude oI the line voltage at the terminals of the souce. e) Calculatethe magnitudeof the phasecurent in the load. f) Calculafe the magnitude of the phase current in lne soutce. 11,20 The impedance Z in the balanced three-phase circuir in Fig.P11.20is 600 + j450 o. Find a) tAB.IBc, and IcA, b) laA,IbB,and lcc, c) Ib,, tb, and Iac. tigureP11.20 a

69[40: kV

I

"

{

A

11,21 For the circuit shown in Fig. P11.21,find ""t' a) the phasecurrentslAB,IBc, and IcA b) the line currentsIaA,lbB,and \c when Zl : 4 8 + j1.4A. 22: 1.6 jDA, 2 3 : 2 5 + j 2 5[ 1 .

arl'd

tigureP11,21

72OD2y

S€ct'ion11,5 1112 A balanced three-phase sourceis supplying 90 kVA at 0.8 lagging to two balanced Y-connected paiallel loads The distribution line connecting the source to the load has negligible impedance.Load 1 is purely resistive and absorbs 60 kW Find the per-phase impedanceof Load 2 if the line voltageis 415.69V arein sefies. andlhe impedance componenls 1L23 The total apparentpower suppliedin a balanced, three-phaseY-A systemis 3600VA. The line voltage is 208 V If the line impedance is negligible and the power factor angleof the load is 25".determine the impedanceof the load. 11.24 In a balanced tbiee phase system, the source has an abc sequence, is Y connected, and The sourcefeedstwo loadq both van = 1204qv are Y-connected. The impedanceof load 1 o{ which j6 power for the a-phase + O/{. The complex is 8 2 is 60013!iy4. Find the total complex of load power suppliedby the sourc€. 11.25 A three-phase positive sequ€nce Y-conrected source supplies 14 kVA with a power factor of 0.75 lagging to a paralel combination of a Y-connected load and a A-connectedload.The Y-connectedload uses9 kVA at a power factor of 0.6laggingand has an a-phasecurrent of 101:lq A. a) Find t}le complex power per phase of the A-connectedload. b) Find the magnitudeof the line voltage.

Prcbtenrs 461 11,26 Calculate the complex power in each phase of the unbalancedload in Problem11-21. 1127 Threebalancedthrce-phaseloadsare connectedin parallel.Load 1 is Y-conrectedwith an impedance ol 300 + j100 O/d; load 2 is d-connectedwitl an impedanceof 5a0[ - P700glf; and load 3 is 112.32+ /95.04kVA. The loadsare fed from a distribution line with an impedanceof 1 + i10 O/d. The magnitude of the line{o-neutral voltage at the load end ofthe line is 7.2 kV. a) Calculatethe total complexpower at tlte send ing end of the line. b) What percentageof the averagepowei at the sendingend of the Lineis deliveredto the loads? 11.28 a) Find tlle rms magnitude and the phase angle of IcA in the circuit shownin Fig.P11.28. b) What percent of the averagepower delivered by the three-phasesourceis dissipatedinthe threephaseload? Figure Pl1,28

5031I) j1380O

t6o

b 3()

5031r)

j24a B 5031() O 11380

1,1,000r12E v j6{I

j1380O

c 3O

j 2 4A

0.80pf lag,and L = 168kW and 36 kVAR (magnetizing). The magnitude of the Linevoltage at the teminals of the loads is 2500\6 V. a) What is the magnitudeof the line voltageat the sendingend of the line? b) Wlat is ttre percert efficiency of the distribution line with iespectto averagepower?

11.31The three pieces of computer equipment described below are installed as part of a computation center. Each pieceof equipmentis a balancedthree phase load rated at 208V Calculate(a) the magnitudeof the line cuirent supplyingthesethree devicesand (b) the power factor ofthe combinedload. . Hard Ddve:5.742kW at 0.82pflag . CD/DVD drive: 18.566kVA at 0.93pf lag . CPU:line current 81.6A, 11.623kVAR

tL32 A

three-phase line has an impedance of 0.5 + j4 oft. Tl:'e Line feeds two balanced tl eephaseloads connectedin parallel.The first load is absorbinga lotal ol 6q1.2 kW and delirering 201.6 kVAR magnetizingvars. The second load is A-connected and has an impedance of 622.08+ j181.44A/6. The line to,neutral voltage at the load end of the line is 7200V What N ue magnitude of the line voltage at the source end of the line?

1133 At full load, a commercially available 200 hp, threephase hduction motor opemtes at an efficiency of 96% anda powerfactoro{ 0.9lag. The motor is supplied from a three-phase oudet with a line-voltage rating of 208V a) What is the magnitude of the line current dmwn fiom the 208V outlet?(1 hp = 74614'; b) Calculare the reactive power supplied to the motor.

11,34 The line-to-neutral voltage at the teminals of the 11:9 Show that the total instantaneouspower in a balancedtbree-phasecircuit is constantand equal to \NhereU- and 12 represent the max1.5U^I- cos0a2, imum amplitudesof the phasevoltage and phase curent, respectively. 11.30 A balancedtbree-phasedistribution line has an impedanceof 1 + j5 O/d. This line is usedto sup ply three balancedthree-phaseloadsthat are con, nected in parallel. The tiree loads are Lr = 75 kVA at 0.96 pf lead, lz = 150kVA at

balancedlhfee-phase load in rhe circuilsho$n in Fig-P1l.34 (page462) is 480V At this voltage,the load is absorbing60 kVA ar 0.8pflag. a) Use VAN as the reference and express I.o in polar form. b) Calculate the complex power associatedwith the ideal thrce'phasesource. c) Check that the total averagepower delivered equals the total averagepower absorbed. d) Check that the total magnetizing reactive power deliveied equalsthe total magnetizingreactive Powerabsorbed.

462

Batanced Thr€€-Phase Circuits

1L38 The total power delivered to a balanced three-

FigurcP11.34

02o

v.n

/0.4o

-js o

11.35 A

balanced three-phase source is supplying 1800kVA at 0.96pf lead to two balancedY-connected panllel loads The distribution line connecting t}te souice to the load has negligible impedance.The powei associatedwith load 1 is 192 + i1464 kVA. a) Detenine the impedanceper phaseof load 2 iJ the line voltageis 6400V3 V and the impedance componentsale m senes, b) Repeat (a) with t}le impedance components in parallel.

11.36 A balancedthree phaseload absorbs190.44kVA at a leading power factor of 0.8 when the line voltage at rhererminakor rheloadis lJ.800vrind fourequivalent circuits that can be used to model this load.

tL37 The output of tlle balarced posilive'sequence three-phasesourcein Fig. P11.37is 78 kVA at a leadingpower factor of 0.8.Theline voltageat the sourceis 208\5 v. a) Find the magnitude of the line voltage at the load. b) Find the total complex power at the terminals of the load.

phase load when operating at a line voltage of 6600V3-V is 1188 kW at a lagging power {actor ol 0.6.The impedanceof the distibution line supplying the load is0.5 + j4 o/d. Under theseoperating conditionsr tle drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive.To compensate, a bant of A-connectedcapacitorsis placed in parallel with the load. The capacitor bank is designed to furnish i920 kVAR of magnetizing reactive power when operated at a line voltage of 6600\,€ v. a) w1lat is ttre mag tude of the voltage at tie sendingend of the lhe when the load is opemt ing at a line voltage of 6600\4 V and the capacitor bank is discoDnected? b) Repeat(a) witl the capacitorbank conn€cted. c) What is the averagepower efficiency of the line in (a)? d) wllat is tlle averagepower efficiercy in (b)? e) If the systemis operating at a ftequency of 60 Hz, what is the size of eachcapacitor in microfarads?

11.39 A balanced bank of delta-connected capacitom is connectedin paralel witl t]le load describedin AssessmentProblem 11.9.The effect is to place a capacitor in parallel with the load in each phase. The line voltage at the terminals of the load thus remains at 4160V The circuit is operating at a frequency of 60 Hz.The capacitors are adjusted so that the magnitude of the line current feeding t})e paiallel combinationofth€ load and capacitorbank is at its minimum. a) What is the sizeof eachcapacitor in miqofarads? capacitois b) Repeat(a) for wye-connected c) Wiat is t}le magnitude of the line current?

F l g u rP e1 1 . 3 7

Section11.6 11.40 Dedve Eq$ 11.56and 11.57.

11.41 The two-wattmeter method is used to measure the power at the load end of the line in Example 11.1. Calculatethe readingof eachwattmeter.

Probtems 463 11.42 The two wattmete$ in Fig. 11.20 can be used ro computethe total reactivepower ofthe load. a) Pro\e rhi. sralement by .houiog rhal

^''3(W- W\: ^/3v'1'"i"0,r.

b) Compute the total reactive power from the wattmeter readirgs for each of the loads in Example 11.6.Checkyour computationsby calculating the total reactive power directly from the given voltage and impedance.

11,43 In the balanced three-phase circuit shown in Fig.P11.43,thecurrentcoil ofthe wattmeteris con, nected in line aA, and the potential coil of the wattmeter is connected across lines b and c_Show that the wattmeter rcading multiplied by 1,6 equals the total reactivepowei associatedwilh the load. The phasesequenceis positive.

11.46 The balarc€dtbree-phaseload showninFig.P11.46 is fed from a balanced,positive-sequcnce, rnreephase Y-connected source. The impedance of the line connecting the souce to the load is negligible. The line-to-neutral voltage of the source is 4800V a) Find the readingofthe wattmeterin warts. b) Explain how you would connect a second wattmeter in t}le circuit so t]tat the two wattmeterswould measurethe total power. c) Calculatethe readingof the secondwacmeter. d) Verify that the sum of the two wattmeter readings equals the total averagepower delivered to the load.

figureP11.46

it ----

Figure P1r,43

- I

- 1 -

600kvA <{B 0-96pf lag

w-

---{c

11,47 a) Calculatethe readingof each wattmeterin the circuit shol*tr in Fig. P11.47.The value of Zo is

601j!:4. b) Verify that the sum of the wattmeter readings equalsthe total averagepower deliveredto the A-connectedload. 1L44 The line,to-neutral voltage in the circuit in Fig.P11.43is 720V the phasesequenceis posirive, and the load impedance1s96+ j72 A/0. a) Calculatethe wattmeterreading. figurc Pt7-47 b) Calculate the total reactive power associate{:l with the load.

4801qv 11.45 a) Calculatetlrc complexpower associatedwith each phaseof the balatced load in Problem 11.20. b) Ifthe two-wattmetermethodis usedto measure ure averagepower deliveredto the load,specify the readingof eachmeter.

oi . n 480/-120"V

i w ,

464

circuits Thre€-Phase Batanced

11,48 a) Find the reading of each wattmeter in the circuit sho\rn in Fig. P11.48 tt 2^=60/-30"{l'

ZR- 24lyq

andZc = 80A O.

b) Show that the sum of the wattmeter readings equals the total average power delivered to the load unbalancedthree-Phase figureP11.46

trtolrzo"v!---

11.49 The wattmeten in the circuit in Fig. 1120 read as w' folows: u1 = 114,29164 w, and w, = 618,486.24 V. is 7600v3 voltage of the line The magnitude 2,. posiri!e. Fnd is Thephasesequence fl50

a) Calculate the reading of each watfmeter in the circuit sho\an h Fig P11.50 when z = 216 - j2M A. b) Check that the sum of the two wattmeter readings equals the total power delivered to t]Ie load c) Check that \4(W1 - W2) equals the total magnetizing vars delivered to the load.

figureP11.50

c) Show that the sum of the two wattmeter readings equals the total power delivered to the unbalancedload.

Sections11.1-11.6 1L52 Refer to the Pmctical Perspective example: ,lli|l#h a) Constru"t u power triangle for the substation Ioad before the capaciton aie connected to the bus. b) Repeat (a) aftel the capaciton aie connected to the bus c) Using ttre line-to-neutral voltage ar the substation as a reference, construct a phasor diagram that depicts the relationship between VAN and Ve before the capacitors are added. d) Assume a positive phasesequenceand construct a phasor diagram that depicts fhe relationship between Vas and Vu6. 1L53 Refer to the Practical Perspectiveexample.Assume mamrrthe fteouenc! of the utilitYis 60 H2. a) What is tle /-!F mting of each capacitoi if the capacitors are delta-connected? b) What is the pF rating of each capacitor if the capacito$ are wye-connected? 11.54 In the Pmctical Perspective example,what happens volldge level al lhe generalingplant if lhe -rg11a; "_"'"" to the ar Il8 kV the subslation subslalion is maintained and the added capacitor to zero, load is reduced connected? bank remains

I I I I

6900 L 12Ev 6900 &t2c v 11.51 The two-wattmeter method is used to measure the power delivered to the unbalanced load in Problem 11.21.The current coil of wattmeler I rs placed in line aA and that of wattmeter 2 is placed in line bB a) Calculate the reading of wattmeter 1 b) Calculatethe readingof wattmeter2.

11.55 In the Practical Perspective example, calculate the kw beforeand after tbe c?pacitors ;fffl#l toral Uneloss_in are connecteclto tle suDstallonDUs 1L56 Assume the load on tle substation bus in t]le mamdtPracticalPersDecli\eexampledrops lo 240kW and """"*' o00tugn.Liting kvAR Al;o assu;e lbe capacilofs remain connected to the substation a) What is the magnitudeof the line-to-line voltage at the generating plant that is required to maintain a line+o-line voltage of 13 8 kV at the substation? b) Will this power plant voltage level cause problems for other customem?

Prublems 465 11.57 Assumein Problem11.56ttrat when the load drops ,ll|ffSt! to 240 kW and 600 magnetizilg kVAR the capacitor bank at tlle substation is discomected. Also assume tlat t]le line-toline voltage at the substation is maintained at 13.8kV a) What is the magnitude of the line+oline voltage at the generating plart? b) Is the voltage Ievel found in (a) withh the acceptablerange of variation?

c) What is the total line loss in kW when the capacitors stay on line after the load drops to 240 + j600kVA? What is tle total iine loss in kW when the capacitofi are removed after the Ioad drops to

240+ i600kvA?

e) Based on your calculatiom, would you recommend disconnectingthe capacitors after the load drops to 240 + j600 kVA? Explain.

Introduction to the Transform LapLace 12.1Definitionofth€ L.plaeeTranstotilp.467 12.2TheSteprunctionp. 468 p. 420 12.3TheImputseFunctjon p. 474 12.4Fundionat Transforms p. 475 12.5opehtionaL Transforms p- 481 12.6Apptying th€ L.placeTransfolr't p. 482 12.7Inverse Transforms 12.8?olesandzercsof rlst p.494 p. 495 12.9InitiaL-andFinal-Vatue Theorems

transformofa 1 Beableto catcutate th€ Laptace functjofLrsing the defifjtioi of Laptace tmnsfom,the Laptace tlanslomtabL€,and/ora tabteof opentionat tB nsforms2 B€abteto calcuLate the inve6e LapLace transformusiig pattialfiactionexpansioiafd the Laptace transform tabLe.

3 Underctard andknowhowto usetheinitiaL vatuetheorcm aid thefinalvaLue theorcm-

We now introduce a powerful analytical technique that is widely used to study the behavior of lineal, lumped-paramctcr circuits.The methodis basedon the Laplacetransform,wlrichwe define mathcmaticallyin Section12.1.Betbre doing so,we need to expiainwhy another analyticaltechniqueis needed.Fimt. we wish to considcrthc transientbehaviorof circuitswhoseclescribing equationsconsistofmore than a singlcnode voltaEleor meshcurrentdillerelltial equation. In other words,we want io considcr multiplc node and multiple-meshcircuits thal are describedby selsoI linear differentialequations. Second.rve \vish to determinethe transientresponseof cir cuits whose signalsourccsvary in ways more complicatedtha11 the sinple dc level jumps consideredin Chaptcrs7 and 8. Third, wc canusc thc Laplacetranslbrmto introducethe colcept ol the transfer function as a tool for analyzingthe steady-statesinusoidalresponseoI a circult when the frequencyof the sinusoidal sourccis varicd.Wc discussthe transferfunclion in Chapter 13. Finally,we wish to relate,in a svstematicfashion,thctime domain behaviorof a circuit lo its ftequency-domainbehavior-. Using the Laplacetransformwjllprovide a broadcrunderstandingolcircuit functions. In this chapter,we introduce the Laplace tlanslorrn.discuss ils pertinent characteristics, and presenta systcmaticmethod lbr transformingfrom the liequencydomainto the tirne domain.

12.1 D€finjtion TGnsform 467 ofthe Laplace

12.1 Detinitionofthe Laptaee Transform The Laplace lmnslbrm of a lUnction is given by the expression

s\f(t)]

-I

f(t)e-"'dt,

(12.1) { Laptacetransform

wherethesymbolg{f(/)} is read"th€ Laplacetransformof/(r)." TheLaplacetransfbrmof/0) is alsodenotedF(r): thatis,

F(0: s{/o}.

\12.2)

This notationemphasizes that when the integralin Eq. 12.1hasbeeneval uated,the resultingexpressionis a functionofs.In our applications, t rep resents the time domain, and, becausethe exponent of e in the integral of Eq. 12.1must be dimensionless, s must have the dimensionof reciprocal lime, or ftequen+ The Inplace transform transforms tlre problem from lhe time domain to the frequency domain. After obtaining the frequencydomain expressioflfor the unkrrowl we inverse-transform it back to the time domair, II the idea behind the Laplace transform seemsforeign, consider anotherfamiliar mathematicaltransfom. Logaritbmsare usedto change a multiplication or division problem, such as A : BC, into a simpler addition or subtraction problem: log A = log BC = logB + log C. Antilogs are usedto caffy out the inverseprocess.Thephasoris another transform;aswe knowftom Chapter9, it convertsa sinusoidalsignalinto a complex number for easier,algebraiccomputation of circuit values. After determinirg the phasorvalueof a signal,wetransformitback to its time-domainexpression.Both of theseexamplespoint out the essential feature of mathematicaltmnsforms:They are designedto create a new domain!o make the mathematicalmanipulationseasier.After finding the unknownin the new domain,weinversetransformit back to the original domain.In circuit analysis,we use the Laplacetmnsfoim to transforma setol integrodifferenlialequationsfrom the time domain 10a setof alge braic equationsin the ftequencydomain.W()thereforesimpiify the solution for an unknom quantity to the manipulationof a set of algcbraic equations. Beforewe illustratesomeof the impo ant propertiesof the Laplacc transform,somegenemlcommentsare in order,First,note thal the irte gral in Eq.12.1isimproperbecausethe upper limit is iniinite.Thuswe are confronled inrmediately with the question of whether the integral converges. In other words, does a given /0) have a Laplace transform? Obviously,the lunclions of primary interestin engineednganalysishave

468

Transfom to th€Laptace Intoduction

Laplace transforms; otlena'rse we would not be intelested in the tfansform. In Linear circuit analysis,we excite circuits with sources that have Laplace tiansforms. Excitation functions such as I or ?', which do not have Laplace transforms, are of no interest here. Sec;nd,becausethe lower limit on the integral is zero' the Laplace lransform ignores /(r) for negative values of t Put another way, F(s) is aleterminedby the behavior of /(t only fdr positive values of t To emphasizethat the iower limit is zero,Eq 121 is frequentlyreferred to as the otr€-sialed,or udlatenl, Laplace transform.In the two-sided' or bilateral' Laplace transform, tle lower limit is co We do not use the bilateral folm herel hence F(.r) is understood to be the one-sided transform Another point regarding the lower limit concerns the situation when origin as, /(t) has a discontiouity at the odgin lf /(D is continuous at the ioi example,in Fig.12.1(a) /(0) is not ambiguous.However'iff(r) hasa finite discontinuity at the origin-as, for example, in Fig 12.1(b)-the question adses as to whether the Laplace transform integral should i;clude or exclude the discontiouity ln other words, should we make the lower limit 0- and include ttre discontinuitv, or should we exclude the discontinuity by making the lower limit 0+? (We use th€ notation 0- and 0- to denote values of I just to the left and right of the origin, respectively.) to For reasoDs we may chooseeither aslong aswe are consistent, Actually, (b) (a) be eiDlainedlater.we choose0.' asthe lower limil. function anddisconlinuous Figuru12.1A A continuous Becausewe are using 0- as the lower limit, we note immediately that atthe orisin. attheoriqin.(a)/(r) k continuous the integation from 0- to 0+ is zero The only exception is when the dis_ (b)/(r) k discontinuous aithe oriqin. continuity at the origin is an impulse function, a situation we discussin Section 12.3.The important point now is that tlte two Iunctions shown in Fig. 12.1have the sam€unilateralLaplacetransformbecausethere is no impulse function ai the origin. The one-sided Laplace transform ignor€s /(t) foi t < 0 what hap_ pens pdoi to 0- is accounted for by the initial conditions Thus we use the Laplace tiansform to predict the rcsponseto a disturbancethat occurs after initial conditions have been established. In the aliscussionthat follows, we divide fhe Laplace transfoms into two tlpes: functional transforms and operational transforms.A funcdonal trsnsform is the Laplace transfom of a specific function, such as sin ol. I' t", and so on.An op€rationaltransformdefin€sa genelalmathematical property of the Laplace ffansform, such as finding the tiansfom of the derivative of /(t). Before considering functional and operational tiansfoms, however, we need to inftoduce the step and impulse functions

72.2

'lhe StepFunction

we mayencounter lunclionslhal havea discontinuil)or iump ar lhe origin. For example,we know froD earlier discussionsof transient behavior that switching operations create abrupl changesin currents and voltages We accommodatethese discontinuities mathematically by introducing the steDand imDulse functions.

12.2 Th€StepFunction 469

Figuie 12.2illustrutes the step function. It is zero for . The symbol for the step function is fr(4. Thus, the mathematical definition of the step funotion is

ru0)=0, t<0, K u ( t ) =K , t > 0 .

(123)

rigure rz.z.l trrestep runctton.

If1(is 1,the functiondelinedby Eq.12.3is the unil step. The stepfunctionis rot defined at I = 0. In situations wherc we need to define the rransition between0 and 0+.we assumethat it is linear and that

ru(0) - 0.s1(.

(r?.4)

As befoie, 0 and 0r represent symmetric points arbitrarily close to the left and dght of tle origin. Figure 12.3illustrates t}le linear traNition lrom 0- to 0'. A discontinuity may occLrrat some time other than I : 0; for example, in sequentialswitching.A step that occuis at 1 = a is expiessedas (r.,(t d).Thus Ku(t-a):0,

t
Ku(t-a)-K,

t>a.

\12.5)

If 4 > 0, the step occurs to the ght of the origin, and if a < 0, the step occun to the left of the origin.Figure 12.4illustratesEq. 12.5.Note that the step function is 0 when the argumentI - /I is negative,and it is K when the argumentis positive. A step function equal to Kfor t < d is written as Ku(d - r). Thus Ku(a

t):K.

t
Ku(a-t)-0,

t>a.

The discontinuity is to t})e left of the origin when a < 0. Equarion 12.6is shownin Fi9.12.5. OIIe application of the step function is to use it to wdte t]le mathe matical expressionfor a function that is nonzero for a finite duration but is defined for all positive time. One example useful in circuit analysis is a finite-width pulse, which we can create by addirg two step furctions_ Th€ tunction -K[,i(l 1) ,,0 - 3)] has rhe value r for 1< I < 3 and rbe value 0 everfwherc else,so it is a finite-width pulse of height fa itritiated at I = 1 and teminated at 1 = 3. In defining this pulse using step functions, it is helpful to think of the step tunction !.(l - 1) as ,'rurning on,'the constafltvalueK at I = 1, and the stepfunction -&(t 3) as,,tuning off,the constant value l. at t : 3. We use step functions to tum olr and turn off linearfunctionsat des ed timesin Example12.1.

Figue12,5A A stepfunctronKu\d - t) fot a > 0.

470

lranstorm Introduction to theLaplace

to Represent a Fundionof FiniteDuration UsingStepFunctions Use step functions to write an expression for the functionillustratedh Fig.12.6.

+ 2 r 8 , o n a t t = 3 , o f f a t t - 4 - Thesestraight line segments and their equations Fig.12.7.Theexpressionfor /(l) is f(t) :2tlu(t) &(

- u(t - 1)l + ( 2t + 4)lu(t- L) 3)l + (2t - 8)["(r - 3) u(t 4\1.

forExampte 12.1. tigure12.5A Thetunctjon

Solution The function shown in Fig. 12.6is made up of Linear segmentswith brcak points at 0, 1, 3, and 4 s.To con stiuct this function, we must add and subtract lhear functions of the proper slope.We use the step function to initiate and terminate these linear segments at ttre proper times.In other words,we usethe step function to tum on and turn off a straight line with the following equations:+2t, on at t = 0, off at t = 1; -Zt + 4, on at t - 1, off at t =3t and

tuned Figure12,7 a Definiiion ofthethreeljnesegmentr formthetunctjonshou/n 0nandoff withstepfunctionsto in Fig,12.6.

NOTE: Assessyour unde/standing of stepfrnctians by trJinq Chapter Prcbkns 12.1 and 12.2.

12.3 TheImpulseFunction When we have a finite discontinuity in a function, such as that illustrated in Fig.12.1(b),the derivativeof the funclion is not defined at the point of the discontinuity. The concept of an impulse ftrnctionl enablesus to define the dedvative at a discontinuity, and thus to define the Laplace tiansform of tlat derivative. An imp se is a signal of infinite amplitude and zero duiation. Such signals don't exist in nature, but some circuit signals come very close to approximating tlis definifion, so we find a mathematical model of an impulse useful. Impulsive voltages and currents occul in cir_ cuit analysiseither becauseof a switchitrg operation or becauset]le circuit is excited by an impulsive source. We will analyze these situations in Chapter 13,but here we focus on defining the impulse function generaly

i The inpulse function is aho known as the Dirac delta iunctio..

12.3 TheImpulse Function 471 To define the derivative of a function at a discontinuity, we first assumethat the funcfior varies linearly aqoss ttre discortinuity, as shown in Fig. 12.8,where we observe that as €+0, an abrupt discontituity occurs at the origin. Wher we dilferentiate the function, the derivative between € and +€ is constant at a value of 1/2€. For I > €, the derivative is aa'(' (). Figure 12.9 shows these observations graphically. As € approaches zero,the valueof /'(t) betweenf€ approaches infhity. At the same time, the dumtion of this large value is apprcaching zero. Furthermore,the areaunder/'(t) between+€ remainsconstantas € +0. In this example, the area is unity. As € approacheszero, we say that the funcrion betweenre approachesa unit impulsefunclion. denored5(). Thus the derivative of/(/) at the origin approachesa unit impulse function as € approacheszerc, o{ /'(0) -6G)

:r + 0.5

Figure12,84 A magnified vjewofthe disco'rtjnujty in -€ Fis.12.1(b), assuftring a lineartransjtjon between

f'(! )

as€ +0.

If the area under the impulse function curve is other thar unity, the impulse tunction is denoted K6(t), where ( is the area.1
Figure 12.9t' Th€dedvativ€ ofth€tunctjon shown jn Fig.12.8.

Many differelt variable-parameter functions have the aforementioned characteristics. In Fig. 12.8,we useda linear tunction /(t) = 0.5r/€ + 0.5. Another example of a variable pammeter function is the exponential function:

/(,)

: Ln'r, 2F-

\12.7)

As € apprcacheszero, the function becomesinJinite at the origin and at the same time decays to zero in an inJiniresimal le gth of time. Figue 12.10 illushates the character of /(,) as € + 0. To show that an impulse function is created as € + 0, we must also show thal the area under the function is indeDendentof €. Thus

K

t'ea= | !a''ar+ I !"-''a, z. Ja.e

.tn

=' ' G l ' ' e-t/' l-* 1"1, K

-

e,/.lo

K

K

2+

Z=

K

K.

tigure12,10A A vadabLe-panmeter tunction used to generat€ animputse tunctjon. (12.8)

which tells us that t]Ie area under the cuve is constant and equal to tr units. Therefore,as € + 0, /(1) -16(r).

472

Iniroduction to theLaptace lransform Malhematically, the impulse futrctiotr is defined K6(t)dt : K;

11,2.e)

60) : 0, r + 0.

f(, K6(/

a)

(12.10)

Equation 12.9states tlat the area under the impulse function is constant. This area represents the strength of the impulse. Equation 12.10 states that the impulse is zerc everywhere except at t : 0. An impulse that occnrs at r = l' is denoted K6(t - a). The graphic slmbol for the impulse function is an alrow The strength of the impulse is given parenttreticaly next to t]le head of the arrow Figure12.11showsthe impulsesf6(t) and (6(t - d). An important prcperty of the impulse function is the sifting propefty, which is expressedas

f(t)6(t-a)dt-f(a),

tigure12.11i A qraphic representation oftheimpube 16(,) andr5(t a).

112.11)

where the function /(t) is assumedto be continuous at t : d; that is, at the location of the impulse.Equation 12.11showsthat the impulsefunction sifts out everything except the value of /(t) at I - a. The validity of Eq- 12.11follows from noting that 60 ") is zero everywherc except at r = a, and hence the integral can be written I:

/I f(t)8(t a)dt:

Ja

I

.l "-,

f(t)6(t

- a)dt.

(12.1?)

But becausef(t) is continuousat d, it takeson the value/(a) as t *.r, so

I :

td+.

ta+€

f(a)6(t a)dt= f(a).1. .1". ,6(t : f(a).

f (.t)

atdt

We use the sifting prope y oI the impulsefunction to lind its Laplace transfornl

= / u1,;"a,= [" a|)a,=t. r1u1,;1

(b) Figurc12.12l Thefint dedvative of theimpulse function. functionusedto G) Theimpuke-genenting (b)Thefilst definethefirstdenvative of ihe impulse. dedvatjve 0f theimpulse-generatjng function that apprcaches 6'(t) as€+0.

(12.13)

112.14)

whichis animportantLaplacetmnsformpair that we makegooduseof in circuit analysis. We can also define the derivativesof the impulsefutrction and the Laplace transform of these derivatives.we discussthe first derivative, along with its transform and then state the resulr for the higher-order derivatives, Thefunctionillustratedin Fig.12.12(a)generatesan impulsefunction as € + 0. Figue 12.12(b)showsthe dedvativeof tlis impulse-generating function,whichis defitredasthe dedvativeof the impulse[6'0)] as€ - 0.

12.3 TheImputse Function The derivative of the impulse function sometimes is referred to as a momenttunction,or unit doublet. To find the Laplace traNform of 6'(r), we simply appty rhe defining integral to the function shown in Fig. 12.12(b)and, after inregmting,ter € + 0. Then

: s[f. .],,", [(-i)*,"] .r,,r,rr 1im

s-?'. + ,t-e '_

2s

(12.15) In derivingEq. 12.15,we had to usel'H6pital's rule twice to evaluatethe indeterminateform 0/0. Higher-orderderivativesmay be generatedin a manner similar to that used to generatethe first derivative (see Problem 12.11),and the defining integral may then be used to find its I-aplace transfom. For the ,th derivativeof the impulsefunction,we find that its Laplacetransform simplyis s'; that is,

rr{6('\t} : s'.

112.16)

Finally,an impulsefunction canbe thoughtof as a derivativeof a sfep function;that is,

ut,t= o"!t)

0 \12.!7)

f'lt) Figure12.13presentsthe graphicinterpretarionofEq.12.17.Thefuncrion shownin Fig.12.13(a)approachesa unit stepfunctionas € +0. The furction shown in Fig. 12.13(b)-the dedvarive of the Iunction ir Fig.12.13(a)-approachesa unit impulseas € +0. The impulse function is an extremely useful concept in circuit analysis,and we say more about it in tlle following chapte$. We introduced t]te concepthereso that we car includediscontinuitiesat the odgin in our definition of the Laplace transform. NOTE: Asse.Js)Joul undentanding of the impulse function by nying ChapterPrcblems12.5 12.7.

ftgsr€12.13A Thejmpulse functjon asthedenvabve (a)/O J r(r) as€-0, and 0rth€stepfunction: (b)/'(r)+ 5(r)as€+0

474

Intmduction totheLaptace T€nstornr

12.4 FunctionaI Transforms A furctional transform is simply the Laplace lransfom of a speciJiedfunction of t. Becausewe are limiting our intoduction 10 the unilateral, or onesided,Laplace tansfom, we defire all functions to be zero for t < 0 . We derived one lunctional transform pair in Section 12.3,wheie we showed that the Laplace transform of the udt impulse function equals1;seeEq. 12.14.A secondillustration is the unit step function of Fig. 12.13(a),where

t 1 , q 4 1= [ y 6 n . o , = [ - , " " 0 , :;".:;

(12.18)

1

Equation 12.18shows that tle Laplace transform of the unit step function is 1/s. The Laplace tmnsform of the decaying exponential function shown in Fig.12.14is

tl""'l -

Pigure12.14^ A decaying e{ponentiat

/

-

ede4dt= Jo Joe''"'ar-,

l

o.

o?.re)

In derivingEqs.12.18and 12.19,we usedthe fact that integrationacross the discontinuity at the origitr is zero. A tlird illustratiotr of findhg a functional transfom is the sinusoidal function shownin Fig. 12.15.The expressionfor l(l) for t > 0- is sin dl; hen€ethe I-aplace transform is

e{sind1}:

l"

ut.un"' o,

= [("'';"'^)-'.

Fiquer2.r5 A A sinEoidat tunction fo' I > 0.

I t l

= - t -

dt

2j

t

-

r \ " + lr)

112.20) Table 12.1 gives an abbreviatedlist of Laplace transform pairs. It includes the functions of most interest in an introductory coune on circuit aDDlications,

12.5 0peBtionat Tnnsfoms 475

Tlre

,f(, c> o-) 6(r)

(imp'ilse)

u(t)

rG) 1

1 1 ;'

(ramp)

I

(exponenlial)

(cosine)

1

(danped ranp) (damped sine)

G+;Fia

12.5 0perationalTransforms Operational transforms indicate how mathematical operations oerformed oDeirher/(ri or F(,) areconvenedinto lheopposiridomajn.iteoperations of pdmary interest are (1) multiplication by a constantt (2) addirion (subhaction); (3) differentiatio{ (4) integmrion; (5) rranslarion in the time domain; (6) translation in the hequency domain;and (?) scalechanging.

Muttiptication by a Constant Fiom the defining integral, if

e{f(t)J = F(s),

s tKf (t)j : KF(r.

(12.21)

476

Introduction tothe Laplace T6nsfom

Thus,multiplicationof /0) by a constantcorespondsto multiplying F(s) by thesameconstant,

Addition(Subtraction) Addition (subtraction)in the time domair rranslatesinto addition (subtraction)in the ftequencydomain.Thusif

s{l(r} : r(r), s{/r(r)} = rr(r), s{/3(r)}= r3(t, then g{ft(t)

+ f2O - /:(4}

= r(")

+ Fls) - F3(s),

(12.2?)

which is dedved by simply substituting the algebraic sum of time-domain functions into the defining integral.

Differentiation Differentiation in the lime domain conesponds to multiplying F(r) by r and ttren subtracting t}le initial value of f(t-that is, /(0 )-fton this prcduct:

' { + } = r F ( s ) - r { o) .

112.23)

which is obtahed directly from the definition of the Laplace transfom, or

'{+\fl+)"",,

lrz.24)

We evaluate the integral h. Eq. 12.24 by integrating by parts. Letting u = e ", ajnddt = ld f (t)ldtldt yletds

.\+\ : n'',rl*I*',u,*''0,

\12.25)

Because we areassuming that /(t) is Laplacetransformable, the evalua tion of e-Y0) at I : oois zero.Thereforethe right-handsideof Eq. 12.25

f((D + sr

rr(r) /(0-). ftr)e'-.crt:

12.5 operationat Transtorms477 This observation completes the derivation of Eq. 12.23.It is an imporant result becauseit states that differentiation in the time domain teduces to an algehaic operation in the s domain. We determine the Laplace tmnsform of higher-order dedvatives by using Eq. 12.23 as the starting point. For example, to find rhe Laplace transtbm of the second dedvative of /(t), we first let

. .

df(t) 112.26)

dt

NowweuseEq.12.23to write G(r) = rF(s)

/(0 ).

\12.27)

But because ds(t) _ d2f(t\ ' dt dtz

"{ot:.")

- "{t'!!)

l d t )

l d f

)

-scrr) - B(0).

(1?.28)

Combhing Eqs.12.26,12.27,and12.28gi'tes I atrr,tl .t O I ri=rj'i - "'rr',- ,tro-t "'),'.

1tz.zs:

We find the Laplace ffansfom of the rth derivative by successively applying the preceding process,which leads to the general result

* l o ' t ' ! \ = s , F r , ) r , , / { o1 - , ^ z l l t o t I d," I

d!

J-,-3af9-) ,--

d" lfg )

'---:-,

Lrz,ru)

Integration Integration in the time domain correspondsto dividing by r ir the s domain. As before,we establishthe relationship by the defining integral:

'\lro^| =Ill[to,,,1,"""

(12.31)

478

Intrcductjon t0 theLaplace Transform We evaluatethe integralon the right-handsideofEq.12.31 by integrating by partq first letting

Then

: f(t)dt,

The htegratior-by-padsfomulayields

-;f'r,r,,l;. .{[:o,*1= f 1,,r0,,,,, The first telm on the right'hand side of Eq. 12.32is zero at bot}l the upper and lower limits. The evaluation at t]le lower limit obviously is zero, whereasthe evaluation at the upper limit is zero becausewe are assuming that /(l) has a Laplace tiansfolm. The secondierm on the dght-hand side of Eq. 12.32is F(s)/s; t}leiefore

'\lrta*j:?,

(12.33)

which rcveals that the operation of inte$ation in the time domain is transfomed to the algebraic operation of multiplyiDg by 1/r in the r domain. Equation 12.33and Eq. 12.30form the basis of the earlier statement that ttre Laplace transform translates a set of integrodiffeiential equations into a set of algebraic equations.

Translation in the TimeDomain IJ we start witll any function /(t),l(t), we can represetrt the samefunction, tra0slated in time by rhecon
:t{f(t

o)u(t a)}:e'"F(s), a>0.

(12.34)

For example,knowingthat

s{ta(t)}

1

, Notethat throushoulwemultiply anyarbirra.yfnDcrion by the unii steplmctjon ,G) l(r) to ensurethat the resultinsfunctionis definedio. all Dositivetine.

12.5 0peratjonat lransfolms 479

Eq. 12.34 permits writing the Laplace tmnsform of (t directly:

a)u(t - a)

9{(t - a)u(t a)}=t". Tbeproofof Eq.12.34 follows fromthedetining integrai: ].* I u ( t- a ) f ( t a ) e - " t d t

sIG-a\u(t-a)\:

.ln'

=

, o)n"a,.

l-

(12.35)

In writing Eq. 12.35,we took advantage of r(t d) = 1 for r > /i. Now we change t]le va able of integration. Specifically,we let r = r - lr. Then r : 0 when r = a, .r : co when I = co, and !r.r - dt. Thus we write the integralin Eq.12.35as

a)u(t- 4j =

sIf(t

|

ftOe {*o a,

="-"'J*rav*a' : e_"F(s)' which is what we set out to prove.

TEnslationin the Frequency Domain Tlanslatiotr in tlle frequency domain coresponds to multiplication by an exponential in the time domain: 9[e-"'f(t)]

= F(s + a),

(1236)

which follows from the defining integral. The derivation of Eq. 12.36is left to Problem12.16. We may use the relationship in Eq. 12.36 to dedve new transfom pai$. Thus,knowing that i/ I cosorl we useEq.12.36to deducethat j/t?

-

co(@l}

\ ' a (t + a)2 + e|'

Scate Changing The scale-changeproperty gives the relationship between /(t) and F(r) when the time variable is multiplied by a positive constant:

sua,*:in(i),,, o,

(12.37)

480

Ttnsform to theLaplace Introduction

property the derivationof whichis left to Problem12.22.Thescale-change where time-scale work, especially in experimental particularly useful is are madeto facilitatebuilding a modelof a system. chaDges use We Eq.12.37to formulatenewtransformpails Thus,knowingthat g{coit r} :

l+1'

we deducefrom Eq. 12.37that sla a(sl(,)z+L f +&]'

1 E{cosot} :

Table12.2givesan abbreviatedlist of operationaltransforms NOTE: Assessyour unilentanding of theseoperutionaltransformsby trying ChaptetPrcbkns 12.24and 12.25.

Operarior

J\r'l

r(s)

Multiplication by a constant

KT(I)

(r(r)

Additioi,/subtraction

l(, + l,(r) - ,fd4+ ...

F(r)+F,(s)-F(s)+'

first derivative (tine)

df(t) dt

srG) /(o-)

a'fG)

df(r) lrG) - ,,f(o ) -

second de vative (time) nth derivative (tilne)

d'f(.t,

"'FG)

s"flf\ "

Tme integal

lf(rtdx

- df,tO ) tu2

F(r)

f(t - c)a(t

F(r + a)

Translation in hequency Scile €hanging Ft r derivative (r)

tf(.t,

"th deiivative (s)

{f(t\

r integral

ro

aF6)

, ,,.{!!)

t,-,,,..

^df(t\ { |i d' \ftD-l d{-l

1 2 . 6 A p py i n qt h eL a pa c eT , a t r f o , m 481

objective1-ge abl.eto catcutate the Laptace transform of a functionusingthe Laptace transform tableor a tableof oPerational transforms 12.2 Use Lheappropriateoperationaltransfbrm fiom Table 12.2to find the Laplacetransfbm of eachfunction:

2 Answer: (u) --; + ts aJ(b)

a:) Pe-'t b) :(r dl

Bs (s + d)'

'' sinh6rJ. (c)

Bz'

(r' + o')'

NOTE: Al\o trJ,ChaptetPrcbtems1L 17 and 11.18.

12.6 Ag:p[yimg tl"l* l-eptaeefraslsforn'] We now illustraleholv to usethe Laplacetransformto solvethe ordinary integrodiffercnlial equations thal descr.ibethe bchaviol of tumped parametcrcircuits.Considerthe circuit shorJnin Fig. 12.16.We assunrc that no initial eneryyis storedin the cfcuit at the instanl\r,henthe switch, which is shor'lirgthe dc current soulce.is opencd.Theproblem is ro find tiguE12.16r A paratletfrao ort. the timc domainexpressionfor ,(/) when I > 0. We bcgin by writing thc integr-odifferenrial cqualior1that ,(/) rnusl satisf,v. We need only a singlc node-volrageequationto describethc cir. cuit- Summingthe currentsawaylrom the top node iD the circuit gcner

+.! l,'^,to' +(

- = I ^ . u- -( t ) .

dr

(12.38)

Nole that in writing Eq. 12.38.we indicaredthe openingof the switctrin the stepjump ofthe sourcecurrenrfrom zero to 1d.. After dcriving the integrodiffcrenlialequations(in this example.jusr one),\\'e transfbrmthe equationsto the r domain.Wc lvill Dot go through the stepsofthc transformationjn dclail,becausein Chapter13we will dis coverhow to bvpassthem and gcneratethe s-domainequationsdirecdy Briefly though.we use three opcralioDaltransformsand one functional tralrsformon Eq. 12.38to obtain

?.;+

-,rr1: r.(i), +c['r(r)

(12.39)

an algebraicequation in wlich y(s) is thc unknown variabtc.We are assunring lhat the circujtpararnererR.l-.and C,aswell asrhe sourcecurrent 1dc,are known; the inilial volrage on thc capacitor (0 ) is zero becausethc iDitial enere)' sturcd in the cjrcuit is zero-Thus we ha\,e reducedthe problemto solvingan algebraicequation.

482

Tmnsform Introduction to the Laptace

Next we solvethe algebraicequations(again,justone jn this case)for the unknowns.SolvingEq.12.39for y(s) gives

ur,r(|.-!.,.) =*, Ia./C

C+(llRc)s+(llLC)

\12.44)

To find r(t) we must inverseftansform the expressionfor v(s). We denotethis invene operation

., (t): s1|v(s)\.

\12.41)

The next steD in the analvsisis to find the inverse tmnsform oI th€ r-alomainexpiession;thisis the subjectof Section12 7.In that sectionwe also present a final, critical step: checkiDg the validity of the resulting The need for suchcheckingis not unique to the time-domainexpression. Laplace transfom; conscientious and prudent engineers ahvays test any dedved solution to be sure it makes sensein terms of known system behavior. Simplifying the notation now is advantageous.We do so by droppirg the parentheticalt in time-domainexpressionsand the parenfheticalr in frequency-domain expressions.We use lowercase letters for all timedomainvariables,and we representthe correspondings-domainvariables wiih uDDercaseletlers. Thus

Y{u}:Y or u=rr{v!, E{i}:1

or i::FU},

9{f) - F

ot f:

e'{F},

,Assessyour understanding of this matetial by trying Chapter NO?t Ploblem 12.26.

12.7 InverseTransforrns The expressionfor I/(s) in Eq. 12.40is a ntional function ofr;that is,one that can be expressedin the form of a ratio of two polynomials in s such that no nonintegral powers of s appear in the polynomials ln fact, for linconslanllhe circuitsqhosecomponenl\aluesare ear.lumped-pdr0meler voltaget curren(t are al$a)\ and for lhe un-knoPn r'domarnexpfession( workmg (You by verify this observation may rational functions of s. of s, mtiorlal lunctions Problems12.2712.30.)If we caninverse-tmnsform

12.7 Inverse Innsforms 483 we can solve for the time-domain expressionsfor the voltages and currents.The purpose of this section is to present a straightforward and systematic technique for finding tlle inverse tansform of a rational {unction. In general, we need to find the inverse transfom of a function that has the folm

rc)=#

a n s n + a nf n r + . + a 1 s + s o b ^ s ^ + b ^ - 4 f d n1 + . + 4 s + b o -

\12.4?)

The coefficients a and b are rcal constants,and the exponents m and n are positive integers.fhe ratio N(s)/r(s) is called a proper rationd furctior lf m > n, and an improp€r rational fir€tion if m < ,?. Only a proper iational function can be exparded as a sum of partial ftactions. This restriction posesro problem, as we show at the end ol this section.

PartiaIFractionExpansion: ProperRationa[Functions A prcper rational function is expanded hto a sum of partial fraclions by writing a telm or a sedes of tems for each rco1 of D(s). Thus D(s) must be in factored folm before we can make a partial fraction er?ansion. For each disainctroot of ,(r), a shgle term appearsin the sum of partial fraction$ For each multiple root of D(r) of multiplicity r, the expansion contains r tems. For examole.in t}re rational function

s + 6 r(r+3)(s+1)' ' the denominatorhasfour rcots.Tho of theseroots are distinct-namely, atr = 0ands - 3. A multiplercot of multiplicity2 occursat s = -1. Thusthe pa ial fraction expansionof this functiotrtakesthe form Kz _ = Kr tl ,r . . l

s(r+3)(s+1),

Kr

K4 r ' l s r t l - '

1 1 2 1 3 )

The key to the partial ftaction technique for finding inveBe transforms lies in recognizing the /(t) coresponding to each telm in the sum of partial fractions. Frcm Thble 12.1you should be able to ve fy that

s+6 .^ | + 3Xr + 1)' [s(s 3)(r = 1Y, + Kre-3'+ K{e | + K4et)u(t).

(12.44)

All that remains is to establish a technique for detemhing tlle coefficients ((1, 1<2,-K3,. . .) genemted by making a pa ial fraction expallsiorlThere are four general forms this problem can take. Specifically,the roots of D(s) arc either (1) real and distinct; (2) complex and distinct; (3) real and repeated;or(4) complexand repeated.Beforewe considereachsituation in turn. a few seneml comments are in oidei,

484

Inhoduction to ihe Laptace Transform

We used the identity sign = in Eq. 12.43to emphasizethat expanding a mtional function into a sum of partial ftactions establishesan identical equation. Thus both sides of the equation must be tle same for all values of the va able r. A1so,the identity relationship must hold when both sides are subjected to t]le same mathematical operation. These chamcteristics are pertinent to determining the coefficients, as we will see. Be sure to vedfy that the rational function is proper. This check is imporranr because nolhingi0 rbepfocedure lor findi;g ihe \ driousKs will ale you to nonsenseresults iJ the mtional function is imprcper. We presetrt a procedure for checking the fs, but you can avoid wasted effort by forming the habit of asking youisell "Is F(s) a pioper rational function?,'

PartialFractionExpansion: DistinctRea[Rootsof ,(s) We first consider determining the coefficients in a partial fraction er?ansion when all the roots of ,(s) are real and distinct. To find a I< associated with a term that arisesbecauseof a distinct root of D(s), we multiply both sides of the identity by a factor equal to the denominator beneatl the desired l(. Then when we evaluate botl sidesof the identity at the ioot corresponding to tne multiplying factor, the right-hand side is always rhe desiredl(, and the left-hand side is alwaysits numerical value. For example, -

a6(' - s)(s t 12) r ( , - 8 X rl o ,

Kt r

K) cr8

K1 r-b

To find the valueof 1K1, we multiply both sidesbys and then evaluateborh sidesats=0:

a 6 ( r 5 ) ( r+ r 2 ) l (,-8)ir or I.o

'

,

K," I K.sI 8 1 . , ' ^ 1 .o '

-96(5)412) := K 1: 1 2 o .

112.46)

To find the valueof &, we mulriply borh sidesby s + 8 ard thenevaluare borhsidesat s = -8:

96(s+sxr+12) "(" + 6l i"= s

:-

-(rts+ 8)l

+ s)l - ^,, ' -. ft:ds r "* 6 )l , = - " ' l , =.

"

96(-3)14)

(-8X-2)

= K-:

1)

\12.47)

Therl ri is 96(r+5)ts+i2)l I r(( + r
,lbp'

-2. "')ult).

|?..0)

0bjedive2-Be ableto catcutate the inverseLaptace transformusingpartialfractionexpansion andthe Laplace transform tabte 12.3 Find/(r)if

12.a Find/0) if 6s2+2or+26 (r+1l(r+2Xs+3)

Answer: /(r) = (3e1 + 2e'''1+ e-3)u1). NOTE: Aho ty ChltptetPrcblems12.10(a) and(b).

7s' +63r+134 (s+lt(r+4)(s+5r' Answer: /(r) : (4e 3t + 6e t - 3e st)u(t).

486

Introduction totheLaptace Tlansfom

PartiaIFractionExpansion: DistinctComplex Rootsof D(s) The only difference between finding the coefficients associatedwith distirct complex roots and finding those associatedwith distinct real roots is that the algebra in tle former involves complex numbers We illustrate by expanding tie rational function: _ "'"' -

t00(s | 3) ," * o1i ' o' + :1

( 1 25 1 )

We begin by noting that F(s) is a proper rational function. Next we must find the roots of the quadratic telm s2 + 6r + 25: f + 6r + 25 = (s + 3 - j4)(r +3 + ja).

Oz.5zl

With tlle denominator in factorod fom, we proceed as before: 100(s+ 3)

(r+6)(f+6s+25) K I

K . K . r-3 /4 r+3-/4

b

To find i:1, &, and i(3, we use the same processas beforel

'K- _

100{i| 3) |

- _

t'

fr + b)(,

J

''

t 1

2s

o s ' 2 s 1 ,.. l00rr- Jr

,,'

too{ 3r

t _ -

too1in)

I t4)l. -r. ,.

{3

/4( 18.,

: 6 _ i8 = 10s 153.13,

02.55)

r00rr- Jr | ( r + 6 X r 3 t 4 r l . -, r "

'-' :6

t \ . . t \

100{i4) 13 j4\-j8l

+ j8 = 10erj:t3.13.

02.56.)

Then

-12 100(s.f 3) (r+6Xl+6.t+25) s+6

r0/ 53.1s" s+3- j4

10/s3.73.

12.7 InvereTnnstornrs487 Again, we need to make someobseNations. First,in physicallyreaLizable circuits, complex roots always appear in corjugate pairs. Second,the coefficients associatedwith these conjugate pairs are themselves coniugates.Note, for example,that i(3 (Eq. 12.56)is the conjugareof f2 (Eq. 12.55).Thusfor complexconjugateroots,you actuallyneedto calculate only half the coefficients. Before inverse{mnsformingEq. 12.5?,we checkthe partial fraction expansionnumerically.Testingat -3 is attractjvebecausethe leffhand sidereducesto zero at this value:

f lr)--i

t 0/ - 5 J . 1 J .

t2

t0 - - - / 5J.l].

i,

= -4 + 2.5/36.a'7"+ 2.5/

36.87.

: -4 + 2.0 + j1.5 + 2.0 , j1.5= 0. Wenowproceedto inversetransformEq.12.57: J

.' i[

l o o r s +- ' 3 r :-:-.'

I

| = r- t2r 6

l { r + 6 ) ( s ' +6 r + 2 s ) l

l'c 't r4rl lge r1

+ 10?,,53.13.€_(3+14), )r,(l).

(12.58)

ln geneial, having t]te function it t]le time domain contain imaginary components is urdesirable. Fo unately,becausethe terms involving imaginary components always come in conjugate pairs, we can eliminate the imaginary componentssimplyby addingthe pairs: 10e t53.13 e (3 i4' + l\e js3.r1 a(1+i4)l :10a1t(eit4t 5313')+ e-t({ '.rr")) = 20€ircos(4r 53.13"),

\12.59)

which eMbles us to simplifyEq. 12.581

1 0 0 (+r 3 ) ," ,1 I + 6r 6)(sz + + 2s)J [(r : I 72eat+ 2].]tcos(ar

53.13')lu0).

112.6A)

Becausedistinct complex roots appearftequently in lumped-parameter linear circuit analysis, we need to sulnmarize these results with a new

488

lntroduclion to the Laphce T€nsforn

transformpair. Whenever D(s) containsdistinct complexroots-that js. jB)(r + d + ip)-a pair ot rermsof rhe form factorsof lhe form (s + a

I< s+a-jB

K" s + d +j B

(12.61)

appearsin the partial Iractior expansion.where the partial fractjoncoeffi cientis,in gencral,a complexnumber.ln polar form.

K - lKleto= lI<11A,

(2.62)

where lK I dcnotesthe qlagnitudeof the complexcoefficient-Then

K':

Klett= KLA.

(12.63)

thc conplex conjugatepair in Eq.12.61atwaysinvcrsc Lransformsas

t.l

K [r+a-jp

,

K'

\

s+d+jpj

=2Ke"tcos(fu+a).

\r2.64)

In applyingEq. i2.64it is importantto nole thai (is definedasrhe coefficiert associatcd with the denominarorterms + a - i6. and K' is delined .'slhe coerhcrenr a..oiiaredsirh rhedcnomi||xlor, o iB.

obiective2-Be abteto catculate the inveBeLaptace transform usingpartiatfractionexpansion andth€ Laplace transform tabte

Ansrver:/(r) = (10e5/- 8.33e{r sin12r"t/.

12.5 Find/0) it 10(r'+ 119) (r+5)(r2+lur+1oo) NOTE: AIto tly ChaptetPrcblems12.40(c)and (d).

PartiatFraction Expansion: Repeated Rea[Rootsof ,(s) To find the coefficientsassociated with the terms generatedby a mukiplc root of multiplicity /,wc mulliply both sidesof the identity by the multiplc rool raisedto its /th power.Wefind the L appearingover the factor raised to lhe fth power by evaluatingboth sidcso{ the identit}' ar rhe mulriple roo1.To find the rcmaining(J' 1) coefiicients,we differentiateboth sides of the identity (,' 1) times.Ai the end of cachdiffereDtiation.we evalu ate bolh sidesof thc identity at the multiplc roo1.The right-handsidc is

12.7 In!€rselransfoms 489

always,thedesired,
=^ "'-ti#|"=;='.Y?"

(1?.66)

To find K2, we multiply both sidesby (s + 5)3 and then evaluateboth sidesat -5i

loo(s+ 2s)| _

+ -K,+ K3(s+ tl"= s

l

' t . ' + Xi(s + 5),1

,

F2.67)

tPfl' = ) - - \ o, o x2 K3 o+ K4xo -)t {

(12.68) , r:.:

.:., i,a :l'

I r . - r . J I . : : i r i i ; .. . t 1 1t . ' , 1 . ) l j , - .

To find tr3 we first mustiiriUliipli both sidesdfiEql 1255by (si+ 5)lr'Next

' !,o-,

-

,,., : , i :r ',*.#tri(j+,5Fi"=_,, (12.69) '*[-{i2],=-

= K3= -100.

,r . (12J0)

490 Intmduction to the Laptace Tnnsfonn To find fa we fint multiply both sidesof Eq. 12.65by (r + 5)r. Next we differertiate both sides twice with respect to:r and then evaluate both sides at r = -5. After simplifying t]le first derivative, the second deriva-

,4[-3]",:..,*[!+ s),(2s s2

+ o+

'1",

+ + s)]"=,, *lr3l"= s r4[zro{"

_40 = 2K+

(12.77)

SolvingEq.12.71for Ka gives K4 = -20.

112.t2)

100(r+ 25) _ 20 _ 400 100 _ 20 (s+ 5)3 (r+5), r+5 s(s+s)3 s

112.13)

Then

At this point we can check our expansion by testing both sides of Eq.12.73 at r = -25. Noting both sidesof Eq. 12.73equal zero when r: 25 gives us confidencein the corectness of the partial fiaction expansion. The inversehansform of Eq. 12.73yields

+ 25)\

-! e, 1 1 0 0 ( s l , 1 s + s 1Jr -

" 120- 200r'e

'rlllvl. ' luor? - 20r

\12.74)

12.7 InvereTnnsfoms491

PartialFractionExpansion: Repeated Complex Rootsof ,(s) We handle repeated complex roots in the sameway that we did repeated real roots: the only difference is that t]le algebra involves complex numbels Recall tlat complex roots always appear ia conjugate pairs and that the coeff;icients associatedvrith a conjugate pair are also conjugates,so t]tat only haiJ tle trs need to be evaluated. For example,

' "'

768 112.15)

G t +* r z 5 7

After factoring the denominator pollnomial, we [aite

Kr K2 (s+314)' -t+3-j4

(r+3l/4)' r+3-14

.

t12 76\

Now we need to evaluate only K1 and (2, because/
,768

'

| j4fl._",,4 tr+3

768 \12.77)

( i8f The value of K2 is

R2

-*L

168 [ . s + 3 + j4)'

_ G2Q68)

r3+j4

l=.,*,.

rl,

2(768)

11?.78)

492

Introduction to the Laptace TGnsfor'n

FromEqs.12.77 and12.78,

Ki-

12,

(12.79)

Ki=i3=31y.

(12.80)

We now group the partial ftaction expansion by conjugate tems to obtain -t "^ -12

rG): [; LTJ+3

- j4), + (s+ 3 + j 4 \ ' )

| 3 / -90' \ { + 1

3 /c0'

i 4

\

s + l + a /

(12.81)

We now write the inverse transform of F(r): 3t 3'cos(4r- 90')]r(r). f(t) = [-24te cos4t + 6?

(12.8?)

Note that if F(r) has a real root a of multiplicity r in its denominator, the term in a partial fraction expansion is of the form

(r + d)' The inve$e tmnsform of this term is

=ffi", ,,{.fo}

(12.83)

If F(r) has a complex root of a + iB of multiplicity r in its denominator, the telm in pafiial ftaction expansion is the conjugate pair

(s+d- jBY (s+a + jB)' The iDversetraNform of this pail is :P1{

lG+'

"

K' iB)' (r+d+jB)' +

I : f 2Kn,-," "''o'rn'* alla'1' rx [;

(12.84)

Equations 12.83and 12.84are the key to being able to inverse-tmnsform any partial fraction expansiotr by inspection. One fu trer note regarding t-hesetwo equations: In most circuit analysisproblems, r is seldom greater tlan 2.Therefore. the inverse transform of a mtional function can be handled wit}l four transform pairs. Table 12.3lists these pairs.

12.7 Inverse Transtorms493 TABLE12.3

Fo|JrUsefulTransfornPairs

Pair Number

Natue of Rools

T(' K K

Kte rtu(t)

K K " s+a jP s+d+ jB K K ' (\+d jpr (r+d+ jB).

2 (Ld ''cos(pt + t ),(r) h KIe-'t.os(.tu+ a)u|)

N , 2 r I n p a i r s t a r d 2 , , < n a r c a l q u n t i t y , q h e r a s i n p a n s 3 a n d 4 . , < i s l n e c o n p l ,c<x Iq u a n t i t r

Obiective 2-Be abteto catcutate the inverseLaptace transform usingpartia[fractionexpansion andthe Laplace transform table Answer:f(r) : (-20te 2tcost+ 20e"sintr0).

12;7 Findf0) it -F 'r"c' r :

"

lr2+ 4t + 5r2'

NOTE: Also tty Chdpter Problem 12.41(e).

PartiaIFraction Expansion: Improper RationaI Functions We concludethe discussionof partial fraclion expansionsby relurning ro an observationmade at the beginning of this section, namely, that improper rational functions pose no sedousproblem in finding inverse tmnsforms.An improper rational function can alwaysbc expandedinto a polyromial plus a proper rational function. The polynomiat is then inversctransformedinto impulse functions and derivativesof impulse functions.Theproper rational functionis invene-transformedby the tech niques ouuined in this section.To iltustrate the procedure.we use thc lunction

ta + 13r' + b6s2+ 20ur+ 300 s' +9s+20

(12.85)

Dividing the denominatorinto the numerator until the remainderis proper rationalfunclion gives

F(r):r' +4s+10+ sherc lhe refmrJos

IUU'r,

30s+ 100 s ' + 9 s+ 2 0 '

- r)s | '0) r. lhe renrrindef.

(12.86)

494

Intbduction to ihe Laptace Tnnsfom

Next we er?and the proper rational function into a sum of partial ftacriorls:

30s+ 100

f+9s+20

-20 30r + 100 = (s+4Xr+5) -+-

50

-.

l\287\

Subsrituting Eq.12.87 inroEq.12.86 yields . n I r r )- s z 4 r - 1 0

5 r"'?

0 rr-"-

(12.88)

Nowwecaninverse-tmnstorm Eq.12.88byinspection. Hence ".. e5() .d;t!) t(tl--t | 4-, I106(r) (20e 4t

st)est)u(t).

(12.89)

12.8 PolesandZerosof F(s) The mtional function of Eq. 12.42also may be expressedas the ratio oI two factored polynomials.In other wordsJwe may write F(s) as

F t s )-

K f r + . r ) ( r+ . 2 ) . l s + . , ) (s+pr)(r+pr.(s+p-)'

(12.90)

where 1{ is the constart a/6_. For example,we may also write rhe function

8s2+120r+400 2s4+ 20s3+ 70s,+ 1O0r+ 48

r("):

8(s' +15s+50) 2(r4+ 10s3+ 351 + 50r + 24) 4(s+s)(i+10) (r + 1Xr + 2)(r + 3)(s+ 4)'

(12.91)

12.9 Initial-andFinalvalue Theorems495 The roots of the denomimtor polynomial,thatis,-pr, -p2. h,..., pu , are calledthe polesof F(s) rhey are rhe valuesof s ar which F(s) becomesinfinitely large.In the function describedby Eq.12.91,thepoles of F(r) are -1, 2, -3, and -4. fhe roots of the numerator polynomial, that is, - zr, - 22, 21,. . . , zn, are calledthe zeros of F(s); they are the valuesof s at which _F(r) becomeszero.In the function descdbedby Eq. 12.91,the zerosof tq(r) aJe-5 and 10. In what folows, you may find that behg able ro visualize rhe poles and zeros of F(r) as points on a complex r plane is helpful. A complex planeis reeded becausethe roots of the pol)'nomialsmay be complex.In t]le complex r plane, we use the hodzontal axis to plot tlle real values of s potes Figure 12.17A Ptotting andzeros onthesptane. and the vertical axis to plot the imaginary values of s. As an exampleof plotting the poles and zerosof F(r), considerthe function

'-'

10(r+ 5)(r + 3 l4i{r + 3 + i4} (r+10.)(s+6 l8r(s+6+j8)

\12.9?)

The poles of F(r) are at 0, 10, -6 + j8, and -6 - j8. The zeros are ar -5, -3 + J4,and -3 - i4. Figuie 12.17showsthe polesand zerosplottedon the s plane,whereX's rcpiesentpolesand O's representzeros, Note that the polesand zerosfor Eq.12.90are locatedin the finite r plane. F(s) can also have either an /th-order pole or an J'th-order zero at infinity. For example, the tunction descdbed by Eq. 12.91 has a secondorder zero at infinity, becausefor large values of r tle function reduces to 4/r', and F(s) : 0 when r = oo.In this text,we are interestedin rhe poles and zeros located in the finite r plane. Therefore, when we refer to the poles and zeros of a mtioml function of s, we are refering to the finite polesand zeros.

12.9 Initiat- andFinat-Value Theorems The idtial- and final-value theorems are useful becausethey enable us to determin€ from F(s) the behavioi of f(1) at 0 and c<).Hence we can check the initial and final values of /(t ro seeif they conform with known circuit behavior, before actually finding the inverse trarNform of F(r). The initial-value theorem statesthat

(1293) < Initial valuetheorem and the final-value theorem statesthat

(12.94) < Finalva(u€theorem The initial-value theorem is basedon ttre assumptionthat /(r) contains no impulsefunctions.ItrEq.12.94,\4'e must add the restriclionthat the theorem is valid only if the poles of F(s), exceptfor a fint-order pole at the odgin,lie in the left half of t}le s plane.

496

Intrcductjon Tnnsform t0 theLaptac€ To prove Eq. 12.93,we stafi with the operational transfom of the first derivative: ( sr\

_ r { L f = , r r s r l r oI lat )

t6,tr

l - i dl e"tlt.

Jo-

{r?.e5J

oo:

Now we take the limit as r-

lim t,r{\) - t(0 )t = rim [*!" s.aln

'0,.

dt

rr2.e6)

Observe thattheright-handsideof Eq.12.96may bewrittenas I fo dt ^ ljm :e'tu

I I , .d\Jo

l*dt | :e"'dIl.

dr

.lo. dl

"- 0; As s- &, (df/dt)e hence tle secondintegal vanishesir the limit.The first integralreducesto /(0') /(0-), which is indeperdentofr. Thus the iglt-hand sideof Eq.12.96becomes

in

r! "-',h - f(0 )

I dt s_6 ln

/(0 ).

lLz.st)

Because/(01is irdependentofr, the left-handside of Eq.12.96may be

[mfsFrs)- /r0 )l - um[rFrr)]-/(0 ).

(12.e8)

From Eq$ 12.97and 12.98,

lirnsFG): /(0') = lja-f(r), whichcompletesttreprcof of the initial-valuetheorem. The prcof of the fiml-value theoremalsostartswith Eq. 12.95.Here we takethelimit asr + 0:

lirqtrFG)-/(r)l=t\^"(fX-'")

(12.ee)

The integrationis witll respectto t and the limit operationis with respect to s,sotheright-hand reduces to sideof Eq.12.99

!L"'a,\ = If'd

r^( I r .o\..i. .,

l

t i,d' .tn

(l2 roo)

Becausethe upperlimit on tle integial is iniidte, this integal may alsobe wiiiten asa limit process:

fff"-' |^4.,,.

(r2.r01)

12.9 Initiat-andFinal-Vatue Theorems 497

where we use ]|, as the symbol of integration to avoid confusion with the upperlimit on the integral.Canying out the integrationprccessyields

,\*t/(/)

.f(o)l -,'1.*t/0)l- f(o-).

(72.102)

Suhstituling Eq. 12.102 into Eq. 12.99gives

riqlrrq(s)l - /(0-) =,tij*f(r)l

"f(o ).

(12.103)

Because Eq. 12.103reducesto the final-valuetheorem, /(0 ) cancels, namely, lim sF(r) " : lim f(r).

r-0

The fiml-value tieorem is useful orly if /( oo) exists.This condirion is true only if all the polesof F(r), exceprfor a simplepole ar the origin,tie in the left half of the s plane.

TheApplicationof Initial- andFinal-Va[ue Theorems To illustmte the applicationof the initial- and finalvalue theorems,we apply them to a function we used to illustrate partial fraction expan sions.Considerthe transform pair given by Eq. 12.60.The initial-value rneoremgrves lim sF(s) = lim

100"[ + (3/r)]

,:;s'[1 + t6lr)]U+ (6/r.)+ (2sls1l

,4,f

: -12 + 12= 0. (r : l-12 + 20cos(-53.13')l(1)

The fural-value theorem gives l00r(s + 3) . l u n \ / t , l t j m - , : - 0 , {-lr J-n (j + 6)(s. + 6r + 25)

,Iij*/(l)

= limi-12€ 6! + 2Oer,cos(4r

5:.r:")lU(r) - o.

In applying the theorems to Eq. 12.60,we already had rhe timedomain expressionand were merely testingour understanding.But the real valueof the initial and final-valuetheoremslies in beingable to test the r-domain expressionsbefore working out the inve$e transform. For example,considerthe expressionfor V(s) given by Eq. 12.40.Alrtough we cannotcalculater'0) until the circuit parametersare specified,wecan checkto see if y(s) predictsthe corect valuesof r.r(0+)and z,(oo).We know from the statementofthe problemthat generatedy(s) that ?r(0+)is zero.We alsoknow that ?)(6) must be zero becausethe ideal inductot is a perfectshort circuit acrossthe dc current source.Finally,we know that the polesof y(s) must lie in the lefr hau oI the r plane becauseR. a. and C are posi!i!econqtanrs. Hencethe polesot rl (r) al.o lie in the teft hall of the s plane.

498

IntrodLr.t on to the Lapta(e Tmn+orm

Applying lhe inilial valuelhcor'cnlyiclds lim jY(:!)

=

5(/dc/c)

"r:;r2[1 + 1/(^ct + 1/(...cr11

Applying the fhal-value theoremgives

l'r,"Yt'l= ilii

s(1d./c)

,2 + (rl RC) + (rrlC)

'llre

derjvedexpressionfor v(s) correctl]'predictsthe initial and final valuesof r(l).

objective3-Understandandknowhowto usethe initial vatuetheorenandthe finat vatuetheorem 12.10 Use the initial and linal valuelheorems1()fird the inilial and Iinal valuesofl0) in Assessment Problems12.4,12.6, and 12.7.

Answer:7,0;4.1;and0,0.

NOTE: AIso ttj Chaptet Prcblen 12.17.

in } r5 $U{_ni The Laplace transform is a tool for convertingtimedomarnequ"rion. ilro lrequency-donJ n equdrion.. accordingto the following generaldefinition:

:tll!\:

I

.lo

I()e"'tt= F(s),

where l(r) is the tine-donain erpression,and F(s) is (Seepaee467.) the frequeDcy-domain expression. ("(r) The sl€p frnction describesa function that experiences a discontiruit]' from one constant level to anotherat somepoint intime. 1(is the magnitudeofthe iunp; if Il = 1, d"(t) is the l|nit step function. (See page469.) The inpnlse funclion (5(r) is defined

I rtrt*: r .J 60) : 0, r + 0. r is the strengthof the nnpulse;if r : 1, K6O is the u t inpulse function.(Seepage171.)

A functional transform is the Laplace transform of a specificfunction. Imporiant functional transformpairs arc summarizedin Table12.1.(Seepage471.) Opcrutionaltr:rnslbrmsdefinethc gcncralmathematical propelies of the Laplace lransfomr.lmportant opera' tiondl trdDsformpairs are sur1lnarizedin Tabtc 12.2. (Seepage47s.) ln linear lumped-parametercircuits,I(s) is a ratronal functionof .r.(Seepage182.) II F(r) is a proper rational function.the inversetransfbrm is found b], a partial fraction expansion.(See pagc,l[33.) IfF(r) is nn inproper ftlional iunction,il canbe inverselranslomred by lirsl exparding il into a sum ol a poly' nomialard a properralionalfunclion.(Secpagc493.) FG) can be expressedas the ratio oftwo factoredpoly' nonjals.The rooh of the denominatorare calledpoles and are plotted as Xs on the complex.!plane.The roots of the numeratorare calledzerosand are plotted as 0s on the complexs plane.(Seepase49s.)

499

The initial-value theorem states that

The theorem is vaiid only if the poles of F(s), exceprfor a first-order polo at the origin, Iie in the left half of fhe s plane.(Seepage495.)

lirn- /(.) = tin sF(r) The theorem assumes that /(t) conrains no impulse lunclions.(Seepage495.) The final-valuetheoremstatestlat

iim /(r) = lim sF(s).

. The inifial- and final-value theorems allow us to predict the initial and final values of /(t) from an s-domain (Seepage497.) expression.

Probtems S€clion 12.2 12.1 Use step functions to write the expressionfor each function shom.

12.2 Use step functions to \r,Titethe expression for each of the functionsshowr inFig.P12.2.

Figure P12.1

Figue?12,2

/o

J('

-20

-10 (a)

fo

rG)

12.3 Make a sketchof /0) for 25s < r < 25s when /(l) is givenby thefollowingexpressior: f(t):

(20t+ 400)u(t+ 20) + (40r+ 400)"(r+ 10) + (400- 40r)r(r 10) + (20t - 400)u(t 20)

500

Inhoduction to the kptaceTransfom

12,4 StepIunctionscanbe usedto definea pirdop function. Thus (t 1, u(t 4) defines a vrindow 1 unit high and3 unitswide locatedon the time axis between1 and4. A function/(t) is definedasfollows

f0) : 0,

tL8 Er?lain why the folowing lunction generates an impulse function as € - 0:

t
:30t,

0< t = 2s

: 60,

2 s= t < 4 s

/\ = oocos[;' - rJ.

4s <, < 8s:

: 301- 300

8s = l = 10s

12,9 In Section 12.3,we used the sifting property of the impulse tunction to show that lt{6(t)} = 1. Show that we can obtain the same result by finding the Laplace hansform of ttre rectangular pulse that exists between +€ in Fig. 12.9 and then finding the limit of tlis transform as € + 0.

Lr.10 a) Showthat

'ta,: l* rr,n'r, a) Sketchl(t) over the interval 2s < t < 12 s. b) Use the concept of the window function to write an expressionfor /(1).

t',,t.

(Hizt: Integiate by parts") b) Use the formula in (a) to showthat E{6' (,)} = r.

Section 123 12.5 a) Fhd the area under the function shown in Fig.12.12(a). b) What is the duiation of t]Ie function when € = 0? c) What is the magnitudeof/(o) when € - 0?

x2.11The triangular pulses shoM in Fig. P12.11 are equivalentto the iectangularpulsesin Fig.12.12(b), becausethey both enclose the same area (1/€) and they both approach infirlity proportional to 1/€2 as € + 0. Use this triangular-pulse representation for 5'(.) to find the Laplacetransformof6"(t).

12.6 Evaluate the following integmls: t 4 " a) / - / (r' a)[5n) a5G 2)]/r.

Flgure P12.11

6'(r)

t4-

b) 1: / lt6(r)+ 6G+ 2.5)+ 60 s)ldr. 12.7 Findl(r) if

fro: ! [ rpten'a-, and

12.1i, Show tlat 3+ io F(d)=,.

.

n'(d).

E{6'r1(t} = sr.

Prcblems501 Sections12.+12.5

l r l

Lr.13 Find t]Ie I-aplace transform of each of t]le following functions: a) f(t) = tu-d' b) /(t) = sinot'

c) Find iP {:=l

I

}.

vr)

d) Checkthe resultsofparts (a), (b), and (c) by first differentiating and then transforming.

12.19 ^) Find tle Laplace tmnsform of the futrction illus-

c) "fG) = sin (or + 0)'

d) ,f0) : r; e) /(r) : cosh(r+ 9). (I{,rl: SeeAssessment Problem12.1.)

XLl4 ^) Find the Laplacetmnsformof re-'r.

tated in Fig.P12.19. b) Find the Laplace tansfolm of the first derivative of the function illustrated in Fig. P12.19. c) Find the Laplace tmnsform of the second derrvaliveoI Lhefuncrionilluslraledin Fig.P 12.t9. Figur€ P12.19 f (!,

b) Usethe opentional transformgivenby Eq. 12.23 to find the Laplacetransformof:

i

LIT

(/?-'' ).

c) Check your result in part (b) by fhst differenti, ating and then transforming the resulting expression. Find t]te Laplace transform (when €-0) of tle derivative of the exponential function illustrated in Fig. 12.8,using each of the following two merhods: a) First differettiate the functiotr and tlen find the tansform of the resulting function. b) Use the opemtional tmnsform given by Eq.12.23.

I by first integmtingand tlen transforming.

Showthat

sle-"' f(t)j = F(s + a).

t".r,^t,t ot{l u^0,}. Fkdsl

Lr.20 a) Find the Laplacetransformof

( r

I

I ydy I.

Checkthe resultsof (a) and (b) by firsr integrar ing andthen transfomhg.

tr.tro .rur{f,.r.,}. o,"ro r {4.or.,}. [d.

)

b) Check the result obtained in (a) by using rhe operationaltransformgivenby Eg. 12.33 1221 Find the Laplacetransformof eachof the following functions: a) f(r): 2oe5t' 2)u( 2). b) /(r) : (8r - 8)lu(t 1) u(t - 2)l + (24 qt)Iu(t - 2) u(t - 4)l

+ (8r- a0x,(t 4) l,(r- s)1.

12.22 Showthat

eIf(at)j :

:'(il

502

Intmductjon to theLaplace Tmnstorm

lL23 Fhd the Laplacetransform{or (a) and (b). a, f u t:

12.27 The switch in the circuit in Fig. P12.27 has been open for a long time. At t : 0, the switch closes. a) Derive the integrodifferential equation tlat govems the behavior of the voltage o, for 1 > 0. b) Showthat

+ l e a ts i n u t l .

b) ,f0) = / .a'cnsax dx. c) Verify the rcsults obtained in (a) and (b) by first carrying out the indicated mathematical opemtion and then finding t]le Laplace transform.

uJll :

udc/RC

s2+(1lRC)s+(llLC)

c) Show that L2.24 a) civen that F(s) : g{/(t)}, sho\i'that

11"): dF(")

=.rr.r,,,.

ud.Q/RLC)

sls' +(1/RC)s + (l/ZCr'

ligwe Pr2.27 Showthat

t-I\'; c)

= Yt( ftttl.

Usetheresultof (b)to findg{rs}, g{r sinBr},

andg{te ' cosht}.

D2A There is no energy stored in the ckcuit shown in 12.8 a) show that if FG) = s{f0)}, Laplace-[ansformable,t]ren

and {/(t)/t} is

1,-'r"":-{*} (H/r?rrUse the definingintegral to write

: l"-(l) ro*^")* 1"",r,0" andthenrevercetheorderofintegration.) b) Startwith theresultobtainedin Prcblem12.2(c) for g {tsinBt} and usethe operationaltransform given in (a) of this pmblem to find :r I sin Bt].

Fig.P12.28al the time the switchis opened. a) Dedve the integrodifferential equation that govems the behavior of the voltage o,. b) Showthat

Ia"/c %G): t 2 + ( 1 l R c ) s + ( I I L C ) c) Showthat

1ls) =

s' +(URC)S+(IILC)

P12.28 Fjgur€

SectionLl,6 12.26 In the circuit shown in Fig. 12.16,the dc current sourceis ieplacedwitl a sinusoidalsouicethat deliversa curent of 5 cos10tA, Thecircuit componentsarc -R: 1 (), C : 25mF, and a : 625mH. Findthenume calexpression for y(r).

12.29 There is no energy stored in the ciicuit shown ir Fig. P12.29at the time t}le switch is opened. a) Derive the integrodifferential equations that govem the behavior of the node voltages u1

b) Show that

Figore PI2.31 11"(r)

VG)= c [ s 2 + ( R l L ) s + ( 1 l L c ) l ligtrrc?12.29 12.32 a) Wdte tle two simultaneous differential equatrons that de,scdbetle circuit showl in Eg. p12.32 in rermsol lbe meshcurren15 4 dnd/r. b) Laplace-transformthe equarionsdedved in (a). Assume that the initial eneryy stored in the cilcutt rs zero. c) Solvethe equationsin (b) for 1r(r) and /r(s).

1230 The switch in tlle circuit in Fig. P12.30has been in

position a for a long time. At , - 0, the switch movesmstantaneously to positionb. a) De ve the integrodifferential equation tlat govems t]le behavior of tlle current ia for t > 0+.

tigureP12.32 60()

25H

l .1 1-0 0-" -0 )lvI) \"l slH Tl

t -"'/

|

)40o it-"t

b) Show that

.fdclr+ (1/RC)l

Is' + (llRc)s + (llLc\l

Seclion 12.7 12,33 Find 0(t) in Problem12.26.

Figure P12.30

1),jolo

R

= C L

12.31 The switch in tlle circuit in Fig. P12.31has been in

position a for a long time. At , = 0, the switch moves instantaneously to position b. a) Derive the integrodifferential equation rhat governs the behavior of the voltage r,, for 1>0-.

1i1.34 The ctucuit parameters in the circuir in Fig. P12.27 6nr! are tR:10ko; a:800mH; and c = 100nF.If rdcis 70 V find a) o,(1)for I > 0 b) t,(r) for t > 0

t1,35 The c cuit parameters in the circurr seen rn E d Fig P12.28have the followirg values:n : 4 kO, L - 2.5H,C = 25 nF, and 1d. = 3 mA. a) Find 0,(t) for 1 > 0. b) Find to0) for I > 0. c) Does your solution for i,(r) make sensewhen 1 = 0? Er?lain.

b) Showthat Ydcls+ (R/1.)

[s' + (RlL)s + (llLC)]'

11.36 The circuit pammeters in the circuit in Fig. P12.29 *pt E are R : 2500 A. L : 500 InH; and C : 0.5 pF. If

ts(r): 15r,G)mA,find,'lt.

504

Intrcdudion to theLapLace lransfbln

12,37 The circuit pammetersin the circuit in Fig. P12.30 arc n:50O, L : 37.25Ix'H, and C = 2 pF. ll = 1d" 100mA, find i,(t) for 1 > 0.

12.42 Fhd /(t) for each of the following tunction$

10r' +85s+95 s2+6r+5

a) The circuit parameten ir the circuit in Fig. P12.31 a r e R = 5 0 0 0O , l , : 1 H , and C:0.25t1F. If Yd. = 15 v, find 0,0) foi t > 0.

Use tlle results ftom Problem 12.32 and the circuit shownin Fig P12.32to a) Find 4(1) and i,(/). b) Find t1(6) and tr(co). c) Do the solutions for i1 and i make sense? Explain.

5(r' +8s+5) F(r) :

s' +4s+5

0

s' +25r+150 F(,) = s+20

12.43 Find /(1) for each of the following functions.

100(r+ 1) r,(s, + 2s + 5)'

a)

12.40 Find/(t) for eachof the followingtunctions: a)

18s' +66r+54 (r+1)(s+2)(r+3) 8r3+8912+311r+3oo

b)

r("): ( r20c +1f

0

s(s+2)(r' +8r+1s) c.)

o,,

17s2+172t+700

(r+2xr' +12r+100) 56s'+112r+5000 s(s' +r4r+625)

12.41 Find /(r) for each of the following functions.

4

s(r' -ss+s0) rG)=

d)

e.)

sG+42 s4(r + 1)

12.,14 Derive ttre transfom pair given by Eq. 12.64.

12.45 a) Derive the transformpair givenby Eq.12.83. b) Derive the translormpair givenby Eq.12.84.

s'(s+ 10)

10(3s' +4s+4) r(r + 2)' ")

.7)

40G+ 2) r(s + 1)3

Sections12.8-0.9 12.46 a) Use the initial-value theorem to find the initial valueof ?,in Problem12.26.

13 6s' + 15s+ 50 r' (f+4s+5)

b) Can tlre final-value theorem be used to find th€ steady-state valueof r'?W}ly?

f+os+s (s + 2)l t6s3 + 12tz + 216s

(s' +2s+5)'

728

Apply t}le hitial- and final-value theorems to each transform pair in Problem 12.40.

.

12.8 Apply the inifial- and finalvalue theoremsto each transfom pair in Problem12.41.

12.49 Apply the initial- and finalvalue theorems to each tmnsformpai in Problem12.42.

probLems 505

12.51 Use the initial- and final-valuetheoremsto check the initial and final values of the current and vol! agein Problem12.28.

!2.52 Use the initial- and final-value theoiems to check the initial and final values of the current in Problem12.30.

12.50 Use the initial- and fhal-value theoremsto check the initial and final values of the curent and vol! agein ?roblem 12.27.

Apply t}le initial- ard final-value theorems to each transformpair in Problem12.43.

Transform TheLaplace in C'ircuit Analysis 13.1Circuitttenentsin thes Donainp. 508 in the 5 Domainp. 511 13.2CircuitAnatysis p. 512 13.3AppLications p. 526 13.4TheTransfer Function Funciion in PartiatFnction 13.5TheTransfer p. 528 Expansions Funchon andthe Convotution 13.6TheTransfer Integratp. 5J1 tunctionandthe 13.7lhe Transfer p. 5i7 SinusoidalResponse Steady-State in Circuit 13.8TheImputseFuncbon

1 Beabteto tGnsforma circuitinto the 5 donain usingLaptace t€nsforms;be sureyou howto rcpresent the initiat undeGtand in the conditjons on energy-stoEge €tements 2 Knowhowto anatyze a circujtin the s'domain ard be abLeto tlanfom an s'domainsotution backto the tine domain. the definitionandsiqnifica.ceof 3 Unde6tand ih€ transferfunctionandbe abt€to catcutate the hansferfurctior for a circujtusjng technjques. s-domain 4 Knowhowt0 usea cncuifstansferfunctionto jis calculate th€ cjrcuj{surjt inpulseresponse, andits steady-state unit steprcsponse, input. rcsponse to a sjnusoidal

506

The Laplace transform has two characteristicsthat make it an attractivetool in circuit analysis.Fi$t,it transformsa setof linear constantcoefficient differential equations into a set of linear polynomialequations,which are easierto manipulate.Second,it automaticallyintroducesinto the polynomialequationsthe initial valuesof the current and voltage variables.Thus,initial conditions are an inherent part of the transform process.(This contrasts with the classicalapproachto the solution of diflerential equations,in which initial conditions are consideredwhen the unknowncoefficientsare evaluated.) We beginthis chapterby showinghow we car skip the stepof w ting time-domainintegrodifferentialequationsand transfbrming them into the r domain. In Sectio[ 13.1,we'll develop the r-domaincircuit modelsfor resistors,inductors,and capacitorsso that we can write .r-domainequationsfor all circuits directly. Section13.2reviewsOhm's and Kirchhoff'slawsin the contextof the,t domain.After establishilgthesefundamentals,weapply the Laplace transform method to a variety of circuit problems in Section13.3. Analytical and simplificatior techniquesfirst introducedwith resistivecircuits-such as mesh-curlentand node-voltagemeth ods and sourcetransformations-can be usedin the s domain as well. After solving for the circuit responsein the s domain,we inversetransformback to the time domain,usingpartial fraction expansion(asdemonstratedin the precedingchapter).Asbefore, checkingthe final time-domainequationsin terms of the initial conditionsand final valuesis an impo ant step in the solution process. The s-domaindescriptionsof circuit input and output lead us, in Section13.4,to the colcept of the transferfunction.Thetransfer function lor a particular circuit is the ratio of the Laplace transformof its output to the Laplacetransform of its input. In Chapters14 and 15,we'll examinethe designusesof the transfet function, bu1 here we focus on its use as an analyticaltool. We continuethis chapter with a look at the role of partial fraction

PracticaI Perspective Surge Suppressors personal Withthe adventof home-based computers, modems, fax machines, andothersensitiveeLectfonic equipment, it is necessary to provideprotectjonfromvoLtage surgesthat can occurjn a househotd circujtdueto switchinq. A comnefcialtyavaiLabte surgesuppressor is showf jn the accompanyjng figure.

Howcanfljppinga switchto turn on a ijght or turn off a hair dryercausea voltaqesurge?At the end of thjs chapter, we witl answerthat questjonusjng LapLace transformtech niquesto analyzea circuit.WewjlLitlustratehow a voltage surgecanbe createdby swjtchingoff a resistjveloadin a circuit operatingin the sinusojdaI st€adystate.

507

508

lhe tapa.eTEnsfo,m in CkcLrit Anatysis

er?ansion(Section13.5) and the convolution integral (Section13.6)in employingthe transfer function in circuit analysis.We concludewith a look at the impulsefuflctionin circuit analysis.

in the s Domain 13.1 CircuitElements The procedurelor developingan.r-domainequivalert circuit foi eachcir' cuit element is simple.First, we write the time-domain equation that relatesthe teminal voltage to the terminal culrent. Next, w.3lake the Laplace transform of the time-domainequation.This step generatesan algebraicrelationship between the s-domain current and voltage.Note that the dimension of a tmnsfomed voltage is volt-seconds,and the dimension of a transformed curent is ampere-seconds.A voltage-to-curenl ratio in the r domain caries the dimensionof volts per ampere.An impedance in the s domair is measuredin ohms.and an admittanceis measuredin siemens.Finally,we constructa circuit model that satisfiesthe relation ship betweenthe s-domaincurrent and voltage.We use the passivesign convention in all tie deivations.

A Resistor in the s Domain We begin$riththe resistanceelement.From Obm's law, 1) :

Ri.

(13.1)

Because R is a constant, ttre Laplace transfom of Eq. 13.1 is

V=RI.

?

?

v:e.{D}

(13.2)

and I=gIi}.

"j n , ' t ! n j t I

Equation 13.2statesthat the r domain equivalentcircuit of a resistoris simply a resistanceof R obms thal caries a current of I ampere-seconds and hasa terminal voltageof yvolt seconds. (a) (b) Figure 13.1 shows the time- and frequency-domaincircuits of the resistor. Note that going froD the time domain to the frequencydomaiD Fig(re13.1Alhercsjstance ehnrent. G)rimedomain. doesnot changethe resistanceelemcnt.

t

i

. Ii

t tlllo ri

J

figure13.2A AnjnductoroflhenrXs canying an initialcunent of 10ampercs.

An Inductorin thes Domain Figure 13.2showsan inductor carrying an initial current of /o amperes. The time-domainequationthat relatesthe terminal voltageto the termi-

"=,#.

(r3.3)

13.1 Circuit Etenents in ther Domain 509 The LaplacetransformofEq.13.3 gives

Y = rlsl

t(u )l = sLI

Lto.

(13.4)

TWo different circuit configurations satisfy Eq. 13.4.The first consists of an impedance of .rl, ohms in serieswith an independent voltage source ofalo vollseconds,asshownin Fig. 13.3.Note that the polarity marks on the loltage sourcealo agreewith the minus sign in Eq. 13.4.Note also that 110 carriesits own algebraicsign;that is, if the initial value of j is opposite to the reference direction for i, then 10has a negative value. The secondr-domainequivalert circuit that satisfiesEq.13.4consists Figure 13.3A Iheseries €quivatent cjrcujttor an oI an impedanceof r-L ohms in parallel with an independentcurent inductor of L henrys canying aniniijatcunent of 10 sourceof 1o/r ampere-seconds, as shownin Fig. 13.4.We can dedve the alternativeequivalentcircuit shownin Fig. 13.4in severalways Ore way is simplyto solveEq.13.4for the currentl aIld then constructthe circuitto satisfythe resultingequation.Thus

', :

v+LIo

"z

= JV ' ; I o

!!

Two otier waysare:(1) {ind the Norton equivalentof the circuit shownin Fig.13.3and (2) startwith the inductor currert as a function ofrhe inductor voltage and then fiird the Laplace tmnsform of the resu]ting integral b equation.Weleavethesetwo approachesto Problems13.1and13.2. figure13.4 paraltelequivalent circuittor an Ifthe initial energystoredin the inductor is zero,that is,if10 - 0, rhe jnductorofA Th€ a henrys canying aninjtiatcurcntof10 s-domain€quivalentcircuit ofthe inductorreducesto an inductorwith an impedanceofsa ohms.Figure 13.5showsthis circuil.

*1 )

A Capacitor in thes Domain An initially charged capacitor also has two r-domain equivalent circuits. Figure 13.6showsa capacitorinitially chargedto y0 volts.The terminal current is

vlsLtl

I tigure13.5A TheJ-domain circuitforaninductor rvhen theinitiatcurentisz€ro.

,

'Iiallsforming Eq. 13.6yields

? ,

'-t"

1 : C[sY - ?,0 )]

I Figure13.5A A capaciior ofC farads injtiattycharged to ro votts.

I=sCV-CVa,

(13.7)

which indicatesthat the r-domain current1is the sum of two bmnch currents (Jnebranchconsistsof an admittanceofrC siemens. and the second branchconsistsof an independentcurrentsourceofCyo arDpereseconds. Figure13.7showsthis parallelequivalentcircuit.

510

IheLaptace Iransfom in circuitAnatysis

we derive the seriesequivalentckcuit for the charyedcapacitorby solvingEq.13.7for y:

(-t)'.7

thc

b

rigure13.7A ThepaEltetequivaL€nt cjrcujt fora capacitor iniiialty charu€d to vovotts.

.r.

ri

hc

"T'

A Y

(13.8)

Figure 13.8showsthe circuit that sarisfiesEq. 13.8. In the equivalent circuits shown in Figs. 13.7 and 13.8, yo carries its own algebraic sign. In other wordi if the polarity of % is opposite to tlre reference polarity for o, yo is a negative quantity. If the initial voltage on the capacitor is zero, both equivalent c cuits reduce to an impedance of 1/rC ohms,asshowr in Fig.13.9. In this chapter, an important first problem-solving step will be to choose between the parallel or series equivalents when inductors and capacitors arc present. With a little forethought and some experience,the correctchoicewill often be quite evident.Theequivalentcircuitsare summarizedin Table13.1.

va/'

figure13.8a Theseries equivateni circuiifora jnjtiatly capaciior chaqed io v0votts.

*1, , *, t,,

FREQUENCYDOMAIN

TIME DOMAIN

.1" rl

,:R

.1" tl r5R

t,

,ln

V=RI

?=Ri

I

I b

Figure 13.9a Thei-domain cjrcujttor a capacitor js zerc. when theinitiatvothge

.1" il ,lzlro

rl

I,

._r

i =!f" ,a' * 4

V=sLI-LIt'

* * T" ,, . +cyo

.l

" =lIo ia' + vo

lsc thc

A

I

Y

'

r

1

C

_. vr ZI as

V

r

"

13.2 Cncuit Anatysis in thesDomain511

13.2 CircuitAnalysis in the s Domain Befbreillustratinghow to usethe r-domainequivalentcircuitsin analysis, we needto lay somegroundwork, F st, we know that if no energy is stored in the inductor or capacitor, the relationshipbetweenthe teminal voltageandcurrentfor eachpassive element takes the form:

V=ZI,

(13.9) < ohm'stawin theJ-domain

whereZ refersto the.r-domainimpedanceof the element.Thusa resistor hasan impedanceof R ohms,an inductor has an impedanceofra ohms, and a capacitorhas an impedanceof 1/sC ohms.The relationshipcon tained in Eq. 13.9 is also contained in Figs. 13.1(b), 13.5, and 13.9. Equation 13.9is sometimesreferredto as Ohm'slaw for thes domain. The reciprocal of the impedanceis admittance.Therefore, the s domain admittance of a resistor is 1/R siemens,an inductor has an admittance of l/sa siemens, and a capacitorhasan admittanceol sC siemens The rulesfor combiningimpedancesand admittancesin the.r domain are the sameas thosefor frequency-domaircircuits.Thus series-parallel simplificationsand A-to-Y convetsionsalso are applicableto s-domain In addition,Kirchhoff'slawsapplyto s-domaincurrentsand voltages. Their applicability stems from the operational transfom stating that the Laplacetransformof a sum of time-domainfunctionsis the sum oI the transformsof the individualfunctions(seeTabte12.2).Becausethe algebraic sum of the curr€ntsat a node is zero in the time domain.the alsehrdic.um or rbe rran.formedcuffentcis dlso,/ero.A simrtarsrateminr holds for the algebraic sum of the transfomed voltages around a closed palh.Thes-domain verrionor Kirchhoifsla$ri.

alg>1 = 0,

(13.10)

alg>Y : 0.

(13.11)

Becausethe voltageand currentat the terminalsofapassiveelement are related by an algebmicequation and becauseK chhoff's laws still hold, all the techniquesof circuit analysisdcvelopedfor pure resistive networks may be used in r domah analysis.Thus rode voltages,mesh curents! sourcetransformations,and Th6venin,Nortonequivalentsare all valid techniques,even when erergy is storedinitially in the inductors andcapacitors. Initially storedeneryyrequiresthat we rnodifyEq.13.9by simply adding jndependentsourceseither in seriesor parallel with the element impedances.The addition ol these sources is governed by Kirchhoff'slaws.

512

jn CjrcujtAnatysjs Th€Laptace Tnnsform

transforms obiective1-Be abteto transforma circlit into the 5 domainusinqLaDtace 13.1 A 500 O resistor.a 16 mH inductor,and a 25 nF capacitorare connectedin parallela) Expressthe adniltance oI lhis parallelcom' biMtion ofelementsas a ratioral function b) Computethe n mericalvaluesof tlle zeros and poles. Answer: (a) 25 x 10 e(s2+ 80,000r+ 25 x 103)/sr (b) z1 : 10,000 j30,000; : 40,000+ j30.000rpr = Lr. z2

13.2 The parallelcircuit in Asses$nenlProblem13.1 is placedin serieswith a 2000f) rcsistoI. a) Expressthc impedanceofthis seriescombinalion as a lational function ofj. b) Computethe Dumedcalvalueso[ thc zcros and poles. Answer: (a.)2000(.r+ 50,000)'/(." + 80.000s + 25 x 103); (b) .zr = -?, = -50,000; -40'000 - j30,000, Pr = pr= 40,000+ /30,000.

NOTE: Also try ChapterProblems13.1and 13.5.

13.3 App!ie*t'i*iis we Dolvilluslralehow to usethc Laplacctransformto determineihe transient behavior oI severallinear lumped paramctcr circuits.We start b]' analyzingfamiiiar circuilsliom Chaptcrs7 and 13bccanscthey representa simple slarling piaceand bccauscthcy show that the Laplacetransform the easeof manipuapproachyieldslhe sameresulls.In all thc cxan1ples. lating algebraicequalions inslead of difibrcntial equations should be apParent.

TheNaturalResponse of an RCCircuit

Figurc13.10.,tThecapacitor djscharge circuit.

vo

we first revisit the natural responseof an -RC circuit (Fig. 13.10)via Laplace transform techn'ques.(You may want to review the classical analysjsofthis samecircuit in Section7.2). Thc capacitoris initially chargedto vovolts.and we are interestediD the lime domain expressions ibr i and r.wc start by finding L ln transferring the circuitin Fig.13.10to thc r domain,wchavea choiceofrwo equivaienl circuilslor the chargedcapacitor.Bccrusewe are interestedin ihe culletrt.thesedesequivalenlcircuiIis more attractivc:it rcsultsin a singlenesh circuil in thc ircqucnc!,domain.Thuswe constructthe s-domaincircuil shownin Fig.13.11. Summingthc voltagcsaroundthe meshgeneratesthe expression

+:+,

+ RI.

fiqure13.11.ir AnJ domajn equivatent circuitfortlre circuitshownin Fig.13.10.

( 1 3r .2 )

Sol\ingEq.13.12for / yields I =

CV|I

uulR

R C s + 1 .' + (fRC)

(13.13)

1 3 . 3 A p p l i c a t i o n s5 1 3

Note that the expressionfor 1is a proper rarionatfunctionoI s and canbe inverse-transformed by inspcctioni i =

'iR':uO, ;e

(13.14)

which is equivalentto thc expressionfor the currentderivedby the classi cal mcthodsdiscussed in Chapter7.ln rhal chapter,rhc currentis givenby Eq.7.26,where7 is uscdin placeofRC. Aller we havc fouid i, the easiestwav to dctcrmine t) is simpllr to apply Ohm'slawrthat is,Irom the circuil, Rt = V,f 'R'ult).

u:

(13.15)

We Dowillustratca way to find o from rhecircuitwirhoutfirstfindingi. In this alternativeapproach,we retum to the original circuit of Fig. 13.10 cYo and lransferjt to the s domainusingthe parallelequjvalentcircuit for the chargedcapacitor.Using the parallel equivalentcircuit is atrractivenow becausewecandesc be the resultingcircuitin rcrmsof a singlenode volf Figure11.12s Anr-domain equivalent circuitforihe age.Figure13.12showsthe new.r domain equivalcntcircuit. circuitshownin Fiq.13.10. The node-voltageequationthat describesthe ncw circuit is

t

lo, "c, = cu".

$ n

(13.16)

SolvingEq. 13.16tor Y gives

,,

uo s + (l/nc)

113.71)

Inverse-transformiDg Eq.13.17lcadsto the samecxpressionfor 1)givenby Eq.13.15, namciy. o = Ve 'tRc = Vae'tu(,t). (13.18) Our purposein deriving by direct use of the transformmethodis to showthat thc choiceof which.! domain equivalentcircuit io useis influerced by rvhichresponsesignalis ofinterest.

objective2-l(now howto analyze a aircuitin ther domainandbeableto transform anJ domainsotutionto the trmedomain 13.3 The swilchin thc circuitshownhasbeenin positior a for a long lime.At I : 0, the smtch is thrown to positionb. a) Find 1,I4, and y, as rationalfunctionsof r.

(b) t = 20? l'5oruG)mA, o1 = 80e r'5o'a(r)V, u2 = 20e 125o'u(t) Y.

b) Findlh
Answer:(a)1- 0.02/(r+ 1250), Yr:80/(s+1250), V=20/(s+1250]\; NOTE: Ako trr ChaptetProblems13.9nnd 13.10.

r00v

5ko

514

Thetaptace Transform in Cncuit Anatysis

TheStepResponse of a ParallelCircuit

Fig(re13.13A Thenep response of a paratteL RLrcircuit.

ldc

1

sC

R

Y sLII!

Next we analyzethe parallelRIC circuit.shownin Fig.13.13.thatwe first analyzedin Exanple B.7.Theproblemis to tind the expressionfor iz after the constantcurrent sourceis switchedacrossthe parallel elements.The initial energy stored in thc circuit is zero. As before,we begin by constructingthe r-domain equivalentcircuit shownin Fig.13.14.Note how easilyan independentsourcecan be [ansformed from the time domain to the frequency domain. We transform the sourceto the r domair simplyby determiningthe Laplacetnnsform ofits time-domainfunction.Here,openingthe switchresultsin a stepchangein the currentapplicdto the circuit.Therelorethe s-domaincurent sourceis g{ 1d"u(1.)}, or /dJr. To lind 12,we filst solvefor y and then use

v Figurc13.14A Thei-domajn equivatent circuitforthe circuitshownin Fig.13.13.

sL

(13.19)

to establishthe s-domainexpressionfor 1r Summing the currents away ftom the top node generatesthe expression SCV

'N'i:+

\13.?0)

SolvingEq.13.20for Y gives

toJC i+(llRc)s+(llLC)

(13.21)

SubstitulingEq.13.21intoEq. 13.19gives I,JJ LC

rlrr+tr/Rc)r+(r/.c)l

(\3.2?)

Substitutingthe numericalvaluesoflR, a, C, and 1d"into Eq. 13.22yields

384x r05 s(s' +64,i)oor+16x10)

(13.23)

Before expandingEq. 13.23irto a sum of partial fractions,we factor the quadraticterm in the denominator: IL

384 x 10. (13.24) r(r + 32,000 j2a,000)(.r + 32,000+ j24,000)'

Now, we can test the r-domain er?ression for 1r by checking to see whether the final-value theorem predicts the correct value for tr ai 1 = cir.All the polesof 1r, exceptfor the firslorder pole at the origin,iie in the left half oI the s plane,so the theoremis applicable.We krow from the bchavior of the cir€uit that after the switch has been open for a long time, the inductor will shorFcircuit the current source.Therefore, the final valueoI i, must be 24 mA.The limit of 11, ass+Urs lim 11' :

384x 1d 16 x 108

(13.25)

(Currents in the s domain cary the dimension of ampere seconds,so the dimension of s1/_will be amperes.)Thus our r-domajn expression

13.3 AppUcatjons 515 We now proceedwith the partial fraction expansionofEq.13.24: K2

K1

s + 32,000 i24,000 K; r+32,000+j24,000

(13.26)

The partial fraction coefficients are

384x 105: 2 4 16 x 103

10r,

(13.27)

384 \ ld ( 32,000+ j24,000)(j48,000)

: 20x 10"1J48

(13.28)

Substituthgthe numedcalvaluesof 1<1and ii2 into Eq.13.26andinversetnnsfoming the resulting expressionyields iL : IA + 4Oe324ffit cos(24,000r+ 126.87')lr(r)mA.

(13.2e)

The answer given by Eq. 13.29 is equivalent to the answer given tor Example 8.7 because 40cos(24,000t+ 126.87'):

24cos24,0001 32 sin 24,000,.

If we weren't using a previous solution as a check,we would tesr Eq. 13.29to make surethat i.(0) satisfiedthe giveninitial conditionsand ir(oo) satisfied the known behavior of the circuit.

516

lhe taptace Tmnsfom in CjrcujtAnatysjs

TheTransientResDonse of a Parattet RLCCircuit Anoiher exampleof ushg the Laplace transfom to find the transient behaviorof a circuit arisesftom rcplacirg the dc currentsouicein the circuitshownin Fig.13.13 with a shusoidalcunentsource. TheDewcurrent sourceis (13.30) where 1u = 24 mA and {, = 40,000 rad/s. As before, we assumethat the hitial elergy stored in the cncuit is zero. The r-domain expressionfor the source curent is s1-

(13.31)

The voltage aooss the paralel elements is (I!:/C)s

s' +(i/RC)s+(r/rc.)

(13.32)

Sub(lilulingL.q.lJ.3l inlo Eq. 11.32rerullrin

QJC\5,

+ (11LC)l' G'+ '1t"' + (JIRC)S from which IL

(.t^/LC)s

sr- 1s' + u)[r' + (l/nc)s + (r/LC)]

(13.34)

Substituting thenumerical valuesof 1-, (1), R,a,and C intoEq.13.34gives 384X 10ss

(s'+ 16x 1o3XC+ 64,ooos + 16x 103)

(13.35)

We noq wrile the denominalorrn facloredform: 184 r 105s

IL

G - joxr + j.,,)G+ d - jB)G+ d + jB)'

(13.36)

whereo : 40,000,a = 32,000,andB : 24,000. Wecan'ttestthe final valueof iz with the final-valuetheorembecause 1r hasa pair of poleson the imaginaryaxis;that is,polesat +i4 x 104. Thuswe must first find i, and then checkthe validity of the expression from knowncircuit behavior. Whenwe expandEq.13.36 intoa sumof partialfractions, wegenerate theequation '

K, KI K2 + j40.000 s-/40.000 s s+32,000-j24,000

+

r+32,n00+j24.n00

(13.37)

13.3 Appticaiions517 The numerical values of the coefficients Kt and 1(2are

Kr

384x 105040,000) + j16,000)(32,000 + 164,000) 080,000)(32,000 =7.5x1o3/ 9(,,

K2

(13.38)

384x 101-32,ooo + j24,ooo) (-32,000 j16,000)(-32,000 + j64,000xj48,000) = 12.5x 10 .12[!.

(13.3e)

Substituting the numericalvaluesfrom Eqs.13.38ard 13.39inro 8q.13.37andinvene-transforming theresultingexpression yields tr = [15cos(40,000, 90") + 25e-3xm.cos (24,0001 + 90')lrnA, : (15sin40,tD0, 25erxom,sin 24,000i)aomA.

(13.10)

Wenow testEq. 13.40to seewhetherit makessensein termsof the giveninitial conditionsandthe knowncircuit behaviorafter the suritchhas beenopenlor a long time.For I : 0, Eq. 13.40predictszeroinitial current, whichagrceswith the initial energyof zeroin the circuit. Equation13.40 alsopredictsa steady-state curent of rz,. : 15sin40,000tnA,

(13.41)

whichcanbeve fiedby thephasormethod(Chapter9).

TheStepResponse of a MuttipteMeshCircuit Until noq we avoided circuits that required two or more node-voltage or mesh-currentequations,becausethe techniquesfor solvingsimultaneous differentialequatioN are beyond the scopeof this text. Howevet,using Laplacetechniques,we can solve a problem like the one posedby the multiple-meshcircuit in Fig. 13.15. Here we want to find the branch currents 4 and i2 that arise when t}le Figue13.15A A muttiph-mesh Rlcircuit. 336 V dc voltage source is applied suddenly to r})e circuit. The inirial energystoredin the circuitis zero.Figure13.16showsthe r-domainequiv8.4r 10r alert circuitof Fig. t3.15.Thetwo mesh-currentequationsare

336 -

0:

336 (42 + 8.4s)11 4212,

42IL+ (90+ 10r)1r.

\13.42)

l)f') l"/i\

48r}

tigure13.16.r'Therdomain equjvatentcircuittorthe jn Fis.13.15. (13.43) circuit shown

Theil,aptacelTran3bh in CircuitAnatysis

UsiDgCramer's ftethodltdsolvefotrtririild'/t:rneobta i'.

: .'":-a .1 . lrlaihrb..il m"iiirrl'' "

iirl;,.1r,ifaF:.laif,iz)l i

: 84(f + 14s+ 24)

..

"

12)i.,.' E+fu+,?)(,f',1 _y1'1y1ir: r ,xraj.:.i jr1:/r.i:, I

: , j i i r . " l . i) 1 i r i ] i ] . i : , i r

''

1336/s - 42 | 90+l0rl l0

..

l+z- e.a"r:ol'l

i 1lt'

/v2- | .t

- 1 ,2.

'.ii, : .:..1..,..\!'4i ra.!4!1i'

.

,)

:it '

" u

I I

,,ji i.r frr'Jll

: ....

\"

,.

it,

J /,.r,irri r.i d.1!rj:,:,;

(13,441

,: t,t

(13.46)

"

Bi{iad.,6tiEdc.15.hjB:4*,|r;jr, i.: Ji.:r i.: i,. ,jl'!.:1i].,:l]:],;li1):..1.''.;|i].J.i,.rj]|':j.ii

-.' 1

Nr 40G+e) A s G- 2 X s + l 2 ) -

\73.47)

t , . i , 1 t ;r ,. . . r r , _ , :

N2 168 '., : - ar (J- tG { 12)'

(13.48)

Expanding 11aDgl?,i$o$surtr9!Pll-rg9!,gq9tio,ns_.9Le,q.,. .. . ,

rrr.ii

I l, ii rl]rt:i

i, : ltsi - t'it-z - trblultS4,

(13.5r)

ir= Q - &4e-a + 1.4:et})u\t) L,

(13.52)

13.3 Apptications519 Next we test the solutionsto seewhetherthey mate sellseiD termsofthe circuit.Becauseno energyis slorcd in the circuit at the instantthe switch is closed,both t1(0 ) and i(0 ) must be zero.The solutionsagreewith theseinitial values.After the switch hasbeen closedfor a ]ong time, the two inductorsappcaras short circuits.Therefore,the final valuesof i1 and

i 1 ( c o ) :M q

t:(6):

= isA,

1544-2) =7A. ^

113.54)

One final testinvolvesthe numericalvaluesof lhe cxponentsand cal culatingthe voltagedrop aooss the 42 O resistorby threedifferent methods.From the circuit,the vollageacrossthe 42 J) resistor(positiveat the top) is D- 42tt, i,t

JJh

dl 8.4 4bt, ,

l

-

d

Inl'.

{13.55,

l

You shouldverify that regardlesso{ which form of Eq. 13 55 is used,the voltageis It : (3s6

23s.2ea - 100.80""')u0) v.

We are thus confidentthat the solutionsfor i1 and t, are correct.

ans domainsotutionto the anatyze a circuitin the s domainandbeableto trensform

.(;) { - IsG + 3)l/t(s + 0.5)G+ 2)1, + 6)l/b(r + o.s)(r+ 2)l; v2: 12.5(sz = (1s (b) ;, ]e{5' + ler')u1r.tv,

13.5

u2=(15- + "4:' + + en')"(t)v:

a) Oeflu. tt'. "-aornainexpressionsfor yL

- 2.5 v; (c) 1)r(0*). 0, ?.'2(0') : = (d) t'r ?,, 1s V

andv2.

NOTE: Also try ChapterPtublems]3'26 and13'27.

.

520

TheLaptace Tlansfonn in CircuitAnatysis

TheUseof Th6venin's Equivalent

b Figur€13.17a A circuittobeanalyzed using T h 6 v e n ienqsu i v a h n ttihnes d o m a i n .

20{)

60f)

a

In this sectionwe showhow to uscThdvenin'sequivalentin the.r domain. Figure 13.17showsthe circuit to be analyzed.The problem is to find the capacitorcurrentthat resultstom closingthe switch.The energystoredin the circuitprior to closingis zcro. To find ic, we first constructthe s domainequivalentcircuit and thcn Iind theTh6veninequivalentof this circuitwith respectto the terminalsof the capacitor.Figurc 13.18showsthe r-domaincirc tTheTh6veninvoltageis the open-circuitvoltageacrossterminalsa.b. Under open-circuitconditions,lhereis Do voltage acrossihe 60 O rcsis 1or.Hence

(a80/sx0.002s) 480

za.t0' b

20 + 0.002s

s+ 104

(13.56)

The Th6veninimpedanceseenfrom terminalsa and b equalsthe 60 O resistorin sedeswith the parallelcombination of the 20 O resistorandthe 2 mH induclor.fhus

figure13.18a Ther-domajn modetofthe circuit s h o winn F i g1. 3 . 1 7 .

^ . 0.002s(20) 80(s+ 7s00)

-

.! + lOa

(13.57)

UsingtheTh6veninequivalent, we reducethe circuitshoNr in Fig.13.18 to the one shownin Fig. 13.19.It indicatesthat the capacitorcurenl 1. equalsthe Th6venin vollage divided by the total seriesimpedance.Thus,

80 (r + 7500) s+104

a80/(r+ r0a) [ 8 0 ( \+ 7 . 5 u 0G) + 1 0 4 )+] [ ( : . t o s l s ] b fiqurc11.19A A simptified versjon ofihe cjrcuit shown in Fig.13.18,usinga IhCvenjn equivalent.

(13.58)

Wcsimplify Eq.13.58 to -'

+ 25 x 106 (s + 5ooo)r' r, + lo,ooos

(13.59)

A partial fraction expansionofEq.13.59 generates

r. -"

3o,ooo + 6 (s + 5000)' s+5000'

(1r.60)

the inverse transform of which is ic : ( 30,0001e-sm0, + 6e sm9u0) A.

(13.61)

We now tesl Eq. 13.61to see whether it makes sensein terms of known circuit behavior.From Eq.13.61,

ic(D : 6 A.

(I].62)

This result agreeswith the initial current in the capacitor,ascalculatedlrom the circuit in Fig. 13.17.The initial inductor current is zero and the iDitial capacitorvoltageis zero,so the initial capacitorcurrentis 480/80,or 6 A. The final valueof the currentis zero,which alsoagreeswith Eq.13.61.Note also from this equation that the current rcvenes sigD when r exceeds 6/30.000,or 200ps. The lact that ic reversessign makes sensebecause, when the switch first closes,the capacitor begins to charge.Eventually this charge is reduced to zero becausethe hductor is a short circuil at r = ,ro. The sign reversal of ic reflects the chargingand dischargingof the capacitor.

13.3 Apptjcations 521

Let's assumethat the voltage drop acrossthe capacitoroc is also of interesl-Once we krow lc, wc find ,. by integmrionin the time domain; that is, 0 c = 2 x 105/ (6

'm'rx. 10.000.r)€

(13.63)

Allhough the integrationcailed for in Eq. 13.63is not difficult, we may avoidit altogetherbylirst finding the r domainexpressionfor % and rhen finding t,c by an inversetransform.Thus ,v, . : i r cI :.

2x105 6s s (r + 5000),

12x tos (s + 5000)''

(13.64)

lrom which

0c : 12 x 105te5mora(r)

(13.65)

You shouldverify that Eq. 13.65is consistentwirh Eq. 13.63and that ir alsosupportsthe observations madewith regardto the behaviorofic (sec Problen 13.33).

obiective2-Know howto anatyze a circuitin thes domainandbeabLeto transform ans-domain sotutionto the time domaln 13.6 The initial chargcon lhe capacitorin ahecircuit \ho\rn i\ 7ero.

(h)/,b: [20(s+ 2.4rl[(({+ 3)(r+ 6r.

ar find rher-domdinfte\enin
[20(:r+ 2.4)]/[s(s + 2)], 2 . 8 )t \ 2):

'IOTE: Al'a trt (hartcr Prcblrn 13.3).

A Circuitwith MutuaIlnductance The next exampleillustrateshow to usethe Laplacetransformto analyze the transient tesponseof a circuit that contains mutual inductance. Figure 13.20showsthe circuit.Thc make-before-break swirchhasbeenin positiona for a long time.At , = 0, the switchmovesinstantaneously to positionb.Tho problemis to derivethe time-domainexprcssionfor i.

0.5F

522

TheLaptace Innsfomin circujtAnatysj,

Figure13.20A A cjrcuitcontaining nagneticatly coupted coils.

(\ 30

M\ 0H

(Lz M) 6H 2()

IM)J2H

r0O

We begin by redrawingthe circuit in Fig. 13.20,with the switch in position b and the magnetically coupled coils replaced with a T-equivalent circuit.l Figure 13.21showsthe new circuit. We now transform this circuit to the s domain.In so doins. we note that

rr(o)-g=sA,

(13.66)

i(r) = 0.

\13.67)

jn Fjg.13.20, figur€13.21 Thecircujtshown wiih thenagnetjcatty coupted coitsrcplaced bya T-equivatent

30

2A

10r} 10 Figure 13.22l Ther-domain equivatent circujt forih€ circuit shown in Fjg.13.21.

Becausewe plan to use meshanalysisin the s domain,we use the se es equivalent circuit for an inductor carrying an initial cwent. Figuie 13.22 shows the r-domain circuit. Note tlat tlere is only one independent vol! age source.This source appearsin the vefiical leg of the tee to account for the initial valueof t}Iecurent in the 2 H inductorof 4(0-) + ir(o ), or 5A. The branch carrying i1 has no voltage source becausea1 M : 0. The two r-domain mesh equations that describe ttre circuit in Fiq.13.22 arc

(3+2s)Ir+2s12=10

(13.68)

2 s I r+ ( 1 2 + 8 s ) 1 2 = 1 0 .

(13.69)

Solvingfoi 12yields -'

(r + 1)(r + 3)

(13.70)

ExpandingEq. 13.70into a sumof partial ftactionsgenemtes I2

1.25 s + 1

1.25 s + 3

(13.71)

Then, i, = (i.25€ l

r.25e-t')ult)A..

113.72)

1 3 , 3 Applications 523

Equalion 13.72reveals thal i, increasesfion zero to a peak value of 481.13n1A iD 519.31ms atlcr the slvitchis movedto posirionb. Ttrereafter, l, decrcases exponentiallytorvardzero.Figurc 13.23sho$'sa plot oIi, versus1.'lliis responsemakcssensein terms of the known physicalbehavior of thc nagneticallycoupledcoils.A currcnl caDerist in thc l,2 inductor I (nis) only if there is a timc varyirg currenr in rhc a1 inducror.As ir decreases trol11i$ initial valuc oi 5 A, L increascstrom zero and thcn approaches Figur€13.23.i Theptotoft: versus tforthe circuit zclo as il approachcsZcro. shownin Fig.13.20.

objective2-Know howto anatyzea circuit in the s domainand be abteto transforman s-domainsotutionto the hme domain 13.7 a) Verify from Eq.13.72that t2reachesa peak valueof48l.l3 mA at, = 549.31nrs. b) Fin.l4. toi I 0. tor rhecircuir.hu$n in Fig.13.20. c) Compuledir/dr when t2is at its peak value. d) Expressl, as a funclion ofdir/dr whcn 12is al its peak valuc. e) Use the resultsobtainedin (c) and (d) to calculatethe pcak valueof i2.

A n s w e r :( a t J r . / r / /= 0 w h e n t : l l n 3 l t i ( h l t l : 2 . 5 ( c ' + t t t ) u Q )A l

(c) -2.89 A/s; (d) i2 =

@di/dt)l121

(e) 481.13 nA.

NOTE: Also ty ChapterProblens 13.36and 13.37.

TheUseof Superposition Becausex'e arc analyzinglinear lunped-parametcrcircuits,we can use superpositionto divide the responseinto componentsthat can bc idenlified with parlicular sourccsand iDitial condiriorls.Disringuishingthese componcnlsis criticalto bcing ableto usethc lransferfunction,lvhichwe introduceil1the Dextsection. Figure 13.24shows our illustrative circuil. We assumcthal at the instantwhen the two sourccsare appliedto thc circuir,rhe induclor is carrying an niidal current of p amperesand that thc capaciroris catr!,ingan initial voltageof ,y volts-Ihc dcsiredresponseof the circuit is the voltage acrossthc rcsislorR2.labeled1r2. Figurc 13.25showsrhe i domain eqnivalcnrcircuit.We optcd lor the parallcl cquivalentsfor /- and C becauser,c anlicipatedsolving lor % usingthe node-voltagemclhod. To lind y, by supcrposition,we calculalethe componcnloI % resulting holn eachsourccacling alone,and rhen we sum thc conlponents_ We bcgin with y! acting alote. Opening cach oI the threc current sources deacti!alesthem. Figurc 13.26sho\rsthe rcsulling circuit.Wc addedrhe nodc vollageyj to aid the aDah,sis.The primesoD la and y2indioaierhar

figure13.24,i A circuitshowing the useofsuper, position in r donainanatysis.

rC

Figure13.25:l Thes-domain equivahntfor thecircuit af lig.73.24.

TheLaptace Transtorm in (ircuitAnaLysis they are the components of 14 and % attributable to 4 acring alone.The lqo equatioDsthat describelhe circuitin Fis. 13.2bare

(+r+.,,),i,,,;:!,

(13.73)

Figur€ 13.26 lhecircuiishown in Fig.13.25 withy,

"cu1* (] * "c )u1= o. \^:

/

\13.74)

For convenience,we int.oduce the notation 1

1

4r-^+-+sC;

(13.75)

YD: -scl

(13.76)

vr,=! + "c.

\13.77)

Substituting Eqs.13.75-13.7 into Eqs.13.73 and13.74gives YIlV't+ \2Vi:

VclRb

Yr2v' r+Yr2Vi=O.

(13.78)

\13.79)

SolvingEqs.13.78 and13.79fot Vigives -Yi/R,

-

YrY,-Yl

'

(r3.80)

Witl the current source 1s acting alone, the circuit shown in Fig. 13.25 ieduces to the one shown h Fig. 13.27.Here, yi and yi are the components of I,i and y2 resulting from Is. If we use t]te notation irtroduced in Eqs.13.75-13.77, the two node-voltageequationsthat descibe the circuit lr]'Fig.13.27 are

Figure13.27l Thecircujtshown in Fjg.13.25, with Is actingalone.

YrVi+YnVi=0

(13.81)

Yr2Vi + YtVi = 1,.

(13.82)

and

SolvingEqs.13.81 and13.82for yi yields " =

E1 Y.Yr- yit *

(13.83)

13.3 Apptications 525

To find the component of f2 resulting from the initial energy stored in the inductor (yi), we must solvethe circuit shownin Fig.13.28,where

YiVi' + Yr2Vi : -pls,

(13.84)

Yr2vi'+Y22v!=0.

in Fig.13.25,with (13.85) Figore13.28A Ihe circuitshown the energized indlctoractingatone.

Thus

v \- ' :

\o Y.tYr yiz

(13.86)

FIom the circuit shown in Fig. 13.29, we find the component of y, (yi\ resulting from the initial eneryy stoied in the capacitor.The nodevoltageequationsdescribingthis circuit are

\rvi'

+ Y12vi"=yC,

(13.87)

Ylzvi'+Y22vi': -'tC.

(13.88) figurc13.29A Thecircuitshownin Fig.13.25,rvith theeneqjzed capacitor actingatone.

Solvingfor Yi' yields

v!'

(&r + Ylr)c Y,tYzz Yiz

v.

(13.89)

The expfessionfor Y, is

vr:vi+ vi+vi' +vi" -

(\tl R) ,, - y2n'c

YIIYD

Y-2/s Yh2

YD

Ytr Y11Yn yrD'e -ctYt-Y) |

^ YtY)-Yi)

v.

(:1.90)

We can find y2 without usingsuperpositionby solvingthe two nodevoltage equations that describe tie circuit shown in Fig- 13.25.Thus

v

Y n v t+ Y t 2 V=2+ +

,

"

yC-!.

Y.U+Y.U:I--\C.

(13.91)

/.13.e2)

You should verify in koblem 13.43that the solution of Eqs. 13.91and 13.92for y2givesthe sameresultas Eq.13.90.

526

ThetapLace Tnnsfomin CircuitAnaLysis

objective2-Know howto anaLyz€ a circaitin the 5 domainandbeableto transform ans-dornain sotutionto the timedomain 13.8 The energysroredin the circuit shownis zcrc at the instantthe two sourcesare turned on.

Ahswer:(a) (r00/3)c'r (100/3)e+'lu0)V; (b)L(soJr, " (50J/. 3lr/(, \ (c) [soe'' - 5o€4']r(t v.

a) Find the comporentof o for I > 0 owingto the voltagesource.

2A

b) Fhd the compoDentof t, for t > 0 owingto the currentsource. c) Find the expressionfor ? when I > 0. NOTE: Also try Cha et Prablem 13.42.

13.4 TheTransfsrFuele{t** The transf€rfuncrionis definedasthe s-domainratio of the Laplacetransform of the output (response)to the Laplace lransform of the inpul (source).In computingthe transferfunction,we restrict our attentionto circuitswhere all initial conditionsare zero.lf a circuii hasmultiple independentsources, we canfind the transferfunction for eachsourceand use superyositionto find the responseto all sources. The transferfunctionis

"r"r:.H.

Definitionof a transferfunction ,r

(13.93)

wherc v(r) is the Laplacelral1slormof the oulput signal,aDd,Y(s) is ihe Laplace lranslorm o[ the input signal.Note that the transfer function dependsoD what is defined as the output signal.Consider.for example. the seriescircuit sholvn in Fig. 13.30.If the current is defined as the responsesignalof the circuit. Figure 13.30/r Asenes,?lf cncuit. 5C

H(s)

R+sL+]/sC

t z L C+ R C t + 1

\r3.94)

In de ving Eq. 13.9,1, we recognizedthat l correspondsto the output y(s) and vs correspondsto the input,Y(r). Ifthe voltageacrossthe capacitorisdefinedasthe outputsignalofthc circuitshownin Fig.13.30,the transferfunctionrs

rlG)

v

1/'C

R+sl,+l/sc

rrac+RCs+l

(13.95J

Thus,becausecircuits may have multiple sourcesand becausethe definition of the output signal of interest can vary a single circuit can genente many transfer functions, Remember that when multiple souces are hvolved, no single transfer function can reptesent tlle total output transfer functions associatedwith each sourcemust be combined using supeiposition to yield the total response.Example 13.1 illustrates the computation of a transfer function for knoM numerical values of 4 L, and C

Derivingthe Transfer Fundionof a Circuit The voltage source ?s drives the circuit shown m Fig.13.31.The responsesignalis the voltageacross the capacitor, rl,.

1000o

a) Calculatethe numericalexptessionfor thetransfer furction. b) Calculatethe numedcalvaluesfor the polesand zerosof the transferfunction,

1000r)

Figur€13.32 A Ther-domain equivalent circuitforthecjrcuit shownjn Fig.13.31.

Solving for % yields

''

+ s000)Ys 1000(s s2+ 6000s+ 25 x 106

Hencethe transferfunctionis Figure13.31A Thecircuittor txamph 13.1.

v" uE 1000(s + s000)

Solution

s' +6000r+25x106

a) The first step itr finding the transfer tunction is to construct the r-domair equivalent circuit, as shown in Fig. 13.32.By definition, the transfer function is the mtio of y,/ys, which can be computed from a single node-voltage equation. Summing the curents away from th€ upper node genemles

u"s 1000

250+ 0.05r 106

b) Thepolesol H(s) aretherootsof thedenominator polynomial.Therefore pr =

3000 J4000,

ft:

3000+ j4000.

Thezerosof l/(.r) arethe rootsof the numerator pol],nomial; thusF1(s)hasazeroat zr :

5000.

528

TheLaptace lransfornin CircuitAnatysis

Objective3-Understandthe definition and significanceof the transferfunction;be abLeto derivea trensferfunction 13.9 a) Dcrive the numericalcxpressionfor the transferfunction %/1s for the circuit shown.

Answer: (a) (b)

a(s) = to(s + 2)/(s/+ 2r + 1o); p 1: - 1

+ j3, pz=

1- j3,

b) Give the nurncricalvalueof eachpole and zeroora(s).

NOTE: Also try ChaptetProblem 13.19

TheLocationof Potesandzerosof lr(s) For lincar lumped-paramctercircuits,H(r) is alwaysa rational funclion oI r. Complex poles and Tcros alwatrsappcar in conjugale pairs lhe poles of H(.f) must lie in thc left half of the r plane if the responseto a boundedsource(one whosevalueslie wirhin sornefinite bounds) is to be boundcd.ThezerosoI A(r) may fie h eilher the right half or thc lett half of the .! planc. in mind, lYe ncxt discussthe rolc With lhesc generalcharacl€r'istics plays response funclion We begin with thc in determirirg thc that 11(s) for finding partial traclion expansiontechnique l(/).

Fc!net'isn 13.5 {h* Transfer in F*rtialFraeti**Fxpn*si*ns From Eq. 13.93we car witc the circuit oulput asthe producl of the transfer lunclion and the driving iunction:

YG) : H(s)x(s).

(13.96)

We have alreadynoted thai -47(s)is a rational function ofr. Reference1() Table13.1showsthat X(s) alsois a rationalfunclion ofJ for the excilalion fuDclionsofmost interestin circuit analysis. Expandingthe right-lund sjde of Eq.13.96into a sum of Pdr'lialtactions produccsa term for each polc of H(s) and X(r). Rememberfrom Chapter12 that polesare lhe roots of the deDominatorpolynomial;zeros are the rools of thc numerator polynomial.'fhe lerl11sgeneratedby thc polcs of H(s) give rise to the transientconponent o[ the total response, whcreasthe terms gencratedby the polesof X(s) give rise to the steadyresponse.wc mean the slate componeDtoI lhc rcsponse.By steady-state responsethat exislsatter the traDsientcomponertshave bccomenegligi ble. Example13.2illustratesthesegeneralobservations.

13.5 TheTBnsfer Function in Partiat FGctjon Apansions 529

Analyzing the Transfer Function of a Cir.uit The circuitin Example13.1(Fig.13.31)is driven by a voltage source whose voltage increases linearly $rith time, namely,?rs: 50t (l). a) Use the tmnsfer funclion to find ?.,,. b) Identily the transient component of the rcsponse, c) Identily the steady-state componert oI the response, d) Sketch0oversusrfor 0 < t < 1.5ms.

(4000r+ 79.70") r,, = [10\6 x 10 ae3moicos + 10r 4 x 10 al"(r)V. b) The ffansientcomponentof 0, is 10\6 x 10-aerm'cos(40001+ 79.70'). Note that this term is generated by the poles ( 3000+ j4000)and (-3000 j4000) of the tansfer function, c) Thosteadystatecomponent ofthe rcsponse is

Sotution a) From Example13.1, H(") =

Thetime-domainexpressionfor r, is

+ s000) 1000(s s2+6ooos+25^tu6

The transfom of the ddving voltage is 50/s2; therefore, t}le s-domain expression for the output voltage is

1000(s + 5000) s0 _. %:G'+6ooor+25x1017'

(10r 4x104),0). Thesetwo tems are generatedby the secondorderpole(K/s') ofthe drivingvolrage. d) Figure 13.33showsa sketchof o. versust. Note that the deviationftom the steady-state solution 10,000t- 0.4mV is imperceptiblealter approximately1 ms.

Thepartialtaction expansion of % is

v

Kr

16

s + 3000 j4000

l4

K 0 0 i0 + j K4 0 0z0 ' 7 K -, s + 3 -;

z

We evaluate the coefficients (1, 1{2, and K3 by usingthe techniquesdescribedin Section12.7:

1
(1(i: s\5 x 1001 12-J!,

1,2 10 8 6 4 2

K2 = 10, K3-

4 x 10 4.

Figure13.33A Th€g€phofr, ve6us r fortuampte 13.2.

ThetapLace Transfom in CircuitAnatysjs

objective4-(now howto usea circuit'stransferfunctionto ca(culate the circsits imputseresponse, unit step response. andsteady-state response to sinusoidat input 13.10Find(a) theunit stepand(b) theunjt impulse aesponse oI lhe circuitshownin Assessmenl Problem13.9. Answer: (a) [2 + (1013)ercos (31+ 126.87')]z(4 v; 'cos O) r0.s4€ (31 18.a3'),Ov.

NOTE: Also try ChapterPrcblens 13.76(a)and (b).

13.11 The unit impulscrcsponseofa circuitis -''cos(240r P, \ r"ti 10.000P wheretan d *. a) Find the transterlunclion of the circuit. b) Find the unil stepresponseof the circuit. Answer: (a) 9600s/(r2+ 140s+ 62,500); (b) 40e ?o/sin240r V.

observations onthe Useof H(s)in CircuitAnalysis Example 13.2clearlyshowshow the transferfunction H(r.) reiatcsto the responseof a circuil through a partial fraction expansion.Holvever,the exampleraiscsquestionsabout the practicalityofdriving a circuil with an increasingramp voltage that generatesan incrcasingramp r'esponse. Eventualiythe circuil componentswill fail under thc stressoI exoessive voltage,and when that happensour linear model is no iotger valid.The ramp responseis ofinterest in practicalapplicationswherethe ramp function increasesto a maximumvalue over a linitc lime inleNal. If the rime takcn to reach this maximumvalue is long conrparedwith the time constanlso[ lhe circuit,the solutionassumingan unboundedramp is valid for this tinile time interval. We make two additionalobservationsrcgardingEq. 13.96.First, let's look al the responseof ihe circuit duc to a delayedinpur. If the inpur is delayedby a seconds. Y.\:((t

a)u(.!

,)] = " ^xG).

and,from Eq.13.96,the responsebecomes y(J) : H(r)x(r)e ," If l(r) : I

\73.91)

'{11G)x(r)}, rhen,fron Eq. 13.e7. y(t

a)u(t - d) - e rIH(s)x(.s)e"'j.

(13.98)

Therefore,dela]'ingthe inpul by., secoDdssimply delavsthe responsc functionby d seconds.Acircuil thal exhibilsthis characteristic is saidto bc time invariant. Second.if a unit impulsesourcedrivesthe circuit.the responseof the circuitequalsthc inverselransformofthe transferfunction.Thusii

.rG)= 60), then-r.(s): I

Y(r)= ,ry(t.

(13.99)

13.6 Tle TclsterFuldiooanothe(o,votLtio.Int"gr'

531

Hence,fromEq.13.99, y(t) - h(t),

(13.100)

wherc the inverse transform of the tansfer function equals the unit impulse rcsponse of the circuit. Note that this is also the natural response of the circuit becausethe application of an impulsive source is equivalent to hsfantaneouslystorh)genergyin the circuit(seeSection13.8).The subsequent release of this stoied energy gives Iise to the natural iesponse (seePioblem 13.86). Actually,the unit impulseresponseof a circuit,ft(t), contains€nough inJormation to compute the responseto any souce that drives the circuit. The convolution integFl is used to exhact the responseof a circuit to an arbitrarysourceas demonstratedin the next section,

13.6 TheTransfer Function andthe Integral Convolution The convolutionhtegral relatestle output y(t) of a lhear time invariant circuit to the input r(t) of the circuit and the circuit's impulseresponse ft(l). The integralrelationshipcanbe expressed in two ways: .y(r):

l'or^t,t,^ t a ^ : l *

n,, nl,l^ta^. {13.101)

We are interested in t}le convolution integral for several reasons.First, it allowsus to work entirely in the time domain.Doing so may be beneficial in situations where r(t) and ,(t) are known only thrcugh experimental data, ln such cases,the transformmethod may be awkward or even impossible,as it would requte us to computethe Laplacetransfom of expedmentaldata. Second,the convolutioninte$al introducesthe conc€pts of memory and the weighting function into analysis.we will show how the concept of memory enablesus to look at the impulse response(or rhe weighting function) ,(t) and predict, to some degree,how closely the output waveform replicates the input waveform, Finally, the convolution integral provides a formal pioc€dure for finding the inverse transform of products of Laplace transfoms. We basedt]Ie de vation of Eq. 13.101on the assumptiontlat ttre circuit is linear and time invariant. Becausethe cir:cuit is linear, the principle of superyositionis valid, and becauseit is time hvariant, the amount of the responsedelay is exacdy the sameas that of the input delay,Now consider Fig. 13.34,in which the block containing fi(t) rcprcsents any linear Figuer3.34 a A blockdhgramofa generat ci'cLrit. time-invadant circuit whose impulse rcsponse is known, r(t repiesents tlle excitation signal and J(t) iepiesents the desired output sigllal. 'We assume that r(t is the general excitation signal shown in Fig. 13.35(a).For conveniencewe also assumethat i(t = 0 for t < 0 . Once you seethe derivation of t}le convolution integral assumingx(t) = 0 for t < 0', the exteNion of tlle integral to include excitation functions tllat exist over all time becomesapparert. Note also that we pemit a discortinuity ir r(r) at the odgin, that is, a jump between 0 and 0-.

532

TheLaptaceTransfom in CjrcuitAnatysjs Now we approximate i(1) by a sedesof rectangular pulses of unifom width Ai, as shownin Fig.13.35(b).Thus

.!(r)- ro(t +.!r(,) +.

+ i,(r) +

.,

(13.102)

where ri(l) is a rectangular pulse that equals r(,\i) berween ,\, and ,\,+r and is zeroelsewhere. Note thal the ith pulsecanbe expressed in termsof step lunctions; that is,

r(t - a(^,X,G- ^,) rlr

(I + ^,\)l].

The next step in the approximation of r(t) is to make A,\ small enoughthat the ith componentcan be approximatedby an impulsefunction of strengthi(i')Al. Figure 13.35(c)showstle impulse represenration, with the strength of each impulse shown in bracketsbesideeach arrow. The impulse representation of i(t) is -r(t) = x(,\o)d,\6(t - ,\) + '(,\1)A,\60 - ,\1) + 5

s tcl Flguer3.35 Theexcitation signatofx(r). (a)A genelaL €xcitation signat. (b)Approxjmatjnq r(r) witha senes (c)Appronmatjng ofputses. r(r) witha series of

+ r(DAI6(t - ^) + ..

(r3.103)

Now when r(t) is represented by a series of impulse functions (which occur at equally spaced intervals of time, that is, at ,\, ,\1,lr, . . .), tlre responsefunctior )(1) consistsof the sum of a seriesof unifomly delayed impulseiesponses. The strengthof eachresponsedependson tle strength of the impulse driving the circuit. For example,let's assumethat the unit impulse response of the circuit contained it t]le box in Fig. 13.34 is the exponentialdecayfunction shown ir Fig. 13.36(a).Then the approximarion ofy(l) is the sum of the impulsercsponsesshownin Fig.13.36(b). Analytically, tle expressionfor l(1) is

)(r) - r(,\o)A.Ir(r-,\o) + r(Ir^^ti(

+ r(,\t^^rt

,\1)

,\t +

+ r(,l'j)A,in(t D +..

(13.104)

As AI+ 0, the summation in Eq. 13.104approaches a continuous integlatio4 or

it^,,r,, - ^,)^^- /-.r(^)ft(r ^)d^.

(13.105)

Therefore, t)-

(b) Figure 13.36A Theapprcximation ofy(r).(a)Ihe jmpulse response oftheboxshown in Fjg.13.34. (b)summjng theimpube rcsponses.

|

x(^)h(t - ^)d^.

(13.106)

If r(l) existsover all time, then the lower limit on Eq. 13.106becomes ,x: thus,in geneml,

ttt: l',{\nt, t)at

(13.107)

11.6 TheTEnsfur Function andtheConvotution Inteqrat 533 which is the secondform of the convolutionintegral givenin Eq. 13.101. We dedve the first folm of the htegral ftom Eq. 13.107 by making a change in the vadable of integmtion. We let ,, : I ,l, and then we note Ihat du = -di,ll : -oo when ,\ : ca, and !r: +co when .d = -co. Now we can write Eq.13.107as

4h(u)((1u),

tfO: l-,ft

: y1t1 l*,(t

,7n@Xa,1.

(13.108)

t(r)

But becausea is just a symbol of integration,Eq. 13.108is equivalentto the first form ofthe convolutionintegral,Eq.13.101. The integral relationshipbetweeny(0, ,0), and r(r), expressedin 8a.13.101.often is written in a shorthandnotation: y(t) : h(t) * x(t) : r(t) * ft(t),

(13.109)

where the asterisk signifies the integral relationsbjp between ft(t) and r(t). Thusfi(t) * r(t) is readas"&(t) is convolvedvrith r(t)" andimpliesthat h(1)' Y(t)'

/I h(^\Y(t ^)d^. '/--

(a)

,(r) M 0 (b)

whereasr(t) + ft(t) is readas"r(t) is convolvedwith h(t)" and impliesthat

(D. rv,

t'

I rl^)hQ ^)d^ '/a 11 0

The integals in Eq. 13.101give t}le most general rclationship for t]le convolution of two functions However, in our applications of the convolution integral, we can changethe lower limit to zero and the upper limit to t. Thenwe car write Eq.13.101as

t

r

Jn

,lo

y ( t )= l t ( t ) t ( r - ^ ) d ^ =

lx(^)h(t- ^)d^.

G) .r(r - I)

(!.rrol (d)

we change the limits for two reason$ Fftst, for physically realizable circuits,t(t) is zero for t < 0. In other words, there can be no impulse responsebefore an impulseis applied.Second,westart measuringtime at the instantthe excitationi(l) is turned on;thereforeiro) : 0 for t < 0-. A gmphic interpretation of the convolution inte$als contained in Eq.13.110is importantin the useof the integralas a computationaltool. We begh with the fi^t integral. For purposes of discussion,we assume that the impulse responseof our ckcuit is the exponential decay function shown in Fig. 13.37(a) and that the excitation function has the wavefom shownin Fig. 13.37(b).In each of theseplots,we replac€dt with ,[, the symbol of integation. Replacing ,trwith -,\ simply folds the excitation function over the vertical a-{is, and rcplacing i with t i slides the folded function to the dght. See Figures 13.37(c) and (d). This folding

l'(IF(r

I)

(e)

inteQetation ofthe tigure13.37A A grap.hic jntesmt convolutjon /;n(^)-v( .\)d^. (a)rhe (c)Ihe (b)Theexcitation function. impulse rcspons€. folded€xcitation function.(d)Ihe fotdedexcitatjon tunctjon disptaced t unjtr.(e)Theprcduct t(I)x( - ^).

534

Th€Laptace I'ansform in ti(uitAnaLysis operation gives dse to the term conrolution. At any specified value of r, the response function )(r) is tlle area under the product function ft(,\)r0 - ,t), as showl] in Fig.13.37(e).It should be apparenrfrcm tlfs plot why the lower limit on the convolution itrlegral is zero and t]le upper limit is t. For ,\ < 0, tle product ,(i)"r(r - I) is zero because,(,\) is zero. For,\ > r, the prcductl'(,l')r(t l) is zero becausex(r I) is zero. Figure 13.38shows the second form of the convolution htegral. Note that tlle product furction h Fig. 13.38(e) confirms the use of zero for the Iower limit and t for the upper limit. Example 13.3 ilustrates how to use the convolution integral, in conjunction with the unit impulse rcsponse,to find the rcsponseof a circuit.

r'(I)

.x(^)

t( r)

l?(

r)

ll(r rF(r)

tigurc13.38 A gmphic interyeiation oftheconvotu(a) tjonjnteqmL ^)'(,\)d,\. The impuke /;n(r (b)Theexcitatjon r€sponr€. function. k) Therotded impulse response. (d)Thefolded imputse response displrc€d, units.(e)Theproductr(r i)x(I).

13.6 TheTnnsfer Function andtheConvotution Inteqrat 535

Usingthe Convolution Integralto Findan outputSignal

,(r)

The excitation voltage ri for the circuit shown in Fig.13.39(a)is sho*n in Fic.13.39(b). a) Use the convolution integml to find o,. b) Plot u, over the rangeofo < r < 15 s.

0 1H

6 ,n: Y G) figure13,39L Thecncuit andexcitation vottage tor Exampte 13.3.(a)The (b)Th€excitaijon cjrcuit. vottaqe.

Figure13.40 A lhe impukercspons€ andthetotd€dercjtation function for ExampLe 13.3.

Sotution a) The fust step in using the convolution integral is to find the unit impulse response of the circuit. We obtain the erpression for % from ttre r-domainequivalertof the circuitin Fig.13.39(a):

i(\)

u

%: -10). w})en or is a unit impulse function 6(t),

r , 0- ^ )

%=h(t)=e'u(t). fiom which h(^) = e ^u(^). Using the first form of the convolution integral in Eq.13.110,weconstructthe impulser€sponse and folded excirarion funcrion shoun in Fig. 13.40,which are helpful in selecting the limits on the convolution integral. Sliding the folded excitation function to the right requires brcaking the integration into three intervals: 0 < I < 5 ; 5 < t < 1 0 ;a n d 1 0 < t < o o .T h e breaksin t]le excitatior function at 0,5, and 10 s dictate ttrese break points. Egure 13.41 sho\qs ttre positioning of the folded excitation for each of these intenals. The analytical exprcssion foi 0; in the time intenal0 < t < 5 is

t,.=4r,0
?]r(r

t)

20

Figlre 13.414 Thedjtptac€rnent of oi(t

^) tor thrce

536

]he Laplace in (ircuitAnaLysk Transfom

Hence, the analytical er?rcssion for the folded excitation function in the interval r - 5 <,1
i\=4(t-,1),

r

r, (v) 20 18 T6 14 1,2 10 8

5
We can now set up the three integral exprcssions for oo,For 0 < 1< 5s:

": I'o'1' ;'1"^1^

2

= 4(e'+ r - 1)V.

Fors< t < 10s, 1).=

20e^d^ +

Jo

=4(s+e-t

e\l

l' ,a{,

t)" ^at

Figur€13.42 A ThevoLtage rcsponse ve6ustjmetol Exampte 13.3.

') v.

Andforl0=r
: 4(e-t

^\.^di

Jt-s

I

1,.47

2

4.54 8.20 12.0'7 16.03

3

e-r 5) + 5d (' 10))V

b) We have computed ?r. foi 1 s intervals of time, using the apprcpriafeequalion.The resulfsare tabulatedin Thble 13.2and showngraphicallyin Fi9.13.42.

5 6 ,7

18.54 19.56 19.80

19.93 10 ll t2 1l l4 15

19.9'7 7.35 2.70 0.99 0.37 0_13

NOTE: Asse.ss,our understandingof conolution by trying Chapter Problems 13.57and 13.58.

TheConcepts of Memoryandthe WeightingFunction ,/(r - I) rurure luirrrrppenr| $ i---\ I "e | t

\ p c c r{ h r s h a o D ( n < J ) \

t-it,r:fiI-,-^ Figur€ 13.43A Thepast,prcsenl andfutur€ vatues of

We mentioned at the beginning of this section thar the convolution integral introduces the concepts of memory and the weighting function into circuit analysis.The graphic interyretation of the convolution integnl is the easiestway to begin to grasp these concepts.We can view the folding and slidingof the excitationfunction on a timescalecharacterizedaspast, present, and Iuture. The vertical axis, over which the excitation function 'I(t) is folded,representsthe presentvalue;past valuesof i(1) lie to the dght oI tne vertical axis, and future values li€ to the left. Figure 13.43 showsthis descriptionof r(t). For ilustrative purposes,we usedthe excitation functionftom Example13.3. When we combinet]le past,present,and future viewsof.r(r r) with the impulse response of the circuit, we see that the impulse response weightsr(l) accordingto prcsentand pasl values.For example,Fig.13.41 showsthat the impulseresponsein Example13.3giveslessweight to past valuesol r(t) than to the prcsentvalue of r(t). In other words,the circuit retainsless and lessabout past input values.Therefore,in Fig. 13.42,o. quickly appmacheszero when the present value of the input is zero (that

13.7 TheTransfer Function andtheSteady-state Sinusoidat Response

is,when / > 10 s). In other words,becausethe presentvalue of the input receivesmore weight than the past values,the output quickly approaches the present value of the input. The multiplication of r0 - i) by t(l) gives rise to the practice of relerring to the impulse response as the circuit weighting ftrctioD. The weighting function, ir turn, determires how much memory tlle circuit has. Memory is the extent to which the circuit's responsematchesits input. For example' if the impulse respome, or weighting function, is flat, as shov,n in Fig.13.44(a),it givesequal weight to all valuesof r(t), past and present. Sucha cbcuit has a pefect memory-However,if the impulseresponseis an impulse function, as shown in Fig. 13.44(b),it gives no weight to past valuesofi(t). Sucha c cuit hasno memory.Thus the more memory a circuit has,the more distortion tiere is between tie waveform of tlle excita, tion function and the waveform of the responsefunction. We can show this relationship by assuming that the circuit has no memory, that is, ,(t : ,a6(1),and ther noting from the convolutioninte$al that

/, =

Jah(^\x(t

Figurc 13.44A Weishtins functions. (a)turf€ct nremory.(b)Nomemory.

^)d^

- n)d^

t",46(^)r(r = Att\t).

(13.111)

20 18 t6

\/

1,2 Equation 13.111showsthat, if the circuit has no memory,the output is a scaledrcplica of the input. The c cuit shownin Example 13.3illustratesthe distofiion between input and output for a circuit that has some memory. This distortion is clear when we plot the input and output waveforms on the samegraph, as inFip.13.45.

13.7 TheTransfer Function andthe Steady-State SinusoidaI Response Once we have computed a circuit's traNfer function, we no longer need to peform a separatephasoranalysisof the circuit to detemhe ils steadystateresponse.Instead,we use the transferfunction to relate the steadystateresponseto the excitationsouice.Fi$t we assumethat

r(r) : ,4cos((,t + d),

(13.112)

and then we use Eq. 13.96to find the steady-statesolution of ),(r). To find the Laplacetmnsformofr(t), we first write x(t) as x(t) : ,4 cos(,t cosd

Asin.,,tsird,

(13.113)

8 2 810

12 14

, (s)

Figure13.45 A Theinputandouhut waveforms for Exampte 13.3.

538

lhe taptace Transfom in CncuitAnalysis

from which

x(") =

(,4cosd)s _ (,asind)o s2+ ,& s2+.2 ,4(scosd o sind) f+0.?

(13.114)

Subsrituting Eq. 13.114 into Eq. 13.96gves the.r-domain exprcssion for

Y(s) = //G)

,4(,'cosd

&,sin d)

(13.115)

The number We now visualizethe paitial ftaction expansionof Eq. 13.115. of terms in the expa$ion depends on the number of poles oI H(r). necause a(s) is not specified beyond being the transfer function of a physicallyrealizablecircuit,the expansionofEq.13.115is Ki sila

KL s-l,)

...

+ )

terms generatedby the polesoflt(s).

(13.116)

In Eq.13.116,thefiISt two termsresultfrom tle complexconjugatepoles of tle ddving source; that is, ,r2 + r,,2= G - jr)G + jo). However, the terms generated by tle poles of ll(s) do not contdbute to t}le steady-stat€ responseof y(t), becauseall tlese poleslie in the left half of the s plane; consequendy,the corresponding time-domain tems apprcach zero as t incleases.Thus the first two terms on the right-hatrd side of Eq. 13.116 determine the steady-state response.The problem is ieduced to finding the partial fraction coefficient .Kr:

".-' :

H(r),a(rcosd- usin,r)|

,+i.

l,=,,

- aslr'O) , H(ja)A(j@cos6 2ja H ( J d ) ,( c o s d + j s i n 0 ) - 1--. ._,, , tHuu\Ac".

(lJ.l17l

In general, H(Jd) is a complex quantity, which we recognize by x'dting it in polar form;thus HAa') = H(io) eiat!').

(13.118)

Note ftom Eq. 13.118that both the magnitude,fl(J(,)|, and phaseangle, 9(o), of the tmnsfer tunction vary with t}le ftequency ,,. When we subslitute Eq. 13.118into Eq. 13.117,t}le er?ressionfor l(r becomes

K1:

+

H(jq eitE,,+i']

(13.119)

13.7 lheTnnsfer Function andtheSteady-Staie SinusojdalResponse 539 We obtain t]le steady-staresolution for y(t) by inveise-transfoming Eq.13.116and,in the process! ignoringt}le termsgeneratedby the polesof H(s).Thus

r13.t20) < Steady-state sinusoidat response comput€d usinga transferfunction which indicates how to use the transfer function to find the steady state sinusoidal iesponse of a circuit. The amplitude of the responseequals the amplitude of the souce,,4, times the magnitude of the transler function, H(yz,r)1.Ihe phase angle of the response,d + d(o), equals the phase angle of the source,d, plus the phase angle of t]te transfer function,0(d). We evaluate both lH(J(,)l and d({r) at the frequency of the souce, o. Example 13.4 illustrates how to use tlle transfer function to fhd the steady-state sinusoidalresponseof a circuit.

Usingthe Transfer Function to Findthe Steady-State Sinusoidal Response The circuit fmm Example 13.1is sholl,'rlin Fie. 13.46. The shusoidal source voltage is 120cos (5000t + 30') V. Find the steady-stateer?ression for o,.

1000r}

The frequency oI the voltage souice is 5000 rad/s; hencewe evaluatel1(s) at 1{(J5000):

1i(j5000)=

1000(5000 + J5000) 25+ 106+ J5000(6000) + 25 x 106 i + i 1_ 1 - j 1

Figure13.46A Thecircuit forExanrpte 13.4.

Solution FromExample13.1, IlG) =

\n

Then,ftom Eq.13.120,

,," = c3*4-,,rooor + 30. 4s")

1000(r+ 5000)

r' + 6000s+ 25* 106

= 20\4 cos(5000r- 15")v.

The abiLity to use the transfer Iunction to calculate the steady-statesinusoidalresponse ofa circuit is important.Notethat if we know H(t{r), we also know H(s), at leasttheoretically.In other words,we canrevene the process; insteadof using .41(r)to find H(lo), we use H(Jo) to find }1(r). Once we know I1(s), we can find the responseto other excitation sources.In this application,we determine1l(J-o)experimentallyand then constructH(s) ftom the data.Practically,this experimentalapproachis not alwayspossible; however,in somecasesit doespiovidea usefulmetlod lor derivinglt(s). In theory,the rclationshipbetweenH(s) and H(/o) providesa link betweenthe time domain and the frequency domairl The transfer function is also a very usefultool in problemsconcerningthe frequencyresponseofa circuit,aconceDtwe introduce in the nexl chaDter,

540

jn CircuiiAnatysis lhe Laptace Iransform

objective4*Know howto usea circuifstransf€rfunclionto calculate the circuit'simpulseresponse, unit step response. andsteady-state rcsponse to sinusoidal input 13.12 The currcnt sourcein the circ it shownis delivering 10cos4r A. Use the tlansferfunctionto compulethe steady-state expressionfor o,,

b) Replacethe 50 kO resistorwith a variablc resistorand computethe value ofresistance necessary to cause?r,to lead os by 120'. 10ko

1 0k l l

- 63.43')V. Answer: 44.7cos(4r 13,13 a) For the circuitshown,find the steadystate expressiorfor l,, when ?'s= 10cos50'0001v'

+ 90') V; Answer: (a) 10cos(50,000r (b) 28,867.s1 O.

NOTE: Also ny Chqter Ptublem\ 13.74and 13.75.

13.8 Y[ietrmpu{se Funaticn i* {ireuit &nal.ysf s Impulsc lunclions occul in circuit analysiseither becauseof a swilchil1g operation or becausea circuit is excited by an impulsive source.The Laplacelranslorlncan be usedto predictthe impulsivccurrcnlsand volt agcscrealedduring switchingand the responseof acircuit to an impolsive soulce.We begin our discussionby showinghow to create aD impulse Iunction with a switchingoperation.

Switching 0perations figurc13.47., A circuitshowing the oeationofan

We usetwo diffcrcnt circuils10illustratehow an impulsefunction canbc createdwith a s\,i.itching operalioD:a capacitorcircuit.and a seriesinduc-

Capacitor Circuit

Figure13.48| Theldomainequivalent cncuitforthe circuiishown in Fig.13.47-

In the circuit shownin Fig. 13.47,the capacitorCr is charged1oan initial vollage of % at the time the switch is closcd.Thc inilial chargeon C, is zcro.Theproblem is to find the expressionfor,(r) as R -0. Figure 13.48 shorvslhe s domaiDequivalentcircuit.

13.8 TheImpulse tunction in CircuitAnatysis From Fie. I3.48,

uo/" R+(1/rcr)+(l/scr) -

Va/R s + (1/RC")'

R:'Rt

\13.r21)

where tllle equivalent capacitanceCrC2l(Cr + Cr) is replaced by C". We invene-transformEq.13.121by inspectionto obtain

': (&.,-*.)*r,

(13.122) tigule 13.49A lhe plotof i(r) versus,fortwo different vatues of R.

which indicates that as R decreases,the initial curent (yo/R) increases and the time constant (RC") decreases.Thus, as R getssmaller, t]te curent starts from a larger initial value and then drops off more rapidly. Figue 13.49showsthesecharactedstics ofi. Apparently i is apprcaching an impulse function asR approacheszerc becausethe initial value of i is approaching infinity and the duration of i is approaching zerc. We still have to detemine whether t}le area under the current function is independent of R. Physically,the total area under the i venus r cwve representstlle total chargetransfe ed to C2after the svritch is closed.Thus

-

|]f;",^,'0,voc.'

113.1?3)

which saysftat the total charge transfered to C2is indep€ndent of n and equalsyoc" coulombs.Thus, as R approacheszero, the current approaches an impulsestrengthYoC";that is,

i - uoc.6(t).

(13.721)

The physical interpretation of Eq. 13.124 is that when R : 0, a finite amount of charge is transfened to C2 instantaneously.Making R zero in the circuit shown in Fig. 13.47showswhy we get an instantaneoustansfer of charge.With R : 0, we create a contradiction when we close the switch: that is,we applya voltageacrossa capacitorthat hasa zeroinitial voltage. The only way to have an instantaneous change in capacitor voltage is to have an instantaneous transfer of charge.Wlen the switch is closed, the voltage acrossC2 does not jump to y0 but to its final value of (13.125)

We leavethe derivationofEq.13.125to you (seeProblem13.78). If we set R equalto zero at the outset,t}le Laplacetransformanalysis will predict the impulsive current response.Thus,

uo/" (1/sC1)+ (1/rcr)

= VC,C,U" fr=c'uo

( 11 32 6 )

542

TheLaplace T€nsform in CircuitAnatysis In writing Eq.13.126,weusethe capacitorvoltagesat, = (f. The itrvene transfonn of a consfant is the constanl times the imDulse function: therefore.fiom Ea.13.126.

i : c"ua6(t\.

1t3.127)

The ability of the Laplace transform to predict correctly the occurrence of ar impulsiveresponseis one reasonwhy the transformis widely used to analyze the transient behavior of linear lumped parameter time-invaiiant

SeriesInductorCircuit The circuit shown ir Fig. 13.50 illustiates a second switching operation that produces an impulsive rcsponse. The pioblem is to find the timedomain expression for o, after the switch has been opened. Note that opening the switch forces an instantaneous change in the curent of 42, which causes?,oto contain an impulsive component, Figure 13.51 shows the .r domain equivalent with the switch open. In deriving this circuit, we recognized that the cuirent in t]le 3 H inductor at r : 0_ is 10 A, and the crllrent in t}le 2 H inductor at r : 0 is zero. Using the initial conditions at t : 0 is a direct consequenceof orrr using 0 as the lower limit on t}le defining integral of the Laplace tmnsform. We dedve the expression for % ftom a single node-voltage equation. Summingthe curents awayftom the node betweenthe 15 O resistorand the 30V sourcegives

u

% - [ ( 1 0 0 / s ) + 3 0=] 0. 3s+10

(13.128)

Solving for % yields

.. 40(r+ 7.s) l2(r + 7.s) % = " ( . + t ' , + 5

(13.r29)

We anticipatethat o, will contain an impulseterm becausethe second termon thedght-handsideofEq.13.129 is animproperrationalfunction. We can expressthis improperfraction as a constantplus a rational function by simplydividingthe denominatorinto the numerator;thatis,

12(r+ 7.s)

(13.130)

r+5

.-

Ll

r=0

figurc13.501 A circuitshowing th€ creatjon of anjmpulsive voLtage.

I 100

Figure13,511Thes domain equivaleni circuitforthe cjrcuitshownin Fig.13.50.

13.8 TheImputse Functjon in CircuitAnatysis 543 Combining Eq. 13.130with the partial fiaction expansion of t}le fint rerm on t}le dght-handside ofEq.13.129gives

6ll

)n

ln

60

10

(13.131)

-'---;,

trom which oo : 126(.) + (60 + 10? 5')4t) v.

1t3.132)

Does this solutio[ make sense?Before aNwering that questio4 let's fint dedve the expressionfor the curent when r > 0-. After the switch has been opened,the cuffent in a1 is the sameas the cunent in 14. If we reference t}Ie current clockwise arolurd the mesh,t}Ie s-domain exTressionis

000/s)+ 30

20

5s+25

,("+t

6 r + 5

4 = ;4 - " + 5 * " +65

=t

2 {13.133)

Inverse+ransforming gives Eq.13.133 i - (4 + 2e 5)u(.i)A.

(13.134)

Before the switchis opened,t]le cu ent in al is 10A, and the current in la is 0 At from Eq. 13.134we know tlat at t : 0+,the currentin at and in Iz is 6 A. Then, the current in Zl changesinstantaneouslyfrom 10 to 6 A, while the curent in a2 changeshstattaneously from 0 to 6 A. From this value of 6 A, t]le current deqeases exponentially to a final value of 4 A. This final value is easily verified from the circuit; that is, it should equal 8 100/25, or 4 A. Figure 13.52shows these charactedstics of ir and i2. How catr we verify that these instantaneousjumps in the inductor current make sensein tems of the physical behavior of the circuit? Fkst, we 4 note tlat the switching operation places the two inductors in series.Any . 2 impulsive voltage appearhg acrossthe 3 H hductor must be exactly balancedby an impulsive voltage auoss the 2 H inductor, becausethe sum of the impulsive voltages around a closed path must equal zero. Faraday,s tiguE13.52l Theinductor currents ve6us fforih€ law states that the induced voltage is proportional to tlle change in flux circujt shown in Fjg.13.50.

544

TheLaptace Iransfomin circujtAnatysjs

linkage (?J: l^/dt). Therefore,the changein flux linkage must sum to zero.ln other words,the total flux linkageimmediatelyafter switchingis the same as that before switching.For the circuit here,the flux linkage betore sNitcbinsis i : Lrir + L2i2 = 3(10) + 2(0) : 30 Wb-tums.

(13.135)

Immediatelyafter switching,it is

,\:(Lr+LrXo*):5(0).

(13.136)

gives Combining Eqs.13.135 and13.136 (0) = 30/s:6,4..

(13.137)

Thus the solutionfor i (Eq- 113.1341) agrees$ith the principleof the conservation of flux linkage. We now test the validity of Eq. 13.132.First we checkthe impulsive jump of i2 from 0 to 6 A at 1 = 0 givesdse terrn 126(r).The instantaneous to an impulse of strength 66(l) in the dedvative of ;2. This impulse gives rise to the 126(t)in the voltageacrossthe 2 H inductor.For I > 0+,di2/ll is 10?-5rA/s; therefore,the voltaget', is b. = 15(.4+ 2e 5t) + 2(-1oe sr) = (60 + 10e-5/)&(r) v.

(13.138)

Equation 13.138agees with the last two terms on the righlhand side of Eq. 13.132;thuswe have confirmedthat Eq. 13.132does make sensein terms of known circuit behavior. We can also check the instantaneous drop from 10 to 6 A in the current i1.This drop gives se to an impulseof -46(t) in the de vativeof t1. Therefore the voltage acrossar contains an impulse of -1260) at the o gin.Thisimpulseexacdybalancesthe impulseacrossa2;that iq the sum of the impulsivevoltagesarounda closedpathequalszerc.

ImpulsiveSources Impulsefunctionscan occurin sourcesas well aslesponses:suchsoulces arecalledimpulsiv€sources. An impulsivesouce driving a circuit imparts a finite amountoI energyinto the systeminstantaneously. A mechanical analogyis strikinga bell with an impulsiveclapperblow.After the energy basbeentransferedto the bell,the naturalresponse of the bell determines the tone emitted (that is, the frequencyof the rcsulting sound waves)andthetone'sduration. vo6(4 In t]le circuit shownin Fig.13.53,animpulsivevoltagesourcehavinga strengthof y0 voltsecondsis appliedto a sedesconnectionof a resistor and an inductor.When the voltagesourceis applied,the initial energyin Figure 13,53l AnRtcircujt €xcited byanimputsive the inductoris zero:therefoiethe initial currentis zero,Theieis no voltage drcp acrossR, so the impulsivevoltagesourceappearsdircc0y across-L,

13.8 TheImputse Function in Circuit Analysl 545 Ar impulsive voltage at the termimls of an inductoi establishesan instantaneous cullent. The curent is

,: | ['roato.

(13.139)

Given that the integral of 6(t) over any hterval that includes zero is 1, we find that Eq. 13.139yields

(o:f"

(13.140)

Thus,in an inlinitesimal momentj the impulsive voltage souice has stored

;"(+)'=:+'

113.111)

in the inductor. The current yo/L now decays to zerc in accordancewith the natural responseof the circuit;that is,

i =fe'nu1t;,

113.\42)

where r = Z/R. Remember ftom Chapter 7 tlat the natuial iesponse is attibutable only to passive elements releasing or storhg energy,and not to the effects of sources.When a c cuit is driven by only an impulsive source, the total responseis completely defined by the natural respoNe; the duiation of the impulsive source is so irfinitesimal that it does not contribute to any forced response. We may also obtain Eq. 13.142by direct application of the Laplace transform method. Figure 13.54shows the r-domain equivalent of the cir, cuit in Fig.13.53. Hence

u"

vo/L

R + s a - s + (R/L)'

i:b"<^ttt, :\"-tr,

(13.143)

(73.744)

Figue13.54 Ther'domain equjvahnt circujtforih€ circuitshownin Fig.13.53. 10()

3H

Thus t}le Laplace transfom method gives the conect solution for i > 0+. Finally, we consider the case in which intemally generated impulses s06(r) ,: o and exlernally applied impulses occur simultaneously.The Laplace transform approach automatically ensures the corect solution for t > 0+ if inductor curents and capacitor voltages at t = 0 are used in consttucting the s-domain equivalent circuit and if externally applied impulses are represented by ttreir transforms. To illustmte, we add an impulsive voltage figu|e13.55.| Thecircuitshownin Fjq.13.50with source of 506(t) in series witl the 100 V source to the circuit shown in anjmputsive votiasesource addedin sefieswiih the Fig.13.50.Figure13.55showsthe new arangement.

546

TheLaptace Tlansfom in CircujtAnaLysjs

50

lSS

I

15(}

Att: [, i1(0):10Aandir(0 ) : 0 A. TheLaplace tiansform of 506(t) : 50. If we usethesevalues,the r-domainequivalentcircuit is as sho\rnin Fig.13.56. Theexpressionfor I is 50+(100/r)+30 25+5r

%

Figure13.56 Theedomainequivatent cjrcujtforthe circuiishown in Fjg.13.55.

: - +16s+5 =

20 s ( r +s )

1

6 =+

1

2

-4

4 -

4

(13.145)

ftom which ilt)

ll)ea'

)ult) A.

(13.146)

The exprcssion for % is

,

' = 3^^l 2 t, r +

\ -^

3Z(s.7.5) . 40fr 7.5) rr: r(r + ).)

2.s \

- t +

r + )/

60

60

20

r+5

60

113.147)

from which

,. = 326(.)+ (60e5' + 60)r(r,V. tt,izlA) t6 l4 . t 2

(13.148)

Now we test the resultsto see whethertley make sense,From Eq. 13.1216, we seethat the currentin a1 and-L2is 16A at I : 0*. As in t]le previouscase,the switchoperationcausesi1 to decreaseinstantaneously from 10to 6 A and,at the sametime,causes12to increasefrom 0 to 6 A. Superimposed on thesechangesis the establishmentof 10A in -L1and 12 by theimpulsive voltagesource;that is,

6

.

z

Figule 13.57A Theinductor cunentr versus iforthe circuit shown in Fig.13.55.

i : ;f

/'soat'ra*: roa.

(13.149)

Therefore i1 inffeases suddenly lrom 10 to 16 A, while i, increasessuddenly ftom 0 to 16 A. The final value of i is 4 A. Figure 13.57shows t1, rr, and i gaphically.

13.8 TheImpulse Function in Cnuii Anatysis 547 We may alsofhd the abrupt changesin i1 and i without usingsuperposition. The sum of the impulsive voltages affoss l,1 (3 H) ard f, (2 H) equals505(1).Thus rhe changein flux linlage must sumto 50;that is, A , \ 1+ A , l " ' = 5 0

(13.150)

Becausei : li, we expressEq.13.150as

3Air+2Ajz=50.

(13.151)

But becauseil and i2 must be equal after the switching takes place,

r1(0)+Alr:r,(0)+air.

113.15?)

Then,

10+Ait=0+4i2.

(13.153)

SolvingEqs.13.151 and13.153 for Ai1andAi2yietds (13.154)

Ai = 16A. Theseexpressionsagree with the previous check. Figure 13.57also indicates that the de vatives of j1 and j2. will contain an impulse at t = 0. Specificaly, the dedvative of i1 will have an impulse ot b5(/). and the derivativeof i2 sill ha\e an impul.e of Io51r1. FigurelJ.58tar.{ b r.fespeclively. illuslrare thederivarj\eaof 4 and12. Now let's tum to Eq. 13.148.Theimpulsivecomponent326(r)agrees with the impulse of 166(t) of di/dt ar $e oigin. fhe terms 60t5r + 60 agreewith the fact that for 1 > 0+. --_ ^dt D , , :l ) r + I , . We test the impulsive component of di/dr by noring rhat it produces an impulsivevoltageof (3)66(t),or 186(1),acrossar. This voltage,along with 326(r) across42, adds to 5060). Thus the algebraic sum of rhe impulsive voltages around the mesh adds to zero. To summarize,the I-aplace tmnsfom will correctly predict the creation of impulsive curents and voltages tlat arise ftom switching.However, the s-domain equivalent circuits must be based on initial conditions at , = 0-, that iq on the initial conditions that exist prior to the disturbancecausedby the switching.The Laplace transform will correctly predict the iesponse to impulsive driving sources by simply representing these sources in the s domain by their correct tiansforms NOTE: Assessyow understanding of the impuhe function in circuit analysisby trting Chapter Problems 13.83and 13.84.

Figure13,58A Ih€ denvative oftl and,r.

548

lheLaptace lranrform in tircuitAnatysis

Practical Perspective Surge Suppressors As mentioned at the beginningof this chapter,voltagesurgescanoccurin a circujtthat is operatjngin the sinusoidat steadystate.our purposeis to transformjs usedto determine the creatjonof a surge showhowthe LapLace in voLtage conductors of a househotd circuit betweenthe line and neutraL whena toadis switchedoff dudngsinusojdaL steady-stat€ operation. Consider the circuitshownin Fig.13.59,whichmodetsa househo[d cifoneofwhichis swjtchedoff at time r : 0. Tosinptit/ cuitwith threeLoads, the anatysis,we assumethat the line-to-neutralvottage, V., is 120A V (rms), a standardhousehoLd voltage,andthat whenthe loadis switchedoff at l : 0, the vaLueof Va doesnot change.Afterthe switchis opened,we canconstructthe s-domaincircuit,asshownin Fig.13.60.Note that because the phaseangLeofthe voitageacrossthe inductiveloadjs 0', ontythe inducthe initiaLcurrentthrouqh the inductiveLoadis 0. Therefore, in the tancein the line hasa non-zeroinitial condition,whichis modeted i-donain circuit as a vottagesourcewith the vatue 2110,as s€en r'n Fjg.13.60. Just beforethe switchis openedat I : 0, eachof the toadshas a !vrt1apeakmdgnilude ol 120v2 lbcT\. steady-stare sinJsoida. vo,tage Alt of the currentflowingthroughthe line from the vottagesourceys is dividedamonqthe threeloads.Whenth," switchis openedat t = 0, atl of resistiveLoad.Thisis the currentin the tinewjt[ ftowthroughth€ remaining because the currentin the inductiveloadis 0 at I - 0 andthe currentin an lherefore,the voLtage dropacross inductorcannotchangejnstantaneously. the remainjngloadscan experience a suryeas the Linecurrentis directed throughthe resistive[oad.Forexampte, if the initral currentin the Lineis 25 A (rms) and the impedance of the resistivetoadis 12 O, the voLtage - 42A-3V dropacross Lneresistor sLrges from lbq.7v ro ()5X\tXl2l whenthe switchis opened.Ifthe resistiveloadcannothandlethis amount of voLtage, it needsto be protectedwith a surgesuppressor suchas those shownat the beginningof the chapter

F i g u r e1 3 . 5 9d L i r 'r i L \ e d o ' f l o d u e d . i i r , I i 1 9 " u ' g F v o L . a g 4

Figure13.60A SyTrbotic r domain cjrcuit.

yaur undestandingaf this PradicaLPeRpediveby trying Chaptel N,TE: Assess PrcbLems 13.89and13.90

549

5ummary We can representeach of the circuit elemerK as an r-domain equivalent circuit by Laplacetransforming the voltage-cu ent equationfor eachelement:

. A time.invariant circuit is one for which, if the input is delayed by a seconds.the responsefunction is also delayedbya seconds.(Seepage530.)

. ResistorY = nI . Inductor:Y : sLI

LIo

. Capacitor:Y : (I/|C)I

+ U"/s

In theseequations,y : 9{u}, t =:S{i}, 10 is the ini tial curaent thrcugh the inductor, and y0 is t}le initial voltageacrossthe capacitor.(Seepages508-509.) We can pedorm circuit analysisin the s domain by replacing eachcircuit(lement$ ith il. r.domainequiva lent circuit. The resulthg equivalent circuit is solved by writing algebraic equations using the circuit anaiysis tecbniquesfrom resistivecircuits.Table 13.1 summarizes the equivalent circuits for resislors, inductors, and capacitonin the s domain.(Seepage510.) Circuit analysisin the r domainis pa icularly advantageousfor solvillgtransientresponseproblemsin linear lumped parameterc cuits when initial conditionsare krown. It is also useful for problems involving multiple simultaneousmesh-currentor node-voltageequationq becauseit reducesproblems to algebraicrather than differentialequations.(Seepages517-519-)

. The output of a c cuit, y(l), can be computedby con, volvingthe jnput,i(t), with the impulseresponseofrhe circuit,i(t):

y(t)= h(t)* *t4: = x(t) + h(t) =

I'

lo

ntt|,1, tlat

x!)h(t

^)d^.

A graphicalinterp{etationof the convolutionintegral often providesan easi€rcomputationalmethod to generate]]r(t).(Seepage531.) We can usethe transfer function of a circuit to compute ifs steady-state responseto a sinusoidalsource.To do so. make the substitutions = ./r, in H(s) and representthe resultingcomplex number as a magnilude and phase angle.If

The transfer function is the r-domain ratio of a circuit's output to its input.lt is rcpresentedas

x(t): A cns(at+ 4),

)'G)

H(j@\ - \H(ja) eje(.'),

-{ (s.) where y(s) is thl3 Laplace transform of the output sig nal, and X(s) is the Laplacetransformof the input signal. (Seepage526.)

then

The partialfraction expansionofthe productfl(r)-lf(s) yields a term for each pole of H(i) and -Y(s). The ll(s) terms correspondto the transientcomponentof the total response;the ,Y(r) terms corespond to the steady-state component.(Seepage528.)

(Seepage539.)

If a circuit is ddven by a unit impulse,r(1) : 50), rhen the responseof the circuit equalsthe inverseLaplace transform o{ the tiansfer function, }(r) = g-tlH(s)} : n(t). (Seepages530-531.)

Laplace transform analysis correctly predicts impulsive curr€nts and voltages arising from switching aDd impul, sivesources. You must ensurethat the r-domainequivalent circuits are basedon initial conditionsat I : 0.r, that is,pdor to the switching.(Seepage540.)

- A Htiut co.ldt | .b | 0\d). ) li

550

Th€Laptace Iunsformin cjrcujtAnatysjs

Problems Section 13.1 13.1 Find the Norton equivalent of the circuit shown in Fig. 13.3. 13.2 Dedve the s-domain equivalent circuit shown in Fig. 13.4 by expressing the inductor current i as a function of the terminal voltage o and tien finding the Laplace tiansfom of this time-domain inlegral equation.

13.8 Find the polesand zerosof the impedance seen looking hto the terminalsa,b of the circuit shown in Fig.P13.8.

FigffeP13.8

0.5H

13.3 Find the Th6venin equivalent of the circuit shown in Fig.13.7.

25O

=20mF

S€ctiotr 13.2 13.4 A 10 kO resistor,a 5 H inductor,and a 20 nF capacitor are in seies. a) Express t]le r-domain impedance of this series combination as a iational function, b) Give the numerical value of the poles and zeros of the impedance. 1jl.5 A 5 kO resistor, a 6.25 H inductor, and a 40 nF capacitor are in parallel. a) Express the s-domain impedance of this parallel combirution as a mtional function. b) Give the numericalvaluesof the polesand zercs ot the impedance. 13.6 A 1 kO resistor is in sedeswith a 500 In}I inductor. This se es combinarion is in parallel with a 0.4 liF

Section 13.3 13.9 The switch in the circuit shown in Fig. P13.9 has Er.r been in position x for a long time. At ,:0, the ropo.irionl. swirchmo\esin.ranrdneously a) Constructans-domaincircuit for 1 > 0. b) Find %. c) Find 1J,.

Figure P13.9

a) Expressthe equivalent.r-domainimpedanceof theseparallelbranchesas a mtional function. b) Determine the numedcal values of the poles and zeros, 13.7 Find the poles and zeros of the impedance se€n looking into the terminalsa,b of the circuit shown in Fig.P13.7. FigureP13.7

13,10 The svritch h the circuit in Fig. P13.10has been in 6nc position a for a long time. At I :0, the switch movesinstantaneously to positionb. a) Construct the r-domain circuit for t > 0. b) Find %. c) Find 11. d) Find zl" for t > 0. e) Find iL for I > 0.

Pmbt€ms551 Figure P13.10 10nF

13.13 The switch in the circuit in Fig. P13.13 has been strd closed for a long time before opening at t = 0. a) Conshuct the r-domain equivalent circuit for

r>0. b) Find%. /1b.r=0

c) Find ?], for r > 0.

tigureP13.13 13.11 The make-before-break swilch ir the circuit it EIIG Fig.P13.11hasbeenin positiona for a long time.At t : 0, it moves instantaneously to position b. Find ?)ofort> 0-

50f)

100

Figure P13,11

5H 15kO

l3.1rl The nake-before-break switch in the circuit seen in tstrG Fig. P13.12 has been in position a for a long time before moving instantaneouslyto position b a = 0. a) Construct the s-domain equivalent circuit foi t>0. b) Find 1, and t,. c) Find % and z',.

13.14 The switch in the circuit seen ir Fig. P13.14has 6Ec beenin positiona for a long time.At , : 0,itmoves instantaneously to position b. a) Find %. b) Fhd o..

FiguIe P13.12 tigureP13.14

!,, 5ILF

Iransfom in circujtAnatysjs 552 ThetapLace 13.15 Thereis no energystoredin the circuitin Fig.P13.15 Eoe at thetimetheswitchis closed.

FigureP13.18

a) Find lJ, for t > 0. b) Does your solution make sense in terms oI known circuit behavior? Explain. figurePl3.15

13.19 The switch in the circuit in Fig. P13.19has been EDft closedfor a long time. At r : 0, the switch is opened. a) Find ?).(t) for t > 0. b) Find j,(t) for t > 0. Fig(reP13.19

13.16 The switchin the circuit in Fig. P13.16has been in 6trft positiona for a long time.At, - 0, it movesinstartaneously from a to b. a) Constructthes-domaincircuit for I > 0.

+

b) Find 1"(r). c) Find i,(t) for t > 0.

1)o

Figurc Pl3.16

i

0.5 H

13.20 Find % and o, in the circuil shownin Fig. P13.20if EPla the initial energy is zero and the switch is closed

Figure P13.20 L2A

5/"F

l20v

20mH 13.17 Find x'" in the circuit shown in Fig. P13.17 if "PIc ts - 154(t)A.There is no energystoredin fhe circuitatr:0. tigureP13.17

200mH

13.21 Repeat Problem 13.20if the initial voltageon the 6trc capacitor is 20 V positive at the lower terminal. 13.22 a) Find the s-domain expression for % in the circuir in Fig.P13.22. b) Use the r-domain expressionderived in (a) to predict the initial- and final-values of r,,. c) Find the time-domainer?ressionfor ?,,. Figure P13.22

7ko 13.18The switchin the circuitin Fig. P13.18has been 6r.r closedfor a longtimebeforeopeningat I : 0. Find ?),fort> 0,

.6u(t)mA.-T-3.2nF

13.23 Find the time-domain expiession for the curent in 64c Urecapacitorin Fig. P13.22.Assumethe reference direction for ic is down.

FlgufeP13.26

30tu(r)v 1324 There is no energy storcd in the capacitors in the E ic circuit in Fig. P13.24ar rhe time the switch is closed. a) Construct the s-domain circuit for t > 0_ b) Find 1r,I{, ard Yr. c) Find ir,01, and t2. d) Do your answersfor i1, 01,and 1,2make sensein terms of known circuit behavior? Explain.

13.2? There is no eneryy stored in the ctucuitin Fig. P13.27 6trG at the time the sourcesare energized. a) Find 11(s)atrd1,(s). b) Use the initial- and 6na1-valuetheoiems to check t}le initial- and fhal-values of llc) and tr(t. c) Find 4(1) and ir() for r > 0.

figureP13,24

Figur€ P13.27 4OO

---*lr

2.5rF 1OH

5qTF

1s4(r)A

60,(r)V

13.25 There is no eneryy stoied in rhe circuir in Fig. p13.25 6rlc at the time the voltage sourceis energized. a) Find % and 1". b) Find ?)oand i, for, > 0. 1328 There is no eneryy stored in the circuit in Fig. P13.23 EPr.r at the time the voltage souce is tumed on, and 'os= 75u(t) Y a) Find % and 1,. b) Find 0" and to.

Figure P13.25 400()

50,000r"3or(r)

+

5H r,

c) Do tle solutions ior 1). a'nd,i. make sense in tems of knowl circuit behavior? Explain. figllreP13.28 4mF

13.26 There is no energy stored in the circuir in Fig. p13.26 a) Use the node voltage method to find 0o. b) Findrhetimedomaine\pression lor 1,. c) Do your answeNin (a) and (b) make sensein tems of known circuit behavior? Explain.

554

jn Circuit Analvsis Transtom TheLaptace

1319 The initial energy in the circuit in Fig. P1329 is xnc zero.Theideal.voltagesourceis 600&(t)V

tiglre P13.31

a) Flnd %(r). b) Use the initial- and final-value theorems 10 find t'o(o') and 0,(oo). c) Do the values obtained in (b) agree with kno*r circuit behavior? Er?larn. d) Find 0,(4. Figurc Pt3.29 at 4

2oH

1oo __L_--

-

lT"*

'

140 f,l

d) Find i,(1)

13.30 Thereis no eneryystoledin t}Iecircuit ir Fig P13.30 Edc

att:0

13.32 There is no eneryy stored in t}!e circuit in Fig. P13 32 Btrc at the time the current source turns on. Given that js = 100'(, A: a) Find 1,(s). b) Use the initial- and final_value theorems to find i(0") and,.(co) c) Detemine if the results obtained ill (b) agree vrith known circuit behavior.

.

a) Find %. b) Fhd t',. c) Does your solution for 1'omake sensein tems of kDown circuit behavior? Er?lain

P13.32 Figure 20 i4

;

25H

zs()

P13.30 Figure

13,33 Beqinninawith Fq. l3.b5,.bowlhal lhe capacitor

3&(r)A

15a(,)v

cu;e01 ; the cjr;uil in Fig. 13 L9 is posidte lor 0 < I < 200ps and negahvefor I > 200ps Also show that at 20Ops, the current is zero and that t\is correspondslo when due/dr is teto

13.34 The switch ir the circuit shown in Fig. P13 34 has 13.31 There is no energy stored in the circuit in Fig. P13 31 Ed( at the time the current souce is energized. a) Find 1aand Ib. b) Find i. and tb. c) Find %, Vb,and %. d) Find t)., tb, and ttc. e) Assume a capacitor will break down whenever its terminal voltage is 1000V How long after the current soluce tulns on will one of t]re capacltors break down?

been open for a long time. The voltage of the sinusoidal sourceis o, = y..io1tt + d) The switch closesat t = 0. Note that tlte angle d in the voltage exprcssion detemines the value of the voltage at thi moment when the switch closes, tllat is, 0s(0) - Y- sin C. a) Use the Laplace tmnsform method to find ifort > 0. b) Using the expiession derived in (a), write the expression for the current after the switch has beenclosedfor a long time.

Probtems 555 c) Using the e4ression derived in (a), write tle expressiotrfor the transient componelt of j. d) Find the steady-state expression for i using the phasor merhod.Verify that 'our eypressionis equivalentto that obtainedir (b). e) Specify the value of d so tlat the circuit passes directly hto steady-state operation when the switch is closed.

13,37 a) Find the cunent in t]le 4 kO resistor in the circuit in Fig. P13.36.The reference direction for the current is dowll through the resistor. b) Repeat part (a) if the dot on the 12.5H coil is revefsed,

13.38 The magneticaly coupled coils in the circuit seen trPI" in Fig. P13.38 carry initial currents of 3 and 2 A,

Figu|e Pt3.34

r:o

a) Find the initial energy stored ir the circuit. b) Find 11and 12. c) Find il and iz. d) Find the total energy dissipared in the 600 and 1350O resistors. e) Repeat (a) (d), witl the dot on the 90 mH inductor at the lower teminal.

rJ

13.35 The two switchesin the circuit shown in Fig. P13.35 E?'.r operate simultaneously.There is no energy stored ir the circuit at the instant the switches close.Find ilt) fot t > 0' by first finding the s-domain Th6venin equivale[t of the circuit to the left oI the terminalsa,b.

tigureP13.38 30nH

f

|/-\ I r / r' ) 4 0 m H )

6ooot

I

Figurc P13.35

.]r-\l-. 1

.]l | /---\ | ( 9 o m r t rr , ) l

|

,'"1 \,L

/4r:soo

I

13.39 In t})e circuit in Fig. P13.39,switch 1 closesat I : 0, o"a and the mate-before-break switch moves instantaneously lrom position a to position b. a) Construct the s-domain equivalent circuit for l>0. 1336 Therc is no ene4y stored in the circuit in Fig. P13.36 BtrG at t]le time the switch is closed. a) Find 11. b) Use the initial- and final-value theorems to fhd ,1(0*)and 4(co). c) Find i1.

b) Find 1r. c) Use the initial- and final-value theorems to checklhe initialandtinai!aluesoi i2. d) Find i2 foi t > 0*. Figure P13.39

Figure P13.36 1.5H 0.5H

1ko

1H

556

TheLaptace TEnsforn in Cncuit Anatysk

13.40 The make-before-break svritch in t}le circuit seenin 5m Fig.P13.40hasbeenin positiona for a long time.At t = 0, it movesinstantaneouslyto position b. Find t,forl > 0.

13.44 Find rJ,(t) in t})e circuit shown in Fig. P13.44if the Erc ideal op amp operates within its linear range and .,s = 400u(r)mV.

figureP13.44

Flgure P13.40

4nF

13.41 The switch in the circuit seen in Fig. P13.41has 6flft beenclosedfor a long time beforeopeningat, = 0. Use the Laplace transform method of analysisto find i,. 13.45 The op amp in t}le ciicuit showr in Fig. P13.45 is aac ideal. There is no eneigy stored h the circuit at the time it is energized.If 1," = Z6,06gtr1t,t, t.a (a) %, (b) ?,.,(c) how long it takesto satuiatethe operational amplifier, and (d) how small the rate of inqeasein 0s must be to preventsatumtion.

FigureP13.41

72Y

2H

50f,)

Figure P13.45 10nF

13,42 There is no energy stoied in the circuit seen in 6PIa Fig. P13 42 at the time the two so[ces are energized. a) Use the pinciple oI superpositionto find %. b) Find t), foi t > 0.

FigrreP13,42 2ko 60a(r)V

50 -

1H

soonF ( I )l2(rlnA 4 k o

13.92for /2 13.43V€rifythatthesolutionofEqs 13.91and yieldsthesameexpression asthatgivenby Eq.13.90.

13.46 The op amp in the circuit shown in Fig. P13.46 is E rc icleal.Thereis no energystoredin the capacitorsat the instantthe clrcuit is energized. a) Find 2,,if Isr = 16!10)V and z)s,: 8u(r) V. b) How mary millisecondsafter the two voltage saturate? sourcesare turned ondoesthe op aIl1lp

figureP13.46

Seciions 13.+-13.5 10 nF

13.49 Find the numerical expressionfor the transfer func, tion (y"/y, oI each circuit in Fig. P13.49 and give the numerical value of the poles and zercs of each traNferfunction. figureP13.49 25kO

800

r--l

15kO

13,47 The op amps in the circuit show]) in Fig. P13.47are sPIt! ideal. There is no energy stored in t}le capacito$ at t : 0-.If0s = 180r(t) mV, how manymilliseconds elapsebefore an op amp saturates?

5nH

Figure P13.47

800ko 400k()

i 13.48 The op amp in the ctucuit seen in Fig. P13.48 is tsdc ideal.There is no energy stored in the capacitors at the time the circuit is energized. Determine (a) %, (b) r,,, and (c) how long it takes to saturatethe operational amplifier. Figure P13.48 100nF

13.S0The operational amplifier in the circuit in Fig. P13.50 isideal. a) Find tlre numerical expression for the transfer tncrion H(s) : U"/Vs. b) Give rhe numericalvalue of eachzeio and pole of 11(s). Figure P13.50

100nF

250nI

l0 ko s09,nF 400ko

6u(r)V

400ko

1 2 . 5V

i

t

;

Transform in CircujtAnatysjs 558 Th€Laptace

13,51 The opemtion amplifier in the cncuit in Fig. P13.51 is ideal. a) Derive the nume cal er?ressionof the transfer funcrion H(r:U/vs for the circuit in Fig.P13.51. b) Give the numedcal value of each pole and zero of H(s). tigureP13,51

Figure P13.53

400ko + 62.5nF 400ko 800kf)

1354 In the circuir of Fig. P13.54 i, is ttre output signal

8nF

and ?rsis the input sEnal. Find the poles and zeros of tlre traNfer tunction.

62.5kO

Flgure P13.54

;

13.52 The operational amplfier in the circuit in Fig. P13.52 is ideal. a) Find fie numerical expression for the ttansfer lnncnon H(s) - WVg b) Give the numedcal value of each zero and pole of 11(r). rigureP13.52 Cr=,lnF

Rr:25kO

13.55 There is no energy storcd in the circuit in Fig. P13.55 Epr.r at uhetime the switch is opened.The sinusoidalcurient sourceis generatingtho signal60 cos4000t mA. The responsesignalis the current L. a) Find the tiansfer function 1,/1s. b) Find 1,(s). c) Descdbethe naturc ol the transientcomponent of i,(t) without solving for to0). d) Desoibe the mture of the steady-statecomponent of io(r) without solvhg for t"0). e) Verify tle observations made in (c) and (d) by finding i,(r). Figure P13.55

13.s6a) Find the transfer function /o//s as a function of 13.53 a) End the numerical exprcssionfor the tiansfer tunction H(s) = U.lU for the circuit tn Fig.P13.53. b) Give the numericalvalueof eachpole and zero of 11(s).

o

/r.for the circuit seen in Fig. P13.56. Find the largest value of p that will prcduce a bounded output signalfor a bounded input signal. Find i" for rl = -0.5, 0, 1, 1.5, and 2 if is : r0&(,) A.

Pobtems559 FigreP13.56 0.5k()

t ) '" Srko

10H

c) Repeat(a) when ft(l) changesto the rectangular pulsesho*n in Fig. P13.60(c). d) Sketch ](r) venus I for (a)-(c) on a single graph. e) Do lhe skercbe. iD(d) makesen5e? E\plain. Figure P13.60

o(') I

Section 13.6 13.57 A rcctangular voltage pulse ?'i - [!r0) - u0 - 1)] V is applied to the circuit in Fig. P13.57.Use rhe convolution integal to find 0o.

orl

-F---1

4 0 r

4 0 t (a)

Figure P13.57 1H

13.58 Interchange

the inductor and resistor in Problem 13.57 and again use the convolution inte$al to find I'o.

13.59 a) Find ,(t) * r(r) when ,(r) and r(t are rhe rectangularpulsesshownin Fig.P13.59(a). b) Repeat (a) when r(t) chatrgesto the rectangular pulseshov,nin Fig.P13.59(b). c) Repeat(a) when n(t) changesto tle iectangular pulseshom in Fig.P13.59(c). Figur€ P13.50

13,61 The voltage impulse iesponseof a circuit is showr in Fig. P13.61(a).The input signal to the circuit is the rectangulai voltage pulse sho$,nilt Fig. P13.61(b). a) Dedve the equations for the output voltage. Note the range of time for which each equation is applicable. b) Sketch ?J,for 0 < t < 27 s. figurcP13.61

/'(t)(\t (a)

v,(^)(v) 13.60a) Giveny0) : nG) * "0), fi"d y(r) whent(r) and r0) are the rectangularpulses shown in Fig.P13.60(a). b) Repear(a) whenft(t) changesto the rectangular pulsesho$,nin Pig.P13.60(b).

560

lhe taptace T€nsfom in Circuii AnaLysis

13.62 Assume the voltage impulse response oI a circuit can be modeled by the tdangula-r wavefom sho\ln in Fig. P13.62.The voltage input signal ro fiis circuit is the step function 4r0) V. a) Use the convolution integral to derive the expiessionsfor the output voltage. b) Sketch tle output voltage over the intenal 0to25s. c) Repeat parts (a) and (b) if the area under the voltage mpulse responsestaystlle samebut the width of the impulse responsenarrows to 5 s. d) Which output waveform is closer to replicating ttre input waveform: (b) or (c)? Explah. Flgure P13.52

13.63 a) Assume the voltage impulse responseof a circuit is h(t):

{l;,'

tigureP13.54

I (pt

13.65a) Repeat Problem 13.64,given that the resistor in the circuit in Fig. P13.49(d)is increasedto 400 O. b) Does inqeasing the resistor ircrease or deqease the memory of the circuit? c) mich circuit comes closest to transmitting a replica of the inpur volrage?

l:t.66 a) Use the convolution integml to find i" in tlle cilcuit in Fig. P13.66(a)iI is is the pulseshown ln Fig.P13.66(b). b) Use t}le convolution integral to find z,o. c) Showrbaryoursolulions for r, andi, areconsistent by calculating o. a'r.d i" at 1 ms, 1- ms, 4- ms,and 4- ms. Figure P13.55

t<0;

Use the convolutioninte$al to find the output voltageif the input signalis 10r(r) V. b) Repeat(a) if the voltageimpulsercsponseis h(t) :

tl.

r<0; - 2.),0=r=0.5s; I > 0.5s.

c) Plot the output voltage versus time for (a) and ( b )f o r o < I < 1 s .

!.64 a) Use the convolution hteglal to find the output voltage of the circuit ir Fig. P13.49(d) if the input voltage is the rectangular pulse shown in Fig.P13.64. b) Sketch o,(t ve6us t for the time itrterval 0=t<100ps.

13s7 The sinusoidalvoltage pulse shown in Fig. P13.67(a) is applied to the circuit shown in Fig. P13.67(b).Use the convolution irtegral to find the value of o, at t = 2.2s.

Probt€ms 561 fig!r€ P13,67

1,70 The current souce in the circuit shown in Fig. P13.70(a)is generating the waveform shown in Fig. P13.70(b).Use the convolutionintegral to find oa^tt = 7 ms.

xt(t)

625kO

Figure P13.70 80nF

(b)

13.68 a) Find t]le impulse response of the circuit sho*n in Fig. P13.68(a)if ?s is the input sipal and i, is t])e output signal. b) Given that oB has the wavefom shown in Fig. P13.68(b),use the convolutionintegral to find to.

is(mA)

c) Does lo have the samewavefom as os?Why?

2 3 4

5

6

7

Flgure PI3.58

60 ].71 The input voltage in the circuit seenin Fig. P13.71is bt = 10lu(t\ - ?(, - 0.1)l v.

600140 f}

a) Use tle convolution integral to find o,. b) Sketcho,for0 < r < 1s.

13.69 a) Find the impulse rcsponse of the circuit seen in Fig. P13.69if rs is t.heinput signal and 0o is t}le output signal. b) Assume that the voltage souce has the wavefo1m shown in Fig. P13.68(b). Use the convolution integral to find o,. c) sketch 2)"for 0 < r < 1.5 s. d) Does ?o have the samewaveform as os?Why? Flgure P13,59 40()

Figure P13.71 4()

+

l:1.72 Use the cotrvolution integral to find ?Jdin the chcuit seeninFig. P13.72if t" : 75u(t) V. FigunP13.72 50()

125tnF r.

10()

400nH

2H

562

jn CircuitAnalysis TheLaplace Transtorm

13.73 a) Show that if )(t) = r(t) *.'(t), tlen

Figure P13.76

r{G)xG).

l0 nF

b) Use the result given in (a) to find /(t) if

s(s + a)'

Section 13,7 13.74 The transfer function for a linear time-invariant circuit is

u. v,

104(r+ 6000) s2+875s+88x106'

If zJs: 12.5cos8000r V, what is th€ steady-state expression for t,?

13.7 When an input voltageof240u(t) Visappliedtoa circuit,the respo$e is knoM to be 01],: (75

100e-eo'+ 25e 32ffi1)u(i)v.

What will tlle steady-state response be 0s = 40cos16,000V?

if

13.75 The operational amplifier in the circuit seen in ErG Fig. P13.77is ideal and is operatingwithin its lina) C.alculatethe transfer function y,/ys. b) If os = 2001r''10cos8000t mV, what is tlle steady-stateer?ression for r',?

Section 13.8 13,78 Show that after yoc" coulombs are translefied from C1 to C2in the circuit shownin Fig. 13.47,the voltageacrcsseachcapacilorisCrUol(Cl + C2\.(Hint: Use the conservation-of-charge principle.)

Figure P13.75

47 k(l

13.76 The op amp in the circuit seenin Fig. P13.78is ideal. rs'tt a) Find the transferf$clionv.lvs. b) Find z).iJ ts = 10"(1) V. c) Find the steady-state er?ression for r', if Ds: 8 cos2000tV'

13.79 The parallel combination of R2 and C2 in the circuit shownin Fig. P13.79representsthe input cbcuit to a catlode-ray oscilloscope(CRO). The parallel combination of R1 and C1 is a circuit model of a compensatinglead that is used to connect the cRo to the source.Theie is no eneigy stored in C1 or C2 at the time when the 10V souice is coturectedto the CRO via the compensating lead. The circuit values are Cr = 5 PF, C2 = 20 PF, Rl : 1 MO, and

R::4Mo

a) Find r,. b) Find t . c) Repeat(a) and (b) givenCl is changedto 80 pF.

Flgu€P13.79

Figure P13.83

+ ;

(}.80 Show that if RrCl : R2C2in the circuit shown in Fig. P13.79,r,, will be a scaled replica of the source voltage.

13.81 The inductor 4 in the circuit shownin Fig. P13.81 is carryingan idtial current of p A at the instant the switchopel1s. Find (a) z(t); (t) i1(t);(c) i(r); and (d) ,\(r), wheie ,\(r) is the total flux linkageir the circuit.

]R

13.82 a) I-et R - co in the circuit shown ir Fig. P13.81, and use the solutionsderived in Prcblem 13.81 to find ?,(1),ir(l), and r2(r). b) Let R : oo in the circuit shown in Fig. P13.81 and use the Laplace transform method to find ,(t),4(t), and t(1).

l:}.83 The switchin the circuit in Fig. P13.83hasbeenin position a for a long time. At , = 0, the switch movesto positionb. Compute(a) o(0 ); (b) ?r(0-); (c) tdo-); (d) (D; (e),1(0*)t (f) zl0*); and

k),r(o-).

figureP13.84

0.8ko rr+

tigureP13.81

r'

13.84 The switch in the circuit in Fig. P13.84 has been closed for a long time. The switch opens at t : 0. compute (a) i1(0 ); (b) n(0+); (c) i,(0-); (d) 4(0+); (e) i(t);(f) ir(t);and (s) D(,).

13.85 There is no energy stored in the circuit in Fig. P13.85 at the time the impulsive voltage is applied. a) Find r"(t) for t > 0. b) Does your solution make sense in terms of known circuit behavior? Explain.

Flgure P13,85 100O 5 rnH

564

lhe Laptace TBnsform in Cncuii Anatysjs

(}.86 The voltage source in the circuit in Example 13.1is changed to a unit impuise; that is, ?rs: 5(1). a) How much energy does the impulsive voltage source store in the capacitor? b) Ho\q much energy does it store in the inductor? c) Use the transfer tunction to find ?,o(,). d) Show that the responsefound in (c) is identical to the response generated by fLst charghg tle capacitor to 1000 V and then releasing the charge to the circuit, as shown in Fig. P13.86.

13.88Therc is no energy stored in the circuit h Fig- P13.88 at the time the impulsive current is applied. a) Find ?Jofor t > 0+. b) Does your solulion make sense in tenns of known circuit behavior? Explain.

Figure P13.88 40nF

Figure Pt3.85

Seciioff 13.1-13,8

t3a7 fhere is no energy stored itr the circuit itr Fig. P13.87 at t]le time the impulse voltage is applied. a) Find ir for t > 0*. b) Find i for t > 0+. c) Find 0, for t > 0-.

1:1.89Assumethe line-to-neutralvoltageV, in the 60 Hz p:llflljl}lcircuit of Eg. 13.59is 120l!| v(rms). Load R, is absorbhg1200W; load Rr is absoibing1800W and load -Y, is absorbing350 magnetiziq VAR. The inductivereactanceof the line (X1) is 1 O. Assume Vsdoesnot changeafter the switchopens. a) Calculatethe inirial valueof t,0) and i,(t). cirb) FindV,, o,(t),and?,(0*)usingthes-domain cuitof Fig.13.60. c) Testthe steady-state componentof z', usingphasor domainanalysis, d) Using a computerprogramof your choiceplot 1lovs.tfor0 = t = 20ms.

d) Do 'our solurioD.for ir. r). and ,o makesensein termsofkDoqDcircuilbehariorltyplain.

FlgureP13.87

mD(r) v

13.90Assumethe switch in the circuit in Fig. 13.59 opensat the instant the sinusoidalsteady-state J#;lI#jE voltage oo is zero and going positive, i.e., uo : 120\/t sin 120ntV . a) Findto0) for t > 0. b) Using a computerprogiam of your choiceplot r,"(t)vs.tfor 0 < I < 20ms. c) Conpare the disturbancein the voltage in part (a) with tlat obtained in part (c) of Problem13.89.

Problems 565 13.91 The purpose of this problem is to show that the pi$fi*iiEline-to-reutral voltage in the circuit in Fig. 13.59 can go directly into steadystate it the load R6 is disconnectedfrom the circuit at precisely the right time. Let ?, = V^cos(I2jnt d') V, where y- = 120\4. Assume ?s doesnot changeafter R, is disconnected.

a ) Find the value of 0 (in degrees) so tlat o, goes

directly into steady-state operation when the load Rbis discomected. b) For the value of d fourd in part (a), fird ?,,(1)Ior t>0. c) Using a computer program of your choice plot, on a single graph, for -10ms < l < 10ms, o,(t) befoie and after load R1,is disconnected.

Introduction to Frequency )elecuveLrrcurts 7

14.1 SomePretiminari€sp. 568 14.2 Low-Passlilters p. 57A 14.3 tligh-PassFilters p. 577 14.4 Bandpass fiLters p. 582 14-5 Bandrdectfitters p. 59-t

1 Kiowthe ffr and8f circujtconfiguratjons that act as tow'pass fitteruandbe abteto desigi Rl andRtcncujt component vatuesto meeta specifi ed cutofffr€quency. 2 (nowthe {L andRf cncuitconfiguntjoisthat act as hjgh'passfittersard be abteto desjgn Rl andff cjrcuitcomponent vatuesto meeta specifi ed cutofffrequency. 3 Xnowthe Rrf cncuitconfig!rations ihat act as bandpass fitters,understand the definjtlonof andr€tationship amofqthe certerfiequency, cutofffrcqLrencj€s, bardwidth,andquatity ractorof a bafdpassfitter,andbe abteto desiqnRifcncuit compofentvatuesto meet designspecificatjons. 4 Knowthe RrCcncuitconfigurations that act as bandreject fitters,understand ihe definitionof andrctationshjp amongthe certerfrequency, cutofffieqlencj€s,bandwjdth, andquaLity ractorof a bafdrejectfitter,and be abteto designRtrcircuit componeitvaluesto me€t desiqnspecifications.

|

I

.

t^.

Up to this point in our analysisofcircuitswith sinusoidalsources, the sourcehequencvwas held constant.In this chapter,we analyze the effectof varyingsourcefrequencyon circuit voltagesand currents.Thc rcsult of this analysisis the frequencyresponseof a clrcurl. We've seen in previous chaptcrs that a circuit's response dependson the typesof elementsin the cjrcuit, the way the elements are connected. and the impedance of the elements, Although varying thc frcquencyof a sinusoidalsourccdoesnot changethe elementtypes or their connections,it does aLtel-the impedanceofcapacito$ and inductors,becausethe impedanceof theseelemcntsis a function offrequency.Aswe will see,thecareful choiceof circuit elenel1ts,their values,and their connections to other elementsenablesus to constructcircuitsthat passto the output only those input signalsthat residein a desiredrange of fr-equencies. Such circuits are called frequency.selective circuits. Many devicesthat communicatcvia clcct c signals,suchas tclcphones, radios, televisiolls,and satellites,employ fiequencyselectivecircuits. Frequency-selective circuits ale also calledfilters becauseof their ability to liller oLltcertair input signalson the basisof frequellcy.Figure 1,1.1on page 568 representsthis ability in a sim plistic way.To be more accurate,we shouldnote that [o practical liequency selectivecircuit cafl pefcctly or completelyfilter out selectedftequencies.Rather,filter-sattenuate-that is,weakenor lessenthe effect of-any input signalswith tiequenciesoutside frequenciesoutside a particular frequency band. Your homc stereosystcmmay havea graphicequalizer,which is an excellent exampleof a collectionof lilter circuils.Each band ill the graphic equalizeris a lilter that amplifiessounds(audiblc ftcqucncics)in thc flequencyrangeof lhe band and attenuateslrequenciesoutside of that band. Thus the graphic cqualizcr cnablcs you to changcthe soundvolume in eachfiequencyband.

PracticaI Perspective Pushbutton Telephone Circuits In this chaptet we examinecircuitsin whichthe sourcefrequencyvaries.The behavjorof these cjrcuitsvariesas the sourcefrequency varj€s,because the impedance of the reactive components is a functjonof the sourcefrequency. These frequency dependent circujtsare caLLed filteB ano are useo ir manycommonetectricaldevices. In radios,fitte15areused to setectoneradiostation'ssignalwhilerejectjfgthe signats from otherstransmjttingat differentfrequencjes. In stereo systems, fittersareusedto adjusttheretativestfengthsofthe Low-and high-frequeicy components of the audjosignaL. Fr'tters areatsousedthroughouttetephone systems. prcducestonesthat you hear A pushbuttonteLephone whenyou pressa button.Youmayhavewonder€d aboutthese tones.Howareth€y usedto tetl the tetephone systernwhjch button was pushed? Whyare tonesusedat att?Whydo the tonessoundmusr'cal? Howdoesthe phonesystemtel[ the dif ferencebetweenbuttontonesandthe normatsoundsof peo pletatkingor sjnging?

The tetephonesystemwas designedto handleaudio sjgnats-thosewith frequencjes between300 Hz ard 3 kHz. Thus,alt signalsfrom the systemto the user have to be audibte incLuding the diattoneandthe busysignal.Similarly, att signabfrom the userto the systemhaveto be audjbLe, incLuding the sjgratthat the userhaspressed a button.It js importantto djstjngujshbuttonsignalsfromthe normaL audjo (DT[4F) 5ignat,so a duat-tone-muttipte-frequency d€signis employed. Whena numberbuttonis pressed, a uniquepairof sinusoidaL toneswjth ve[/ preckefrequencies is sent by the phoneto the telephone systern. TheDTI4F frequency andtiming specifrcations makeit unLikety that a humanvoicecouLdpro ducethe exacttone pairs,evef jf the personweretryjng.In the centralteLephone facitjty, electriccjrcuitsmonitorthe audiosignat,ljsteningfor the tonepairsthat signaL a number. In the Practical Perspective exampte at the endofthe chapter, we witLexamine the designof the DTMF filtersusedto determinewhjchbuttonhasbeenpush€d.

t

568

Introduction to Frequency S€lective Cjrcuits

We begin this chapter by anallzing circuits fiom each of the four major categoriesoffiltels: low pass,high pass,bandpass,and band reject. The tmnsfei function of a circuit is the starting point for the frequency figure14.1A Theactionofa fitieronaninputsiqnat responseanalysis Pay close attention to the similarities among ttre transtesuLrs in anoutputsignat. fer functions of circuits that pedorm the same filtedng function. We will employ these similarities when designing filter circuits in Chapter 15. Inout sgnrr

t--l

ourout srgnaL

14.1 SomePretiminaries Recall from Se€tion 13.7that the txansferfunction of a cicuit provides an easyway to compute the steady-stateresponseto a sinusoidalinput. Therc, we consideredonly fixed-ftequency souices.To study the ftequency response of a circuit, we replace a fixed-frequency sinusoidal source with a varyingfrequencysinusoidalsourc€.The hansfer function is still an imnensely useful tool becausethe magnitude and phaseof the output signal depend only on the magnitude and phaseof the transfer function fl(j.r). Note that tlte approach just ouUined assumesthat we can vary tlle frequency of a sinusoidal souce vrithout changing its magnitude or phase angle.Therefore, the amplitude and phase of the output will vary only if those of the transfer function vary as the frequency of tle sinusoidal sourceis chalged. To turther simplily this first look at frequency-selective circuits, we will also restrict our attention to caseswhere both the input and output signals are sinusoidal voltages,as illustrated in Fig- 14.2.Thus, the transfer function of interest to us will be the ratio of the Laplace transform of the ouQut voltage 1o the Laplace tramform of the input voltage, or - %(s)/I4(s).We shouldkeepin mind,however,thatfor a particular Figure 14.2 A circujt withvottage inputandoutput. H(s) application, a current may be either the input signal or output signal of intercst. The signals passedfrom the irput to the output fall withit a band of frequencies called the passbatrd.Input voltages outside this band have their magnitudesattenuatedby the circuit and are thus effectivelyprevented ftom reaching the output terminals of the circuit. Frequencies not in a circuit's passbandare in its stopband.Frequency-selectivecircuits arc categodzed by tlle Iocation of the passband. One way of identifying the t)?e oI frequency-selective circuit is to examhe a ftequency responseplot. A frcquency responseplot showshow a ckcuit's transfer function (both amplitude and phase) changes as the sourceftequency changesA frequency responseplot has two parts. One is a graph of l1t(io)l versus frequency d. This part of the plot is called the magnitude plot.The other part is a graph of du(o) ve$us frcquency .). This part is called the phas€ angle plot. The ideal ftequenry responseplots for tle four major categoriesof filtels are shown in Fig. 14.3.Pa s (a) and (b) ilustrate the ideal plots for a low-pass and a high-pass filter, rcspectively. Both filten have one passband and one stopband, which are defined by the cutofi ftequency that separatesthem. The names low pass and high pasr are derived from the magnitude plots: a low.pass filler passessignals at frequencies lower than the cutoff frequency from the input to the output, and a high-pass filter

14.1 SomePretiminanes569

H(i.)l I

elia) o(i.") 0"

I

e0",)

o(i.)

o(i0\) 0'

0'

e(ia")

Figure plotsofthefourq,pes 14.3& Ideatfrequency r€sponse offiLter circujts. (a)Anideatlow+ass fitter.(b)Anjdeathigh-pass fitter.(c)AnideaLbandpass fittel (d)Anideatbandrej€ct fitter. passessignalsat frequencieshigher than the cutoff frequency.Thus the termslon and ft8, asusedhere do not refer to any absolutevaluesof frequency,but rather to relative values with respect to the cutoff frequency. Note from the graphs for botl these filters (as well as those for the bardpassand bandrejectfilters) that the phaseaflgleplot for an ideat fil tervades lin€arlyin the passband.Itis of no inlerestoutsidethe passband becausetherethe magnitudeis zero.Linear phaseva ation is necessary to avoidphasedistortion(seeChapter16,pp.770-773.) The two remainingcategoriesof filters eachhavetwo cutoff frequencies.Figure 14.3(c) illustrates the ideal ftequency responseplot of a bandpassfiller, which passesa source voltage to the output only when the sourcelrequency is within the band defined by the two cutoff frequencies. Figure 14.3(d)showsthe ideal plot of a bandrej€ctfflter, which passesa sourcevoltageto the output only when the sourcefrequencyis outsidethe band dclined by the two cutoff frequencies The bandreject filter tius rejects,or slops,the sourcevoltagefrom reachingthe output when its frequency is within the bard defined by the cutoff frequencies. In specifying a realizable filter using any of the circuits from this chapter,itis importartto note that the magrlitudeandphaseanglecharacterislics are not irdependent.In other words,the charactedsticsof a circuit that result in a particular magnitude plot will also dictate the form of the phaseangle plot and vice versa.For example,once we selecta desired fbrm for the magnitude responseof a circuit, the phase angle iesponse is also determined.Alternatively,if we selecta desiredlbrm lor the phase

570 Intrcduchonto Frcquency Setective Cncuits angle response,the magnitude responseis also determined.Although there are some ftequency-selective c cuits for which the magnitude and phaseanglebehavior can be independentlyspecified,these circuits are not presentedhere, The next sections present examples of circuits from each of the four Iilter categories.They are a few of the many circuits that act as filters. You should focus your attention on tying to identify what properties of a circuit determine its behavior as a filter. Look closely at tle form of the transfer function for circuits that perform the same filtedtg functions Identifying tie folm of a filter's transfer tunction will ultimately help you in designing filtering circuits for particular applications. All of the filteis we will consider in this chapter are passivefiilters, so calledbecausetheir filtering capabilitiesdependonly on the passiveelements:rcsistors,capacitors,and inductoN.The largestoutput amplitude such filters can achieve is usualy 1, and placing an impedance in se es with the sourceor in parallel with the load will decreasethis amplitude. Because many pmctical lilter applications require inoeasing the amplitude of the output, passivefilters have som€ significant disadvantages.The only passivefilter desoibed in this chapter that can ampMy i1s output is the seriesRLC resonant filter. A much greater selection of ampliJying filters is found among the active filter circuits, the subjecl of Chapter 15.

14.2 Low-Pass Filters Here, we examinetwo circuitsthat behaveas low-passfilters,the series R-L circuit and ttre series RC circuit, and discover what characteistics of lhesecifcujlsdelefminerhecurofffrequenc).

L

R

(a) L

R

(b)

Figure 14.4a (a)A series Rl low-pass fitter.(b)rhe equjvatent cncuitat o : 0. andG)Th€€quivatent

TheSeriesRl.Circuit-QualitativeAnalysis A series RI, circuit is shown in Fig. 14.4(a).The c cuit's input is a sinusoidal voltage source with varying frequency. The circuit's output is defined as the voltage across the resistor. Suppose the frequency of the sourcestartsvery low and increasesgradually.We krow that the behavjor of the ideal rcsistorwill not change,becauseits impedanceis independent of frequency.But consider how the behavior of the inductot changes. Recall that the impedance of atr inductor is joa. At low ftequencies, the inductor's impedanceis very small compared with the rcsistor\ impedance, and the inductor effectively functiors as a short citcuit. The term low frequencies thus refen ro any frequencies for which ol, << R. The equivalent circuit for o = 0 is shown in Fig. 14.4(b). In this equivalent circuit, the output voltage and tlre nrput voltage are equal both in magni tude and in phase angle. As the hequency increases,the impedanceof the inductor increasesrel, ative to tlat of the resistor. Increasing the inductor's impedance causesa correspondhg ircrease ir the magnitude of the voltage drop across t]le inductor and a corresponding d€creasein the output voltage magnitude. Incieasing the inductor's impedanc€ also itfoduces a shift in phase angle betweenthe inductor's voltage and current.This results in a phaseangle diJference between the input and output voltage.The output voltage lags the input voltagg and as the frequency increases,this phaselag approaches90'.

571 At high frequencies,the inductor's impedance is very large compared with the resistor's impedance,and the inductor thus functions as an open circuit, effectively blocking the flow of curent in the circuit. The term ,,ig, frequenciesIhus rcfefi to any frequencies for which (da >> n. The equiv, alent circuir for (d = oo is shown in Fig. 14.4(c),where the ourpur volrage magnitude is zero. The phase angle of the output voltage is 90' more negative than that of the input voltage. Based on t]le behavior of the output voltage magnitude, this seiies Ra a0,) circuit selectively passeslow-frequency inputs to the output, and it blocks high-frequency inputs from reaching the output. This circuit's responseto 1.0 varying input frequency thus has the shape shom in Fig. 14.5.These two plots comprise t}le frequency response plots of the se es Rl, circuit in Fig. 14.4(a).The upper plot shows how ld(j@) vades with frequency.The lower plot showshow 9(j(o) vaiies as a function of hequency.We present a 0 more formal method for constiuctiry these plots in Apperdix E. e(i.) We have also supedmposed the ideal magnitude plot for a low-pass 0" filter from Fig. 14.3(a)on the magnitudeplot of the Ra filter in Fig. 14.5. There is obviously a difference between the magnitude plots of an ideal filter and the frequency rcsponse of an actual Ra filter. The ideal filter exhibits a discontinuity in magnitude at the culoff frequency, (,., which 90' createsan abrupttansition hto and out of the passband. While thisis,ideally, how we would like our filte$ to perfom, it is not possible to use real pLotforthe response componentsto constnrci a circuit that has this abrupt tansition in magni- Figurc14.5A Thefrequency senes ft circuitin Fig. 14.4(a). tude. Circuits acting as low-pass filters have a magnitude response that changesgraduallyfrom the passbandto the stopband.Hence the magnitude plot of a real circuit requires us to define what we mean by the cutolf ftequency, o..

Definingthe CutoffFrequency We need to define the cutoft frequency, da, for realistic filter circuits when the magnitude plot does not allow us to identify a single frequency that divides tie passbandand the stopband.The definition for cutoff frequencywidely usedby electricalengineersis the frequencyfor which the transfer Iunction magnitude is decreased by the factor 1/\,, from its

(14.1) < Cutofffrequenc!' definition where }1-ax is the maximum magnitude of the transfer function. It follows from Eq. 14.1 that t}le passband of a realizable filter is defined as tle range of lrequencies in which tle amplitude of the output voltage is at least ?0.7% of the ma-\imum DossibleamDlitude. The constant1/\4 usedin defining ttri cutomtequency may seemlike an arbitrary choice.Examining another consequenceof the cutoff liequency will make this choice seem morc reasonable.Recall from Section 10.5that the averagepower delivered by any circuit to a load is proportional to Vl, where yr is t}te amplitude of the voltage drop aqoss the load: 1V? 2 R

(11.2)

572

rntroduction to Frequ€ncy Setectiv€ Cjrcujts ffthe circuit hasa sinusoidalvoltagesource,V(jo), then the load voltage is also a sinusoid, ard its amplitude is a function of the frequency (r. Define P-"" as the value of tlre averagepower deliveredto a load when t})e magnitude of the load voltage is maximum:

1 v 1.^u* 2

R

(14.3)

If we vary the frcquencyof the sinusoidalvoltagesource,I40o),the load voltage is a maximum when the magnitude of the circuit's transfer furction is also a maximum: (14.4)

Now considei what happens to the average power when the frequency of the voltagesouce is o.. Using Eq. 14.1,we determinethe magnitudeof the load voltaseat o^ to be

vLtj@)l = lH(tu)lUl

=Lu ,, \,5-'^*' 114.5) SubstitutingEq. 14.5into Eq.14.2,

P(*,)=:v+d I

\iw*)

2

R

1 v"- J2 2 P-* 2

(r1.6)

Equation 14.6 shows that at the cutotr frequency o., the average power delivered by the circuit is one half the maximum averagepower.Thus,r.r.is also called the halfTow€r ft€quency.Thereforc, in the passband,tle average po\ter deliveredto a load is at leasr50o. ot rhe maRimumaveragepower. sL

Y,(r)

TheSeriesnl Circuit-QuantitativeAnalvsis R

%(,

Figure14.6A Ther-domajn equivatentfor thecircuit in Fig.14.4{a).

Now that we have defined the cutoff lrequency for real filter circuits, we can analyzethe seriesna circuit to discoverthe relationshipbetweenthe component values and the cutofl frequency for this low-pass filter. We beginby constructingth€ r-domainequivalentof the c cuit in Fig.14.4(a), assuminginitiat conditions of zero. The resulting equivalent circuit is shownin Fig.14.6.

14.2 Low-Pass Fitters 573 Tbe volraeetransferfunctronlor thiscircuil,,

H(,) =

R/L

s+R/L

\14.7)

To study the frequencyresponse,we make the substitutions - jd in Eq.14.'7. R/L jo + R/L

(14.8)

We can now separateEq. 14.8into two equations.The fi^t definesthe transfer functior magnitude as a function of frequency; the seconddefines the hansfer function phase angle as a function of frequency:

n(i,) =

RlL

G+GB'

d(ia,) = - tan

\n/

(14.e)

\r4.\a)

Close examination of Eq. 14.9 provides the quantitative support for the magnitudeplot shownin Fig.14.5.W}lenl, = 0, tlre denominatorand the numeratorare equal and lH(i0)l = 1. This meansthat at o = 0, the input voltage is passed to the output terminals without a change in the voltagemagnitude. As the ftequencyincieases,the numentor of Eq. 14.9is unchanged, but the denominator gets laiger. Thus l}I(ia)l decreasesas the frequency increases, as shown in the plot in Fig. 14.5.Likewise,as the frequency increases, the phaseanglechangesfrom its dc valueof0', becomingmore n€gative,as seenfrom Eq.14.10. When a,: c., t}le denominator of Eq. 14.9 is infinite and lH(ico) : 0, as seenin Fig. 14.5.At o : oo, the phaseanglereachesa limil of -90', as seenfromEq.14.10 and the phaseangleplot in Fig.14.5. Using Eq. 14.9,we can computethe cutoff ftequency,od. Remember that .r. is defined as the frequency at which lH(Ja ") : (I/\n)H ^ ". Fol the low-passfilter, l1max= lH(jo) , asseenin Fig.14.5.Thus,foi the circuit in Fig.i4.4(a),

: jx1u11.,1

R/L

\,/A;snj

(14.11)

Solvirg Eq.14.11for o., we get

(14.12) { Cutofffrequency for ru fitters Equation 14.12provides ar impo ant result. The cutoff ftequency, o", can be set to any desiied value by appropdately selectingvalues for R and L. We can therefore design a low-passfilter wilh whatever cutoff frequency is needed.Examplel4.l demonshatesthe designpotentialofEq.14.12.

574 Intoductionto Frcquency Setective Cncuitr

Designing a Low-Pass Filter Electiocardiology is the study of the electric signals produced by the hea{. These signals maintain the heart's rhltlmic beat, and they ate measuredby an instrument called an electrocardiograph.This instru, ment must be capable of detecting pedodic signals whose frequency js about 1 Hz (the normal hearr rale is 72 beats per minute). The instnrment must operate in the presenceof sinusoidalnoise consisting ot signalstrom rhe suffoundingelecrical environ ment! whose fundamental fiequency is 60 Hz the frequency at which electdc power is supplied. Choosevalues for R and I, in the circuit of Fig. 14.4(a)suchthat the resultingcircuit could be used ir an electrocardiographto filter out any noise above 10 Hz and pass the electric signals from the hea at or neat 1 Hz. Then compute the magnitudeof % at 1 Hz, 10 Hz, and 60 Hz to see how well the filter pefolms.

Solution The problem is to select values for R ard L that yield a low-pass filter with a cutoff frequency of 10 Hz. From Eq. 14.12,we seethat R and a cannot be specified independently to generate a value for o.. Therefore, let's choose a conmonly availabl€ value of 1,, 100 mH. Before we use Eq. 14.12to compute the value of R needed to obtain the desired cutoff frequency, we need to convert the cuto{f frequencyfrom hertz to radiansper second: o, = 2r(L0) = mn r^dls.

We can compute the magnitude of % using the equarion% : H(ja) .lul

u"(.) :

R/L

\/"j +1.R8 20t

\,ll + cw.;

lu tut.

Table 14.1 summarizesthe computed magnitude valuesfor the frequencies1 H4 10 Hz, and 60 Hz. As expected,the input and outputvoltageshavethe samemagnitudesat the low hequency,becausethe circuit is a low-pass filter. At the cutoff frequency, the output voltagemagnitudehasbeen reducedby U\D. ftorr. the unity passband magnitude.At 60 H4 the output voltage magnitude has been reduced by a factor of about 6, achieving the desiredattenuationof the noisethat could corrupt the signal the electrocardiographis d€signedto

NoW solve for the value of R which, together with a : 100mH, will yield a low-passfilter with a cuf off frequencyof 10 Hz: R:a,L

: (20?r)(100 x 10-3) : 6.28().

I tL) 60

1.0 t,0 1.0

0.995 0.707 0.164

A SeriesRCCircuit

figuE14.7A A sedes fr tow-pass fitter.

The series RC circuit shown in Eg. 14.? also behavesas a low-pass filter. We can verify this via the same qualitative analysis we used previously. In fact, such a qualitative examination is an important problem-solving step that you should get in the habit of performing when analyzing filters. Doing so wlll enable you 10 prcdict the filtering charactedstics (low pass,high pass,etc.) and thus also prcdict the general form of the lransfer function. If

the calculated transfer function matches the qualitatively predicted folm, you have an important accuracycheck. Note that tle circuit's output is defined as the output acrossthe capacitor. As we did in the previous qualitative anal)ris, we use three frequency regions to develop the behavior of the setiesRC circuit in Fig. 14.7: L Zero ftequency (a = 0): The impedance of the capacitor is infinite, and the capacitor acts as an open circuit. The input and output voltagesare thus the same, 2. Frequenciesincrcasingfrom zeroi Tf\e impedance of the capacitor decreases relative to the impedanceofthe resistor,and the soutce voltage divides betweenthe resistiveimpedanceand the capacitive impedance.Theoutput voltageis thus smallerthan the source voltage. 3. Infinite frequency (.,, = oo): The impedance of the capacitor is zero,and the capacitoiactsas a short circuit.The output voltageis tlus zero, Based on this anal)sis of how tle output voltage changesas a function of trcquency,t}le seriesRC circuit functions as a low-passfilter. Example 14.2 exolorcs this circuit ouantitativelv

Designing a SeriesfC Lory-Pass Fitter For the seriesRC circuit in Fig. 14.7: a) Find the transfei function between the soutce voltage and the output voltage. b) Determine an equation {or the cutoff frequency in t]le sedesRC circuit. c) Choose values for R and C that will ield a low' passfilter with a cutoff fiequency of 3 kHz.

Sotution a) To de ve an expressionfor the transfer function, we first constructther-domain equivalentofthe cicuit in Fig. 14.7,as showr in Fig. 14.8. Using .r-domain voltage division on the equivalent circuit, we find I

H(r) :

RC I RC

Now' substitute r : jo atrd compute the magnitude of the resulthg complex expression: I RC =

laU,)1

(".-t)'

x(r)

Y,(J)

Figure14.8^ Th€s'domain equivatentforthe circuit jn Fig.14.7.

b)At the cutoff frequency (,c, lg(jro)l is equat to (1,1\/2)HM. For a low-pass filter, H-"- - H(jo), and foi t}le circuit in Fig- 14.8, H(J0) - 1. We can then descdbethe relationship amongthe quantitiesR, C, and o.:

lH(ia,\=+0\

1 RC / 1 \ 2

l__: I \ nc,i

Soi,vingthisequationfor (,", we get

576

Inttoduction to Frequenq Setective Cjrcujts

c) From the resultsin (b), we seethat the cutoff frequency is determined by the values of R and C. BecauseRand C cannotbe computedindependently,let'schooseC : 1 pF. Given a choice,we will usualiy specify a value for C first, rather than for R or a, because the number of available capacitor values is much smaller than the number of resistor or inductor values, Remember

that we have to conve the specified cutoff ftequencyftom 3 kHz to (2r)(3) krad/s:

-'^

1 a"C

(2,X3 x 103)0x 10) : 53.05 {).

Figure 14.9summarizesthe two low-passfilter citcuits we have examined. Look carcfully at the tansfer functions. Notice how similar in form they are-they differ only in the terms that specifJ the cutoff ftequency. In fact, we can state a genenl form for the tmnsfer functions of these two lowpassfilten:

Transfer functionfor a low-Dass fitter >

'L

+ vi

R

R/L ,trt=,+NL

v"

+ , r ( r ' : r +Ir/RC fcv" o,= rlRC Figure 14.9A Twotow-pass fiLters. thes€ries nLand thesed€s RCtogetherwith thenhansfer tunctions and

114.t3)

Any circuit with the voltagemtio in Eq.14.13would behaveas a low-pass filtei wiih a cutoff ftequency of (,c.The problems at the end of the chaptei give you other examples of circuits with this voltage ratio.

Retatingthe Frequency Domainto the TimeDomain Finally, you miglt have roticed one orher important relationship. Remember our discussion of the natural responsesof the fust-order Rt and RC circuits in Chapter 6.An important pammeter for these circuits is the time constant, 7, which characterizes the shape of the time response. For the Ra circuit,the time constanthasthe value a/R (Eq.7.14);for rhe RCcircuit,the time constantisnCGq.7.24). Comparethe time consranrs to the cutoff fiequencies for these circuits and notice that 114.14) This iesult is a direct consequence of the relationship between the time rcsponse of a circuit and its frequency response,as revealed by the Laplace transform- The discussion of memory and weighting as representedin the convolutjonhtegral of Section13.6showsthat as (,.+,]., rhe filter hasno memory, and the output approachesa scaledrcplica of the input; that is,no filtedng hasoccu ed.As o. + 0, the filter hasincreased memory and the output voltage is a distortion of the input, becausefilter-

1 4 . 3 Higlr-Pds Fitters 577

objective1-Know the 8l andRCcircuitconfigurations that actas Low-pass filters 14.1 A seriesllclow,pass filter iequircs a culolf frequcncyof 8 kHz. Use R - 10 kO and computc the valueof C required.

14.2 A seriesRZ low-passfiher with a cutofl frequcncyof2 kHz is needed.tlsing n = 5 kO, compure(a) I;(b) H0o) ar 50 kH4 and (c) d(id) at 50 kHz. Answer: (a) 0.40Hr (b) 0.{)4; (c) 87.71'.

NOTE: Also try ChaptetPtubkns11.1and 11.2.

14.3 lirglh-Fass Filt*rg We rext examinc two circuils that function as high-passfiliers. Oncc again,they are thc scriesRa circuit and the scriesRC circuir.We will see that the samcscricscilcuitcanact aseithcr a low passor a high-passfilrcr, dependingon Nhcre lhe output voltageis dctined.Wewill alsodeterminc the relationshipbellveenthe componentvaluesaDdthe cutoff frequcncy of thesefilters.

TheSeries RCCircuit-Qualitative Anatysis A scdesRC circuit is shorn in FiS.f4.f0(a).In contrast10 ils low pass counlerpartin Fig.14-7,tho outpul voltagehere is detincdacrossthe resislor,not the capacitor.BecauscoI this.the effectofthe changingcapacitive impedaDce is clifferenrthan it wasit the low-passconfiguration. A1 d = 0, the capacitorbehaveslike an open circuil. so there is no currentflowingin thc rcsislor.Tlis is illustratedin the equivalentcircuitin Fig.14.10(b).In this circuit.thereis no voltagcacro,isthe resistor,and thc circuit filters out fie low lrequencvsourccvoltagebefore it reachesthc

G) C

As the frcqucnc) of the voltage sourccincreases, the impedanccoi the capacitordecrcascsrclalive to tire impedanccoI the resistof.and rhe C sourcevoltageis now divided betlveenthe capacilorand the resistor.fhe oulput voltagemagnitudcthusbegiDsto increasc. R When the frequcncyoI lhe sourceis infinitc ((, = c.), the capaciror behavesasa short circuil.and thusthereis no vollageacrossrhe capacitor. This is iliustratcdin the equivaientcircuir in Fig. 14.10(c).In this circuii, the input voltageand output voltagearc the sane. Thc phaseaDgledifferencebctx'cen the soulce and output voltages Figure 14,10t (a)A se esrr high-pass fitter(b)the alsovaricsas lhe Irequencyof the sourcechaDges.For o = 6, th c outpu! equivatent ciruii at o - 0: and(c)theequivatent voltagcis thc sameas the input voltagc,so the phaseanglediffcrcncc is

578

Introduction to Frequency Setective cncuits zero. As the fiequency of the source deoeases and the impedance of the capacitor hdeases, a phase shift is infoduced between the voltage and the current in the capacitor. This ueates a phase difference between the source and oulput voltages.The phase angle of the output voltage leads that of t])e source voltage. When a = 0, this phase angle difference reachesits maximumof +90'. Based on our qualitative analysis, we see that when the output is defhed as fie voltage acrossthe resistor, the seriesRC circuit behavesas a high-pass filter. The components and €onnections are identical to the Iow-pass series RC circuit, but the choice of output is differcnl. Thus, we have confimed the earlier obse ation that the filtedng charactedstics of a cfucuit depend on the definition of the output as well as on circuit comDonents.values.and comections. Figure 14.11shows the ftequency responseplot for the seriesRC high-pass filter. For reference, the dashed lines indicate the magnitude plot for an ideal high-passfilter. We now tum to a quantitative analysisof this samecircuit.

H0.)l 1.0

TheSeriesfiCCircuit-QuantitativeAnalysis 0

To begin, we conshuct the r-domain equivalent of the c cuit in Fig. 14.10(a).This equivalentis shown in Fig. 14.12.Applying s-domain voltage division to the circuit, we *Tite the transfet function:

o(i.) +90.

s + I/RC Making the substitution s = io results in Flgue14.U l. Thefrequency ptotforthe response series fr cncuit in Fig.14.10(a).

L JC

Yi(s)

+ Y,(r)

fltt,)t=

(11.15)

Next,we separate Eq. 14.15into two equationsThefirst is the equation describingt]le magdtudeof the tmnsferfulctioq tlle secondis the equation describingthe phaseangleof the transferfurction:

nA,) Figure14.12a Thei-domajn equjvatent ofthe circuii jn Fig.11.10(a).

jd+ IlRC

' \E +nlRcf

o(i.) :90"

tan loRC.

(14.16)

114.17)

A closelook at Eqs. 14.16and 14.17confirms the shapeof the frcquencyresponseplot in Fig. 14.11.Using Eq. 14.16,we can calculatethe cutoff frequency for the sedesRC high-passfilter. Recall that at the cutoff frequency, the magnitude of t}le tra$fer funcrion is (l/\4)Itmax. For a high-passfiltei, H* = lZl(t.).=- - fl(ico)i, as seenfrom Eg. 14.11. We can construct an equation for a)" by setting tlle left-hand side of Eq. 14.16 to (y\'A)lH(j'x'\, noting that for this series RC circuir, Ifi(jco) = 1:

t2

\Gll!RC),-

(14.18)

1 4 . 3 N i q hP a 5tsi t e 6

579

SolvingEq.14.18for (,",we get 1

(74.7e)

Equation 14.19presents a familiar result.The cutoff ftequencyfor the seriesRC circuit has the value 1/RC, whether tle circuit is configured as a low-passfilter in Fig. 14.?or as a high-passfilter in Fig. 14.10(a).This is perhapsnot a surprisingrcsult,as we have alreadydiscovereda connection between the cutoff ftequency, (,., and the time constant,r, of a circuit. Example 14.3 analyzes a sedes Ra circuit, this time configured as a high-passfilter. Example14.4examinesthe effectofadding a load resistor in panllel with the inductor.

Designing a SeriesRl High-Pass Fitter Show that the seriesRa circuit in Fig. 14.13also acts like a high-passfiltefl a) Derive an expression for the circuit's transfer function, b) Use t}le result frorn (a) to determine an equation Ior the cutoff frequency in tlre seriesRL circuit. c) Choosevaluesfor R and I that will yield a highpassfilter wirh a cutoff frequency of 15 kHz. R

R Y(s)

v,(t

Flgur€14.14 A Ther-domain equivatent ofthecjrcuitin Fig.14.13.

b) To find an equation for the cutoff frequency,first compute the magnitude of H(/.0): L

Figure14.13a Thecircuittor Example 14.3.

Solution a) Begin by constructing tle .r-domain equivalent of the seriesri, circuit, as shown in Fig. 14.14. Thenuser-domainvolragedir i.ionon rheequiralenl circuit lo conslrucllhe lransler[unction:

s+R/L Making the substjtution r : jd, se get

nUot :

sL

ja + RIL'

Notice that this equation has the same form as Eq.14.15for the seriesRChigh-passfilter.

E\i.):

\/"1+ gnj

Then, as before, we set the leffhand side of this equation to (11\5\H^*, basedon the defmtion of the cutoff frequency .,,.. Remember that 1/mx = lH(J@)l for a high-pass filter, and for the seriesRL circuit,lH(joo)l = 1. we solvethe resulting equation for the cutoff frequencyi \/, This is tle same cutoff frequency we computed for the sedes-Ra low-pass filter. c) Using t}le equation for (,. computed in (b), we recognize that it is not possible to specify values for R and l, independently. Therefore, let's arbitrarily selecta value of 500 O for R. Remember to convert the cutoff frequency to radians per second:

,

n

5

0 0 "'"- -(2,,)(15,000)

580 Intrcductjon to Frequency S€lective Cjrcujll

Loading the Seriesfl High-Pass Fitter Examine the effect of placing a load resistor in parallel with t}le inductor ir the Rl, high-pass lilrer showninFig.14.15: a) Detemine the transfei furction for the circulr m Fig.14.15.

magnitude is (1)(112)= 112,andthe curofffrequencyis (15,000X1/2)= 7.5kHz. A sketchot the magnitudeplots of the loadedand unloaded circuitsis shownin Fig.14.17.

b) Sketch the magnitude plot for the loaded Rl, high-pass filter, usitrg the values for R and a from the circuit in Example 14.3(c)and letting R/ - R. On lhe ramegraph..kercbfie mag0itude plot fo{ the unloaded RL high-passfilter of Example14.3(c).

R L'

RL

Figure14.15A Thecircuit forExanrpte 14.4.

Solution a) Begin by transfoming the cbcuit in Fig. 14.15ro the s-domain,as shownin Fig.14.16.Use voltage division across the parallel combination of irductor and load resistorto computethe transfer function:

RLSL

ll(r) =

Rz+sa

^*^3? R+Rr

/ R , \ \R+R/./ R.

'- - \ /R + R , /\ R t

, a.:

Kt r+@.'

KR/L.

Note that (). is the cutoff frequency of the loadedfilter. b) For the unloaded Ra high-passfilrer from Example 14.3(c),the passbandmagnitudeis 1, and the cutoff hequencyis 15 kHz. For rhe loadedRa high-passfilter, R = Rr = 500O, so 1( = 1/2.Thus,for the loadedfilter, tlle passband

R

+

v,(s)

SLi

RL Y"(,)

Figure14.16a Th€r-domain equivat€nt of thea,cuitin Fig.14.15.

r

a 1 2'r2

0

30 40 f",ro f" 2D Frequency (kHz)

50

flgure14.17A Thenagnjtude ptotstortheunloaded Rl high,pass fiLter of Fig14.13 andthetoaded Rthiqh-pass filter of Fig.14.15.

14.1 Hjsh-Pass Fjtte6

581

Compadng the transfer functions of t}le unloaded filter in Example 14.3 and t})e loaded filter in Example 14.4is uselul at this point. Both transfer functions are in the form: '''"

t(J

r

_!_

A ( i RL )

JC

with /( : 1 for the unloaded filter and ( : nz/(R + Rz) for t})e Ioaded filter. Note that the value of ( for the loaded circuit reduces to the value of 1
+ vi

''

d l r r= r

+ Vac

@,= 1/RC

R

+ vi

,'r(r)=

sL

Figuc14.18A Twohighpass filters, thes€n€s Rfand 81,together theseries wjththeirtransferfunctions and

(14.20) < Transfer funationfor a high-pass fitter Any circuit with the transfer function in Eq. 14.20would behave as a highpass filter witl a cutoff frequenry of o". The problems at the erld of tl1e chapter give you other examplesof circuits $rith this voltage mtio. We have drawn attention to another important relationship. We have discoveredthat a se es RC circuit has the samecutotf frequency whether it is configured as a low-passfilter or as a high-passfilter. The sameis tiue of a series RZ circuit. Having prcviously noted the connection between the cutofl ftequency of a filter circuir and the time constant of that same circuit, we should expecf the cutoff frcquency to be a characteristic parameter of the circuit whosevalue dependsonly on the circuit components, their values.and the wav thev are connected,

lntroduction to ftequency Sehctive Circuits

objective2-Know the f,l andnCcircuitconfigurations that actas high-pass fitters 14.3 A seriesRL high passlilter has R = 5 kf,) and l- - 3.5mH. Whal is d. for this filter? Answer: 1.43Mrad/s. 14.4 A seriesRC hlgh passfilter hasC : 1 pF. Computcrhc cutoff frequencyfor the lbllowing valuesof R:(a) 100();(b) s kor and(c) 30 kO. Answ€r: (a) l0 krad/sr (b) 200md/s; (c) 33.33rad/s.

14.5 Computcthe lransferfunction of a sericsnC iow'passlilter tiathas a load resistor,R,in paraltclwith its capacitor.

Answer: H(r) = :.

1 pr I

.+-KRC

uhere K

R+RI

NOTE: Ako tt! Cha?terProbk,ls14.12and 14.13.

14.4 S*ndpass Fi(ters The next filters rvc examiDeare thosethat passvollageswithir a band ot liequencics1o the oulput while filtering our vollagesat frequenciesour sidethis band.Thesefillers are somewhatmore complicatedthan the 1ow passand high-passfilters of the prcviousseclions.Aswe havealrcadvseen in Fig. la.3(c).ideal bandpassfiltcrs have lwo cutoff frequencics.o,r and o,.r.which idenlify the passband. For rcaljsticbandpassfilters,rheseculoI fiequcncicsare againdefinedas the frcquenciesfor which the magnitudc of thc translerfuncrionequals(1/\/: )r/mu.

CenterFrequency, Bandwidth, andQualityFactor Thcre are three other importanl palanelers that characterizca bandpass filtcr. The first is the center fiequency,a,,, defired as the frequcncyIoI which a circuit\ transferfunctionis purelyreal.Anorhernamcfbr lhe centcr lrequencyis the resonlnt lrcqucncy.Thisis the samenamcgiven1(lthe frequencvthat charactcrizes the naturalresponseof thc second-ordercircuitsin ChaptcriJ,becaulethey are the samefrcqucnciesI When a circuiris driven at ihe rcsonantftequency,we sa],that thc cjLrcuit ]s in resonance, becausethe frcquencyoIlhe forcingfunctionis thc sameasthe naturalfrequencyof thc circuit.flle centerfrequencyjs thc gcometriccenterof the passband,that is.d, : r6;,ar. For bandpassfilters,thc nagnitudeofrhe lraNfer functionis a maximull at the centerfrequency( H-,, : 1II(ja") ). The secondparamcleris the bandwidth,B. which is the widlh of the passband.Ihe final paraneter is the qrality factor,which is the ratio of the centerfrequcncJr1()lhe bandNidth.'Ihequalitylaclor givesa measure ofthe width of the passband, ildependent ofits locatioDoDthe frequency axis-It also dcscribesthe shapeof the magnitudeplot, independentof Although there are five differcnt paramelersthar characterizcthc bandpassfiller .,.1,@.r,o,. p, and O only two of the five can be speci fied independcnlly.lnother words,oncewc arc able to solvefor any two

583 ofthese parameterqtheother thrce canbe calculatedfrom the dependent relationships among them. We will defhe these quantities more specifically once we have analyzeda bandpassfilter. In the next section,we examine two RIC circuits which act as bandDassfilters. and then we deri!eexPre.cionc for dll ol lheircbaraclefiitjc pafamerers.

TheSeriesRLCCircuit-QualitativeAnalysis Figure 14.19(a)depicts a sedesRaC circuit. We want to coNider the effect of changing the souce ftequency on the magnitude of the output voltage. As before, changesto the source ftequency result in changesto the impedance of the capacitor and the inductor. This rime, the qualitative analysisis somewhatmoie complicated,becausethe circuit has both an inductor and a capacitor. At (, = 0, the capacitorbehaveslike an open circuit, and the inductor behaveslike a short c cuit. The equivalent circuit is shoM in Fig.14.19(b).The opencircuitrepresentingthe impedanceof the capacitor prevents curent from reaching the resistor, and the resulting output voltageis zero, At (, = co, the capacitorbehaveslike a short circuit, and the inductor behaveslike an open circuit. The equivalent circuit is shown in Fig. 14.19(c).Theinductor now preverts current ftom reachingthe resis tor, and againthe output voltageis zero. But what happens in the frequency region between ar :0 and o = or? Betweenthesetwoextremeqboth the capacitoiand the hductor have finite impedances.In this region, voltage supplied by the source will drop aooss both the inductor and the capacitor, but some voltage will ieach t]le resistor. Remember that the impedance of th€ capacitoi is nega, tive, whereas the impedance of the inductor is positive. Thus, at some fiequency,the impedance of the capacitor and the impedance of the inductor haveequalmagnitudesand oppositesigns;thetwo impedances cancelout, causirg the output voltage to equal the source voltage. This special frequencyis the centerfrequenry,o,. On either sideof o,, the output voltage is lessthan the sourcevoltage.Note that at .d,, the sedescombinationof the inductor and capacitor appea$ as a shoit circuit. The plot of the voltage magnitude mtio is shown in Fig. i4.20. Nore that the ideal bandpassfilter magnitude plot is overlaid on the plot of the seriesRaC transferfunction magnitude. Now consider what happens to the phase angle of the output voltage. At the ftequency where the source and output voltage are the same,the phaseanglesare the same.As the frequencydeffeasei the phaseangle contribution from the capacitor is larger than that from the inductor. Becausethe capacitorcontributespositivephaseshift,the net phaseangle at the output is positive.At very low frequencies, the phaseangle at the output maximizesat +90". Conversely,if the frequency increasesfiom the ftequency at which the source and the output voltage are in phase,the phase angle contribution from the inductor is larger than that from the capacitor.The inductor con tributesnegativephaseshifi, so the net phaseangleat the output is nega' tive. At very high frequencies, the phase angle at the output reaches its negativemaximum of -90'. The plot of the phaseangledifferencethus hasthe shaDeshowninFis.14.20.

tcl Figure 14.19A (a)Asedes Rtrbandpass fitter;(b)the for@= 0; andG)theequivahnt equivatent circujt

H\i-) 1.0 -,1.

a

e(j"'j 90'

90" pLot Figure 14.20& Thefrequency response forthe sefies RtCbandpass fittercncLrit in Fig.14-19.

584

Introduction to Frcquency Setective Circuits

TheSeriesRl.CCircuit-QuantitativeAnalysis We begin by &awing the s-domait equivalent for the se es RaC circuit, as sho\ln in Fig. 14.21.Use s-domain voltage division to \4Titean equatior for the transfer function:

4(s) Figur€14.211 lhe s-domain equivalent forth€ circuit in Fis.14.19(a).

(RlLts s' + (RlL)s+ (LlLC)'

\14.21)

As before, we substitute r = jo into Eq. 14.21ard produce the equations for the magnitude and the phase angle of the tmnsfer function: a(R/L)

laj.)l :

d(r-) = 90"

(14.22)

.f uR/Lr f tan 'l -*L l(1/LCl @' )

\14.?3)

We now calculate the five patamete$ that characte ze this RZC bandpassfilter. Recall that the center frequency,oo, is defined as the frequency for which the circuit's ffansfer function is purely real. The transfe. func, tion for the RaC circuit in Fig. 14.19(a)will be real when rhe frequency of ure voltage source makes t}le sum of the capacitor and inductor imped-

1 t."t - -i"rc

u.

(11.24)

Solving Eq. 14.24for o", we get

Centerfreqoency >

\14.?5)

Next, calculatethe cu1offftequencieq(,.t and oa2.Remember that at the cutoff frequenciesjthe magdtude of the rransfer tunction is (1/\,t )Anax. Becauseil-,, = lHQo") , we can calculareflnax by subsrituting Eq. 14.25 ir\to Eq.1.4.22. Htu

= H(ja,)

oo@/L)

\IOtrcl -;,*+AAn \,(!lraFtL) ' / - - . '

V ltl,lLC) tr/LC)1, - +|\/t| L

:1.

LCxR/L\|

14.4 Bandpass Fitters 585

Now set the left-hand side of Eq. 14.22to (U^VDH^ 1/V2) aDdprepareto solvefor (,":

(which equals

a,(R/L) \/,

vrctn ,. ;O"nUj tIt-zn;

on,nc;l + t

\14.26)

We can equate the denominato$ of the two sidesof 8q.14.26 to get

L +. l,- . . ^

1

114.27)

ReanangingEq. 14.27resultsin the folowing quadmticequation:

'lt+-,n- yc =o.

(14.28)

Thesolutionof Eq. 14.28yieldsfour valuesfor the cutoff frequercy.Only they identwo of thesevaluesare positiveand havephysicalsignificance; of this filter tifv the Dassband

(14.29)

< cutofffrequencies, seriesruCfittec (14.30)

We cair use Eqs. 14.29and 14.30to confirm that the center frequercy, &),, is the seomeldcmeanol the iwo culoll freouencies:

< Retationship between centerfrequenay andcutofffrcquencies

/l R

r r n r--\rrtcr,t/ tl lnz r' t r n v r r r t * Vf-zr V\:rl V\:r/ \icll (14.31)

586

(iftuiis IntrodLrction to Frcquency Setective

Recall that the bandwidth of a bandpassIilter is defined as the difference between the two cutoff ftequencies, Becaused.2 > ().1 we can compute the bandwidthby subtractingEq.14.29from Eq.14.30: Relationshipbetweenbandwidthand cutoff freouencies>

fl ^"- + l

/ n Y-

\n)

R

(+)[ + .\n) - (r"t) /nY

i'

\14.32)

The quality factor, the last ol the five characteristic parameters,is defined as the ratio of centerfrequencyto bandwidth.Using Eqs.14.25and 14.32:

ouatityfactor> (1/LC) (R/ L) L

(14.33)

cP

We now have five parametenthat characterizethe seriesRaC bandpassfilter: two cutoff frequencies,a,.1and d.2, which delimit the passband; the center fiequency, do, at which the magnitude of the transfer function is maximum;thebandwidth,B, a measureofthe width of the passband;and the quality factor,O, a secondmeasureof passbandwidth.As previously noted, only two of these patameters can be specfied independently in a design. We have already observed that the quality factor is specified in terms of tie center frequency and the bandwidth. We can also rewrite the equations for the cutoff frcquercies in terms of the center frequency and the bandwidth:

_ p 2

: L + 2

(!)'

\14.34)

(BY

\t)

114.35)

Alteinative foms for these equations express the cutoff ftequercies in terms of t]le quality factor and the center frequency: ..t:

o.'

(+) l

I - . tro Ir

/ r \rl ( V t r a lI

114.36)

I

Also seeProblem14.17at the end of the chaptei.

\r4.37)

I4.4

Bandpas5 Fjlte6

587

The examples that follow illustrate the design of bandpass filters, introduce another RLC ciicuit that behaves as a bandpass filter, and examine the eflects of souce iesistanc€ on the characteristic parameters ofa seriesRaCbandpassfilter.

Designing a Eandpass Filter A graphic equalizer is an audio amplifier thar allows you to select different levels oI amplification within differe0t frequenc) fegionsUsing the series RaC circuit in Fig.14.19(a),choosevaluesfor R, L, and C that yield a bandpasscircuit able to select inputs witiin the 1-10 kllz frequency band. Such a circuitmight be usedin a graphicequalizerto select this ftequency band from the larger audio band (generany 0-20 kHz) prior to amplification.

Sotution We need to compute valuesfor R, a, and C that producea bandpas\liler qirh curoff frequencie( of 1 kHz and 10 kHz There are many possible appioachesto a solutior. For instancq we could useEqs 14.29and 14.30,which specifyd.1 and (d"2 ir terms of R, a, and C. Becauseof the form of theseequations,the algebraicmanipulationsmight get compiicaled.Illstead,we will use the fact that ttre center frequency is the geometric mean of the cutoff fiequencies to compute .rd, and we will then use Eq. 14.31 to compute a and C from a,,. Next we will use the definition of quality factor to compute O, and last we will use Eq. 14.33to compute .R.Even though this approach involves more individual computational sfeps, each calculatron rs fairly simple. Any approach we choose will provide only two equations hsufficient to solve for the three unknowns because of the dependencies among the bandpassfilter charactedstics.Thus, we need to select a value for eithei R, a, or C and use the two equalionswe ve cho<ento calculalelhe remaining componentvalues.Here, we choose 1/rF as the capacitor value, because there are stricter limitations on conmercially available capacitors than on inductors or rcsistom, We compute the center frequency as the geometric mean of the cutoff frequencies:

Hz. f .: \/Trt; = {m00xm!00) = 3162.28

Next, compute the value of L using the computed center fiequency and the selectedvalue for C. We must .emember to convert the center ftequency to radiansper secondbeforewe car useEq.14-31:

= 2.533InH.

itc

lzn(3162.28)12(1tr4)

The quality factor, O, is defined as tlle ratio of the center ftequency to the bandv.idth. The bandwidth is the difference between the two cutoff frequency values.Thus,

f. fr-

fA

3162.28 10,000- 1000

NowuseEq.14.33to calculate rR:

'^ r.,,1T cQ,

= 143.24 A.

00-6)(0.3s14)'

To check whetler tnese component values produce the bandpassfilter we want, substitutethem into Eqs 14.29and 14.30.Wefind that @.1= 6283.19rad/s (1000Hz), dc, : 62,831.85 radls (10,000Hz), which are the cutoff frequencies specified for the filter. This example rcminds us that only two of the five bandpass filler parameteis can be specified independently. The other three parameterc can always be computed from the two that are speci fied. In turn, these five paramerer values depend on the three component values,R, a, and C, of which only two can be specified independently.

588

lntroduction to Frequency Sehciiv€ Cjrcuits

Designing a Parattet RICBandpass Fitter a) Show thar thc RLC circuit in Fig. 14.22is also a bandpassfilter by derivingar expressionlbr the transfcrf unction 1-1(s). b) Computethe centerfrequency,oo. c) Calculatethe cutofi trequencies, @.rand o.2,the bandwidth,B, and the quality factor,Q. d) Computevaluesfor R and a to yield a bandpass filter with a center frequencyof 5 kHz and a bandwidthof200 Hz, usinga 5 AF capacitor.

b) To find the center frequency, , we need to calcu late where the transfer function magnitude is maximum.Substitutingr : jain H(s),

H(i,))=

RC

,tl ,Itr. r V\rc --l * \ Rci I

' .

(.^"

+)' -R/

L

C

The magnitude of this transfer function is maxrmum when the term /

Figure14.22,i. ThecircujttofENamph 14.6.

1

\zc

Solution

'/

\2

is zero.Thus,

a) Beginby drawingther-domain equivalentofthe circuit in Fig.14.22,as shownin Fig.14.23.Using voltagc division, we can compute the lransfer function lbr the equivalent circuit if we first compule the equivalentimpedanceot the parallel combinationof l, ard C, identjfied as Z.q(r) in Fig.14.23:

a . ," '

arcl H(ja)l = 1.

H-*:

of the o At the cutoff frequencies,the magnitude =

transfer funcrion is (1/\4)11-,, l/\,4. Substifutingthis constanlor tlle left-handsideofthe magnitudeequationand then simplifying,weget

L

z.e!):

' " = l ilT

1 .'C

[| * ' A '' oll : . '

_l RC

Squaringthe left-handsideof this equationonce againproducestwo quadraticequatioDsfor the cutoff frequencies,with four solutions.Only two ofthem are positiveand thereforehavephysical significanc€:

I

z*(,) R

r,(") +_ )

.

rtsc

^

_

1 ,"t: -ZnC.

r,(") .,r. :

F i g u c 1 4 . 2 3B l h er - d o mra" q l i v d l Ip o l _ h ef ( u 1 i l Fig.14.22.

1

znc

.

&.Cutofffreouencies for oarattet BlCfitters

ta.a 8"-dpdrfiltes We compute the bandwidth from the cu! offfrequencies:

d) Use the equation for bandwidth in (c) to compute a value for R, given a capacitance of 5 I!F. Remember to convert the bandwidth to the appropnate umts:

1 -RC

1 BC

Fhally, use the definitionof qualityfactor to calculateO:

(22X200)(s. 1ir6)

a=olq R2c L Notice that once again we can specily the cutoff frcquencies for this bandpass filter in terms of its center frequency and bandwidth:

599

Using the valueofcapacitanceand the equation for center frequencyin (c), computet}le inductor value:

,

1

- " : ' -BV \n' /p'i- -; '

^':t.^Kil..?.

x 1o{) [2,(5ooo)]'(s = 202.64 u.H.

Detennining Effectof a Nonideal VoltageSource on a ilc Bandpass Filter For eachof the bandpassfilte$ we have constructed, we have alwaysassumedan ideal voltage source,&at is, a voltage source with no series r€sistance.Even though this assumptionis often valid, sometimesit is not, as in the case where the filter design can be achievedor y wilh \ aluesot R. L. dndCwboseequivalent impedarce has a magnitude close to the actual impedanceof the voltage source.Examine the effect ol assumjnga noDzero source rcsistance!R,, on the charactedsticsof a sedesRLC bandpassfilter.

Solution a) Begin by transforming the circuit in Fig. 14.24to its r domain equivalent,as shown in Fig. 14.25. Now use voltage division to construct the transfer function: R

x u. (4;4), - i I

a) Determine the transfer function for the circuit in Fie.14.24. b) Sketch the magnitude piot for the circuit in Fig. 14.24,usingthe valuesfor R, a, and C from Example 14.5ard setting& : fi. On the same $aph, sketchthe magnitudeplot fot the circuit in Example14.5,whereRi - 0.

Yis)

Figur€14.24 A Th€circujtfor Examph 14.7.

y,('

Flgure14.25 a Ther-oorainequvdlenlor lfe.'(Lit il Fig.14-24-

590

Introductionto Frequency SeL€ctjve Cjrcuits

Substitute r = j,) and calculate the tmnsfer function magnitude:

Finally, the quality factoi is computed from the center fiequency and the bandwidth:

n

a0,\ =

\TIrc

t

,t; .t' , ^.R+ V\.c "i *l'' . ,/

The center frequency, oo, is the frequency at $bich this rran.ferlunclioomagnitudeis marimum, which is

From this analysis, note tiat we can write tle transfer function of the seriesRIC bandpassfil' ter with nonzerosourceresistanceas

KB" s' +Bs+4'

LC At the cenler frequency,the maximummagnitudeis

H-,, = lH(/-,)l:=

t2

-.

R+Ri

The cutoff fiequencies can be computed by set ting ttre tansfer function magnitude equal to (1./tr2)H ^,.:

,"=

/R+R,\r.

I

\ rL )-rc

The bandwidth is calculated from the cutoff frequencies:

^

I'=

Bs

r1(")-

R+R, ."r= + Zr R+Ri , Zf

Nole that when 4 = 0, I( - 1 and the tmnsfer function is

R +& L

2500

5000

b) The circuit in Example 14.5 has a center ftequencyof3162.28Hz and a bandwidthof 9 kHz, and H-* = 1. lf we use the same values foi R, a, and C in t}le circuit in Fig. 14.24 and let Rr = R, then the center frequency rcmains at 3162.28 kHz, but B : (R + R)lL = 18kHz, and Hnax - R/(R + Rr) : 1/2. The rransfer function magnitudes for these two bandpassfilters are plotted on the samegraph in Fig.14.26.

7500 10000 12500 15000 17500 20000

f (.Hz\

Figlre14.26.rThemagnitude ptotsfora s€ries nlrbandpass frlterwitha zerosource resistance anda nonz€ro source resistance,

14.4 Eandpass Fitterc 591

If we compare the charactedstic parameter values for the filter with R, = 0 to t}le values for the filter with & + 0, we seethe followingl . The cetrter frequencies are ttre same. . The maximum transfer function magnilude for the filter with & + 0 is smallerthdnfor rhe tiiterq ith R, - 0. . The bandlridth for the filter v.ith Ri ta 0 is larger than that for the filter with Ri = 0. Thus, the cutoff frequencies and tie quality factors for the two circuits are also different.

I JC

vi

/t(s) =

(R/z)r + (R/L)s+ rlLC

The addition of a nonzero source resistanceto a seriesRIC bandpassfil@.= ,,/-tlLc B= R/L ter leaves the center frequency unchanged but widens the passband and rcduces the passbandmagnitude. R Here we seethe same design challenge we saw with the addition of a load resistor to the high-passfilter, that isJwe would like ro designa bard! pass filter that will have the same filtering properties regardless of any vi S L v. JC intemal resistance associatedwith tle voltage source. Udortunately, filtels consfucted fiom passive elements have their filtering action altercd with the addition of source resistance.In Chapter 15,we will discover that i/nc ' active filters are insensitive to changesil source resistance and thus are st + slRC+ 1/LC better suited to designsin which this is an impofiant issue. p = 1,/RC a"=\,-1/LC Figure 14.27summadzestle two RaC bandpassfilten we have stud, ied. Note that the expressionsfor tle circuit tmnsfer functiom have the Figure 14,27A TwofLCbandpass fitters, together wjth same form. As we have done previously, we can create a geneml form for €quatjons forih€innsf€rfunctjon, center frequency, the transfer funclions of these two bandpassfilters:

(14.38) < Transfer functionfor RlCbandDass fitter

Any circuit witl the transfer functior in Eq. 14.38acts as a bandpassfilter with a center frequency (1)oand a bandwidth B. In Example 14.7,we saw tlat t}le transfer functior can also be written in the folm

KB" s2+Bs+0t'

114.39)

where the values for lK and B depend on whetler the sedesresistance of the voltage souce is zero or nonzero,

Retatingthe Frequency Domainto the TimeDomain We can identify a rclationship between tle parameteis that characterize t]le frcquency responseof RaC bandpassfilters and the parameters that characterize the time response of RLC circuits. Cnnsider tle sedes RaC circuit in Fig. 14.19(a). In Chapter 8 we discovered tlat the narural responseof tlis cir€uit is characterized by tle neper frequency (a) and the resonant frequency (oo). These paramete$ were expressedin tems of the

592

Introduction to Frequency Setective Circuih

crrcul componcntsin Eqs. 8.5Nrnd 8.59,which are repeatedhcrc lor

R 2L

114.4A)

114.4t) We seethat thc sameparameteridois usedto characte ze both thc time responseand thc frequencyresponse. Thafs why the centerfrequcncyis also called thc resonantfrequency.The bandwidlh ard the ncper fre, queDcyare relatedby the equation

P:2".

114.42)

Recallthat the naturalresponseofa seriesR/,C circuitmay bc underdamped,overdamped.or critically damped.The transition from overdampedto underdampedoccurswhen oj = a'. Considerthe rclalionship betweena and p from Eq.14.,12and the definirionofrhe qualjryfaclor O. The transilion from an ovcrdanped to an underdampedrcsponseoccurs when Cl = 1/2.Thus,a circuit whosefrequcncyresponseconrainsa sharp peak at d,, indicating a high 0 and a narrow bandwidth,will have an underdampednatural rcsporNe.Converscly,a circuit whosc liequency responsehas a broad bandrvidrhand a low Q will have an overdamped

objective3-(now the Rlf circuitconfigurations that actas bandpass fitters 14,6 Usjng the circuit in Fig.11.19(a),compurethe valuesof R and l, to give a bandpassfiller with a centerfrequencyof 12kHz and a quality factor of 6. Use a 0.1pF capacitor.

14,8 Recalculatethe componentvaluesfor the cir cuit in Exampie14.6(d)so that lhe frequency responsooflhe resultingcircuit is unchanged usinga 0.2pF capacitor.

Answer: | : 1.76mH, -R = 22.10O.

Answer: a = 5.07mH.n = 3.98kO.

14.7 Usjng the circuit in Fig.14.22,compurethe va1uesofl, ard C to give a bandpassfilter wlth a center ftequency of 2 kHz and a bandwidrh of 500Hz. Use a 250 O resistor.

14,9 Recalculatethe componcntvaluesfor the circuit in Example14.6(d)so that the quality iactor of the resnltingcircuil is unchangedbut the cenlerfrequencyhasbeennoved ro 2 kHz. Use a 0.2pF capacitor.

Answ€r: a : 4.9?mH,C : 1.21pF.

Answer: R = 9.95kO,L = 31.66mH.

NOTE: Alxo try ChaptetPrcblems14.21and t4.22.

11.5 BandrejectFitters593

14.5 Bandreject Filters We turn now to the last of the four filter categories-the bandrejectfilter. This filter passessourcevoltagesoutside the band betweenthe two cutoff ftequenciesto the output (the passband),and attenuatessource voltagesbefore they reach the output at frequenciesbetween the two cutoff frequencies(the stopband).Bandpassfilte$ and bandreiectfilrers thus pedolm complementaryfunctionsin the frequetrcydomain. Bandreject filters a.rechamcterized by the same parameters as band passfilters: the two cutoff ftequencies,the center frequency,the bandwidth, and the quality Iactor. Again, only two of tlese five parameters can be specified independently. In the next sections,we examine two circuits that function as bandreject filters and then computeequationsthat rclate the circuit componerlt values to t]re characteristic parameters for each circuit.

TheSeriesRLCCircuit-QualitativeAnalysis Figure 14.28(a) shows a seiies RaC circuit. Although the circuir componentsand connectionsaie identicalto thosein the seriesnac bandDass filrerin Fig.l4.lora).the circuirin trg. t4.28{a)hasan imporranrdiiterence:the output voltage is now defined acrossthe inductor-capacitor pair. As we saw in the case of low- and high,pass filters, rhe same circuit may perform two different filtering functions, depending on the definition of the output voltage. We have already noted that at o : 0, t]le inducto. behaveslike a short circuit and the capacitor behaveslike an open circuit, but at ,, = oo, these roles switch. Figure 14.28(b) presents the equivalent circuit for o : 0; Fig.14.28(c) Fesents the equivalent circuit for o : c). In both equivalenr circuits! the output voltage is delined over an effective open circuit, and (c) thus the output and input voltages have the same magnitude. This series RaC bandreject filter circuit then has two passbands-{ne below a lower Fiqure 14.28A (a)Asenes Rr bandrcject filter. cutoff frequency,and the other above an upper cutoff fiequency. (b)The equjvatent cjrcuit for@= 0. k) Theequivalent Between these two passbandsrboth tlle inductor and the capacitor have finite impedances of opposite signs.As the frequency is increased from zero, the impedance of tle hductor increasesand that of the capacilH(ia) tor deqeases.Therefore the phase shift between the input and the output approaches-90' as ot approaches1/dC. As soon as ol, exceeds1/oC, 1.0 the phaseshiJtjumps to +90" and then approacheszero as o contitrues to !

\z

At some frequency between the two passbands,t]le impedancesof the inductor and capacitor are equal but of opposite sign. At this frequency, 0 the seriescombhation of th€ inductor and capacitor is that of a short circuit, so the magnitude of the output voltage must be zero_This is the cen- e(i') 90" ter liequencyof this seriesRIC bandreiectfiltei. Figure 14.29presents a sketch of the frequency responseof the series RIC bandrejectfilter from Fig. 14.28(a).Note rlar rhe magnirudeplot is 0" overlaid with tiat of the ideal bandreject filrer from Fig.14.3(d). Our qualitativeanalysishasconfirmedthe shapeof the magnitudeandphaseangle 90' plots. We now turn to a quantitative analysis of the circuit to confirm this tiequency iesponse and to compute values fot the parameters that charac- Figure 14.29A Thefrequency ptotforihe rcspons€ terize this response. jn Fis.14.28(a). senes RrCbandrej€cifittercjrcujt

594

Introduction to Frequency Setective Circuits

TheSeriesRICCircuit-QuantitativeAnalysis Aftei transforming to the s-domair, as shom in Fig. 14.30,we use voltage division to constnct an equation lor the transfer function:

Y,(r) H(r) :

Figure14.30l. Ther-domain equivaLeni of th€cjrcuit in Fig.14.28(a).

-

sr+] 1 l{+rL+-;

. R s'+tr+

1

1 /r

(14.43)

Substitutej(.) for s in Eq. 14.43ard generateequationsfor the transfer functionmagnitudeandthe phaseangle: | 1 H(ia)l =

e0,,):

li-*l

"l

ttt ^' r-r.''' V(.c--l.\t/

\14.44)

(r4.45)

Note ttrat Eqs. 14.44and 14.45confirm the frcquency responseshape pictured in Fig. 14.29,which we developed based on the qualitative analysis. We use t]le circuit in Fig. 14.30 to calculate the center frequency. For the bandreject filter, the center frcquency is still defined as the fiequency for which tlle sum of the impedancesof the capacitor and inductor is zero. In the bandpassfilter, the magnitude at the center frequency was a maximum, but in the bandreject filter, this magnitude is a minimum. This is becausein the bandreject filtei, ttre center frequency is not in the passband; rather, it is ir the stopband. It is easy to show that the center frequency is given by

(14.46)

SubstitutingEq.14.46into Eq.14.44showsthat lnja")l : O. The cutoff frcquencies, the bandwidth, and the quality factor are defined for the bandreject filter in exactly the way they weie foi the bandpass filtels. Compute the cutoff frequetrcies by substituting the constant (1/V2)HD!; for the lelt hand side of Eq. 14.44 and then solving for o.r

14.5 Bandreject Fitters 595 and o.2. Note that for the bandreject filter, Itiw = 1{(i0) : and for the sedesRaC bandrejectfilter in Fig.14.28(a),Hh* :

R

/ / R \ ' ,- i 'r

n-,,1\n) R 2L

//R\, V \2L,i

1 LC'

fl(i*)i, 1.Thuq 114.47)

(14.48)

Use the cutoff frequencies to generate an expression for t]le bandwidth,B: (14.49)

Finally, the center frequency and tle bandvridth produce an equation for the quality factor, O:

,:E;.

(14.50)

Again, we can rcpresentttre expressionsfor the two cutoff ftequencies in lermsof the bandwidthandcenterfrequency, aswe did for the bandpassfilter .", _

lJ ,.

@'

\14.51)

(14.52)

Alternative forms for theseequationsexpressthe cutoff frequenciesin termsof the quality factor andthe centerfrequency:

*,:*l-i.,t{#-]

(14.53)

\14.54)

Example14.8presentsthe designofa seriesRaCbandrejectfilter.

596

Circujts S€tective Intrcductjon t0 Fcquency

Designing a SeriesRICBandrejedFitter Using the seriesRaC circuit in Fig. 14.28(a),compute the componentvaluesthat yield a bandreject filter with a bandwidthof 250Hz and a centerftequetrcyof 750Hz.Usea 100nF capacitor.Compute (1).2, and8. valuesfor R,l,, .d"1,

Use Eq.14.49to calculateR:

R: PL = 27(250)(450 x 10 3) : 70'7A.

Solution We begin by ushg the definition of quality factor to compute its value for this filter:

The valuesfor the centerfrequencyand band widttrcanbe usedin Eqs.14.51and 14.52to com pute the iwo cutoff frequencies:

,":-t*

Q-a./P=3. to conUseEq. 14.46to computea, remembering (rd per vert to radians secold:

= 3992.0rad/s, ."=;,

I

x 104) [2,(?50)]1100

: 5562.8rad/s. The cutoff frequenciesare at 635.3Hz and 885.3Hz Their differcnce is 885.3 6353 = 250H2, cor.' firmjng lhe specilied bandwidth The geomerfic - 750H/. conLrminglhe meani. v/i65)(885j) specified center frequen+

As you might suspectby now, another configuration that produces a bandreject filter is a parallel RaC circuit. Whereas the analysis details of the paiallel RZC circuit are left to hoblem 14 32, the results are summadzed in Fig. 14.31,along with the series RIC bandreject fifter. As we did for other categories of filters, we can stale a general form for the tansfer functions of bandieject filte$, replacing the constant terms with B and .d":

fitter > fundionfor f,LCbandreject Transfer

(14.55)

Equation 14.55is useful in filter design,becauseany circuit with a transler function in this form can be used as a bandreject filter'

14.5 Bandr€ject Fitters 597

vo

L

sc s, + t/Lc s'+(R/L)s+1/LC @.=.v/1/LC

a.= yI/LC

B=R/L

p= 1/Rc

rigure14.31 Twoilr bandreject filt€rs,together wiihequations forthetnnstur tunction, cent€r frequency, andbandwidth ofeach.

598

Introduction to Frequency Setective Cncujts

PracticaIPerspective Pushbutton Tetephone Circuits In the PracticatPerspectjve at the stat of this chapter,we described the (DTMF) duat-tone-muttipLe-frequency systemusedto signalthat a button hasbeenpushedon a pushbuttontelephone.A keyetem€ntof the DTMF systemjs the DTMF receiver-a circuitthat decod€s the ton€sproduced by pushinga buttonanddetermines whichbuttonwaspushed. In orderto designa DTI4F reciever, we needa betterunderstanding of the DTlvlF systefir.As you canseefrornFig.14.32,the buttonson the tetephoneareorganized into rowsandcolunns.Thepair of tonesgenerated by pushinga buttondepends on the button'srowandcotumn.Thebutton'srow determjnes its low-frequency tone, andthe button'scotumndetermines its "6" pressing high-frequency tone.1Forexampte, the button produces sinusoidattoneswith the frequencies'770Il2 and14'77HL At the teLephone switchjngfacitity, bandpassfilters in the DTI4F receiverfirst detectwhethertonesfrom both the Low-frcqu€ncy and highgroupsare simuttaneousLy present.Thistest rejectsmanyextrafrequency neousaudiosignaLs that are not DTI4F. If tonesare presentjn both bands, otherfiLtelsare usedto setectamongthe possibl€tonesin eachbandso jnto a uniquebuttonsignat.AdditionaI that the frcquencies canbe decoded testsareperformed to preventfatsebuttondetection.Forexarnpte, ontyone tone per frequencybandis attowed;the high- and low-bandfrequencies muststartandstop within a few firitliseconds of oneanotherto be consideredvatid; and the hjgh- and tow-bandsignalamplitudesmust be suffi' cienttyctoseto eachother. You may wonderwhy bandpassfiLtersare used instead of a high passfjtter for the high-frequency group of DTI4F tones and a Low-pass gfoup of DTI4F fitter for the tow-frequency tones.The reasonis that the tetephonesystemusesfrequenciesoutsideof the 300-3 kHz band for purposes, othersjgnating suchas dngingthe phonekbetL.Bandpass fitters preventthe DTlvlF receiverfrom enoneousLy detectingthese other signals.

v *v 7.

t'l

Figue14.324 Tones generated bythercwsand pushbultons. columns oftehphone

NjfE: Assessyaur undestanding of this PracticdLPespective by hying Chapter Problen s 14.43-1 4.45.

l Afourth hiqh frequency toneis Eservedat 1633Nz.Ihi5toneis usedinfiequenttvand is not produced bya standad12-buttontelephone.

Probtems599

5ummary . A frequency s€lective circuit, or fflt€r, enablessignalsat certain ftequercies to reach the output, and it attenuates signals at other frequencies to prevent them from r€aching the output. The passband contains the fre, quencies of those signals that are passed;t}le stopband contahs the frequencies of those signals that are atten, uated.(Seepage568.)

. Bandpassfilters and bandreject lilten eachhave two cut off foequencies,(d.1and rrr. Thesefilters aie further characterizedby their center ft€qu€ncy (.,,,), bandwidth (B), and quality factor (O). Thesequantities are defined as o"= t-a1n6 p = o.2

0+1,

o: alp. The cutoff ftequency, o. , identifies t]le location on t}le trequency axis that separates the stopband from the passband. At the cutoff frequency,the magnitudeof the transferfunctior equals(1/1U )H-"* (Seepage571.)

A low-pass ffller passesvoltages at frequencies below o. and attenuatesftequenciesabove .()..Any circuit with the lransfer function fl("):

(see pagesJ65 )6b.) . A bandpassfilter passesvoltages at frequencies within the passband,which is betweer @crand (,.2. It attenuates frequenciesoutside of the passband.Any circuit with the transfer function

Ps

3+Bs+4 functionsas a bandpassfilter. (Seepage591.) . A bandreject fflter attenuates voltages at frequencies within the stopband,which is betweeno"r and o.2. It passesfrequenciesoutsideof the stopband.Anycircuit with the transfer function

functions as a low-pass filter. (Seepage 576.)

A high-passfill€r passesvoltages at frequencies above o. and attenuates voltages at frequenciesbelow o.. Any circuit with t}le transfer funclion

rl(s) = - L Iunctionsas a high passfilter. (Seepage58i.)

sr+-j st+8"+-1, functioN as a bandrejectfilter. (Seepage596.) . Adding a load to the output of a passivefilter changes its filtering properties by altering the location and magnitude of the passband.Replacing an ideal voltage sourcewith one whosesourceresistanceis nonzeroalso changesthe filtering propertiesof the iest of the circuir, again by alterhg the locatjon and magnitude of the passband.(Seepage589.)

Problems Section14.2 14.1 a) Find the cutoff frequency in hertz for the Rl, filter shownin Fig.P14.1.

250mH

CalculateH(id) at o.,0.3o., and 3o". c) If o, : 50 cos ol V, D.rite the steady,stateexpres-

sion for ?o when .r:

figurcP14,1

(r", o - 0.3(d., and

600

Inhoduction to Freq0ency Sehctive Cjrcuitr

l|2 Use a 25 mH inductorto designa low-pass, -Ra,passive lilter with a cutoff frequency of 2.5 kHz. !1!B!!!1 a) Specify the value of the iesistoi. b) A load havinga resistanceof ?50O is connected acrossthe output teminals of tle filter. What is the corner, or cutofl lrequency of the loaded filter in hertz?

14,5 A resistoi denoted as Rr is connected in paiallel with the capacitor in the circuit in Fig. 14.7. The loaded low passfilter circuit is shown in Fig. P14.5. a) Derive the expression for ttre voltage transfer function V,/I4. b) At what frequencywill the magnitudeof H(Jo) be maximum? c) What is the maximum value of the magnitude ot H(jo)? d) At whar ftequency wi]l the magnitude of g0o) equal its maximum value divided by \2? e) Assumea rcsistanceof 300ko is addedin parallel with the 4 nF capacitor in the circuit in Fig. P14.4.Find a., nQI), nUQ, aQ0.za,). and 1i(j8a").

143 A resistor, denoted as Rr , is added in serieswith the inductor in the circuit in Fig. 14.4(a).The rew lowpassfilter circuit is shown in Fig. P14.3. a) Derive tle expression for d(r) wher€

H(s): WV.

b) At what hequencywill the magnitudeof fl(Jd) be maximum? c) Wlat is the ma-\imumvalue of the magnitude oflt(io)? d) At what frequency will tlle magnitude of H(jo) equal its maximum value divided by \U? e) Assume a resistance of 75 O is added ir series witl the 250 mH inductor in the circuit in Fig. P14.1.Find a,, H(j0), H(ja.), H(j0.3a,\. and 1t(j3.d.).

Figure P14.5 R

t4s li!-rM!

Figure P14.3

&

L

1|,4 a) Find t}le cutoff frequency (in heitz) of the lowpassfilter sho\lll in Fig. P14.4. b) CalculaleH(io) at .r,,0.2o., ar.d8a". €) If ?,i:48ocosotmv, write the steady-state expressionfor ?,,when o = d.,0.2o., and 8o.. Figure P14.4

20ko

Use a 25 nF capaciror ro design a low-pass passrve filter with a cutoff frequency of 160krad/s. a) Specifythe cutoff tequency in hertz. b) Specify tle value of the filter resistor. c) Assume tlle cutoff frequency cannor increase by more than 8o/".What is the smallestvalue of load resistancethat can be connectedauoss the output teminals of the filter? d) If the resistor found in (c) is connected across the output terminals, what is the magnitude of H(/l,)wheno=0?

14,7 Design a passiveRC low passfilter (seeFig. 14.7) with a cutoff fiequency of 500 Hz using a 50 nF capacitor. a) What is the cutoff frequency in rad/s? b) What is the valueofthe resistor? c) Draw your circuit, labelingthe componentvalues and output voltage. d) What is the transfer function of the filter ir palt (c)? e) ff the filter ir part (c) is loaded witl a resisror whose value is the same as the resisror part (b), what is the transfer tunction of this loaded filter?

Problens601 f) What is the cutoft fiequency of the loaded filrer ftom part (e)? g) Wlat is the gain in the passband of rhe loaded filter from part (e)?

14.8 Study the circuit shown in Fig. P14.8 (wirhout the load resistor). a) As ., - 0, the inductor behaveslil(e what circuit component?what valuewill the output voltage 0ohave? b) As (,+ co, the inductor behaveslike whar circuit component?Whaf value will the output voltageoohave? c) Basedon parts (a) and (b), what rypeof filrering doesthis circuit exhibit? d) What is the transfer function of the unioaded filter? e) If R : 1 kO and I, = 20 mH, what is rhe curoff tequency of the filter in rad/s?

S€ction 14.3 14.10 a) Find the cutoff frequency (in hertz) for rhe highpassfilter shownin Fig.P14.10. b) Find H(j(d) at o., 0.1.d.,and 10o.. c) If o, = Soocos(()tmv, write the steady-state expressionfbr oo when o : .d., (1)= 0.1o., and tD = 10ac. FigureP14.10 2.5 DF

:

-

.

14.11 A resistor,denoted as R., is connectedin series with the capacitorin the circuirin Fig.14.10(a).The new high,passfilter circuitis shownin Fig.P14.11. a) Derive the expression for l1(s) where

H(") = vJU.

b) At what fiequencywill the magnitudeof H(jo) be maximum? c) What is the maximum value of the magnitude ot H(ja)? d) At what frequencywill the magnitudeof t110rr) equalits maximumvaluedividedby \4? e) Assume a resistanceof 10kO is connectedin serieswith the 2.5 nF capacitorin the circuit in Fis. P14.10.Calculate (,., H(ja,), H(io.l.,.,), and H(j10o.).

Figure P14.8 I I R

: I

_l

RL

l1(,):#

Flg(reP14.11

14.9 Supposewe wish to add a load resistorin parallel with the resistorin the circuit shownin Fig.P14.8. a) What is the transfer lunction of the loaded filter? b) Comparethe transferfunction of the unloaded filter (part (d) oI Problem 14.8)afld the trans fer function of the loaded filrei (part (a) of Problem14.9).Arethe cutoff frequenciesdifferent? Are the passband gains different? c) What is the smallestvalueof load resistalcethat canbe usedwitl the filter from Problem14.8(e) such that the cutoff frequency of the resulting filter is no more than 10% different from me unloadedfilter?

t4.x2 Using a 20 nF capacitor, design a high-passpassive l!0!!!11

filter with a cutoff frequency of 800 Hz. a) Specifythe valueofR in kilohms. b) A 68 kO resistoris connectedadoss the output terminals of the filter. What is rhe curoff flequercy,in hertz,ofthe loadedfilter?

602

Circuits to Fr€qu€ncy Selective Introduction

14.13 Using a 25 mH lductor, design a high-pass,R-L, p?iTfll{passivefilter with a cutoff frequency of 160krad/s. _;tr4 a) Specify the value of the resistance. b) Assume theJilter is cornected to a purc resistive load. The cutoff frequency is not to drcp below 150krad/s. What is the smallest load resistor lhat can be contecled acrossthe outpul lerminalsof the filter?

14.14 Design a passiveRC higl passfilter (seeFig. 14.10[a]) with a cutoff ftequency of 300 Hz using a 100 nF capacitor. a) What is the cutoff ftequency in rad/s? b) wltat is the value of the resistor? c) Draw your circuit, labeling the comporent values and output voltage. d) Whar is fte transfer function of the filter in Part (c)? e) If the filter iII part (c) is loaded witl a resistor whose value is the same as the resistor h (b), what is tle transfer futrction of tlis loaded filter? I) What is the cutoff frequency of the loaded filter ftom pafi (e)? g) What is t])e gain in the pass band of the loaded filter from part (e)?

14.15 Consider the circuit shown in Fig. P14.15. a) With the input and output voltagesshown in the figurg this circuir behavesiike what t,?e of filte b) what is the transfer function, H(s) : %(s)/I4(r), of t}Iis filter? c) What is t})e cutoff frequency of this filter? d) wlat is the magnitude of the fillei's transfer tunctionats: rar.?

14.16 Supposea 1kO loadresistoris attachedto the filter itrFig.P14.15. a) Whatis the transferfunction,l/(s) = %(r)/I4(s), of this filter? b) What is the cutoff frequencyof this filter? c) How doesthe cutoff lrequencyof the loadedfilter comparewith the cutoff jlequency ot the unloadedfilter in Fig.P14.15? d) What elseis differentfor tlese two filters?

Section 14.4 14.f

Show that the altemative forms for tle cutoff frequencies of a bandpass filter, given in Eqs 14.36 and 14.37,canbe derivedfrom Eqs.14.34and 14.35.

14,18 Calculate the center frcquency, fie baldwidth, and the quality factor of a bandpass filter that has an upper cutoff frequency of 200 krad/s and a lower cutoff frequency of 180 krad/s. 14,19 A bandpassfilter hasa center,or rcsonant,tequency of 80 krad/s and a quality factor of 8. Find the bandwidth, the upper cutoff ilequency, and the lower cut ofI frequency.Expressall answersit kilohertz 1420 Use a 20 nF capacitor to design a seriesRIC bandsu cenliker.asshownal rhetoDofliq. l4.27.The 'lli'Ii Dass ier t-equencyot I he riiref ic 20 kl-t-z'and rhe qualir' factor is 5. a) Specify the values of R and a. b) what is the lower cutoff frequency in kilohertz? c) What is the upper cutoff frequency in kilohertz? d) What is tlle bandvddth of t}le filter in kilohertz? 1421 For the bandpass filter shown in Fig. P14.21,find

"-,

G) .,", (b) t, G) o , @)a"b G) f ,b $) aa, G) f a, and (h) B.

Pl4.t5 Figure 1ko

tiglre P14.21 300 {)

50mH

Probt€ms 603 14.22 Using a 25 nF capacitor in the bandpass circuit shown in Fig. 14.22,design a ftlter with a qualiry fac*lsJPM _tsini tor of 10 and a center fiequency of 50 krad/s. a) Specifythe numericalvaluesofR and a. b) Calculate the upper and lower cutoff frequenciesin kilohertz. c) Calculatethe bandwidthin hertz. 14.23 For the bandpassfilter shownin Fig.P14.23,calcuEtrc late the following: (a) /,; (b) O; (c) /.r; (d) /.,; and (e) B. tigureP14.23

14,24 The input voltage in the circuit in Fig. P14.23is 800cos(,t mV. Calculat€ t}le output voltage when G) (,, : (,.; (b) o - ro"1;and(c) o = -"2. 14.25 Design a seriesRIC bandpassfilter (seeFrg.14.19[a]) with a quality ol 5 ard a center ftequency of 20 krad/s, using a 0.05 I,F capacitoi. a) Draw your circuit, labelingthe componentvaluesand output voltage. b) For the filter in part (a),calculatethe bandwidth and lhe !aluesof the tqo cutolifrequeDcie\ 1416 The input to the sedes RLC bandpass tilter designedin Problem 14.25is 200coso1mV. Find rhe voltage drop adoss the resistor when (a)o - o,; (b)., = ac1:k) a = o,,;(d) o : 0.1o.; (e) o = 10a.. 1417 The input to the series RaC bandpass filter designedin Problem 14.25is 200cosot mV Find the voltage drop acrossthe seriescombinationof the inductor and capacitor when (a) ., = o,; (b) o: o.1; G) o = o,,; (d) 4, = 0.1o.; (e) o : 10o". 1428 A block diagram of a system consisting of a sinusoidal voltagesouice,an RIC sedesbandpassfiltei, and a load is shownin Fig.P14.28.Theinremal impedanceof tle sinusoidalsourceis 36 + J0 O, and the impedanceofthe load is 320 + i0 O.

The RIC series bandpassfilter has a 5 nF capacitor,a center frequencyof 250krad/s, and a quality factor of 10. a) Draw a circuit diagram of the system. b) Specify the rumerical values of a and R for rhe filter section of the system. c) Wlat is t}le quality factor of the interconnected system? d) What is the bandwidth (in hertz) of the inter, connectedsystem? Fig!reP14,28

1429 The purposeofthis problem is to investigatehow a resistiveload connectedacrossthe output terminals of the bandpasslilter shown in Fig. 14.19 affectsthe quality factor and hencethe bandwidth of the filterirg syst€m.The loaded filrer circuit is shownin Fig.P14.29. a) Calculate the transfer function y,/I{ for the cir cuit sbownin Fig.P14.29. b) What is the er?ressionfor the bandwidth of the system? c) What is the expressionfor the loaded bandwidth (pr) as a function of the unloadedbandwidtl (Bu)? d) What is tie expression for the quality factor of tbe system? e) What is the expressionfor the loaded quality factor (Cr) as a function of t}le unloaded quality tactor (qu)? f) What are the expressionsfor the cutoff frequencieso.t and o.2? figurcP14.29 R

604

Intrcductjon to Frequencr Setective Cncuits

14.30 Consider the circuit shown in Fig. P14.30. """ a) Find r,.,.. b) Find P. c) Find O. d) Find the steady-state expiession foi t', when 0, = 750cosoJ mV. e) Show tlar if RL is expressedin megohms the O of the circuit in Fig.P14.30is ^

2

14.33 For the bandrejectfilter in Fig. P14.33,calculate Em (a) o.;(b) fo;(c) O; (a) oa;(e) /a; (0 zoa;G)fa; and (h) B in kilohertz. tigureP14,33

5

u= 1+n5/R, f) Plot O venus RL for 1 MO < RL < 40 MO. Figure P14.30 1.25MO

14,31 The parameten in the c cuit in Fig. P1,1.30are R = 100kO, C=4pF, and a=400pH. The quality factor of tle circuit is not to drop below 9. What is t}le smallest permissible value of tle load resistorRL?

14,34 Use a 100nF capacitorto designa bandrcjectfilter, p?iii6lras shown in Fig. P14.34.The filter has a center frea;i;i quency of 50 kHz and a quality factor of 8. a) Specify the rumerical values of R and -L. b) Calculatethe upper and lower corner,or cutofl tequenciesin kilohertz. c) Calculate the filter bandwidth in kilohertz. FiguleP14,34

100nF

Section14.5 14.32 a) Show (via a qualitativeanalysis)that the c cuit inFig.?14.32is a bandrejectfilter. Supporrlhe qualilari!eanal].isoi(a) b) I nd ng the voltage transfer function of the filter. c) Derive the expression foi the center fiequency of the filter. Derive the expressions for the cutoff frequene) wlat is th€ expressionfor the bandwidthof the filter? f) What is the expression for the quality faclor oI t}le circuit?

14.35 Assume the bandrejectfilter in Problem 14.34is 6dc loadedwith a 932O resistor. a) What is tlle quality factor of the loaded circuit? b) wlat is the bandwidth (in kilohertz) oI the loadedcircuit? c) w}lat is the upper cutoff lrequency in kilohefiz? d) Wlat is the lower cutoff frcquency in kilohertz?

tigureP14,12 14.36 Designan RaC bandrejectfiltei (seeFig. 14.28[a]) with a quality of 2/3 and a center hequency of 4 krad/s,usinga 80 nF capacitor. a) Diaw your circuit,labelingthe componentvaluesand output voltage, b) For the filter in part (a), calculate the bandvidth a0drhevalue.o( lhe l\ o culoii frequencies.

14,37 The input to tlle RIC bandrejectfilter designedin Problem 14.36is 125cosotmV. Find the vottage drop acrossthe seriescombinationof the inductor and capacitor when (a) ,) = aa; $) a = b.1; (c) a, = ar.,;(d) o : 0.1a.t(e) @ : 10!,".

14.38 The input to the RaC bardrejecl lilter designedin Pioblem 14.36is 125cos.rrmV. Find t}Ie voltage drop across the resistor when (a) o = ro,; (b) o - o"1; (c)(d = @d,;(d)o : 0.1a.;(e)o = 1u,t..

14.41 The load in the bandreject filter circuit shown in Fig.P14.39is 36 kO.The centerfrequencyofthefilter is 1 Mrad/s, and the capacitor is 400 pF. At very low and very high frequencies,the amplitude of the sinusoidaloutput voltageshouldbe at least96010 of the amplitudeofthe sinusoidalinput voltage. a) Specify the numerical values of R and a. b) What is the quality factor of the circuit?

Sections14.1-14.5 1439 The purpose of this problem is to investigate how a resistiveload connectedacrcssthe outputterminals of the bandreject filter shown in Fig. 14.28(a)affecrs the behaviorof the filter. The loadedfilter circuitis shownin Fig.P14.39. a) Find the voltage transfer function yo/14. b) Wlat is the expressionfor the centerfrequency? c) W}lat is the expiessionfor the bandwidth? d) What is the er?ression foi the quality factor? e) Evatuatefl(jo.). f) Evaluatea(jo). g) EvaluateH(/co). h) Wlal are the er?ressions for the corner frequencies1,.l and o.r? Figurc P14.39

14.40 The paramerers in rhe circuir in Fig. p14.39 s P r Ga r c R : 5 k O , t = 400mH, c:250pF, and Rr = 20ko a) Find,r,, B (in kilohertz),and O. b) Find fl(iO) and l1(jco). c) Find f., and /.r. d) Showtiat iJ RL is expressedin kilohmsrhe O of the circuitis

o:8[1 + (s/RL)]. e) Plot C versusRL for 2 kO < RL < 50 kO.

14,42 Giventhe following voltagetransferfunction:

v H@_i 4x106 r' +500r+4x106 a) At what frequencies(in radiansper second)is the ratio of Yo/I4equal to unity? b) At what ftequency is the ratio maximum? c) What is the maximumvalue of the ratio? 14.,fi} Designa seriesRaC bandpassfilrer (seeFie. 14.27) p?$l,jii, for detecting t.helow-frequency rone generated by iE.;a; pushinga telephonebutton as shownin Fig.14.32. a) Calculatethe valuesof l, and C that place the cutoff frequencies at ihe edges of the DTMF low-frequencyband.Note that the resistancein standardtelephone circuils is alwaysR = 600 O. b) What is the output amplitude of this circuir at eachof the low-bard frequencies, relativeto the peak amplitude of the bandpassfilter? c) W}Ial is the output amplitude of this circuit at the lowestofthe high-bandftequencies? 14.44 Design a DTMF high-bandbandpassfilter simitar liglffitr to the low-band filter design in Problem 14.43.Be lone. .0.,.n sure lo iDcluderhe lourlh hiph-trequenc) r{tJJH,/.rDyourdesign. whar is lhe response amplitude of your filter to t}le highest of the lowfrequencyDTMFtones? 14.45 The 20 Hz signalthat drgs a telephone'sbell hasto t:l|;Ill#r have a very larye amplitude to produce a loud enoughbell .igoal.Ho$ muchlargercan rhe ring .!-r.. '' iDg sienalamplituLle be. relari\e lo lhe loq-bank DTMF signal,so that the rcsponseof the filrer in Problem 14.43is no more than half as large as the largestof the DTMF tones?

ActiveFilterCircuits 15.1 fnst-OrdefLow-Pass and High-Pass 15.2 S.aling p. 612 15.3 0p Anp Sandpass and Bandreject Hlters p. 615 15.4 Higherord€r0p Amp FiLtersp. 622 15.5 NarowbandBandpass and Bandreje.t fi|terc p. 636

I Knowthe op ampcjrcuitsthat behaveasfi6tandhish-pass fittersandb€ abL€ orderLow-pass componentvatues for thesecircuits to catcuLate of cutofffrcquency and t0 m€etspecifications B€abl€to desigffiLtercjrcuitssia'tjngwith a prototypecircuitandusescatingto achjeve desnedfrequency Esponsecharacterktics and howto us€cascaded fi6t- and Undelsland secondoder Butt€rwothfitten to imptement afd bandreject tow pass,high pass,bandpass, to cakuLate 8e abteto usethe designequations for prototypenarcwband, component vaLues nLtetsio meetdesjred bandpasr, ard bandreject

lJp to this point, we haveconsideledonly passivefiltcr circuits, that is,filter circuitsconsistingof resistors,inductors,and capaci1ols.ll1ere are areasof application,however,where active circuits,those lhat employ op amps.have cefiain advantagesovcr passivefilters.Fo1instancc,activecircuitscan producebandpass and bandrejestfilters without using inductors.This is desirable becauscinductoft are usuallylarge,heavy,costly,and they may introduce electromagneticfield effccts that compromise the desiredhequencyresponsechalacteristics. Examinc the transtertunctionsof all the lilter circuits from Chapter 14 and you will noticc that the maximum magnilude docs not exceed 1. Even though passiveresonant filters can achievevoltage and curent amplification at the resonant liequency,passiveliheN in genelal are incapableof amplification, bccausethe output magnitudedoesllot exceedthe input magni tude.This is not a surpdsingobscrvation,as many of the transfer tunctions in Chapter 14 were derived using voltage or curcnt division.Active filtcrc provide a control over anpiification noi availablein passivelilter circuits. Finally, rccall that both the cutotTfrequency and the pass band magnitudeof passivefilters wcre altered with the additio of a resistiveload at the output oI the filter. This is not thc case with activefilters,due to thc propertiesof op amps.Thus,we use activecircuitsto implemeDtlilter designswben gain,load variation, and physicalsizc are impofiant parameteLsin the design specifications. ln this chapter,we examinea few of the many filtcr circuits that cmploy op amps.As you will see,theseop anlp circuitsovcr come the disadvantagesof passivcfilter circuits.Aiso, we will show how the basic op amp filler circuils can be combincd to achievespecificftequencyrcsponsesand to attain a more nearlv ideal tilfer response.Note that throughout this chaptcr l{e assumcthat cvcry op amp behavesas an ideal op amp.

606

PracticaI Perspective Bass Votume Control In this chapter,we continueto examinecircuitsthat arefre, quencys€tedjve.As described in Chapter14,this meansthat the behavjorof the circuit dependson the frequencyof its sinusoidal input. Mostof the circuitspresented herefatLjnto oneofthe four categorjes identifiedin Chapter14-tow,pass fitters,high-pass fitters,bandpass fitters,and bandreject fitters.But whereas the circuitsin Chapter14 wereconstructed usingsources,resjstors,capacitors,and inductors,the circuitsin thjs chapterempLoy op amps.Weshatlsoonseewhat advantages areconferred to a fitter cifcujt constructed using op amps. Audjoelectronjcsystems suchasradios,tapeptayers, and ptayers CD often providesepafatevotumecontrotslabeled "trebte"and"bass." TheseconkoLspermitthe userto select

the votumeof high frequencyaudiosjgraLs("treble")independent of the volume of tow frequencyaudio signats ("bass").Theability to independentty adjustthe amountof amptification(boost)or attenuatjon(cut) in thesetwo frequencybandsaLtows a tistenerto customjze the soundwith moreprecisionthan wouldbe providedwith a singtevoLume conkol. Hencethe boost and cut control circuit is aLso referred to as a tone controlcircuit, The PracticalPerspective exanrpteat the end of this chapterpresents a cifcuit that impLements bassvotumecontrot usinga singLeop amptogetherwith resistors andcapacitors. An adjustabLe resistorsuppfiesthe necessary controt overthe ampLification in the bassfrequency range.

607

15.1 First-Order Low-Pass Fitters andlligh-Pass

fittel Figurer5.r A A filst orderLow-pass

Consider ttre circuit in Fig. 15.1. Qualitatively, when the ftequency of the sourceis varied,only the impedarceof the capacitoris afrected.At very low frequencies,the capacitor acts like an open circuit, and the op amp circuit acts like an amplifier with a gain of R /-Rr.At very high frequencies, the capacitor acts like a sho circuit, thereby connecting the output of the op amp circuit to gound. The op amp circuit in Fig. 15.1thus functions as a low-passfilter wittr a passbandgain of -Rr/Rr. To confirm this qualitative assessment,we can compute the hansfer function H(r) = %(r)/I1(s). Note that the circuit in Fig.15.1has the general form of the circuit shown il Fig. 15.2, where the impedance in the input path (Z) is the resistorR1,and the impedancein the feedbackpath (Zf) is the parallel combination of the rcsistor R2 ard tlle capacitor c. The circuit in Fig. 15.2 is analogous to the inverting amplifier circuit from Chapter 5, so its trarsfer functjon is Zt/Zi. Therefore, the transfer tunction for the circuit in Fig. 15.1is

zl Z,

-" (;"1) R1

Figure15.2A A genemLop ampcjrcuit.

(15.1)

R.

(15.2)

and

1 Note that Eq. 15.1has the sameIorm as the generalequation for low passfilters given in Chapter 14, with an irDportantexception:Thegain in the passband,i(, is set by the ratio RzlR1.The op amp low-passfilter thus permits the passbandgain and the cutoff ftequencyto be specified independentry

Plots A NoteAboutFrequency Response Frequencyresponseplots,introduced in Chapter 14,provide valuableinsight into the way a fiIter circuit functions Thus we make extensive use of frequency responseplots in this chapter, too. The frequency responseplots il Chapter 14 comprised two separateplots a plot of the transfer function mag!.itude versusfrequercy, and a plot of the transfer functior phaseanglg in degreeqversus frequency.w}len we use both plots' they are normally stackedon top of one another so that they can sharethe sameilequency a,\is.

lc.l

F i J - 0 d q l o w - P dar n dP : g h - P af ist r e $

609

In this chaptor, we uso a special type of frequency response plots called Bode plots. Bode plots are discussedin detail in Appendix E, which includes detailed information about how to construct these plots by hand. You will probably use a computer to consrructBode plots,so here we summarize the special features of these plots. Bode plols differ from the frequency rcsponseplots in Chapter 14 in two impoitanl ways. First, instead of using a linear axis for the fiequency values, a Bode plot uses a logarithmic axis.This permits us to plot a wider turge of frequencies of hterest. Normally_we plot three or four decadesof frequencies,say from 1o'rad/s to 10"iad/s, or 1 kHz to 1MH4 choosingthe frequency mnge where the tmnsfer function charactedstics are changing. If we plot botl tle magnitude and phase angle plots, they agair share the frequency axis, Second, instead of plotting ttre absolute magnitude of the transfer function ve$us fuequency,the Bode magnitude is plotted in decibels (dB) versus the log of the frcquency. The decibel is discussedin Appendix D. Bdefl$ if t})e magnitude of the transfer function is lH(/'o) , its value in dB is given by ,4dB: 20 loglo H(jir)|. It is important to remember that while lli(jo) is an unsigned quantity, ,4dBis a signed quantity. When 2{dB: 0, the transfer function magnitude = 0. When 2{dB< 0, ttre traNfer function magniis 1, since201og1o(1) tude is between 0 and 1, and when ,4dB > 0, the transfer function magnitudeis greaterthan 1.Finally,note that : 20log1oL1/\41

3dB.

Recal tlat we deline tlle cutoff frequency of filters by determining t}Ie ftequency at which the maximum magnitude of the tmnsfer function has been reduced by 1/rZ. If we translate this definition to magnitude in dB, we define the cutolf frequency of a filter by determidng the frequency at which the maxjmum magnitude of the transfei function ill dB has been reduced by 3 dB. For example, if the magnitude of a low-pass filter in its passband is 26 dB, the magnitude used to fi[d the cutoff frequency is 26 3:23d8. Example 15.1 illustmtes the desigr of a fi$torder low pass filter to meet desircd specifications of passband gain ard cutoff frequency, and also illustrates a Bode magnitude plot of the filter's transfer function.

Designing a Low-Pass 0p AmpFilter

Solution Using the circuit shown in Fig. 15.1,calculate values for C and R2that,togetherwith n1 : 1 O, produce a low-pass filter having a gain of 1 in the passband and a cutoff fiequency of l rad/s. Construct the transfer function foi this filter and use it to slclc! a Bode magnitude plot of the filter\ frequency response.

Equation 15.2 gives tle passband gain in terms of Rl and R2,so it allowsus to calculatethe required valueof R2: -R2: iiR1

: (1X1) =1O.

610

ActiveFitterCncuits

Equation15.3then permitsus to calculateC to meetthe specifiedcutoff frequency: ^'

1 &a. I

(1X1) = 1F. The tansfer function for the low passfilter is givenby Eq.15.1: fl(s) = _

r1 s + 1

TheBodeplot of l}I(jd)l is shownin Fig.15.3.This is the so-calledprototne low-passop amp filter, becau.e it u.esa resisror \alueof I O anda capaci tor valueof 1 R and it providesa cutoff frequency of 1 rad/s.As we shallseein the nextsection, prototypefilteN providea usefulsta ing pointfoi the designof filters by usingmore realisticcompotrent valuesto achievea desiredfrcquencyrcsponse,

vi

f*

-1

0.5

1.0 @GadA)

5.0

IO

ptotofthetow-pasr Flgue15.3 TheBod€ masnitud€ filterfrom ExampLe 15.1.

You may have recognized the circuit in Fig. 15.1 as the integrating amplifiei circuit introduced in Chapter 7. They are indeed the same circuit, so integralion in t]Ie time domain correspondsto low-pass filtering in the fiequency domain. This relationship berween integralion and low-pass filterilg is further contumed by the operational I-aplace transform for integrationderivedir Chapter12. The circuit in Fig. 15.4is a fi$torder high-passfilter. This circuit also hasthe generalform of the circuit in Fig. 15.2,only now tle impedancein the input path is the seriescombination of R1and C, and the impedance in ttre feedback path is the resistor R2.The transfer function lor the circuit in Fig 15.4is thus nltt: _

-7. _ zi R

2

^

Flgure 15.4 Afirslorder hjgh-pass fitter.

l 115.4)

-,,-

Rz Rr'

(15.5)

and a . : *

I

(15.6)

15.1 First-orde. Low-Pass andHigh,Pass Fitters 511 Again,the form of the transferfunctiongivenin Eq.15.4is the sameasthal givenin Eq.14.20,the equationfor passivehigh-passfilrers.And again,the active filter permits t}le design of a passbandgain greater than 1. Example 15.2 considers the design of ar active high-passfilrer which must meet ftequencyresponsespeciJications from a Bode plot.

Designing a High-Pass 0p AmpFitter Figure 15.5 shows the Bode magnirudeplot of a high-passfilter. Using the active high passfilter c1rcuit in Fig-15.4,calculatevaluesof R1 and R2 that produce the desired magnitude response.Use a 0.1pF capaciror.Ifa 10 kO loadrcsistoris addedto thisfilter, how wil the magnituderesponsechangc?

30 20 10

Solution Begin by writing a transter function that has the magnitudeplot shownin Fig. 15.5.To do this,note that the gain in the passbandis 20 dB; therefore, ( : 10.Also note that the 3 dB point is 500rad/s. Equatiofl 15.4is the tmnsfer function for a highpassfilter, so tle traNfer function that has the magnitude responseshownin Fig.15.5is givenby

10r s+500 We cancomputethe valuesof R1and R2neededto yield this transfer function by equating the transfer functionwith Eq. 15.4:

---'

-(Rzln1)r l0r s + 5 0 0 s + (1/R1C)

3

tll

20 30

-10i

5 10

50 100 o (rad/t

s001000 500010,000

Figure15,5,L TheBodemagnitude ptotofthe high-pass fiLterfor

Equating the numerators and denominatorsand then simplifying,we get two equations: R,

10:;.

l

s 0 0 =_ .

Using the specifiedvalueof C (0.1pF),we find Rr = 20kO, R2 : 200kO. The circuit is shownin Fig.15.6. Becausewe have made the assumptionthat the op amp in this high-passfilte{ circuir is ideal, the addition of any load resistor,regardlessof its resistance, has no effect on the behavior of the op amp.Thus,the magnituderesponseof a high-pass filter with a load resistor is the same as that of a high-passIiltei with no load resistor, which is depictedin Fig.15.5.

200ko

Figure15.6 r. Thehjgh-pass filterfor Exampte 15.2.

612

ActiveFitterCiLcuits

andhigh-pass fi(tersal|dbeableto asfirst orderlow-pass objectivel-Know the op ampcircuitsthat behave values catcutate theircornDonent 15,1 Computethe valuesfor R?and Cthat ]'ield a high-passfiller with a passbandgain of 1 and a cutoff frequency of 1 .ad/s if Rl is 1 O . (Notdr This is the prololype high-passfilter.)

15.2 Compuiethe resisiorvaluesneededfor the low-passfilter circuitin Fig.15.1to producethe lmnsler function 20.000

.'1G)= .,+ sooo

Usea5l[Fcapacilor. Answer: -R1: 10O, n, : 40O.

A n s w e rR: 2 : 1 O , C = l F . NOTE: Alro bt ChaptetPrcblens15.1and 15.7.

15.2 Scal.5ng Il1thc dcsignand analysisof both passiveand activefilter circuits,working lvilh elcmcnt valuessuch as 10, 1 H, and 1 F is convenient.Although Lhesevalucs are unrealistic for specifyingpractical components,they greatlysinrplili computations. After nal
R' : k,"R, L' = ft_t.

and c, : c/k_.

(15.7)

Note that k,,,is by definilior a posilivereal numberthat canbe eithcr less than or greaterthan 1. lr frequencyscalirg.we changcthe circuit paramete$ so that at the new frequency,the impedanceofcach elementisthe sameasit wasat the originalfrequency.Becauselcsistancevaluesare assumedto be independ_ eni of frequency,resislorsare unaffectedby frequencyscaling.Itwe let kt denotethefrequencyscaleIactor,bothinductorsand capacilorsarc multiplied by 1/fr1.Thus for frequencyscaling. R'=R,

t,':L/kr,

a l r . dC ' = C / k f .

(15.8)

The frequencl,scalefactor kt is alsoa positivereal numberthal canbe lcss than or greaterthan unity.

15.? Scaling 613

A circuit can be scaledsimultaneouslyin both magnitude and frequency.The scaled values (primed) in terms of the odginal values (unprimed)are ,R, = k^R, L'

{ Component scatefactors C'

l

-

(15.e)

TheUseof Scatingin the Designof 0p AmpFilters To use the conceptof scalingin the designof op amp filters,fi$t select lhe cutoff frequency,(,r., to be 1 nd/s (if you are designinglow- or high, passfillels), or selectthe center frequency,(ro, ro be l rad/s (if you are designingbandpassor bandrejectfilters).Then selecta 1 F capacitorand calculatethe valuesof the resistorsneededto give the desiredpassband gain and the 1 rad/s cutoff oi certer frequency.Finally, use scalingto computemore realisticcomponentvaluesthat give the desiredcutoff or centei ftequency. Example 15.3 illustrates the scaling process in general, and Example15.4illustratesthe useofscalingin the designof a loiv passfilter.

Scating a SeriesRICCircuit The seriesRIC circuit shownin Fig.15.7hasa cen: 1 rad/s, a bandwidthof ter frequencyof VflC R/a : l rad/s, and thus a quality factor of 1. Use ssalingto computenew valuesofn and a that yield a circuitwith the samequality factor but with a center frequency of 500 Hz. Use a 2 ,uF capacitor-

before scaling,whereas the pimed variablesrepre, sentvaluesafter scaling.

= 2'!oo): :r+r'se' Now, use Eq. 15.9to computethe magnitudescale factor that, together with the frequency scalefactor, will yield a capacitorvalueof 2 AIF:

,

1 C (3141.s9)(2 x 10-6)

Use Eq. 15.9againto computethe magnitude,and frequency-scaled valuesofn and a: tigure15.7A Th€senes RlCcjrcuit forExanpte 15.3.

n' : r-R : 159.155 O, kL'=lL=50.66m}l.

Solution Begin by computing the ftequency scale faclor thar will shifi the center frequency ftom l rad/s to 500 Hz. The unprimed vaiiables representvalues

WitI these component values, the center fiequency of the sedes RLC circuit is = 3141-61rcdlsor500Hz, and the band\/!i7 widrh is R/a : 3141.61rad/sor 500Hz; thus rhe quality factor is still 1.

614

A.tiveFitterCircuits

Scalinga PrototypeLow-Pass 0p Amp Fitter Use the prototype low pass op amp filler from Example15.1.alongwith magniludeand frequency scaling,to computc thc r€sislor valuesfor a lowpassfilter with a gain o[ 5, a cutoff frequencyof 1000 Hz, and a ltcdback capacitor of 0.01pF. Constructa Bode plol ofthe resultingtransferfunc tion's magnitudc.

Thc final componcnlvaluesare Rr :3183.1 O. _Rr: 15,915.5 I), C : 0.01pF. Thc lransferfunctionof the fi]ter is givenby 31,415.93

s + 6283.185 The Bode plot ofthe magnitudeof this transfcr iunctior is shownin Fig. 15.8.

Solution To begin.usefrequencyscalingto placc ihc culoff ficquencya!1000Hz: kr:

a',/a, = 2'(1000)/1= 6283.185.

where lhe primed variablehas the ncw value a d the unpdmedvariablehasthe old value oI the cutofl frequercy.Then compute thc magnilude scale Iactor that. togetherwith t/ = 6283.185,will scale the capacitorto 0.01pF:

1 C k, C'

= 15,q15.5. (b283.18s)(10 1

Sincercsislolsare scaledonly by using magnitudc scaling. = 15,915.s Ri = .Ri = [_R : (15,915.5X1) O. Finally, we need to meei thc passbandgain specification.We can adjust thc scaledvalues of either Rr of Rr, becauseK = nrl,4r. If we adjust nl, we will changethe cutott ticqucncy.because o. = 1/RrC. Therefore,we can adjusl the value of R1to alter only the passbandgain:

= 3183.1 R] : X'lr - (1s,91s.s)/(5) O.

2Ll

15 10

5 lL)

-15

--10

50 r00

5001000 flHz)

5m0 10,000

ptotof tlretowpss fitterfrom Fjgure15,8..'.TheBod€magnitude Erample 15.4.

0bjective2-8e abteto designfitter circuitsstartingwith a protot!'pe andusescatingto achieve desiredfrequenq, responseand componentvatues 15.3 Wh"r magnirudc rnd lrequenc).caletacror. will transformthe prototypetrigh-pass filter into a high passfilter with a 0.5pF capacilor and a culoff frequencyof 10kHz? NOTE: Ako try ChapterPrcblems15.13and 15.11.

Answer: ty = 62,831.85. ft,,,= 31.831

15.3 0p AmpEandpass andBandreject Fitte6

15.3 0p AmpBandpass andBandreject Filters We now tum to the analysisand designof op amp circuits that ac1as bandpass and bandreject filte$. While there is a wide variety of such op amp circuits, our initial approach is motivated by the Bode plot construction shownin Fig. 15.9.Wecan seeftom the plot that the bandpassfilter consistsof tlree separatecomponents: 1. A unity-gain low-passfilter whose cutoff frequency is oa2,the larger of the two cutofl ftequencies; 2. A unity-gah high-pass filter whose cutoff ftequency is a,.1, the smaller of the two cutoff frequencies;and 3. A gain componert to provide the desired level of gain in the passband, Thesethree componentsate cascadedin sedes.They combineadditively in the Bode plot construction and so will combine multiplicatively in the r domain. lt is important to note that this method of constructing a bardpassmagnitude responseassumesthat the lower cutoff ftequency (o.1) is smaller than the uppei cutoff frequency (ar.2).The resulting filter is called a hoadband bandpassfilter, becausethe band of ftequenciespassedis wide. The formal definition of a broadband filter requjres the two cutoff frequenciesto safisfy the equation

l0 20

\ l l

10

10 2D

A,il|

)'ll

30

-r0i

10

50 100

5001000 500010.000

Figure15.9l Constiucting ptoiof a bandpass the Bodemagnitude fitrer.

615

616

AciiveFitterCircuits

As illustratedby tlle Bode plot constructionin Fig. 15.9,we require the magritude of tle high-pass filter be unity at the cutoff hequency of the low-passfilter and the magnitude of the low-passfilter be unity at t}le cut off frequency of the high-passfilter. Then the bandpassfilter will have the cutoff frequercies specified by the low-pass and high-passfilters. We need to determine the relationship betweer o.r and o., that will satisfy the requirementsillustmtedin Fig.15.9. We can construct a circuit that provides each of tle three components cascading by a low-passop amp filter, a high-passop amp filter, and an inverting amplifier (see Section 5.3), as showll in Fig. 15.10(a). Figure15.10(a)is a form of illustrationcalleda block diagram.Each block representsa componentor subcircuit,and the output of one block is the input to the next,in the djrectionindicated.Wewish to establishthe relationship between (Da and aa that will permit each subcircuit to be designedindependently, \rthout concernfor the other subcircuitsin tlle cascade. Then the designof the bandpassfilter is reducedto the designof gain a unity first order low passfilter, a unity-gain first-order high-pass filter, and an invertingamplifier,eachof which is a simplecircuit. The transfertunctionofthe cascadedbandpassfilter is the productof the transfcrfunctionsofthe thre€ cascadedcomponents:

H(r)

u.

u

-R') " *,)/ / \/ \{+u.r/\s+d.r/\ R, / - K0t.2s Ka,zs s2+ (@"1+at)s + a.pa'

(b)Ihe cjtcujt. Figure15.10a A casaded opampbandpass fitter (a)Ihe btockdjasram.

(15.10)

15.3 0p AmpBandpas5 andEandrcject Fjtte6

We notice dght away that Eq. 15.10is not in the standard form for the transfer function of a bandpassfilter discussedin Chapter 14,namely,

Bs

Hsp

l+Bs+t]'

In order to convert Eq. 15.10irto the form of the standard transfer func tion for a bandpassfilter, we requirethat (15.11)

WtenEq.15.11holds, \.at+ad=oa ard the transfer function for ahe cascaded bandpass filter in Eq. 15.10

lJ(r) :

-Kaas s2+ar2s+o"7a4

Oncewe confirm that Eq.15.11holdsfor the cutoff tequenciesspecified foi the desired bandpassfilter, we can design each stage of the cascaded circuit independently and meet the filter specifications. We compute the values of Rr and Cz in t}le low-pass filter 1o give us the desired upper cutotf frcquency, oa:

1 '" - rq;c,

115.12)

We compute the values of ftr and C.r in the high-passfilter to give us the desired lower cutoff fiequency, o.1: RnCp

(15.13)

Now we compute the values of 1Rjand nf in the inverting amplifier to provide the desiredpassbandgain.To do this,we considerthe magnitude of the bandpassfilter's tiansfer function, evalualed at the center frequency,.ro: - Ka.2Qa"\

H(i,.)t :

Aa)2+aeja)+a.pa =

K,'t.z

\15.14)

Recall from Chapter 5 that the gain of the inverting amplifier is Rl/Ri Theiefore, Rr

wad:8.

(15.15)

Any choice of resistorstlat satisfiesEq. 15.15will produce the desired passbandgaiD. Example 15.5illustratesthe dosignprocessfor the cascadedband passfilter.

617

618

Activ€FitterCndritr

Designing a Broadband Bandpass 0p Anp Fitter Design a bandpassfilter for a graphic equalizer to provide an amplification of 2 wirhin rhe band of ftequenciesbetween 100 and 10,000Hz. Use 0.2pF capacitors.

Nexr.we rurnro lhe high-pass srage. I rom Eq. l5.lJ. RrlCs

: 2n(100), 1

x 10-6) [2,(100)x0.2

Solution We can design each subcircuit in the cascadeand meet the specified cutoff frequency values orly iJ Eq.l5.ll holds In this casej,).2 = 100(0.1, sowe can saythat (,.2 >> o.t. Beginwith the Iow-passstage.From Eq.15.12,

RtCt

! 7958O. Finally,we need the gain stage.From Eq. 15.15,we see there are two unknowts, so one of the rcsistors can be selectedarbitrarily. Let's select a 1 kO resistor for Rj. Then,from Eq.15.15, Rf = 2(1000)

: 2,(10000),

:2000f}-2ko. 1

[2i00000)](0.2x 10-6) ! 80O.

The resulting circuit is shown in Fig. 15.11.We leave to you to verily that t]le magnitude of this circuit's transfertunction is rcducedby 1/\4 at both cutoff frequencies, veiifying tle validity of the assumplrono.2 >> (r.1.

O.2 ttF

tigure15.1r Thecascaded opanpbandpass fitrerdesjsned in Exampte 15.5.

We can use a component approach to the design of op amp bandreject filters too, as illustratedin Fig. 15.12.Lile rlre bandpassfilrer, rhe bandreject filter consists of three sepamte components.There are important differences,however: 1. The ur ty-gain low-passfilter has a cutoff frequency of .r"1,which is t]le smaller of the two cutoff frequercies. 2. The unity-gain high-pass filter has a cutoff frequency of o.2, which is lhe largerol the lwo culofffrequencies. 3. The gain component provides the desired level of gain in the passbands.

15.3 0p AmpBandpass andBandreject Filrels 619

30 20 10

/1 )ll

e

t

\

20

tl

Hi g h

30 -40

5 10

l{|l

s0 100 5001000 500010,000 o (rad/9

Figure ptotofabandrejedfitter. 15.12A Constructing iheBode magnitude

The most importantditterenceis that thesetluee comporentscannot be cascaded in serieqbecause they donot combineadditivelyonthe Bode plot- Instead,we use a parallel connectionand a summingamplifier,as shownboth in block diagramform and asa circuitin Fig.15.13.Again,it js assumedthat the two cutoff fiequenciesare widely separated, so that the resultingdesignis a broadbandbandrejectfilter, and o.2 >> d.1. Then eachcomponentof the paralleldesigr car be createdindependently, and tlle cutofl tequency specifications will be satisfied. The transler function ofthe resultingcircuitis the sumof the low-passand high-passfilter trans fer functions.From Fig.15.13(b),

n , . ' -' I ff i)l l' L r + -;,,(

-, 't

'

--l

(,"(r +.,.r) + r(r + o.r)\

G +..,X" + ..')

/ (15.16)

;,( (r + o.r(r + d.,) / Using the samerationaleas for the cascadedbandpassfilter, the two cutoff ftequenciesfor the transferfunction in Eq. 15.16are .r.r ard d., only if .r.2 >> {r" |. Then the cutoff ftequencies are given by the equations I

R t Ct '

1 RsCs

(15.17) (15.18)

620

rigure15.13r. A pa€iLet opampbandreject fitter.(a)IhebLock djasEm. (b)rhecncuit. In the two passbands(as s + 0 and r + oo), the gain of the transfei function is R/ /R,. Therefore, (15.19)

As vrith the design of the cascadedbandpassfilter, we have six unknowns and three equationsTlpically we choosea cornrnerciallyavailable capacitor valuefor Cr and Cr. Then Eq$ 15.17and 15.18permit us to calculate]Rt and Rr to meet tle specified cutofl ftequencies.Finally, we choose a value for eitler Rf or & and tlen useEq.15.19to computethe otherrcsistance. Note the magnitudeof the transferlunction ir Eq. 15.16at the center rrequency,oo - "\/oc1,ocz:

l R ,/ + 2a"1(ja.) + r1(r,"): n' (j@.)z l;( \ 0.)" + ( o d + a d l j o . ) Rt

o,pe

)l

+ o"1@a

2r.,

Rr 2a-.

(15.20)

Ifo.2 >> o.1,rhen H(jr,)l << ZRflR,(aso.r,/(r.2<< 1),so rhemagnitudeat the centerfrequencyis muchsmallerthant}Iepassband magnitude,

15.1 0p AmpBandpars andBand'eject Fitte's 621

Thus t}le bandreject filter successfullyrcjects ftequencies near the center frequency,again confirming our assumption that the parallel implementa tion is meant for broadband bandreject designs. Example 15.6 illustrates the design piocessfor the parallel band r€jectfilter.

Designing a Broadband Bandreject 0p AmpFitter Designa circuit basedon the paralel bardrejectop amp filter in Fig. 15.13(b).The Bode magnitude responseof this filter is shown in Fig. 15.14.Use 0.5&F capacitorsin your design.

15

D

6.54 5

Sotution From the Bode magrlitudeplot in Fig. 15.14,we see that the bandreject filter has cutoff frequencies of 100rad/s and 2000rad/s and a gain of3 in the pass bands.Thus,,r.2 = 20d.r, so we make the assumption that o.r >> o.t. Begin with the prototype low-passfilter and usescaling to meet the specifications for cutoff frequency and capacitor value. The ti€quency scalefactor,tf is 100,which shifts the cutofTfrequenc)from I rad s lo 100rdd/.. tte magnitude scalefactor f,, is 20,000,which pemits the use of a 0.5pF capacitor.Using these scale factors resultsir the following scaledcomponenlvalues: R' : 20 kO' C1 = 0.5pF. The resulting cutoff frequency of the low passfilter I RLCt 1 (20 x 10)(0.s x 10 6) : 100radls. We use the same approach to design thc high passfilter, startingwiti the prototpe high-passop amp filter. Here, the frequency scale factor is ,t/ - 2000, and the magnitude scale factor is ,ta : 1000,resultingin the following scaledcom ponentvaluesi R/r : 1 kO, Cg : 0.5s.F. Finally, because the cutoff frequencies are widely separated,we can use the ratio -Rl/Rr to

0

l

".)

a - s

-15 20 25 -3 t1

50 100

5001000

500010,000

ptotforth€ Figure15.14"{ TheBode magnitud€ circujtto bed€signed jn Exampte 15.6. establisht}Ie desiredpassbandgain of 3. Let\ choose & = 1kO, as we are aheadyusingthat resistance R. R,tor RH. Then R/ JkQ. and ( : parallel op amp band3000/1000 3 The resulting rejectfilter circuitis sho$nin 8g.15.15. Now let's check our assumplion that a).2>> @.1by calculatingthe actual gain at the specifiedcutoff frequenciesWe do tlis by making the substitutionss : i2,(100) and r : J2T(2000) into the tmnsferfunctionfor the parallelbandreject filter,Eq.15.16and calculatingthe resultingmagnitude. We leave it to the reader to verify that the magnitude at the specitied cutoff ftequetcies is 2.024, which is less than the magnitude of 3l\a = 2-12 that we expect.Therefote, our rejecting band is somewhatwider than specifiedin the problem staiement-

622 0.5r.F

in Exampte 156 bandrcject fittercjrcuild€sisned rigure15.15A Theresuttins your undelstandingof thismaterialbr trying ChapterPrcblems15.30and 15.31. NOTE: Assess

15.4 Higher0rder0p AmpFil.ters You have probably noticed that all ol the filler circuits we have examined so far, both passiveand active, are ronideal. Rem€mber tiom Chaptei 14 that an ideal filter has a discontinuityat the point of cutofl which sharplydividesthe passbandand the stopband.Allhough we cannot hope to constructa circuit with a discontinuousfrequencyresponse, we canconstructcircuitswith a sharper,yetstill continuous,transitionat the cutoff frequency.

ldenticalFilters Cascading How can we obtain a sharpei transition between the passband and the stopbandl One approachis suggestedby ttre Bode magnitudeplots in Fig. 15.16.This figure showsthe Bode magnitudeplots of a cascadeof idenaicalprototwe low passfiltersand includesplots ofjust onefilter,lwo It is obviousthat asmore threein cascade. and {our in cascade. in cascade. transition ftom the passbandto the filters are addedto the cascade,the lor constructing Bode plots (fiom stopbandbecomessharper.The rules filter, the [ansition occun with an Appendjx E) tell us that with one (dB/dec). pei Because circuitsin decade asymptoticslopeof 20 decibels plot, two filters a cascade with cascadeare additiveon a Bode magnitude = + 40 dB/dec; for ol 20 20 has a transition with an asymptoticslope four filters, it is is dB/dec, and for tbree filters. the asymptotic slope 60 80 dB/dec, asseenin Fig.15.16.

15.4 Higher0rder 0p Anp Fitt€rs

20 10 3 \

lLl

I

I {irst

i\

20

\

.g 30

I

\

-40 ird o

-50

\

r..l"nl.ol)

60 70 80 0.1

0.5

1

5

l0

tigure15.15A TheEodemagniiude ptotof a cascad€ ofidentjcat prototypend-order fi tt€rs.

In general,an n-element cascadeof identical low-passfilten will tmnsitionfrom the passbandto the stopbardwith a slopeof 20l, dB/dec. Both the block diagram and the circuit diagram for such a cascadearc shown in Fig. 15.17.It is easy to compute the transfer function for a

C

risure 15.17 A Acascade t"*p.:l)nft',.{"1Ihe blockdiag€m.(b) Ihe cjrcujt. oria"",i*r*i v-q,rrl

623

cascadeof ,, prototwe low-passfilten-we just multiply the individual transfer functions:

'-"'

/ - t \ / - t

\

/

\.l*1/\r+1/

r \

\s+1/

= (r(-ry

(r5.21)

+ 1)'

The order of a filter is determined by tlle number of poles in its tmnsfer function.From Eq. 15.21,we seethat a cascadeof first-orderlow-pass filters yields a higher order filter. In fact, a cascadeof n first-order filters produces an ,th-order filter, having ,1 poles in its transfer function and a final slopeof 20n dB/dec in the transitionband. There is an important issue yet to be resolved, as you will see if you look closelyat Fig.15.16.Asthe order of the low-passfilter is indeasedby adding protot]?e low-passfilters to t}le cascade,the cutoff frequency also changes. For example,in a cascadeof two first-orderlow-passfilters,the magnitudeofthe resultingsecond-orderfilter at.r" is -6 dB,so the cutoff ftequenry oI the second-order filter is not .d". In fact, the cutoff ftequency As long as we are able to calculate the cutoff ftequency of the highei order filten formed in tbe cascadeof first-order filten, we can use frequency scaling to calculate component values that move the cutoff frequency to its specified location. If we start with a cascadeof n prototype low-pass filteis, we can compute the cutoff ftequency for t]te resulting nth-order low passfilter. We do so by solving for the value of o"" that results

in H(ja) : Y\nl

HG)=(,*rr€,

= : t,(*",)t la.: t), +, \r2'

.7"+ |

\\a)

'

95 = ,,1^+ r, (t5.2?) To demonshatethe use of Eq. 15.22,let's compute the cutoff f.e' quency of a fourth-order unity-gain low-passfilter constructed from a cascadeof four prototypelow-passfiltels:

V{4-

= o.+:s,ua7..

(15.23)

ThuE we can design a fourth order low-passfilter with any arbitrary cutoff frequency by starting with a fourth-order cascadecoflsisting of protot!?e

15.1 Nisher order 0pAmpFitters 625 low-passfilters and then scaling the componentsby k, =.,16435 to placethe cutoff frequencyat any valueofo. desired. Note that we can build a higherorder low-passfilter with a nonunity gain by adding an inverting amplifier circuit to the cascade.Example 15.7 illustratesthe designof a fourth order low-passfilter with nonunitygain.

Designing a Foufth-order Low-Pass 0p ArnpFitter Desigr a fourth-order low-pass filter vrith a cutoff frequemy of 500Hz and apassbandgain of 10.Use 1 t[F capacitors. Sketch the Bode magnitude plot for this filter.

Solution

Finally, add an irverting amplifier stagewith a gain of Rf/Ri = 10.As usual,we can arbitrarily selectone of the two rcsistor values. Becausewe a.re akeady using138.46O resistorsslet R; = 138.46O; then, R, : 10tRi= 1384.6O.

We begin our designwith a cascadeof four prototype low-pass filters. We have already used Eq. 15.23 to calculate t}le c$toff frequency for the resultingfourth-orderlow-passfilter as0.435md/s. A frequency scale factor of /./ : 7222.39will scale the component values to give a 500 Hz cutoff frequency.A magnitudescalelactor of fr", = 138.46 permitsthe useof 1 f.F capacitors. The scaledcomponentvaluesare thus

R : 1 3 B . 4 6 O ;C = 1 p F . 1pF

The circuit for this cascadedthe fourth,order low-passfilter is shown in Fig. 15.18.It has the traNfer function

n(s):

t r0l

rrrr ro ::::=:

lr l

The Bode magniludeplot for this transferfunc, tion is sketchedin Fig.15.19. 1pF

tigure15.18A Thecascade circujtforihe fourthorderlow-pass fiLterdesjgned in Lyampte 15.7.

-30 t0

50 100

5001000 I (Hz)

500010,000

figure15.19A TheBodemagnjtude ptotforthefouth,order tow,pass fitt€rdesigned in txampte 15.7.

By cascadingidentical low-passfiltels, we can increasethe asymptotic slope in the transition and control the location of the cutoff frequency,but our approach has a serious shortcoming: Th€ gain of the filter is not constant between zerc and the cutoff ftequency (d". Remember that in an ideal low-passfilter, tie passbandmagnitude is 1 for all fiequencies below t]le cutoff frequency.But in Fig. 15.16,we seethat the magnitudeis less tlan 1 (0 dB) for frequencies much lessthan rhe curoff ftequency. This nonidealpassbardbehavioris best undeistoodby looking at the magnitudeofthe transferfunctionfor a unity-gaifllow-pass,th-order cascade.Because

H(t :

(t + o.")''

the magnitudeis givenby

aQa)l I t/o' + a'- | 1 /

-\

n'

115.21)

(V(d/o.,f + 1J As we can see liom Eq. 15.2, when o << a.,, the denominator is appioximately 1,and the magnitude of the tansfer lunction is also nearly 1. But aso - o.,, the denominatorbecomeslarger than 1,so the magnitude

15.1 Higherord€r 0pAnrpFitters 627 becomessmaller thatr 1. Because the cascadeof low-pass filters results in this noddeal behavior in the passband,otler appioaches aie taken in the desigr of higler order filteis. One such approach is examined next.

ButterworthFilters A unity-gainButterworlh low-passfilter has a transfer function whose m,miirdP

i< oiven hv

lnQa)l =

.,/14t.F'

\\5.25)

where ,?is an integer that denotes the order of the filter.r Whenslud)ingEq. 15.25. norethe touoqing: 1. The cutoff frequency is o. rad/s for all values of n. 2. II n is large enough,the denominatoris alwayscloseto unity when 3. In the exprcssion for lfl(jo) L the er"onent of o/o. is always even. This last obsenation is impo ant, becausean even exponetrt is rcqu ed for a physically realizable circuit (seeProblem 15.24). Given an equation for the magnitude of the trarsfer function, how do we find I/(sX The dedvation for H(s) is gready simplified by using a prototype filter- Therefore, we set da equal to 1 rad/s in Eq. 15.25.As before, we will use scaling to transform the prototype filter to a filter that meets the given filtering specfi cations 'To find ,H(s), filst note that if N is a complex quantity, ther ' N : NN", whereN' is the conjugateofN.Ir followsthat

aQa\ 2 = nQa\n(

ju\.

(15.26)

But becauser = jo, we canwdte

lH(/(,)1': flG)a(-t. Now otlselve that s' =

\15.?7)

o'. lhus.

lHU,)l'= 1

1 + @')' 1

1.( rry I

1 + (-1fs,' H(r)H(-r) =

1+ ( 1)'sh

(15.28)

I Tlis filte. wasdevelopedby lhe Britisn engineerS.Butterwortnandreportedin ltrrulst Easiaeetiag1 (1930):536-54L

629

Active Fitter Cncurts The procedurefor finding fl(s) for a givenvalueofn is asfollows: 1. Find t})e roots of the polynomial | ( - l)'r1 L. 2. Assign the left-half plane roots to 1{(s) ard rhe dghfhalf plane roots to E(-s). 3. Combine terms ifl the denominator oI r/(s) to form first- and second-order factors. Example15.8illustratesthis process.

Calculating Butterworth Transfer Functions F i n d r h e B u l l e r q o f l h l r a n s f e rl u n c t , o n \ l o r f l and,r = 3.

2

For 11= 3, we find the roots of the polynomial 1+(-1)ri6:0. Reanarging tems,

Sotution For n : 2, we find the roots of the pol],nomial 1+(

1),s= { 0.

Rearrangingterml we find

sa= 1=1L1![. Therefore, the four roots are

sr: 11]{- 1lt2 + jl^/r, s,-11J!q:= U\n+ jb/2, st: 1/41: -11\/5+ jl^e,

s6 = 11!: = 1/360' . Therefore, the six roots are

\ = 11!:: r, s2= r1g::

112+ j\,3/2,

x - 11]41: -112+ i\f312, s a = 1 1 1 p : =1 + i 0 , ss= 1/::!41: -112+ j\".312, sa= 113!4l: 112+ -i\4312. Roots sr , 14, and 15are in the leffhall plane.Thus,

+ -j^n. E - 11:fA= 11.\a (s + 1)(r + 1/2 jtSlz\(s + 112+ j.!512) Rootss2and s3are in the left-halfplane.Thus,

I

(r + 1)(s,+ s + 1)'

G+ y\a

jl"rt)G+ rl\,a+ j/\./z)

(s' +!t2r+1)

w€ note in passing that the roots of the Butteiworth pol).nomialare alwaysequally spaced around ttre unit circle in the s plane. To assistin the designof Butterworth filters, Table 15.1 lists the Butterworthpolynomialsup to ,? : 8.

15.4 Higher0rder 0p AmpFitte* TABLE r5.1 Normatized(so that or:

629

r rad/r) ButteMorthPotynohialsup to the Eighthorder ,th-Order ButteNoit[ Pol]Tomial

(r+1) (r' +V2r+1) (r+1Xs'+r+1) (r' + 0.765s + 1)(r' +1.848I + 1) (r + 1)(r' + 0.618r + lxs' + 1.618r + 1) (r2+ 0.518s + 1Xr2+ \4 + 1xr' + 1932s+ 1) (r + 1xr'+ 0.445r+ 1)(r2+ 1.247r + 1) + lxsr + 1.802s (r?+ 0.390r + 1)(l + 1.11is + r)(l + 1.6663r + l)(r' + r.962! + 1)

I 2 3 5 1 8

ButterworthFilterCircuits Now that we know how to specifythe tiansferfunction for a Butterworth filter circuit (either by calculatingthe poles of the transfer function directlyor by usingTable15.1),wetum to the pioblem of designhga circuit with such a transfer function. Notice the form of the Butterworth pol]'nomialsin Thble15.1.They are the pioduct of first- and second-order factorsi theiefore. we can conslruct a circuit whose transfer function has a Butterwoitl polynomial in its denomhator by cascadingop amp circuits, each of which provides one of t}le neededfactors A block diagram of such a cascadeis shown in Fig. 15.20, using a fifth-order ButteNorth polynomial as an example. All odd-order Butterworth polynomialsinclude the factor (s + 1), so all odd-order Bulterworth filter circuits must have a subcircuit that provides the transfer function 11G) = VG + 1) This is the tmnsfer furction of the protot)?e low-passop amp filter from Fig. 15.1.So what remainsis to fird a c cuit that providesa transfertunction of the form ..1(s)= 1/(s'+ bls + 1). Sucha circuiris shownin Fig. 15.21.Theanalysisof this circuit begins by writing the s-domain nodal equations at the noninverting ierminal of the op amp and at the node labeledY,:

u--u .'i,

tr" - %)rcl +

u - - v -= jRr 0, u . u

YJC'+---=n' R

115.?9)

(15.30)

v.ffi' figure15.20a A cascade of first-andsecond-order circujts withthejndjcated transfer yietdinq functions a fifth-order low-pass Buiielwodh fitterwitho. = 1 rad/s.

Figure15.214 A circuitthatprovjdes thesecond order iransfer function forihe Butteftorthfiltercascade.

630 Simplifying Eqs.15.29 and15.30yields (1 + RC6)U = U, U+ (1+ RC2!)U.:0.

12+ RC$u"

(15.31) (15.32)

UsingCramer's rulewith Eqs.15.31 and15.32,we solvefor %:

4

2+RC1sUl 1 0l

lz+nc,, (1+RcrOl 1

1+RCrr I

R2crc2s2+2RC2s+1

(15.33)

Then,rearrangeEq. 15.33to write the rransferfunction for the c cuit in Fig.15.21:

_u_ .,..,

PCIC2 j

2

1

(15.34)

Finally.set R : 1 Q in Eq. 15.34rrhen

1 CrCz

(15.35)

Note thatEq.15.35hasthe form requiredfor the second-ordercircuit in the Butterwo h cascade.In other words, to get a tmnsfer function of the form

s' +brs + 1' we uselhe circuitin Fig.15.21 dndchoosecapacilorvaluesso lhal b1 = ;

2

L1

a n d1 = - . L]L2

(15.36)

We have thus outlined the procedure for designing an ,?th-oider Butterworth low-pass filter circuit with a cutoff liequency of o. : 1 rad/s and a gain of 1 in the passband.We can use frequency scaling to calculate revised capacitor values that yield ary other cutoff fiequency, and we can use magnitude scaling to provide more realistic or practical component values in our design.We can cascadean inve ing amplifier circuil to provide a gain other than 1 in the passband. Example15.9illustratesthis designprocess.

11.4 Nigher 0rder 0pAmpFitters 631

Designing a fourth-orderLow-Pass Butterworth Filter Desigr a fourth-order Butterworth low-passfilter wilh a cutoff frequencyof 500 Hz and a passband gain of 10.Use as many I kO resiston as possible. Compare the Bode magnirude plot for this Butterworth filter with that of the identical cascade filter in Example15.7

The precedingvalues for Cj , Cr, Cj, and Ca leld a fourth-order Butterworth filter with a cutoff frequencyof 1 rad/s. A frequencyscalefactor of kt:3141.6 will move the cutoff frequency to 500Hz.A magnitudescalefactor of k- = 1000will permit the use of 1kO resisto$ in place of 1 O resistors,The rcsultingscaledcomponentvaiuesare R : 1kO,

Sotution From Table 15.1, we find that the fourth-order Butterworthpolynomialis (r, + 0.765s+ 1)(s, + 1.848s+ r). We will thus needa cascadeof two second-order fil ters to yield tle fou h-oder transferfunction plus an inverting amplifier circuit for the passbandgain of 10.The circuitis shownin Fig.15.22. kt the first stage of the cascadeimplement the transfer function for the polynomial (r' + 0.765s+ 1). FromEq.15.36,

Cr = 831nI' C, - 121nF. C3 - 344 nFF, Ca:294nF' Finally,we need to specifythe resistorvaluesin the inverting amplifiei stageto yield a passbandgain of 10.Let R1 : 1kO;then Rt : 10R1: 10kO.

CI:261F' C2:038F' Let the second stage of the cascadeimplement the transfer function for the polynomial (r' + 1.848r+ 1). From Eq.15.36, C3 = 108 F, c4: 0.924F.

gajn. Figure15.22A A fouth'orderSutteftorthfitterwithnon'unjty

Figure 15.23 compares the magnitucle responsesof the fou h-order idenricalcascadefilter from Example 15.7 and the Butterwofih filter we just designed.Note that both filters provide a passbandgain of 10 (20 dB) and a cutoff ftequercy of 500Hz, but the Butterworth filter is closerto an ideal low-passfilter due to its flatter passbandand steeper rolloff at the cutoff ftequencl Thus, the Butterworth designis preferredover the identical cascadedesign.

632

\

-

I

\

o h

\ l0

l0

50 100

5001000 (Hz) f

500010,000

r€sponses for a figure15,23& A compafison of the magnitude andButteftodh fourth-order Low-pass fitterusingth€jdentica t cascade

TheOrderof a ButterworthFilter

H(ja\

dB

It should be apparcnt at this point that the higher rhe older of the comesto that of Butterwo h filter, the closerthe magnitudecharacteristic the magnitudestays an ideal low-passfiltor.In other words,asn incleases, close to unity in the passband,the transition band narrows, and the magni tude staysclos€to zero in the stopband.At the sametime, as the order increases,the number of circuit components iroeases. It follows then tlat a fundamental problem in the design of a filter is to determine the smallest value of r?that will meet the filtering specfications. are usually In the designof a low-passfilter, the filte ng specifications given in terms of the abruptness of the transition region, as shown in Fig. 15.24.Once Ap, ap, A,, and o, are speciJied,the order of the Butter,vorthfilter canbe determined. For the Butterworthfilter.

, , : ,n,-"----"" l

",ri4

: . regionfor a Figule15.24a Defining thetransjtion

10logr(1+.d?), r^'^-

=

(15.37)

I

+.',,3'). 101ogro(1

(15.38)

15.4 Higher 0rder0pAmpFjlte6

It follows fmm the definition of the logarit}lm tlat 10414"=l+.,];,

(15.3e)

10{.1n,= 1 + o2'.

(15.40)

Now we solve for di and ol and then folm the ratio (d/or)".

1/{0-nt,

1 V l O o . 1 \ _r

/,," \' \uet

', oe

We ger

1t5.41)

where tlle symbols.r" and o, have been introduced for convenience. From Eq. 15.41we can wdte n loEn@, / a p) = loglr(ooloo),

togn@,/.r,) togln@Jror)'

115.42)

We can simplify Eq. 15.42 if (,? is the cutoff frequency, becausethen ,4p equals-20logro14, and o, : 1. Hence loglo{'"

' =

Lcte'J%\

(15.43)

One furthei simplification is possible.We are using a Butterworth filter to achieve a steep transition region. Therefore, tie filtering specification will make 10{ra" >> 1.Thus -

..

1^ 0 05'1

loglo{/r ! -0 05"4'

(15.44) (15.45)

Therefore, a good appmximation for the calculation oI z is

" =

0.05/" Ll,'r\.J%)

115.16)

Note that (o/(dp : /,/fp, so we can work with either radians per second or hertz to calculaten, The order of the filter must be an integer; hence, in using either Eq. 15.42or Eq. 15.46,we must selectthe neatestinteger value greater t]lan the result given by the equation. The following examples illustrate tle usefulnessof Eqs. 15.42 ald 15.46.

633

634

ActiveFitierCncuits

Determining the orderof a Butterworth Filter a) Determinethe order of a Butteiworth filter that has a cutoff frcquercy of 1000 Hz and a gain of no more than -50 dB at 6000Hz.

10{r( 50)is much greaterthan 1. Herce, we can useEq.15.46with confidence: , =

b) What is the actualgain in dB at 6000Hz?

r-0.05)r 501

: 1 -) -l 10g10(6000/1000)

Therefore,we need a fourth-orderButterworu filter. b) We can useEq.15.25to calculatethe actualgain at 6000 Hz. The gain in decibels will be

Solution a) Because the cutoff ftequency is given, we krow r, = 1. We also rote ftom the specification that

t K - )n los o{

1 ,_-

\ |

\ v 1 + 6 ", /

02.25dB.

An AtternateApproach to Determining the orderof a Butterworth Filter a) Determine the order of a Butterworth filter whosemagnitudeis 10 dB lessthan the passband magnitude at 500 Hz and at least 60 dB lesstllan the passbandmagnitudeat 5000Hz. b) Detemine the cutoff frequency of the filter (in hertz). c) What is the actual gain of the filter (in decibels) at 5000Hzl

Therefore we need a third-order Butterworm filter to meet the specifications. b) Krowing tllat the gain at 500Hz is - 10 dB, we can detemine t}le cutoff frequency.From Eq. 15.25we -10log1o[1 + (ald")61 :

10,

whereo : 10002rad/s.Therefore 1 + (a10,,)6: r0,

50lution L

a) Becausethe cutofr ftequencyis not given,we use Eq. 15.42to determinethe order of the filter:

.,-."40'ditTf

ffi

: 2178.26 ftdls.

r=:, It followsthat

", - 14o d(aIJ

= tooo,

: 10, t',Jop: f,/f p: 5000/s00 locro(1000/3) loglo(10)

f, = 34668 H2. c) The actual gain of the filter at 5000 Hz is

-l0losro[1+ (s000/346.68)6]

r: =

69.54dB.

15.a Njsher 0rder0p AmpFjtters 635

ButterworthHigh-Pass, Bandpass, andBandreject Filters An nth-order Butterworth high-passfilter has a transfer function with the ,th-order Butte$orth polynomial in the denominator,just like the ,rth-order Butterworth low-passfilter. But in the high-passfilter, the numerator oI the tansfer lunction is r', whereas in the low-pass filter, tie numerator is 1. Again, we usea cascadeapproachin designhglhe Butterworthhigh-pass filter. The fiist order factor is achieved by including a prototwe high-pass Iilter (Fig. 15.4,with Rr : R, = 1 O, and C : 1 F) in the cascade. To produce the second-order factors in the Butterworrh polynomial, we need a circuit with a transler function of the form

H(r) :

s2+rtr+1

Sucha circuit is shown in Fig. 15.25. This circuit has the transfer function

u. r)+ R C s2+ R r R 1l SettingC = lFyields Figure15.25AA second-oder ButteMo'th hjgh-pass fittercircuit.

1 ' )- & " 2* R , &

(15.48)

Thuq we can realize any second-order factor in a Butteiwo h polynomial of the form (s' + b1s+ 1) by includingin the cascadettre second-order circuitin Fig.15.25witl resistorvaluesthat satisfyEq.15.49:

br:+ano r :

1

RF;.

(15.49)

At this point, we pauseto make a coupleof obsenationsrelalive to Figs. 15.21 and 15.25 and their prototlpe tmnsfer functions 1[G2 + bf + 1) and r'/G' + 6rs + 1).'These obsenations are important becausethey are true in geneml. First, the high-pass circuit in Fig. 15.25wasobtainedfrom the low-passcircuitiII Fig.15.21byinterchanging resistonand capacitorsSecond,the prototypetransferfunctionofahighpassfilter can be obtainedfrom that of a low-passfilter by replacings in the low-passexpressionwith 1/r (seeProblem15.46). We can use ftequency and magnitude scaling to design a Butterworth high-pass lilter with practical component values and a cotoff frequency other than l rad/s. Adding an irverting amplifier to the cascade will accommodatedesignswith nonunity passbandgains.The problems at the end ofthe chapterincludeseveralButterworthhigh-passfilter designs.

636

ActiveFitterCircuits

Now that we can design both ,th'order low pass and high-pass Bufterworth filters $ilh arbitrary cutoff frequenciesand passbandgains, we can combinctheseliltels in cascade(aswe did in Section15.3)to producerth-ordcr Bulterworth bardpassfilten. Wc cancombiDethesefilters in paralleiwith a sunming amplifier (again.as we did in Section15.3)to producerth order Butterworthbandrejectfiltcrs.This chapter'sproblems alsoincludeButterworthbandpassand bandrejectfilter designs.

objective3-Underctandhowto usecascaded first- andsecond-order ButteMorthfitt6rs Answer: R1 = 0.?07O, R, : 1.41O.

15.4 For the circuit in Fig. 15.25,find vnluesof-R1 and R) $al )ield a .econdorderprrJrol)?e Buller$ortl high-pacc liltcr. NOTE: AIso try ChapterProblems15.33,15.36and 15.37.

15.5 &lxrrowfon*ci S**dpnss al:d Rar,4r,*i.:r: iii tr,..: The cascadeand parallel componentdesignsfor synthesizingbandpass ard bandrejectfilters from sirnplerlow-passand high-passiillers havethe restrictionthat only broadband.or low-Q. filters wiU rcsull. (The Q. of course,standsfor qrdliry /a.il7r.) Tlis limitation is due principally1l)the fact that the transfcrfunctioft for cascadedbandpassand parallel baDdreject fillers have discrelereal poles.The synthesistechniqueswork best for cutoff frequcnciesthat are widely separatedand therelore yield the lowest quality factors.But the largestquality factor we can achievewith discretereal poles ariseswhen the cutoff frequcncies,and thus the pole locations.are the same.Considerthe transfcrfunclion that resulrs: -' \ -.\/ / \s + d /\s + u /

s' +2.,s+.1 0.58s s2+Bs+tf,'

(15.50)

Eq.15.50is in the standardform ofthe transferfunction oI a bandpassfilter, ard thus we can determine the bandwidth and cenler frequency directly:

B:2,-

(15.51)

.Z="t

\15.52)

From Eqs.15.51and 15.52and thc detinilion oI Q, we seethat

8 2 a , 2

(15.53)

1..5

Id,owbdrdBardodss filte$ andBandrciect

637

Thusv.it} discretereal poleqthe highestquality bandpassfilter (oi band= 112. rcjectfilter)we canachie,reha'sQ To build activefilte^ with high quality factor valueEwe needan op ampcircuit that canproducea transferfun€tionwith complexconjugate poles.Figure15.26depictsonesuchcircuit for us to analyze.At the invening input of the op amp,we sumthe currcntsto get

v,,

u _-u

l/sc

fi3

figure15.26A Anactivehigh-Obandpass fittel

Solvingfor %,

',% - _u

(15.54)

"n.C-

At the node labeled a, we sum the crrlrents to get

u u =v ^ u- 1 Auc -"& v ^ 1Ac

&

Solvingfor V , U - 0 + 2sRtC+ RIl R2)U- sRrCU".

(15.55)

SubsrituringEq. 15.54into Eq. 15.55and then rearranging,we get an er?ression for the transfer function %/14:

s RrC

r1("):

1

2

(15.56) 1

'

s-+Rrcs+R"oR.c,

Reo

R,Rl ^ ^, R2

.. RrllR, -

Since Eq. 15.56is in t}le standard form of the transfer function foi a bandpassfilter, that is,

KP"

?+Bs+of,,' we can equate terms and solve for the values of the resistoN, which will achievea specified center hequency (o,), quality factor (O), and passband

sain(to:

^ -

2 RrC' 1

KF: _; ,

"

{15.57) (15.58)

1 R.qR3C'

(15.59)

638 At this point, it is convenientto define the prototype versionof the circuit in Fig. 15.25 as a circuit in which do : 1 rad/s and C = 1 F. Then the expressions for 1R1,R2, and Rr can be given in terms of the desired qualitylactor andpassbandgain.We leaveyou to show(in Problem15.38) that tbr the piotot)?e circuit, the expressionsfor R1, R, , and R3 are

&: Q/K. R2_ QIQQ, K), Rz : 2Q. Scalingis usedto specifypracticalvaluesfor the circuit components. This designprocess is illusrrared in Example15.12 Designinga High-o Bandpass Filter Design a bandpassfilter, usingthe circuit ir Fig.15.26, which has a center frequency of 3000 Hz, a quality factor of 10, and a passbandgain of 2. Use 0.01 ,oF capaciton in your design.Compute the transfer func tion of youl circuit, and sketcha Bode plot of its mag nitude rcsponse.

Solution SinceO:10 and lK = 2, the valuesfor Rl, R2, and R3in the prototypec cuit are

Fisur€15.27A lhehigh-0 bandpass fitt€rdesisned in Examph 15.12.

R] : 10/2 : 5,

I

5 dB (sa;

R, - 10/(200 2) = r0l1e8, R3-2(10):20. The^ scaling facton are kt = 60002 and 4- : 106/kf.After scaling,

T T

Rt = 265kO, R, - 268.0O, R3 : 106.1kO. The circuitis shownin Fig.15.27Substituting the values of resistanceand capacitarce in Eq. 15.56givesthe transler function for this circuit:

H(s) :

3770s s2+t885.0s+355^1ob

It is easyto seethat this transfer function meets the specification of tlle bandpass filter defined in the example. A Bode plorof its magnirude response is sketchedin Fig.15.28.

-

T

-20 -25 30

t

-35 100

5001000

{ 500010,0m 50,000100.000 f(H")

Figlre15,28A TheEodemagnitude ptotforthe high-Q bandpass fitter desjgned in Exanpte 15.U.

15.5 Narowband BandDass andBandreiect Fitters 639 fhe parallel implementation of a bandreject filter that combines lowpassand high-passfilter componentswith a summingamplfier hasthe same low-O restdction as the cascadedbandpassfiltor. The ciicuit in Fig. 15.29is an aclive high-Q bandrejectfilter ktro*n asthe twin-T notch fflt€r becauseof the two T-shapedparts of the cicuit at the nodeslabeled a and b. We begin the analysis of this circuit by summitrg the currents away from node a:

(u ursc +(va*ry=o u)sc ulzscR+ 2I - U[scR+ 2ol : sCRU.

(15.60)

Summing the currents away from node b yields

v u u* " u *^ (Yb R

R.j

o%)2sc = 0

vhl2+ 2RCsl Vo[l+ 2oRcsl: Vt.

(15.61)

Summing the currents away frcm the no nverting input termiml of the top op amp gves

1a-aY"*!ii\

=o

,RCU-Ub+(sRC+1)%:0.

I

\15.62)

I "C

JC

v,

0-')n 4 2

oR

tigure15.29^ A hish,oactive bandrcjectfiLter

640

ActiveFitterCircuits From Eqs.15.60-15.62, we car useCramer'srule to solvefor %:

2(Rcr + 1) 0 JCRI4 0 2(Rcr + 1) U R C s 1 0 (RCs+2') O l2(RCr+ 1) r 2 o R c+ r l) u 2rRCr l) I -RCs -1 RCr+ I | (Pc's' + I)U RzC2s2 + 4Rc(1

(15.63)

o)s + 1

Rearranging Eq.15.63,we cansolvefor thetransferfunction:

ITG):

(t.**a)

n

(15.64)

which is in the stardard fom for the transfer function of a bandreiect filter r

T@0

r'+Bs+d6

(15.65)

EquatingEq$15.64and15.65gives 1 -a

1 R2C2'

(15.66)

^ a $ c ) ' R c

115.61J

In this circuit, we have tbiee paiameters (R, C, and a) and two destn constraints((," and p). Thus otreparameteris chosena$itrarily;it is usually the capacitor value becausethis value t'?ically provides the fewest commercially available options. Once C is chosen,

_

1

(15.68)

and

-- = -' - L = ,- l 4-"

4Q'

(15.69)

Example 15.13illustrates the designof a high-O active bandreject filter.

1".\

Na'owbard Bardp4,andBardrcject rilte \

641

Designing a High-oBandreject F'itter Design a high-o active bardrejecr filrer (based on the circuit in Fig. 15.29) with a cenler ftequency of 5000 rad/s and a bandwidth of 1000rad/s. Use 1 pF capacitorsin your design.

Therefoie we need resiston with the values 200 O (n),100 o (R/2),190o (rR), and 10 o [(1 - ')R]. The filtal designis depicledin Fig. 15.30,and the Bode magrlitudeplot is sho*'IrinFig.15.31.

Sotution In the bandreject prototwe filter, oo : l rad/s, R : 1 O, and C : 1 F. As just discussed, otrueo, and O aie given, C can be choser arbitrarily, and R andl' canbe fourd from Eqs.15.68and 15.69.From the specifications,O: 5. Using Eqs. 15.68 ard 15.69,weseethat

R:200(), (' = 0.95.

Figure 15.30a Thehish-Oactive bandrcjectfiLrer desisn€d in [xampte 15.13.

10

I -10

20 1000

5000 10,000

50,000 100,000

rigule15.31A TheBodemagnitude plotforih€ hish-Oactive bandrej€ct fiti€rdesiqned in Example 15.13.

642

Active Fitter Circuits

0biective4-Be abteto usedesignequations to calculate vatuesfor prototype narrowband, component bandpass. andbandreiect filters 15.5 Designan activebandpassfilter with O = 8, K - 5 . a D d d "- 1 0 0 0f a d / sU . .e lr,l cafaci tors,and specifythe valuesof all resistors-

15.6 Designan activeunity-gainbandrejectfiller with o, = 1000rad/s and I : 4. Use 2 tuF caplrcilorsin your design,and specifyihe values of R and.'.

Answer: ,R1: 1.6kO, R, = 65.040, Rr = 16kO.

. Answen R : 500O , ,7 : 0.9375

NOTE: AIM try Chapter Probkm 15.58.

PracticaIPerspective Bass Volume Control Wenowlookat an op ampcircuitthat canbe usedto controL the ampLjfication ofanaudjosignaI in thebassrange. Theaudjorangeconsists ofsjgnats havingfrequenci€s from20 Hz to 20 kHz. Thebassrangeinctudes frequenciesup to 300 Hz.Thevotumecontrotcircuitandjts frequency response are shownin Fig.15.32.Thepartjcularresponse curvein the famjiyof response curvesis setected by adjustingthe potentiometer settingin Fjg.15.32(a). In studyingthe frequency response curvesin Fjg.15.32(b)notethe fot' Lowing.First,the gain jn dB canbe eitherpositiveor negatjve.Ifthe gair is positivea signaLin the bassrangeis anrptifiedor boosted.If the gain is negativethe signaljs attenuatedor cut. Second, it is possjbte to seLecta jn response characterjstic that yieldsunitygair (zerodB)for alt ftequencies the bassrange.Aswe shallsee,if the poteftiometeris set at jts nridpojit, the cjrcuitwitl haveno effecton signabin the bassrange.FinatLy, as the frequency increases, aL[the charactestic responses approachzerodB or unity gajn. Hencethe voLume controlcircuitwjtl haveno e{fecton signats ir rhe rppe endor lrebleranqFo'LhedLdiotreqrenLiF..

-dBr -dBr iBi

(a)

(b)

Figufe15.324 (a) Bass votume controtcircuit;(b) Bass votume contrclcircuit frequency response.

PracticatPerspectjve 643

Thefirst stepin analyzing the frequency response of the circuitin js to caLcutate tig. 15.32(a) the transfer functjon%/I4. Tofacititate this equivaLent cjrcuitis qiv€nin Fig.15.33.Thenode catculation thes-domain jn thecircujtto support nodevottag€ vottages % andVbhavebeentabeted is determined by the numerical Thepositionof the potentjometer anatysis. vatueof d, asnotedin Fig.15.33. f!nctionwewritethethreenodevottage equations Tofindthetransfer for thevottaqe ratio that describe the circuitandthensolvethe eq0ations The node vottage are eqlations %/%.

u ^ v ' ( v "lbr\( =u: '"

u

(1- a)R,

X'

-v- : + ( u4-

y - )-5 . 4

"4^ " - 0^ ;

(1 - a)R,

t v o - n-' oR,

Thesethreenode-vottage equations canbe sotvedto find % asa functionof 14and hencethe transferfunctjon/t(r): H(") =

V_

u"

-(a, +an,+RrR,Crr) R1+ (1

a)R, + -RrRCrr

It fottowsdirecttythat

nQq -

(&+ dR2+ joRE2Ci [R1 + (1

d.]R, + jo,RlR C1l'

functionwiL[generate the famjlyof Nowlets verifothat thistransfer in Fig. 15.32(b). First note that when frequency response curves depicted : is for atl frequencies, i.e. d 0.5themagnitude of AUar) unity

laj,)l :

Rr+0.5R2+JoRrR2C1 :1. lRl + 0.5R, + jdRrRrcl

Wheno=0wehave &+ aR2 111(j0)= Rl + (l a)Rz' of ]11(10)at d = 0. that is observe that Il(io)l at a : 1 is the reciprocal nt to)o

RrtR2 R,

t It, /o). o-.

Wjth a tjttte thoughtthe rcadercan seethat the reciprocalretationship notjust o - 0. ForexanpLe o = 0.4 andd = 0.6 hotdsfor all frequencies, aresymmetrjc with a = 0.5 and -(Rr + 0.4Rr) + t('RrRrcl

EQ4"=od=

(Rr + 0.6R,) + joRrR2cl

whiLe H(Jb)"=o.o=

(R1+0.6R) + jaR$2C1

(R1+ 0.4lRr)+ ja&R2cl

Flgure15.33 4 Thes'domain cncuitfortheba5s voLume controL. Notethata detemin€r the potentiom€t€r seitjng,so0 < a < 1.

644

Active Fitter Circuits

1 Hllu)" a.a- ,, n\ta)o aa It fotiowsthat depending on the vatueof a the votumecontrolcircujtcan eitherafiplifo or attenuatethe inconingsignat. ThenLlmerical vaLues of R1,Rz,andCt arebasedon two designdeciThe first ampLification or attenuation sions. designchoiceis the passband in the basslange(aso+ 0). Theseconddesignchoiceis the frequency at whichthis passband ampufication or attenuationis changedby 3 dB. The component vaiueswhjchsatis{ythe designdecisions are caLcuLated wjth d equalto either1 or 0. gainwitl be (Rr + Rr)/Rr Aswe havealrcadyobserued, the maxjmum maximum wiL[ be Rr/(R1 + R2). If we assume and the attenuation (Rl + nr)/Rl >> 1 then the gain (or attenuation)witl differ by 3dB fromits maxir,rum valuewheno : 1/R2Cr.Thiscanbe seenby notingthat

l"(,#)1",

l R 1+ R 2 + j R 1

lRr + iRll

R 1+ R 2

&

* /,'l l

r /n, + n,\

L 1+ j 1 and

"(,#)..

v2\

R1 /

R1 + jR1

Rr+Rr+jR1l 1+j1l

=",(#E)

yourundetstanding N'TE: Assess of thisPtddicalPerspedive by trying Chapter Problems 15.59and15.60.

Sumrnary Active filters coNist of op amps,resistors,aIId capacitors, They can be configued as low-pasEhigh-pasqbandpass, and bandreject filte$. They overcomemany of the disadvantagesassociatedwitl passivefilters. (Seepage 608.) A prototype low-passlilter has componentvaluesof R 1 = R 2 : 1 O a n d C : 1 F , a n d i 1 p i o d u c eas u n i t y passbard gain and a cutoff frequency of 1 rad/s. The prolotype high-passfilter has the same component valuesand also producesa unily passbandgain and a cu! off frequencyof l rad/s. (Seepages609and 612.)

Magnitude scalhg can be used to alter component val ueswithout changingthe frequency responseof a circujt. For a magnitude scalefactor of t ,, the scaled (pimed) values of resistance,capacitance,and inductance are R' = k^R,

L' : k^L,

and C' : Clkn.

(Seepage612.) Fr€qu€ncy scaling can be used to shift the frequency responseof a circuit 1()another frequency region without changhg the overall shape of the frequency response.

Summary645

For a frequency scale factor of k/, the scaled (pirned) valuesof resistance,capacitance,and inductance are R' = R, L' = L/kJ,

nnd C' : Clkr.

(Seepage612.) Componentscan be scaledin both magnitudeand f.equency,wilh the scaled (primed) component values grver by R' : k_R, L' = (k_lkf)L.

a

c' = c/(k_kf).

(Seepage613.) The dcsignof activelow'passard high-passfilters can beginwith a prototypefilter cilcuit. Scalingcan then be applied to shift the frequencyresponseto the desired cutoff frequency,using component values thal are commerciallyavailable.(Seepage613.) An active broadband bandpassfilter can be constructed usinga cascadeofa low passfilterwith the bandpassfiI, ter's upper cutoff frequency, a high-pass filter with the bandpassfilter's lower cutoff frequency,and (optionally) an inverting arnplilier gain stage to achieve nonunity gain in the passband.Bandpassfilters implementedin this fashior must be bioadband filters (o.2 >> o.r), so that the elementsof the cascadecan be specifiedinde pendentlyofone another.(Seepage616.) An active broadband bandreject filter can be conslructedusinga parallelcombinationof a low-passfilter with the bandeject filter's lower cutoff frequency and a high-passfilter witl the bandrcjectfilter's upper curoff {fequenc}.Ihe ourpursare then ted into a summing amplifier, which can produce nonunity gain in the passband.Bandrejectfilten implementedin this way must be broadbandfilters (o"2 >> (r"r), so that rhe low-pass and high-passfilter circuitscan be designedindependently ofone anotler. (Seepage620.) Highcr order activefillers have multiple polesin their transfer furctions, resulting in a sharper transition from the passband to the stopband and thus a more nearly ideal frequencyresponse. (Seepage622-) The tiansfer function of an nth-order Butterworth lowpass filter with a cutoff frequencyof 1 rad/s can be determined from the equation

H(s)H(-r) -

1 + (-1)'sh

by ' tinding the roots of the denominatorpolynomial . assigningthe left-half plane roors ro fl(r) . writirg the denominatorof H(s) asa productof firsrand second-orderfactors (Seepage627-628.) Tlle fundamental prcblem in the design of a Butterworthfilter is to determinethe order ofthe Iilter. The filter specification usually delines the sharpnessof the transitionband in termsof the quantities,4",o_,A,, and (,". From thesequantiti€s,we calculatethe'smlles't integer large. than the solution to either Eq. 15.42or Eq.15.46. (Seepase633.) A cascadeof second-orderlow pass op amp Iilters (Fig.15.21)with 1 O resistorsand capacitorvaluescho, sen to prcduce each factor in the Butterwortl polynomialwillFoduce an even-orderButterworthlow-pass filter. Adding a prototype low-passop amp filter will prc duce an odd-o.der Butterwo h low-pass filter. (See page629.) A cascadeof second-orderhjgh pass op amp filters (Fig.15.25)with f F capacitorsand resisrorvaluescho, sen to produce each factor jn the Butterworth poly, nomial vill produce an even order Butterwo h high-passfiltei. Adding a prorotype high-passop amp filter will producean odd-orderButterworth high pass filter. (Seepage635.) For both bigh and low-passButterworth filters, frequency and magnitude scaling can be used to shift the cutoff frequency from 1 rad/s and to include realistic componentvaluesin the design.Cascadingar inverting amplifier will produce a nonunity passbandgain. (See page630.) Butterworth low,passand high,passfilters can be cas cadedto produce Butterworth bandpassfilte$ o{ any order . Buttelworth low-passand high-passfilters can be combined in parallel with a summing amplifier to producea Butterworth bandrejectfiller of any order r. (Seepage635.) If a high-Q,or nar(owband,bardpass,or bandrejectfilter is needed,the cascadeor parallel combinationwill not work. lnstead,the circuitssho$n in Figs 15.26and 15.29are used with the appropdatedesignequation$ Typically, capacitor values are chosen from those com mercially available,and the designequationsale used to specilythe resistorvalues.(Seepage637.)

646

ActiveFilterCircuits

Probtems Section 15.1

tigurcP15.3

15,1 Find the transfer function %/q for the circuit shown in Fig. P15.1if Zt is the equivalentimpedance of tle feedback circuit, Zi is the equivalent impedanceof the input circuit,and the opemtional amplifieris ideal.

Figure P15.1

r5,4 a) Using the circuit in Fig. 15.1,designa low-pass filter with a passbandgain of 15 dB and a cutoff frequenc)ot In kH,,.A..ume a 5 nf capaciror is available. Diaw the circuit diagram and label all components.

152 a) Use the resultsof Problem15.1to find the tmnsfer lunclionol lhe crrcuilqhownin fig. P15.2. b) What is the gain of the circuit as o - 0? c) What is the gain of the circuit as o + N ? d) Do your ans$erslo (b) dnd (c) makesensein terms of known circuit behavior?

15.5 Designan op amp-basedlow passfilter with a cutofl frequency of 500 Hz and a passbandgain oI 10 using a 50 nF capacitor. a) DIaw your circuit, labelingthe componentvalues and output voltage. b) II the value of the feedback resistor in the filter is changed but the value of the resistor m the forward path is unchanged, what charactenshc of the filter is changed?

FigureP15,2

15.6 The input to the low pass filter designed in Problem15.5is 200cosolmV a) Supposethe power suppliesare +%.. What is the smallestvalue of %. that will still causethe op amp to operatein i1sLinearregion? b) Find the output voltage when (, : (,.. c) Find the output voltagewhen o = 0.1.10. d) Find the output voltagewhen o : 10o0.

RepeatProblem15.2,ushg the circuit shown in Fig.P15.3.

I5.7 a) Use the circuit in Fig. 15.4to designa high-pass filter with a cutoff lrequenry of 40 kHz and a passbandgain of 12 dB. Use a 680 pF capacitor in the design. Draw the circuit diagram of the filter and label all the components.

Ploblenis647

15.8 Desgn an op amp-basedhigh passfilter with a cutoff frequency of 300 Hz and a passband gain of 5 using a 100 nF capacitor. a) Draw your circuit, labeling the component values and the ou4)ut voltage. b) If the value of tle feedback resistor in the filter is changed but the value of the resistor it the forward path is unchanged, what chaiacteristic of the filter is changed?

15.9 The input to the high pass filter designed in Problem15.8is 150cos/dtmv a) Suppose ttre power supplies are fy.d. Wiat is t]le smallest value of y.. that will still causethe op amp to opemte in its linear region? b) Find tle output voltage when a, : oc. c) End the output voltage when o - 0.1a)0. d) Flnd the output voltage when (, : 10(,\).

15.11 The voltage tlallsfer function for either high-pass prololypefiltef sho$Titr Fig.Pl5.l I is rItrl = ---:_Show that if either circuit is scaled in botlt magnitude and frequency,the scaled ffansfer function is (s/ kr\ H ' ( s ): . i - . ls/ Kl | + l

Flgsre P15.11 C=1F

!-t-

R=1O

Section152 15.10 The voltage tiansfer function of either low-pass prototypefilter showntu Fig.P15.10is I/(s) _

I

_.

Show that if either circuit is scaled in both magnitudo and frequency,the scaledtansfer funclion is l H'(s) : t s / K l t+ | Figule P15.10 R=1O

15.12 The voltage tamfer function of the prototype bandpassfilter shown in Fig. P15.12is

a(r):

G)"

u.(i)".'

Show that if the circuit is scaled in both magnitude and ftequencv. t}le scaledtansfer function is

(r(;)

("')'. (iXi).' L=lH

Flgu'€P75.72 1Ft=1H

648

AciiveFiltercircuits

15.13 a) SpeciJyt}Ie component values for t}Ie prototype passivebandpassfiltei describedh Pmblem 15.12 iflhe qualiryfactoroftbe filleri.25. b) Specifythe componentvaluesfor the bandpass filrer describedin Problem 15.12if the quality factor is 25; t}le center,or resonart, frequency is 100krad/s; and t}le impedanceat resonance i s 3 . 6k O c) Draw a circuit diagram of the scaledfilter and label all the components. 15,14 An altemative to the prototype bandpass filter ilustrated h Fig. P15.12is to make o, = l rad/s, R = 1O, and I = I henrys. a) Wlat is the value of C in the prototype filter circuit? b) w}lat is the transfei funcfion of the prototype filter? c) Use the altemative prototlpe circuit just described to designa passivebandpassfrlter that has a quality factor of 20, a center frequency of 50 krad/s, atld an impedanceof 5 kO at resonance. d) Draw a diagramof the scaledfilter and label all the components. e) Use the iesults obtained in Problem 15.12to wdte the transfer function of the scaled circuit,

15.15The passivebandpassfilter illustratedin Fig. 14.22 has two prototype circuits.In the first prototype c i r c u i t ,o o : l r a d / s , C : 1 F , l,-1H, and R-Oohms. In the second prototype circuit, od=1rad/s, R:1O, f a r a d s ,a n d C=O L: (1lQ)henrys. a) Use one of these prototype circuits (your choice)to designa passivebandpassfilter thal has a quality factor of 16 and a center frequency of80 krad/s.The resistorR is 80 kO. b) Draw a circuit diagram of the scaledfilter and label all components.

Show that tle obse ation made in Problem15.16 with respect to the transfor function for the circuit in Fig. 14.28(a)also appliesto the bandrejectfilter circuit (lower one) in Fig.14.31.

15.f

15.18 The passive bandrcject filter illustrated in Fig. 14.28(a) has the two prototlpe circuits shom in Fig.P15.18. a) Showthar for both cncuitE the transfer function is 12+1

u .( i ) ' . ' b) Wdte the transfer function for a bandreject filter that has a center trequency of 50 kmd/s and a quality factor of 5. Fig!reP15.18

't o! "" "

15.19The two prototype venions of the passive band reject filter shown in Fig. 14.31(lower circuit) are shownin Fis.P15.19(a)ard (b). Show that the transfer function for either s2+1 (;),.' FigureP15.19

'a'

,11--

15.16 The tiansfer function for tie bandreject filter shownin Fig.14.28(a)is

(+)

(f)".(#) Showthat if the circuit is scaledin both magnitude and ftequency,the transfer function of the scaled circuit is equal to the transfer function of the unscaled circuitwith r replaced by (s/,tl),where,tl is the frequencyscalefactor.

E.m

Scalethe bandpassfilter in Problem 14.21so that the cenferfoequencyis 250kHz and the qualityfactor is 7.5,using a 10 nF capacjtor.Determine the valuesofthe resistor,theinductor,ard the two cuf off frequenci€sof the scaledfilter.

probtems649 15.21 Scalethe bandrejectfilter in Problem14.33to get a center tequency of 500krad/s, using a 50pH inductor.Determine the valuesof the resistor,the capacitor, and the bandwidttr of the scaled flltei. 15.22 The circuit in Fig. P13.26is scaled so that the 4 kO resistoris replacedby a 20 kO resistorand the 5 nF capacitor is replaced by a 100pF capacitor. a) What is tle scaledvalueofZ? b) W}Iat is the er?ression for io in the scaledcircuit? 15.23 Scalethe circuit in Problem 13.29so that the 10 O resistor is increased to 1 k0 and the frequ€ncy of the voltage responseis increasedby a factor of 1000.Find t,0). 15.24 a) Show that iJ tlle low-passfilter cilcuit illustrated in Fig. 15.1is scaledin both magdtude and frequency,the transfer function of the scaledcircuit is the sameas Eq- 15.1with s replacedby s/k/, where q is the frequency scalefactor. b) In the prototrye version of the low-pass filter circuit in Fig. 15.1, rr" = l rad/s, C = 1F, Rz = 1O, and Rr = 1/tri ohms. Wlat is the transfer function of the prototype circuit? c) Using the result obtained in (a), derive the tmnsferfunctionofthe scaledfiltei. 15.25 a) Show that if the high-pass filter illustrated in Fig. 15.4 is scaledin both magnitude and frequency, the tmNfer function is the same as Eq. 15.4wirh r replac€dby r/t , where &l is the Irequency scalefactor. b) In the prototwe version of the high-passfilter circuit in Fig. 15.4, (,,. : l rad/s, Rl - 1 O, C = 1 F, and n2 : tri ohms.Wlat is the transfer function of tlle prototype circuit? c) Using the result in (a), derivethe transferfunction of the scaled filter. Section 15.3 15J6 a) Using 20 nF capacitors,design an active broadband first-order bandpassfilter that has a lower ,'-fj:i" cutoff frequency of 2000 IIz, an upper cutoff fre;;iai quency of 8000 Hz, and a passband gain of 10 dB. Use prototype versionsof the low-pass and high-pass filten in the design process (see Problems15.24and 15.25). b) Write the transfer function for the scaledfilter. c) Use the transfer function derived in part (b) to find li(/o.), where.d, is the centerftequencyof the filter.

d) What is ttre passbandgain (in decibels) of the file) Using a computerprogramof youi choice,make a Bode magnitudeplot of the filtei.

E.n

a) Using 5 nF capacitors, design an active broadband first-order bandrejectlilter with a lower cutoff frequency of 1000IIz, an upper cutoff frequency of 5000 Hz, and a passband gain of 10 dB. Use t]le prototpe filter circuits inttoducedin hoblems 15.24and 15.25in the design process, b) Draw tle circuit diagmm of the filter and label all the components. c) What is the transfei function of the scaledfilter? d) Evaluatethe transferfunction derived in (c) at the center frequency of the filter. e) What is the gain (in decibels) at the center fiequency? f) Using a computerprogramof your choice,make a Bode magnitude plot of the lilter transfer lunction.

1528 Show that the circuit in Fig. P15.28behavesas a bandpass filter. (Hr?t find the transfer tunction for this circuitand showthat it hasthe sameform as ttre transfer function for a bandpassfilter. Use the resultfrom Problem15.1.) a) FiDd the center ftequency, bandwidth and gain for this bandpassfilter. b) Find the cutoff frequencies and the quality for this bandpassfilter. figuruP15.28 10 pF

650

ActiveFiLterCircuits

8.29 For circuits consisting of resistoG capacitoq inductorE and op amps,IH(i,) ' involves only even powers of (d.To illustmte this, compute lH(Jo) l' for the tlree circuits in Fig. P15.29when V H(') : -:.

Pl5,29 Figure R

Sectiotr 15.4 15.32 The puryose oI this problem is to illustrate the advantageof an,?th-order low-passButterworttr filter over the cascadeof r identical low-pass sections by calculating the slope (in decibels per decade) of each magnitude plot at the comer frequency o.. To facilitate the calculation, let y represent the magnitude of 1ie plot (in decibels), and let "1 = logld,. Then calculate d/dr at (o"for each plot. a) Show that at the coner ftequency ((," : 1 rad/s) of an ,?th-order low-passprototype Butterworth filter, dy

-

londB/dec.

Show that for a cascadeof r identical low-pass protot)?e sections,the slopeat o. is 'l"

')(tr()t n

1\

dB/dec. 2r/" c) Compure dy/dx for each type of filter for n:1,2,3,4,and.n. d) Discussfhe significanceof the resultsobtained tII paft (c). 15.33 a) Derermine the order of a low-pass Butterworth filter that has a cutoff frequency of 1000 Hz and a gain of at least-40 dB at 4000H2. b) What is the actualgain,in decibels,ar 4000It? 15.34 The circuit in Fig. 15.21 has the tansfer function given by Eq. 15.34.Show that if ttre c cuit in Fig. 15.21 is scaled in both magnitude and frc quency,the transfer function of tle scaledciicuit is PC1C,

H'(") =

/ s \ -2 , 2 / r \- , 1530 Design a unity-gain bandpassfilter, using a cascade connection, to give a center ftequercy of 50 krad/s and a bandwidtl of 300 krad/s. Use 150 nF capacitors. Specify /.1, /.r, Rz, ard Rl'.

15.31 Design a parallel bandreject filter with a center frcquencyof 5 kHz, a bandwidthof 30 kHz, and a passband gain of 4. Use 250 nF capacitors,and speciJy all resistor values.

\krl

1

Rc,\4/ Rtc'r;

15,35 a) Write the hansfer function for the prototype lo$'-pass Butterworth filter obtained in Problem15.33(a). b) Write t}le traNfer function for the scaled filter in (a) (seekoblem 15.34). c) Check the expression derived in pafi (b) by using it to calculate tle gain (in decibels) at 4000 Hz. Compare your result with that found in Problem15.33(b).

probLems 651 15.36 a) Using 2 kO resisto^ and ideal op amps,design a circuit that vdll implement the low-pass Butterworth filter speciJiedin Problem 15.33. The gain in the passbandis one. b) Construct the circuit diagram and label all componentvalues, 15,37 a) Using 25 nF capa€itors and ideal op amps, design a high-pass unity,gain Butter,vorth filrer ,ffiTi$ with a cutoff frequency of 5 kHz and a gain of at least 25 dB at 1kHz. b) Draw a c cuit diagram of t]Ie filter and label all component value$

t5.42 a) Use 300 pF capacitors in the circuit in Fig. 15.26 to design a bandpassfilter with a quatty factor of 20, a center frequency of 8 kHz, and a passband gain of 40 dB. b) Draw the c cuit diagram of the filter and label all the components.

15.43 Show that if o. = 1 rad/s and C : 1F in the circuir in Fig. 15.26,t.heprototype values of Rl, R2, and R3are

^

o

a

15.38 Veri!, the entries ir Thble 15.1 for rl = 5 and

2Q2 15.39 The circuit in Fig. 15.25 has the transfer function given by Eq. 15.47.Show tlat if the circuil is scaled in both magnitude and ftequency, the transfer function of the scaled circuit is ("')' d'(,)

=

2 /r\ 1 - n,c \4/ R,Ra' \4/ /s\'

Hence the hansfer function of a scaled circuit is obtained from ttre transfer function of an unscaled circuit by simply replacing s in the urNcaled transfer function by r/&r, where ,t, is the frequency scaling factor.

15.40 a) Using3 kn fesistors andidealop amps.designa low-passunity-gain Butterwonh filter that has a cutoff frequency of 20 kHz and is down at least 25 dB at 100kHz. b) Draw a circuit diagram of the tiltei and label aU tlte componenfs,

15.41The high-pass filter designed in Problem 15.37 is cascaded with the low-pass filter designed in Problem15.40. a) Describe the twe of filter fomed by this interconnectiotr. b) Specify tlre cutoff fiequencieq tle midfrequency, and the quality factor of the filter. c) Use tlle resultsof Problems15.33and 15.38to dedve the scaledtmnsfer futction of the filter. d) Check the deiivation of (c) by usingit to calculare It0oo), wheie .r" is the midhequencyof the Iilter.

K'

Rz = 2Q'

15.44 a) Design a broadband Butterwonh bandpassfilter with a lower cutoff frequency of 1000TIz and an upper cutolf frequency of 8000 Hz. The pass, band gain of the filter is 10 dB. The gain should be down at least 20 dB at 400 Hz and m kHz. Use 50 nF capacitors in the high-passcircuit and 5 kO resistors in the low-pass circuit. b) DIaw a circuir diagram of the filter and label all the components.

15.45 a) Derive the exptession for the scaled fiansfer tunction for the filtei desigred in Problem 15.44. bJ U.ing rheexpressioD deri!edin fa L find thegain (in decibels)at 500H2 and 5000Hz. c) Do the values obtained in part (b) satisi, the fil" tedng specifications given in Problem 15.44?

15.46 Derive the prototype transfer function for a fifthorder high-pass Butterworth filter by firct writing the transfer function for a fifth-order prototype low-pass Butterworth filter and then replacing s by 1/r in the low-passexpression.

15.47 The fifth-order Butterworh filter in Problem 15.46 is used in a system where the cutoff frequency is 10krad/s. a) Wlat is ttre scaledtransfer function for the filter? b) Testyour er?ressionby fhding the gain (in decibels) at the cutoff frequency.

15.48 The purpose of this problem is to guide you plfJfltr through the analysis necessaryto establish a design procedure lor derermining t}le circuit components in a filter circuit. The circuit to be analyzed is shown in Fig.P15.48.

Fiqure P15.48

a) Analyze the circuit qualitatively and convince yourselfthat the circuit is a low-passfilter with a passbandgainof RzlR1. b) Support your qualitative analysisby deriving the transfer function U.IU. @int: In deriving the transfer function, represent the resistors witll theirequivalentconductances,tlatiqGr: 1/Rr, ard so forth.) To make tlle transfer function useIul in terms of the entriesh Table 15.1,put it in the form

Kb.

c) Now obse e that we have five ciicuit components Rt, iR2,R3,C1,and C2 and three tlansfer function constraints-(, 61, and b,. At first glance, it appears we have two free chorc€s among the five component$ However, when we hvestigale ttre relationships between the circuit componentsand the transfer function constraints, we seethat if C2is chosen,there is ar upper limit on C1 in order for R2(Gr) to be realizable.With tlis in mind, show that iJ C, = 1F, the three corductancesare given by the expressioff Gr:

part of a third-order low-pass Buttenvorth filter havinga passbandgainof B. a) If C2 = 1F in the prototype second-ordersection, what is the upper limit on C1? b) If tlre limiting value of Ct is chosen,what are the prctotype valuesof Rr, Rr, and R3? c) If the comer frequencyof the filter is 50 kHz and C2 is chosen to be 250 pE calculatethe scaledvaluesofC1, R1,rR2,and Rr. d) SpeciJythe scaledvaluesof the resistonand the capacitor in the firslorder section of the filter. e) Construct a circuit diagam of the filter and label allthe componentvalueson the diagam.

KG2l

c, = (g\c,, tt, + t/ttt,

t41t + r1q

2(r+ r() For G2 to be rcalizable,

-'

15.49 Assume the circuit analyzedin Problem 15.48is

h? 4b.(1+ K)'

d) Basedon fie resultsobtainedin (c), outline the design procedure for selecting the circuit componentsoncer, b., ard b1are known.

15.50 Interchange the Rs and Cs in the circuit in p?::TixFig.P15.48;thatis, replaceRl with Cr, R2 with C2, R3with C3,Cr with R1,and C2with R2. a) Describe the type of filter implemented as a resultof the interchange. Confirm the filter type descibed in (a) by deriving the transfertunction %/ q. Write tle transler lunction jn a form that makesit compatible with Table15.1. c ) Set C2 : C3 = 1 F and derive the expressions for C1,Rl, and R2in tems ofl., bl, and b,. (See Problem15.48for the definition of br and bo.) d) Assumethe filter describedin (a) is usedin the sametype oftlird-order Butterworth filter that has a passbandgain of 8. WitI C, = C3 = 1F, calculatethe prototypevaluesof Cl, R1,and R2 in the second-ordersectionof the filter.

ftobhns 653

15.51a) Use the circuitsanalyz€din Problems15.48and 15.50to implementa broadbandbandrejectfilter having a passbandgajn of 20 dB, a lower cornei hequency of 800 IIz, an upper corner frequencyof 7200Hz, and an attenuationof at least 20 dB at both 1500Hz and 13.5kHz. Use 50 nF capacitors whenever possible. b) Draw a circuit diagram of the filter and label a the components, a) Derive the transfer function Ior the bandreject filter describedin Problem15.51. b) Use the transfei function derived in part (a) to Iind the attenuation(in decibels)at the center Irequency of the filter.

15.53 The puryose of this problem is to develop the designequarions ior rhecircuiliD Fig.Pl5.)1.(see Pioblem 15.48for suggestions on the development of designequations.) a) Basedon a qualitativeanalysis, describ€the type of filter implementedby the circuir. b) Vedfy the conclusion rcached in (a) by deriving the transfer tulrction V"IU. Write the tralrsfer function in a form that makes it compatible with the entdesin Table15.1. c) How many ftee choicesare there in the selec tion of tlle circuit components? d) Derive the expressions for the conductances G1 : 1/n1 and Gz = llRz in terms of C1, Cr, and the coefficientsb, and b1. (See Problem 15.48for the definition of b, and d.) e) Are thereany restictions on Cl or C2? f) Assume the circuit in Fig. P15.53is used to design a fourth order low-pass unity gain Butterworth filtei. Specify the prototype values of Rl ard R2 in each second-order section if 1 F capacitorsare usedin the prctotypecircuit. figurcPr5.53

15.54 The fourth-order low-pass unity-gain Butterworth ,?nlfll{ filter in hoblem 15.53is used in a systemwhere the cutoff frequency is 25 kHz. The filtei has 750 pF caPacitors. a) Specify the numedcal values of Rr and R2 in each section of the filter. b) Draw a circuit diagramof the filter and label all the components.

15.55 Interchange the Rs and Cs i]) the circuit in Fig.P15.53,that is, replaceR1 with C1,R2 with C2, plisJ:Jl{ anclvrcevema. a) Analyze the circuit qualitatively and predict the type offilter implementedby the circuit. b.) Verify the conclusion reached in (a) by deiving the transfer function U.IU. Writ€ the transfer functior in a form that makes it compatible with t}le enrriesin Table15.1. c) How many free choicesare there in the selection of the circuit components? d) Find R1and R2asfulctions of 6., 4, C1,and C2. e) Are there ary restrictions on C1and C2? f) Assume the circuit is used in a third-order ButteNortl lilter of the t}pe found in (a). Speciry the prctotype values of Rt and n2 in the secondorder sectionof the filter it Ct = gt : 1 , .

15.56 a) The c cuit in Problem 15.55is used in a thirdorder high-passButterworthfilter that hasa cul ,liTiJ" off lrcquency of 40 klIz. Specify the values of Rl and -R2if360 pF capacitorsare availableto con struct the filter. b) Specifytlre valuesof resistanceand capacitance in the finforder section of the filter . c) Draw th€ circuit diagram and label all the componenlS, d ) Give the numerical expressionfor the scaled lransfer function of the filter. e) Use the scaledtransferfunction de ved in (d) to find the gain in dB at the cutolf frequency.

654

ActiveFitterCircuit5 rigureP15.62

S€ction 155 1557 a) Show that tle transfer function for a prorotype bandrejectfilter is

1lsCI

12+1

s' +([lQ\s+1

(1 o)Rz aRz

(l-d)R,

Y

dR,

b) Use tlle result found in (a) to find the transfer tunction of the filter designedin Example 15.13 R4+ 2R3

1s.s8a) Using the circuit shown in Fig. 15.29, design a na[ow-band bandreject filter having a center frequency of 4 kHz and a quality factor of 15. Basethe designon C : 150nF.

0

L rq

q

L rC1

Rt [-r ] t".rrlri

Draw the circuit dia$am of the filter and label all component values on the diagram. What is the scaledtnnsfer function of the filtei?

nr

S€€{iotrs15.1-15.5 1559 Using tle circuit in Fig. 15.32(a) design a volume pll$l*ii, control c cuit to give a maximum gain of 20 dB and dfii a eaiDol 17 dB al a ffeoue0cvof 40 Hz. Use an "*"" tt-t t o resisror and a lOdko potenriometer. Iest your design by calculating the maximum gair at a, = 0 and t]le gain at rr: 1/R2C1 using the selectedvalues of n1, R2,and C1, 15.60 Use the circuit in Fig. 15.32(a)to design a bassvolft;Ijii€ ume control circuit that has a mardmum gain of ,e.o*, 13.q8dB that drops otf 3 dB ar 50lt. 15.61 Plot the maximum gain in decibels versus d when = 0 for the circuitdesigredin Problern1s.59.I-et tifflHlEo a vary from 0 to 1 in incrementsof 0.1. 15.62 a) Show that the circuitsin Fig. P15.62(a)and (b) are eoui!alent. b) Show that the points labeled x and y in

Fig.P15.62(b)are alwaysat the samepot€ntial. c) Using the information in (a) and (b), showthat lhecircuitin l-ig.I5.33cdnbedrdwna(shownin Fig.P15.62(c). Showtlat the circuit in Fig. P15.62(c)is ir the form of the circuit in Fig.15.2,where Rr+(1 - d)R2 + n1R2C1s 1 + Rrclr

zf

Rl+dR2+RrlR2Cls 1 + R2C1s

15.63 An engineeing project manager has received a ;ffmi, proposalfrom a subordjnale\rho claimsrhe circuil shoqnin Fig.P15.63 couldbe u.edas a rreble!olume control circuit if Ra >> Rt + R3 + 2R2. The subordinatefu her claimsthat the voltagetranster function for the circuit is

vItG) : ; -{(2R3+ n4) + (1

B)& + n.l(BRa+ ntc,s}

{ ( 2 R r + R 4 ) + ( 1 B ) & + R 3 l ( B R+a R ) C r r } where R, = R1 + R3 + 2R2. Fortunately the projecf engineer has an electdcal engineering undergraduate student as an intern arld therefore asks t]le student to check the subordinate's claim. The student is askedto check tle behavior of tlle fansfer function as (,+0; as o+oo; a]Id the behavior when (.) = oo and B variesbetween0 and 1. Based on your testing of the hansfer function do you tlink the circuit could be used as a heble volume control? Explain.

Prcbtems 655 figuruP15.63 Rr R2

'0 C,

c) Is Rasignificantly greater rhan R,? d) When B = 1, what is the boostin decibelswhen a : L/&c2'l e) When B - 0, whar is the cut in decibels when a : 1/&C2? 0 Based on the results obtained in (d) and (e), what is the significance of rhe frequency 1/R3C, when R4 >> Ro?

(1-B)n4I pR4 -R3

15.64 In the circuit of Fig. P15.63the componenrvalues plglfiiir are n1 = Ru = 20 kO, R: = 5.9kO, Ra : 500kO, and C2 = 2.7 nF . a) Calculatethe maximumboostin decibels. b) Calculate the maximum cut in decibels.

15.65 Using the component values given in pr#flI#hProblem 15.64,plot the maximum gain in decibels versus B when a = 0. LeL P vary from 0 to 1 in incrementsof0.1.

Fourier Series 16.1 Fo!rier SeriesAnatysis: An oveffie$ p. 658 16.2 The FourierCoefficientsp. 659 16.3 TheEffectof Symmetryon the Fourier Coetfi.ients p. 662 16.4 An AlternativeTrigonom€tric form of the FourierS€riesp. 668 16.6

AnAppti.ationp. 6/0 Average-Power [at(utations with Periodic

In the preceding chapters,we devoteda considerableamount oI discussionto steady-statesinusoidalanalysis.One reasonfol this interestin the sinusoidalexcitationfunction is that it allows us to find the steady-stateresponseto nonsinusoidal,but peri odic, excitations.A periodic funct'ionis a lunction that repeats itself every I secondsFor example,the triangular wave illustrated in Fig. 16.1on page 657 is a nonsinusoidal,but periodic, break waveform. A periodiclunction is one that satisfiesthe relationship f(t) = ftt x nr1,

16.7 Therms VaLueoI a Periodic fun.iion p- 678 16.8 TheExponentialFormof the Fourier 16.9 Amplitudeand PhaseSpecvap. 682

(16.1)

\rhere '? is an integer (1.2,3, . . .) and I is the period.The function shownin Fig.16.1is pe odic because

f(rd = ,f(ro- T\ : f(ta + T) = f(to + 2r) : 1 Beabteto catcutate the trigonometric formof the Fouriercoefficients for a periodicwaveform lsjng the definitionofthe coeffjcieitsandthe sjmptificationspossibte if the wavefomexhibits oneor morctypesof symmeht 2 Krowhowto analyze a circLrjys response to a periodicwaveform usingFouriercoefficjents powerdeLivered 3 BeabLeto estimatethe average numbe'of Fourier to a r€sistorusinsa smaLL fom of the 4 Beabteto catcutate the exponentiaL Fouriercoefficjents for a penodicwaveform and magnitude usethemto generate andphase ptotsfor that waveform. spectrum

for any arbitrarily chosenvalue of a0.Note that Zis the smallest time interval that a periodic tunction may be shifted (in either direction)to producea function that is identical1()itsell Why the il1terestin periodic functions?One reason is that periodicuave man)clcctricalsourceiol practical\alLregenerale forms.For example,nonfilteredelectronicrectifieN driven from a sinusoidalsourceproducesrectified sinewavesthat are nonsinusoidal,but periodic.Figures16.2(a)and (b) on page657showthe waveformsof the full-wave and half-wavesinusoidalrectifiers, respectivcly. The sweepgeneratorusedto control the electronbeam ol a producesa periodictriangularwavelike cathode-rayoscilloscope the one shownir Fig.16.3. Electronicoscillators,which are usefulin laboratorytestingof equipmenl, are designed to produce nonsinusoidal periodic waveforms.Function generators,which are capableof producing wavefbrms, square-wave, t angular-wave, and rectangular-pulse are found in most testinglaborato es.Figure 16.4illustratesq.pical waveforms.

Another practical problem that stimulates intercst in periodic functions is that power generators,althoughdesignedto producea sinusoidal waveform,cannotin practicebe made to producea pure sine wave.The distortedsinusoidalwave,however,is periodic.Engineersnatuially a.re interest€din ascertainingthe consequences of exciting power systems with a slightly distorted sinusoidal vollage. T h + 2 7| Interest in pe odic functions also stemsfrom the geneml obsenation that any nonlinea ty ir an otherwiselinear circuit createsa nonsinusoidal pedodic function. The rectifier circuit alluded to eartier is one example of Figule16,1,6 A periodkwavdo this phenomenon.Magnetic saturation,which occus in both machines and transfomen, is another example of a ronlinearity that generates a nonsinusoidalperiodicfunction.An electronicclippingcircuit,which uses transistorsaturation,is yet anotherexalnple. Moreover, nonsinusoidal pedodic functions are important in the analysisof nonelectricalsystems.Problemsinvolving mechanicalvibration, fluid flow, and heat flow all make use of Deriodic functions. In fact. the study and analysis of heat flow in a metal iod led the French mathe maticianJeanBaptisteJosephFourier (1768 1830)to the trigonometric seriesrepresentationof a periodicfunction.Thisseriesbearshis nameand of a nonfiltercd sinuis the startirg point for findhg the steady-statercsponse to periodic exci- Figure16.2A 0utputwaveforms soidatrectjfier. Futlwave rcctjficabon. (b) Natf-wave G) lationsof electriccircuits.

tigule 16.3A ThetdanguLar waveform of a cathode'my generatoL oscitloscope sweep

vn

produced Figurc16.4a Wavetorms byfunctjon used in taboratory testing.(a)Square wave s€nelator (b)Irjanqutar wave.G) Rectansutar pulse.

16.1 FourierSeriesAnatysis: An Overview What Fourier discovered in investigating heat-flow problems is that a periodic tunction can be rcpresented by an infinite sum oI sine or cosine functionsthat are harmonicallyrelated.In other words,the period of any trigonometric term in the infinite series is an integral multiple, or harmonic, of the Iundamental period Z of the pedodic function. Thus for peiodic/(r), Fouriershowedthat/(t) canbe expressedas

Fourier seriesrepresentation of a periodic fundion >

(16.2)

where,?is the integersequence1,2,3, . , , . 1n8q.76.2, a., a", and b" are kno*n as the Fou er co€fficients and are calculated ftom /(t). The telm (.)o(which equals 2?r/7) represents the fundamentelfrequ€ncyof the periodic function f(l). The integral multiplesof d0 that is,2d0,3o0,4.r0,and so on aie knorpnas the harmonic frequenciesof/(t). Thus 2o0 is the secondharmonic,3.d0 is the third harmonis and nrro is the nth harmonic of f(t). We discuss the detemination of the Fourier coefficients in Section16.2.Beforepursuingthe detailsof usinga Fouder seriesin circuit analysis, we first need to look at tlre process in general terms. From an applicarions point of vie\ we can er?ress all the periodic functions of interest in terms of a Fourier series.Mathematically, the conditions on a periodic function /(1) that ensure expressing/(,) as a convergent Fourier series rknown as Dirichl€t's conditions) are that 1. /(t) be single-valued, 2. f(t) have a finite number of discontinuities in t})e periodic inteival,

3. /(r) havea finite numberof maximaand minimain the periodic interval, 4. theintegral I

f(t)ldt

Anypedodic functiongenemredby a physicallyrealizablesourcesatisfies Dirichlet'scondjtions.Theseare sufficientconditions,not necessaryconditions.Thus if f0) neets these requircments,we know that we can expressit asa Fouder series.However,if/(t) doesnot meettheserequ ements,we still may be able to expressit as a Fouder series.The necessary conditionson /(t) are not known. After we have determined /(t) and calculated the Fouder coefficients (d,, a,, and 6,), we resolve the periodic source into a dc source (ao) plus a sum of sinusoidal sources(r, and 6"). Becausethe periodic souce is driv ing a lhear circuit, we may use the principle of superposition to find the steadystateresponse,Inparticulai,we fust calculatethe rcsponseto each sourcegeneratedby the Fouriersedesrcprcsentationof/(t) and then add the individual rcsponsesto obtain the total response.The steady-state

16.2 lhe Fourier Coefficients 659

response owing to a specific sinusoidal source is most easily found with the phasormelhodofaraltsit The procedue is straightforward and involves no new techniques of circuit analysis. It produces the Fouder series repiesentation of lhe steady-stateresponse;coirsequently, the actual shape of the response is unknom. Futlemrore, the responsewaveform can be estimated only by adding a sufricient trumber of tems together. Even though.the Fouier seriesapproach to finding the steady:state.respolsedoes have some drawbacks,it introduces a way of thinking about a problem that is as important as getting quantitative results. In fact, the conceptual picture is even more important in some rcspectsthan the quantitative one.

16.2 TheFourierCoefficients AJter defining a periodic function over its fundamental period, we determine the Fourier coefficients from the rcIationships

(16.3)

(16.4) < Fouriercoefficient5 (i6.5)

In Eqs. 16.4 and 16.5, the subscdpt k indicates the kfi coefficient in the integersequence1, 2,3, . . . . Note that a, is the averagevalueof /(t), dr is twice the averagevalue of /(1) cos,/rol, and 6k is twice the avemge value of /(r) sin toot. We easily derive Eq$ 16.3-16.5fiom Eq. 16.2by recalling the following hteglal relationships,which hold when m and n are intoge$:

J,,,

s]|nme)otdt = 0.

Ior all fi,

(16.6)

'o"^'$ o' =

0,

for all m,

\16.7)

0,

for all m and n,

(16.8)

'"o"^ '"'on'ro' : f','" f,,^"

"io^,0, "n no,n, a,

fotallm+n,

T 2'

(16.e)

660

' ,o' ^-,,,,,or r.o, a!

f'

o. - t ,T

to(a)tm - n. . I o fm = n .

(16.10)

We leaveyou to vedfy Eqs 16.6-16.10inProblem16.5. To derive Eq. 16.3,we simply integrateboth sidesoI Eq. 16.2over one pedod: f,. r

J,.

,,,,0, -

for,

J"

\""

6

lio+?

= I

: \ )a,cosn-6r- b.sinn-o!ldt

a,&+>l

/h+?

\ a , c o sn d a t b a s i n n u d ) d r

= a;f + O.

(16.11)

l-qualronl6.J followsdirecllytrom tq. 16.IL To dedve t}le expiession for the ft1h value of a,, we first multiply Eq. 16.2by coskol and then irrtegiate botl sidesover one period of /(t): l'"rr I

f(t)coskaotdt:

J

J

>/

lh+r axcoska& dt I

h

| at cosnutt!coskuol + bnsinnuutcoskdutldt

= 0 + a // (Tt\r - 0 .

r i o z. )

SolvingEq.16.12for ak yieldsthe er?ressioninEq.16.4. We obtail the expression for the frth value of b, by filst multiplying borh sidesof Eq. 16.2by sin /..rot and then integrating each side over one period of /(t). Exampte16.1showshow to useEqs.16.3-16.5to find t}le Fouder coefficients for a specific periodic function.

Findingthe FourierSeriesof a Triangular Waveform with NoSymmetry Find the Fouder series for the periodic voltage shownin Fig.16.5.

Figure16.5 A Thepeiodicvottage for Example 16.1.

Solution When usingEq$ 16.!16.5 to find a,, dr, and b*, we may choose the value of t0. For the periodic voltage of Fig. 16.5,the best choice for to is zem. Any other choice makes the rcquired integntions moie cumbe^ome.fhe expiessionfor o(t) between0 and fis

*r:(?)

16.2 ]lreFoudef Coeffici€nts661 The equationfor a, is

Thc equationfor the kth valueof D, is

,.:il'(?)'"=in"

b k= :

I

l+)L\i\k-,tdl

2v^l

1 = __I_--:_cln/.dur

Thisis cle"rlylh<J\sfrge \ dlueot lhe $dvetor11 in Fig.16.5. The cquationfor the kth valueof d,,is

: :ilt | 0 /'\

2U_/ | -coskd^1 T' \ k,-6

1c,). k-a

-

_

|

."r-rt)l;

2,k I /

- v ^

"^=: I l+ )rcosr-o,.ir l J a \ t / = -l

\K_di

/-

1tk

' *f;,-r..,r)

The Fourier seriesfor 1](r)js v

utrt: ll t

')l:o

for all /,.

=

v

N l

l)l sil r- r n;=tn

v - v ^srn. d0/ ^

v ,sLn , .^ldr,/

^

v :a sin3rant-

objective1-Be ableto catcutatethe trigonometricfonn of tfte Fouriercoefficientsfor a periodicwaveform 1 6 . 1 D e r i \ e l h e e y p r e s s i o n < l o r r . . l 1 , . atnodr lbh{e pefiodicrolragerunc(ionshownit l- - qr V

1_t 3

Answer: d! - 21.99V, zr =isrn i

v.

Dk=i{1 - cos-} v. ^OrE: At\o t, ChdpterPrcbt?n' lo.l lo ,.

p rf o b l e mt o . t . 1 6 . 2 R c f e rr o A . s e . s m e n . a) What is rhe aleragevalueofthe periodic voltage? b) Computethe numcricalvaluesof a1 o5 and | , bs. cr ltf - l2s.ol.m\.whatislh t uen d a m c n l r l frequencyin mdiansper second? d) wllat is thc.frcquency of the third hamonic in hertz? e) Wdte the Fourier seriesup to a]ld includilrg the fifth harmonic. Answen (a) 21.99Y ( b ) 5 . 2V . 2 . 6 V 0 V - L J .a n dl . 0 r V : (tV 4.5V 0V 2.25V and 1.8V; (c) 50 rad/si (d) 23.87Hz; (e) 1,(t) = 21.99- 5.2cos50r + 9 sin 50r + 2.6cos100r+ 4:5sin 100r 1.3cos200r+ 2.25sin20Or+ 1.04cos250r- l.8.in.)5tr V.

Findingthe Fouriercoefficients,in general,is tedious.Thereforeanything that simplifiesthe taskis beneficial.Fortunately,a periodictunction that possesses ce ain typesof symmetrygreatly reducesthe amountof woik involvedin finding the coetficients.In Section16.3,we discusshow s),rnmelry affect5rbecoefficienls i, a Fourierserieq.

16.3 TheEffectof Svmmetrv

Coefficients onthe Fourier Four types of symmetry may be used to simplify the task of evaluating the Fourier coefficients: . . . .

even-tunctionsymmetry, odd-functionsymmetry, half-wave synmetry, quarter-waveslmmetry.

The effect of each type of symmetry on the Fourier coefficients is discussed in the following sections.

Even-Function Symmetry A lunction is defined as ever if (16.13)

Ev€nfunction>

Functions that satisfy Eq. 16.13 are said to be even becausepolyromial tunctions with only even exponents possesstlis cha-racreristic.For even Denodic functions. the eauations for the Fouder coefficients reduce to

"':i.1" rt'ta''

(16.14)

sk:

(16.15)

coskaotdt, ; Jo fQ)

bk : 0,

for all k.

(16.16)

Note that all the b coefficients are zero iJ t}le periodic function is even. Figure 16.6 ilustrates an even periodic tunction. The derivations of Eqs. 16.14-16.16 follow directly from Eqs. 16.3-16.5.Itr each dedvation, we select ,o = -?/2 and tlen break the intenal of integration into the rangefrom T/2 to 0 and' to T/2, or 1 lrl2

"-+I flgure15.61An ev€npeiodicfunction,

f(t) = f(-t).

f(t\ dt

:+t-,t+1" f(t\ dt

f(t) dt.

(16.17)

16.3 TheEff€ct ofsymnetry oniheFouier Coefficjents663 Now we change the variable of integmtion ir the first integral on the right-hand side of Eq. 16.17.Specifically,we le1 t = -i and note that /(r) : /(-r) - /(x) becausethe function is even.We also observethat x: T/2whent: T/2anddt: dx.Tl\en l0

la

fr,z

f(t\t dr\ - I I Jrlz JO

I litat J-fl2

t(x\dx.

(16.18)

which showsthat the integrationfrom f/2 to 0 is identicalto that from 0 to f/2; thereforeEq. 16.17is tlre sameas Eq. 16.14.The deiivafion of Eq.16.15proceedsalongsimilarlines.Here,

ak =;

f(t) cnskoatdt

I

+

cosk'otdt, i.1,, f(t)

(16.19)

but to I f(t)co'A.or,lt

J rl2

ta | l(r)cos(-k-ax)('dx)

J7:/2

: I,,,

(16.20)

As before, the integmtion from f/2 to 0 is identical to that from 0 to ?/2. Combining8q.16.20with Eq.16.19yieldsEq.16.15. A11the b coefficients are zero when f(t) is an even periodic tunction, becausethe integration frcm Z/2 to 0 is the exact negative of the integration from 0 to f/2; thar is,

P

t')

I J0\'ink-a!dt - / /(yrsinl rud)( .1.r) J TP JTz

= -.1" f @sint<.otat

\16.21)

When we use Eqs. 16.14and 16.15to find the Fourier coefficients,the interyalof integation must be between0 and I/2.

0dd-Function Symmetry (16.22) < oddfunction

664 Functionsthat satisfyEq. 16.22are said to be odd becausepolynomial functionswith only odd er?onentshave this characteristic. The exprcssions for the Fourier coefficients are lt6.?3) a* : 0, bk::

Figffe16.7A Anoddperiodic funcirion

f(t) = f( t).

for all k: I

1t6.21)

flt ) sinkdar,lt.

\t6.?5)

Note that all the d coefficientsare zero if the periodic function is odd. Figure16.7showsan odd pedodicfunction. We use the sameprocessto derive Eqs.16.23-16.25 that we used to deriveEqs.16.1416.16.Weleavethe derivationsto you in Problem16-6. The evenness, or oddness,of a periodicfunction can be destroyedby shifting the function along the time axis.In other words, the judicious choice of where I : 0 may give a periodic function even or odd symmetry. For €xample,the triangularfunction shownin Fig. 16.8(a)is neither even nor odd. However, we car ma&e the function even. as shown in Fig.16.8(b),orodd,as shownin Fig.16.8(c).

Half-Wave Symmetry A periodic function poss€sseshalf wave symnetry if it satisfies the (a)

t(t)

l(t) = f(L r/2).

(r6.26)

Equation 16.26statesthat a periodicfunctiol hashalf-wavesymmetryif, after it is shifted one half period and inverted, it is identical to the original function.For example,the functionsshownin Figs.16.7ard 16.8havehalfwavesymmetry,whereasthoseinFigs.16.5and 16.6donot.Note that halfwave s],mmetry is not a fuflction of where t = 0. lfa periodicfunctionhashalf-wavesymmetry,both irk and bk are zero for evenvaluesof k, Moreover,dr also is zero becausethe averagevalue of a function with half-wave slDmetry is zero. The expressions for the Fouder coefficients are 116.27) 116.28)

ak=:

I

G) Figure16.8A Howthechoice of wh€r€I = 0 can makea periodjc function (a)A even,odd,or neither. perjodic iianguhrwaveihaiis neitherevennorodd. of(a) madeevenbyshiftinsthe {b)Thetrianqutarwave functjon alonqtheI axis.(c)Ihetrianguhrwaveof G) nadeoddbyshiftingthefunction atonsihe t axjs.

f c ) c o s k d r td r .

for ,t odd;

b*=0,

1.1" to"n*wat,

(16.29)

(16.30)

for k odd.

(16.31)

1 6 . 3 TheEffectof Symmetry ontheFouiefCoefficjents 665

WederiveEqs,16.27-16.31 by starting$.ith.Eqs..16.3=16.5laod choois' ing the interval of integranoaas -T 12to T/2. We then divide this range into the inte als -Tl2 to O and 0 to T/2. For example,the derivation for a* is ak:

; J,^

f(t)cosk!,otdt

2 frp ,fO)cosL0&dt T JI .r/2

?, T.

+:l

dt. flt) coskor,t

' : ! - - r l )

tt = T/2,

when I : 0;

wrcat = -1/2,

dt = dt.

Werewrite the first integal as tn

|

J1/2

fttcosk-ntdt

-

.TP

I

J0

tr" - T/2)coskuDtx T/2tdx. (16.33)

Note that

fG-r14=-f(x).

Therefore Eq.16.33becomes

l)do"o"r,.o,

a,-

*"o..oskaaxrix(16.31) lo''', ,ut

IncoryoratingEq.16.34into Eq.16.32gives ak =

1 ;(

- coskt)

rT/2 f(t)cos ka$ dt. .lr

(16.35)

But cos/rir is 1 when,t is evenand 1 whenf is odd.ThereforeEq.16.35 generates Eqs.16.28 and16.29. We leaveit to you to verify that this sameprocesscan be usedto dedveEqs.16.30 and16.31(seeProblem16.7). Wesummarizeour obse ationsby notingthat th€ Fourierseriesrepresentationof a pe odicfunctionwith hall wavesymmetryhaszeroaverage. of dc.!alueandconrainr onl)oddharmonict

Symmetry Quarter-Wave The telm quarteFwave symmetry descdbes a pe odic function that has half-wave symmety and, in addition, symmetry about the midpoint of the positive and negative har-cycles. The function ilustrated in Fig. 16.9(a) has quarter-wave s]'nmetry about the midpoint of the positive and negative half-cycles.The function in Fig. 16.9(b) does not have quarter-wave symmetry,although it does have half-wave synmetry. A periodic tunction that has quarter-wave symmetry can always be made either even or odd by the proper choiceof the point where t = 0. For example,the function shown in Fig. 16.9(a)is odd and can be made evenby shiftingthe function I/4 units either right or left alongthe 1axis. However,the function in Fig. 16.9(b)can never be made either even or odd.Totake advantageof quater wavesymmetryin thecalculationof the Fouriei coefficients, you must choose the point where I : 0 to make the function either even or odd. If the funclion is made even. then d,. = 0, becauseof half-wavesymmetry; (b) Figure 16.9A (a)A function thathasquaderwave quarteF symnretry. (b)Afunction thatdoes nothave

a* = 0, for k even, becauseof half-wave symmetry;

* = | l''o fa\"o,0,r0,,

roraodd;

br = 0, for a ,t, becausethe function is even.

(16.36)

Equations 16.36 result from the function's quarter-wave syrnmetry in addition to its being even. Recall that quarter-wave symmetry is superimposedon half wavesymmetry,sowe can eliminatea! and 4* for k even. Comparingthe expressionfo. ar, & odd,in Eqs.16-36with Eq.16.29shows that combining quarter-wave symmetry \.ith evennessallows the shortening of the range of integration from 0 to 7/2 to 0 b f/4. We leave the derivation ofEos.16.36to vou in Problem16.8.

16.1 TheEffectofsymnretry onthe Fourier Coeffi.jents

Ifthe quarterwavesyrnmetdcfunction is madeodd, 4! = 0, becausethe functionis odd; dr = 0, for all k, becausethe functionis odd; br = 0, for k eveq becauseofhalf wavesymmetry; h

8 /r4 t (t)sin kartdt. ; I

lor ( odd.

(16.37)

Equations16.37are a direct consequence of quarter-waveslmmetry and quarter-wavesymmetryallowsthe shorteningofthe inter oddness.Again, val of integrationftom 0 to T/2 to 0 to T/4.We Ieavethe deivation oI Eqs.16.37to you in Problem16.9. Example16.2showshow to usesynrmetryto simplifythe task of findins the Fourier coefficients.

Findingthe FourierSeriesof an odd Function with Symmetry Find the Fouder seriesrepresentationfor the curient waveformshownin Fig.16.10.

In the intervai 0 = t < Tl4, the expiession for i(r) is

i(t)=;t.

Thus

o^=] 1,"!,"^u-,,0,

Figure16.10,! Theperiodic wavefom for Example 16.2.

Solution We beginbyiooking for degreesof symmetryin the waveform.We lind that the function is odd and,in addition,has half-waveand quarter-wavesymmetry Becausethe function is odd, all the a coefficientsare zero;that is, ar = 0 and l'k = 0 for a f. Because the function has half,wave symmetry, br = 0 for even valuesot l. Becausethe function has quarter-wave s]rynmetry the expression for ,* for odd values01& is 8-filt) = -

321^ l sinL-ot r" \ k"oi :

i(r)sintroordr.

8t-(

r)

t/.tsoo(

The Fouder se es representationof i(r) is $

>

I

az

-sin-SD

2

f ,=ns. n"

+I''^

81rt -;kn r, I stn

k"h

|

ru.I

^ ' q sln Jdol

|

-

15 slD )dnl

I 49

_

.ln /dnr

for a periodicwaveform coefficients objective 1-Be abteto catculatethe trigonometricform of the Fourier 16.3 De ve the Fourier seies for the pedodicvoltagesnown.

12Y,, : - _ Answer: f,(/) " r' "-i1:.

sin(r'rlll

NOTE: Also try ChaptetPrcblens 16.10and 16.11

Forwt Tnigonornetric 16.4 AveAtterslative sf the FsurierScries In circuit applicationsof the Fourier series,we combinethe cosineand Doing so allows sin.3termsinthe seriesinto a singleterm for convenicnce. thc representationofeach harmonicof ?,(t)ori(r) asa singlephasorquartity.The cosineand sine lerns may be mergedin eilher a cosineexpression or a sine expression,Becausewe chose the cosine format in the phasormethod of analysis(seeChapter9), we choosethe cosineexpression hcre for ihe alternativelorm of the series.Thuswe write the Fourier seriesin Eq. 16.2as

l(t) = a,-+ >A,,cos(naot 0"),

(16.38)

where,4,,andd,,aredefinedby thecomplexquanlity

o^ jt,, = {d, + 41_!" = t,t -a".

(16.39)

We dcriveEqs.16.38and 16.39usingthe phasormelhodto add the cosine and sinc terms in Eq. 16.2.We begin b]' expressilgthe sine functionsas cosinefunctions;thatis.we rewdte Eq.I6.2 as f(.i):

a,+ )a,cosn,,'t + b,,cos("oo1 90").

(1640)

Addlng the termsunder lhe summationsign by usingphasorsgives E{a,cos naot} : a, p

0{b, co.1r-er q0)} : 0,,l lq =

(16.41)

1n,,.

116.42)

Then 2l{a,,cos(naxt+ ,, cos(roi1

90')} : a,

jb"

(16.13)

Senes 669 Tisonometric Form 0ftheFouder 16.4 AnAtternahve When we invene+ransfom Eq.16.43,weget

- tt',:= + b,cos(nont ancosn.,at ^,)2rr'^l r,r. nr.*, SubstirutingEq. 16.44into Eq. 16.40yields Eq. 16.38.Equation 16.43 cofesponds to Eq. 16.39.If the periodic function is either even or odd, ,4,,reducesto either a, (even) or 6i (odd), and 0, is either 0" (even) or 90'(odd). The denvation of the altemative form of the Fouder seriesfor a given periodic function is illustrated in Example 16-3.

voltage fourierSeriesfor Periodic calculatingFormsof the Trigonometric a) Dedve the expressionsfor a&and ,l for t}le peri odic functionshowninFig.16.11. b) wrire lhe first lour rerm. ol lhe Fourierseries representation of ?J(t) using the format of Eq. 16.38.

and )

bk:

Irl4

;.lr

v-sinka.atdt coska6t r/a\

: r2V\ /

k , " " )

=( rF\ ( ' - .t 9/ ) b) The averagevalueof t(l) is

T r 3f 4

2

1

T sT 3f '7f 2r 1 2 4

for ExampLe 16.3. fundoon Figure15.11A lhe pedodic

u^g/4)

jbklor k:1,2,and3arc

Thevaluesof ar

..

vr,

',Evv-.-45', - - t .v-

a,

tD

t

ib -0-

Solution a) The voltage z,(r) is neither even nor odd, nor does it have half-wave symmetry Therefore we useLqs.lo.4and 16.5lo frnd,, andb . Choo.ing toaszero,we obtain

a , - i..b -

a t-9O

t''

-v^ ^

.v^ \a2v-

i:-

^

/

t3s".

Thus lhe lirsl four term. in lhe Fourier.eries representationof ,0) are

a,] a,* l',,{o)"o"r,.,,, *: |ll"'''r^"'"0.", rr - "' ,.,1r1

4

t

' "/iw -2c651ud n

+ lijilqrs(3@ot

^5')

:: co5(2@0/ q0')

135') + ...

670

FoderSenes

0bjediv€1-Be abteto catculate the trigonometric formof the Fouriercoefficients for a periodicwaveform 16.4 a) Conputc /1 ,4sand dL 0\ for the pcriodic functionshownif Y,, = 9f; V. b) Using the format of Eq. 16.38,wdte the Fourier seriesfor o(r) up 1()and including the fiftir harmonicassuningZ : 125.66ms. Answer: (a) 10.4,5.2,0,2.6.2.1V and 120", 60', not defincd, 120', -60'; (b) 0(, = 21.99+ lo.4cos(5or 5.2cos(100t 60")+ 2.6 cos(200t 120') + 2.1cos(250t 60")V.

120') +

NOTE: Ako tt)' ChaptetPrcblem16.18.

16.5 An A*ulieatioll Now wc illuslrale how lo usea Fouricr seriesreDreseDtution of a neriodic c \ . ' , r l r o nt u n ( r i u lno l i n dr h c c r c a o\ y( a l er e \ p o ; . c u at i n e a r c . r i u iIr}.e RC circuil shownin Fig. 16.12(a)wjll provide our example.'$c circuil is cncrgizedwith the periodic squarcwave voltageshown in Fig. 16.12(b). Thc vollageacrossthe capacjioris the desiredresponse. or output,signal. The fint step in finding thc steadyslareresponseis to represcntthe pcriodic excitationsourcewith ils For.der series.After noting that the sourcehas odd,half-wave.and quarter \vavesymmetrywe know !ha! Lhe Fouder coefficientsreducoto ,r, \rilh k restrictedto odd integervalues:

s y'ia bk: ; I V,,sinkart dt t,ln AV

= + iDl Figure16.12?, AnRf circuitexcited bya penodic (a)Ihe8r series v0ttaqe. cncuit.(b)Ihe squae-wave

(r is odd).

(16.45)

Then the Fouricr seriesrepresentationof % is

,"

''in,.,.

AV

_* >

\16.46)

Wriling the seriesin expandcd[orm. lve have 4V- .

4V^. ^ it

4V* + -sin

4l_ 5dl + _ sinTdnr+

11,6.47)

The voltagc sourceexpressedby Eq. 16.:17is the equivaleDtof infidtely many sericsconnectedsinusoidalsourccs,cach sotrce having its orvn amptitudeand ficquercy.To fird the contributionof cach sourceto lhe output voltage,wcusethe principleof superpositiorl.

16.5 AnApplicatjon 671

For any one of the sinusoidal souces, the phasor-domain expression for the output voltage is "

(16.48)

1+ i@RC

Al1 the voltage sources are expressedas sine functions, so we interpret a phasor in terms of the sine instead of the cosine.In other wordq wher we go from the phasor domain back to the time domain, we simply write the time-domain expressionsas sin(o, + d) instead of cos(ol + d). The phasor output voltage owing to the fundamental ftequency of the sirusoidalsourceis

v- , =

(4u,JrJ /O'

(16.49)

1+ i@oRC

Writing Vl in polar folm gives

v,r :

$u^)/ - 9r

(16.50)

nY1 + u6R'C'

91 :

(16.51)

tan-'ooRc.

From Eq. 16.50,the time-domain expressionfor the fundamentalfrequency component of ?J.is

trv 1 + @6R'C'

in(@d

Br).

\16.52)

We derive the tlird-hamonic component of the output voltage ir a simiphasorvolrageis lar manner. The rhird-barmonic (4v-/3n\,t0" "' I + 73oxRC

4V.

(16.53)

/-Ft,

3,\h + s"?rfrC 13oonc. 83 : tan

(16.54)

Thetime-domainexpressionfor the t]Iird-harmonicoutput voltageis ,- - ----!h sin(Jdnr- P r' . 3:'V | + guiR'C'

(1655r

Hence the expressionfor the kth-harmoniccomponentof the output voltageis

4V^

sin(koot k ^,4TF&Fe

Br:

tan 'koonc

- Bk)

(r is odd).

(k is odd), (16.56)

(16.57)

We now write down the Fou er se es represertationof the output voltage: "

. .

4Y- : sin(n@o/ B,) t ,=(ls...nt/t + (rdoRc)2

(16.58)

The derivation oI Eq. 16.58was not difficult. But, althoughwe have an analyticexpressionfor the steadystateoutput,what?r,G)looks lile is not immediately apparcnt from Eq. 16.58.As we mentioned ea.rlier,this shortcomingis a problem with the Fourier sedesapproach.Equation 16.58is not useless, however,becauseit givessomefeel for the steady-state waveform of x'o(t), if we focus on the frcquency response of the circuit. For example,if C is large, 1/r?ooc is small for the higher order harmonics.Thus the capacitorshort circuitsthe high-frequencycomponentsof tle input wavelbrm,and the higher order harmonicsin Eq. 16.58are negligible comparedto the lower order harmorlics.Equatior 16.58reflecasthis condition in that.for larue C.

"-#"-e+"in('oo1 eo")

"#. F.;-""'"'

(16.59)

Equatior 16.59showsthat the amplitudeof the harmonicin the output is decreasingby 1/n', comparedwith 1/n for the input harmonics.IfC is so large that only the fundamentalcomponentis significant,then to a first

Do(r, n

_cosoiy',

(16.60)

and Fouiier analysistells us that the square-waveinput is defomed into a sinusoidaloutput. Nowlet's seewhathappensasC- 0.The circuit showsthat ooard ?s are the same when C = 0, b€cause the capacitive branch looks like ar open c cuit at all frequencies.Equation 16.58predictsthe same result because, as C + 0, x^:-

>

srnndnr.

(16.61)

But Eq.16.61isidenticalto Eq.16.46,and theieforeod- rs asC +0. ThusEq. 16.58hasprovenusefulbecauseit enabledus to predictthat the output will be a highly distorted replica of the input waveform if C is large,and a reasonablereplica if C is small.In Chapter 13,we looked at the distortionbetweenthe input and output in tems of how much memory the systemweightingfunction had.In the frequencydomain,we look at the distortion betweenth€ steadystate inpul and output in terms of how the amplitude and phaseof the harmonicsare altered as they are transmitted through the circuit. w}len the network significantly alters the amplitude and phaserelationships among the harmonics at tlle output relative to that at the input, the output is a distortedversion of the input. Thus, in the frequencydomain, we speak of amplitude distortion and phasedistortion.

16.5 AnApptication 673

For the circuit here,amplitude distortion is prcsent becausethe amplitudesof the input harmonicsdecreaseas 1/n, whereasthe amplitudesof the output harmonics decreaseas

" \/1 + (rr"Raj' This circuit also exhibits phase distortion becauset}le phase angle of each input hamonic is zero, whereasthat of the flth harmonic in the output sig nal is - tan ' ,?ooRC.

An Applicationof the DirectApproach to the Steady-State Response For the simple RC circuit shown in Fig. 16.12(a),we can derive the expression for the steady-state responsewithout resortingto the Fourier series representation of the excitation function. Doing this extra analysis here adds to our understanding of the Fourier seriesapproach. To find the steady-state er?ression for o" by straightforward circuit analysis,we reason as follows. The square-wave excitation function alternates betweer charging the capacitor toward +ya and y-. After the circuit reaches steady-stateoperation, this altemate charging becomes pedodic.We know from the analysisof the singletime'constantRC circuit (Chapter7) that the rcsponseto abnpt changesin the driving voltageis exponential.Thus the steady-statewaveform ol the voltage aooss the capacitorin the circuitshownin Fig.16.12(a)is asshownin Fig.16.13. The analyticexpressionsfor o.(t) in the iime htelvals 0 < t < Z/2 andT /2 = | = T arc

Toward+I4,,

Figure16.13/\ Thesteady-state wavefom ofr, forthe

q = v- + (vt - V-)e1txc,

( 1 6 . 6 2 )c j r c u i t i F n i q1. 6 . 1 2 ( a ) .

, t ) .=

(16.63)

O=t=T/2; V - + ( V 2 + U ^ ) e V ( t 2 ) ) / R CT, l 2 < t < 7 .

We dedveEqs.16.62and 16.63by usingthe methodsofChapterT, assummarized by Eq. 7.60.We obtain the valuesof y1 and /2 by noting ftom Eq.16.62that

v2:v^+

(V - v;ert2Rc

(16.61)

andfromEq.16.63 that rt2Rc U : v- + (v2+ v.\e SolvingEqs.16.64and16.65for Vrandy2yields v-1t e-r 2Rc 1 vt: -V=

(16.65)

(16.66)

Substituting Eq.16.66 into Eqs.16.62 and16.63gives

u": v^-

t)
4-"'tRc,

(16.67)

and o.-

t-

-.:1-*-1t-'t

2 ^

t)-t
|6.681

Equatiom 16.6?and 16.68indicate that ?,o(r)has half-wave symmetry and that therefore the averagevalue of z, is zero.Thisresult agreeswith

Toward+Y",

674

lourier Senes

Figue15.14A Theetrectofcapacjtor rizeonthe

the Fourier seriessolution for the steady-stateresponse namely,that becausethe excitation function has no zero ftequency component,the responsecan have no suchcomponent.Equations16.67and (16.68)also show the effect of changingthe size of the capacitor.If C is small, the exponeltial ftrnctionsquickly vanish,oo : f- between0 and Z/2, and 1). : U^betweer T /2 and T. In other words,o, + os as C - 0. If C is large, the output waveform becomestdangular in shape,as Fig. 16.14 rholrs.Nore lhar tor large C. \ e ma) appro\imalelhe expoDenlial 'P' lerms p and p- r-''lR' by lhe linear terms | - (/ RC) and (T/2\llRC], respectively.Equation 16.59 gives the Fourier 1. {[t seriesof this triangular waveform. Figure 16.14summarizesthe results.The dashedline in Fig. 16.14is the input voltage, the solid colored line depicts the output voltage when C is sma , and the solid black line depictsthe oulput voltage when C is large. Fhally, we veriry' that th€ steady state response of Eqs. 16.67 and 16.68is equivalentto the Fouder seiiessolutionin Eq.16.58.Todo so we simply derive the Fouder series rcpresefltation of the periodic function desffibedby Eqs.16.67and 16.68.Wehavealreadynoted that the periodic voltage response has half-wave symmetry. Therefore the Fourier series containsonly odd harmonics.For * odd,

"r=| 1,"(o- , *41i|)*,o-", o, -8RCV^

I[1 + (/rooRc)']

il"& 4U^

-

(,t is odd),

(16.69)

2l -e i Fc y dl |+ et ' R()srnkd

8k.DU,R2c2

T[1 + (kooRc)'1

(r is odd).

(16.70)

To showthat the resr ts obtainedftom Eqs.16.69and 16.70are consistent vith Eo.16.58.wemust Drovethat

4U,,, Yai + bi =

1

tor a,! 11.*i,nif '

(16.71)

and that ak bk

=

kaoRc.

176.72)

We leave you to verify Eqs. L6.69 1.6;72iu Problems 16.22and 16.23. Equations16.71and 16-72are usedwith Eqs.16.38and 16.39to de ve the Fourier sedesexpressionin Eq- 16.58;we leave the details to you in Problem16.24. With this illustrative circuit. we showed how to use the Fourier series in conjunctionwith the pnnciple of superpositionto obtain the steadystate responseto a periodic driving function. Again, the principal short comingof the Fourier seriesapproachis the difficulty of ascefiainingthe waveformof the response. However,by thinking in termsofa circuit'sfte' quency rcsponse,we can deduce a reasonableapproximation of the steady-stateresponseby ushg a fhite number of appropriate terms in the (SeeProblems16.28and 16.29.) Fourier seriesrepresentation.

Functions 675 16.6 Averag*Pows Catcutations !!ith P€riodic

a circuit'sresponse to a periodicwaveform obiedive2-Know ftowto anatyze 16.5 The periodictriangular-wavevoltageseenon the left is appliedto the circuit shownon the righl.Deriv( ihe lirst lnrccnon/erolerm. in the Fou er sedesthat iepresentsthe steadystarevoltagez1o if %, = 281.257rmV and rhe period of the input voltageis 2uih ms. Answer: 2238.83cos(10/- 5.71') + 239.46cos(3ot - 26.5?")+...mV 16.70')+ 80.50cos(501 16.6 fhe pedodic squarc-waveshownon the left is appliedto the circuit shownon the right. a) Derive the first four nonzerotermsin the Fourierseriesthat representsthe steadystate voltage ?J,if yu = 210?rV and the pedod ofthe input voltageis 0.2rrmsb) Which harmonicdominatesthe output vollage?Explain why

100k{)

+ 88.81')+ Answer: (a) 17.5cos(10,000t - 95.36")+ cos(30,000r 26.14 + 168cos(50.000, + 98.30')+ 17.32cos(70,000t

.V;

(b) Thefifth hamonic,at 10,000 rad/s, filter because thecircuilis a bandpass rad/s with a centerfrequency ot 50,000 quality factorof 10. anda

l0 kt)

NOTE: ALsotry ChaptelPrcblens 16.27and 16.28.

16.6 ;kerage-Fower eateutatiems with Feriod'lcF$netiosls o[ the voltagcand curenl at Ifwe have thc Fourier seriesrepreseniatioD a pair of tcrminals in a linear lumped-parametercircuit, we can easily expressthc averagepower at the terminalsas a function of the harmonic vollagesand currents.Using the higonomelriclbrm of the Fourier series enpressedin Eq. 16.38,we write the pe odic vollage and current at the telminals oI a network as 'o = Ydc+ >4cos(,oot

el,),

(16.73)

t = 1d. + )/,cos(trdor

d,").

176.71)

676

Fourier Series The notationusedin Eqs.16.73and 16.74is definedasfollows: yd. = the amplitudeof the dc voltagecomponent, y, - the amplitudeofthe nth-harmonicvoltage. 0d, = the phaseangleofthe nth-harmonicvoltage, 1d. = the amplitude of the dc current component, 1r - the amplitudeofthe nth-harmoniccurrent, 4, = the phaseangleofthe nth-harmoniccurrent. We assumethat the cment reference is in the direction of the reference voltage drop acrosstbe termhals (using the passivesign convention), powerat the terminalsis ?i.The averagepoweris so that the instantaneous

+1"""+t"

(16.7a)

To find the expressionfor the averagepower,we substituteEqs.16.73and 16.74into Eq. i6.75 and integrate.Atfirst glance,this appearsto bc a for midable task, becausethe product ,)i requ es multiplying two jnfinire series.However,the only tems to surviveintegrationare the productsof voltage ard current at the samefrequency.A review of Eqs.16.8 16.10 should convince you of the validity of this observation. Thetefore Eq.16.75reducesto P-,v,tl"t

-r]|l+r v . I , c o \ t n a- a ot . n l r>-/ r=tr J\

x coslnoat

ei)dt.

(16.76)

Now' usingthe trigonometricidentily co\aco(P-

t 2coi{o' P)

l cos(o B}. 2

we simplify8q.16.76to P - vd.td. :>

-,'

- 0., + cos(2r?a,ol

lcosro., n...1

/

qi')lrlt.

(16.1t)

The secondterm under the integralsig! integratesto zero,so

+ i?*,(pP : v,r.Id.

e,,1.

(16.78)

Equalion 16.78is particularly important becauseit statesthat in the case of an interactior betlveena periodic voltage and the correspondingperiodic current! the total averagepower is the st1lltof the averagepowers obtained from the interaction of currents and voltagesof tlle samefrequency.Currents and voltages of different frequencies do not interact to produce avetage power. Therefore, in average-powercalculations involving periodic furclions, the total average power is the superposition of the average powe$ associatedwith each harmonic voltage ard curent- Example 16.4illustrates the computation of avcragepower involving a periodic voltage.

Functions677 16.6 Averaqe-Power Catculations withPenodic

Average Power for a Circuitwith a Periodic VoltageSource Catculating Assume that the periodic square-wave voltage in Example 16.3is applied aqoss the terminalsoI a 15 O resistor. The value of y- is 60 Y and that of f is 5 ms. a) wrile lhe {irstfivenonTerolermso[ thefourier cefle\ represenrarion of r(r). Lse rhe rrigonometric form given in Eq. 16.38. b) Calculate the average power associatedwith eachterm in (a). c) Calculate the total avemge power delivered to the 15 O resistor. d) Wlat percentageof the total power is delivered by the first five terms of the Fourier series?

Thus,using the first five nonzeroterms of the Fourierseries, ?r(r): 15 + 27.01cos(400,r 45") + l9.l0cos(800,r 90') + 9.00cos(1200rt- 135') + 5.40cos(20002r 45') +.

V.

rsatpliedrolhelefminal. oI a resi' b) tte \ olrage tor, so we can lind tlre power associatedwith eachtermasfollows: 152

P,,.===lsW. l5

",=|ff="o.t"*.

Solution

119.10'= 12.16 W, 2 t 5

a) The dc component of ?,(t) is

(60)(z/4) d,=

T

:15V. P3:

From Example16.3we have

A2: 60lr = 19.10v,

;1s

= 2.70W.

'p' - = : i : a : 2 1 5

A1 - .,A6olt = n.01v, dr : 45"

l q 2

n 0 7w

c) To oblair the total average power delivered to the 15 O resistor, we fi$t calculate the Ims value of z'(t):

0z: 90''

ltrr)'Q/4) : v900: 30v.

4=20\f2ln:9.00v, The total averagepower deliveredto the 15 O

d3 = 135" At=0,

3n2

P' = -:

or= o',

t5

d) The total power delivered by the fint five nonzerotermsis

ds:45"' 21 do: t:

b0w.

2r(1000)= a U Urt a d / s . 5

P:

Pa"+ Pr+ P2+ P3+ Ps : 55.1W 5 .

This is (55.15/60X100), or 91.9270of the total.

16.7 ThermsValueof a Periodic Function Tbe rms value of a periodic function can be exprcssed in tems of the Fourier coefficients; by definition,

; I

(16.7sJ

fl)'dt

Representing f(t) by its Fouder sedesyields

n ",-

--lt t"l

Vrl,

N

2

(16.80) >A.costndae r "t )dt.

f"

The integral of the squared time function simplifiesbecausethe ody terms to survive integration ovei a pedod are the product of the dc term and the harmonic products of the samefrequency.All other products integrateto zero.ThereforeEq.16.80reducesto - ?

' /

\

;\dr + >t,All t:4 sl:1l

I

?'\\,2.)

'

(16.81)

Equation 16.81states that the rms value of a periodic function is tlle squareroot of the sum obtainedby addingthe squareofthe rms value of each harmonic to the squa.reoI tle dc value.For example,let's assumethat a periodicvollageis fepresenled by lhe linileseries r' - 10 + 3Ocos(rrd- 01)+ 2'cos(2aot - a2) + 5 cos(32,,0, 93) + 2cos(5dor- 05). The rms valueof this voltageis

1G + Qol\/-/)' + (2V\r42 + (5/\"r2)2+ (2/^"2\2

\,564s: n.65v.

16.8 TheExponentiat Fornrofthe FourierSeries 679 Usually,infinitely many terms are requiredto representa pedodic function by a Fourier series,and thereforeEq. 16.81yieldsan estimateof the true rms value.We i ustratethis resultir Example16.5.

Estimating the rmsVatueof a PeriodicFunction Therefore,

UseEq.16.8110estimatethe lms valueof the voltage h Example16.4.

Solution

,v, ' n [ ^ , 1 3 1 -1 1 ' u r , 1 ' - r i o o ] ' - , r . , o l ' V- \rr/ \'r/-\'z/'\':/

FromExample16.4, Ydc: 15V, U = 21.01/\,4v,

= 28.76v.

the Ims valueof the fundamenlal From Example 16.4,the true rms value is30VW.3 a p p r o a crhh i s v a l ube\ i n c l u d i n g m o r e am nd oreharmon l n ( r m s \ a l u e o l h e l h r f dh a r m o n r c i ; . i n F q . r b . 8 r . F oeri a m p l e . i f w e i n c l u d e r h e h a r m o n i c i throughk : 9, the equationyieldsa valueof 29.32V the rms valueof the fifth harmonic.

V2= 19.101^'ay, the rms valueof the secondharmonic, v -quu v2v V. = 5.401",4V,

NOTE: AssessJow understandins ofthb materisl by trying Chaptet Prcblems 16.36and 16.37.

Form 16.8 TheExponential of the FourierSeries The exponential form of the Fourier series is of interest becauseit allows exDonentialform ofthe seriesis us to exDressthe seriesconciselv.The

f(t) = > c"et*"t,

c,-;

(16.82)

1 t4'r fo)e-nbr'dt I

(16.83)

To deriveEqs.16.82and 16.83,we return to Eq. 16.2and replace thl] cosineandsinefunctionswith their exponentialequivalents: (16.84)

2

(16.85)

2j SubstirutingEqs.16.84and 16.85into Eq.16.2gives

f(t) = '" + ;?G,^' , e, '1, \1",'"

= ".

n, 1

* ('" +rjb.)",^, oa.wt 2(";'o^)r.^,

Now we define C" as - tt^

.

"4

A i^\ -

t"n,

) "n

'-/

ab

n _ t.2.3.

(16.8i)

) 4

Fromthe definition of c", tf)

c.

t'ot

+l+ z L t JIn :;

1 fh+r

I , Jr, 1

=+ I r

tttr+T

Jtr

r

r"

I - i', It h ftrtsin r-,rdrI tutco,nunrdr tJt I

f1)@osn..iat- isinna{)d.t

ft)e i""'rdt,

(16.88)

which completes the deivation of Eq. 16.83.To complete the derivation of Eq.16.82,wefiist observeftomEq.16.88 that co=;

1 ftr+T ,,i(t)dt = a.. I I

(16.8e)

Jtt

Next we note that C-.'

I t " ' I - ib,t. Jtttc"-d d! C) - :ta. -r I

(lo.eo,

J,,

yields 16.89, and16.90inroEq.16.86 Substituting Eqs.16.87, f (r) = Ca + >C"d^r

+ Cie j"'"t)

= >Creitu] + >C,iitu|.

(16.e1)

Note that the secondsummationon the dght-handside of Eq. 1691 is C,e/t'dfrom 1to-c<,;thatiq equivalent to summing >c'netnad = >cnenb&.

(16.e2)

Becausethe summationfiom 1 to -co is the sameas the summation ftom o<,to 1,we useEq.16.9 to rewriteEq.16.91: -

1

f (t) = >CFt^c + >chetnad n=0 = >c"er*4, which completes the derivation of Eq. 16.82.

(16.e3)

'6.8

rormoI lhe loJderSe:e. T\e kponeoridl

681

We may also expressthe Ims value of a pedodic function in terms of the complexFourier coefficients. From Eqs.16.81,16.87,and 16.89,

(16.e4)

\/a;-4 Ld :

d;'

(16.95)

(16.96)

Substituting Eqs. 16.95 and 16.96 into Eq. 16.94 yields the desired

(16.97)

Example 16.6 lllustrates the process of finding the exponential FourierseriesreDresentation ofa Deriodicfunction,

findingthe Exponential Formof the FourierSeries Find the exponential Fourier sedesfor the pedodic voltageshownh Fig.16.15.

= J:!-\c

t"-r, 2 - et"u, 2)

= 2U-. n-o'1t n-Psln

-rl2

0r/2

T ./2 T T+r/2

Figure16.15Alhe pedodicvottage forExample 16.6.

Here,because o(t) hasevensymmery,bn = 0 for all n, and hencewe expectC, to be real. Moreover,tbe amplitudeof C, foljowsa (sin.v)/.r indicated distribution,as whenwe rewrite -

Lr:

Solution Using ,2 as the starting point for the integiatior\ we have,from Eq.16.83, a

=11

vh T

v a-in-vnl

sln (naor/2) V-r ----------t-

-1

nanrtL

We saymore about this subjectin Section16.9. represenralron The exponenl'alseries oi rC) is

44\'in ("0"/2)/",," ,,,,, ' = S / T t na\rtt2 ,,="-\ = 1u_,) S \ r 1,1

sjn(nott/2r,^u ' n-or/2

682

objective4-Be abteto catcuhtethe exponential formof the Fouriercoefficients for a periodicwaveform 16,8 Derive the cxlressionfor the Fburier coclli cienisCi fbr the pedodicfunction shown. llirr:Takc advanlageof symmetryby usjngthe fact that C, = (a" - jb")i2.

16,9 a) Calculatethc rms valLreof the periodiccnrreni in AsscssmentProblem16-8. b) Using Cr-Cn, estimatethe rms valuc. c) What is the perccntageo[ error in the value obtainedin (b),basedon thc true value found in (a)? d) For this periodicfunction,couldfewer terms be usedto estimatethe rms valueand siill insurcthc error is lessthan l%?

I (ns)

Answer: (a) \'/34 A; (b) s;7'7'7 Al -0.93 (c) %! ( d ) y e . : i r( I ( e a r eu \ c d . l h e f f o fi . 0.9870.

Answer: C, = -j-!(1 +3cos?).nodd NOTE: Ako trt ChapterProblens 16.11and16.45.

16.9 Asnglituete andFiras*Spe*trs Aperiodicrime functionis detiiedbyits Fouriercoefficientsand ils period. ID other words.$,hcnwc know d,, d,, 0,,,and Z. we can construcl/(r), at leasttteoretically.When we know d,,and b,,,we alsoknow the amplitude (A,) anclphaseanglc( d,,)of eachharmonic.Again,wc cannol,in genelal. visualizewhat thc pedodicfunctiontooks like in thc lime domair from a description()1lhe coefficienrsand phaseangles;ncvcrlheless, we recognize that thesequantiliescharacterizeihe periodic funclion oompletelnThus. rlith sufficicntcomputingtime, we can synthesizcthc lime domain wavelorm from thc amplitudeand phaseangledata.Also,whena periodicdriving tunction is exciiirg a circuit that is higtrly hequeDcyseleciive.the Foudel seriesof tlle stead\,-statc rcsponseis dorninatedbyjust a fe$Jtcrms. Thus the descriptionof the rcsponsein tems of anplitude and phascmay providean uDderstanding ofthc outpul $aveform. We can presentgraphicallythe descriptionof a periodic function in tcrnrsor the anpiitude and phascanglcol eachterm in the Fourier series of ft).'Ihe plot of the amplitudc of cach term versusthe frequencyis callcdthc lmplitrde speclnlmof/C), and lhe plot oI the phaseangleversusthc frequercyis calledthe phasespectrumol f(r). Becausethe amplitudc and phaseangledata occur al disclelevaluesof the frequency(that is,at rr0.2rd0.3o0, . . .),lhcsc plols alsoare Ieferled to asline spectra.

An lltustrationof AmplitudeandPhaseSpectra Amplitude ard phasespectraplol\ are basedon either Eq.16.3E(,4,,and 0,) or Eq.16.82(C,,).we focuson Eq.16.82aDdleavethe plots basedon Eq. 16.38to Problemi6.4ll.To illuslrale the amplitudeand phasespcctra,

16.9 AmpLitude andPhase Spectra 683

which are based on the exponential form of the Fourier sefes, we use the pe odic voltage of Example 16.6.To aid the discussion,we assumethat V- = 5Vandr = f/5. Frcm Example16.6, (^ r :

sin(naaal2)

-vhr ; ---= I noar/z

(16.98)

which for the assumed values of y- and r reduces to

(r6.9e) Figure16.16illustrates theplot of themagnitude ofC, fromEq.16.99lor valuesof r rangingfrom -10 to +10.The figue clearlyshowsthat the ampliludespectrumis boundedby the envelopeof the (sinr)/.rl function.Weusedthe orderof the harmonicasthe frequencyscalebecausethe numedcalvalueof Z is not specifled.lvhen we know I, we alsoknow o0 andthe trequency correspo0ding lo eachharmonic. Figure 16.17providesthe plot of ( sinr)/rl versusr, wherc r is in mdians It showsthat the function goesthrough zero wheneverr is an Flglre16.16A TheptotofC, versus u whenr = I/5, irtegral multiple of n'. FromEq. 16.98, for thepenodjc vottaqe for Exampte 16.6.

"-.o

(16.100)

From Eq. 16.100,we deduce that the amplitude spectrum goes ttrrough zero whenever n4r is an integer. For example,in the plot, "/Z is 1/5, and thereforethe envelopegoesthrough zero at n : 5, 10,15, 10, 15,and so on. In othor words, the fifth, tenth, fifteentl,... harmonics are all zero. As 2n l.sn -n 0.5n0 the reciprccal of /f becomesan increasingly larger integer, the number of harmonics between every ,7 radians inueases If nr/f is not an inreger, (sinr/, versus 16.17 lhepLotof '. the amplitude spectrumstill follows the l(sinr)/xl envelope.However, Fiqure the envelope is not zerc at an integral multiple of oo. Because C, is rcal for all ,, the phase angle associatedwith Cn is either zeroor 180",dependhgon the algebmicsignot (sin .''l5)/(n1tl5). For example.lhe phaseangleis zero for /? - 0. - l. +2. -3. and -14.Il i< not defined at ,r = +5, becauseC+sis zero.The phaseangle is 180' at n = t6, t7, t8, and +9, and itis not definedat +10.Thispattem repeaK itself as,r takes on largei integei values.Figure 16.18showsthe phase angleofC, givenby Eq.16.98.

180' 90' 1 5 1 3 1 1 9 7 5 3 1 -74 -72 -1,0 -A 6 4 2 Flgure16.18A Thephase angteof C,.

3 5 7 911 13 15 2 4 6 8 t O 7 2 1 4t 6

684 Now' what happcns 1o lhe amplitude and phase speclm if t(t) is shifted along the time axis?To Iind out, we shift the periodic vollagein E r a m p l (1 6 . 6 , un n l \ t o r h . r i g h l B ) h ) p o r h e . i \ .

,,()

: s

(16.101)

Therefore

,-(r

to):

>

C,p)no\( ti =

t

C,,e t"qt'Ej"qt,

(16.102)

which ildicatesthat shiftingthe odgin hasno effecton the ampliludespec-

(16.103)

However. referenceto Eq. 16.87revealsthat the phase spectrumhas changcdto (An+ no0t0)rads.For example,lel's shift the periodicvoltage in Example16.1r/2 units to thc right.As before,we assumethat r = 7/5: thcn the new phaseangle81,is

ai=

( 0 h +n r / s ) .

(16.104)

We havepio$cd Eq. 16.104in Fig.16.19fou rangingfron -8 ro +8. Note that no phascangleis associated with a zeroamplitudecoefficient. You may lvonder why we have devoted so much attenrion ro the amplitudcspeclrumof the periodic pulsc in Example 16.6.The reasonis that this partic! ar periodicwaveformprovidesan excellentway io illus irate thc transitionfrom the Fouricr seriesrepresentationof a periodic lunction to the Fouder transfom representarionof a nonpcdodicluncFigure16.19r! Theptoiof0;,versus n for Eq.16.104'. tion.We discussthe Fouricr translonnin Chaprer17. "

objettive4-Be ableto calc!latethe exponential forrnof the Fouriercoefficients Jora periodicwaveform 16.10 The fuDctionin Assessment Problem16.8is shifl
16.48and 16.49.

r(t:+ :

Lg * 2.""[V

tjr/2xn+teiha{A.

5unmary 685

Summary A periodicfunction is a function that repeatsitself every Z seconds,

. A period is the srnallesttirne interval (f ) that a periodic functioncanbe shiftedto producea functionidentical to itser. (Seepage656.)

. The Fourier seriesis an infinite se es used to represent a periodic function.The seriesconsistsof a constant telm and infinitely many harmonically related cosine and sineterms.(Seepage658.)

The fundamental ftequency is the frequency determined by the fundamentalperiod (/0: l/T or an:2tf). (Seepage6s8.)

The harmonic flequency is an integer multiple of the fundamentalfrequency.(Secpage658.)

The Fo$rier coeffici€nts are the coNtant term and the coefficientof each cosineand sine term in the series. (SeeEqs.16.3-16.5.) (Seepage659.)

Five typ€s of symmetry are used 1{)simplify the computation of the Fouier coefficienls: . ?r,€&in which all sine terms in the seriesare zero . od4 in which all cosineterms and the constantterm . halil'aye, in which all even harmonics are zero

. ln the alternativeform of the Fourier sedes,eachharmonic represenledby the sum of a cosine and sine term is combined into a single term of ihe form A, co(".rol 0,). (Seepage668.)

For steady-stateresponse,the Fourier series of the response signal is determined by first finding the responseto each componentof thc jnput signal.The individual rcsponsesare added (super-imposed)to form the Fourier series of the responsesignal.The rcsponseto the individual lerms in the input se es is found by €itherftcquencydomain orr domain analysis. (Seepage670.)

The waveform of the respoNe sigml is difficult to obtain without tlre aid of a computer. Sometimesthe frequency respoNe (or filtering) characteristics of the circuit can be used to ascertain how closely the output waveform matchesthe input waveform.(Seepage672.)

Only hamonics of the samefrequencyinteract to produceaveragepower.The total averagepowerls the sum of the averagepowersassociatedwitl each frequencl (Seepage675.)

The rms value of a periodic function can be estimated (SeeEqs.16.81,16.94.and lrom the Fouder coefficients. 16.97.)(see pase678.)

The Fouier series may also be written in exponential form by usingEuler'sidentity to replacethe cosineand sine terms with their exponential equivalents.(See pase679.)

' qusrter-wave, hslf-wave, ewn, in which the series coDtainsonly odd harmoniccositreterms . qudnprwav?.hau-wavp,odd. in which thc serie.containsonly odd hamonic sineterms (Seepase662.)

The Fourier series is used to predict the steady-state responseof a systemwhen the systemis excitedby a periodicsignal.The seriesassistsin finding the steadystateresponseby transferringthe analysisfromthe tjme domainto the frequencydomain.

Probtems FiguePr6.2

Sedions 16.1-16.2 16.1 For each of the periodic functions in Fig. P16.1, specfy a) o, in radiaN per second b) /o in hertz c) the valueof z. d) the equations for at and bt e) ,(t) as a Fourier series

Figur€ P16.t

r@t

16.3 Dedve the Fourier sedes for the periodic voltage shownir Fig.P16.3,giventhat

T

ott| : 20ocos= tv .

T

I

r (/,r)

D(r) -

lOOcos- lY. I

T 4

3

Flgure Pl6.3

r,(r)v

16.2 Find the Fourier series exprcssionsfor the peiodic voltage functions shown in Flg. P16.2. Note tlat Fig.P16.2(a)illustrates the squarewave;Fig P16.2(b) illustrates the full-wave rectified sine wave, where a(t) : Uhsir\Qr/:l)t,0 < I < 7; and Fig. P16.2(c) illustrates the half-wave rcctified sine wave, where D(t) - Vnsin(zrlT)t,o < t < T12.

T 3T/4 T/2 T/4 0 T/4 r/2 3r/4 r

T 4 '

Probtems 687 16.4 Derive Eq.16.5.

a) What is the fundame[tal frequency in hertz? b) Is the function even?

16.5 a) Vedfy Eqs.16.6and 16.7.

c) d) e) f)

b) Vedfy Eq. 16.8.Hirf Use the tdgonometric identity cosd sin B : ; sin(a + B) -;sin(a B). c) VeriJy Eq. 16.9.Hintr Use the trigonometdc identity sin a sin B : ;cos(d - B) ; cos(a + B). d) Veffy Eq.16.10.flinr Use the trigonometric identity cosa cosB = icos(a B) +;co(d + B).

Secrion 16.3

l6.tl

16.6 Derive the expressionsfor the Fourier coefncientsof an odd periodic function. Itirrl. Use the same techque asusedin t})e text in deriving Eq$ 16.14-16.16. 16.7 Show t]lat if /0) : /(r - ?/2), tlle Fourier coef, ficients bt are given by tle expressions

Is the function odd? Does the function have half-wave symmety? Does the function have quarter-waves)'mmetry? Give the l1ume cal expressionsfor a!,4*, and bt.

It is given tlat o(r) = 20rcos0.25ztv over the inte al -6=t=6s. The function tlen repeatsilself. a) What is the fundamental frcquency in radians pei second? b) Is the furction even? c) ls the tunction odd? d) Does t]le functior have half-wave slmmetry?

bk=;.lo

fO sink,D,tdt,

for t odd.

Airrt: Use the same technique as used in the toxt to dedveEqs.16.28and 16.29. 16.8 Dedve Eqc 16.36. Hint: Start $.irh Eq. 16.29 and divide ttre inteflal of integration into 0 to Z/4 and ?/4 to f/2. Note that becauseof evennessand quarter-wave symmetry,f(t) = -f(T 12 - t) nt the mtelvalT/4=t
i(,) : 4000r A,

0 < r < 1.25ms;

(t:sA,

1.25ms < t < 3.75ms;

i0) = 20 - 4000tA,

3.75ms < l < 6.25ms;

i(t) = -s .q.,

6.25ms = 1< 8.75ms;

i(t) = -40 + 4000, A

8 . 7 5 m=s 1 < 1 0 m s .

16.12 Find the Fouder series of each periodic function showr)in Fig.P16.12.

rigureP16,12

v -T

T/2 0

-v

r / 2l r

16.13 a) Dedve the Fourier series for the pedodic volF agesho$n in Fig.P16.13. b) Repeat (a) if the vertical ieference axis is shifted f/2 units to the left.

16.14 tt is given ttrat f(t) = o.al -5 < r < 5 s .

4

rigureP16.15

fo

owr the inteflal

Construct a pedodic function that satisfies this /(t) between 5 and +5 s,has a period ol20 s, and has half-wave s)'mmety.

b) Is the function even or odd? Does the function have quarter-wavesymmetry? Derive the Fourier sedesfor /(/). Wite the Fourier sedesfor /0) if /(t) is shifted 5 s to the right.

16.15 RepeatProblem 16.14given that /(t) = 0.4t3over t h ei n t e r v a l 5 < t < 5 s .

16.16 The periodic function shown in Fig. P16.16is even and has both half-wave and quafier-wave symetry a) Sketch one ful cycle of the tunction over the idrtewal -T/4 < t < 3T/4. b) Derive the expression for the Fourier coeffi' c) Wdte the first thiee nonzeio terms in the Fouder expansionof f(l). d) Use the tust tbree nonzero terms to estimate

f(r/8).

16.17 It is somerimespossible to use syrnmetry to find the Fouder coefficientq even though the original function is not symmetricall With this thought in mind, considerthe function in AssessmentProblem16.1. Observe that o(/) can be divided into the two functions illustrated in Fig. P16.17(a) and (b). Furthermore, we can make o2(l) an even function by shifting it Z/6 units to rhe right.This is illustrated in Fig. P16.17(c).At this point we note that o(t) = r,(0 + o2(t) and that the Fourier seriesof o1(t) is a single-tem seriesconsisthg of y-. To find the Fouiier se es of o2(0, we first find the Fourier sedes of o2(t - T16) and then shift tlis series f/6 units to the left. Use the technique just outlined to veriJy the Fourier sedes given as the answer to Assessment Problem16.2(e).

FigurePl6.13

sT/4 r 3T/4 Tl2 T/4 0 r/4 T/2 3r/4 T 5T/4

PmbLems 689 FlgurcP15.17

16.19 For each of the periodic tunctions in Fig. P16.1, derivethe Fourier sedesfor ?r(t)ushg the form of Eq.16.38.

L 2 L r ! L 5 ! 3

3

3

1,7

3

16.20 Dedve the Fourier series fot the periodic function described in Problem 16.10,using the form of Eq. 16.38.

16,21 Derive the Fourier seriesfor the periodic function constucted in Problem 16.14,using the fom of Eq.16.38. T 2 T T 4 T 5 T 2 T )V^

a

t

r

t

I

S€ction 16.5 rz(t

T/6)

16.22 Dedve Eq$ 16.69and 16.70.

16.23 a) Derive Eq. 1.6.71.Hint: Note tlat ,j. 4VJrk + ka,Rcak. Use this expressionfor bk to find a7+ b? in terms of ar;.Then use the explession for /'t to deriveEq.16.71. b) DeriveEq.16.72. Section 16.4 16,18 a) Dedve the Fourier series for the periodic function shom in Fig. P16.18 wher y, : 3782 V. Write the seriesin the form of Eq. 16.38. b) Use the fircl five nonzero terms to estimate og/8). Figure PI6.18 ,(r)

16.24 Showthat when we combhe Eqs.16.71and 16.72 witl Eqs.16.38and 16.39,the rcsultis Eq. 16.58. Ilrtr Note ftom the definition of Bk that a b*:

^ -k rarrPk'

and from the definition of r9tthat tan9k: -cotBk Now use the tigonometric identity

tanr : cot(90- r) to showthat0e= (90 + Bk).

16.25 a) Show that for large values of C, Eq. 16.67can be approximated by the er?rcssion

' /rl:

u-f 4RC

25 rnH

u+rr

RC'

Note that this expression is the equation of the triangular wave for 0 < t < I/2. Hilrts, (1\ Let e 4Rc x 1- (t/RC) and arpRc = 1 - QI2RC1; Q) put the resolting expression over the common delominator 2 - (T/ZRC); (3) simplify the numerator; and (4) for large C, assumethat r/2RC is much lessthan 2. b) Substitutethe peak valueof the triangularwave irto the solution for Problem 16.12 (see Fig. P16.12(b) and show that the result is Eq.16.59.

(b)

16.28 The periodic square-wave voltage described in tstrc Assessment Problem 16.6 is applied to t]1e circuit shownin Fig.P16.28. a) Derive the filst four nonzero terms in the Fouder seriesthat representsthe steady-state voltaget),.

16.26 The square-wave voltageshownin Fig.P16.26(a)is tstrc appliedto the circuit shownin Fig.P16.26(b). a) Find the Fourier series representatjonof the steady-state currenti, b) Find the steady-stateexpr€ssionfor i by straight forward circuit analysis,

b) Which frequency component in the input voltage is eliminated from the output voltage? Explain why.

Figure P16.28 20mH

figureP16.26

(a)

1629 The full-wave rectified sine wave voltage sho\rn rn 1627 The periodic squarewave vollage seen in Fig. P16.27(a) is appliedto lhe circuirshown iD Fig.P16.27(b).Derive t}le first three nonzeroterms ir the Foudei seriestlat representsthe steady-state voltage ?,, if U^ = 601rV and the period of fte input voltageis ?' ms.

Fig. P16.29(a) is appliedlo lhe circuil shown in Fig.P16.29(b). a) Find the first four noizero terms in the Fouder se es representationoft]o. b) Doesyour solutionfor oomake sense?Explair.

Pmblems691 Flgore P15.29 ,s(V)

Figure P16.31

10ko

l0H

Section 16.6

+ {

5ko

2.5rrF

(b)

16.30 The periodic current shown in Fig P16.30(a)is used to energize the circuit shown in Fig. P16.30(b)Wdte the time-domain expression for the fiftlharmonic current in the exptession for io-

16.31 A periodjcloltage hariog a periodol 0.Ir me i. given by the following Fourier sedes: \

l"z _- g =.+),=Ptr. t

8. nr srn tcos

1632 The periodic voltage across a 81 q 2 kO resistor is shownin Fig.P16.32. a) Use tlle first four nonzero terms in the Fourier series representationof o(t) to estimate the averagepower dissipated in the resistorb) Calculate the exact value of the average powet dissipatedir the 81 n2kO resistor. c) What is the percentage error in the estimated valueofthe averagepower dissipated?

Figure P16.32

v(/)v

nd'l v'

This periodic voltage is applied to the circuit shown in Fig.P16.31.Find the amplitudeand phaseangle of the component of o, that has a lrequency of 300krad/s.

Figure P16.30

(b)

692

FourierSeri€s

1633 The tdangular-wave voltage sourceis applied to the circuit in Fig. P16.33(a).The triangular-wave voltage is shown in Fig. P16.33(b).Estimate t}le average power delivered to tle 20 kO resistor when the circuit is in steady-state operation.

Section16.? 16.35 The voltage and current at the terminals of network are + 45') + 60sin1500r o : 80 + 200cos(5001 V, j = 10 + 6 sin(500t+ 75') + 3cos(1500t + 30")A.

Figur6 P15.33 20mH 50pF

20 ko

The current is in the direction of t]le voltage drop acioss t]!e terminals. a) Wflar is $e averagepower at t}le terminals? b) whal is lhe rms\alueof rhevollage? c) Wlat is t]le Ims value of the curlent?

(a)

'" (v)

t 0,t

16.36 a) Find the rms value of the voltage shown in Fig. P16.36 foi V-:100V. Note that the Fourier series for tlfs periodic voltage was found in AssessmentProblem 16.3. b) Esrirnate the rms value of the voltage, using ttre lirst three nonzero terms in the Fouder sedes representationof or(r). Figure P16.35

16.34 The periodic currcnt shown in Fig. P16.34is applied toalkOresistor. a) Use the fiist tlree nonzerc terms in the Fouder sedesrepresentation of i(, to estimate the average power dissipated in tle 1 kO resistoi. b) Calculate the exact value of tlle average power dissipatedin the 1 kO resistor. c) What is the percenrageof error in the estimated value of the avemge power?

figoreP16.34

t(mA)

16.3? a) Estimate the rms value of the periodic squarewave voltage shown in Fig. P16.37(a) by using the [irsr five non?eroterms in rbe fourier series representationof ?,(1). b) Calculate the percentageof error in the estimation if

% " * " restimated : I value

r]

roo.

c) Repeat parts (a) and (b) if the periodic square\aa\e \ ol(ageis replacedby the pefiodicl riangu lar voltaseshownin Fig.P16.37(b).

Pobtems693 Figure P16.37 2 (v)

c) U<ethe firsr tourDonzerorermso[ t-bee\pression derivedir fb) ro eslirnarelhe rms\alue otis. d) Find the exact rms value of is. e) Calculate the percetrtage of elror in the estimated rms value,

?,(v)

t (ns) Figure P16.39 i,

16.38 a) Estimate the rms value of the full-wave rectified sinusoidalvoltage shown in Fig. P16.38(a)by using the fiIst three nonzero terms in Ure Fourier seriesrepresentation of o(r). b) Calculate the percentage of error in the estimation (seeProblem16.3?). c) Repeat (a) and (b) iJ the tull-wave rectfied sinusoidal voltage is replaced by the half-wave recti fied sirlusoidal voltage shown in Fig. P16.38(b). Flg!reP16.38 1'(v)

16.4'0 a) Use the first four nonzero terms in the Fourier sedes apprcximation of the periodic voltage sbownin Fig.P16.40ro esrimale its rms!alue. b) Calculate the true lms value of !e voltage. c) Calculate the percentage of error in the estimatedvalue, t (n0 FigureP16.40

vs z'(v)

0

-v_/2 16.39 a) Derive the expressionsfor the Fouier coetfrcients for the periodic flrrent showl in Fig. P16.39. b) Write the first four noMero terms of the seies using the alternative trigonometric folm given by Eq.16.39.

-v-

T/8 T/4 3r/8r/2 5T/837/417/8 r

t

l

tL__l[ - lL__ll

694

Fouder Senes

t6Al Assume the periodic function described in Problem 16.16is a current is with a peak amplitude of2 A. a) Find the Ims value of the curent. b) If this current exists in a 54 O resistor, what is the averagepower dissipated itr the resistor? c) If ts is approximated by usiry just the fundamental frequencytem of its Fourier sedes,what is the averagepower delivered to the 54 O resistorT d) what is the percentageof error in theestimation of the power dissipated?

Sedion16.E 16.43 Use the exponentialfolm of the Founer senesto write an expressionfor the voltage shown in Fig.P16.43.

Figure Pr6.43 1,(,)

16,42 The Ims value of any periodic triangular wave having the form depicted in Fig. P16.42(a)is indepedent of ta and lr, Note tlat for the function to be '[1ae single valued, tn < tb. mrs value is equa] to vpl\,4. venry dtis obse ation by finding the fim value of the ttrrce waveforms depicted in Fig.P16.42(b)-(d).

Figure P16.42 lJ (v)

? (v)

x,(v)

? (v)

Probtems 695

16.,9 Derive the er?ression for ttre complex Fourier coefficients for the periodic curent showr in Fig.P16.44.

rigneP15,45 50ko 100mH =

6.25nF

Figure Pt6.44 (,

T/2

3r/4

16,45 a) The pe odic cunent it Problem 16.44is applied to a 60 O resistor.If1- = 20 A what is the averagepower delivered to the resistor? Assume (0 is approximatedby a truncated exponential folm of the Fourier seriesconsisting oI rhe firsl se!en nonlero lefms, lhal i\ n : 0, 1,2,3, 4, 5, 6 andT.Wlat is t}le rms value of the current, using this approximation? c) ff the approximation in part (b) is used to represent i what is the percentageof errcr in the calculatedpower?

16.47 a) Fhd the rms value of the periodic voltage in Fig.P16.46(b). b) Use lhe compler coelficienrsdefi!ed in Problem 16.46(b)to estimate the rnts value of os. What is the percentageol eror in the estimated rms value of zrs?

o

Section 16.9

16.46 The periodic voltage sourcein the circuit shown in Fig. P16.46(a)ba. rhe \ aveform sho$n in Fig.P16.46(b). a) Derive the expressionfor Cn. b) Fird t.he values of t]le complex coefficients C", C-L C1,C 2, C2,C4, C3,C a, and Cafor t}le input voltage os if y- = 72tV an T = 50t ps. c) Repeat(b) for oo. d) Use the complex coefficients found in (c) to esri mate the avenge power deliveredto the 200kO resistor.

16.48 a) Mal@ an amplitude and phase plot, based on Eq. 16.38,for the periodic voltagein Example 16.3. Assume y- is 40 V Plot both amplitude and phase versusfur., wheie,r = 0, 1,2,3, . . . b) Repeat(a),but basethe plots on Eq.16.82.

16.49 a) Make an amplitude and phase plot, basedon Eq. 16.38. for the periodic volrage in hoblem 16.32.Plot bolh amplitude and phase v e n u s n o o w h e r e:n 0 , 1 , 2 , . . . b) Repeat(a),but basethe plots on Eq.16.82.

696

16.50 A pedodic voltage is rcpresentedby a huncated Foudersedes.The amplitudeandphasespectraare shownin Fig.P16.50(a) and(b),rcspectively. a) Write an expressionfor the pedodic voltage usingtheform givenby Eq.16.38. b) Is the voltagean evenor odd function of 1? c) Doestle voltagehavehalf-wavesymmetry? d) Doestlre voltagehavequarter-wavesynmetry?

Figur€ P15.51

(17.64\

Figure P16.50

0 9Cf

Sectiotrs16.1-16.9 16.52 The hput sigral to a third-order low-pass Butteiworth filter is the periodic triangular-wave voltageshownin Fig P16.52.The cornerfrequency of the filter is 1 rad/s.Wite the first three termsof the Fou er se es that represents the steady-state output voltageof ttre filtei.

16.51A periodic function is representedby a Fourier seriesthat has a finite number of terms.The amplitude and phasespectraare shownin Fig.P16.51(a) and (b), respectively. a) Wdte the expressionfor the periodic current usingthe form givenby Eq. 16.38. b) Is the current an even or odd function oI t? c) Doesthe curent havehalf-wavesymmetry? d) Calculate the rms value of the curent in milliamperes. e) Wfirerheexpooential formof the Fourierserie'. f) Make the amplitude and phase spectra plots or the basisof tlre exponential series.

figureP16.52

\

(.n

/

il \

V;I V

, / t

rG)

Probtems 697 16.53 The input signal to a second-order low-pass Butterworth filter is a fu[-wave rectified sine wave with an amplitude of 2.5?rV and a fundamental ftequency of 5000 md/s. The comer ftequency of the lilter is l krad/s. Write the first two rerms in th€ Fourier series that rcpiesents the steady-stateoutput voltage of the filter.

16.54 The transfer funcnon (VIV) for the narrowband bandpassfilter cftcuit in Fig.P16.54(a)rs

f+Bs+a!,'

4

Find (,, B, ard 03 as functions of the circuit palametersR1,R2,R3,C1,and C2.

b) Wdte the first thee terms in tle Fourier senes thatrepresents ,, if rs i\ the periodic\ollagein Fig.P16.54(b).

Figure P15.54

Transform TheFourier 17.1 Th€ D€rivationof the Fourier Itanttorm p. 699 of the Founer 17.2 TheConvergence Lntegtal p. 741 to tind Fourier :17.3UsingLaplaceTransfonns

in the Limit p. 206 Tradsforms 7 7 . 4Fourier p. 708 Properties Someli'lathematicat p. Transforms Zlo 1 7 . 6operationat p. 214 77.7 CicuitApplications p. 717 Theorem r 1 . 8 Parseval's

of a the Fou ertransfornr 1 BeabLeto catcuLate ofthe roLtowing: functjonusirg anyor aLL . Thedefinitjorofthe Fouriertnnsform; . Laplace transforms; .

properties ofthe Fourier l4athematical

.

0pentionaitransioms.

Knowhowto usethe Fouiertransfolnto find of a circuit. the response theotemand be abteto Parsevafs Undedafd aboutthe energy useitto answerquestions bands. containedwithin specificfrequency

698

In Chapter 16, we discussedfhe representationof a periodic function by mealls of a Fou er series.This seies representation enables us 10 describethe periodic function in terms of the frequencydomajnattributesof amplitudeand phase.TheFourier transtbrmextendsthis frcquency-domaindescriptionto lul1ctions that are not periodic.Throughthe Laplacetransform,we already introducedthe idea of transformingan aperiodicfunction from the timc domainto the frequencydomail You may wonder,then, why yet another type oI transformation is necessary.Stdctly speaking,thc Fouriertransformis not a new transform lt is a special caseof the bilatetal Laplacetransform,with the real pait of the complex frequencyset equal to zero. However,in terms ol physicalinterpretation,the Fourier transformis better viewed as a limiting caseoI a Fourier series.Weprcsentthis point of view in Section17.1,wherewe derivethe Fouier transformequations. The Fourier fransformis more useful than the Laplacetrans_ form in certaincommunicationstheory and signalprocessingsltuations.Although we canlot pursue the Fourier transfolm in depth, its introduction here seemsappropriatc while the ideas underlyingthe Laplacetransformand the Foufier seriesare still freshin your mind.

17.r TheDerivation oftheFouderTnnsfom699

17.1 TheDerivation of the Fourier

Transfornn We begin the derivation of the Fourier tmnsfom, viewed asa limiting case of a Fourier series.with the exDonential form of the seriesi

r,,':

Cr:;

r " i,*,

!

(17.1)

1 ITP f(t).t'qt dt. I

\17.?)

In Eq.17.2,weelectedto startthe integrationat t0 = -?/2. A owing the fundamental period Z to hclease without limit accomplishes t]le transition from a periodic to an apedodic function. In other words, if Z becomesiniinite,.the function never repeats itselJ ard hence is aperiodic. As f increases,the separation between adjacent harmonic frequenciesbecomessmallerand smaller.In paiticular, 2n n @ o :@ o = i .

Ad=(r+I)do

(17.3)

andas f getslargerandlarger,the inuemental separationAo approaches a differentialseparationd.r. From Eq.17.3,

;---

as/l ..@.

\17.4)

As the period increases,the frequency moves from being a discrete vari, able to becoming a continuous vadable, or noo+o

as7+

co.

(17.5)

In terms of Eq. 1?.2,asthe period increases,the Fourier coefficients C, get smaller.In t}le limit, C,+0 as f +co. This rcsult makes sense, becausewe expect the Fourier coefficients to vanish as t]le functior loses its pedodicity. Note, however,the limiting value of the product C,7; tlat is,

,.,- f-r,,* j''

dt

asT -

@.

(17.6)

In writing Eq.17.6we took advantageof the relationshipin Eq.17.5. The integral in Eq. 17.6is fte Fouri€r transform of f(t) and is denoted

(17.7) < Fourierfansform

700

IheFouri€r Iransfom We obtain an explicit expressionfor the inve$e Fourier trallsfolm by invesligatingthe limiting form of Eq. 17.1as Z - co. We begin by multiplying and dividingby I:

f(t) :

,ir.,n,^"(1)

(17.8)

As f -o., the summalion appioacheshtegration, C,f -F@), naa- a,andllT - dal2r.Thus ir the limit,Eq.17.8 becomes

InverseFouriertransformts

tv)= ! [- '61"'' o-'

117.e)

Equations17.7and 17.9definetlle Fouriertransform.Equation 1?.7transforms the time-domainexpression/(r) into its correspondinefrequencydomain expressionF(o). Equation 17.9definesthe inverseoperationof transformins-F(o) nrto /(t). Let's now derivethe Fouriertransformof the pulseshownin Fig.17.1. Note that tlis pulsecorespondsto the periodicvoltagein Example16.6if we let Z + oo.The Fouriertansfolm of z'(r)comesdirectlyftom Eq. 17.7:

v(0') = putse. Figure17.1d A voltage

t:;

v,itatdt

v , ,/ ^ . . - , \

(17.10)

t /

/ o \

whichcanbe put in the form of (sinr)/r by multiplyhgthe numerator anddenominator by r.Then,

v(a) - v^r

sin or /2

.o42

.

171.1r)

For the pedodic train of voltage pulsesill Example 16.6-the expressionfor the Fourier coefficients is V^r sin ntiar 12

T

naor/z

117.12)

CompadngEqs. 17.11and 17.12clearly showsthat, as the time-domain function goes from periodic to apenodic,the amplitude spectrumgoes from a discreteline spectrumto a continuousspectrum.Futhermore, the envelopeol the line spectrumhasthe sameshapeasthe continuousspec trum.Thus,asZincreases,thespectrumof linesgetsdenserand the amplitudes get smaller,but the envelopedoesn't changeshape.The physical interpretation of the Fourier transfom y(rr) is therefore a measure of the fiequencycontent of o(l). Figore 17.2ilustrates theseobservations. The amplitudespectrumplot is basedon the assumptionthat ? is constantand Iis increasins,

17.2 TheConveqence oftheFouri€r Integnt 707

(a) ct

v(^)

Figure 17.2i Tlansition oftheamptiiud€ specbum asf(t) Ca\e,\Js roapeiodk.ld) h"t. T t-5; soe\aor pedodic (b)c, ve6us ,1a!,z/r = 10;(c)y(o) ve$us @.

77,2 TheConvergence of the Fourier Integral A function of time /0) has a Fourier transform if the integral in Eq. 17.7 €onverges.If /(t) is a $/ell-behaved function that differs ftom zero over a finite interval of time, convergerce is no problem. We -behaved implies that /(t) is single valued and enclosesa finite area ovei the range of integration. ln pmctical terms, all the pulses of finite duration ttrat interest us are well-behaved functions.The evaluation of the Fourier transfom of tle rectangular pulse discussedin Section 17.1illustrates this point. If /(t) is different ftom zero over an infinite interval, the convergence of tlle Fouder integial dependson the behavior of /(r) as i+co. A single valued function that is nonzero over an infinite inteflal hasa Fouder hansform iJ t]le integral

l-,,r,to'

exists and if any discontinuitiesin /(r) are finite. An example is the d€cayingexponentialfunction illustrated in Fig. 17.3.The Fou er transfonn of /(t) is

rot - l- r*u-''' a,: l" o"'" .' a, Figure 17,1AThedecaing exponentjattunctjon K."'u(t).

K

1t7.13)

A third important group of functions have great practical interest but do not ir a strict sensehave a Fourier tmnsform. For example,the integral in Eq. 17.7doesn'tconveryeif /(t) is a constant.Thesamecan be said if /(t) is a sinusoidalfunction,cosoot,or a step function,(!(t).These functions are of geat interest jn circuit analysiEbut, to include them in Fourier analysiq we must rcso to som€ mathemalical subterfuge.Fi$q we create a function in the time domain that has a Fourier transform and at the same time can be made arbitrarily close to the function of interest. Next, we find the Fourier ffansfom of the approximating function and ther evaluate the limiting value of F(ir) as this function approaches/(1). Last, we define the limiting value of F(d) as the Fourier fansform of /(r). Let's demonstrate this technique by finding the Fourier transform of a constant.We can approximate a coNtant with the exponential function

f(t): Ae't, e>0.

lr7.u)

gmphically. As € -0,f(t) -,4. Ficure17.4showsthe approximation The Fouriertransformof /(1) is

r{,) =

l"-et'"

i^at*

o" "n'' 0,. fo-

117.15)

Canying out ttre integrationcalledfor in Eq.17.15yields Figure17.4A Theapproxjmation ofa constant withan

A

f(d)=

E

A

2

. + '' €rla la

=

e

A

---. e'+ u'

(17.16)

The tunction given by Eq. 17.16generatesan impulse function at o : 0 as € +0. You can veriff this result by showingthat (1) F(o) approaches infinity at d = 0 as € -0; (2) the dumtion of F(o) approacheszcro as € -0; and (3) tlre area under F(o) is independentof €. The area under F(@)is the strengthofthe impulseand is

['

-"o . o- = 4,A l- ^!-

J ,€'+

-'

.lo e'+ o'

= 2*A.

\17.71)

17.3 Using Laplace Tnnsformsto FindFouner Transtorms703 In the limit. f0) approachesa constant,4, and F(r,.')approachesan impulsefuDction2r,45(o). Therefore,ihe Fouriertransfom of a cofftanr A is defhed as 2,,45(o).or i+{A} = 2tA6(a).

(17.18)

In Section 17.4, we say rnore about Fourier transforms defined through a limit process.Before doing so.in Scclion17.3r,e showhow to take advantageof the Laplacetransform!o lind the Fouier traDslorn of functionsfor which the Fouder integralconverges.

objective1-Be ableto calcllatethe Fourier transform of a fundion 17.1 Usethedefiningintegralto find theIouner transformof the following functio s: a) f0): A, -rl2=t <01 fl.):A, f(r) - 0 b ) . i ( r )= 0 , "', f(t) = te

0< t=r/2; elsewhere. r<0; t >- 0,a > 0.

nn'""', rur-,1 - \ 4. , /ll\ i

17.2 The Fouder transfom of/(t) is givenby

r(,,) - 0, r@) : 4, F(a\ : L. F(a) - 4, tq(l,) = 0,

-3
Find /(1).

"".911. 2/

1 ,.. (D) ------_ _--,

Answer:/(r) = !10 ,in:,

(s + j.r'

NOTE: Also tty Chapter Prcbtens U2 and 17.3.

17.3 lJs'ing Laplaee Transfernrs ts FinciFounier Transfcrms We can usea table of unilateral,or one-sided.Laplacetransformpairs to IiDdthe Fouriertransfbrmollunctions forwhich the Fourierintegralcon verges.The Fourierintegralconvergeswhen a1lthc potesoI F(s) lie in the left half of the s plane.Note that if I(r) haspolesin the right half of the r p^laDe or alongthe imaginaryaxis,l (r) doesnot saiisfythe constrainithat dl exists. /;ll0) The follo\,r'ing rulcs applyto the useof Laplacclransformsto find the Fouriertransformso{ suchfunctions. 1. If l(r) is zero for I < 0 , we obtain the Fouricr rraDslormof i(t) from the Laplacetransfom oI/(t) simplyby replacingr by lo.Thus

?tIf (t)j : :t I f (t\j,= j".

(17.19)

,t .in 24.

For example,say that

.f0) - 0,

t<0-;

e'cosaot,

f(t\:

t>0'.

Then

I

otlltt\t-

|

=-

i - ,

1s- al r ",11.t., \ju I al I ..li' 2. Becausethe raDgeof integration on the Fourier integral goes from oo to +,ro, the Fouder transform of a negative-time function exists A negativelime futrctiotr is noizero for negative values of time and zero for positive values of time. To find the Fouder tmnsfolm of such a function, we proceed as follows. First, we reflect tle negative-time function over to the positive+ime domain and then find its one-sided Laplace tmnsform. We obtain the Fouder tlansfolm of the original time functiotr by replacing r with -io. Therefore,when/(t) = 0 for t > 0*,

s{f (r)J: e{f ( t\\ F t_.

(17.20)

For example,if

.f(r): o,

(for r > 0-);

fQ) : e'cosaat,

(fort=0).

/(-r) : 0,

(fort<0);

then

/(-t):

t"cosoor,

(for t > 0*).

Both /(t) and its milror imageare plotted in Fig.17.5. The Fouder transfom of /(t) is

s {f (t)} : e {f (-t\},=_i.

Figue 17.5 a Th€r€fl€ctjonof a negatjve'time functjon overto the positive-time domain.

r+a I

1 s +o)' a .+1 ,'ll ,+",fr1"=1.

(- ja + a:)2+ b4' Functions that are nonzero over all time can be resolvedinto positiveand negative{ime tunctions.We use Eqs 17.19and 17.20to find the

u.3

Using LptaceTnnsb'ms to FindFouderTcnsfons705

Fourier traDsford of rbe posilive- and mgative,time functioDs,respecI ively.The Fourier ranslorm ol the orighal functioD is tbe surnof the two transfoms, Thus if we let

/"-(r)=/(r) (forr>0). f (t): f (t)

(for r < 0),

f(t)=f+(t)+.f-(t) and

s{f (t)} = s{fr(t)} + slf (t)} - Y { / - r l ) } , . /-. r t [ - \

t t ] " -p .

0;.2,

Atr exampleof usingEq. 1?.21involvesfinding the Fouder transform of e-". For the original function;the positive andnegativetime functionsare f*(t) = "-"

and f (t) = dt.

g t l ' ( r )| = : I r +a' gtl

1

(-t\l -

fterefore.tromEq.17.21.

..

st?."|'tt

r I !-

r I ____Ll

|

als_p s+al!__j-

1 ja+a

l -j6,+a

=7+7' If /(t) is even,Eq. 17.21reducesto oilflt\\

:


! . t J-t t ,

', ')l - ;-'

lL7'221

If l(t) is odd,then Eq. 17.21becomes

s{f(t)} : s.If @}"=i.- s{f (t)},- i,.

,17.23')

706

objective1-ge ableto catcutate the Fourier transform of a function 17,3 Find the Fouriertansform ofeach function.In eachcase.,i. a po\iri\erealcoo\laol.

4 /G):0,

.f(r)= e'tira,ot, b) f(t) = 0.

t>0. r>0,

c) f(4 - te-'',

r>0, r<0.

Answer: (a)

(a+jat),+a&' 1

,-.

NOTE: Abo try Chapter Prcblen 17.1.

17.4 Fourier'!-ransfornns in the i-intit As we pointed out in Section17.2,lheFouriertransformsof sevcralprac tical functionsmust be definedby a limit process.Wenow return to these typesoffunclions and developLheirtransforms.

TheFourier Transform of a Signum Function We showed that the Fourier transform of a constanlA is 2r,46(0) in Eq. 17.18.Thenext function of interestis thc signumfunction,defined as +l lbr t > 0 and -1for t < 0.The signumfunctionis denotedsgn(1)and canbe expressedin termsofunit'stcp functions,or sgn(r)

sgn(t)= (t) -,(-r.

\r7.?4)

Figure17.6showsthe functjon graphicaily. To find the Fourier lransformoI lhe signumfunction.we first crcatea Iunctionthat approachesthe signumfunctionin the limit:

sgn(i)= liqk ''u(t) FiguE17.6A Thesignum functjon.

I\4

171.25)

The function insidethe brackets.plotted in Fig. 17.7,has a Fourier transibrm becauscthe Fouder integralconverges. Becausc/(r) is an odd function. we useEq.17.23lo find its Fourier transform:

s{/(,)}:i;1" ,-*1"=,, I

Fisure17,7A A function thatapproaches sgn(l) as

e"'u( r)1, € > 0.

1

-2j. \11.26)

17.4 Folrier Transfoms in theLimit 707

As € - 0,/(1) - sen(t),ands\f(t)I + 2lia.'I}Jercrorc,

sfte'(r)]: 3.

\17.27)

TheFourierTransform of a Unit StepFunction To find the Fourier tmnslorm of a ud1 step function, we use Eqs.17.18and 1'7.21.We do so by recognizing tlBt the unit-step function can be

I

I

'l{r)=t+tssn(r).

117.2s)

Thus,

="{1} . *{f"*,,,} *r,,,,r = .a@), !.

l.17.29)

TheFourierTransform of a CosineFunction To fiDd t]re Fourier transform of cosdot, we retum to the hvene{ransform integal ofEq.17.9 and obse ethatif F(a):2r6(a-ad,

(17.30)

then

rc =|l-v,a<.-,"1"u'a.

(17.31)

Usingthe siftingpropeftyof the impulsefunction,wereduceEq. (17.31)to f(t) : ei''t '

117.32)

Then,frcm Eqs.17.30and 17.32, s{eta$} : 2r6(.,-(|,i.

(17.33)

We now use Eq. 17.33to find the Fourier transfom of cosoot, because cosaot:

ej@d + a-irt 2

(11,34)

Thus,

s{cosa&} = =

r@{ei"t}

|e"N,

+ s{e-,qt})

- ao)+ br6(a+ ao)l

= a@ - ofi + '.6(0 + (,,o),.

(17.35)

TheFouriertransformof sin.ootinvolvessimilarmanipulation,which we leavefor Problem17.5.Thble 17.1presentsa summaryof the hanslorm pajrsoftbe imponanlelemenla-ry functioDs, Wenowtum to theproperdesof the.Fouriertransfom that enhatceour ability to desciibeaperiodictime-domainbehaviorir termsof frequencydomainbehavior.

TJe

to

F(@)

impulse

6C)

t

A sgrtr)

2iA6(@) n6(a) + 1,/jt

positive-time ex?onedial

e--L(tt

negative-time exponential

e'u( t,

t1!o i.1,o>o 2a/(az+e]),a>o

positive' and negativeline eponential

2n6(a - @o) ,[6(@+ed+6(0-ab)] j,[6(@ +.qr) - 6(0 - aro)]

17.5 SomeMathematical Properties The first mathematical property we call to your attention js that F(o) is a complei'quantity and can be. e4)ressed in eithet rectangular or polar form. Thus from the defining integral, f(t e i't dt

F(') :

= ./

/(tXcos,t - i sin'",,t)dt

-i : l-)ot.*,,., l-

fosinatdt.

(17.36)

17.5 Some l,4aihemahcat Prcpedies

e1,y=

"o",t at l- 71t1

n*,1:

l-y6si^,tat.

(11.31)

(17.38)

Thuqusingthe definitionsgivenby Eqs.17.37and 17.38in Eq.17.36,weget F(o) = A(a) + jB(o\ -

F(,o)leia\o.

\17.39t

The following observations about F(.d) are pertinent: . The rcal part of F(rr)-that iq,4(o)-is an even function of d;in other words,A(o) = ,4(-/d). . The imaginary pa of F(@)-that is, B(o)-is an odd turction oI o; in otler wordq B(o) : A( (,). . The magnitude of F(.d) that is, V,4'(o) + B2(.d)-is an even func. The phase angle of F(/,,)-that is, 0(t:,) : tan lB(o)lA(a);s an odd function of .",. . Replacing .,, by -o generatesthe conjugate of F(o); in other words,

F(-4 = r'@).

Hence, if /(r) is an even function, F(o) is real, and if /(t) is an odd function,F((o) is imaginary.If/(1) is even,from Eqs.17.37atrd17.38,

4.) =zl

t|)cosatat

\17,40)

and ,(@) : o.

117.41)

A(o) = 0

\17.4?J

ff f(t) is an odd tunction,

and

oa,t: -z

I

ttn"in.ra,.

117.43)

derivationsof Eqs. 17.40 17.43for you as Problems17.10

and17.11.

709

7X0

IheFounerTransforn If /(r) is an even function, its Fouder tnnsform is an even function, and if f(l) is an odd function,its Fou er transformis an odd function. Moreover,if/(t) is an evenfuflctior, from the inverseFourier hregral, r

tV)

r

-

1

1

tt

Jr.

:

*f',',,*"-'

-l/

6

l

;a-)P"'d-

t_6

+ is".dt\d'n

r 1 . y c o s , r a+, o

: =

Figure 17.8A A reciangutar frequenq spectrum.

f

tvr''d--^

^ l zn

I

Ala) cos-t d-.

117.11)

Now compareEq. 17.44with Eq. 17.40.Note that, exceptfor a factor of 1/22, these two equations have the same folm. fhus, the waveforms of A(d,) a'rd f(4 become interchangeable if f0) is an even function. For example,we have alreadyobservedthat a rectangularpulse in the time domain produces a frequency spectrum of the form (sin o)/zo. SpecificanX Eq. 17.11e4ressesthe Fourier transformof the voltagepulse shownin Fig.17.1.Hencea rectangulaipulsein the frequencydomainmustbe generated by a time domain function of the form (sin r)/i. we can illustrate tllis requirement by finding tie time-domain function f(/) corresponding to the frequencyspectrumshownin Fig.17.8.From Eq. 17.44,

-

tU):^

2 tt

I znJr

zMt\inutJ"

M c o s u t d u =- l - I tn\

|

| a

:+1.*y) =*(-^\#)

\11.45)

We say more about the ftequency spectnm of a rectangular pulse in the time domain versusthe rectangularftequency spectrumof (sint/t after we introduce Parseval'stheorem.

17.6 0perationaI Transforrns Fourier transforms,like Laplace tmnsforms, can be classified as functional and operational. So far, we have concentrated on the functional tnnsform$ We now discusssomeof the importantoperationaltmnsforms.With regardto the Laplacetransform,theseoperationaltransformsare similar to thos€ discussedin Chapter 12. Hence we leave tieir proofs to you as Problems17.12-17.19.

17.6 operationat Tnnsfoms 711

Multiplicationby a Constant From the defidng integral, if

s{f o} = F(a\,

slxf(t\j = rr@)

'46t 117

Thus,multiplication of /(t) by a constant conesponds to multiplying F(o) by that sameconstant.

Addition(Subtraction) Addition (subtraction)in the time domaintanslates into addition (subtmction)in the frequencydomain.Thusif s{f t(t)} = Fla), tt{f2(t)} : F2@), s{f 3G)}: F3@),

s{f{t)

f2(t) + /3(r} = r(.d) - Fr(a) + F3G)),

117.47)

which is derivedby substitutingthe algebraicsum of time-domainfunctronsinto the definhg integal.

Differentiation The Fou er transfoft of the firclderivative of /(t) is

cl!+l

I d . J

- i.r@t.

tt7.1l)

The/ltl derivativeof/(t) is !j\

I a' "',;' trti I d t

I )

ti@YF(@).

Equations17.48and 17.49are valid if/(l) is zero at +co.

Integration

sa\= l';ata,,

(17.1e)

tlten

=+. s{s(tr\

(17.50)

EquatioD17.50is valid if

f(x) drc: 0.

ScateChange Dimensionally, time and frequency arc rcciprocals. Therefore, when time is shetched out, frequency is compressed(and vice versa), as reflected in the functional transform

e t l { d / r-l l r ( 1 ) . a \a /

,' -u.

(17.51)

Note that when 0 < d < 1.0,time is shetched out, whereaswhen a > 1.0, time is compressed.

Translation in the TimeDomain The effect of translating a function in the time domaitr is to alter the phase spectrum and leave the amplitude spectrum untouched. Thus - a)} : e-r" e@).

s{f(t

117.52)

ffl' is positivein Eq. 17.52,the time function is delayed,and if d is negative, the time furction is advanced.

Tnnstationin the Freguency Domain Tfanslation in the frequency domai[ coresponds to multiplication by the complex exponential in rhe time domain: ei{ej.&fG)} : F(a, - at).

(17.53)

Modulation Amplitude modulation is the process of varying the amplitude of a sinusoidalcarier. If the modulatingsignalis denoted/(t), the modulatedcarder becomes/(1) cos.dot.The amplitude spectrum of this carier is one-half the amplitude spectrum of /(t) centered at tl')o, that is. .}{/fr) co( dor} -

I 2Ftu--r\

I 2l-td

uot.

Ot.r|)

Convo[ution in the TimeDomain Convolution in the time domain coresponds to multiplication in tie hequency domain. Ir other words, )(t)

-

/I r(^)h(t ^)d^

J -

17.6 0perationatlransfoms 713

e{)(t)} = Y(o) : x(a\H(a).

117.55)

Equation 17.55 is important in applications of the Fou er transform, becauseit states that the transform of the response function y(.r) is the product of tle input ffansform X(l,)) and the system function l1(@). We saymore aboul this relatiotrship in Section 17.7.

Convotution in the Frequency Domain Convolution in the ftequency domain coresponds to finding the Fouder transform of the product of two time functions.Thus if

f(tt : f{t)f2(t),

' 6 1= j f ' ' , 1 , y r 7 . - a o ,

(17.56)

Table 17.2 summarizesthese ten opemtional transforms and another operational transform that we introduce in Prcblem 17.18.

^n

n(@)

Kf(t)

KF(o)

/n, - ,f,t) + R(r)

F1@)

.t^f(t)/dt'

0o)"F(o)

flat)

1p{q1.,'o

Fr(@\ + Fx@)

t(-at 1 ;F(a

/-,t^rrt,

^ir^

1 ad +;F(a

+ ao)

x(@)H(@)

flt)ho

i J *FI@F{'

{f(t\

u)'i;

')d"

714

TheFourier Transfom

obiective1-Be abteto calcutate the Fourier transform of a function 17.4 Supposcf(r) is definedasfollows:

11= t 14 t + e .

l=

t = tt,

0
f(t)=-4t+A,

a) Find the secondderivativeofl(t). b) Fnld the Fou er transform of the second derivalive. c) Use the resultobtainedin (b) to find thc Fouder transformof the functionin (a.). (Ir,ir Use the operationaltransformof differentiation.)

= 4 u (' ,

t)

tllat ls,

1'(,) :

f(t) = 0,

Answer: (a) -

17.5 The rectangularpulseshowncan be expressed as the difference betweer t\vo step voltages;

n*('.l) - u"(,- i)v

Use the opemtional transfom for translation in the time domain to find the Fouiier translorm of 1r(r).

j5(r)

""1,(' -;),

4Al -r (.o ) , \cost 4A / (c) , {t d-r \

-r /\ :

ur\ co!; J. L/

Answet fT(o): v"t,

sin(orl2) @r/4

NOTE: 41ft 0) ChaplerProblen l7.ln.

17.7 €ireur't Applieations The Laplacetransformis ltsedmore widely to find the responseof a circuit than is the Fouriertransform,for two reasons. First,the Laplacetransform integralconvergesforawider rangeof driving functions,and second, it accommodatcs initial condiljons.Dcspitcthc advantages ofthe Laplace transfom. we canusethe Fouricr transformto find thc rcsponse.Thefundancntal rclationshipunderlyingthc uscolthc Fouricr transformin transient analysisis Eq. 17.55,which relatesthe translorm oI the responsc y(o) Lothe transformofthe input X(d) al1dthe lransferfunclion 1{(rr) of the circuit.Nole that l1(.r) is the familiar H(r) wilhr replaccdbyjo. Example17.1illustratcshow to usc thc Fouricr transformto find the rcsponseoI a circuil.

U.7

C i ( L rA itppti.atiom 715

usingthe FourierTransform to Findthe Transient Response Use the Fouder transformto find L(t) h the cncuit shown in Fig. 17.9.The current sourceis(t) N the signumfunction20 sgn(r)A.

Evaluating1ir and /(, gives

..=+ 10.

&=-= 3()

r,(,),i

rnl

,r0)

Therefore

1H t'\u)=-

10 ld

1 4r

10

: lu)

tigur€ 17.9 6 Theci,cuitfor Ecmpte17.1. The response is

i.(t) = s 1lr"(@)l = ssgno- roea'r(r).

Sotution The Fouriertransformof the driving sourceis

/s(o) = 9{20 sgn(r)}

:^(1) 40 Ja The transfer function of the circuit is the ratio of Id to I3;so

a@\ =

ro

TheFouriertransformof i"(t) is

Figure17.10showsthe response.Does the solution make sensein terms of known circuit behavior? The answeris yes,for the following reasons. The current sourcedelivers 20A to the circuit between-m and 0. The resistancein eachbranch govems how the -20 A divides between the 1wo branches. Ir particular, one fourth of the -20A appearsin the to branch;therefore l, is 5 foi r < 0. When the currenl source jumps fiom 20A to +20A at l:0, i, apprcachesi1s final value of +5 A exponentiallywith a tim€ constantof i s. An important charactedsticof the Fouder transform is that it directly yields the steady-state response to a sinusoidal ddving function. The reason is that the Fourier transformof cosoot is based on the assumptionthat the function existsover all time-Example17.2illustiatesthis feature.

r"(a): r s@)H(a) 40 ia(1 + i,o) Expanding1,(o) into a sumof pailial fractionsyields t.\1))

Kt

=-- (+la

K2 +rta

Figure17.10A lhe plotof i,(r) ve6usr.

Usingthe FourierTransform to Findthe SinusoidaI Steady-State Response 'Ihe

currcnt sourcein the circuit in Exarnplc 17.1 (Fig. 17.9)is shaDgedto a sinusoidalsource.The oxprcssionfor the currentis l{(l) : 50 cos3r A. Ilse the Fouricrtransfom nethod to find l,(r).

Sotution The translbrmoI the ddving functionis 1s(o) - s0f;[6(o 3) + 6(&,+ 3). A. before,rherrdn(re.tuncliunol lhe circbitic n\d)

=

I 4rla

Becauseof thc siftingpropertyofthe impulsefunction,lhe casicstway to fiDdthe inversetransformof ,1,,(o)is by the inversionintegral: i"(t): ?t '{1.(a)}

*[ -:) - son / a(r,, + 6 ( d + 3 ) 2;.1-l 4 )",',-\4+lt

4

^- / pil.

\ .

The transformof the currentrcsponsethen is -. -- 6(- t)+6(d+t) l,(d)=)uu . +rJa

ti) I rrr/1ro3' \

,;6r'

s

5

/

:5[2cos(3i

36.87')] = l0cos(3t - 36.87"). We leaveyou ro verify that thc solutionfor i,(r) is identicalto that obtaincdby phasoraDalt'sis.

objechve2-Know howto usethe Fouriertransform to find the response of a circuit 17.6 The cunent sourcein the circuit showndeliven a currert of 10sgn (i) A.The responseis the voltageacrossthe 1 H iDductor.Compute (a) 1s(.,,);(b)H(jo);(c) %(a,);(d) o,(r); (e) t1(0 );(f) r,(0+):(g) rl0 );(h) i(0+); (i) r"(0 );and (j) 0,(0+). 1f} i1

i.) ,t|+o 1 H Answer: (a) 20lj.r; (b) aj@/(5+ jo); (c) 80/(s + ja); (d) 8oe5' (r) v; (e) -2 A; (f) 18A:

(g)8A; NOTE: ALro lty ChapterProblems17.22,17.28,and 17.32.

( h )8 A ; (i) 0Y (j) 80v 17.7 Theloltagesourcein tlle circuitshownis gen eralingthevoltage = % e'u(. t) + u(t)Y. a) UsetheFouriertraNformmcthod1ofind ,,. b) Compute?,,(0.),?),(0*), ard ?),(oo).

A n s w e r :( a ) o ,

t .eu(-r) -

* j

O); v'

|

t 'utt\ ^?

r

"g"p;v,

i",i"

2

6

17.8 ParevattTheorem 717

17.8 Parseval's Theorem Parseval'stheoremrelatesthe energyassociated with a time-domainfunction of finite energyto the Fouriertransfom of the function.Imaginethat the time-domainfunctionf(r) is eitherthe voltageaqossor the curent in a 1 f,, resistor.The enersvassociated with this functionthen is wto - [

f'G)at.

117.51)

Parseval'stheoremholds that this sameenergycan be calculatedby inte$ation in rhe frequencydomain,or specifically,

l'ru>0,:jf 'o1'0.

(17.58)

Therefore the 1 O ene4y associatedvrith /(r) can be calculated either by integrating the square of /(l) over all time or by integrating 1/2r' times the square of the magnitude of the Fourier transfom of /(t) over all frequencies.Parseval'stheorem is valid if both integials exist. The average power associaled with time-domain signals of finite eneigy is zero when aveiaged over all time, Therefore, when comparing signalsof this type, we reso to the energy content of t]Ie signals Using a 1 O resistor as the base for the energy calculation is convenient for comparing ttre energy content of voltage and current sigmls We begin the dedvation of Eq. 17.58 by rewriting tlle kernel of the htegral on the lefthatrd side as f(l) times itself and then expressingone /(t) in telms of t]le inversion integral:

l*rato':l* turnoo, a.lat. l-1"atl|l^, asea'

(17.59)

We move /(l) inside the interioi integral, becausethe integiation is with respect to ..,, and tl1en factor t]le constant 1/2r' outside both integrations. ThusEq.17.59becomes

ftrr"

= irLll^F@)roei",ta,]dt. (17.60)

We reverse the order of integation and in so doing recogize that F(o) can be factored out of the ittegntion with respect to I Thus

o,lr.. l* r',,,r,r1, l* n,,,f*,,,n'''

(17.61)

The interior integml is F(-o), so Eq.17.61reducesto

.10.. l' ro* : !z n . l a[- op1o1

.la

1t7.62)

In Section 17.6, we noted that F( a) = F1a). Thus the product F(d)ir(-o) is simply the magnitudeof lq(o) squared,and Eq. 17.62is equivalentto Eq.17.58.Wealsonotedrhar F(d)l is an evenfuncrionof o. Therefore, we can also wr:ite Eq. 17.58as

= o, l-;'ato, ) l* vo,1r

(17.63)

A Demonstration of Parseval's Theorem We can best demonstratethe validity of Eq. 17.63with a specificexample.If

f(t\ : .*', the left-handsideofEq. 17.63becomes

* l,-"--o, l'"-"o': l--a"',, - "-. , 1^0 r -e 1 -^ 1 -l za 14

: r r1

zalo

1

1

117.61)

2o=;.

The Fourier transform of /(t) is 2a and thereforethe dght-handsideofEq.17.63 becomes Ao' 4a t t d t.-tl-t l [ ' , . d u - 11 . \l ' ' - t a n ' ltlo l t Jn (a' uf\ ,t a 2at az d/

'{o-

\

,\

J

" /

nl )1a - o

:1

\11.65)

Note that the resultgivenby Eq.17.65is rhe sameastlat givenby Eq.17.64.

TheInterpretationof Parseval's Theorem Parseval'stheorcm givesa physicalinterprctationthat tlle magnitudeof the Fouder transfom squared, F(o) ', is an energydensity(in joulesper hertz).To seeit, we write the right-handsideof Eq. 17.63as 1 r"'

ar

l rntll *" Ja I ) r t 2 " l t ) r n d r = 2,lo

dl.

{17.06)

where lF(2rrf)l'dl is the energyin an infinitesimalband of frequencies (d/), andthe total l O energyassociared with/(r) isthe summation(integlarion)of F(znf)\'llf over all fiequencies.We can associarea portion

1i.8

Pa6evaLt Theo€m 719

of the total eneqy with a specified band of frcquencies. In other words, the 1 O energy in the frequency band ftom.q to ('2 is

lF(a)|'d@. wa = ; J.,

(77.67)

Note that expressingthe integation in the ftequency domain as

*l*,,-,'oinstead of

Fr.i'

1 , " -

:

F('i)"]a

I

allowsEq.17.67to be wdtten in the form I

f-

wo - ; I -n rv

,

rt-tlz (t-

I

lq

lrotl' ,t-. -,_ n JI-

(17.68)

Figure17.11showsthe graphicinteryretationof Eq. 17.68. Examples17.317.5illustrate calculationsinvolving Pa$eval's theoiem.

Figure17.11A Thegraphic interprehiion of Eq.17.68.

AppMngParsevalt Theorem The curent in a 40 (} resistor is i = 2\eatu(t\ A..

,,,

wao(l==

What percentage of the total eneryy dissipated in the resistor can be associatedwith the frequerry band0
40 /i . t |n

400

-:--:--4ra

M,@0 /1 f

\ 2

t1b

,o

-\

l o l

= !99911\: ,non,

Solution The tolal energydissipaledin the 40 O resistoris ].wllri,= 40 | 400e* dt JO

: 1 6 , 0 0o0 4, l - : 4 o o 0 J .

n \2/ w]th the frequenc)band The energya.qocraled 0 < (' < 2v3rad/sis

40 4oo ... " * n : ; [' o1 , . , ,*4_ ,

r -,.1,fr\ 'tl" : ro,ooo/ " lt*' I \ . /

We can check this total energy calculation witl Pa6eval'stheorem:

8000/r\ , ' \ 3 /

F@):#i

8000 3 '

Hence the percentage of tle total energy associated with this range of frequencies is

Therefore

e@\ =

20

\,4 + ,.?

8000/3 ":

4fJ00

" L00 = 66.67ok.

ApptyingParsevat's Theorem to an ldea[Bandpass Filter The inpul\ohagero an idealba0dpass lilterrs

The total 1 O energy available at the output of the filter is

D(t) = 120e-2atu(\v' ._, = 1 /43 14,400 The filter passesall frequercies that lie between 24 and zE rad/s, without attenuation, and completely iejects all frequencies oulside this passband. a) Sketch ly((,r) ' for the filter inpur volrage. b) Sketch %(o)l'for rhe filrer output voltage. c) What percentageof the total 1 O energy contert of the signal at the input of the filter is available at the output?

"'

; J, si6* -'"'

:

600

;rn

, -]*

ilzo

:!t'^"" ,"'-''):T(#-;) : 61.45J. The percentage of ttre input energy available at the output is

Solution a) The Fourier transform of the filter input voltage is

, : lof ' no:2o.4.vo.

120

Therefore

v\-)l'

v(..)1": 14,400 576+ o]'

Fig.17.12sho*s the sketchof ly({r)1, versus(r. b) The ideal baDdpa.sfilrer rejectsali lrequencies outside the passband,so the plot of yo(o) , versus(oappearsas shownin Fig.17.13.

Figureu.12,| ly(o)|'versus o forExample 17.4.

c) The total 1 O energy available at tlle input to the filter is

1 --. * ' : ; . 1 , , /-

14,400 sib+.,d-

600 r' = 300J. 1 1 2

v"(@:)|' 25

zo

/ r = 14,400 tan' 9 1 I " \; 2 4 l o/

15 10 5

Figure17.13l

y(d) ' veuusa for Exampte 17.4.

17.8 ParevahTheocn 721

ApptldngParsevat's Theorem to a Lotv-Pass Fitter Parseval'stheorem makes it possibleto calculate the energyavailableat the output of the filter even if we don't know the time-domain expressior for r,,(t). Supposetle hput voltageto the low-passRC filter circuit shown in Fig. 17.14is 0(t) : 15e 5'&(t)V.

10ko

Figure17.14 A Thetow-pAsRCfitterfo, Lamph 17.5.

a) What percertage of t}le 1 O energy available in the input signal is available in the output signal? b) Wlal percenlageol lbe outpur eDer$ is associatedwith the hequency range 0 < o < 10 nd/s?

We can easilyevaluatethe integral by expanding the kemel irto a sum of partial fractions;

22,500 (2s+.'x100 + o')

Solution

w,

I

J0

100+

Then

a) The 1 O energy i the input signal to the filter is ,.F

25 + a2

--10/ l''t rtsc dt - 225' ,.rl = 22.sJ ru

-"=+{[

25+az

l0

3 0 0[ 1 / u \ r s\2/

TheFoufiertranstorm oI rheourpulvollagei.

uo@)- u@)H@),

r /"\l 10\2/

_- _

The energyavailableir the output signaltherelote is 66.67'h of the eneryy availablein the inputsignal;thatis,

q(,)=*

15

rt:;(100\:66.67%.

L/RC I/RC + ia

10 10 + jo

Hence

u"(.): ( s + i o x 1 0 + j . , ) 22,500 (25+d')(100+@') The 1 O energy available in the output sigml of th€ filrer is

w^: = I

.1" 1oo+ ,"?J

22,500 (25+@1(100+d' )

b) The output energy associatedwith the frequen(ry range0 < d = 10 rad/s is

soo( tto a.

W" ^ = - \ l - l - l

' (./o 25+uz

= Jn\/t ,r0 " \j'an T-

/'"

,/.

I

Ja 100+@t) I

' a ,n, t 0 \ ro ro/

3 0 /2 z

,\

"\z.sa ^)

: 13.64J. The total 1 O energyin the output signalis 15I so lhe percentageassociated wit]I the frequency range0 to 10rad/sis90.9770.

TheEnergy€ontainedin a Rectangular VoltagePulse We conclude our discussion of Parseval's theotem by calculating the energy associatedwitl a rcctangular voltage pulse. In Section 17.1 we found the Fouder transform of the voltase Dulseto be

v(.) :

v(.)

sinor/2

(17.69)

To aid our discussion,we have redrawn the voltage pulse and its Fourier tansform in Fig. 17.15(a)and (b), respectively.These figures show that, as the width of the voltage pulse (r) becomessmaller, the dominant po ion of the amplitude specfum (that is, the spectrum from 2T/r to 2n/r) spreadsout over a wider range of frequencies.This result agrceswitl our earlier comments about the operational tmnsform involving a scale changg in other words, wher time is compressed,frequency is stretched out ald vice vefia. To transmit a single rectangular pulse with reasonable fidelity, the bandwidth of the system must be at least $.ide etrough to accommodate the dominant podion of the amplitude spectrum.Thus the cutotf frcquency should be at leasr2T lr radls, or 1lr Hz. We can use Palseval's theorem to calculate the ftaction of the total enefgy associatedwitlr r'(a) that lies in the frequenc.yfinge 0 < a < znfr. From Eo.17.69.

*:||,,,.r,^+ffi^.

(17.70)

(b)

Figuro pukeandits lT.t5 A Therectangular vottaqe Fouri€rtrandom. (a)Iherectangubr pulse. vottage (b)TheFouder transtornr ofr(,).

To carry out the integation called for in Eq. 17.70,we let

177.71) nodng $at

a,=ia.

\17.72)

and that when a : 2tlr.

\17.73)

Ifwemakethesubstitutions givenby Eqs,17.71-17.73, Eq.17.70becomes --. W

ZV'^t f" s;n'x - l ^ d x nJo x'

.

(7i.14)

We can inlegrate the integral in Eq. 17.74by parts. If we let (71.75) (11.16)

r7.8 Pa6evats Theorem

du : 2sinxcost dx : sit2r dx,

\17.77J

and I

o = -

.

117.78)

Hence

/'sin' x

| -or= Ja x'

s j r l- rl ' .r

:o+

f" r ^ | I --smz.rd.r r lo Jo

d, 1"":@!

111.79)

Substituting Eq.17.79intoEq.17.74yields -.. = - av1^rf"sin2x w

t

I -ax. .lo 2t

(17.80)

To evaluate the integral in Eq. 17.80,we must first put it in the form of 21r sin /y.We do so by letting ) : 2t and noting that dy = 2 dr, and ! : when .r z.Thus Eq.17.80becomes

,

zv2_r fz"ya a,. T .lo

(17.81)

Y

The value of the integral in Eq. 17.81can l,e found in a table of sine intesrals.rIts value is 1.41815.so

w = 3(1.41815).

117.82)

The total 1 O energy associatedwitt ?(t) can be calculated either ilom the time-domain integration or the evaluation ofEq.17.81 with the upper limit equal to iDJinity.In either case,tlle total energy is (17.83)

The fraction of the total energy associatedwitlt the band of frequencies between 0 and 2r'h is

w

' w 2V2^r(1.41815\

"(vk ) : 0.9028.

117.84)

Therefore, approximately 90% of the energy associatedwith o(t) is contained in the dominant poition of the amplitude spectnm. ' lt. e-u.**it" unoL sr.goL,Eandbook of Mokehaticolaa.nb,r (NewYork:Dover, 1965),p.244.

obiedive3-understandParsevat's theoremandbeabteto useit 17.8 The voltageacrossa 50 l) resistoris

17.9 AssumeLhatthe magniludeof the Fou er transformof r(t) is as shown.Thisvoltageis appliedto a 6 kO rcsislor.Calculatcthe total energydeliveredto the resistor.

D = 4te'u(t)Y.

v (.ia)

Whar perccntageofthe rotat eDergydissipated ir the resistorcan be associated with thc frequencyband0 < (, < t,/3rad/s?

o (rad^)

Answer: 94-23%. NOTE: Abo ty Chqter Problent7.39.

Sunnm'lny* The Fourierlrunsformgivesa frequencydomaindcscriptionofan aperiodictimcdomaintuncrior.Dependingon the nature of rhe time domain signal, one of three approaches 1ofindingitsFourierrranslormmay bc used:

. The Fouriertransfbnnof a responsesignal _r(r)js Y(o) = x(.o)H(a). where-Y(o) is the Fourier rranslormofthe jnpur signal .r(.),and H(o) is the transferfunction H(s) cvaluaredar r = jo. (Seepage713.)

. If the time-domainlignal is a wctl-behavedputseof linitc duration,thc integral thal definesthe Fourier translormis used. If the ore-sidcdLaplacetransformof/(r) exisrsand all the polesot I(s) lie in the left half of lhe s planc. F(s) ma]' bc usedto find F(d). If l0) is a conslaDr,a signumfunction,a slep function,or a sinusoidalfunction.the Fouder lraNform is tound by usinga limit process. (Seepage699.)

Funclionaland opcrationalFourierrransformslhat are usefulin circuiranalysisare tabulalediD thblcsl7.1and 17.2.(Seepages7()Uand 713.)

.

fhc Fourier transform accommodalesborh negativetimc and positivetime fuDciionsand lhereforeis suiled to problemsdescribedin terms of eventsthat start al I = co.In contrast.the unilateralLaplacetransformis suited to problemsdescribedin terrns of iritial condi tions and elents that occul for I > 0.

. The magnitudco[ rhe Fourier ttansform squaredis a measureof thc energydensity(joules per hcrrz) in rhe Irequencydomain(Parseval's theorem).Thuslhe Fourier transformpemits us to associatea fracrionof the total energycontainedh fO with a specifiedband of frequeDcies. (Sccpage717.)

Probtems 725

Problems Figure Pl7.3

S€ctioDs17.1-f.2 1?.1 Use the defining integral to find the Fouder transfolm of the following functions:

a) f0) : 'A- a" ^' ! , )''

2=t<2:,

/(r) - 0, b) flrr : 1. f():

:t

+ A. + A.

f(t) : 0,

-1 -, -n. 2 u
Sections17.3-17.5 171 a) Find the Fourier transform of the function shown .r'F1g.P17.2. b) Find F(a,) when o : 0. c) SketchlF(d)l versusa when A : 2 and r = 7. Hint: Lvaluate r(-)l al d - l. -12.-13.. . . . +15. Then use the fact that lt (o)J is an even function of (,. tigurcPu.2

17.4 Find tle Fouriertanslorm of eachof the following functions.In all of tle functionEd is a positivercal constant and-c! < t < c<), a) /(t) = ltlcnr'r' b) f(t) : le aa; c) /(t) : e-'r4cosoet; d) /(1) = e 'r4sin {,ot' e) /(t) : 6(t

d.

17.5 Dedve S{sin .,,0t} 17.6 If /(t) is a rcal function of t, show that the inveNion integral reduces to

r,,t

,l.l-),t,-tro"/

a(a)sin drld..

17.1 1I f(t) is a real, odd function of 1, show tlat the irvenion integral reduces to

17.3 TheFoudertransformof /(t) is shownin Fig.P17.3. a) Ftnd/(t). b) Evaluate/(0). c) Sketch/(r) for 10 < t < 10s when ,4 = 2?7. andao: 2radfs.Hint: Evaluate/(r) at t = 0, 1, 2,3, - . . , 10s andthen usethe Iact that /(t)

/c)= : [*n@)";",,a.. 17.8 Use the inversionintegral (Eq. 17.9)ro show tlat g- \ {2lja} - sen(t).Hin: Use Problem17.7. 17.9 Find g{cos ,}0t} by usingthe approximating function " coso'or' f(r)=" where € is a positive real constant.

726

TheFouder TE.sfom

17.10 Show that if /(r) is an even funclion, A(a):2

/-

f.16 Given ty(t) = I x(^)h(t - ^)d^,

fO cosatdt,

Jo

,.t{

show that v(.r) : g{y(t)} = X(.o)a(.,,), where X@) = er{x(ol and H((,) = sF{r(r)}. fltnr. Use the definingintegralto r\,Tite

B(r,) = 0 17,11 Show that if /(,) is an odd function,

/-[ /-

g;ry(t\l

l

eut : -z f@s:n .r ar. I

Next, reve$e the order of btegration and then make a changein tlte variable of integation; that is, let u = t i. :

f1.r7 Given f(t) Section 17.6

(112n\ I

: ja,F(.,,), where f17.12 a) Show that g{df(t)ldt} = F(a) s {f(t)}. Hi:nt: Use the defining integrat and integrate by parts. b) Wlat is the rcstriction on /(r) if the rcsulr given in (a) is valid? = (/o),I(o), where c) show that !rId"f(t)ld{} F(a) = s{f(t)}. r.13

r

t

1

-

,

c t ,I ^ , M , 1 = , , , . t . (./-lh) )

where F(o) : S{/(r)}. fltnr: Use the defining integral and integmte by pad$ b) What is the restriction on /(r) if the resutt given in (a) is valid? c) If f(te) - e ^u(x), can rhe opentional rransfolm in (a) be used?Explain. f.14

a) Show tlat s[f(a!)I:

| a

Fl

/ , , ,\ l, a>0. \a /

b) Given that f(ot) = "-" for d > 0, skerch F(a) = s{f(at)} for d : 0.5,1.0,and 2.0. Do your sketches rcflect the obsenation that compressionin the time domain co espondsto rtrercliirg in the frequency domair? 17.15 Dedve eachof the following operational tansfoms: a) s{f(t - a)} : e i* F(a)i b\ s{d"t f(t)) = F(o

ai;

c) s{/(r)cos (,0r} = jr@ - a) + iF(a + o,\i).

f{t)f21),

show that I(o)

:

F1Q)F2(a- u)dr. ffr ri Firsr, use the

defining integral to expressF(.r) as /I ftt)f)(t)e-tr tu.

F(d):

.,t F

Second,use the inversion integral to *Tite f ,u) ::-

a) Showthat (

I

I I I x\^\hv ^td^ le id dr. J d L J _

A\0,) = 0,

1 t *

l F,tu\ej'' du. z n l ' '

Third, substitute the expression for /1(t) hto the defining integral and then interchange the order of integration. 17.18 a) Show that

f s,r,..,,1

: s{!f\t)J.

Itrl +

L " l b) Use the result of (a) ro find each of the followirg Fouder transforms: s{te '' u(t)},

s {lt)e-''}, 'q$e a'I' : l1(t/,(r),where 17.19Supposethat/O /1(r) = cosoo,, f2G):1,

-r/2
lr0) :0,

elsewhere.

a) Use convolution in the frequoncy domain to find F(o). b) What happens to F(d) as the width of /r(r) increases so that f(r) includes mote and more cycleson /(4?

Probtems 727 17:5 RepeatProblem17.24exceptieplaceo,(l) with i.(t. f.20

a) Use tlle Fouder transfom method to find ro(r) in the circuit shown in Fig. P17.20.The initial value of t,(t) is zero, and the souce voltageis

12su(t) v. b) Sketch i,(t) versus t. Figure P17.20 10()

1726 a) Use the Fourier transformto find o, in the circuit h Fig.P17.26il ts = 2 sgn(r)A. b) Does your solutionmake sensein terms of knowncicuit behavior?Explain. FlgoreP17.26

+

ouJ;

----L(r)

40()

+

i.(t

1)

5oo:

t'"(r)500nH

17.21 RepeatProblem17.20if the input voltage (os) is """ changed to 125sgn(t). 1:,.22 a) Use tle Fourier transform method to find 2,"(t) in tlle circuit shown h Fig. P1'7.22if r,s = 20sgn(r)V. b) Does your solution make sensein terms of knowncircuit behavior?Explain. tigure P17.22 40k{)

f.23

Repeat Problem 1'7.22except rcplace,ua?)\\-tth io1).

fr4 The voltage source in the circuit in Fig. P17.24 is 8nft given by the eryression I,s : 15 sgn(t)V.

1727 Repeat Problem 17.26except replace i, with o,.

17"28 a) Use the Fourier tiansform to find oo in the circuit in Fig.P17.28if ?)8equals30t5t A. b) Find ,,"(0-). c) Find r.',(0*). d) Use the Laplace transfom mettrod to find u, fort > 0. e) Does the solution obtained in (d) aeree with r', for 1 > 0+ ftom (a)? rbureP17.28 10f,)

a) Find ?,(r). b) Wlat is tle valueof 0.(0-)? c) Wlat is tlle value of 1J.(0*)? d) Use the Laplace transform method to find o,(t) for t > 0-. e) Does tle solution obtained in (d) agree with o.O for I > 0+from (a)? Figure Pt7.24

50ko

0.1F

1729 ^) Use t]Ie Fou er tiansform to find r, in the circuit in Fig.P17.28if ?,sequals30t5! A. Find t,(0-). c) Find i,(0). Use the Laplace transform method to find L fort>0. Does the solution obtained in (d) agee witl ," forI>0'ftom(a)?

o

17.30 a) Use the Fourier tnnsform method to find 2" in the circuit in Fig. P17.30if u" = 125"o", tt U . b) Check the amwer obtained in (a) by finding the steady-state expression for 0, using phasor domain analysis. Figure P17.30 5H

40J) 20H

x,

figureP17.33 2.5() 5 rnl

17.34 a) Use the Fourier hansfolm method to find 2,"in the circuit shown h Fig. P17.34. The volrage sourcegenemtesthe voltage uE= g\e 4Jaty'

360r}

b) Calculare ,"(0 ). ,,(0'). and ,,(co). c) Fhd l1(0 );ir(0*);?,c(0 );andoc(0+). d) Do fie results in part (b) make sensein rems of known circuit behavioi? Er?lain.

17.31 a) Usp the Fourier transformmelhodto find odin thecircuitin Fig.P17.31 when is : 45efftu(-t) + 45eaauu(t) A.

Flgure Pt7.34 20pF

b) Findt.(0-). c) Findi(0*). d) Do t])e answen obtahed in (b) and (c) make sense in terms of known circuit behavior? Explain. tigureP17.31

5ko

+1.2spF

I)

1735 a) Use the Fouier transform method to find oo in rhe circuit in Fig. P17.35when u, : 6t)e5' u(.-t) + 9o0re-5'u(r)v.

5H

b) Find t',(0-). c) Find o,(0+). 1732 Use the Fourier tnnsform method to find i, in the ""( cncuit in Fig. P17.32if ?s = 200 cos2500, v.

Flgure Pl7.35 DA

Figore Pl7,32 400nF

f.36 1733 The voltage source in the circuit in Fig.P17.33 a& generatingthe signal Dc= 18e4tu(-t) - 12u(t)Y. a) Find 0D(0 ) and t',(0+). b) Find i"(01 and ,o(0+). c) Find z)".

Wlen t]Ie input voltage to the system shown Fig.P17.36 i. 8rO) V. lhe oulput\ oltageis ao = [60 - 40e-5t+ 20e 2u]u(r)v. What is the output voltage if oi : 8 sgn(t) V? Figure P17,35 \to

@

f,l_-l

D"\,

Problems 729 Figurc P17.40

Seclion 17,8 17.3? It is siven.hat F(a) = e'u( a) + e-'u(a). a) Find /(r). b) Fird the 1() energy associatedwirh /(t) via time-domainintegration. c) Repeat (b) using frequency-domain integration. d) Find the value of (,1 if /(l) has 90% of the energy in the ftequency band 0 < o < {rr. 17.38 The input current signal in the circuit seen in Fig.P17.38is ic: 3etstA,

)no -.

v,td)=-

1 0 0 ' a d s d l ' 2 0 0 f a sd:

V@) : O,

elsewherc.

r > o*.

W}tatpercentageofthe total l O energycontentin the output signallies in the ftequencyrarge 0 to 10 rad/s? tigurePr7.38

f r o o i,,l

t)

17.41 The amplitude spectrum of the input voltage to the high-passRC filter in Fig-P17.41is

200mH

17.39 The input,voltage in the circuit in Fig. P17.39is z'g= 60trrrll0) v. a) Find ?,,(1). b) Sketchlvs(o) foi 10 < o < 10 rad/s. c) Sketch %((,,)lfor -i0 < o < 10 radls. d) Calculate tlle 1 O energy content of 2s. e) Calculate the 1 O energy content of o,. f) Wlat percentageof the 1 () energy content in 0s lies ir the frequency range 0 < .) < 10 rad/s? g) Repeat (f) for o,. Figure Pr7.39 20()

a ) Sketchll4(o)I'for 300< (, < 300rad/s.

Sketchi%(.,,)]'for-300 < (.)< 300iad/s. ") Calculatethe 1 O energy in the signal at the input of the filter. o , Calculate the1 O energyinthesignalat theoutput of the filter. FigurePI7.41 0.5tF

.---l

17,42 The input voltage to the high-pass RC filter circuit in Fig.P17.42is ti(t) : A.''u(t). Let a denotethe cornerfrequencyofthe filter,that iq a : 1/RC. a) W}Iat percentage of the energy ir t]le sigul at the output of the filter is associatedwjth the frequencyband 0 < o = aif a: a'l b) Repeat(a), giventhat d : V3a. c) Repeat(a), siventlata = alt/1.

17.40 The circuit shown in Fig. P17.40is driven by the Flgure Pl7.42 i" = 30e 't u(t\ pA. What percentageolthe total l O energycontentin the output voltage oo lies in the ftequency range 0<@<4rad/s?

r l

Two-Port Circuits 18.1 TheTerminalEquationsp. 73? 18.2 TheTwo-PortParamete6p- /32 18.3 AnaLysis of the Terminated Two-Port 18.4 Interconnected Tl^io-PotrCltcuits p. 747

Beableto catcutate anys€t oftwo-port pa€metenwith anyof the foLLowing m€thodsl .

l4easurements madeon a circuit;

.

Converting fromaroih€rset oftwo-por. pacm€tersusjngTabLe 18.1.

Beabteto analyze a ieminatedtwo-portcircuit to find cufferts,vottages, impedances, aid ratiosof interestusirg Table18.2. Krowhowto anatyze a cascade interconnectiof

730

We have frequently focused on the behavior of a circuit at a specified pair of lerrninals. Recall thal we introduccd the Th€veninand Norton equivalcntcircuitssolel]'to simplify circujt analvsisreiativeto a pair of terminals.ln analyzingsomeelectri cal systems, tbcusingon lwo pairsof tcrminalsis alsoconvenicnt. particular, ln this is helpful when a sigral is led into one pair of tefininals and thcn, aftel being proccssedby the system,is exlr'actedat a secondpair of tcrminals. Becausethc tcnninal pairs represent the points where signalsare cither fed in or extracted,thev arc rcfer-r-ed to as the pofts of the system.In this chapter.wclin1itthe discussionto circuitsthal haveone input ancl one output por-t.Figure 18.1on page731illustratesthe basictwo porl building block- Use of this building block is subjectto several rcstrictions.Fint, thcrc can be no ener€iystorcd within the circuil. Second,thcrc can be no iDdependentsourceswithin the cir'cuit:depcndentsources, howevcr,are permilted.Third, the current into the port must cqual the cuuent out of the port; that is, tl = ri aDdl, : li. Four-th.all externalconnectionsrDustbe madc to eithe. the input port or the output port:no suchconncctionsare allowedbetweenports.that is,betweenterminalsa and c, a and d, b arrdc,or b and d-Theserestrictionssimplylimit the rangeof cir cuit problernsto whichthe two-port formL ation is applicable. The fundamcntalprincipleunderlyingtwo-port modelingof a systemis that o y the terminal variables(ir, U1.12, and r.r2)are of interest.We haveno interestin calculatingthe currentsand volt agesinsidcthe cilcuit.We havealreadystlessedtenninal behavior il1the analysisof operationa]amplifiercircuils.ln this chapter,we formalizethat approachby introducingthe two port parameters.

1 8 . l T h eT e , m i nEaqt u a t i o 6 7 3 1

18.1 TheTerminat Equations Inviewinga circuitasa two'port netwoik,we are interestedin relatingthe curent and voltage at one port to the curent and voltage at the other port. Figure 18.1showsthe relerencepolarities of the terminal voltages ropur .l"' Ourput and the rcference directions of the terminal cu[ents. The references at port eachpo are symmetricwith respectto eachother;that is.at eachport the curent is dhectedinto the upper terminal,and eachport voltageis a dse from the lower to the upper terminal. This symmetry makes it easier to generalizethe analysisofa two-port network and is the reasonfor its uni Figure18.1i Thetwo pot buitding block. versalusein the literature. Themost geneialdescriptionofthe two-port network is caded out in the s domain. For purely resistive networks, the analysisreducesto solving problemscan be solvedeither by resistivecircuits.Sinusoidalsteady-state lirst fhding the appropriate r-domain er?ressions and then replacing s wifi io, or by direct analysis in tlle frequency domain. Here, we wite all equationsin the s domain;resistivenetworksand sinusoidalsteady-state solutionsbecomespecialcasesFigure18.2sho$sthe basicbuildingblock in tems oI the s-domainvariables1r, 14,Ir. and yr. Figure18,2A Thei-domain two'portbasic OI liese four terminal variables, only two are independent. Thus for bujldjngbtock. any circuit, once we specify two of the variables, we can find ihe two remaining unknowns. For example, knowing y1 and % and the circuit wilhin the box, we can determineIt and 12.Thus we can descibe a twoport networkwithjust two simultaneousequations.However,thereare six differentwavsin which to combinethe four variables:

vr :

z[Ir + ar2l2,

v2= z2rlr + a?212;

(18.1)

Ir : yrvr + Jr2U2, I2

: Y21vr+ Jnu2;

(18.2)

Vr: a1rV2 at2l2, Ir = a2rvz a2zl2; u2- bru

b 1 2I r

12 = b2tU

b22li

(18.3)

(18.4)

U=hxl1+hr2v2, 12: h\Ir + hLV2;

(18.5)

11: gltU + gnl2, V2: g2tvt + 92212.

(18.6)

732

Two-Port tlcuits

Thesesix setsof equationsmay alsobe consideredas three pai$ of mutuallyinve$e relations,The fiIst set,Eqs 18.1,givesttre input and output voltagesasfunctioDsof the input and outpu! cwrents.The secondset, Eqs,18.2,givestle inverserelationship,that is, the i[put and output currents as iuDctionsof the input and output vokagesEquations18.3and 18.4areinvene rcIations,asare Eqs.18.5and 18.6. The coefficientsof tle current and/orvoltagevadableson the righthandsideof Eqs.18.1-18.6are calledthe parametersof the two-port circuit.Thus,whenusingEqs.18.1,we refer to the z parametersof the circuit. Similarly,we refer to the , parameters,the d parameters,the b parameters, tlle ,, parameters,andthe g parametelsof the network.

18.2 TheTwo-Port Parameters We can detemine the parameteft for ary circuit by computation or measurement, The computation or measurement to be made comes directly from the parameter equations. For example, supposethat the problem is to find the I parameters for a circuit. From Eqs. 18.1,

-"

u 1l

u ztz: ;l

l

ul z"r= il ul

z.::l

o,

(18.7)

O,

(18.8)

o,

(18.9)

o.

(18.10)

Equations18.7-18.10revealthat the four a parameten can be described as follows: . z1ris the impedance seen looking into port 1 when port 2 is open. . zD is a transfer impedance. It is tle mtio of the port 1 voltago to the port 2 curellt when polt 1 is open. . 221is a traNfer impedance.It is the ratio of the port 2 voltage to t]le port I cullent when port 2 is open. . 222is the impedance seenIooking into port 2 when port 1 is open. Therefore the impedance parameterc may be either calculated or measued by first opening po 2 and determinhg the mtios 14/11 atrd y2/1r, and then opening po 1 and detemdning the ntios U/12 ^nd VrJ12. Example 18.1illustmtes ths detemhation of the z pammeters for a resistive circuit.

18.2 TheTwo-Pori Pa€mete6 733

Findingthe z Parameters of a Two-Port Circuit Find the z parameters Ior the circuit shown rn Fig.18.3.

!-

5rr

and therefore

v

'''

-h

0.75u

I, t =o

= - : 7 5 O

Vr/to

When 11 is zero, the resistanceseen looking into p o r l " i c l h c l 5 O r e . i . l o ri n p a r a l l e$li r b l b e \ e r i e s combinationof the 5 and 20 O resistors. Therefore y,

(15)(25) ^^__ ^

ligure 18.3 A ThecjrcuitforE\ampL€ 18.1.

When port 1 is open,11is zero and the voltage Y1is

Solution The circuit is purely resistive,so the r-domam cllcuit is alsopurely resjstive. With port 2 open,that is, 12 : 0, theresistanceseenlooking into port 1ls the 20 O resistorinparallelwith the seriescombination ofthe 5 and 15 () resistors. Therefore

u

V:::-(20)=0.8I/'. l+lt)

With port 1 open,the currentinto port 2 is

(20x20)-^ ^

r1

"

rr lt=rl Wlen 12is zero, Y2is

9.3'ts'

Hence

u,:rr{31tr;:n.tru,,

-"

Equations 18.?-18.10and Example 18.1 show why the paramete$ in Eqs. 18.1are called the z paramete^. Each parameter is the ratio of a voltageto a curent and therefore is an impedancewith the dimension of ohms. We usethe sameprccessto determinethe remainingportparameters, which arc either calculatedor measured. A port parameteris obtainedby eitheropeningor shortinga por1.Moreover,a port parameteris an impedance.an admittance.or a dimensionless ratio. fhe dimensionless mtio is the ratio oI either two voltagesor two clrlrents.Equations18.11-18.15 sunmarizetheseobservations.

Irl

]n=;l

Y 2= r;

u

T^

s,

s)

tt=;l

r.l

S

(1e.11)

v.,l lzlt -a

o.sy2 Vztq.315

734

Two-Pot Circuits

ul

UI aD= -i

al1=;l

a^ =;

t-l

S,

u . l t2l , D2r: ;l

u

o,

/,1

-

I z l v= o

(18.12)

ul b-,. = - - l O. rrl! =n

+,,="

(18.13)

J.

ul

hD: -l

+1.,"

no::l

s.

(18.11)

ll

(18.15)

v2lt,=o

, I 412

;

u

-, I

8r,-,

The two-port parameten are also desffibed in relation to t}le re€iprocal sets of equations,The impedance and admittance paramelefi are grouped into the inmittance parameters.The term immiflance denotes a quantity that is eithei an impedance or an admittance.The a and b pammeters are called the traftmission parameten becausethey descdbe the voltage and current at one end of the two,port network in terms of the voltage and current at the other end.The immittance and transmissionparameters are the natual choicesfor relating t]le port vadables.In othei words, they relate eithei voltage to cwent variables or input to output vaiables. The i and g paiamete$ relate cross-variables,that iE an input voltage and output currcnt to an output voltage and input curreDt.Therefore the ,t and 8 parameters are called hybrid parameters. Example 18.2illusfates how a set ofmeasuemenls made at the rerminals of a two-po circuit can be used to calculate the /' parameters.

18.2 TheTwoPoar Parameterc 735

Findingthe d Pararneters fron Measuremenrs fhe following measurements pcrtaii 1l)a two-pofi circuit operating in the sinusoidal steady state. Withport2 oper avoltageequaltol50cos4000r V is applied to port 1. The culrent into pofi I is - 15") A, and the port 2 voltage is 25cos(40001 100cos(40001+ 15') V. With port 2 shortcircuired, a vollageequalto 30cos4000/V is appliedto port 1. Thc current into port 1 is l-5cos(4000/+ 30') A, and the current into port 2 is 0.25cos(4000r + i50') A. Find thel? paramelelsthai can dcscribe the sinusoidalsteadystatebehaviorofthe circuil.

FromEqs.18.12,

Solution

Therefore

,' =

2518 :0.2s1 "': n)'=,: @:s. 1001]t I1

The secondsetofncasuements gives

T:3OAV, vr:0V,

rL= ].slfqA, rz = 0.2s1f1! A,.

301!: l .,,= vil,. "=0.2s1p:

The first sc1oI measurements gives

v 1= l 5 0 a v ,

1so1! = :151 8:, v, ,,=o 10011t:

\ = 251_E A,

o,,:

vr : 100111v. r , = 0 A .

I,l

1201j!:4.

1.5,/30"

i,,=o=68:uLg.

objectiv€1-Be abteto catculateany set of two-port parameter5 18.1 Find the lr paramelerslor the circuitin Fig.18.3. Answer: )1r = 0-25S, )l, = /rr = 02S, y22:t5 ) 18,2 Find tbe a andl?parameterslor the circurtm Fig.18.3. Answer; g = 0.1St gD = -0.75; gr1 : 0.75; . \ - 5 ( ) :i - 4 Q : , r 0.8. I h^= 0.8;h22= 0.10615

18.3 The following measurements $'eremadeon a two port resistivecircuit.With 50 mV applied to port 1 and port 2 open,thc currenl into port 1 is 5 /rA, and the voltagcaqossport 2 is 200mV With port 1 sbort circuitedand 10 mV appliedto port 2,the currcnt into po.t 1is 2 r!A, and the currentinto por! 2 is 0.5pAFird the I parametersof fte network. Answer: 8r1 = 0.1mS 82t = 4l

szz: 20ko".

NOTE: Ako try Chaptet Probkms 18.2,18.4,arul 18.5.

Relationships Amongthe Two-Port Parameters Becauscthe six setsof equationsrelale 1()the samevariableqthe parametersassociated rvithanypair ofequationsmustbe relatedto the parameters ofall thc other pairs In otler wordEiI we know one setof parameters, we candedveall the other setsfrom the knowDset.Becauseof thc amountof aigebrainvolvedin thesederivations. wc nerelv list rhe resulrsin Table18.1.

J n a i b n L h l A,y a27 b21 hn Bt

'"

Y1) aY lzt A'r,

A.r a)1

h1

I

__

qzL

hzr

Ab bz

hzz

1

_ 422 _ b t

Yr

P) AL

1

-A:v. - 4 , , Yr2

I

b=-: zr2

h t =

s11

,

Pl

8r2

z1t

y11

att

Aft

1

alr

Yr2

La

h )

ei

A?

|

:

b2

Ag

h.

a., bD = - : : : . - : = aD D|

hr

s)j A8

zt2

C12

91r a/2 lrr = Lz

Y12

_ 422 :

=

4t

1

Ag

btz

hL

9tz

.tr2

A.a

t

aD

bLz

z2l

1

a12 211

--

z22 zt2

822

Lb

h2r

82r

bn

ht

gz2

1

Ay

bzt Lb

hzz - 8n hr czr

br A,

1 a g hzt 9zt r

a

J12

!12

-ht

:

.?1

b., br

aj1

Yn

b brr

Ag

A8

b z r: L h bn

1

= 1i2

bt

221

All

= -ho,

8zr

hr - 8zz lht Bzt

1

l,l

A

an

._-_ 1 _Ly _at

n"

212

be: Ab

Yr Y2t

Ay 1 h" ) = - : : : : - : z,2z Y11

1, gn

1

z2l

z2z z)t

hn

bzz _ _A.h Ab hzt

221

1

hn _ 8 \ 2

al\ _bn - Ah !11.. bD hr\

lr1

1

8

A

z

1

bp:

ht

Lz=zrrzn-zDz2\ LY: Yiin

ir2Y2r

LaI : al1a22

aDa2I

A,b = bft12-

bDb21

L,h = hnh22

heh21

Lg = 811822-

8r282r

Althougl we do not deriveall the relationshipslistedin Thble18.1,we do derive those between the z and y parameters and between the I and a parameten These derivations illustrate the geneial processinvolved in relatirlg one set of parameters to another. To find the z pammeters as functions of the J, parameteN, we fi^t solve Eqs. 18.2 for I4 and y2. We

18.? TheTwo:Port hi?mete$ 737

lr11

11.1,1,; : i-';,,,i., r,r :: r,i;.

r'l= -!41, * LLr. u- - lY."-ay Ly

ay' ,", ,-,,;, ComparingEqs.18.16and 18.17with Eqs 18.1shows

t , rrrt= 3,,' Ay' 221 = --=,

^y

lrr Lv

-i'

,

To find the I pammelersas functiotrsot the c Darame€rs.we rearranqe Eqs 18.3in tb; formof Eqi"t8.I andthio compare coet6cienrs. Fromtf,e secondequationin Eqs.18.3,

y,- = J-1, , !2.1,. a\-

A2i

'.

(18.?2)

Therefore,substitutingEq. 18.22htq the fLst equationof Eqs.18.3lelds ,,

ana)t

( aipn - a t'-,/ t \t t-)- . turzr \

i. ': , l.

nA,8\

FromEq.18.22, 1

(18.26) (18.27)

Example 18.3illustates tlle usefulnessof t}le parameter conversion table.

Findinglr Parameters from Measurements andTabte18.1 Two setsof measurements aie madeon a two-port iesistive circuit. The first set is madewith port 2 opeq and the secondsetis madewith port 2 shortciicuited.The resultsarc asfollows: Port 2 Open I,i = 10mv 1r = 10pA v2= 4I)Y

The d parameters afe

10 x 10-3 ul "':.'..1,,=o:-4{) = -0.25 x

Port 2 Short-Circuited \ = 24mv Ir = 20 pA

"

Find the li parameters of the circuit. u12

i,l

l 0 x 1 0 6= -0.25 x 10{S, 40 vz t'=o

ul ,l

24 r 10-1_ _r, n l

LI

20xlo6=-2ox10r.

t0

Solution we can find /l11 and ll21 directly from the shortcircuit test:

--

ul

ht:;l

10 3,

rz lr .=o

10'

The numericalvalueof Aa is

2 4 x 7 0 3: 1.2kO, 20x106

Aa = araz2

r^l

=5x10-6

nzr:7

10, 20 x 10n The pammetem hr2 and h22 cannot be obtained dnecdy from the open-circuit test. However, a check of Eqs 18.7-18.15 itrdicates tlat t]Ie foui d paramete$ can be derived from the tesl data, T'herelorc, h12ar.d h22can be obtaitred through t]le conversion table. Specifically,

ava21 6x10:6:

Thus

106

20^10'

0.25x 10 6 -20x103

10-6.

739

Reciprocal Two-Port Circuits If a two-port circuit is reciprocsl, the following relationships exist amotrg tlle Dort Darameters: (18.28)

Yr2: Y2r,

(18.29)

atan - a72a2r: La = 1.

(18.30)

b11b72 bpb21 = Ab = 1,

(18.31)

ho = - htt,

(18.32)

(18.33) A two-port circuit is reciprocal if the interchalge of an ideal voltage souice at one port witl an ideal ammeter at the other port produces the sameatnmeter rcadhg. Consider, for example,the resistive circuit shown in Fig. 18.4.w}len a voltage source of 15V is applied to port ad, it produces

figure18.4A A reciprocaLtwo-port cjrcuit.

a current of 1,75A itr the ammeter at Dort cd.The ammeter current is easily determined once we know the volage ybd.Thus ubd . ubd 15 . uba .. *20='' 60' io

(18.34)

and Ybd: 5 V. Thereforc

r-.2n+(Jj-1.1s^.

(18.35)

If tle voltage sourceand ammeter are interchanged, the ammeter will still read 1.75A. We verily this by solving the circuit shown iII Fig. 18.5: Vod . Ubd . Vod

$'3tt'

n

15

=''

(18.36)

ybd- 7.5V. Thecunent/adequals FromEq.18.36,

r , u = f f + .:1r 1. r s e .

(18.37)

A two-port circuit is also reciprocal if the itrterchange of an ideal current souce at ol1eport with an ideal voltmeter at the other port produces the same voltmeler readiDg.For a reciprocal two-pofi circuit, only three calculationsor measurementsare neededto determhe a set of parameters A reciprocal two-port circuit is symmetric if its ports can be interchanged without disturbing the values of the teminal currents and voltages.Figure 18.6 shows four examples of symmetric two-port circuits. In

jn Fig.18.4,wiihihe vottage tigur€18,6a Four Flgure18.5A Thecircuitshown (a)Asymmetric examptes of symm€tric two-port circuits. tee. sou(eandammeter interchanged. (b)A symm€tdc pi.(c)A symmetric bddsed tee.(d)A symmeiric taiiice.

8 . 1 A l a r r . io r f h"

-p

mhdleT d w oP o . i1 ' ( u |

suchcircuits,the foliowing addirionJlr
br - bn hyh22 8rr822

-

(18.41)

heh21 = Lh = 1,

118.42)

812821= Lil = I

(18_43)

For a symmctricreciprocalnetwork. only 1wocalculationsor mcas urementsare neccssary10determineall the two po parameters.

0bjective1-Be abteto catcutate anysetof two-portparameters 18.5 The followingmeasulements were madeon a resistivetwo-po network that is syrnmelric and rcciprocalWith poft 2 open,y1 : 95 V and 1i = 5 A;with a short circuit acrossport 2,

the \: 7152y andIz: -2.12A. Calculate ofthe two-portnetrvolk, z parameters Answer: rrl : zz2= tg A, z1z: a2\= 77A.

NOTE: AIsotr! ChapterProblem 18.12.

18.3 ,4nalysis sf the Terminatsd

Two-Fort eircuit ln the typicalapplicationof a two port model.the circuit is driven at porr 1 and loadedat pofi 2. Figure18.7showsthe r-dornaincircuit diagramfor a lypically terminatedtwo port model.Here, Zg representsthe intemal impedanceof lhe sourcc,ys the internalvoltageof the source,and Z/ th e load impedance.Analysisof this circuit involvcsexpressingthe terminal culrerts and vollagesas functions of the two-porl parameters,ys. Z€. and Z L. Si-\characteristics of thc terminatedtwo-port circuit defineits termi- Figurc18,7s A ternrinated two-podmodet. nal behavior: . thc inpul impedanceZiD = y1/11,or the admitta ceYjr: IlVr . the output current /2 . the Th6veninvoltageand impedancc(yrh, Zri with respecrto port 2 ' thc currenl gain1y'1l . the voltagegainYy'f . lhe voltagegain Yzlfs

741

742

Two-Port Circuits

in Termsof the z Parameters Thesix Characteristics To illustrate how these six chamctenstics ale derived, we develop the expressionsr:sing the z parameten to model the two-port portion of tlle circuit. Table 18.2 summarizesthe expressionsinvolving the ), a, ,, r, and g pammeters

Z ; o =z u -

YDYzlzL

4 . = Y r ' - 7 + yzzz L

zn+ ZL

!zrVe

ztv"

+ yrtzL + yrzs + LyzszL - y21vs

lzr+Zsl(zn+ZLi-zDz2l

yt + LtZc 1+ vttz. Zlh = z2x

4r+ Zs

'tt

Iz= II zzz+ ZL

L=

V,

u2

u

z2rzL

vr

4jZL + Lz

v2 vs

II

zzrZt

kr+

z)(zz2 + z) - 42221

yr + L,tZ,

yln+ LyZL -

J^ZL 1 + yl2ZL

uz vg

YztZt

tnttZ,ZT

- 1t + yrz)(l

+ vzzzL)

b2zzL + bE bztzL + b11 -\^b

vs

bizs+ brlzszl+ b?2zL+ bD

a r Z L + a a + a z l z a Z L +a 2 2 z s

vsLb bzz+ b21zs

vs

brzs + brz b^zs + b!) Ir= I1 v2=

1 ZL

Iz= b|t + b^zL II

v z _ A.bzL

VI

u

ZL u2 vs (a|+ az\z)ZL+ af + a22zs

LbzL uz_ + b?2zL+ bzrzszL br+ b11zs vs

b\z + br2zL

18.3 Anatysis oftheT€minated Two-Port Circuit 743

hDhnz L

X.=grr-

1,+ hzzL (1 + hrxz)(h1t + z) - hDhzrzL

C1,821

8n+ Zt -gtVc

Q + s 1 1 2 ) ( s n + z L-) StzCrlzs

_hrVs

B"us 7 + EnZs

z"+ hr

CDC^ZE

| + Blrzs

h_

I2

I1

7 + hrzL

r1

grZL + LE

v2

-hrZt

u2

LhzL + hi

v1

SzZt 8zz + Zt

u

uz us

httZ t

(h11+z)(r + h2zz)

hDrhtzL

u2 SnZL vs (1+ srz)(szz+ zLt snszrzs

The dedvation of any one of the desired expressions itrvolves the algehaic manipulation of the two-poft equations along wilh the two constraint equations imposed by the tem)iMtions If we use the z-parameter equations,the four tlat describe the circuit in Fig. 18.7are

Vr= zrJr + zDI2,

118.44)

U2=2,lr11+22212,

(18.4s)

Vr = Vs I&s,

(18.46)

V2:1227.

(18.47)

Equations 18.46 and 18.47 describe the constraints imposed by the terminations. To find the impedarce seenlooking hto port 1, that i\ Zi = Vr/ 11,we proc€edas follows.Itr Eq. 18.45we replacey, with 122, and solv€the resulting expressio[ for 12:

.

"

zztlr Zr.'. ztt

(18.48)

744

Two-Pot Cilcuits We then substitute thls equation into Eq.18.44 and solve for Znr: 212221

-

zn t

(18.49)

Lt.

To find the teminal current 12, we fhst solve Eq. 18.44 for 11 after replacing l,i with tlle right-hand side of Eq. 18.46.The result is Us - z1zl2 {18.50)

211I Z,

We now substituteEq. 18.50into Eq. 18.48and solvethe resultingequation for 12:

- z2rvs (zt+Z)(zD+Z)-zr2z21

(18.51)

The Th6venin voltage with respect to port 2 equals y2 when 1, = 0. With 12 = 0, Eqs.18.44ard 18.45combineto yield Z 2 1 jI :

v2 t. ':

vr Zzt-.

(18.52)

B]It Vr : Vs I1ZE, and Ir : Vsl(Zs + z:'J,);therefore substituting t])e rcsults into Eq. 18.52yields the open-circuit value of y2:

Vzt,=o - Vrn

Lvc

(18.53)

The Thdvenin, or output, impedance is the mtio yr/12 when % is replaced by a sho circuit. When Vs is zero, Eq. 18.46reduces to

Vr:

IrZs.

(18.54)

Eq.18.54into Eq.18.44gives Substituting ZnIz z1I Z,

(18.55)

We now useEq.18.55to rcplace11in Eq.18.45,withthe resultthat ,,1 U = z- r h - 2 2 2 z )z2r r, 2l l v " - a z- t - Z s

(18.56)

The currentgain 12/11comes directlyfiom Eq.18.48:

Iz

- zzr

11

21 l 722

(18.57)

18.3 Analysk oftheTeminated Two-Pot Circujt 745 To derive the expressiotrfor the vottage gaitr yy'y1, we sta by replacing 1, in Eq.18.45with its valuefrom Eq.18.47;rlus

/ -u\

v r : z z l L+ z > \ j

(18.58)

).

NextwesolveEq.18.44for 1r asa functionof I,i andy2:

/ -u-\ ,tr!t=vr- rrr\i)

.

V

znvz

zi

aizr

(18.59)

We now replace 11 in Eq. 18.58with Eq. 18.59and solve the resulting ei?ression for Yz/l,i:

v2

u

ziZL + 41222- zr2z2r

zztZL znZL + L,z'

(18.60)

To derivethe voltageratio yr/ys, we first combineEqs.18.44,18.46, and18.47to find 11asa function of y2and y':

. 'r:

us xovz. * zr.,+ 4 4k" + z)

(18.61)

Wenow useEqs.18.61and18.47in conjunctiorwith Eq.18.45ro dedve an expressioninvolvingonly y2and 4; that is, Znzt2V2

r,

'

4t, n 4

",

t. Ie

1 j : ), ,

,,. .2" zr"'

(18.62)

which we can manipulate to get tlle desired voltage ratiol

zzrZt

%

(zr1+ Z")(zzz+ Z)

znzzt

(18.63)

The first entries in Table 18.2summadze the expressionsfor these six' arlflbuteso[ lhe terminatedh^o-porl circujl. Also lieled aJe lhe correspondhg expressionsin terms of the y, 4, ,, ,, and g parameters. Example 18.4 illustmtes the usefulnessof the relationships listed ir Table18.2.

Two-Poft Analyzing a Terminated Circuit The two-port circuit shown in Fig. 18.8 is described in termsof its b oarameters.the valuesof which are bL = -20' br : -2 mS,

612= 3000O, b22: -02.

b) The averagepower deliveredto the 5000 O load is 2$.162 : 6.93W.

2(s000)

a) Find the phasorvoltag€V2. b) Fnd the averagepower deliveredto the 5 kO load. c) Fmd the averagepower deliveredto the input polt. d) Find the load impedance for maximum average powertransfer. e) Find the maximum averagepower deliveredto the load in (d).

c) To find the average power delivered to the input pofi, we first find tlre input impedance Zn . Frcm Table18.2, bnzL + bp

b^zL + br ( 0.2xs000)- 3000 2 x 10 3(s000) 20

:$= t.r.r:t

Solution

Now 11followsdirectly:

a) To find Vr, we have two choicesftom the entries in Table 18.2.We may choose to find 12and then find V?from tle relationship V : -IrZr, or we may find the voltage gain Vr/Vs and calculate V, from the gain.Let's use the latter approach.For the 6 parametervaluesgiven,wehave

Ab = (-20x-0.2) - ( 3000x 2 x 103) -4 6: 2.

Ir:

The avemgepower deliveredto the input po is n 7xq472 Pr = r-(133.33)

= 41.55w

d) The load impedance for maximum power transfer equals the conjugate of the Th6venin impedanceseenIooking inro po 2. From Table18.2,

From Table 18.2,

v2

500 ='789.47 mA. 500 + 133.33

brzR + b12

L,bzL + br2+ brlzs b22zL+b2\zszL

b^zE + b22

( 20x500) 3ooo

(-2Xs000)

+ [-2 3000+ (-20)500+ (-0.2)s000 10 19 Then,

n,=(lf),*:,u.."zau

(-2 x 10 3)(500) 0.2

r0 \500xs000)]

11n6n ::=10,833.3jO. L t

metelore zL: z+r,:10,833.33 {1. e) To find the maximum averagepower delivered to Zr, we fiNt find V2from tlle voltage-gain expresa. thisgainis sionvrlv!. when ,/L ;c 10.8J3.33

v"! :

0.83J3.

Thus vr:

(0.8333) (500): 416.61 V, 1

Pr(maxinurn) = ^ -= Figue18.8A ThecircuittorEftmph18.4.

6 612

= 8.01W.

18.1 Interconnected Two-Poft Circuits 747

objective2*Be ableto analyze a terminated two-portcircuitto find currents. vottages. andratiosof interest 18.6 Thc 4 parametenof the two-port network ' l ' o $ n a r ( d -, 5r l { r ' . , r 2- l 0 Q = d2r 10+ S, and 422= 3 x 10-'.The nef wolk is driven by a sinusoidalvoltagesource having a maximum amplitude of 50 mV and an ir1lernalimpedanceof 100 + i0 ().It is tetmi natedin a resistiveload of 5 kO. a) Calculatethe averagepower deliveredto the load resistor. h) Cxl(ularerhe loddrcsr\lalcetor miximum averagepower. c) Calculatethe maximumaveragepower deliveredro thc resisrorin (b).

Vs

Answer: (a) 62 5 mW: (b) 70/6kO; (c) 74.,1mW.

NOTE: Also try ChapterPrablems18.30,18.32, snd 18.37.

18.4 !*t*rqs*fi ected'i"1ir$-F$yt {ir{{.xits S),rthcsizing a large,complex syslemis usuallymadceasierby first designing subscctionsof the system. lntercoDnecting thesc simpler, easier-todesign unitsther completesthe syslem.Ifthe subsecriorls are modeledby two port circuils,s},lrthesisinvolves the analysisof interconnecledrwo-porr circuils. Two-port circuits may be interconnecrcdIive ways: (1) in cascade, (2) in series.(3) in parallel,(4) in series-paratlel, and (5) in paratlel-series. Figure18.9on pagc748depictsthesefivc basiciDterconnections. We analyzeand illustraleonly the cascadeconnectionjn this section. However.iflhe four olher connectionsmeel certainrequircmenlqwecan obtainthe paramelersthat describcthe irterconnectedcircuitsby simply addingthc individualnetwork parameters. In parricular.lhea parameters describethc seriescornection,lhc ) parametersthc parallelconnection. the I paramele$the series-parallcl conDection, and lhe I parametersthc l parallel-scries connecrion. -fhe cascadecomrectionis imporlant becauseit occursfrequenily in the modelingoI large systems.Unlike the other four basicinterconnections,there are no rest ctionson using the parametcrsof the individual two-port circuits1l] obtain the parametersof the intcrconnectedcircuits. The a parametersare bestsuitedfor describingthc cascadeconnection.

rAdctailerl disc!$ion of thcsclour inter.onnertionsis prcscntedi. Henry Runon and JosephBordogn", t/er fn I]\Ie/rorkt: Fuh.ttoas, Filt(rs, Anausf (Ne{ York: Mccraw-Hill,

748

Two-PodCircuits

tigure18.9a lhefivebasicinterconnections oftwo'poi! circuits. palaLteL. (b) (c) (d) (e) ParatLel'serie5. cascade. series. %rattet. series G) We anal)ze the cascade connectior by using the circuit shown in Fig. 18.10,where a singleprime denotesa parametersin the fiIst circuit and a doubleprime denotesa paramete$in the secondcircuit.The output voltage and curent of the fint circuit are labeled yi and /i, and the input voltageand currentofthe secondcircuit are labeledyi and Ii. The problem is to derivethe d-parameterequationsthat relate y2 and /2 to y1 and 11.In other words,we seekthe pair of equations U=artu2-ar2l2

(18.64)

It=a2tv2-anl2,

(18.65)

where the d parameters are given er?licitly in terms of the d parameters of the individual circuits.

Circuit 1

Circuii 2

Figlre18.10i. A cascade connectjon.

18.4 Interconnecied Two-Port Cncuits 749 We beginrhederivarion b] noti0gtrom fig. 18.I0tbar

v1: aiyv'2- ai2l\,

u8.66)

11= a^Vi

(18.67)

aizti.

The interconrectionmeansthat yi: Vi ^nd Iitheseconstaintsinto Eqs.18.66andEqs.18.67 yields

Ir. Substituting

V1:ai1V\+a!el\,

(18.68)

It:airv'r+abli.

(18.69)

The voltage Vi and the cunent 1i are related to y2 and 1, thrcugh the d parameters of the secondcircuit:

vi= aitv2- ahl2,

(18.70)

I'r = ai1U2 abl2.

(18.71)

Wesubstitute Eqr 18.70and 18.71hto Eqs.18.68and 18.69ro generate therelationships betweenVr, I1 andVz,I2i + a\2ab\Ir, U = (aiai1 + a:r2ai)U2-@11ai2

(18.7?)

I r = (!4{h + ai2ai)u2-@^ai2+ anah)r2.

(18.73)

By compadngEqs.18.72ad 18.73to Eqs.18.64and 18.65, we ger rhe desiredexpressions for the d parametersof the interconnectednetwotks, a!:a\1ah+alr2ai1,

(18.74)

ar2: aiab + a:Dab,

(18.75)

at:d2ra:L+4241,

{18.76)

an= 41ai2 + abah.

118.77)

If more than two units are connected in cascade.the a Darametersof tbe equivalenllwo port circuit can be fou0d b' successrvely reducingrhe odginal set of two-port circuits one pair at a time. Example 18.5illustrates how to use Eqs. 18.74-18_77to analyze a cascade connection with two amDlifier circuils.

750

Two-Poir Cncuits

circuits Tryo-Port Analyzing Cascaded Two identical amplifien are connected in cascade, as shown in Fig. 18.11.Each amplifier is desciib€din terms of its ft parameters.Thevalues are ftrl = 1000O, 1,12= 00015, h:r = 100, and i,2 = 100 pS. Find the voltage gain Vz/Us.

: 10.25x 10 6, arz= aipl2 + ai2a:n = (5 x 10r)( 10)+ ( 10x-10") = 0.095O, an-4ra:t+4zdzt lo 4r | {-o.ol)( lo ")

- (-lo{r5

: 9.5x 10 eS,

185. Figure18,114 Thecircuitfot Exampte

an= aip\z + dnd22

Solution

= (-10nx-10)+ (-10')'

The first step in t]|J]di'JgVrlvc is to convert fiom ft parameters to l' paiameters. The ampliliels aie identical, so one set of a parameten describes the ampiifiers:

: 1.1x 10-4.

hzt

:*F='*'o-0,

FromTable18.2,

u2 Ys

ZL

(dr1 + a21Zs)ZL+ ap+ s22ZE

,i,=f:ff= 'on,

104

+ 0.095+ 1.1x 10-4(500) [10.25x10-6+9.5x10-e(500)]101

ai=

h", hr:

-100 106 : lou

ru'5'

104

0.15+0.095+0.055 -t

ab:

lb1:

I

lw:

10''

Next we use Eqs. 18.74 18.77 to compute the d parametefi of the cascadedamplifiers: arr= a\ra\1+ a\2d2r

:25 x 10-6+ (-10X 10 )

10s 3 = 33,333.33 Thus an input signalof 150/,V is ampiified to an output signalof 5 V For an altemativeappioachto finding the voltagegain yr/ys, seeProblem18.41

751

0biective3-Know howto analyze a cascade interconnection of two-portcircuits 18.7 Eachelementin the symmctricbridged-tee cilcuil showDis a 15 O resislor.Two()1urese bridgedteesare connectcdin cascadebetween a dc voltagesourceand a resistiveload.'nre dc vollagesourcehasa no load voitageof 100V and an internal resisranccot 8 O.The load r c , i . l o ri \ h d i r . r e dun r r lm e \ i m u mp u $ e ri , deliveredto the load.Calculate(a) the load (b) the load vollage,and (c) the resistance, load power.

NO'|E: Aho try ChapterPrcblem t8.38.

Answer:(a)14.44 O; ( b )i 6 v (c) 17.73 w

5iur"*rxary Thc lwo-pon nod€l is used!o desc be the performancc of a circuit in terms of thc vollage and current a1 rts inpul and output ports.(Secpage730.) The model is linriled !o circuitsin which . no indcpendentsourcesare insidethe circuitberween thc ports: . Doenergvis storedinsidethe circuitbetweenthc ports; . the currentinto the pofi is equalto thc currentout of theport;and . no cxternalconnectionscxist belweenthe input and ourporpo s. (Scepage730.) Two of the four tcrmiial variables(f . ,/1,yr, 1, are independent;thcrefore. only tNo simultaneousequationsinvoh,irg the tourvadablesare neededto descdbe lhe circuit.(Seepagc731.) The six possiblcscls o{ sinultaneousequationsinvolving the four tcrminalvariablesare callcdthe:-,_v-,d-.b-. ,-. and g paramelerequations.Scc Eqs.18.1-18.6. (See page731.) Theparameterequationsa1ewrirtenin the s domain.The dc valuesof tlle paramctcrsareobtainedby settingr = 0, and the sinusoidalsteadyslate valuesare oblainedbv settingr = j@.(Seepage731.)

ADy setof paramelersmay be calculatedor measuredby invokingappropdaleshortcircuit and open circuit conditionsat thlJinpul md outputports.SceEqs.18.7-18.15. (Seepages732and73,1.) The relationshipsamongthe six setsof paramctcrsare giveDin Table18.1.(Seepage736.) A two porl circuit is reciprocalif the inlerchangeof an ideal voltagesourceat one port with an ideal ammeter at thc olher port producesthe sanreannneterreadingThc ei1eclof reciprocityon thc lwo pofi paramerersis givenby Eqs.18.28 18.33.(Seepage739.) A reciprocal two port circuit is sJm etric iI its porrs can be inlcrchangedwithout disturbing the values of the tenninal currerts and voltages.The added effeci of symmetry on the two-port parametersis given by (Seepages?10ard 741.) Eqs18.38 18.43. Thc pcrlor-manceof a two-port circLil cornectedto a Thivenfu equivalentsourceand a load is summarizedby the relalionshipsgivenin Table18.2.(Seepage742.) Large net\rorks can be divided into subnelworksby meansof iilerconnectedtwo-port models.Thecascade connectiorlwils used in this chaptcr 1{)illustrate the analvsis ol inlerconnected two porl cilcuits. (See page747.)

752

Two-Port Circuits

Probtems 18,7 Find the i parameters for the circuit in Fig. P18.7.

Sections18.1-182 18.1 Find the l' atrd I parameters for the circuit in Example18.1.

FlgoreP18.7

!\

611

5rr

lor rhecircuilin Fig.Pl8.2. 18.2 Findlhe z parameter. Figure P18.2

fr- so

60rt

L

+ 20()

vz

18.8 Find t}le b parameteN for the circuit shown in Fig.P18.8. tigun P18,8 .\-

5()

18,3 Use the results obtained in Problem 18.2 to calcuIate the ) paiameters for the circuit in Fig. P18.2.

18.4 Find rhe z paramelerstor lhe circuil sbo!\n iD Fig.P18.4. Figure P18,4

lE.9 Select the values of Rr, n2, and R3 in the circuit 10f)

in Fig. P18.9 so that /'11=1.2. atz=340", a21= 20 mS, ar.d ar2 : 1.4. Fig{rcP18.9

185 Find the b parameters for the circuit in Fig. P18.5. Figu€P18.5

fl-

+m

v | 2 x 1 Or V 2

18.10 Find the i Darameteisof the t\qo'Dort circuif shown in Fig.P18.10. Flgure P18.10

1

18.6 Use the results obtaitred in Problem 18.5 to calculate the 8 parametemof the circuitin Fig.P18.5.

2 5 oj l o o

1f)

PmbL€ms 753 18.11 The following direct-current measurements were madeon the two-port networkshowr in Fig.P18.11. Pon I Op€n

Port 1 Sho(-Ciroit€d

Y 1: 1 m V

1 1= - 0 5 p A

u, = 10v

/z = 80rA u2=5v

Iz = 200pA

18.16 Dedve the expressions for the functions of the a parameters. 18.17 Derive the expressions for the y paramelersas functions of the , parameters.

Calculaaethe lr parameters for the network. Flgure P18.11

18.1E Derive the expressionsfor the I parametersas functionsof the z parameters. 18,19 The opemtional amplilier h the circuit shown in Fig. P18.19is ideal. Find the 8 parametersof the circuit.

18.t2 a) Use the measurements given in Problem 18.11 to find the 4 parameters for the network. b) Check your calculations by finding the a pammeters dircctly from the i parameten lbund in hoblem 18.11. 18.13 Find the frequency-domain values of the ) parame, tersfor the two-pofi circuit shoM in Fig.P18.13. Figure P18.13 lr

18.20 Find the s-domair expressionsfor the l parameters lz

loo

ot the two-poit circuit shown ir Fig. P18.20. Figure P18.20

ma

*

l2so

18.14 Find the , paramete$ for tlte two-pon circuit shownin Fig.P18.13.

18.21 Filrd the r-domain expressionsfor the a pammeters 18,15 Find tle l, pammeters for the operational amplfier circuit shownir Fig.P18.15.

of the two-polt circuit shownin Fig.P18.21. Figure P18.21

Pigurc Pl8.l5

1F 8000

754

Two-Pot Circuits

18.22 Is the two-port chcuit shown in Fig P18.22s]'rnmetric? Justify your answer.

b) Use the a parameters to derive tle Th6venn equivalent circuit witl respect to the terminals of the load. I ime-domdin c ) Derivelhe sready*tare eypression

Figure P18.29

10 mII

160mH

1823 a) Uqelhe definingequa on. ro tind rher-domain expressionsfor tl1e ft parameters for the circuit h Fig.P18.23. b) Showthat tlre resultsobtainedin (a) agee witn the ,-parameter relationships for a rcciprccal syrimetric retwork.

18.30 The y pammeters of the amplifier in tle circuit shownir Fig.P18.30are lrr : 25 mS:

yt2 :

y21:

= -40 mS. ]}]22

250mS;

Figure P18,23

1 mS;

Find the mtio of tie output power to that supplied by t}le ideal voltage source. figureP18.30 |

100 1/U V Section18.3

!L-r''""'l

G-t

100{)

1814 Derive the er?ression for the input impedance (.2i" : U/ I j) ol the ci,rcuitin Fig. 18.7in terms of the b parametem. 1825 Dedve the expiession for tlle current gain 1zl/1 of the circuitinFig.18.7in termsof the I parameters"

18.31 The 8 parameters for the two-port circuit in Fig.P18.31are

1826 Derive the exprcssion for the voltage gain yz/yl of the circuitin 8g.18.7 in lerms of the ) parameters

s

1817 Derive the eryression for the voltage gain yy'ys of the circuitin Fig.18.7in terms ofthe ft parameters.

g.

1828 Find the Thdvenin equivalentcircuit with respect to port 2 of the circuit in Fig. 18.7ir tems of the z parameten. 18.29 The linear transformer in the circuit show[ in Fig.P18.29hasa coefficientof couplingof0.65.The transformer is driven by a sinusoidal voltage source whoseintemal voltageis os = 100cos2000tV.The intemal impedanceof the sourceis 10 + J0 O. a) Find the frequency-domain l' parameters of the linear trunslbrmer,

-

l

t

l S;

n5

10.5:

f!)--0.5-/0.5; gz2 1.5 12.5O.

The load impedance Z/ is adju.redtor ma\imum averagepower transfer to Zr. The ideal voltage sourceis generatinga sinusoidalvoltageof x's = 42\4 cos50001V. a) Find the Ims valueof Y2. b) Find the aveiagepower deliveredto Zr. c) W}Iat percentageof the averagepower developed by the ideal voltage sourceis delivered by ZL?

Probtems 755 FigureP18.31

FigureP18,34

18.32 The ), parametersfor the two-port power amplifier circuirin Fig.P18.32are

J'r1= 100ms;

)r, =

2/..S;

rz -

50 pS.

18.35 a) Find the z parametenfor the two-port network in Fig.P18.35. b) Fird o, for 1 > 0 wh()n t'! = 50"(r) V. Figure Pr8.35 t'-

to

The internal impedanccof the sourceis 2500 + JOO, and the load impedanceis 70,000+ l0 O.The ideal voltage sourceis generatinga voltage

i--;

,"--'^

I

I

l

-T--trO

r:

0s = 80\rcos 4000rmV. a) Find the rms valueof Yr. b) Find the averagepower deliveredto Zr. c) Find the avcragepower developedby the ideal

Figure P18.32

r!

18.36 The followingmeasurements were madeon a resistive two port network.With port 2 open ard 100V appliedto port 1, the port 1 currcnt is 1.125A, and the port 2 voltage is 104V. With porl 1 open and 24V appljcd1()port 2, the port 1 voltageis 20V and the port 2 curreDtis 250 mA. Find the maximum power (mi iwatts) that this two port circuit can deliver to a resistiveload at port 2 when port 1 is driven by an ideal voltagesourceof 160V dc. 18.37 The following dc measurements were made on the r e s i \ r i ! ne e r $ ^ r k\ h o $ ni D F i g . P l ir ? I1 = 25V

Il = 41v

Fig.P18.32,find

vr:0v

a) the value of Zr for maximum averagepower transferto Z, b) the maximumaveragcpower deliveredro Zr c) the averagepowcr developedby the ideal vollage sourcewhenmaximun poweris delivcredto Zr.

1z =

vz: 20v 1, = 0A

18.33For lhe terminated two-pofi amplifier cfcuir in

500nA

A variable resistor Ro is connecledacrossport 2 and adjustedfor maximum power liansfer to R.. Find the maximumpower. Figurc P18.37

18.34a) Find the.r-domainexpressions for thel?parameters ofthe circuil in Fig.P18.34. b) Pon 2 in Fig.P18.34i. rermrndrcd in u fesi.rance of 800(), and port 1 is driven by a stepvoltage sourceo(r) : 45"(1)V. Find ol0 lor r > 0 if C : 0.1/-iFand a : 400mH.

756

Two-Pod Circuits

Section18,4 18.38 The g and l, parameters for the resistive two porls in Fig.P18.38are givenby 8l1 : 10mS;

ftr1= l50O; h)2:

0.05:

82\ : 20;

h2t =

0.70

g?2: 24 kA;

h22: I 00 pS;

18,40 The networks A and B in tlre cncuit in Fig. P18.40 arc reciprocaland symmetric.For network A, it is known that .ri1 : 4 ^nd a\2 : 5 A. a) Find the d parametersof network B. b) Find vy'V when I, : 0. Figur€ P18.40

Calculate0, if?s : 1095 V dc Figure P18.38 200

tr'l

)

lko

Isl

S€ctions18,1-18.4 18,41 a) Show that the circuit in Fig. P18.41is an equiva' lent circuit satisfiedby the i-parameter equations. b) Use the ,-parameterequivalentcircuit of (a) to find the voltage gaill V2JV i(r the circuit in Fig.18.11.

18.39 The z parameteN of the fint two-port circuit in Fig.P18.39(a)are

FigureP18.41

.1l = 200(I:

zr2 = 20 dl;

.21=

222= 40kO

16 MO:

The circuit in the secondtwo-port circuit is shown in Fig. P18.39(b),where R = SkO. Find x'. if 0s : 15 mV dc. Figure P18.39 5000 )

sko

lzl b

1

d

d (a) R

2

f

18.42 a) Show that the circuit ir Eg. P18.42is an equivalent circuit satisfiedby t}le z-parameter equations. b) Assumethat the equivalentcircuit in Fig.P18.42 is ddven by a voltagesourcehaving an internal impedanceof Zs ohms.Calculatethe Thdvenin equivalentcircuit with respectto port 2. Check your results againstthe appropriate entries in Table 18.2.

Probtems 757 18,43 a) Show that the circuit in Fig. P18.43is also aD equivalentcircuit satisfiedby the z-pammeter equations. b) Assumethat the equivalentcircuitir Fig.P18.43 is teminated in an impedanc€ of Z, ohms at poit 2. Find the input impedancey1/11.Check your results against the appropriate entry in Thble18.2.

tigur€P18.43

18.44 a) Derive two equivalent circuits that are satisfied by the y-parameterequations.llr?tr Start with Eqs. i8.2. Add and subtract )Iy, to the fi^r equation of the set. Construct a circuit that satisfies the resulting set of equations,by thinking in terms of node voltages,Derive an alternative equivalent circuit by first altering the second equationin Eq. 18.2. b) Assume tlat port 1 is driven by a voltage source havingan intemal impedanceZs, and port 2 is loaded wittr an impedance Zr. Find the current gain 1r,/1r.Check your results against the appropriate entry in Table 18.2. 18.45 a) Derive the equivalent circuit satisfied by the 8-pammererequauons, b) Use the 8-parameter equivalent ciicuit derived in part (a) to solve for the output voltage in koblem 18.39.Hrdr Use Problem 3.64to simplily the secondtwo-port circuit in Prcblem 18.39.

The Solutionof Linear nppena,xf\

lJ\ Simultaneous Equations Circuit analysis hequently involves t]le solutior of linear simultaneous equations.Our purpose here is to review the use of deteminants to solv€ such a set of equalions.fhe theory of determinants (witl applications) can be found in most intemediatelevel algebratexts.(A particularlygood r€ference for engheering students is Chapter 1 of E.A. Guillemin's Tie Mathematics of Circuit Anaryrtr [New York Wiley, 1949]. In our review here,we will limil our discussionto the mechanicsof solvirg simultaneous equations with determinants

A.1 Pretiminary Steps The first step in solving a set of simultaneous equations by determinants is to wfite the equationsin a rectangular(square)format.In other words,we anange the equations in a vertical stack such tlat each variable occupies tlle same horizontal position in every equation. For example,in Eqs.A.1, tllre variables 4, i2, and h occupy the first, second, and third position, respectively,on the left-hand side of each equation: 21ir-9i2-12i3=-33, -3i1+6i2-21=3, -8i-

(A.1)

4iz + 22i, = 50.

Alternatively, one can descdbe this set of equations by saying that il occupiesthe first column in the array,i2 the secondcolumn, and 13the third column. If one or more variables arc missing from a given equation, they can be inserted by simply making their coefficienl zero. Thus Eqs.A2 can be "squaredup" as shownby Eqs.A.3:

402+31\=76,

(A.2)

1q+2\=5; 2ut-D2+0r\=4, 0\+4t2+3\=16,

(A.3)

'hr+0t2+2?3=5. 759

760

TheSoLuiion ofLinearSimultaneoLrs Equation!

A.2 Crame/s Method The value of each unkno*n vadable in the set of equations is exprcssedas the ratio of two determinant$ If we let N, with an appropriate subscdpt, representthe numefator determinantand a representthe denomhator determinant.then the tth unknown r, is

.

N

o

(A.4)

The denominator determinant A is the same for evelv unknown variable aDdi. calledlhe charaderistic d€terminant of lhe ir ot equarions. The numerator deteminant Nr varieswith each unknown.Equation A.4 is rcferredto as Crameasmethodfor solvingsimultaneousequations.

A.3 TheCharacteristic Determinant Once we have organizedthe set oI simultaneousequations into an orderedaray, as illustratedby Eqs.A.1 and A.3, it is a simple matter to form the charactedstic determinant. This determinant is the square aray madeup from the coefficientsof the unl(nowr variables.For example,the characteristic determinantsof Eqs.A.l and A.3 are

121 e ^-13 6

121 -21

4

221

18

2 A = 0 1

L 0 4 3 , 0 2

(A.5)

(A.6)

respectively.

A.4 TheNumerator Determinant The numeratordeterminantNr is fomed from the characteristicdeterminant by replacingthe ,tth column in the characteristicdeterminantwith the column of valuesappearingon the right-handside of the equations.

A.5 TheEvatuation ofa Deteminant761 For example, the numerator determinants for evaluating i1, 12,and t3 in Eqs.A.l are

-e -12 6 21,

l-33 N,:l 3

I s0 -4

I':

-;

-11

(A.7)

221 -r )l

;l

t,; '; ;f

(4.8)

and

ltt

e

331

6

31.

&= l-3 l-8 -4

(A.e)

501

The numemtor deteminatrts for the evaluation of ?)1,,2, and ?)3in Eqs.A.3 are

14

-1 0

l s

0 2

&=116 4 31,

^ ,'1;l li

(A.10)

(A.11)

and

l" N,=

l0 17

-t

4l

4 161. 0 sl

(A.12)

A.5 TheEvaluation of a Determinant The value of a determinant is found by expatrding it itr tems of its minors. The mfuor of any element in a deterrninant is the determinant tlat remains after the row and column occupied by the element have been deleted. For example,the minor of the element 6 in Eq. A.7 is

| ::

1zl

I s0 221'

762

TheSotution of tjnear sjmutianeous Equations whilelhe minorol lhe element22 in Eq.A.- r\

l-33 |

3

-el 6l'

The cofactor of an element is its minor multiplied by the signcontrolling factor 1{i+'' where i and i denotethe row and column,respectively, occupjedby the element.Thus the cofactorof the element6 in Eq.A.7 is

33 rz 'r{' +also zz| I and the cofactorofthe element22 is

-l '1 1 : *| : 1 13- 3 3 6 The cofactor of an element is also rcferrcd to as its siened minor. tte sign-conrfoltiDg laclor l" / will equal l;r I depending on whetheri + j is ar evenor odd integer.Thusthe alg€braicsignof a cofac tor alternatesbetween+1 and 1 aswe move alonga row or column,For a 3 X 3 determinant,theplus and minussignsform the checkerboardpat' tern illustratedhere:

+ +

+ + -

+

A determinant can be expanded along any row or column. Thus tle fiISt stepin making an expansionis to selecta row j or a columnl. Once a row or columnhasbeenselected,each elemert in that row or columnis multiplied by its signedminor, or cofactor.The value of the determinantis the sum of theseproducts.As an example,let us evaluatethe deteminant in Eq. A.5 by expandingit along its first column.Following th€ rul€s just explained,wewrite the er?ansionas

' ' l- , , , , 1 ' 1 1, 0 . ' , , ^ -' r r' ' io 2.2] -, , - , ,o ' 2t I -l 4 l-4 -

fhe 2 x 2 determinantsin Eq. A.13 can alsobe expandedby minors. The minor of an elementin a 2 x 2 determinantis a singleelement.Itfollows that the e).pansionreduces to multiplyhg the upper-left element by the lower-rightelementandthen subtmcthgftom thisproducttheproduct

A.5 TheEvatuation ofa Determinant763

of the lower-leftelementtimesthe upper-rightelement.Usingthis observation,we evaluateEq.A.13 to A, = 21(132- 8) + 3(-198 - 48) - 8(18+ 72) = 2604-'t38 - 120: 1146.

(A.14)

Had we eleded to expandthe deteminant alongthe secondrow of ele' metrts,we would havewritten

-,'l

-rzl

l-o 1.' r , ,',l Jz r ^: -3(-1)l , ::l+6(+1)l-: 8 l-4 221 l-u 221 = 3(-198 - 48) + 6(462 96) + 2( 84 :

138+ 2196 312:1146.

-ol 4l

72) (A.15)

The numericalvaluesof tle determinantsN1, N2, and N3 given by Eqs.A.7,A.8,andA.9 are N1 = 1146,

(4.16)

N2 = 2292,

(4.17)

& = 3438.

(A.18)

and

It folloss ftom Eqs.A.15 tbroughA.18 that the solutionsfor il, ,2,andi3in Eq.A.1are

-

A

.

A

and

;,:f; =:e.

(4.1e)

764

lhe 50tuti0n Equafions of Linear simuttaneous

We Ieave you to veriry that the solutions for 01,02,and ,3 in Eqs.A.3 are 1 ) ,= - :

9.8V,

I t,t l,, : :- = -23.6 V.

(A.20)

and

u] = =

: 36.8 v.

A.6 Matrices A system of simultaleous linear equations can also be solved using matrices. In what folowq we biefly review matrix notation, algebra, and terminology.I A matdx is by definition a rectangular array of elements;thus

(4.21)

is a matdx with m mws and n colunns.We describe A as being a matdx of oder mby n,or m x ,1,where ,1 equalsthe number of rows and r' the number of columns.We always specify t}le rows fint and the columlls second. The elemerts of tllre matrix-atu aD, aB, . ..----can be real numbels, complex numbers,or tunctions. We denote a matrix with a boldface capital letterThe aray in Eq. A.21 is frequently abbreviated by writing

a : l"t)-",

\4.?2)

wherc.rt is the elementin the ith rcw and thelth column. If m : 1,A is calleda row matrix,that iE A:

Ia!

ar2 aB

'

a1,].

(4.23)

' An ercelenl int.oduclorylevel texl in nahix applicatioN 10circuil a&lysis is Lawene P, Huelsnd, Ct !tu, Matnta! an.l Liheo.vectot Spa.es INN York; Mcc.aw-Hill. 1963).

Atgebn 765 A.7 Matrix If

"l:]

: 1. A is called a column metlix, that is,

(4.21)

If m : r, A is caled a squale matrix. For example,if m : n = 3, il\e souare3 bv 3 matrix is

ap o'.) fa1 A= a 2 2a r l l aa 2 r 3r q2 a\l

(A.25)

I

Also note ttrat we use brackets [] to denote a matrix, whereaswe use verticallines I to denotea deteminant.It is important to know the difference.A matrix is a rectangular aray o{ elements A determinarl is a function of a square aray of elements.Thus if a matrix A is squarc,we can define the deteminant of A. For example.if

1,1 o = 1 2 tt]' Lo

*'" : l1 D - ]l l = 3 0 - 6 = 2 4 . l''

A.7 MatrixAlgebra The equality, addition, and subtraction of matrices apply or y to matric€s of the same order. Tko matdces are equal if, and only if, their correspon_ ding elementsare equal. h other words, A = B if, and only iJ,dij = bii for all i and i. For example,the two matrices in Eqs.A.26 and A.27 are equal because a1 : b1,a12= bD,a21: b2r,andan: bnr

^-[': ?:] . =f: -?:l

(4.26)

(A.21)

766

lhe soLution of Linear Simultaneous Equations

If A and B arc of the same order. then

C:A+B

(A.28)

implies (4.2e) For example,if

^ = Lf4s r6z -1ol ,{l'

(A.30)

and

': | ;: ': ?:l then

,:l-1i; -?i)

(A.31)

(4.32)

The equation

D:A_B

(A.33)

implies Q1 : a4

b11

(A.34)

For the matdcesinEqs.A.30 and A.31,we would have

":l-:?"i ill

(A.r5)

Matrices of the same order are said to be confo nabl€ for addition and subtmction. Multiplying a matrix by a scalar & is equivalent to multiplying each element by the scalar.Thus A = /.8 if, and only if, ai, = ,t4r. It should be noted tlat t may be real or complex. As an example, we wil multiply the matrix D in Eq. A.35 bv 5.The result is

" =l;fl -i3l33l

(A.36)

Matrix multiplication can be peformed only if the number of columnsin the ftst matrix is equalto the numberof rowsin the second matdx.In otherwords,the productAB requirestlie numberof columnsin A to equalthe numberof rowsin B.The order of the resultingmatrixwill be the number of rows ir A by the number of columnsin B. Thus if C = AB, whereA is of orderm X pandBisoforderp X r,thenC will be a matrix of order,n x ,r.Whenthe numberof columnsin A equalsthe numberof rowsin B, we sayA is conformableto B for multiplication. An elementin C is sivenbv the formula c..: ).a.b, .

(A.37)

The formula given by Eq. A.37 is easy to use if one remembers that matrix multiplicationis a row-by-columnoperation.Henceto get tle ith, ith term in C, eachelementin the ith row of A is multiplied by the corresponding element in the ith column of B, and the resulting products are summed.The following example illustmtes the procedure.We are askedto find the matdx C when

.

1 6 3 2 ' l L] 4 6l

(4.38)

and

B =

[?;]

(A.3e)

Fint we notethat C will be a 2 x 2 matrix andthat eachelementin C will requiresummhgthree products. Tofind C11wemultiplythecorresponding elementsin rcw 1 of matxixA with the elementsin column1 of matdx B and then sumthe products.We can visualizethis multiplication and summingprocessby extractingthe corespondhgrow andcolumnfrom eachmatrix andthen lining themup elemenlbv element.Soto find C,, we have RowlofA Column 1 of B

Cn:6x4+3x0+2x1=26. Tolind C12wevisualize RowlofA Colurm 2 of B

768

IheSoLution of Linear Simuttaneous Equatjons

thus cn : 6 x 2 + 3 x 3 + 2 x (-2) = 17. For C2rwe have Row 2 ofA Column 1 of B and

C21=7x4+4X0+6X1=10. Firally, for C2 we have Row2ofA Columr 2 of B from which

C 2 2 : 1 x2 + 4 x 3 + 6 > < ( 2 ) : 2 . It followsthat

'N ":""=;i:

(A.40)

In general, matrix multiplication is not cormutative, that isr AB * BA. As an example, consider the product BA for the matrices in Eqs.A.38 and A.39. The matrix genemted by this multiplication is of order 3 x 3, and each telm in the resulting matdx rcquhes adding two products Therefore if D : BA, we have

ze zo ,ol I D=l 3 12 t8 L4

5 - r 0l

(A.4r)

Obviously,C + D. Weleaveyou to verify the elementsin Eq.A.41. Matrix multiplicationis associativeand distributive.Thus

(AB)C: A(Bc),

(4.42)

A(B+C):AB+AC,

(A.43)

(A+B)C:AC+BC.

( .14)

A.7

IIr Eqs.A.42,A.43,andA-44,we assumethat the mafficesare confomab]e for addition and multiplication. We have already noted that matrix multiplication is not commutative. There are two other properties of multiplication in scalar algebra that do not carry over to matrix algebra. First, the matrix product AB : 0 does not imply either A : 0 or B : 0. (NoterA matrix is equal to zero when all its elemeflts are zero.) For examDle.if

^ = 3[ i-] . " = l : : l to ol - 0 . 00

AB

Hence t}le product is zero, but neitler A nor B is zero. Second,tlre matrix equation AB : AC do€s not imply B : C. For example,if

"-[;3]":[;:]''."=l;:1, AB 4C

fi

4l

;s.

b uBr , c .

The fanspose of a matdx is formed by interchanging the rows and columN For example,if fr > rl

| /,il

l= qio.,r,*l',iul

lzssl

l:osl

The transpose of the sum of two matices is equal to the sum of the rransposes.lhal is. (A + B)r = Ar + B7.

(A.i5)

fhe transpose of the product of two matrices is equal to the product of the transposestaker in reverse order. In other words, IABII:

BrAr.

(A.16)

llatrixALgebG 769

770

TheSolutjon of Lin€ar Sinruttaneous Equations

Equation A.46 can be extended to a product of any number of matri-

IABCD]I = DrCrB/Ar.

(4.47)

If A : Ar, tne marrix is saidto be symm€tric.Only squarematrices canbesynmetric.

A.8 ldentity,Adjoint,andInverse Matrices An idenlity rnatrix is a square matrix where at : 0 for i + j, aJJdaii : I for i = j. In other words, all the elements in an identity matrix are zero exceptthosealorg the main diagonal,wheretley are equalto 1.Thus

0 1 0 0

0 0 1 0

t ; i lt i : l l " - [ i are all identity matrices.Note that identity matricesare alwayssquare.We will usethe slmbol U for an identity matrix. The adioint of a matrix A of orde X ,?is defined as

adjA: tArrl"x,,

(A.48)

where Aij is the cofactor of a,;. (See Section A.5 for the definition of a cofactor.) It follows from Eq. A.48 that one caII t]rinl of firding the adjoint of a square matdx as a two-step pmcess.First construct a matrix made up of the cofactors of A, and tlen transposethe matrix of cofacton. As an examplewe will find the adioint ofthe 3 x 3 matrix

[ 1 2 3 l A=l 3 2 11. L 1 1 5l The cofacto$ of the elements in A are

A 1 r- 1 ( 1 0 1 ) : 9 , A r , : 1 ( 1 5+ 1 ) : 1 6 , a13:1(3+2):5, Azr=-1(10-3):-7, Arr:1(5+3):8, Ar: 1 ( 1+ 2 ) = 3 , : 431 1(2 6): -4, A3' = 1(1 9) = 8, 433=1(2-6)=-4.

A.8

The matrix of cofacton is

[e R=l-7

16 sl 8 - 31 . 8

L-4

4l

It follows that the adjoint of A is

l s - t - + l

a d j A = B r =l - 1 6 8 8 1 . l s - 3 - 4 1 One can check the arithmetic of finding the ad.ioint of a matdx by using the tleorem adjA.A:

detA.U.

{A.4e)

Equation A.49 tells us that the adjoint of A times A equals the determinant of A times the identity matrix, or for oul example, detA : 1(e) + 3(-7) - 1(-4) = -8. If we let C = adj A . A and use the technique illuslrated in Sectiotr A.7, we 6nd the elements oI C to be

q r = 9 2 1+ 4 = - 8 , q2=18-74-4=0, cB=27-'7-20:O, c2r--16+24-8:0, c22=-32+16+8=-8, cx=-48+8+40=0, c a =l 5 - 9 + 4 : 0 , caz:10-6-4-0, ca3:15-3-20= 13.

Therefore

o ol

o o'l

[8 [t c = l 0 - 8 o l : - 8 1o 1 o l L 0 o - 8 1 L 0 o 1 l = detA'U. A squarematdx A has an inve$e, denoted as A-r, if

A-1A=AAr=u.

(4.50)

Identjty,Adjoint,andInve6ellaticet

77!

772

lhe Solutjon of Linear Simulianeous Equations

Equation A.50 tells us that a matrix either piemulripLied or posrmultiplied by its inverse generates the idertity matrix U. For the inve^e matrix to exist,it is necessary that the determinantofA not equalzero.Only square matriceshaveinverses,and the inverseis alsosquare. A foimula for findina the inverse of a mat x is ,

adi A

(4.51)

The formula in Eq. A.51 becomes very cumbersome if A is of an order laiger than 3 by 3., Todaythe digital computereliminaresthe drudgeryoI havingto find the inverseof a matrix in numericalapplicationsof matdx algebra. It foilowsfrom Eq.A.51 that the inverseofthe matrix A in the prev! ousexampleis

: "'[-'i_il 7 8 -3

: l

o.s;s I r.rzs ) 1 | -o.orio.rri

:rl

You shouldveriJythatA tA= AA r:U.

A.9 Partitioned Matrices It is often corvenient in matrix manipulations to partition a given matrl\ into submatrices.The o ginal algebraicopemtionsare then carriedout in terms of the submatdces.In partitioning a matdx, the placement of the partitionsis completelyarbitrary,with the oire restrictionthat a partition must dissectthe entire matrix.In selectingthe pa itions,it is also necessary to make sure the submatdcesare conformableto the mathematical operationsin which they are involved. For example, consider using submatrices to find the product C : AB.where

ti L:

2 3 4 3 0 2 1 - 1 2 1

;:l

? You can learn altemarive ncthods ior inding ihe inveEe in dy introdudory text on natrix theort. See,lor eranple,Fnnz E. Hohn- Elcmetuat! Mari,,lkbla (Ncw york:

I'latrices 773 A.9 Patitioned and

" [ : ]

Assume that we decide 10 partition B into two submatdceq 811 and Bzt;thus

":fn'l

"

[B",.]'

Now sinceB hasbeen paftitioned into a two'row column matrix, A must be partitioned irto at least a two-column matrix; otheiwise the multipiication ca rot be pedomed. The location of the vertical pa itions of the A matdx will dependon the definitions of 811and Bn. For example,if

T r_- l 8 ", , : l

|

|

f1l

|

ttJ

0 l a n d 8 ',- , = 1 , .

L lt

then Att must coltain tlree columns, and A12 must contain two colurms, Thus the partitioning shown in Eq. A.52 would be acceptable for executins the Droduct A3:

l 5t r4 c=l_10

3

2

3

i:l I

2

L:;-i

.

(A.52)

3 0

Il on the other hatrd. we DartitioD the B matrix so that

"". ""=[i] "":[3] then A11must contain two columns, and AD must contain thrce columns In this casethe partitloning shown in Eq. A.53 would be acceptablein executins the Droduct C : AB:

iil

[11 131

c=l-1 0

2 0 I 3 0

(4.53)

774

TheSolution of Linear Simulhneous Eqoations For puiposes of discussion,we will focus on the partitionhg given in Eq.A.52 and leaveyou to verify that the partitionirg in Eq.A.53leads to the sameresult. From Eq. A.52 we can write

- tB"l A118.. 4'12821.

g

1l ' e'rl

la.,'J

(A.54)

It followsfrom Eqs.A.52 andA.54 that

"^:l-i 1i]nl"i] -":[.i:]'li] and

"[j] The A matri{ could also be pa itioned horizontally once the vertical partitioning is made consistent with tie multipiication operation. In this simple problem, the horizontal paltitions can be made at the disdetion of the analyst. Therefore C could also be evaluated using the partitionhg shownin Eq.A.55:

1 5

c=

2 4

3 3

4 2

5 1

2 0 I

-

1 O 2 l - 3 1 0 1 - 1 1 0 1 0 2 1 1 2 0

3

.

(A.55)

A,9 Fdftitioned l,latrices t75

crr-ArlBi+A,rB1. c2t-a2tilll-ArB2.

You shouldverif] thal

't

tr 2 1|

c , ' = l;; ; l l o l 'f l4; s, lll lll ; l L-al

_ f - ' l + f u l _I 1 1 l L 7l

[-'

L 6l

o

L13l'

2l[ 2l

r'l

[-i ll c , r - ol | - , l l o l l 9 ' l ; ] L o 2 1 l L - 1 1L 2 o . l ' t-rl t t

t t sl t t -r:'l t l

= l 1 l + l0 l : l 1 1 . L-11 L-61 L 7l and

'

[".]

I 13'l c = l - 1 31 .

L-;l

776

Ihesotution of Ljnear Simuttaneous Equations

A.10 Apptications The following examplesdemonstrate some applications of matdx algebra in circuit analysis.

Use the matdx method to solvefor the node volt ages?]1and 02in Eqs.4.5and 4.6.

Solution

It follows frcm Eq. A.62 that the solutions for x'r and 1,2are obtain€d by solving for the mat x productA-rI. To find the invene of A, we first find the cofactorsof A.Thus

^,r-( 1)10.6):0.6, ^1r=( 1f( o.s):0.5,

The first stepis to rewrite Eqs.4.5and 4.6in matix notation. Collectingthe coefficientsof r,1 and t2 and at the same time shifting tlle constant terms to t}le right-handsideofthe equationsgivesus 1.7rr

0.522: 10, (A.57)

d,1 : ( 1)3( 0.5) : 0.5,

Ln= ( r)\1.1):1,.i. The matrix of cofacton is

. 13i ?tl

0.5ttt+0.622:2. It followsthat in matrix notation,Eq. A.57 becomes

I r.7

o . s tl u , l t l o l

L u . s o . oLl o l L z l ' AV=1,

(A.5e)

lnh rrrl a d' j A - B ' - l - . ,:. LU.) 1./l

A:

v:

L U.J

U.OJ

n - - sl l r 1 . 7 , { 0 . o r , 0 . 2 5 -r 0 . - ? . v ul

(A.60)

From Eqs"A.65 and A.66, we can write the nverse of the coefficient malrix, that is,

t ,' .,t .

^ ,-

tb2)

trol t = l: .

1 fo.o o.sl olz os tz

To fiDdrhe elemenrs ot lhe v malfi\. we premultiply both sidesof Eq. A.59 by the inverseof A; thus (A.60)

EquationA.60 reducesto

o.sllrol o ',' : z1oo[o.o o . sr . z z l 1oo[7] 77 8.4I

| e.oel ' I i0.91

(A.68)

It follows direcdythat (4.61)

It,l

l,')

v = A-rL

(467)

Now the productA-1I is found:

t z

Uv=A1I,

I t7 | ".

.._1,

A'Av=Art.

(4.65)

The determinantof A is de'A

0.51

(A.64)

and the adjoint ofA is (A.58)

| 1;7

(A63)

(A.62)

I e.oel

110.911' or ?)1: 9.09V and z)2= i0.91v.

(A.6e)

A.10 Applicatjons 777

1)'(-45 - 80): 12s,

Usethe matdxmethodto find the threemeshcur rentsin thecircuitin Fig.4.24.

Ar1:(

Sotution

^,r = (-1)5(-1oo 2s) : 12s,

t}le me.h-cuffenr equationrlhat de\cribelhe cifcuit in Fig.4.24are givenin Eq.4-34-Theconstraint equationimposedby the currentcontrolledvoltage sourceis givenin Eq.4.35.Wlen Eq.4.35 is substituted into Eq. 4.34, the following set of equations

^i = (-1f(20 + 200\:220,

L.n : ( 1)4Q2s- 1oo): 125,

ar:

( 1)'( 1oo-1oo)=200,

r\I

( 1 1 6 ( 2 -5205 ) - 2 2 5 .

Thecofactormatrix is 25ii

5i2

20i1= 50,

6s 7ol

| ,o = B | 125 125 r2s L220 200 22s)

5ii+10ir-4t3=0, -5i1- 4i2+9l:Q

( A7 0 )

from which we can write the adjoint of A:

In matrix notation,Eqs.A.70 reduceto AI = V,

r,

fra (A.71)

rrs

d d i A- B / - | " t

ltnl

, r i , o o|

ps ns)

ln

5 -2ol

(A.14)

,o,n

The determinant of A is 2U 25 -5 5 : 10 4 detA

I A=l s 10 -4, e) L s 4

s

ti'l

r= ; .

4

9l

: 25(90 16)+ s(-45 - 80) s(20+ 200)= 125.

;,

It followsfrom Eq. A.73 that

and

12s 2zol

. f74 A r = ;l 6s 12s 2001. --- '70 125 225)

['l v-l 0l L 0l It follows from Eq. A.71 that the solution lbr I is

t=A'v.

(4.72)

We find the inverseofA by usingthe relationship , A-'=

adi A -:----.. det A

(A.73)

To find the adjoint of A, we first calculate tie cofac toIS of A. fhus All:(

1 F ( 9 0- 1 6 t ) = ' 7 4 ,

Arz: ( 1)3(4s - 20):6s, dr3:(-1)4(20+50):70,

(A.76)

The solutionfor I is

,|

-+ r:s )ol so | [2,].oo'l

I - r r { l b s 1 2 5 , , n 0| '-"1-o

0l=

r2s )251L ol

2 6 . 0 0] . ( A . 7 / )

L28.oo.l

The meshcunents follow directly from Eq. A.77.Thus

I i;l Izo.ol ,, ,r

126.0 | 128.01

{A.78)

or i1 = 29.6 A.,i2: 26 A, and13: 28 A. ExampleA.3 illustrates the application of the matrix method when the elements of the matrix are complex

778

Ihe sotution of Linear sinrultaneo6 Equations

Usethe matrix methodto fhd the phasormeshcur, rents11and 12in the circuit in Fig.9.37.

The cofactormatrix B is B

Solution

26 [( t 02

i13) I 27 - lrb)l j16) (13 lia) ]

(A85)

Summingthe voltagesaroundmesh1 generatesthe

The adjoint of A is

(1 + j2)rl + 02 -- j16)(rr r) = 1501!:. (A.7e)

- B / f l : : t : l : l l : -- l : : ' ( A 8 o ) adiA L( )7 jr6) (r3 jr4r)

Summingthe voltagesaroundmesh2 producesthe equation

The delerminantofA is

(12 y'6xr,- 11)+ (1 + j3)r' + 39r,:0.(A.s0) The current controlling the dependent voltage

t, : (rr

Iu).

fl3 - illlf2h | ,lJr

(A.81)

AJter substitutingEq. A.81 hto Eq. A.80, the equationsare put into a matix format by lirst co ecting,in eachequation,t}le coefficientsof 11and Ir; thus fll li4)l r 2 7- i l b r l

.","=lflit',ill,[i.';,2; : 60 - j4s.

AI=V,

(4.83)

-l12 jt6rl "^ _ f D / 1 4 127+ tt6 (26+ jrll l' fr , l | = l;'1. andv: Lr2.l

L

u

It followsfrom Eq.A.83 that (A.s4)

The inverse of the coefficient matdx A is fourd usingEq.A.73.In this case,thecofactorsofA are ^u=( 1 ) , ( - 2 6- j r 3 ) : 26- j13, Liu= e\3(.2'7 + j16) = 2'7 - j16, ^I,l : (-1)i(-12 + j16) = 12 - j16.

Ar, = (-1f03

j14)= t3

(60 - i45)

14.

(4.88)

EquationA.88 canbe simplifiedto

6 0+ i 4 s f t - 2 6 l 1 J ) 0 2 l l b ) l 'a- r 5 b 2 sL l 2 7 j l o ) ( r 3- j 1 4 ) l I | 65 - ,r3o 96

t28l

:'sl -oo lr+s o+, rr;]

(A8e)

SubstitutjngEq.A.89 into A.84 givesus

I

I = A-rV.

(A.87)

[(-26 113) 02 - i16)l t(-27 j16) (13 j14))

-

l 'r' s. -n1/.r r l

jtb)

Theinve$e of the coefficientmatrix is

rll l l o r t /= l 5 n 0 " . , ^ - . 1 6o / i t 2 b- j l r l ) n.

No\using matrix notation.Eq.A.82 is written

I l2 - jtn)Q-

t l , l - I f ( 0 5- 1 1 3 0 ,( q o 1 2 8 )l l s o 0 r.:l :-S f-oO llasl (qa- lt7) U ] l( 26 - js2\l (A.e0) l(-24 jsst)' It followstrom Eq.A.90 that j 5 2 \- 5 8 . 1 4 , l l o . 5 7 ' A . lt - ( )6 ,r o , ( 2a - /58) - r.2.'7, 122.48'A. ' -'' l

In the fiist three exampleEthe matrix elementshavebeennumben-real numbersin ExamplesA.1 and A.2, and complexnurnbersin Example A.3. It is also possiblefor t}le elementsto be fimctions Example A.4 illustrates the use of matrix algebrain a circuit problem where the elementsin the coefficient matrix are functions

A.10 Apptkations779

Use the matrix method to derive expressions for the node voltagesy1and y, in the circuitin Fig.A.1.

-

Sotution Summingthe curents away ftom nodes 1 and 2 gereratesthe followingset of equations:

u v " - j31 K

+ llsc + (V - y,)sc :0.

u + (v2 v)sc + (/, ;

/'4's2)

Letting C = 1/R and collecting the coefficients of Y1and Y2gives us

(G+2sc)vr-scvz:Gvs, sCU+ (G + 2sc)v2= scv{

(A.e3)

Writing Eq.A.93 in matrix notationyields

AV:t.

fG + 2sc

A=L

rc

A.4. tiguruA.1 d Ihe circLrit for E/"ampte

Ys)sc:0

(A.e1)

Thedeteminantof A is rC I lC+2sC derA | ^ ^ ^ ^ l - G ' + t : C C + 3!C2. t s L (A.98)

The inverseof the coefficientmatrix is

[c + 2rc

sc I

G+zlcl L rC (G2+4scc+3s2c2)

sc I

c+2{c]'

(A.ee)

It foLlowsfrom Eq. A.95 that

n =Lh) f l l . " ' a r =fL.r( cyr- ll i l

lG+ 2sc

It followsfrom Eq.A.94 that

v=ArI.

ttl:

L sc

(4.e5)

As before, we find the inverse of the coefficient matrix by first finding the adjoint of A and the determinant of A. The cofactors of A are All=( lflc + 2sCl : 6 1 2tg. ^r,=(-1)3(-sc)=rc, lzr=(-r)3(-sc):sc, 4,, : (-1)4[6 + 2scl = G + 2sC. .a I ^ "+ ' , ^lsLl . Cr

(G' +4rcc+Ji2c2)

(A.100)

Carryiflgout the matdx mu]tiplicationcalledfor in Eq.A.100 gives

1 tvll = l(e +2scc + Cc'\v"1 q'cc + tCCll 1z"cc+ z?c'\v, )' lv,] fc, + (A.101)

Now the expressions for l{ and y, can be written d ectlyftom Eq. A.101;thus

The cofactoi matdx is l t-: + ) " a B= ^-"' L .rc

sc llcysl

G+2sc)lsc\)

lG2+z\cc+?c' lvs (A.102) ' (.G2 + 4sCG+ 3s2cz)

\A.96J

and therefore the adjoint of the coefficient matrix is

and

2(sCG+ ?c2)vs (G'+4tcc+3"tC\

(A.103)

780

TheSolutjon of Lin€ar Simultaneous Eq0ations

In o r filral example,we illustralehow matrix algebracan be usedto analvzethe cascadeconnectionof two two-oc,ncrrcuns.

Show by meansof matrix algebrahow the input variablesyr and 1r can be describedas functiom of the output vadablesy2 and 12 in the cascadeconnectionshowr in Fig.18.10.

Theseconstraintrelationshipsare substitutedillto Eq.A.104.Thus

lrlt:;: Illl l s:n a\zllvrl latl at I| r\ )'

Solution We begin by expressing, ir matri\ notation, the relationshipbetweenthe irput and outputvariables of eachtwo-po circuit.Thus

(A.107)

The relationshipbetweenthe hput variables(14,1r) and the output variables (y2, 1, is obrained by substitutingEq.A.1O5into Eq.A.107.Theresultis

f"'l : fai, ,i,lf,i, -,i,1f",l { A ' 0 8 )

trl

_ l^i., ,:,,11v51 (4.104) L.t d.ll ril

trt .at

at) lai

, , ' , 1- / . , -

After multiplying the coefficient malrices, we have

tlll

lAli]

(A'05)

Now the cascadeconnection imposesthe constraints (4.106)

ttl

-(a\1ai,+ a:labJlv,] + l(a\pL o\d!,) + abqh t(41a'i1 ) U,) @iph + ai2ah)

(A.10e)

Note that Eq. A.109 correspondsto writing Eqs.18.72 and 18.73in matrix form.

Appendix

Complex Numbers

Complex numben were hvented to pemit the enraction of the squarcroots of negative numben Complex numbers simplify the solution of prcblems tlat would otherwise be very difficult. The equation I + 8x + 41 : 0, for example, has no solution in a number system that excludes complex numbe$. Thesenumbeis, and the ability to manipulate them algebraically, are extremely useful in circuit allalysis.

8.1 Notation There are two ways to designatea complex nurnbei: wit]l the cartesian,or rectangulai,form or with the polar, or trigonometriq folm.In the r€ctarg ar Iornr, a complexnumber is written in tems of its real and imaginary components;hence n:a+

jb,

(8.1)

whered is the real component,b is the imaginarycomponent,andj is by definitionVl.1 In tlle polar form, a complex number is written in terms of its magdtude (or modulus)and angle(oi argument);hence (8.2) where c is the magnitude, d is ttre argle, e is the base of t]le natural logaithm, and, as befoie,j = V-1. In the literature,the symboll:q is frequentlyusedin placeof erd;thatis,the polar form is written

,t = c1!.

(8.3)

Although Eq. B.3 is more convenient il printing text material, Eq. B.2 is of pfmary importance in mathematical opemtions because the rul€s for manipulating an exponential quantity are well known. For example,because * O)" = l', then (et1' = pto; lgsausey = 1/r,, $et: e-ig - 1/ei0; and so foith. Becausethere are two ways of expressingthe samecomplex number, we need to relate one form to the other. The transition from the polar to the iectangularform makesuseofEuler's identity: (8.4)

r You may be nore fanilid wirh the notatior ; = \/=- In electrical engineering, i is used as the s)dbol lof curent. and tence in eleclrical engineering liierature,l is Ned to derote

781

742 A complex number in polar form can be put in rectangular form by writing cel:

c ( c o s d+ j s i n d ) = ccosd + icsind)

(8.5)

The tmnsition from rcctangular to polar form makes use of the geometry of the dght tdangle, namely, / _\ a+tD=lva'+t1'lep \ /

(8.6)

tan9 : b/a.

(B.7)

It is not obvious ftom Eq.8.7 in which quadranfthe angle p lies.The ambiguity can be resolvedby a graphical represertation of the complex number.

8.2 TheGraphicaI Representation of a Comptex Number A complex number is represented graphically on a complex number plane, which usesthe horizontal axis for plotting the real component and tigure8,14 ThegEphical rcpresenrarion of' + i, lhe \erricala"\istor plollinglhe imdginar)componeDt.The angleot rhe when , andb arcbothpositive. complex number is measuredcounterclockwise ftom the Dosifive real axis. Ile graphicalplot of lhe complexnumber, - a + 1i t 7o",;f *e assume rhata andb arebolh positive. is sho\D i0 Fig.B.t. This plot makesvery clear tlle relationship betweenthe rectangular and polar foms. Any point in the complex-numberplane is uniquely defhed by giving eithei its distance from each axis (that iE a and b) or its iadial dis tance from the origin (c) and the angle of the mdial measurementp. It followsfrom Fig.B.l that d is in the first quadrantwhen d and 6 are = 4+j3 5tr0!?" 4+j3:5/143.13. borhpo
B.l

component.Thusthe conjugateof d + lb is l' - jb, and the conjugateof .i+ jbis-a - Jb.Whenwewrite a complexnumberin polar form, we fbrm its conjugate simply by reversing the sign of ttre angle d. Therefore the conjugate of c/q] is cAz:. The conjugale of a complex number is designatedwith an astedsk.In otler words,'1. is underctoodto be tle conjugateof r. Figue 8.3 showstwo complexnumbersand their coniugatesplotted on the complex-numberplane. Note tlat conjugation simply rcflects the complex numbers about the

nz:

Arithmehc operatjons 783

a+jb=cle.)

n: = a- jb=<:I 01 tisuc 8.3A lhe co'iptexnumbeE,1and,r amdthen conjugates,iandzi.

B.3 Arithmetic0perations Addition(Subtraction) To add or subtract complex numbers,we must expressthe llumbels in tectangular folm. Addition involves adding the real parts of the complex nunbe$ to lolm the real part of the sum, and the imaginary parts to folm lhe imagina4pan of the .um.Thus. i[ $e aregiven t\=8+

j16

n2: 12

j3,

and

tllen nr + n2 = (8 + 12) + j(16 - 3) = 20 + j13. Sublraction follows the samerule. Thus nz- \=

(12

8)+ j(-3 - 16):4 -r19.

If the numbers to be added or subtracted are given in polar form, tley are first converted to rectangular form. For examplg if

ry= 10/:!i and ,12=5/-135', then

\ + n 2 = 6 + i 8 3 . 5 3- 5i 3 . 5 3 5 : (6 3.53s)+ i(8 3.53s) - 2.465+ j4.465:5.10/6L.L0",

784

Comptex Numben

and ( 3.535 i3.535)

+i8

nr-n2:6

- 9.535+ i11.535 : 14.966 /50.42'.

Muttiptication(Division) Multiplicationor divisionof complexnumberscanbe carriedout with the numberswritten in either rectangularor polar form. However,in most cas€s,the polar form is more convenientAs an example,let's find the prcduct /,1/r2when nr = I + j10 and n, : 5 j4 Using the rectangllar nlnr: (8 + j10)(5

j4) = 40

j32 + jso + 40

=80+j18 : 82/12.68'. If we usethe polar form, the multiplicationthn2becomes ntnz : (2.81'/ 51.31')(6.40/

3866")

- 82/12.68' = 80 + i18. Thefirst stepin dividingtwo complexnumbenin rectangularfolm is to multiply the numeratorand denomhatorby the conjugateof the denomi nator.This reducesthe denominatorto a real number.We then divide the real numberinto the newnumeiatorAs an example,let's find the valueof n1= 6 + i3 andn2 = 3 Jl we have nlfn2,where n 1_ 6 + i 3 _ ( 6 + i 3 X 3 + i l ) (3-i1X3+i1) n2 3-i1 18-j6 ia 9 + 1 =

t -l - + ! ; 1_5 -

t0

't \'

l

il 5

= 2.121!t:. In polar form, the division of /'l by ,t2 is h. n-

6.11/ 26.5'7' 1tA /

1141.

= 1.5+ i1.5.

-_-'

-

8.5

8.4 UsefulIdentities In working with complex numbers and quantities, the following identities are very useful: (8.8) /- tv i\ :

1

(8.e)

. 1 I = -l . , "+it = -1 otit/2

Civen lhat,t = a + Jb - cll.

=

+

(8.10) (8.11)

i

(8.12)

ir toUowslhal

nn'=d+8=&,

(8.13)

n+ n'=2a,

(8.14)

n

(8.r5)

n:i2b,

nln' : 1.12!:.

(8.16)

8.5 TheIntegerPower of a Complex Number To raisea complexnumberto an integerpower&,it is easierto first \\.rite the comDlextrumberin Dolarform,Thus nk:1a+p;k = (cei\k : ckeikr

= l( cos&d + j sink0). For example, (2ei12)5 = 25ei&J' : 32ei6v

= 6 + jn.1\ and (3+ j4)1: (5eis313)4= 51eP1232' :625epr252"

= -s27 j336.

Ihe IntegerPower ofa Complex0mber

786

8.6 TheRootsof a Complex Number To find the ,tth root of a complex number, we must re€ognizethat we are solviry the equation xk

cejo : 0,

(8.17)

which is an equation of the tth degiee and therefore has & roots. To find the ,t roots, we fint note that ceia = cet(!+2r) = cei(!+4a) =

--

(8.18)

It follows fiom Eqs. B.17 and 8.18 that

kei\1/k = cltkeio/k, = cr/ke7a+2nvk, hIcej(a+2n)lr/k = crlkela+4nYk, t : Icejla+4n)lr/k \=

(8.1e) (8.20) (8.21)

We continue the process outlined by Eqs. B.19, B.20, and B.21 until the roots start repeating. This will happetr when the multiple of 7I is equal to 2f. For example,let'sfind the four roots of 81ei60'. We have \

- q]1/aeidJla - 3ei1s',

r Y2 - 8l lelr6o-1601 - JP o<', rr 8 l a e r ' 6 07 u l 4 J ? r ' o + . 81r/4ei(60+r$0)/4= 3eJr35', 4: = r. 8 t l a p i r 6 0l a o ' a - V r r - r " = j e i r , Heie,.r5 is the same as xt, so the roots have started to repeat.Therefore we krow the four roots of 81et60'are the values given by -xr,.x2,.r3,and ra. It is worth noting t}lat the rcots of a complex number lie on a circle in tlre complex-number plane. The radius of the circle is cll*. The roots are ur Jormly distdbuted around the circle, the angle between adjacent roots j60' bel{],g equal to 2T fk ftdians, or 360/A degrees. The four roots of 81e are

showr plotted in Fig.B.4.

Figure8.4A Th€fourrootsof 8le ls'

More MagneticaLl,y on ,^ ooo'"0"L Coupl"ed Coil,s andIdeaI lranslormers

C.1 Equivatent Circuitsfor MagneticaLLy Coupled Coils At times,it is convenientto model magneticallycoupled coils with an equivalentcircuit that doesnot involve magneticcoupling.Considerthe two magreticaliycoupledcoils sho$,nin Fig. C.l. The resistances R1 and R, representthe windirg resistanceof cachcoil.The goalis to replacethe magneticallycoupledcoils insidethe shadcdareawith a sel of inductors that are not magneticallycoupled.Beforc derivitg the equivalentcircuits, we mustpoint out animportant restriction:Thevoltagebetweentermimls b and d must be zero.In othcr words.if lerminalsb and d can be shoried Fig(reC.1A Ihe cncuitusedlo devetop an€quivalent togetherwithout disturbjngthe vollagesand currentsin the original cir- cncuitfor magnetjcalty coupted coits. cuit, ihe equivalentcircuits deived in the maierial that fotlows can be used to model the coils.This resl{ction is imposedbecause,while the equivalentcircuitswe developboth havefour terminals,two ofthose four terminalsare shortedtogether.Thus,the samerecluirementis placedon the originalcircuits. We begin dcvelopingthe circuit modelsby writing the two equations that relate the terminal voltagesor and 02 to the terminal currentstr and i2.For th(rgivenreferencesand polarity dots,

,,- ,r!" *ft

{c.1)

and ,,*.

TheT-Equivalent Circuit

(r.2)

a L t M

LzM

c

R2

To arive al an equivalentc cuit for thesetwo magneticallycoupledcoils, + - i we seek an arrangementof inductorsthat can be describedby a set of 1 \ M equationsequivalentto Eqs.C.1and C.2.Thekey to finding the arange, ment is to regardEqs.C-1andC.2asmesh'currcntequationswilh it and i2 as the meshvariables.Thenwc nced one meshwith a total irductanceof Ir H and a secondmeshwith a total jnductanceof l,2 H. Furthermore,the two meshesmust havea commoninductanceoIM H. fhe T-arrangement FigureC.2A TheT-equivalent circuitforthemagnetiof coilsshownin Fis. C.2satisfiestheserecuiremcnt$ cattycoupted coiLs of Fig.c.1.

788

andldeallransfomer !1ore onMasnetjcalty Coupt€d Coils You should verify that the equations relating q and ?)2to t1 and i2 reduce to Eqs.c'1 and C.2.Note the absenceof magletic coupling between tle inductors and the zero voltase between b and d.

Ther'-Equivalent Circuit we can derive a r-equivalent circuit for tlle magnetically coupled coils shownin Fig.C.1.Thisdedvatior is basedon solvingEqs.C.1 and C.2for the derivatives dild/ and di2/dt a]J'dthen rcgaldlng the resulting expressions as a pair of node-voltage equations.Using Cramer's method for solving simultaneous equations,we obtain expressionsfot dildt and dirldt:

di.

ar

l a 1M l lr2 L2l

L"

L t L )- v 2 " '

lt , u

M

LtLz M"

'

(r.3)

l Mr " l

l v' t " I

di ol lvl ... --L r d,= Lr; 14' L L. - Ir'u Lth

^ur. M' -

rc.4)

Now we solve for 4 and i, by mulriplying botl sidesof Eqs C.3 and C.4 by dt and then integrahng:

a: t.rot,7f 61"'ua,- LrL2M-

M2

I' ,0,

(c.5)

and

iz = iz(O)-

M

t' L, I' I D.dt+-I u'dr. LtL'z M'.Jo

L1L2 - M2 J"

{C.6)

If we regard 11 and t'2 as node voltages,Eqs. c.5 and C.6 descdbe a circuit of the folm shown in Fis. C.3.

FigureC.3A Thecjrcuitusedto derjvethez-€qujvatent circuitfor magneticaLty coupted coils.

c.1 Equivatent forllasneticatly Cncuits Coupled toits 789 All that remairBto be done in dedvingthe z-equivalentcircuit is to find ZA,lB, andZc asfunctionsof Zl, L2,and M. Weeasilydo so by writing the equationsfor il and i2 in Fig. C.3 and then comparingthem with Eqs,C.5andC.6.Thus t l il - irfo) , la..h LAJO

,

t t t llo\

LBJU

or)di

=,,,0), (/--|")[,,* i"l'",,

(c.7)

and

- -l.l',.0, l"l"'," " r* ,,- ,,t0,

=i,ro1, fl',,a,,(L. *) I',*

(c.8)

Then 1 LR

LtQ

(c.e)

M2'

7 _ L2- M LA LrL2 - M2'

(c.lo)

1 Lc

(c.11)

L 1 LrL2

M Ml'

Wher we incorponte Eqs.C.9-C.11into rhe circuit shownin Fig.c.3, the rr-equivalentcircuit for the magneticallycoupledcoilsshownin Fig.C.1is asshowr iDFis. C4.

Figure C,4a Ther-equivalent circuittorthe magneticaLty coupted coikof Fig.C.1.

790

M o r eo n l v d 0 n e l i cl voC o - o l e Ido ' f d r d l d p . l , d 1 J o r r "

UL,

tt'

M

rigureC.5.d The'-equivatent cjrcuitused for sinLrsoidat steady-state analysis.

Nole that the initial valuesofjl and i, are explicitin the r'-equivalenl circuitbut implicit in the T-equivalentcircuit.We are focusilg on the sinu soidalsteady-state behaviorof circuitscortaining mutual inductance,so we can assumethat the injtial valuesof ir and i2 are zero.We can thus elimhale the curent sourcesin the 7-equivalentcircuit, and the circuit shownin Fig.C.4simplifiesto the one shownin Fig.C.5. The mutual inductancecaries its own algebraicsign in the T- and ,-equivalent circuits.In other words,if the magneticpolarity of the coupled coils is reversedfrom that given jn Fig. C.1,the algebraicsign of M reverses.Areversalin magneticpolarity requiresmovingone polarity dol without changingthe referencepolarities of the terminal currents and voltages. ExampleC.l illustratesthe applicationof theT-equivalentcircuit.

a) Use the T-equivalent circuit for rhe magneticaly coupledcoilsshownin Fig.C.6to find the phasor currcnrcI, and 1.. tle .ource trequenc)i. 400nd/s. b) Repeat(a), bul with the polarity dot on the sccondarywindingmovedto the lorverferminal. Figure C.7a TheT-equivahnt circujtforihe maqneticatty coupted coilsin trampte C.1.

Sotution

j2400

j100

.|r'cxffie' a) for lhe polafit)dol\ 5bosDin Fig.C.6. a value of +3 H in the T-equivalert cilcuit. Theretoferhe threeiDductaDces in lhe
M=9

3:6Hl

L2

M=4

3:llli

modet0fihe equivatent Figurc C.8a Ihefrequency-domain 5000 jlooo 20oo j2400o

j400o 1000

M = 3H. Figure C.7 showsthe T-equivalentcircuit. and Fig.C.8showsthe frequency-domain cqujvalcnt circuit at a ftequencyof400 rad/s. Figure C.9 shows the frequency-domarn circuitfor the original system.

IL

300O"V

-\" F i g u EC . 9a c i h r ! o f t i g .C . 6a. i L hr h et r a q r e r i c a t . , coilsrcptaced bythenl-equivatent circuit. coupted

500() j100o a 200O j1200o t00O

300r!l'v

Fiquc(.6.

i

+

'

vr j3600o

Tle f,"ql"ny-do-.i1"q r\d q |

j2500o

| Lil fo f drple1. .

/ 1200O

r o r ' v a , €(rit' c r i \ I o . l v a q n e t i l aC.ol yr p l e( o i k

C.l

Here the magneticallycoupled coils are modeled by the circuit shown i]) Fig. C.8.To find the phasor curents Ir and 12,we first find the node voltageaooss the 1200O inductivereaclance.l[ne userhe lo\ er nodeas the refereDce, the singlenode-voltageequationis

v-300

v

700 i.500

v

1i200 o00 j2 t00

As before,we first find the node voltage aqoss the centerbralch, which in this caseN a capacitivereactanceof J1200O.If we usethe lower node as reference, the node-voltage equauonls

v-300

11200

= 56.s7/

Then , lr

300 (136- t8) - 61.'5 /-71 57"mA (rm, -^^,.^/uu + Jlfuu

I1

_

-',;^

/ZIUU

- iq.oJ/b3.,11"mA rfmsl.

b) Wlen the polarity dot is moved to tlle lower terminal of the secondarycoil, M cardesa valueof I H in the l.equivalent circuit.Beforecarrying out the solution with the new T-equivalentcircuit, we note that reversingthe algebraicsign of M hasno effect on the solution for 11and shifts 12 by 180'.Thereforewe anticipatethat lr = 63.25 /-71.57.

98.13.V (ms).

300 ( 8 js6) 700+ j4900 : 63.25/-71.57'InA (rms)

and

tr=ffi : 59.63 (rms). / 116.57'mA

14800o

j2800o

116.57'mA (rns).

We now proceed to find these solutions by using the new T equivalent circuit. With M : 3 H, the three inductanc€sin the equrv alent circuit are L1

M=9

( 3)= 12H;

L2

M=4

( 3)=7H;

M:

'

mA OrJJs)

and

I, : 59.63/

900 + 1300

Then

and t, - * YW

v

v:-8-ls6

136 j8 = 136.24 / 3.37"Y (rms).

V:

v

700 + 14900

Solving for Y gives

Solvingfor Y yields

figureC.10d Thefreou"n.y ,ncrt -ol donail eqrivdlFnt 3 H a n d o = = M 4 0 0r a d / s .

500O /100O 200O j4800o

j2800o 100()

3}j.

At an operating ftequency of 400iad/s, rhe ftequency domain equivalent circuit requires two inducror. d0da capacrlof. asshownin fig. C I The resulting frequency-domain circuit for the originalsystemappearsin Fig.C.11.

791

FigureC.1rA Theftequenc] domain equivaLent cncuitfor Exampte c.1(b).

792

andIdeat TGnsformers l,lore onJ'4asneticatty Coupted Coits

in C.2 TheNeedfor IdealTransformers

Circuits the Equivalent The inducto$ in the T- and ?r-equivalent chcuits of magnetically coupled coils can have negative values. For example, if -L1= 3mH, L, = 12 m}f, aud M : 5 mH, th€ T-equivalent circr.lit tequiies an inductor of 2mH, and the r-equivalent circuit requires an inductor of 5.5mH. Thesenegativeinductancevalues are not troublesomewhen you are using the equivalentcircuits in computations.However, if you are to build the equivalentcircuitswith circuit components,the negative inductorscan be bothersome,The reasonis that wheneverttre ftequency of the sinusoidalsourcechanges,you must changethe capacitorused to simulate the negative reactance,For example, at a frequetcy of 50 kmd/s, a 2 mH inductor has an impedanceof i100 O. This imped ancecan be modeledwith a capacitorhaving a capacitanceof 0.2rrF. If the frequency changesto 25 krad/s, the 2mH inductor impedance changesto j50 O. At 25 krad/s, ttris requires a capacitor with a capaci tance of 0.8&F. Obviously,in a situationwhere the ftequencyis vaiied continuously,the use of a capacitorto simulate negativeinductanceis practicallyworthless. You can circumvent the problem of dealhg with negative inductances by intioduchg an ideal transformer hto t}Ie equivalent circuit. This doesn't completely solve the modeling problem, becauseideal hansformers can or y be approximated. However, in some situations the approximatior is good enough to warant a discussionof using an ideal bansformer in the T- and ri-equivalent circuits oI magnetically coupled coils.

M

L

M

atL, Ma t,- uo

" , I1 rl l -;

LtLz

il rdeal a(LtL, - M' ) M

M2

ldeal

cncuit formaqneticatty coupl€d coils. Figure c.l2 a Thefourways ofusjnq anjdeattEnsfomer in th€T andz-€quivaLent

c.2 TheNeed forldeatlransforme6 in theEquivatenr Circuits 793 An ideal translormet can be used in two different wavs in either the L z M , y Tequi\alenl or lbe r-equivaleDtcircuir.FigureC.l2 .ho\ ( lhe rwo arrangementsfot each tlpe of equivalent circuir. + l Verifying any of the equivalent circuirs in Fjg. C.12 requhes showing ."'l1f'1? only that, foi any circuit, the equations relating \ arrd rz lo ilifdt aad dir/dr are identicalto Eqs.C.1andC.2.Here,wevalidarethe circuitshown IdealL: in Fig.C12(a);we leaveit to you to ve fy the circuirsin Figs.C.12(b),(c), and (d).To aid the discussion, we redrew the circuit showt in Fie.C.12fa) G) asFig.C.13,addingthe variablesi0 and zh. FisureC.13A ThecircuitofFiq.C.12(a)wiih t0andro From this circuit.

ill

,,=

l_

M\di.

\u

;)*+

( r,

u\ aio

M d

i(i1

+ia\

(c.12)

and

\",;)at

M d + ii. *\io

(c.13)

The ideal transformer imposesconstraints on ?o and io: (c.14) 4 = at2'

(c.15)

Substituting Eqs.C.14andC.15inroEqs.C.12andC.13gives ur:

Lr

.Iit

M d + dt tlt{di2\

(c.16)

and

Y =j { ' " ' t * l !

(c.17)

FromEqs C.16andC.17, di. di. 1 - L t . + M : at /11

(c.18)

,,: u!r1l + u4. dt

(c.1e)

and

EquationsC.18and C.19are identicalto Eqs.C.1 and C.2:thus,insofaras terminalbehaviorisconcerned,the circuirshownin Fis_C.13is eouivalent ro the magnerically coupledcorl<showninsrderbeUoiin fie. C.t.

194

l'4ore Coupted Cojkandldeallrunsformeff on Magneiically

In showing that ttre circuit in Fig. C.13 is equivalent to the magnetically coupled coils in Fig. C.1, we placed no restiictionson the turns ratio d. Therefoie, an infinite number of equivalenl circuits arc possible. Fu hermore, we can always find a tums ratio to make all the inductances Dositive.Three values of d are of Darticular interest:

"(; i

"

I

M Lt'

(c.20)

Ideal L2

(t

(c.21)

kz)L, and

' =\ t;^ t figffe C.14A Two€quivatent circuiiswhen a = M/LI.

Lt(r E)

(a)

r.r1 |

Ideal (D)

FigureC.15a Twoequivahnt circuitswhen a = L,IM.

(c.22)

The value gf a given by Eq. C.20 eliminates the inductances L1 Mla a1:da'Lr - dM from the T-equivalent circuits and the inducaM) aM, and a2(L1L2- M2)l(dLt rances (ara,- rP)l(dL. r-equivalenr circuil".Thc \alue ot d ai\en b\ Eq. (-."1 from the etfmrnares rhe irductances(Lzr-) lU a) and L, - ai rrom rne rLrl, - V2, tLaMt dnd f-equi!alenrcircuir\and tbe induclances M')lL, dM) fiom the ,'-equivalentcircuits a'(44 Also note that when d = M/L, the circuitsin Figs.C.12(a)and (c) becomeidentical,andwhen a = L2/M,the circnitsin Figs.C.12(b)and (d) becomeidentical.FiguresC.14and C.15summarizetheseobservations. In deriving tle expressions foi the inductances there, we used t}le Expressing the hductances asfunctions of t]le rclationship M = krZE. self-inductancesL1 and a2 ard the coefficient of coupling k allows the values of d given by Eqs.C.20 a C.21not only to rcduce the number of inductancesneededin the equivalentcircuil,but alsoto guaranteethat all t}le inductanceswill be positive-We leaveto you to investjgatethe consequencesof chooshg the value of d given by Eq. C.22. The values of d given by Eqs. C.20-C.22 can be detemined expedmertally. The ratio M/41 is obtained by driving the coil desigrated as having Nr turns by a shusoidal voltage source.The source ftequency is set high enough t}lat r)11 >> R1, and the N, coil is left open. Figure C.16 showsthis arrangement. With the N" coil oDen.

lr Y2 = jtiMlr Noqas lolr FigueC.16a Experimentatdetemjnation ofthe ftna M /Lr

(c.23)

>> Rl, the curent 11is

\:

j,h

(c.24)

jn theEquivaLent C.2 TheNeed forldeatTmnsform€rs Circuits

SubstitutingEq. C.24into Eq. C.23yields

M

/V\

(c.25)

\v/,,=.:t'

in which the ralio M/l,r is the terminal voltage ratio corespondilg to coil 2 beingopenithat is,12: 0. We obtain t]te ratio a/M by rcversing the procedurc; that is, coil 2 is energized and coil 1 is left open. Then

b

tu\

M

\v,), =o

(c.26)

Finally, we obsefle that the valu€ of a given by Eq. C.22 is the geometdc mean of these two voltage ratios; thus

Mi \L1 M

E;

!t

lc.?1)

For coils wound on nonmagnetic cores, the voltage ratio is not tlre sameas the turns ratio, asit very nearly is for coils wound on feromagnetic cores Because the self-inductancesvary as tle square of the lumber of tums, Eq. C.27 reveals tlat the tums ratio is apprcximately equal to the geometric mean of the two voltage ratios, or

[i

x.

r!) /!)

(c.28)

Appendix

TheDecibeL

Telephoneengineerswho were concernedwith the power lossacrossthe cascadedcircuitsusedto transmit telephonesignalsintroducedthe deci, bel.FigureD.1 definesthe problem. There,pi is the powei input to the system,?1 is the power output of circuit A, p2 is the power output of circuit B, and p, is the power output of the system.Thepower gain of eachcircuit is the ratio of the power oul to the power in, fhus

o^: 4. P

Figure0.1 A Three cascaded circuits.

ana o" = L r - r )

The overallpower gain of the systemis simply tie productof the individual galns,or Pn=

The multiplicationof power ratios is convertedto addition by meansof the logarithm;thatis, p, , logrnj . log ooA

log o6

loe,no6.

This log mtio oI the powers was named the bel, in honor of Alexander GrahamBell. Thus we caiculatethe overallpower gain,in bels,simplyby summingthe powergains,alsoin bels,of eachsegmentof the transmission system.In practice,the bel is an inconvedentlylarge quantity.Onetenth of a bel is a more useful measureof power gain;hencethe decibel.The numberofdecibelsequals10 timesthe numbei ofbels,so Numbefotdecibels l0loA,. P''

When we usethe decibelas a measureof power ratios,in somesituations the resistanceseen looking into the circuit equalsthe resistance loadingthe cjrcuit,asillustratedin Fig.D.2. When the input resistanceequalsthe load resistance, we caDconvert tie powerratio to eilher a voltageratio or a curtent ratio:

p.

R. = Rr.

oZ",ln' ,i,/R-

(;)'

Pigrre0.2 A A circuitin whichth€jnputresjstance equats the toadresistance.

797

798

TheDecibet

P.=

t3*Rr

(;l

Theseequationsshowthal the numberofdecibelsbecomes g Number of alecibels- 20 logro

: 20 loglo!s

TABLE 0.1 soniedB-Ratio iairs dB 3 6 r0 15 20

Ralio 1_00 1_41 2.00 3.16 5.62 10.00

dB 30 40 60 80 100 1,20

Ratio

3r-a 100.00 1or 1o' los 106

(D.1)

The definitionof the decibelusedin B odediagrams(seeAppendixE) is borrowed from the results expressedby Eq. D.1, since these rcsults .rppl)to aoy rransferluncrioninrolvinga vollageratio.a currenlr,lio. a ratio.You shouldkeep the voltage-to-cunentratio,or a current-to-voltage origiMl definition of the decibelfirmly in mind becauseit is of fundamental importancein many engineeringapplications. Wher you are working with transfer function amplitudes expressedin decibels,having a table that translates the decibel value to the actual value of the output/input ratio is helpful. Table D1 gites some useful pai$ The ntio corresponding to a negative decibel value is the reciprocal of the pos' itive ratio. For example, 3 dB corresponds to an output/input ralio of 1/1.41,or 0.?07.Interestirgly, 3 dB correspondsto the half-powerfre quencjesof tle filter circuits discussedin Chaplen 14 and 15. The decibelis alsousedas a unit of powel when it erpressesthe ratio of a known power to a referencepower.Usually the relerencepower is "decibelsrela1 mW and the power uflit is witten dBm, which standsfor tive to one milliwatt." For example,a power of 20 mW correspondsto i13 dBm. AC voltmeterscommonlyprovidedBm readingsthat assumenot only a 1 mW referencepower but also a 600O referenceresistance(a value commonlyusedin telephonesystems).Sinc€a power of 1 mW in 600O corespondsto 0.7746v (rms),that vollageis read as0 dBm on the meter. For analogmeten, there usually is exactly a 10 dB differencebetween adjacentrarges.Although the scalesmay be marked 0.1,0.3,1,3, 10,and so or,in fact 3.16V on the 3 V scalelinesup with 1V on the 1V scale. Somevoltmeters provide a switch to choosea reference resistance(50, 135,600,or900O) or to selectdBm or dBV (decibelsrelativeto one volt).

o**o* E BodeDiagrams As we haveseen,the ftequencyresponseplot is a very important tool lbr analyzing a circuit's behavior. Up to this point, howevei, we have shoM qualitative sketches of t}le frequency responsewithout discussinghow to create such diagams. The most efficient method for generating and plotting the amplitude and phase data is to use a digital computer; we can rely or it to give us accumtenumericalplots of H(j(r) and o(jo) versus(d. However, in some situations, preliminary sketches using Bode diagmms can help ensure the intelligent use of the computer. A Bode diagram, or plot, is a graphical technique that gives a feel for the frequency responseof a circuit. These diagrams are named in recogdtion of the pioneering work done by H. W. Bode.l They are most useful for circuits in which the poles and zeros of H(s) are reasonablywell sepaiated. Like the qualitative frequency response plots seen thus fai, a Bode dia$am consistsof two separate plotsr One shows how the amplitude of A0r,) vades with ftequency, and the other shows how the phase angle of l1(J(,) varies with ftequency. In Bode diagrams,tlre plots are made on semilog graph paper lor greater accuracy in representing the wide range of frequencyvalues.Inboth the amplitudeand phaseplots,the ftequency is plotted on the horizontallog scale,and the amplitudeand phaseangle are plottedon the linoar verticalscale.

E.1 Real,First-Order PotesandZeros To simplify the development of Bode diagmms, we begh by considering only caseswhere all the poles and zerosof li(r) are real and first order. I-ater we will present caseswith complex and repeated poles and zeros. For our purposes,having a specific expression for H(s) is helpful. Hence we basethe discussionon ((s + zr) s(s + A)

--.

(E.1)

from which n l l' a l : - .

K(ju + z\) l @ \ l @+ P t )

(E.2)

The first step in making Bode diagrams is to put the expression for H(jo) in a standard form, which w€ derive simply by dividing out the polesand zeros:

Kzr( + ja/z)

nU.) = pt(ja)(l I see H. w Bode. Ncr,"//.,1,

+ jalp)

(E,J)

allsi antl Fee.lba.k Desig, (Nev York Van Nostrand, 1945).

799

800

BodeDjagrams

Next we let 1<, represent the constant quantity Kzr/pb ard aI Lhe sametime we expressll(./o) in polar forml

aQ.) =

r.t + ia/zl1L al 129:lr+ jalptl 1A K" I + idlzl + tulp\

;# u |

/u,. _ 90. B.).

(E.4)

From Eq.E.4,

v04:

K.lI + jalzlt

e@\:!tr-eo" - 81.

(E.5)

(E.6)

By definition,thephaseangles'y'randBl are (E.7) (E.8)

l:t

The Bode diagramsconsistof plotring Eq. E.5 (amplitude) and Eq. E.6 (phase)asfunctionsof o.

E.2 Straight-Line AmplitudePlots The amplitude plot irvolves the multiplication and division of facton associated with the polesand zercsof.It(s). We reducethis multiplicatior and division to addition and subtraction by expiessing the amplitude of d(j(d) i]) terms of a logadthmic value: the decibel (dB).2 The ampiitude of H( i(,) in decibelsis ,4dB: 20loglol1(i{r)|.

(E.e)

To give you a feel for the unit of decibels,Table E.1 provides a tnnslation betweenthe actualvalue of severalamplitudesand their valuesin decibels.E\pressing Fq. F.5 in lermsof decibels gi\ es 0 3 6 10 15 20

1.00

30

1.47

40 60

2.00 3.1,6 5.62 10.00

80 100 120

31.62 100.00 103

,4dB= 20logro

Koll + jti/zr

- r + j-lp1l

= 2olog1oKo + 20log1o1+ ja/zrl

1o'

- 20logroo 20log;loll+ jalp1.

105 106 )\e

A D p c i d i \D r o r m o r er o t o m a r i o ne g r r dn e t h e d e c i b e l .

(E.10)

E.? Straight-Line Amptitude Ptots 801

The key to plotting Eq. E.10 is to plot each telm in the equation separately and then combine the sepamte plots giaphically. The individual factors are easy to plot because they can be approximated in all casesby straight lines. The plot of 20 logto tr, is a horizontal straight line because1lo is not a function of frequency.The value of this term is positive for l., > 1, zerc for tr, = 1, and negat'vefor fa < 1. Two straight lines approximate the plot of 20 log101 + j.,r/at . For small valuesof o, the magnitude 11+ jo/zr l is approxirnarely 1, and rherefore 20logrol1+ jzolz1l+0 aso+0.

(E.11)

For large values of o, the magnitude 1 + ia/ zrl is approximately .r/21, and therefore

20logn1 + ja/ ztl-20log1s(o/:l

as o-

m.

(E.12)

On a log frequencyscale,20log10(o/zr)is a stxaighrtine with a slope of 20 dB/decade (a decade is a 10,to-1 change ir toequency).This straighr line inte$ects the 0 dB axis at (, = ar. This value of .,, is called tle corner frequency. Thus, on the basis of Eqs. E.11 and E.12, two shaight lines car approximale the amplitude plot of a first-order zero, as shown in Fig. E.1. The plot of -201og10o is a straight line having a slope of _20 dB/decadethat intersectsthe 0 dB axis at., = 1.Two stmight lires approximatethe plot of 20logn1 + j.,,lprl.Here the two srraighrtines inte$ect on the 0 dB a-\is at o = Pl. For large values of ar, the straight line 20logo({r/pJ has a slope of 20 dB/decade.Figurc E.2 showsrhe stmight-line approximation of the amplitude plot of a first order pole.

20

0loero i; .), 20dB/de

1t Decade

1

2

3

4 5678910

20

3 0 4 05 0

@ (Iad^)

tigureE.1A A staighftjneapprcximation ptotof a ofthe amptiiude

802

Bode Dids'ans 5 0

tPl"".ifr

-5 'uu

1C

Yt-

-lo )0 15

Pi.r1.=\

B,

20 1

2

3

4 5678910 @Gad/t

20

3 0 4 05 0

ptotof a first-order pote. figurcE.2l A sinjghltine approximation ofiheampLitude

Flgure E.3 showsa plot of Eq. E.10 for 1<, = l4d, z1 : 0.1 rad/s, and pt = 5 rad/s. Each term in Eq. E.10 is labeled on Fig. E.3, so you can verifl' that the individual tems sum to create the resultant plot, labeled 20 losn H (ja)|. Example E.1 illustrates the construction of a straightline amplitude plot for a tatrsfer function chamctenzed by first-order poles and zercs.

:!tl]ll|r,],

'fli'oiI 0 30

-?J

2Olosror1{/o tti L

t l l

20

ill

l0 Em (

t0 0

019s11+ j

10 -_

u1 005 0 r

0.5 1.0 5 l0 @Gad/t

l.l

50 100

pLot forEq.E.10. Figore E.3 A stmjght-lin€ approxjmatjon oftheamptitude

500

E.2 Straight'Line Amptitude Ptoh

For the circuit in Fig.E.4:

803

10mF

a) Computethe tmnsferfunction,H(s). b) Construct a straighcline approximationof the Bode amplitudeplot. c) Calculatezologlol/t(j(.)) at o = 50 rad/s and o = 1000rad/s. d) Plot the valuescomputedin (c) on the straightline graphtand e) suppose that ?,i0) : s cos(s00r + 15") V. and then use the Bode plot you constructedto prc dict the ampiitudeof ?J.(, in the steadystate.

FigureE.4A Thecircujtfor ExampLe E.1.

0.11(J50)

..1(j50)=

( 1+ j s x l + l 0 . s )

Sotution

0.9648/ 15.25",

a) Translormine the circuit in Fig. E.4 into the .r-domainand then usings domainvoltagedivislon glves

20logloIi(i50) = 201ogr00.9648 -0.311d8;

(R/L)r 0.11(j1000)

r' +(R/r)s++

+y'o) 0 + y'00x1

Substituting the numerical values from the circuit,we get

110r

0.1094/ 83.'72" I

l10s

-19.22dB.

s'+ 11or + looo (s + 10Xs+ 100) b) We beginby writing l1(lo) in standardform:

n04 =

40 30 20

0.i1j.,

t1+ i(d/10x1 + j(..1100)l'

ldB

AdB: 20lognlH(ja)

= 20109100.11 + 20logroliol

:or.g.lr i inl,,.r,1,,,ft| I

r

U

|

|

l

U

Figurc E.5 shows the straightline ptot. Each lerm conrributingto the o!erall amplirudei. identilied.

v

10

The expiession for the amplitude of H(jo) in decibelsis

10

zologto1 . 1 1 2t) za 30 -,10 -50 60

t0 g

t$l,; l

-.fu

t t

F

( 12.5)

t).,'',:

t -

20logro 1+jlft 20logro r l l

t0

50 100

5001000

o (md^)

ptotforthetransfer functionof tlgureE.5A Thestraight-tine amptjtude the circuitin Fig.E.4.

804

Bode Diaqrams

d) SeeFig.E.5. e) As we canseefrcm the Bode plot in Fig.E.5,the value of /dB at a) = 500 rad/s is approximately -12.5 dB.ThereIore, = O.24 Al : 10\ \25t20)

We can compute the actual value of H(io)l by substituting(, = 500 inro the equation for H(i@\l:

= r1(js00)

0.11(js00) : -17.54" . (1 + js0)0 +i5) 0.22/

Thus, the actual output voltage magnitude for the specified signal source at a ftequency of 500rad/s is

= 1.19V. v-.: Av,n = (0.24xs)

: 1.1v. u_a= lA)u^t: (0.22xs)

E.3 MoreAccurate AmplitudePlots We can make the slraightline plots for fi$forder poles and zeros more accurate by corecting the amplitude values at the comer frequency, one half the corner fiequency, and twice the coner frequency. At the comer frequency,the actual value in decibels is ,4dB.: i20log1oll + j1l | "0log o\2 ! +3 dB.

(E.13)

The actualvalueat one halfthe cornerfrequencyis

I = 2l togrolt Ads 't2I ) * |

- -20log VS4 s ! J1dB.

(E.14)

At twicethe cornerfrequency,the actualvaluein decibelsis

A.rBt= +201o9rc1+ j2 = t20logro\6 !

t7 dB.

(8.15)

In Eqs E.13 E.15, the plus sign applies to a tust-order zero, and rhe minus sign applies to a fi$t-order pole. The straightline apprcximation of the amplitude plot gives0 dB at the comer and one half the corner frequencies, and +6 dB at twice tlle comer frequency.Hence the corections are +3 dB at the comer frequency and al dB at both one half the comer ftequency and twice the corner frequency,Figure E.6 summarizesthesecorrections A 2+o-1 change in frequency is called an octave. A slope of 20 dB/decade is equivalent to 6.02dB/octave, which for graphical purposesis equivalentto 6 dB/octave.Thus the corecuons enumeraredcorresponclto one octave below and one octave above the corner frequencv

E.4 Strajght LinePhase AnqtePtots 805

25

15 t

lI

'11;: 7 5 l0

Idu

'' r,io

ldl IB

\

15 20 25

2c

Figure pLots E,6A Corrected ampLitude forafirst,orderz€ro

If the poles and zerosof 11(r) are well sepamted,inserting theseconections into the overall amplitude plot and achieving a reasonably accumte culve is relatively easy.However, if the poles and zeros are close together, the overlapping corrections are difficult to evaluate,and you're better off ushg the straight-line plot as a first estimateof the amplitude characteristic. Then use a computer to refine the calculations in the frequency range of

E.4 Straight-Line PhaseAnglePlots We can also make phaseangle plots by using straighlline approximations. The phaseangle associatedwith the constanttr. is zero, and the phase angleassociated with a first-orderzero or pole at the origin is a constant + 90'. For a firsForder zero or pole not at the origin, the stmight-line approximations are as follows: . For frequencieslessthan one tenth the comer frequency,the phase angleis assumedto be zero. . For frequenciesgealer than 10 times the comer frequency,the phase angleis assumedto be 190". . Betweenone terth the cornerfrequencyand 10 timesthe comer ftequency,the phaseangleplot is a straightlirle that goesthrough0" at one-1enththe corner ftequency,145' at the corner frequency,and i90'at 10 timesthe cornerfrcquency. In all thesecases,t}le plus sign applies to the fiNt,order zero ard tlle minus signto the filst-orderpole.FigureE.7 depictsthe straight-lineapproximation for a first-order zerc and pole. The dashedcurves show the exact variation of the phase angle as the frequency varies. Note how closely the

806

BodeDjaq'aris

EtL-t-f1)|l l 4 / 1= t a n - ' ( o / ? r )

60'

Straightline approxinar

I

e la)

t illiltl

| i l Ii l 1

0

:xiiti*ltn

30" -60. -90" 4l1l pl10

zr

pI

10.1 7Op1

o Gadi0 Figure E.7a Phase plotsforafirst-order angte zeroandpote.

straight-line plot approximates the actual variation it phase angle. The maximum deviation between the slmightline plot and t}le actual plot is approximately 6". Figure E.8 depi€tsthe shaight-Iitre approximation of the phaseangle of the hansfer functior given by Eq.8.1. Equation 8.6 gives the equation for the phaseangle;the plot correspondsto al - 0.1 iad/s, ard pl - 5 rad/s. An illuslration of a phase angle plot using a straightline approximation is givenin ExampleE.2.

90" 6Cf

30' d(d)

0

-h=

ta^-\ (0,1p)

-3Cf -60" -9ff

0.01

0.1

0.51.0 (radis) o

5 10

tigureE.8^ Astraightline appBximation ofthephase ptotforEq.8.1. angle

E.5 BodeDiag,ams: complex Poles andZeros 807

a) Make a straight-linephase angle plot for the transferfunctionin ExampleE.1. b) Compute ttre phaseangle p(o) at o : 50, 500, and 1000rad/s. c) Plot the valuesof (b) on the diagramof (a). d) Using the resultsfron Example E.1(e) and (b) of this example,compute the steady-stateout put voltage if the source vottage is given by

and

0(j1000):

83.72".

c) SeeFig.E.9. d) We have

v^o: H(js,})v_t : (0.22)(10)

(s00r- 25')V. r'(t = 10cos

Solution 0.=a(@)+0t

a) FromExampleE.1, H(.j.,) =

0.11(/o) [1 + l(d/I0)[1

+ (d/100)]

:

7'7.51' 25"

-

t454".

Thus,

0.11[o

/(t'1 11+ i(o/10)11+ la,/100)l Therefore, 0(.): ,!r Ft - Bz,

B,

p).

where tl = 90', Pr : tan-1(o/10), and Br: tan 1(o/100). Figure E.9 depicts the straight-line approximationofd(o). b) We have

"o0) - 2.2 cos(500t 102.54')v. t, = 90"

a

H(j50)- 0.e61_ll4:,

-30"

H(jsoq)= o.221_JJlf ,

60'

fl(i1000)= 0.r1/ 83.72' .

90"

Thus, 1s.25"

0(js00)-

'77.54" ,

)2 =

tan'(oi100) aLO)

120"

d(j50):

I

o (a)

510

5)

"t

-77.54)

( - 8: -72\

t.U I 50100

5001000

FigureE.9a A stra'9ht-line app'o.mdron of0(d) lor f"anplef.2.

E.5 BodeDiagrams: Complex Potes andZeros Complex poles and zeros in tlle er?ression for 1{(s) rcquire special attention when you make amplitudeand phaseangleplots.Lefs focuson the contribution that a pair of complex poles makes to the amplitude and phase angle plots. Once you understand the rules for handling complex poles,their applicationto a pair of complexzerosbecomesapparent.

808

BodeDiaqrams The complex polos and zeros of H(r) always appear in conjugate pairs The first stepin makingeither an amplitudeor a phaseangleplot of a tmnsfer function that contains complex poles is to combine the conjugate pair into a single quadratic term. Thus,for

K j B ) ( r +a + j B ) ' we filst rewritetheproduct(s + d - iBXs + d + jB) as fl(r) :

(s+a

(s + d)2 + 92 : 12 + 2o.r+ a2 + 82.

(E.16)

(E.17)

Wlen making Bode diagmms,we write the quadratic term in a more convenient folm:

s2+ 2as + otz+ 92 = i + z{o"s + a7,. A direct comparisonof the two formsshowsthat

a|:&+P'

(E.18)

(E.1e)

and (E.20)

The term o, is the cornerfrcquencyof the quadraticfactor,and ( is the dampingcoefficientof the quadraticterm. The cdtical value of { is 1. If ( < 1,the rootsofthe quadraticfactorarecomplex, andwe useEq.E.18 to rcpresentthe complexpoles.If 4 > 1, we factor the quadraticfaclor into (s + plxs + p2) and then plot ampliludeand phaseir accordance previously.Assumhg withthediscussion that4 < 1,wercwdteEq.E.16as K

s' +21drs+otr'

(E21)

We then write Eq. E.21 in standard form by dividing through by tl1epoles and zeros.For the quadratic telm, we divide through by o,, so

K 11G): -i 1+ \sl-.)t + zgtl-")'

(E.22)

H(io):

(E.23)

ftom which | - |..' |.',) + J\2ad/un)'

, . K .,. ,?. Beforediscussing the amplitudeand phaseanglediagramsassociated with Eq.E.23,for convenience we replacethe ratio (,/(,, by a newvariable,!r.Then H(.ja') : Now we write lt(ta)

Ko

I

u2+ j2tu

IE.24)

in polar form:

n0,) =

lr

u' + \ i2tu1A'

(E.25)

E.6 Amptitude PLoti 809

from which AdB : 20 lognlH(ja) : 21logaK" - 201og1ag- u,) + j2ta,

(E.26)

and

Bt- - t^n-'4+.

o(.) -

\E.27)

E.6 Amptitude Plots The quadrariclacrorconrribulesLolhe amplirudeo[ Ht ld) by meaosol rbe rerm -20loglolt l., - 121fi1. Beciu.eu - u a,.u-j as o -0, and & + co as o + co. To seehow tlre term behavesas o mngesfrom 0 to 6,

-2}tosrcIt u't - j2(u

zotogd\fr :

i\

t tfu,

l1logralu4+ 2u2(2{, 1) + 11,

10logl]olu4+ 2uz(2t2- 1) + 1l - 0, -101og16[aa + 2]e{

D +11-

- 40log10rr.

(E.2s)

(E.2e)

(E.30)

From Eqs.E.29 and E.30,we conclude that the approximate amplitude plot consists of 1wo straight lines. For .", < on, the straight line lies along the 0 dB ax[ and for o > (,,, the straight line has a slope of 40 dB/decade. These two straight lines join on tte 0 dB axis at u - 7 ot o = @r. Figure E.10 shows the straight-line approximation for a quadratic factor with 4 < 1.

20 10 0 \

t0 20

40 da/decade

30 40 -50 @"

70@"

o (rad/t

FlgureE.10 TheamptjtldepLotlorapajrofcompLex potes.

810

Bod€ Diaqrams

E.7 Correcting Straight-Line Amptitude Ptots Correcting the straight-line ampLitudeplot for a pair of complex poles is rot as easyas correctinga fi$t-order real pole, becausethe corrections dependon the dampingcoefticient(. FigureE.11showsthe effectof l on the amplitude plot. Note that as ( becomesvery small, a large peak in the amplitude occurs in the neighborhood of the comer frequency o,(" = i ). When t > 1/',/2, tle corrected amplitude plot lies entirely below the straighfline approximation.For sketching purposes,tle straightline amplitude plot can be corrected by locating four points on the actual curve. These four points correspond to (1) one half the corner frequency, (2) the frequency at which the amplitude rcaches its peak value, (3) the corner frequenry and (4) the frequency at which the amplitude is zero. FigureE.12 showsthesefour points. At one half the comer frequency(point 1), the actualamplitudeis

AB@,J2)-

10los1o(f'+0.s62s).

(E.31)

The amplitudepeaks(point 2) at a ftequencyof

.,

-,Vt

2('

(E.32)

andir basa peakamplitudeof AdB(dp):

lo logro[4f,(1

f,)].

(E.3t

At the comer frequency(poht 3),the actualamplitudeis

AdR@,r: 20loEn2t.

(E.34)

Thecorrected amplitudeplot cross€s the0 dB axis(point4) at

..: ."\/2O tO : .,,2.,.

(E.35)

The deiivations of Eqs. E.31, E.34, and E.35 follow ftom Eq. E.28. yieldsEqs.E.31 EvaluatingEq. E.28 at & = 0.5 and r, = 1.0,respectively, and E.34 Equation E.35corespondsto finding tlre value ofIl that makes ua + 2u2(212- 1) + 1 = 1.The derivationof Eq. E.32 requiresdifferentiating Eq. E.28 with respect to ll and then finding the value of ,l where the dedvativeis zero.EquationE.33is the evaluationofEq. E.28 at the value ofIl found in Eq. E.32. Example E.3 i ustrates the amplitude plot for a transfer function with a pair of complex poles

E.7 Correcting Shajsht Lin€Amptjtrde Ptots 811

10 I

(r t0 IO

.3

\

-20

-30

-50 o (rad/s) FigureE.lt A Theeftuctof 4 ontheampftude ptot.

3 2

i

1

a

\

0 I

Aae

2 -3

-5

7 a"D

o (rad^) Figure points t.t2 l Four onthecorrected ampLitud€ ptorfo.a panof

812

BodeDiasGms

Compute the transfer function for the circuit sho$n in Fig.E.13. a) Wlat is the value of the corner frequency in radiansper second? b) Wlat is the value of /(,? c) What is the value of the damping coefficient? d) Make a straighrline amplitude plot ranging from 10to 500md/s. e) Calculate and sketch the actual amplitude in decibelsat (r&/2,(r?,.d,,andcr.. f) From the straight-Line amplitude plot, describe the type of filter represented by tie circuit in Fig. E.13 and estimate its cutoff frequency,(,)..

50mH

1f)

Figure E.13A Thecircuittor ExampL€ E.3.

d) SeeFig.E.14. e) The actual amplitudes are AdR@12) = -1o loglo(o.6o2s): 2.2 dB, ap - 50\6.92 : 47.96:/,:d,ls, AdB(o") =

(0.96)- 8.14dB, 10 loslo(0.16)

Adr(o,) =

20loglo(0.4):7.96 dB,

@" = \420,r, = 6i.82 ftd I s, ,4dB(o,) = 0 dB. FigureE.14showsthe corected plot. 0 It is clear from the amplitude plot in Fig. E.14 tlat this circuit acts as a low-pass filter. At the cutoff frequency, the magnitude of the transfer luncrion.H(i@.,.is 3 dB lesslhanlhe ma\;mum magnitude. Fiom the corect€d plot, the cutolT frequency appears to be about 55 rad/s, almost the same as that predictedby the straightline Bode diagram.

Sotution Tiansform the circuit ir Fig. E.13 to the s-domam andI henuses-domain\ olragedi! isionro ge. 15 t0 5 0 5 10

LC

f+(f)"+,.! !alues, thecomponent Substituting

2500 s2+20.r+2500 a) Fromt}leexpression for lt(s),.ri = 2500;there= fore,o, 56t667.. b) By definition,r, is 2s\0lol,or r. c) Thecoefficientof r equals2ro, ; therefore

t =f,= o.zo.

/dB

)r

I0) \.

15

-20 25 30 35 40 _45 2 ptoifor ExampLe E.3. Figuret.14 A Theamptjtude

E.8 Phase AngLe Plots 813

E.8 PhaseAnglePlots The phaseangleplot for a pair ofcomplex polesis a plot ofEq. E.27.fhe phaseangleis zero at zero frequencyand is -90" at the cornerfrcquencv. - 180'as @1Il It approaches ) become.large.As in rbecaseot lhe amplitude plot, 4 is important in determining the exact shapeof the phaseangle plot. For smallvaluesof f, fte phaseanglechangesrapidly in the vicinity of the coner frequency. Figure E.15 shows the effect of 4 on the phase ange plot. We can also make a straightline approximation of tlle phase angle plot for a paii of complex poles.We do so by dmwhg a line tangent to the phase angle curve at the comer frequency and extending this lhe until it intersectswith the 0" and -180" lines.The line tangentto the phaseangle curveat -90'has a slopeof -2.3/f rad/decade(- i32l1 degrees/decade), and it intersectsthe 0' and 180' lines at ,1 : 4.81 { arlclu, = 4.91t, respectively.Figurc E.16 depicts the straighrline approximation for 4 = 0.3 and showsthe actualphaseangle plot. Compaing the srraighrline approximationto the actualcurveindicatesthat the aDoroximationis reasonable iD lhe vicinit) ot lhe comef tfequency. Houerer, in Lheneiehbor hoodof rr and,r. lhe error i. quileIarge.In ExampleF.4.$ e .ummarrze our discussiorof Bode diaqrams.

0 15' 15" -30' 60' 75"

.t

l

.t

Nu

_30" 60'

).3

07\

0(a) -90" 120"

105. -120"

_15ff

135' 150' -165" -180"

I

-l 0.2 0.4

1.0

2

'1J 'l ] 4

rigurc8.15a Iheeffect of{ onthephase ptot. ansle

180" 1.0 2.0 o Gadis) f i g u r eE . l 6A A s l z i o ' . r j r ea D p . o r i . a t ioor t e p l a s F anstefor a panofcomptex potes.

814

Bode DiasBms

a) Conpure the transfer function for the c cuit shownin Fig.E.17. b) Mak€ a straightline amplitude plot of

from which H(j,i)

20losrol1/(l(,,) . c) Use the straighlline amplitude plot to determine the type of filter represeflted by this circuit and then estimate its cutoff frequencf. d) What is the actual cutoff ftequency? e) Make a straightline phaseangleplot of t1(jo). f) wlat is the value of d(o) at the cutoff frequency ftom (c)? g) What is the actual value of 0({r) at the cutoff frequency?

11+ jal25l1!) t- (.,lt0)2+ j0.4(0110)l1a

:

Note that for the quadratic factoi, u = o/10. The amplitudeof -&(jo) in decibelsis AdB:z}Iogrcl

+ ja/25

,,",,,[], 1;;'.r-(..)]l andthe phaseangleis 0(ot)= h

Fr

250mH

,(a/2s), trl = tarr . 0.4(t/1o)' B . '- , = t a n 1 (d/10)'

Fisure E.17A Thecircuit forExampte E.4.

Solution

Egllfe E.l8 showsthe amplitudeplot. plotin Fig.E.18, c) Fromthesfraigltt-line amplitude this circuit actsas a low passfilter. At the cutoff ftequency,t}le anplitude of li(jo) is 3 dB less than the amplitudeirl the passband.From the plor.$e predictlhar lhe culoff trequency is approximarely13rad/s.

a) Tlansfom ttre circuitir Fig.E.17to the r-domain andthenpeform s-domainvoltagedivisionto get H(.):

,r2+r4s+r"!

Substituting the component values from the clrcuit gives

4(s + 25)

s' +4s+1oo b) The fust slep in making Bode diagrams is to put H(j.d) in standard form. Because 11(s) contains a quadraric factor, we first check the value of (. We find that I = 0.2 and o,, = 10,so

s/25 + 1 1+ (s/10f + 0.4(s/10)'

mr"e,,[lr (# +r0.4# ll -" 80

I

510

50100 5001000 o (rad/s)

ptottorExample E.4. Figure t.t8l Theamptjtude

E.8 PhareAnqLe PLotl 815 d) To solve for the actual cutoff frequency,replace s with j(d ir fl(r), compute the expressionfor

Computing the phase angle,we see 0(a,) = 0(i16) = -125.0'

serlH(j.i"\l = g.va) H_ _ = ll\n, lH(jaJ)|,

and solvefor o.. First,

Note the large elror in the predicted angle. In general, straight-line phase angle plots do not give satisfactory results in the frequency band where the phaseangle is changing.The straightline phaseangieplot is usefulonly in predicting the generalbehaviorof the phaseangle,not in estimatingactualphaseanglevaluesat particular ftequencies.

+ 100 nud = ( r . o4Uo) f+40o)+1oo Then,

llr(i,Jl =

\/@S;tn v@-AF + t4;j

\/r'

Solving for o. gives us a, = 16 rad/s. e) Figure E.19 shows the phase angle plot. Note that the straightline segmentof d(a,) between 1.0and2.5rad/s doesnot havethe sameslopeas the segmentbetween2.5and 100iad/s. t) FIom the phasearyle plot in Fig. E-19,we estimate the phase atrgle at the cutoff frequency of 16 rad/s ro be -65'. g) We can compute the exact phase angle at the cutoff frequency by substituting r : j16 into the transferfunctionii(r)i

H(j16):

4(j16+ 2s)

Ut6),+4(ir6)+1oo

o (-)

0' 45' 90" 135' 180'

510

50100 5001000 @(rad/s)

tigun E,19^ Ihe phase anqlepLotfor E€mph[.4.

An Appendi, AbbreviatedTableof E

I Trigonometric Identities

1. sia(a+ B) = sindcosP+ cosasinB T sindsinp 2. cos(a+ B) : cosacosB " + B d P ^=^ r. srna+. srnp I srn cos 2 2

4 .s i n o s i n B , . " ' 1 ^ ; u ) " ' " ( '

,')

5.cosa+ coso='*(-r)"-(l/) ^ ^ l o B \ . ( a ' n PJ o . c o s a c o s p= - r . t n ( - J . t n ( , 7. 2 sioasinp = co(d - P)

cos(a+ B)

8. 2 cosacosB= cos(d- B) + co(o + B) 9. 2 sillocosB: sin(a + B) + sin(d- p) 10.sin2a: 2 sinacosa

1 1 +;cos2d 12. cos'o: ; t

1

lJ,sln_d=---cosla

tand + tanB 14. tan(a + B) = 1 + tana tanP 15. tan2a :

2 tand

811

nppenai,. f An Abbreviated Tab[e

U of Integrals

L

I xe- dx = .(ax - 1.)

2. I x'e"^ ax : 1.\'P

zor + z1 I "o,o"

3 . /,.ino" a" = 1"i,'or J a '

a

4. [, "o"or,1'= 1coso, + 1"ioo'

a'

.l

5.

In,,ro" o, - $ a :r+ab' i' n b , l-

6. l e ' c o s b x d t = .t

7. 8. 9.

tco.o,1

.l

f

dt .l r'+ a' I

dx

J lx2+i)'

oat

)- -(acosbr 0-+D-

1 a

b.inbY.,

.r 1/ tt za' \x' + a'

1 a'-

sin{a2ta btt sin(dl--::

sin(d b)r | . vnd'rsrnDf*d'r ' 2\a - b) /

sin(a I 10. / cosdr cosD,tll,t - -;:tg-D)

13.

a * x

,,, , , .1 / b

cos(a- - - r1r | b)r - i - - - -con.? ,.

-

^ t\0

J

12

b)-r h)

z\a'D)

J

. 1 1 . /t s r n a Y c o q D r a -r

.-r\ aI

o)

) t\a - D)

a"/

b

sin2dx

t . .

lstn-atat: 2t " x cos'axdn:t+ .l

ta sir2ax 4"

(+,o.0: I

A A X

/ ' ' : ro a'+ r'

t :

l u , d : u ; [+.".0

[:,o. o; " ' : l" ^ 1 7 . , .o /-sinar

t r 16. | ? sinaxdx = ="indr a,l

a'

819

nnoeno,x Ll Answers

I I to Setected Problems

Chapter 1 110giga-watt hous a) 111.6seconds b) 2480bytes 1.6 0.10mm 1,9 6 sin4000t mC 1.12 a) 600WftomAtoB b) 2000W fromB to A c) 2400W from B to A d) 4800W fromA to B 1.17 a) 3.1mW h) 1.24pI .) 21.6'7pI 1.24 a) r=0.634s b) s.196mw (delivered) c) r = 2.366s

2,14 a) A 5 A currentsourcein parallelwith a 20 O

1.1 1,3

b) 8 0 w 2 . 1 7 a)

(v) 80 '70 60 50 40 30 20 10 0

d) 5.196nw (extracted) e) 0mJ,4mJ,4mJ,0mJ 1.26 770mW

Chapter 2 2.2

1700W

2.3

Not valid,due to 4 A and 5 A cunent sourcesin the rightmostbranch

2.6

Norvalid, sincet}le voltagedrop betweenthe top and bottom nodes is different due to different voltage sourcesin tle left and dght branches

2.8

Notvalid, sincethe voltagedrcp betweenthe top and bottom nodes is different due to different voltagesin the left and right branches

2.10 8 kO resistor 2.11 4 kO resistor

400 225 300 500 (-A) \ b) A 75 V sourcein serieswith a 200 O resistor 150

c) 125mA d) 375mA €) 500mA

0

A linear model cannotpredictnonlinear behavior

2,18 a) 2 A b) 0.5A c) 4 0 v d ) 25W,80W,20W e) 125W 2.19 a) 2 . 1 4 , 1 . A 6 b) 192V c) Powerdeveloped anddissipated is 768W 2.24 a) 22.22 W,33.33 W,11.11W,16.67 W,0 W b) 83.33W is 83.33W c) Powerdeveloped anddissipated 2.28 a) 4.5V anddissipated is 741mW b) Powerdeveloped

821

822

to SeLected ProbL€ms Answ€rs

2.34 j = 385mA, so a warningsignshouldbe posted

c) 40O,50 O and60O,45 O and30O;simplified

circuit is

and piecautions taken

A

150rl50v(

)

75()

12o and20O,28o and21O; simplified circuitis 72o-

200mA 2.36 a) Pdn = 59.17w;4"s = 29.59W; P,,.k = 7.40W

1) 7.5o

18()

b) 30 O and 5 O,9 O and 18 O;simplifiedcircut ls

b) ram : 1414.23s; rbs =1071.13s\ tdnk : 10'4/2 54 s

40()

c) All values are much greater thar a few minutes

20r}

8V 6A

4.2857 A

2.37 a') 40V b) No,12 V/800 O : 15 mA will causea shock

100kO and300kO,75kO,50kOand150kO; simplifiedcircuit is 75kO

2.38 3000V

Chapter 3 3.1

0.5v

a) 3 kO and8 kO,5 kO ard 7 ko;simplified circuit is

25 kO

11kO

3.5

I ) 10ko

6 ko:

3.6 circuit is 240A

lE(] {}

l0v 340[}

a) 6ko

b) 5ooo c) s0o

12ko

b) 180O and100a. 140O aDd200O:simplified

25 kO

a) 6O b) 14o c) 125kO 3,13 a) 66v b) 1.88W,1.32W 4,12,408 A ., 1'7,6'72 3.15 a) 60kO,15ko b) 1/8w 3.21 257.5A,rc30A,82404,41.2k{L

Answe6toSetected Probtems 823

3.22 a) b) c) d) 3.23 a) b) c) ?rn

1.2mA 12V 3.2V 1.33V 3.2Y 13.33mA 20mA (25/12\

rt t

:

50 - (25 l2)

Chapter4 4.1

| "ne6'

b, 112s00 3.33

3.48

3.52 3.53

3,54 3.71

3,72 3.73

c) Yes a) 99,950 O b) 4950O c) 50J) a) 900O b) 25mA c) 800O,180mW d) 900o,90 mw 48v,24v a) A-connected R -R3-Ra become Y-connected 3 O-18 f)-6 O;equivalent resistance is 50O b) Y-connectedR2-Ra-Rs becomeA-connected 90O 270O 45 O;equivalent rcsistance is50O c) Conve the delta connectionRa-R5-R6 to its equivalentwye.Convef tle wyeconnection R3--Ra-R6 to its equivalentdelta. 99f} A,Rz = 1.1435 O, R: = 1.2O, & = 1.03'72 Rr = 1.1435 O, R: : 1.0372 O, Ra = 0.0259 O, : : = Rb 0.0068 O, R. 0.0068O, Rd 0.0259 O = : Pdi,. 624W Pa.1 a) Rl = 0.6106 A, Rz: 0.'7122 O, R3 : 0.768O, Ra = 0.7722 A, R5 : 0.6106 O,R" : 0.0244 A, = = = Rb 0.0066 O, Rc 0.0066 o, Rd 0.0244 o = b, i1 17.52A,i1&: 187.5w or 15oWn; i2 = 16.23A,3R2 = 187.5W or 150Wn; ib = 33.75A, iSRb= 7.5w or 150Wm; Pder= 997.5 W = Pdis"

a) 5 b)3 c ) - t s + t 1+ i 2 = 0 , - t 1+ i a + i 3 : 0 , ts-tu-t::0 d)2 e) R1i1+ R3i3- n2i2= 0, R3i3 + R5i5

4.2

-Raia= 0

al 11 b) 10 c) 11 d) 10 e)5

0 s d 7 4.3

a\ 2 b)5 c\ '7 d) r,4.'7 4.4 a) 10 b)4 c) 4 d) Avoid the threemesheswith curent souices 4.6 10V 4.9 25y,90y 4.1O a) 4 A,2 A,2 A,3 A.,-1 A. b) s82w 4.79 375W 4.20 a) 165W 4.21 4.26 4.27 4.31

b) 16sw 3.2v 200v,1.2w 26v a) 9.8A, -0.2A, 10A b) 1.72 A, 1.08 A,2.8A

4.32 a) 1140W, developed b) 1140W absorbed

824 Answers tosetected Prcbtenrs 4.37 98W 4.38 2J00W 4.47 a\ 264336W w b) 6847.36 c) checks 4.42 al 5.2ml. b) 200mW c) 2.912mW 4.47 99W 4.50 a) 5.7A, 4.6A,0.97A, -1.1A, 3.63A b) 1319.685 W 4.54 a) Meshcwent methodrcquiresfewerequations b) 4nW c) No-you cansolvefor the voltagefrom the meshcurrents d) 200mw 4.56 a) Theconstraint equations areeasierto tormulateifthe nodevoltagemethodis used b) 180w 4.59 a ) 3 m A b) 3mA 4.62 a ) 1 A b) 1A 4.63 4 8 V , 1 6 ( ) 4.66 8 mA (down),10kO a) 45.28V b, s.6'7a/" 4.71 86.40V,43.2kO 4.77 0 v , 8 f , ) 4.79 16.67 f),150{) 4.80 a) 12{) b) 48W 4.91 a ) 3 8 V b) 72.2W 4,92 3 0 v 4.105 r,1= 39.583 V, t2 = 102.5V 4,706 I'r = 37.5V, r,, - 105V V,1J2= 117.5V 4,707 01= 52.083

Chapter 5 5.1

a) Positi!e ln\crbng N

Po$er

qupll)

inp,, I ---.._ or,ou, ) .-' Non-in\errng Dy'

i.put

Neg:live porer supply

b) The hput resistance;i,,: 0 c) The openloop voltagegain;(r,,,- r,,,) : 0 d) tr, : 4V 5.2

a\ -12Y b)

18 v (saturatet

c) 10V d) -14 v e) 18v Gatumtes) t) 2.8125Y< tr" < 7.3125 V 3.1mA a) Non invertingamplifier b) 4.5V a) Manypossible designs; oneusesa single10kO inputresistorandlour series-connected 10kO resistorsin the feedbackpath

b) r10 v 5.12 a) Invertingsummirg amplifier

b) 6v c) -7.5V
6V

b) 8.4V < t]b < 13.2V 5.75

V"

An5we6 Probte'ns825 to Setected

5.17 a) Non-inverting ampMier b) 3.7sD, 2.4V = 0, = 4Y .l

b) ? r = 0 o- 6V 0 = 0

5.18 a) 7.56V b) -3.97V = os = 3.97V c) 35kO

p:0 p - 961 4.8W p:0

s.24 20ka 5.25 a) 15.95V b) -638mV < rb < 962mV

w-0 u:48t21 u : 4 8 t 2 - 4 . 8 t+ o . 1 2 l u:0

5,26 a) 56.25mV kO b) 114.3 c) 80ko

5.33 32.89kO < R: < 33.11kO 5.42 a) 19.9844 bl 736.1pY 6) 5003.68o d) 20,0v, s000o

5 4 3 2 I

oo,lr:6.74

Chapter6 i-16t4 i=0.8-16rA

t:0

0
rr(A)

30.98 999.smv,999.87mv 36'7.94pY 836.22pA 3 1 , 1 V , 0v , 0 A

a) i=0

r<0

t,d

5.48 a) 2kO b) 12mO

6.1

,<0 0 < 1< 25ms 25 < t < 50ms 50ms< t

30 25 20 15 10 5

5.32 a) 19975 b) -0.0s .) 399.5

5.43 a) b) .) d) e)

t <0 0 < t < 25ms 25
t
60 40 20 0 0 20 -40 60

2

rG)

826 Answers toSetected Prcbtems 6.15 a) -200 x 103r+ 40V b) 4 x 1051 20V c) 100v

6,43 2.8nWb/A,3.6nWb/A 6,47 a) 18.5J b) 18.5J c) 6.5J d) 6.5J 6.48 a) 5 A b) No

d)

z,(v) 120 100 80 60

I

6.4e ', =;u"0) + u(0) 6,51 a\ 2C

100

0

1m

400

500 t(pt

6.21 20H 6.22 a) -12e 2uA h) -9e-'o' + 13A c) 3? 'o' 13A d) s40J e) 390 J f) 3380J l

^

S ) , ( 1 0 ) ( 1 3 )- ,

Chapter7 7.7

7.2

t

f 3 0 ) fl 3 y = J J 8 0 J( c h e c k 5 r

b, l8e-xaa+2'7y c, 12e25au- 2'7Y

7.3

d\ s.4 pI el 23.625pJ 18.225 pJ t -

: 0

b) 3c4g: o

6.26 5 nR initial voltageis -10V 6.27 a\ 3Oe2nuV

1

d1

,

1

" , { 2 7 1= s ) ; ( 2 0 1 0 4 r 2 7 y ' ; r 3 o. l o 18.225pJ di" di. = d-5j t +t 90i, 6.34 a)' 40= d b) 40[-€ , + 11.25e225t1+g)[e-t 50e t

5e1r5\ -

105e I + 56.26e2.25tV d ) -48.75 V , which is consistent witl the circuit's

ct

behavior 6,42 a) 160mH,1.25 b) 0.25x lorwb/A,o.2s x 10 6wb/A

a) b) c) dr a) b) c) d) e) a) b) c) d) e)

5A a0ms 5e ' r' A, -4ooe"Y-5ooe "V 9.020/" 0.2mA,0.2mA 0.2mA, -0.2 mA o.2e10"mA 0.2e10mA The currentin a resistorcanchange hstantaneously 0A 62.5mA 87.5mA 62.smA 150nA

0 0A

g) 62.5eamrmA

h) 0v D -12.sV i) 0v k) t) 7.21 a) b) c)

-12.5a4m0' V 150 62.5e{om'mA 125'+15v,-15€ 15!+ 15v 15er25tmA,60s s62spI 1125pJ,4500sJ

Answersto Setected Probtems 827 7.24 al 39.6e2MtmA b) 14.0s% 7.33 9 + 3aso'om' mA, -60e soooot V 4oau y 7.34 a' 4 + 2e A,76 76e-14tu

b) 40v0v 7.35 a) 3 - 19e 2mr A b) 15 + 285e-2omrv 7.50 a) 40 40e 5mo'V b) 2 - -5mo'mA c) 2 + 4e-sooo' mA 5om'mA d) 8 4e e) 6mA 7.51 15

105e-4oo'V

7.92 25ms 7.93 30,000t 30mY10+ 5e' ooo'mY 30,0001 40 5e '{m'mV 7.103a) 1.091MO b) 0.29s 7.104a) 8.55flashes/min b) 5s9.3ko 7.105a) 24.3flashes/min b) 99.06nA peryear c) $43.39

Chapter 8 8.1

7.53 a) 120V

b) -1s0v c) 2ms

7.54

7.65

7.67

d) -5.4mA e) -150 + 270?-soo'V f) 5.4e5m'nA a) -30 V b) 25V c) 2.5ms d) 1.97ms a, 0.2 0.2eloooorA b) 15dloooo! v c) 0.25- 0.25e-1o,mo'A d) 50 + 50elonmrmA €) Yes a, 4t) 40esmo'mA b) 10dsom!V c) 16 16a5mo'mA d) 24 24e 5tna'mA e) Yes -559.12mV

7.12 7.76 r,.:100Y0 < t < 250ms; 0o= 100s1mo(.-o2tv250ms< I < 7.86 a) 2 ps b) 529.83 7.47 fi3.23 ps

8.2

8,3

8,4 8.5

8,19 8,20 8.21 8,25 A.26 8.27 8.45

a) b) c) d) e) a) h)

-20,000rad/s 5000rad/s, overdamped 7812.5O 8000+ j6000rad/s,-8000 - j6000rad/s 6250O 800(),200mH,12,500 rud/s,10,000 mdls -6.25a564t+ 25e2oooot mA, 5e 5000r_ 5d 20,000! mA, 1.25e 5000r_ 20e 20,0rrr mA

a) 800 mH

b) c) d) e) a) b) a)

2500o 125V 12.5mA t4m0'(12.5 cos3000r+ 68.75sin3000,mA 10kO,12.5nF, 5 x 10sV/s,25V (2s,000r 7.s)rlmo'mA 100rad/s,80rad/s,6.2sH,2s i,4 20mA, 5mA b) 5e 160'+ 5?4'v qtnA c) -25a164t+ 25e d) 5" 160'- 20e4ormA 140?-2000 + 200e-3mo V 60?{00'cos3000t 320d{00'sin 3000tv (60 - 132x 10ar)e-1onoo' V 'oo'nA 15 40a50i 5e 15 - 45e3orcos 601 10e30'sin60rmA 15 1500re-100,45e-100, mA 40 40a5m0lcos 5000t 40esm0'sin5000tv

828 Answers toSetect€d Probtems - 40e 5000! 8.46 40 - 200,000tt5000' v 8.47 4tJ 53.33e4w+ 1333t300tu v < = 8.58 a) 0 r 0.2s125t2v, 4rY 0.2s < r < rsar: + 12.5t 1.25V,r- 1V 6.25t2 b) 2.844s 8 . 5 9 0 < r = 0 . 2 s i 2 5 5 O e+ 1 25e-2tv2 , +2e1tyl t>O.zst 6.25+ 22.38a-(a2) 15i1t' (/ 02)v, V 05 166e-2(-0.2) ps 8.64 a) 55.23 b, 262.42v v c) tnax: 53.63ps,1]0n) : 262.15 8.65 a) 40mJ b) 27,808.04 V c) 568.15 V

9.72 a) 50Hz

90' c) 40() d) 127.32m}f b)

e)

i40o

9.13 a) 251,32J.41rad/s b) 90' c) -19.89o d) 0.2pF

j19.89O

e) 9.14 a)

51lZ {r

b) 50ps 9.15 a) 1()

Chapter 9 9.1

9.5

a) b) c) d) €) f) g)

120IIz 8.33ms 100v '70.'71v 45',0.7854 rad 1.042ms 3.125ns 170V 60Hz 3'7'7 radls 1.05rad -60'

a) h) c) d) e) f) 16.67ms g) 2.78ms h) -170sin120'1V

20t20" A

9.9

+ 2 cos(400t+ 23.13) A a) 1.84e53333i 53333t -1.84e h, A,2cos(400r + 23-13")A c) 133.61mA d) 2 4,,40Orad/s,23.13' el 36.8'7"

j2 tl

x 10ar+ 34.46")V c) 46.4cos(5 9.16 a)

500t6tr'v

9.21

v;

+ j20o

bl 46.413!gv

i) 25118 ms j) 2519rns 9.8

20o

9.26 9.27 9.28 9.34 9.4O e.41

bl 1/23.13"A + 23.13)A c) l cos(8000r a, 223.6/2657" mS b) 200mS c) 100mS dl 2.24A s000nd/s (5000r+ .{5")V 33.94cos 32cos(8000t+ 90') V 2/3 A 60/ 36.a'7" v,a.64 + jL1.52A jzs A 81 f!q:A,50

j400o

Answers io S€lected Probtens829 204,0.4 - j1.2A 15.81/18.43" V 72 + j96 = 1201pEv 12cos50001V 11.31cos(5000t45")V - 56.31)A, a) 18.03cos(10,000r - 180')A 5 cos(10,000r b) 0.5 c) 112.5mJ,37.5 mJ 9.73 a) 0.5657 b) 1.sA 9,77 800+ j600o 9.81 a) 241 + i7.25: 247.111]gv D - j32 4,241.+ j8 :241.131f2yv c) -j26.90O

9.47 9.51 9.56 9.58 9,61 9.72

9.42 I

\nr -.. -.\

iv^P v^ \\t"= (v^p\ _ tjR,

9.85 a) 11= 241YA,h:2.041l|A, r1 = 21.961!A,r4 = 19.4o1yA, rs= 4.61!A,\ = 2.551!A b, 0.421y4 9,86 a) 0A b) 0.436A c) Wlen the two loadsare equal,morecurrentis drawnfrom t}le pdmary

10 Chapter 1O.2 a\ b) c) d) 10.3 a) b) 10.13a) b)

A04.16W(abs),2404.16 VAR (abs) 1s5.29w(abs),-s79.s6vAR (del)

-80 w(del),60vAR, 100VAR Balances(80w) Balances(80VAR) 0.qlagging. 0.43:0.43 leading.-0.q: -0.82 0.57leading, b) 0.94tagging, 0.343 ='75.891_l!l!f0 j24 70,22a, 72 b) 0.9487leading 10.41a) 30 + j10k0 mw b) 16.87s 10.44a) 1404 V(rms) b) 280w .l 9.720/" 10.45a) 3240W b) 6480W 10.49a) 9W b) 20O,8 mH €) 17.31w'yes d) 18.7sW e) 30O,9 InH 0 Yes o,18.26w 10.50a) 8 mH,31.62 h) Yes c) Yes 10.s8 160W 10.59a) 8 b) 250w 10.66a) 28.8O b) 28.8O c) Yes

10.17a) b) c) 10.21a)

v2

10,67a) Pr = -

v' E v2v2

v' rR.+ n,)

P , , : - =

R'xz

(del) 427.53W(del), 1174.62VAR -307.82w (del),845.72vAR (abs) Yes Yes 60v(ms) 300w

Pu:

PL

lv,

I \PL

_

P1,

Pu b) 1125W 10.6836(),24 o

P"

v ' \I /-vl ' \ PM)\PM

)

830

Answe6 to SeL€cied ProbLems

Chapter 11

W = VLILlcos(o 30") cos(0+ 30")l = 2Y,1' sin0 s n 30" : VJ fiin9 Thus,\6(w, w1\ = \f3vLI Lsnq = Qr

11.42 a) W,

11,1 a) abc b) acb 11.2 a) Balanced,positivephasesequence b) Balanced,negativephasesequence

b) 11.52a,

41'72.'79 VAR 1.2MVAR

c) Balanced. posirj\e pbasesequence d) Balanc€d,negativephasesequence e) Unbalanced,due to unequalamplitudes f) Unbalanced,due to unequalphaseangle separation 11.8 !'AB : 13,799.25cos(ot 30")V, ,tBc : 13,799.25 cos((ot + 90') V, ((,,r- 150")V z,c"a: 13,799.25cos 11.9 a) 32.84A(rms)

b) 12,8as.9aVGms) 11.10a) I,A = 241_fSEA,rbB - u1p!!:A,

t c = a1_14!4lA

b) vab: 8313.841:fqV,Vbc: 8313.8419q V, vca= 8313.84l,150. V c) vaN: 46s9.9611!!:v, vBN: 4659.961115.63. V, vcN = 4659.961_4fl1Y d) y^B : 8071.2811!4Y, vBc: 8071.28M{I v, ycA = 8071.28/ -154.37. y 11,14a)

2005(f

v G.0

b) 0.23l1s6.87'AGmt c) 35.3/116.63'V(rrns) 11.15159.5/29.34" V(rms) 11.1610.81,/71.34' A(ms)

11.2461201)55!vA 71.25a) 1833.46132V A b) 519.62V(rns) 11.47 19'7 .29W,4't6.63W

1.2MW

G) D' l -

I2MW

11.53 a) 16.71&F b) s0.14/,F 11,54 V"bl = 12,548.8 V, so the voltageis belowthe acceptablelevel of 13 kV Thus,wher the load at the substationdrcps to zero,the capacitorbank must be switchedofi 11.s5 Ploerorc)= 81.66kW, PL("rl.,)- 40.83kW

Chapter 12 12.1 a) (120+ 30r[(&G + a) "0)] +(120:0r)[u0)- u0 - 8)] + ( 360+ 30rlu(r8)-u(r-12)l / - \ / - \ b ) t 5 0 s i o" r l , t ) - ( 5 0 s r n , /) l / z l \ . z l \ .) (30- 3tltu(l) - (r 10)l 12.2 a) (2.st+ 50)u(t+ 20)- 2.stu(t)+

4)

(sr so)u(r 10) 0 , 5(r | g\u(t- q) 5(r o),(/ | or - 5t 3)u0+ 3) + 5(1 3)40 3) + s0 6)u(t 6) - s(r - e),0 9) 72.5 a) 1.0 b) 0 c) 12.6 a) 52 b) 6.25

Answe8 to Seteded Prcblems 831

12.7

318

12.25 1,F(u)du: l,-ll-ra;-.1t*

1 (s + a)"

12.13

r-T ,"I I I I l()e-'' du ldt

Jo- LJ,

b)

('cos0+ rsir0

d)

I

?

=;,,rl-)''

siDld + slcosh0]

G'- 1)

b)

='\+j

1 I

r

b) Y{rsinBr} :

.) check 12.26

12,18a)

200s'

(r2+4or+64)(r,+1oo)

r) [5e-" + 10e-6icos(& s3.13")]r(r) d) [8 + 50?-7cos(24t + 16.26')]u(t)

c) 2 d) check

" , ' " ' : 1 1 I n , v 'a' ,l ds dsLJr' j :-

tf(t)e "'dt

I

dF(s) So jrtll(/)] = -rb)

,+t

s-+p-

12.40 a) I3et+ 6e-L + 9e-3tlu(t) b) [10 + se '1 - 8e 3t + e 5t]u(t)

b)

12.24a)

e-a dudt

=;,,,111,i),,

I

12.17al

fQ) |

I

":.':': ":.':': So

I

r l f l t s r -d"lt

I

-lflu)e"'dt

:i:

( tf I t"f(t\e"'at

72.41a, b) c) d, e) 72.42c\ 72.47al b) c) d)

rorlr(r) l40t - 8 + 16e 110 40rc-2t+ 2oe-'tlu(i) "cos(t + 53.13")] G) [10t 5 + 10e 2tlu(t) + I)e lQt l.st' L50r€'co(|2l l{r.'oo)| 20P-'cosf)r 36.87)l&(r) 6'(t) + s6(t + 50€ 'oru(r) f(0') = 18,/(oo) - 0 /(0*) = 8,/(co) = 10 /(0) - 11,/(oo):0 /(0*) : s6,/(€) : 8

: ( r),e{t"l(t)}

s.tfi-ry g{r sh Br} : e{e 1 cosht}

13 Chapter 2Bs

G' + B'\' s2+2s+2 s'(r + 2)'

13.4 a)

5lr' +2000s+1071 J

b) Zeros at -1000 + j3000rad/s and 1000

i3000mdls;pole at 0

832

Answerc to Setected PrcbLems 13.16 a)

25 X 106s

a)

s2+5o0o.r+4x106 Z€ro at 0;polesat -1000rad/sand b) 4000rad/s

n0lsv 120lsV

13.9 a) 2000()

0.8sO

1l-x-10'

0-8 X 106^

b)

0.8 x 106o

1 2 x 1 03 V

b)

12,000

r' +25oor+106

c) (-8e-sm' + se 'ooo!)ll0)v 13,10a)

10ko 103ho :1cF A

100(r+ 104) r' +104r+25x106 -0.01(s+ 7s00) c) (s + 5000)' d) 15x 105tesm, + 100d5ooo'lrc) v e) [25,000r + 10]€5om'r(r) mA b)

13,15a) [5 6e 5'+ ae xlu(t) y b) Comparesolutionat t = oand r : ooto

circuit at t : 0 andt = co

3(r + 2s00)

s2+2ooos+107 (3000r- 26.57')u(.i)A c) 3.35e-10m'cos 13.26 a) (256e {mo' - 4e-10,m.)&(r) v 1o,lroo, ao,'$o)u(r) b) (?5 + 5e 80e nA = c) Comparesolutions at t 0and1= c€to circuit at t = 0andr = c. 13.27 al

15r' + 15s + 24 12s2+ 63s+ 24

, G + r ) G+ 3 ) ' , ( , + r X !+ ,

b) Initialvalues: 15A, 12A; finalvalues:4A,4 A c) (4 - 2'7e't + 38e-3')u(t)A,

(4 + 21e-t 19e 3\uO A _ 5e rooo,l"(1) 13.35 [5 5000r?rooo, mA 2'76.25(s + 4OI)) 13.36 a) s(s+200)(r+8s0) b) Initial valueis 0, final valueis 650mA c) 650 - Ase 2Nt 225e-35u )u(t) mA ,00, 13.37 a) (51? 51?-3s9u(r) mA 'm' 350!)rO + 51e b) ( 51e mA 6x 104(r+ 4000)+ 96 x 106 73.42a) r(s+2000)(r+3000) b) (56 - 108emo' + 52e-{0)r0) v jL. 13.49 a) no zeros. ooteat so rad/s J+JI)

b) -i=:. zeroat 0, pole ar 50 rad/s r+5tJ , t3.

J " ru-raos t0".r.rortr.Potear

Answers to Setecied Probtems833

O

I ^ r"n,6n o . * t . t o s . p o l e a l

,-., 100 e) -=;.

13.57 (e

3

106rad.

poleat -125 rad s no zeros.

400

l)e-'V

300

13.58 (1 - e)€ IV

zo0

13.74 50cos(8000t+ 36.87) V

100

-16 x 10as

13.75 a)

(s+8000)(r+16,000) - 161.57") b) 4 cos(8000t V

13.76a)

s(r + 9000) (r+2000Xr+4000)

- 2se-4urk\u|) V b) (35e-2c.rJt + 30.96)V .) 11.68cos(2000r 13.83a) 8 0 v b) 20v c) O V d) 326(r)&A e) 1 6 V

0 g) 20v

et o.6e-2"re'u(t) A -0.6 e 2x16tu(t) A

[ 1.6x 10-36(r)] 17200e2\ra"'u(t)lv 13.89 a) i,(o ) = i,(o+): oA; iz(o-) : rr(0+): 35.36A 1440r(122.cn\r%- 3O]U1\D ,, b)

'.) =

+

G + 147s")G'+ 14'+ool) 3O0\D

s + 1475n bo = 252.89e185"t + 772.62cos(120nt+ 6.85) V Do(O+): 424.26Y c)

200

13.90a) b)

20.58e-Bb"'+ 172.62cos(120Tt83.15')v

'.(v) 200 150 100 50 0 -50 100 -150 200

10

12.5

15

t(ns) c) Voltagespikesin Problem13.89but not here

13.84a) 0.8A b) 0.6A c) 0.2A d ) -0.6 A f)

0

vo = 122.06l1ql v(rmt

14 Chapter 14.1 a) 954.93H2 h) H( ja") = 0.701111t:, H( j0.3d.)= -L6;70",H ( j3a,\ = 0.9s78,r -'71.5'7' o.3162/ c) 0,((,") :35.36cos(6000t 45")Y : 47.89cos(1800116.70')y o,(0.3(.)") : o.(3o.) 15.81cos(18,0}0t '71.57")v 14.2 a) 392;70A Hz b) 1640.85 14.12a) 9.95ko b) 917.03 Hz 14.13a) 4kO b) 60ko 14.21a) 1 Mrad/s b) 159.15 kHz c) 7.5

834 Ans\j/ers toSetected PmbLems d) 935.sskrad/s e) 148.90kH2 f) 1068.89 kmd/s s) 170.12kll2 h) 21.22kH2 14.22a) 8 kO,16mH h) '7.s'7kH2,837 kllz cl 195.77 LIz 14.33a) 8 Mnd/s b) 1.2"tM}]z c) 16 d, '7;7sMrad/s €) 1.234MHz f) 8.25Mrad/s g) 1.31Mrad/s h) 79.58kHz 14.34al 254.654,fi132 pH b) 46.9'7kHz,53./2 k}jz .) 6.25kHz 14.43a) 0.39H,0.10r!F h) lus,") - veals,l: 0.7071v0."r1, lVnos,l: Vszu) = O.94slvn""tl c) l\21qH,l= O.344Vrat 14.44L - 0.D5H.C 0.057t!F.0.344vp*l 14.4563.7timesaslargeastlle DTMF tones

15.7 a) R1 = 5.85kO, R2 = 23.29kO b)

15.13a) 1H,1F,0.04 o b) 0.9H,0.11nF,3.6 ko c) 0.11nF

1lL = llQF $lQ)s b) r' +(1/O)s+1 c) 5 k o , 2 H , 0 . 2 n F

75.14 a't

d)

Chapter15 15.4 a) n1 = 0.5?kO, Rz : 3.18kO b) €)

2500r

s' +25oor+25x103

15.30fct : 1291.4Hz,fc2 = 49,037.85 Hz, = 21..64 A, RE : a2L.64 A 15.31 Rr = 7U.6dt,RH = 20.7O; if Rr : l kO tnen

Rr:4kO 15.33a) 4 b) -48.16dB

koblems 835 Answe6 to set€ded nF,30.44nF;secordstage: 15.36 a) First stage:208.05 86.12nE 73.53nF b) 86 nF

16 Chapter ndls md/s,.d,b: 785,398.16 16.1 a) rad = 69,813.17 : = 11,111 11H2,f,b 12skHz b)," c) dla : o,dlb : 18.75V d) ak" = 0lor all k; bku: 0 for k even;

c o .T l v r o r i o, o a r

t,"='\l z nkL

+

t

j'*

.r(b:

I

kf 5 0I . t"l-." s,Dt + s,ntJvrorau(i k,l

brb : 0 for all k e) 0"(, =

O 15.37a) 900.32 O,1800.63 b)

i:: t

t

t

/

:1 2

, = t " t - , s ,n \

--\

v siou@"/ cos= j,i l

,'b0)= 18.7s+

Aili,i,!i*snT)*,,,rv ';in\

76.2 a,

15.58a) R - 26sA.oR = 261O, (1 o)R : = 300nF 4.4O, C : 150nFF,2C b)

%

t,/

!o",-",.t

t v f I cosr.^t lV b) ::!!| | + 2 ! i.' l ' L 4n i1l

.t

u^ u^.

L ziiiL

i

4

4.4o 16.3

i

I "*",.,"

300

+ 50 cosrr,/

cos(n,/2) cos(/l.,,,t) v ,=oros @2 l')

600 S

16.10 a) 100Hz

s2 + o4 lo6rr r'+ 533.33os+ 64 x 10612 15.59ChooseR1 : 11.1kO,X, : 100k.().Then, C = 39nF,lH(Jo)|."" : 20.01dB, lH(jlR2c) :17.04d8 15.60Choose-Rr= 100kO, thenR2 = 400kO, C1 - 7.96nF

b) no c) yes d) yes e, yes f) do - 0, itk = 0 for all 4, b* : 6 1o. I "u"n' k1t" 80 .. D * = J J S r n- t o r / a o o o n-k'

836 Answers toSetected Probtems 16.11a) grad/s

16.48 a)

,{r (v) b) ro c) yes d) no 16.18a)

20 18 t6 I4 12 10 8

Ahcos(na;t 0,) wherc

>

2 o1
AkLs, =::,-" | ; K

\nk

\

jt lmV,rorodd*

0 1 2 3 4

)

b) 888.92 mV

b)

16,27214.66 cos(2000t 26.57')+ 44.38cos(6000r+ - 68.20)V 123.69')+ 17.83cos(10,000r 16.28a) 839.82 cos(10,000r 1.19) + 278.78 cos(30,000r + 174.64") + 118.74 cos(70,000r 171.70.) V b) The5th harmonicat 50krad/s is etiminatedby the bandrejectfilter whosecenterftequencyis 50kmd/s

'7 5-3-t

1 3 5 7

16.49al

16.32a) 28'7.06W b) 300w .l -4.3L0/"

)A*l

500

16.3341.52mW 16.36a) ?4.5356v(rms) b) 74.5306v(rms) 16.37a) 77.9578v(ms) b) -2.5so/" c) 46.1880V(rms),0.0156%

0

76.44Cr = a. T

C"=;|l2co'tnr

2t - nt stntnlt2)

21.

n : l:1,+2,. 16.45a) 4000W bl 7.72A c) 10.57%

3."

Prcbt€ms 837 Answe6 to Setected 17.3 a) =[@,t

1c,l 500 f

cos(a.tlz) 2sin(a"tlz)l

b)0 c)

4oor

,*l

/(t)

1.0 0.8 0.6 0.4 0.2

200l-

Lool-

0.0F;

-0.2 -0.4 0.6 0.8 -1.0

r

2

3

4

5

2(a' - a') (d + 0?)2 ..^ (o' -') b) lqtao':----: -

17.4 a)

77 Chapter

c)

2Al^ 17.2 al r lrcos=o-f t b)

t

. M + d,rsrn ^

-l zl t

d)

a2 + (.

..;'

d+(.+-.\'

d + (a

ot.)2

a' +(a+,t")2

e)

t sinl(d do)'/21 r str[(a + a.blz] + t 2 (a - a.)(rlz) t (a + o")(rl2\ b) r((d) +,t6(o - o,) + d(.,,+ a,o)l 17.22a, msgn(t)- 40e5or0) V b) Yes,checkthe idtial conditionsandfinal values i.za a, sestu(-t)+ (12.5et 7.5est\u(t)v

c,

1 7 . 1a9,

F(,)l 1.0 0.9 0.8

o.1 0.6 0.5 0.4 0.3 0.2 0.1 ':15

b) 5v

.) sV d) (r2.s.t '|.ses'\u(i)v - 10

e) Yes + 90")nA 17.32166.67cos(2500r

838

Aiswers to Set€.-ted Probt€ms

17,39 a) 15eat - se.5'lu(t) +. r1dtu1t)v

0 %.95"/" s) 96.93"/" ChapterlS

20

l%(') 4 3 2 I

-r0 -8 -6 -4 -2 d, 720r e) 28J.

o

2

4.

6

8

10

18.11lr1r: 1000O; i12 : 1 x 1!a:h2r = 40 hn=20x101.5 18..12a, ai -, -4 x 1O1;aD = -25 A, . a2r= -5:x 10-' S;ar, = -0.025 . 9l ,tt =, -4 x 1c-a;ar, = -25 gtr' i | . Pzr=' -5 x 10' S;ar" = -6,925 7A30 1s.625 18.32a) 281180"V(rms) b) 11.mmw c) 2.88pW 18.377.5W 18,383.88V

Index A

B

a-phasevoltage,434 Active filter c cuits,s€eFrequenry'selective circuits Admittance, impedance and, 350 Ammeters, 68,69-70 Amplifiercircuits, 45, 103-104,110-111,125-126 Kirchoffs lawsappliedin, 45 mesh-currentmethodanalysisof, 110-111 node-voltagemethodanalysisof, 103 104 Ohm's law applied in, 45 Thdvenin equivalent usedin, 125 126 Amplifiers, 154-185,26V263,312j16 integrating,260-263,312316 operational(op amp), 1s4-18s Amplitude modulation, 712 Amplitude plots, 682-684,800-804,804-805,809,810-812 Bode plots, 800 804,804 805,809,810812 complexpoles,corecting for, 810-812 coner frcquency and, 801-802,804 805 quadmtic factor and, 809 straighlline, 800-804,804 405, 810412 Amplitude spectrum,defined, 682 Analogmeters,68 Analysis, ree Circuit analysis;Laplace transform; analysis Sinusoidalsteady-state power, Apparent 401-402 AppliaDc€ ratings, 397 398 Arcing current, 188-189 Attenuation,defined, 566 Average power, 394-398,406, M5446, 452454, 675-678 defined, 394 measuringin three-phasecircuits,452+54 periodic functions, calcularions with, 675-6?8 power factor, 396 sinusoidalsteady-state analysisand, 394-398,406 three-phasechcuits, 445446, 452454 two-wattmetermethod, 452454 wye (Y) connectedloads,in, 445 446

b phasevoltage, 434 Balancedthree-phasecircuits,432-465 conditionsfor balanced,437 electdcalpower,transmissionand distdbutionol 433, 455 456 line current, 440 line voltage, 440 measuringaveragepower in, 452-454 phasecurrent, 440 phasevoltage,440 power calculations in, 445-451 voltage, 435-436 souces,three-phase 1wo-wattmeter method, 452454 voltages,43H35,435 436 wye-delta (Y-I) circuit, analysis of, 442 445 wye-wye (Y-Y) cncuit, analysis of, 436:142 Bandpassfilters, 569,582-592,615618,636638 bandwidth, 582-583,586 broadband,615-618,636 Butterworth, 636 centerft equency,582-583,584,585 cutoff frequency, 585,588 frequency domain-time domain relationship in, 591-592 narrowband,636-638 op amp, 615-618 parallel,RLCcircuit aq 588-589 passivefilter circuits, 569,582 592 qualitativeanalysisof seriesRLC circuit, 583 quality factor, 582 583,586,636638 quantitativeanalysisofseriesRaccircuit, 584 587 resonant frequency, 582 seriesRIC circuit as, 583,584 587 transfer function for, 591 Bandrejectfilters, 569,593-597,618622,636,639441 broadband,618-622 Buiterworth, 636

839

840

Index

narrowband, 639 541 op amp, 618-622 passivefilter circuits,569,593-59? qualitativeanalysisofseriesRaccircuit,593 quantitative analysisof seriesRaC circuit, 594-597 se esRaccircuitas,593,594-597 transfer function, 596 twin-Tnotch,639 Bandwidth,bandpassIilters. 582-583,586 Bassvolume control, 607,642-644 Black box, defined, 59 Block diagrams,616 Bo{teplots,609-611,?99-815 circuitsand, 609-611 activefiequency-selective amplitude,800 804,804-805,809,810 812 complex poles and zeros, 807-809 corner frequency, 801-802,804-805 phaseangle,805 807,813-815 real,firslorder polesand zeros,?99-800 stmightline, 800-804,804805,805-807,810812 Branch, defined, 94 Broadbandfilters, 615-618,618622,636 bandpass,615-618 bandrejecr,618-622 quality factor and, 636 Butterworthfilters, 627-636 bardpass, 636 bandreject, 636 design of circnits, 629 632 high pass, 635-636 low-pass, 627-634 order of, 632-634 tansfei function, 628-629

r' c-phasevoltage, 434 capacitance,defined, 186 Capacitive circuitq 395-396 Capacitors,195-199,200-2\3,2fi 2n 309,509-51'0, 540 542 curent in, 196 displacemenl cullent, 195 energy in, 197 powerin, 196-197 s'domain,in, 509-510

sedes-parallelcombinations for, 200-203 switching operation for impulse function ol 54[!-542 voltagein, 196,309 Cascadeconnection,312-313,622-52J identical op amp filters, 622-627 integratingampliJiers,312-313 Center frequercy, bandpassfilte$, 582-583,584 Characteristicequation,Raccircuits,28T-288,308 Circuit analysis,10-11,4246.92 153,170-1?2,506565, 6'70475,714716 amplifier circuit, 45 circuitswith realisticresistors,93,133 136 conceptualmodelfor, 10 dependentsources,42 40 Fouder seriesand, 67M75 Fouder transform, using, 714-716 ideal circuit components, 10 inverting-amplifiercircuit usingrealisticop-ampmodel, 1'70 1'71 Kirchoffs lawsappliedto, 43 '15 Laplace transform in, 506-565 matlematicalmodelEuseof, 10 ma-\imumpower transfer, 12G129 mesh-currentmethod, 105 107,107-109,109112, 112 116 node'voltagemethod, 97 100,100-101,101-104, 112 1'16 noninve ing-amplifier circuit using rcalistic op-amp mode| 1'71 1'72 Norton equivalent, 119-120,121 Ohm's law applied to, 43 45 ove iew of, 1G-11 physical prototype, 10 planar circuits, 94 sensitivityanalysis,93,133-136 simultaneous equations, 9546,96 9'7 source transfomationE 116-119,121' 122 superposition, 129 132 techniquesof, 92 153 terminology for, 94 97 Th6veninequivalent,119 121,122-123,123126 unknown voltage, finding, 44 C cuit theory 5-6 electdcal circuit, 5 electdcal engineeing, 5-6 frequency,6 system,5 lumped-parameter

Index 841

cinc[irs, 2 2L, 22-55,56-91,92-153,228-2$, 244-329, . Seealso 39+396,432465,566-605 ,'730J5'7 ,606-655 Models; Circuitanalysis; Idealbasiccircuitelements; operationalamplifier activefilter, 606-655 :126 amplifier, 45,103-104,110-L11.,125 three-phase, 432-465 balanced blackbox, 59 capacitive,powerin, 395-396 curent and, 11-12,12-13,24-28,36-3'7,38:39 curent-divider, 6ft5 delta-to-wye(A-to-Y) equivalent,73-76 dependentsourcein, 42-45 divisionof voltageand current, 65 67 electrical, defined,5 and,2,3-a eleitricalengineering resistance, 28-32 electrical electrical safery,23,46-4'7 energyand, 14-16 first-order, 228 283 fiequencyselective,566 505 idealbasicelements, 12 13,2255 inductive,powerin, 395 intemationalsystemof units (SI) for, 8-9 Kirchoffs laws, 3642, 4345 measurement of voltageandcullenr, 65-67 model,constructionof, 10,32-35 Ohm'slaw,28 32,4345 open,33-34 pamllel-connected elements,59 60 pi-tot-tee (,{o-T) equivalefi,'73-:74 plaltar, 94 power and, 14-16 realisticresistort 93,133-136 rear windowdefroster,57,7G79 resistive,powerin, 3921395 resistorsand, 2a.3O-32,5862 responsesof, 228-283,284-329 secord-order, 284-329 series-cornected elements, 58-59 sedes-pamlelcombined,6M2 short, 33-34 sourcesand, 22128,56 teminology for, 94 97 two-porq 730-757 variables ot 2-55

voltageand, 11 12,12-13,24-24,3'l:i9 voltage-divider, 62 64 wheatstone bddge, 71 72 Closed path, defined, 37 Coefficient of coupling, 211 212 Common mode input voltages, 166 167 Common mode rejection ratio (CMRR), 168 168 Communication systems,electrical enginee ng, 3 Complex frequencies, 289 Complex numberE 781-786 a thmetic operations for, 783-785 graphical representation of, 782-783 identitiesol 785 integer power of, 785 notation, 781-782 roots of, 768 Complex poiver, 401402, 403405, 446 447 alternate forms of, 404-405 apparent po\qer, 401 calculatiol of, 402 defhed, 401 sinusoidal steady-stateanalysisand, 401-402,403 V'I relationships and, 403-404 wye (Y) connectedloads,44947 Computer systems,electrical engineering, 3 4 Conceptual model for engineerhg design, 10 Conductance,29,31,350 CorNtantsources,56 Cortrol systems,electdcal engineedng, 4 Convolution, Fourier transform of, 712 713 Convolution integral and the tmnsfer function, 531-537 Comer frequency, amplirude plots, 801 €02, 804-{05 Cosine Iunction, Fouder transfom of a, 707 708 Cramer'smethod, 760 Critically damped response, 289,299-300,309 capacitorvoltagestepresponse,309 currentnaturalresponse,309 voltage,289,299-300 Current and voltage (V-I) relationships, 342-345, 403404 capacitors, 344-345 inducto{s, 343-344 passive in lhe frequency domain. circujlelements 34 :45 power calculations, 403 404 rcsistors, 342-343 analysisand, 342-345 sinusoidalsteady-state

442 Currentratioq 368-369,369-370 determination of, 368-369 dot convention,369 ideal transfomers ard, 368i69,369:3'71 polarity of, 369-370 CrlllenI, 11-12,12-13,U 28, 36-3?, 38-39,56,65-67, 68 70,156-161,188 189,19u-191,195,196,230132, 309,332134, 342345, 341 arcing, 188-189 capacitonand, 195,196 cotrstant, 56 defined, 11 dnect (dc),56 directionreference,12-13 displacement, 195 division, 65-67 electdcalchargeand, 11-12 inductorsard, 190 191 input constraint,159 Kirchoff 's law, 3637, 38-39, 347 measuring,68 70 natural response,deriving expressionfor, 230 232 operational amplifier (op amp) terminals, 156-161 passivesign corvention, 13 Ra circuits, natural responsein, 230-232 RLC sedescircuits, natural responsein, 309 sinusoidalsource,332-334 sinusoidalsteady-sta1e analysisand, 332-334, 342-i45,34'7 sourceE24 28,56 V-l relationships, 342-345 Current-divider circuit, 64-55 Cutoff frequency,568,571 572,573,575,585 586,588 bandpassfiltels, 585 586,588 bandwidth, rclationship to, 586 center liequency, telationship to, 585 defined, 568,571-572 half-power ftequency, 572 low-passfilters, 573,575 paratlelRaC filterq 588 rearnrrerolcuttq )/l-)/t seies nC filters, 575 sedesRl, filters, 573 seriesRaC filters, 585

D Dampediadian frequency,295296 Dampiq coefficient, 296 Damping factor, 296 d'Alsonval ammeter, 69-70 Decibel, unit of, 797-798 Delta (A) cornected load, power calculations in, 447-448 Delta (A) interconrect, 73 Delta-to,wye(I-to-Y) tansformarions!13J6,352155 equivalent circuits, 73'76 impedarces, 352-355 Determinants,160'764,765 chancteristig 760 evaluation of, 761-764 marrices,765 numerator, 760-?61 Differcnce-amplifier circuit, 165 169 commonmode input voltages,166 167 commonmode rejectionratio (CMRR),168-168 differential mode input voltages, 16G167 measuringperibmance, 168-169 Differentialmode hput voltages,16G167 Difterenlialion in lime domain.47{r477,7l. Digital meterE 68 Dirac delta function, r€s Impulse function Direct curent (dc) sources,56 Direction references,12-13 Dirichlet'sconditions,658 Division of voltageand current, 65-67 Dot convention,20+20'7,369 Dual irline package(DIP), 156

E Effective value, 399 Electrical circuits, s€€Circuits Electrical engineedng, 2, 3-8 circuit theory, 5 5 communication systems, 3 computer systems, 3-4 control systems,4 overview of, 3-8

Index 843 problem solving, 6 8 role of, 2 sigEl-processing systems, 4-5 Electrical power, tansmission and distdbution of 433, 455456 Electrical safety, 23,46-47 Electricalsources, re?Sources Ercrgy, L4 16,192.197,212214,717 :7A. Seeako Power calculation of in rectangulai voltage pulse, 722 723 calculations,212 214 capacitors and, 197 circuirs and, 14-16 inductors and, 192 mutualinductanceand, 212-214 Paneval'stheoremand, 717 724 Equivalent circuits in magnetically coupled coils, 787 791 Exponential form of Fourier s€ries, 679 682

F Faraday'slaw, 207 Feedbackresistorqintegratingamplifierswith, 314-316 Filters,seeFrcquency-selectivecircuits Finafvalue theorem, Laplace transform, 495 498 Firslorder circuits, 228 283 defined,228,230 general solution for, 248-254 integratine amplifier, analysisot 260-263 natumlrcsponseof, 228,230-236,236-240,248-254 resistor-capacitor (RC) circuits, 228,236-240,U6t48, 248-254,257258,258259 (na) circ]uils,228,230:36,A7 246, resistor-inductor 248-254,255-256 responseol 228-283 sequenrial switching, 254-258 step rcsponseoI, 228,240-248,248-254 unboundedresponseof, 258-259 Firsforder filtels, 608-612 Bode plots, 609-611 high-pass,61M11 lowpass, 608 610 op amp circuils, 608-612 Flashinglight circuit, 229,263:264

Fourier coefficients, 658,65944,662468 even-funclion symmetly, 662 663 half-wave symmetr$ 66,k66 odd-function symmetry, 663 {64 periodic tunction and, 658,659-64 quarter-wave symmetry, 666-667 s'.mmetry,effectof on, 662-668 Fourierintegml convergence of, 701 703 Fourier series, 65ft97 amplitude, 682 684 analysis, 658 659 appiication ol 670'675 average-powercalculations, 675 678 coefficientq 658,659-662,662-668 Dirichlet'sconditions,658 exponential form of, 679-682 fundamental frequency, 658 hamonic frequency, 658 introduction to, 65{q57 pedodic functions, 656,6154'78, 6784'79 Phasespectra, 682-684 rms value, 678-6?9 steady-stateresponse,finding witl, 67tl-675 symmetryand, 662-668 trigonometric form of, 668-670 Fouder transform, 698 :729. Seeabo Opelati,or6'l tmnsforms circuit applicationsol 714 716 cosine function, of a, 707 708 defined, 699 derivationoq 699-701 elementary tunctions, 708 functional, 706-708,710-714 inverse,700 Laplace transforms used to find, 703 706 limits ol 70G708 mathematical properties of 708 710 operational, 710-714 Parseval'stheoren, 717-724 signum function, of a, 70G707 unit step function, of a, 707 Frequency,defined, 6 Fiequencydomain, 338,342-346,346-348,34u352, 352 355,355 360,360 361,467,479,51G57'7, 591-592,712,'713

844

Index

convolutionin, 713 defined, 338 delta-to'wye(A-to-Y)transformations,352-355 Fourier transform and, 712,713 high passfilters, relationship to time domain in, 591-592 impedanceand, 345,348 352,352-355 Kirchoff's laws, 34G348 Laplacetranslormand, 46?,479 low-passfilters! relatiotrship to time domain i4 57G577 mesh-currentmethod, 360-361 node-voltagemethod, 359-360 Noiton equivalent circuit, 355 passivecircuit elements in, 342-346 reaclance.145 .inusoidal sleady-slale analysis aDd,J42 J6l sourcetranslbrmations and, 355-358 fh6venir equivalent circuit, 355,357,358 time domain, relationship to, 516 577,591-592 translationin, 479,712 V-I relationships, 342-345 Frequency response,defined, 566 Frequencyresponseplots,568,608-611 Frequency scaling, 612 Frequency-selective circuirs,566-605,605-655 active filter circuits, 606-655 attenuation,566 bandpass filler\ 5bi1.582-592. ol5 ol8. b3o-ol8 bandrejecl tilteri 5oq.5qJ-597,o18 o22,o3o,blg{41 bassvolume control, 607,642444 block diagams, 616 Bode plotq 609-611 brcadbandfilters, 615418,61a-422,636 ButteNortl filters, 627-536 cascadingidentical filters, 622,627 cutofftequency, 568,571-572,573,575,585-586,588 defined, 566 filters, 566,567 tirst-orderfilters, 608-612 tiequency responseand, 566 frequencyresponseplots, 568,608-{11 higher-order,622-636 high-passfilteN, 568 569,5'77-582,610411,635-.636 low-passfilters, 568,570-5'7'7.608-610,62:7634 magnitude plot, 568

narrowbandfilters. 63ffi42 opampfilters, 608411,613414,615-622,622436 passband,568 passivefilter circuirs, 566-605 phase angle plot, 568 pushbufton telephone circuits, 567,598 scaling,612514 sropband, 568 rransfertuncrions,5j6,581,591,596,628-429 Functional transforms, 468,474475 Fundamentalfrcquency,defined, 658

lJ Gain, defined, 157 General solution for responses,248 254,286-290 charactedstic equarion, 287 288 complexftequencies,289 f rt-order circuirs, 248 254 neper frequency, 289 resonantmdian ftequency,289 second,order circuits, 286-290

H Harmonic frequency,defined, 658 Heating appliances, 391,41'7418 Higher-order op amp filten, 622-636 High-passfil1ers,568-569,577-582,610 611,635-636 acrive fi1ter circuirs. 610 611 Butterworth. 635 636 defined. 568 569 firsrorder. 610 511 opamp, 610 611 passivefilrer circuirs,568-569,577-582 qualitarive analysisof se es RC circuir, 577 578 quantitativeanalysisof seriesRC circuit, 578-582 seriesRC circuirsaq 577-578.578582 tmnsfer funcrion for. 581 Househotd disrdburion circuir, 331,375 Hybrid parameters,two-poft c cuits, 734

Index 845 I

Ideal basic circuit elements, 12-13,22-55J8G227 active, 25 capacitors, 195 199,200 203,217 227 c!Itenl,7z 73,24 28 direction reference, 12 13 dot convention, 204 207 electiical resislance, 22, 28 32 in series,38 inducton, 188-195,200-203,21?-227 mutualinductance,188.203-2U,209-214 rode, 36 passive,25,188 passivesignconvention,13 polarity reference,12 13 proximiryswirches,1a7,214216 iesisto^, 28,30 32 self-inductarce, 207-209 series-parallel combinations,200-203 sources,24-28 \oltage, 72-73,24-28 Ideal transformer,361,365 371,792:795 curent mtios, 368-369,369 370 d€fined, 365 equivalent circuits, in, 792-795 impedancematching, use of for, 371 timiting values, 36G368 polarir)ofratios.l6q-J70 sinusoidalsteady-stateanalysis,361,365371 voltageratios, 368-369,369370 Ignirion circuit, 2a5,31'7320 Inmittance, defined, 734 k[pedance,345,348-352,352:355,363-365,3'71 admittance,350 combi ng in seriesand parallel. 348-352 conductance, 350 defined,345 deltato-wye (A{o-Y) transformations,352 355 linear transformers, 363-364 matching,ideal tmnsformeruseof for, 371 parallel, 350 passivecircuit element of, 345 reactance,345 reflected impedance, 363-364

series, 348 349 susceptance, 350 Inpioper rational function, 483,493494 Impulse function, 47M73,54LI-547 de[ircd,4'72 derivatives oI, 472-473 impulsjvesources,544-547 Laplace transform circuit analysis and, 540 547 sifting property, 472 strength ol 471 swilchingoperations,540 542,54 544 unit, 471 variable-parameterfunction,4Tl Impulse,defined, 470 lnductance,186,188,2031M,24114 defined, 186 Faraday'slaq 207 mutual, 188,203207,209 214 self, 207J08 Inductive circuits, 395 Inducto$, 188 195,200203,21721,54 541 curent in, 190-191 energy in, i9 power in, 191-192 combinationsfor, 200-203 series-parallel switchjngoperationfor impulsetunction of, 542-544 voltage in, 188 189 Infidte ftequency, 575 Initial-value theorem, Inplace transform, 495 498 Inpurcohlrainrs.158.159 Instantaneouspower,392-393,448449 Instantaneous real power, 394-398 Inte$als. 531-537,701-703,819 InteFating amplifiers, 260 263 ,312 316 cascadecoinections,312313 feedbackresistors, with, 314 316 first-ordercircuitsand, 260-263 response,analysis ol 260-263,312:316 secord-ordercircuitsard, 312-316 sequentialswitching,analysisof with, 262 Integratior in time domai]tr, 4'7'7-4'78,'711J12 Intemationalsystemof units (SI), 8-9 InverseFouder transform, 700 Inverse Laplace transfoims, 482 494 impropermtioral functions,483,493+94 partial fraction expansion, 483-494

846 propei ralional functionq 483-484 rational funclion, 482-483 transform pails for, 493 Inverting-ampliJier circu\I, 161 1.62,170-17L

K Kirchofs laws, 3642, 4345, 346 34a, 511 amplifier circuit, applying io, 45 circuit analysis,appliedto, 43-45,511 circuit model using, 41 42 closedpath, 37 current, 3G37,38-39,34? electrical circuits and, 36-42 elements in series, 38 frequency domain, in the, 346 348 loop, 3'7 node, 36 s-domain,usingin, 511 sinusoidalsteady'stateanalysisard, 34G348 unknown current, finding using, 40 unknown voltage,applyingto find, 44 \ oltage, 3'7,39, 346-34'l

L Lagging power factor, 396 Laplace traNform, 46G505,506 565,'703J06. Seealso Operatioml transfoms application ot 4814a2, 512-526 circuil analysisand, 50G565 defined, 467 468 final-value theorem, 495-498 Fouder tiansforms, used to find, 703-706 frequercydomainand, 467,479 functional transforms, 468,47447 5 improper rational functionE 483,493-494 impulse tunction, 4'70473, 54U54'7 initial-value theorem, 495 498 inverse, 482 494 mutual inductance,analysisof a circuit with, 521-523 natuml responseusing, 512 513

one-sided,468 opemtional transforms, 468,4'l 5481 pai,rs,474-4'75,493 partial fraction expansion,483+85,528 531 polesof F(s), 49a-495 proper rational functionsr 483-484 rationalfunction, 482483 s'domain, 467,506,508-510,511-512,520 521,,523-526 step function, 468-470 stepresponseusing, 514-515,517-519 .uperposition. u.e of in \-domarn.523 526 surgesuppressors, 507,548 Th6venin equivalent, use of in s-domain, 520-521 time domainand, 467,476'479 transler function, 526 52a, 528-531,531-531,53'7 540 transientresponseusing, 516-517 unilateral, 468 zerosof F(s), 494-495 Leadingpower factor, 396 Line currerlt, defined, 440 Line spectra,682-684 Line voltage,defined, 440 Linear equationqsolutionof simultaneous,759-780 Linear transformer,sinusoidal steady-state analysisof, 361-365 Loop, defined, 37,94 Low-pass filters, 568,5'70-57'7,608-610,627 629 active filter circuits, 608-610 Butterwofih, 62?-629 cutoff ftequency,57L 572,573,575 defined, 568 fintorder, 608 610 frequency domainrimedomaiDrelarionship in. 576 5'77 fiequency increasing from zero, 575 op amp, 608-610 passivefilter circuirs,568,570 577 qualitativeanalysisof seriesRZ circuit, 570-571 quantitative analysis of seiies Ra circuit, 572-574 seriesRC circuits as, 574 576 seriesRl, circuits as, 5'7U5'71,572-574 transfer functions, 576, 628-629 zercftequency,575 Lumped-parametersystem,5

M Magtretically coupled coils, 787-795 equivalent circuits for, 787 791 ideal translormersin, 792 795 z-equivalentcircuit, 788 790 T-equivalent circuit, 787 788 Magnitudeplot, 568 Magnitudescaling,612 Matiices, 764-780 '7'70-:7'71 adjojjlt, '165J'10 algebra, applicationsof, 77G780 column, 765 conformable,766 defined, 7& determinant of, 765 identity, 770 '7'71-:772 lr'\efie, partitioned, '772J'75 square,765 transposeof, 769 Maximum power tmnsfe\ 126 129, 410417 absorbed,412 circuit analysis and, 12G129 conditionfor, 411 restiicted, 412-413 power calculations,410+17 sinusoidalsteady-state Measurement,8-9, 35,68-70,71 73,168 169 ammeterq 6B,69-70 commonmode rcjectionmtio (CMRR), 168-169 delinedquantitieESI unitsr 8 derived SI urits, 9 difference-amplfierperformance, 168-169 prefixes, SI units, 9 resistance,?1-73 SI units of, 8-9 terminal,modelsbasedon, 35 voltageand current, 65-67 voltmeter, 613 Memory, concept of in transfer function, 536-537 Mesh,defined, 94 Mesh'curent method, 105-10?,107-109,109-112, 112 116,360-36L amplifiercircuit,analysisof, 110 111 circuit analysisusing, 105-1W,107109,109 112, 112-116 defined, 105

dependent sourceq useofwitl, 107-109 frequency'domainc cuits, 360-361 introductionto, 105-107 node-voltage method,comparison o! 112-116 sinusoidal steady-state analysis using,360 361 specialcases of, 109 112 supermesh, 110 Models,10,3235,170-173 analysisof circuitsusingrealisticop amp, 170-173 circuir,10,3235 conceptual, 10 constructionof 10,32-35 electdcalbehavior, 35 flashlight,33-34 mathematical, 10 operational amplilier(opamp),170173 terminalmeasuements, basedon, 35 Multiple meshcircuit,stepresponseof usingLaplace tmnsform,517 519 Multiplicationby a consta\t, 475-476,'711 Mutualinductance, 188,203-2U,209-214,521-5 coefficient of coupling,211-212 conceptot 209 211 defined,188 dot convention, 20+207 energycalculations, 212 214 Laplacetransfom analysisof a circuit with, 521-523 self inductance,relationshipto, 211-212

N Narrowband filrers, 636-642 bandreject,639 641 bandpassfilters, 63fu38 twin-T no1ch,639 quality factor and, 63fu38 Natuml response, 228,230-236,236-240,248-254, 286-291,291-301,308-311,512-513 cdtically damped voltage, 289,299-300 current in RIC seriescircuits, 309 defined, 228 forms of for voltage of RLC cilqtit, 297 :01. general solution for, 248 254,2a6 290 Laplacetransform,using, 512-513 methodof calculating,234,238,250 overdamped voltage, 289,291-294 parallelRaC circuit, 286)9L,291-30I

848

Index

(RC) circuits,23G240,248-254, resistor-capacitor 512 513 iesistor-inductor (Ra) circuits, 230-236, 248-254 (RaO circuits,286-291, resistor-inductor-capacitor 291-301 seriesRIC circuits,308-311 time constant,232134,237 underdampedvoltage, 289,295298 Negativefeedback,defined, 158 Negative(acb)phasesequence, voltages,434 Neperfrequency,289,308 Neutral terminal, 435 Node,defined, 36,94 Node-voltagenetllod, 97-100,100-101,101-104, 112-116,359160 amplifier circuit, analysis of, 103 104 circuitanalysisusing, 97 100,100 101,101 104, 112-116 defined, 98 dependentsources, useof with, 100-101 fiequency-domair circuits, 359-360 introduction to, 97-100 mesh-cufientmethod,comparisonol 112-116 sinusoidalsteady-state analysisusing, 359 360 specialcasesot 101-104 supernode,102-103 Noninverting-amplifier cilcnjl, L64 165.17L 172 Norton equivalentcircult, 119 120,121,123,355 ciicuit analysis of, 119 121 defined, 121 frequency-domain version, 355 sinusoidalsteady-state analysisof, 355 sourcetransformationsusedto derive, 121,355

0 Ohm'slaw,29,43-45,511 amplfier circuit,applyingin, 45 circuit analysiqappliedto, 43-45,511 electrical resistance and,29 r-domain, usingin, 511 unknoM voltage,applyingto find, 4,1

One-sided Laplacetransform,468 Op anp,seeOperational amptifier(op amp) Opencircuit,33-34 608-611, Operational amplifier(op amp),154-185, 613414,61502,622436 bandpass filters,615618 bandreject filters,618522 ButteNorttr filte6, 627-636 cascadingidenticalfilte$, 622-627 curents,156-161 ditrerence-amplifier c cuit, 165-169 (DIP), 1s6 dualin-linepackage filters,608-611, 613614,615422,622436 higher-orderfilters, 622-636 high'pass filters,610 611 inveiting-amplifiercircuit, 161 10,170 171 low-pass filterE 608 610 noninve ing-amplfier circnit, 764 L65,7771'72 realisticmodelfor, 17G-173 scaling, useof ir desigrof filte$, 613-614 straingages,155,1'7 3-l'74 summing-amplifier circuit,163-164 terminals,156 161 voltages,15G161 Operational transforms, 468,475481,71|.1714 addition,476,711 amplitudemodulation,712 convolution,712-713 defined, 468 differcntiation, 176-417,111 Fouder,710 714 frequency domajn,in the, 479,713 integ ation, 47'7 478, 771:712 Laplace,468,475-481 multiplication by a comta , 4'754'76,711 scalechanging,4'79480, 712 subtraction,476,711 timedomain,in the, 41*479,712113 translation,478479 typssof, 480,713 289,291-294,309 Overdamped response, voltage capacilor stepresponse,309 currentnaturalresponse, 309 voltage,289,291-294

Index 849

P Parallel-connectedcircuit elements, 59-60 Parallel RaC circuits, 2a4,2a6)91, 291-30L,301-3A characteristic equation, 287-288 naturalresponse!2U,286:291,291107 neper frequency, 289 resonanlmdian frequency,289 stepresponse,284,301 307 Parseval'stheorem, 717-724 applicationsofto filten, 718 719 demonstration of, 718 energy contained in rectangular voltage pulse, calculation of using, 722-723 inter?retation,718-719 Panial fraction expansion, 483494,528 531 circuit analysisusing H(s), 530-531 distinct complex roots of D(s), 48G488 distinct real rcots of r(r), 484-485 improper rational functions, 493ts494 prcper rational functiotrs, zE3-493 repeatedcomplexroots of , (s),491-493 repearedrcal roorsofD(r), a88-a90 time invadant, 530 transfer function in, 528-531 Passband ftequencieqdefined, 568 PassivecircuitelementE25,1a8,342-346 frequency domain, in the, 342-346 ideal basic circuits and, 188 impedance, 345 reactance,345 signal ir phase, 343 sirusoidal analysisand, 342-346 source,25 V-I rclationships, 342 345 Passivefilter circuits, seeFrequency-selectivecircuits Passives8n convention, 13 Path,defined,94 Period of sinusoidal function, 332 Periodicfunctions,656,658-662,675-678,678 679 average-power calculationswith, 675 678 defined, 656 Fourier coefficients in, 656,658 662 Fourierseriesand, 656,6756'7a,6'7a-6'79 rms value ot, 678-679 Phaseangle,defined, 332

Phaseangleplot, 568,805-807,813-815 Bode plots, 805-807,813-815 complexpolesand, 813-815 passivefrequency-selectivecircuits and, 568 staight-line, 805 807 Phas€current, defhed, 440 Phasespectra plots, 682-6821 Phasespectrum,defined, 682 Phasevoltage,defined, 440 Phasordiagramq 372-374 Phasors,337-342,342-345,403-404 defined, 337 frequencydomain, 338 inverse transfom, 338 340 power calculations, 403-404 representation,338 transfom, 338 V-I relationships, 342-345, 403404 Physical prototype, use oI in circuit analysis, 10 7-equivalent circuit, 788-790 Pi (,) interconnect,73 Pi-to-tee (z-to-T) equivalent circuits, 73 76 Planarcircuits,94 Polarity of curent and voltage ratios, 369-370 Poladtyrefeiences,12 13,15 Poles,494195, 528,799-800,807-809 Bode plots and, 799-800,807-809 complex, 807-809 F(s),locationof in Laplacetransform, 494 495 H(s), location of in transfer function, 528 real,first-order, 799€00 Ports,defined, 730 Positive (abc) phasesequence,voltages! 43.{ Powei, 14-16, 126-129,191-192, 196-197,3E2-i98, 401402,41041'7. Seeals, Power calculations algebraic sign of, 15 apparent,401 402 applianceratings,397 398 average,394 398 capacitive circuits, 395 396 capacitorsand, 196 197 circuit analysis, 126-129 circuitsand, 14-16 complex, 401-402 defined, 14 factor, 396

850

lndex

inductive circuits, 395 inductorsand, 191-192 instantaneous, 392 393 instantaneous rcal, 394 398 maximum power tiansfer, 126-129,410-411 polarity reference,15 reactive, 39,[-398 resistive circuits. 394-395 sinusoidalsteadystateanalysis,392 398,401-402, 410417 Power calculations, 39M31, 445451,675-67A averagepower, 394 398,406,445446,6'75-678 complex power, 401402,404405,446-447 delta loads(A),balanc€d,447-448 heating appliances, 391, 41'7418 instantaneouspower, 392-393, 448449 instantaneousreal power, 39zl-398 ma-{imumpower transfer, 410 417 periodic functions, average-powercalculations with, 675-6'78 reactivepower, 394 398,406 rootmean-square (rms) value, 399-400 sinusoidalsteady-state,390-431 thre€'phase circuits, 445-451 V-I relationships and, 403-404 wye (Y) loads,balanced,M5446,446447 Powerfactor, 396 Pioper rational functions, 483 484 Proximity switches, 181,214 216 Pspicesensitivityanalysis,135-136 Pushbutton telephone circuitsr 567,598

a 636638 Qualiryfacror,582-583,586, bandpass filters,5B2-583, 586 broadband bandpass filterE 636 nairowband bandpass filters,631k38

R Rational functions, 482 494. Seedlso Partial fraction er?ansion defined, 482 483 improper, 483,493494

partial ftaction expansion of 483-494 proper, 483 493 RC circuits, se?Resistor-capacitor (RC) chcuits Reactance,passivecircuit element of, 345 Reactivefactor, 396 Reactivepower, 394 398,406 Rear window defroster,57,7G79 Reciprccal two-port ciicuits, 739-741 Reflectedimpedance,363-364 Resistance,22,28 32,71-72.Seedb, Ohm'slaw basic circuit elemenq as a, 22 conductance,29,31 defined, 28 measuring,71-72 Ohm's law, 29 resistors.28,30 32 symbol for, 28 Wheatstonebddge circuit, 71-72 Resistive circuits, 39,1-395 Resistor-capacitor (RC) circ[its, 228,23G40, 246148, 248-254,257158,258-259, 512,513, 574 576, 577 5a2 cutoff frequency, 575 defined, 228 generalsolutionfor responses of, 248-254 high-passfilters,analysisof as, 577-578,578-582 Laplace transfolm, using for analysisof, 512-513 lowpass filteis, arlalysis of as, 574-5'76 natuial responseof, 236-240, 512-513 qualitative analysis of series, 577-578 quantitative analysisof serieE 578 582 sequentialswitching,257 258 se es circuitsasfilters, 514 576,577 578,5'7q5az stepresponse,24G248 time constant,237 unboundedresponse,258-259 voltage, deriving exprcssion for, 237 238 (RI) circvits, 228,230236,A1 246, Resistor-inductor 248-254,255-256, 570 5'71., 572 5'74 curent, derivingexpressionfor, 230 232 cutotr frequency, 573 defined, 228 generalsolutionfor responses of, 28-254 low-pass filteis, analysis oI ^s, 5'70-5'71,5'72-5'74 naturalresponseof, 230-236 qualitativeanalysisof series,570 571 quantitative analysis of series, 572 574 sequentialswitching,255 256 seriescircuitsasfilters, 510 571,572574

Index 851 steady-state response, 233 stepresponse, 21-246 time constant,232-234 hansientrcsponse,233 (RLC) circ\\its,284 329, Resistor-inductor-capacitor 514-515, 516-517, 583,584 587,588-589, 593,594-597 bandpassfilterq analysisof aq 583,584-587,588-589 bandrojectfilten, analysisof as, 593,594-597 criticallydampedresponse,289,299-300 cutoff frequency,585,588 ignition circuit, 285,317-j2O Laplacetransform, analysis of using,514 515,516517 naturaliespotrse of, 286191,291-301, 308 311 overdampedresponse,289,291294 palallel,284,286-291,291 301,301307,514-515, s16-517,588-589 qualitative analysis of sedes,583,593 quantitativeanalysisof series,584-587,594-597 series,284,308311,583,584-587, 593,594-597 stepresponse of, 301-307, 308-311, 514-515 hansientresponse of 516-517 underdampedresponse,289,295298 Resistorq28,30-32,58-62,73 corductance of, 31 curentin,30 defined,28 delta(A) interconnect, 73 pamllel,in,59.60 pi (z) interconnect,73 powerin, 30-32 series,in,58-59 series-parallel combined,6{162 tee(T) interconnect, 73 voltagein, 30 31 llTe (Y) interconnect,73 Resonantfrequency, bandpassfilte$, 582 Resonantradianilequency,289,308 Response, 22&283,284-329,335 337,537-540,566, 670.6'74 first-ordercircuits, 228-283 flashingtight circuit, 229,263264 Fourierseriesapplicationto, 6'70-674 frequency,566 genemlsolutionfor, 248-254 ignition circuit, 285,317-320 integratingamplifie$, 260-263,312216 natJJJ al, 228,230-236,23GAO, 248254,2a6-291, 291-301,308-311

(RC) circults,228,236240,246-248, resistor-capacitor a8:254 resistor-inductor(Rf ) ci,tcuits,228,230236,247-246, 248-254 resistor-inductor-capacitor (Rl,C) circuits, 284-329 second-ordercircuits, 282[-329 sequentialswitching,25zl-258 sinusoidal,335-337,537-540 solutionsfor, 248-254 steady-state, 233,335:337, 670475 step,228,240-248,U8-254,301 307,308-311 transferfunction and steady-slate sinusoidal,537-540 transient, 233 unbounded, 258 259 R-Lcircuits,se€Resistor-inductor(Rl) circuirs (RaC) RICcircuits,see Resistor-inductor-capacitor c1Ictuts (Ims) value Ims value,re€Root-mean-squa.re (rms)value,333,335,399 Rootrnean-square 400, 678-6'79 effectivevalue, 399 periodictunctions,678 679 powercalculationsand, 399-400 sinusoidal sources, 333,335

5 Scalechanging,479480,'712 Scaling,612614 componentscalefactors, 613 designhgop ampfi1te6,useo! 61!614 ftequency,612 magnitude,612 r-domain, 467,506,508 510,511-512,520-521,523-526 capacitor in, 509 510 circuitanalysis in, 511-512 circuit elementsin, 508-510 equivalent circuits,510 inductorin, 508-509 introduction to, 506 Kirchoff'slawsard, 511 Laplacetansform function of, 467 Ohm'slawh, 511 resistorio,508 superposition, useof itr, 523-526 Thdvenioequivalent,useof in, 520-521

852

Index

Second-order circuils, 28,[-329 defined, 286 general solution for responsesof, 286-290 ignition circuit, 245,31'7-320 Integmting amplifierq 260:263,312116 naturalresponseof, 286.291,291301,308311 parallel RLC, 284,28G29L,291 j01,301 j07 rcsistor-inductor-capacitor (R-LC) circuits, 284-329 seriesR-LC,284,308 311 stepresponseol 301 307,308 311 voltage rcspoNes of parallel RaC circuits, 291-301 Self-indnctance, 2U -208, 211-212 Sensitivityanalysis,93,133-136 Sequential switching, 254-258 Series-connectedcircuir elements, 58 59 Series-parallel combinations,6M2,200 203,348-352 admittance, 350 capacitoin 200 203 circuit elements, 60 62 impedances, 348-352 inductorq 200-203 resistorq 60-62 susceptance, 350 SeriesRLC circuitE 284,308 311 capacitorvoltageir, 309 characteristic equation, 308 natural rcsponse, 308-311 nepei hequency, 308 resonant radiar frequency, 308 stepresponse,308-311 Shortcircuit, 33-34 STunirs.rp?[nrernarionals]srem o{ units(Sl): Measurement Sifting propert$ impulse function, 472 Sig0al-pfocessiDg s)slems.eleclricalengineering.4-5 Signum ftrnction, Fourier tansform of a, 70G707 Simplifyingtechniques,ieeTransfomations Simultaneous equationE 95-96,96 97,759:780 circuit analysisand, 95-96,9G97 linear, solution of, 759-780 numberof,95-96 systematicapproach using, 96 97 Sinusoidalresponse,335-337,537-540 steady-state compotrent,33G337 steady-state sinusoidalanalysisand, 335-337 transfer function and, 537-540 traNient component,335

Sinusoidalsources,332-335 cu enr, 332-334 period, 332 phaseangle, 332 voltage, 332 334 Shusoidalsteady-state analysis,330-389,390-431 cutent-to-voltage (V-I) relationships, 342-345 delra{o-\1]'e (A{o-Y) tansformationq 352-355 frequency donain, 338,342-346,346-348,348-352, 352-355,355-360,360-361 householddistributioncircuit, 331,375 ideal transfomer, 365 371 impedance,345,348-352,352-355 Kirchoff's laws, 346 348 mesh-currentmethod, 36G-361 node-voltage metlod, 359-360 Norlon equivalent c cuit, 355 passivecircuit elements, 342-346 phasor diagmms for, 372-374 phasors,337-342 powercalculations,390 431 response,335 337 source transformations, 355-358 sources! 332-335 Th6veninequivalentcicuit, 355,357-358 transformen, 361:365, 365:3'71 Source transformations, 116-119,121-122,355158 circuit analysiq 116 119 defined, 116 frequency-domain circuits, 355 358 No on equivalentc cuit,usingto derivea, 121,355 sinusoidal alalysis and, 355-358 steady-srate analysisand, 355-358 Th6veninequivalentcircuit,usingto derivea, 121-122, 355,357358 Sources,24-28,56,332-335,435-436 activeelement, 25 constant,56 controlled, 25 dependent, 24 direct cuffert (dc), s6 ideal cufient, 24 idealvoltage, 24 independent,24 neutralterminal, 435 passiveelement, 25 sirusoidal, 332-335

symbolsfor, 24 25 testing interconnections of, 26 27 three-phasevoltage, 435-436 Steady-state response,233,335-33'7,6'7c-675 component, 336-337 defined. 233 direcr approach to, 673-674 Fourier series,finding with, 670-675 sinusoidalanalysisand, 335 337 Step function, Laplace transform, 468-470 Stepresponse,22a,24024a,248-254,301-307,308-311, 514-515,517519 capacitor voltage in seriesRIC circuits, 309 defhed, 228 direct approach,303-304 geneml solution for, 248 254 indirect approach, 302-303 Laplacetransform,analysisof usirg, 514-515,517-519 method of calculating, 250 multiplemeshcircuit, 517-519 parallel-Rlccircuits,301-307 resistor-capacitor (RC) cncrits, 24GA8,248 254 (RL) circu\ts,241 246,248 254 resistor-inductor (RLC) circuits,301-307, resistor-inductor-capacitor 308-311,514-515 seriesRIC circuits,308-311 Stopbardftequencies, defined, 568 Straingages,155,1731'74 Summing-ampliJiei circuit, 163-164 Supermesh, conceptoi 110 Supemode,conceptoi 102-103 Supeiposilion,129-132,523-526 circuit amlysisand, 129-132 defined, 129 l-aplace transform analysisusing, 523-526 t-domair, useof in, 523-526 Surge suppressom,507,548 Susceptance,impedance and, 350 Switches, 187,188-189,214-216 Switchingoperations,540-542,542-544 capacitor circuit, 540-542 impulsefunctionsand, 540 5A,542 544 se es inductor circuit, 542 544 Symmetry,662 668 even'function, 662-663 Foudei coefficients,effects of on, 662-568

half,wave, 66M66 odd function, 663-6& quarter-wave, 66ffi67

T T-equivalent circuit, 787-788 Tee (T) inlerconnect,73 Teminals, 156-161 current for op amps, 15c161 operational amplifiers (op amps), 156 voltag€for op amps, 15G161 Terminated two-port cir lls, 741 :747 Th6venin equivalent cilcujts, 119 121,722 123,123-126. 355,357:58,5m 521 amplifiei circuil, using in the, 125-726 circuit analysisof 119-121 defined, 120 finding, 120-121 frequency-domain venion, 355,357 358 independentsoutces,derivingwith, 123 125 Laplacetmnsform,analysisof using, 520 521 .r-domain,useofir, 520 521 shusoidalsteady-state analysisol 355,357-358 sourcetransformationsusedto derive, 121-122, 357-358 Three-phasecircuits,seeBalancedthree-phasecircuits Timeconstant,232234,5'76 Time domain, 467,4764'79,5765'77,591592,'112:713 convolutionin, 712 713 differenriarion in, 476:177 Fourier transform and, 712-713 ftequencydomain,relationshipto, 576 577,591 592 high-passfiltem, relationship to frequency domain in, 591-592 integrarion in, 477-478 Laplacetransfomand,467,4764'79 low-passfiltels, relationship to frequency domain in, 57G577 opemtional Laplace transfoims, 476-479 translationin, 47&419.'712 Time invariant. defined, 53 Transducers. 155

854

528 531,531-531 Tlansferfunction,52G528, ,537-540, 576,581,591,596,628 629 Butterworthfilters, 628-629 circuit analysisusingH(r), 530 531 convolutionintegraland, 531-537 defined, 526 derivationof in a circuit, 527 circuitE576,581,591,596,628529 frequency-selective Laplacetransformcfucuitanalysisand, 526-540 memoryand, 536-537 pa ial ftaction er?ansion,useol in, 528-531 polesand zercsof fl(r), 528 response and,537 540 steady-state sinusoidal time invariant, 530 weightingfiinction, 563-537 Tfansformations,73 :76, 11G119,121-122,352355, 355-358 delta-to-wfe(A-to-Y) equivalentcircuitq 73-76 delta-to-wye(A-to-Y) impedance,352-355 355-358 frequency domaiq 352-355, pi-to{ee (z-to-T) equivalentcircuits, 73 74 358 sinusoidal analysis and, 352-355,355 121-122, 355-358 source,116-119, Tiansformers,361 365,365371,'792J95 defined, 362 equivalentcircuits,in, 792 795 95 ideal, 361,365-371,792-:7 linear,361-365 primarywinding, 362 reflectedimpedance,36!364 secondarywiadirg, 362 30l-3b5.365 17l siDusoidal steady-slale anal)si.. 336 Tlansientcomponent,sinusoidalresponse, 233,516517 Tlansientresponse, Translation, 47&79, 712 Fouriertransform,using, 712 ilequencydomain,479,712 Laplacetiansfoms,using, 478-479 timedomain,478479,112 Tiigonometdcform of Fourierserie$ 668-670 Tiigonometricidentities,table of, 817 Twin-Tnotchfilter, 639 'l\ o-poflcircuirs.730-757 analysisof, 741-?47 conversionof parametem,736 hybdd paiameters,734 immittance,734 interconnected,747-751

paftmetels oI, '732:741 '739:741 rcciprccal relarionshipsamong. 735-7Jg symmetric reciprocal, 740 terminalequatioN foi, 731 732 tefinlnaled, 741J41 transmission pammeters, 734 z parameters,732133,'742J 46 Two-wattmeter method, three-phasecircuitS 452-454

U response, 258-259 Unbounded 3B respoDse, 249,295.:298, Underdamped capacitorvoltagestepresponse,309 cwent naturalresponse,309 dampedradianfrequenc$ 295-296 damphg coefficient,296 dampingfactor, 296 voltage,289,295298 UnilateralI-aplacetransfom, 468 Unit impulsefunction, 471 Unit stepfunction, 469,707

V Var (VAR),unitof reactivepower,395 Vaiable-parameterfurction, 471 V-I relationships, r€sCurrentandvoltage(V-I) relalionships Virtual short condition, 158 volt-amps(vA), unit of complexpower, 401 Voltageratios, 368-369,369j70 ol 368-369 determination dot convention,369 idealtransfomenand, 368i69,369:3'71 poladty of, 369-370 39,56,65-67,6u10, voltage,11 12,72-13,24-2a,37, 196,23148, 291-iO1, 15G161, 16G167,188-1a9, 332-334,342345,346-34'7 , 434435, 435436 a-phase,434 balancedthree-phase,434435,435436 b-phase,434 capacitorsand, 196 commonmodeinput, 16G167

Index 855 constant,56 c-phase, 434 critically damped response, 289,299 300 defined, 11 differentialmode input, 166 167 direct (dc), 56 division, 65-67 elect cal chargeand, 11-12 gain, 157 inductorsand, 188-189 input constaint, 158 Kirchoff'slaw, 37,39,346 347 measudng,68 70 natuml response, 237 238,291-301 negative (acb) phase sequence, 434 negative feedback, 158 opemtionalampMier(op anp) terminals,156-161 overdamped response, 289,291194 poladty reference, 12-13 positive (abc) phase sequenca, 434 RC circuits, deriving expressionfor natural responseol

231-238 RaC circuit,naturaliesponseformsof for parallel, 291 iO1 second-oidercircuit iesponse,291-301 smusoroal source,JJZ-JJ4 sinusoidal steady-state analysis and, 332 334,342345, 346-34 sources, 24-28,56,332 334,435436 underdampedresponse,289,295-29a V-I relationships,342-345 virtualsho condition,158

Voltage-divider circuit, 62-64 Voltmeter, 68

W Watt (W),unit of averagepower, 395 Weighting function, concept of, 563 537 Wleatstone bddge circuit, 71 72 Wye (Y) connected loads, 445446,44644'7 Wye (Y) htercormect, 73 Wye-delta (Y-A) circuir, analysisoi 442-445 Wye-wye (Y-Y) circuit, 436-442 analysis o{, 436-442 conditions for balanced three'phase circuit, 437 single-phase equivalentcircuit, 438

z '732 :746 z parameters, :733,742 determinationog 732 733 terminatedtwo-port circuit rsir.g, 742J46 two-po circuits, 732-:733,742-:7 46 Zero fiequency,575 Zeros, 494-495,528,799-800,807-809 Bodeplotsand,799+00,807+09 complex,807+09 F(s),location of in Laplacetransfom,49rH95 li(r), locationof in tratrsferfunction, 528 real,fi^t-order, 799 800