Linear Control Theory: The State Space Approach

r i'

Linear Contro l Theory

Linear Control Theory The State Space Approach

Frederick Walker Fairman Queen's University, Kingston, Ontario, Canada

John Wiley & Sons Chichester • New York· Weinheim ·Brisbane· Singapore· Toronto

Copyright

1998 John Wiley & Sons Ltd, Baffins Lane, Chichester. West Sussex P019 IUD, England 01243 779777 National International ( -r 44) 1243 779777

e-mail (for orders and customer service enquiries): cs-books(c!;wiley.co.uk Visit our Home Page on http:/ jwww.wiley.co.uk or http://www. wiley .com All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means electronic, mechanical, photocopying , recording, scanning or otherwise, except under the terms of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, 90 Tottenham Court Road, London WIP 9HE, UK without the permission in writing of the Publisher. Other Wiley Editorial Offices

John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, USA Wiley-VCH Verlag GmbH, Pappelallee 3, D-69469 Weinheim, Germany Jacaranda Wiley Ltd, 33 Park Road, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 2 Clementi Loop #02-01, Jin Xing Distripark, Singapore 129809 John Wiley & Sons (Canada) Ltd, 22 Worcester Road, Rexdale, Ontario M9W ILl, Canada

Library of Congress Cataloguing-in-Publication Data

Fairman, Frederick Walker. Linear control theory : The state space approach ; Frederick Walker Fairman. p. em. Includes bibliographic al references and index. ISBN 0-471-97489-7 (cased: alk. paper) I. Linear systems. 2. Control theory. I. Title. QA402.3.F3 1998 97-41830 629.8'312-dc21 CIP British Library Cataloguing in Publication Data

A catalogue record for this book is available from the British Library ISBN 0 471 97489 7 Typeset in part from the author's disks in 10jl2pt Times by the Alden Group, Oxford. Printed and bound from Postscript files in Great Britain by Bookcraft (Bath) Ltd. at least This book is printed on acid-free paper responsibly manufacture d from sustainable forestry, in which two trees are planted for each one used for paper production.

To Nan cy for her unt irin g sup por t

crt,,

Contents xiii

Preface

1

Introduction to State Space 1.1 1.2

1.3 1.4

1.5

1.6

1.7

1.8

1.9

1.10 1.11

2

Introdu ction Review of Second Order System s Pattern s of behavi or 1.2.1 1.2.2 The phase plane Introdu ction to State Space Modeli ng Solving the State Differe ntial Equatio n The matrix expone ntial 1.4.1 1.4.2 Calcula ting the matrix expone ntial 1.4.3 Proper and strictly proper rationa l functio ns Coordi nate Transfo rmation Effect on the state model 1.5.1 1.5.2 Determ ination of eAt Diagon alizing Coordi nate Transf ormatio n Right-e igenve ctors 1.6.1 1.6.2 Eigenv alue-ei genvec tor proble m 1.6.3 Left-eig envect ors 1.6.4 Eigenv alue invaria nce State Traject ories Revisit ed Straigh t line state trajecto ries: diagon al A 1.7.1 1.7.2 Straigh t line state trajecto ries: real eigenv alues 1.7.3 Straigh t line trajecto ries: comple x eigenv alues 1.7.4 Null output zero-in put respon se State Space Models for the Comple te Respon se Second order proces s revisite d 1.8.1 1.8.2 Some essent ial feature s of state models 1.8.3 Zero-s tate respon se Diagon al form State Model Structu re 1.9.1 1.9.2 Proper ties 1.9.3 Obtain ing the diagon al form state model Compu ter Calcula tion of the State and Output Notes and Refere nces

State Feedback and Controllability 2.1 2.2 2.3

Introdu ction State Feedba ck Eigenv alue Assign ment Eigenv alue assign ment via the contro ller form 2.3.1

1

2 5 7 9 9 10 12 12 13 14 15 16 17 19 20 21

22 23

24 25 26 26 28 29 32 32 33 35 37 39

41 41 42

44 45

viii

2.4

2.5

2.6 2.7

3

2.3.2 Realizing the controller form 2.3.3 Controller form state transformation 2.3.4 Condition for controller form equivalence 2.3.5 Ackermann"s formula Controllability 2.4.1 Controllable subspace 2.4.2 Input synthesis for state annihilation Controllable Decomposed Form Input control of the controllable subspace 2.5.1 Relation to the transfer function 2.5.2 Eigenvalues and eigenvectors of A 2.5.3 Transformation to Controllable Decomposed Form Notes and References

State Estimation and Observability

47 49 50 52 55 56 59 60 61 62 63 64 66

67

Introduction Filtering for Stable Systems Observers Observer Design Observer form 3.4.1 Transformation to observer form 3.4.2 Ackermann's formula 3.4.3 Observability A state determination problem 3.5.1 3.5.2 Effect of observability on the output Observable Decomposed Form 3.6.1 Output dependency on observable subspace 3.6.2 Observability matrix 3.6.3 Transfer function 3.6.4 Transformation to observable decomposed form Minimal Order Observer 3.7.1 The approach 3.7.2 Determination of xR(t) 3.7.3 A fictitious output 3.7.4 Determination of the fictitious output 3.7.5 Assignment of observer eigenvalues Notes and References

67 68 69 71 72 73 74 75 75

Model Approximation via Balanced Realization

91

4.1 4.2 4.3 4.4

91 91 94 96 96 98 99 101 104 107 108 109 111 111 112 114

3.1 3.2 3.3 3.4

3.5

3.6

3.7

3.8

4

~

Contents

4.5 4.6 4.7

4.8

4.9

Introduction Controllable-Observable Decomposition Introduction to the Observability Gramian Fundamental Properties of W0 4.4.1 Hermitian matrices 4.4.2 Positive definite and non-negative matrices 4.4.3 Relating E0 to -\[W0 ] Introduction to the Controllability Gram ian Balanced Realization The Lyapunov Equation Relation to the Gramians 4.7.1 Observability, stability, and the observability Gramian 4.7.2 Controllability Gramian Revisited The least energy input problem 4.8.1 Hankel operator 4.8.2 Notes and References

77 78 79 80 80 81 82 82 83 84 85 86 90

Content s

"-

115

Quadratic Control

5

5_1 5.2 5.3

5.4

5.5

5.6 5.7 5.8

6

6.3

6.4 6.5

6.6 6.7

7

Introd ucti on Observe r Based Controlle rs Quadrat ic State Feedbac k Control Motivatin g the problem 5.3.1 5.3.2 Formula ting the problem 5.3.3 Develop ing a solution Solving the OCARE Stabilizin g solutions 5.4.1 5.4.2 The Hamilton ian matrix for the OCARE 5.4.3 Finding the stabilizin g solution Quadrat ic State Estimatio n Problem formulat ion 5.5.1 5.5.2 Problem solution Solving the QFARE Summar y Notes and Referenc es

Introduc tion LOG State Feedbac k Control Problem Problem formulat ion 6.2.1 6.2.2 Develop ment of a solution LOG State Estimatio n Problem Problem formulat ion 6.3.1 6.3.2 Problem solution LOG Measure d Output Feedbac k Problem Stabilizin g Solution The Hamilton ian matrix for the GCARE 6.5.1 6.5.2 Prohibiti on of imagina ry eigenval ues 6.5.3 lnvertab ility of T11 and T21 6.5.4 Conditio ns for solving the GFARE Summar y Notes and Referenc es

7.3

Introduc tion Time Domain Spaces 7.2.1 Hilbert spaces for signals 7.2.2 The L 2 norm of the weightin g matrix 7.2.3 Anticaus al and antistabl e systems Frequen cy Domain Hilbert Spaces 7.3.1 The Fourier transform 7.3.2 Converg ence of the Fourier integral 7.3.3 The Laplace transform 7.3.4 The Hardy spaces: 1-£ 2 and 7.3.5 Decomp osing £ 2 space 7.3.6 The H2 system norm The H:x. Norm: SISO Systems 7.4.1 Transfer function characte rization of the Hx norm 7.4.2 Transfer function spaces 7.4.3 The small gain theorem The Hoo Norm: MIMO Systems 7.5.1 Singular value decompo sition

Hi

7.4

7.5

147 149 149 150 153 154 155 157 158 158 159 162 165 166 166

167

Signal and System Spaces 7.1 7.2

115 116 119 120 121 122 127 127 130 133 137 137 140 143 145 145

147

LQG Control 6.1 6.2

ix

167 167 168 170 172 173 173 175 176 177 178 179 181 181 183 184 185 185

Contents

X

7.6 7.7

8

System Algebra 8.1

8.2

8.3

8.4

8.5

8.6

8.7 8.8

9

Introduction Parallel connection 8.1.1 8.1.2 Series connection System Inversion Inverse system state model 8.2.1 SISO system zeros 8.2.2 MIMO system zeros 8.2.3 Zeros of invertible systems 8.2.4 Coprime Factorizatio n Why coprime? 8.3.1 Coprime factorizatio n of MIMO systems 8.3.2 Relating coprime factorizatio ns 8.3.3 State Models for Coprime Factorizatio n Right and left coprime factors 8.4.1 Solutions to the Bezout identities 8.4.2 Doubly-cop rime factorizatio n 8.4.3 Stabilizing Controllers Relating W(s) to G(s), H(s) 8.5.1 A criterion for stabilizing controllers 8.5.2 Youla parametriza tion of stabilizing controllers 8.5.3 Lossless Systems and Related Ideas All pass filters 8.6.1 Inner transfer functions and adjoint systems 8.6.2 Summary Notes and References

H00 State Feedbac k and Estimation 9.1 9.2

9.3

9.4

9.5 9.6 9.7

10

7.5.2 Induced 2-norm for constant matrices 7.5.3 The L.x. Hx norm for transfer function matrices Summary Notes and References

Introduction Hoo State Feedback Control Problem 9.2.1 Introduction of P_ 9.2.2 Introduction of G~(s) 9.2.3 Introduction of J-inner coprime factorizatio n 9.2.4 Consequen ces of J-inner coprime factorizatio n Hoc State Feedback Controller Design equations forK 9.3.1 On the stability of A+ B 2 K 2 9.3.2 Determinat ion of 6. 9.3.3 Hx State Estimation Problem Determinat ion of Te(s) 9.4.1 Duality 9.4.2 Design equations for L 2 9.4.3 Sufficient Conditions Summary Notes and References

H00 Output Feedbac k Control 10.1 10.2

Introduction Developme nt

186 189 190 191

193 193 193 195 196 197 198 199 200 201 202 204 205 206 207 209 212 213 214 215 217 219 220 221 223 223

225 225 227 229 229 230 231 234 234 236 239 242 242 243 244 245 246 246

247 247 248

Cont ents

10.3

10.4

10.5 10.6

10.2.1 Refor mulat ion of P" 10.2.2 An Hx state estim~tor ack 10.2.3 Introd ucing estim ated state feedb Hx Outpu t Feed back Contr ollers 10.3.1 Centr al contr oller 10.3.2 Contr oller param etriza tion 10.3.3 Relat ion to Youla param etriza tion Hx Sepa ration Princ iple 10.4.1 A relati on betwe en Hami ltonia ns 10.4.2 Relat ing stabil izing soluti ons 10.4.3 Deter minat ion of L0 Summ ary Notes and Refer ences

Linear Alge bra

A

A.1 A.2 A.3 A.4 A.5

Multi ple Eigen value s and Contr ollabi lity Block Uppe r Trian gular Matri ces Singu lar Value Deco mpos ition (SVD) Differ ent Form s for the SVD Matri x Inver sion Lemm a (MIL)

B

Reduced Ord er Model Stability

c

Problems C.1 C.2 C.3 C.4 C.5

D

Probl ems Relat ing to Chap ter 1 Probl ems Relat ing to Chap ter 2 Probl ems Relat ing to Chap ter 3 Probl ems Relat ing to Chap ter 4 Probl ems Relat ing to Chap ter 5

MATLAB Experiments 0.1

0.2

D.3

0.4

State Mode ls and State Resp onse 0.1.1 Contr oller form 0.1.2 Seco nd order linear beha vior 0.1.3 Seco nd order nonlin ear beha vior 0.1.4 Diago nal form Feed back and Contr ollabi lity 0.2.1 Contr ollabl e state mode ls 0.2.2 Unco ntroll able state mode ls Obse rver Base d Contr ol Syste ms 0.3.1 Obse rver based contr ollers vior 0.3.2 Obse rver based contr ol syste m beha State Mode l Redu ction ble syste ms 0.4.1 Deco mpos ition of uncon trolla ble and/o r unob serva ty 0.4.2 Weak contr ollabi lity and/o r obser vabili rvabi lity 0.4.3 Energ y interp retati on of the contr ollabi lity and obse Gram ians 0.4.4 Desig n of reduc ed order mode ls

xi 248 251 253 254 255 256 260 261 262 267 269 269 270

271 271 272 274 276 277

279

283 283 285 287 288 290

293 293 293 293 295 296 297 297 298 299 301 303 303 304 305 306 307

References

309

Index

313

Preface g control This book was written with the intent of prmidin g students and practicin system control use to needed theory engineers with the basic backgro und in control those of nt treatme detailed a with begins design software more productively. The book the of er remaind the in needed are that aspects of the state space analysis of linear systems manner: text. The book is organized in the following n \ia • The first four chapters develop linear system theory including model reductio balanced realization. • Chapter s S and 6 deal with classical optimal control theory. Hx control • The final four chapter s are devoted to the development of suboptim al theory. needed The mathem atical ideas required in the development are introduc ed as they are beyond venture to reader the e motivat to using a "just-in -time" approac h. This is done ic contror ', .the usual topics appeari ng in introduc tory undergr aduate books on .. automat leYel books duate postgra to d restricte been to more advanced topics \Yhich have so far titles. their in control" "robust and having the terms "mathem atical control theory" the final at course ester two-sem or one a This book can be used as the text for either lewl. duate postgra g beginnin the at course r year undergr aduate level or as a one semeste " or S}Stems and ''signals either in course Students are assumed to have taken a basic state the of ge knowled tory introduc an . "autom atic control ... Althoug h not assumed would space analysis of systems together with a good understa nding of linear algebra book. this in d presente ideas benefit the reader's progress in acquiring the t Yiew of Ideas presented in this book which provide the reader with a slightly differen are as ks textboo other reading by d obtaine control and system theory than would be follows: and 1or • The so-called PBH test \Yhich is usually presented as a test for controllability ues in eigenval obserYability is used through out the present book to characterize output and;or control problems involving eigenvalue assignment by state feedback injection. deYelop• An easy to underst and matrix Yariational technique is used to simplify the LQG and ic ment of the design equation s for the time invarian t, steady-state. quadrat controllers. ment of the • The relatively simple idea of the L 2 gain is used as a basis for the develop H 00 controller.

xiv

Preface

Concerning the style of the book, the beginning section, "Introduction", for each chapter contains motivational material and an overview of the ideas to be introduced in subsequent sections in that chapter. Each chapter finishes with a section called "Notes and References", which indicates a selection of other sources for the material treated in the chapter, as well as an indication of recent advances with references. I would like to thank the following colleagues in the Department of Electrical and Computer Engineering at Queen's University for proof-reading parts of the manuscript: Norm Beaulieu, Steve Blostein, Mingyu Liu, Dan Secrieu and Chris Zarowski. Special thanks go to my former research student Lacra Pavel for proof-reading and advice on Chapters 6, 9 and 10 as well as to Jamie Mingo in the Department of Mathematics and Statistics at Queen's University for his help with some of the ideas in Chapter 7. Thanks go also to Patty Jordan for doing the figures. Finally, I wish to acknowledge the contribution to this book made by my having supervised the research of former research students, especially Manu Missaghie, Lacra Pavel and Johannes Sveinsson. The author would appreciate receiving any corrections, comments, or suggestions for future editions should readers wish to do so. This could be done either by post or e-mail: < [email protected] >.

1 Introduction to State Spa ce

1.1

Introduction

A well known behavioral phenomen on of dynamic systems is the appearanc e of an output in the absence of an input. This effect is explained once it is recognized that the internal storage of energy in the system at the beginning of the response time will produce an output. This kind of behavior is referred to as the system's zero-input response. Alternative ly, the production of an output caused solely by an input when there is no energy storage at the start of the response time is referred to as the zero-state response. These two classes of response are responsibl e for all possible outputs and in the case of linear systems we can always decompose any output into the sum of an output drawn from each of these classes. In this chapter \Ye will use the example of a second order system together with both the zero-input response and the zero-state response to introduce the reader to the use of the state space in modeling the behaYior oflinear dynamic systems.

1.2

Review of Second Order Systems

A commonly encountere d physical process which we will use in the next two sections to introduce the state modeling of linear dynamic systems is the electric circuit formed by connecting an ideal constant resistor Rc, inductor Le, and capacitor Ce in series in a closed loop as shown in Figure 1.1 Suppose the switch is closed at t = t, < 0 so that there is a current flow i(t). t ::.0: 0. and a voltage across the capacitor y(t), t ::.0: 0. Then applying Kirchhoff' s voltage law yields

Rei(t)

+Led~~)+ y(t) =

0

where the current in the circuit depends on the capacitor voltage as

"( ) =

l

t

ce dy(t) dt

Combining these equations gives a second order differential equation in the capacitor voltage, y(t), d l'(t) d\·(t) (1.1) -·-+a2 y(t) = 0 -·-+a I dt dt 2

2

Introduction to State Space

i(t)

L,

Switch

C,

Figure 1.1

y(t)

Electric circuit with charged capacitor. Switch closed prior to t = 0

where

and we refer to the capacitor voltage as the system's output.

1.2.1

Patterns of behavior

The differential equation (1.1) is said to gm;ern the evolution of the output, y( t), since it acts as a constraint relating y( t), d~~t) , and ddt~r) to each other at each instant of time. We will see now that once the initial conditions, i.e., the values of initial output, y(O), and initial derivative of the output, y(O), are specified, the differential equation, (1.1 ), completely determines the output, y(t), for all positive timet E (0, oo). We obtain y(t) as follows. Suppose we have y(t) such that (1.1) is satisfied. Then denoting the derivatives of y(t) as dy(t) = g(t) dt

d2y( t) = h( t) dt 2

we see that equation (1.1) becomes ( 1.2)

Now the only way this equation can hold for all t > 0 is for h(t) and g(t) to be scalar multiples of y(t) where g(t)

=

sy(t)

( 1.3)

Otherwise equation (1.2) can only be satisfied at specific instants of time. Therefore with this assumption assumption (1.2) becomes p(s)y(t) = 0

( 1.4)

Review of Second Order Systems

3

where p(s) is the second degree polynomial pi,sj =

l- + a 1 1--:- a2

Finally, equation (1.4) holds for all time, when y( t) is not zero for all time, i.e., the trivial solution. if and only if sis any one of the roots. {..\, : i = 1, 2} of p(s),

(1.5) Returning to the requirement that y(t) and its derivatives must be constant scalar multiples of each other, equation ( 1.3), the function that has this property is the and has series expansion exponential function. This important function is denoted as

est=~ (st)i ~

i=O

.,

( 1.6)

/.

where i!. (factorial i). is the product i! = ( i) ( i - 1) (i - 2) ... ( 1)

i i

=1

>0 =0

Notice that a bit of algebra shows us that the derivative of ~.Y. equation (! .6 ). has the desired property of being an eigenfunction for differentiation, dest

-=se dt

sr

Now we see from the foregoing that est satisfies equation (1.1) when s = )q or .A 2 . Therefore any linear combination of e)'~t and e>-,t satisfies equation (1.1) so tharthe output y(t) is given in general as

(1.7) where the k;s are constant scalars chosen so that y(t) satisfies the initial conditions. We can be do this by solving the equations which result from setting the given values for the initial conditions, y(O) and y(O). equal to their values determined from equation ( 1. 7), i.e., by solving ( 1.8)

for k1, k2. Notice that we can do this only if .A 1 t- A.c- In orderto proceed when A. 1 = .A2 we replace equation (1.7) with ( 1.9)

4

Introduction to State Space

and determine the k;s from the appropriate equations to ensure that the initial conditions are satisfied. Returning to the behavior of the physical process that is under analysis, notice that since R"' Le, and Ce are real, the a;s are real. As a consequence the roots >.; of p(s). equation (1.5), are both real or are both complex. Moreover when these roots are complex they are conjugates of each other, i.e., ).. 1 = >.2. More generally, if all the coefficients of a polynomial of any degree are real, each complex root must be matched by another root which is its complex conjugate. This property is important in the context of the behavior oflinear physical processes since the parameters of these processes, e.g., mass, heat conductivity, electric capacitance, are always real so that the coefficients of p(s) are always real. Now a plot of the output, y( t), versus time, t, reveals that there are two basic patterns for the behavior of the output depending on whether the >..;s are real or are complex conjugate pairs. If the A;s are real, we see from equation (1.8) that the k;s are also real and the output y(t): t E (0, oo) is given as equation (1.7) or (1.9). In this case we see that the output voltage y(t) exhibits at most one maximum and decays without oscillation to the time axis as t tends to infinity. Notice from equation (1.5) that the A;s are real provided the parameters Re, L"' Ce have values such that C'i/ ~ a2 . < a2 , then we see from (1.5) that Alternatively, if the >..;s are complex, i.e., if >.. 1 = X2 and from (1.8) that k 1 = k;. Thus k 1 e>'~ 1 and k 2e>- 21 are complex conjugates of each other and their sum which gives y(t), equation (1.7), is real. Incorporating these conjugate relations for the >..;sand the k;s in equation ( 1. 7) allows us to write the output as a damped oscillation

C1/

(1.10) where k 1 = Re[kt]

+ jim[kt]

_ _1 (Im[ktl) e -tan Re[ki] Thus we see from (1.1 0) that the output voltage across the capacitor, y( t), swings back and forth from its initial value to ever smaller values of alternating polarity. This behavior is analogous to the behavior of the position of a free swinging pendulum. The capacitor voltage (pendulum position) eventually goes to zero because of the loss of heat energy from the system resulting from the presence of Re (friction). In this analogy, voltage and current in the electric circuit are analogous to position and velocity respectively in the mechanical process. The inductance Le is analogous to mass since the inductance resists changes in the current through itself whereas the inertial effect of mass causes the mass to resist change in its velocity. In addition, notice from equation ( 1.1 0) that the frequency of the oscillation, Im[).. J], as well as the time constant associated with the decay in the amplitude of the oscillation, (Re[>..J])- 1 , are each independent of the initial conditions and depend on the system parameters, Re, Le, Ce only, i.e., on a1, az only.

Review of Second Order Systems

5

The previous discussion leads to the following characterization of the zero-input · response of dynamic processes whose behavior can be modeled by second order differential equations with constant coefficients. 1 (i) The zero-input response, y(t): t > 0, depends on the set of signals {e'\; : i = 1, 2} referred to as modes of the system where the constants >..il (system eigenvalues), are roots of the polynomial p(s), (characteristic polynomial). (ii) The steady state zero-input response is zero, i.e., limy(t) = 0, for any initial conditions if and only if all the >..;s are negative or have negative real part, i.e., Re[>..;] < 0, i = 1, 2 . In this situation we say that the system is stable. (iii) We have Re[>..;] < 0, i = 1, 2, if and only if a 1 > 0 and a 2 > 0. More generally, the condition a; > 0, i = 1, 2, · · · n for systems whose behavior is governed by differential equations in the order of n > 2, is necessary but not sufficient for the system to be stable, i.e., is necessary but not sufficient for all >..;s to have negative real part t~oo

1.2.2

The phase plane

We have just seen that, when there is no input, a second order system having specified a;s has output,y(t), which is specified completely by the initialconditions,y(O) andy(O). This important observation suggests that the same information concerning the behavior of the system is contained in either (a) a plot of y(t} versus tor (b) a plot of y(t) versus y(t). Thus if we make a plot of y(t) versus y(t), the point representing y(t),y(t) in the y(t) versus y(t) plane traces out a curve or trajectory with increasing time. The two dimensional space in which this trajectory exists is referred to as the state space and the two-element vector consisting of y(t) and y(t) is referred to as the state, denoted as x(t) where

x(t)

=

[~(t)] y(t)

This approach to visualizing the behavior of a dynamic process was used by mathematicians at the end of the last century to investigate the solutions for second order nonlinear differential equations, i.e., equations of the form (1.1) but with the a;s functions of y(t) and/or y(t). The term phase plane plot was used to refer to the state trajectory in this case. Since, in general, the dimension of the state space equals the order of the differential equation which governs the output behavior of the process, the state space cannot be displayed for systems of order greater than two. Even so, the mathematical idea of the state space has become of great practical and theoretical importance in the field of control engineering. Referring to the previous section, we see that the state trajectory for a dynamic process whose behavior can be modeled by a second order differential equation with constant coefficients; can exhibit any one of the following four fundamental shapes. (i) If the A;s are complex and Re[>..;] < 0 the system is stable and the state trajectory spirals inwards towards the origin. (ii) If the \·s are complex andRe[>..;] > 0 the system is unstable and the state trajectory spirals outwards away from the origin.

6

Introduction to State Space

y(t)

b d

y(t)

y(t)

a

Figure 1.2

Plot of y(t) vs. t and y(t) vs. y(t) when A;s complex

(iii) If the A.;s are real and both A;s are negative the system is stable and the state trajectory moves towards the origin in an arc. (iv) If the A.;s are real and one or both are positive the system is unstable and the state trajectory moves away from the origin in an arc. Notice that state trajectories (ii) and (iv) do not occur in the present example of an electric circuit. This results from the fact that the parameters R"' Le, Ce are positive. Thus the coefficients, a; : i = 1, 2 of the characteristic polynomial, equation (1.4) are positive so that the A;S are negative or have negative real parts. This implies that we are dealing with a stable dynamic process, i.e., state trajectories tend to the origin for all initial states. So far in this chapter we have used an electric circuit as an example of a system. We used the character of the behavior of this system in response to initial conditions to

Introduction to State Space Modeling

7

introduce the concept of the state of a system. In the next section this concept is made more specific by introducing the mathematical characterization of a system referred to as a state model.

1.3

Introduction to State Space Modeling

We saw in the previous section that once a second order system is specified, i.e., once the a;s are given numerical values, the zero-input response is determined completely from the system's initial conditions, y(O),y(O). In addition, we noted that the second derivative of the output is determined at each instant from y(t) and y(t) through the constraint (1.1). These facts suggest that it should be possible to obtain the zero-input response by solving two first order differential equations involving two signals, x 1 ( t), x 2 ( t), which are related uniquely to y(t),y(t). One straightforward way of doing this is to identify y(t) with x 1 (t)and y(t) with x 2 (t), i.e.,

y(t)

= XJ (t)

(1.11)

y(t) = x 2 (t)

( 1.12)

An immediate consequence of this identification is that at every instant the derivative of x 2 (t) equals x 1 (t)

x2(t)

( 1.13)

= x1 (t)

Moreover, rewriting the second order differential equation, (1.1), as

:t (y(t)) =

~a 1 y(t) ~ a2y(t)

andusing equations (1.11-1.13) gives us the differential equation for x 1 (t) ail(1.14)

Thus we see from equation (1.13) and (1.14) that the derivative of each of the x;s is a (linear) function of the x;s. This fact is expressed in matrix notation as ( 1.15)

x(t) = Ax(t) where

A=

[ ~al I

x(t)=

[xi(t)] x2(t)

with the vector x(t) being referred to as the state, and the square matrix A being referred to as the system matrix. In addition we see from equation (1.12) that

y(t) = Cx(t)

(1.16)

Introduction to State Space

8

where

c

=

[0

1]

t matrix. with C being a row vector referred to as the outpu ion (1.1) is equivalent to the vector equat ential differ In summ ary the second order (I.l6) . These equat ions, (1.15, 1.16) differential equat ion (1.15) and the outpu t equat ion system in the absence of an input. constitute a state mode l for the second order d by a block diagr am involving the Alternatively, the state model can be represente integrators, and scalar multipliers interc onnec tion of blocks which opera te as summers, tor 1/sis used to indicate integration. on the comp onent s of the state. The Laplace opera to obtai n a state model for the More generally, we can use the foregoing procedure sses as follows. zero-input response of higher order dynam ic proce 1h order process is gover ned by n an of nse respo Supp ose the zero- input ( 1.17) where

fy comp onent s of the state with Then we proce ed as in the second order case to identi derivatives of the outpu t as

(t) = y(n~l)(t) x2(t) = Y(n~2)(t)

X]

(1.18)

equat ion (1.15) having a system Thus using (1.17, 1.18) we obtai n a vecto r differential

x.(t)

i,(t)

x,(t)

x,(t)

-a,

-a,

Figure 1.3

Block diagra m repres entati on of the state model

·g

Solving the State Differential Equation

matrix A given as -a I 1

A=

-a2

0

-a3 0

-an

0

0

0

0

0

0

0

0 (1.19)

and output equation, (1.16), having an output matrix C given as

c=

[0

... 0

1]

The pattern of zeros and ones exhibited in A, (1.19), is of particular importance here. Notice that the coefficients of the characteristic polynomial

appear as the negative of the entries in the first row of A. Matrices exhibiting this pattern are referred to as companion matrices. We will see shortly that given A in any form, the characteristic polynomial is related to A as the matrix determinant

p(s)

=

det[s/- A]

This fact is readily seen to be true in the special case when A is in companion form.

1.4

Solving the State Differential Equation

Recall that the solution to a scalar differential equation, e.g., (1.1), involves the scalar exponential function, e>.t. In this section we will show that the solution to the state differential equation, (1.15), involves a square matrix, eAt, which is referred to as the matrix exponential.

1.4.1

The matrix exponential

Suppose we are given the initial state x(O) and the system matrix A, either constant or time varying. Then we obtain a solution to the state differential equation, (1.15), by finding (t), the square matrix of scalar functions of time, such that

x(t) = (t)x(O)

(1.20)

where (t) is referred to as the transition matrix. Since the state at each instant of time must satisfy the state differential equation, (1.15), the transition matrix is a matrix function of the system matrix A. In this book A is constant. In this case the dependency of (t) on A is captured by the notation

(t)

=eAt

(1.21)

10

Introduction to State Space

where the square matrix eAt is referred to as the "matrix exponential of At" since it can be expressed as an infinite series reminiscent of the infinite series for the exponential of a scalar. (1.5), i.e., ( 1.22)

In order to show that the transition matrix given by (1.22) solves the state differential equation, (1.15), we differentiate the foregoing series expansion for the matrix exponential of At to obtain 2A 2 t 3A 3 t 2 At deA 1 ~=A+--+--Ae

2!

dt

=

3!

4A 4 t3

+--+··· 4!

AeAt

Then using this relation to differentiate the assumed solution x(t) =eAr x(O)

( 1.23)

yields

x(t) = AeA 1x(O) = Ax(t) and we see that (1.23) solves the state differential equation, (1.15).

1.4.2

Calculating the matrix exponential

There are many ways of determining eAt given A. Some of these approaches are suitable for hand calculation and others are intended for use with a digital computer. An approach of the first kind results from using Laplace transforms to solve the state differential equation. We develop this approach as follows. We begin by taking the Laplace transform of (1.15) to obtain ( 1.24)

sX(s) - x(O) = AX(s)

where A is 2 x 2 we have x;

Xi(s) =

j xi(t)e-' dt 1

0

Then rewriting (1.24) as (sf- A)X(s)

= x(O)

( 1.25)

we see that provided sis such that (sf- A) is invertible, we can solve (1.25) for X(s) as X(s) = (sf- A)- 1x(O)

( 1.26)

Solving the State Differential Equation

11

Now (sl- A)- 1 can be expressed using Crammer's rule as (

sl -A

)

adj[s/- A] - det[s/ -A]

_1 _

where when A is an n x n matrix, the adjugate matrix, adj [s/ - A], is an n x n matrix of polynomials of degree less than n and det[s/ - A] is a polynomial of degree n. Finally, taking the inverse Laplace transform of (1.26) yields

x(t) = £- 1 [(s/- A)- 1)x(0) and we see, by comparing this equation with (1.23), that

Now in the case where A is the 2 x 2 matrix given by (1.15), we have (sl _A)-I= adj[s/- A] det[s/- A]

=

[

s+ai -1

a2]-I

(1.27)

s

/

where

adj[s/- A]=

[s s -a2] +a 1

1

Notice from the previous section that det[s/- A]= p(s), (1.4), is the characteristic polynomial. In general any n by n system matrix A has a characteristic polynomial with roots {>..i : i = 1, 2 · · · n} which are referred to as the eigenvalues of A. The eigenvalues of the system matrix A play an important role in determining a system's behaviq.r. Returning to the problem of determining the transition matrix for A, (1.15), we apply partial fraction expansion to the expression for (sl- A)- 1 , (1.27), assuming det[s/- A] has distinct roots, i.e., >.. 1 -1 >..2 , to obtain (1.28) where

12

Introduction to State Space

Finally, taking the inverse Laplace transform of (1.28), we obtain the transition matrix a~~ -)q)•'l(e>" 1 -.\2e;.,,

-

e.>.'')]

+ AJe;.,,

(1.29)

We will show in the next section that there are other ways of modeling a dynamic process in the state space. This non-uniqueness in the state model representation of a given dynamic process results from being able to choose the coordinates for expressing the state space. In the next section we will use this fact to simplify the determination of eAr by working in co-ordinates where the state model has a diagonal A matrix.

1.4.3

Proper and strictly proper rational functions

Before continuing to the next section, notice that when A is ann x n matrix, adj [sf - A] is an n x n matrix of polynomials having degree no larger than n - 1. Thus, since the characteristic polynomial for A, det[sl- A], is of degree n, we see from (1.27) that 1 is ann x n matrix of strictly proper rational functions. (sfIn general a rational function

Ar

n(s) r(s) = d(s) is said to be; (i) strictly proper when the degree of its numerator polynomial is less than the degree of its denominator polynomial, i.e., deg[n(s)] < deg[d(s)] (ii) proper when the degree of its numerator polynomial equals the degree of its denominator polynomial, i.e., deg[n(s)]

=

deg[d(s)]

In subsequent chapters we will see that this characterization of rational functions plays an important role in control theory.

1.5

Coordinate Transformation

In Section 1.3 we saw that the zero-input response for a system could be obtained by solving a state vector differential equation where the components of the state were identified with the output and its derivatives. In this section we examine the effect of changing this identification.

.

..,.,.

~··.

Coordinate Transformation

1.5.1

13

Effect on the state model

Referring to the second order process used in the previous section, let x( t) denote the state obtained by setting

= v[x1(t)] [.Y(t)] y(t) x (t)

(1.30)

2

where Vis any invertible (nonsingular) 2 x 2 matrix of constants. In the previous section V was the identity matrix. Now we see from (1.11, 1.12, 1.30) that the state x(t) used in the previous section is related in a one-to-one fashion to the state x(t) as

x(t)

(1.31)

Vx(t)

=

where we say that x(t) is the state in the old or original coordinates and x(t) is the state in the new or transformed coordinates. Then the state model parameters in the old coordinates, (A, C), are transformed by a change of coordinates to (A, C) in the new coordinates as

v

- -

(A, C) ~ {A, C)

(1.32)

where

We can develop this relation as follows. First using (1.31) in (1.15) we obtain

V x= AVx(t) which, since V is invertible, can be multiplied throughout by

x (t) = Ax(t) where

Again, using (1.31) in (1.16) we obtain

y(t) = Cx(t) where

v-I

to give

14

Introduction to State Space

Notice that the transition matrix, eAt, which applies in the new coordinate s is related to the transition matrix, eAt, in the original coordinate s as

(1.33)

1.5.2

Determination of eAt

The flexibility provided by being able to choose the coordinate s for the state model representa tion of a dynamic process is often of considerab le use in the analysis and design of control systems. We can demonstra te this fact by using a change of coordinate s to calculate the transition matrix. Suppose we are given a two dimension al system matrix A having a characteris tic polynomia l, det[s/- A], with distinct roots (eigenvalues), i.e., )q -=f. >. 2 . Then we can always find a coordinate transforma tion matrix V so that the system matrix A in the new coordinate s is diagonal and

i (t) = Ax(t)

( 1.34)

where

with entries along the diagonal of A being the eigenvalues of A. Now when the system matrix is diagonal, the correspond ing transition matrix is also diagonal. We can see this by noting that the state differential equation in these coordinate s, (1.34), consists of two scalar first order differential equations which are uncoupled from each other

x 1(t)

=

.A 1x 1(t)

x2(t)

=

>-2x2(t)

so that their solution can be immediate ly written as

x1 (t) = /· 1 t x1(0) x2 (t) = e>.'tx 2 (0)

( 1.35)

which in matrix form is ( 1.36)

Diagonalizing Coordinate Transformation

15

Thus we see that the transition matrix is indeed diagonal

Having determined the transition matrix for A, we can use (1.33) to determine the transition matrix for A as

(1.37) with V being the coordinate transformation matrix which makes A diagonal. Now we will see in the next section that, in general, the coordinate transformation matrix V needed to make A diagonal depends on the eigenvectors of A. However in the special case when A is a 2 x 2 companion matrix, (1.15), with >.. 1 -=1- >..2, the required coordinate transformation matrix is simply related to the eigenvalues of A as

(1.38) We can see that this coordinate transformation gives rise to a diagonal system matrix by / . using

(1.39) to obtain

Then since J + a1s + a2 = (s- A.I)(s- >..2) we have a 1 = -(>.. 1 + >.. 2) and""a2 = >..,>..2. Therefore the foregoing expression for A reduces to

(1.40) Finally, the expression obtained for eAt using V, (1.38), in (1.37) equals (1.29) which was obtained at the end of Section 1.4 through the use of Laplace transforms. The foregoing approach to the determination of the transition matrix requires the determination of a coordinate transformation matrix V which diagonalizes the system matrix A. We will see in the next section that the columns of the coordinate transformation matrix required to do this are right-eigenvectors for A.

1.6

Diagonalizing Coordinate Transformation

As mentioned previously, the roots of the characteristic polynomial for a square matrix A are called the eigenvalues of A. In this section we will see that corresponding to each of A's

Introd uctio n to State Spac e

16

igenvector. More over we will see that eigenvalues there is at least one right and one left-e transf orma tion V requi red to make when the eigenvalues of A are distinct, the coord inate A. [n addit ion we will see that A diagonal has columns equal to the right-eigenvectors of of A. ectors v-I has rows which are the transpose of the left-eigenv

1.6.1

Righ t-Eig env ecto rs

matri x in comp anion form, (1.15), Cons ider the special case when A is a two-b y-two g the chara cteris tic polyn omial as having unequ al eigen value s{.\: 1,2}. Then writin At= -a 1 Ai- a 2 we see that i

= 1, 2

(1.41)

or

i = 1, 2

( 1.42)

where

i = 1, 2 the i1h eigenvalue, right-eigenvector Notic e that (1.42) is a general expression relating vi is said to be the right-eigenvector pair (Ai, vi) for the any squar e matri x A, where a major role in the state analysis of corre spond ing to the eigenvalue Ai· These pairs play prese nt instan ce is a result of A being systems. The partic ular dependence of vi on Ai in the in comp anion form. v- 1A V diagonal, we combine the Conti nuing with the const ructio n of V to make n equat ions given by (1.42) fori= 1 and i = 2 to obtai

AV= VA

( 1.43)

where V, A are given as

A=[A01 A2OJ A , V is invertible and we can preNow when A has distinct eigenvalues, i.e., A1 i= 2 multiply (1.43) by v- 1 to obtai n

matri x requi red to make A diagonal. Thus we see that Vis the coord inate trans forma tion necessarily in comp anion form, x, More generally, suppo se A is any n x n matri not out that this condi tion of turns it Now j. which has distin ct eigenvalues, ,\ i= A1 : i ic of A to be indep enden t, i.e., vi and vi distinct eigenvalues is sufficient for the eigenvectors enden t colum ns and is theref ore point in different directions. There fore V has indep

Diagona lizing Coordina te Transfor mation

17

invertible where (1.44) i

=

1, 2,- · · ,n

( 1.45)

and v- 1AVis diagonaL In the special case when A is in compani on form, (1.19), its eigenvalues, {>..- : i = 1, 2, · · · , n} are related to its eigenvectors, {vi : i = 1, 2, · · · , n} as ,n-1 v iT -:_ [ /\" I

.A;

1)

In order to see that this result holds, set the last entry in vi equal to one. Then taking A in compani on form, (1.19), solve the last scalar equation in (1.45) and use the result to solve the second to last scalar equation in (1.45). We continue in this way solving successive scalar equation s in (1.45), in reverse order, until we reach the first scalar equation. At this stage we will have all the entries in vi. These entries satisfy the first scalar equation in (1.45) since A; is a root of the characteristic polynomial whose coefficients appear with negative signs along the first row of A. Jn general, when A is not in any special form, there is no special relation between the eigenvalues and the corresponding eigenvectors. Thus in order to determine the eigenvalue, right-eigenvector pairs when A is not in any special form we need to determine (.A;, vi) : i = 1, 2, · · · n so that the equations i = 1,2, · · ·n

(1.46)

are satisfied.

1.6.2

Eigenvalue-Eigenvector problem

The problem of determining (.A;, vi) pairs which satisfy (1.46) is referred to as the eigenvalue-eigenvector problem. There are well established methods for solving this problem using a digital computer. In order to gain additional insight into the nature of the eigenvalue-eigenvector problem we consider a theoretical approach to finding eigenvalue-eigenvector pairs to satisfy (1.46). To begin, suppose we rewrite (1.46) as (AI- A)v = 0

(1.47)

where 0 denotes the null vector, i.e., a vector of zeros. Then in order for the solution v to this equation to be non-null we must choose .A so that the matrix .AI - A is singular, i.e., does not have an inverse. Otherwise, if AI- A is invertible we can solve (1.47) as

~nd the only solution is the trivial solution v = 0. However when (AI - A) is not Invertible, (1.47) can be satisfied by v f= 0.

Introduction to State Space

18

Now from Cramme r's rule for matrix inversion we have (AI _ A )_ 1= adj[M- A]

det[M- A] an Therefor e )...] -A does not have an inverse when det[>..I -A] = 0, i.e., )... = >..; eigenvalue of A. Next recall that singular matrices have dependen t columns. Therefor e A;!- A has dependen t columns so that we can find scalars {vk : k = 1, 2, · · · , n} not all zero such that n

L[(\1 - A)]ku~

= 0

( 1.48)

k=l

be where [(>..J- A)]k: k = 1,2· · · ,n denote columns of >..J- A. Notice that (1.48) can where vi = v and >..; = )... rewritten as (1.47) with

Since we can always multiply (1.48) through by a nonzero scalar a, the solution, vi, to (1.48) or (1.47) is not unique since m/ is another solution. More generally, we say that the eigenvectors of a given matrix are determin ed to within a scalar multiple, i.e., the directions of the eigenvectors are determin ed but their lengths are arbitrary . Assumin g that A has a complete set of (n independ ent) right-eigenvectors, we can decompo se any initial state as n

x(O) = l:a;vi = Va

(1.49)

i=l

where

with the a 1s being found as a=

v- 1x(O)

This decompo sition of the state into a linear combina tion of right-eigenvectors of A plays an importan t role in the analysis of system behavior. This is illustrate d in the next of section where we will use the eigenvectors of A to reveal certain fundame ntal properties state trajectories. Unfortun ately, when an n x n matrix A has multiple eigenvalues, i.e., when det[>..I - A] does not have n distinct roots, the number of eigenvectors may or may not be equal ton, i.e., A may or may not have a complete set of eigenvectors. When A does not ed have a complete set of eigenvectors, other related vectors, referred to as generaliz a provide to tors eigenvectors, (Appendix), can be used together with the eigenvec it case this in However (1.49). decompo sition of the state space in a manner similar to

Diagonalizing Coordinate Transformation

19

is impossible to find a nonsingular matrix V such that v- 1AVis diagonal. Then A is said to be not diagonalizable. In summary, the condition that A has distinct eigenvalues is sufficient but not necessary for A to be diagonalizable. Most of the time, little additional insight into control theory is gained by discussing the case where A does not have a complete set of eigenvectors. Therefore, we are usually able to avoid this complication without loss of understanding of the control theory.

1.6.3

Left-Eigenvectors

Suppose A is any n x n matrix having n distinct eigenvalues. Then we have seen that the matrix V having columns which are right-eigenvectors of A is invertible and we can write (1.50) where Avi =>..;vi: i = 1, 2, ···nand /

Now suppose we post-multiply both sides of (1.50) by

v- 1 to obtain

v- 1A = Av- 1 Then if we denote

(1.51)

v- 1 in terms of its rows as

and carry out the matrix multiplication indicated in (1.51) we obtain

A] [)..>..2w2T w Tl

w:T A ws [

..

1

1

.

...

wnT A

)..n wnT

Therefore it follows, by equating corresponding rows on either side of this equation, that i = 1,2, · · · ,n

(1.52)

20

Introduction to State Space

which transposing through out gives i = 1, 2, · · ·, n

( 1. 53)

of AT Thus we see from equatio n (1.53) that the column vector w; is an eigenvector left the on with corresponding eigenvalue .A.;. However, since the row vector wiT appears it ish distingu to side of A in equatio n (1.52), w' is referred to as a left-eigenvector of A from the correspo nding right-eigenvector of A, namely, v;. rs just Notice that, since v-I V is an identity matrix, the left and right-eigenvecto defined are related to each other as

=0 i=fj} = 1 i =i

(1.54)

In addition notice that any nonzero scalar multiple of w;

satisfies (1.52), i.e.

Therefore zi is also a left-eigenvector of A and we see ftom equation (1.54) that the left and right eigenvectors are related as ziT v

1

=0 = 'li =I 0

i

=I j

i

=j

in general

nality This basic characteristic ofleft and right-eigenvectors, referred to as the orthogo r of behavio the to property, is responsible for a number of fundam ental facts relating state models.

1.6.4

Eigen value invari ance

eigenvalues Before we go to the next section, it is importa nt to note the basic fact that the invertible of A and of A are the same whenever A is related to A as A = v-I A V for any v-I and matrix V. We can see this by premultiplying both sides of equatio n (1.45) by inserting vv-I between A and vi, viz.,

Then setting

ti

= v-I vi and taking v-I A v =A gives

Thus the which implies that the eigenvalue, right-eigenvector pairs for A are (A.;, t;).

State Trajectories Revisited

21

eigenvalues of A equal the eigenvalues of A independent of the coordinate transformation matrix V. Alternativelv. another way we can see this fact is to carry out the following manipulations det[s/- A]= det[sl- V- 1AV] = det[V- 1 (s/- A)Vj

v-I det[s/- A1 det V

=

det

=

det[s/- A]

Thus A and A have the same characteristic polynomial, and since the roots of a polynomial are uniquely dependent on the coefficients of the polynomial. A and A have the same eigenvalues. Finally, since the differential equations modeling the behavior of dynamical processes must have real coefficients, we can always work in coordinates where the state model parameters, (A, C), are real. As a consequence, if (.\,vi) is a complex eigenvalue-eigenvector pair for A, then (,>,j. l,tx) is also an eigenvalue-eigenvector pair for A.

1. 7

State Trajectories Revisited

We saw in Section 1.6.2 that assuming A has a complete set of eigenvectors. any initial state can be written in terms of the eigenvectors of A, (1.49). In this section this fact is used ~to gain additional insight into the nature of a system's state trajectories and zero-input response. More specifically, under certain conditions on the matrix pair, (A, C), a system can exhibit a null zero-input response, y(t) = 0 for all r > 0, for some non-null initial state, x(O) -=1- ¢.When this occurs we say that the state trajectory is orthogonal (perpendicular) to cT, denoted cT __Lx(t), since the output depends on the state as y(t) = Cx(t). Two vectors a, f3 are said to be orthogonal if aT

f3 = 0

When the state space is n dimensional, a state trajectory which produces no output lies in an n - 1 dimensional subspace of state space which is perpendicular to the vector cT. For example, if the system is second order, this subspace is a straight line perpendicular to cT: if the system is third order. this subspace is a plane perpendicular to cT Thus, in the second order case, we obtain a null output if we can find an initial state such that it produces a straight line trajectory

x(t)

=

1(t)v

satisfying Cv

=0

where !(t) is a scalar function of time and vis a constant two-element vector. When n > 2,

Introduction to State Space

22

straight line any trajecto ry orthogo nal to cT can be decomp osed into a sum of re an Therefo cT. to nal trajecto nes all lying in the n - 1 dimensi onal subspac e orthogo property the of nding understa an understa nding of straight line trajectories is essential to states. posed by certain systems of having a null zero-inp ut response to certain initial

1.7.1

Straig ht line state trajectories: diago nal A

a straight line Suppose A is a 2 x 2, real, diagona l matrix. Then the state trajecto ry is can see this We axes. ate whenever the initial state lies only on one of the two coordin immedia tely as follows. Conside r the initial states

x1 (0) =

[

.X1 (0)]

0

1.5.2 that where x1 (0) and, x2 (0) ~re any real nonzero scalars. Then recalling from Section the state that (1.36) from see we l, the transitio n matrix eAt is diagona l when A is diagona correspo nding to each of these initial states is

x(t)=

[.xJ(OJe,x'r]

x(t) =

[.x2 (~e,\2 t J

for x(O)

= x1(0)

for x(O)

=

(1.55) x 2 (0)

ates The foregoin g suggests that the trajecto ry for any initial state in these coordin

x(O)

=

[~I (0)] x2(0)

can be written as

( 1.56) where

/ :k

=

1, 2

are columns from the 2 x 2 identity matrix, viz.,

ry More generally, when A is a real, diagona l, n x n matrix, the state trajecto

(1.57) results when the initial state

State Trajectories Revisited

23

has components which satisfy

.X;(O)

=0 #0

fori# k

fori= k

where ik is the k 1h column from the n x n identity matrix. The foregoing result implies that the zero-input state response for any initial state can be written as n

x(t)

= .L:(xk(O)/'k 1) / k=I

when n

x(O)

=

L .xk(O)ik k=I

In summary, the state trajectory is the vector sum of state trajectories along coordinate axes where each coordinate axis trajectory depends on one of the system modes in the set ). I . of _system modes, {e k : k = 1, 2 · · · , n}. 1 Now in order to generalize the foregoing result to the case where A is not diagonal, we suppose, in the next section, that the foregoing diagonal case resulted from a coordinate transformation from the original coordinates in which A is given.

1.7.2

Straight line state trajectories: real eigenvalues

Recalling that V, equation (1.44), is the coordinate transformation needed to diagonalizeA and taking x(O) as in (1.57) we have (1.58) where Avk = >.kvk. Then, using the series expansion for the matrix exponential, (1.22), we see that when x(O) is given by equation (1.58) we have

(1.59) Now with >.k real we see from equation (1.59) that the point representing the state moves along the eigenvector, if, towards the origin when >.k < 0 and away from the origin when >.k > 0. The case where >.k is complex is taken up in the next subsection. More generally, assuming A has a complete set of eigenvectors, we can write any initial state as n

x(O)

= 2:: "(;Vi i=I

(1.60)

Introduction to State Space

24

where 'Y

=

v-t x(O) and

Then the state trajectory can be written as n

x(t) =

2,:,)'Y;e.\')v;

(1.61)

i=l

Now recall, from Section 1.2, that a system is stable if lim x(t)

=

for all x(O)

0

( 1.62)

t--+'XJ

Therefore assuming A has real eigenvalues, we see from equation (1.59) that the system is stable if and only if i = 1, 2, · · ·, n

It should be emphasized that a system is stable if and only if equation (1.62) is satisfied. Therefore if we need to restrict the initial condition in order to ensure that the zero-input state trajectory goes to the origin, the system is not stable. For example, referring to the expansion (1.60), any initial state, x(O), which is restricted so that its expansion satisfies

ri = o

when\ 2': 0

has a state trajectory which goes to the origin with time even though A has some nonnegative eigenvalues.

1.7.3

Straight line trajectories : complex eigenvalue s

Since the differential equations governing the input-output behavior of the physical processes we are interested in controlling have real coefficients, we can always choose to work in coordinates so that A, Care real matrices. Then since the eigenvalues, if complex, occur in conjugate pairs we see that the corresponding eigenvectors are also conjugates of each other, i.e., if(>., v) is a complex, eigenvalue-eigenvector pair, then(>.*, v*T) is also an eigenvalue-eigenvector pair of A. Now if the initial state is any scalar multiple of the real or imaginary part of v, the resulting state trajectory lies in a two dimensional subspace of state space composed from the real and imaginary parts of v. More specifically, suppose ( 1.63)

where 'Y is any real scalar and v

= Re[v] + jlm[v]

25

State Trajectories Revisited

Then frow (1.59) we obtain

x(t)

=

O:re(t)Re[v]

+ O:;m(t)Im[v]

( 1.64)

where

a,e(t) =

1'eRe[.X]t cos(Im[A.]t)

O:;m(t) =

-1'eRe[.X]t

sin(Im[A.]t)

The foregoing generalizes the observation made at the beginning of the chapter that, for second order systems, spiral shaped trajectories result when the roots of the characteristic equation consist of a complex conjugate pair. In the present case where the system order n > 2, the spiral shaped trajectory lies in a two-dimensional subspace of the n-dimensional state space. Notice that the initial stat~ was chosen to ensure that the state x( t) is real. However if we choose the initial state as

/'

with the real scalars o:; satisfying

. then x(O) would be complex and the resulting trajectory would lie in a 2-dimensional subspace of a complex state space.

1.1.4

Null output zero-input response

Having discussed straight line state trajectories, we return to the problem $.tated at the beginning ofthis section concerning the possibility of having initial states which produce null outputs. Suppose the output matrix happens to satisfy (1.65) for some eigenvector, v\ of A. Then it follows from (1.59) that if x(O) = 1'kvk then

and we see the output is null for all time when the initial state lies along vk. This effect of the existence of non-null initial states which do not affect the output, is related to a property of the state model's A and C matrices which is referred to as the system's observability (Chapter 3). More is said about this matter in Section 3.5.2. In order to obtain some appreciation of the importance of this effect consider a state model with A having all its eigenvalues, except A.k, in the open left-half plane. Then this state model is internally unstable since x( 0) = vk produces a trajectory which moves away

Introduction to State Space

26

equa tion (1.65) from the origin. How ever if C satisfies outp ut a;Jd in this case we have

this traje ctor y has no effect on the

for all x(O)

lim y(t) = 0

t-v:yc

and the system is exte rnal ly stable. beha vior mod el is inter nally unst able its outp ut This dem onst rate s that while the state ical process, ly impossible to exactly mod el a phys is stable. How ever , since it is practical r only and is resp onse to initial state s exists on pape the fore goin g stab ility of the outp ut in on, we say not robu stly outp ut stable. For this reas referred to by saying that the system is open left-half A matr ix has all its eigenvalues in the that a system is stab le if and only if its plane. which uctiv e to cons ider cond ition s on C Before we leave this section, it is instr t and leftfied. Recall, from Section 1.6.3, that righ guar ante e that equa tion (1.65) is satis eigenvalues are orth ogon al, eigenvectors corr espo ndin g to different

of eigenvectors so that The refo re supp ose A has a com plete set of the left-eigenvectors of A,

we can expa nd C in terms

n

C

( 1.66)

L...a ;wiT = "'"' I

when ak = 0 fied whe n Cis inde pend ent of wk, i.e., The n we see that equa tion (1.65) is satis pter 3 to acte riza tion of C will be used in Cha in equa tion (1.66). This stru ctur al char h relat e to their observability. develop prop ertie s of state models whic

1.8

lete Response State Space Models for the Comp

section el a syst em's zero -inp ut response. In this So far we have used the state space to mod zeroem's syst a of g conn ectio n with the mod elin we take up the use of the state space in state response.

1.8.1

Sec ond ord er pro ces s rev isit ed

age source Sect ion 1.2, supp ose we conn ect a volt Retu rnin g to the electric circuit used in (cap acito r t) al equa tion gove rnin g the outp ut, y( as show in Figu re 1.4. The n the differenti voltage), beco mes d2y(t) +a dy(t ) +a2y(t) 1 dt dt2

and where u(t) is the inpu t (sou rce voltage)

alla 2

= b2u(t)

are as in (1.1) with b2

( 1.67)

= a2 .

State· Space Models for the Complete Response

27

i(t)

u(t)

Vollagc Souroc

y(t)

Figure 1.4 Electric circuit with voltage input

Suppose, as in Section 1.3, that we choose the components of the state as [ xx 1 (t)] 2 (t)

[y(t)]

(1.68)

y(t)

so that (1.69)

"Then we see from (1.67) that (1.70)

and from (1.68-1. 70) that the state differential equation and output equation are given as

x(t) = Ax(t) + Bu(t)

(1.71)

y(t)

(1.72)

=

Cx(t)

where

A= [

-a, 1

B=

[~]

C= [ 0

with B being referred to as the input matrix. This state model is complete in as much as it can be used to obtain the output caused by any specified combination of initial state, x(O), and input, u(t). We will see that the matrix product of the initial state and transition matrix, which gives the zero-input response, is replaced in the calculation of the zero-state response by an integral, referred to as the convolution integral, of the i:nput and transition matrix. Before showing this, consider the following modification of the foregoing state model.

Introduction to State Space

28

Suppose we rewrite (1.67) as

where

and proceed in the manner used to get equation (1.71, 1.72). This gives the alternative state model

+ Bcu(t)

:<:(t) = Acxc(t) y(t)

=

(1.73)

Ccxc(t)

where Ac = A and Be= : 2 B =

[~]

Cc=b 2 C= [0

b2 ]

This state model, equation (1.73), is an example of a controller form state model. Controller form state models are characterized by having Be as the first column of the identity matrix and Ac in companion form, (1.19). The controller form state model is used in the next chapter to provide insight into the dynamics of closed loop systems employing state feedback.

1.8.2

Some essential features of state models

First notice that when we change coordinates by setting x(t) = Vx(t) for some constant invertible matrix V, the resulting state model in the new coordinates is given by .X (t)

=

Ax(t) + .Bu(t)

y(t) = Cx(t) where the parameter s for the original and transformed state models are related as (A, B, C)

where

v f-+

- - (A, B, C)

l

State Space Models for the Complete Response

29

Second, notice that there are physical processes which transmit the effect of the input directly to the output. In this situation the output equation for the state model has an additional term, Du(t), i.e., the output equation is

y(t) = C:dt) + Du(t) Notice that, unlike the other state model parameters (A, B, C), the D parameter is unaltered by a state coordinate transformation. Third. some dynamic processes have more than one scalar input and or more than one scalar output. For instance a dynamic process may have m scalar inputs. {ui(t): i E [I.m]} and p scalar outputs. {r,(t): i E [l,p1}. In this situation the system input, u(t), and system output, y(t), are column vectors of size m andp respectively

z?(r)

=

[u 1 (t)

u2 (t)

rT (r) =

[Y! (t)

Y2(t)

lim ( t)

]

Further, in this case, the state model has an input matrix B with m columns and an output ma!rix C with p rows. More specifically, the general form for state models used here is given as

x(t)

Ax(t)

+ Bu(t) }

y(t)

Cx(t)

+ Du(t)

(1.74)

where x(t). u(t) and y(t) are column vectors of time functions of length n,m and p respectively and the state model parameters, (A, B, C, D), are matrices of constants of size n x n, n x m, p x n, and p x m respectively. Finally, systems having p > 1 and m > 1 are referred to as multiple-input-multipleoutput (MIMO) systems and systems having p = 1 and m = 1 are referred to as singleinput-single-output (SISO) systems.

1.8.3

Zero .. state response

Recall from Section 1.3 that the zero-input response, Y:,(t), depends on the transition matrix and initial state through multiplication as Yzi(t) = CeAr x(O). In this section we will show that the zero-state response, Yzs(t), depends on the transition matrix and the input through integration as

where the integral is known as the convolution integral. We begin the development of the foregoing relation by assuming that the initial state is null, x(O) = o, and that the state model parameters. (A, B, C, D) are known. Then taking Laplace transforms throughout the state differential and output equations, ( 1. 74 ),

lntf'oduction to State Space

gives sX(s)

=

AX(s)

+ BU(s)

( 1.7 5)

Yzs(s)

=

CX(s)

+ DU(s)

(1.76)

Next solving (1.75) for X(s) yields X(s) = (sf- A)- 1 BU(s) and substituting this expression for X(s) in (1.76) yields Yzs(s) = G(s) U(s)

( 1. 77)

G(s) = C(sf- A)- 1B + D

(1.78)

where

Now, recalling from Section 1.4.3 that (sf - A) - 1 is a matrix of strictly proper rational functions, we see that Gsp(s) is strictly proper where

so that

=0

lim Gsp(s) S--->00

Thus G(s), (1.78) is given by G(s)

= Gsp(s) + D

( 1.79)

and lim G(s) = D S--->00

Notice that two state models which are related by a coordinate transforma tion have the same transfer function. We can see this by calculating the transfer function using the state model in the transformed coordinates, (A, ii, C, D) -

-

-

1-

-

G(s)=C( sl-A)- B+D and substituting A=

v- 1AV, ii = v- 1B, C = CV, I>= D to obtain G(s)

= CV[V- 1 (sf- A) vr 1 v- 1B + D = C(sf- A)- 1B + D = G(s)

State Space Models for the Complete Response

31

Next recalling the following general property of Laplace transforms

J 00

Yzs(s) = F,(s)Fz(S)¢=}-Yzs(t) =

J,(t- T)fi(T)dT

0

we see that setting

F1 (s) = C(sl- A)- 1 so that

fi (t) = CeAt

f 2 (t) = Bu(t)

gives the inverse transform of (1. 77) as

Yzs(t) =

C

1

00

eA(t-r) Bu(T)dT

+ Du(t)

Notice that when Dis null and the input is an impulse, u(t)

(1.80)

= 8(t), (1.80) gives

/

( 1.81)

In addition recalling from section 1.4.2 that

we see that Yzs(t), (1.81), has Laplace transform equal to Gsp(s) as we expect since the Laplace transform of the impulse response equals the transfer function. In addition we see from (1.81) that the zero-input response equals the impulse response when the initial state is x(O) = B. Now we need to define eAt as a null matrix when t < 0 in (1.80, 1.81). This is done to match the mathematics to the physical fact that the future input, u(T) : T > t, does not affect the output at time t, i.e., y 2 s(t) is independent of future values of the input. This property of the transition matrix, i.e., (t) = 0 for all t < 0, is referred to as the causality constraint and applies when we use the transition matrix in connection with the zero-state response. Thus the causality constraint forces the integrand in (1.80) to be null forT> t and enables (1.80) to be rewritten as

Notice that when the transition matrix is used in connection with the zero-input response we can interpret¢( -t) fort> 0 as the matrix needed to determine the initial state from the state at timet, i.e., x(O) = ¢>( -t)x(t), which implies that¢( -t) is the inverse of (t). At this point we can see, by recalling the principle of superposition, that when a system is subjected to both a non-null initial state, x{O), and a non-null input, u(t), we can write

32

Introduction to State Space

the output as y(t)

= Yz;(t)

+ Yz (t)

(1.82)

1

where Yz;(t)

= CeA 1x(O)

Yzs(t)

=Clot

eA(t-T) Bu(T)dT

+ Du(t)

Notice that: (i) y 2 ;(t), the zero-input response, is caused solely by to x(O) (ii) Yzs(t), the zero-state response, is caused solely by to u(t). In this section we have developed several essential relations involving the state model of a dynamic process. These important relations provide the complete response of a system to any specified input and any specified initial state.

1.9

Diagonal Form State Model

In subsequent chapters we will encounter a number of fundamental properties of state models in connection with their use in the design of feedback control systems. A simple way of beginning to appreciate these properties is to consider state models in coordinates where the system matrix A is diagonal. This type of state model is referred to as a diagonal or normal form state model. We encountered this model earlier in Section 1.5.2 in connection with the determination of the matrix exponential, and in Section 1. 7.I in connection with straight line state trajectories.

1.9.1

Structure

Suppose the state model for a given nth order SISO system has an A matrix which is diagonal. Then the state vector differential equation decomposes into n scalar equations

x;( t)

=

.\x;( t)

+ b;u( t)

i = I, 2, · · ·, n

(1.83)

with the output being a scalar multiple of these components 11

y(t)

=

L c;x;(t) + Du(t)

( 1.84)

i=l

We can visualize these equations as a block diagram of the sort introduced in Section 1.3 to model the zero-input response. Alternatively, when we view this state model in the frequency domain so as to obtain the plant's transfer function, we see that since the state model's system matrix, A, is

Diagonal Form State Model

Figure 1.5

33

Block diagram representation for diagonal form state model

di~onal, we have (sf- A)- 1 diagonal,

0 0 (1.85)

0

0

This fact simplifies the dependency of the transfer function on the elements in the Band C matrices since we have G(s) = C(sl- A)- 1 B + D

C·b· =~ L-'-'-+D i=1

s- .A;

(1.86)

where

1.9.2

Properties

Notice from (1.83) that the evolution of the components of the state, xk(t), are dccoupled from each other, i.e., xk(t) is independent of {x;(t): i E [O,n], i # k}. This fact plays an important role in the following discussion.

34

Introduction to State Space

We begin by noting that this decoupling of components of the state from each other implies that any component of the state, say xk(t), depends on the input through the kth component of the input matrix, bkl only. Therefore, if one of the components of B is zero, the corresponding component of the state is not affected by the input. Alternatively, notice from (1.86) that when bk = 0, the transfer function, G(s), does not have a pole at ).,k and the zero-state response does not involve the mode e>.k 1. These phenomena are related to a property of state models known as controllability. Roughly speaking, a system is said to be controllable if its state can be forced, by the input, to equal any point in state space at some finite time . Therefore if bk = 0, the evolution of the kth component of the state, xk(t), is independent of the input, u(t), and the state model is not controllable since there is no way of manipulating the input to make xk(t1 ) equal some specified value after the passage of a finite amount of time, t1 . We will see in the next chapter that controllability is a necessary and sufficient condition for being able to arbitrarily assign the closed loop system matrix eigenvalues by using state feedback. Unlike the diagonal form state model, the controllability of a state model in other coordinates, depends on both the A and B matrices. The foregoing observations were obtained by viewing the system from the input side. We can obtain dual observations by viewing the system from the output side involving the output instead of the input. We develop these observations as follows. First, since equation (1.83) decouples the evolution the components of the state from each other, we see that the output, y(t), (1.84), depends on the kth component of the state, xk(t), through ck only. Therefore, if one of the components of Cis zero, the corresponding component of the state does not affect the output. Alternatively, notice from (1.86) that when ck = 0, the transfer function, G(s), does not have a pole at )..k and the zero-state response does not involve the mode e>.kt. The foregoing phenomena are related to the property of state models known as observability. Recall that we encountered this property in the previous section in connection with the system's zero-input response. In the present instance we see that the observability of a system also relates to the system's zero-state response. Notice that unlike the diagonal form state model, the observability of a state model in other coordinates, depends on both the A and C matrices. In summary, we see from equation (1.86) that if either bk or ck is zero for some integer k E [0, n], then the system transfer function G(s) does not have a pole at Ak and the zerostate response does not involve the mode e>.kt. Moreover, under these conditions, if we transformed coordinates and recomputed the transfer function from the parameters of the state model in the new coordinates, i.e.,

we would find that the both the numerator polynomial, Cadj(s/- A)B, and denominator polynomial, det(s/- A) (or det(s/- A) since the characteristic polynomial is invariant to coordinate transformation), would have a zero at )..k· Since s- Ak would then be a common factor of Cadj(s/- A)B and det(s/- A), it would cancel from the ratio of these polynomials and be absent from the transfer function. Thus the eigenvalue of A at )..k is not a pole of G(s) and has no effect on the system's input-output. Given a transfer function, any state model that is both controllable and observable and has the same input-output behavior as the transfer function, is said to be a minimal

Diagonal Form State Model

35

realization of the transfer function. Notice that this terminology reflects the fact that lack of either controllability or observability increases the dimension of the state model over the minimal dimension needed to match the zero-state behavior of the state model with that of the transfer function. As we will see, this fact play's an important role in control theory.

1.9.3

Obtaining the diagonal form state model

The relation between the form for the transfer function, equation (1.86) and the diagonal form state model suggests a way of determining a diagonal form state model for a given SISO transfer function, G(s), having simple poles. This is done as follows. We begin by using one cycle of division when the given transfer function is proper, to separate out the strictly proper part of the transfer function. Thus suppose we are given a transfer function which is proper n

L Gp(s)

=

!

.

f3n-iS 1

i=On-1 sn + L an-isi i=O

Then dividing the denominator into the numerator enables Gp(s) to be written as the sum of a strictly proper transfer function plus a constant, i.e.,

where

i = 1,2, · · ·n .,..

Then assuming the strictly proper part, Gsp(s), can be expanded in partial fractions as n

Gsp(s) =

k·

L s_ 1

i=l

'·

( 1.87)

/\l

we can obtain the diagonal form state model by factoring the k;s as k;

.l

; ~: '

'

= C;b;

and referring to equation (1.86). Notice it is only possible to obtain a diagonal form state model if the transfer function can be expanded as equation (1.87), i.e., if the system can be decomposed into first order systems. When Gsp(s) has simple poles we can always expand Gsp(s), as equation (1.87). Notice that if Gsp(s) has gjmple poles some of which are complex, the diagonal form

Introduction to State Space

36

state model will be complex. In some situations we may want the benefits of the diagonal form state model, i.e., decoupled state differential equations, but require the state model to be real. In this case a reasonable compromis e is to combine complex conjugate pairs of terms in equation (1.87) so that the resulting realization has an A matrix which is block diagonal with each pair of complex conjugate poles correspond ing to a 2 x 2 real block on the diagonal of A. More specifically we expand Gsp (s) as

G,p(s)

=

G,pr(s)

+ G,p,(s)

(1.88)

where n,.

Gspr(s)

=

k·

L s _'A i=l

A; real 1

with the nc complex pairs of poles of Gsp(s) being the poles of Gspc(s), i.e.,

Now we can obtain a state model from (1.88) as

A=

[A1

A3

c

= [

A2] A4

cl c2]

where (A 1 ,B 1 , C 1) constitute a diagonal form state model for Gspr(s), i.e., A 1 is diagonal with diagonal elements equal to the real poles and the elements of B 1 and C 1 satisfy cub; 1 = k; fori= 1, 2, · · · n1 . Both off diagonal blocks of A, namely A 2 and A 3 are null. The remaining blocks constitute a state model for Gspc(s) as A4I

0

0

A42

0 0

0 =

B22 B2 =

A4 =

c2

B21

0

[ c21 c22

A4n,

B2n,

C2n,]

where {(A 4 ;,B2;,C2 ;): i= 1,2,···nc} are 2 dimensional real parameter state models which we obtain from { Gci : i = 1, 2, · · · nc} with

Computer Calculation of the State and Output

37

One such real parameter state model is given as A4;

= [ -~il

C2;

= [hi]

-a,.2l 0 J

i

1.89)

bi2)

Finally notice that we encountered a restricted version of the state model (1.89) in Section 1.8.1. In the next Chapter we will make extensive use of this form which is referred to as a controller form. At that time we will provide a means for its determination from a given transfer function.

1.10

Computer Calculation of the State and Output

A fundamental problem encountered in using digital computers to analyses control problems is created by the fact that the dynamic processes needing to be controlled operate in continuous time whereas digital computers perform operations in discrete time. In this section we indicate how this fundamental mismatch is overcome. In order to be able to use a digital computer to calculate the system output y(t) from (1.82) we need to employ a piecewise constant approximation of the input u( t) owr the time interval of interest, [0, tN]· This is done by partitioning the interval into N equal nonoverlaping contiguous subintervals each of time duration T. Then the piecewise constant approximation, u( t), of the input is given by u(t)

= u(t;.;)

(1.90)

where

tk = kT: k = 0, 1, 2, · · · N. The set {u(tk) : k = 0, 1, 2, · · · N} is referred to as the discrete-time equivalent of the continuous-tim e input, u(t), and the values, u(tk), of the discrete-time equivalent input are referred to as the sampled values of the input. The continuous time "staircase like" function u(t) generated by this approximation is referred to as the sample and hold equivalent of the input. Figure 1.6 shows a plot of a typical input and its sample and hold equivalent. The output, y( t), at the sample times, tk : k = 0, I, 2, · · . , N, is computed by repeating the calculation indicated in equation (1.82) for each subintenal with kT considered as the start time for the computation of y([k + 1] T). This is carried out in a cyclical fashion with the state that is computed at the end of a given subinterval being used as the initial state for the computation of the state at the end of the next subinterval. More specifically, the computations are started by specifying the initial state, x(O), and the input. u( t), over the interval of interest, [0, t\;. The computer program then decides on the sample spacing T. This is done so that the sample and hold version of the input is a "close" approximation to the actual input in the time interval [0, tN]· Then taking the

Introduction to State Space

38

u(t)

ii(t)

Figure 1.6

Typical input and its piecewise constant approximation

input as u(O) over [0, T] and using (1.82) we compute y(T) from T

x(T)

=eAT x(O)

+

j

eA(T-T) Bu(O)dT

0

y(T)

=

Cx(T)

+ Du(T)

which can be rewritten as x(T)

= Fx(O) + Gu(O)

y(T) = Cx(T)

( 1.91)

+ Du(T)

where

T

G

=

j

eA(T-T) EdT

0

Then to compute y(2T), we replace x(O) by x(T) and u(O) by u(T) in (1.91). This gives x(2T)

= Fx(T) + Gu(T)

y(2T) = Cx(2T)

+ Du(2T)

This procedure continues with the output at the (k + l)th sample time being given as

+ I)T] = Fx(kT) + Gu(kT) y[(k + I)T] = Cx[(k + l)T] + Du[(k + l)T]

x[(k

r-···

Notes and References

-~-'

39

There are several ways in which the computation ofF and G can be carried out. A simple way of doing this is to truncate the series expansion for the matrix exponential, (1.22), with the upper limit, n0 , of the sum being chosen large enough to achieve some specified accuracy. Thus F, Gin (1.91) can be expressed as

G=

(lo

0

T

no A;

i

)

L~da B+Eo i=O

l.

where the error matrices Ep and E 0 are made negligible by choosing n0 "large enough" so that we can take F and G as

/

Notice that the choice ofthe number of subintervals needed to cover the main interval, i.e., [O,tN], also plays a role in the accuracy of the computation, with an increase in accuracy being achieved at the expense of more computation time i.e., the accuracy goes up the smaller T and the larger N.

1.11

Notes and References

There have been a number of excellent textbooks expounding on state space analysis of systems. Some of these books are [5], [4], [23], [6], [9], [15]. Throughout the eighties the book by Kailath [23] was used widely as a text for courses on state space analysis of linear systems. Although Brockett's book [5], is quite old and was . written for a more mathematically sophisticated reader, the first half of the book does give the reader a nice exposure to the more general nature of the subject. The treatment of state modeling in the brief book by Blackman [4], shares the view with the present book that understanding basic ideas in the state space can be facilitated by thinking geometrically. Concerning books that provide insight into matters pertaining to linear algebra, Strange's book [39] gives an excellent introduction. The book by Golub and Van Loan [17] provides a more advanced treatment which is more orientated towards computer computation. In addition the book by Brogan [6], plus the appendix to Kailath's book also provide useful information on linear algebra.

2 State Feedba ck and Controllability

2.1

Introduction

An important and well known consequence of feedback is behavioral modification.

F~edback is present when the input to a system depends on that system's output. When feedback is used to modify a system's behavior the resulting closed loop system is referred to as a feedback control system. For the purposes of discussion we can separate the control system into a plant and a controller. The plant consists of the physical process to be controlled plus any transducers, actuators or sensors, needed to interface the physical process output to the controller and the output of the controller to the input of the physical process. While controllers were implemented in the past by a physical process such as an electric filter, their implementation now is often done by a digital computer. Figure 2.1 shows the general setup for a control system. Ut(t)

Yt(t)

-·

PLANT

u2(t)

Y:(t)

CONTROLLER

Figure 2.1

Block diagram of the feedback control scheme

external input controlled input

desired output measured output

where u 1 (t), u2 (t),y 1(t),y2 (t) have dimensions mh mz,PI ,p2 respectively.

State Feedba ck and Controllability

42

The general form of the state model for the plant is ,"\:(t) = Ax(t) + B 1 (t)u 1 (t) + B2 u2 (t) } YI(t) = C1x(t) + D 11 u1(t) + D 12 u 2 (t) Y2(t) = C2x(t) + D21u 1(t) + D 22 u2 (t)

(2, 1)

the This chapter focuses on the effect of using a feedback signal compos ed from compon ents of the plant's state as the controlled input, (2.2) input. This where K has m 2 rows, one row for each scalar signal making up the controlled nts ingredie two of one is and k feedbac state as to referred is importa nt kind of feedback er. controll the of needed to complete the design it is The need for something in addition to state feedback can be appreciated when state use to are we if re Therefo n. unknow usually is recognized that the plant state ng the feedback as a basis for the design of a controller, we must have a means for estimati is which r, estimato state a as to referred model, state plant state. This is done by a state plant the on tion informa The model. state plant's designed from knowledge of the signals is containe d in measurements of all or some of the plant's input and output measurethese to equal input r's estimato the supplied to the state estimato r by setting There plant. the of state the mimics state its that so ments. The state estimato r is designed about made ions assumpt the on ng dependi r are several different types of state estimato exact, e.g., ments, measure their and signals the relation between the actual plant chapters ent subsequ In n. unknow partly or corrupte d by random measurement noise, ance of the we will study both the design of these estimators and the effect on the perform feedback. state with tion conjunc in use their by feedback control system that is caused so that we known is model state plant's the of In this chapter we assume that the state in using goal Our (2.2). equation by given can use it to generate the state feedback signal input zero and state zero the which in system state feedback is to produce a closed loop which to extent The use. plant's the to al behavior is changed in some way which is benefici a system's the plant's controlled input can affect the plant state, is referred to as es determin model state plant the of ability controllability. We will see that the controll plant's the from differ can system loop the extent to which the state model of the closed model state model when we use state feedback. The importa nt property of state chapter this out through guises t differen controllability will be encountered in several and the rest of the text.

2.2

State Feedback

state model Applying state feedback, equation (2.2), to the plant, equation (2.1) gives the for the feedback system as x(t) y 1 (t) Y2(t)

= = =

(A+ B 2 K)x(t) + B 1 (t)uJ (t) } (C 1 +D 12 K)x(t) +D 11 u 1(t) (C2 + D 22 K)x(t) + D21u1 (t)

Figure 2.2 gives a block diagram interpre tation of state feedback.

(2.3)

State Feedback

43

YI(t)

u1(t)

PLANT u2(t)

x(t)

L__

Figure 2.2

CONTROLLER

1.-----

Block diagram representation of state feedback.

Notice that the main effect of using state feedback is the transformation of the system matrix from A, when the plant stands alone, to A + B 2 K, when the plant is embedded in a state feedback loop. Thus we see that the main purpose of state feedback is the assignment of eigenvalues to the closed loop system matrix A + B 2K to achieve a set of specified performance criteria for the behavior of the closed loop system from the external input u1 (t) to the desired output y 1 (t). This fact leads immediately to the question: can we choose K so that the eigenvalues of A+ B2K are assigned to any specified set of values? The answer to this question is obviously of great importance to the use of state feedback in the design of feedback control systems. The answer to this question, which is given in the following theorem, involves the use of the left-eigenvectors of A. Since this property applies in general to any state model (A, B, C, D) we will drop the subscript on B, i.e., B2 will be replaced by B. Theorem 2.1 Whenever A has a left-eigenvector w;, i.e., wiT A= >.;wiT, such that

the corresponding eigenvalue of A, ,X;, is invariant to state feedback, i.e., >.; E >.(A + BK) for all K. Proof Suppose the condition in the theorem is satisfied. Then multiplying the closed loop state feedback system matrix, A+ BK, on the left by wiT gives

wiT(A +BK) =wiT A +wiTBK However since wiT A= >.;wiT and wiT BK = 121, the previous equation becomes

and we see that >.1 is an eigenvalue of the system matrix for the state model of the feedback control system, since>.; E >.(A+ BK) for all K. • It turns out that when there are no left-eigenvectors of A which satisfy B = 0, the eigenvalues of A are all assignable to any specified values by state feedback. This is shown later in Theorem 2.4. Notice that when we have the condition B = 0 so that the corresponding eigenvalue >.1 cannot be assigned by state feedback, we refer to >.1 as an

wT

wT

State Feedb ack and Controllability

44

ion wf B 1- 0 we refer to uncontrollable eigenvalue. Conversely, when we have the condit state feedback, as a conthe corresponding eigenvalue A; which can be assigned by trollable eigenvalue. llable (uncontrollable) Definition 2.1 An eigenvalue >..; of A is said to be a contro A;~ >.[A+ BK] eigenvalue of the pair (A, B) if we can (cannot) find K so that invari ant to changing Uncon trollab le (controllable) eigenvalues for a pair (A, B) are coordinates. This can be seen as follows. have Suppose A; is an uncontrollable eigenvalue of (A, B). Then we (2.4) (2.5)

n wiT and A, in equation Now insert rr-l between wiT and Bin equati on (2.4) and betwee by T. Then we obtain (2.5). In additio n, postm ultiply both sides of equati on (2.5)

where

of the state model in the Thus we see that A; remains an uncontrollable eigenvalue transfo rmed coordinates. have no eigenvalues Reflection on the foregoing reveals that plant state models which ty that state feedback of A which are both uncontrollable and unstable have the proper having this proper ty are can be used to make the closed loop system stable. State models is not stabilizable when referred to as stabilizable state models. Conversely, a state model impor tant proper ty since it has uncon trollab le eigenvalues which are unstable. This is an condit ion for a control the stability of a contro l system is almost always a necessary system to be of arty use. if all its uncontrollable Definition 2.2 A plant state model is said to be stabilizable eigenvalues are stable. feedback control Recall that the poles of the transfer function for an SISO output the present case of state system are constr ained to lie on branches of the root locus. In matrix A + BK are system loop closed the of feedback, the locatio n of the eigenvalues nal flexibility additio This llable. contro are they that , uncon straine d provided, of course one adjustable param eter of state feedback over output feedback arises from there being ck, whereas there is only in K for each compo nent of the state when we use state feedba ck. one adjustable param eter when we use consta nt output feedba

2.3

Eigenvalue Assignment

has the same eigenvalues Recall, Section 1.6.4, that a state model's system matrix . In addition, notice that independent of the coordi nates used to express the state model

Eigenvalue Assignment

45

if the state, x(t), in the original coordinates i& related to the state, x(t), in the new coordinates as

x(t) = Tx(t) then the state feedback matrix is transformed as

K = KT since

u(t) = Kx(t) = KTx(t) This fact provides us with the flexibility of being able to carry out the determination of the state feedback matrix in coordinates which are the most suitable for this computation. We saw in the previous section, that we can choose K to achieve a specified set of closed loop eigenvalues, {JL; : i = 1, 2, · · ·, n }, provided that each uncontrollable eigenvalue of the pair (A, B) equals one of the desired closed loop eigenvalues. Assuming this condition is satisfied, the state feedback matrix, K, which is needed to make the set of closed loop eigenvalues equal the specified set of eigenvalues can be obtained by equating the characteristic polynomial for A+ BK, denoted a(s), to the characteristic polynomial 1(s) having roots equal to the specified set of eigenvalues, {JL; : i = 1, 2, · · · n}, as /

a(s)

=

')'(s)

(2.6)

where

()= II( n

/

S

S -

JL;

)=

S

n

+ 'YI S n-I + 'Y2S n-2 + · · · + /n

i=l

Notice that the coefficients of a(s) are functions of the elements of K. Thus we see that equation (2.6) gives rise to n equations in the unknown elements of K, i.e., a;=/;

i = 1,2, · · ·n

In general, these equations in the elements of K are coupled in ways which depend on the A and B matrices. Therefore it becomes difficult to set up a computer program to form and solve these equations for all possible (A, B) pairs. However if the plant's state model is in controller form these equations are uncoupled and therefore easy to solve.

2.3.1

Eigenvalue assignment via the controller form

A state model for an nth order SISO system is said to be in controller form when A = Ac is a companion matrix and B = Be is the first column of the n x n identity matrix. More

46

State Feedback and Controllability

specifically the controller form state model has parameters ~a I

Ac =

~a2

~an-I

~an

0

0

0

0

0

0 0

0 B

~

c~

0

0

0

0

with D and C being whatever is needed to model the plant's behavior in these coordinates. Use of the controller form state model greatly simplifies the calculation of the state feedback matrix Kc needed to make Ac + BcKc have a specified set of eigenvalues. This results from the fact that in these coordinates the state feedback system matrix, Ac + BcKn is also in companion form ~a 1

+ kc1

~a2

+ kc2 0

Ac +BcKc =

0 0

0

~a3

+ k3

~an-I+ kn-1

~an+

0

0

0

0

0

0

kn

0

0

with the kc 1s being the elements of the state feedback matrix

Recall from Chapter 1 that since Ac + BcKc is a companion matrix, the coefficients of its characteristi c polynomial appear, with negative signs, along its first row. Therefore the coefficients of the characteristi c polynomial for the closed loop system o:(s), equation (2.6), are related to the elements of the state feedback matrix K as

i = 1, 2, · · · n

(2.7)

Then if we want the closed loop system matrix, Ac + BcKn to have eigenvalues at {p,1 : i = 1, 2, · · ·, n }, the o: 1s are obtained from

and the elements {kci : i from equation (2. 7) as

=

1, 2, · · · , n} of the feedback matrix K are easily determined

i

=

1, 2, · · · n

(2.8)

Before showing how to transform coordinates so that a given state model is transformed to controller form, we consider the problem of determining a controller form state model from a given transfer function.

Eigenvalue Assignment

2.3.2

47

Realizing the co-ntroller form

Having seen that the assignment of eigenvalues is easy to do when the system state model is in controller form, we turn now to the problem of getting a controller form state model for a system specified by a transfer function or differential equation. Theorem 2.2 Suppose we are given the a;, b; parameters in the differential equation or transfer function model of some plant, i.e., n-1

n-1

Y(n)(t) + I:>n-iY(i)(t )

=

Lbn-iU(i)(t ) i=O

or

(2.9)

where

Y(s) = Gsp(s)U(s)

= 0 for all i = 0, 1,2,· ·: ,n- 1 Then the controller form state model for this plant has parameters (Ac, Be, Cc) given by

with y
-a I

Ac

Cc

=

=[

-a2

-a3

-an-I

-an

0

0

0

0

0

0

0

0

0 Bc-

0

0

0

0

0

0

bl

b2

b3

bn ]

(2.10)

Proof Factor the transfer function as Gsp(s) = N(s)D(s) where n-1

N(s) = Lbn_;Si i=O

D(s) =

--n--1--

sn +

L: an-is;

i=O

Then we see that

Y(s)

=

N(s)Z(s)

(2.11)

Z(s)

=

D(s)U(s)

(2.12)

JB

State Feedback and Controllability

where Z(s) is the Laplace transform of the signal z(t), i.e. z(t) =

.c- 1 [Z(s)]

Since (2.12) gives the zero state response, z(t), of a system having transfer function D(s) and having u(t) as input, we see that (2.12) implies that n-1

z(n)(t)

+ L.>n-;Z(i)(t)

u(t)

=

(2.13)

i=O

and setting i

=

1, 2, · · ·, n

(2.14)

in (2.13) gives

where the matrices Ae, Be are as stated in the theorem. Finally, since the initial conditions are all assumed zero, we see from (2.11) that n-1

y(t) = l::)n-iz(il (t) i=O

and using the relation between the componen ts of the state and the derivatives of z(t), (2.14), we see that the foregoing expression for the output can be written as

• where the matrix Ce is as stated in the theorem. Notice that since Gsp(s) is strictly proper, De = 0. Alternatively when the given transfer function, Gp(s), is proper, we can proceed as in Section 1.9.3 to obtain its controller form (Ac, Be, Cc, De) with Ae, Be, Cc as in Theorem 2.2 and D = {30 where n

2:: f3n-i Si GP (s) =

0- - -''-.=-'--

n-1

sn

+ "'a 0

n-1

si

i=O

with i

= 1, 2, · · · n

In summary, Theorem 2.2 shows that the controller form state model is easily obtained when we are given either the transfer function or the differential equation governing the process behavior. In some cases we may be given a state model for an SISO plant which is in a form other than the controller form. In this situation it may be possible to simplify the determina tion of the state feedback matrix K by:

Eigenvalu e Assignme nt

49

(i) determinin g the coordinate transforma tion matrix, Tc, to transform the given plant state model to controller form (ii) choosing Kc using (2.8) to assign the eigenvalues to the controller form state model (iii) using the coordinate transforma tion matrix Te to obtain the require state feedback matrix asK= KeT-; 1 . This approach, while not of practical significance when applied as just described, does enable the developme nt of a practical computati onal algorithm for determinin g K. This algorithm which is referred to as Ackerman n's formula will be developed in Section 2.3.4. At this point we need to develop both a method for determining Te and conditions on the given state model which enables its transforma tion to controller form.

2.3.3

Control ler form state transformation

Suppose we want to find a coordinate transforma tion matrix, Te, which transforms a given state model, (A, B, C), not in controller form to a state model, (A, ii, C), which is in controller form. Then assuming, for the time being, that such a transform ation exists, we can solve this problem by proceeding as follows. Since, in general A, and A= T- 1AT have the same characteris tic polynomia l, we can use the coefficients of the characteris tic polynomia l for the given A matrix, det[s/ - A]. to construct the controller form system matrix A = Ae from Theorem 2.2. In addition, since Be is the first column of the n x n identity matrix we have ii = Be without doing any calculation. Therefore neither Ae nor Be requires the determina tion of Tc. However Tc is needed to determine C = CTc = C,. In what follows we use Ac- Be and their relation to . Te to develop an algorithm for determinin g Tc, one column at a time. To begin with, notice that the first column of Tc can be obtained immediate ly from inspection of the relation Tcii = B when T e is written in terms of its columns.

r

0

where ti is the irh column of Tc. Having determined the first column of Tc as B, we can see how to determine other columns of Tc by inspection of the relation TeA= ATe with A set equal to Ac and Te expressed in terms of its columns,

0

0

0

-an-1

-an

0

0

0

0

0

0

0

0

=

A[t 1

t"

t"

l

:Ju

;,(ale

l"eeaoacle and Controllabili ty

Thus equating like positioned columns on either side of this equation gives -a 1 t 1 +t 2 =At 1 -a2 t 1 + t 3 = At 2 -ant 1 +

)

(2.15)

t~ = Atn-!

(2.16) and we see that the remaining columns of Tc can be generated successively from (2.15) by starting with t 1 = B and proceeding as t 2 =a 1t 1 +At 1 t 3 = a2 t 1 + At 2

)

(2.17) tn

= an_ 1t 1 + Atn-l

To recap, we can generate the columns of Tc by (i) obtaining the coefficients of the characteristic polynomial for the given A (ii) setting the first column of Tc equal to the given B. (iii) using (2.17) to determine the remaining columns of Tc in succession. Notice that so far we have no assurance that the matrix Tc which results from using (2.17) will be invertible. However T c must be invertible if it is to be a coordinate transformati on matrix. We show now that only those state models which have the property of controllability will produce a nonsingular Tc matrix as a result of applying the foregoing algorithm. Before doing this notice that we did not need (2.16) to get (2.17). This extra equation will be used in Section 2.3 .5 to derive the Cayley-Ham ilton theorem.

2.3.4

Condition for controlle r form equivale nce

In order to determine conditions on the given state model which guarantees that it can be transformed to a controller form, i.e., to insure that the matrix Tc which is generated by (2.17) is invertible, we expand (2.17) by successive substitution to obtain

(2.18)

Eigenvalue Assignment

51

Then writing these equations in matrix form gives (2.19)

where

fl= [B

AB

A 2B

...

An-Is'

R=

al

az

an-I

a!

an-2 an-3

0

1

0

0

0

0

0

Now since the product of two square matrices is invertible if and only if each matrix in the product is im ertible. Tc is imertible only if both \2 and R are invertible. However since R is an upper-triangula r matrix having nonzero elements along its diagonal, R is invertible for any given state model. Therefore T, is invertible if and only iffl is invertible. The matrix n, which depends on the interaction between the given state model's A and B matrices, is referred to as the given state model's controllability matrix. For systems having only one input the controllability matrix is square and the state model is said to be controllable if its controllability matrix is invertible. However when the input consists of m scalar inputs, n. has nm columns and only n rows. In this case the state model is said to be controllable if its controllability matrix is full rank, i.e., all its rows are independent or equivalently, n of its nm columns are independent. More generally. the property of controllability of a given state model is preserved under coordinate transformation. We can see this by noting that the controllability matrix n of the state model in the transformed coordinates is related to the controllability matrix n of the state model in the original coordinates as

n = [JJ

.4.e ...

= [T- 1B

=

;p-Iel

T- 1ATT- 1B

···

T- 1An-IB]

r- 1n

(2.20)

Therefore since D has fewer rows than columns we have

for some a -1 0 only iffl has dependent rows, since ,BT = implies that

aT T- 1

= 0T only if a

= 0.

This

rank[n] = rank[T- 1D] = rank[D] where the rank of a matrix equals the maximum number of its columns which are independen1. However since, for any matrix, the number of independent columns equals the number of independent rows. we see that, in general rank[D] .::::; n

52

State Feedback and Controllability

since n is ann x nm matrix for any n dimension state model having m scalar inputs. This fact will be used later in the fourth section of this chapter. When the state model is single input and controllable, m = 1 and n, 0 in (2.20) are square and invertible. Therefore we can obtain the coordinate transformation matrix, T, from (2.20) as

r

=

nrz-l

(2.21)

Then comparing this expression with (2.19) we see that rz-l = rl;:- 1= R. Moreover the controllability matrix for a state model in controller form is readily calculated as

A~- I Be]

nc = [Be AcBc -a I

0

2 al-

a2

-a I

0

0

0

0

* * *

(2.22)

0

which, being upper-triangular with fixed nonzero entries along the diagonal, is invertible. Thus all controller form state models are controllable. To recap, a given single input state model can be transformed through an appropriate coordinate transformation to a controller form state model if and only if the given system is controllable. Moreover we have just seen that an uncontrollable state model cannot be made controllable by a coordinate transformation and that any controller form state model is controllable. In addition, we will see that there are several other aspects of system controllability which play important roles in control theory. A more basic view of system controllability is developed in the fourth section of this chapter.

2.3.5

Ackermann's formula

As mentioned at the end of Section 2.3.1, we could calculate K to assign the eigenvalues of + BK, for a given single input controllable system by transforming the given system to controller form , finding Kc, and transforming Kc back into the original coordinates through the relation K = KcT;:- 1• We will see now that we can use this idea to develop a computational algorithm for obtaining K which is known as Ackermann's formula. In order to do this we will need the following result. Theorem 2.3 (Cayley-Hamilton) Any square matrix A satisfies its own characteristic equation. Before proving this theorem, the following explanation may be needed. Recall that the characteristic equation for A, A

(2.23) is a scalar equation which is satisfied by the eigenvalues of A.

53

Eigenvalue Assignment

Therefore Theorem 2.3 tells us that if we replace the scalars in (2.23) by the matrix A, we obtain the result

We can readily show the Cayley-Hamilton theorem as follows. Proof Premultiplying (2.18) by A and using (2.16) yields (An+ a 1 An-I

+ a2An- 2 + · · · anl)B = 0

(2.24)

Now (2.24) holds for any B. Therefore letting B = I in the foregoing equation yields An+ a 1An-I

+ a2An- 2 +···ani= 0

•

Theorem 2.4 (Ackermann' Formula) Given a single input controllable state model (A, B, C, D) and the desired characteristic polynomial, a(s), for the closed loop system

a(s)

=

det[si- (A+ BK)]

then the required feedback matrix K is given by K

= -qT[a(A)]

where qT is the last row of the inverse of the controllability matrix, 0, for the given pair (A, B). Proof For convenience, and without loss of generality, suppose the given plant is of dimension n = 3. Then the plant state model's system matrix A has characteristic polynomial

a(s) = det[si- A] = s 3 + a 1s 2 + a2 s + a3 Let the desired characteristic polynomial, a(s), for the state feedback system matrix be denoted as

Then using the coordinate transformation matrix Tc developed in Section 2.3.3 we transform the given state model for the plant to controller form

so that the state feedback system matrix in these coordinates is given by

State Feedback and Controllability

54

where

Therefore the desired characteristic polynomial can be written by inspection from the companion form for Ac + BcKc as

a(s) = s3 + (al- kc1)i

+ (a2- kcl)s + (a3- kc3)

= a(s)- (kc1i + kc2s + kc3)

(2.25)

Next recalling that A and T- 1AT have the same characteristic polynomial for any invertible T, we see that a(s) is the characteristic polynomial for both A and A c. Therefore from the Cayley-Hamilton theorem (Theorem 2.3) we have

a(A) = a(Ac) = 0

(2.26)

Therefore we see from (2.25, 2.26) that

(2.27) where I is the 3 x 3 identity matrix

However we can readily verify that

Therefore we see that pre-multiplying (2.27) by the third row of the identity matrix gives

-3T a (A c ).

l

= -

k ell·IT

-

k c2l-2T

-

k c3l·3T

(2.28)

Then inserting the identity matrix T;; 1 Tc between i 3 T and a(Ac) on the left side of (2.28) and post-multiplying throughout by r;; 1 yields (2.29) Moreover using the facts that

(2.30)

Controllability

and that KcT-;: 1

55

= K we see that (2.29) can be rewritten as i 3TT;: 1a(A)

= -K

(2.31)

Finally recalling (2.19) we see that

~

"' R-•n-•

~ [~ ~

}-•

.(2.32)

where

and* are elements of R- 1 which are not needed. Thus using (2.32) in (2.31) gives

and the theorem is proved. • Notice that since only the last row of n- 1 i.e., qT, is required when we use this result to compute the feedback matrix, K, we need only compute q which satisfies

with in being the last column of the n x n identity matrix. This avoids having to do the more intensive computation of n- 1 .

2.4

Controllability

So far we have encountered the effects of system controllability twice, once in connection with the condition for eigenvalue assignment by state feedback, (Theorem 2.1 ), and again in connection with the existence of a coordinate transformation matrix which transforms a given state model to controller form, (Section 2.3.4). In this section we encounter system controllability in the context of the basic problem of the input's ability to manipulate the state of a given state model. The following definition of a state model's controllability is made with this problem in mind. Definition 2.3 A state model (A, B, C, D), or pair (A, B), is said to be controllable iffor every possible initial state, x(O), we can find at least one input u(t), t E [0, t1 ] and some finite final time t1 < oo so that x(t1 ) = !ll, i.e., so that the state is annihilated by the input in a finite time. . It is important to notice in the foregoing definition, that t1 is required to be finite. This IS done to prevent all stable systems from being considered to be controllable. More specifically, since stable systems have the property that lim x(t) = !ll

1--+oo

for any x(O) when u(t) is null

56

State Feedback and Controllability

all stable systems would be controllable if the final time, tr, in Definition 2.3 were allowed to be infinite. · In order to develop the implications of the foregoing definition we need to recall, from the first chapter, that the state x( tr) which results from having an initial state, x(O), and an input, u( t), is given by

J It

x(tr)

+

= eA 11 x(O)

eA(tr-T) Bu(T)dT

(2.33)

0

Then if the state is annihilated at time t1 , i.e., if x(tr) u(t), must be chosen so that

=

0, we see from (2.33) that the input,

J It

eA 11 x(O)

=-

eA(It-Tl Bu(T)dT

(2.34)

0

However from the series expansion for

eAt,

(Section 1.4.1) we see that

and so that we can simplify (2.34) as

J It

x(O)

=-

e-AT Bu(T)dT

(2.35)

0

This equation is the basic constraint which must be satisfied by any input which drives the system state from x(O) to the origin in state space in finite time, t1 < oo. Notice that the following three questions are immediately evident: 1. For each initial state, x(O), is there at least one input, u(t), which satisfies (2.35)? 2. If it is not possible to satisfy (2.35) by some input for each initial state, how should the initial state be restricted to enable (2.35) to be satisfied by some input u(t)? 3. Iffor a specific initial state, x(O), it is possible to satisfy (2.35), what is the specification of the input u(t) which does so? Notice that when the answer to the first question is in the affirmative the system is controllable in the sense of Definition 2.3.

2.4.1

Controllable subspace

We will show now that a criterion for a given state model or pair (A, B) to be controllable in the sense of Definition 2.3 is that the controllability matrix be full rank, i.e., rank[S1] = n, when the state model is n dimensional where

Controllability

57

Notice that when rank[fl] = n, n has n independent columns so that we can always find a constant vector, 1, for any initial state, x(O), such that x(O) = !:11

(2.36)

However, if rank[OJ < n, then n has fewer than n independent columns, and it is not possible to find 1 to satisfy equation (2.36) for some x(O). However, those x(O) for which we can find 1 to satisfy equation (2.36) are said to lie in the range or image ofn, denoted as x(O) E range[fl] We we will show, later in this section, that those initial states, x(O), satisfying x(O) E range[fl], are those initial states for which we can find u(t) to satisfy equation (2.35). This fact answers question 2 and leads to the following definition. Definition 2.4 Initial states, x(O), for which we can (cannot) find u(t) to satisfy equation (2.35) are said to lie in the controllable subspace Sc (uncontrollable subspace Sc) of the state space. Thus we see from the answer just given to question 2 that Sc = range[fl]

An important property of the controllable subspace is given in the following theorem. Theorem 2.5 If x(O) Erange[n] then Ax(O) Erange[fl]. Proof Suppose x(O) E range[fl]. Then we can find a constant vector, 1, to satisfy (2.37) where the I;S are constant vectors having the same length as u( t). However, we know from the Cayley-Hamilton theorem, (Theorem 2.3), that n

An

= -

L a;An-i i=l

where the a;s are coefficients of the characteristic polynomial

Therefore multiplying (2.37) by A and using the Cayley-Hamilton theorem gives Ax(O) = AB11 +A 2 B12 + · · ·An-lBin-l

+

(t

-a;An-iB )In

=fl')'

(2.38)

where ···

ln-2 - lna2

In- I

-

lnaJ]

State Feedback and Controllability

58

and we see from (2.38) that Ax(O) E range[n]

•

Notice that this theorem implies that if x(O) E range[n], then Akx(O) E range[n] for all integers k. This fact enables us to show that rank[D] = n is a necessary and sufficient condition for S, to be the entire state space so that there is an input satisfying (2.35) for all initial states. Thus rank[D] = n is a necessary and sufficient condition for a state model to be controllable in the sense of Definition 2.3. This fact is shown now in the following theorem. Theorem 2.6 If a state model (A, B, C, D) or pair (A, B) is controllable in the sense of Definition 2.3 then rank[D]

=

n

where

Proof Suppose (A, B) is controllable in the sense of Definition 2.3. Then substituting the series expansion for eAT in (2.35) yields

= Bro + AB11 +A 2B12 + · · · =

n,o +Ann/! + A2nn/2 + ...

(2.39)

where I

kT

=

[

T rkn

T rkn+l

k

= 0, 1' 2 ...

with i=0,1,2···

However from Theorem 2.5 we have k

=

0, 1, ...

(2.40)

Therefore the columns on the right side of (2.39) span the entire state space if and only if rank[D] = n. Otherwise when rank[D] < n there are initial states x(O) ¢:. range[D] so that (2.39) is impossible to satisfy by any u( t) and the state model is not controllable in the sense of Definition 2.3. This implies that rank[O] = n is necessary for the state model to be • controllable in the sense of Definition 2.3.

59

Controllability

2.4.2

Input synthesis for state annihilation

The following theorem shows that rank[n] = n is sufficient for a state model (A, B, CD) or pair (A, B) to be controllable in the sense of Definition 2.3. Theorem 2.7 Ifrank[n] = n then we can satisfy (2.35) for any x(O) by choosing u(t) as

where

Proof Suppose W is invertible. Then substituting the expression for u(t) given in the

theorem into the right side of (2.35) gives

tf e-ArBu(T)dT= ~ }otf e-Ars [-BTe-A T]r dTW- 1x(O) - }o /

=

ww- 1x(O)

=

x(O)

which is the left side of (2.35). Thus we have shown that the state model is controllable in the sense of Definition 2.3 when W is invertible. In order to show that W is invertible, suppose rank[n] = n and W is not invertible. Then we can find a =f. 0 such that (2.41) Now since the integrand in the definition of W is quadratic, i.e.,

where

we can only satisfy (2.41) if for all '

T

E [0, tf]

(2.42)

However using the power series expansion of the matrix exponential we see that

60

State Feedback and Controllability

where kT _

{3

-

l"

(-T )kn

(kn)!

( -T )kn+l

(kn

+ 1) !

(-T)(k+l)n-1

]

([k + l]n- 1)!

Thus (2.42) is satisfied only if (2.43)

However since rank[r!] = nand n is n x nm, we see that n has independent rows and only a= 0 satisfies (2.43) and (2.41). This proves that W is invertible when rank[r!] = n. • In the next section we will see that when rank[r!] < n we can still use the foregoing approach to provide a method for determining an input which annihilates an initial state x(O) in a finite time provided x(O) Erange[r!]. This is achieved by using a controllable decomposed form for the system's state model so that the controllable part of the system is immediately evident.

2.5

Controllable Decomposed Form

So far we have considered two special types (canonical forms) of state model, the controller form and the normal or diagonal form. In this section we introduce another form for the state model of an n dimensional system. The purpose of this form of state model is to simplify the specification of the controllable and uncontrollable subspaces, denoted as Sc and S 2 respectively, by aligning them with subsets of the coordinate axis. More specifically, when a state model is in controllable decomposed form the state is given by

where x;(t) is of length n; and

x(t) ESc

if x 2 (t)

x(t) ESc

if x 1 (t) =0

= 0

Then any state, x(t) can be decomposed as (2.44)

where xc(t) ESc and xc(t) E S 2 with ac, ac being scalars. Notice that Scl_Sc A state model whose state space is orthogonally decomposed in the foregoing manner is referred to as being in controllable decomposed form. The state model for a system to be in this form is defined as follows. Definition 2.5 A state model having m scalar inputs is said to be in controllable

61

Controllable Decompose d Form

decomposed form when its A and B matrices have the following block matrix forms

B=[~]

A=[~

(2.45)

where (All B!) is a controllable pair; All A 2 , A 4 and B 1 have dimensions: n 1 x n 1 , n 1 x n2 , n 2 x n 2 , n 1 x m respectively with n = n 1 + n2 being the dimension of the state space.

2.5.1

Input control of the controllable subspace

We begin the discussion of the effect of an input on the state when the state model is in controllable decomposed form by noticing that the product of upper-triang ular matrices is upper-triang ular, and that the coefficients of the series expansion of eAt involve powers of A. Therefore the transition matrix, eAt, for a state model in controllable decomposed form, (2.45), is upper-triang ular, At_

~

e

l

f[A~ * ~ i=O

0

A~

il

where the block marked * is not of importance here. Next recall, from the previous section, that if u(t) drives the state from x(O) to the origin in finite time, t1 , then (2.35) is satisfied. Therefore since the eAt is upper-triang ular and B has its last n2 rows null when the state model is in controllable decomposed form, we see that (2.35) can be written as

x'(O)l [~(0)

=-

lot (oo2:.)-l)i [A;Bl. ~Ii) U(T)dT 1

I

i=O

0

0

I

(2.46)

l.

or

x 2 (0)

=

0

Thus we see that (2.46) can not be satisfied by any input u(t) ifx2 (0) f= 0. Alternatively , 1 since (A 1 , B!) is a controllable pair, we can always find u(t) to force x (tr) = 0 for any 1 x (0) and ~(0). More specifically, when the state model is in controllable decomposed form, (2.45), we see that

so that

State Feedbac k and Controllability

62

Therefore, proceeding in the&ame manner as was done to obtain (2.35) we see that if x 1 (tr) is null then we must choose u( t) to satisfy the following equation (2.47) where

Then using the result given in Theorem 2.7 we see that one input which satisfies (2.47) is

where

To recap, we have just shown that we can always find an input to annihilate the projection of the initial state onto the controllable subspace. More specifically since any initial state can be written as (2.44) with t = 0, we can find u(t) so that xc(tf) = 0 for any tf

< oo.

2.5.2

Plelation to the transfer function

Concerning the input-ou tput behavior, suppose the state model has A,B as specified by (2.45) and C partition ed as

with ci beingp X ni, i = 1' 2 where pis the number of scalar outputs. Then from the block upper-triangular form for A, (2.45), we see that (sf- AJ)- 1A 2 (sl- A4)- 1 ] (sf- A 4 )- 1

Therefore the system transfer function is given as G(s)

=

C(sl- A)- 1B

(I _- C IS

-

C 1adj[s/- AI]B1 A )-IEI _ I det[s/- AI]

(2.48)

Notice that the order of the transfer function is no more than n 1 • This reduction in the order of the transfer function results from the eigenvalues of A4 being uncontrollable. We

tion in the numb er of eigenvalues of A that will see in the next chap ter that a furth er reduc funct ion may occu r as a resul t of the go over as poles of the corre spon ding trans fer order to bette r appre ciate the fact that the inter actio n of the state mode l's (A, C) pair. In the following. eigenvalues of A 4 are unco ntrol lable , consi der

2.5. 3

Eig env alue s and eige nve ctor s of A

ectors of A when the state mode l is given Supp ose we attem pt to deter mine the left-eigenv to solve in contr ollab le deco mpos ed form. Then we need WT A= AWT

struc ture of A, (2.45), we see that the for the pair {.:\, w}. How ever from the block , foregoing equa tion expa nds to two equa tions (2.49) (2.50) where

ied when we let w1 = with w; being oflen gth n;. Then (2.49) is satisf

o and (2.50) becomes (2.51)

2 1 .:\ is an eigenvalue of A 4 • w is the Ther efore for this choice of w , we see that = [ 0 wf} is the corre spon ding leftcorre spon ding left-eigenvector of A 4 , and wT s of A4 are eigenvalues of A. Notic e that eigenvector of A. Thus we see that eigenvalue in this case we have

H'T B

= [0 w

n [:

1

l

=0

s of A 4 are unco ntrol lable . so that eigenvalues of A which are also eigenvalue values of A 1 . We can see this by notin g Now the rema ining eigenvalues of A are eigen of A 1 and w1 is the corre spon ding leftfrom (2.49) that if w 1 i- 0 then .:\ is an eigenvalue mpos ed form the pair (AJ>B!) is contr oleigenvector of A 1 . Since in contr ollab le deco s of A I are contr ollab le. More over since lable, eigenvalues of A whic h are also eigenvalue s of any pair (A, B) are disjo int \Ve have the contr ollab le and unco ntrol lable eigenvalue 1 is invertible. Ther efore when w is a leftthat if.:\ E .:\[Ad then .:\ ¢:. .A(A 4 ] and .AI- A 4 2 red w 2 to make wT = [ w 1T w T J a lefteigenvector of A I, we can deter mine the requi eigenvector of A from (2.50) as w2T = w!T A2(A I-

A4)-1

A is imme diate ly evide nt when one The foreg oing prop erty of the eigenvalues of state feedback when the state mode l is in attem pts to use state feedback. Thus if we use

64

State Feedback and Controllability

uncontrollable decomposed fDrm, the closed loop system matrix is

where the state feedback matrix is partitioned as (2.52)

with K 1 , K 2 being m x n 1 and m x n2 . Now we just saw that block upper-triangular matrices have the property that their eigenvalues equal the union of the eigenvalues of their diagonal blocks. Thus the eigenvalues of the state feedback system matrix satisfies

and A+ BK has a subset of eigenvalues, .A[A 4 ], which is clearly unaffected by state feedback. However since (A 1B 1 ) is a controllable pair, all the eigenvalues of A 1 + B 1K 1 can be assigned by K1 . Notice also, from the definition of stabilizability, Definition 2.2, that a system is stabilizable if, when its state model in controllable decomposed form, the A 4 partition of A is stable. Finally notice that Ackermann's formula, which we developed to assign the eigenvalues of single input controllable state models, can also be used to assign the controllable eigenvalues of uncontrollable systems. This is done by transforming coordinates to put the given state model in controllable decomposed form. Ackermann's formula is then applied to the controllable pair (A 1 , B 1) to determine the K 1 partition of K, (2. 52), with K2 being chosen arbitrarily. Finally, we can obtain the feedback matrix K 0 in the original coordinates from K 0 = KT- 1 where Tis the coordinate transformation matrix needed to transform the given state model to controllable decomposed form. In the next section we consider how this coordinate transformation matrix can be determined.

2.6

Transformation to Controllable Decomposed Form

In this section we will indicate the role played by the controllability matrix in constructing a coordinate transformation matrix to put any state model in controllable decomposed form. We begin by noting that when a state model is in controllable decomposed form with the (A, B) pair being given by (2.45), the last n2 rows of the controllability matrix consist of nothing but zeros, i.e.,

0

[i i] where

(2.53)

Transformation to Controllable Decomposed Form

65

Notice that the columns of 0 2 depend on the columns of 0 1 , i.e., we can always find a constant matrix 8 to satisfy Therefore recalling that the subspace spanned by the independent columns of n is the controllable subspace, Sc, we see that range[O]

=range([~])= Sc

This fact together with computationally robust methods for finding a basis for the range of a matrix, e.g., singular value decomposition, (Chapter 7), provides a means for constructing a coordinate transformati on matrix T which takes any state model to controllable decomposed form. The following theorem provides some insight into why a coordinate transformati on based on the controllability subspace is able to achieve this result. Theorem 2.8 A coordinate transformati on matrix T transforms a given state model (A, B, C, D) to controllable decomposed form, (Definition 2.5) if range[TJ] = range[O] = Sc where

n is the controllability matrix for the given state model and

with T 1 , T2 being n x n 1 , n x n2 and with T invertible where rank(O) = n1. Proof Let QT denote the inverse of T, i.e.,

[~n [ T1

QTT=l

Tz] = [

~ ~]

(2.54)

Then the transformed A, B matrices are given by

Recall that for the transformed state model to be in controllable decomposed form, (Definition 2.5), we need to show that A3 and B2 are each null. To show that B2 = 0, notice from (2.54) that QJ T 1 = 0 so that the columns of T1 are orthogonal to the columns of Q2 • However range(TJ) = Sc so that we have Qfx=!ll

when

X

ESc

(2.55)

State Feedback and Controllability

66

Now since B appears as the first m <:;olumns of rl, range[B] c range[n]

=

S,

(2.56)

and we see from (2.55) that the columns of Bare orthogonal to the columns of Q 2 so that

B2 To show that A3 since

=

QJB = 0

= 0, recall from Theorem 2.5 that if x ESc than Ax E Sc- Therefore range[TJ]

=

Sc

we have range[ATt] C Sc

(2.57)

Therefore it follows from (2.57, 2.55) that the columns of AT1 are orthogonal to the columns of Q2 so that

•

Notice, in the foregoing proof, that (2.55-2.57) imply that we can find constant matrices 8 1 , 8 2 such that B= T181 AT1 = T182

and since QJ T1 = 0 we achieve Q 2 B = 0 and Q 2 AT1 = 0 as required for T to transform the given system to controllable decomposed form.

2. 7

Notes and References

The Ackermann formula was one of the first attempts to develop an algorithm for assigning the eigenvalues to a single input state model by state feedback. More recent work on this problem has concentrated on developing algorithms which are least sensitive to errors caused by the need to round off numbers during the execution of the algorithm using finite precision arithmetic on a digital computer. Further information on this problem can be obtained by consulting [43]. The Cayley-Hamilt on theorem plays an important role in calculating matrix functions and can be used to provide an efficient method for calculating eAt once the eigenvalues of A are computed. Modern computer oriented algorithms rely on methods which truncate the infinite series expansion for eAt after having put A in a certain computationall y beneficial form known as real Schur form, [17]. The approach to the problem of transforming a state model to controllable decomposed form which was discussed in Section 2.6 forms the basis for the command CTRBF in MATLAB.

3 State Estimation and Observabi lity

3.1

Introduction

In the previous chapter we saw that state feedback could be used to modify basic aspects of a plant's behavior. However, since the plant state is usually not available, we need a means for obtaining an ongoing estimate of the present value of the state of the plant's state model. In this chapter we will see that this can be done by using measurements of the plant's input and output. The problem of determining the state of a state model for a plant from knowledge of the plant's input and output is referred to in general as the state estimation problem. There are two classical types of state estimation: deterministic state estimation and stochastic state estimation. Deterministic state estimation, which is the subject of this chapter, assumes that the system input and output are known or measured exactly. The goal of deterministic state estimation is the determination of an estimate of the state having error which tends to decrease with time following the initiation of the..estimation procedure. However, in stochastic state estimation the input and output signals are not assumed to be known exactly because of the presence of additi\"e stochastic measurement noise having known statistics. In this situation the state estimation error is always present. The goal of stochastic state e-stimation is the determination of an estimate of the state so that, in the steady state, the average or expected value of the state estimation error is null while the expected value of the squared error is as small as possible. Stochastic state estimation is discussed in Chapter 6. Both deterministic and stochastic state estimation are further subdivided according to when the input and output signals are measured relative to when the state estimate is needed. If the plant's input and output are measured over the time interval [0, T] and if we need the state estimate at time te we have 1. a prediction problem if te > T 2. a filtering problem if le = T 3. a smoothing problem if le < T

The state estimation problem which is of concern in connection with the implementation of state feedback is a filtering problem since we need an ongoing estimate of the

68

State Estimation and Observability

plant's state at the present time. Prediction and smoothing problems, which are not discussed it1 this chapter, arise, for example, in hitting a moving target with a projectile by aiming at the target's predicted position and in estimating the true value of data obtained from experiments done in the past when the measured data are corrupted by noise. The computer implementation of a solution to the filtering problem, either deterministic or stochastic, takes the form of a state model which is referred to in general as a filter. The filter's input consists of both the plant's input and output and the filter's output is the state of the filter, i.e., the filter is designed so that its state is an estimate of the plant model's state. In the deterministic case the filter is referred to as an observer. In the stochastic case the filter is referred to as a Kalman filter. The Kalman filter is taken up in Chapter 6.

3.2

Filtering for Stable Systems

Suppose we know the parameters (A, B, C, D) of a plant state model as well as the plant's input and output, { u( t), y( t) : t E [0, tel}. Then, from Chapter 1, we can express the plant state at any time tc in terms of the initial state and input as (3.1)

However since the initial plant state is usually unknown we are only able to calculate the zero state response. We use this fact to form an estimate of the plant state at time tc as (3.2) where from (3.1) we see that the plant state, x(te)) and plant state estimate, x(te)) differ by the plant state estimation error, x(te), (3.3)

Now if the plant's state model is stable, we see from (3.3) that the state estimation error, x(te), approaches the null vector with increasing estimation time, te, lim x(te) = 0

for all x(O)

=

x(O)

lc---+00

In this situation the state estimate, x(tc), given by (3.2), can be a reasonably good approximation to the plant model's state, x( te), provided te is large enough to ensure that the effect of the initial state estimation error x(O) is negligible. In this case the estimate is said to be an asymptotic estimate of the state since the state estimate approaches the state being estimated asymptotically with time. Alternatively, we can view the state estimate obtained in the foregoing approach as the output, Ye(t), from a system called a state estimator having state model (Ae, Be, Ce)· More

Observers

69

specifically we have

x (t) = Aex(t)

Ye(t)

=

+ Beu(t)

(3.4) (3.5)

Ce_x(t)

Then since the state of the plant model is governed by

x=

Ax(t)

+ Bu(t)

(3.6)

we see by subtracting equation (3.4) from equation (3.6) that the state estimation error is independent of the plant input and is governed by the differential equation x (t)

= Ax(t)

(3.7)

where x(t)

= x(t) - x(t)

Therefore the state estimation error is independent of the plant input and is given by (3.8) Now if the initial state of the plant were known we could take x(O) = x(O) as the initial state for the estimator, equation (3.4). Then .X(O) = 0 and we see from equation (3.8) that we would have the desirable result of exact state estimation, x(t) = x(t), for all time. Usually we don't know the initial plant state. In this case we could set x(O) = 0 in the state estimator, (3.4), so that x(O) = x(O) and provided A is stable, the state estimate would approach the actual plant state asymptotically for any initial plant state. However, unstable plants are encountered quite frequently, and must be stabilized by feedback, e.g., the feedback control of satellite rocket launchers. In these cases the estimate obtained from the state estimator, (3.4), diverges from the plant st~te for any x(O) -#0 since eAt becomes unbounded with time and the state estimation error given by equation (3.8) grows indefinitely. Notice that it is impossible to know a physical parameter such as the initial plant state exactly. Therefore in practice it is not possible to set x(t) = x(O) so as to obtain x(t) = 0 from (3.8). Thus whenever the plant is unstable, we are unable to use (3.4) as an asymptotic state estimator. Not knowing the initial state and needing to estimate the state to implement stabilizing state feedback for unstable plants, forces us to seek another approach to state estimation.

3.3

Observers

Notice that the foregoing simple approach to the plant state estimation problem ignored the additional information on the plant state which is present in the plant output, i.e.,

y(t)

=

Cx(t)

+ Du(t)

(3.9)

In this section we show how to use this information to obtain an asymptotic state estimator for unstable plants.

70

State Estimation and Observability

Suppose, for the moment, that the plant output, y(t), is unknown. Then, assuming we have an estimate of the plant state, x(t), we can use it to obtain an estimate of the output, ji(t), by substituting x(t) for the plant state in (3.9), to get

y(t) = Cx(t)

+ Du(t)

(3.10)

However, we assume in this chapter that the actual plant output, y(t), 1s known. Therefore we can define a plant output estimation error as

(3.11)

y(t) = y(t) - Y(t)

Now since we can form.Y(t) from u(t),y(t) and x(t) all of which are known, we can use y(t) to indicate the accuracy of the plant state estimate since y(t) t 0 is an indication that the state estimate differs from the actual state. More importantly, we can use y(t) to correct future state estimates by letting jl( t) affect x (t) by subtracting Ly( t) to the right side of (3.4) to obtain

x= Ax(t)

+ Bu(t)- Ly(t)

which using (3.10, 3.11) can be rewritten as

.

x= Fx(t)

+ G [u(t) y(t)

(3.12)

l

(3.13)

where G= [B+LD

F=A+LC

-L]

When F is stable, the system represented by (3.13) is referred to as an observer. In order to determine the effectiveness of an observer in obtaining an asymptotic estimate of the state, consider the differential equation for the state estimation error, x(t) = x(t)- x(t). We obtain this differential equation by subtracting (3.13) from (3.6) to obtain

.X (t)

=

Fx(t)

(3.14)

where F =A +LC

Then the state estimation error is given by

Notice that unlike the simple approach to state estimation presented in the previous section, we now have the possibility of getting an asymptotic estimate of the state, even if the plant is unstable. We do this by choosing L to make F stable. Questions regarding the possibility of doing this are answered by examining certain properties of the righteigenvectors of A.

Observer Design

71

Theorem 3.1 Whenever A has a right-eigenvector v; satisfying

the corresponding eigenvalue A; of A is an eigenvalue of A + LC for all L. Proof Suppose the condition of the theorem is satisfied. Then multiplying A the right by v; gives

+ LC on

However, since Cvi = 0 and Avi = A;Vi we see that the foregoing equation becomes

• Thus A; is a fixed eigenvalue of the matrix A + LC for all L We will see that if vi is an eigenvector of A such that Cvi =f. 0 then we can choose L so that the corresponding eigenvalue, A;, of A can be assigned to any desired value as an eigenvalue of A + LC. This leads to the following definition. 'Alefinition 3.1 An eigenvalue >..; of A is said to be an observable (unobservable) eigenvalue for the pair (A, C) if we can (cannot) find L such that A; is not an eigenvalue of A+LC. Reflection on the foregoing reveals that we can design an observer for a given state model, i.e., we can find L so that A + LC is stable, provided the plant state model has no eigenv~lues which are both unstable and unobservable. Plant state models having this property are referred to as being detectable. Definition 3.2 A plant state model is said to be detectable if all its unobservable eigenvalues are stable. Thus we can only determine an observer for a given state model if that state model is detectable. This is of obvious importance in the implementation of state feedback using •· an observer for the stabilization of an unstable plant. Recall, in the case of state feedback, that the controller form state model facilitates the calculation of K to achieve a specified set of closed loop eigenvalues. In the present situation there is an analogous form for the plant state model which facilitates the calculation L to assign a specified set of eigenvalues to the observer. This form, which we encounter in the next section, is referred to as an observer form.

3.4

Observer Design

Suppose we are given a state model (A, B, C, D) for the plant and we want to determine L so that the observer eigenvalues, >..-[A + LC], equal a specified set {!1; : i = 1, 2 · · · n}. Then provided any unobservable eigenvalues of the pair (A, C) are in the specified set, we can determine L so that the observer eigenvalues coincide with the specified set. This could be done by equating coefficients of like powers on either side of the equation a(s) = 7(s)

72

State Estimation and Observabi/ity

where n(s) = det[sl- (A+ LC)]

( ) II( 11

I s =

S -

f-Li

)

= S

n

=

s"

+ o 1.1n-I + n 2 s"- 2 +···+On

+ II S11-l + I2Sn-2 + · · · + In

i=l

The resulting equations, li = ai: i = 1,2, · · · n, in the elements of L are, in general, coupled in a way that depends on the plant's A and C matrices. This makes it difficult to set up a general procedure for determining L from these equations. However, when y(t) is a scalar and the state model for the plant has all its eigenvalues observable, this difficulty can be overcome by using coordinates which put the state model for the plant in observer form.

3.4.1

Observer form

Recall that any state model (A, B, C, D), for a plant is related to the plant transfer function G(s) as G(s) = C(sl- A)- 1B + D Then a system having transfer function GT (s) has state model (A, B, C, 15) where

State models (A, B, C, i5) and (A, B, C, D) are said to be duals of each other when their parameters are related in this way. When G(s) is symmetric, the dual of any state model for G(s) is also a state model for G(s). In the SISO case G(s) is a scalar so that GT (s) = G(s). Therefore, in this case, we can obtain a canonical state model referred to as an observer form, by forming the dual of the controller form state model. Thus if the state model for G(s) is in controller form, (A"' Be, Ce, De), the dual state model is in observer form, (Ao =A'{, B0 = C'{, Co= B'{, D0 =De), or more specifically 0

-a!

-a2

0

-an-2

0

0

-an-I

0

0

-an

0

0

0

0

bl

0

0

b2

Eo

0

...

0

0

0

bn-2

0

I

bn-l

0

0

bn

OJ

Do

D

,j

l

Observer Design

73

An indication of the importance of this form is evident from the ease with which we can assign observer eigenvalues through L. Thus when the plant is in observer form, F 0 = A 0 + L 0 C 0 is in the same form as A 0

+ lol -az + lo2

0

-al

0

0 0

/1

0 0

/2 L o-

Fo-

0

+ lon-2 -an-I + lon-1

0

0

0

0

0

-an+ lon

0

0

0

-an-2

ln-2 ln-1

0

ln

Therefore if we want the eigenvalues ofF to satisfy a specified characteristi c polynomial, 'Y(s), we have n

det[s/ - F] =

sn

+

L

')';Sn-i

i=1

and the required elements of L 0 , {10 ; : i = 1, 2, · · · n} are easily obtained from loi

=a,.-')';

i=l,2···,n

In order to preserve the eigenvalues of A + LC under a coordinate transformati on we see that if L assigns a desired set of eigenvalues to A + LC in the transformed coordinates then we need to replace L by L = T L to assign the same set of eigenvalues to A + LC in the original coordinates. This is seen by noting that

3.4.2

Transformation to observer form

In the previous chapter, (Section 2.3.3), we developed an algorithm for generating a coordinate transformati on matrix T so that any single input controllable state model is transformed to a controller form state model. This algorithm can be adapted to provide a coordinate transformati on matrix so that a single output observable state model is transformed to an observer form state model. This adaptation is made using the duality between the controller and observer forms. Thus by replacing A, Bin the algorithm given in Section 2.3.3 by AT, cT we obtain a matrix T which transforms the pair AT, cT to controller form. Then r-T is the coordinate transformati on matrix needed to transform the given state model to observer form. Just as the transformati on to controller form is only possible if (A, B) is a controllable pair so here the pair (AT, cT) must be controllable or equivalently the pair (A, C) must be observable, i.e., rank[U] = n

(3.15)

74

State Estimation and Observability

where

The matrix U is called the observability matrix. Notice, in general, that U is a pn x n matrix where pis the number of elements in y(t). Also in general we have rank[U] :<:; n since the rank of any matrix is never greater than its smallest dimension. Notice that if a pair (A, C) is observable, i.e., rank[U] = n, we can assign all the eigenvalues of A + LC by choosing L. Otherwise if we are not able to assign all the eigenvalues, one or more of the right-eigenvectors of A satisfies Cv; = 0 (Theorem 3.1) and we can show that (3.16)

Therefore at least one of then columns ofU is dependent so that rank[ U] < nand the pair (A, C) not observable. Alternatively, if all the eigenvalues of A+ LC can be assigned, all the right-eigenvectors of A, v; satisfy Cv; =f. 0 and (3.16) is not satisfied by any right-eigenvector of A. Then assuming A has a complete set of eigenvectors, any vector q having n elements can be written as n

q

= LO!;Vi i=l

and

implying that the columns of U are independent and rank[U] = n. Thus all the eigenvalues of A+ LC are assignable if and only if (A, C) is an observable pair.

3.4.3

Ackermann 's formula

Observability

3.5

75

Observabil ity

So far we have encountered two effects of system obsenabilit;. Both the capability of assigning all the eigenvalues to A + LC and the capability of transforming coordinates so that the state model is put in observer form requires that the given state model be observable, i.e., rank:uJ = n where U is pn x n observability matrix, (3.15), with p = l. It is important to notice from the discussion so far, that we can still design an observer to estimate the state of a given state model when the state model is unobservable, i.e., when rank[U] < n. provided the given state model is detectable, Definition 3.2. Thus observability is sufficient but not necessary for the existence of an observer for a given plant state model. In order to gain further insight into the meaning of observability, we are going to consider the problem of determining the state at some estimation time from the derivatives of the input and output at the estimation time. Notice that in practice we avoid using signal differentiation since noise acquired in measuring a signal can appear gt;eatly enlarged in the derivatives of the measured signal. Therefore our intent in discussing the problem of determining the state from the derimtives of the input and output is to provide additional insight into the theoretical nature of observability. We will show that we cannot solve this problem unless the state model is observable.

3.5.1

A state determinat ion problem

Suppose we are given the state model, (A, B, C, D), and the derivatives of the input and output at some estimation time, te, i.e., { y i) ( te ) 'u i) ( te ) .. l. --

0 ' 1' 2 '

...

'n - l }

Then from the output equation (3.17) we have p equations in then unknown components of x(te), where pis the number of elements in y(t). Now if p = n, Cis square and if the rows of Care independent. or equivalently if the elements in y( t) are not related by constants, i.e., are independent, then we can solve (3 .17) for the state at time te as

without requiring derivatives of the input and output signals. However usually p
76

State Estimation an~ Observability

Suppose p = 1. Then we have one equation, (3.17), which we can use in the determination of then components of the state x(tc)· We can generate a second equation for this purpose by taking the derivative of (3.17) (3.19) and using (3.18) to obtain (3.20) We can generate a third equation for determining x(te) by differentiating (3.20) and substituting (3.18) for the derivative of the state. This gives

Continuing in this way we can generate additional equations in the state until we have · the required n equations. These equations can be collected and given as (3.21) where

c

y(te)

z = Y-

ru

/

Y=

CA

1\te)

U= CAn-1

/n-1)(te)

f=

U=

D

0

0

0

CB

D

0

0

CAB

CB

D

0

CAn- 2 B

CAn- 3 B

CAn- 4 B

D

Now since we developed (3.21) from a consideration of the effect of the state and input on the output, we have z Erange[U]so that we can always find x(te) to satisfy (3.21). When (A, C) is an observable pair, all n columns ofU are independent, rank [U] = n, and UTU is invertible SO that when p > 1 we have

and when p

=

1 we have

Observability

77

However when rank[ U] < n, U has dependent columns and we have (IS=

(3.22)

Ux(te)

for some non-null x(te) which we denote by x 0 (te)· Solutions to (3.22) when rank[U] < n are said to lie in the null space of U denoted when Hence if x(te) is any solution to (3.21) then we have x(te) + x 0 (te) as another solution to (3.21). Therefore we see that if (A, C) is not observable it is impossible to determine the true state, x( le), from the derivatives of the system's input and output. In summary we have shown that the observability of the state model is a necessary and sufficient condition for being able to determine the actual state of the plant state model at any time from derivatives of the input and output at that time.

3.5.2

Effect of observability on the output

Continuing the discussion begun in Section 1.7.4 we see that when (A, C) is an unobservable pair with right-eigenvector satisfying Cvj =(IS then when x(O) = v; we obtain the zero-input response as y(t)

=

CeA 1x(O)

=

C [I+

A

;t

=

1 I C[I +At+ 2 ! A 2 t 2 + 3! A 3 t3 + · · ·]x(O)

,2 2 1 ,3 3 ] ; >-.-t C ; + 21! /\; t + 3! /\; t + · · · v = e ' v =

(IS

(3.23)

We can illustrate this important effect by considering the unobservable state model

C= [+1 +1] Then since A is in companion form, (Section 1.6.1 ), we can express its right-eigenvectors as A1 =

-1,

A2 =

-2}

and any initial state given by x(O) = av 1 =

[+a] -a

gives rise -to a trajectory which lies in the null space of C for all time so that y(t) = 0 for all t.

State Estimation and Observability

78

Notice also that the transfer function corresponding to the foregoing state model is given by

G(s)

=

s+1

s 2 + 3s + 2

s+2

and the unobservable eigenvalue of A at -1 is not a pole of the transfer function. This effect was encountered in Section 1.9.2 in connection with diagonal or normal form state models. We will see, in Chapter 4, that this property of having certain non-null initial states which produce no output when a state model is unobsevable leads to a characterization of observability in terms of the rank of a matrix called the observability Gramian. The observability Gramian is involved in determining the energy in the output caused by a given initial state .

. 3.6

Observable Decomposed Form

Recall, from the previous section, that we are unable to determine x 6 (t) Enull[U] from knowledge of a state model's input and output derivatives at timet. We refer to null[ U] as the state model's unobservable subspace, denoted S 6 , i.e., S 6 =null [U]. Moreover we refer to range[ UT] as the state model's observable subspace, S 0 • Now it turns out that any solution to (3.21) can be written as x(t)

= x (t) + x 6 (t) 0

where (i) x 6 (t) Enull[U] = S6 and is arbitrary otherwise; (ii) x 0 (t) Erange[UT] = S0 and depends uniquely on z. Moreover the observable and unobservable subspaces are orthogonal, S 6 ..LS0 , i.e., (3.24) for any x 0 (t) Erange [ ur] and any x 6 (t) Enull[ U]. This can be seen by using the fact that

if and only if we can find w to satisfy

so that we have

In this section we show that any unobservable state model can be transformed, using a change of coordinates, so that in the transformed coordinates the observable and

Observable Decomposed Form

79

unobservable subspaces align with subsets of the coordinate axes. More specifically, when a state model is in observable decomposed form the state is given by

x(t)

=

[:~~~n

(3.25)

where xi(t) is oflength n; and

x(t)

E Sa

if x 2 (t) =Ill

x(t)

E

S0

if x 1 (t) =Ill

Notice that (3.24) is satisfied when the state is given by (3.25). Now the structure required for the state model to be in observable decomposed form is given in the following definition. Notice that this structure is dual to the structure of the controllable decomposed form, Definition 2.5, in the sense described in Section 3.4.1. Definition 3.3 A state model having p scalar outputs is said to be in observable decomposed form when its A and C matrices are in the following block forms

where (AhC 1) is an observable pair and AhA 3 ,A4 , C 1 have dimensions n1 x n 1, n2 x n 1, n 2 x n2 and p x n 1 respectively with n = n 1 + n 2 being the dimension of the state space.

3.6.1

Output dependency on observable subspace

?•

One way of seeing that a state model in observable decomposed form decomposes the state space in the manner specified in equation (3.25) is to note that, in these coordinates, the transition matrix is lower triangular,

so that the zero input response is given by

(3.26) Thus the zero input response is independent of x 2 (0) and S 0 is as specified in equation '(3.25). In addition since (All CJ) is observable, we could determine x 1 (t) from derivatives of the output. Thus Sa is also as specified in equation (3.25).

80

State Estimation and Observabi lity

3.6.2 Observ ability matrix Alternatively, we can also see that the decomposi tion of the state space given by equation (3.25) is achieved when the state model is in observable decomposed form by inspection of the observability matrix.

0 0

c1 CIA

U=

C An 1 -l 1 1 C1A~ 1

0 0

C1Ai- 1

0

~]

[u1 u2

where

rank[U] = rank[UJ] = n 1

and we have

x E null[U]

if and only if

x

= [ :2 ]

(3.27)

for any x 2 of length n - n 1 .

3.6.3

Transfe r function

Again, notice that the observable decompose d form gives the transfer function as G(s)

= [C 1 0] [

(s/-AJ)- 1

* with order equal to or less than n 1 depending on the controllability of the pair (A 1, B1). Notice also that the eigenvalues of A 4 are not poles of the transfer function since they are unobservable. This latter observatio n is readily verified from the structure of (A+ LC) in these

Observable Decompose d Form

81

coordinates since

and since A

+ LC is block lower triangular we have

with >.[A4 ] being eigenvalues of A (A, C).

3.6.4

+ LC for all L, i.e., being unobservabl e eigenvalues of

Transformation to observab le decomposed form

Now we saw in Chapter 2 that the controllable subspace is A-invariant, i.e., if x Erange[n] then Ax Erange[n], (Theorem 2.5). The analogous result here is that the unobservabl e subspace is A-invariant, i.e., if x Enull[U] then Ax Enull[U]. When the state model is in ot>servable decomposed form we can see this fact directly from (3.27) and the block structure of A. Alternatively, we can see that this result holds more generally in any coordinates by noting that we can always express the last block row of UA using the Cayley-Ham ilton theorem, Theorem 2.3, as

Therefore the last block row of UA depends only on block rows of U and thus if Ux = 0 thenUAx = !1). Now the A-invariance of the unobservabl e subspace provides a means for determining the coordinate transformati on matrix needed to put a given state model in observable decomposed form. Theorem 3.2 A coordinate transformati on matrix T transforms a given state model (A, B, C, D) having observability matrix U with rank[U] = n 1 to observable decomposed form if ?·

range[T2 ] = null[U] where

with T~o T2 being n x ni> n x n """- n 1 such that r- 1 exists. Proof Let QT denote the inverse ofT, i.e., (3.28)

··~

State Estimation and Observability

82

Then the transformed A and C matrices are given by -

A= T

_1

AT=

[Qf AT1 Qf AT2] QJ AT1 QJ AT2

C=CT=[CT1

CT2J= [cl

C2]

Now since C appears as the first prows ofU and T 2 is chosen so that UT2 = 0, we have CT2 = 0. Finally since range[T2 ] = S 8 and QT is the inverse ofT we see from (3.28) that

C2 =

when x E S 0

Q[x = 0

Then since AxE Sa if x E Sa, we have range[AT2 ] = S 0 and thus

3.7

Qf AT2 =

0.

•

Minimal Order Observer

Recall from Section 3.3 that use of the plant output enables the construction of an asymptotic state estimator for unstable plants. This estimator, referred to as an observer, has dimensions equal to the dimension of the plant state model. In this section we further recruit the plant output into the task of estimating the plant state by using it to reduce the dimension of the observer. More specifically, if the plant output y( t) consists of p independent scalars, {yJt): i = 1,2, · · ·p}, then y(t) contains information about the state in the form of a projection of the state onto a p dimensional subspace of state space. This fact allows us to concentrate on the design of an estimator for the remaining part of the state consisting of its projection on the remaining n - p dimensional subspace of the state space. In this way we will see that we can obtain a plant state estimator having dimension n- p which is referred to as a minimal or reduced order observer. We can develop this reduced order observer as follows.

3.7.1

The approach

Suppose the model for the plant whose state is to be estimated is given as

x(t) = Ax(t) y(t)

=

Cx(t)

+ Bu(t)

(3.29) (3.30)

where

Notice that there is no loss of generality in assuming that the state model has D = 0, since when D =1- 0 we can replace the left side of (3.30) by q( t) = y( t) - Du( t) and use q( t) in place of y(t) everywhere in the following. Now we assume throughout that the components {yi(t) : i = 1, 2, · · · ,p} of the plant

Minimal Order Observer

83

output y( t) are independent , i.e., we assume that p

l::O';Y;(t) = 0

for all t E [0, oc)

only if n;

=

0

for all i E [Lp]

i=i

This implies that C has independent rows. As mentioned earlier in this chapter, if p = n then Cis invertible and we can solve (3.30) for x(t) as x(t)

= c- 1y(t)

Since we usually have p < n, Cis not invertible. However, even in this case, we can still use y(t) and (3.30) to obtain information on the state. More specifically, we can determine xR(t),the projection of the state on range [cT], from (3.31) where c# = cT ( ccT)- 1is referred to as aright inverse ofC since cc# =I. Then we can wrjte x(t) as (3.32) where XN (t) E

null[CJ

Now since we can be obtainxR(t) from(3.31), we see from (3.32) that we can determine the complete plant state provided we can develop a method for estimating xN (t). Before doing this we show that xR (t) can be determined from (3.31 ).

3.7.2 Determination of xR(t) Notice that the independenc e of the rows of Censures that CCT is invertible. We can see this as follows. Suppose ccT is not invertible. Then we can find y # 0 such that

where w = cTy. However this is possible only if w = 0 and therefore cT has dependent columns. Since this contradicts the assumption that C has independent rows we must have CCT is invertible. Now the expression for c#, (3:31), can be derived as follows. Since xR(t) Erange[CT] it follows that (3.33) for an appropriate w(t). Then multiplying this equation through by C and using the

~-/--

84

State Estimation and Observabilit y

invertability of CCT enables

US

to determine w(t) as (3.34)

However from (3.32) we see that

y(t) = Cx(t) = CxR(t) and we can rewrite (3.34) as (3.35) Finally, premultiplyi ng this equation by CT and using (3.33) yields (3.31).

3.7.3

A fictitious output

Having determined the projection, xR(t), of x(t) on range [cT] along null[C], we need to develop a method for estimating xN (t). This is done by introducing a fictitious output, yF(t), and using it withy(t) to form a composite output, yc(t) as

y c (t)

=

y(t) [ YF (t) ]

=

c ] x(t) [ CF

=

Ccx(t)

(3.36)

where C F is chosen so that Cc is invertible, i.e., C F has n - p independent rows each independent of the rows of C. Then we can solve (3.36) for x(t) as (3.37) Before showing how to generate YF(t), it should be pointed out that we can ensure the invertability of C c by choosing C F so that range[CJ]

= null[C]

(3.38)

We can see this by using the general result for matrices which is developed at the beginning of Section 3.6 in terms of the matrix U, i.e., range [ uT] l_null[ U] Therefore we have range [cJJ l_null[C Fl and it follows from (3.38, 3.39) that null[C] n null[CF] = {0}

(3.39)

Minimal Order Observer

85

so that only if

X=0

which implies that Cc is invertible. Now assuming we chose C F to satisfy (3.38) we can readily show that (3.40) where

This enables (3.37) to be rewritten as (3.41) I

3.7.4

Determination of the fictitious output

Having obtained the state of the plant model in terms of the known plant output and the fictitious plant output, (3.41 ), we need a means for generating the fictitious output. This is done by using the plant state differential equation to set up a state model which has the plant's input and output, (y(t), u(t)), as input and has state which can be used as an estimate of the fictitious output. The derivation of this state model is given as follows. We begin by multiplying the plant state differential equation (3.29) by CF to obtain (3.42)

Then differentiating x(t), (3.41), and substituting the result in (3.42) gives

However from (3.38, 3.40) we have CFC# written as

=

0 and CFC# =I. Therefore (3.43) can be

(3.44) Now equation (3.44) suggests that an estimator for yF (t) has differential equation (3.45) where YF(t) will be an asymptotic estimate of y(t) provided CFAC# is stable. This follows from the differential equation for the estimation error, yF (t), which we can

86

State Estimation and Observability

determine by subtracting equation (3.44) from equation (3.45) to obtain

where

Thus we have

so that for all

yF (0)

(3.46)

if and only ifF R is stable.

3.1.5

Assignmen t of observer eigenvalue s

Now the eigenvalues ofF R = CFACt are fixed by the plant state model. Therefore when some of these eigenvalues are unstable we cannot satisfy (3.46). Recall we encountered the same sort of difficulty in Section 3.2 in connection with trying to estimate the state of a plant state model without using the plant output. We can overcome the present problem when the state model, whose state is to be estimated, is detectable by replacing CF everywhere in the foregoing development by (3.47)

where LR is used in the same way that Lis used in Section 3.3 to assign the eigenvalues of an observer's system matrix, A+ LC. This replacement is effected by replacing yF (t) by z(t) = Tx(t) in (3.36) to obtain [ y(t)] = Crx(t)

(3.48)

z(t)

where Q= [ I LR

0] I

Notice that since Cc is invertible and Q is invertible, independent of LR, Cr is invertible for any LR. Therefore we can determine x(t) from (3.48) for any LR as x(t)

=

My(t)

+ Nz(t)

(3.49)

Minimal Order Observer

87

where (3.50 i

Now by forming C rCi 1 =I we can show that .'vi and X must satisfy CN=0 TN=!

(3.51)

Therefore recalling from (3.38. 3.40) that

[cFc] Lc~

(3.52)

#

we see that the constraints (3.51) on Jf and X are satisfied when S=Ct

In order to obtain a differential equation for z(t), we follow the approach used to get a differential equation for yp(t). (3.44). Therefore multiplying equation (3.29) by T and substituting for x(t) from (3.49) yields :i(t) = TAXz(t)

where z(t)

=

+ TAMy(t) + TBu(t)

(3.53)

Tx(t) and TAN= CpAC1

+ LRCAC1

TAM= CpAC#

+ LRCAC#- CpAC1LR- LRCAC1LR

This differential equation suggests that an estimator, having the same form as an observer, (3.13), can be formulated for z(t) as

z· (t) = Fri(t) + Gr [u(t)] y( t)

(3.54)

where Fr =TAN

Gr=[TB

TAM]

Then the estimation error for z(t) is given by

I(t)

=

(eFT 1)i(O)

where i(t) = z(t)- z(t) and we have an asymptotic estimate of z(t) if and only ifF r is " stable. Thus provided F Tis stable, we can usez(t) generated by equation (3.54) in place ofz(t)

88

State Estimation and Observabilit y

in (3.49) to obtain an asymptotic estimate of x(t) as x(t)

=

My(t)

+ Nz(r)

(3.55)

Notice that the dependency ofF T = TAN on LR, (3.53), implies that we are only able to choose LR so that TAN is stable if the pair (CpAC'J;, CAC'J:) is detectable. We show now that this detectability condition is satisfied when the plant state model or pair (A, C) is detectable. Theorem 3.3 The pair (CpAC#, CAC#) is detectable if and only if the pair (A, C) is detectable. Proof Suppose (>., v) is an eigenvalue, right-eigenvector pair for A. Let r be a coordinate transformati on matrix given as

r = [ c# c#]

(3.56)

Then premultiplying the eigenvalue-e igenvector equation Av rr- 1 between A and v gives

=

AV by r-l and inserting

Aq = >.q where q =

(3.57)

r- 1v and

Then from (3.52) we see that Cis transformed as

c = cr = C[ c# c# ] =

[1

0]

Now suppose A is an unobservable eigenvalue of the pair (A, C). Then recalling that unobservable eigenvalues are invariant to coordinate transformati on, we see that q, (3.57), satisfies

Cq =

0

which from the form for C requires q to have its first p elements zero

where q2 has n- p components. Then using the form just obtained for eigenvalue-e igenvector equation, (3.57) becomes

A, we see that the

Minimal Order Observer

89

implying that ,\is an unobservable eigenvalue of the pair (A 4 , A2 ) which equals the pair (CFAC1, CAC1) specified in the statement of the theorem. This shows that the observability or detectability. (if Re )] ~ 0 I. of (A. C) is necessary for the obsen ability or detectability of the pair (C FAC1, CAC1). Sufficiency can be shown by reversing the • foregoing argument. Notice that when the plant state model's (A, C) pair is observable the coordinate transformation r, (3.56), just used in the proof of Theorem 3.3, can be used to determine LR, (3.47), to achieve some specified set of eigenvalues for the minimal order observer's system matrix, F T, (3.54). One way of doing this \Vouid be to use Ackermann's formula, introduced in the previous chapter, on the pair (A 4 . A 2 ). (3.57). The proof of Theorem 3.3 suggests that if we work in coordinates where the state model's C matrix is

C= [I 0] then a minimal order obsener to estimate the state in the transformed coordinates can be determined by taking T, (3.47), as

where Q is any nonsingular matrix of appropriate size. This ensures that CT, (3.48), is invertible independent of LR. Notice that Q plays the role of a coordinate transformation matrix for the coordinates of the minimal order observer and therefore Q does not affect the minimal order observer eigenvalues. Therefore referring to (3.54, 3.55) and taking Q =I for simplicity. we obtain a minimal order obsener for the estimation of the transformed state as

ij(r) = My(t)

+ ~vz(t)

z (t) = T ANz(t)

+

T AMy(t)

+ T Bu(t)

where N=

TAN = A4

[~]

+ LRA2

TAJ1 = LRAI- A4LR- LRA2LR

+ A3

In summary, we have seen that a minimal order observer for ann-dimension al plant state model is (n- p )-dimensional, where p is the number of independent scalar plant outputs available for use in estimating the plant state. As in the case of the full order observer introduced in the first few sections of this chapter, unobservable eigenvalues of the plant state model's pair (A, C) are fixed eigenvalues for the minimal order clJserver's system matrix FT, (3.54). Thus as in the case of the full order observer we are unable to design an asymptotic observer when the plant state model is not detectable.

90

State Estimation and Observabilit y

3.8

Notes and Referenc es

The obse1 ver was originally proposed by David G. Luenberger in his Ph.D. thesis in 1964. For this reason the observer is sometimes referred to as a Luenenberge r observer. Previously, in 1960, the use of a state model having the same structure as an observer was proposed by R. E. Kalman for the purposes of least squares signal estimation in the presence of additive noise. We will encounter the Kalman filter in Chapter 6. For further discussion of observers the reader is referred to Chapter 4 of [23] Finally, the term " output injection" is used by some authors to refer to the system signal manipulatio n we use to obtain an observer having system matrix A + LC. This terminology should be compared with the use of the term "state feedback" in connection with the formation of a system having system matrix A + BK.

4 Mod el Approximation via Bal anc ed Rea liza tion

4.1

Introduction

In previous chapters it was noted that whenever a state model (A, B, Ci is either not cotitrollable or not observable the transfer function resulting from the calculation C(sl- A)- 1B has pole-zero cancellations. Thus system uncont;oll ability and/or unobservability gives rise to a transfer function with order less than the dimension of the state space model. Therefore systems that are "almost uncontroll able" or "almost unobservable", should be able to be approxima ted by a transfer function model of order less than the dimension of the state model. Reduced order model approxima tion can be of importanc e in the implement ation of feedback control systems. For instance, model order reduction may be beneficial when the large dimension of the full state model gives rise to computati onal problems during the operation of the control system. This chapter treats the problem of extracting a subsystem of a given full order system which works on that part of the state space which is most involved in the input-outp ut behavior of the original system. This is accomplished by introducin g t\m m;trices called the observability Gramian and the controllab ility Gramian. It is shown that these matrices can be used to quantify the energy transfer from the input to the state and from the state to the output. In this way it is possible to identify which part of the state space is controlled most strongly and which part is most strongly observed. Then in order to construct a reduced order model by extracting a subsystem from the given system. it will be shown that the strongly controlled subspace must coincide with the strongly observed subspace. To achieve this coincidence we change coordinate s, if necessary, so that the Gramians become equal and diagonal. The resulting state model is said to be a balanced realization. The reduced order model is then obtained by extracting a subsystem from the balanced realization.

4.2

Controllable-Observable Decomposition

The idea for model order reduction being pursued in this chapter grew out of the basic fact that only the jointly controllable and observable part of a state model is needed to model the input -output behavior.

Model Approximation via Balanced Realizati on

92

The controlla ble and observable part of the state space can be viewed as one of four possible subs paces for the state space of any given state model. More§pecifically, consider the four dimensional state model in diagonal form which is shown in Figure 4.1. Inspection of the block diagram reveals that the four dimensional state model has a is transfer function which is only first order since only the first compone nt of the state which space state the of part that Thus output. the affects both affected by the input and is is both controlla ble and observable forms a one dimensional subspace Sea which specified as if x 1(t) = 0 fori= 2, 3,4 x(t) E Sea where

On further inspection of Figure 4.1 we see that there are three other subspaces sci5J SCOJ Sea specified as

Sea x(t) E Sea x(t) E Sea

x(t)

E

if x,(t) = 0 fori= 1,3,4 if x 1(t) = 0 fori= 1,2,4 if x 1(t) = 0 fori= 1, 2, 3

where Sea' Sea' Sea' Sea are referred to as follows.

Sea Sea Sco

Sea

is the controllable and observable subspace is the controllable and not observable subspace is the not controlla ble and observable subspace is the not controlla ble and not observable subspace

More generally when we are given a higher dimension state model, (A, B, C), which is

~~--~

u(t)

~ -~ Figure 4.1

Example of controllab le-observa ble decompos ition

Controllable -Observable Decomposit ion

93

not in any special form, we can extract a subsystem from the given system which has its state space coincident with the controllable and observable subspace. This can be done by using two successive coordinate transformati ons as described now. First we change coordinates so that the state model is in a controllable decomposed form . (4.1) with (Ac1, Bc1) being a controllable pair. The transformati on matrix to do this was discussed in Chapter 2. Notice that, in the new coordinates, we have the full n dimensional state space split into controllable and uncontrollable subspaces, Sc, Sc. It remains to split Sc into observable and unobservable subspaces, i.e., to split Scinto Sco and Sco· Next we carry out a second coordinate transformati on which affects only the controllable part of the controllable decomposed state model, (4.1). This is done using a coordinate transformati on matrix, T, which is constructed as I

with To being chosen so that the controllable state model (Ac1, Bc1, Cc1) is transformed to observable decomposed form, i.e., so that

0]

(4.2)

with (Aol, Col) being an observable pair. The transformati on T0 to do this was discussed in Chapter 3. Since coordinate transformati on preserves controllability, the.state model (4.2) is controllable. Therefore the subsystem of the observable part of the ~tate model (4.2), i.e., (A 01 , B01 , C01 ) is both controllable and observable. Thus the transfer function for the entire system has order equal to the dimension of this subsystem and is given by

G(s)

,~;

-

= Col (sf -

-

Ao!)

-1-

Bol

Notice that the foregoing model reduction procedure applies to full order state models which are either not controllable or not observable or both not controllable and not observable. More important, the reduced order model we obtain has exactly the same input-outpu t behavior as the full order model. In this chapter we are interestedin obtaining a reduced order model when the full order state model is completely controllable and observable. In this situation any reduced order model will have input-outpu t behavior which differs from the input-outpu t behavior of the full order state model. Our goal is to develop a method for achieving a reduced order state model having input-outpu t behavior which approximates the input-outpu t behavior of the full order state model. We will see that we can develop an approach for solving this model approximati on problem by replacing the concepts of controll-

Model Approximation via Balanced Realization

94

ability and observability by concepts of energy transfer. We will see that the distribution of energy among components of the state is related to properties of Gramian matrices which replace the controllability and observability matrices in this analysis. Just as coordinate transformations are important by extracting the controllable and observable part of a state model, so too are coordinate transformations importanUo being able to separate out that part of a controllable and observable state model which is most important in the transfer of energy from the system's input to the system's output. The state model in coordinates which enable the identification of this part of the system is called a balanced realization.

4.3

Introduction to the Observability Gramian

In order to define the observability Gramian we need to consider the related idea of output energy. Recall that if the state model (A, B, C) is known, then the zero input response, y(t), caused by any specified initial state, x(O), is given as

y(t) = CeA 1x(O)

(4.3)

The output energy, E 0 , for each initial state, x(O), is then defined as the integral of a nonnegative, real, scalar function v( t) of y( t), i.e., v(t) ;:::: 0 for all t E [0, oo)

Notice that since v(t) is non-negative real for all t, the output energy E 0 for any initial state x(O) must satisfy for all x(O)

E0

;::::

E0

= 0 only if v(t) = 0 for t;:::: 0

0

A simple way to choose v(t) so that E 0 has these properties is as follows. if y(t) is a real scalar if y( t) is a complex scalar

then choose then choose

if y(t) is a real vector if y(t) is a complex vector

then choose then choose

v(t) = /(t) v(t) =I y(t)l2 v(t) = YT (t)y(t) v(t) = y*(t)y(t)

Notice that when y(t) is a scalar, y*(t) denotes the complex conjugate of y(t) and IY( t) 12 = y* (t)y( t). Alternatively when y( t) is a complex vector, y* (t) denotes the transpose of y(t) and the complex conjugation of each entry in the resulting row vector. Thus if y(t) is a complex vector of length p then p

y*(t)y(t)

=

LIYz(t)l 2 1=1

More important, notice that each of the foregoing choices is a special case of the last

Introduction to the Observabillty Gramian

95

choice. Therefore, to avoid special cases, we define the output energy E 0 , in general as

Eo

=

looc y* (t)y( t)dt

(4.4)

so that after substituting from (4.3) we obtain

Eo= x*(O) r

=

(loao eA'rc*ceA dt)x(O) 1

(4.5)

(4.6)

x*(O) W 0 x(O)

where the constant n x n matrix W 0 is called the observability Gramian. Notice that in order for the foregoing integral to converge, or equivalently for W 0 to exist, the observable eigenvalues of (A, C) must be in the open left half plane. This can be seen by referring to section 7 of Chapter 1. Alternatively, if (A, C) is observable W 0 exists only if A is stable. Notice also that W 0 must be 1. an Hermitian matrix, i.e., W o = W ;, 2. a non-negative matrix, denoted W 0 ~ 0. I

Requirement number one results from the integral defining W 0 , (4.5). Hermitian matrices have entries in mirror image positions about the diagonal which are conjugates of each other, i.e., (Wo)ij = (W 0 )j;. When W 0 is real we require W 0 = W~ and W 0 is referred to as a symmetric matrix. Hermitian (symmetric) matrices have eigenvalues and eigenvectors with some rather striking properties which we will develop in the next section. Requirement number two results from the output energy being nonnegative for all initial states since a non-negative matrix, W 0 , has the property that

for all vectors x(O) of appropriate length. Now it turns out that we can always find a matrix M such that any Hermitian, nonnegative matrix W 0 can be factored as W 0 = M* M. This fact allows us to write x(O) W 0 x(O)

=

x(O)* M* M x(O) n

= z*z =

L)d2 i=l

where z = Mx(O). Since I: 7= 1 lz;l 2 ~ 0 for any z we have W 0 ~ 0. Moreover since L %:1iz;l 2 = 0 if and only if z =¢,we have x(O)* W 0 x(O) = 0 for some x(O) -=f 0 if and only if M and hence W 0 is singular, i.e., if and only if W 0 has one or more zero eigenvalues. More is said about this in the next section. Finally the observability of a state model is closely tied to the nonsingularity of its observability Gramian. This can be seen by noting from (4.4) that E 0 = 0 if and only if y(t) = 0 for all time. However we saw in Chapter 3 that the zero input response is null, y(t) = 0, for some x(O) -=f 0 if and only if (A, C) is unobservable. Therefore it follows that

Model Approximation via Balanced Realization

96

(A, C) is unobservable if and only if W 0 is singular. Further insight into this fact is given in subsequent sections where it is shown that W 0 can be obtained by solving a linear matrix equation which is referred to as the Lyapunov equation and which involves A and C in a simple fashion.

4.4

Fundamental Properties of W0

In the previous section we saw that the observability Gramian is a member of the class of matrices which are characterized as being Hermitian and nonnegative. In this section we give some of the essential features of these kinds of matrices.

4.4.1

Hermitian matrices

Recall that a matrix W 0 is said to be Hermitian if it equals its conjugate transpose, W 0 = W~. This implies the ijth and jith entries are conjugates of each other. The eigenvalues and eigenvectors of Hermitian matrices have some rather striking properties which are given as follows. Theorem 4.1 If W 0 is Hermitian then: (i) W 0 has real eigenvalues (ii) W 0 has right eigenvectors which satisfy v;*v 1 = 0, when er 0 ;

#- er J 0

Proof (i) Let (er 0 , v) be any eigenvalue-eige nvector pair for W 0 , i.e.,

(4.7) Then premultiplying (4.7) by v* gives (4.8) and taking the conjugate transpose throughout this equation gives v* W~v

= er~v*v

(4.9)

Then subtracting (4.9) from (4.8) we obtain v*(W 0

-

W~)v

=(era- er~)v*v

(4.10)

.

Now since W 0 is Hermitian, the left side of (4.1 0) is zero. Therefore either er o = er~ or v* v = 0. However v* v = 0 holds only if v = 0, and since vis an eigenvector of W 0 , we have v*v #- 0. Therefore (4.10) can only be satisfied if er 0 = er~ which shows that era must be ~.

era!

Proof(ii) Let (er 0 ;,v;: i = #- ero2· Then we have

1,2) be two eigenvalue-eige nvector pairs for W 0 with (4.11) (4.12)

97

Fundamen tal Properties of Wo

Next operate on these equations as follows: (i) premultipl y (4.11) by 1·2* (ii) postmultip ly the second form for (4.12) by v 1 (iii) subtract the result of (ii) from (i) This yields (4.13)

However, since H" 0 is Hermitian , the left side of (4.13) is zero. Moreover we are assuming • that 0'01 =fc 0'~ 2 . Therefore (4.13) is satisfied only if vtv 2 = 0. different to ing correspond rs eigemecto , symmetric .. i.e reaL is Notice that when W 0 . .. v iT v j = 0 : 1. 1_;_ ],. 0'0 ; 1_;_ O'oJ· Iues are orth ogona,I I.e . e1genva Recall that if a matrix has distinct eigenvalues, it has a complete set of eigenvectors and can be diagonaliz ed by a matrix haYing columns made up from its eigemecto rs. Thus when Wa has distinct eigenvalues we can use its eigenvectors {v; : i = 1, 2, · · · n} to form a matrix V 0 \vhich diagonaliz es W 0

v- 1 w v 0

0

-2: 0

(4.14)

0-

where

:E 0

=

["~1

0

0

(l'o2

0 =

.

0

0 V 0 = [v 1

v2

diag[0'0 J, O'o2'

· · ·, O'on]

O'on vn]

HoweYer if the eigenvectors are normalized ,

then we see from Theorem 4.1 that

( 4.15)

1 Therefore V~ 1 = V~ so that V~W 0 V = 2: 0 . Matrices satisfying the condition V~ are referred to as unitary matrices. Alternatively, when W 0 is real then (4.15) becomes

=

V~

V~Va =I and therefore V~ 1

=

V~ so that V~ W 0 V 0 = 2: 0 • Matrices satisfying the condition

98

Model Approximation via Balanced Realization

v,;

are referred to as orthogonal matrices. It can be shown that any Hermitian V;,- 1 = matrix car1 be diagonalized by a unitary matrix even if the Hermitian matrix has multiple eigenvalues. To recap, we have encountered the following matrix types. Definition 4.1 If M is complex then M is Hermitian

if M

=

M*

M is unitary

if

M- 1 =

M is symmetric

if

M = M

M is orthogonal

if M- 1 = M

M*

If M is real then

4.4.2

7

7

Positive definite and non-negative matrices

Suppose that we change coordinates, viz., x(O) = Tz(O), with T nonsingular. Then for each initial state x(O) in the old coordinates, there is a unique initial state z(O) in the new coordinates and vice versa. Thus we see from the expression for the output energy, (4.6), that the output energy can be rewritten in terms of the initial state in the new coordinates as (4.16) where W0 = T*W 0 T. Now if we choose T as the unitary matrix V 0 , (4.14), then

and (4.16) becomes 11

E0

= z*(O)~oz(O) =

L

(4.17)

CT0 ;IzJO)I 2

i=1

However since there is a unique z(O) for each x(O) we have that E 0 2 0 for all x(O) if and only if E 0 2 Oforallz(O). Thusweseefrom(4.17)thatE 0 2 Oforallx(O) ifandonly if all eigenvalues of W 0 are non-negative, i.e., croi

20

i= l,2,···n

Otherwise, if W 0 were to have some negative eigenvalues, we could choose z(O) so that z1(0) c/c 0 if cr01 < 0

and

z1(0) = 0 if cr 01 > 0

i = 1, 2, · · ·, n

which would give E 0 < 0, for the initial state x(O) = Tz(O) and the requirement for Eo to be non-negative for all initial states would be violated.

Fundamental Properties of Wo

99

Finally, we can show by a similar argument that E 0 > 0 for all x(O) of 0 if and only if all eigenvalues of W 0 are positive. To recap, we have defined the matrix properties of positive definiteness and nonnegativeness as follows. Definition 4.2 W 0 said to be:

(i) (ii)

when x*W 0 x > 0

positive definite, denoted as W 0 > 0 non-negative, denoted as W 0 ~ 0

when x* W 0 x ~ 0

for all x for all x

of 0

Moreover these properties impose certain restrictions on the eigenvalues of W 0 namely

4.4.3

>0

(i)

then

U0;

(ii)

then

U0 ; ~

0

for

U0;

for

U0; E

E

>.[W 0 ] >.[W0 ]

Relating E0 to .A[W0 ]

In order to begin to appreciate the role played by the eigenvalues of W 0 in achieving a lower order state model approximation we consider the problem of finding the initial state, Xmax(O), which produces the most output energy, Eomax• when the initial state is constrained so that X~ax(O)Xmax(O)

=1

(4.18)

When x(O) is constrained in this way we say x(O) is normalized to one. Notice that this constraint on the initial state removes the possibility of obtaining an unbounded output energy by choosing an initial state with an unbounded component. We begin the development of a solution to this problem by noticing that if we use a coordinate transformation matrix T which is unitary, the constraint (4.18) is preserved, i.e., if T satisfies T* T = I then we have ?·

x*(O)x(O)

=

z*(O)T*Tz(O)

=

z*(O)z(O)

(4.19)

Therefore if x*(O)x(O) = 1 so does z*(O)z(O). Now we have seen that using the unitary matrix T = V 0 as a coordinate transformation matrix makes the observability Gramian diagonal in the new coordinates. Therefore our problem can be solved by first solving the problem in coordinates in which the observability Gramian is diagonal and then transforming the solution, Zmax(O) in these coordinates back into the original coordinates. Suppose we are in coordinates where the observability Gramian is a diagonal matrix, E 0 • Suppose also that we have reorder the components of the state in these coordinates so that the largest entry on the diagonal ofE 0 is first, i.e., u01 ~ CT0 ; : i = 2, 3, · · · , n. Then we see from equation (4.17) that the output energy satisfies n

n

Eo=

Luo;iz;(O)I i=l

2

::=;

U0 1

L i=l

I

z;(O)

2

1

(4.20)

100

Model Approximation via Balanced Realization

However since we are assuming that x(O) is normalized to one, we see from (4.19) that z(O) is also normalized to one n

z*(O)z(O)

=

L

I

z;(O)

12

= 1

(4.21)

i=l

Therefore under this constraint we see from (4.20) that the output energy, E 0 , is bounded by the observability Gramian's largest eigenvalue, i.e., (4.22) Notice that if we choose

then (4.21) is satisfied and the output energy becomes n

Eo=

L D"o;lz;(O)I

2 = D"oJ

i=O

and we achieve the upper bound on E 0 , (4.22). Thus the normalized initial state in the original coordinates which maximizes the output energy is obtained as (4.23)

1 0

0

Therefore our problem is solved by setting the initial state equal to the eigenvector of W 0 corresponding to the largest eigenvalue of W a when that eigenvector is normalized to one. Moreover, it is not hard to see from the foregoing that, in general, we have

when x(O)

= vk

The foregoing analysis indicates that the relative importance of the components of z(O) in producing output energy depends on the size of an eigenvalues of W 0 relative other eigenvalues of W 0 . We can use this observation to suggest how we are going to be able to obtain a reduced order state model approximation. We do this as follows. Suppose we choose the coordinates for the state space so that the eigenvalues of W 0 are in descending size, i.e., a 01 ?: a 02 ?: · · · ?: a 0 n with W 0 diagonal. Suppose we consider z(O)

_.{.,;:;.

".

Introduction to the Controllability Gramian

to have all its elements equal, z;(O)

=

101

zi(O). Then the output energy is given by 0

0 0: 0

1z(O) = I~ 11

!Joi

!Jon

where

Moreover, we can always choose an integer n 1 E [0, n]large enough so that (4.24)

where

n,

Eo]

= ~.

L i=l

n

IJc;

Eo2 =

~.

L

!Joi

i=n 1+1

Therefore in this situation a good approximation to the output energy is

Since the last n- n 1 components of the initial state are less important to the zero input response than the first n 1 components under the foregoing condition of an equal element initial state, a reduced order state model in these coordinates could be formed as (AhB 1 , C\) where A1 is the n 1 x n 1 partition in the top left corner of A and .8 1, C\ are corresponding partitions of B, C. However, our interest in this chapter is in obtaining a reduced dimension state model whose input-output behaYior approximates the input-output beha\io;· of the full dimension model. The foregoing argument for neglecting the last n- n 1 components of z(t) focuses only on the zero input response. Therefore what is missing from the previous argument, if we are interested in modeling the input-output behavior, is the effect the input has on the neglected components of z(t). The effect of the input on different components of the state is related to properties of a matrix known as the controllability Gramian.

4.5

Introduction to the Controllability Gramian

The controllability Gramian. which we introduce in this section, is im olved in characterizing the energy delivered to the state when the input is a unit impulse and the initial state is null. Since the Laplace transform of the zero state response to this input is the given system's transfer function the distribution of energy among the components of the state in this situation is appropriate to our goal of obtaining a reduced order transfer function approximation. We begin the dewlopment of the controllability Gramian as follows.

102

Model Approxim ation via Balanced Realizatio n

Suppose we are given a state model, (A, B, C). Then the state in response to a unit impulse at time zero when the initial state is null is given by

Then the energy, Ec, transferred from this input to the state is given by (4.25) Notice that the integrand obtained here is strikingly different from the one obtained in the integral for the output energy, (4.5). We proceed now to express Ec (4.25) as an integral which appears as the dual of the integral expression of E 0 • More specifically, we show now that Ec can be expressed as the trace of the integral obtained earlier for E 0 , (4.5) with A replaced by A* and C replaced by B*. The trace of a matrix is a scalar which is defined as follows. Definition 4.3 The trace of any square matrix M of size n, is defined as the sum of its diagonal entries, n

= 2: m 11

trace[M]

i=l

where mu is the ijth entry in M. In order to proceed we need the following result involving the trace of the product of two, not necessarily square, matrices Theorem 4.2 Given two matrices M, N of size p x n, and n x p then trace [M N] Proof Let M N

trace [N M]

=

= Q. Then p

trace[MN] =

2: q

11

i=l

where q 11 is the result of multiplying the

ith

q,, =

row of M by the

ith

column of N,

n

2:munii j=l

Therefore we can rewrite the trace of the matrix product as p

trace[MN]

n

= 2: L

mun;;

i=l j=l

Next interchang ing the order of the summation s gives n

trace[MN]

=

p

L L n mu 11

j=l i=l

·.~

Introductio n to the Controllab ility Gramian

103

and we see that

• where SjJ is the /h entry on the diagonal of S = N M, and the result follows the is matrix square any of trace the that is theorem this of ce consequen An immediate 1 sum of its eigenvalues. More specifically if JJ = 1"- \Y, then n

trace[M] = trace[V- 1AV] = trace[Avv - 1] =

2:..:::\ i=l

where>..; E .\[Jf1 • Returning to the problem of re-expressing the energy En (4.25), we use Theorem 4.2 with

to obtain

Ee

= trace[Wc]

(4.26)

\Vhere We. the controllab ility Gramian. is given by (4.27) Notice that We is Hermitian and that We is non-negative since the integrand in (4.27) is quadratic. Notice also that if we replace A by A' and B by C* in the foregoing definition of the controllab ility Gramian. We. (4.27), we obtain the obsenabil ity Gramian, W 0 , (4.6). Thus all the properties discussed in connection with W 0 hold for WeMore specifically. since We is Hermitian , Tre has real eigemalue s and can be diagonalized by a unitary matrix, V 0 i.e., (4.28) where V;:- 1 = v; and E, is real and diagonal. Notice that Vc has columns which are normalized right eigenvectors of We and the diagonal entries in the diagonal matrix Ec are the eigenvalues of rr,. In addition recalling the discussion following the proof of Theorem 4.2 that the trace of a square matrix equals the sum of its eigenvalues, we have

Now we will see in the next section that we can always transform coordinate s so that the controllability Gramian is diagonal. Assuming we have done this transfon11ation, we permute componen ts of the state, if need be. so that the entries along the diagonal of Ee appear in descending size, i.e., uc; 2': uc(i+I) : i = 1, 2, · · · n- 1. Then we can always

104

Model Approximat ion via Balanced Realization

choose n 1 so that

where with

Eel =

L"'

n

aci

i=l

and with

Ec2 =

L

ac;

i=n 1+1

This inequality implies that the first n 1 components of the state in these coordinates receive most of the energy from the input. If in addition, in these coordinates, the observability Gramian happened to be diagonal with entries on the diagonal being in descending size, the first n 1 components of the state would also be relatively more important than the last n- n 1 components in the transfer of energy to the output. We could then obtain an approximati on to the full order transfer function for the system as

where the full state model in these special coordinates has its system parameters, (A 6 , Bb, Cb), partitioned to reflect the partitioning of the state, i.e.,

In the next section we will see that it is always possible to transform the coordinates so that any controllable and observable state model has controllability and observability Gramians which are equal and diagonal, i.e., so that in the new coordinates

with L:b having entries in descending size along its diagonal. A state model having this property is referred to as a balanced realization. A reduced order approximati on, Gapx(s), to the full order system is obtained from the balanced realization in the manner just described. Finally, we are assuming the pair (A, B) is controllable so that every component of the state receives some of the energy supplied by the input. Therefore, from the foregoing, we have all eigenvalues of We bigger than zero or We> 0 when (A, B) controllable. In addition, it turns out that rank[Wc]

4.6

=

rank[D]

Balanced Realizati on

Suppose we are given the controllability and observability Gramians, W"' W 0 , for a controllable and observable state model, {A, B, C, D}, of the system to be approximate d. Then in this section we will show how to construct a coordinate transformati on matrix so

~;

Balanced Realization

105

that the state model in the new coordinates is a balanced realization. Just as in Section 4.2 we saw that a succession of two state coordinate transformations are required to achieve a controllable-observable decomposed form, so too in this section we will see that we need a succession of two transformations to achieve a balanced realization. The first coordinate transformation gives a new state model with controllability Gramian equal to an identity matrix. The second coordinate transformation keeps the controllability Gramian diagonal and diagonalizes the observability Gramian obtained from the first transformation so that the resulting Gramians are equal and diagonal. To develop these transformations we need the fact that, if the coordinates are changed using coordinate transformation matrix T, then the Gramians (We, W0 )for the state model in the new coordinates are related to the Gramians ( W c• W 0 ) for the state model in the original coordinates as (4.29) where x(t) = Ti(t). These relations can be obtained by replacing (A, B, C) in the integrals for ~lie and W0 by (T- 1 AT, T- 1 B, CT). We use (4.29) now to develop a the sequence of three coordinate transformations T; : i = 1, 2 and P so that T b = T 1 T 2 P transforms the given controllable arid observable state model to a balanced realization, where Pis the permutation matrix required to put the diagonal elements of the Gramians in decreasing size. We will show now that these coordinate transformation matrices are determined from the Gramians for the given state model. In the next section, we will show that the Gramians for the given state model can be determined by solving two linear matrix equations known as Lyapunov equations. For the time being suppose we are given the controllability and observability Gramians, We, W 0 for the state model we wish to approximate. Then we form the first coordinate transformation as

1

1

r,..~

where v; WcVc = Ee with v; = V;:- 1and E~ > 0 is diagonal with mE~= Ec· Then using (4.29) we see that the Gramians which result from the first coordinate transformation are given as

Next, we take the second coordinate transformation as Tz

-

=

- _!

VoEo4

where V(; W0 V0 = E0 and E~ > 0 with (E~) 4 = E 0 . Then using (4.29) we see that the Gramians which result from the second transformation are given by

-I

Wo

=

TiWoTz

and we have the Gramians equal and diagonal.

=E~

106

••••

•

Model Approxim ation via Balanced Realizatio n

'

It remains to permute the componen ts of the state so that the entries on the diagonal of

the Gramians are ordered in decreasing size. This correspond s to a third coordinate transforma tion, this time using a permutatio n matrix Pas the coordinate transforma tion matrix so that

where

and the ub;s are referred to as the Hankel singular values for the system. More is said about this in Section 4.9 At this stage we can obtain the balanced realization, (Ab, B1, Ch), from (T- 1AT, T- 1B, CT) with T = T 1 T 2 P. Then partitionin g the balanced realization as (4.30)

with n 1 large enough so that n1

n

i=1

i=n 1+1

I:: ubi » I::

ubi

we obtain the reduced order state model as (Ab 1 ,Bh 1 ,Ch 1,D) and the reduced order transfer function model as

Recall that we are assuming that the full order state model is stable, i.e., Ab is stable. Therefore we get a bounded output for every bounded input to the full order, unreduced system. Thus if Gapx (s) is to be a meaningfully reduced order approxima tion to the full order transfer function, we expect Gapx(s) to be stable. Otherwise the behavior of the reduced order system would be dramatical ly different from the behavior of the original full order system. The following theorem specifies conditions which guarantee that the reduced order model is stable. Theorem 4.3 If in (4.30) we have ubn 1 'I uhn 1 _,_ 1 then Ab 1 is stable. Proof Appendix In the next section we show that the Gramians for a given state model can be determined by solving linear matrix equations called Lyapunov equations. In this way the calculation of the integrals in the definition of the Gramians is avoided. We will also see that further relations between the Gramians for a state model and the observability, controllability, and stability of that state model are uncovered by considering the Gramians to be solutions to Lyapunov equations.

I

The Lyapunov Equation

4.7

107

The Lyapunov Equation

We begin this section by generalizing the output energ~ as follows. Recall that in the absence of an input the observability Gramian gives the output energy caused by any initial state as

Now suppose we replace this constant quadratic function of the initial state by e0 (tl, a time varying quadratic function of the state given as (4.31)

e0 (t) = x*(t)Px(t)

where Pis a positive definite Hermitian matrix and xit) function is always positive for all t

E

= eA 1x(O).

(0, oo) and all x(O) #

"Kotice that this

(4.32)

0

Next assume that e0 (t) has a derivatiYe \\hich is a negative quadratic function of the st;rte de 0 (t) · _ = -x*(t)Qx(t) < 0 dt

for all t

-

E

(0, oo) and all x(O)

#

0

(4.33)

where Q is a positive definite Hermitian matrix. Then we see from (4.32, 4.33) that e0 (t) is a positive, decreasing function of time which is bounded below by zero. This implies that (4.34)

However since Pis invertible we have e0 (t) lim x(t)

= 0

= 0 only if x(t) = 0 and (4.34) implies that for all x(O)

(4.35)

t-+00

Therefore equations (4.31). and (4.33) imply that the state model is internally stable, i.e., the state model's A matrix has all its eigenvalues in the open left half plane. NmY if we differentiate (4.31) and use :\"(t) = Ax(t) we can express the time derivative of e0 (t), as

e (tl 0

=

x*(r)(A*P~ PA)x(tl

These observations lead us to formulate the following theorem concerning the stability of A. Theorem 4.4 If for any Hermitian Q > 0 we obtain P > 0 as a solution to the linear matrix equation A*P+ PA = -Q

then A is stable.

(4.36)

108

Model Approximation via Balanced Realization

Proof Suppose (.A, v) is any eigenvalue, right-eigenvector pair for A. Then pre and post multiplying (4.36) by v* and v respectively gives v*(A*P+ PA)v

= -v*Qv

which recalling v* A* = .A*v*becomes (.A*+ .A)v* Pv = -v*Qv

(4.37)

However since v* Pv > 0 and -v*Qv < 0 for all v # 0, we see that (4.37) is satisfied only if (.A*+ .A)= 2Re[.A] < 0. This shows that all eigenvalues of A lie in the open left half plane and therefore A is stable. • The linear matrix equation (4.36) is called a Lyapunov equation after the Russian mathematician who originated this approach in the study of stability.

· 4.7.1

Relation to the Gramians

We show next that when we take Q = C*C the solution P to the Lyapunov equation, (4.36), is the observability Gramian, i.e., W 0 satisfies the Lyapunov equation (4.38)

Analogously, the controllability Gramian, We, is the solution the Lyapunov equation (4.39)

The foregoing relations between the Gramians and the solutions to Lyapunov equations are shown in the following theorem. Theorem 4.5 IfF is a square stable matrix and if (4.40)

then

FW+ WF* = -Q

(4.41)

Proof We begin the proof by defining the matrix function

Then we differentiate Z(t) to obtain

Next recalling the series expansion of er 1 , we see thatF* and eF' 1 commute. Therefore

The Lyapunov Equation

109

the foregoing equation can be written as dZitl

---crt= FZI t) + Z( t)F

(4.4.:2)

Finally, since A is being assumed stable, we hcne Zix) = 0. Therefore integrating (4.42) from 0 to oo gives

lax dZ(tl =lax [FZ(tl Z(t)Fjdt Z(oo)- Z(OJ =

F[laoo

-Q=FW+

eF1 QeF' 1dt]

+[lax

eF1 QeF'rdt]F*

wr

•

No\Y there are two important observations we can make when \Ye compare the Lyapunov equation (4.36) with the Lyapunov equations for the Gramians (4.38, 4.39). First, notice that unlike Q on the right side of(4.36), C*C and BB* in (4.38, 4.39) are only non-negative. Second, (4.38, 4.39) are linear equations in the elements of the matrices W 0 , We. Therefore we may be able to solve (4.38, 4.39) for matrices W 0 and rv, when A is not stable. However since the observability and controllability Gramians (4.5, 4.27) are only defined when A is stable, the matrices W 0 , We obtained by solving (4.38, 4.39) are system Gramians only if A is stable. The implications of these observations are taken up next.

4.7.2

Observabil ity, stability, and the observabil ity Gram ian

What follows are theorems which describe how the concepts named in the title of this section are interrelated. Theorem 4.6 If (A. C) is observable and if P > 0 satisfies the Lyapunov'equa tion AXP+PA = -C*C

(4.43)

then A is stable. Proof Suppose, contrary to the theorem, that A is not stable, i.e., Av = ,\v with Re[.A] :::0: 0 for some .A E .A[A]. Then proceeding as in the proof of Theorem 4.4 we obtain 2Re[.A](v*Pv) = -v*C*Cv

(4.44)

Now since (A, C) is observable we have Cv = ~ i=- 0 and L'·c~ Cc· > 0. However we are assuming that Re[.A]::;:: 0. Therefore (4.44) is satisfied under these conditions only if we have~·· Pv < 0 which contradicts the condition P > 0 specified in the Theorem. Therefore • our assumption that A is unstable is false. is C) (A, and unstable is A when that proof, Notice, in connection with the foregoing have we observable (4.45)

110

Model Approximation via Balanced Realization

is unbounded and the observability Gramian, W 0 , is undefined. However this does not mean that we are unable to find P to satisfy (4.43) in this case. The foregoing theorem has the following corollary. Theorem 4.7 If A is stable and Pis positive definite and satisfies (4.43) then (A, C) is observable. Proof Suppose, contrary to the theorem, that (A, C) is not observable. Then we must have a(?., v) pair, Av = >.v and Cv = 0. Therefore proceeding as in the proof of Theorem 4.6 we obtain 2Re[>.](v' Pv) = -v*C*Cv = 0

(4.46)

Now in order to satisfy (4.46) we must have Re[>.] = 0 or v* Pv = 0 or both. The condition Re[>.] = 0 is impossible since we are assuming that A is stable. The condition v* Pv = 0 is also impossible since we are assuming that P > 0. Thus (4.46) can not be • satisfied and therefore (A, C) must be an observable pair. To recap, the foregoing pair of theorems state that A is stable when (4.43) is satisfied by P > 0 if and only if (A, C) is an observable pair. Since the observability Gramian W 0 , (4.45) is defined when A is stable and positive definite when (A, C) is an observable pair, if the solution P to (4.43) satisfies P > 0 when (A, C) is not an observable pair Pis not the observability Gramian. The following theorem characterizes the effect of having a P > 0 as a solution to the Lyapunov equation (4.43) when the pair (A, C) is not observable. Theorem 4.8 If the pair (A, C) is not observable and P > 0 satisfies the Lyapunov equation (4.43) then A has imaginary axis eigenvalues. Proof Since the pair (A, C) is not observable, we can choose a (>., v) pair so that Av = >.v and Cv = 0. Then we see from (4.46), that 2Re[>.](v* Pv) = 0

(4.47)

However since P > 0, (4.47) is satisfied only if Re[>.] = 0. • To recap, we have shown in the foregoing theorems that A is stable when the Lyapunov equation A*P+ PA

= -C*C

is satisfied by P > 0 if and only if the pair (A, C) is observable. Alternatively, if this Lyapunov equation is satisfied by P > 0 when (A, C) is not observable, the unobservable eigenvalues of A lie on the imaginary axis. Notice from the proof of Theorem 4.7 that eigenvalues of A that are not on the imaginary axis are observable and lie in the left half plane. Analogously, A is stable when the Lyapunov equation AP+PA*=-BB*

is satisfied by P > 0 if and only if the pair (A, B) is controllable. Alternatively, if this Lyapunov equation is satisfied by P > 0 when (A, B) is not controllable, the uncontrollable eigenvalues lie on the imaginary axis and the controllable eigenvalues are all in the left half plane.

Controllabil ity Gramian Revisited

4.8

111

Controllability Gramian Revisited

In this section we develop another use for the controllability Gramian which provides us with a different way of viewing the process of model reduction and a new way of thinking about a system's input-outpu t behavior.

4.8.1

The least energy input problem

An alternative use for the controllability Gramian, We, arises in connection with the solution to the problem of finding the input u0 (t) : t E ( -oo, 0] which expends the least energy in taking the null state at time t = -oo to some specified state at time t = 0. In order to solve this problem we need to recall, from Chapter 1, that the basic equation relating the initial state and input to the final state is

x( t1 )

=

eA 1f x( -oo) +

l

lf

-oo

eA(tr -T) Bu(T)dT

(4.48)

Then, in the present situation, we have x( -oo) = !Zl, and t1 = 0 so that (4.48) becomes x(O) ·=

/

1:

e-AT Bu(T)dT

(4.49)

Now we want to choose u(t) to satisfy (4.49) so that the input energy, EcUl given by (4.50)

achieves its minimum value, Ecua, determined from

Ecua

=

min Ecu u(t)

It turns out that one input that solves this problem is given as (4.51)

where We is the controllability Gramian, (4.27). Notice that this input satisfies (4.49) and has energy determined as Ecu

=

1:

u*(T)u(T)dT

(1:

=

x*(O)W~ 1

=

x*(O)W~ 1 x(O)

=

e-AT BB*e-A'TdT ) W~ 1 x(O) Eeuo

Notice that u0 (t) exists provided We is invertible. This requirement that the (A, B) pair be controllable is a necessary condition for being able to satisfy (4.49) for any x(O), (see Chapter 2 Section 2.4). Recall that the coordinate transformati on matrix, Ve, (4.28), that diagonalizes We is unitary, V~ 1 = v;. Thus we see that this coordinate transformati on also diagonalizes

112

~

Model Approximation via Balanced Realization

} I

w;:- 1 since Therefore we can use this fact to rewrite E,uo as

(4.52) where

z(O)

=

v:x(O)

Then if the components of z(O) are equal and have been permuted so the entries along the diagonal of ~c are ordered, abi 2': ab(i+ 1l : i = 1, 2, · · · n - 1, we see from (4. 52) that most of the energy in the input is needed to achieve components in the lower partition of z(O). This implies that the lower partition of z(t) is harder to control than the upper partition. Now in the following we make use of the foregoing alternative interpretation of the controllability Gramian to provide further insight into the use of a balanced realization to achieve model reduction. Recall that a balanced realization has both Gramians equal to the same diagonal matrix with the entries encountered in descending the diagonal forming a nonincreasing sequence. Then we saw, in Section 4.4.3, that most of the energy supplied to the output by an initial state with equal component values comes from components in the upper partition of the initial state. However, we have just seen that the components in the upper partition of the state are more easily manipulated by the input than components in the lower partition. Therefore, in balanced coordinates, the subspace of the state space corresponding to the upper partition of the full state is more involved in the input-output behavior than the subspace corresponding to the lower partition of the state. This suggests that we can capture most of the full order system's input-output behavior by using a reduced order state model formed by truncating the balanced realization in the manner given at the end of Section 4.6.

4.8.2

Hankel operator

We can obtain further insight into a system's input-output behavior by using the present interpretation of the controllability Gramian. This is done by noting that: (i) the controllability Gramian is used to characterize the system's transformation of u(t) : t E ( -oo, OJ into x(O); (ii) the observability Gramian is used to characterize the system's transformation of x(O) into y(t) : t E [0, oo) when u(t) = 0: t E [0, oo). This suggests that, in addition to the concept of a system implied by its transfer function, G(s), i.e., that the system transforms or maps system inputs, u(t) : t E [0, oo ), to system outputs, y(t) : t E [0, oo ), we can view the system as a map from past inputs, u(t) : t E ( -oo, OJ to future outputs, y( t) : t E [0, oo ). This map is referred to as the

n '

Controllabil ity Gramian Revisited

113

system's Hankel operator and is denoted as r G· When a system is stable its Hankel operator has an operator norm which is referred· to as the Hankel norm. Operators and operator norms will be developed in more depth in Chapter 7. For the time being we want to show that the Hankel norm of a system equals the largest entry on the diagonal of the Gramians when the state model for the system is a balanced realization. Suppose we are given a strictly proper transfer function, G(s), and minimal state model, (A, B, C), for some stable system. Then the Hankel operator, r G' for this system has output y(t) : t E [0, oo) produced by input u(t) : t E ( -oo, OJ when x( -oo) = 0 and

u(t) = 0: t E [O,oo).

Now we define the system's Hankel norm, denoted IIG(s)IIH as

IIG(s)IIH

=

(max

u(~12..

y*(t)y(t)dtl)~ [It' tao u*(t)u(t)dt

(4.53)

IE(O,oo) x(-oo)=l'l

where (-)! indicates the positive square root. There is a close connection between a system's Hankel norm and that system's balanced realization as shown in the following t~orem.

Theorem 4.9 The Hankel norm IIG(s)IIH.'for a stable system having transfer function G(s) is given as (4.54) where ab 1 is the largest entry on the diagonal of the Gramians in balanced coordinates. Proof Recalling (4.29) we see that the eigenvalues of the product of the controllability and observability Gramians are invariant to a coordinate transformati on since we have

ww c

0

=

r- 1 wc r-•r·w

0

T

=

r-'wc w r 0

(4.55)

Thus when Tis the coordinate transformati on which puts the system in balanced form we have

where {ab; : i = 1, 2, · · · n} are referred to as the Hankel singular values. Next recall from Section 4.4.3 that when the state model is a balanced realization, the ~aximum output energy, Eomax> produced ?Y ~n initial state, z(O), constrained to satisfy z (O)z(O) =a , when u(t) = 0: t E [0, oo ), 1s giVen by Eomax

[Eo]= = max "(0)

[z",.x(o)] *:Eb[zmax(O)] = a 2ab 1

z' (O)z(O)=a2

where [~ax(O)f =[a

0

· · · OJ

114

Model Approxim ation via Balanced Realizati on

Alternatively, recalling the solution to the least energy input problem, we see from produce (4.52) that using the least energy input u0 (t) : t E ( -oo, 0] when z( -oo) = 0 to zmax(O) gives the least energy as )

E cmin

=

· [Ecu ] mm

=

[z max ( 0 )] * :Eh-I z max ( 0 )

ull) IE( -x.O]

Cl=(}hI

.:;(-'X)=O

:(Oj=:max (O)

Therefor e we see that the maximum value of the ratio of energy ~ut to energy in is given as Eomax

~

=

2

,2

(Tbi =/\max

[

we wo ]

cmm

which from (4.53, 4.55) yields (4.54 ).

4.9

•

Notes and References

Hermitia n matrices play an importan t role in many branches of science and engineering, e.g., quantum mechanics. In general a square matrix having repeated eigenvalues may not be able to be diagonalized by a similarity transform ation, i.e., we may not be able to find V such that v-I MV is diagonal. However a Hermitia n matrix can always be diagonalized even if it has repeated eigenvalues, [39]. The term "Gramia n" arose originally in connectio n with the problem of determining if the vectors in a set of vectors {v; : i = 1, 2, · · · , n} were linearly independent. The Gramian G was defined as the scalar whose value was determined as G = det[M], where (M)u = v; *v1. The independence of the v;s required G to be nonzero, [5]. As used in system theory, the term Gramian describes a matrix, not a scalar, which is formed in manner reminiscent of the foregoing matrix G. Balanced realization was first used in the area of digital filters to combat inaccuracies (roundoff) resulting from computa tion errors brought about by the requirem ent that calculations be done using finite precision arithmeti c [29]. The use of a balanced realization to obtain a reduced order model approxim ation and the method given here for calculating a balanced realization was originally presented in [27]. Theorem 4.8 and the proof of Theorem 4.3 (Appendix) were originally given in [34]. The idea of a balanced realization was extended to unstable linear systems, [22] and to a class of nonlinea r state models, [38]. Finally, we need to take note of the following importan t aspect of model reduction. We will see in the remainde r of this book that optimal controller design techniques, e.g., quadratic , Gaussian , and H x feedback control theories, make use of an observer as part of the feedback controller. Since the observer order matches the plant order, there can be problems in implementing controllers for use with high order plants. Unfortun ately, the stability of the closed loop system may not be preserved when the controlle r designed using the optimal control theory is replaced by a reduced order model. A good summary of approach es to this problem up until 1989 is given in [2]. A more recent approach to solving this problem, involving H X! robust control theory is given in [30] for linear plants and in [32] for a class of nonlinea r plants.

5 Qua drat ic Control

5.1

Introduction

In the preliminar y part of this chapter. we introduce the basic approach to feedback control known as observer based feedback control. In this scheme the plant state x(t) in the state feedback control signal u(t1 = Kx(t) + v(r) (Chapter 2) is replaced by the estimated plant state i(r) obtained from an observer (Chapter 3). As we will see in the remainder of this book, this scheme has enabled the developme nt of several optimizati on based controller design techniques . In addition to quadratic control treated in this chapter, we will see that observer based feedback control is used in linear quadratic Gaussian, LQG, control and in HX! control. Following this, we take up the use of quadratic control to determine the K and L matrices in an observer based feedback control setup. Quadratic control arises\\ hen there is a need to maintain the control system's steady-stat e output constant in response to a constant control input. In this situation the control scheme is said to operate as a regulator. In practice, impulse like disturbanc e signals, i.e .. signals having large amplitude and short time duration. enter the control loop and cause the output to jiggle about its desired level. A well known example of this is the regulation of the pointi~g direction of an outdoor dish antenna. Unfortuna tely the dish acts like a sail and experience s torque disturbanc es because of wind gusts which tend to change its pointing direction. Therefore we need to choose K and L so as to alter the closed-loo p system dynamics. i.e .. assign the eigenvalue s of the closed-loo p system matrix, in a way which reduces the effect on the output of these impulse like disturbanc e signals. In quadratic control this choice of K and Lis done so as to minimize the energy in the output signal caused by the disturbanc e. We will see that the determina tion of the optimal K and L matrices needed to achieve this minimum requires the solution to two nonlinear matrix equations known as algebraic Riccati equations. Special solutions to these equations, referred to as stabilizing solutions are required. We will develop conditions on the plant and optimizati on criteria which will ensure the existence of such solutions. We will encounter the Riccati equation again in connection with other optimal control schemes developed in this book. More specifically, w_e will see that the algebraic Riccati equation is invohed in the solution of the di~turbance attenuatio n problem when the disturbanc es are viewed as rand::>m signals With known statistics (Chapter 6) and when the disturbanc e signals are restricted to the so-called L 2 class of signals (Chapters 9 and 10).

116

5.2

Quadrati c Control

Obser ver Based Contro llers

Although errors in the estimated plant state in an observer based control system produce effects at the control system's output, we will see that the dynamics of the observer do not contribut e to the control system's zero-state response. More specifically, we will show that the control system's transfer function equals the transfer function for the closed loop feedback system when exact state feedback is used, i.e., when the estimated state equals the plant state. Suppose the plant state model paramete rs (A, B, C, D) are given. Then recalling the development of the observer, (Chapter 3), we see that the-plant state, x(t), and the observer state, x(t), are governed by the following vector differential equation s

x(t)

=

Ax(t)

+ Bu(t)

x (t) =(A+ LC)x(t)

(5.1)

+ (B + LD)u(t) - Ly(t)

(5.2)

Now recall from Chapter 2 that to apply state feedback we set u(t) = Kx(t) + v(t). a Lacking exact knowledge of the plant state we replace it by its estimate, .X( t), to get control input

u(t) = Kx(t)

+ v(t)

(5.3)

Then substitut ing (5.3) in (5.1, 5.2) and the plant's output equation , we see that the control system is a 2n dimensional system having state model given as

Xc(t) = AcXc(t) + Bcv(t)

(5.4)

+ Dcv(t)

(5.5)

y(t) = Ccxc(t) where

Ac =

BK

[A

- LC A + LC + BK

Cc = [C

v(t) +

]

x(t)] Xc(t) = [ x(t)

DK]

u(t)

~

y(t)

PLANT

+

~

STATE FEEDBACK

x(t)

OBSERVER

1<---

l Figure 5.1

Setup for observer based feedback control

Obsen/er saseci cOiitroriers,

111

sion n, the closed-loop Notice that since the plant and observer are each of dimen state mo~el is minimal, plant the if even ver system state model is of dimension 2n. Howe , (5.4, 5.5), IS not controli.e. observable and controllable, the closed-loop state model the resulting state model is in lable. We can see this by changing coordinates so that _ as follows. controllable decomposed form, (Section 2.5). We do this to Be matrix input ormed Recall that controllable decomposed form requires the transf a , halves equal oned into two have a lower partit ion which is null. Since Be is partiti given by coord inate transf ormat ion matrix, T, which does this is

the contro l system state Therefore, after using this coord inate transf ormat ion matrix model parameters, (5.4, 5.5), become Ac =

[Acl ~c2] 0

Ac4

Cc = [C+D K

[

,

x(t) x(t)

l

(5.6)

-DK]

where Aci =A+ BK

Acz = -BK

Ac4

= A+L C

x(t) = x(t) - x(t) plant state estimation error, Notice, from the null block structure of Ac and Be, that the er function is given as x(t), is uncontrollable and that the contro l system transf Gc(s)

= Cc(s l- Ac)-I Be+ De = [C+D K -DK ][(sl -Acd -I 0

=

(C + DK)[ sl- (A+ BK)r 1B + D

*_ (s/- Ac4 )-

I]

[B] +D

0 (5.7)

i.e., (A, B) controllable and Therefore provid ed the plant state model is minimal, (A + BK, C + DK) observable (A, C) observable, we have (A + BK, B) controllable and ing transfer function and the forego the in s so that there are no pole-zero cancellation e this contro l system transfer contro l system and plant have the same order. Notic we used exact state feedback. function is the same as we obtain ed in Chapt er 2 when outpu t or zero-state behavior !here fore we can conclude that the contro l system's inputer. However, the contro l system's zero-input IS unaffected by the presence of the observ follows. response is affected by the observer. We can show this as the differential equations that see we (5.6), l, mode state 's system From the contro l

118

Quadratic Control

governing the plant state and plant state estimation error are ~\'(t) = A,lx(t)

.X (t)

+ Ac2~'X-(t) + Bv(t)

(5.8)

= A, 4 ~x( t)

Then the state estimation error is independent of the input and depends only on its initial value as

x(t)

=

e(A+LC)tx(O)

(5.9)

Therefore substituting this expression for .X(t) in (5.8) yields

x(t) =(A+ BK)x(t)- BKe(A+LC)rx(O)

+ Bv(t)

(5.10)

and we see that the evolution of the plant state depends on the state estimation error. However, this effect which is caused by the presence of the observer in the feedback loop, disappears with time. More specifically, recall from Chapter 3 that L must always be chosen to make A+ LC is stable. Therefore we see from (5.9) that lim.X(t) = 0

for any .X(O)

(-'t(X;-

Thus, for some time following the initiation of the feedback control, the plant state, x(t), and plant output, y(t), are both affected by the plant state estimation error, .X(t). Eventually the term A, 2 x(t) can be neglected in comparison to other terms on the right side of (5.8) so that for large t the derivative of the plant state can be approximate d as

x(t) =(A+ BK)x(t)

+ Bv(t)

(5.11)

Notice that (5.11) is the differential equation for the plant state when exact state feedback is used. Finally, recalling that the eigenvalues of the system matrix A, are preserved under coordinate transformati on, (Section 1.6.4), and that block upper-triang ular matrices have eigenvalues which are the union of the eigenvalues of the diagonal blocks, (Appendix), we see from (5.6) that

>.[A,] = >.[A,] = >.[A + BK] u >.[A + LC] Thus half the closed loop system eigenvalues depend on K and are independent of L, and the other half depend on L and are independent of K. This spliting of the closed loop eigenvalues is responsible, in part, for being able to break the optimal control problems treated in this book into two simpler problems: 1. the state feedback control problem 2. the state estimation problem This fact, referred to as the separation principle, has been a major theme running through the development of linear control theory since its discovery in the 1960s.

Quadra tic State Feedba ck Control

119

state In summar y, the use of an observer-estimated plant state in place of the true plant having system control k feedbac a to leads ration in a state feedback control configu as we would transfer function (steady-state. input-ou tput behavior) which is the same system, control the of start-up the at initially r. Howeve . feedback obtain using exact state the affect will state plant true the from state ted r-estima observe any departu re of the state ted control system output. This effect diminishes \Vith time as the observer-estima approac hes the true plant state. ues for Recall, from Chapter s 2 and 3. that we needed the specification of eigenval control k A + BK and A + LC in order to determine K for the design of the state feedbac , we are only and L for the design of the state estimato r, respectively. However, in practice re, in this given some desired goals for the behavio r of the control system. Therefo control the situation, we need methods for turning specified goals for the behavio r of and in system into values for the K and L matrices. In the remaind er of this chapter, for goal subsequ ent chapters , we will show in each case that by interpre ting the specified develop can the control system's behavio r as a mathem atical optimiz ation problem , we equation s for determi ning the required K and L.

5.3

Quadratic State Feedback Control

mandato ry There are many possible control system design objectives in addition to the ents arises requirem al addition these of One requirem ent that a control system be stable. control the entering signals nce disturba ration when there are large-amplitude, short-du the design to need we n situatio this In t. constan loop and we want to maintai n the output , possible as much as , diminish to and system controller both to stabilize the closed-loop optimal the to solution The output. plant the on the effect of the disturba nce signals L both to quadrat ic control problem attempt s to achieve this goal by determining K and control the of re departu the in energy the e minimiz stabilize the closed-loop system and to \\e section, s previou the in ed mention As value. tate system output from a desired steady-s lems: subp.rob simpler two into problem this can use the separati on principle to split l. the quadrat ic state feedback control problem and 2. the quadrat ic state estimati on problem with the solution to 1. determining K, and the solution to 2. determining L. ic state In this section we develop the design equation s forK which solves the quadrat of the sum the s, equation feedback control problem . When K satisfies these design ed. minimiz is signal, nce energies in the output and feedback signal caused by the disturba ic quadrat the as known One of these design equation s is a nonline ar matrix equation to needed und backgro cal theoreti control algebraic Riccati equatio n (QCAR E). The solve this equatio n is provide d in the next section, Section 5.4. ic state !n S~ction 5.5 we develop the design equation s for L to solve the quadrat c algebrai an of solution the involves this es!Imatlon problem . Again we will see that the as known is and QCARE the to related Riccati equation . This equatio n is closely to the quadrat ic filtering algebraic Riccati equatio n (QF ARE). The required solution the solving for 5.4 Section in ed develop res procedu using QFARE can be obtaine d ~ QCARE .

Quadratic Control

120

5.3.1

Motivating the problem

Let the disturbance input to the plant be denoted as u 1 ( t); let the controlled input involved in implementing the state feedback be denoted by u2(t). Then the plant state model is given by the equations

x(t)

=

Ax(t)

+ B1 u1 (t) + B2u2(t)

(5.12)

y(t)

=

Cx(t)

+ D 1u 1 (t) + D 2 u2 (t)

(5.13)

and setting u2 (t) = Kx(t) + v(t), where v(t) is the command input, we obtain a state feedback control system having state model given as

x(t)

=

Ax(t)

+ Bl

+ B2v(t)

(5.14)

y(t)

=

Cx(t)

+ D 1u 1 (t) + D 2v(t)

(5.15)

Uj

(t)

. where

disturbance input controlled input

y(t) v( t)

control system output control system command input

However, if (i) the disturbance input is null, u 1 (t) = 0 (ii) the control system command input is constant, v(t) = v5 , (iii) the feedback matrix K makes the control system stable so that all eigenvalues of A lie in the open left half plane, then the constant steady-state control system state, x" is obtained from the control system state differential equation, (5.14) by setting x(t) = 0. Then using x, in (5.15) gives the steady-state control system output y 5 as

u,(t)

PLANT v(t)

+

y(t)

u,(t)

~

+

I+--CONTROLLER

Figure 5.2

If

Setup for disturbance input attenuation

Quadratic State Feedback Control

121

where Xs =

limx(t)

t-x

Now suppose the control system is operating at this steady-state level at time t = t 1 when a short-duration disturbance signal, lasting T seconds, arrives at the disturbance input. Then the disturbance input can be written as u 1 (t)

#

for t E h

o

t1

+ Tj

otherwise

=o

Therefore, shifting the time origin to the instant immediately following the end of the disturbance, i.e., to time t 1 + T. we are left \vith the state immediately following the end of the disturbance being given by

However since A

+ BK is assumed stable we have lim xd(t) = o {-H:X!

and the state of the plant returns to its steady~state Yalue with time, limx(t) = x, {-H)O

Our goal in choosing K is to decrease the effect of the disturbance input on the control system output during the return of the output to its steady-state value. Since superposition applies. we can assume, without loss of generality, that the control system command input is null, v( t) = vs = 0. Then the desired steady-state output is null and any deviation of the output away from being null is caused by the presence of the disturbance input. Thus we want to choose K so that the action of the feedback signal attenuates these deviations as much as possible.

5.3.2

Formulatin g the problem

Recall, from the previous chapter, that we used the observability Gramian and its relation to the output energy to obtain a measure of the zero-input response. We use this idea here to reformulate the problem just discussed as the problem of choosing K to minimize the ~ffect of a non-null initial plant state on the sum of the energy in the output and the energy In the feedback signal, i.e.,

where lQc =

=

laoo (y'(t)y(t)- pu2(t)u2(t))dt la

00

[x*(t)C*Cx(t)

+ u;(t)[pJ]u 2 (t)]dt

(5.16)

122

Quadrati c Control

with p being a positive scalar and JQc being referred to as the cost or the performa nce index for the quadratic control problem. Notice that the reason for including the feedback signal energy in the performa nce t) index, i.e., for including u 2 ( t) in the integrand , is to limit the feedback signal u 2 ( t) = Kx( given is this of on explanati detailed more A to a level that can be physically implemented. as follows. When we choose K to minimize J QCo we discover that the smaller p the larger the elements inK. If we were to let p be zero so that there would be no contribut ion to the to cost, lQc, from the feedback control signal, u2 (t), we would find that the K required feedback required the case this in Thus entries. minimize lQc would have unbound ed control signal u 2 ( t) would be an impulse. This physically unrealizable requirement prohibits the use of p = 0. However, while increasing the size of p has the beneficial on effect of reducing the size of the required feedback signal, the effect of the disturban ce between ise comprom a pis of size the for choice the output is increased. Thus a beneficial limiting the size of the control signal, u2 ( t) and limiting the effect of the disturban ce on the outpu~.

In order to simplify the development to follow, we will assume that there is no direct feed through of the feedback controlle d input u 2 ( t) to the plant output, i.e., we assume D 2

=0

In addition, since it is always possible to choose coordina tes for the state space so that the paramete rs of the state model are real, we will assume that these paramete rs are real. Finally, for generality in what follows, we will replace C* C and pi in (5.16) by Qc and Rc respectively so that the performa nce index becomes ( 5.17) all where Qc, Rc are real, symmetric matrices with Qc ?: 0 and Rc > 0 implying that for x(t) and u(t) we have

x*(t)Qcx(t)?: 0 u).(t)Rcu2(t) > 0 These restrictions on Qc and Rc are needed to ensure that the contribut ions to lQc from both u2 (t) and x(t) are never negative and so that R, is invertible. In summary , we have formulate d the problem of designing a state feedback controlle r to combat the effect of unknown short-du ration disturban ce inputs as an optimiza tion problem. The solution to this optimiza tion problem requires that K be determin ed so that is a quadratic performa nce index in the disturbed state and the controlle d input minimized.

5.3.3

Develo ping a solution

From the discussion earlier in this section, we suppose that the comman d input is null, has v( t) = 0, and that the time origin is taken at the instant a short-dur ation disturban ce

Quadra tic State Feedba ck Contro l

just ended. This leaves the plant state at some non-null value, x(O) · have the plant state given by

= xd

123

and thereafter we

( 5.18)

and the controlled input given by (5.19)

Thus using (5.18, 5.19) in the quadratic control cost, (5.17), yields ( 5.20)

where Qc = Qc + K*RcK

J. =A +B2K -

I integral, (5.20), Now since we are assuming that K is stabilizing, i.e., A is stable, the converges to a real symmetric matrix PQc, ( 5.21)

and the performance index, (5.20), can be written as (5.22)

ov ¥AUation Recall from Chapte r 4 that PQC• (5.21), is the solution to the Lyapun (5.23)

to this Lyapunov Therefore, since A, Qn (5.20), depend on K, so does the solution, PQc, writing PQc and by fact this denote We equatio n and the performance index, JQc, (5.22). JQc as PQc(K ) and JQc(K, xd) respectively. From the foregoing we conclude that our goal is to find K such that (i) K is stabilizing, i.e., A + B 2 K is stable, and all initial states, xd . (ii) the solution, PQc(K ), to (5.23) minimizes JQc(K, xd), (5.22), for satisfies (i) and .. In the following development we use K 0 to denote the value of K which (u). Thus K 0 satisfies (5.24)

for all disturbed initial states, xd·

124

Quadratic Control

Now we are going to use a matrix variational approach in connection with the Lyapunov equation, (5.23) to develop an equation for P QCo, where P QCo denotes the matrix PQc which solves (5.23) when K = K 0 • We begin by noting that any real and stabilizing K can be written as ( 5.25) where Eisa positive real scalar and b K is a real matrix having the same dimension as K with E(bK) being referred to as a perturbation on K,n the desired value forK. Since P QC depends on K through the Lyapunov equation, (5.23), the perturbation, E(bK), on K 0 produces a series expansion for PQc about its optimal value PQco, ( 5.26) where HOT (higher order terms) stands for terms in the expansion involving powers of E which. are 2 or greater. The term (b1P QC) is a symmetric matrix which depends on (b K) in a manner to be determined. Next we substitute the expansion of PQc, (5.26), into the expression for the cost lQc, (5.22). This gives an expansion for lQc about its minimum value lQco (5.27) where

and again HOT stands for terms in the expansion involving powers of E which are 2 or greater. Now since the terms indicated by HOT in both (5.26) and (5.27) depend on {Ei: i=2,3,···}weseethat lim[HOT] E-->0

=

0

( 5.28)

E

Therefore it follows that the scalar, (b 11Qc), in (5.27) and the matrix, (b 1PQc), in (5.26) can be expressed as (5.29)

( 5.30) Notice that (5.29) is reminiscent of the derivative in calculus. This fact together with ideas we encounter in using calculus to obtain minima of a scalar function of a scalar variable enables us to visualize the present problem of minimizing a scalar function J QC of a matrix K as follows.

Quadratic State Feedback Control

125

Suppose, for simplicity we assume u2 ( t) is a scalar. Then K is a row vector with n scalar entries. Then we can think of JQc together with the components of K as forming ann+ 1 dimensional space. Imagine making a plot of the scalar JQc on the "vertical axis" vs. K with K being represented by a point in a "horizontal" hyperplane, or plane when n = 2. The resulting surface in this space would appear as a bowl with the bottom of the bowl as the "point" {JQCo• K 0 }, i.e., the optimal point. At this point (8tJQc), (5.29), is zero both for all8 K and for all initial disturbed states xd. This is analogous to the derivative of a scalar function of a scalar variable being zero at points where the function has a local minimum. Notice from the dependency of (8 1JQc) on (8 1PQc), (5.27), namely

that (81J QC) is zero for all xd if and only if (81PQC) is a null matrix. Therefore the optimal value forK, K 0 , makes (8 1PQc) null for all (8K) which maintain K, (5.25), stabilizing. Therefore in order to determine K 0 we need to determine an equation for the matrix (8 1PQc) which involves (8K). We can do this by substituting the expansions forK and for PQc (5.25, 5.26), in the Lyapunov equation, (5.23). The results of doing this are detc;rmined as follows. begin by denoting the result of using the expansion of K, (5.25), in A and Qc, (5.20), as

We

A= A + EB2(8K) 0

Qc = Qco + E[(8K)* RcKo + K~Rc(8K)] +HOT

(5.31)

where

Then substituting these expressions for A, Qc as well as the expansion of P Q~' (5.26), in the Lyapunov equation, (5.23), yields A~PQCo + PQCoAo + Qco + E(M + M*) +HOT= 0

(5.32)

where

Notice that the first three term~ in (5.32) sum to a null matrix since PQco satisfies the Lyapunov equation, (5.23), when K = K 0 • Thus (5.32) becomes

E(M + M*) +HOT = 0

(5.33)

Then dividing (5.33) through byE, letting Ego to zero, and using the limiting property,

126

Quadratic Control

(5.28), yields M

+ M* = 0

Finally, substituting for M from (5.32) yields the following Lyapunov equation relating (8 1PQc) to (8K) (5.34)

Recall, from discussion earlier in this subsection, that K = K 0 when matrix for al18K which maintain K, (5.25) stabilizing. Therefore when matrix, (5.34) becomes

(8 1PQc) (8 1PQc)

is a null is a null

( 5.35)

which· is satisfied for all stabilizing perturbations, 8 K, only if M 1 is a null matrix. Therefore we see from the definition of M 1, (5.32) that (5.36)

where we have used the assumption Rc > 0 to insure that Rc is invertible. Notice, at this stage of the development, that the state feedback matrix K 0 is readily determined from (5.36) once we have P QCo· It turns out that P QCo is a solution to a nonlinear matrix equation known as the quadratic control algebraic Riccati equation (QCARE). This equation can be developed by setting K = K 0 , (5.36), in the Lyapunov equation (5.23). After carrying this out, we get the QCARE as A* P QCo

+ P QCoA -

P QCoRQcP QCo

+ Qc = 0

(5.37)

where

Having determined the QCARE it remains to solve it for PQca such that A+ B2Ka is stable when PQco is used to determine K 0 from (5.36). We can recap this section as follows. We began by introducing a matrix variational approach and using it to determine an equation for the optimal feedback matrix, Ka, (5.36). This equation involves P QCo• the matrix which solves the Lyapunov equation, (5.23), when K = K 0 . Then we saw that by eliminating K 0 from the Lyapunov equation by using (5.36) we obtained a nonlinear matrix equation in PQCo referred to as the QCARE, (5.37). In the next section we will be concerned with conditions on the plant state model parameters and the parameters of the performance index, Rc and Qc, which ensure the existences of the so-called stabilizing solution, P QCo• to the QCARE, i.e., the solution which makes A- B 2 R; 1B* PQco stable.

127

Solving the QCARE

5.4

Solving the QCARE

We have just seen that the determin ation of the state feedback matrix K 0 , (5.36), which g minimizes the performa nce index JQc, (5.17), requires the determin ation of a stabilizin establish to is section present the of goal main The (5.37). , solution to the QCARE condition s on the plant state model paramete rs and weighting matrices Qc and Rc which ensure the existence of a stabilizing solution to the QCARE. We will be aided in this task by resorting to a 2n x 2n matrix called the quadratic control Hamilton ian matrix. Before doing this we consider aspects of the closed loop stability which are brought out using the fact the PQco is the solution to the Lyapuno v equation , (5.23), when K = K 0 •

5.4.1

Stabilizing solutions

Consider the following theorem and its proof. Theorem 5.1 A0 is stable if (i) (A, B2 ) is a stabilizable pair and (ii) P QCo satisfies the QCARE with P QCo 2': 0 and (iii) (A, Qc) is a detectabl e pair. wh(;e A0 =A+ B2 K 0 with K 0 given by (5.36) and PQco satisfying (5.37). Proof The necessity of (i) is obvious since we have, from Chapter 2, that a pair (A, B) is we stabilizable if we can find K such the A+ BK is stable. Therefor e assuming (i) holds, stable. be to A for sufficient 0 will show that (ii) and (iii) together are Notice from (5.36, 5.37) that

Therefor e adding and subtracti ng PQcoRQcPQCo on the left side of the QCARE , (5.37), gives the Lyapuno v equation , (5.23), with K = K 0 , (5.38) where

We proceed now to use the ideas from Chapter 4 concernin g stability in connectio n with the solution of Lyapuno v equations . First we rewrite (5.38) by expandin g Qc0 , (5.31), using K 0 , (5.36). This enables (5.38) to be rewritten as (5.39)

~en pre and post-mul tiply this equation by

Yields

vi* and vi respectively where

A vi

= >..vi

0

(5.40)

128

Quadratic Control

Next recalling that R, > 0 and that Q, 2: 0 we have (5.41) and the right side of (5.40) is either negative or zero, i.e., either (a) or (b)

Suppose (a) holds. Then we see from (5.40) that (5.42) and neither Re[A;] nor vi* PQcoVi can be zero. Therefore we see from condition (ii) in the theorem that v'* PQcov' > 0. Thus (5.42) is satisfied only if Re[A;] < 0. Alternatively, suppose (b) holds. Then we see from (5.41) that both and This implies that and

(5.43)

and we see that ' i /\;V

=

A~ o V ;

=

(A

~

1 B 2 Rc B*P 2 QCo ) V ;

=

AV i

(5.44)

Thus (A;, vi) is an eigenvalue, right-eigenvector pair for A satisfying Qcv; = 0 implying that\ is an unobservable eigenvalue for the pair (A, Q,). However (A, Q,) is detectable, condition (iii). Therefore if (b) holds we have Re[\] < 0. To recap, since the right side of(5.40) is either negative or zero and we have shown that in either case Re[A] < 0 when the conditions in the theorem hold. Thus we have shown that the conditions in the theorem are sufficient for A0 to be stable. • We will show in the next subsection that condition (iii) in the statement of the foregoing theorem is not necessary for A0 to be stable. Instead, we will show that only condition (i) in Theorem 5.1 and the condition that any imaginary axis eigenvalues of A be observable eigenvalues for the pair (A, Qc) are needed to ensure the existence of a stabilizing solution to the QCARE. Some appreciation of this weakening of condition (iii) can be obtained by reconsidering the foregoing proof as follows. First, notice that when case (b) in the proof holds, (5.40) implies that

Solving the QCARE

129

and either (bl): Re[.\] = 0 or (b2): PQcoV; = 0

However, recall from the proof that when case (b) holds the eigenvalue).., of A0 is also an unobservable eigenYalue for the pair (A, Q,). Therefore if we impose the condition that (A, Qc) has no unobservable imaginary axis eigenvalues, then case (bl) is impossible. Second, though not apparent at this stage in the discussion. we will see in the next subsection that if both condition (i) in Theorem 5.1 and the condition that (A, Qc) haYe no unobservable imaginary axis eigenvalues are satisfied, then eigenvectors of A can only satisfy the condition for case (b2) to apply if the corresponding eigenvalue of A is stable. Finally, notice in the proof of Theorem 5.1, that case (b) is impossible when the pair (A, Qcl is observable and thus PQcu > 0 \\hen (A. Qc) is obseryable. Again, though not obvious now, we will show that we can replace this condition for P QCo > 0 in case (b) by the condition that (A, Q,) have no stable unobservable eigenvalues. Therefore this condition together with condition (i) in Theorem 5.1 and the condition that (A, Q,) haye no imaginary axis eigenvalues are the necessary and sufficient conditions for PQco > 0.

The uniqueness of the stabilizing solution is shown in the following theorem. Theorem 5.2 The stabilizing solution to the QCARE is unique, when it exists. Proof Suppose P QCI and P QC are two stabilizing solutions to the QCARE. This enables us to write

+ Qc = 0

A*PQcl

+ PQClA-

A* PQc2

+ PQc2A- PQc2RQcPQc2 + Q, = 0

PQciRQcPQc!

where A; = A - RQcP QCi : i = 1, 2 are each stable. Next taking the difference between these two equations yields A*(b.PQc)

+ (b.PQc)A- PQclRQcPQcl + PQc2F·QcPQc2 =

0

where (b.PQc) = PQCI- PQc 2. Then adding and subtracting P QCl RQcPQc 2 enables this equation to be rewritten as (5.45) Now pro_cee~ing in the same manner as was done to prove Theorem 4.5, we can shO\v that when A 1 , A 2 are stable the matrix equation "4i(~PQc)

+ (!:::,.PQc)A2 =

has solution !:::..PQc which can be written as the integral

.\1

130

Quadratic Control

However since the right side in (5.45) is null, M is null and the integrand of the foregoing integral is null for all time. Therefore we have !::,.PQc = 0 which implies that • PQci = PQc 2 and the stabilizing solution to the QCARE is unique. Having examined certain properties of stabilizing solutions to the QCARE, we turn next to questions concerning the existence and calculation of such solutions. An important tool for answering these questions is the Hamiltonian matrix.

5.4.2

The Hamiltonian matrix for the QCARE

In this section we will show the relation between the solutions of the QCARE and properties of a 2n x 2n matrix made up from the parameters of the QCARE, (5.37). This matrix, called the quadratic control Hamiltonian matrix, is defined as HQc =

[AQ,

( 5.46)

where RQc = B 2 K; 1B;. Notice that the blocks of H QC depend on all the matrices in the QCARE, (5.37), except its solutions P QC· Recall that we are interested in the stabilizing solution denoted P QCo· In the following theorem, we give a property of H QC which is important in establishing conditions for the existence of a stabilizing solution to the QCARE. Theorem 5.3 It is not possible to find a stabilizing solution to the QCARE, (5.37), if H Qc, (5.46), has imaginary axis eigenvalues. Proof Define the 2n x 2n matrix T as (5.47)

where P QC is any Hermitian matrix. Notice that Tis invertible for all Hermitian matrices, P QC, having inverse given by

r-I =

[P~c ~]

Next let HQc be related to H QC as -HQc =

T -1 HQcT =

[Az-

R-- QAc*

l

where A= A- RQcPQc Z =A* PQc

Notice that

+ PQcA- PQcRQcPQc + Q,

( 5.48)

•

Solving the QCARE

131

independent of P QC· Therefore if we choose P QC so that Z is null, H QC is block uppertriangular and (5.49) >.[HQc] =>.[A] u >.[-A'] However since complex eigenvalues of A occur in conjugate pairs, we have A[A*] = >.[A]. This together with the fact that >.[-A] = ->.[A] allow us to rewrite (5.49) as >.[HQcl =>.[A]

u -A[A]

(5.50)

This shows that the eigenvalues of H QC are mirror images of each other across the

imaginary axis, i.e., if 'TJ E >.[H Qcl then-ryE A[HQcl· Now when we choose PQc so that Z is null, PQc is a solution to the QCARE, (5.37). Moreover, recalling the expression for the feedback matrix K 0 , (5.36), we see that A, (5.48) can be rewritten as

when PQc = PQco, the stabilizing solution to the QCARE. However, if HQc has an imaginary axis eigenvalue, (5.50) shows that this eigenvalue must also be an eigenvalue for Xfor z!l Hermitian matrices PQC which satisfy the QCARE. Therefore it is impossible • to choose P QC to satisfy the QCARE, i.e., to make Z null, so that A is stable. To recap, the 2n eigenvalues of H QC split into two sets of n eigenvalues each such that each eigenvalue in one set has a corresponding eigenvalue in the other set which is its mirror image across the imaginary axis. Thus if PQC is a stabilizing solution to the QCARE, the eigenvalues of A, (5.48), equal n of the eigenvalues of H QC which are in the · open left half plane. However, if H QC has imaginary axis eigenvalues, H QC does not have n eigenvalues in the open left half plane. Therefore, in this situation, a stabilizing solution to QCARE does not exist since the eigenvalues of A must include at least one imaginary axis eigenvalue of HQc for all solutions, PQc, to the QCARE. We need now to establish conditions which ensure that H QC does not have imaginary ?~ axis eigenvalues. This is done in the following theorem. . Theorem 5.4 The quadratic control Hamiltonian matrix H QC• (5.46), is devoid of Imaginary axis eigenvalues if (i) (ii)

(A,,RQc) has no uncontrollable eigenvalues on the imaginary axis and

(A, Qc) has no unobservable eigenvalues on the imaginary axis

Proof Suppose HQc has an eigenvalue atjw. Then we have

(5.51)

HQcv=jwv

[ A .

Qc

R.Q: ] [ v 1 ] = jw [ v 1 ] -A

Vz

Vz

or (jwl - A )v 1 = RQcVz (jwl + A*)vz

=

Qcvi

(5.52) (5.53)

Quadratic Control

132

Next, we premultiply (5.52) by

v2 and (5.53) by vr

to obtain

v}(jwl- A)v 1 = v.2RQcv2 v[(jwl

+ A*)v2

=

v[Qcvl

(5.54) (5.55)

Then taking the conjugate transpose on either side of (5.55) and recalling that Q,. 1s Hermitian enables us to rewrite (5.55) as ( 5. 56) Now we see, by comparing (5.54) with (5.56), that (5.57) How~ver Qc and RQc are each non-negative. Therefore the only way (5.57) can be

satisfied is for each side to be zero. Thus we have (5.58) (5.59) and (5.52, 5.53) become (5.60) (5.61) This shows that jw is an eigenvalue of A and that v 1 , v 2 are the corresponding right and left-eigenvectors of A. However if v 2 f 0, (5.58, 5.61) imply that (A, RQc) has an uncontrollable eigenvalue on the imaginary axis and condition (i) is not satisfied. Therefore if (i) is satisfied, we must have v 2 = 0. Alternatively, if v 1 f 0, (5.59, 5.60) imply that (A, Qc) has an unobservable eigenvalue on the imaginary axis and condition (ii) is not satisfied. Therefore if (ii) is satisfied, we must have v 1 = 0. From the foregoing we see that if conditions (i) and (ii) are satisfied, (5. 51) holds only if v = 0. Thus, contrary to the assumption we made at the beginning of the proof, H QC has • no imaginary axis eigenvalues when the conditions of the theorem are satisfied. 1 Notice that since R, is non-singular and RQc = B 2R-; B2, we have only if Therefore condition (i) in the foregoing theorem is equivalent to (i)

(A, B 2 ) has no uncontrollable eigenvalues on the imaginary axis.

We will show in the next subsection that, though sufficient for H QC to have no

Solving the QCARE

133

iJDaginary axis eigenvalues, the conditions in the foregoing theorem are not sufficient to ensure that the QCARE has a stabilizing solution.

5.4.3

Finding the stabilizing solution

The proof of Theorem 5.3 suggests a way of determining the stabilizing solution to the QCARE. More specifically, we saw t~at when the stabilizing solution to the QCARE exists, we can find T, (5.47), such that HQc, (5.48), is block upper-triangular with the top diagonal block being stable. This suggests that the stabilizing solution to the QCARE can be determined from the eigenvectors of H QC corresponding to the stable eigenvalues. However, it turns out that we can avoid having to compute eigenvectors. Instead we can use Schur decomposition on H QC to determine the stabilizing solution to the QCARE. In order to see how to do this we need the following basic result form linear algebra. Theorem 5.5 If (5.62)

MU=UW

where M and Ware square with W being invertible and U having independent columns, then/ (i) each eigenvalue of W is an eigenvalue ofM, i.e., .A(W] c .A(M] (ii) range[U] is an M-invariant subspace corresponding to A.[W], i.e., range(U] is spanned by the eigenvectors of M corresponding to the eigenvalues of W. ~oof ~et (.A;, vi) be any eigenvalue-eigenvec tor pair for W. Then premultiplying . Wv' = .A;v' by U and using (5.62) we obtain

Mw=A.;w

.

where w = Uvi and we see that A; is also an eigenvalue of M. This completes the proof of ~

.

Next assuming the eigenvectors of W are complete, we can express any vect~r q of the same dimension as these eigenvectors as

n, q= LO'.;Vi i=i

so that post-multiplying (5.62) by q gives

Ms=r where

n, r= Up

p = L(A.;a;)vi

i=i S=

Uq

We see that M maps range(U] into itself, i.e., range[U] isM-invariant.

134

Quadratic Control

Finally, notice that if V 1 is a matrix whose columns are the eigenvectors of M corresponding to the eigenvalues of M which are also eigenvalues of W, then MVI =VIAl

where A 1 is diagonal with the eigenvalues of W along its diagonal. Then post-multiplying this equation by the non-singular matrix e such that U = V 1e and inserting ee-l between V 1 and A 1 gives MU=VW

where

Thus .A[W] =.-\[AI]. Moreover since

e is non-singular, we have

range[U] = range[V 1e] = range[VI] This shows that range[U] is spanned by eigenvectors of M corresponding to the • eigenvalues of W. In order to develop a relation between the quadratic control Hamiltonian matrix and the QCARE which we can use as a basis for solving the QCARE for its stabilizing solution, we proceed in the following manner. Suppose we have the stabilizing solution, P QCO" Then the QCARE, (5.37), can be written as

and we see that (5.63) where HQc

=

[A

Qc

Now since A- RQcPQco is stable, we see from Theorem 5.5 that (5.63) implies range [ _ :QCo ] is the H Qc-invariant subspace corresponding to the stable eigenvalues of H QC· The standard way of computing this subspace, when it exists, is to use Schur decomposition. This numerically reliable method computes an orthogonal (unitary when H Qc is complex) matrix T such that the transformed quadratic control Hamiltonian matrix, HQc, (5.48), is block upper-triangular with first diagonal block, H11, being

Solving the QCARE

135

stable, i.e., ( 5.64)

Notice from (5.48) that H 11 =A- RQcPQco = "4o. the stable system matrix for the state feedback control system. Now since Tis unitary, T* = T -t. we see that premultiplying (5.64) by T and equating the first n columns on either side of the resulting equation gives (5.65) where Tis partitioned into n x n blocks, T;i, with T 1 denoting the first n columns ofT, i.e.,

Then we see from Theorem 5.5 and (5.65) that (i/ >.[Hn] c >.[H Qcl (ii) range[Td is an HQcinvariant subspace corresponding to >.[H 11 ]. Therefore since H 11 is stable, the columns ofT 1 span the stable eigenspace of H QC• i.e., the columns of T 1 can be expressed in terms of the eigenvectors of Hoc whose corresponding eigenvalues lie in the open left half plane. Recall that range[Td =range[T 18] for any non-singular matrix 8 of appropriate dimension. Hence if T 11 is non-singular the columns of T 1T]} also span the stable eigenspace of H QC· Therefore assuming T 11 is invertible, we postmultiply (5.65) by T]} and insert T]}T 11 between T 1 and H 11 on the right side of(5.65) to obtain HQc[T21IT]/]

= [T21/T)/](TuHIIT)/)

Then comparing this equation with (5.63) we see that PQco is given by PQco

=

-T21TJ11

(5.66)

and we have the stabilizing solution to the QCARE. Notice that in order to use the foregoing approach to solYe the QCARE for its stabilizing solution, not only must H QC have no imaginary axis eigenvalues, (Theorem 5.3), but in addition T 11 must be invertible. Notice also that when T 11 is invertible we have P QCo > 0 only if T 21 is also invertible. The following theorem establishes conditions ?n the parameters of the plant ahd control cost which ensure that T 11 and T 21 are Invertible. Theorem 5.6 Suppose we can find a unitary matrix T so that H 11 is stable and

Quadratic Control

136

where

Then we have the following results: (i) T 11 is invertible if (A, B 2 ) is stabilizable; (ii) if T 11 is invertible, then T 21 is invertible if (A, Qc) has no stable unobservable eigenvalues. Proof (i) From (5.65) and the blocks in H QC and H

+ RQcT 21

11

we have

= T 11 Htt

( 5.67)

QcT11- A*T2l = T21H11

( 5.68)

AT 11

letting let (,\, v) be any eigenvalue-right-eigenvector pair for H 11 and postmultiplying (5.67, 5.68) by v gives Th~n

AT 11 v + RQcT 21 v = ,\Tllv

( 5.69)

QcT11v- A*T2lv = ,\T21 v

( 5.70)

Now suppose TIIv = 0. Then T 11 is not invertible and (5.69, 5.70) become RQcw = 0 A*w

=

-,\w

(5.71) (5.72)

where w = T 21 v. Then if w = 0 we would have

T ]v=0 [T21 11

implying that T has dependent columns and is therefore not invertible. This is impossible since we are assuming that Tis unitary. Therefore 1v -:10. Now since H 11 is stable, (5.71, 5.72) imply that (A, RQc) is not stabilizable, and since RQc = B 2R-; 1B2 with R-; 1 non-singular, we have (A, B 2) is not stabilizable. This contradicts the condition given in (i). Therefore no eigenvector v, of H 11 satisfies T 11 v = 0 when (A, B 2 ) is stabilizable. Thus assuming the eigenvectors of H 11 are complete, the foregoing condition shows that T 11 is invertible when (A, B 2 ) is stabilizable. Proof (ii) Suppose TII is invertible and T 21 v = 0 for an eigenvector v of H11· Then (5.69, 5.70) become As= ,\s

(5.73) (5.74)

where Tllv = s. Therefore since H 11 is stable and,\ is an eigenvalue of H 11 , we have that (A, Qc) has a

Quadratic State Estimation

137

Stable unobserva ble eigenvalue. This contradict s the condition given in (ii). Therefore no eigenvector v ~f H 11 satisfies T 21 v # 0 if (A! ~c) has no stable unobs~rvable e_i~envalues. Finally assummg H 11 has a complete set of eigenvectors. the foregomg cond1t1on shows • that r 21 is non-singul ar when (A, Qc) has no stable unobservable eigenvalues. Although more involved, it can be shown that the foregoing theorem holds even when done using generalized H 11 does not have a complete set of eigenvectors. This is . . . . eigenvectors. In summary, in thts sectwn, we ha\e considered the problem of solvmg the QCARE for its stabilizing solution. The so-called Hamiltoni an matrix was introduced for this purpose. By using this matrix it was shown that _the comp~ta:ionally ~seful technique of Schur decompos ition could be employed to obtam the stabthzmg solutiOn to the QCARE when it exists. In addition, the Hamiltoni an matrix enabled the developme nt of important conditions on the plant and performan ce index parameter s which are necessary and sufficient for the existence of a stabilizing solution to the QCARE.

Quadratic State Estimation

5.5

So far we have solved the quadratic control problem under the assumptio n that the state of th7 plant is known. In this case the controlled input is I

However, when the plant state is unknown, we must resort to the use of an observer in an observer based feedback control scheme as discussed at the beginning of this chapter. Doing this and using the optimal quadratic state feedback matrix K 0 , developed in previous sections, gives the controlled input as (5.75)

where

x(t) = x(t)- x(t) Since u2s(t) = K 0 x(t) is the controlled input \vhich minimizes the quadratic state fee~back control cost, JQc, (5.17), the departure of the controlled input, u2E(t), from its optimal value, u25 (t), causes the cost JQc to increase. Our goal in this section is to design an observer to minimize this increase in the quadratic control performan ce index. We approach this goal by solving the optimizati on problem of choosing the observer's L ~atrix so as to minimize the energy in the state estimation error caused by both an un~uls~ disturbanc e in the measurem ent of the plant output and an initial plant state esttmatwn error.

5.5.1

Problem formula tion

Recall, from Section 3.3, that ideally the observer estimates the state of the plant hased on the true plant output, y(t) = Cx(t), where for simplicity, we assume, as in the quadratic state feedback control problem, that the plant is strictly proper. However, in practice,

138

Quadratic Control

only a measured plant output is available and this output differs from the true plant output by an additive measurement noise signal, w(t). Thus ify 111 (t) denotes the measured output we have Y111 (t) = y(t)

+ 1r(t)

It should be emphasized that, unlike the quadratic state feedback control problem where the disturbance is at the plant input, the disturbance here is at the input to the observer.

v(t)

PLANT

+

u,(t)

:E

y(t)

+

~ CONTROLLER

Figure 5.3

Setup for output measurement noise attenuation

output measurement noise

y(t)

control system output

controlled input

v( t)

control system command input

Now applyingy 111 (t) to the input of an observer, we obtain an estimate of the plant state from

x= Ax(t)

.Y(r)

=

+ Bu(t) + L[.Y(t) - Ym(t)]

cx(r)

and we see that the plant state estimation error, x(t) differential equation

.x (r) =

Ax(t)

+ LHi(t)

=

x(t)- x(t), is governed by the

A= A +LC

(5.76)

As in the state feedback control problem, we assume the unknown disturbance is a largeamplitude, short-duration signal which we model as an impulse, w(t) = 15(t- to)· Suppose the control system has been in operation with a constant command input and no measurement noise, for a time which is great enough so that the plant state estimate has converged to the steady-state value for the plant state, i.e.,

x(t)

=

x,

x(t)

= 0

Quadratic State Estimation

139

. Then shifting the time origin to t 0 , the time where the output measurement noise impulse occurs, we have w(t) = 8(t), x(O) = 0 and we see from (5.76) that the state estimation error becomes (5.77) Now we want to return the state estimation error to its null value as rapidly as possible following the occurrence of the measurement noise impulse. Therefore (5. 77) implies that L = 0 is the best choice for this purpose since then x(t) = 0 and the disturbance would have no effect on the state estimate. However, this approach to setting up the quadratic state estimation problem does not take into account effects on the state estimation error which arise from the initial plant state being unknown. For instance, if the plant is unstable, we have seen in Chapter 3 that choosing L null causes any estimation error present at start-up to grow without bound. To overcome this problem, we reformulate the foregoing optimization problem by assuming that the initial state estimation error has an additional component not caused by the output m~asurement noise, w(t), i.e., we take x(O) as

..x(o) =

.xd

Thet'we see from (5.76) that the plant state estimation error, x(t), caused by both x(O) = xd and w(t) = 8(t) is given by (5.78)

Now we want to choose L to decrease the effect on the plant state estimation error of having both a non-null initial state estimation error, and an impulsive output measurement noise. One way of doing this is to solve the problem of choosing L so as to minimize the energy in the state estimation error, i.e., find L such that

where JQE

=loco x*(t)x(t)dt

However recalling the use we made of the trace of a matrix in connection with the deve!opment of the controllability Gramian in Chapter 4, we see here that JQE can be rewntten as

Which using (5.78) becomes (5.79)

Quadratic Control

140

where A= A +LC

Now in order to make the optimization problem more tractable we replace M by

(2e,

where Re, Qe are real symmetric matrices with Qe 2' O,Be > 0. Then our problem is to choose L to minimize J QE such that A is stable, where· ( 5.80)

with A =A+LC

The development of equations for doing this proceeds along lines similar to those used to develop the equations for K 0 in Section 5.3.

5.5.2

Problem solution

We begin the development by noting that the required stability of the observer ensures that the integral, (5.80), converges to a real symmetric matrix P QE so that (5.81)

where PQE =

l

(X)

-

At-

e Qee

-

A't

dt

( 5.82)

Recall, from Chapter 4, that PQE' (5.82), solves the Lyapunov equation (5.83)

Notice that since A and Qe, (5.80), depend on L so does PQE and JQE· We use the notation PQE(L) and JQE(L) to denote this fact. Therefore our goal is to find L so that (i) A(L) is stable, and (ii) the solution, PQE(L), to (5.83) minimizes JQE(L), (5.81). We denote the value of L that satisfies (i) and (ii) as L 0 . Thus L 0 , the optimal value of L, satisfies

Quadratic State Estimation

141

Now we employ the matrix variational technique used in Section 5.3 to develop L 0 • Thus suppose Lis perturbed away from L 0 as L

=

L0

+ E(fJL)

(5.84)

where Eisa small positive real scalar and (8 L) is any matrix having the same dimensions as L such that A is stable. Then we can write the expansion of PQE about PQEo as (5.85) where HOT (higher order terms) stands for terms in the series involving powers of Ewhich are 2 or greater. Next we obtain the expansion of J QE about its minimum\ alue, J QEo, by substituting (5.85) in (5.81). Doing this and using the fact that the trace of sum of matrices is the sum of the trace of each matrix gives JQE

= trace[PQ£] = trace[PQEo] + Etrace[(8 1PQ£)] +HOT (5.86)

= JQEO + E(SIJQE) +HOT Notice that

( 5.87) Now comparing the present development of equations for L 0 with the development used to obtain equations for K 0 , Section 5.3. we see that the optimal value for L, L 0 . makes (S 1JQE) = 0 for all8L. Therefore we see from (5.87) that an equation for (8 1P QE) is needed. W~ begin the determination of an equation for (S 1PQE) by using (5.84) to express "4 and Qe , (5.80), as

A= A + E(8L)C 0

Q" = Qo -t- E[L

0

Rc(8L)*-'- (8L)ReLo]-r- HOT

where

Then substituting these expressions for Lyapunov equation, (5.83), yields

A, Q and

the expansion for

PQE•

(5.85), in the

(5.88)

142

Quadratic Control

Now the first three terms on the left side of (5.88) sum to a null matrix since P QEo the Lyapunov equation, (5.83), when L = L 0 • Thus using this fact and recalling from Section 5.3 that satisfie~

l.im ,~o

[HOT] = 0 E

we see that dividing (5.88) byE and letting Ego to zero gives (5.89) Notice that (5.89) is a Lyapunov equation in (8 1PQt} Therefore, since A0 is required to be stable we must have (81 PQE) ;::: 0. Thus no eigenvalue, <J;, of 81P QE is negative, i.e., { <J; ;::: 0 : i = 1, 2 · · · , n}. Moreover, referring to the development in Section 5.3, we see that J QE is minimized by choosing L so that 81J QE = 0 for all 8 L. Thus after rewriting (5.87) as n

81JQE

= trace[(8IPQ£)] =

L<J;

i=1

we see that 81J QE = 0 for all 8 L only if all the O";S are zero for all 8 L which implies that (8 1PQE) is null for all8L. Now referring to (5.89) we see that if (8 1PQ£) is null then

(8L)Mt + M1 (8L)* = 0 which holds for all 8 L only if M 1 is null. Therefore L 0 must satisfy

which since Re

> 0 can be solved for

L 0 as

(5.90) Notice that in the same way that K 0 depends on P QCo• the stabilizing solution to the QCARE, so L 0 depends on P QEo• the stabilizing solution to an algebraic Riccati equation commonly referred to as the quadratic filtering algebraic Riccati equation (QF ARE). This equation is obtained by substituting L 0 , (5.90) in the Lyapunov equation, (5.83). Thus after some algebra we obtain the QF ARE as (5.91) where RQE = C* R;: 1 C. This completes the development of the equations for the optimal value of L. In summary, the same variational approach which was used in Section 5.3 to determine equations for the optimal state feedback matrix K 0 was used to determine equations for the optimal observer matrix L 0 • In the next section we will find PQEo so that the QFARE,

Solving the QFARE

143

I), is satisfied and A - PQEoRQE is stable by resorting to ~ Ham_il_t~nian m~trix approach similar to the approach used to solve the QCARE for Its stabthzmg solutiOn.

5.6

Solving the QFARE

In this section we exploit the dual nature of the quadratic state feedback and state estimation problems. We begin by recalling the QCARE, (5.37) and QFARE, (5.91)

QCARE: QFARE: where

RQE- C*R-e 1 C Then comparing these equations we obtain th correspondences given in Table 5.1. Recall that if (A*, C*) is stabilizable then (A, C) is detectable. Therefore the foregoing corf!Spondences together with Theorem 5.1 in Section 5.4.1, imply the following sufficient conditions for the existence of a stabilizing solution to the QFARE. Theorem 5.7 A0 is stable if (i) (A, C) is a detectable pair and (ii) P QEo satisfies the QFARE with (iii) (A, Qe) is a stabilizable pair.

P QEo :::::

0 and

where A0 =A+ L 0 C with L 0 given by (5.90) and PQEo satisfying (5.91). Next recalling the quadratic control Hamiltonian matrix, H QC• (5.46), we see from the foregoing correspondences that the quadratic estimation Hamiltonian matrix, H QE• needed to solve the QFARE for its stabilizing solution should be defined as

_[A*

HQE-

Qe

RQE]

(5.92)

-A

where RQE = C* R;;'c Then substituting PQE in place of PQc in the similarity transformation matrix T, (5.47)

Table 5.1 Estimation A C*

QCARE-QFARE Symbol Correspondences Control A*

Re Qe

B2 Rc Qc

pQEo

PQco

144

Quadratic Control

gives the transformed quadratic estimation Hamiltonian matrix as T

-1

-

H QE T = H QE =

[A-*y

where A=A-PQERQE

Y = PQEA*

+ APQE- PQERQEpQE + Qe

Notice that H QE and fi QE have the same eigenvalues for all T. Therefore we can see, by choosing T so that Y is null, that H QE has eigenvalues which are mirror images of each other across the imagin~ry axis. This means that any imaginary axis eigenvalues of H QE are also eigenvalues of A for all P QE that make Y null, i.e., that satisfy the QF ARE. Thus when H QE has imaginary axis eigenvalues the QF ARE does not have a stabilizing solution, cf., Theorem 5). Conditions which ensure that H QE has no imaginary axis eigenvalues are determined by analogy with Theorem 5.4. Theorem 5.8 The quadratic estimation Hamiltonian matrix, H QE• (5.92), is devoid of imaginary axis eigenvalues if (i) (A, C) has no unobservable eigenvalues on the imaginary axis and (ii) (A, Qe) has no uncontrollable eigenvalues on the imaginary axis Assuming a stabilizing solution to the QF ARE exists, we can use the Schur decomposition of H QE to determine this solution as was done in Section 5.4.3 to determine the stabilizing solution to the QCARE. Therefore referring to Section 5.4.3, we use the first n columns ofT, the unitary or orthogonal matrix which transforms H QE to a block upper triangular matrix with the first diagonal block stable. Then the stabilizing solution to the QF ARE is given as PQEo

= -T21T]}

where

Conditions required for the existence of the stabilizing solution to the QF ARE can be obtained by analogy with Theorem 5.6. Theorem 5.9 Suppose we can find a unitary matrix T so that H 11 is stable and

where HQE= [

A*

Qe

Summary

145

we have the following results: (i) Tn is invertible if (A, C) is detectable; (ii) if TIl is invertible, then T 21 is invertible if (A, Qe) has no stable uncontrollable eigenvalues. Finally we can show that the stabilizing solution to the QF ARE is unique, when it exists, by proceeding in the same fashion as was used to prove Theorem 5.2. This completes the discussion of the stabilizing solution to the Q FARE.

5.7

Summary

We began this chapter with a description of observer based state feedback and its properties. This was followed by the derivation of desig_n equations f~r both an optimal quadratic state feedback controller and for an optimal quadratic state estimator (observer). In each case the design equations were nonlinear matrix equations referred to as algebraic Riccati equations, i.e., the QCARE and QF ARE. The requirement for the state feedback system and observer to be stable restricted the use of the solutions of each of these equations to their so-called stabilizing solution. Conditions for the existence of these stabilizing solutions were developed, in each case, in terms of the parameters of the planyand performance index. This was done through the use of a related four-block matiix known as a Hamiltonian matrix. We saw, in Theorems 5.4 and 5.6, that the QCARE has a stabilizing solution if and only if (A, B2 ) is stabilizable and (A, Qc) has no unobservable imaginary axis eigenvalues. Alternatively we saw in Theorems 5.8 and 5.9 that the QFARE has a stabilizing solution if and only if (A, C) is detectable and (A, Qe) has no uncontrollable imaginary axis eigenvalues.

5.8

Notes and References

The algebraic Riccati equation has played a role of great importance in the evolution of control theory. An extensive view of this subject is given in [3]. The use of the Schur decomposition of the Hamiltonian matrix to solve the quadratic state feedback control problem, is implemented in the MATLAB Control System Toolbox under the '"dommand lqr. More detailed information on the Schur decomposition is given in [17]. The matrix variational technique used here to develop the design equations for the optimal K and L matrices was used previously to reduce the effect of inaccurate modeling ofthe plant on the performance of a quadratic control system [28]. For more recent uses ofthis approach see [13] and the references therein. Since the stabilizing solutions, P, to either of the algebraic Riccati equations ~countered in this chapter are uniquely determined from the corresponding Hamiltontan matrix, H, we have H f---+ Pis a function. This fact has lead to the use of H Edom(Ric) to d~ote those H that admit a stabilizing P and P =Ric(H) to denote the stabilizing . ~lution P determined fr~m H_. Thi~ notation has become standard in the control · terature. For a more detailed discussiOn see Chapter 13 of [47]. When either the plant is time-varying, or when the control cost is evaluated over a finite of time, the optimal controller is time-varying and is determined by solving two matrix differential equations. The development of these conditions is beyond the ofthe present text. The reader is referred to [I] and [1 O]for more information on this

6 LQ G Control

6.1

Introduction

metho ds for choosing the This chapte r continues the development, begun in Chapt er 5, of As in Chapt er 5, our goal K and L matrices in an observer based feedback contro l system. on the contro l system's here is to comba t the effect that unkno wn disturbances have t loss of generality, that output. Thus, as discussed in Section 5.3.1, we can assume, withou l system output is caused the contro l system's comm and input is null so that the contro durati on charac ter of the solely by the disturbance input. The assumed sporadic, short characterized as being disturbances used in Chapt er 5, is replaced here by disturbances a linear system produce an persistent random signals which when applied as an input to average power if output with bound ed average power, where y(t) has bound ed lim T->oo

2_ fT y*(t)y (t)dt < oo T }o

energy unbou nded. Notice that the persistence of the disturbances makes the output the output energy use longer Therefore unlike the quadra tic contro l problem, we can no to attenu ate ability 's system l caused by the disturb ance input as a measure of the contro mance perfor a as used is disturbances. Instea d the steady-state average power output measure. zero mean Gauss ian We will assume throug hout that the disturbances w;( t) are random vectors with covariance given as (6.1)

148

LQG Control

and E[·] is the expectation operator from probability theory. Moreover we will assume that the output caused by these random disturbances is ergodic so that the steady-state average power at the output is given as lim _!._ {T y*(t)y(t)dt T }o

=

T -we

lim E[y*(t)y(t)]

(6.2)

t-->cc

Now we will begin by assuming that the plant to be controlled has state model specified as

x(t)

=

y 1(t)

=

Y2(t)

=

+ wo(t) + B2u2(t) I C 1x(t) + w1 (t) + D12u2(t) C2x(t) + w2(t) + D22u2(t)

Ax(t)

(6.3) (6.4)

(6.5)

where disturbances

controlled output

controlled input

measured output

and u2 ( t), y 1 (t), y 2 (t) have dimensions m 2 ,p 1 ,p2 respectively. Our goal here is to choose K, Lin an observer based feedback control system so that J cc is minimized where

fcc= {-'>CC lim E[y7(t)yJ (t)]

(6.6)

Since the disturbance signals are assumed to be zero mean Gaussian random vectors, this type of optimal control is known by the acronym LQG (Linear Quadratic Gaussian). More recently this optimal control problem was interpreted as a minimum norm problem and the term H 2 optimal control was also used. More will be said about this in Chapter 8 following the development of ideas involving signal and system spaces. w,(t)

~ y,~) -~~~------------~~+~ wo(t)

PLANT u,(t)

CONTROLLER

Figure 6.1

LQG control configuration.

LQG State Feedback Control Problem

149

As in the quadratic control problem, the separation principle allows us to obtain the solution to the LQG control problem by combining the solutions to two subproblems: (i) the LQG state feedback control problem (ii) the LQG state estimation problem. The LQG control problem is also referred to as (iii) the LQG measured output feedback control problem. In what follows we assume that all matrices are real so that we can write (·)Tin place of(·)*

6.2

LQG State Feedback Control Problem

In this section we dewlop the design equations for the optimal state feedback control matrix K which minimizes lee, (6.6). We will do this using the \ariational approach introduced in the previous chapter. We will see that the determination of the optimal K requires the solution of an algebraic Riccati equation.

6.2.1

Problem formulation

Suppose the plant state, x(t). is known. Then we want to determine the state feedback matrix, K, involved in generating the controlled input, u2 (t) = Kx(t), so that1cc, (6.6), is minimized. Notice from (6.3. 6.4) that under state feedback the system relating the disturbances to the controlled output is given as x(t) = Ax(r)

+ 1r0 (t)

(6.7)

Yt (t) = Cx(t)

+ w 1(t)

(6.8)

where

and w0 (t) and 1r 1 (t) are unknown zero mean Gaussian random wctors having covariances given by (6.1). Now in order to proceed \Ye must have H 1 (t) = o. This requirement can be seen by noting from (6.1) that when w1 (t) cf0 the performance index, fcc, (6.6), is unbounded for all K because of the impulsive nature of the covariance of w1 (t). Therefore in order for the LQG state feedback control problem to be properly posed we must have w1 ( t) = o. Therefore under this assumption the statement of the LQG state feedback control problem becomes: minJcc K

150

LQG Control

given that lim E[yf (t)yl (t)] lee= 1---+x x(t) YI (t)

=

Ax(t)

+ w0 (t)

= Cx(t)

(6.9) (6.10) (6.11)

with A stable where (A, C) are given in (6.7, 6.8) and w0 (t) is characterized from (6.1) as

The variational method used in the previous chapter is used now to develop equations forK to solve this problem.

6.2.2

Development of a solution

Since A is stable and lee, (6.9), involves the steady-state behavior, the performance is unaffected by the initial state of the plant. Therefore only the zero state controlled output response is needed. Thus referring to (6.1 0, 6.11) we see that this response is given by (6.12) Then using (6.12) we see that the expectation needed in the cost, (6.9), can be written as

Now we can express the right side of this equation by employing the trace relation used in the development of the controllability Gramian in Chapter 4. This relation is restated here as (6.14) for any compatible vectors vi and matrix 8. Therefore identifying w0 ( T 1 ) with v 1 and w0 ( T2 ) with v 2 enables (6.13) to be rewritten as

Next evaluating the expectation using (6.1) reduces (6.15) to

LQG State Feedback Control Problem

151

Finally, use of the sifting property of the impulse gives

E[yf (t)y 1 (t)] =trace and substituting T = t index, JGe• (6.9) as

T2

[1 W 1

and allowing t

00 e"F(t-T2 )CT CeA(t-T2 )dT2

--->

J

(6.17)

oo enables us to express the performance

J 6 e = trace[WooPGcl

(6.18)

where

Now recalling Theorem 4.5 in Chapter 4, we see that P 6 e satisfies the Lyapunov equation (6.19) and isfthe o.>servability Gramian for the pair (A" C). Thus since (A, C) depends on K so does PGe· Now the equations which determine K so that J 6 e, (6.9), is minimized can be developed by following the variational approach used in the previous chapter. Thus if K 0 denotes the optimal value forK we have

and expanding he and PGe about K 0 yields

he= heo + t:(l5,he) +HOT = trace[WooPGeol

+ t:trace[W00 (8 1P 6 e)] +HOT

(6.20)

where

~d

t:

is a small positive real scalar with l5 K being any real matrix having the same

~ension as K such that A is stable. In addition, recall that HOT denotes higher order 'Fms in E and therefore satisfies

limHOT =0 <-->0

E

(6.21)

we see from (6.20, 6.21) that (6.22)

152

LQG Control

Recall from the previous chapter that lee is minimum when (o 1Jcc) = 0 for all bK. Thus we see from (6.22) that we need a relation between (61 Pcc) and 0 K. This relation is developed by substituting the expansions for K and Pc;c as given in (6.20) into the Lyapunov equation (6.19). This gives (6.23) where

A

0

=

A+B2Ko

Co = C 1 + D12Ko

Notice that the first three terms on the left side of (6.23) sum to a null matrix since P GCo satisfies (6.19) when K = K 0 . Therefore dividing (6.23) through byE, letting Ego to zero, and using (6.21) yields (6.24) where

and we see that (o 1PGc) is governed by a Lyapunov equation. Now in order for (o 11Qc), (6.22), to be zero for all OK, we must have W 00 (o 1PGc) null for all OK. However W 00 is independent of OK. Therefore we can have W 00 (o 1PGc) null for all oK only if (8 1PGc) is null for all8K. Thus we see from (6.24) that if (8 1PGc) = 0 for all8K then we must have M 1 = 0. Therefore assuming that the columns of D 12 are independent so that Df2D 12 invertible, we can solve M 1 = 0 for K 0 as (6.25) where

As in the case of the optimal quadratic control problem, the matrix PGco is obtained as the solution to an algebraic Riccati equation. We develop this equation by substituting, Ko, (6.25), in the Lyapunov equation, (6.19). Thus after substituting for A, C from (6.7, 6.8) in (6.19) we obtain

AT PGco

+ PGcaA + K"[; Qo + Q(; K + Cf C1 + K"[; Df2D12Ko = 0 0

which after substituting for K 0 from (6.25) becomes

AT PGco

+ PGer,A- 2QT; (Di;D\2)-l Qo + cf cl + QT; (Df2Dl2)- 1Qo = 0

(6.26)

LQG State Estimation Problem

153

Finally substituting for Q0 , (6.25), gives [AT~ C[ Dn(D[zDtz)- 1 BfjPGco

~PGcoBz(Df2Dl2)- 1 BzPGco

+ PGco[A ~ B2(D[zDn)-I D[zCJ]

+ CfCt

~ CfDJ2(DfzDn)- 1DfzCt

=0

which we can write more compactly as (6.27) where At= A~ Bz(D'fzDn)- 1DTzC 1 Rt = Bz(DfzDI2f 1B[ QI = C'f[I ~ DI2(D'fzDu)- 1D'fzJCt

Notice that

Now the algebraic Riccati equation (6.27) is referred to as the Gaussian control algebraic Riccati equation, (GCARE). Thus we obtain the optimal K, K 0 , by solving the GCARE for its stabilizing solution, i.e., the solution P GCo which makes A + B 2 K 0 stable. As in the previous chapter, we will show that the GCARE can be solved for its stabilizing solution by using an appropriate Hamiltonian matrix. This matter is considered further after taking up the state estimation problem. Before leaving this section, notice that the requirement that D[2D 12 be invertible is similar to the requirement we encountered in Section 5.3.2 that Rc be invertible. Both Rc and D[2D 12 play the role of introducing a penalty on the use of the controlled input in "~ minimizing the control costs JQc and lac respectively.

6.3

LQG State Estimation Problem

In this problem we assume that the plant state x( t) is unknown and that we want to design an observer to estimate it from the controlled input, u2 (t) and measured output, Yz(t), (6.3, 6.5). Unlike the problem just treated, the controlled output y 1 (t), (6.4) is not used. Our goal now is to choose the observer matrix L so that JGE> the steady-state average power in the state estimation error, i(t), is minimized where JGE

=lim E[.XT(t)i(t)] =lim trace[E[i(t)ir(t)]] (----700

(6.28)

t-oo

Notice that this performance index involves the steady-state, state estimation error covariance matrix P GE where

154

LQG Control

n of the In order to proceed we need to review the role played by L in the evolutio estimati on error. This is done as follows.

6.3.1

Probl em formu lation

having the Recall, from the beginnin g of this chapter, that the state model for the plant as given is output only its measure d output, Y2(t), as

+ wo(t) + B2u2(t)

x(t)

=

Ax(t)

Y2(t)

=

C2x(t)

( 6.29)

+ w2(t) + D22u2(t)

(6.30)

vectors where the unknow n disturba nces w0 (t) and w2 ( t) are zero mean Gaussia n random that an see we 3, Chapter to referring Then (6.1). with covarian ce matrices as specified by as observer for this state model can be cast

+ B 2u2(t) + L[y2(t) - Y2(t)]

.X (t)

=

Ax(t)

.Y2(t)

=

C 2x(t)

+ D22u2(t)

(6.31) (6.32)

where we have replaced the unknow n disturba nces by their expected values. on Again as in Chapter 3, we can obtain a differential equatio n for the state estimati gives This 6.32). (6.30, using and (6.29) error by subtract ing (6.31) from

i

(t)

=

Ax(t)

(6.33)

+ Bw(t)

where

w(t) =

[wo(t)l

x(t) = x(t)- x(t)

w2(t)

A= A+ LC2

B=

[I

L]

nce vector, Notice from (6.1) that the compos ite zero mean Gaussia n random disturba w(t), has covarian ce (6.34)

where

that A is Recall from Chapter 3, that in the absence of disturbances, we chose L so hes the approac which estimate state a s generate r observe stable. This ensures that the l beneficia this nately Unfortu error. on estimati state plant initial plant state for any t persisten of asympto tic behavio r of the state estimato r is not possible in the presence

··--..... ·~'!".'"""''-<'1"1?:'11'.

toG st~,;; E:fdimiliion-:liriiiii"~'rir

155

.·disturbances like those being considered here. However the expected value of the state estimation error will be null in the steady-state when A is stable and the expected values of the disturbances are null. Estimators which have this desirable property are said to be unbiased. Therefore since we are assuming zero mean disturbances, we obtain unbiased estimates of the plant state provid~d our choice of L to minimize h£, (6.28), is constrained by the requirement that A be stable. This plus the fact that the performance index, hE, (6.28), concerns the steady-state behavior of the estimator, implies that we can ignore the state estimation error response due to any initial state estimation error. Therefore using (6.33) we obtain (6.35)

Now this relation can be used to develop an expression for the state estimation error covariance matrix PGE· We do this by following the steps we used to develop PGc in the previous section. Doing this yields

I

(6.36)

T

Finally, we can obtain the steady-state error covariance matrix, = t- T 2 and taking the limit as t tends to infinity. This yields

PGE•

by setting

(6.37)

Notice from Chapter 4, (Theorem 4.5), that PGE• (6.37), is also the controllability Gramian for (A, B) and satisfies a Lyapuno v equation (6.38)

where P GE depends on the observer matrix L since both A and B depend on L. Thus we see from (6.28) that we need to adjust L so that A is stable and so that the solution to (6.38), PGE, has minimum trace. A solution to this problem is obtained now by using the duality lletween the present problem and the LQG state feedback control problem which we ~lved in the previous section.

Problem solution

i

begin by using the dependency of and C, on K, (6.7, 6.8) to expand the Lyapunov (6.19) and by using the dependency of A and Bon L, (6.33) to expand the

156

LQG Control

Lyapunov equation (6.38). This gives AT Pee+ PeeA

+ KT sf Pee+ PecB2K

+Cf C1 + Cf D12K + KT Df2C1 + KT D[2D 12 K APeE

=

0

+ PeEAr + LC2PeE + PeEcf Lr +Woo+ LW2o

+ Wo2LT + LW 22 LT = 0

(6.39)

(6.40)

In order to compare the symbols in these Lyapunov equations we make use of the fact that covariance matrices are symmetric. Therefore we can factorize W, (6.34), as (6.41)

so that

Notice that these factorizations allow us to relate w0 (t) and w2 ( t) to a zero mean Gaussian random vector u 1 (t) as (6.42) where

Now we can use the foregoing factorizations, (6.41), to rewrite the Lyapunov equation (6.40) as

+ LC2PeE + PeEcT Lr +B1Bf + LD21Bf + B1Df1LT + LD21Df1LT = 0

APeE + PeEAr

(6.43)

Then comparing terms in (6.43) with terms in (6.39) we obtain the symbol correspondences for the LQG state feedback control problem and for the LQG state estimation problem given in Table 6.1. Thus we see that the formulas for the optimal Land PeE, i.e., L 0 and PeEo' can be obtained by substituting from these correspondences in the formulas for K 0 , Peeo given in (6.25, 6.27)). Doing this yields (6.44) (6.45)

Table 6.1

GCARE -GFARE Symbol Correspo ndences Control

Estimatio n

------

PGc

PGE

AT KT

A L

sr

Cz

cf

Bl

Dfz

D21

where

A2 =A- BtDf1 (DctDfJ )- 1 C 2 R2

=

Cf(D21 Dft)- 1 C2

Q2 = B 1 ~I- DJ1(D 21 Df~)- 1 D 21 ]Bf Notice that

A2- pGEoR2 =A+ LoC2 filtering Now the algebraic Riccati equation (6.45) is referred to as the Gaussia n (6.44) in PGEo ting substitu by d algebraic Riccati equatio n, (GFARE ). Thus L 0 is obtaine makes L resulting the 0 since where PGEo is called the stabilizing solution to the GFARE stabilizing A+ L 0 C 2 stable. As in the previous chapter, the GFARE can be solved for its 6.5. Section in this show will We solution by using an appropr iate Hamilto nian matrix. on estimati state LQG the to solution Before going on, notice that the foregoing dent indepen have must D that 21 problem requires that D 21 D[1 be invertible. This implies (t) are rows. Thus when this conditio n is satisfied the elements in the random vector·w 2 s in element the of subset a that sense deterministically indepen dent of each other in the n conditio this When exactly. (r) w in 2 w2(t) can not be used to determine another element . singular being as to is not satisfied, the estimati on problem is referred

6.4

LQG Meas ured Output Feedback Problem

estimati on Having solved the LQG state feedback control problem and the LQG state d output problem , the separati on principle allows us to determine the LQG measure feedback controll er by combini ng these solutions as follows. in the First, replacing the plant state by the observe r's estimate of the plant state expression for the controll ed input gives

r. Then \Vhere Ka is obtaine d from (6.25, 6.27) and i(r) is the state of the observe (6.31, n, equatio tial differen state r's observe the in t) ( u for on 2 substitu ting this expressi

1

158

LQG Control

6.32), yields the state model for the controller as .X (t)

=

u2(t)

= K 0 .X(t)

A,cx(t)

~

L 0 Y2(t)

( 6.46) (6.47)

where Ace =A+ Lu( C2

+ DnKa) + B2Ko

and L 0 is obtained from (6.44, 6.45). This completes the formulation of the controller which solves the LQG control problem. It remains now to develop conditions on the plant state model paJameters which ensure the existence of stabilizing solutions to the GCARE and the GFARE.

6.5

Stabilizing Solution

Since stabilizing solutions to both the GCARE and GF ARE are needed to implement an LQG controller, sufficient conditions for the existence of these solutions, expressed as conditions on the plant parameters, are of considerable importance to designers of these controllers. As in the quadratic control problem, the stabilizing solutions to the GCARE, (6.27), and to the GFARE, (6.45), can be computed, when they exist, by using appropriate Hamiltonian matrices. We will use this fact, in the same manner as was done in the previous chapter, to develop conditions on the parameters of the plant which ensure the existence of stabilizing solutions for the GCARE and GF ARE. The duality of the GCARE to the GF ARE allows us to obtain these results for both by treating the GCARE in depth and extending the results obtained to the GF ARE through the use of duality.

6.5.1

The Hamiltonian matrix for the GCARE

Referring to Section 5.4 we see that the GCARE has a stabilizing solution if and only if the associated Hamiltonian matrix Hac has (i) no eigenvalues on the imaginary axis, (ii) T 11 invertible where

T,=[~~:J is any 2n x n matrix whose columns span the Hac corresponding to the stable eigenvalues of H GC·

invariant subspace of

where Hac is composed from the GCARE, (6.27), in the same manner as H QC is composed from the QCARE in Section 5.4.2. Thus we see that (6.48)

Stabilizing Solution

159

with AI =A~ B:(D[2DI2)- 1 Df~CI

R1

= B2(Df2DI2)-I BJ

QI

= Cf[/ ~ DciDf2Dic)- 1Df:JCI

Also recall. from the previous chapter, that condition (i) is necessary and sufficient for the existence of a nonsingular 2n x 2n matrix T, (which can be orthogonal), so that H cc can be transformed to block upper triangular form such that then x n matrix H II is stable where (6.491

Then condition (ii) is required to construct the stabilizing solution to the GCARE from

and we see that Pcco > 0 requires that T 21 be invertible. In what follows we obtain necessary and sufficient conditions on the plant parameters so that (i) and (ii) hold. We do this by modifying theorems obtained in the previous chapter taking into account the differences in the dependency of the Hamiltonian matrices H Qc and H cc on the plant parameters. Conditions on the plant parameters which ensure that P ceo > 0 will also be dewloped.

6.5.2

Prohibition of imaginary eigenvalue s

Recall that in the quadratic state feedback control problem we obtained conditions on the plant and performance index which ensured that H QC has no imaginary axis eigenvalues. These conditions were given in Theorem 5.4. Therefore using Theorem 5.4 and comparing H QC with H cc, we see that H cc, (6.48), has no imaginary axis eigenvalues if: (a) the pair (A 1 • RI) has no uncontrollable imaginary axis eigemalues and (b) the pair (A 1 , QJ) has no unobservable imaginary axis eigenvalues. Notice that these conditions are not immediately applicable to the given data of a plant state model and performance index since they do not explicitly state how these parameters are involved in contributing to the presence of imaginary axis poles for H cc· This deficiency is overcome in the following theorem. Theorem 6.1 H cc, ( 6.48), has no imaginary axis eigenvalues if (i) (A.B 2 ) has no uncontrollable imaginary axis eigemalues, (ii) __,;E(-:x:.oo)

where B 2 is n x m2

160

LQG Control

Proof (i) Referring to the proof of Theorem 5.4, we see that smce Df2 D 12 1s nonsingubr we have

if and only if and therefore A is an uncontrollable eigenvalue of (A 1 ,R 1 ) if and only if A is an uncontrollable eigenvalue of (A, B2 ). Thus condition (i) here is equivalent to condition (i) in Theorem 5.4. Proof (ii) We begin by considering the pair (A 1 , Q1 ) where A, =A - B2(Df2Dt2)- 1Df2C1

Thus if jw is an unobservable eigenvalue of (A 1 , Q1 ) we have

which can be rewritten as (A- jwl)v- B2(Df2D!2)- 1Df2C,v = 0

(I- D!2(Df2D12)- 1Df2)C,v = 0 Now these equations can be written as (6.50)

where 8= [

A- jwl

B2

C,

D12

~]

]

Since .6. is invertible, 8.6. has independent columns only if 8 has independent columns. However (6.50) implies that 8.6. has dependent columns. Therefore 8 must have dependent columns and rank[8] < n + m 2 This shows that condition (ii) is sufficient for (A 1 , Q1 ) to have no imaginary axis eigenvalues. Now we want to use this fact to show that condition (ii) is sufficient for (A 1, Q 1 ) to have no imaginary axis eigenvalues. In order to proceed we need to be able to factor D, as De=

(

') T

D~

D21

( 6.51)

Stabilizing Solution

161

where here

However, we can only factorize De in the foregoing manner if De is symmetric and nonnegative. The symmetry of De is obvious. We can show De~ 0 as follows. Since D 12 has independent columns, we can solve the matrix equation D 12 u

= y,

y,. E range[Dd

as

Therefore it follows that

and (6.52)

However since any p 1 dimensional real vectors y can be written uniquely as Y

= Yr + Y1.

(6.53)

where y, E range[Dn]

YJ. E null [Dfz]

we see from (6.52) that (6.54)

and D c is said to be a projector. Finally, since in general y"[ y .1.

=

0, we see from (6.54, 6.53) that (6.55)

for all real p 1 dimensional vectors, y. Therefore since the right side of the foregoing equation can not be negative we have shown that De ~ 0 and therefore we can factorize De as (6.51) and we can proceed to show that when condition (ii) is satisfied (A 1, QJ) has no imaginary axis eigenvalues. Suppose condition (ii) is satisfied. Then we showed earlier in the proof that any eigenvector-imaginary axis eigenvalue pair (v,jw) of A 1 satisfies (6.56)

where

162

LQG Control

Thus we must have

so that

or

This shows that

for any eigenvector v of A 1 associated with an imaginary axis eigenvalue which implies that the pair (A 1 , Q 1) has no unobservable imaginary axis eigenvalues. • The foregoing proof reveals an important property of Dc which we will make use of in Chapter 9 and 10 in connection with the Hx feedback control problem. Not only is D, square and non-negative but in addition is a contraction since it has the property that UT

u:::; YTY

for any p 1 dimensional vectors u, y which are related through De as u =DeY· This can be seen by noting from (6.53-6.55) that T

Y Y

=

T

Yr Yr

T + Y1Y .L

A matrix or operator having this property is referred to as a contraction. In addition, notice that matrices which are contractions have eigenvalues which are bounded above by one, i.e.,

Amax[Dc] :S: 1 This property can be seen by refering back to Chapter 4 and the discussion of the significance of the largest eigenvalue of a non-negative matrix (Section 4.6.3). Finally, notice that this constraint on the eigenvalues of De is also a consequence of De being a projector since De projects from a higher to a lower dimenional space.

6.5.3

lnvertability ofT 11 and T21

Having established conditions on the plant state model which ensure that H cc has no eigenvalues on the imaginary axis, it remains to determine conditions on the plant which ensure that T 11 , (6.49), is non-singular. We can do this by appropriately modifying Theorem 5.6 as follows. Theorem 6.2 If H cc, (6.48), has no imaginary axis eigenvalues so that we can find T, (6.49), so that H 11 is stable where (6.57)

Stabilizing Solution

163

then we have , (i) T 11 invertible if i A, B2 ' is stabilizable: DzC . (A if invertible is T 1 : has no stable unobservable then 1 21 (ii) if T 11 is invertible, eigenvalues where

A, =A- B2(Df2DJ2)- 1Df2C1 D, = I - DdDf2D12)- 1Df2

=

(nz)

T Dz

Proof (i) We begin by recognizing that (6.57) implies that

which we can expand as (6.58) (6.59) Now suppose H 11 has an eigenvalue-eige nvector pair(.>., v) such that

Then post-multiplyin g (6.58. 6.59) by v we see that (6.60)

-Afw = >.w

(6.61)

where w = T 21 v. NO\Y ifw =owe would haYe

implying that T 1 and hence T has dependent columns. This is impossible since T is invertible. Therefore w -=f. 0. Finally since H 11 is stable we have Re[)..] < 0 and therefore (6.60, 6.61) imply that (A 1 , R 1 ) is not stabilizable. This is equivalent to (A, B 2 ) not being stabilizable since R1 = B2 (Df2 D 12 )- 1Bf, Therefore no eigenvector, v, of H 11 satisfies T 11 v = 0 when (A, B 2 ) is stabilizable. Thus assuming the eigenvectors of H 11 are complete we haw shown that T 11 must be invertible when (A, B 2 ) is stabilizable. If H 11 lacks a complete set of eigenvectors. the proof is completed by resorting to generalized eigenvectors. Proof(ii) Suppose T 21 v = o with s = T11 u 'I 0 for some eigenvector,''· of H 11 . Then

164

LQG Control

post-multiplying (6.58, 6.59) by v we obtain A]S

(6.62)

=AS

(6.63) where I

M=D'cC1 However since, in general, we have null[Mr]l_ range [M], we see that (6.63) is satisfied only if s Enull[M]. Therefore if (6.63) is satisfied then (6.64) I

Moreover since H 11 is stable, (6.62, 6.64) imply that (A 1, D;C 1) has a stable eigenvalue. Therefore no eigenvector, v, of H 11 satisfies T 21 v = 0 when (A 1 ,D~CJ) has no stable unobservable eigenvalues. Thus assuming ~he eigenvectors of H 11 are complete we have shown that T 21 is invertible when (A 1, D;.c J) has no stable unobservable eigenvalue. Again as in the proof of (i), if H 11 lacks a complete set of • eigenvectors, the proof is completed by resorting to generalize~ eigenvectors. Notice that when (A, CJ) is not observable neither is (A 1, D~C 1 ). Therefore when the GCARE has a stabilizing solution, a necessary condition for T 21 to be non-singular so that Pcco > 0 is that (A, C 1) have no stable unobservable eigenvalues. However this condition is not sufficient for P ceo > 0. This can be seen from the following considerations. First, reviewing (6.51-6.56) we see that unobs~rvable

where null [Di J is of dimension m 2 and it may turn out that C 1v # 0 but Dt.c 1v = 0. Df2 D 12 = RD is symmetric and invertible it can be factored and its Second, since I square root Rb is invertible. Hence we can write (6.51) as

(D;.1)T D'c1=I- D12 (Rb1)-1( Rb1)-T Df2 which implies that

[ (Rb)d

-T

D"(;

l

has orthonormal columns. Therefore it is impossible to have a non-null vector q such that I

D'cq =

0

and

(Rb1)

-T

D'{;

g

=

0

Stabilizing Solution

Thus if vis an eigenvector of A which satisfies C 1 v fc 0 but (Rb)-T D[2 C 1v # 0 so that

I

D~C 1 v = 0,

165

then we have

and vis an eigenvector of A but not of A 1• Notice in this case that unlike the eigenvectors of A, the eigenvectors of A 1 depend on B2 implying that the positive definiteness of the stabilizing solution depends on B2 . In summary, the GCARE has a stabilizing solution, Poco: if (lc) (A, B 2 ) is stabilizable; (2c)

rank [[

A- jwl wE

cl

(-oo,oo)

Finally if a stabilizing solution, Poco, exists then !

(3c) Poco> 0 if (A 1 ,D~C 1 ) has no stable unobservable eigenvalues.

6.5.4

Conditions for solving the GFARE

In order to obtain conditions on the plant parameters which ensure that the GF ARE has a stabilizing solution, we can use the correspondence between matrices in the GCARE and the GF ARE given in Table 6.1. Therefore substituting these correspondences in (lc, 3c) we see that the GFARE has a stabilizing solution, PoEo if (le) (AT, Cf) is stabilizable; (2e)

rank [[

AT-jwi T

Bt

cf]] = T

D21

n + P2

wE

(-oo,oo)

.,~

Moreover, if a stabilizing solution, P GEo, exists then 1

(3e) POEo

> 0 if (AJ, D~Bi) has no stable unobservable eigenvalues, where

However these conditions can be restated in a more convenient form by using the facts that the stabilizability of any pair (A, B) is equivalent to the detectability of the pair (AT, BT), and that the rank ofamatrixequals the rank of its transpose. Thus the GFARE has a stabilizing solution, PoEo• if (le) (A, C2 ) is detectable; (2e)

rank [[

A- jw/

Cz

B1 D21

]]

=

n +p 2

wE

(-oo,oo)

166

LQG Control

Moreover, if a stabilizing solution,

PG£o,

exists then

(3e) Pc;eo > 0 if ( A2, B 1Df) has no stable uncontrollable eigenvalues.

6.6

Summary

In this chapter we have given a derivation of the design equations for the LQG optimal feedback controller for a linear time-invariant continuous-time plant. The performance criterion or cost function which is minimized by the optimal LQG controller is the steadystate expected or average value of a quadratic form in the output vector to be controlled (the controlled output) when the disturbance input is a zero mean Gaussian random vector. Necessary and sufficient conditions were given for being able to design the LQG controller.

6. 7

Notes and References

Again, as in the quadratic control problem treated in the previous chapter, we have seen that the algebraic Riccati equation plays a central role in the design of optimal controllers. For other instances where the algebraic Riccati equation arises in control theory see Chapter 13 of [47]. The LQG optimal control problem was recently referred to as the H 2 optimal control problem. The reason for this will be given in Chapter 7. Assuming the appropriate existence conditions are satisfied, the LQG controller can be calculated using the command h2lqg in the MATLAB Robust Control Toolbox. The observer obtained by solving the LQG state estimation problem was originally given by Kalman as an alternative to the Wiener filter for extracting a desired signal from an additive combination of the desired signal and noise. This aspect of the LQG problem has had a great impact on a wide variety of industrial problems. There are many books which deal with this subject. For example, informative treatments are given in [7, 10, 25].

7 Signal and System Spaces

7.1

Introduction

In the previous two chapters we were concerned with the problem of designing controllers to attenuate the effects of disturbance inputs on the output of a feedback control system. We assumed that the disturbance input was either an impulse (Chapter 5) or a random signa1with an impulsive covariance (Chapter.6). In the next two chapters we develop ideas needed to solve the disturbance attenuation problem for a broader class of disturbance input. Signals in this class have finite energy and are denoted by £ 2 [0, oo) where £ is used in recognition of the mathematician H. L. Lebesgue, pronounced "Lebeg", the subscript 2 is used in recognition of the quadratic nature of energy, and the bracketed quantities indicate the time interval over which the signals are not always zero. Therefore f( t) E £ 2 [0, oo) if (i) f(t) has bounded L 2 norm (finite energy),

[lao J*(t)J(t)dt]!= IIJ(t)ll2<

00

(7.1)

I

where [·] 2 denotes the positive square root andf*(t) denotes the conjugate transpose of

f(t). (ii) f(t) is null for negative time

f(t) = 0

for all

t E ( -oo, 0)

(7.2)

In this chapter we will use the foregoing idea of the L2 norm for signals to develop system norms defined in both the time and frequency domains. This is made possible by Parseval's theorem which relates the L2 norm in the time domain to the L2 norm in the frequency domain.

7.2

Time Domain Spaces

In classical control theory there was no apparent need to consider signals defined for negative time since control problems usually have a well defined start time. However the

168

Signal and System Spaces

need for more sophisticated mathematics to deal with new approaches to control problems requires that we enlarge the domain of definition of signals to include the negative time axis and to consider the operation of systems in both positive and negative time.

7.2.1

Hilbert Spaces for Signals

Consider a signal vector ,f( t), which unlike signals in .C 2 [0, oo), is not zero for all negative time but still has finite energy over the time interval ( -CXJ, CXJ), i.e.,

(7.3) Any f( t) which satisfies (7 .3) belongs to the normed space .C 2 ( -oo, oo). Alternatively, another related normed space, denoted .C 2 ( -oo, 0], consists of finite energy signals which are null for all positive time. Notice that .C2 [0, oo) and .C 2 ( -oo, OJ are each subspaces of .C 2 ( -oo, oo ), i.e.,

Now from the foregoing we see that if[(t) E .C 2 ( -oo, oo) it is always possible to write

f(t) =f+(t)

+ f_(t)

(7.4)

wheref+(t) E .C2 [0, oo) andf_(t) E .C 2 ( -oo, OJ with

f+(t) = f(t)

t

>0

j~(O)

= ""f(O)

f_(t) =f(t)

t

<0

f_(O)

=

(1- "")f(O)

where"" E [0, lJ. Notice that once"" is specified,f+(t),f_(t) are uniquely dependent onf(t). Moreover since f+ (t) and j"_ (t) are nonzero on disjoint intervals, we see that

l:f~(t)f_(t)dt =

(7.5)

0

Now (7.4) together with (7.5) enables the L 2 norm of any f(t) decomposed in the following fashion

=

[LCXJ [f~(t)f+(t)Jdt +

=

[Ill~ (t) II~ + III- (t) II~]~

1:

E

.C 2 ( -oo, CXJ) to be

I

[f*_(t)f_(t)Jdt],

which is reminiscent of the Pythagorean theorem for right-angled triangles.

(7.6)

Time Domain Spaces

169

In the foregoing signal analysis we encountered an integral involving two signals in £ 2(-oo, oo ). In general the quantity

1h

a*(t)p(t)dt

(7.7)

is referred to as the inner product of a(t) and p(t) for any equal length vectors a(t),p(t) E £ 2 (a,b) and is denoted as< a(t),p(t) >,i.e.,

< a(t), p(t) > =

1h

a*(t)p(t)dt

(7.8)

Notice that, unlike the norm which involves only one signal and is always a real nonnegative scalar, the inner product involves two signals and is not restricted to be positive or real. Notice that the norm and the inner product are related as I

lla(t)ll2 =[J2

(7.9)

I

where again [·J 2 denotes the positive square root. J\· signal space with an inner product is referred to as inner product space. Under additional technical constraints (completion), an inner product space is referred to as a Hilbert space. Hilbert space has found widespread use in applied mathematics, physics and engineering. In addition to being normed spaces, £ 2[0, oo ), £ 2( -oo, oo ), and £ 2( -oo, OJ are each Hilbert spaces with inner product defined by (7.7). An important consequence of inner products is the property of orthogonality. Two , signals are said to be orthogonal or form an orthogonal pair if they have an inner product which is zero. Thus we see from (7.5) thatf+(t),J_(t) E £ 2(-oo, oo) are orthogonal. The classical example of signals which are orthogonal arises in connection with the Fourier series decomposition of periodic signals. This decomposition exploits the periodicity and orthogonality of the following signal pairs: {cos(wt), sin(wt)}; {cos(nwt), cos(mwt)}; {sin(nwt), sin(mwt)}; where n, m are unequal integefs and the interval of integration used in the inner product is the period~ of the periodic signal being decomposed. Two spaces S 1 , S 2 are said to be orthogonal, denoted S 1 _L S 2 if

< a(t),p(t) >= 0

for all a(t) E S 1 and all p(t) E S2

Thus we see that £ 2( -oo, OJ _L £ 2[0, oo ). More important, £ 2(-oo, OJ n £ 2[0, oo) = 0 so that £ 2(-oo, OJ and £ 2[0, oo) are orthogonal complements of each other, denoted by £2(-oo,OJ = £t[O,oo) and the decomposition of any signalf(t) E £ 2( -oo, oo) given by (7.4) is captured by the following relation between the involved Hilbert spaces (7.10)

where EB is referred to as the direct sum of the spaces involved.

170

7.2.2

Signal and System Spaces

The L 2 Norm of the Weighting Matrix

Recall, from Chapter 1, that the zero state response of a single-input, single-output system can be calculated from its state model as

(7 .11) with impulse response denoted y 1 ( t) being given as

(7 .12) In addition recall from Section 1.8.3 that the causality constraint requires

YJ(t)

=

0

for all

t

<0

Therefore the square of the L 2 norm of YJ(t) for stable systems can be determined by recalling the observability and controllability Gramians, W 0 , We from Chapter 4. Thus we have

(7.13)

Alternatively, since YJ(t) is a scalar, we can determine the square of the L2 norm ofy1 (t) as

(7.14)

Thus we see from (7.13) or (7.14) that the impulse response of a stable system has finite L2 normandhence YJ(t) E £ 2 [0,oo). The extension of the foregoing result to multi-input, multi-output systems is more involved since, in this case, CeA 1B is a matrix referred to as the system's "weighting matrix". This terminology is used in recognition of the fact that CeA 1B applies differing weights to the components in the system's input vector in the calculation of the zero state response

y(t)

=lot CeA(t-T) Bu(T)dT

Notice that the Laplace transform of a system's weighting matrix is the system's transfer function. Notice also that the weighting matrix is the impulse response when the system is single-input, single-output. In what follows we will use the sum of the energies in each entry in the weighting matrix to generalize the L2 norm of the impulse response to the multi-input, multi-output case.

r

Time Domain Spaces

171

l.[.

Suppose u(t) is a vector of length m and y(t) is a vector oflengthp, i.e., the system has m scalar inputs and p scalar outputs and the weighting matrix, CeAt B, is p x m. Then if the input is

u(t) = u~(t) = I~8(t) where I~ is the ith column of them x m identity matrix, then the output is

where Ii is the i1h column of the n x m matrix B. Now we want to calculate the sum of the energy in each of the outputs in the set of outputs {y}(t) : i = 1, 2, · · · m} resulting from the application of each input in the set {u} : i = 1, 2, · · · m} respectively. Clearly, we can write the desired energy as

(7.15) where y1 (t) is the pm dimensional vector

y}(t) yi{t)

y'J'(t) Now we can rewrite (7.15) as

IIYI(t)ll2

=

(trace[fooc (CeA 1B)*CeAtBdtJY

(7.16)

=

(trace[B*W0 B])!

(7.17)

Alternatively, recall (Theorem 4.2) that for any matrix M we have trace[M* M] Therefore setting M

=

= trace[MM*]

CeAtB, we see that (7.16) can be rewritten as

(7.18) Finally, since II.Y1 (t)ii 2 , (7.16), is the sum of the energy in each scalar signal in the p x m

172

,

Signal and System Spaces

matrix CeAr B, we define the L 2 norm of the p x m matrix CeAr B as

(7.19) which we see from (7 .17, 7 .18) can be written in either of the following two forms IICeA 1BII 2 = (trace[B*W 0 B])~

(7 .20)

= (trace[CWcC*])~

(7.21)

We will see in Section 7.3.6 that the foregoing time domain L 2 norm of the weighting matrix equals an appropriately defined frequency domain L 2 norm of the system's transfer function which is referred to as the H 2 norm.

7.2.3

Anticausal and antistable systems

We begin by considering the single-input, single-output case. Suppose u(t) E £-2( -oo, 0]. Then substituting negative t in (7 .11) and considering t decreasing, i.e., reversing the direction of integration in (7 .11 ), we see that the output for negative time is given as t E

[O,oo)

or t E ( -oo,

OJ

(7.22)

Notice that this system operates in negative time, i.e., runs backward in time. Thus an impulse input at the time origin, u(t) = b(t), affects the output at earlier (negative) times. This effect is contrary to nature where dynamic processes have the general behavioral property that the output now is independent of the future input and is solely caused by past initial conditions and the cumulative effect of the input over the past interval of time. Natural systems having this behavior, where the cause precedes the effect, are referred to as causal systems. Artificial systems having the predictive behavior exhibited in (7.22), where the effect precedes the cause, are referred to as anticausal systems. Now if we let u(t) = b(t) then we see from (7.22) that the impulse response of an anticausal system is given by

YJ(t)

= -CeA 1B

Yr(t)

=

0

t E ( -oo,

OJ

(7.23)

for all t > 0

Notice that y r ( -oo) is zero only if- A is stable, i.e. only if A has all its eigenvalues in the open right half-plane or equivalently, only if no eigenvalue of A is in the closed left half-plane. Systems that have this property are referred to as being antistable. Recall that

173

Frequency Domain Hilbert Spaces

a system is unstable if at least one eigenvalue of its A matrix lies in the closed right halfplane. Thus we see that antistable systems are a special class of unstable systems. The introduction of this class of system allows us to express the transfer function of an unstable system, provided it has no imaginary axis poles, as the sum of two transfer functions, one for a stable system and one for an antistable system. We can obtain this sum decomposition by expanding the given transfer function in partial fractions. The foregoing readily extends to the multi-input, multi-output case as follows. If an anticausal system is antistable the system's weighting matrix satisfies lim W(t)

l-----+-00

=0

and the system's A matrix has no eigenvalues in the closed left half-plane. Moreover, the system's zero state response is given by

y(t)

=-lot

W(t- T)U(T)dT

W(t) = -CeA 1B

t E (-oo,O]

so that 0

II W(t)llz=

l

(trace[["" W*(t) W(t)dt])

2

Finally, in summary, we have shown that the weighting matrices for causal systems which are stable satisfy

whereas the weighting matrices for anticausal systems which are antistable satisfy

W(t) E C2( -oo, OJ

7.3

Frequency Domain Hilbert Spaces

We begin this section by relating the time domain Hilbert space C2 ( -oo, oo) to the frequency domain Hilbert space £ 2 . This relation is made possible by Parseval's theorem in the general theory of the Fourier transform. Following this we introduce two frequency domain Hilbert spaces, 1i2 and Ht which constitute an orthogonal decomposition of C2 and are known as Hardy spaces.

7.3.1

The Fourier transform

Recall that the Fourier transform and the inverse Fourier transform are defined as

F[.f(t)] .f(t)

=

F(jw) = l:.f(t)e-Jwtdt

I =27f

loo -oo

.

F(jw)e1wt dw

(7.24) (7.25)

174

Signal and System Spaces

Also recall the following well known sufficient condition for the convergence of the integral defining the Fourier transform of[(t), (7.24)

1: [f*(t)f(t)]~dt

<

X

(7 .26)

I

where again [·Pis to be interpreted as the positive square root. Notice that whenf(t) is a scalar, (7.26) reduces to f(t) being absolute value integrable, i.e.,

1:

lf(t) I dt <X

(7 .27)

Signals satisfying (7 .26), or (7 .27) if appropriate, are said to be in the Lebesgue space .C 1 ( -x, x).

Now we are interested in signalsf(t) E .C 2 ( -x, x ). Therefore we need to be concerned about the convergence of the Fourier integral, (7.24) whenf(t) E .C 2 (-x,x),J(t)~ .C 1 ( -x, x). It turns out that the Fourier integral of signals in this latter subspace converges for allmost all w except for, what is referred to as, a set of measure zero. This means that Parseval's theorem

!

00

_

00

1!00

J*(t)f(t)dt = 2 71"

-oc

F*(jw)F(jw)dw

(7 .28)

applies to allf( t) E .C 2 ( -oo, oo) independent of the values assigned to F(jw) at points w where the Fourier integral does not converge. This fact is discussed further in the next subsection. The important conclusion to be drawn from the foregoing is that the L2 norm of a signal in the time domain, llf(t)/1 2, equals an appropriately defined L 2 norm in the frequency domain, IIF(Jw) ll2, llf(t) ll2

=

IIFUw) ll2

(7.29)

where llf(t)ll2 = [J:!*(t)f(t)dtr I IIF(Jw)/12 = [ 271"

Joe F*(jw)F(jw)dw]~ oc

(7.30)

(7 .31)

Notice that the definition of the frequency domain L 2 norm, (7.31) includes the scaling factor (27r)- 1 which is missing from the definition of the time domain L 2 norm, (7.31). Any F(jw) satisfying

//F(Jw)// 2 <

x

is said to belong to the L 2 frequency domain space denoted .C2. Notice that the frequency and time domain L 2 norms are denoted by .C 2 and .C 2 ( -x, oo) respectively.

Frequency Domain Hilbert Spaces

175

Now Parseval's theorem, (7.28), allows us to calculate the time domain inner product of/1(t),J2 (t) E .C2(-oo, oo) in the frequency domain as

<JJ(t),J2(t) > = < F1(jw),F2 (jw) > where F;(Jw)

(7.32)

= F[fi(t)] and = 1:ft(t)f2(t)dt

(7.33)

(7.34)

Thus .C2 is an inner product space which can be shown to be complete. Therefore .C2 is a Hilbert space with inner product defined by (7.34). The reason which make it possible for us to extend Parseval's theorem from .C 1( -oo, oo) to .C2(-oo, oo) arises from the theory of Lebesgue integration which is beyound the scope of this book. However the following subsection is included in an attempt to provide some insight into this matter.

7.3:2

Convergence of the Fourier integral

Supposef(t) is specified as

f(t)

=

(t + l)- 1

t::=:o

=0

(7.35)

t
'Then

f(t) tt.CI ( -oo, oo) or more specifically

lo

1

oo

--ldt = t+

00

However we can determine the Fourier transform of (7.35) at all frequencies except 0. To see this we need to show that

w=

lo

oo

1

-1-sin(wt)dt < oo +t

lo

and

oc

1

-1 -cos(wt)dt < oo +t

We can do this by noting that

lo

1

oc

=L 00

-1 -sin(wt)dt +t

ck

k=O

where fork= 0, 1, 2 ···we have

J

tk+l

ck =

k

krr

tk=-

w

1

-1 -sin(wt)dt

+t

(7.36)

176

Signal and System Spaces

Then since f( t) is a positive monotonical ly decreasing function, the foregoing series is an alternating series satisfying lim

ck =

0

k

=

0, 1, 2 ...

k---+·XJ

which is known to converge. In a similar manner we can show that the integral involving coswt converges. Therefore in this example (7 .24) defines F(jw) unambiguou sly for every w except w = 0. Now since F(jO) is undefined in this example, suppose we set it to n. Then we would find that the L 2 norm IIF(Jwll~ is independent ofn. This fact is stated mathematica lly, in this case, by saying that the Fourier transform of[(t) specified as (7.35) is uniquely determined for all w E ( -oo, oo) except w = 0 which is a point of measure zero. The foregoing example demonstrate s the fact that, in general, whenf(t)~.C 1 ( -oo, oo) andf(t) E .C 2 ( -oo, oo) the Fourier transform F(jw) is defined uniquely as an element of the Hilbert space .C2 but is only determined "almost everywhere" as a point function of jw, i.e., is uniquely determined except on a set of measure zero. Therefore the Fourier transform is a unitary operator between Hilbert spaces .C 2 ( -oo, oo) and .C 2 since for any f 1 (t),J2 (t) E .C 2 ( -oo, oo) we have (7 .37) where the inner product on the left is defined by (7.34) and the inner product on the right is defined by (7 .33). Hilbert spaces which are related in this way are said to be isomorphic and the unitary operator between the Hilbert spaces is called an isomorphism. In the present case the isomorphism is the Fourier transform and .C 2 ( -oo, oo) is isomorphic to .C 2 .

7.3.3

The Laplace transform

In order to characterize the properties of the Fourier transforms of causal and anti causal signals in .C 2 [0, oo) and .C2 ( -oo, OJ, respectively, we need to define two Laplace transforms, the usual one denoted .C+[f(t)] for causal signals and one denoted .C_[f(t)] for anticausal signals. These transforms are defined as .C+[f(t)] =

~~1~ f(t)e-' dt

Re[s] >a+

(7 .38)

.C_[f(t)] =

~~1~f(t)e-' 1 dt

: Re[s]
(7 .39)

1

where a, a+, and a_ are real scalars with a being positive. Recall that the existence of the Laplace transform, .C+ [f(t)], requires that[( t) have a lower bounded exponential order, i.e., there must exist a finite real scalar a+, called the abscissa of convergence , such that lim f(t)e-at = 0 t~oo

= 00

(7 .40)

I, I

Freque,.cy Domain Hilbert Spaces

177

This implies that .C+[f(t)] has no poles to the right of a line called the axis of convergence which is parallel to the imaginary axis and which cuts the real axis at the abscissa of convergence, s =a+. Now it turns out thatf(t) E .C2 [0, oo) thenf(t) satisfies (7.40) with a+ = 0. Therefore the integral (7.38) converges for Re[s] > 0 when f(t) E .C2 [0,oo). This implies that .C+[f(t)] is devoid of poles in the open right halfplane whenf+(t) E .C2 [0, oo ). We can use a similar argument to show that .C_ [/( t)] is devoid of poles in the open left half-plane whenf(t) E .C2 ( -oo, 0] Now suppose we are given a causal signalf+(t) E .C2 [0, oo) and an anticausal signal f_(t) E .Cz(-oo,O]. Then we have

f+(t)e-a+t E .C 1 (-oo,oo)

a+ E (0, oo)

f_(t)e-a_t E .C,(-oo,oo)

a_ E (-oo,O)

and

F[f+(t)e-a+t]

= l:f+(t)e-a+te-Jwtd t = F+(a+ + jw)

F[f_(t)e-a_t] = l:f-(t)e-a_te-Jwtd t = F_(a_

+ jw)

for any real scalars a+ > 0, a_ < 0. Recall that the Fourier integral, (7.24), converges for almost all w when f(t) E .C2 ( -oo, oo ). Therefore if we take a+ and a_ zero we have the Fourier transforms , forf+(t) andf_(t) as

= .C+[f+(t)]ls=jw F_(jw) = .C_[f_(t)]ls=jw

F+(jw)

(7 .41)

with {F+(jw) :wE S+}, {F_(jw) :wE S_} being sets of arbitrary finite elements where S+,S- are sets of measure zero. Then we see thatf(t) E .Cz[O, oo) (j(t) E .Cz(-oo,O]) has a Fourier transform almost everywhere on the imaginary axis if and only if the Laplace transform .C+[f(t)] (.C_ [f(t)]) has no poles in the closed right (left) half-plane. Functions having these properties are said to belong to the Hardy space 'Hz ('H~-).

7.3.4

The Hardy spaces: 1{2 and

Hi

In this section we will introduce frequency domain spaces denoted 'Hz and 'Ht so that we can write any F(jw) E .Cz as

where

These spaces are called Hardy spaces after the mathematician G. H. Hardy who carried out extensive studies on these and other related spaces, e.g., 'H00 space.

,

Signal and System Spaces

178

The definition of the Hardy spaces H 2 and Ht involve ideas from the theory of a complex functions of a complex variable. A function of this sort is said to be analytic at are exist not does e derivativ point, if it has a derivative at that point. Points where the referred to as singular points. The only singular points a rational function has are its poles. H 2 consists of all complex functions of a complex variable which are analytic at every point in the open right half-plane and have a finite L 2 norm, (7.31). Alternatively, the Hardy space Ht consists of all complex functions which are analytic in the open left halfis plane and have a finite L2 norm, (7.31). The analytic requirement in these definitions either in function No spaces. these for norm needed to ensure that (7 .31) is a well defined H 2 or Ht can have imaginary axis poles. In the rational case, H 2 consists of all strictly proper functions which have no poles in the closed right half-plane, whereas Ht consists of all strictly proper functions which have no poles in the closed left half-plane. In summary, if F(s) is real rational then (i) F(s) E H 2 if and only if F(s) is strictly proper and has no poles in the closed right half-plane. (ii) F(s) E Ht if and only if F(s) is strictly proper and has no poles in the closed left halfplane. (iii) F(s) E £ 2 if and only if F(s) is strictly proper and has no poles on the imaginary axis. Notice that if F(s) is the transfer function of some system, then in case (i) the system is stable, and in case (ii) the system is antistable.

7.3.5

Decom posing £ 2 space

Recall that the L 2 time domain Hilbert spaces decompose as

Then since the Fourier transform maps £ 2[0, oo) onto H 2 and £ 2( -oo, OJ onto Ht and the same inner product is used for £ 2 , H 2 and 1-if, we have

(7 .42) and the Fourier transform is an isomorph ism such that

£2 ( -oo, oo) £2[0,oo) £2( -oo, 0]

is isomorph ic to is isomorph ic to

£2 H2

is isomorph ic to

'Hf

As an illustration, suppose we are given the following signal f(t)

where a, bare real and positive.

=e-at

t 2::0

Frequency Domain Hilbert Spaces

179

Then we can writef(t) as

f(t)

= f+(t) + f_(t)

where

f+(t) =e-at f_(t) = 0 f_(t) =it f+(t) = 0

fort~

0

fort :s; 0

Since the Laplace transforms of these signals are given by £+[1'-..(t)]

.

= - 1-

Re[s]

s+a

1 C_[f_(t)] = - b s-

>-a

Re[s] < b

we see that both C+[f+(t)] and C_[f_(t)] are free of poles on the imaginary axis. Thereforef(t) has a Fourier transform which can be written as 1

1

F[f(t)] = -.- b + -:-------+a JW-

JW

However £+[f+(t)] and C_[f_(t)] are each zero at w = oo with £+[/+(t)] being analytic in the closed right half-plane and£_ [J_ (t)] being analytic in the closed left halfplane. Therefore we have C_[f_(t)] E Hf

F[f(t)]

E

£2

The foregoing example demonstrate that, in general, systems with transfer function ,~ satisfying G(jw) E £ 2 can be written as

G(s) = G 1 (s)

+ G2 (s)

where

G2(s)

7.3.6

E

Hf

The H2 system norm

Recall from Section 7.2.2 that when the state model for a given system is minimal, the time domain L 2 norm of the weighting matrix is finite only if A is stable and Dis null. Therefore in this case the system's transfer function

G(s) is strictly proper, i.e., G( oo)

=

=

.C+ [CeAt B]

= C(sl- A)-I B

0 and has no poles in the closed right half-plane. Thus we

Signal and System Spaces

180

have G(s)

E

7-i 2 and using Parseval's theorem we have

[[CeA 1B[[ 2

=

(trace[lx (ceA 1 B)*(ceA 1B)drJY

=(trace [L 1: G*(jw)G(jw)dw] Y= [[G(Jwll2

(7 .43)

Thus the time domain L 2 norm of the system's weighting matrix equals the frequency domain L2 norm of the system's transfer function, G(s). Moreover, since in this case we have G(s) E 7-i 2 , this norm is usually referred to as the system's H 2 norm. Notice that if G(s), U(s) E 7-{ 2 with G(s) rational, then G(s) U(s) E 7-{ 2 since each term in the product is analytic in the open right half-plane has no poles on the imaginary axis with the product being zero at infinity. Thus the zero state response from a stable causal system having a strictly proper transfer function satisfies Y(s) E 7-{ 2 or y(t) E £ 2[0, oo) for any input u(t) E £ 2 [0, oo ). In order to extend this result to the case where the transfer function is only proper we need another Hardy space. In the next section we will see that transfer functions of stable systems which are only proper, i.e., G(oo) -=f. 0, lie in a Hardy space denoted 7-iCXJ. Thus if U(s) E 7-{ 2 and G(s) E 1ix then the product G(s)U(s) is analytic in the open right halfplane and has no poles on the imaginary axis. Clearly, in the case when U(s) is rational we have G( oo) U( oo) = 0 and Y(s) E 7-i 2 or y(t) E £ 2 [0, oo) for any input u(t) E £ 2[0, oo ). It turns out that this result holds in the more general case when U(s) is irrational. Thus we have

Y(s)

=

G(s)U(s)

E

(7.44)

H2

when

G(s)

E

7-i 00

Just as there is a frequency domain norm for functions in either 1i2 or Ht, namely the L 2 norm, we will show in the next section that there is a norm for functions in Hem called the H 00 norm. Thus 7-{ 00 is a normed space. However, unlike 7-{ 2 or Ht there is no inner product defined on 7-i 00 . Therefore H 00 is not a Hilbert space. Now the space of rational transfer functions that have no poles in the closed left halfplane and are finite at infinity is denoted 7-i~. Notice that we do not use 1i~ to denote this space since the concept of orthogonality is missing. Therefore in addition to (7.44) we have the dual result

Y(s)

=

G(s)U(s)

E

Ht

(7.45)

when G(s)

E 7-i~

U(s)

E

Ht

We will see in the next three chapters that the Hardy spaces, 1i00 , 1i~, 1i 2 and Ht, that

The H 00 Norm: SISO Systems

181

we are examining in this chapter are essential to the solution of the Hoo control problem. This solution consists of a feedback controller which stabilizes the plant and constrains the Hx norm of the plant's transfer function, from disturbance input to desired output, to be less than a specified scalar. The Hx norm is introduced now as follows.

7.4

The H00 Norm: SISO Systems

At the end of the previous section we introduced the H 2 norm of a system's transfer function, IIG(s)lb by relating it to the time domain L2 norm of the system's impulse response, CeAt B. In this section we introduce another type of frequency domain system norm referred to as the "H infinity norm" denoted IIG(s)lloo- We do this now by relating IIG(s)llx to an equivalent time domain norm referred to as the "L2 system gain". Suppose we are given a stable single-input, single-output system having zero state response y(t) when the input is u(t). Then we define the L 2 system gain, /'o, as /'o

= sup lly(t)ll2 . u(t)E£2[0,oo) llu(t)ll2

(7.46)

u(f),'O

where "sup" is the abbreviation for supremium or smallest upper bound of a set, which in this case is the set of positive real numbers generated by the ratio of the L 2 norms in (7.46) as u(t) is varied over £ 2 [0, oo). Notice that "sup" is used instead of"max" in case the set over which the search is carried out does not contain a maximal element. In addition, notice that we do not consider the null input to avoid dividing by zero. Now the use of a signal norm to define a system norm as in (7.46) is referred to by saying that the signal norm induces the system norm. Therefore the L 2 system gain, (7 .46), is induced by the L 2 signal norm. Notice that, for a given system, this norm is, in general, not equal to the L 2 norm of the system's impulse response, CeAt B. Since we are assuming the system is stable, we saw in the previous S€ction that y(t) E £ 2[0,oo) when u(t) E £ 2[0,oo). In addition we can use the fact that £ 2[0,oo) is isomorphic to 1t2 to rewrite the L 2 system gain, (7.46) in the frequency domain as l'o =

sup II Y(s)ll2 U(s)EJt2ll U(s) ll2

(7.47)

U(.<),'O

Therefore comparing (7.46) and (7.47) we see that the L2 system gain in the time domain is equivalent to the frequency domain system norm induced by the H 2 norm. This norm is referred to as the "H infinity norm", denoted as IIG(s)ILx:· Notice that this norm and the H 2 norm (Section 7.3.6), for a given system are, in general, not equal.

7.4.1

Transfer function characterization of the H 00 norm

Suppose we are given a stable system and that we can choose the system input so that the ratio in the definition of the L2 system gain, (7.46), is close to being maximized. More

Signal and System Spaces

182

specifically suppose the zero state response, Yopt(t), caused by the input

U 0 p1 (t)

satisfies

(7 .48) where Eisa small positive scalar in the sense that E « lo· Then recalling that linear systems satisfy the principle of superpositio n, we have ay 0 p1 (t) when au 0 p1 (t) for any constant a. Therefore (7.48) can be written as

where Ynor(t) = O:Yopt(t) This shows that we can obtain the L 2 system gain, /o, (7.46), or H 00 system norm, IIG(s)lloo, (7.47), by restricting u(t) or U(s) to satisfy the following conditions:

(i)

u(t)E.C 2 [0,oo)

or

U(s) E H 2

(ii)

llu(t)ll2 =I

or

IIV(s)ll 2 =I

Thus (7.46, 7.47) become 'Yo= sup II y(t)ll2 =

sup II Y(s)ll2

(7 .49)

Notice that

(7 .50) where O'max

=

sup IG(Jw)l wE(-oo,x)

(7.51)

and 0' max is referred to as the Lx norm of the function G(jw). However, since U(s) is restricted in (7.49) as IIU(s)ll2= (

-1 1

27r

00

-oo

~ 2 IU(Jw)l dw) = 1

(7 .52)

we see that (7.50) becomes -1 27r

1

00

-oo

IG(Jw)U(jw) l 2 dw :S: O'~ax

(7.53)

r

The H 00 Norm: SISO Systems

183

Now ittums out that we can choose U(jw) in (7.53) subject to (7.52) so that the left side of (7 .53) is arbitrarily close to a~ ax. This implies that the supremium in (7.49) is D'max and we have lo

=

O'max

IG(J.,.:)! = IG(Jw)llx

}UP

1

(7.54)

~ec.-x,x)

Notice that sup IG(iw)l < oo wE(-oo,oo) provided G(s) has no imaginary axis poles. However we have sup u(t)E£2[0ov) u(t)#O

lly(t)ll2 < llu( t) ll2

00

only if G(s) has no poles in the closed right half-plane, since y(t) is unbounded for some u(t) E £ 2[0, oo) othenvise. Therefore IIG(Jw)lloo is referred to as: (i) the L 00 norm of G(s) when the system is unstable and G(s) has no poles on the imaginary axis including s = oo; (ii) the H 00 norm of G(s) when the system is stable and G( :x) < oo. In addition, transfer functions that have bounded H 00 (L 00 ) norm form a normed space denoted 'H 00 (£ 00 ). Notice that 'Hoo C £ 00 • More is said about this in the next subsection. Now there are two different ways of interpreting IIG(Jw) lloo· One arises from defining IIG(I.u)iloc as

IG(J..v·)ll

= 00

sup II Y(jw)ll2 U(iw)E£2II U(jw)ll2 U(Jw)#O

Then in this case we are considering G(jw) to be an operator which maps C(j;_,)) E £2 to Y(jw) E £ 2 and IIG(Ju.:)lloo to be the operator norm induced by the frequency domain L2 norm or, if the system is stable. by the time domain L2 norm. Alternatively, if we take (7.54) to be the definition of IIG(Jw)lloc then we are considering G(jw) to be a function of w and IIG(Jw) lloo is a function norm.

7.4.2

Transfer function spaces

Since any proper rational transfer function with no imaginary axis poles has a finite Loo norm, we can define three normed transfer function spaces depending on the location of the transfer function's poles. Therefore assuming the rational function G(s) is proper we have (i) G(s) E Hoc when all poles of G(s) are in the open left half-plane (ii) G(s) E 'H~ when all poles of G(s) are in the open right half-plane (iii) G(s) E L 00 when G(s) has no poles on the imaginary axis.

184

Signal and System Spaces

Notice that when G(s) E Lx we can use the partial fraction expansion (after one cycle of division) to write G(s) as

'Nhere

Then assuming some scheme for making G+(s), G-(s) unique, e.g., we can insist that c- (s) be strictly proper, we can capture this spliting of a transfer function in L 00 into the sum of stable and antistable transfer functions as a direct sum decomposition of the spaces involved (7 .55) Alternatively, suppose two stable systems having transfer functions G1 (s), G2(s) E H)Q form the components of a cascade connection. Then the transfer function of the composite system, G 1 (s)G 2 (s), is stable since its poles are contained in the poles of the component transfer functions. In addition, the transfer function of the composite system is proper or strictly proper since the product of proper rational functions is proper independent of any pole-zero cancellations that might occur. Therefore G1 (s)G 2(s) E Hoo ifG 1 (s),G2(s) E rt 00 • Now we can readily show, from the definition of the H 00 norm, (7.47), that the H 00 norm of the transfer function of the cascade connected system is related to the H 00 norms of the transfer functions of the component systems as

(7.56) This result plays an important role in control theory. An example of its use is given in the following subsection.

7.4.3

The small gain theorem

Prior to the use of the H 00 norm in control theory, the H 00 norm appeared in a number of different guises in various engineering applications. For example, the resonant gain used to characterize the performance of an electric filter is the H:XJ norm of the transfer function relating the filter's input and output voltages. Another instance of this occurs when the Nyquist stability criterion is applied to determine the stability of a feedback system having transfer function T(s)

G(s) T(s) = 1 + G(s)H(s) Then when G(s), H(s) E Hoc the distance from the origin to the point on the polar plot of G(Jw)H(Jw) which is furthest from the origin is IIG(s)H(s)lloo· This geometrical observation leads to the following result which is known as the small gain theorem. Recall that when G(s)H(s) E 1i 00 , the Nyquist criterion states that the feedback system is stable if and only if the polar plot of G(Jw)H(jw) does not encircle the point

The H 00 Norm: NIIMO Systems

185

-1 + jO. Therefore since the foregoing geometrical interpretation of IIG(s)H(s)lloo implies that the polar plot of G(jw)H(jw) cannot have any encirclements of -1 + jO if

IIG(s)H(s)l\x< I

(7.57)

we have (7 .57) as a sufficient condition for the feedback system to be stable when G(s)H(s) E 'Hoc· This result, referred to as the small gain theorem, is important in connection with robust stabilization where a single fixed controller is required to stabilize any one plant in a set of plants. A simplified example of the application of the small gain theorem in this context is given as follows. Suppose we are given a fixed controller having transfer function H(s) E 'H 00 which satisfies

IIH(s)llx:s; a

(7.58)

Then using (7.56) we can show that (7.57) is satisfied if 1

IIG(s)lloo
(7.59)

;

Therefore any stable plant satisfying (7.59) is stabilized by the given fixed controller. Notice that by interchanging the role of the controller and plant, the foregoing development leads to a characterization of a set of stable controllers any one of which can be used to stabilize a given fixed stable plant. This completes the development of the H 00 (L 00 ) norm for single-input, single-output systems. We need now to proceed to develop this norm for multi-input, multi-output systems.

7.5

The H00 Norm: MIMO Systems

We have just seen that the H 00 norm of a single-input, single-output systefh is the H 2 induced norm of the system's transfer function, G(s). In order to extend this idea to the multi-input, multi-output case, where G(s) is a matrix, we need to introduce the induced 2-norm for constant matrices, denoted as IIG(jw)ll (for fixed w). Then we can obtain the H 00 norm of the matrix G(j w) as the largest induced 2-norm as w varies over (-oo, oo ), i.e.,

IIG(jw)lloo=

7.5.1

sup (IIG(jw)ll) wE(-oo,oo)

Singular value decomposition

Before we can develop the induced 2-norm for a constant matrix we need to introduce the singular value decomposition (SVD) of a constant matrix. This is done as follows. Recall, from Chapter 4, that Hermitian matrices have real eigenvalues and orthogonal eigenvectors. Moreover notice that the product matrices M* M and MM* formed from any complex matrix M, are Hermitian and nonnegative. Therefore these product

186

Signal and System Spaces

matrices have orthogonal eigenvectors and real nonnegative eigenvalues. This general observation leads to the matrix analysis technique known as SVD. The following theorem, which is proved in the appendix, defines the SVD of a matrix. Theorem 7.1 Any p x m matrix of complex constants, M, which has rank r can be decomposed as M = UI:V*

(7.60)

where U, and V are p x p and m x m unitary matrices with

I:=[~ ~] U1 is p

X

U2 is p

r

V1 ism x r

X

V2 ism

p- r X

m- r

and I: 0 is an r x r, real, positive definite, diagonal matrix denoted as a!

0

0

0

a2

0

0

a,

= diag[a 1, a 2, ···,a,]

I:o = 0

with diagonal entries referred to as singular values and ordered so that

i = 1, 2, · · ·, r- 1 Proof See Appendix The SVD of a matrix has become ever more useful since the establishment of practical methods for its computation in the early 1970s. The SVD is of practical use wherever we require the solution of problems involving matrices having small rank. One problem of this sort, which we encountered in Chapter 4, involves the determination of a lower order state model approximation of a state model having controllability and/or observability matrices which are almost rank deficient, i.e., the state model is almost uncontrollable and/or unobservable. We were able to obtain a solution to this problem without requiring the SVD by using the state model's Gramians. Since the Gramians are nonnegative Hermitian matrices, they enjoy the nice properties which make the SVD so useful, namely, real and nonnegative eigenvalues and mutually orthogonal eigenvectors.

7.5.2

Induced 2-norm for constant matrices

We now proceed with the development of the constant matrix norm induced by the 2norm for constant vectors. This is done in the proof of the following theorem by employing the SVD and its properties.

;~

The H 00 Norm: MIMO Systems

187

Theorem 7.2 Any constant p x m matrix J1 has a norm induced by the 2-norm for constant vectors'' hich equals a 1 • the largest singular value of .\1. Proof The induced 2-norm for a matrix M, ' \1 , is defined as IIMII =sup IIMxll J!xll xEC

(7.61)

xi=-

where em is the vector space of all vectors of length m having constant complex entries and llxll equals the positive square root of the sum of the squared absolute value of each entry in x, i.e., (7.62)

Notice that unlike the L 2 norm for a time varying vector which we used earlier, the 2norm for a constant vector has no subscript, i.e .. lll 2 denotes the L 2 norm whereas 11·11 denotes the 2-norm. Next since the transformation J;fx is linear. multiplying x by a constant scalar changes both the numerator and the denominator in (7.61) by the same amount so that the ratio in (7.61) remains unchanged. Therefore we have the alternative definition of the induced 2norm of M given by IIJ1II =sup' ,Vfxll

(7.63)

Since them x m matrix V in the SVD of M, (7 .60), is invertible, we can use its columns { 1./ : i = l, 2, · · · m} as a basis for em. Thus we can express any x E em as m

x= La;v' = Va

(7.64)

i=O

where with a E em. Now, since Vis unitary we have

111

= a*a = Lla;l 2 = llall 2

(7 .65)

i=O

and the 2-norms of x and of a are equal. In addition since Vis invertible there is a unique a for each x and vice versa so that supiiMxll = supiiMVaJJ xECm

cxECm

11<11~1

llall~l

(7.66)

188

\~

Signal and System Spaces

Then using the SVD of M, (7.60) we can write

IIMVall =

(a*V*M*MVa)~

* ~2 ~-(a I:ma)--

where

(

2 2) r ~O"iln;l

(7 .67)

I:; is them x m diagonal matrix given by I:o2 = d"Jag [0"1,2 0"2,2 · · ·, O",2].

Therefore from (7.63), (7.66) and (7.67) we have

II

Mil~ b~t.aila;l 2 r

(7 .68)

Now the summation on right side of the foregoing equation satisfies the following inequality r

L O"flail

r

2 ::;

O"T Llail 2

i=l

(7.69)

i=l

where 0" 1 is the largest singular value of M. However since a is restricted in (7 .68) to have unit norm, we see from (7.69) that (7.70) Notice that if the components of a satisfy i?2 then

llall =

1 and r

L O"flail = O"T 2

i=l

which determines the supremium in (7.68) as

II Mil= 0"1

•

Notice, in the foregoing proof, that the supremium is achieved for a = I 1, the first column of them x m identity matrix. Therefore, letting a= I 1 in (7.64) gives the result that the unit 2-norm x that gives the supremium in (7.63) is the right singular vector of M, v 1, corresponding to the largest singular value of M, 0" 1 . Thus

IIMxll

~=0"]

,.

189

The H 00 Norm: MIMO Systems

when x = (3v 1 for any complex scalar, (3. Further consideration of this fact leads us to the important conclusion that

!IMxll :S 0"1 llxll

for all

X

E Cm

(7.71)

with equality onlywhenxisa scalar multiple ofv 1 . Finally, notice that ' JIII = M is a scalar. 1

7.5.3

'Jfl when

The Lx, H 00 norm for transfer function matrices

Having developed the idea of the induced 2-norm for constant matrices, we are now in a position to give the definition of the Lee norm of a transfer function matrix, G(jw). This norm is defined as the supremium of the induced 2-norm of G(j..,:) over all wE ( -oo, x)

IIG(j,.-.:) lloo=

sup

O"J (i~)

(7.72)

~'C(-oo.:x.

where 0" 1(jw) is the largest singular Yalue of G(j..v·). As in the SISO case, when an MI:\10 system is stable its transfer function matrix G(s) has finite L 00 norm, which is referred to as the Hoc norm and G(s) is lies in the normed space denoted by Hx. Alternatively. when the system is antistable, its transfer function matrix G(s) has finite Lee norm, (7.72), and G(s) lies in the normed space denoted H:;,. Finally notice that, as in the SISO case, all G(s) with finite Lx norm are denoted b:; G(s) E £ 00 . Thus Hcc and H~ are each subspaces of £ 00 which are disjoint and complete the £ 00 space so that (7.55) holds for MIMO systems. In subsequent deYelopment we will need the following result which is an immediate consequence of the foregoing. Theorem 7.3 A real rational transfer function G(s) E Lx has an L 00 norm less than a finite positive scalar ~., i.e.,

]G(j..v·) lloo < ! if and only if F(jw) is positive definite for all w, i.e., 'Vw E ( -oo, oo)

F(jw) > 0 where

F(jw) ="-/I- G*(jw)G(jw) Proof Let the SVD (Theorem 7.I) of the p x m matrix G(j w 1 be given as

G(jw) = U(jw)L,(jw) V*(jw)

where U(jw). V(jw) are p x p and m x m unitary matrices respectively and L,(jw) = [

L,o(.fw) 0

~]

L,0 (jw)

=

diag[O"J (jw), 0"2 (jw), · · ·. O",(jw)]

190

Signal and System Spaces

~..

with r = rank[G(jw] and cr 1(jw) ~

crJjw)

i

>1

(7.73)

Then using the SVD of G(jw) we can write F(jw) as (7.74)

where L:~,(jw) is them dimensional diagonal matrix

L:~7 (jw) = diag [crf(jw), cr~(jw), · · · , cr~ (jw), 0, · · · , 0] Next pre and post multiplying (7.74) by V*(jw) and V(jw) respectively gives

V*(jw)F(jw) V(jw)

= L:~

(7.75)

where

Now since V(jw) is unitary and therefore invertible, we have

F(jw) > 0

V*(jw)F(jw)V(jw) > 0

if and only if

Thus we see from (7.75) that

F(jw) > 0

if and only if

[·-?- crf(jw)] > 0

(7.76)

However, since the L 00 , norm for G(s) is defined as IIG(Jw)lloo=

sup

wE( -oo,x)

cr1 (jw)

we see from (7.76) that

F(jw) > 0 for all wE ( -oo, oo)

if and only if

IIG(Jw)llx< I

and the theorem is proved. • In the next chapter we will use the foregoing theorem to obtain necessary and sufficient conditions on the state model parameters, {A, B, C, D}, of a system which ensures that the system's transfer function has L 00 norm less than a given finite real scalar.

7.6

Summary

We began by formalizing the idea, used earlier in Chapters 5 and 6, of relating the size of a signal to the energy in the signal. This led to the time domain Hilbert space £-2( -oo, oo) and its orthogonal subspaces £ 2 ( -oo, OJ and £ 2 [0, oo ). Following this we used the Hilbert space isomorphism from £ 2 ( -oo, oo) in the time domain to £ 2 in the frequency domain to introduce the frequency domain spaces H 2 , H~ known as Hardy spaces.

Notes and References

191

Next we introduced the notion of the size of a system. This was done by using the induced operator norm to characterize a system's ability to transfer energy from its input to its output. As with signal norms we saw that there is an equivalence between a time domain system norm known as the system's L 2 gain and a frequency domain norm known as the Hx norm. Unlike the Hardy spaces, H 2 , H} for signals which are equipped with an inner product, the Hardy spaces, 'H00 , 1{~ for systems are normed spaces only.

1.1

Notes and References

There are many texts on Hilbert space. A readily accessible treatment of this subject which includes material related to the control problems being considered here can be found in [46]. Unfortunately books on Hardy spaces require a considerable background in functional analysis and complex analysis. Some of these references are given in [14, p. 13}. An interesting treatment of complex analysis as it applies to Laplace and Fourier transforms is given in [26]. The example following (7.27) in Section 7.3.2 which is used to discuss the Fourier transform of signals in £ 2 that are not in £ 1 is taken from [45, p. 272]. The convergence of the series (7.36) is discussed in [36, p. 71]. The applicability of Parseval's theorem for functions in £ 2 ( -oo, oo) is discussed in [37, p. 185]. An excellent reference for the SVD is [17]: The small gain theorem is used to solve a wide variety of robust control problems including both linear, [47] [18] and nonlinear, [42] [21] [33], systems. Finally a good basic introduction to the ideas in this and the next chapter can be found in [11].

8 System Alge bra

8.1

Introduction

In the previous chapter we were interested in characterizing signals and systems in terms of normed spaces in both the time and frequency domains. This led to the introduction of the frequency domain spaces H 2 , Hi and £ 2 for signals and 'H00 , H;;, and £oc for systems. In the present chapter we consider a number of operations involving systems in these spaces. We will be especially interested in obtaining state models for the systems which result from these operations. To facilitate this endeavour, we use the compact equivalence relation

G(s)

s

=

[AC DB]

(8.1)

to denote that

G(s)

=

C(sl- A)- 1B + D

.,. We begin the discussion of these operations by considering the effect of connecting systems in parallel and in series.

8.1.1

Parallel connection

One of the simplest examples of operations with systems consists of connecting several systems in parallel so that all systems have the same input and the output from the connection is the sum of the outputs from each system. The connection of the first order component systems resulting from a partial fraction expansion of the system's transfer function is an example of the parallel connection of systems. The input -output constraints which ensure that r component systems are connected in parallel are u(t)

= u;(t)

y(t)

= L:Y;(t)

i = 1, 2, · · · r

(8.2)

r i=l

(8.3)

194

System Algebra

u(t)

Figure 8.1

Parallel Connection

so that

Y(s) = G(s)U(s) where r

G(s)

=

L G;(s) i=l

with G;(s) : i = 1, 2 · · · r being the transfer functions of the component systems. Thus assuming we know state models for each of the component systems as

G;(s) ~ [ A' C;

Bl

' D;

i = 1, 2, .. · r

(8.4)

we want to determine state model parameters {A, B, C, D} for the composite system, i.e.,

G(s)

~ [;

;]

We do this as follows. First we write the state equations for each component system as

+ B;u;(t) = C;xi(t) + D;u;(t)

xi(t) = A;xi(t) Y;(t)

(8.5)

where i = 1, 2, · · · r. Next we use (8.5) together with the constraints imposed by (8.2, 8.3) to give a state model for the composite system as x(t)

=

y(t)

=

+ Bu(t) Cx(t) + Du(t) Ax(t)

(8.6)

Introduction

195

where AI

0

0

A2

Il

A= 0

C=[Ct

0

c2

C,]

m

XI (t) x 2(t)

x(t) =

B=

x'(t)

;

D=LD; i=l

Notice that the dimension of the state model for the composite system is the sum of the dimensions of each component system. MoreoYer. MIMO systems can be connected in parallel if and only if each of the systems has: (i) same number of inputs, and (ii) the same number of outputs. Notice in the case of the partial fraction expansion of G(s), the A;s are the poles of G(s). Finally notice that if we are given G(s) E Lx then we can use partial fraction expansion to decompose G(s) into the parallel connection of two systems such that G(s) = GI (s) + G:c (s J \\here GI(s) E n 00 and C:c (s) E H~ with the eigenvalues of AI (A 2 ) being in the open left (right) half plane. This fact was given in the previous chapter as the direct sum decomposition of the spaces involved, i.e., £ 00 =Hoc EB 7-i~.

8.1.2

Series connection

Next consider a series connection of two systems, often referred to as system composition. The constraints imposed by this connection are

YI(t)=r(t)

(8.7)

so that

Y(s) = G(s)U(s)

(8.8)

where

T\ow we can de\elop the state model for the composed system, by imposing the constraints (8.7) on the state models for the component systems, (8.4), with r = 2. The result is as follows.

xi(t)

= A 1x 1 (t)

x2 (t) =

A 2 x:c(t)

y(t) = C 1x 1 (t)

_u_(t)-------i>!)l

G2(s)

Figure 8.2

+BIC:cY 2

+ B2u(t) + DI C 2x 2(t) + D1D 2u(t)

1-----0>1)1

G,(s)

Series Connection

1--Y(_t)_

196

System Algebra

which can be rewritten as

y(t)

= [ C1

and we see that the composed system has state model

(8.9) where

Notice that the constraint u1 (t) = y 2 (t) implies that the number of inputs to the system labeled 1 must equal the number of outputs from the system labeled 2. Alternatively, we could connect the systems in reverse order if the number of inputs to the system labeled 2 and the number of outputs from the system labeled 1 are equal. In this case the transfer function for the composed system would be G2 (s)G 1 (s). Notice that we get the same composed system independent of the order in which the component systems are connected in series only if the matrices G 1 (s) and G2 (s) commute so that G 1 (s)G 2 (s) = G2 (s)G 1 (s). Thus in the SISO case the result of composing systems is independent of the order in which the component systems are connected. In the remainder of this chapter we will take up the problem of determining the state model for several other types of system operation. One of these problems concerns the determinatio n of the state models for the systems needed in a series connection so that the resulting composed system has a specified transfer function. This problem is known as the system factorization problem. The requirement that the factors have certain properties gives rise to a number of different classes of system factorizations, several of which are of great importance to the development of control theory.

8.2

System Inversion

Before we begin considering system factorization we need to consider the related problem of finding a system inverse. Suppose we are given two SISO systems having transfer functions b m-1 b m + · · · + bm G (s) = os + 1s 1 s n + a1sn-1 + ···+an

(8.1 0)

sn+a1sn- 1 +···+an G (s) - - - - - ; - - - - - - , -

(8.11)

2

-

boSm

+ b ]Sm-1 + · · · + bm

er function which is unity, i.e., Then the series conne ction of these systems gives a transf Thus it would appea r that the input. its to equal the series connected system has its outpu t system labeled 1 has the system labeled 2 as its inverse. sinusoidal steady state gain However, physical processes have the prope rty that their G ( oo) and G2 ( oo) to be finite. canno t be unbou nded at infinite. Therefore we require 1 m). There fore we can only have Now G 1 ( oc) ( G2 (oo)) is finite if and only if m ~ n (n ~ a given physical process having both G 1 ( oo) and G2 ( oo) finite if m = n. This means that prope r (not strictly prope r) or is (s) transf er function G 1 (s) has an inverse if and only if G1 D matrix. ro nonze a equivalently only if the state model for G1 (s) has has an inverse if and only if (s) G on 1 In the MIMO case, a system having transf er functi systems having the same only Thus its state models have aD matrix which is invertible. for the inverse system of model state numb er of inputs as outpu ts can have an inverse. The s. a given invertible system can be developed as follow

8.2.1

Inverse system state mod el

for an MIM O system Suppose we are given the transfer function and state model

~ [~· ~]

G(s)

D- 1 exists

so that state equat ions for the system are

+ Bu(t)

(8.12)

= Cx(t) + Du(t)

(8.13)

x(t) = Ax(t) y(t)

Then solving (8.13) for u(t) gives u(t) = -D- 1Cx(t)

+ D- 1y(t)

(8.14)

which when substi tuted in (8.12) gives

(8.15) outpu t u(t), input y(t), and Thus we see from (8.14, 8.15) that the inverse system has state model given as

G- 1 (s)b

[~:

Bx] Dx

(8.16)

where Ax =A- BD- 1C

ex=

-D- 1 C

= BD- 1 Dx = D- 1

Bx

useful in developing results in As we will see, the state model for the system inverse is contro l theory.

1

System Algebra

198

Notice that the transfer functions for aninvertible SISO system and its inverse (8.1 O, 8.11) are reciprocals of each other. Therefore the poles of the given system equal the zeros of the inverse system and vice versa. A similar statement can be made for invertible MIMO systems. To see this we need interpret the idea of a system zero in terms of a system's state model. This will be done by first considering system zeros for SISO systems.

8.2.2

S/SO system zeros

Recall that an SISO system has a system zero at s = s 0 if its transfer function, G(s), is zero for s = s0 or

U(s 0 )

Y(s 0 ) = 0

-=/=

(8.17)

0

where

Y(s) = G(s)U(s) Notice that (8.17) implies that G(s 0 ) = 0. Now in order to arrive at an interpretation of system zeros which we can use in the MIMO case, we consider a system's zeros in terms of a state model for the system. Suppose we are given a controllable and observable state model for the system, s G(s) =

[AC DB]

(8.18)

Then taking x(O) = 0 since system zeros are defined in terms of the system's zero state response, the Laplace transform of the state equations for the system yields

[0 ] -B] [X(s)] Y(s) U(s)

[sfC -A

=

(8.19)

D

Therefore if s0 is a system zero we have

[ sC0 ! - A ] X(so)

+

[-

D

B] U(so)

=

U(s 0 )

0

-=/=

0

(8.20)

which implies that the column

is dependent on the n columns in

Therefore s = s0 is a system zero if and only if the (n side of (8.19) is not invertible or rank [

s0 1- A

c

+ 1)

-B] < n+ D

1

x(n + 1) matrix on the left

(8.21)

199

System Inversion

As illustration, suppose we are given a second order system in controller form with D zero. Then (8.19) is [

s+ a 1 ~1

which after eliminating X 1 (s) becomes

(s 2 + a 1s + a2)X2 (s) ~ U(s) = 0

(8.22)

(c 1 1 + c2 )X2 (s) = Y1s)

(8.23 I

Notice from (8.22) that if C(s0 ) i= 0 then X(s 0 ) 7'= 0. Consequently (8.23) implies that = 0 for U(s 0 ) cJ 0 only if s0 is a zero of the polynomial

Y(s0 )

However

which shows that s0 is indeed a zero of G(s). By proceeding in the same fashion. we can show. when D cJ 0. that the foregoing state model has a system zero at s0 which is a root of the polynomial D(s 2 -L a 1s- a2 ) + c 1 s + c2

which is the numerator of G(s). The foregoing ideas are used now to extend the definition of a system zero systems.

8.2.3

to

MIMO

MIMO system zeros

Suppose the system denoted by (8.18) has m inputs and p outputs \Yith m :.:; p and with all m columns of B independent. Then a complex number s0 is a system zero if G(so) C(s0 )

=o

for some Us 0 )

#

o

(8.24)

Notice that unlike the SISO case \Yhere (8. 17) implies that G(s 11 ) = 0. (8.24) does not imply that G(s0 ) = 0. Now with appropriate modification, (8.20) implies that s = s0 is a system zero if at least one of the m columns of

[-:J

200

[1

System Algebra

\1 I

is dependent on the columns of

Therefore, recalling that the rank of a matrix cannot exceed its smallest dimension, we see from (8.21) that s0 is a system zero if rank [

-B]
s0 I- A

D

c

(8.25)

Alternatively, when m > p the matrix in the foregoing inequality cannot have rank greater than n + p so that (8.25) is satisfied for all s0 and (8.25) is meaningless. In order to overcome this problem we define s0 to be a system zero if rank [

s0 I- A

c

-B]

D


where p =max ( rank [

sf- A

sEC

C

We want now to apply the foregoing ideas to invertible systems. Since invertible systems are square, i.e., m = p, we can use (8.25) to define the system zeros for this class of system.

8.2.4

Zeros of invertible systems

Suppose G(s) is invertible. Then we see, in this case, that the condition for s0 to be a system zero, (8.25), is equivalent to det [

s 0 I- A

c

-B] =0

(8.26)

D

Notice that if the state model is not controllable the foregoing condition is also satisfied when s0 is an uncontrollable eigenvalue of the system since we have an eigenvalue-left-eigenvector pair (.A, w) of A which satisfies wT (M-

A) = 0 and

wT B

= 0

Therefore

(8.27) and (8.26) holds for)\= s0 . Thus, in this case, s 0 is both a system zero and a system pole. Alternatively, if (A, C) is an unobservable pair a similar argument can be used to show that at least one of the eigenvalues of A is both a system zero and a system pole.

·~

t

Coprime Factorizatio n

201

However, if the state model is both controllable and observable with Dis nonsingular, then there is no system zero which equals a system pole. In addition, each system zero of G(s) is a system pole of G- 1 (s) and vice versa. To see this suppose s0 is a system zero. Then (8.26) is satisfied and is equivalent to det [[

s0 l - A

(8.28)

c

where M is any nonsingular matrix of appropriate dimension. Now suppose we choose M as

~] Then the left side of (8.28) becomes det [[

sol- A

-:]

c =

det[s0 / - A+ BD- 1 C] det[D]

However since G(s) is invertible, we have det[D]

(8.29)

=I 0 and (8.29, 8.28) imply

where

and Ax is the system matrix for G- 1 (s), (8.16). Therefore s0 is a pole of G-l.(s) as well as being a zero of G(s). Finally, we will see that invertible systems which are stable with stable inverses play an important role in control theory. These systems, which are referred to as "units", form a subspace U 00 of 1{00 and are characterized in terms of their transfer functions G(s) as

G(s)

E

Uoc

if and only if

G(s), G- 1 (s) E 1i00

In the case of an SISO system, this definition implies that a system is a unit if it has a transfer function which is proper with no poles or zeros in the closed right-half plane.

8.3

Coprime Factorization

Problems in system factorization are concerned with the determination of two systems, called factors, whose series connection has the same input-outpu t behavior as ihe system being factored. Additional constraints on the factors determine different classes of factorization.

202

System Algebra

Coprime factorization can be characterized as a series connection of a stable system and the inverse of a stable system with no unstable pole-zero cancellations between factors. As an example consider the system with transfer function G(s) where G(s) -

(s+l)(s-2)

_..:...._~..:...._-----'--_

- (s

+ 3 )(s -

4) (s

(8.30)

+ 5)

Then a coprime factorization of this system is indicated as follows:

where

Nt(s)=(s-2) (s+l) (s + 3)a(s)

Mt(s) = (s-4)(s+5) a(s)

with a(s) being any polynomial of degree 2 having no zeros in the closed right half plane. Notice that this choice of degree for a(s) ensures that M 1 (s) is invertible whereas the restriction on the location of the zeros of a(s) ensures that a(s) is not involved in any unstable pole-zero cancellations between N 1 (s) and Mj 1 (s). Thus even if a(s) has zeros at s = -1 and/or -5 the resulting pole-zero cancellations in N 1(s) and Mj 1(s) are allowed since they are stable. More important, notice that the closed right-halfplane zero of N 1 (s) at s = 2 and of M 1 (s) at s = 4 are different so that there are no unstable pole-zero cancellation between N 1 (s) and Mj 1 (s). Stable systems which do not share any system zeros in the closed right-half plane are said to be coprime. Notice, from the foregoing example, that the form of coprime factorization, N(s)M- 1 (s), bears a striking resemblance to the expression for a transfer function in terms of its numerator polynomial N(s) and denominator polynomial M(s). Of course the "numerator" and "denominator" in the present context are not polynomials but are each rational and can therefore be thought of as transfer functions for "numerator" and "denominator" systems.

8.3.1

Why coprime?

In the foregoing example the numerator N 1 (s) and denominator M 1 (s) are coprime. In order to better appreciate this fact we give a second factorization of G(s) in which the factors are not coprime Suppose were-express G(s), (8.30), as (8.31)

where N 2 (s) = (s- 6)(s- 2)(s + 1)

(s

+ 3)f3(s)

() _ (s-4)(s+5)(s -6)

M2 s -

jJ(s)

with f3(s) restricted to be any degree 3 polynomial with no zeros in the closed right half plane. Although this factorization has the desired property that N 2 (s), M 2 (s) E H 00 with

I

····-

'·'·"~·-~-~~--.

··· c~~rin: ''ciOr:;i•uon·

203

(s) and Mz(s) be Mz(s) being invei:tible, it does not satisfy the require ment that N 2 is not a coprim e (8.31) Thus 6. = s at zero coprim e since they share a right-h alf plane unstabl e polehaving not of ance import the factoriz ation of G(s). We can demon strate . follows as factors zero cancell ations betwee n having one of its n Suppos e that we have a strictly proper scalar transfe r functio n G(s) - 1 poles being in the poles, n > 2, at s = s0 on the positive real axis with the remain ing n as open left-hal f plane. These specifications imply that G(s) can be written

G(s)

=

q(s)[s - (s0 +E)] (s- soJP(s)

0 :::;

E

<

00,

So

>0

no zeros in the where the degree of q(s) is less thanp( s) with q(s0 ) f. 0 and p(s) having is stable only if E = 0 closed right half plane. Now the system having this transfe r functio n inspect ion of the so that there is an unstabl e pole-zero cancell ation. However, closer rely, in practice, on consequences of such a pole-zero cancell ation reveals that we cannot This becomes unstabl e pole-zero cancellations to make an unstabl e system stable. 1 e, £:j: [G(s)] = g(t) immediately evident when we conside r the system's impulse respons

g(t) = Koe501 + r(t) where

Ko =

-E

q(so) p(so)

with r(t) bounde d for 0 :::; E < oo. e response is Notice that if E = 0, K 0 esot is missing from g(t). Then the system's impuls 501 is present in g(t) and the e K bounde d and the system is stable. Howev er, if E f. 0, then 0 is unstabl e. Thus the system's impuls e respons e tends to infinity with time and the system system's stability is catastro phicall y sensitive to E. 01 witlftim e and the Conversely, if s0 is in the open left-ha lfplane then K 0 e" tends to zero system is stable for all E. ro cancellation, Compa ring the effect on the system behavi or of the two types of pole-ze y to unstabl e stabilit 's system a of stable and unstabl e, we see that the lack of robustn ess nt to reliable detrime a ation pole-zero cancell ation makes this type of pole-zero cancell factors in a the of y stabilit the control system design. We will see that the coprimeness and control lers ining determ for coprim e factoriz ation can be exploited to provide a means additio n in but stable utput which not only make a feedback control system input-o that the so plant and ler preven t unstabl e pole-zero cancellations betwee n the control closed loop system is interna lly stable. system zeros of Return ing to coprim e factoriz ation, we saw earlier in this chapte r that see now, that we re, Therefo . an invertible system are poles of that system 's inverse 1 if and only if ted preven are (s) Munstabl e pole-ze ro cancell ations betwee n N(s) and i.e., if and plane, alf right-h closed the M(s) and N(s) have no commo n system zeros in only if M(s) and N(s), are coprime. 1 G(s) is unstabl e. In additio n, since N(s) is always stable, M- (s) is unstabl e if are the only G(s) of poles e unstabl the e, coprim Moreov er, since M(s) and N(s) are

204

System Algebra

unstable poles of M- 1 (s). Finally, since M(s) is stable, M- 1 (s) has no zeros in the closed right-half plane. Therefore the closed right-half plane zeros of G(s) are the only closed right-half plane zeros of N(s).

8.3.2

Coprime factorizatio n of MIMO systems

Moving on to MIMO systems, we need to distinguish between two types of coprime factorization depending on the ordering of the factors. Thus a p x m real rational transfer function matrix G(s) can be coprime factored as

G(s) = N(s)M- 1(s)

(8.32)

G(s) = M- 1 (s)N(s)

(8.33)

or as

where N(s), N(s) E 7-f. 00 are both p x m and M(s), M(s) E 7-f. 00 are m x m and p x p respectively. In addition both pairs {N(s), M(s)}, and {N(s), M(s)} must be coprime. Notice that the two varieties of factorization (8.32, 8.33) are referred to as right and left coprime factorizations since the denominator is on the right in (8.32) and on the left in (8.33). In the SISO case these two varieties coincide since products of scalars commute. We give now, in the theorem to follow, an alternative characterization of coprimeness. We will see later that this characterization is also useful in the problem of determining controllers which make the closed loop control system internally stable. Theorem 8.1 The factors M(s), N(s) E 7-f.DC {M(s), N(s) E 7-f.oo} are right {left} coprime if and only if there exists X(s), Y(s) E 7-f. 00 {X(s), Y(s) E 7-f.oo} which satisfy the following Bezout, (pronounced "bzoo") identity

X(s)M(s) {M(s)X(s)

+ Y(s)N(s)

(8.34)

=I

+ N(s) Y(s) =I}

(8.35)

Proof (it) Consider the SISO case. Suppose M(s), N(s) are not coprime, i.e., M(s 0 ) = N(s 0 ) = 0 for some s0 in the closed right-half plane. Then since X(s) and Y(s) are stable, X(s 0 ) < oo, Y(s 0 ) < oo. Therefore we have X(s0 )M(s0 ) = 0, Y(s 0 )N(s0 ) = 0 and (8.34) cannot be satisfied. This shows, in the SISO case, that satisfaction of the Bezout identity is sufficient for M(s),N(s) to be coprime. The MIMO case is treated as follows. Rewrite the Bezout identity, (8.34), as

(8.36)

L(s)R(s) =I where

L(s) and L(s), R(s) E

7-{ 00

=

[X(s)

Y(s)]

are m x (m + p) and (m

R(s)

+ p)

=

[ M(s)] N(s)

x m respectively.

I I .

j

Coprime Factorization

205

Next suppose that M(s), N(s) are not coprime with s0 being a zero of M(s) and of N(s) which lies in the closed right-half plane. Then s0 is a system zero of R(s) and we see from (8.24) that there is a complex vector U(.111 1 # 0 such that

However since L(s)

E

Hoc all elements of L(s0 ) are finite. Therefore we have Uso)R(.Io) U(.1o I = 0

and the Bezout identity (8.36) is not satisfied. Therefore we have shown that, in the MIMO case, if the Bezout identity, (8.36) or (8.34), is satisfied then M(s), N(s) are coprime. We can show, in a similar way. that (8.35) is sufficient for Jf(s),N(s) to be coprime. Proof (only if) This will be done in the next section by using certain state models we will determine for the transfer functions· JJ(s), S·(s), M(sl. N(s). X(s). Y(s), i(s),

•

¥w.

Before going on to the development of state models for the various systems involved in coprime factorization, we need to consider the possibility of there being more than one coprime factorization for a given system. This is done now as follows.

8.3.3

Relating coprime factorizations

The coprime factors in a coprime factorization of a given system are not unique. Moreover the factors in two different coprime factorizations of the same transfer function are related by a unit, (Section 8.2.4). This fact is shown in the following theorem. Theorem 8.2 Suppose G(s) has right and left coprime factorizations

Then G(s) has right and left coprime factorizations given by

G = S 2 (s)M2 1 (s) =

Jf2 1 (s)S.c(s)

if and only if there exists units R(s). L(s)

E

U00 such that

N 2 (s) = N 1 (s)R(s)

N 2 (s) = L(s)N 1(s)

Jf2(s) = M 1(siR(s)

jf2 (s) = L(spf 1(s)

(8.37)

Proof (if) If R(s) E Hx then we see from (8.37) that N 2 (s), M 2 (s) E Hoc since the composition (series connection) of stable systems is stable. In addition we see from (8.3 7) that (8.38)

206

System Algebra

However since N 1 (s), M 1(s) are coprime the Bezout identity (8.34) is satisfied. Therefore pre and post-multiplying the Bezout identity by R- 1 (s) and R(s), and using (8.37) we obtain (8.39) where

Finally if R- 1 (s) E Hoc then we have X2(s), Y2(s) E Hoc and M 2(s) and N 2(s) are 1 coprime from Theorem 8.1. Therefore we have shown that N 2 (s)M2 (s) is an additional right coprime factorization of G(s) if R(s) E Ucc. 1 We can show, in a similar fashion, that if L(s) E Uoc then MJ. (s)N2 (s) is a second left coprime factorization of G(s). Proof(only if) Consideration of(8.38) shows that the invertability of R(s) is necessary. 1 Therefore suppose R(s) is invertible but R(s) '/:. U00 • Then R(s) '/:.Hoc and/or R- (s) '/:. Hx. Suppose R(s) '/:. H 00 • Then since N 1 (s)M\ 1 (s) is a coprime factorization of G(s), we see from (8.37) that either N 2 (s), M 2 (s) '/:. H 00 or N 2 (s), M 2 (s) E H 00 . Suppose N 2 (s), M 2 (s) '/:. H 00 • Then N 2 (s)M2 1 (s) is not a coprime factorization of G(s). Alternatively, suppose N 2 (s), M 2 (s) E H 00 • Then we see from (8.37) that all unstable poles of R(s) are cancelled by zeros of N 1 (s) and M 1 (s). This implies that N 1 (s) and M 1 (s) share closed 1 right-half plane zeros and therefore, contrary to assumption, N 1 (s)M\ (s) is not a coprime factorization of G(s). Therefore R(s) E H 00 Alternatively, if R- 1 (s) '/:. Hoo then R(s) has closed right-half plane zeros. Thus we see from (8.37) that N 2 (s) and M 2 (s) share closed right-halfplane zeros and are therefore not coprime. Thus N 2 (s)M2 1 (s) is not a coprime factorization of G(s). The foregoing shows that N 2 (s)M2 1 (s) is an additional right coprime factorizations of G(s) only if R(s) in (8.37) is a unit. In the same way we can show that M2. 1(s)N2 (s) is a second left coprime factorization • of G(s) only if L(s) in (8.37) is a unit. always is it since proper always is M(s) that section this In summary, we have seen in invertible. Alternatively, N(s) is proper when G(s) is proper and strictly proper when G(s) is strictly proper. We will see in the next section that Y(s) is always strictly proper so that Y(oo) = 0. Therefore the product Y(s)N(s) is always strictly proper so that Y( oo )N( oo) = 0 no matter whether G(s) is proper or strictly proper. Thus we see that 1 a necessary condition for satisfying the Bezout identity, (8.34), is X( oo) = M- ( oo) which implies that the state models for X(s) and M(s) must have D matrices which are inverses of each other. Similarly, state models for X(s) and M(s) must have D matrices which are inverses of each other.

8.4

State Models for Coprime Factorization

In what follows we develop, from a given minimal state model for the system being factored, minimal state models for its coprime factors, M(s), N(s), M(s), N(s) as well as the factors X(s), Y(s), X(s), Y(s) needed to satisfy the Bezout identities, (8.34, 8.35).

%07

8.4.1

1

Righ t and left copr ime factors

state model for the system Suppo se we are given a contro llable and ob~ervable (minimal) to be factore d G(s) =s

B]

[A

(8.40)

C D

Then applyi ng state feedba ck (8.41)

u(t) =Kx(t )--H(t )

ing state equati ons for the with K chosen to make A + BK is stable. yields the follow closed loop system

x(t) =(A+ BK)x( t) + Bv(t)

(8.42)

y(t) = (C 7 DK)x (t), Dv(r)

(8.43)

r functio ns S(s), .H(s) Now we can interp ret (8.41- 8.43) as two systems haYing transfe with each system having v(t) as input

Y(s) = N(s) V(s)

(8.44)

M(s) V(s)

(8.45)

U(s)

=

where

"V(s)

s

=

[AC +BK + DK

B]

M(s) =s

D

[A+ BK B] K

(8.46)

I

n noti~ that Jf(s) is Notice that N(s), M(s) E Hx since A+ BK is stable. In additio y matrix . This allows us to invertible since its state model has aD matrix which is an identit solve (8.45) for V(s) and substit ute the result in (8.44) to obtain

Y(s) = N(s)A r 1 (s) U(s) relatio n of the foregoing The following simple example provid es an illustra tion of the state model s, (8.46) to coprim e factori zation . Suppo se we want to obtain a coprim e factori zation for s- 1 G(s)

= s(s- 2)

in contro ller form. so that We begin by obtain ing a minim al state model for G(s). say G(s) =s

[AC

B]

D

208

System Algebra

where A=

[~ ~]

C= [1

-1]

Next we need to choose K so that A + BK, is stable. Suppose we do this so that A has a multiple eigenvalue at -1. Then K is K = [ -4

and A+ BK, and C

-1]

+ DK in the state models for the factors, -2

A +BK = [ l

+ BK

(8.46) are

C+DK=C=[1

-1]

Finally, we get the transfer functions for the coprime factors from the state models, (8.46), as

M(s)

= K[sl- (A+ BK)r 1B +I=

s2

1

s

1

N(s)=(C+DK)[sl-(A+B K)r B=

(s + 1)

-

2

2s

(s + 1)

2

and we see that N(s)M- 1 (s) = G(s) The required coprimeness of N(s), M(s) in this example is seen by inspection. We can show that the factors, M(s), N(s), (8.46), are coprime in general by using the definition of a system zero given in Section 8.2.3 of this chapter. This is done as follows. We begin by noting that the state models for the factors, (8.46), have the same system matrix, A + BK, and the same input matrix, B. Therefore the states of these systems are equal for all time if their inputs are equal and if their initial states are equal. Recall from Section 8.2.3 that the system zeros are defined in terms of the zero state response. Therefore in order to investigate the coprimeness of the factors specified by (8.46), we take the initial states of these state models null, i.e., equal. Moreover, we note from, (8.44, 8.45) that the inputs to these state models are the same being equal to v( t). Therefore the states of the state models for the factors M(s) and N(s), (8.46), are assumed equal in the following investigation of the system zeros of the factors. Suppose s0 is a common system zero of the factors M(s) and N(s) given by (8.46). Then we see from (8.20) that the following conditions must be satisfied [sol- A - BK]X(s0 ) [C + DK]X(s 0 )

-

BU(s0 ) = 0

+ DU(s0 )

KX(so)

(8.47)

0

(8.49)

+ U(s 0 ) = 0

(8.49)

=

: i

~')

~. ..

State Models tor Coprime Factorization

209

Thus substituting for U(s 0 ) from (8.49) in (8.47, 8.48) yields [so/- A]X(s0 ) = 0 CX(s0 ) = 0

which implies that the pair (A, C) is unobservable. This contradicts our assumption that the state model for G(s) is minimal. Thus s0 cannot be both a zero of N(s) and a zero of M(s) which is sufficient for N(s), M(s) to be coprime. Notice, from the foregoing demonstration of the coprimeness of M(s) and N(s), (8.46), that the detectability of the state model used for G(s), (8.40), is both necessary and sufficient for the coprimeness of these factors, i.e., for there to be no common closed righthalfplane system zeros for M(s),N(s). In the next section we will provide additional evidence for the coprimeness of the pair M(s),N(s), (8.46), by finding state models for X(s), Y(s) to satisfy the Bezout identity, (8.34). Before doing this we determine state models for some left coprime factors of G(s) by utilizing the fact that the right coprime factorization of GT (s), i.e., (8.50) becomes the desired left coprime factorization of G(s) after transposition, i.e., G(s)

=

M- 1(s)N(s)

We do this as follows. From (8.40) we have

Then from (8.46) we see that the right coprime factors of GT (s) are given a~.

where L T replaces K in (8.46). Notice that as K was chosen to make A + BK stable in (8.46), so must L T be chosen now to make AT + CT L T stable. Therefore after doing the required transposition we obtain the left coprime factors for G(s) as -

s

M(s) = Notice that we have N(s), M(s)

8.4.2

E

[A+ LC ~] c

(8.51)

1-£00 with M(s) being invertible.

Solutions to the Bezout identities

In this section we will see that there is an important connection between the coprime factors for a controller in an observer based feedback control system (Section 5.2), and

~, v

.,ysre m Atgeo ra

the pairs {X(s) , Y(s)} and {X(s), Y(s)} in the Bezou t identities, (8.34, 8.35). This connection is used here to establish state models for these pairs. We will see later that this connection can be used to characterize all contro llers which impar t stability to a closed loop contro l system. In order to obtain a coprime factorization for the contro ller, we need to recall that the relevant equati ons defining an observer based contro ller for the plant, (8.40) are ~ (t)

= A-~(t) + Bu(t) + L[Y(t )- y(t)]

y(t) = Cx(t)

u(t)

=

+ Du(t)

Kx(t)

Then we see from these equati ons that the state equations for the contro ller are

x (t)

=

Ax(t)

u(t)

=

Cx(t)

+ By(t) (8.52)

where

A = A+ BK + LC + LDK so that the contro ller transf er function H(s) is given by

Notice that the contro ller is strictly prope r. Notice also that since we are assuming that the plant has m inputs and p outpu ts, the contro ller hasp inputs and m outpu ts. Therefore H(s) is an m x p matrix. Next using (8.46, 8.51) we see that the right and left coprime factorizations of the contro ller are

where N y(s) b [

Ny(s )b with K and choose

i

~ + BK

[~+ic

!l !]

M y(s) b [

~ + BK ~

l

My(s)b[~+ic ~]

being chosen so that the factors are stable. One way of doing this is to

K=

C+D K

i

=

-(B+ LD)

State Models for Coprime Factorization

211

where K, L were used in the previous section to obtain state models for the left and right coprime factorization of G(s), (8.46, 8.51). Notice that this choice provides the required stability since the A matrices for the state models of the right and left coprime factors of H(s) are given as

Therefore using this choice forK and L we see that the coprime factor state models for the controller become Nn(s) -

s =

[A +BK

s

[A+ LC

Nn(s) =

K

Mn(s)

s

=

( ) .!_ Mn s-

K

[A +BK C +DK

[A + LC

(8.53)

;(B+LD)]

K

(8.54)

We show now that the Bezout identities, (8.34, 8.35), can be satisfied by making use of the foregoing factorizations of the observer based controller for G(s). Theorem 8.3 If a system has transfer function G(s) with minimal state model specified as (8.40) and coprime factorizations specified as (8.46, 8.51), then the Bezout identities, (8.34, 8.35), are satisfied by X(s)

=

Mn(s)

Y(s) = -Nn(s)

(8.55)

X(s)

=

Mn(s)

Y(s) = -Nn(s)

(8.56)

where Mn(s),Nn(s ),Mn(s), and Nn(s) are given by (8.53, 8.54) Proof We need to establish that Mn(s)M(s) - Nn(s)N(s) =I

(8.57)

We can do this by recalling the state model for systems connected in series, (8.9) and using the state models for Mn(s), M(s), N n(s), N(s) (8.54, 8.46) to obtain

[ [A +LC Mn(s)M(s) ob 0

[ K

Nn(s)N(s) ob

[[A+LC 0 ( K

-(B+LD)K ] A+BK

[-(B;LD)l]

(8.58)

K -L(C+DK )] A+BK 0

[-fl]

(8.59)

Then changing coordinates for the state representation of Nn(s)N(s) by using the

212

System Algebra

coordinate transformation matrix T T

=

[

~ ~]

yields

NH(s)N(s)cb

A+ LC 0 [ K

-(B+ LD)K] A+BK

[[

K

(8.60)

]

Finally comparing the state models given by (8.60) and by (8.58) we see that

which shows that (8.57) is satisfied. We can show that (8.56, 8.51, 8.53) satisfy the Bezout identity, (8.35) by proceeding in • a similar fashion.

8.4.3

Doubly-coprime factorization

The foregoing right and left coprime factorizations of G(s), (8.46, 8.51) constitute what is referred to as a doubly-coprime factorization. This type of factorization is defined from the following observations. First recall, in general, that any right and left coprime factorizations of G(s) and of H(s) satisfy

Therefore it follows that

M(s)N(s) - N(s)M(s) = 0

(8.61)

0

(8.62)

MH(s)NH(s)- NH(s)MH(s)

=

Next recall, from Theorem 8.3, that if we use the particular coprime factorizations for G(s) and H(s) given by (8.46, 8.51, 8.53, 8.54) then we have the following Bezout identities

MH(s)M(s)- NH(s)N(s) =I

(8.63)

M(s)M H(s) - N(s)N H(s) =I

(8.64)

Finally, notice that (8.61-8.64) can be written as the single matrix equation

[ Mf!(s) -N(s)

-i!_H(s) M(s)

l

[M(s) N(s)

N H(s)] MH(s)

=

[I0 0] I

Now we use this relation to define a doubly-coprime factorization as follows.

(8.65)

Stabilizing Controllers

213

Suppose a system with transfer function G(s) has left and right coprime factorizations

G(s)

= _M-I

(s)N(s)

= N(s)M- 1 (s) Then these factorizations taken together constitute a doubly-coprime factorization of G(s) if there exists a set of transfer functions, {MH(s),NH(s),MH(s),NH(s)} E 1ix, · which satisfies (8.65). doubly-coprime the between relation foregoing the use In the next section, we factorization of a plant and the stabilizing controller for the plant to provide a means of characterizing all controllers which stabilize a given plant, i.e., all controllers which render the feedback control system internally stable.

8.5

Stabilizing Controllers

Recall, from the discussion in Section 8.3.1, that achieving stability through unstable pole-zero cancellations is not robust. Therefore we cannot rely, in practice, on unstable pole-zero cancellations between the m;1thematical models for the plant and controller to lead to the implementation of a stable feedback control system. To avoid this pitfall we need a controller with model H(s) which when used with a plant having model G(s) makes the state model for the closed loop system internally stable. Controller models, H(s), having this property are said to stabilize the plant model, G(s), or, alternatively, are said to be stabilizing. Alternatively, the internal stability of the state model for the closed loop system is equivalent to the input-output stability from {u 1(t),u 2 (t)} as input to {y 1(t),y 2 (t)} as ·output for the interconnection shown in Figure 8.3. Satisfaction of this condition implies that 8(s) E 'H00 where (8.66) We will show that 8(s) E 7t00 is equivalent to W(s) E 7t00 where

[ E 1 (s)] = W(s)[Ut(s)] U2 (s) E 2 (s)

(8.67)

In order to establish conditions on G(s) and H(s) which insure that W(s) E 'Hoo we need to determine relations between the blocks of W(s) and G(s), H(s). Before doing this consider the following fundamental aspect of feedback control systems. Recall that transfer functions of physical processes must be bounded at infinity. Therefore to be physically meaningful G(s), H(s) and W(s) must be either proper or strictly proper. Assuming that G(s) and H(s) are proper or strictly proper, the feedback control system is said to be well-posed if W(s) is proper or strictly proper. We will see that W(s) is proper or strictly proper if and only if

[/- G(=)H(oo)r'

exists

System Algebra

214

~----e_,(0__7 ~---Y~I(_t)__~

J

+

1

u,(t)

y,(t)

Setup for Plant Stabilization Criterion

Figure 8.3

Notice that if either G(oo) or H(oo) is null, i.e., strictly proper, this condition is satisfied and the feedback control system is well-posed. Recall that H(s) is strictly proper, independent of whether G(s) is proper or strictly proper, in the case of an observer based controller, (8.52). Therefore observer based control systems are always well-posed. We proceed now to establish expressions for the partitions of W(s) in terms of G(s) and H(s).

8.5.1

Relating

W(s) to G(s), H(s)

We begin by noting from Figure 8.3 that

E 1 (s)

=

U1(s)+H(s)E2 (s)

(8.68)

E2 (s)

=

U2(s)

+ G(s)E 1(s)

(8.69)

or

U(s)

=

w- 1 (s)E(s)

where W

I -'()=[ -G(s) s

-H(s)] I

Recalling that G(s),H(s) arep x m and m xp respectively we see that w- 1 (s) and W(s) are each (m + p) x (m + p). Now the determination of W(s), (8.67), in terms of G(s), H(s) proceeds as follows. First substituting (8.69) in (8.68) and solving for E 1 (s) as well as substituting (8.68) in (8.69) and solving for E 2 (s) gives E 1(s) = [I- H(s)G(s)t 1 U1 (s) +[I- H(s)G(s)t 1H(s) U2 (s)

(8.70)

E 2(s) =[I- G(s)H(s)t 1G(s)U 1 (s) +[I- G(s)H(s)t 1 U2 (s)

(8.71)

Alternatively, substituting for E 1 (s) from (8. 70) in (8.69) gives

E2(s)

=

G(s)[I- H(s)G(s)t 1 U1 (s) + [I+ G(s)[I- H(s)G(s)t 1H(s)] U2 (s)

(8.72)

Stabilizing Controllers

215

Then comparing (8.71, 8.72) yields the following equalities

[I- G(s)H(s)t 1G(s) = G(s)[I- H(s)G(s)t 1

(8.73)

[I- G(s)H(s)r' =I+ G(s)[I- H(s)G(s)t 1H(s)

(8.74)

Alternatively, by substituting for E 2 (s) from (8.71) in (8.68) and comparing the resulting expression for E 1(s) with the expression for E 1(s) given by (8.70) leads to equations (8.73, 8.74) with G(s) and H(s) interchanged. The foregoing results, which can also be obtained using the matrix inversion lemma, (Appendix), enable W(s), (8.67), to be expressed in terms of G(s), H(s) in the following ways W(s) = [ W 1 (s) W 3 (s)

=

[

Wz(s)] W4(s)

[I- H(s)G(s)r' [I- H(s)G(s)r 1H(s) G(s)[I- H(s)G(s)r 1 I+ G(s)[I- H(s)G(s)r 1H(s)

[I+ H(s)[I- G(s)H(s)r 1G(s) [I- G(s)H(s)r'G(s)

l

H(s)[I- G(s)H(s)r']

(8.75 )

(8.76 )

[I- G(s)H(s)r 1

Thus we see from the dependency of the blocks of W(s) on [I- G(s)H(s)r 1 that W(s) is proper or strictly proper if and only if [I- G(s)H(s)r' is proper or strictly proper which requires that I- DcDn be invertible where G(oo) =De and H(oo) = Dn. Finally, notice from (8.67, 8.75, 8.76) that we can write 8(s), (8.66), as

e(s)= [W3(s) W 1(s)- I Therefore we have 8(s) E Hoc is equivalent to W(s) E Hoc.

8.5.2

A criterion for stabilizing controllers

We showed, in the previous section, that W(s) E Hoc is equivalent to the corresponding closed loop system being internally stable. In this section we give a more useful criterion for deciding if a controller is stabilizing. We will use this criterion in the next section to develop a characterization of all controllers which are capable of stabilizing a given plant. The criterion to be developed here is stated in the following theorem. Theorem 8.4 A controller having transfer function H(s) with right coprime factorization H(s) = N n(s)M!/ (s) stabilizes a plant having transfer function G(s) with left coprime factorization G(s) = M(/ (s)Nc(s) if and only if (8.77) where

216

System Algebra

Proof Recall that 6(s) E Uoc is equivalent to 6(s), 6- 1 (s) E Hx. Since 6(s) is defined as the sum and product of transfer functions in Hx, we have 6(s) E Hoc irrespective of the internal stability of the closed loop system. Therefore we need to show that 6- 1 (s) E Hx is necessary and sufficient for W(s) E H:c In order to show that 6- 1 (s) E Hx is sufficient for W(s) E Hcxc we assume 6- 1 (s) E Hoc and proceed as follows. From the given coprime factorization of the plant and controller we have --1

-1

-

G(s)H(s) =Me (s)Nc(s)N H(s)MH (s) or

Mc(s)G(s)H(s)M H(s)

=

Nc(s)N H(s)

Therefore 6(s), (8.77), can be written as

6(s)

=

Mc(s)[/- G(s)H(s)]MH(s)

This allows us to express[/- G(s)H(s)r 1as

Then we can use this expression to rewrite W(s), (8.75, 8.76) as follows

w( s )

= [/

+ N H(s)6 - 1(s)Nc(s) 1

-

M H(s)6- (s)Nc(s)

N H(s)6 - 1 (s)~c(s) MH(s)6 - 1(s)Mc(s)

l

(8. 78)

Therefore, assuming that 6- 1 (s) E H 00 we see that W(s) E H 00 since each element of W(s) consists of products and sums of transfer functions in H 00 • This shows that (8.77) is sufficient for the controller to be stabilizing. In order to show that 6 - 1 (s) E Hoc is necessary for W(s) E Hoo we assume W(s) E Hoo and proceed as follows. Since H(s) = N H(s)M}/ (s) is a coprime factorization, the following Bezout identity is satisfied

for some X H(s), Y H(s) E H 00 . Then post-multiplying this Bezout identity by 6 Mc(s) and by 6 -I (s)Nc(s) gives the following two equations X H(s)M H(s)6 -I (s)Mc(s) X H(s)M H(s)6 -I (s)Nc(s)

+ Y H(s)N H(s)6 -I (s)Mc(s) = + Y H(s)N H(s)6 -I (s)Nc(s)

where Q 1 (s) = 6- 1 (s)Mc(s) Q2 (s)

= 6- 1 (s)Nc(s)

Q 1 (s)

= Q 2 (s)

-I

(s)

217

Stabilizing Controllers

However we can rewrite these equations using W(s), (8.78), as

+ Y H(s) W2 (s) = Q 1 (s)

(8.79)

+ Y H(s)[W1 (s)- I] = Q2 (s)

(8.80)

X H(s) W4(s) X H(s) W3 (s)

Therefore, since all terms on the left sides of (8. 79, 8.80) belong to 'Hoc, we have (8.81) However the factors of an 'Hoo function need not be in 'Hoc· Therefore we are not able to use (8.81) to conclude that Ll- 1 (s) E 'H 00 • Since G(s) = A-rc; 1 (s)NG(s) is a coprime factorization, the following Bezout identity is satisfied

MG(s)XG(s)

+ NG(s) YG(s)

=I

for some XG(s), YG(s) E 'H 00 • Then pre-multiplying this equation by Ll- 1 (s) yields

and we see that Ll- 1 (s) E 'H::xo since all terms on the left side of this equation belong to 1t00 • This shows that (8. 77) is necessary for the controller to be stabilizing. • By interchanging the role of the plant and controller in the foregoing theorem, we see that a left coprime factored controller, H(s) = Mii\s)NH(s), stabilizes a right coprime factored plant, G(s) = NG(s)M(/(s), if and only if

Li(s) E U=

(8.82)

where

In summary, we see that we can use either (8.77) or (8.82) to determi;e if a given controller stabilizes a given plant. We can use this fact together with a doubly-coprime factorization of the plant to provide a parametrization of all the plant's stabilizing controllers.

8.5.3

You/a parametrization of stabilizing controllers

We have just shown that if there are coprime factorizations of the transfer functions for the plant and controller which makes Ll(s), (8.77), or b.(s), (8.82), a unit, i.e., Ll(s) E U 00 , or Ll(s) E U 00 , then the corresponding closed loop system is internally stable and the controller is said to be stabilizing. However, if the coprime factorizations of the plant and controller happen to also make both Ll(s) and b.(s) identity matrices, then not only is the controller stabilizing but, in addition, the left and right coprime factorizations of the plant constitute a doubly-coprime factorization, (Section 8.4.3), of the plant. The foregoing observation is used now to show that we can construct a set of stabilizing controllers by using a doubly-coprime factorization ofthe plant. The elements of this set are generated by varying an m x p matrix, Q(s), over 1()().

218

System Algebra

Theorem 8.5 Given a doubly-coprime factorization of a p x m plant transfer function

matrix

G(s)

=

--1

-

Ma (s)Na(s)

=

-1

Na(s)Ma (s)

(8.83)

where

(8.84)

then either of the following forms for the m x p controller transfer function matrix stabilize the given plant

H(s)

Nc(s)M-;: 1(s)

(8.85)

H(s) = M; 1(s)Nc(s)

(8.86)

=

for all m x p matrices Q(s) satisfying Q(s)

E

Hoc where

Mc(s)

=

+ Ma(s)Q(s) MH(s) + Na(s)Q(s)

Nc(s)

=

NH(s) + Q(s)Ma(s)

Nc(s) = NH(s)

Mc(s) = MH(s)

(8.87)

+ Q(s)Na(s)

Proof Recall that for (8.85) to be a coprime factorization of H(s), we must have

(i)

(ii)

Nc(s), Mc(s) E Hoc Nc(s), Mc(s) coprime

Since Nc(s), Mc(s), (8.87), are each dependent on sums and products of matrices in Hoc, condition (i) is satisfied. In order to show that condition (ii) is satisfied, recall that Nc(s),Mc(s) are coprime if X(s), Y(s) E Hoc satisfy the Bezout identity

X(s)Mc(s)

+ Y(s)Nc(s) =I

(8.88)

Now substituting for Mc(s), Nc(s) from (8.87) and choosing

X(s) = Ma(s)

Y(s)

=

-Na(s)

yields the following equation

where 81

= ifc;(s)MH(s)- Nc(s)NH(s)

82

=

[Ma(s)Nc;(s)- Nc(s)Ma(s)]Q(s)

(8.89)

Lossless Systems and Related Ideas

219

Then we see that 8 1 = I since the plant transfer function factorization is doublycoprime, (8.84). Moreover we have 8 2 = 0 from (8.83). Therefore (8.88) is satisfied and condition (ii) is satisfied. Thus (8.85) is indeed a coprime factorization of H(s). Finally, in order to show that the controllers, (8.85), are stabilizing. notice that we have just shown that

Therefore, referring to Theorem 8.4, we have ~(s) = IE Ux so that the controllers given by (8.85) are stabilizing. This completes the proof that the controllers given by. (8.85), stabilize the given plant. The proof that the controllers given by (8.86) are stabilizing can be done in a similar • fashion.

8.6

Lossless Systems and Related Ideas

A linear time-imariant system is said to be lossless if it is stable and if it has zero state response y(t) corresponding to any u( t) E .C 2 [0, oc) such that (8.90) Notice, from Sections 7.4 and 7.5 of the previous chapter, that a lossless system has L 2 gain equal to one. However it is important to note that this condition is necessary but not sufficient for a system to be lossless, i.e., for (8.90) to be satisfied for all u( t) E .C2 [0, oo ). In contrast, recall from the previous chapter that .C2 fO. oo) is isomorphic to .C 2 so that the L 2 norms in the time and frequency domains are equivalent. Therefore a system is lossless if its transfer function G(s) E HCXJ maps U(s) E H 2 to Y(s) E H 2 such that (8.91) However since

I Y(Jwll~= -2j

!X

71'

U*(Jw)G*(jw)G(jw)U(jw)dw

-x

we see that (8.91) is satisfied for all U(s) E

G*(jw)G(jw)

=

Im

7-{ 2

if and only if

for all wE (-oc, oc)

(8.92)

where G(jw) is p x m. Notice that p ;:::: m is necessary for (8.92) to be satisfied. Mathematically, we can consider lossless systems to be unitary operators. Recall, from Chapter 7, that the Fourier transform is an isometry or unitary operator between the time domain Hilbert space .C 2 (- oc, oc) and the frequency domain Hilbert space .C2 . In the present instance (8.90) implies that an m-input p-output lossless system is an isometry or unitary operator between the m-dimensional input Hilbert space, .C 2 [0. oc ), and pdimensional output Hilbert space, .C 2 [0, oc ). Alternati\d~, (8.91) implies that a lossless system is a unitary operator between them-dimensional input Hardy space, 7( 2 , and the p-dimensional output Hardy space, H 2 •

zzo An important property of lossless systems can be seen as follows. Notice that if we connect two systems with transfer function G 1 (s), G2 (s) E Hoo is series with G1 (s) being the transfer function of a Iossless system, then we have IIG1 (s) lloo= 1 and from Section 7.4.2 we see that (8.93) However (8.92) implies that the L 2 norm of the output from the composite system is independent of G1 (s). Therefore the inequality (8.93) becomes an equality

In the next chapter we will see that this type of invariance of the H 00 norm plays an important role in the development of a solution to the H 00 control problem. Now since the linear models of the time-invariant physical processes being considered here, e.g., plants, have transfer function matrices whose elements are ratios of polynomials having real coefficients, we have

G*(jw) = GT(-jw)

for all w E ( -oo, oo)

and (8.92) is equivalent to

GT ( -s)G(s) = Im

for all complex s

(8.94)'

We can gain additional insight into these matters by reviewing the earlier development of these ideas in connection with electric filters.

8.6.1

All pass filters

Recall that a stable time-invariant linear SISO system having transfer function G(s) has a steady-state output given as

y(t) = A 0 sin(wt + 8o) when its input is given by

u(t)

= A 1 sin(wt)

where

A 0 = A;yiG*(jw)G(jw)

=

IG(jwiAi

e 0 =tan _1 [Im[G(jw)]] Re[G(jw)]

This system is referred to as an all pass filter if

IG(jwl =I

for all w

E ( -oo,

oo)

(8.95)

Lossless Systems and Related Ideas

221

or if G( -s)G(s) = 1

for all complex s

(8.96)

so that the amplitude A 0 of the steady state output equals the amplitude A; of the input for all frequencies. Thus comparing (8.95, 8.96) with (8.92. 8.94) we see that all pass filters are lossless systems. Moreover if G(s) is the transfer function of an all pass filter, then we see from (8.96) that G(s) is proper, with G(O) = 1, and has poles which are mirror images of its zeros across the imaginary axis, i.e., ifP; is a pole of G(s) then -p; must be a zero of G(s ). For example, the first order system having transfer function

(s- 1) G(s) = (s+ 1) is an all pass filter An all pass transfer function is a generalization of the concept of an all pass filter. This generalization is done by relaxing the requirement of stability. Thus G(s) is said to be an all pass transfer function if (8.96) is satisfied. Notice that the mirror imaging of poles and zeros implied by (8.96) means that when G(s) is an all pass transfer function it has no imaginary axis poles, i.e., G(s) E £ 00

8.6.2

Inner transfer functions and adjoint systems

In order to distinguish between stable and unstable all pass transfer functions, stable all pass transfer functions are referred to as being inner. Thus a transfer function is inner when it is the model for a lossless system. In the case of a transfer function matrix the following generalization is made. If G(s) E 7t00 is p x m, then G(s) is said to be inner if (8.97) is satisfied or co-inner if (8.98) is satisfied where (8.97) G~ (s)G(s) = Im if p 2m G(s)G~ (s) = IP

if p::::; m

(8.98)

for all complex s with

Notice that p 2 m (p ::::; m) is necessary for G(s) to be inner (co-inner). Also notice that if the transfer function is square and inner then G- 1 (s) = G ~ (s) E 7t~. Moreover, since dimensionally square systems have the property that the poles of G- 1 (s) are the zeros of G(s). (Section 8.2.4), we see that in this case that all system zeros of G(s) lie "in" the open right half plane, i.e., lie in the region of the complex plane where G(s) is analytic. This is the origin of the terminology "inner". However in the nonsquare case this concentration of zeros in the open right half plane is not necessary for a transfer function to be inner, e.g ..

G(s)

·.1

=--

s+Vi

[s+l] 1

is inner without having any zeros in the right half plane.

The system whose transfer function is G- (s) is referred to as the adjoint sys!em (relative to a system having transfer function G(s)). Notice that the state model for G (s) is related to the state model for G(s) as

G(s) =s

[AC DB]

(8.99)

since

Properties which the parameters of a minimal state model must satisfy in order for a system to have an inner transfer function are given in the following theorem. Theorem 8.6 G(s) E Hoc is inner if it has a minimal state model

G(s) =s

[AC DB]

with parameters that satisfy the following conditions:

DTD=l BTWO +DTC = 0 AT Wa

+ WoA + cT c

=

0

Proof From the state model for G- (s), (8.99), and the rule for composing two systems, (8.9), we obtain

Next transforming coordinates using the coordinate transformation matrix

gives

- Ar G-(s)G(s) de [ [ 0 [ C, where A2 = -A 7 X- XA- CT C and

AA2 ] C2]

[BB21 ]

1

DTD

Jl ·f.

-~

Summary

223

Finally, applying the conditions given in the theorem yields

X= W

0

A2 = 0

Bl =0

C2=0

DTD=l

and the state model for G (s)G(s) becomes

~]

C G- (s)G(s) =s [A where

A=

r-~T ~]

C=[

c1

0]

B= jj

[!]

=I

Thus we have G- (s)G(s)

= C(sl- A)- 1B + jj =I

•

and the theorem is shown. In developing a solution to the Hoc state feedback control problem in the next chapter, we will extend the foregoing characterization of state models so that the resulting system is J-lossless with J-inner transfer function.

8.7

Summary

In this chapter we have established state models for certain interconnections and operations with systems. This included the ideas of system inversion, system zeros, system coprimeness and lossless systems. We will see in the next two chapters that the operation of coprime factorization coupled with the requirements on a system's state model for that system to be lossless, plays an important role in developing a solution to the H 00 feedback control problem.

8.8

Notes and References

The treatise by Vidyasagar [44], on the consequent ramifications of the coprime factorization approach to feedback control places coprime factorization in the mathematical setting of algebra known as ring theory. The term "unit" refers to elements in a ring that have an inverse in the ring. The development of state models for the coprime factors used here is treated in [14].

9 H 00 State Feedb ack and Estimation

9.1

Introduction

In the previous two chapters we developed ideas for characterizing signals and systems. In the next two chapters we use these ideas to develop the design equations for a controller which is required to (i) stabilize the closed-loop control system and (ii) constrain the L 00 norm of the closed-loop transfer function from disturbance input, u 1(t), to desired output, y 1 (t), to be no greater than/, a specified positive scalar. Figure 9.1 gives the setup for this problem. Before going on to consider this setup in more detaiL we make the following more general observations on this control problem. First, we should recall that to avoid having a system's input-output stability dependent on the physically unrealistic requirement of pole-zero cancellation, we have used the term "stability" to refer to a system's internal stability. Thus when requirement (i) is satisfied so that the closed-loop system is internally stable, the L 00 norm mentioned in requirement (ii) is an H 00 norm. Therefore controllers satisfying (i, ii) are said to solve the H 00 control problem. AlternatiYely. recall from Chapter 7, that the H 00 system norm is induced by the H 2 signal norm in the frequency domain or the L 2 signal norm in the time domain. Therefore when requirement (i) is satisfied, (ii) can be interpreted as the requirement that the closedloop control system's L 2 gain from u 1(t) to y 1 (t) be no greater than f. Thus the H 00 control problem is sometimes referred to as the L 2 gain control problem. In addition, notice that satisfaction of requirement (ii) does not imply that requirement (i) is satisfied since the L 00 norm is defined for both stable and unstable systems provided only that the system's transfer function have no imaginary axis poles. This differs with both the quadratic and LQG control problems since the performance indices, lQc and lGc, are infinite when the closed-loop system is unstable. Concerning the setup for the Hx control problem, it has become customary to replace the actual plant in the feedback control configuration, Figure 9.1, by a composite system referred to as a generalized plant. The generalized plant consists of the plant to be controlled plus any other systems which the control system designer may want to connect

226

H 00 State Feedba ck and Estimat ion u 1(t)

u,(t)

y,(t) GENERALIZED PLANT

CON1ROLL ER

y,(t)

~

Setup for Hex, Control Problem disturbance input y 1 (t) = desired output controlled input y 2 (t) = measure d output Figure 9.1

u 1 (t)

u2 (t)

where u 1 ( t), u2 ( t), y 1 ( t), y 2 (t) have dimensions m 1 • m 2 . p 1 . p 2 respectively. in cascade with the plant (referred to as weights or filters). For instance. the filters can be used to take into account any known spectral characteristics of the disturba nce signals, e.g., narrow band noise. Then when the feedback control system is impleme nted, the controller determi ned by applying the control algorith m to the generalized plant is used with the actual plant. In what follows the signals {u;(t). Y;(t) : i = 1, 2} shown in Figure 9.1 are assumed to be the various inputs and outputs associated with the generalized plant which has realrational transfer function G(s) with given real-par ameter state model

D~~] 1

B

D22

l

(9.1)

As in the LQG and quadrat ic control problems, a solution to the H output feedback 00 control problem takes the form of an observer-based controll er and involves the stabilizing solutions to two algebraic Riccati equations. Thus in order to develop a solution to the Hoc output feedback control problem we need first to consider two simpler problems namely, the Rx state feedback control problem, and the Hoc state estimati on problem. These basic Hx problems are taken up in this chapter with the R)C output feedback control problem being addressed in the next chapter. As we progress through the development of solutions to these problems. we will see that the generalized plant's D matrix plays a pivotal role. However, only certain parts of the D matrix play a role in each problem . For instance, the existence of a solution to the Hx state feedback control problem requires D 12 to have indepen dent columns and only the first p 1 rows of D, denoted by D 1, are involved in the solution to this problem where

However, the existence of a solution to the H.cx. state estimati on problem requires Del to have indepen dent rows and only the first m 1 columns of D, denoted by D • 2 are involved

Hoo State Feedback Control Problem

227

in the solution to this problem where

, are involved in In the next chapte r we will see that the last p 2 rows of D, denote d by D 3 where problem the solution to the H 00 output feedback control

9.2

H00 State Feedback Control Problem

want to choose the Assuming that the state of the generalized plant, x(t), is known we state feedback matrix, K 2 , so that setting the controlled input as (9.2)

n from u 1(t) to makes the closed-loop system internally stable and the transfer functio 'Y· t constan positive d specifie a y 1(t) have H 00 norm no greater than the plant as the Since the measured output, Y2(t), is not needed in this problem, we take of G 1(s) are ters parame the assume we Thus . missing generalized plant, (9.1), with Y2(t) known where

(9.3) with

where

k control problem The relation between the signals and systems in the H 00 state feedbac going to be used are x(t) K and (t) 1 are shown in Figure 9.2. where the additio nal signals u1 . problem control k feedbac in the development of a solution, K 2 , to the Hx state Now after implementing state feedback we have (9.4)

where

and we want to find K 2 such that and

(ii): IITc(s)lloo:C:: 'Y

(9.5)

228

H 00 State Feedbac k and Estimatio n ii, (t)

~

u,(t)

y,(t)

PLANT

u,(t)

K,x(t) STATE CONTROLLE R

K1 x(t)

Figure 9.2

~J

Setup for the Hx State Feedback Control Problem

Next recall from Chapter 7 that the H 00 norm is defined by

1/Tc(s)//CX)=

O"max[Tc(jw)]

sup w;o(-oo,x)

where O"rnaxfTc(jw)] is the largest singular value of Tc(jw). Now since lim Tc(jw)

w-.oo

=

D11

we see that it is not possible to satisfy condition (ii), (9.5), if where

In subseque nt sections we will see that the more restrictive condition (9.6)

must be satisfied in order to solve the H 00 state feedback control problem. In addition, we will see that this condition is also needed to solve the H 00 state estimatio n problem. We can summari ze the strategy we will employ now to develop the design equation s for K2 as follows. l. Start by assuming stabilizing feedback, i.e., assume K 2 achieves condition (i), (9.5). 2. Define a scalar quadratic performa nce index, P.1 , in terms of y 1 (t) and u (t) such that 1 P 1 :S: 0 implies condition (ii), (9.5), is satisfied. 3. Use G1 (s), (9.3), to introduce a transfer function G1(s) relating { u ( t), y (()}as output 1 1 to { u1 ( t), u2 ( t)} as input under the constrain t of stabilizing sta.te feedback. i.e., u2(t) = K2x(t) such that G1(s) E H 00 . This enables P1 to be rewritten as a quadratic in u1 (r) and u2 (t) with U 1(s) E H 2 and U2 (s) restricted to stabilize G1 (s). 4. Convert P1 to a quadratic in { V1(s). V 2 (s)} E H 2 where { V (s). V (s)} arise as 1 a 2 consequence of using J-inner coprime factoriza tion on G1 (s). We will see that the design equation s for K 2 result from carrying out the last step.

Hoo State

9.2.1

Feedback Control Problem

229

Introduction of P1

see from Chapte r 7 Suppose K 2 is chosen so that conditi on (i), (9.5), is satisfied. Then we that if

YI(t) E .Cz(O,oo)

and we can interpr et conditi on (ii), (9.5), in the time domain as a constra

int on the L 2 gain (9.7)

Next define P1 as

(9.8) Then conditi on given in (9.7) is equival ent to (9.9) Now since £ 2 [0, oo) is an inner produc t space we can rewrite P1 as

(9.10)

where

, we see that P 1 , However, recalling from Chapte r 7 that £ 2 [0, oo) is isomor phic to 1t2 as (9.10), can be rewritte n in the frequency domain

P,

= j [ U1 (s)] ,J"'mp [ U1(s)] ) \

=

Y1 (s)

·

Y1 (s)

-1' 2 IIUI(s)ii~+IIYI(s)ii~

(9.11)

and conditi on (ii), (9.5), is satisfied if (9.12)

s stabilizing state It is import ant to remem ber that the foregoing result assume feedback.

9.2.2

Introduction of G1 ( s)

that we can relate Recalling from (9.3) that Y 1 (s) depend s on U 1 (s) and U 2 (s), we see

z;ro

H00 "Blll'e~'i!CIBiil:Tsit'a"'tlf/ihi;\ton · ·.

{ Ur (s), Yr (s)} to { U 1 (s), U2 (s)} throu gh the use of a transf er functi on Gr (s) as

(9.13) where

Now under the constr aint that the state feedback is stabili zing we have for

(i): U1 (s) E Hz (ii): U2 (s) = K2 X(s)

where A.[A + B 2 K 2 ] are all in the open left half plane, and X(s) is the Lapla ce transf orm of the closed-loop system state. There fore we can use (9.13) to rewrite the inner produ ct defining P1 , (9.11 ), in terms of {U1(s), U2 (s)} as

(9.14) where U 2 (s)

= K 2 X(s)

In what follows, it is impor tant to remem ber that even thoug h G1 (s) may possibly be unstable, the restric tion that the contro lled input, U (s) = K2 X(s), must be stabilizing 2 maint ains P1 , given as either (9.8) or (9.14), finite.

9.2.3

Introduction of J-in ner copr ime facto rizat ion Recall, from Chapt er 8, that G1 (s) E Lx has a right coprim e factor izatio n given by -

-

--1

G1 (s) = N(s)M

(s)

(9.15)

where N(s), M(s) E H~ with H(s), i\f(s) being coprime. Now we are going to make use of a special type of right coprime factor izatio n in which lUs) in (9.15) satisfies (9.16) j

. (li1f'

=

[ (l)-~,2 I

/,,,

A system having transf er function N(s) E fix which satisfi es (9.16) is said to be J-inne r and a right coprim e factorization (9 .15) which has H(s) J-inne r is referred to here as a J-

H 00 State Feedback Control Problem

231

inner coprime factorization . The utility of using this type of coprime factorization here can be seen immediately as follows. Let G\ (s) have a J-inner coprime factorization as denoted by (9.15, 9.16). Then using the fact that N(s) is J-inner we see that P'Y' (9.14), can be rewritten as

where

U2 (s)

= K2X(s)

Next recall, from Chapter 8, that since M(s), N(s) E 7t00 we can obtain { U 1(s), U2(s)} and {U1 (s), Y1 (s)} from V1 (s), V2 (s) E 'H2 as (9.18)

(9.19) Then since N(s) E 1-loo we have

(9.20) Finally, by using (9.18) in (9.17) we can rewrite P7 as

p'Y = ( [ =

~:~;~

l

,J'Ymm [

~:~;~

l)

(9.21)

--/I IV, (s) II~+IIV2(s) II~

Notice that unlike U2 (s), which is restricted to a subspace of7t 2 since it must provide stabilizing state feedback, V1 (s), V2 (s) are not restricted to any subspace of7t 2 . We will use this freedom in the choice of the V,.s to ensure that P1 :::; 0.

9.2.4

Consequences of J-inner coprime factorization

Suppose we have a J-inner coprime factorized for G1 (s), (9.15, 9.16). Then to proceed we need to consider state models for these coprime factors. Therefore, recalling (8.46) and Section 8.3.3 we see that state models for the coprime factors, M(s), N(s), (9.15), can be determined from the state model for G1 (s), (9.13), as M(s)

I'

~

[A+BK

Bfl-']

K

fi.- 1

(9.22)

(9.23)

232

~.,,.~,-'~""""'""'"''.,...."'~-"''~~"'~~~"~-,.,.~

H 00 State Feedbac k and Estimatio n

where K, 1:1 are chosen so that M(s), N(s) E H_ 00 with N(s) satisfying (9.16); 1:1 is a constant nonsingu lar (m 1 + m 2 ) x (m 1 + m 2 ) matrix which along with K is partition ed so as to conform with the partition ing of Band 15 1 , viz.,

We show now that the Hoc state feedback control problem is solved when K and 6. are chosen so that the foregoing coprime factoriza tion is J-inner. The equation s needed to determin e K and Do are develope d in subseque nt sections. We begin by using (9.18) to express V1(s), V 2 (s) in terms of U 1 (s), [h(s) as (9.24)

Then using the state model for the inverse system develope d in Chapter 8, we see that the state model for the inverse of M(s), (9.22), is

~]

(9.25)

Notice that the system and input matrices for the generaliz ed plant, G (s), (9.3), and 1 for M- 1 (s), (9.24), are the same. Therefor e the states of these systems are equal for all time if their initial states are equal. The initial state of each system is the same since each is null. This is because we are basing our developm ent of the Hx state feedback controlle r on transfer functions which relate the system output to the system input under the assumpti on the initial state is null. The foregoing observati on together with the state model for M- 1 (s). (9.25). allows us to write the V;s as

(9.26) where X(s) is the Laplace transform of the state of the state model for G (s), (9.3). 1 Now suppose we choose U1(s) and U2 (s) as (9.27) where D1 (s) E 7-{ 2 is an external disturban ce input whose effect will be seen to be equivalen t to the actual disturban ce input U1 (s) E H 2 . Then substituti ng (9.27) in (9.26) yields (9.28)

We show now that P,! <:; 0 for U 1 (s)

E 7-{ 2 ,

(9.12). This is done by first showing that (9.28)

implies that P'Y :S 0 for 0 1 (s) U1 (s) E 1-lz. Recall that

E

for 1-£2 and then showing that this implies P7 :S 0

I

{oo

II Vi(s)lli= 27r 1-oo V7(Jw )V;(jw )dw so that using (9.28) in P'Y, (9.21) can be writte n as (9.29)

Therefore we see from (9.29) that if

[-1' 2 ~i~l +~3~3] < 0

then

P'Y < 0

for ~11 non-n ull 0 1 (s) E 1-£2

p'Y = 0

for

ul (s) =

} (9.30)

(/)

that the H 00 state feedback contr ol Howe ver recall, from the beginning of this section, will show in Secti on 9.3.3 that probl em has a soluti on only if Du satisfies (9.6). We

}

if O"max[DIJ] < '/' then

-1' 2 ~t~~

(9.31)

+ ~3~3 < o

There fore (9.31, 9.30) imply that if O"max[Dn] < '/' then P7 < 0 for a~l non- null

p'Y = 0 for

ul (s) =

}

0 1(s)

E 1-{2

(9.32)

(/)

see this notice from (9.27) that We need next to show that (9.32) implies (9.12). To (9.33)

ack, u2 (t) = K 2 x(t) stabilizes the Now we will show in Section 9.3.3 that the state feedb for any K = (t) u 1x(t). Thus X(s) E 1-{ 2 generalized plant by itself witho ut requi ring 1 E H2 (s) 0 that 1 (9.33) from see we fore input Ut (s) E 1-{2 provi ded u2 (t) = K 2 x(t). There i.e., ied, satisf is (9.12) that s implie for all U1 (s) E 1-{ 2 . This fact toget her with (9.32) and U1 (s) -1- K 1X(s) }

for

U 1(s)

for

U1 (s) = K 1X(s)

E 1-{2

(9.34)

the soluti on of the H"" state feedback as is requi red to satisfy requi remen t (ii), (9.5), for contr ol probl em.

Notic e that the distur bance input U (s) = K X(s) maximizes P.,. There fore U1 (s) 1 1 K1X( s) is referr ed to as the worst distur bance input for the Hoc state feedb

=

ack contr ol system. In summ ary, we have show n that a soluti on to the H= state feedback probl em is obtai ned when (9.6) is satisfied and when we choos e K and L1. so that the right copri me factor izatio n of G1 (s), (9.13), has nume rator N(s), (9.23), which is J-inner. The equat ions forK which make N(s) J-inn er are the design equat ions for a soluti on to the H 00 state feedback probl em. These equat ions are developed in the next section.

9.3

H oo State Feedback Controller

In order to obtai n a J-inn er coprime facto rizati on of G1 (s), (9.13), we need condi tions on the state mode l param eters of a given system which ensure that the system is J-inner. These condi tions are given in Theo rem 9.1. A simila r set of condi tions was given in the previous chapt er in conne ction with inner system s, (Theo rem 8.6). Once we have Theo rem 9.1 we can apply it to the state mode l param eters for S(s), (9.23). In this way we determ ine an equat ion relating 6. to D and an expression for K involving the 1 stabilizing~solution Xoc to an algeb raic Riccati equat ion.

9.3.1

Des ign equ atio ns for K

Notic efrom (9.l3 ,9.l5) that.! V(s)i san(m +pi) x (m 1 +m 2 )matr ix.No wfor lV(s) tobe 1 J-inner, the condi tions given in the following theor em must be satisfied. Theorem 9.1 A system having an (m + pJ) x(m 1 1 + m 2 ) transf er funct ion N(s) is Jinner if N(s) E Hx and the param eters of a minim al state mode l for the system

satisfy

(i) '

where

J·,mp' l.ymm

J ")lnYfl

(ii)

0

(iii)

0

are defined as

Proof The proof is similar to the proof of Theo rem 8. 7. Here we need to obtai n a state mode l for NT ( -s)J., 111pN(s) and chang e coord inates with coord inate transf orma tion matri x T given by

T=lII x1 ] LO

H 00 State Feedback Controller

235

The resulting state model then reveals that in order for NT ( -s)J-ympN(s) = J-ymm we need • to impose the conditions in the theorem. Jis N(s) that so (9.23), N(s), for Now we want to choose~ and Kin the state model 9.1. Theorem applying and inner. We do this by equating N(s) to N(s) To begin, recall that the D matrix in a state model is unchanged by a change of coordinate s. Therefore from N(s) = N(s) we have (9.35)

Then using condition (i) in Theorem 9.1, we obtain (9.36)

where

Notice that D Jc is square. We will show later in this section that D Jc is invertible if D 12 has independe nt columns and D 11 satisfies (9.6). This inverse is needed to determine K. Continuing with the development, we see from Theorem 9.1 and (9.23) that further consequences of setting N(s) = N(s) are AN= A+BK BN

= B~-1

eN=

(9.37) (9.38)

c, + i5 K 1

(9.39)

Then using (9.35, 9.38, 9.39) in condition (ii) ofTheorem 9.1 we obtain

which after multiplying on the left by ~ T becomes (9.40)

Now we see from the definitions of C1 and

15 1, (9.13). that (9.41) (9.42)

Thus using (9.41, 9.36) and assuming D 1, is invertible enables us to solve (9.40) forK as (9.43)

236

H= State Feedback and Estimation

Notice that K depends on the state model parameters for the given generalized plant, (9.3), and on X. We will show now that X is a solution to an algebraic Riccati equation. To begin, we substitute from (9.37, 9.39) in condition (iii) of Theorem 9.1 and use (9.41, 9.42) to obtain ATX +XA +KTBTX +XBK +KTDfe!

+ eTD!K +KTDJcK +ere!= 0

(9.44)

Next grouping terms involving K and KT and using (9.43) to substitute -DJcK for BT X+ Di e 1 enables (9 .44) to be rewritten as

(9 .45) Finally substituting forK from (9.43) gives the equation

(9 .46) where Ax! =A - BD]c1Df el Roc!= BD]/BT

Qool

=

ef(I- DJD]/DT)e l

with

Notice that (9.46) is an algebraic Riccati equation like the GCARE (Section 6.2). We refer here to (9.46) as the HCARE. Since A001 - R 001 X =A+ BK we have K as a stabilizing state feedback matrix if the solution X to (9.46) makes A 001 - Roc 1X stable. As in the quadratic and LQG state feedback control problems, we reserve a special notation for these solutions which in this case is X oc· Notice that X 00 ?': 0. In addition, notice that unlike Df2 D 12 , D Jc is not sign definite, i.e., neither nonnegative nor non-positive. Thus unlike R 1 in the GCARR (section 6.2.2), K" 1 in the HCARE is not sign definite. This makes the determinatio n of conditions on the state model for the generalized plant which ensure the existence of a stabilizing solution to the HCARE more involved than in the case of the QCARE.

9.3.2

On the stability of A+ B2 K2

We have just developed equations for the determinatio n of K to effect a J-inner coprime factorization of G1(s), a system which is made up from the given generalized plant. Thus A + BK is a stability matrix. However the H:xJ state feedback controller, (9.2), affects only u 2 (t) through the state feedback equation u 2 (t) = K 2 x(t). In addition, recall from Section 9.2.4, that condition (ii), (9.5), is satisfied provided K 2 is the last m 2 rows of the matrix K

' .•.I

I JI .

I

of G1(s). There fore it remains to which is neede d in the J-inne r coprim e factor izatio n is determ ined in the foregoing K when ed establish that condi tion (i), (9.5), is satisfi manne r. has a stabilizing soluti on X 00 , More specifically, assuming that the HCAR E, (10.96), to show that A+ B 2 K 2 has all its and that K, (9.43), is determ ined using X= X 0 , , we need m rows of K. This is done in the eigenvalues in the open left half plane when K 2 is the last 2 proof of the following theorem. 0 is a stabilizing soluti on to Theorem 9.2 Ifthe pair (A, B 2 ) is stabilizable and X 00 2: half plane where left open the HCAR E, (10.96), then >.[A+ B 2K 2] are all in the

with K given by (9.43). A Proof Suppo se X 00 is a stabilizing soluti on to (9.46). Then 001 where

-

R 001 X 00 is stable

(9.47) with 12 = A + B2K2. B ) is necessary for 1 2 to be Notice, from Chapt er 2, that the stabilizabilty of (A, 2 is detect able if we can find L C) (A, stable. In additi on, recall from Chapt er 3, that a pair is detect able all unstab le C) (A, such that A+ LC is stable. Alternatively, when then Cv -I 0. There fore >.v, = Av eigenvalues of A are observable, i.e., if Re[>.] 2: 0 when Thus if K 1v = 0 where able. detect since A2 +B1Kh (9.47), is stable, we have (A 2 ,Kt) . proof the in 1 2v = >.v then Re[.>.] < 0. We will use this fact later to and from (9 .45) to obtain Next we add and subtra ct both X 00 B 2 K 2 and its transp ose

(9.48) where

8 = XooB2K2

+ K[ BfXoo + KrD, cK- CfCt

that this fact plus the Now we are going to show that e ~ 0. Then we will see (A2, K 1) is detect able implies that 1 2 is stable. We begin by expan ding KT D1cK

KTD .K = [KT1 Jc

KT] [Mt 2

fact that

M2] [Kt] K2

M[ M4

= K{ MtKt + K{ M2K2 + K[ Mf Kt + K[ M4K2 where we see from (9 .46) that the M;s are given as

Then 8, (9.48), becomes

8

2- Cf C1 = (XocB2 + K{ M2)K2 + K{ (Bf Xoc + M[ Kt) + K{ MtKt + K{ M4K

(9.49)

238

H= State Feedback and Estimation

Next after rearranging (9.43) as

we see that (9.50) Therefore using (9.50) in 8, (9.49), gives

8

+ Df2C1)T K2- K!(M4K2 + Df2C1)

=

-(M4K2

=

-KJ M4K2- C{ D12K2- K'{ D{2C1

-t-

+ K[ M1K1

K[ivf1K1

-

+ K'{M4K2- C{ C1

cf C1 (9.51)

Finally, r-ecall from (9.6) that we require all the eigenvalues of D{1D 11 to be less that 1 2 since this is a necessary condition for the solution of the H 00 state feedback control problem. We can readily show that under this condition we have A1 1 < 0. Thus vve see from (9.51) that 8 :S 0. We can now proceed to show that A2 is stable. Pre and post multiplying (9.48) by,,* and t' where A2 v = .Xv gives 2Re[.X]v*X00 v = v*8v

(9.52)

Now since we have just shown that 8 :S 0 we have either v*8v < 0 or v*8v = 0. If v*8v < 0 it follows from the condition X 00 2: 0 that v* Xocv > 0 and therefore Re[.X] < 0 If v* 8v

=

0 we see from (9 .51) that (9.53)

and since M 1 < 0 we see that (9.53) requires and

(9.54)

Finally, recalling from the beginning of the proof that the pair (A 2 , K 1) is detectable, we see that if K 1v = 0 then Re[.X] < 0. • To recap, we have shown that the state feedback matrix K 2 which solves the H 00 state feedback control problem can be determined as the last m 2 rows of the matrix K, (9.43), when X= Xx the stabilizing solution to the HCARE, (9.46). A remarkable feature of this solution, shown in Theorem 9.2, is that while X)C makes A+ BK stable we also have A + BK2 stable. Before discussing conditions on the generalized plant which ensure the existence of a stabilizing solution to the HCARE. we pose and solve the H x state estimation problem.

. !1·.

9.3.3

Dete rmin ation of ~

Recall that in Section 9.2.4 we made use of the condit ion

We will shmv now that this condit ion is satisfied when O"max[Dil]
and

Df2 D 12 is invertible

of D 11 , D 12 so that (9.36) is We do this by first developing an expression for~ in terms satisfied. This is done in the following theore m. Theorem 9.3 If and then 6 can be taken as

where

with (9.55)

where ],1mm is giYen in (9 .16) and

giYen by Proof Consid er the consta nt symmetric matrix factori zation

~]

(9.56)

from (9.56) that Then identifying the blocks of D Jc with the Mis, (9.55), we see (9.57)

where

r' e are defined in the theorem.

240

H 00 State Feedback and Estimation

Next assume for the time being that r, 8 can be factored in the manner specified in the theorem. Then we can write

where Jymm was given in (9.16) and

Finally refering to (9.57) we have

so that D Jc can be factored as (9.55) provided 8, r have square root factors. Recall fmm the proof of Theorem 6.1 that a matrix has a square root factor provided it is symmetric and non-negative. Now 8 and r are seen to be symmetric by inspection. However unlike 8 it is not evident that r is non-negative. To show that r is non-negative notice that it can be expressed as (9.58)

where De was used in Section 6.5.2

Next recall from Chapter 7 that the singular value vectors vi, u; for D 11 are related as

a;

and corresponding singular

where

Therefore noting that T

i

Dllu = a;v

i

we see from (9.58) that (9.59)

However we saw, following the proof of Theorem 6.1, that De is a contraction so that (9.60)

Hoo State Feedback Controller

241

Therefore from (9.59, 9.60) we have v

2

iTr ;

2

v ?."!-a;

and

for all eigenvectors, vi : i = 1, 2 · · · p 1 of Df1D 11 provided the singular values of D 11 satisfy i

= 1, 2 · · ·p 1

Finally, since Df1D 11 is symmetric, its eigenvectors, v\ are complete and it follows that if

then

for any p 1 dimension al vector v. Therefore r ?. 0 when the necessary condition (9.6) is • satisfied, and we can factor r in the manner stated in the theorem. from see we Thus (9.31). prove to used be now can The foregoing expression for 6 Theorem 9.3 that

and (9 .31) follows. Finally, recall that K, (9.43), requires D 1c to be invertible. We can show that D 1c is indeed invertible, by using the factorizati on of D 1 c given in Theorem 9.3 to obtain an expression D]c1 as -! = 6 D Jc

-lj-1

A-T

rmmL.l.

(9 .61)

Then we see, from the lower triangular structure of 6, (9.55), that .11-1 =

(9.62)

where r. e are given in (9.55). In the next chapter we will use the form for 6 given in Theorem 9.3 to develop the solution to the Hx output feedback control problem. Before we can do this we need to develop a solution to the H 00 state estimation problem.

242

9.4

H 00 State Feedback and Estimation

H 00 State Estimation Problem

Recall from the introduction to this chapter that our ultimate goal is the determination of an Rx controller which uses the measured output to determine the controlled input to the generalized plant so that the resulting closed loop system is internally stable and has transfer function from the disturbance input to the desired output, T0 (s), with H 00 norm no greater than r. In this section we treat the problem of determining an estimator or observer for the generalized plant state, x(t), from knowledge of the measured output, y2 (t), and controlled input, u2 (t). We need to do this so that we can use this state estimator in conjunction with the RXJ state feedback controller determined in the previous two sections to obtain the desired Hoc controller. Notice that since the state of the generalized plant depends on the disturbance input uJ(t), \vhich is unknown. it is not possible to design an observer to give an asymptotic estimate of the generalized plant state. Instead we design the observer so that the L2 gain from the disturbance input to desired output estimation error, _v 1(t), is no greater than{, i.e., so that (9.63)

where (9.64)

and .h (t) is the estimate of the desired output, (9.65)

Alternatively, we must determine the observer so that the transfer function Te(s) from u 1 (t) to ji 1 (t) satisfies

and

(9.66)

As in the quadratic and LQG state estimation problems where the solution was obtained by dualizing the solution to the corresponding state feedback control problem, we will see now that the solution to the Hx state estimation problem can be obtained by dualizing the solution we obtained in previous sections for the Hoc state feedback control problem.

9.4.1

Determination ofT e ( s)

Recalling the development of observers given in Chapter 3. we see that an observer which estimates the state of the generalized plant (9 .1) using u21r) and .1· 2 (r) is constructed as (9.67)

where

243

H 00 State Estim ation Prob lem

d plant , (9.1), we see that However, from the state mode l for the generalize (9.68) alized plant state estim ation error , Thus using (9.67, 9.68) we see that the gener l equa tion i(t) = x(t) - x(t). is the solution to the differentia (9.69) the desired outp ut estim ation error , In addit ion, we see from (9.1, 9.64, 9.65) that distu rbanc e input as y1 (t) = )' 1 (t)- y 1 (t), depends on the state estimation error and (9.70) fer funct ion Te(s) from u1(t) to There fore we see from (9.69, 9.70) that the trans has state mode l given as (B! -+- L2D21)] -Dll

y1 (t)

(9.71)

from this state mode l for Te(s) that Recalling the discussion following (9.5), we see condition (ii). (9.66), cann ot be satisfied if

h we requi red to solve the H 00 state Thus the same necessary cond ition, (9.6), whic solve the Hx state estim ation probl em feedback contr ol probl em. is needed now to namely

(9.72) for the deter mina tion of L 2 so that We will see now that the design equa tions ned by using the fact that the prese nt conditions (i, ii), (9.66). are satisfied can be obtai the Hx state feedb ack probl em solved H')(; state estim ation probl em is the dual of previously in this chap ter.

9.4.2

Dua lity

Te(s) is defined as Recall, from Chap ter 7, that the HOG norm of

))Te(s))lx=

sup

amaxUw)

c<:E(-X .e
Teljw). where a 111 a,(jw ) is the largest singular value of the positive squa re root of the largest is (jw) More specifically, recall that a max (ju.-·) Te(j;.,.,·) or Te(I.;.;) T,~ (jw). Ther efore eigem alue of either of the Herm itian matri ces r; largest singu lar value of r;(Jw ). As the largest singu lar value of Te(jw) equa ls the spon ding to a system having real-r ation al discussed in Chap ter 8, the adjoi nt system corre

244

H= State Feedback and Estimation

transfer function Te(s) has transfer function T[ ( -s) and satisfies

r;(s)

ls=jw=

T[(-s)

ls=jw

Therefore condition (ii), (9.66), is equivalent to the condition

IIT[(-s)llx ~I

(9.73)

where we see from (9.71) that (9.74) Notice that condition (i), (9.66), is satisfied if and only if T[ (-s) is antistable since Te(s) E 1i00 (s) is satisfied if and only if T[ ( -s) E 1i~(s). Thus we see from the foregoing that the H 00 state estimation problem and the H 00 state feedback con.trol problem are dual problems in the sense that a solution to the H 00 state estimation problem requires that

(i): T[(-s)

E

1i~

and

(ii): IIT[(-s)iloo~ I

(9.75)

whereas a solution to the H 00 state feedback control problem requires that (9.76)

9.4.3

Design Equations for L2

In order to exploit the foregoing dual nature of the H 00 state feedback control problem and H 00 state estimation problem, (9.75, 9.76), we need to establish a symbol correspondence between these problems. This can be done by comparing the state models for Tc(s) and TT ( -s), (9.4, 9.74). This gives the following correspondences between their parameters. Recall that we determine K 2 in the H"" state feedback problem by first solving the HCARE, (10.96), for its stabilizing solution, X 00 • Then K 2 is the lastm 2 rows of the matrix K which we determine by setting X= X 00 in (9.43). Therefore, proceeding in an analogous manner we determine L 2 as the last p 2 columns of the matrix L obtained by using the stabilizing solution Y oo to an algebraic Riccati equation. More specifically, the design equations for determining L are obtained by replacing parameters in (9.46) and (9.43) using the parameter equivalences in Table 9.1. Thus doing this and also replacing -X with Y gives (9.77)

and (9.78)

Table 9.1

State mode l Para mete r Corr espo nden ces

Control

Estimation

A

-AT

-c!

Be

LI

K2

-ci

BI

-Bf

cl

-D~

Dl2

-Df1

D11

where A 00 2

=A - B1D f DJ,} C

Roo2

=

CT D}ol C

Qoo2

=

B1

[I- D[ D]01D2JBf

with

) equa tion like the GFA RE, (Section 6.3.2 Noti ce that (9.77) is an algebraic Riccati es mak h whic Y tion HF ARE . The solu which is refer red to here as the and as the stabilizing solu tion to the HFA RE to red A 2 - YR 2 =A + LC stabl e is refer from ed rmin dete is K n as A + B 2 K 2 is stable whe is deno ted Y oo and satisfies Y 00 2 0. Just g Y oo le when Lis dete rmin ed from (9.78) usin stab C (9.43) using XX) so too is A+ L 2 2 \Vhere c is of D 12 and cond ition (9.72) ensu re that D 1 Just as the inde pend ence of the colu mns re ensu ) (9.72 ition cond of the rows of D 21 and invertible so too does the inde pend ence that D Je. is invertible. be back prob lem we requ ired (A, B2 ) to Finally, recall that in the H')C state feed be to K B 2 A+ 2 for r solu tion X)() 2 0 in orde stabilizable and the HCA RE to have a to RE HFA the need state estim ation prob lem we stable. We see by anal ogy that in the Hex, e. stabl be to C L A+ 2 2 dete ctab le in orde r for have a solu tion Y 00 2 0 and (A, C 2 ) to be

9.5

Sufficient Conditions

s in ralized plan t mus t satisfy certa in cond ition Recall that a given state model for the gene 6). pter Cha of stabilizing solu tions (Section 5 orde r for the GCA RE and GF ARE to have the to tions solu re the existence of stabilizing A similar set of cond ition s which ensu

246

H= State Feedback and Estimation

HCARE and HF ARE are (Al) (A2)

(A, B 2 ) and (A, C2 ) are respectively stabilizable and detectable pairs.

[

(A3)

(A- jwl)

cl

and

[

(A- jwl)

c2

have respectively independent columns and independent rows for all real w. D 12 and D 21 have respectively independent columns and independent rows.

Taken together, (Al-A3) constitute a sufficient condition for the solution of both the H 00 state feedback control problem and the H 00 state estimation problem. Notice that (Al,A2) ensure that the Hamiltonian matrices corresponding to the HCARE and HF ARE have no imaginary axis eigenvalues. This can be seen by noting the similarity of the HCARE with the GCARE in Chapter 6 and referring to Theorem 6.1 in Section 6.5.2.

9.6

Summary

In this chapter we have given a derivation of the design equations for solving the H 00 state feedback control and state estimation problems for a given linear time-invariant, continuous-time plant. Unlike the solutions to the corresponding quadratic or LQG problems obtained in Chapters 5 and 6, solutions obtained here for the H 00 problems do not minimize any performance criterion or cost. Rather the solutions to the H 00 problems achieve a specified upper bound on the L 2 gain from disturbance input to desired output with the resulting closed loop systems being stable. As in the quadratic and LQG control problems, algebraic Riccati equations still play a central role in the solution. In the next chapter we will see how the solutions obtained in this chapter can be used to provide controllers which use the measured plant output instead of the plant state to achieve the just mentioned Hoc norm bounding and closed loop stability.

9.7

Notes and References

A great deal of work has been done on the Hx control problem since its inception over a decade ago. One of the first approaches solved this problem by beginning with a Youla parametrization of all stabilizing controllers for the given plant and finished by having to solve a model matching problem. This work is the subject of [14]. This and other early approaches which are reviewed in [8] and [20] are quite complicated and have controllers with unnecessarily high dimension. Indications for a simpler approach came from research on stability robustness [35], which lead to the two-Riccati-equation solution given in this chapter [12, 16]. Following this similar results were obtained using J type factorizations, e.g., J spectral, Jlossless coprime, [19], [41], [20], [40]. Discussion of this approach in connection with the chain-scattering formulation of the generalized plant is given in [24].

10 H 00 Output Fe ed ba ck an d Control

10.1

Introduction

state prob lems of state feedback cont rol and Having seen how to solve the basic Hoc a ing rmin we come now to the prob lem of dete estimation for the given generalized plan t, that: inpu t from the measured outp ut such controller which generates the controlled e and (i) the closed loop system is internally stabl to desired outp ut is no grea ter than a given t inpu ce rban (ii) the H 00 norm from the distu positive scalar J. ut feedback cont rol problem. The setup for This prob lem is referred to as the H 00 outp 9. given previously at the start of Cha pter the outp ut feedback cont rol problem was fer trans that the generalized plan t has know n Again, as in that chap ter, we are assuming function G(s) and state model, i.e.,

, [A B]

G(s) "=

D

(10.1)

C

where l./1 (s)

C2(s)

disturbance inpu t controlled inpu t

desired outp ut mea sure d outp ut

that there are man y controllers which solve In the development to follow, we will see rs Moreover. we will show that these controlle the H:x. outp ut feedback control problem. ng havi m syste the for rs meterized controlle constitute a cons train ed subset ofY oula para Theorem 8.5, that Youla parameterized from ll, Reca ut. outp Ll:,(sl as inpu t and Y 2 (s) as to ition (i) is satisfied. We will show how controllers are stabilizing. Therefore cond (ii). constrain this set so as to achieve cond ition

248

H 00 Output Feedback and Control

10.2

Development

Recall, from Section 9.3.2, that the partition K 2 of K which effects a J-inner coprime factorization of G\ (s) stabilizes the generalized plant. In addition, recall from Section 9.2.4 that P.., ~ 0 for all 0 1 (s) E H 2 and that this implies that P1 ~ 0 for all [' 1 (s) E 7{2 where I 10.2)

This fact is used now to develop a solution to the H 00 output feedback control problem. The plan for doing this is given as follows. 1. We begin by assuming that the disturbance input U1(s) is given by (10.2) with K 1 determined in Section 9. 3.1. 2. Then we show that the signals V1 (s). V2 (s) can be replaced in the expression for the quadratic performance index P1 , introduced in Section 9.2.1, by the "external" disturbance input 0 1 (s) and an output-like signal, Y0 (s). This gives P. = P. 0 where P 10 is us.\(d to denote the new expression of L. The output-like signal, Y 0 (s). depends on 0 1(s), U2 (s) and the state of the given state model for the generalized plant. 3. Next we determine an estimator for the generalized plant state which uses y 0 (t) and achieves Y 10 ~ 0 for all 0 1 (s) E H 2 . This is accomplished by adapting the H 00 state estimator obtained in the previous chapter. 4. Following this we replace the actual generalized plant state in the Hoc state feedback controller determined in Chapter 9, by the estimate of this state which is generated by the state estimator developed in the previous step. This leads to a controller which achieves P., 0 ~ 0 for all 0 1 (s) E 7{ 2 but requires J 0 (t) as an input in addition to the measured output y 2 (t). 5. Since y 0 (t) is not available for use as an input to the controller, the remainder of the development concentrates on eliminating the need for y 0 (t) as an input to the controller.

10.2.1

Reformulation of P1

Recall from the development of the solution to the H 00 state feedback control problem (Section 9.2.1) that P., ~ 0 for all u 1 (t) E £ 2 [0, :x) implies that condition (ii) is satisfied. We are going to use this fact to solve the Hx output feedback control problem. Also recall, (Section 9.2.3), that P,1 was expressed in terms of the signals Vds). V 2 (s), which resulted from a J-inner coprime factorization. i.e.,

(10.3) and recalling (9.25) we have

need to the H 00 state estim atio n prob lem we Now in orde r to inco rpor ate the solu tion state the g min assu ed plan t state. We do this by tore -exp ress P 1 in term s of the generaliz can L\ that and 9.3.1 ion man ner specified in Sect feedback mat rix K is dete rmin ed in the be taken, from The orem 9.3, as ( 10.4)

where

8 =

eT/ 2 8 1/ 2

= D{2D12

[ = rTI 2[ 112 =-/ Im,

- D{1(Ip, - D12 8- 1Df2)D11

generalized ) we see that the state mod els for the The n subs titut ing for U1- (s)1 from (10.2 become plan t G(s), (10.1), and Ar (s), (10.3),

(1 0.5)

(10.6)

where

els for s and inpu ts to the foregoing state mod ::\otice that the state differential equa tion these of each of s notice that the initial state G(s) and i;j-l (s) are identical. In addi tion uts outp to ts g tran sfer functions to relate inpu models are identical because we are usin e thes of s state each mod el be null. The refo re the which requires that the initial state of the n whe t plan l to the state of the generalized models are identical for all time and equa equa tion obse rvat ion toge ther with the outp ut distu rban ce inpu t is given by (10.2). This t on the nden depe 1 ), reveals that V 1 (s), V2 (s) are of the state mod el for iU- (s), (10.6 n as generalized plan t state X(s) and are give (10.7) V 1 (s) = i:\ 1 G\(s) (10.8) U2(s) V2 (s) = -L\4 K2X (s) + L\3U1 C1) + L\ 4 the L\is. Vi(s)s together with the relat ions for We can use these expressions for the . rem ner given in the following theo (10.4) tore -exp ress PT (10.3), in the man

250

H 00 Output Feedback and Control

Theorem 10.1 If the state feedback matrix K is determined as in the previous chapter so that its partition K 2 solves the H 00 state feedback control problem, then P'Y, (10.3), can be re-expressed as P'Y = P'Yo where

(10.9) with

and X(s) being the state of the generalized plant. Proof From (10.7) we see that

-;

2

/

2

2'

2

~

2i

~

J I (s)llz= -~ ~~~~ U1 (s)ll2= -( \~1 U1 (s). ~~

~.

[I

(s))

(10.10)

where

Similarly from (10.8) we see that

II V2(s) II~ = II~4K2X(s) II~+II~4U2(s) 11~+11~3 ()I (s) 11~-2(~4K2X(s), ~4 U2(s)) - 2(~4KzX(s), ~3 ()I (s)) + 2(~4U2(s), ~3 ()I (s)) However from the dependency of the T

~~ ~~ = lm,-

i

-2

T

D11Dll

~is

(10.11)

on D 11 and D 12 , (10.4), we have

T ( T )~! T + "Y -2 D11D12 D12D12 D12D11

~f ~3

=

Dt!D12(Df2D12r 1DfzD11

~r ~3

=

Df2D11

~r ~4

=

Df2D12

Therefore using these relations in (10.10, 10.11) gives

( 10.12) and

- 2(D 12 K2X(s), D 12 U2(s))- 2(D 12 K2X(s), D11 ()! (s))

+2(D 12 U2

.D 11 U1 (s)i

Finally substituting (10.12) and (10.13) in P. ,(10.3), gives P.

il0.13) =

P'Yo• (10.9).

•

I

j

ralized plan t state is available for state Notice that, in the ideal case when the gene Y (s) = D 11 0 1 (s) and P"~0 ' (10.9 ), becomes feedback, we can set U2(s) = K2 X(s) so that 0

P"~o = (Oh (-"-/ lm, +Df 1Du )U1) we are assuming that O"max[DIJ] ~ "'(. which satisfies P10 ~ 0 for all U1 (s) E 'H 2 since prese nt case, we use its estimate, X(s), to Alternatively, since X(s) is unkn own in the replacing U2 (s) by K2 X(s) in Y0 (s) and generate the cont rolle d input. Therefore substituting the result in P10 , (10.9), yield (10.14)

where

Y0 (s) = Y0 (s) - Y0 (s) = D12K2X(s) -

D11

[h (s)

X(s) = X(s) - X(s) P ~ 0 for all 0 1 (s) Thus we see from (10.14) that in orde r for 10 develop an estim ator for x(t) so that

(i) : Tac(s) E 'Hoo

(ii): IIToc(s)lloo~ "!

E

'H 2 we need to (10.15)

where (10.16)

We obta in a solut ion to this problem by which is an H 00 state estimation problem. ation prob lem obta in in Section 4 of adapting the solut ion to the H 00 state estim Chap ter 9.

10.2.2

An H 00 state est ima tor

prob lem just pose d with the H 00 state A comp ariso n of the Hoc state estim ation ls that the role played by the desired estimation prob lem solved in Chap ter 9, revea d by y0 (t), the error in estim ating y0 (t). outp ut estim ation error , y1(t), is now being playe using u1(t) in place of u 1(t). In addition, in the present situation, we are the Hoc state estim ation prob lem given in to ion Therefore in orde r to use the solut a system which relates u1 ( t), u2 ( t) as inpu t Chap ter 9 we need to develop a state model for as the generalized plan t. We can obta in to Yo(t),y 2 (t) as outp ut and has the same state ut, Y1 (s) from G(s), (10.5), by Y 0 (s), (10.9). such a system by replacing the desired outp fer function by G0 (s), we obta in a system Thus doing this and denoting the resulting trans which is specified as (10.17)

252

H 00 Output Feedback and Control

where

Now we could develop a solution to the present H 00 state estimation problem by obtaining T0 c(s), (1 0.15), from the state model for G0 (s), (1 0.17) and proceeding as we did with T0 in the previous chapter in the development of an Hx state estimator for the generalized plant. However, since our goal here is to determine a stable estimator for the state of G0 (s) subject to the requirement that (10.18)

we can obtain the design equations for the desired estimator by replacing the parameters in the design equations for the H::xJ state estimator for the generalized plant state model, G(s), (10.1), by the corresponding parameters in the state model for Ga. (10.17). Inspection of the state models for G(s) and G0 (s) shows correspondences between the state modelparameters as given in Table 10.1. Therefortthe H 00 state estimator for G0 (s), (10.17), has differential equation (10.19)

Notice that we are assuming temporarily that y 0 (t) is known and can be used in the state estimator, (10.19). This assumption will be removed later since the final controller we are seeking has only y 2 (t) as input and zdt) = K 2.-\:(t) as output. Notice also that L 0 in (10.19) corresponds to Lin the H::xJ state estimator for the generalized plant given in Section 9.4.1. Now the design equations for L 0 are obtained by replacing the parameters in the HFARE and the equation for L which we obtained in Section 9.4.3 by the corresponding parameters as indicated in Table 10.1. Thus L 0 is obtained (when it exists) by finding the solution Yx 3 to the following ARE such that A 0 + L 0 C0 is stable Ax3 Y oo3

+ Yoo3A~3

- Y oo3Roo3 Yoo3 + Qcc3 = 0 La= -[Yoc3ci + BtDJ]D}ri

where Ax3

=

Aa- BtDf D]ot Co

Rod= C"[;DjjCu Q:xc3

with

=

Qx2

=

Bt (I- Df D.!a1 D2)Bf

(10.20)

Table 10.1

the gen Par ame ter corr espo nde nce s for (s). G ized plan t, G(s), and 0

A B

C C, C2

D

eral-

A0 B

C0 -D1 2K2 C2 + D 21K1

D

tion, Y 003 , to the blem treated in Chapter 9, the solu As in the H 00 state estimation pro s Y003 ~ 0 and is sfie sati le, A 0 + L 0 C0 stab Y 003 R 003 = A es 003 mak ch whi , ameters of ARE, (10.20) ation (10.20). Notice tha t the par equ to tion solu ing iliz stab the problem. referred to as the H 00 state feedback control to tion solu the e olv inv , .20) ends on the the ARE, (10 solution to equation (10.20) dep ing iliz stab a of e tenc exis the Therefore ation suggests tha t the k control problem. This observ solution to the H 00 state feedbac m and the H 00 state ble pro back control H the 00 stat e feed h bot to s tion solu tence of a existence of plant, does not guarantee the exis ized eral gen n give a for m ble trasts with estimation pro problem for tha t plant. This con trol con k bac feed put out H h the LQ G solution to the 00 which has a solution whenever bot m, ble pro trol con k bac feed put the LQ G out solutions. state estimation problems have state feedback control and LQG H 00 out put feedback the e of a solution to tenc exis the t tha 10.4 tion Sec in e stabilizing We will see h the HC AR E and HF AR E hav bot t tha s uire req y onl not m control proble solutions satisfies the addition the pro duc t of these solutions, X 00 and Yom but in following inequality

10.2.3

feedback Introducing estimated state

pla nt under the state estimator for the generalized erate the state So far we have developed an H 00 gen to generalized pla nt is being used assumption tha t the state of the (t)• j! (t)\u 2 and of Y2(t), In this case the estimate of y 0 (t), 0 feedback signal, u2 (t) = K2 x(t). are seen from (1 0.17) to the generalized plant state, x(t), Yz(t)iuz(t) based on the estimate of be (10.21) ) Yo(t)lu 2 (t) = -Dl2Kz.X(t) + D12uz(t (10.22) D22u 2 (t) Yz(t)lu 2 (t) = (C2 + D21K1)x(t) + is generated as However, if the controlled input

(10.23)

254

H 00 Output Feedback and Control

we see from (10.21, 10.22) that (10.24)

Yo(t) = Yo(t)- Yo(t) = -yo(t)

h(t) = Y2(t)- Y2(t) = (C2 + D21K1

+ D22Kc)i( t)- Y:>(t)

(10.25)

where we have y0 (t), (10.21), null when the controlled input satisfies (10.23). This fact plays an important role in the developme nt of the controller. Continuing , we see that use of(l0.24, 10.25) in (10.19) yields (10.26)

where

and the controlled input is generated by a controller with transfer function f(s) as (10.27) However, since y 0 (t) is unknown we need to eliminate it as part of the input to this controller in a manner which preserves P'Y0 ::; 0 for all U 1 (s) E H 2 . We will do this in the next section. Before we go on to develop these J 0 (t)-free controllers, we summarize what has been achieved so far in the development of a solution to the H 00 output feedback control problem. 1. We began by incorporat ing the solution to the Hoc state feedback control problem in the performan ce index, P,, used in the previous chapter. The expression we obtained was denoted P 1 ()' It was shown that P 10 depends on YoU), ( 10.9). an output from a system, (1 0.1 7), whose state equals the state of the generalized plant. 2. Next, in Section 10.2.2, we used the solution to the H 00 state estimation problem obtained in Chapter 9 to obtain an H 00 state estimator based on the output y 0 (t) so that the error in estimating Yo(t), i.e., Yo(t), satisfies IIYa(t)ll2< 1llu1 (t)ll2· 3. Finally, in Section 10.2.3, we showed that by using the estimated state to generate the state feedback signal as u2 (t) = K 2 .x(t) and the H::x: state estimator based ony0 (t) gave E H2 as a controller which satisfied p 'YO ::; 0 for all cl (s) E H2 or equiYa1ently all ul problem. control feedback output Rx; the solve to required

1 0.3

Hex Output Feedback Controllers

As mentioned before, the controller given by equation ( 10.27) cannot be used as it has the unknown output-lik e signal, y 0 (t), as an input. In this section we will shO\\ hO\\ this unknown signal can be removed from the input to the controller, (10.27).

~~~

nw •n• •. 7f~biiijARHNOIMOH vw

to as the cen tral s to a single con trol ler referred One way of doi ng this which lead urb dist anc e, (Section dist urb anc e inp ut is the wor st controller, is to assu me tha t the 9.2.4), so tha t 0 1 (s) is null. em. This app roa ch relate y 0 (t) to u 1 (t) thro ugh a syst Ano ther way of doi ng this is to feedback con trol put out Hxof which solves the one any lers trol con of ily fam yields a ame triz ed stabilizing ily is a sub set of the You la par fam this t tha see will We . problem d in Cha pte r 8. ized plan t which we enc oun tere controllers for the given general

10.3.1

Ce ntr al Co ntr oll er

Yo(t) satisfies tha t the erro r in esti mat ing ion, sect ious prev the Recall, from e inde x P'Y 0 ' (10.14) as re we can write the per form anc 51 0 (t) = -y 0 (t), (10.24). The refo (10.28) and the rela tion (10.16) as (10.29) t ned the esti mat or, (10.19). so tha However recall tha t we dete rmi (10.30) 10.30) tha t Thus we see from (10.28, 10.29,

put feedback con trol (ii) in the stat eme nt of the H 00 out as is requ ired to satisfy con diti on problem. led inp ut, u2 (t), r sup pos e we generate the con trol Now 0 1(s) is unk now n. Howeve anc e inp ut is the wor st er the assu mpt ion tha t the dist urb und ), 0.27 (1 ler, trol con the from is null. The n in this und er the assu mpt ion tha t u1(t) i.e., 4), 9.2. n ctio (Se e, anc urb dist out put , (10.27), is is null and ther efo re the con trol led 29), (10. ). YJ~ t tha see we n atio situ given as

-

Lo2]

0

(10.31)

where

the cen tral controller. This con trol ler is referred to as eralized plan t und er is an Hx stat e esti mat or of the gen In essence the cen tral con trol ler the con trol led inp ut e inp ut is the worst dist urb anc e and the restriction that the dist urb anc the Bx state feedback u (t) = K 2.'{(t), where K 1 solves depends on the stat e estimate as 2 is free from Yo(t). of obt aini ng a con trol ler which con trol pro blem . Thi s is one way

H 00 Output Feedback and Control

10.3.2

Controller parametrization

Suppose thP unknown output Y 0 (s) which is present in the controller, (10.27), is replaced by Yq,(s) where

Yq,(s)

= ~(s)U 1

(s)

(10.32)

with (i) : ~(s) E

Hoc

(ii) : jj~(s) 1\oc::; I

Notice that when U1(s) E 1i 2 we have Yq,(s) E Hz and jjYq,(s)jjz::; ri\UJ(s)llz· Thus when Y0 (s) is replaced by Yq,(s) in (10.28) we have

which implies that P10 ::; 0 for U1 (s) E 1i2 as required to satisfy condition (ii) in the statement of the Hoc output feedback control problem. Next supp(;se we have W(s) E Hrx; such that (10.33) Then replacing Y 0 (s) in (10.33) by

and solving (10.33) for U2 (s) gives the desired controller as (10.34)

where

with ~(s) constrained as specified in (10.32). The specification of W(s) is given in the following theorem Theorem 10.2 The state model for the transfer function W(s), (10.33), satisfies W(s) E 1ix and has state model

( 10.35)

where

. . 'JtilB .

• ••

•••

feed bac k con trol in the solu tion of the H 00 stat e Pro of Since K was dete rmi ned • H E ) 00 W(s (Section 9.3.1), we hav e pro blem so that A+ BK is stable, 34) becomes (10. that so null ) W (s) we cho ose
cop rim e dete rmi ne W 2 (s) and W4 (s) by H E 00 , we can (s) W (s), 4 W e 2 sinc Now and W 4 (s) = Mr(s) lly, we cho ose W2(s) = Nr(s) factorizing f 2 (s). Mo re specifica s) cop rim e. Mr( s), 1 Nr(s), Mr(s) E H 00 and Nr( where f 2(s) = Nr( s)M i (s) with as oted den is (s) stat e mod el for r 2 Recall from Sec tion 8.4.1 that if the

rim e fact ors off 2 (s) are given as then stat e mod els for the righ t cop

Nr(s) ~ [Ar +B rKr Cr +D rKr

Br] Dr

Mr(s) E H 00 • where Kr is cho sen so that Nr(s), Now from (10.31) we hav e Dr =0 gives the cop rim e fact ors as so that cho osin g Kr = (C2 + D 3 K)

s Nr(s) = Wz(s) =

[A+ BK

-Lo2]

Kz

0

[ A + BK

-

s Mr(s) = W 4 (s) = Cz

+ D3K

(10.38)

Lo2] I

of the Hoc stat e K was dete rmi ned in the solu tion where Nr(s), Mr(s) E 'Hoo since s com plet es the Thi le. stab 9.3.1) so that A+ BK is feedback con trol pro blem , (Section W (s) and W 4 (s). dete rmi nati on of stat e mod els for 2 that we recall from (10.17) and (10.33) (s) Nex t in ord er to dete rmi ne W 1 (10.39) Y0 (s) = Gal(s)Dl (s) + GoZ(s)Uz(s) (10.40) s)U (s)

Uz(s)

=

WJ(s)Ya(s)

+ W2(

1

258

H 00 Output Feedback and Control

Now if the system inverse,

6;;21 (s), were to exist we could solve (10.39) for U2 (s) as (10.41)

and we would have W1 (s) = 6;;21 (s) from a comparison of (10.41) with (10.40). However we see from (10.17) that 6a2 (s) is not invertible in general since its state model,

has aD matrix, D 12 , which is not invertible in general, (Section 8.2). This difficulty can be overcome by recalling that we are assuming D[2 D 12 is invertible since this is a necessary condition for the existence of a solution to the H 00 state feedback control problem. Thus G02 (s) = D'[2 6 02 (s) is invertible where (10.42)

Therefore we rewrite (10.39, 10.40) as

Y (s) =Go! (s)Ut(s) + Go2(s)Uz(s)

(10.43)

WI (s) Y0 (s) + Wz(s) 0 1 (s)

(10.44)

0

U2(s)

=

where

Then solving (10.43) for U2 (s) (10.45)

and comparing (1 0.44) with (10.45) yields ( 10.46)

Finally, using the form for the state model of a system inverse, (Section 8.2), we see from (I 0.42) that ( 10.47)

and (10.47) yields ( 10.48)

DLD 12 =I, given as where DL is a left inverse of D 12 , i.e.,

(10.49)

T D t12 = (DT12D12 )-l D12

(s). This completes the determination of W 1 and in the second block equa tion s in (10.33) ite rewr (s), W e 3 In order to determin (l0.1 7)as (10.50) Yz(s) = W3(s)Y0 (s) + W4(s)01(s) (10.51) Yz(s) = G0 3(s)01(s) + G0 4(s)U2(s) 0) in (10.51) Then substituting for U2 (s) from (10.4 ( 10.52) als that and comparing (10.50) with (10.52) reve

the (10.17), and W 1(s), ( 10.48) toge ther with The n using the state models for G04 (s), in tion 8.1 ), we obta rule for combining systems in series, (Sec

ugh the out affecting W 3 (s). We can do this thro Nex t this state model can reduced with dina te coor the let dinates. Therefore if we use of an appr opri ate change of coor tran sfor mati on matr ix be

dina tes is the state model for W3 (s) in the new coor [

T- 1 A w T

r- 1 B w

CwT

Dw

[A +0 BK

A

r [ C + D3K 2

where

l1

= ~ w J3 w LCw

1

Dw J

~2~2KJ

[B2;j l]

D22K2;

D=zDi 2

2

260

H 00 Output Feedbac k and Control

Finally, notice that W3 (s) is given by W3 (s)

-

-

l -

= Cw(sl- Aw)- Bw + Dw = (C2 + D3K)[si - A- BKr 1B2DL + D22 DL

which implies that W 3 (s) has a state model given as

•

In summary , in this section we have determin ed a family of controlle rs, H(s), (1034. next 10.35), any one of which solves the H 00 output feedback control problem. In the ized parametr oula Y the of subset a to rs controlle section we relate this family of controllers.

1 0.3.3

Relatio n to You/a param etrizat ion

rs. Recall, from Theorem 8.5, that the Y oula parametr ization of stabilizing controlle 8.4.3). (Section H(s), for a plant, G(s), involves the doubly coprime factoriza tion of G(s), These controlle rs are given by (10.53)

where

with

-~H(s)l [~G(s) MG(s)

;\G(s)

Now in the present situation , the H'Xl output feedback controlle r is connecte d between the generalized plant's measured output, Y2 (t) and controlle d input, u 2 (t). Moreove r, this controlle r must stabilize the generalized plant for all disturban ce inputs satisfying (Section 9.2.4). Therefore. ul (s) E H2 including the worst disturban ce, Uj (t) = Klx(t) assuming the disturban ce input is the worst disturban ce, we see that the controlle r must stabilize a plant G(s) = Gn(s), where from (10.5) we have (10.54)

systems involved in a doub ly coprime Thus recalling the state models for the has a doub ly coprime factorization given by factorization, (8.46, 8.53), we see that G22 (s)

Nn(s )

~

[ Ae

~=eKe ~ J

Nc(s) ~ [ Ac + BeK e Ce+ DeK e

Be] De

Mn( s) d: [ Ae + BeK e Ce+ DeK e Mc(s )

~

L1c] (10.55)

~e]

[ AG +;:GKG

as specified in (10.54) Then substituting the state model para mete rs and Le = -L02 in (10.55) gives

and setting Ke

= K2

(10.56)

in the general , we can repla ce Q6 (s) However, since 2 Q6 (s) E 1(::o if Qe(s) E 'H 00 (s) so that DL

D1


(10.57)

where

the generalized plan t and have the same Thus controllers satisfying (10.57) stabilize determined in the previous section, (10.34, form as the H 00 outp ut feedback controllers n
10.4

H00 Separation Principle

the controller which solves LQG outp ut Recall, from Chap ter 6, that we obta ined corresponding solutions to the basic LQG feedback cont rol prob lem directly from the cont rol problems. We referred to this fact plant state estimation and plan t state feedback case, being able to solve the correspondas the separation principle. However in the Hx estimation problems is only a necessary ing basic Hoc state feedback cont rol and state ut feedback cont rol problem. condition for being able to solve the H 00 outp l condition needed to ensure the existence In this section we will show that the addi tiona rol problem is of the solution to the Hoc outp ut feedback cont (10.58)

262

Y

H 00 Output Feedback and Control

'

..0

where X 00 , Y 00 are stabilizing solutions to the HCARE and HFARE, (9.46, 9.77). Before doing this we show now that Y = 3 and L 0 , (10.20), needed to construct the controllers, (10.34, HU5) can be calculated directly from the stabilizing solutions to the HCARE and HF ARE as follows: ( 10.59)

ZL 2 ]

(10.60)

where

Z=(I- 1 - 2 Y DC X CXJ )- 1

with L being the solution to the H 00 state estimation problem for the giwn generalized plant (9.78). The approach used to show these results relies on a similarity relation between the Hamiltonian matrices associated with the HFARE. (9.77), and the ARE. (10.20).

10.4.1

A relation between Hamiltonians

We are going to make use of the following ideas from Chapter 5. The Hamiltonian matrix Hp associated with the ARE

AP+PA 7

-

PRP+ Q = 0

110.61)

1s g1ven as

(10.62) The eigenvalues of H p exhibit mirror image symmetry across the imaginary axis.

A[Hp]

=

A[(A- PR{]

A -(A-PR)]

This fact is seen from the following similarity transformation of H p

fl

= P

rl H T P

= [

T

(A-PR) AP+PA*-PRP+Q

-(A~ PR)]

where

T with

=

[

I

-P

Hp being block-upper triangular when

0]I

P satisfies (10.61).

(10.63)

<

263

H00 Separation Principle

Notice that equating the first n columns on either side of (10.63) yields (10.64) so that if Pis a stabilizing solution to (10.61) then

is the stable subspace of H P· Now the relation between the Hamiltonian matrices associated with the HFARE, (9.77), and the ARE, (10.20) which is given in the following theorem will be used to show (10.59, 10.60). Theorem 10.3 The Hamiltonian matrix H 2 correspondin g to the HFARE (9.77) and the Hamiltonian matrix H 3 correspondin g to the ARE (10.20) are similar and related as (10.65) where i

= 2,3

Proof Suppose (10.65) is valid. Then carrying out the matrix multiplicatio n which is indicated on the right side of (10.65) enables (10.65) to be rewritten as

H2

= [

(Aoo3

-~- 2 Qoo3Xoc) T

r- 2 (A;;:,3xoo + XocAoo3) -~- 4 XooQx3Xoo + Rx3] -(Aoc3 -I- 2 Qoo3X"")

Qoo3

(10.66) Then equating like positioned blocks on either side of (10.66) yields (10.67)

Q""2 = Qoc3 Aoo2 =

Aoo3 -!- 2 Q""3Xoo

R""2

I- 2 (A~3Xao

=

(10.68)

+ XooAoc3)

-~- 4 XxQoo3Xoc

+ Roo3

(10.69)

Now we show that (10.65) is satisfied by showing that (10.67-10.69) are satisfied. Recall from (10.20) and the HFARE, (9.77), that

A=2

A - B1 Df D.!a1 C

C~D7t}C0

R=2

c 7 v--Jo1c

B1(I- D[Dj}D2)B f

Q=2

B, (I- D2T D}a1 D2 ) B,T

Ax3

A0

Rx3 Qoo3

-

81 Df D_j} Co

(10.70)

H= Output Feedback and Control

264

where Ao

D2

[Dll]

D;o

[ -"/IP~ + D 11Df1 D 21 D 11

A+B 1K 1

D21

DllD~l

Co

D21D21

[ -D12K2

]

C2 +D2,KI

and (10.67) is satisfied. Next in order to show (10.68) we use (10.70) so that (10.71)

Now we are going tore-express the right side of this equation using results from the H 00 state feedback control problem. Recall from the solution to the H 00 state feedback control problem given in Chapter 9, (9.43), that (10.72)

Now writing D Jc as

~]

(10.73)

where

enables (10.72) to be rewritten as

DIITc IK =I2 K,- B,Txoo

(10.74) (10.75)

where C,K=D 1K+C 1

(10.76)

Then K 1 is obtained from (10.74) as (10.77)

Next we can readily show that

H 00 Separ ation Princi ple

265

where

which after substit uting for K 1 from (1 0. 77) becomes

( 10.78)

Howev er writing D 1a. (10.70), as

gtves

and (10.78) can be rewritt en as (10.79) right side of (10.71) gives Return ing to (10.71) we see that use of (10.79, 10. 77) on the

AX!,-

Ao.c2 =; - 2B1 (Bf Xoo + Df1 elK)

-~- 2 B1Df ( [~]elK+ D]a1D2Bf Xoo)

= ~~- 2 (B1Bf X:xc- B1Df Dj} D2Bf Xoc) is satisfied. and we see from the definition of Qoc 3 , (10.70), that (10.68) Finally it remain s to show that (1 0.69) is satisfied, i.e., that (10.80) where

We begin by using (10.67, 10.68) to express

n as

H 00 Output Feedback and Control

266

Then using the definitions for Ax 2 and Qx 2 , (10,70), enables \1 to be written as !.1

=

~r- 2 (AT Xx:

+ Xx:A-

eT MT Xoc- XxMe)

+ 1-4 (X""B1BfXoo- XxMD2BfXoo) (10,81)

where

Next we are going tore-express ATX::x: + X 00 A for use in (10,81), Recall from the development of the HCARE given in the previous chapter that

which we can rewrite using the expression forD In (10,73), as (10,82)

However, from the definition of elK• (10,76), we see that expanding e[KeiK and ef elK gives

which when substituted in (10,82) yields

Finally we can expand this relation by substituting for K 1 from (1 0. 77) to obtain (10.83)

where T

Fcx = eiKeiK-

elT elK-

T

eiKel

Fx = XocBIDfl elK+ eT~D]]Bf XX+

efKDllDfl elK

Returning to (10.81), we can use (10.83) to rewrite (10.81) as (10.84)

This completes the expansion of n for use in showing (10.80) Finally using ( 10. 79) yields (10.85)

Hoc Separati on Principle

267

and from R003 , (10.70), we have Roc3

= Roo2- , -2 [FcK-

cT MTXoo- XocMCJ

+ ,-4 [Fx + XooBtDzD]j DzBf Xoo]

(10.86)

where M,FCK and Fx are given in (10.81, 10.83) Finally we see from (10.86) and (10.84) that (10.80) is satisfied so that (10.69) is • wW.

10.4.2

Relating stabilizing solutions

The relation between Hamilton ian matrices, (10.65), is used now to show that (10.59) is satisfied and that condition (10.58) is required in order to solve the Hoc output feedback control problem. Recall from (10.64) that the Hamilton ian matrix, H 2 , corresponding to the HFARE satisfies (10.87) so that

is the stable subspace of H 2 . 1 Therefore premultiplying (10.87) by
which indicates that range [

/ _

-2 X

~'_ y 0000

y 00

l

is the stable subspace of H 3 , since A 002 - Y00 R 002 is stable. Now since the range of a matrix is unchanged by post-multiplication by a nonsingular matrix, we have 2x

range [ I_,-· oo -Yx provided I-

yx

l [ =range

] I_ ) -I ( 2 -Y00 I-1' XxYxo

,-z X"" Yx is invertible. Therefore we see that

H 00 Output Feedba ck and Control

268

is the stable subspace of H 3 . However we saw in the previous subsection that

is the stable subspace of H 3 . Thus we see that (10.88) ence of and (10.59) is established. Notice in (10.88) that the rightmost equality is a consequ the ensure to the matrix inversion lemma, (Appendix). We show next that ( 10.58) needed from follows existence of the solution to the H 00 output feedback control problem (10.88). we have Recall that Y:xJ 3 and Yex; are each stabilizing solutions to AREs so that ) of(l0.88 on Y 003 , Yoo 2: 0. Thus we see from (10.88) that Z 2': 0. However in the derivati the for required we required Z to be invertible. Therefore we can conclude that Z > 0 is n on existence of a solution to the H 00 output feedback control problem. Now the conditio Y 00 , X 00 which ensure that Z > 0 is given in the following theorem. Theorem 10.4 Z > 0 if and only if (10.89) where Z= (1- ' / - 2 Y 00 X 00 )- 1

> 0 if and only if z- 1 > 0. In addition recall that z- 1 > 0 if and only if Az-• > 0 for all Az-1 E .x[z- 1]. Proof Recall from Chapter 4, that Z

Let Az-1 be any eigenvalue of z- 1 • Then

(10.90) and from the dependency of z- 1 on Y xXx. we have (10.91)

Then solving for Az-' gives (10.92)

and we see that Az-' > 0 if and only if /\YX < largest Finally we see from (10.92) that ,\ 2 > 0 for all .\ 2 E .X[Zi if and only if the • . theorem the in gi~en n conditio the satisfies ooX:xJ], eigenvalue of Y00 X 00 , Amax[Y

Summary

10.4.3

zn

Dete rmin atio n of L0

ished directly form (10.88). The relation between L 0 and (L, Z), (10.60), can now be establ calculated directly from the As mentioned earlier, this relation enables Y 003 to be since L depends on Y 00 and Z stabilizing solutions to the HCA RE and the HFARE depends on both X 00 and Y00 • in (10.20) to be In order to establish (10.60), recall that Lm was determined (10.93) (10.85) yields Therefore substituting for Y 003 and C0 from (10.59) and

where

However from the definition of Z, (10.88), we have

so that (10.94) becomes 2 1 L 0 = -Z(Yo oCT +B1D I)Dj0 -"Y- ZY00 C[K[I

=ZL -"(- 2 ZYooC [K[l

0]

0] (10.95)

where L was obtain ed in Section 9.4.3.

10.5

Summary

ARE developed in the previous In this chapt er we have shown that if the H CARE and HF these solutions satisfy chapter have stabilizing solutions, Xxo, Yoo and provided

m. In additi on we have shown then we can solve the HQO outpu t feedback contro l proble which solve the Hoo outpu t H(s), llers, contro that when these conditions are satisfied the feedback contro l proble m are given by

where

Nc(s) = W 1 (s)
ci>(s) E 1-{""

270

H 00 Output Feedback and Control

with W(s) having state model

and with K, L being determined from

+ Df CI)

K

=

-Dj} (BT Xoo

L

=

-(Y XCT + B!DI)Djj

where X00 , Y 00 are stabilizing solution of the HCARE and HFARE, viz.,

A~1X + XAxl- XRoo1X + Qool = 0

with

=

A - BD}c1Df C 1 BD}/ BT

Ax2 =A- B1DJ D}a1C R:x2 = CT D"Jol C

=

Cf (I- DID)c1Df)Cl

Qoo2 = B1 [I- D[ D"J01D2JBf

Aool

=

Roo I

Qool

and

D;c

D;o

=

[-lIm, + Dft Du

D~D12]

Df2D11

D12D12

[

=

7 -''(lp,

+ D11D1tT T

D21D11

D11D~ D21D21

l

Conditions on the generalized plant state model parameters which ensure that the HCARE and the HF ARE have stabilizing solutions were given at the end of Chapter 9.

10.6

Notes and References

Hx control can be used in conjunction with the small gain theorem (Section 7.4.3) to provide a means for designing controllers to stabilize plants haYing inexactly specified models [47, pp. 221~229]. These ideas were extended recently to plants having a class of nonlinear state models [31]. Additional material on the ideas presented in this chapter can be found in [18].

Appendix A: Linear Algebra

Most of the current "first" books on control systems with terms like "automatic control" or .. feedback control" in their titles have an appendix containing a summary of matrix theory. These can be consulted for this purpose. The topics treated in this appendix are selected to give the reader additional insight into issues alluded to in the text.

A.1

Multiple Eigenvalues and Controllability

When some of the roots of the characteristic polynomiaL det[.A/- A] for ann x n matrix A are repeated, A may not have a complete set ofn independent eigenvectors. Notice that if (A.l) where the .Aks are unique A; # )...i i # j i,j E [I, m], then we say that )...k has multiplicity or is repeated nk times. Since the degree of the characteristic polynomial, (A. I), is n, the dimension of the A matrix, and since an nth degree polynomial always has n roots, counting multiplicities, we have

nk

:\ow there is at least one eigenvector, vk 1, corresponding to each eigenvalue >..k : k E [1, m]. However since )...k is repeated nk times. we obtain a complete set of eigenvectors only if there are nk independent eigenvectors, vki : i = 1, 2, · · · nkl corresponding to each >..k, k E [1, m], i.e., only if

has nk independent solutions v = vki, i = 1, 2, · · · nk for each k E [l, rn]. Any square matrix of size n having a complete set of n eigenvectors gives rise to a

272

Appendix A: Linear Algebra

diagonal matrix under the transformati on

where then columns of V are eigenvectors of A, i.e.,

and A is a block diagonal matrix having n1 x n1 diagonal blocks, A 1 : i ).. 1 at each diagonal position, i.e.,

l

0

1.0

A,~

=

1, 2, · · · m, with

r :,0 ;,0]

[>..,

Matrices having a complete set of n eigenvectors are referred to as diagonalizable matrices. Any matrix whose eigenvalues are simple, i.e .. do not repeat, is a diagonalizable matrix. Alternatively, a matrix A that does not have a complete set of eigenvectors cannot 1 be diagonalized by any invertible matrix V in the transformati on v- A V. In this case A is said to be defective, i.e., defective matrices are not diagonalizable matrices. For more detail the reader is referred to [17, pp. 338-339]. Concerning the controllability of a state model having an A matrix with multiple eigenvalues, it turns out that when A has more than one eigenvector correspondin g to any eigenvalue it is not possible to control the state from one input, i.e., the pair (A, B) is uncontrollable for all B having only one column. The foregoing fact can be shown as follows. Whenever A has two independent righteigenvectors correspondin g an eigenvalue >.., A has two independent left-eigenvectors, w 1 , w2 correspondin g to this eigenvalue. Then any linear combination of these lefteigenvectors is also a left-eigenvector of A, i.e.,

where

with the a 1s being arbitrary, constant, non-zero scalars. Therefore we can always choose the a 1s for any vector B of length n so that wTB= 0

showing that >.. is an uncontrollable eigenvalue for all vectors B.

A.2

Block Upper Triangular Matrices

Block upper triangular matrices play an important role in many different situations in control theory. They are directly related to invariant subspaces. A subspace W of Vis said to be an A-invariant subspace if for all x E W we have Ax E W.

~

'

J

rj

r l

I

Block Upper Triangular Matrices

273

Suppose we are given a square matrix of size n which has a block upper triangular structure, e.g., (A.2)

where All A 2 , A 3 , and A 4 are n 1 x any x of the form

nb

n1 x

n2, n2

x

n~>

and n 2 x

n2

with n

=

n 1 + n 2 . Then

where x 1 is any vector of length nh lies in an A-invariant subspace, since x E W

In this case W is called an eigenspace corresponding to the eigenvalues of A which are also eigenvalues of A 1• To see this consider the eigenvalue-eigenvector equation for A. We can rewrite this equation in the following way A 1v\

+ A2v~ = >.;v~

(A.3)

= >.;v~

(A.4)

A4v~ where

is the partitioned form for the eigenvector vi corresponding to eigenvalue>.;. Suppose that v~ is null. Then we see from (A.3) that must be an eigenvector of A 1 corresponding to eigenvalue >.;. Thus the eigenvalues of A1 are eigenvalues of A i.e. >.(AI) C >.(A) with the corresponding eigenvectors of A being

vi

with A 1v\ =>.;vi Real Schur decomposition [17, p. 362], provides a computationally robust method for obtaining an orthogonal matrix Q such that

has the form (A.2). In addition there are ways of doing this so that the eigenvalues of A 1 are a specified subset of the eigenvalues of A.

274

A.3

Appendix A: Linear Algebra

Singular Value Decomposition (SVD)

In this section we give a proof of Theorem 7.1, viz., Theor.:m 7.1 Any p x m matrix of complex constants, M, which has rank r can be decomposed as Jf

=

U:EV*

(A.5)

where U, and V are p x p and m x m unitary matrices with

U 1 is p x r

U2 is p

Vl ism

V2 ism

X I'

X jJ- I'

m- r

X

and 'Eo is an r x r, real, positive definite, diagonal matrix denoted as O'J

0

0

0

0'2

0 = diag[0" 1 , 0'2 , · · ·, 0',.]

'Eo= 0

0

O'r

with diagonal entries referred to as singular values and ordered so that

i = 1. 2, .. · , r

-

1

Proof Let (O'T, v;) denote an eigenvalue-eigenvec tor pair forM* Man m x m Hermitian non-negatve matrix. Thus we have M* Mv; = O'JV;

(A.6)

with O'T real and positive so that we can always order the eigenvalues as 2 O'J

::0.

2 O'r+l

2 0'2

=

::0. .. • ::0. O'r2 2

O'r+2

= ··· =

>0 2 O'm

= 0

Notice that since rank[M] = r, only r of the eigenvalues of M* Mare non-zero. In addition, since M* l'vf is Hermitian, we can always choose the eigemectors so that they are orthonormal =

1

=0

i = 1, 2, · · · m

(A.7)

1- j

(A.8)

for i

Next we form a matrix V from the foregoing orthonormal eigenvectors of M* }.1 so

.ll

215

that (A.6) can be written as the matrix equations (A.9)

M*MVI = V 1 I;6

(A.lO)

M*MV2 = 0

where V = [VI V1

= [v 1

V2] v2

,..2 "-'O = ···

V2

v,]

d"' Iag [a 21 o Cf 22' ... '

=

[vr+l

Vr+2

2]

rJ r

...

Vm]

Notice that the orthonormal nature of the eigenvectors, (A.7, A.S), makes them x r matrix V 1 satisfy Vj V 1 =I, and the m x m- r matrix V 2 satisfy Vi V 2 =I. Thus we have V* V = VV* =I, i.e., them x m matrix Vis unitary. )Jext, pre-multiplying (A.9) by Vj and then pre- and post-multiplying the resulting equation by I; 01 gives

which we can interpret as

(All) where

Notice from (A.11) that the p x r matrix U 1 has orthonormal columns. Therefore we can always choose a p x p- r matrix U 2 having orthonormal columns so that the p x p matrix U

is unitary. i.e., U' U = UU* = I. We show next that the foregoing unitary matrices. U, V satisfy (A.5) in the statement of Theorem 7.1. We begin by expanding U* MV in terms of its blocks U*MV= [

U*MV I

I

Ui"o\t!VI

U·*.1MV2] U2MVc.

(A.12)

Then we see that the 1,1 and 1,2 blocks can be reduced by substituting for U 1 from (All) and using (A.9, A.10). Doing this yields I;r) 1 VjM*A'fV 1 = I; 01

UjMV1

=

U)MV 2

= I; 01 V) M* MV 2 =

I::6 = I;o

I; 0 Vj V2

=0

p;ext the reduction of the 2,1 block in (A.l2) relies on using the

expres~ion

for Uh

-------

---·~--·u

(A. ll) to ena ble MV 1 to be repl aced by U,E o so tha t

u;M v, = u;u ,E 0 = 0 Fiually in ord er to reduce the 2,2 block in (A.12) we use the fact tha t MV 2 seen by not ing , from the con stru ctio n of V b (A.IO), tha t M* Mv ; =

= 0. This is

0

where v; is any colu mn of V . Thi s implies tha t 2 v;M *M v;= O

and letting z;

= Mv ; we see tha t

which is only possible if z; is null . Thu s ivfv ; mu st be null and sinc e v; is any colu mn of V 2 we have MV 2 = 0. The refo re we see tha t the 2,2 block of (A. l2) becomes

Thu s from the foregoing we see tha t (A. l2) is reduced to U'M V

=

[~ ~]

and we hav e the form for the SVD of M which is given in the theo rem .

A.4

Different Forms for the SVD

•

The re are occasions where con stra ints imp ose d on the SVD by the dimensions of the mat rix being dec omp ose d can be tak en into acc oun t so as to pro vid e a mor e explicit fom1 for the SVD. Thi s possibility aris es from the fact tha t the ran k of a rect ang ular mat rix can not exceed the sma ller of its dimensions. The refo re since Af is p x m, we hav e rank[M]

= r:::;

min (p, m)

and the SVD of M, (A.5), can be expressed in one of the following form s dep end ing on which of the dim ens ion s of M is larger. If p < m, i.e., M has mor e colu mns tha n rows, then (A.5) can be wri tten as

I

~ '\;i

where U, V in (A.5) equal UP and [ Vrl

Vr 2 ] respectively. In add itio n Vr 1 , Vp 2 are

.

.

-

Matrix Inversion Lemma (MIL)

277

m x p and m x (m- p) with :EP being a p x p diagonal matrix :EP = diag[a 1 , a 2, ···,a, 0, · · ·, 0]

In this case it is customary to refer to the p diagonal entries in :EP as the singular values of M and the p columns of UP and of VP 1 as the left and right singular vectors of M respectively. Similarly, if p > m, i.e., M has more rows than columns, then (A.5) can be written as V m2 ] [

:E.m] 0 V*m

where U, V in (A.5) equal [ U mi U m 2 ] and V m respectively. In addition U 1111 , U m 2 are p x m and p x (p - m) with L:m being the m x m diagonal matrix

Again it is customary to refer to the m diagonal entries in :Em as the singular values of M and the m columns of Umi and of V m as the left and right singular vectors of M respectively.

A.5

Matrix Inversion Lemma (MIL)

The following relation has been of considerable importance in the development of control theory. Lemma If D and I: are nonsingular matrices then

where L = r2 R

+ \li I:

= ~.r~-

o-I\li(o-1\li- I:-Ir'o-1

Proof Expanding the product RL yields

(A.l3) where =

o-1\li(o-'\li

+ I;-~~-~

r2=

o-'\li ( rl- 1\li

+ I::- 1) -I o- 1wE

r,

278

Appendix A: Linear Algebra

Then we see that r1

+

r2

=

D~ 1 W(
= 0 ~1\lf(
+ I:-1)-1 [I:~ I+
n- 1WI:

Finally substituting (A.14) in (A.13) gives RL =I.

(A.14)

•

_,_j ~~~~

Appendix 8: Reduced Or de r M od el Stability order model is stable, we require the As mentioned at the end of Section 4.6, since the full the input -outp ut behavior of the reduced order model to also be stable. Otherwise model behavior as the difference reduced order model canno t approximate the full order ls diverges in response to a mode between the outpu ts of the full and reduced order a system matrix A 1 which is a has l common boun ded input. Since the reduced order mode inates, coord ced partition of the full order system matrix Ab, in balan

A has all its eigenvalues in the we need a criterion which can be applied to ensure that 1 es this prope rty was given, without open left half plane. A practical criterion which ensur we are going to prove this theorem. proof, as Theorem 4.3 in Section 4.6. In this appendix entry in 'Ebl is the same as any Theorem 4.3 If in balanced coordinates, no diagonal diagonal entry in Eb 2 then A 1 will be stable where

Proof Suppose A 1 is not stable. Then we have..\

E

..\[Ad such that Re[..\];::: 0

(B.l)

of the theorem, Ab is unstable. Now we are going to show that under the conditions only possible if the given system is ation However this is impossible since a balanced realiz is stable. ions for the controllability and To begin, consider expanding the Lyapunov equat

280

Appendix B: Reduced Order Model Stability

observability Gramians as

1 A2] [~bl

[A A3

A4

0

Then we see that

~bl

0.

~b2

l

+

[~bl

0

~b2

0

l

[Aj.

A)]

A2

A4

=-

[B1] B [Bj

Bi]

(B.2)

2

satisfies the following Lyapunov equations

(B.4) (B.5) Now since ~bl is diagonal with all diagonal entries being positive and real we have > 0. Therefore we see from the proofs of Theorems 4.7 and 4.8 that the assumed unstable eigenvalue, A, of A 1 must lie on the imaginary axis and be an unobservable and uncontromtble eigenvalue of the pairs (A 1 , C 1) and (A 1, B 1) respectively. Therefore the corresponding right and left eigenvectors, (B.l ), satisfy ~bl

(B.6) C1v=O

(B.7)

Next we show that there are many eigenvectors corresponding to this eigenvalue. In what follows we show how these eigenvectors are related. To begin, postmultiply (B.5) by the eigenvector v, and use (B.7) to obtain or

(B.8)

However, A is on the imaginary axis. Therefore -A* = A and we see from (B.8) that is an additional left eigenvector of A 1 corresponding to the uncontrollable eigenvalue A and therefore we have ~b 1 v

(B.9) Next premultiply (B.4) by v* ~b 1 , and use (B.8, B.9) to obtain \ *"2 AV uhl = -V *'<'2 ub1·4*I

or

Thus we see that both v and ~; 1 v are (independent) right eigenvectors of A 1 corresponding to the eigenvalue A. Continuing in this fashion we can generate sets of left and right eigenvectors corresponding to A as

s"

= {~ /,J'L'

:i

=

1. 3, s.... , 2k + 1}

sv = {~hi v : i = 0, 2, 4,

0

0

0 ,

2k}

(B.IO) (B.ll)

.l.~x. ,,

'

281

Appendix B: Reduced Order Model Stability

where k is an integer which is chosen large enough so that we can find a matrix Q to satisfy (B.l2) where M is the n 1 x (k + I) matrix

Notice from (B.l2) that under the transform ation :Eb 1, the range of M is unchanged. This fact plays an importan t role in what is to follow. Now we saw in Theorem 5.5, that (Bl2) implies that every eigenvalue of Q is an eigenvalue of :Eb1

(B.13) This importan t fact will be used to show that A 3M= 0. Notice that since the elements in lSw, Sv are left and right eigenvectors, respectively, of A 1 corresponding to the uncontro have we lable and unobservable eigenvalue.>.., v*:E~ 1 B 1 = 0

CI:E~Iv

=0

i=l,3, .. ·,2k+l

(B.14)

i = 0,2, ... ,2k

(B.15)

~BIB;

(B.16)

~C2,CI

(B.17)

Returning to (B2, B.3) we have

+ :Eb1A:i = A2,:Ebl + l:b2A3 =

A2l:b2

Then premultiplying (B.l6) by (B.l4, B.15) gives "' 2ubiV = ub2 A*"'; A *"'i+l 2L..bl V

=

~

v*:E

hi> and postmultiplying (B.l7) by :Ei 1v

"'i+l V A )L.Jbl

"'

~L.Jb2

i = 1, 3, ... '2k

"'; A 3uhJV

+1

i = 0,2, ... ,2k

and using

(B.l8) (B.l9)

Next we can rewrite (B.l8) as (B.20)

and (Bl9) as I "' 2 :Eblv L..b2 A*[

3 :Eblv

···

"'2 AM 2k+l v ] = ~ub2 3 :Ebl

(B.21)

where :Eb2 was used to premultiply (B.l9) and M was defined previously in (B.l2). Now since the left sides of (B.20, B.21) are equal we have (B.22)

282

Appendix 8: Reduced Order Model Stability

Moreover we can use (B.l2) to rewrite the expression on the left side of this equation. Doing this gives rise to the following Lyapunov equation ~Q- 2:;~2~

=0

(B.23)

where Now it turns out if Q and 2:;~ 2 do not have any eigenvalues in common then~= 0 is the only solution to (B.23). Indeed Q and 2:;~2 do not share any eigenvalues since the eigenvalues of Q were shown to be a subset of the eigenvalues of 2::~ 1 , (B.l3), and from the statement of the theorem we have that no eigenvalue of 2:;~ 1 is an eigenvalue of 2::~ 2 . Thus we have A 3M = 0. Since the columns of M consist of all the right-eigenvectors of A 1 corresponding to the imaginary axis eigenvalue,\ of A 1 , we have that (.X, q) is an eigenvalue right-eigenvector pair for the full system matrix A 6 where q

=

[~]

smce

However, since ,\ lies on the imaginary axis, A 6 is unstable. This contradicts the fact that A 6 is stable. Therefore our assumption at the start of the proof that A 1 is unstable is false and the proof is complete. •

l..

·.·.·.i

Ap pen dix C: Pro ble ms

of the The following problem s illustrate the basic ideas introduc ed in the earlier chapters ment. develop under are es principl d advance more book. Problem s dealing with

C.1 1.

Problems Relating to Chapter 1 matrix Use the fact that any matrices M 2 and M 3 have the same eigenvalues as the M 1 if

ied for any nonsing ular matrix S, to determi ne relations between the unspecif of set same the has matrix each that so 4, 3, 2, 1, = k elements in the matrices Ak : akjs. the for values the specify 3} eigenvalues. If this set is { 1, -2,

AI=

A3=

2.

["i' [!

1

a12 0

0 0

Tl ""]

A2 =

["'' !] a21

a31

A,

a23 a33

[

c

0 0

:

0

a31

a32

n

matrices Find the eigenvalues {Aki : i = 1, 2} for each of the following system for tors, Ak : k E ~1, 6;. Then calculate the correspo nding right eigenvec one. Ak : k = 1, 2 letting the first compon ent of each eigenvector be

vi

AI= [

A4=

~3 ~1]

[~~

j~]

A2

=

As=

[-1 O

3

l

-2j

[-5 -6 ~]

A3

= [ ~2

A6=

~1]

[:2 !s J

284

3.

4.

Appendix C: Problems

Using Laplace transforms, and without calculating numerical values, determine which of the transition matrices corresponding to the Aks. k E [1, 6], given in the previous question have entries dependent on only one mode, e:u. A certain system has a zero-input state response of

x(t)

=

[x1(t)] x 2 (t)

where

5.

6.

7.

Find, without using Laplace transforms, the eigenvalues and eigenvectors of the A matrix for the system's state model. Using the trajectories obtained when the initial state is an eigenvector, sketch the state trajectory for the given initial states x;(O) : i = L 2 and system matrix A. Use your result to determine a relation between the elements of C, namely, c 1, and c2 , so that the system output is bounded for all initial states.

Calculate the eigenvalue left-eigenvector pairs,{.\. 11i: i E [L 2]}, for A given in the previous question. From this result determine an output matrix C such that for any initial state the zero-input response y(t) depends only on the mode e~ 21 • What is the relation between the C matrix you obtain here and the C matrix you obtained in the previous question. Suppose a state model has system matrix, A, which is partially specified as

If the last component of each right-eigenvector is 1, find: (i) the initial state, x(O), which gives a zero input response of

y(t)

=

when C = [ -1

4e- 1

l

1

(ii) the output matrix C which gives a zero input response of

y(t) 8.

=

when x(O) = [ 1 ]

4e- 1

-2

Recall that if ¢(t) is a transition matrix then

(i)

0(0) =I

(ii)

O(t) = Ao(r)

If M(t) is a transition matrix Aflt) = ~]

' , find the

et;s

3

[

I

+2

-21

ole e ct3et - e-2r

a 2ei a4e'

-,

~e

-c l

+ e-2t J

and the corresponding system matrix A.

functions Determ ine diagon al form state models for systems having transfe r

9.

GI (

3s- 1 s) = -,s2'+---:3,--s_+_2

Gz(s)

=

s 2 - 3s+ 2 2 3 2 s + s+

10. If the transfe r functio n for a certain system is s+4 G(s)

= (s + l)(s + 1 + j)(s +I- j)

system where find the real parame ters a 1, a2 , a 3 , c1 , c2 , c3 in the state model for this

D=O

that the state goes to 11. Use the eigenvectors of A to find B (other than the null vector) so the initial state is and the origin with time when the input is a unit impulse, u( t) = c5( t) null. It is known that one of the eigenvalues of A is at + 1 and

, u(t) 12. Ifthe steady-state output is zero when the input is a unit impulse when b and b for 3 values 2 (withou t using Laplac e transfo rms) the

-1

4/3] 8/3

0

3

-3/2

c = [1 C.2

1]

=

c5(t), find

D=O

Problems Relating to Chapter 2

a system matrix, A certain plant is known to have a control ler form state model with for the plant is A, which has eigenvalues at {-1, 1,2,3}. Furthe r, this state model steady state plant's known to have direct feedthrough matrix, D, which is zero. If the x(O) = 0, and 1 = output is y( oo) = 5 when the plant input and initial state are u( t) find the control ler form state model for the plant. If the state model for 2. Suppose we are given the transfe r functio n G(s) for some plant. in each case, so that xi(O), state, the plant is in control ler form, determine the initial

1.

286

Appendix C: Problems

the correspondin g output, y;(t), is as specified. The input is zero.

(s+4)(s+5 )

G (s) = . .,.s-+-'-1-) .( . .,.s-'. (---!--'-2..,...). .,.s-'---~--3...,-) .(

3.

Given that .\[A]= {I, 2, -3} determine left-eigenvectors of A, w;, assuming the first entry in each left-eigenvector is one. Use your result to find a and (3 in B so that x( oo) is null when u(t) = 8(t) and x(O) = 0

A= [ 07 0l 0]1 -6

4.

0 0

Suppose there is a set of three state models each having the same A matrix but different B and C matrices as given. Determine which models if any can be transf2rmed to controller form. Do this without using Laplace transforms. case(i)

A=[; ~] 5.

6.

~1]

c1 = [1

1J

1]

case(ii)

B2 = [

~]

c2 =

case(iii)

B3 = [

~]

C3 = [ -4

[2

3]

Specifyfore achstatemod el,{(A,B;,C ;): i= 1,2,3},given intheprevio usproblem: (i) the controllable eigenvalues, (ii) the uncontrollable eigenvalues. In addition. specify which of these three state models is stabilizable. Given the state models

CI=[1

7.

B! = [

0 0

l]

find, without using Laplace transforms, a coordinate transformati on matrix T in each case so that the state models in the new coordinates are each in controller form. Specify T and the controller form state model, C4, B, C) in each case. Use your knowledge of the controllable decomposed form and the eigenvalues of lower triangular matrices to rapidly determine the polynomials PI (s)

= det[s/- A]

P2(s)

= Cadj[s/-

A]B

Pro'bfiiiiilflfiilatlng 16 Cftapter 3 if the state model paramet ers are 0

7

-6

0

0

0

0

0

0

0

0

0

0

0

-2

-3

-2

0

A=

5

3

4

-4 6

8.

0

0

D=O

2

-1

C= [3

0

B=

zeros, i.e., Determi ne the transfer function which is free from coincide nt poles and whose numera tor and denomi nator are coprime . = 5 for all Determi ne B so that the steady state output is y(oo) = 10 when u(t) positive t and

+2 A=

r+1

0 0]

C=[1

-2

C.3 1.

Problems Relating to Chapter 3

Den(s), of Find the numera tor polynom ial, Num(s), and denomi nator polynom ial, model state g follmvin the to nding the transfer function G(s) correspo

-5 A=

0

0

0

-10

0

-10

0

0

-4

0

0

0

0

OJ

C= [1

0

0

B=

4

0

D= 1

where

G(s) = Num(s) Den(s) matrix but Suppose there is a set of three state models each having the same A can be any if models which ne Determi given. as matrices C differen t B and

288

Appendix c:

Problems

transformed to observer form. Do this without using Laplace transforms.

A=[: ~]

case(i)

Bl=

case(ii)

B2

case(iii) 3.

4.

=

[-;1]

c1

[!]

c2 =

B3 = [

~]

c3

=[I [2

1] 1j

= [ -4 3]

Specify for each state model, {(A, B;, C;) : i = 1, 2, 3}, given in the previous problem: (i) the observable eigenvalues, (ii) the unobservable eigenvalues. Specify which of these state models allow the design of an observer to estimate their state . Suppose a plant state model is in controller form and is 3 dimensional with A having eigenvalues at 1,2,-3. If the C matrix in this model is partially specified as

-2 q]

C =[I

determine values for q which would make it impossible to design an observer which could be used to obtain an asymptotic estimate of the plant state.

C.4 1. 2.

Problems Relating to Chapter 4

If every entry in a square matrix is either 1/2 or -1/2 determine the size of the matrix and the sign pattern of its entries if the matrix is orthogonal. Find real numbers a, b, c 1 , c2 , and c3 such that M is orthogonal where a

= [a

M

a 3.

4.

b 0 -b

c1 ] c2

c3

M 1 and M 2 are square matrices having eigenvalue- right eigenvector pairs as follows:

forM 1:

{0,1}

and

0,2(.[2-1 (}

for M 2 :

{0, 1}

and

{[1,2]T.[2

1]T}

respectively. Is either of these matrices symmetric? If so which one. Justify your answer. For each matrix, specify which of the following properties holds: positive definite, non-negative, indefinite, i.e., not positive definite. not nonnegative. not negative definite. (Hint: use the distinguishing properties of the eigenvalues of each type of matrix) l'vfl =

[ 1 -2] [] 0 0] -2

0 0

5

2 0

0 3

M2 =

[~

~] [01 00 02] 2

0

I

. Jl --~~

5.

Which of the following matrices could be either a controllability or an Gramia n. Give reasons for your answer in each case.

[1-2]

MI=[l 2] 2

observability

2

-1

-1

The observability Gramia n for a system is determined to be

6

w

= [

2 -4]

-4

0

8

the input, Is there an initial state vector, x(O) other than the null vector, which when justify. If If"no" zero? always is which y(t), u(i), is always zero, produces an output, has this which vector, null the than other "yes", give an example, i.e., give an x(O), property. ed such that 7. If, in the previous problem, the initial state vector is restrict e for the possibl value largest the find zero), xT(O)x(O) = 1, (and u(t) is always ion. restrict this under y(t) signal output the integral from 0 to oo of the square of value. this e produc would which Specify x(O) (subject to the restriction) te the observability 8. Determine, in each case, if the pairs (A, C) are observable. Calcula n in each case. equatio ov Lyapun riate Gramia n manually by solving the approp definite. Relate positive is n Gramia onding Determine, in each case, if the corresp bility in each observa the on s finding your to your findings on the positive definiteness case

9.

case(i):

A=[-10 -22]

C =[I

I]

case(ii):

A=[-10 -22]

C= [0

1]

that the r;s are Determine the observability Gramia n for the given system assuming real.

c=

[0

0

1]

D=O

r;s make the What values of the r;s make the Gramia n singular? What values of the . explain so if related? system unobservable? Are these values

Appendix C: Problems

290

10. A certain plant has state model matrices (A, B) and observability Gramian W 0 given as

w

0

=

[31 21]

Given that D = 0 and recalling that the integral defining W o does not exist when A is unstable, find the transfer function of the plant using the Lyapunov equation relating A, C, W 0 to find a and C. 11. Suppose that a certain system is known to have maximum output energy equal to 2 0. Suppose the for all possible initial states satisfying x~(O) + x~ (0) = I, \vhen u(t) this system that suppose addition In ~]. [~, = 0) ( xT is this achieves initial state that satisfies also which state has minimum output energy equal to I for some initial xt(O) + x~(O) = 1 when u(t) = 0. Find the system's observability Gramian and the initial state that gives this minimum output energy.

=

C.5 I.

Problems Relating to Chapter 5

An observer based controller (Fig. 5.1) generates a feedback signal,f(t), from

f(t)

=

Hz(t)

i(0 =Fz(0 +G 1v(0+G2y(0 where

2.

If the plant output equals the first component of the plant state \ector, find: (i) the closed loop transfer function from v(t) to y(t), (ii) the characteristic polynomial for the observer. If in the previous problem the plant state model is in controller form and the plant output is the sum of the first three components of the plant state, determine the closed loop transfer function from v to y for the observer based feedback control system where 4

3.

0

-4

0

7

-13

0

13

If in the observer based feedback control system shown in the Figure 5.1 we are given the transfer functions from v(t) to y(t), Gvy(s), and from y(t) to f(t), Gyr(s), where

f(t) is the feedb ack signal, deter mine the contr oller when GV\(s) . ·

1

~-= -3~~-2 + 2s + 1

s

+ 3s

state mode l matri ces (Q, R, S, T)

-2 ' 3s-? + 2 s + 3 Gyr(s · ) = s·,

with plant mode l param eters given as

and contr oller equa tions

i(t) = Qz(t)

+ Rv(t) + Sy(t)

f(t) = T::(t)

Appe ndix D: MATL AB Exper iment s

0.1

State Models and State Response

The intention of this lab is to introduce some of the ways MATLAB can be used in connection with state models and to provide further experience in thinking in terms of the state space.

D.1.1

Controller form

Given the transfer function model for some system, the command

[a, b, c, d]

= tf2ss(num. den)

can be invoked to find an equivalent state model, (use the help facility), where num and den are entered as row vectors containing coefficients of the numerator and denominator of the transfer function. Try the following examples 1

Gl (s) = s2 G2 (s ) =

+ 3s + 2

s-7

+ 2s +-i

---co-----

3s2

+ 2s + 1

Check your result by using the command "ss2tf', (use the help facility), to convert back to the transfer function models.

D.1.2

Second order linear behavior

Recall that the state at any time is represented by a point in ann-dimension al space and the path traced out by this point is referred to as a state trajectory. Unlike cases where the system order is greater that 2 , we can readily display the trajectory for a second order system as a graph with the two components of the state forming the axes of the graph.

~~-"'-""'""

294

~·- '-_

...,.,-.

Appendix D: MATLAB Experiments

Consider the system whose behavior is governed by the following second order linear differential equation

where

Then the controller state model is

A

= [ - 2~wn

-;~ ]

C= [0 w~]

B

= [ ~]

D=O

Now we are going to investigate the zero-input state response by obtaining the state trajectories for one value of natural frequency, say wn = 1 and different values of the damping ratio, say

( = 0.3, I, 10,-0.3 each for some initial state, say xT (0) = [2 0] One way to do this is to invoke the command, (use the help facility), initial(A, B, C, D, xO)

Then from the resulting graph of the output vs. time that is obtained on the screen, you can decide over what interval of time you will need the state. Suppose you decide you need to know the state in the time interval [0. 5] seconds every 0.5 seconds. We can create a vector, t, of these times by incorporating this data in a vector as start:increment: end. This is done here by typing t =

0: 0.5: 5

and

[y, x, t]

=

initial(A, B. C, D, xO)

This produces a column y, followed by two columns. one for each component of the state. The entries in these columns give the values of the output, and components of the state at 0.5 s intervals from 0 to 5 s. Now you can obtain the graph of the state trajectory by entering

plot(xi:.l),x(:. 2)) where x(:, i) is all rows in the ith column of x. Your graph can be labelled using the

295

same line procedu re (on line help) where you enter each line (there are four) on the separate d by commas . Explain why these plots are markedl y different for different signs on(.

D.1.3

Secon d order nonlin ear behav ior

nonlinea r Consider the system whose behavio r is governe d by the second order , equation Pol der van a as to differential equation , referred /

2

l(t)- (1 -/(t))y (ll(t)

+ y(t) = 0

manner Then if \Ve assign compon ets of the state to derivates of the output in the same of the ents compon the of r behavio the case, linear the in form er as done to get the controll equation s state are seen to be governe d by the following first order nonline ar differential

x1 (t) = (1- x~(t))x 1 (t)- x 2 (t) x2(t) = x 1 (t) ned using Now the state trajecto ry obtained for a specific initial state can be determi and type editor an open file this create MATLA B by creating an m-file called vdpol.m. To

functionyp yp

= vdpol(t, q); = [(1- q(2) * q(2)) * q(l)- q(2); q(l)];

solvers, say Save the file and invoke one of the ordinary (vector) differential equatio n ODE23

[t, q] = ode23('vdpol', [tO,({], qO) as a column where tO is the start time, if is the finish time and qO is the initial state, entered vector by typing

of q. The sample values of the compon ents of the state are returned as the columns states initial the of each for state the getting try Use tO= 0, if= 25 and

x(O)

=

[0.2] 0

and

[~]

Obtain a graph of the state trajecto ry in each case by typing

plot(q(:, I). q(:. 2)) on as Notice that q( :, 1) is all the elements in the first column of q produce d by the simulati state. the of ent compon first the values of the

296

Appendix D: MATLAB Experiments

In the second case where xT (0) = [2 0] it is instructive to concentrate on the latter part of the trajectory, say from time step 40 to the end of the simulation. To find out what this end valt:e is type

size(q) Then if this end value is 200 type

plot(q(40: 200, l),q(40: 200,2)) to obtain the trajectory from time 40 to time 200. Generalize the nature of the trajectories in terms of their dependency on the initial state. Compare the state trajectories you obtain for the present nonlinear system with those you obtained for the linear system in the previous section.

D.1.4

Diagonal form

So far we have only considered the controller form state model. There are several other important canonical forms for the state model of a linear dynamic system. One of these is the diagonal or normal form. This form is characterized by having its system matrix A which is diagonal. This simplifies solving the state differential equation since the derivative of each component of the state is dependent on itself only. Recall that if [A, B, C, D] is a state model for a given system so is [Q- 1AQ, Q- 1B, CQ, D] for any nonsingular matrix Q of appropriate size. where Q is a coordinate transformation matrix. Also recall that if the eigenvalues of A are distinct, we can change coordinates so that A in the new coordinates is diagonal. This is done by letting Q have columns equal to the eigenvectors of A. We are going to use these facts. Use MATLAB to determine the parameters, (A. B, C, D), of the controller form state model for the system having transfer function

G(s)=

2

s -s+20 s 4 + 5s 3 + 5s 2 - Ss - 6

Use the command ·'roots" to decide if all the roots of the denominator are poles of G(s). Calculate the eigenvalues and eigenvectors of A by typing

[Q, E]

=

eig(A)

and the characteristic polynomial of A by typing

poly( A) How do the poles of G(s) compare with the eigenvalues of A, i.e .. diagonal entries in E? Should each of these sets be the same and if not why not? Use Q to transform the controller form state model to a diagonal form state model.

..j. ) ~·/'

Feed back and Cont rolla bility

297

This can be done by typin g

*A *Q inv(Q ) * B

aa = inv(Q )

bb =

cc = CA

*Q

dd= D

the poles of G(s)? How do the diago nal entries of aa comp are with mine the residues, poles and any deter to due" "resi Finally, inYoke the comm and r). This is done by typin g const ant term D (nonz ero only if G(s) is prope [res,poles, d]

= residue(num, den)

s in bb and cc? How do the residues of G(s) relate to the entrie

0.2

Feedback and Controllability

with the tasks described in what follows, In order for you to proce ed as rapid ly as possible us matrices you will be enter ing in the you shou ld plan a subsc riptin g scheme for the vario computer.

D.2.1

Controllable state mod els

Enter the state mode l 5

9

-6

3

0

-3 0

3 2

0

0

0

0

4

a44

0

b2 b3 b4

0

0

5

7

-6

hs

2 0

A=

c=

[1

B=

D=O

1]

s with the following assignment to the variable entrie

bi = 1 : i = 2, 3, 4, 5

and

function using the comm and "ss2z p". Obta in the poles and the zeros of the trans fer that part of the state space is superfluous Are there any pole-zero cancellations indicating to the input -outp ut beha vior? comm and "ctrb " and check its rank Next obtai n the contr ollab ility matri x using the using the comm and "rank ". of A. Then find the state feedback gain Using the comm and "eig" find the eigenvalues s to -1. Try doing this with both the matrix K which shifts the unsta ble eigenvalue Check using the comm and "eig" to see comm and '"place'' and the comm and "ack er". m matrix A + BK are those you specified. that the eigem alues of the state feedback syste

298

Appendi x D: MA TLAB Experim ents

Obtain the poles and zeros of the state feedback system using the comman d "ss2zp". Are there any pole-zero cancellations? What effect does the state feedback have on the transfer function zeros? Do other experiments if need be so that you can give a general reasons for your answers to the foregoing questions. Obtain the feedback matrix which produces a closed loop transfer function which has no finite zeros (they are all cancelled) and poles which are all in the open left half-plane. Does the rank of the controllability matrix change. Explain your answer by using eigenvectors to identify uncontro llable eigenvalues.

D.2.2

Uncontrollable state model s

In this section we are going to examine the use of state feedback when the given state model for the plant is not controllable. One way to obtain an uncontro llable system to enable this study, is to form the input matrix B so that it is linearly dependen t on a subset of the eigenvectors of A. We do this as follows. Invoke "eig" to get the eigenvectors of A. These will be displayed in a square matrix you specify in the argumen t for "eig". Suppose this matrix is V and Dis used as the diagonal .matrix displaying the eigenvalues. Create a new input matrix which we call Bl by combining a subset of the columns of V. i.e., some of the eigenvectors of A. Suppose we do this by combinin g all the eigenvectors correspo nding to the unstable eigenvalues of A with all the weights used to combine these eigenvectors equal to 1. This can be done using the MATLA B subscripting rules (note the use of:). Obtain the controllability matrix ("'ctrb") for (A, B1) as well as the rank of this controllability matrix. Obtain the transfer function for the state model (A, Bl, C, D) and comment , with an explanation, on any pole-zero cancellations. What is the order of the system indicated by the transfer function and does the transfer function indicate that this system is stable? To gain more insight into the nature of uncontro llable state models. invoke "ctrbf' to transform coordina tes so that the state model is in controlla ble decompo sed form, (Section 2.5). In this case we see that the block matrix consisting of the last two rows and columns of A and the last two elements of Bin these coordina tes has special significance. If(Aban Bhar) are the system and input matrices in the controlla ble decompo sed form then we can extract the subsystem correspo nding to these rows and columns by typing AC = AhaA4: 5,4: 5) BC = Bhar(4: 5, 1) CC = Cbm·(1,4: 5)

Then calculate the transfer function using the command ·'ss2zp" for the system (AC, BC, CC) and compare it with the transfer function you obtained earlier for the system in the original coordinates. Explain what you see. Next can you find a state feedback matrix Kin these coordina tes so that the resulting system is internally stable? Repeat the foregoing by making up a new B matrix from a linear combina tion of the eigenvectors correspo nding to the stable eigenvalues of A. Notice that the new AC matrix will be a 3 x 3.

Obse rver Based Control Systems

299

Controllability and Repeated Eigenvalues

s are uncontrollable for any Recall that we showed in Appendix A that single-input system ector associated with eigenv one than input matrix B when the system matrix A has more s. follow as d procee we fact any repeated eigenvalue. To investigate this A. Then use typing by A recall can We lab. this of Assume A is as given at the beginning eigenvectors of A. Are the the comm and "eig" to determine the eigenvalues and typing eigenvalues distinct? Next change the 4,4 element in A by A(4,4 ) = A(4,4 )

+3

matrix. Are the eigenvalues Find the eigenvectors and eigenvalues of the new A input matrix B so that the an ruct Const ? distinct? Is there a complete set of eigenvectors new (A, B) pair is controllable. Explain.

0.3

Observer Based Control Systems

the plant state from measureRecall that.a n observer is used to provide an estimate of will examine the design and ments of the plant' s input and outpu t. In this lab 'YOU estimate of the plant state in behavior of observers and the effect of using an observer's l system. place of the actual plant state in a state feedback contro the tasks described in what In order for you to proceed as rapidly as possible with e you will use in connection follows, read the lab throug h and plan the subscripting schem computer. This is especially with the various matrices you will need to enter in the you will need to make a block impor tant in connection with the second section where t numb ered for use with the diagram of the contro l system with each input and outpu commands "blkb uild" and "conn ect". Observer Estimates of the Plant State

[To Be Done Prior to Using Computer] Suppose you are given the unstable second order plant

for the plant and with a 1 = 0 and a2 = -1. Write the observer form state model observer (Section 3.3)

design an

i(t) = Fz(t) + c[u(t )] y(t) where F=A +LC

G= [B+L D

-L]

with F having eigenvalues at -1, -2. TLAB that the observer state Now you will attem pt to verify experimentally using MA alues of the observer and eigenv and plant state appro ach each other with time. Use the

300

Appendi x D: MATLAB Experime nts

plant state model to determine a suitable length of time over which to let the simulatio n operate. [To Be Done Using Compute r] Begin by entering the observer form plant state model matrices: A. B, C, D. Then use the comman d "acker" or "place" to determine the observer matrix L so that the observer eigenvalues are at -1 ,-2. This is done by typing

where

pis a column vector of the desired observer eigenvalue locations. Construc t F and G and check the eigenvalues ofF using the comman d "eig". Now we are going to check

this observer's ability to track the compone nts of the state of the plant by generating both the plant and observer states when the plant input is a unit step. Invoke the comman d "step" to obtain the plant output, y and state x. Try using a time interval of ten seconds in steps of 0.1 s. In order to do this you will need to create a row vector t having elements equal to the computa tion times. This can be done by typing t=O:O. l: 10

You will need t in the last section of this lab. Next use the comman d "1sim" to obtain the state of the observer when the plant input is a step andy is the plant output in response to a step. This can be done by typing [yl,zl]

= !sim(F,G ,CI>DB, UB,t)

where UB = [uu,y]

DB=[O;O]

and uu is a column vector of 1's of the same length as y withy being a column vector of samples of the plant output you just compute d. The input sequence. uu. could be generated by typing uu

=

ones(IOL 1)

Once you have z l, the state of the observer, (a matrix having two columns of samples of the compone nts of the observer's state) you can compare the observer state, zl. just obtained , with the plant state x by generating the error err=(z l-x)

However recall that since the plant is in observer form, the plant output is the first compone nt of the plant state. Therefore we need only compare the second compone nts of the plant and observer states in order to characterize the performa nce of the observer as a state estimator . In addition in attemptin g to commen t on this performa nce, you might find it helpful to examine the relative error, i.e., z1 (:. 2)- x(:, 2)/x(:, 2)

j

Obser ver Based Contro l System s

301

type For instanc e if we want to compu te this for time 10 we would [zl(lO l, 2)- x(IOI, 2)]/x(I OI, 2) the plant output is null. Now the initial state of the plant in the foregoing genera tion of plant. Theref ore to test the However in practic e we do not know the initial state of the state of the plant and observer's ability to estimate the plant state when the initial state was xT(O) = [1, 1]. observer are not equal, suppos e you though t the initial plant ption, use the comm and Then setting the initial state of the observ er equal to your assum "lsim" with the observer, i.e., type zO

= [1, 1]

[y2, z2] = lsim(F , G, C, DB, UB, t, zO) tion time and obtain a Next obtain the history of the estima tion error over the simula n by typing divisio t elemen by t elemen plot of this history. This can be done using rele

= (z2- x).jx

ts in the first row of the Notice that since the actual initial plant state is null. the elemen (not a numbe r) or oo as its matrix xs are zero, so that the first listing in "rele" has NaN state, e.g., set the initial first entry. Try other guesses for the "unkn own" initial plant ng relative error in the observer state to zO = [100, 100] and obtain a plot of the resulti estimate of the plant state.

D.3.1

Obse rver base d cont rolle rs

[To Be Done Prior to Using Compu ter] section to implem ent a Suppose you want to use the observer studied in the previous study the interac tion which state feedback contro ller. In this section you will begin to connec ted via a feedback takes place between the plant and the observ er when they are matrix K in a closed loop configuration. were used the closed loop Choos e the feedback matrix K so that if exact state feedback model in observer form system would have eigenvalues at -3, -4. Do this for the plant state which you used in previo us sections. the feedback system, i.e., Next referring to Section 5.2, obtain the system equati ons for q for the closed loop system let the plant input u be given as u = Kz + v. The state vector has four entries, viz.,

state for the observer you where xis the state for the plant (in observer form) and:: is the designed in the previous section. [To Be Done Using Compu ter] the plant state model Use the comm and "acker " to find the feedback matri:-; K so that the observer based denote Then -4. -3, (in observer form) has its eigenvalues shifted to

302

Appendix D: MA TLAB Experiments

controller state model as

where F,G, were determined in the previous section and K was just determined. Now you are going to use the commands "blkbuild" and .. connect" to form the observer based feedback control system by interconnecting the various subsystems involved. This is done in two steps. First enter each subsystem, either as transfer functions, (nidi) or as state models (ai. bi, ci, di). In each case i should identify a different subsystem and should be the integers starting with one. In the present case there are three subsystems: the input-controller (i = 1), the plant (i = 2) and the observer-controller (i = 3). The input-controller is trivial and is done by entering its transfer function as nl = 1 and dl = 1. Then the plant and observer-controller can be entered in terms of their state models as [a2,b2,c2,d2] and [a3,b3,c3,d3] respectively. Once you have entered all the subsystems, type nblocks = 3 blkbuild

A message appears "state model [a. b, c, d] of the block diagram has 4 inputs and 3 outputs". You are now ready to proceed to the second step which involves specifying how to interconnect the subsystems you have just identified. This is done through the use of a connection matrix, Q. The connection matrix, Q, specifies how the subsystems inputs and outputs are to be connected to form the observer based control system. Notice that the outputs r 1, y2, and y3 are the outputs from the input-controller. the plant and the observer-controller respectively. The inputs ul, u2, u3 and u4 are the input to the input-controller, the plant and the two inputs to the observer. It is important to keep the ordering of the inputs to the observer consistent with the ordering used in the previous section in forming the state differential equation for the observer. Draw a block diagram showing how all inputs and outputs relate to one another. The connection matrix, Q, has rows corresponding to subsystem inputs with the integer in the first position of each row identifying the subsystem input. Subsequent entries in a given row are signed integers which identify how the subsystem outputs are to be combined to make up the subsystem input. Zero entries are used when a subsystem output does not connect to a subsystem input. The sign associated with each integer entry in Q indicates the sign on the corresponding output in the linear combination of subsystem outputs making up a given input. Since there are three inputs which depend on the subsystem outputs, Q will have three rows. Using your block diagram, enter the matrix Q. Once Q is entered, the inputs and outputs of the feedback control system must be declared by typing inputs= ·1 outputs = [2]

rol er, i = 1, is the obse rver base d feed back cont where here the inpu t to the inpu t-con troll rol cont t, i = 2, is the outp ut from the feedback system inpu t and the outp ut from the plan system. ectio n info rmat ion in the Q matr ix to form At this stage you are read y to use the conn m m from the mult i-inp ut mult i-ou tput syste the observer base d feedback cont rol syste the i.e., de], cc, be, [ac, m The conn ected syste [a, b, c, d) you form ed using "blk buil d". obta ined by typing is m, syste rol cont observer based feedback

ts, outputs) [ac, be, cc, de] = eonneet(a, b, c, d, Q, inpu obta ined in the penc il-an d-pa per work you Check your result with the state mod el you did before start ing this com putin g session. model. calculate the poles and zeros of the Once you have the corr ect closed-loop state side m by using ·'ss2 zp'' (don 't use k on the left transfer func tion for the closed-loop syste s matrix). Are there any com mon pole and if you have alrea dy used it for the feedback of trans fer function with the trans fer function zeros indicated? If so, why? Com pare this uss Disc state feedback (wit hout the observer). the closed loop system assuming exact ms. syste rol cont d base rYer obse of theo ry your obse rvati ons by relating them to the el by mod state loop d close the of ility rvab Finally, check the cont rolla bilit y and obse and com putin g rank s by using "ran k". using the com man ds"c trb" and "obs v"

0.3 .2

avi or Ob ser ver bas ed con tro l system beh

(To Be Don e Prio r To Using Com pute r] for on, and assuming the state s are available For the plan t intro duce d in the first secti state with m syste inpu t for the closed loop feedback, calcu late the outp ut for a step l has poles at -3, -4. Do this for the initia m syste feedback such that the closed loop or asef achc tine earo ugh sket chof theo utpu conditionxO = [O,O]andxO = [10, lO]. Mak using MAT LAB . in com paris on with the results you will obta [To Be Don e Usin g Com pute r] d ut of the state mod el of the obse rver base Using "lsim " obta in the state and outp d" buil "blk g usin prev ious section as a resu lt of cont rol system that you obta ined in the ofO. l Do this over a ten second inter val in steps followed by '·con nect ", i.e., [ac, be, ec, de]. of part first the uu, which you created for use in s (use the t vect or and step inpu t vector, when 10] 0, [1 the plan t state is x(O) = [0, 0] and this lab). Obta in a plot of the outp ut when the observer state is zO = [0, OJ in each case.

0.4

State Mo del Reduction

estimation we saw that cont rolla bilit y and In previous labs on state feedback and state n to (i) assign feedback system poles (ii) assig observability play a key role in being able the tion model of a system has orde r less than observer poles. Recall that the trans fer func ntro lstate model is unobservable and/ or unco dimension of the state model when the ost" vior of a given state model which is ''alm lable. Ther efor e the inpu t-ou tput beha state a ld be able to be appr oxim ated using unco ntro llabl e and; or unobservable shou dimension of the given state model. This model whose dime nsio n is less than the to do in Cha pter 4. In this lab we are going possibility was intro duce d theoretically

some simu lation studies to furth er investigate the effectiveness of this model order reduc tion technique.

D.4.1

Dec omp osit ion of unc ont roll able and /or uno bse rvab le systems

Recall that we can always trans form the coord inate s so that a given state model that is not contr ollab le is trans form ed to contr ollab le deco mpos ed form Ac= [ Acl Ac3

cc =

[eel

0]

Ac4

Be=

[:J

c,2J

where (Ac2, Bc2 ) is contr ollab le (done using "ctrb f"). Alternatively if the given model is unobservable we can chan ge coord inate s so the state mode l becomes

where (A 04 , C02 ) is obser vable (done using "obsv f"') Ther efore when a given state mode l is both unco ntrol lable and unob serva ble we can use the foregoing deco mpos itions to obtai n a state mode l which has a state space which is divided into four subspaces. In the following list of these subspaces C, C indicates contr ollab le and unco ntrol lable , respectively with a similar notat ion for observable and unobservable. Thus the state space splits up into 1 a subsp ace CO of dimension n 1 2 a subspace CO of dimension n2 3 a subsp ace CO of dimension n 3 4 a subsp ace CO of dimension n4 Notice that if the dime nsion of the given state model is n then

An example demo nstra ting the use of MA TLA B to do this deco mpos ition can be carrie d out as follows. Ente r the system state mode l matrices

[ AI " Cl= [O

I :

5

7

-2

0

0

-3

0

0 0

!.] 1l

Iii

[J'

Dl = 0

r

State Model Reduction

305

and determine the corresponding transfer function by entering [zl,pl,kl] = ss2zp(Al,Bl, Cl,Dl)

f_

What is the order of the system? What can be said about the controllability and observability of the state model? We can use the eigenvector tests for controllability and observability to answer this question by interpreting the result of making the following entries [Vl,DRIJ = eig(AI) Cl *VI

and [WI,DLl] = eig(AI') Bl' *WI

As a further check on your results, obtain the rank of the controllability and observability matrices using the commands "'ctrb". "obsY" and '·rank". Use your results to specify basis vectors for each subspace in the decomposition of the state space.

0.4.2

Weak controllabi lity and/or observabil ity

Enter the following system state model matrices

A2~

-1

5

7

:

-2 0 0

0 -3 0

[

I]

.01

C2= [ 0

D2 = 0

and repeat the preyious section for this state model. Contrast the results you obtain now with those you obtained in the previous section. Next obtain the controllability and observability Gramians, (f·Vc, W 0 ), for the system you just entered. Do this by typing

QO = C2'

* C2

QC

= B2*B2'

WO

= lyap(A2'. QO)

we= lyap(A2, QC) (Use "'help" for information on the command '·lyap".l Obtain the eigenvalues of the product of the Gramians. Record these values for future reference. Next use the command "balreal" to obtain a balanced realization by typing [A2b, B2b, C2b,g2, t2]

= balreal(A2, B2, C2)

306

Appendix D: MATLAB Experiments

Determine the eigenvalues of the product of the Gramians you just entered and compare them with the entries in g2. Comment on any correspondences. Next obtain a reduced order model by discarding parts of the balanced realization correspondin g to elements of g2 that are relatively small. For example suppose g2(3) < < g2(2). Then since the elements of g2 are arranged in descending size we can obtain a reduced order model, (A2r, B2r, C2r) by typing A2r

= A2b(l : 2, I : 2)

B2r = B2b(l : 2) C2r = C2b(l : 2)

Use use the command "ss2zp" to determine the poles and zeros of the transfer function for the reduced order system. Is there any obvious relation between the original system transfer function and the reduced order system transfer function?

0.4.3

Energy interpret ation of the controlla bility and observab ility Gramian s

Recall, from Chapter 4, that the sum of the diagonal entries in the controllabili ty Gramian equals the energy transferred into the state over all positive time, from the input when the input is a unit impulse at time zero. We can check to see if this is the case here. Recall from Chapter 1 that the zero state response to a unit impulse is the same as the zero input response to an initial state x(O) =B. Therefore we can use "initial" with the initial state set to B. As before we need to decide on the time interval and sample times for doing this. Suppose we use an interval of 10 s with samples taken every 0.1 s. Then we create a vector of sample times by typing ts=O:O.l: 10 Notice that ts has 10 I elements. Therefore using the command "initial" with B2 in the position reserved for the initial state, we can obtain the resulting state by typing [Y2, X2, ts] = initial(A2, B2, C2, D2, ts)

where X2 has four columns with each column correspondin g to a different component of the state and each row correspondin g to a sample time. Recall that the energy transferred to the state is given by

where

State Model Reduct ion

307

fact we can is the energy transfe rred to the i1h compo nent of the state. Using this an approx imation , EAc, to Ec by enterin g

compu te

eei=X 2(1: 10l,i/* X2(1: lOl,i) with i = 1, 2, 3, 4 in succession and then enter 1

EAc

=

4

-L:ee i 10 i=l

by typing Compa re your result with the trace of the control lability Gramia n

ETc= trace(WC) output from an Recall, from Chapte r 4, that the energy, E 0 , which is transfe rred to the initial state can be determ ined using the observability Gramia n as

te E01 when the Now we can use the proced ure just described for compu ting Ec to compu Y2. Compa re your initial conditi on x(O) = BI by using the data stored previously in result with the result obtaine d by using the observability Gramia n (B2)'

D.4.4

* WO * B2

Design of reduced orde r models

ch to model order In this part of the lab you will examine the balance d realization approa filter. This analog orth reduction. The system to be reduced is a degree eight Butterw of degree inator denom a and system has a transfe r function with constan t numera tor origin. A the from e distanc eight. The poles are located in the open left half-plane at unit list of the poles can be obtaine d by enterin g

[zbw,pbw, kbw] = buttap(8) ing the We can check that the poles are equidis tant from the origin by examin elements of the matrix obtaine d by enterin g

diagon al

pbw * pbw' is obtaine d by first A balance d realiza tion (state model) for the Butterw orth filter to balance d form. it rming transfo then and calculating a control ler form state model Therefo re enter [A3, B3, C3, D3] = zp2ss(z bw,pbw , kbw) and [A3b, B3b, Cb3,g3 , t3]

= balreal(A3, B3, C3)

308

Appendix D: MA TLAB Experiments

Recall that the reduced order model is obtained by discarding the latter part of the state vector in balanced coordinates so that the corresponding discarded part of g3 is negligible in the sum of the elements in g3. Notice that there is a significant decrease in the size of the elements in g3 in going from the fifth to the sixth element. Therefore a reduced order model obtained by discarding the last three elements in the state should give a good approximation to the full order system. The state model for the reduced order approximation can therefore be obtained by entering A3b5 = A3b(l : 5.1 : 5) B3b5 = B3b(l : 5) C3b5

= C3b( 1 : 5)

Notice that this reduced order state model approximation is also balance. This can be seen by generating the controllability and observability Gramians by typing QObS

= C3b5' * C3b5

QCbS = B3b5

* B3b5'

Wob5

= lyap(A3b5'. QOb5i

WebS

= lyap(A3b5. QCbS)

The final part of this lab concerns the performance of the reduced order model. One possible test would be to compare the step responses of the reduced and full order models. [y3b, x3b] = step(A3b, B3b, C3b, D3, 1, ts) [y3b5, x3b5]

= step(A3b5, B3b5, C3b5, D3, I, rs)

Note any marked differences in the plots on the screen and proceed to compare the error between the reduced order model output and the full order model output. This can be done using element by element division, i.e., rete =(y3b5- y3b).jy3b plot( rs, rele)

Repeat the foregoing by deleting more than the last three elements of the state to obtain a reduced order model and use the step response to compare the performance of this approximation with the performance of the fifth order approximation just studied.

r·

References

[I] [2] [3] [4] [5] [6] [7] [8] [9] [10] [II] [12]

Anderson, B. D. 0. and Moore, J. R Optimal Control: Linear Quadratic Methods. PrenticeHall, Englewood Cliffs, New Jersey, 1990. Anderson, B. D. 0. and Liu, Y. "Controller reduction: Concepts and Approaches", IEEE Trans. Automatic Control, TAC-34, pp. 802-812, 1989. Bittanti, S., Laub, A. J. and Willems, J. C. (Eds.) The Riccati Equation. Springer-Verl ag, Berlin, 1991. Blackman, P. F. Introduction to State-variable Analysis. The Macmillan Press, London, 1977. Brockett, R. W. Finite Dimensional Linear Systems, John Wiley and Sons. New York, 1970. Brogan, W. L. Modern Control Theory. 3rded., Prentice-Hall , Englewood Cliffs, N.J., 1991. Brown, R. G. Introduction to Random Signal Analysis and Kalman Filtering. John Wiley and Sons, Chichester, 1983. Chandrasekha ran, P. C. Robust Control of Dynamic Systems. Academic Press, London, 1996. Chen, C. T. Introduction to Linear Systems Theor. Holt Rinhart and Winston, New York, 1970. Dorato, P., Abdallah, C. and Cerone, V. Linear Quadratic Control, An Introduction. Prentice-Hall , Englewood Cliffs, N.J., 1995. Doyle, J. C., Francis, B. A. and Tannenbaum, A. R. Feedback Control Theory. Maxwell Macmillan Canada, Toronto, 1992. Doyle, J. C., Glover, K., Khargonekar, P. P. and Francis, B. A. "State space solutions to standard H 2 and Rx: control problems", IEEE Trans. Automatic Control, AC-34, pp. 831-

847, 1989. Fairman, F. W., Danylchuk, G. J., Louie, J. and Zarowski, C. J. "A state-space approach to discrete-time spectral factorization" , IEEE Transactions on Circuits and Systems-ll Analog and Digital Signal Processing, CAS-39, pp. 161-170, 1992. [14] Francis, B. A. A Course in Hx Control Theory, Springer-Verl ag, Berlin, 1987. [15] Furuta, K., Sana, S. and Atherton, D. State Variable Methods in Automatic Control, John Wiley and Sons, Chichester, 1988. [16] Glover, K. and Doyle, J. C. "A state space approach to H"" optimal control", Lecture Notes in Control and Information Science, 135, pp. 179-218, 1989. [17] Golub, G. H. and Van Loan, C. F. Matrix Computations, 2nd ed., The Johns Hopkins University Press, Baltimore, Maryland, 1989. [18] Green, M. and Limebeer, D. J. N. Linear Robust Control. Prentice--Hall, Englewood Cliffs. New Jersey, 1995.

[13]

310

[19] [20] [21] [22]

[23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44]

References

Green, M., Glover, K., Lime beer, D. J. N. and Doyle, J. "A J spectral factorization approach to Hoc control", SIAM Journal on Control and Optimi::.ation, 28, pp. 1350-1371, 1990. Green, M. "Hx controller synthesis by J lossless coprime factorization", SIAM Journal on Control and Optimization, 30, pp. 522-547, 1992. Isidori, A. "Hx control via measurement feedback for affine nonlinear systems", Int. J. Robust and Nonlinear Control, 4, pp. 553-574, 1994. Jonckheere, E. A. and Silverman, L. M. '"A new set of imariants for linear systemsapplications to reduced order compensator design", IEEE Trans. Automatic Control, AC28, pp. 953-964, 1983. Kailath, T. Linear Systems, Prentice-Hall, Englewood Cliffs, l\. J .. 1980. Kimura, H. Chain-Scattering Approach to Hx Control, Birkhausser Boston, 1997. Kwakernaak, H. and Sivan, R. Linear Optimal Control Systems, Wiley Interscience, New York, N.Y .. 1972. LePage, W. R. Complex Variables and the Laplace Tran.sformj!Jr Engineers, Dover, 1980. Moore, B. C. "Principal component analysis in linear systems: controllability, observability, and model reduction". IEEE Trans. Automatic Control. AC-26. pp. 17-32, 1981. Missaghie, M. M. and Fairman, F. W. "Sensitivity reducing obsen ers for optimal feedback cont.!;_ol", IEEE Transactions on Automatic Control, AC-22, pp. 952-957, 1977. Mullis, C. T. and Roberts, R. A. "Synthesis of minimum roundoff noise in fixed point digital filters", IEEE Trans. Circuits and Systems. CAS-23. 551-562, 1976. Mustafa, D. and Glover, K. "Controller reduction by Hac balanced truncation". IEEE Trans. Automatic Control, AC-36, pp. 668-682, 1991. Pavel, L. and Fairman, F. W. "Robust stabilization of nonlinear plants-an L 2 approach", Int. J. Robust and Nonlinear Control, 6, pp. 691-726. 1996. Pavel, L. and Fairman, F. W. "Controller reduction for nonlinear plants-an L 2 approach", Int. J. Robust and Nonlinear Control, 7, pp. 475-505, 1997. Pavel, L. and Fairman, F. W. (1998). "Nonlinear H" control: a J-dissipatiYe approach", IEEE Trans. Automatic Control, AC-42, pp. 1636-1653, 1997. Pernebo, L. and Silverman, L. M. "Model reduction via balanced state space representation", IEEE Trans. Automatic Control, AC-27, pp. 382-387, 1982. Petersen, I. R. "Disturbance attenuation and Hex- optimization: a design method based on the algebraic Riccati equation", IEEE Trans, Automatic Control, AC-32, pp. 427-429, 1987. Rudin, W. Principles of Mathematical Analysis, 3rd ed., McGraw-Hill Book Company, New York, 1976. Rudin, W. Real and Complex Analysis, 3rd ed., McGraw-Hill Book Company, Ne\\ York, 1987. Scherpen, J. M. A. "Hx balancing for nonlinear systems'', Int. J. Rohust and Nonlinear Control, 6, pp. 645-668. 1996. Strang, G. Linear Algebra and irs Applications, 2nd ed., Academic Press, Ne\\ York, 1980. Sveinsson, J. R. and Fairman, F. W. "Simplifying basic Hx problems", Proc. Conference on Decision and Control, 33,2257-2258, 1994. Tsai, M. C. and Postlethwaite. I. ''On 1 lossless coprime factorization and Hx control", Int. J. Robust and Nonlinear Control, 1, pp. 47-68, 1991. van der Shaft, A. J. "L 2 gain analysis of nonlinear systems and nonlinear state feedback Hoc control", IEEE Trans. Automaric Control, AC-37, pp. 770-784. 1992. Varga, A. "A multishift Hessenbcrg method for pole assignment of single-input system,;", IEEE Trans. Automatic. Control, AC-41, pp. 1795-1799, 1996. Vidyasagar, M. Control System Synthesis: A Factori::.ation Approach, The MIT Press, Cambridge. Massachusetts, 1985.

r

Referen ces [45] [-+6] [47]

311

onal, EngelVidyasag ar, M. Nonlinear Systems Analysis, 2nd ed., Prentice -Hall Internati wood Cliffs, 1993. Cambrid ge, 1988. Young. N. An Introduction To Hilbert Space, Cambrid ge Universit y Press, Prentice -Hall, Control, Optimal and Zhou, K. with Doyle. J. C. and Glover, K. Robust Upper Saddle River, New Jersey, 1996.

Ind ex

Ackerm ann formul a 52, 64, 74 adjoint system 221 adjugate matrix 11, 18 algebraic Riccati equatio n (ARE) 115, 119, 1-!5, !66, 224, 246 GCAR E 153 GFAR E 157 HCAR E 236 HFAR E 244 QCAR E 126 QFAR E 142 system 221 pass all anticausal, antistab le systems 172 asymptotic state estimat ion 69 average power 147 balanced realization I04 Bezout identity 204, 209 causality constra int 31 causal system, signal 172 Cayle:- Hamilt on theorem 52 characteristic polyno mial 9 co inner functio n 22 I compan ion matrix 9 comput er determ ination of state 37, 66 contrac tion 162 controllability 34, 55 Gramia n 101, 112 matrix 5 L 57, 63 control lable decomp osed form 60, 93, 118 eigenvalue 4-t subspace 56

control ler form 47 Hoc-central, parame trized 255 LQG 157 quadra tic 120 convol ution integral 29 coordin ation transfo rmation 12, 28 coprim e factoriz ation 201, 204, doubly 212 J-inner 230 state models 206 covariance 147 decomp osition of a space 169, 178 detectable system 72 direct sum 169 disturb ance 115, 120, 138, 147, 154, 225, 232, 242,24 8 dom(R ic) 145 doubly coprime factoriz ation 212 dual system 72 eigenvalue 17, 96 assignment 43, 64, 74, 82, 86 controllable, observa ble 44, 72 invariance 20 eigenvector 1'7. 96 left-eigenvector 19 right-eigenvector 16 tests (controllability, observability) 43, 71 energy 94, 102. 110, 167 feedback 41 filtering 68

314

Index

filtering (contd.) H.00 242 LQG (Kalman) 90, 155, 166 quadratic 138 Fourier transform 173 Gaussian random vector 147 Gramian controllability 10 I, 112 observability 94, 109 Hamiltonian matrices 130, 158 Hx relation 262 Hankel norm Ill Hardy spaces H2, H.~ 177 H,X)) H.~ 183 Hermitian 111atrix 95 Hilbert space 167 induced norm 181 initial condition response (see zero-input response) inner function 221 inner product space 169 input-output response (see zero-state response) invariant subspace 133 inverse left matrix 259 right matrix 83 system 197 isomorphic, isomorphism 178, 229 Laplace transform 176 Lebesgue spaces time domain: ..CE(-x,oo), £")()(-x,x) 167, 174 frequency domain: £", Lx: 174 L 2 gain 181 linear combination 3, 18 lossless system 219 LQG control requirement for solution 165 state estimation !53 state feedback 149 Lyapunov equation in balanced realization l 07 Hex, control 234

LQG control 151, 155 quadratic control 123-129, 140 matrix Hermitian 95 nonnegative 96 positive definite 98 matrix exponential 9 matrix fraction description (see coprime factorization) matrix square root 160 measurement noise 138 minimal realization 34 reduction to 91 nonnegative matrix 96 norms 2-norm for vectors 94 H 2 signal norm 167 H 2 system norm 172, I 79 H 00 system norm 181, 191 Hankel norm 112 induced 2-norm for matrices 186 L2 induced system norm 181 L 2 time domain signal norm 95, 167 L 00 function norm 182 null space of a matrix 78 null vector. matrix 17. 36 observability 34, 76 Gramian 94, I 09 matrix 73 observable decomposed form 82, 94 eigenvalue 72 subspace 78 observer 70 form 72 minimal order 82 observer-based controllers 116 orthogonal 21 compliment 169, 178 matrix 98 spaces, signals 169 output feedback 149. 157. 254 output injection 90 Parsevars theorem 174 PBH (Popov-Belevich-Hautus) tests

Index (see eigenvector tests) performance index Gaussian 148, 153

H 00

224, 233, 242-252

quadratic 121, 139 phase plane 5 pole placement (see eigenvalue assignment) pole-zero cancellation 34 positive definite matrix 98 projector 161 proper 12 quadratic control 115 requirement for solution state estimation 13 7 state feedback 119

147

range of a matrix (operator) 57, 77, 163 rank of a matrix 51, 59, 76 rational function 12 realization balanced 104 minimal 34,35 Ric(H) 145 robustness (see stability) separation principle 118, 149, 157, 261 singular value decomposition 185, 274 small gain theorem 184, 191 spaces Hardy 179, 185 Hilbert 168, 173 Lebesgue 167 stability internal, external 25 reduced order model 279 robustness 26, 69, 203 stabilizable system 43 stabilizing controllers 213, 215

315

stabilizing solution 128, 135, 158, 227, 267 stable system 6, 109 state computation 37, 66 estimation 67, 153, 242, 251 feedback 41, 120, 151, 227 state model forms controllable decomposed 60 controller 48 diagonal (normal) 32 observable decomposed 78 observer 72 state trajectory 5, 21 strictly proper 12 symmetric matrix (see Hermitian) system factorization 201 system interconnections 193 system inverse 196 system zero 198 trace of a matrix 102, 150 transfer function 34 proper, strictly proper 12 transformatio n to controllable decomposed form 64 controller form 49 observable decomposed form ,81 observer form 73 transition matrix 9, 31 unitary matrix

98

weighting matrix 170 well posed feedback control 214 worst disturbance 234, 259 Y oula parametrization

217, 260

zero-input (initial condition) response

1, 25,

32

zero-state (input output) response

I, 32