Industrial servo control systems: fundamentals and applications

INDUSTRIAL SERVO CONTROL SYSTEMS Fundamentals and Applications

Second Edition, Revised and Expanded

GEORGE W. YOUNKIN Industrial Controls Consulting, Inc. Fond du Lac, Wisconsin, U.S.A.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Marcel Dekker, Inc.

New York • Basel

TM

Copyright © 2002 by Marcel Dekker, Inc. All Rights Reserved.

Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress. ISBN 0-8247-0836-9 This book is printed on acid-free paper. Headquarters Marcel Dekker, Inc. 270 Madison Avenue, New York, NY 10016 tel: 212-696-9000; fax: 212-685-4540 Eastern Hemisphere Distribution Marcel Dekker AG Hutgasse 4, Postfach 812, CH-4001 Basel, Switzerland tel: 41-61-260-6300; fax: 41-61-260-6333 World Wide Wed http://www.dekker.com The publisher offers discounts on this book when ordered in bulk quantities. For more information, write to Special Sales/Professional Marketing at the headquarters address above. Copyright # 2003 by Marcel Dekker, Inc. All Rights Reserved. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA

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FLUID POWER AND CONTROL A Series of Textbooks and Reference Books FOUNDING EDITOR

Frank Yeaple President TEF Engineering Allendale, New Jersey

1. Hydraulic Pumps and Motors: Selection and Application for Hydraulic Power Control Systems, Raymond P. Lambeck 2. Designing Pneumatic Control Circuits: Efficient Techniques for Practical Application, Bruce E. McCord 3. Fluid Power Troubleshooting, Anton H. Hehn 4. Hydraulic Valves and Controls: Selection and Application, John J. Pippenger 5. Fluid Power Design Handbook, Frank Yeaple 6. Industrial Pneumatic Control, Z. J. Lansky and Lawrence F. Schrader, Jr. 7. Controlling Electrohydraulic Systems, Wayne Anderson 8. Noise Control of Hydraulic Machinery, Stan Skaistis 9. Interfacing Microprocessors in Hydraulic Systems, Alan Kleman 10. Fluid Power Design Handbook: Second Edition, Revised and Expanded, Frank Yeaple 11. Fluid Power Troubleshooting: Second Edition, Revised and Expanded, Anton H. Hehn 12. Fluid Power Design Handbook: Third Edition, Revised and Expanded, Frank Yeaple 13. Industrial Servo Control Systems: Fundamentals and Applications, George W. Younkin 14. Fluid Power Maintenance Basics and Troubleshooting, Richard J. Mitchell and John J. Pippenger ADDITIONAL VOLUMES IN PREPARATION

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Preface

In this second edition of Industrial Servo Control Systems, the chapters have been updated and expanded and a new chapter has been added on servo compensating techniques. The book continues to be dedicated to the practicing engineer making the transition from academic theory to the realworld solution of engineering problems related to the application of servo drives to industrial machines. Part I focuses on the evolution and classification of servos, with descriptions of servo drive actuators, amplifiers, feedback transducers, performance, and troubleshooting techniques. Part II discusses system analogs and vectors, with a review of differential equations. The concept of transfer functions for the representation of differential equations is discussed in preparation for block diagram concepts. Discussion of the mathematical equations for electric servo motors has been expanded to include both DC and brushless DC servo motors. The equations for mechanical and electrical time constants are derived with additional analysis on the effects of temperature on these time constants. The representation of servo drive components by their transfer function is followed by the combination of these servo drive building blocks into system block diagrams. Frequency response techniques are introduced because of their usefulness in determining proper servo compensation. Also included are practical formulas for calculating inertia and examples that show how machine components reflect inertia to the

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servo drive motor. Torque-to-inertia ratios for electric servomotors are discussed along with calculations of the reflected inertias for various machine configurations. These reflected inertias are critical when sizing the drive and motor and when determining time constants for a stability analysis. Servo drive analysis and servo compensation techniques include the use of lag and lead compensation for analog-type servo drive amplifiers. This discussion is followed by a description of proportional, integral, and differential compensation, which is very popular with digital-type amplifiers. To classify practical servo performance criteria, the use of servo indexes of performance is discussed for both electrical and hydraulic servo drives. The chapter on servo plant compensation techniques addresses the continuing problems of structural dynamics and resonances in industrial machinery. This type of compensation is different from feedback control compensation in that it uses special electrical devices and software algorithms to correct for machine nonlinearities and mechanical resonance problems. Nonlinearities of dead zone and change-in-gain to overcome such machine problems as stiction and cogging of the servo drive at low velocities are covered. The topic of simple and practical notch networks used to compensate for machine mechanical resonances has been expanded to include frequency selective feedback with suggestions for software implementation. This edition also includes a discussion on the use of a control technique known as feedforward control to minimize the position errors of machine axes with widely varying dynamics. Chapters on servo drive stiffness and drive resolution are an integral part of the analysis of industrial servo drives. Coverage of speed and acceleration has been broadened to include the derivation and application of ‘‘S’’ type acceleration. Machine considerations for ball screw resonances have been expanded to include ball screw critical speed, axial, and torsional resonance calculations, and drive ratio considerations have been expanded to include worm/wheel gear boxes. Friction considerations include the three types of friction and their relation to servo drives on industrial machines. One important aspect of selecting a servo for a given application is sizing the drive, because the servo must be large enough to meet the load requirements. Manual drive sizing forms for both electric and hydraulic servo drives are included in this book. The electric drive sizing forms have been expanded to include both DC and brushless DC drives. Chapter 14 is entirely new and reviews the process of compensating an electric servo drive. It is assumed that this industrial servo drive has a brushless DC motor with a current loop. An example is given for the case in which the motor and current loop are closed with a position loop, and also with a velocity and a position loop. The application of proportional plus

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integral compensation is used throughout Chapter 14 and detailed in the inclusion of machine structural resonances in the position servo loop. Once a servo drive has been properly sized for a given machine, it is necessary to determine whether the machine and servo drive will be stable. There are many classical engineering methods that can be used, but it is also possible to use high-speed computer iterative techniques to simulate the control–servo drive–machine combination. Transient step responses in input are used in this simulation since it is readily possible to duplicate them on an actual machine and compare the results. Machine simulation is used to predict performance before the machine is built. Machine simulation includes the significant parameters of the servo drive: the nonlinearities of feedback; coulomb, static, and viscous friction; machine structural resonance; and machine mass. A number of examples are presented with case histories for comparing simulated response with actual response. My contemporaries encouraged me to pull together a lifetime of servo drive experience and write the first edition of this book. The second edition updates, expands, and increases the usefulness of this book to the practicing engineer. I am sincerely grateful to Dr. Thomas Higgins for providing me, as a student, with the academic tools needed in the field of feedback control. I am also grateful to Dr. John Bollinger and Dr. Robert Lorenz from the University of Wisconsin, for their help over the last 45 years in researching, teaching, and applying principles that have been critical in improving the performance of servo drives. Tom Rehm, my friend of 40 years and a fellow software ‘‘hacker,’’ has been a great help in software technology. In addition, I would like to thank the Giddings and Lewis Machine Tool Company for providing the atmosphere for study and research in the field of industrial servo drives, and my family for their long years of patience as my career progressed. George W. Younkin

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Contents

Preface I

BASICS OF INDUSTRIAL SERVO DRIVES 1 The What and Why of a Machine Servo Just What Are Some of the Benefits of a Servo System? 2 Types of Servos 2.1 Evolution of Servo Drives 2.2 Classification of Drives 3 Components of Servos 3.1 Hydraulic/Electric Circuit Equations 3.2 Actuators—Electric 3.3 Actuators—Hydraulic 3.4 Amplifiers—Electric 3.5 Amplifiers—Hydraulic 3.6 Transducers (Feedback) 4 Machine Servo Drives 4.1 Types of Drives 4.2 Feed Drive Performance

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

5

Troubleshooting Techniques 5.1 Techniques by Drive 5.2 Problems: Their Causes and Cures

6

Conclusion: Machine Feed Drives—An Integral Part of a Machine Control System 6.1 Advances in Technology 6.2 Parameters for Making Application Choices

II

ADVANCED APPLICATION OF INDUSTRIAL SERVO DRIVES

7

Background 7.1 Introduction 7.2 Physical System Analogs, Quantities, and Vectors 7.3 Differential Equations for Physical Systems 7.4 Electric Servo Motor Transfer Functions and Time Constants 7.5 Transport Lag Transfer Function 7.6 Servo Valve Transfer Function 7.7 Hydraulic Servo Motor Characteristics 7.8 General Transfer Characteristics

8

Generalized Control Theory 8.1 Servo Block Diagrams 8.2 Frequency-Response Characteristics and Construction of Approximate (Bode) Frequency Charts 8.3 Nichols Charts 8.4 Servo Analysis Techniques 8.5 Servo Compensation

9

Indexes of Performance 9.1 Definition of Indexes of Performance for Servo Drives 9.2 Indexes of Performance for Electric and Hydraulic Drives Summary

10

Performance Criteria 10.1 Percent Regulation 10.2 Servo System Responses

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Contents

11 Servo 11.1 11.2 11.3 11.4 11.5

Plant Compensation Techniques Dead-Zone Nonlinearity Change-in-Gain Nonlinearity Structural Resonances Frequency Selective Feedback Feedforward Control

12 Machine Considerations 12.1 Machine Feedback Drive Considerations 12.2 Ball Screw Mechanical Resonances and Reflected Inertias for Machine Drives 12.3 Drive Stiffness 12.4 Drive Resolution 12.5 Drive Acceleration 12.6 Drive Speed Considerations 12.7 Drive Ratio Considerations 12.8 Drive Thrust/Torque and Friction Considerations 12.9 Drive Duty Cycles 13 Drive 13.1 13.2 13.3

Sizing Considerations Introduction Hydraulic Drive Sizing Electric Drive Sizing

14 Adjusting Servo Drive Compensation 14.1 Motor and Current Loop 14.2 Motor/Current Loop and Position Loop 14.3 Motor/Current Loop with a Velocity Loop 14.4 PI Compensation 14.5 Position Servo Loop Compensation 15 Machine Simulation 15.1 Introduction 15.2 Simulation Fundamentals 15.3 Machine Simulation Techniques to Predict Performance 15.4 Other Simulation Software 16 Conclusion Glossary Key Constants and Variables

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Appendix: Hydraulic Drive and Electric Drive-Sizing Forms Bibliography

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I BASICS OF INDUSTRIAL SERVO DRIVES

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1 The What and Why of a Machine Servo

The control user should be familiar with servos. The user will understand what a servo is and why it is required in so many applications. This discussion will answer these two basic questions: What is a servo? Why use a servo? Any discussion of servos will have to employ the term ‘‘feedback.’’ Thousands of times every day we require information to be ‘‘fed back’’ to us so that we can perform normal activities. When controlling a car down a highway, feedback is provided to our brain by the gift of sight. How terrifying it would be if we were traveling at 70 mph and we lost the ability to see. Our brain, which is the center of our control system, would have little feedback to help it decide what corrective actions need to be taken to maintain a proper path. The poorer feedback channels still available would be the senses of hearing and touch, which would allow us to ride the shoulder. The result would be a lower speed, poorer control, a very irregular path, and a greater chance for an accident. Inferior feedback on a machine blinds the operator or the control just as it does a driver. When using numerical control and servos, poor feedback can result in inferior parts, poor productivity, and high costs. Essentially, feedback is the retrieval of information about the process being controlled. It verifies that the machine is doing as commanded. There are two types of feedback—negative and

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

positive. Positive feedback, used for instance in radios, is not discussed here. Negative feedback, required to make a servo work properly, subtracts from commands given to the servo so that a discrepancy or error between output and input can be detected. This discrepancy initiates an action that will cause that discrepancy to approach zero. A perfect example of a negative feedback system is a wall thermostat and furnace, as depicted in Figure 1. If, in a 658F room, we set the thermostat for 728F, then 728F can be considered the command. The 658F of the room feeds back, subtracts from the 728F command, and results in a 78F discrepancy or error that instructs the furnace to supply heat. The furnace supplies heat until the negative feedback is sufficient to cancel the command so that no discrepancy exists and no further heat is required. A servo or servomechanism is a system that works on the negative feedback principle to induce an action to cause the output to be slaved to the input. Our thermostat/furnace example was one of a servo which induced the generation of heat. Any servo has two basic elements. These are a summing network and an amplifier. The summing network, as shown in Figure 2, is simply a device that sums the negative feedback (F) with the command (C) to generate an error discrepancy (E). Our driver’s summing network (Figure 3) was his brain. The command would be to keep the car in the right-hand lane. If he was straddling the center line, the feedback through his eyes to his mind would so indicate. If he subtracts this feedback information from the

Fig. 1 Thermostat example.

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Fig. 2 Summing network.

command, he will deduce that an error exists, which can be corrected by moving 2 ft to the right. A machine feed axis drive example (Figure 4) would be to have a command of þ10 in. If, as the machine was executing the command, we took a snapshot of the summing network when the machine reached þ9 in., we would see a command of þ10 in., a feedback of þ9 in., and a resulting error of þ1 in. The error would be the amount of further movement required for the feedback to equal the command. The second main element (see Figure 5) is the amplifier. This is simply a power device that takes a small error (E) and multiplies it by an amplification factor, which is a measure of the muscle or power available to drive the output and thus the feedback device (F). The amplification factor is what is normally referred to as the gain of the system.

Fig. 3

Feedback.

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Fig. 4

Summing network for positioning.

Our driver (Figure 6) mentally computed an error (E) requiring that the car be moved 2 ft to the right. This instruction traveled from his brain through nerves to the muscles in his arms and hands, which turned the steering wheel and with the muscle in his power steering caused the car to move into the right-hand lane. In the machine feed axis example (Figure 7) the error of þ1 in would be in the form of a small voltage, which would cause the servo motor to turn, and axis positioning motion would result. If we connect the two elements together, as shown in Figure 8, a basic closedloop servo system results. A new command will generate an error (E), which will activate the muscle until sufficient movement has caused the feedback (F) to be coincident with the command (C), at which point the error (E) is zero and motion is no longer instructed. The term ‘‘closed loop’’ suggests that after entering a command, signals traveled around the loop until equilibrium is attained.

Fig. 5

Amplifier.

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Fig. 6

Human muscle.

Our automobile and driver (Figure 9) are now a servo system with a command to stay in the right-hand lane. The servo is shown above and the position of the car to the centerline is shown below. When the car is actually straddling the white line, an error is mentally computed, which instructs the driver’s muscle and the car’s muscle to take a corrective action. When this corrective action is taken, the position of the car corresponds to the command, no further error is detected, and no further action is taken. Our snapshot of the machine axis servo feed (Figure 10) showed a command of þ10 in., from which the þ9-in. feedback was substracted. The þ1-in. error instructed the axis to continue moving. As the axis continues on, the error will get smaller and smaller until the feedback indicates that the axis is in position with no resulting error. The earliest axis feed systems used the operator as the summing network that closed the servo loop. The command was located on a part drawing, and the operator’s mind was the summing network. The operator read the scale on the machine, subtracted it from the desired command, and

Fig. 7

Servomotor.

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Fig. 8

Closed-loop servo.

came up with an error. The operator instructed the machine’s muscle by shifting the appropriate levers to engage motion. As the machine neared the commanded point, the error indicated the need to slow down to one or more intermediate geared speeds. Creep speed was used to accurately reach final position. A repeat performance on a second and third axis would complete the operation.

Fig. 9

Car example.

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Fig. 10

Closed-loop position servo.

JUST WHAT ARE SOME OF THE BENEFITS OF A SERVO SYSTEM? Here are six benefits of a machine axis feed servo drive. 1. Shorter positioning time: The servo operates at maximum positioning rate until the ideal time to decelerate, at which point it slows down uniformly to the end point with no hesitation at intermediate feeds. Since it dynamically searches for zero error, variations in machine conditions are compensated for. Positioning time is thus minimized. 2. Higher accuracy: A servo continually homes to the final position so that on January mornings it will continually strive to push the axis toward the end point, whereas on the Fourth of July when the machine might have a tendency to overshoot, the servo error will reverse and force the machine back into position. 3. Better reliability: An outstanding feature of servos is the ability to control acceleration and deceleration so that the mechanical hardware will hold its specification tolerance much longer. 4. Improved repeatability: Repetitive moves to a particular commanded point will show much better consistency. The result is more consistency of parts that are intended for interchangeability. 5. Coordinated movements: Since all axes are closed-loop servos, they are continually responding to the command at all feed rates. Coordinated movements thus require the generation of coordinated commands through employing interpolators with the control. 6. Servo clamping: There is no longer a dependency on mechanical clamps for servoed axes because of the continuous position-

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holding capability of a servo. The stiffness of the servo must be relied on for any contouring movements requiring that both axes be in motion. Properly designed, the servo can also hold the axis very stiffly at a standstill.

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2 Types of Servos

2.1

EVOLUTION OF SERVO DRIVES

The word ‘‘servo’’ is a learned borrowing from Latin where it meant ‘‘slave.’’ Industrial servos had their real beginning during World War II. Some applications were in military gun control and steering control. The evolution of speed and positioning drives since the war is shown in Figure 1. Industrial servos can be classified in four different ways as follows: By use: Cranes and hoists Winders Steering Refrigeration Heating Combustion Machine tool feed and speed Process control Power Guidance By variables controlled: Force

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Fig. 1

Evolution of industrial servo drives.

Tension Level Speed Position Mixture Temperature and humidity Voltage and current Volume Density By type: Chemical: gasoline, paint, etc. Electronic: AFC (automatic frequency control), AVC (automatic volume control), color synchronization Electric drives Hydraulic drives By feedback: Regulators (Type 0) Servomechanisms (Type 1) Acceleration (Type 2)

2.2

CLASSIFICATION OF DRIVES

For purposes of this book, industrial machine feed and speed servo drives are addressed. Machine drives can be classified further as shown in Figure 2.

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Fig. 2 Classification of machine drives.

Mechanical drives are not considered as servo drives, but they are included here as one of the three types of machine drives. Hydraulic servo valve drives have been used extensively as machine servo drives. With hydraulic servo drives it is very important to minimize the trapped oil volume to maintain a stable servo drive. In general the hydraulic resonance of the trapped oil volume should be greater than 200 radians/sec. Thus the servo valve type of hydraulic drive will be used because the trapped oil volume will be minimized by joining the servo valve and motor close together with a manifold. The result will be an increased hydraulic resonance and a more stable servo drive. Servo pump hydraulic drives, on the other hand, have a large volume of trapped oil between the pump and actuator, thus lowering the hydraulic resonance and creating possible stability problems. Hydraulic stepping motor drives had reliability problems and were not used extensively.

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Electric drives are shown in Figure 2 with four of the many types available. The motor starter drives are not servo drives, but they are in the majority since they are used extensively in industrial applications and consumer products such as appliances. Stepping motor drives are used in low-torque, high-speed applications such as tape transports. Direct current (DC) drives have numerous industrial applications. The majority of these drives use a power amplifier and motor. The amplifiers were initially of the Ward–Leonard type using a rotary amplifier. These drives were rugged but had very slow response with frequency responses of a few hertz. The thyratron was a form of an electrical switch that was not reliable and had noticeably large dead times or transport lag. The static amplifier has been used extensively in industrial servo drives. The magnetic amplifier was highly reliable and rugged but was also very slow in response. The silicon controlled rectifier (SCR) is the solid-state version of the thyratron. The SCR is highly reliable and is available in very large current ratings. It has been used extensively in industrial servo drives, both for speed and positioning or feed drives. Like the thyratron, the SCR has a dead time where no current flows. This dead time is described as transport lag causing amplifier phase lag and limiting the DC SCR drive to low servo response. Maximum torque ratings are about 400% of rated, which can be used for forcing (acceleration). For very-high-horsepower industrial servo drives, in the hundreds of horsepower, for example, the DC motor was at one time driven from separately excited generators. These drives have largely been replaced by the SCR drive. Alternating current (AC) industrial servo drives have a number of forms. Those discussed in this chapter are shown in Figure 2. The adjustable-frequency motor drives are used in industrial applications where positioning is not a requirement. An example would be a high-horsepower fan drive. The two-phase AC servo drive has been used in fractionalhorsepower instrument-type servos. For many lower horsepower positioning servo drives the synchronous AC (brushless DC) servo drive is being used extensively. These drives are available in a class of about 30 HP or less. Many of these brushless DC drives are used for positioning or feed drives on industrial machines. These AC drives can be treated as DC drives in the servo application—hence the name ‘‘brushless DC drive.’’ This drive uses transistors in the servo amplifier. The result is a higher performance drive (servo bandwidths of 30– 40 Hz are possible on large machine applications). The disadvantage is that the torque ratings of the drive are limited by the transistor ratings in the amplifier. Characteristically these drives have a 200% maximum torque rating for drive forcing (acceleration). The drive motors have maximum

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torque ratings of about 400%, but the amplifier limits the overall torque rating to about 200% because of the current rating of the transistors. These ratings have been increased with the introduction of the insulated-gate bipolar transistor (IGBT). For high-horsepower servo drive applications, the induction motor with vector control is being used in both speed and positioning drives. These drive amplifiers use transistors which have current limitations. By using IGBTs the horsepower ratings are available upward of 500 HP. These drive classifications are those that will be discussed in Parts I and II of this book. There are other types of drives, but these classifications have been selected for their numerous industrial drive applications.

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3 Components of Servos

3.1

HYDRAULIC/ELECTRIC CIRCUIT EQUATIONS

From a basic hardware point of view it is important to understand the steady-state equations of a hydraulic and electric actuator. The majority of hydraulic positioning or machine feed drives use a rotary actuator with a servo valve attached to the motor to minimize the amount of trapped oil under compression. By minimizing the volume of oil that is trapped between the servo motor and servo valve, the hydraulic resonance will be increased and the drive made more stable. The hydraulic flow equations for a fixed displacement servo motor are shown in Figure 1. The steady-state direct current (DC) motor equations are also shown in Figure 1. The voltage and torque equations include the motor back emf constant (Ke) and the torque constant (KT). These motor constants are used throughout the basic and advanced portions of this book. Therefore it is important to know where to find them and how they are used in the forthcoming servo calculations.

3.2

ACTUATORS—ELECTRIC

It is important from the hardware point of view to be familiar with manufacturers’ motor specifications to find the design parameters such as

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Fig. 1

Hydraulic and electric motor circuits.

the voltage constant and torque constants that are required for servo drive calculations. Sample electric drive manufacturer specifications are included for DC and alternating current (AC) servo drives. DC motor specifications for Gettys motors are shown in Figure 2. AC (brushless DC) Allen-Bradley motor specifications are shown in Figure 3. The motor voltage and torque constants are included in these specifications. For all drive calculations the torque constants for a hot motor (408C) should be used wherever possible.

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Fig. 2 Gettys DC motor data. (Courtesy of Gettys Corp.)

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Fig. 3

Allen-Bradley AC motor data. (Courtesy of Rockwell Automation/Allen-Bradley.)

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In addition to the motor specifications, it is important to be familiar with the speed/torque characteristics of a given drive motor. These characteristics show how much torque can be developed at a given speed to stay within the rating of the drive. It should be noted that the speed torque ratings are for the motor/amplifier combination. The speed torque characteristics for two Gettys DC motors are shown in Figure 4. As the operating point moves to the right (higher torque) the motor will develop increasing sparking at the brushes and thus the drive must be derated. These other operating zones can be used intermittently for drive forcing and acceleration. If the required torque moves further to the right, beyond the operating zones, the motor will be damaged. To prevent this from happening, all commercial servo amplifiers use ‘‘current limit.’’ Speed/torque characteristics for two Allen-Bradley brushless DC motors are shown in Figure 5. These AC motors do not have the commutation limits of the DC motors and therefore can be operated at higher speeds for rated torque conditions. While the DC drive (amplifier and motor) can use 450% rated torque intermittently, for drive forcing, the AC drive package can only operate at about 200% rated torque for drive forcing. As the state of the art in drive amplifiers advances, this 200% limit for the AC (brushless DC) drive should increase.

3.3

ACTUATORS—HYDRAULIC

Most industrial hydraulic servos are of the servo valve classification. Hydraulic servo pump designs usually have a large amount of trapped oil, which results in a low hydraulic resonance and stability problems. Hydraulic servo valve designs using a piston actuator can also have large amounts of trapped oil volume for the longer stroke designs. Therefore most industrial hydraulic positioning or feed drives use rotary actuators. Like electric motors, the hydraulic servo motors have manufacturer specifications. These motors also have a torque constant with units of so many inch-pounds per 100 psi pressure. Hydraulic rotary actuators are generally of the piston type motor or the roll-vane type of motor. A typical specification sheet for a Hartman roll-vane motor is shown in Figure 6. The advantage of the hydraulic servo is the large amount of torque that can be produced. They also have a disadvantage of oil contamination requiring oil filtration of 10 mm. The oil viscosity can change with operating temperature, which can affect the stability of the servo. Oil temperature should be held below 1308F. If the operating temperature exceeds 1408F for

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Fig. 4

Gettys DC motor speed/torque data. (Courtesy of Gettys Corp.)

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Fig. 5 Allen-Bradley AC motor speed/torque data. (Courtesy of Rockwell Automation/Allen-Bradley.)

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Fig. 6

Hartman hydraulic motor data. (Courtesy of Hartmann Controls Inc.)

a long period of time the oil can start to break down, causing the servo valve spool to stick, resulting in a stall or runaway condition. Hydraulic pump noise can also be an annoying condition. From the government Walsh– Healy act, hydraulic pump noise must be limited to 90 dB on the A scale.

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3.4

AMPLIFIERS—ELECTRIC

In Figure 2 for the classification of drives, three classes of amplifiers were shown: rotary amplifiers, electronic amplifiers, and static amplifiers. Of these three types of amplifiers the static solid-state amplifier is predominant. The silicon controlled rectifier (SCR) has been used extensively in industrial applications. The SCR is the solid-state version of a thyratron. As an amplifier component the SCR is very rugged and available in very high current ratings, but by itself it is just a switch that conducts for a part of a half cycle of the AC power. Therefore, for part of the conduction half cycle no current flows. The period where no current is flowing is called the off time of the SCR, which is referred to as the transport lag of the amplifier and is discussed further in Part II. SCR amplifiers are designed in various circuit configurations, with the most popular circuit being the three-phase, half-wave amplifier. For reference, this circuit design for a Gettys DC drive is shown in Figure 7. Each circuit design for the SCR has an efficiency rating, represented by the form factor, which is a measure of the amount of ripple current in its output relative to the average DC value of its output. Thus the form factor is equal to Irms/Idc. For DC SCR amplifiers it is necessary to derate the drive rated torque. The rated torque is divided by the form factor. The derating form factor for various SCR amplifier designs are listed as follows: SCR amplifiers Single-phase, full-wave Single-phase, full-wave/inductor Three-phase, half-wave Three-phase, half-wave/ inductor Pulse width modulation or brushless DC amplifiers

Derating form factor 1.66 1.2 1.25 1.05 1.0

Another class of DC static amplifiers uses pulse width modulation (PWM) techniques. These amplifiers use transistors for current control to the motor. While the PWM amplifier has a high switching rate, allowing for much higher servo performance, it has much lower current ratings than DCSCR drives. In general, these drives have a maximum rating of 200% rated current. The form factor of these drives is one, so no derating is required. A typical block diagram of an Allen-Bradley DC pulse width modulation amplifier is shown in Figure 8. While these drives have a servo bandwidth of about 40 Hz, they are limited in velocity because of the mechanical commutator in the DC motor.

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Fig. 7

Gettys DC SCR electric drive diagram. (Courtesy of Gettys Corp.)

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Fig. 8 Allen-Bradley DC PWM electric drive diagram. (Courtesy of Rockwell Automation/Allen-Bradley.)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Lastly, the brushless DC drives offer higher speeds and higher performance. Most industrial versions of this drive use a synchronous AC motor. A position transducer in the motor measures the armature position and provides a signal to the amplifier to commutate the three-phase armature currents. Thus the motors can rotate at higher velocities than the DC motors. Speeds of 3000–5000 rpm are common. The brushless DC transistor amplifier also has a high switching rate with a 200% limit of rated torque. The amplifier and motor can be treated as a DC drive. A typical block diagram for an Allen-Bradley digital brushless DC drive is shown in Figure 9.

Fig. 9 Allen-Bradley brushless DC block diagram. (Courtesy of Rockwell Automation/Allen-Bradley.)

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3.5

AMPLIFIERS—HYDRAULIC

Hydraulic servo drive amplifiers take the form of a servo valve. Servo pump drives are not being considered. For reasons stated in Section 2.2, to maintain a minimum of trapped oil volume, the servo valve is mounted to the actuator with a manifold. Oil is supplied to the servo valve from a pump through a filter, which usually has a rating of 10 mm. Based on maximum power transfer, the hydraulic pressure at the servo motor through a servo valve is equal to two-thirds of the pressure at the input to the servo valve. In general it is assumed there is approximately a 300 psi line loss between the pump and the servo valve. When selecting a servo valve based on motor speed and required flow, it is important to allow additional oil flow to compensate for system leakage. As an index of performance, 30% more oil flow than required should be used in selecting the flow rating of the servo valve. The typical servo valve force motor (or torque motor) is shown in schematic form in Figure 10. The armature and drive arm are suspended on the flexure tube, which acts as a pivot point and also provides a spring force tending to keep the armature on center between the poles of the ‘‘C’’ section. The force motor has two symmetrical flux paths as shown by the dotted lines. Assuming an electrical current is applied to coil ‘‘A,’’ magnetic flux will be produced in air gap ‘‘a,’’ which tends to draw the armature toward that pole. The armature will move a distance such that the restoring force of the spring suspension (flexure tube) equals the magnetic pull. Thus, the greater the current in the coil, the greater the magnetic pull, and the further the stroke. Since the flexure tube acts as a pivot, the tip of the drive arm strokes in the opposite direction as the armature. The motion at the tip of the drive arm (also known as the flapper) is the force motor output. In summary, the force motor produces a displacement output proportional to differential current. If the coil currents are not the same, then the pull on one side will be greater and the armature will stroke an amount proportional to the difference in the two currents. As the flapper strokes one way, the nozzle pressures A and B will be unbalanced. This causes the spool to move in a direction to reestablish a balance of pressure at the nozzles. As the valve spool moves to a new point of equilibrium, oil will flow through the valve to the motor. Therefore the servo valve can be considered as a positioning servo where the spool is being positioned to allow oil flow. Servo valve spools can be made with overlap (creating a dead zone in output flow), underlap (which can cause system instability), and critical or zero lap. For most hydraulic servo drive applications, a very small overlap is used on the spool. Since the 1970s, advances have been made in hydraulic servo valve and actuator designs to include feedback transducers such as linear differential transducers (LDT).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Pegasus hydraulic servo valve diagram. (Courtesy of Schenck Pegasus.)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

3.6

TRANSDUCERS (FEEDBACK)

One of the important, if not the most important, elements of an industrial servo drive is the measuring device. One of the purposes of the feedback device is to accurately place a tool or workpiece at some desired location, prior to some machine operation. For the control system to know where the tool or workpiece is located, some measuring device must be used to provide this information in a language the control will understand. Measuring devices in general can be called transducers, a device used to transform one form of energy to another form of energy. A simple thermostat is an example of a transducer, where temperature is registered as a physical bending in a bimetallic strip. For some specified deflection of the bimetallic strip as the result of a temperature change, an electric circuit will be activated to run a furnace in the case of a heating control, or a compressor in the case of a refrigerator. There are many examples of transducers in the average home, but the type used with numerical control systems will convert position information from the motion of a machine drive screw or element into electrical signals. These electrical signals are the feedback information

Fig. 11

Feedback devices.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

to the control system in a language the control can understand. A classification of feedback devices is shown in Figure 11. There are two basic types of feedback signals, which also define the control system as to the data information used within the system. The first type of signal is called analog, and the control is usually referred to as an analog system. Analog signals have two characteristic features. They directly represent some physical quantity and are continually varied. A tachometer is an example of an analog measuring device where the output voltage is directly proportional to the velocity of input shaft rotation. The second type of signal is called digital, and the control system is usually referred to as a digital system. Physical quantities are represented through the medium of digits or numbers, which would be discrete electrical pulses in the digital control system. Digital signals or information can be further defined as of two types: incremental and absolute. Incremental signals are merely a train of pulses where each pulse has a specific weighted value such as 0.01 in per pulse. The absolute digital information will take the form of a pattern of pulses. The pattern can have the form of a binary numbering code or some special coding. There are many commercial versions of transducers available. Perhaps the best known device is the tachometer, which is a small generator, sometimes producing AC output voltage but more often a DC output voltage. Tachometers are used to a large extent with velocity regulators. With brushless DC drives, the velocity feedback is synthetically generated. Since these drives have a position transducer as part of the motor (used for commutating motor currents in the amplifier), the position signal is differentiated in the amplifier creating a synthetic velocity signal to close a velocity servo loop. The velocity regulator may not be the complete control system, but it is quite often part of a positioning system. Transducers used for positioning systems have two basic forms as analog devices. Probably the simplest analog device for positioning systems is the potentiometer. As a feedback measuring device, the potentiometer can be a single-turn or a multiturn device. One of the most common forms of feedback measuring devices is the synchro, an AC electromechanical device providing a mechanical indication of its shaft position for some electrical input, or providing some electrical output that is some function of the angular position of its rotor shaft. Another electromechanical feedback device often used is the resolver. The device is similar in appearance to the synchro, but the electrical construction is not the same. Both types are a form of variable transformer, with more or less decoupling between the rotor and stator windings as the armature rotates. A linear version of the resolver is known as the Inductosyn. These devices are discussed later in more detail.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The synchro construction has a single winding on the rotor with three windings on the stator. The stator windings are wound, so that induced AC voltages in each of three windings from the rotor winding are 1208 apart. In application, these units are used as a transmitter or receiver in the transmission of data in position-type servos. A variation of the usual type synchro is the differential synchro. This device has a set of three windings on both the stator and rotor. The differential is used to add another mechanical input to the data transmission system of servo drives. The resolver is a precision induction-type device acting like a variable transformer, with the amount of coupling varying as the sine and cosine of the angular position of its rotor shaft. In control systems, the resolver is used for coordinate transformation in analog computer applications, in position servos as applied to industrial controls, and in other control applications where rectangular conversions to polar coordinate or vice versa are desired. The rotor and stator have two windings, which are wound at 90 electrical degrees with respect to each other. As with the synchro, the resolver can be used as a data transmitter or receiver. The resolver has the advantage that it can transmit data either side of zero degrees (operate in all four quadrants) whereas the synchro can only operate in quadrant one. In addition, the same resolver used to transmit or receive AC signals can be used as a differential unit. An electrical drawing of a resolver is shown in Figure 12. The resolver could be considered as a variable coupling transformer, with the amount of coupling dependent on the angle of the rotor shaft. Considering voltages V1 and V2 as primary voltages, the secondary voltages would be Vr1 and Vr2. Each secondary voltage would be a function of both primary voltages V1 and V2, plus the angle of the rotor shaft. Therefore, the two secondary

Fig. 12

Resolver circuit.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

voltages Vr1 and Vr2 would have the following mathematical relationship to the primary voltages V1 and V2 : Vr1 ¼ V1 cos ya þ V2 sin ya Vr2 ¼ V1 sin ya
V2 cos ya where: ya ¼ actual angle of the rotor shaft in degrees Since the resolver is a rectilinear to polar coordinate transforming device, the voltages V1 and V2 are also a function of the desired rotor angle as follows: V1 ¼ E sin yd V2 ¼ E cos yd where: yd ¼ Desired angle of the rotor shaft in degrees E ¼ rms value of an alternating sinusoidal voltage of a specified frequency When the resolver is used as a measuring device in a continuous positioning system, the position of the rotor shaft is a measure of where the machine element being positioned actually is. The resolver shaft is connected to the movable machine element through a gear train to the drive screw or sometimes by a precision rack and pinion. Therefore, the angular position of the rotor shaft of the resolver is a measure of the linear position of the machine element. A block diagram of the resolver used as a measuring device in a positioning system is shown in Figure 13.

Fig. 13

Servo drive with resolver feedback.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The difference between the actual angle ya and the desired angle yd is the position error. However, the input and output of the resolver windings are electrical voltages, not angles. The input voltages can be a function of the desired angle, as from the equations for V1 and V2 . The output rotor voltage is a function of the difference between the actual angle ya and the desired angle yd . When ya ¼ yd the voltage Vr1 is zero, which is the same as saying the machine slide is in position. This fact can be determined from the preceding equations when yd and ya ¼ 0 : Vr1 ¼ 0 cos 0 þ Eð0Þ ¼ 0 Vr2 ¼ 0ð0Þ
E cos 0 ¼
E

ðnot usedÞ

Note: Only one resolver rotor winding is used. Thus Vr1 satisfies the condition of zero output when yd ¼ ya and it is therefore used; Vr2 is not used. It should be apparent that Vr1 can be made zero (the positioning system null position) by varying either the rotor shaft ya or varying V1 and V2 as functions of the desired angle yd . Most of the resolver devices are used as measuring devices coupled to a machine drive screw. With most industrial machines, the drive screw will have some backlash between its angular position and the position of the element it is driving—a machine slide, for example. For this reason, there has been difficulty in obtaining accurate positioning when measuring from the machine drive screw. Some manufacturers have designed their controls so positioning will always occur from the same direction. In general, measuring from the machine drive screw is not ideal. A considerable effort has gone forth to create a linear measuring device that can be used right at the moving machine slide rather than through a drive screw. These devices are called linear transducers (e.g., Inductosyn) and take the commercial forms of linear resolvers. With the linear analog measuring devices, either the stator or rotor is mounted on the moving part, and the other part is mounted on the stationary machine element. There is an associated attenuation with these devices, and amplification is required. The linear digital measuring devices have taken the form of optical grating devices, and some are magnetic coupling devices that generate a series of pulses, with each pulse being equal to some given distance. Some manufacturers have used precision gear racks on the moving machine element with a rotary measuring device fastened to the stationary element. This technique obviously is better than measuring off the machine drive screw. Each of the three methods of measuring—from the machine drive screw, a precision rack, or a linear transducer—has advantages and disadvantages in cases of application to a machine. In general, accuracy

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

improvements of a factor of 10 can be obtained between measuring from a machine drive screw and a linear transducer. Measuring at the drive screw can provide accuracies of one thousandth of an inch when using a good ballbearing screw. Both analog and digital systems are in use, and equally good results have been obtained with both. Most process-type control systems and numerical contouring systems use a single measuring device. It is characteristic of these types of control systems for the actual position to vary with the varying command position such as to make the position error of the control system very small. A plot of a resolver output voltage vs the rotor shaft angle is shown in Figure 14. With a continuously varying control voltage (V1 and V2 ) the resolver output voltage seldom gets more than a few degrees away from the resolver null position. Digital feedback is used extensively in many industrial servos. The digital feedback transducers can provide improved accuracy over the analog resolver or Inductosyn if the resolution of the pulses count per revolution is high enough (e.g., 500,000 pulses per revolution). For short-travel machine slides, the absolute digital feedback is practical, but in general most digital feedback devices used are incremental pulse generators. For maximizing the accuracy of positioning, the transducer is mounted on the machine slide.

Fig. 14

Resolver output voltage.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 15

Feedback devices.

Therefore for short-travel machine slides there are many linear digital incremental feedback devices that can be used. These linear devices are costly and must be justified. Examples of some of the feedback devices are shown in Figure 15.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

4 Machine Servo Drives

4.1

TYPES OF DRIVES

Positioning systems in general have their most basic form, and also the most complex, in the human being. The brain serves as the summing network that accepts the command for a desired motion or position, the musculatory system serves as the power source or prime mover to cause motion to take place, and the sensory system, such as the eyes, determines the present position. In general, these three things—the brain, the muscles, and the eyes—are analogues to the three basic parts of any positioning system. The brain accepts the command, or reference as it is sometimes called, and compares it to the feedback to answer the question, ‘‘How do we get to where we want to be?’’ The eyes are the feedback device, sometimes called a measuring system, which answers the question, ‘‘Where are we?’’ The difference between the command or the desired position and the actual position (determined from the feedback device) is referred to as the position error. It is this error that makes the prime mover cause motion to take place, resulting in the actual position equaling the desired position and the position error being reduced to zero. Applying these definitions to a position control system, we have three basic parts of a position system, shown in block diagram form in Figure 1. The ‘‘desired position’’ must take the form of a piece of equipment to

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1

Machine servo-drive block diagram.

convert from the language of dimensions to the language of the prime mover. The input may be a calibrated scale in the dimension of inches or degrees with a pointer and handle. The output will usually be some voltage. The ‘‘actual position’’ is usually determined with some measuring device that is generally referred to as a transducer. The output language of the measuring or feedback device must be the same language as the command since the feedback is subtracted from the command signal. The answer is the position error signal. Thus the three basic parts of the servo are the drive, feedback device, and summing junction. The prime mover is usually the ‘‘drive’’ for the positioning system. The position error signal is the input to the drive. When there is a position error, the drive will cause motion to take place. Thus far, the discussion has been around positioning systems. In this discussion there are three types of feedback control systems in common usage. These are referred to as type 0, type 1, and type 2 servomechanisms. The type number refers to the net number of integrators around the loop. A type 0 servomechanism has a constant value of the output, such as speed, and requires a constant error signal under steady-state conditions. One form of a type 0 servo is a regulator, which implies that something is being controlled to produce a desired output, such as speed, despite disturbance conditions. The block diagram of Figure 2 shows a drive used in a speed regulating system where the controlled output is speed. Desired output speed is controlled by the reference input voltage to the drive. The system becomes a regulator when a feedback signal is added with a tachometer generator to measure output speed. As the tachometer input shaft rotates, an output voltage is generated that is proportional to the speed of the tachometer. A summing circuit at the input of the drive will subtract tachometer feedback voltage from the reference input voltage. The resultant

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 2 Speed regulating system.

voltage will be a constant error voltage to the drive under steady-state conditions. An external disturbance, such as loading, or a change in the amount of reference input will cause the regulator to compensate for loading and maintain its original speed or change its original output speed to some new speed if the reference input voltage is varied. For either case, the regulator is acting under transient conditions, and the time required to again reach the steady-state condition, or a new steady-state condition, is the response time. There are times when the reference input is continuously varied, and the regulator output will attempt to follow the change in reference input. A sinusoidal reference input voltage is often used to test the ability of the regulator to follow a changing reference input. As the frequency of the sinusoidal voltage increases, the regulator output will follow the changing reference input until a frequency is reached where the regulator output can no longer follow the varying input. The frequency at which the regulator ceases to follow the sinusoidal reference input is referred to as the cutoff frequency of the servo-drive frequency response. With appropriate compensation (an integrator), a type 0 regulator can be converted to a type 1 system. A type 1 servomechanism will have zero error for any constant value of the controlled output. A constant rate of change of position (velocity) requires a constant error signal under steady-state conditions. This error is usually referred to as the following error in a type 1 positioning system. The positioning machine servo drive is an example of a type 1 servo. Such a feedback control system may include a speed regulator as shown in Figure 3. The feedback signal for this type of servomechanism is an integrated measure of speed (position). Some type of measuring device is used to provide a feedback voltage proportional to position (a resolver for example). The third type of servomechanism is referred to as a type 2 servomechanism, sometimes referred to as a ‘‘zero-velocity error’’ servo.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 3

Positioning servomechanism.

The type 2 servomechanism has a zero error for a constant value of controlled variable (position). These drives also have zero error for a constant controlled variable of speed. A constant rate of change of speed (acceleration) requires a constant error. The type 2 servo drive requires compensation, which results in its being conditionally stable. Changing gains or compensation time constants can produce an unstable drive.

4.2

FEED DRIVE PERFORMANCE

Servo-drive performance is a measure of how well a servo can make rapid changes in velocity, path, etc. and maintain required accuracy in the controlled variable (position). The frequency-response bandwidth is one way to define the performance of an industrial servo drive. To understand the relevance of servo loop bandwidth to servo performance it is necessary to define what is meant by servo bandwidth. ‘‘Bandwidth’’ is a term describing the frequency-response characteristic of a servo drive. In its simplest definition, the frequency response is a measure of how well an output of a servo follows the input as a sinusoidal input driving frequency increases. At low frequencies the output of a servo faithfully follows the input, so the ratio of output to input is 1. This is often referred to as the flat part of the response. At high frequencies, the output of a servo no longer follows the input. The higher the servo bandwidth, the greater the capability of the servo to follow rapid changes in the commanded input. For example, in a machine contouring system, the performance of a servo to make rapid changes in the path (square corners, etc.) is directly related to the servo bandwidth.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

As the frequency increases beyond the flat part of the response, the servo output begins to lag behind the input driving signal. There is a phase lag and an amplitude attenuation between the output and input. When the phase lag reaches 180 degrees the servo is unstable. Thus it is a requirement to establish some performance specification to state how much phase lag is acceptable before instability occurs. Most industrial axis servo drives use a velocity servo inside a positioning servo system. It is critical that the position servo drive have an open-loop phase lag less the 180 degrees to avoid instability of the servo. As an index of performance (IP), a maximum of 135 degrees phase lag has been used as a criterion. This means there are 45 degrees phase lag left before instability occurs, and this 45 degrees is referred to as the phase margin before the servo goes unstable. Thus, the total acceptable phase lag of the internal velocity loop and other components cannot be more than 135 degrees phase lag. In any positioning servo system with an internal velocity servo there will be a required integration of the velocity output of the velocity servo to the position output of the position loop. This integration component takes place in the motor and is measured by the position transducer (often a resolver). Any integration component has a fixed 90 degree phase lag at all frequencies. Thus the total allowable phase lag of 135 degrees for the velocity servo and integration is now reduced to 45 degrees (135790) for the closed-loop velocity servo. Therefore the useful velocity servo bandwidth occurs at a frequency where the closed-loop velocity servo has a phase lag of 45 degrees (often stated as the 45 degree phase-shift frequency). A typical frequency response for a brushless DC servo drive with a load inertia equal to the motor armature is shown in Figure 4. The useful velocity servo bandwidth is 20 Hz. A term to define the performance criteria of a position servo is referred to as the position-loop gain or velocity constant. This gain is actually the open position-loop gain. Referring to Figure 5, the block diagram for a position servo, the open-loop gain is a parameter that represents the product of the individual gains in the position servo loop without the feedback being closed. The closed position-loop gain for this example is position output/ position command. Without the feedback being closed, a position command will cause the drive to have motion. The ratio of output/input will therefore be in units of velocity/position. As an example, if the input command has the units of inches, the open position loop output will be inches per second. Thus the open position-loop gain is Kv ¼ (in.)/(in.-sec) ¼ 1/sec. A position-loop gain, in units of 1/sec, does not have a practical meaning to many machine control people working in industry. By making a units

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Fig. 4 Machine servo frequency response.

conversion of Kv ¼

1 60 sec in: 1 inches per minute 6 6 ¼ 60:06 ¼ sec min 1000 mils sec mil

or ðinches per minute=milÞ616:67 ¼ ð1=secÞ Thus the open position loop gain expressed as some velocity (inches per minute, ipm) divided by distance (mils) has a practical significance.

Fig. 5 Machine servo-drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Typical measured velocity servo frequency response bandwidths for some commercial servo drives are shown in Figure 6. The servo bandwidths are shown versus increasing inertia loads. The Gettys DC silicon controlled rectifier (SCR) drive is a typical three-phase, half-wave DC SCR servo drive. As inertia load is added to the motor, the drop in servo bandwidth is minimal since the motor inertia is 0.437 lb-in.-sec2 compared to the inertia load being added. However, the servo bandwidth is limited since it is a function of the transport lag (dead time in the SCR firing), which contributes phase shift to the servo drive. The DC SCR drive has a forcing capability of 400%, which is an advantage over the transistor type servo amplifiers. If the same DC motor is used with a transistor-type pulse-width modulation (PWM) amplifier (General Electric HI-AK), the servo bandwidth characteristics are much higher since there is no transport lag in the servo amplifier. However, the forcing capability (for acceleration) is limited to 200% because of the transistor current capability. As time passes, newer transistor technology will improve the current capability. With an AC drive (brushless DC Indramat drive) the servo performance is comparable to the DC PWM servo drive and has the same

Fig. 6 Loading effects on servo performance.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

current capacity limitations. The AC (brushless DC) synchronous motor shown has an inertia about half (0.2 lb-in.-sec2) that of the DC motor. The hydraulic (Hartman) servo drive has an armature of about 0.015 lb-in.-sec2. Thus at no load, the performance is over 60 Hz. As a coupling and load inertia are added to the motor shaft, the servo performance drops rapidly. The frequency response is a powerful analytical tool that can be used for design or diagnostic procedures. Industrial servo drives are connected to machines. Since machines can have nonlinearities of backlash, stiction, etc. and mechanical resonances, the frequency response of the servo drive and machine can be used to identify these nonlinearities and structural resonances. An example of a frequency response of a machine slide that has a 40-Hz resonance inside the position loop is shown in Figure 7. The resonant peak is about 25 dB, which results from a structural compliance and an underdamped machine slide (roller bearings). These applications of the frequency response are discussed in further depth in Part II. In addition to the frequency-response method to measure performance of an industrial drive, the transient step response can be used. Measuring the frequency response requires a servo analyzer, which may not always be

Fig. 7 Machine servo frequency response with a resonance.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 8

Second-order transient response.

readily available. A transient step response can readily be made with a simple battery circuit. The step response is used extensively in testing, diagnosing, etc. of industrial machines in the field. A typical transient response for a second-order system is shown in Figure 8. For velocity servo drives an index of performance exists, which recommends that the transient response have one overshoot. Likewise, for positioning servo drives (especially contouring drives) there should not be any overshoot (critically damped).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

5 Troubleshooting Techniques

The time is gone when machines, controls, and drives were ‘‘add-ons,’’ as shown in Figure 1. Early machines used add-on drives. These drives had one thing in common: There was no interaction among the control, drive, and machine. To a large extent these machines were open loop. Today’s numerically controlled machines are tightly integrated machine systems. There is a great deal of interaction between the machine, drive, and control. It is this interaction that results in problems with stability, surface finish on the work part, and accuracy. The actual block diagram for today’s machine systems looks more like the block diagram shown in Figure 2.

5.1

TECHNIQUES BY DRIVE

The primary difference between the machine drives of yesteryear and those of today lies in the number and type of feedback signals that occur with modern control systems. It is true that early drives had load forces fed back to the drive, and these forces could cause a droop in the feed rate with load (called load regulation). Nonlinearities (stiction, lost motion, etc.) would not affect the stability because they were outside the servo loop, but they could cause inaccuracies. By and large, drives were ‘‘hooked’’ onto a machine and in spite of resonances in the mechanical drive, stick-slip, lost motion, etc.,

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1

Machine control system block diagram.

the machines performed with less than acceptable accuracy by today’s standards. With the ever-increasing demand for more accurate machines, more and more of the machine was included in the control/drive feedback loops. Today the machine, control, and drive are an integrated machining system. The existence of resonances between the drive motor and machine slide can make the drive unstable. A stick-slip condition on the machine slide can cause a null hunt (discussed in the next section) in the drive. Poor surface finish can result where load forces are fed back to a drive that has poor servo stiffness. A contouring drive with poor resolution also demonstrates poor low feed characteristics affecting the surface finish. Electrical noise feedback into transducer cables, as the result of poor shielding and isolation, can cause stability and accuracy problems. With just about everything in the machine system being a variable that can affect machine accuracy and stability, where does the technician start to look for the cause of a machine that will not perform properly? It is virtually impossible to diagnose the trouble of a malfunctioning machine with all the control loops closed. Therefore, the control and machine should be

Fig. 2

Machine control system block diagram with nonlinearities.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

separated. With a hydraulic drive the separation should be at the servo valve connector. With a DC electric drive the separation should be at the input to the electric motor. With brushless DC drives the battery box should be connected to the drive amplifier input. The troubleshooting should now proceed in the following sequence.

Closed Velocity Loop The next logical step in the ‘‘setting up’’ or troubleshooting of the servo drive is to close the tachometer or velocity loop and check the performance.

Hydraulic Drive The velocity loop performance can be checked by closing the servo loop external to the control. A battery box connected to the velocity drive input can be used to operate the servo drive independently of the control. By operating the velocity drive back and forth and putting step changes in input voltage to the drive, the servo performance can be observed for stability. Any unacceptable performance must be corrected before proceeding further.

Electric Drive Setup instructions for the electric drive are provided by the drive supplier. This type of drive can also be controlled by a separate variable DC voltage applied to the drive amplifier input terminals. Likewise, by moving the machine slide back and forth and putting step changes in input voltage to the drive, the servo performance can be observed for stability. Unacceptable performance must be corrected before proceeding to close the position loop. In either the hydraulic or electric drive the performance of the velocity loop should be checked with an external DC voltage source. The numerical control voltage sources such as the feed-forward voltage should not be used for trouble shooting. The most obvious test to make with the velocity loop is to make sure the drive can be controlled from standstill to the traverse rate. Whether the feed rates are smooth or jerky, especially at the lower feed rates, should be observed. With either electric or hydraulic drives the velocity loop gain should be set at its highest possible setting. The velocity loop performance is directly related to the gain. With too low a gain the drive will be sluggish and exhibit a speed droop with applied load. One of the prime reasons for poor surface finish with numerically controlled machines is a velocity loop with too low a gain (thus also poor performance and bandwidth). There are engineering evaluation tests that can be used to

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

ascertain the servo loop performance. However, these tests should only be made by a trained analyst. For general use the gain setting can be made by raising the velocity loop gain until the drive goes unstable and oscillates. The oscillation is usually observed as an audible rumble often called ‘‘tachometer growl.’’ When this condition occurs the gain should be reduced until the ‘‘growl’’ ceases. The gain should then be reduced slightly further to allow some margin in reliability of the gain setting. With electric drives the gain setting is often a component part of the setup instructions.

Hydraulic Drives Machine piping should be checked to assure that all return lines run directly to the ‘‘tank.’’ The drain lines should have a loop at the motor to be higher than the motor. Any restriction in the drain line can blow out a motor seal. A blown seal is obvious with a puddle of oil on the floor. Hydraulic pressure and return line pressure can be checked at the servo valve as a rapid way to determine if a piping problem exists. Return line pressure at the motor should be less than 100 psi when the motor is rotating.

Electric Drive Wiring for electric drives does not pose the same problems as hydraulic drives. However, motor armature cables carry currents with very fast rise times from pulse-width modulation (PWM) DC drives and brushless DC drives. These cables should be isolated as much as possible from tachometer feedback cables and position transducer (resolver or Inductosyn scales) cables.

Open Loop Battery boxes can be used for both hydraulic and electric drives.

Hydraulic Drive The battery box connected to the servo valve can be used to move the machine slide back and forth. A normal machine slide will move (break away) with about þ0.004 A or less. This current may not necessarily be symmetrical around zero. The servo valve mechanical null adjustment may cause a slight servo shift if it is not centered.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Electric Drive For DC electric motors, the breakaway current should be approximately less than one-half the rated current. Excessive breakaway currents in either drive indicate a sticking or binding machine slide. Vanes or pistons in a hydraulic motor and armature winding slots in electric motors will cause a cogging effect when these motors are operated as an open loop. The cogging is noticeable at low speeds. It is normal and should not be a cause for alarm. A properly operating tachometer loop will smooth the open-loop cogging action. Machine slide stick-slip can be observed with a battery box in cases of friction slides. With antifriction slides (hydrostatic or roller bearing) stick-slip should not be a problem. In the case of friction slides a breakaway current two or more times the run current can cause a position loop null hunt if the stick-slip is inside the position loop. Hydraulic drives usually have some amount of internal leakage. This leakage is a form of damping for the hydraulic resonance in the drive. It will be of some benefit in damping and will permit higher gains to be used. When the leakage becomes excessive and variable, the low feed rates may become jerky. This is the same resultant effect obtained from varying loads on a lowgain velocity servo loop. The cure is the same for both situations. The velocity loop gain (and thus the bandwidth) should be set to its highest feasible setting. With some hydraulic motors the leakage has been found to be excessive and the motor should be replaced. Some control suppliers make an unnecessary practice of adding cross-port leakage to all hydraulic motors. Cross-port leakage in the form of a drilled set-screw plug in the servo valve manifold should only be used if needed. Cross-port leakage reduces the sensitivity of a drive and if required should be kept as small as possible. The diameter of the drilled plug should be about 0.001 in. With the larger hydraulic motors, experience has shown that a 0.0025- to 0.003-in. diameter orifice will suffice. In general the setup and troubleshooting of electric drives has been found to be simple and reliable. The numerous variables associated with hydraulic drives have proven to be challenging at times.

Closed Position Loop Each control manufacturer will probably have a different technique for closing the position loop. The one thing in common to all controls is that the position loop gain must be set at a predetermined value. The techniques vary for setting this gain. It is quite common to set the gain by measuring the position lag (command position7actual position, called the following error)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

at some given speed. In other cases the output position responses to a step input position may be measured and recorded on a chart recorder. The position-loop gain is then adjusted until the desired response is observed on the chart recorder. Whatever the techniques used to set the position-loop gain, the control manufacturer will have a step-by-step procedure for these adjustments. Once the position-loop gain is set to the prescribed value, the servodriven axis may not perform properly for a variety of reasons. These are some symptoms of faulty position loops: 1. The servo drive may be unstable or it may null hunt at a low frequency. 2. The feed rates may not be smooth. 3. The drive may not be stiff enough. 4. The drive may not position accurately. To relate all the possible faults to the symptoms would result in endless discussion. The next section attempts to relate different kinds of problems to their possible causes.

5.2

PROBLEMS: THEIR CAUSES AND CURES

Table 1 illustrates the causes of problems.

Poor Surface Finish Causes: Other than tooling problems, a poor surface finish can result from a poor low-feed contouring control. If the low-feed-rate contouring control does not operate smoothly it can be defined as a drive with poor drive resolution (the difference between breakaway error and run error is too large). The smaller the control signal required to cause the drive to move, the smoother is the drive and the better the surface finish. The surface finish is also related to the bandwidth of the velocity loop and the position loop. Poor surface finish is very often the end result of a low gain (and thus low bandwidth) tachometer loop. Stiction and high-friction machine ways can also contribute to poor surface finish. A machine way is the surface area where a machine slide makes contact with the structure of the machine. Nonlinearities such as nonuniform leakage, as found in hydraulic motors, can definitely cause surface finish problems. Cures: It is recommended that all feed drives use the highest tachometer loop bandwidth (highest gain) possible. A wide bandwidth

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Table 1

Causes of Problems Problem

Possible causes 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Mechanical backlash Stick-slip Low spring rate Undamped hydraulic resonance Too much hydraulic leakage Too much friction Tach loop gain too low Tach loop gain too high Position loop gain too high Position loop gain too low

Note: X, Definite cause; p, possible cause.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Poor surface finish

Unstable position loop

Null hunt

p p p p X p X p p X

X

X X

Tach loop growl

X X p

X X

Feed rate not smooth

Low stiffness

Insufficient accuracy

p p p p X p X p p X

X

p p p

p p X X

X p X

X

X

tachometer loop will minimize the nonlinear effects of hydraulic motor leakage. The machine slide should have no lost motion (backlash) in the mechanical drive train, and the machine ways should be antifriction (roller bearings, hydrostatic, Rulon way liners, etc.).

Unstable Position Loop Causes: There are several causes for an unstable position loop, with the most obvious being too high a position-loop gain for a given system. When resolver feedback is used at the drive motor, the nonlinear mechanical problems such as lost motion, stiction, and low spring rates will not affect the stability of the drive. However, when direct slide feedback is used (using Inductosyn scales) these mechanical nonlinearities are inside the position loop. Cures: For contouring servo drives the position-loop gain should be in the ‘‘soft servo’’ range (0–2 ipm/mil). The machine dynamic problems can be minimized by using a low position-loop gain when direct machine slide position feedback is used. Some control manufacturers use 0.6 ipm/mil position-loop gain on all axes. In general, most large machines with direct feedback should have position-loop gains of about 1 ipm/mil. It is very important that lost motion (backlash) be absent from drives with machine slide feedback. If a wound-up gear train should lose its windup it will probably result in an unstable position loop.

Null Hunt Causes: A null hunt is a form of instability. A null hunt is a very-lowfrequency oscillation sometimes called a limit cyle oscillation. Lost motion (backlash) can cause a null-hunt condition when combined with large friction forces. Usually a backlash condition inside a position loop results in an unstable drive. The fact that the period of oscillation may be fast or slow is not significant. Stick-slip on a machine slide can definitely cause a null hunt. In general, a null hunt is possible when the breakaway friction is at least twice the running friction. The null hunt associated with stick-slip can also be aggravated by a spring (unstiff drive screw) inside the position loop. With hydraulic drives, excess leakage will cause an unstable position loop, which will oscillate at a low frequency and appear as a null hunt. Cures: For most cases of null hunt caused by backlash, antibacklash gearing (wound-up gearboxes) should be used. When the windup is lost in a wound-up gearbox the position loop is almost guaranteed to be unstable. Friction-type slides should be avoided. For machine ways using roller

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

bearings, hydrostatic ways, or Rulon way liners, null hunt from stick-slip should not be a problem. Excessive cross-port leakage on hydraulic servo drives will cause a null hunt, and care should be taken that if an orifice is required for hydraulic damping the orifice is in place. There have been cases where the orifice was omitted. The orifice is located in the servo valve manifold. Hartman motors have enough internal leakage and do not require additional cross-port leakage.

Tachometer Loop Growl Cause: The instability of a velocity loop will cause an oscillation at a frequency high enough to be audible, thus the term ‘‘tach loop growl.’’ The most significant cause is trying to set the loop gain too high for the existing parameters. Cures: The most obvious cure for an unstable tachometer loop is to lower the loop gain. However, if the loop gain must be lowered the total contouring drive will have reduced performance.

Feed Rate Not Smooth Causes: There are numerous causes for an unsmooth feed rate. Excessive and varying hydraulic leakage can also cause a varying low feed rate in hydraulic drives. Other things, such as the drive nonlinearities of stick-slip, lost motion, low mechanical resonances can affect the drive smoothness. Cures: A prime consideration is to maintain as high a tachometer loop gain (thus bandwidth) as possible assuming other restrictions, such as underdamped hydraulic resonances and mechanical resonances, have been dealt with.

Low Stiffness Cause: Stiffness as referred to here is the servo drive loop stiffness. It is a measure of how many lb–in. of torque is required to rotate the motor shaft a given number of degrees. Too much leakage will reduce the stiffness of a hydraulic drive. On a machine, the stiffness measured at the motor increases by the square of the drive ratio. The stiffness can also be measured at the machine slide where the stiffness increases by the square of the ratio/lead. Drive stiffness is discussed in depth in Part II.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Cures: To maintain adequate stiffness the design of the drive must include the proper sizing. After the drive is assembled the only variables left to adjust are the loop gains.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

6 Conclusion: Machine Feed Drives—An Integral Part of a Machine Control System

6.1

ADVANCES IN TECHNOLOGY

In the post–World War II years, machine and control builders were still provincially minded in building controls and machines as separate entities, believing that a control could be ‘‘hung’’ on a machine with few problems. Advances in technology have shown that the separate elements of control, drive, and machine must be integrated into the total system concept. Each of these elements is dependent on the others. It is the purpose of this discussion to illustrate how feed drives serve the purpose of system interface and prime mover through three areas, namely, selection of the drive, sizing the drive, and finally, evaluating the performance of the drive and total system. In selecting the feed drive, it is appropriate to ask the question, which drive—electrical or hydraulic? In the process of selecting a drive, it is worthwhile to classify the various types of feed drives. Electric drives can be classified into three main types, namely, the DC drive, the pulse-width modulation (PWM) DC drive, and the brushless DC drive. One of the most popular drives has been the hydraulic drive. The greatest share of these drives have been the servo-valve drives, because of the minimal trapped oil volume and lower hydraulic resonance than in hydraulic servo pump drives.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The electric and hydraulic drives have the greatest application for servo drives. A natural question might be, why a servo drive at all? Some of the reasons are: 1. A servo drive will be required for numerical control applications. 2. The servo drive will result in improved drive resolution. 3. Improved drive stiffness will be realized with servo drives, eliminating the need for auxiliary clamps. 4. Positioning time will be greatly reduced. 5. Acceleration can be controlled with the servo drive. 6. The servo drive is easier to apply in different applications because of its modular design. Some historical background would be appropriate prior to discussing specific comparisons in the selection process. In the post World War II era machine feed servo drives were motor-generator electric drives such as the Amplidyne, Rototrol, and Metadyne. On large machines, these drives were limited in performance. Advanced hydraulic drives using high torque-to-inertia ratio servo motors and high-response servo valves quickly overshadowed the rotary electric servo drive. The outstanding advantage of the hydraulic drive was its higher performance, although it had other advantages such as a large motor torque capacity in a small package. As the state of the art of numerical control advanced, with its ever-increasing demands for feed drive performance and accuracy, hydraulic drives seemed to fit the specifications best. Improved hydraulic motors were developed, with less breakaway torque and more uniform low-speed characteristics. In addition, servo valves showed improved reliability in industrial environments. However, the machine user was becoming dissatisfied with fluid maintenance and with the noise problems associated with hydraulic drives. One of the most frequent complaints from the user of hydraulic drives was the noise problem. It grew to such magnitude that enforcement standards were imposed on suppliers to reduce the noise level of hydraulic servo systems. The Walsh–Healy Public Contracts Act, which became effective May 20, 1969, established an upper limit on noise level as a stimulus to noise abatement efforts. The noise problem was probably the greatest deterrent to the use of hydraulic drives. Despite these problems with hydraulic drives, the electric drives’ apparent lack of performance capability made them take a back seat for a considerable time. With the appearance in the 1960s of the low-inertia slotless-armature DC electric drive, however, it looked like electric drives might be due for a revival. These motors had high torque, excellent lowspeed characteristics, and high performance. However, their high torque was

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thermally limited to a 1-sec rating. Performance of the low-inertia drive motors was comparative with their hydraulic counterparts. After several years of evaluation throughout industry, it appeared that the low-inertia DC motor had advantages for machine feed drive applications that require high-performance situations, where many rapid accelerations of low mass loads are the mode of operation. These motors were used briefly on medium-size machines. For the large machine, however, the low-inertia DC servo drives did not provide sufficient feed thrust. The low-inertia DC motor also provided a peak torque at high acceleration rates, which, if not compensated for, could destroy the mechanical components of the feed drive. Therefore, controlled acceleration had to be provided with these drives. Accordingly, it appeared that a DC electric drive with ample torque capacity and less performance than the low-inertia slotless-armature DC electric drive would better suit the machine feed drive market. These motors were referred to as high-torque, low-speed DC motors. Subsequently, during the 1970s and to the present, the silicon controlled rectifier (SCR) DC drive, the PWM DC drive, and the brushless DC drive were introduced to the machine drive market. With this revival of electric drives, making the proper choice for a specific application required detailed knowledge about some of the more important requisites of feed drives and about some of the significant advances in drive motors, amplifiers, and control techniques. Such knowledge can be obtained from a comparative study of a number of the newer types of drive motors. They are compared here for rated torque, maximum torque, inertia, acceleration characteristics, drive stiffness, and thrust requirements.

6.2

PARAMETERS FOR MAKING APPLICATION CHOICES

Rated torques for the drive motors are compared in Figure 1, which shows the difference between several high-torque, low-speed DC and brushless DC electric motors, and several hydraulic motors. The hydraulic motor torque rating is limited by the hydraulic pressure, and these ratings are also continuous. Electric motors have a continuous rating and a higher thermally limited rating. In addition, the electric motors have an absolute maximum rating, which is available for short-duration forcing torques needed in acceleration and deceleration. Relative inertias are compared in Figure 2. The low inertia is directly related to the performances capabilities of drive motors. It is important to note in Figure 3 that performance capabilities are modified as a load inertia

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Fig. 1

Torque characteristics comparison.

is connected to the drive motor. There is a varying reduction in performance capability for a given load, as illustrated. The electric motors have two performance limitations, electrical and mechanical, with the latter being reduced with increased inertia load. Hydraulic drives also have a performance limitation related to load inertia, which is referred to as the

Fig. 2

Inertia characteristics comparison.

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Fig. 3

Loading effects.

hydraulic resonance. The reduction in performance with load, however, is much more pronounced with the hydraulic motors. Therefore, the advantage of high performance with the low-inertia drive motor can only be realized if the motor does not see large reflected inertias at the rotor. As performance improves, the acceleration capacity also increases. Maximum motor acceleration is directly related to the performance and the maximum velocity. Acceptable acceleration rates vary with the application. Some machine builders limit the acceleration to 0.1 G (gravity) or 0.2 G, which is equivalent to a rate of 2320 ipm/sec and 4640 ipm/sec, respectively. If a large machine has a 100,000-lb load, for example, that is being accelerated at 0.1 G or at the rate of 2320 ipm in 1 sec, it is not too difficult to imagine the excessive forces involved. Therefore, feed drives must have some form of controlled acceleration. For the soft servo with low gain and extended error-control features, acceleration is limited. With the highperformance, high-gain feed drives referred to as hard servos, the acceleration must be limited through programming or control techniques. A further requirement is that machine feed drives should have sufficient static stiffness to be insensitive to load disturbances. In addition, a feed drive in a numerical control system must remain stationary or clamped when not in motion. Also, during the standstill period, the axis at rest must resist the load disturbances caused by the reaction forces of the other axes. Relative feed drive stiffnesses are compared in Figure 4, which represents the relative amount of torque that will be developed at the machine drive

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Fig. 4

Drive stiffness comparison.

screw for a load disturbance on the drive screw resulting in an angular deflection of 1 radian (rad). The presence of a high-gain tachometer minor loop adds greatly to the drive stiffness, since the static stiffness is the product of the position loop gain, the velocity loop gain, and the square of the gear ratio when a gearbox is used. A very important part of sizing a feed drive properly is selecting a motor rating large enough to provide the required feed force. For drilling applications, feed forces of 10,000 lb-force are not uncommon. With a surface feed per minute (sfm) of 30, the suggested feed force should be 1000 lb per usable cutting horsepower. The most frequent questions asked by the machine builder concerning drives relate to the choice between hydraulic, DC SCR, DC PWM, or brushless DC electric drives. With the new electric drive motors (DC or brushless DC) it is possible to meet all the requirements of performance and load with electric drives, provided sufficient engineering expertise is brought to bear on the problem. The important criteria in selecting the type of drive is the application of the drive. There are a number of factors the system designer should consider in selecting a drive. Drive motor size may be important. The machine designer may object to the larger physical space requirements of the electric motor. Weight can also play an important role in the selection of a drive motor. Heavy drive motors could be a detrimental factor in the area of machine structural dynamics. Hydraulic motors can operate at a maximum

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

torque continuously, whereas DC electric motors operated from SCR amplifiers must be derated. In general, DC motors must be derated 60% of their rating when used with full-wave, single SCR amplifiers, and 80% of their rating with three-phase, half-wave SCR amplifiers. Brushless DC motors can operate at their rated torque continuously. The DC SCR drive has a 400% rating for forcing or acceleration, while the brushless DC drive has an equivalent 200% rating. Another consideration in selecting a drive is the effects of noise. There are two kinds of noise to consider. While audible noise is generally associated with hydraulic drives, electric noise is an important consideration with PWM DC and brushless DC electric drives. Government regulations have initiated an attack on audible noise pollution, and to meet and surpass these requirements, quiet hydraulic power supplies are available. Electrical noise can cause loss of operation and possible catastrophic failures with digital control systems. Next to diathermy machines, the PWM DC and the brushless DC amplifier are electrical noise generators of the worst kind, and special isolation practices must be incorporated. Standards for the installation of electrical control systems are available from the Institute of Electrical and Electronic engineers (IEEE Std. 518-1982). In addition to the power duration requirements for electric drives, both types of drives have performance limitations. Hydraulic drive performance is usually limited by the hydraulic resonance where the resonance is a function of motor inertial load, motor size, and oil under compression. Electric drives using SCR amplifiers have an on–off timing function. With SCR switching on and off, dead time exists in each cycle of AC power where no power will be applied to the motor from each of the silicon controlled rectifiers. The on–off switching function of the SCR amplifier is referred to as the transport lag of the amplifier and adds phase shift to the servo drive. In addition to the inertial time constant limitation of the DC motor, the transport lag of the SCR amplifier will also limit the performance of the DC drive to about 10 Hz on large machines. The brushless DC drive is not limited by current switching and therefore has performance bandwidths of about 30 Hz on large machines. The DC SCR drive has a current and torque rating of about 400% for the purpose of forcing (acceleration). The brushless DC drive and the PWM-DC drive have an amplifier with transistors, which limits the overall drive to about 200% of rated torque. The second area under consideration is the sizing of the drive. Once the drive has been selected, it is necessary to size the drive properly, and criteria for sizing are based on the proper performance and feed thrust. Sizing a drive requires the expertise of a system analyst; however, the

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technology involved has been reduced to simple computerized programs to be used by design engineers. In summarizing, it would be appropriate to recap some of the more significant points discussed. Today’s industrial machines should be considered as a machine system comprised of a control, drive, and machine. The selection of the type of servo drive should be based on the application. Evaluation of the low-inertia slotless-armature DC motors proved inadequate for industrial servo machine applications. Low-inertia brushless DC motors (e.g., neodymium iron boron or samarium cobolt armatures) have many industrial applications for high-performance applications where it is recommended that the motor armature be matched to the reflected load inertia through a ratio. For an inertia mismatch of up to about four to one (reflected load inertia to motor armature), ceramic magnet armature motors are recommended. Advances in the area of electric drives continue with improvements in amplifiers and motors. The introduction of the insulatedgate bipolar transistor (IGBT) to brushless DC drives and vector-controlled AC drives has increased the torque capacity of these drives by about 50%. New innovations will continue unabated.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

II ADVANCED APPLICATION OF INDUSTRIAL SERVO DRIVES

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7 Background

7.1 INTRODUCTION Part I discussed the basics of industrial servo drives from a hardware point of view. Physical parameters and practical applications were discussed. Part II repeats some of the things in Part I but from a mathematical point of view. The advanced application of industrial servo drives requires the use of differential equations to describe mechanical, electrical, and fluid systems. As applied to servo drives there are numerous academic techniques to analyze these systems (e.g., root locus, Nyquist diagrams, etc.). In working with industrial machinery we live in a sinusoidal world with such things as structural machine resonances. Thus frequency analysis is used in Part II to describe and analyze industrial servo systems. To solve the differential equations describing the physical systems of servo drives, transformation calculus is used to obtain the required transfer functions for the components of servo drives and in analyzing the servo system. There are a multitude of academic textbooks and university courses dealing with feedback control. It is the purpose of Part II to show how the fundamentals of servo drives described in the many academic sources are applied in practice.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

7.2

PHYSICAL SYSTEM ANALOGS, QUANTITIES, AND VECTORS

As a beginning, analogous parameters for an electrical system, a linear mechanical system, and a rotary mechanical system are compared for future reference. In all physical systems there are scalar quantities and vector quantities. Vector quantities can be represented as complex numbers on a complex plane, in polar form, or in exponential form as in Eq. (7.2-1) to (7.2-12).

Scalar Quantities (a) Magnitude only (b) Examples: length of a line, mass, volume

Vector Quantities (a) Magnitude and direction (b) Examples: force, voltage, weight, velocity

Complex Numbers A vector can be represented by its rectilinear components.

Fy Fx cos y ¼ F F Fy ¼ F sin y Fx ¼ F cos y qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jFj ¼ ðFx Þ2 þ ðFy Þ2 sin y ¼

Fy y ¼ tan
1 Fx qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fy ;F
¼ ðF cos yÞ2 þ ðF sin yÞ2 ffy ¼ tan
1 Fx A vector can also be represented on a complex plane.

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(7.2-1) (7.2-2) (7.2-3) (7.2-4) (7.2-5)

F
¼ Fx þ jFy F
¼ ðjFj cos yÞ þ jðjFj sin yÞ pffiffiffiffiffiffiffi j ¼
1

(7.2-6) (7.2-7) (7.2-8)

A vector can be represented in polar form. F
¼ jFjffy qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fy F
¼ ðFx Þ2 þ ðFy Þ2 ffy ¼ tan
1 Fx

(7.2-9) (7.2-10)

A vector can be represented in exponential form. F
¼ jFje jy

y ¼ tan
1

Fy Fx

(7.2-11)

Thus: Rectangular F cos y þ jF sin y ¼

7.3

Polar jFjffy

¼

Exponential jFje+jy

(7.2-12)

DIFFERENTIAL EQUATIONS FOR PHYSICAL SYSTEMS

The differential equations for physical systems can be written for individual servo components such as motors and amplifiers or for complete multiloop servo drives. In actual practice servo drive block diagrams can be put together with a combination of individual transfer functions representing the differential equation of the separate servo drive components. These individual transfer characteristics can, in general, be represented by singleorder or second-order blocks in the overall servo block diagram. A singleorder differential equation or transfer characteristic results from a circuit having a single time-varying parameter. Likewise, a second-order transfer characteristic results from a circuit (mechanical, electrical, or fluid) having two time-varying parameters. Most servo drive components can be represented by either a first-order transfer characteristic (or transfer function) or a second-order transfer function. A transfer function is, by definition, the ratio of the Laplace transform of the output to the Laplace transform of the input. In general a transfer function is a shorthand solution for solving differential equations. The derivation of a single-order electrical circuit transfer function having an inductor as a single time-varying parameter is shown in Figure 1. The steady-state equations for the output voltage, based on a sinusoidal

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Fig. 1 (a) Inductive/resistive circuit. (b) Circle diagram.

input voltage, are 7.3-1 to 7.3-17. Replacing the jo term by the differential operator p or the Laplace transform operator s changes Eq. (7.3-10) to the transform function of Eq. (7.3-19). This transform function can be represented in the frequency response of Figure 2. To illustrate a second-order circuit, the circuit of Figure 3 with two time-varying parameters has the differential equation of Eq. (7.3-20). The output voltage for the unique case of a sinusoidal input voltage is Eq. (7.325). A second-order mechanical circuit for linear translation is shown in Figure 4. Eq. (7.3-30) is the differential equation for this circuit. Assuming a sinusoidal input, the output displacement is Eq. (7.3-35). Lastly, a rotary mechanical circuit is shown in Figure 5. The output angular motion is Eq. (7.3-44). These examples of single-order and second-order circuits are to illustrate that individual servo drive components can be represented mathematically by differential equations, transfer functions, or the absolute case for a sinusoidal input driving source. The circuit shown in Figure 1 is further described for three cases of absolute, vector, or differential analysis followed by the response of this circuit to a step input and a ramp input. ei ¼ iRe þ iXL

(7.3-1)

ei ¼ iZ

(7.3-2)

XL ¼ 2pf ffoL Zffy ¼ Zðcos y þ j sin yÞ

(7.3-3) (7.3-4)

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Fig. 2

Single-order frequency response.

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Fig. 3

Inductive/capacitive/resistive circuit.

Fig. 4 Spring/mass diagram (linear).

Fig. 5 Spring/mass diagram (rotary).

Note: a þ jb ¼ cffy: Z

  Re XL þj ¼ Re þ jXL Z Z

(7.3-5)

ei ¼ iðRe þ jXL Þ ¼ iðRe þ joL Þ ei ei  i¼ ¼  Re þ joL Re 1 þ joL L Re

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(7.3-6) (7.3-7)

L ¼ T1 (7.3-8) R

(7.3-8)

ei ei ¼ Re ð1 þ joT1 Þ Z ei eo ¼ R e i ¼ ð1 þ joT1 Þ

i¼

(7.3-9) (7.3-10)

For 1 þ joT1 ¼ Z eo 1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ei 1 þ ðoT1 Þ2 ffy ¼ tan
1

(7.3-11) (7.3-12) oT1 1

eo 1 oT1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ff
y ¼ tan
1 ei 1 2 1 þ ðoT1 Þ T1 ¼

1 o1

eo 1
1 o ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi ff
y ¼ tan o1 ei o 1þ o 1

(7.3-13)

(7.3-14) (7.3-15)

For o ¼ o1 ; ;y ¼
45 : 1 Amplitude ratio ¼ pffiffiffi ¼ 0:707 2 eo  ; ¼ 0:707ff
45 ei An example of a complex plane plot is:

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(7.3-16) (7.3-17)

eo ¼ Ki ¼

Re e i ei ei ¼ ¼ Re þ pLi 1 þ RLi p 1 þ RLi s e e

eo 1 ¼ ei ð1 þ tsÞ Z di i dt ei ¼ Re i þ Li þ dt Ca Z d 1 ¼p ðf Þt dt ¼ dt p

(7.3-18) (7.3-19) (7.3-20) (7.3-21)

For sinusoidal input, p ¼ jo  ei ¼ i Re þ joLi þ

1 joCa



ei ei joCa ¼ 1 R þ joLi þ joC ½jRoCa þ ðjoÞ2 LCa þ 1 a ei joRCa eo ¼ 2 ½ðjoÞ Li Ca þ joRCa þ 1

i¼

ðjoÞ2 2d Quadratic : þ jo þ 1 2 o sffiffiffiffiffiffiffiffiffiffi o 1 o¼ Li Ca dx dt d2x dx Ma 2 ¼ F
kx
c dt dt d2x dx Ma 2 þ c þ kx ¼ F dt dt

SF ¼ Ma ¼ F
kx
c

(7.3-22) (7.3-23) (7.3-24) (7.3-25) (7.3-26) (7.3-27) (7.3-28) (7.3-29) (7.3-30)

For f ¼ F sin ot, the differential d ¼ jo ¼ p dt

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(7.3-31)

where p is the differential operator. Ma p2 x þ cpx þ kx ¼ F 2

ðMa p þ cp þ kÞx ¼ F 2

½Ma ðjoÞ þ cjo þ kx ¼ F F F=k x¼ ¼ 2 2 ðjoÞ c ðjoÞ Ma þ joc þ k k=Ma þ jo k þ 1

(7.3-32) (7.3-33) (7.3-34) (7.3-35)

For quadratic: "

# ðjoÞ2 2d þ jo þ 1 o2 o sffiffiffiffiffiffiffi k 2d on ¼ ¼ c=k Ma o c o d¼ k 2 ST ¼ Ja ST ¼ J

d 2 y0 dy0 ¼ GT ðy1
y0 Þ
b 2 dt dt

d ¼ p Differential operator dt Jp2 y þ bpy0 þ GT y0 ¼ GT y1 2

½Jp þ bp þ GT y0 ¼ GT y1

(7.3-36)

(7.3-37) (7.3-38)

(7.3-39) (7.3-40) (7.3-41) (7.3-42)

For: y1 ¼ y1 sin ot p ¼ jo ½JðjoÞ2 þ bjo þ GT y0 ¼ GT y1 GT y1 y1 ¼ y0 ¼ 2 2 ðjoÞ b ðjoÞ J þ job þ GT ðGT =JÞ þ GT jo þ 1

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(7.3-43) (7.3-44)

For quadratic: "

# ðjoÞ2 2d þ þ1 o2 o rffiffiffiffiffiffiffi GT 2d b o¼ ¼ J o GT b o d¼ GT 2

(7.3-45) (7.3-46) (7.3-47)

Absolute

Vector

jei j ¼ Ri þ xL i

e
i ¼ Ri þ jxL i

jei j ¼ jijðRi þ xL Þ jei j ¼ jijðR þ 2pfL Þ

e
i ¼ i
ðR þ jxL Þ e
i ¼ i
ðR þ 2pfL Þ

jei j jij ¼ ðRþoLÞ

e
i i
¼ RþjoL

jei j jij ¼ jZj

i
¼ Ze
i

jei j ffi jij ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2

Rffi i ¼ pi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2

Rjei j ffi jeo j ¼ jijR ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2

R
ei e
o ¼ i
R ¼ RþjoL

jei j jeo j ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð1Þ2 þðoRLÞ

e
o ¼

di eiðtÞ ¼ Ri þ ff dt d p ¼ dt eiðtÞ ¼ iðtÞ ðR þ LpÞ e

iðtÞ iðtÞ ¼ ðRþLpÞ

e ff
tan
1 XL

R þðoLÞ

R þðoLÞ

R þðoLÞ

L R

Differential

e
i

ð1þjoRLÞ

Re

iðtÞ eoðtÞ ¼ iðtÞ R ¼ ðRþLpÞ

eoðtÞ ¼

eiðtÞ

ð1þRLpÞ

¼ T1

jei j jeo j ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðT1 oÞ þð1Þ

iðtÞ eoðtÞ ¼ ðT1 pþ1Þ

e
o ¼ ð j oe
iþ1Þ

eoðsÞ ¼ ðT1iðsÞ sþ1Þ

o1

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

e

e
i e
o ¼ ðT1 joþ1Þ

e

Examples Absolute Case ei ¼ 10 V at 60 Hz R ¼ 5O L ¼ 0:2 H ei 1065 50 50 jeo j ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ 0:66 V 25 þ 5679 7513 52 þ ð377Þ2 R2 þ ð2pfLÞ2 ei 10 10 10 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jeo j ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ r ¼ 0:66 V ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼  2 15:1 227 þ 1 0:262pf ðT1 oÞ2 þ 1 þ1 R

Vector Case ei ¼ 10ffy V R ¼ 5O L ¼ 0:2 H f ¼ 60 Hz e
i 6R e
o ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R2 þ ðoLÞ2 ff tan
1 e
o ¼

¼ oL R

10ff0 65 ¼ 0:66ff
86 75:3ff86

10ff0 10ff0 10ff0 e
i ¼L ¼ ¼ ¼ 0:66ff
86 T1 jo þ 1 R 2pfj þ 1 15j þ 1 15:1ff86

Differential Case R ¼ 5O L ¼ 0:2 H eiðsÞ eiðsÞ eiðsÞ ¼ ¼ 0:2 eoðsÞ ¼ ðT1 s þ 1Þ ð0:04 s þ 1Þ 5 sþ1

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

For the Case of Sinusoidal Input eiðtÞ ¼ E sin ot E ¼ 10 V Eo eiðsÞ ¼ 2 ðs þ o2 Þ Eo 25610o eoðsÞ ¼ 2 ¼ ðs þ o2 Þð0:04 s þ 1Þ ðs2 þ o2 Þðs þ 25Þ " # e
25t sinðot
yÞ eoðtÞ ¼ 250o þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð25Þ2 þ o2 o ð25Þ2 þ o2 o ¼ 377



e
25t sinðot
yÞ eoðtÞ ¼ 2506377 þ 142;750 142;440
1 o Phase ¼ y ¼ tan 25 y ¼ 86



E
25t <<< 1 142;750 94;250 eoðtÞ ¼ sinðot
86 Þ 142;440 eoðtÞ ¼ 0:66 sinðot
86 Þ eo ¼ 0:66 V at phase of
86

For Case Where ei(t) ¼ Step ¼ E (Figure 6a) eiðtÞ ¼ E E eiðsÞ ¼ s T1 ¼ 0:04 sec E 25E ¼ eiðsÞ ¼ sðT1 s þ 1Þ sðs þ 25Þ eoðtÞ ¼ Eð1
e
25t Þ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 6a

Step response.

Fig. 6b

For case where eiðtÞ ¼ RAMP ¼ Et .

For Case Where ei(t) ¼ Ramp ¼ Et (Figure 6b) eiðtÞ ¼ Et E eiðsÞ ¼ 2 s eoðsÞ ¼

7.4

25E Eðe
25t þ 25t
1Þ ¼ 25 þ 25Þ

s2 ðs

ELECTRIC SERVO MOTOR TRANSFER FUNCTIONS AND TIME CONSTANTS

In Section 7.3 some simple first- and second-order mathematical descriptions were given to illustrate how the corresponding transfer functions were developed. In Part I, the components of servo drives were described from a physical point of view. These servo-drive components are now described from a mathematical or transfer function point of view. In the analysis of electric servo-drive motors, the equations for the motor indicates the presence of two time constants. One is a mechanical time constant and the other is an electrical time constant. Commercial servo-motor specifications usually list these two time constants. However, it should be cautioned that these two time constants as given in the specifications are for the motor alone with no load inertia connected to

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 7 Ramp response.

the motor shaft. Since these two time constants are part of the motor block diagram used in servo analysis, it is important to know the real value of the time constants under actual load conditions. There are two types of servo motors to consider. The first is the classical DC servo motor and the second is the AC servo motor often referred to as a brushless DC motor. The brushless DC motor is a threephase synchronous motor having a position transducer inside the motor to transmit motor shaft position to the drive amplifier for the purpose of controlling current commutation in the three phases of the motor windings. A derivation of the motor equations and the electrical and mechanical motor time constants will be discussed for the DC motor followed by a discussion for the AC motor. The DC motor equivalent diagram is shown in Figure 8, where: eI ¼ Applied voltage (volts) ia ¼ Armature current (amps) JT ¼ Total inertia of motor armature plus load (lb-in.-sec2) Ke ¼ Motor voltage constant (V/rad/sec) KT ¼ Motor torque constant (lb-in./A)

Fig. 8 Equivalent diagram for a DC motor.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

La ¼ Motor winding inductance (Henries) Ra ¼ Armature resistance (ohms) TL ¼ Load torque (lb-in.) Vm ¼ Motor velocity (rad/sec) a ¼ Acceleration (rad/sec2) The steady-state (DC) equations are: ei ¼ ia Ra þ Ke Vm

ðVoltage equationÞ

T ¼ Torque ¼ ia KT ¼ Ja

ðTorque equationÞ

(7.4-1) (7.4-2)

For the general case, the differential equations are: dia þ Ke V m dt d Laplace operator S ¼ dt ei ¼ Ra ia þ La Sia þ Ke Vm

ei ¼ i a R a þ L a

(7.4-3)

(7.4-4)

ei ¼ ðRa þ La SÞia þ Ke Vm   La S þ 1 Ra ia þ Ke Vm ei ¼ Ra

(7.4-5)

Also : T ¼ KT Ia ¼ JT a ¼ JT Vm S T JT Vm S ¼ ia ¼ KT KT

(7.4-7)

Combining equations gives:   La JT V m S S þ 1 Ra ei
K e V m ¼ Ra KT

(7.4-6)

(7.4-8)

(7.4-9)

Rearranging results in: ðei
Ke Vm ÞKT ¼ Vm La ðR S þ 1ÞRa JT S a

(7.4-10)

This last equation can be represented in block diagram form as in Figure 9. C G The closed-loop equation ðR ¼ 1þGH Þ for this block diagram is: Vm K h  T i ¼ La ei Ra JT S Ra S þ 1 þ Ke KT

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(7.4-11)

Fig. 9 DC motor block diagram.

Rearranging Eq. (7.4-11) gives: Vm KT ¼ 2 L a ei ðRa JT Ra ÞS þ JT Ra S þ Ke KT

(7.4-12)

Dividing by Eq. (7.4-12), KeKT gives: 1

Vm   Ke   ¼ Ra JT La ei S 2 þ Ra JT S þ 1 Ke KT

Ra

(7.4-13)

Ke KT

From the last equation, the motor mechanical time constant, tm, is: tm ¼

Ra JT ½sec Ke KT

(7.4-14)

The total inertia, JT, is the sum of the reflected inertia to the motor shaft plus the motor inertia. The resistance, Ra, is the motor winding resistance plus the external circuit resistance. Thus the motor mechanical time constant is summarized as: P tm ¼

R a JT ½sec Ke KT

(7.4-15)

Also, the motor electrical time constant is: La te ¼ P ½sec Ra

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(7.4-16)

Therefore, the closed-loop motor equation can be expressed as: 1

Vm Ke ¼ ei tm te S2 þ tm S þ 1

(7.4-17)

From the general equation for a quadratic: S2 2dS þ þ1 o2m om where: om ¼

(7.4-18)

pffiffiffiffiffiffiffiffiffiffiffiffiffi 1=tm te

(7.4-19)

The damping factor is: d ¼ 0:5 tm om ¼ 0:5 tm

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 1=tm te ¼ 0:5 tm =te

(7.4-20)

The mechanical and electrical time constants for a brushless DC motor have the same basic equations with some variations. For a brushless DC motor with a wye connected motor, the electrical circuit is as shown in Figure 10. The mechanical time constant is: P RPHASE JTOTAL ½sec (7.4-21) tm ¼ KeðPHASEÞ KT where: KeðPHASEÞ ¼ Motor voltage constant h i ¼ KT ¼ Motor torque constant lb
in: Arms

KeðL
LÞ V
sec RAD 1:73

RMðL
LÞ ¼ Motor resistance [ohms] P RMðL
LÞ ¼ Total motor circuit resistance ¼ 1:35RMðL
LÞ ½ohms

Fig. 10

Equivalent diagram for a WYE connected brushless DC motor.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

P RMðL
LÞ 60:5 ¼ ½ohms RMðPHASEÞ ¼ LL
L ¼ Motor inductance ¼ ½Henries JTOTAL ¼ Motor armature inertia plus the reflected load inertia at the motor shaft ¼ ½lb:-in:-sec2  P

Most manufacturers give the electrical parameters in line-to-line values. Thus some of these values must be converted to the phase values as shown in the preceding text. Summarizing, the mechanical time constant can be computed as: P RL
L Rl
l Jtotalatmotor 2 JTOTAL tm ¼ KeðL
LÞ ½sec ¼ 0:86 (7.4-22) Keðl
lÞ KT 1:73 KT The electrical time constant for the brushless DC motor is computed as: te ¼

Total inductive path LL
L ¼P ½sec Total resistive path RmðL
LÞ

(7.4-23)

Another factor affecting the mechanical time constant is the temperature. Most manufacturers specify the motor parameters at 258C (cold rating). This implies that the magnet and wires are both at room temperature. However, the motors used in industry will operate hotter, which means could reach a magnet temperature of 808C to 908C in a 408C ambient. The winding temperature is considerably more than that. Some means must be used to compensate for the motor parameters rated at 258C. For those manufacturers that offer the hot rating on motor specification parameters, they should be used to calculate the time constants. The parameters of motor resistance, torque constant, and voltage constant should be adjusted, if needed, for the hot rating. The motor resistance will increase; the torque constant and voltage constant will decrease. However, contrary to their implied name both time constants are not of constant value. Rather, they are both functions of the motor’s operating temperature. The electrical resistance of a winding, at a specified temperature, is determined by the length, gauge and composition (i.e., copper, aluminum, etc.) of the wire used to construct the winding. The winding in the vast majority of industrial servo motors are constructed using film-coated copper magnet wire. Based on the 1913 International Electrical Commission standard, the linear temperature coefficient of electrical resistance for annealed copper magnet wire is 0.00393/8C. Hence, knowing a copper winding’s resistance at a specified reference or ambient temperature, the windings at temperatures above or below this ambient temperature is given

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

by: RðTÞ ¼ RðT0 Þ½1 þ 0:00393ðT
T0 Þ

(7.4-24)

where: T ¼ Winding’s Temperature ð CÞ T0 ¼ Specified Ambient Temperature ð CÞ: Using Eq. (7.4-24), a 1308C rise (1558C–258C) in a copper winding’s temperature increases its electrical resistance by a factor of 1.5109. Correspondingly, the motor’s mechanical time constant increases by this same 1.5109 factor while its electrical time constant decreases by a factor of 1/1.5109 ¼ 0.662. In combination, the motor’s mechanical to electrical time constant ratio increases by a factor of 2.28 and this increase definitely affects how the servo motor dynamically responds to a voltage command. In consulting published motor data, many motor manufacturers specify their motor’s parameter values, including resistance, using 258C as the specified ambient temperature. NEMA, however, recommends 408C as the ambient temperature in specifying motors for industrial applications. Therefore, pay close attention to the specified ambient temperature when consulting or comparing published motor data. Different manufacturers can, and sometimes do, use different ambient temperatures in specifying what can be the identical motor. In the same published data servo motors are generally rated to operate with either a 1308C (Class B) or 1558C (Class F) continuous winding temperature. Although motors with a Class H, 1808C temperature rating are also available. Assuming the motor’s resistance along with its electrical and mechanical time constants are specified at 258C, it was just demonstrated that all three parameters significantly change value at a 1558C winding temperature. If the motor’s winding can safely operate at 1808C the resistance change is even greater because Eq. (7.4-24) shows that a 1558C rise (1808C–258C) in winding temperature increases its electrical resistance by a factor of 1.609. Hence, if the servo motor’s dynamic motion response is calculated using the 258C parameter values then this calculation overestimates the motor’s dynamic response for all temperatures above 258C. In all permanent magnet motors there is an additional effect that temperature has on the motor’s mechanical time constant only. As shown in Eq. (7.4-24), a motor’s mechanical time constant changes inversely with any change in both the back EMF, Ke, and torque constant, KT. Both Ke and KT have the same functional dependence on the motor’s air gap magnetic flux density produced by the motor’s magnets. All permanent magnet

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

motors are subject to both reversible and irreversible demagnetization. Irreversible demagnetization can occur at any temperature and must be avoided by limiting the motor’s current such that, even for an instant, it does not exceed the peak current/torque specified by the motor manufacturer. Exceeding the motor’s peak current rating can permanently reduce the motor’s Ke and KT thereby increasing the motor’s mechanical time constant at every temperature including the specified ambient temperature. Reversible thermal demagnetization depends on the specific magnet material being used. Currently, there are four different magnet materials used in permanent magnet motors. The four materials are AluminumNickel-Cobalt (Alnico), Samarium Cobalt (SmCo), Neodymium-IronBoron (NdFeB), and Ferrite or Ceramic magnets as they are often called. In the temperature range,
60 C < T < 200 C, all four magnet materials exhibit reversible thermal demagnetization such that the amount of air gap magnetic flux density they produce decreases linearly with increasing magnet temperature. Hence, similar to electrical resistance, the expression for the reversible decrease in both Ke(T) and KT(T) with increasing magnet temperature is given by: Ke;T ðTÞ ¼ Ke;T ðT0 Þ½1
BðT
T0 Þ

(7.4-25)

In Eq. (7.4–25), the B-coefficient for each magnet material amounts to: B(Alnico) ¼ 0.0001/8C B(SmCo) ¼ 0.00035/8C B(NdFeB) ¼ 0.001/8C B(Ferrite) ¼ 0.002/8C Using Eq. (7.4-25) it can be calculated that a 1008C rise in magnet temperature causes a reversible reduction in both Ke and KT that amounts to 1% for Alnico, 3.5% for SmCo, 10% for NdFeB, and 30% for Ferrite or Ceramic magnets. Like the motor’s electrical resistance, most motor manufacturers specify the motor’s Ke and KT using the same ambient temperature used to specify resistance. However, this is not always true and it is again advised to pay close attention as to how the manufacturer is specifying their motor’s parameter values. Combining the effects of reversible, thermal demagnetization with temperature dependent resistance, the equation describing how a permanent magnetic motor’s mechanical time constant increases in value with increasing motor temperature amounts to: tm ðTÞ ¼ tm ðT0 Þ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

½1 þ 0:00393ðT
T0 Þ ½ð1
BðT
T0 ÞÞ2 

(7.4-26)

Notice in Eq. (7.4-26) that the magnet’s temperature is assumed equal to the motor’s winding temperature. Actual measurement shows that this assumption is not always correct. Motor magnets typically operate at a lower temperature compared to the winding’s temperature. However this conservative approximation is recommended and used. An example will be given to illustrate a change in time constants. To raise the mechanical time constant to a 1558C temperature rating inside a Ferrite magnet motor, for example, the resistance increase will be the following: Rð155 CÞ ¼ Rð25 CÞ þ 0:00393= C6ð155
25Þ6Rð25 CÞ ¼ 1:5109 Rð25 CÞ The voltage constant Ke and torque constant KT will be lowered. Since the magnet material is 108C–158C cooler than the windings the Ke and KT will be the following: Ke ð140 CÞ ¼ Ke ð25 CÞ
0:002= C6ð140
25Þ Ke ð25 CÞ ¼ 0:77 Ke ð25 CÞ The mechanical time constant will increase by the following: tm ð155 CÞ ¼ 1:5109=ð0:77Þ2 ¼ 2:54 tm ð25 CÞ

7.5

TRANSPORT LAG TRANSFER FUNCTION

In Part I, the transport lag was described in the application of silicon controlled rectifiers (SCR). The SCR has a sinusoidal power line frequency applied to it in some form of circuit configuration such as a three-phase, half-wave amplifier circuit. Each SCR will only conduct current in one-half cycle of the line frequency. In addition, the amount of current that will conduct in the conducting half cycle is controlled by the current at the gate of the SCR. Therefore there is a resulting dead time in the conducting half cycle where no current flows. This dead time is described mathematically as the transport lag. Transport lag has the transfer function e
ts ¼ e
jot ¼ e
jy

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

where: y ¼ ot ¼ phase shift of the transport lag y ðdegÞ ¼ ot657:5 o ¼ rad=sec

ðdegÞ

The significance of the transport lag (dead time in the firing of an SCR) is that each type of SCR amplifier circuit will have an increasing phase lag versus increasing frequency. This phase lag adds to the overall phase shift of the servo amplifier, contributing to an unstable servo drive. The transport lags for four different SCR servo amplifier circuits are shown in Figure 11. The relation between the transport lag of the four types of SCR circuits and the phase lag versus frequency is shown in Figure 12. The phase lag of the SCR servo amplifier is a limiting factor in the available frequency response (servo bandwidth) of this type of DC servo drive. The transfer function for transport lag does not have any amplitude attenuation with increasing frequency.

Fig. 11

SCR circuit waveforms.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 12

7.6

Phase-shift characteristics for SCR circuits.

SERVO VALVE TRANSFER FUNCTION

In servo analysis or in system synthesis it is necessary to have a mathematical representation (transfer function) of the servo valve. The usefulness of a linear transfer function to approximate the servo valve response is well established. A servo valve is a highly complex device that has high-order nonlinear responses. The servo valve torque motor can be represented by a second-order transfer characteristic. However, the natural frequency of the first stage torque motor is high (about 700 Hz) and can be omitted since it is much higher than the normal frequencies encountered in practice for hydraulic servos. In practice the useful bandwidth of a servo valve occurs at a phase lag of 45 degrees for the same reasons as explained in Part I, Section 4.2. For

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

industrial hydraulic servo drives, the servo valve can be approximated mathematically with a first-order transfer function as Q=i ¼ Kv =ð1 þ s=ov Þ

(7.6-1)

where: Kv ¼ valve gain, in.3/sec/A ov ¼ valve bandwidth, rad/sec Q ¼ flow, in.3/sec i ¼ current, A Typical values for commercially available Pegasus servo valves are

Model 120 140 160 180

Flow rating (GPM)

Torque motor current rating (A)

Bandwidth (rad/sec)

Half the trapped* volume (in.3)

Valve leakage (in.3/sec/psi)

5 10 15 25

0.2 0.2 0.2 0.2

— 407 300 160

0.17 0.17 0.17 0.17

0.00069 0.00152 0.00193 0.00385

* Add 0.2 in.3 to the valve volume for the valve manifold.

7.7

HYDRAULIC SERVO MOTOR CHARACTERISTICS

The transfer function for the hydraulic servo motor is developed from the flow equations and is discussed in Section 12.3. As a summary, the hydraulic servo-motor transfer function is ym 1 h i ¼ Kv x Dm s22 þ 28h þ 1 oh

oh

where: sffiffiffiffiffiffiffiffiffiffiffiffiffi 2 bD2m oh ¼ V c JT sffiffiffiffiffiffiffiffi KL bJT dh ¼ Dm 2Vc b ¼ bulk modulus of oil ¼ 1 6 105 lb/in.2

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(7.7-1)

Dm ¼ motor displacement, in.3/radian JT ¼ total inertia at motor, lb-in.-sec2 KL ¼ total leakage, in.3/sec/psi Kv ¼ servo valve flow gain, in.3/sec/in. Vc ¼ half the total trapped volume, in.3 x ¼ servo valve displacement from neutral, in. Dm ¼ motor position, radians dh ¼ damping ratio, dimensionless oh ¼ hydraulic resonance, rad/sec2 The mathematical representation (transfer function) of a hydraulic servo motor is a second-order transfer function with an integration term. The second-order denominator has a hydraulic resonance. There is an index of performance for the hydraulic resonance that states that this resonance should be oh ¼ 200 rad/sec or larger. This index of performance is necessary to provide a stable servo drive. The hydraulic resonance is like having a spring inside the servo loop that can cause the servo drive to be unstable. The index of performance for hydraulic drives is discussed further in Section 7.9. The damping factor of the hydraulic resonance is another important consideration. An index of performance for the damping factor is that it should be 0.8–1.0. The damping factor for hydraulic motors is a function of the motor design. Those hydraulic motor designs with minimum internal leakage have lower damping factors. Piston-type motors have low leakage and damping factors. Roll-vane motors usually have higher and nouniform leakage as the motor rotates. In practice it is possible to add cross-port leakage to hydraulic motors and increase the damping factor. The added cross-port leakage is added in the servo valve/motor manifold. The required cross-port hole diameter in the manifold is usually about 0.001–0.002 in. Hydraulic servo-motor damping factors are covered in added depth in Chapter 11. The leakage KL is the total hydraulic leakage of the servo valve, motor, and manifold. In practice the leakage values are not always available from manufacturers’ data specifications. Vigilance may be required in obtaining the leakage values from manufacturers.

7.8

GENERAL TRANSFER CHARACTERISTICS

Thus far mathematical descriptions have been given for the DC motor, transport lag, hydraulic servo valve, and motor. All of these servo components have simple lag-type transfer functions where the output lags the input for a driving input of increasing frequency. Most servo-drive

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

components can be represented by first- or second-order transfer functions. A plot of these transfer functions in the frequency domain is shown in Figure 13. There are some other servo drive components that have not been described as yet. The simplest of these is an amplifier that is described with a transfer function of the amplifier gain K. Next is an integrator represented by a hydraulic piston or DC motor for example. The integrator has the transfer function of K/s, which has a linear reduction in gain at the rate of 20 dB per decade of increasing frequency and a 90-degree phase shift at all frequencies. The gain K occurs at a frequency of 1 rad/sec. The integrator is followed by a differentiator represented by an electric tachometer having a transfer function of Ks. The gain characteristic increases with frequency at the rate of 20 dB per decade of increasing frequency and has a phase lead of 90 degrees at all frequencies. The gain K occurs at a frequency of 1 rad/sec. A simple lead is not a usual component of a servo drive, but it has a transfer function of K(ts þ 1). Both the differentiator and the simple lead have an increasing output with increasing frequency. This condition will amplify any unwanted noise in the machine and servo drive system. Therefore a limit on the differentiation is required. The simple lead transfer function thus becomes Ks/(t1s þ 1) where 1/t1 or o1 provides a limit on the increasing output of the differentiation. The transfer function for the single lead is plotted in the frequency domain as shown in Figure 13a.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 13 (a) Response characteristics—simple lead and lag. (b) Response characteristics—second order and transport lag. (c) Response characteristic— integration and differentiation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 13

(b) Continued

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 13

(c) Continued

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

8 Generalized Control Theory

8.1

SERVO BLOCK DIAGRAMS

Having discussed individual components of industrial servo drives from an operation point of view and a mathematical descriptive point of view, the next step is to combine these components into a block diagram for the complete servo drive. The block diagram is a powerful method of system analysis. In the block diagram each component in a servo system can be described by the ratio of its output to input. Thus the output of one component is the input to the next component. To satisfy an accuracy requirement some form of feedback is required. For a velocity servo drive the output is speed. Without velocity feedback, the speed will vary greatly with changes in load. By providing negative feedback with the use of a tachometer, the feedback voltage is compared with a reference command voltage input by means of a summing junction. The difference in the two voltages (command and feedback) is the error voltage. For a velocity servo drive the error is finite. As load increases the drive slows down, the feedback voltage gets smaller, and the error increases, causing the speed regulator to speed up and maintain the required velocity. To represent the servo drive with a block diagram it is necessary to use block diagram algebra. In its simplest form the block diagram will have a forward loop block, a feedback loop block, and a summing junction as in

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Figure 1. The output controlled variable (C) ratio to the reference input (R) is B ¼ CH E ¼ C=G

(8.1-1) (8.1-2)

R
B¼E

(8.1-3)

R
CH ¼ C=G R ¼ C=G þ CH ¼ Cð1 þ GHÞ=G C G ¼ R 1 þ GH

(8.1-4) (8.1-5) (8.1-6)

The term G represents the total transfer function for the forward loop. The term H represents the total transfer for the feedback loop. Normally the G term is made up of the combined transfer functions of several components. Using Eq. (8.1-6) several examples are given in Figures 2–4 showing the use of block diagram algebra to simplify servo block diagrams, and illustrating their use for a real velocity servo drive having a motor, amplifier, and feedback tachometer. To illustrate the use of block diagram algebra for an electric velocity servo drive having a motor, amplifier, and feedback tachometer, Figure 5 presents an example in the frequency domain, where E ¼ error Ka ¼ amplifier gain; V=V Km ¼ motor gain ¼ 1=Ke ; rad=sec=V KTA ¼ tachometer gain, V/rad/sec R ¼ reference command ta ¼ amplifier time constant, sec tm ¼ motor time constant, sec

Fig. 1

Servo block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 2

Servo block diagram algebra.

Vm ¼ output velocity (controlled velocity), rad/sec The forward loop has the following transfer function: CðjoÞ Ka Km ¼G ¼ EðjoÞ ð1 þ jota Þð1 þ jotm Þ

(8.1-7)

The open-loop transfer function from which the stability is determined is: BðjoÞ Ka Km KTA ¼ GH ¼ EðjoÞ ð1 þ jota Þð1 þ jotm Þ

(8.1-8)

The closed-loop transfer function is: VmðjoÞ Ka Km 6h ¼ RðjoÞ ð1 þ jota Þð1 þ jotm Þ 1þ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

1 Ka Km KTA ð1þjota Þð1þjotm Þ

i

(8.1-9)

Fig. 3

Block diagram algebra.

which can also be written as VmðjoÞ Ka Km ¼ RðjoÞ ð1 þ jota Þð1 þ jotm Þ þ ðKa Km KTA Þ VmðjoÞ Ka Km ¼ RðjoÞ ðjoÞ2 tm ta þ joðta þ tm Þ þ ð1 þ Ka Km KTA Þ VmðjoÞ Ka Km =ð1 þ Ka Km KTA Þ ¼ RðjoÞ ðjoÞ2 ðta þtm Þ þ jo þ 1 ð1þKatmKmtaKTA Þ ð1þKa Km KTA Þ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(8.1-10) (8.1-11) (8.1-12)

Fig. 4 Block diagram algebra.

A general form for a quadratic is: ðjoÞ2 2d þ jo þ 1 o2m om sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ Ka Km KTA om ¼ tm ta

Fig. 5 Electric drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(8.1-13) (8.1-14)

To further illustrate the use of block diagram algebra, a hydraulic velocity servo drive can be used as an example in Figure 6. For simplicity the individual drive components are represented by their respective Laplace transfer functions. A detailed development of a hydraulic servo drive is presented in Chapter 12. The terms in Figure 6 are: Ga(s) ¼ amplifier transfer function Gv(s) ¼ servo valve transfer function Gm(s) ¼ servo motor transfer function KTA ¼ tachometer constant R(s) ¼ reference input Vm(s) ¼ motor output velocity The forward loop has the transfer function VmðsÞ ¼ GaðsÞ GvðsÞ GmðsÞ ¼ GðsÞ EðsÞ

(8.1-15)

The open-loop transfer function is BðsÞ ¼ GaðsÞ GvðsÞ GmðsÞ KTA ¼ GH EðsÞ

(8.1-16)

The closed-loop transfer function is VmðsÞ GaðsÞ GvðsÞ GmðsÞ ¼ RðsÞ 1 þ GaðsÞ GvðsÞ GmðsÞ KTA

(8.1-17)

The open-loop equation by itself is not very useful in determining the system performance, but it can be used to determine the transient performance and the closed-loop performance. In the general equation for

Fig. 6

Hydraulic drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

a feedback control system, CðsÞ ¼

GðsÞ RðsÞ 1 þ GðsÞ

ð1 þ GðsÞ ÞCðsÞ ¼ GðsÞ RðsÞ

(8.1-18) (8.1-19)

the transient solution is found by equating 1 þ GðsÞ ¼ 0

(8.1-20)

and since GðsÞ is a transfer function of the form GðsÞ ¼ K=ð1 þ TpÞ

(8.1-21)

then 1 þ GðsÞ ¼

ð1 þ TpÞ þ K ¼0 ð1 þ TpÞ

(8.1-22)

or expressed in a general form, 1 þ GðsÞ ¼

ðs
r1 Þðs
r2 Þ    ðs
rn Þ ðs
ra Þðs
rb Þ    ðs
rm Þ

(8.1-23)

Therefore the transient conditions are determined from the denominator of Eq. (8.1-18). For a control system with several components in the term GH, the general equation will have the form GðsÞ HðsÞ ¼

K1 ð1 þ a1 s þ a2 s2 þ    þ an sn Þ sn ð1 þ b1 s þ b2 s2 þ    þ bn sn Þ

(8.1-24)

which can be factored into roots: GðsÞ HðsÞ ¼

K1 ð1 þ aa sÞð1 þ ab sÞ þ    þ ð1 þ an Þ sn ð1 þ ba sÞð1 þ bb sÞ þ    þ ð1 þ bn sÞ

(8.1-25)

The open-loop transfer function can be written directly from the block diagram, which will have the form of Eq. (8.1-25), or in a specific example such as Eq. (8.1-8). The nature of the system is determined by the power of sn in the denominator of Eq. (8.1-25). As an example, consider the type 0 regulator velocity system: sn ¼ s0 ¼ 1

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The system is a velocity regulator, and a constant error is required to have a constant controlled variable. For a type 1 positioning system, sn ¼ s1 ¼ s The system will have zero error for a constant controlled variable. For a type 2 acceleration system, sn ¼ s2 The system will have zero error for a constant controlled variable and can maintain a constant velocity with zero error. Block diagram algebra can also be used to find the servo error for a velocity servo drive and a positioning servo drive, as in Figures 7 and 8.

Fig. 7

Velocity loop block diagram.

Fig. 8

Position loop block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Error for a Velocity Servo Drive E ¼R
B

(8.1-26)

B ¼ CH C ¼ EG

(8.1-27) (8.1-28)

E ¼ R
EGH Eð1 þ GHÞ ¼ R

(8.1-29) (8.1-30)

E ¼ R=ð1 þ GHÞ

(8.1-31)

Using Eq. (8.1-8) for an electric servo drive, the error is Ejo ¼

Rjo Ka Km KTA 1 þ ð1þjot a Þð1þjotm Þ

(8.1-32)

For the steady-state case, jo ¼ 0. Therefore, E¼

R 1 þ Ka Km KTA

(8.1-33)

Error for a Positioning Servo Drive E ¼ yD
y0

(8.1-34)

Using Eq. (8.1-10) y0jo ¼ Ejo

KD Ka Km jo½ð1 þ jota Þð1 þ jotm Þ þ Ka Km KTA 

(8.1-35)

Substituting Eq. (8.1-35) into Eq. (8.1-35) yields Ejo KD Ka Km jo½ð1 þ jota Þð1 þ jotm Þ þ Ka Km KTA 

(8.1-36)

Solving for E yields  1 þ KD Ka Km Ejo ¼ yDjo jo½ð1 þ jota Þð1 þ jotm Þ þ Ka Km KTA  yDjo jo½ð1 þ jota Þð1 þ jotm Þ þ Ka Km KTA  Ejo ¼ 1 þ KD Ka Km

(8.1-37)

Ejo ¼ yDjo


For the steady-state case, jo ¼ 0 and therefore E ¼ 0.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(8.1-38)

8.2

FREQUENCY-RESPONSE CHARACTERISTICS AND CONSTRUCTION OF APPROXIMATE (BODE) FREQUENCY CHARTS

Having identified a number of servo drive components mathematically with their transfer functions and placed them into system block diagrams, the next step is to examine how the servo block diagram can be used to analyze servo performance. All industrial servo drives are connected to a machine of some kind, which is often referred to as the ‘‘servo plant.’’ These industrial machines have inherent nonlinearities and structural resonances that are sinusoidal. Therefore the sinusoidal frequency response method is used to analyze the performance of the servo drive block diagrams. For the case of sinusoidal analysis the differential operator (p) or the Laplace operator (s) is replaced by jo. Frequency-response characteristics are plotted on semilog paper. This is to compress the scales to have the important part of the response on an 8.5 6 11-in. sheet of paper. The horizontal or frequency scale is logarithmic. The ratio of output to input of the system transfer function, referred to as the magnitude in db, or attenuation rate, is plotted on the vertical scale. The vertical scale is compressed using decibels. Examples of frequency responses for a single time constant, multiple time constants, and a second-order response follow.

Single Time Constant For a transfer function E0 =Ei ¼ K=ð1 þ jotÞ

(8.2-1)

where: K ¼ 10 t1 ¼ 0:01 sec the frequency response is shown in Figure 9. Note that o1 ¼ 1=t1 . The attenuation rate for a single time constant is 20 dB per decade. From the relation of a change in power level measured in decibels given as 20 log10 ¼ o2 t=o1 t

(8.2-2)

the attenuation for a 10 to 1 change in frequency is 20 dB. A decade is a 10 to 1 change in frequency. A gain to decibel conversion chart is given in

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 9 Frequency response for single-order lag.

Figure 10. The phase lag of the output with respect to the input is given by y ¼ tan
1 ðo=o1 Þ

(8.2-3)

For a single time delay where t1 ¼ 0:01 second, the phase shift (y) is plotted in Figure 9.

Multiple Time Constants A transfer function may have more than one time-varying parameter, such as E0 =Ei ¼ 10ð1 þ jot2 Þ=ð1 þ jot1 Þð1 þ jot3 Þ where: t1 ¼ 2:5 sec t2 ¼ 0:5 sec t3 ¼ 0:025 sec o1 ¼ 0:4 rad=sec o2 ¼ 2 rad=sec o3 ¼ 40 rad=sec

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(8.2-4)

Fig. 10

Decibel conversion chart.

The Bode plot is given in Figure 11. Note that t2 is a lead term and t1 and t3 are lag terms. The total phase shift is computed from y ¼ tan
1 ðo=o2 Þ
tan
1 ðo=o1 Þ
tan
1 ðo=o3 Þ

(8.2-5)

For example, for o ¼ 20 rad=sec, y ¼ tan
1 20=2
tan
1 20=0:4
tan
1 20=40 y ¼ 84
89
27 ¼
32

(8.2-6) (8.2-7)

A phase shift versus frequency chart for a single-order lag is given in Figure 12.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 11

Frequency response for Eq. (8.2-5).

Fig. 12

Bode diagram for single-order lag.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Quadratic Transfer Function For a transfer function of Eo =Ei ¼ 1=ðs2 þ 2don s þ o2n Þ

(8.2-8)

where: d ¼ damping factor on ¼ resonant frequency the attenuation phase diagram is shown in Figure 13. In actual practice most servo drive components can be defined mathematically with first- or second-order transfer functions.

Approximate Frequency Response (Bode Plot) Having discussed the mathematical descriptions of machine servo drive components with their transfer functions and their relation to frequencyresponse characteristics, the subject of approximate frequency-response characteristics (Bode plots) can be addressed. A more detailed discussion of

Fig. 13

Second-order equation frequency response.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

an electric drive is presented in Section 12.3. The simple electric velocity drive example of Figure 5 is used to discuss the application of the ‘‘Rules of Bode’’ in constructing Bode diagrams. Using these rules, the forward-loop, open-loop, and closed-loop responses of the velocity servo drive of Figure 5 can be drawn as in Figure 14 assuming the following constants: Ka ¼ 1000 V=V Km ¼ 2 rad=sec=V KTA ¼ 0:0268 V=rad=sec ð3 V=1000 rpmÞ ta ¼ 0:01 sec tm ¼ 0:02 sec

oa ¼ 100 rad=sec om ¼ 50 rad=sec

The rules of Bode can be summarized as follows to find the closed-loop response: Vmjo =Rjo ¼ 1=feedbackðKTA Þwhen the open-loop response Bjo =Ejo Vmjo =Rjo

is larger than 0 dB ¼ Vmjo =Ejo ðforward loopÞ when the open-loop response Bjo =Ejo is less than 0 dB

For this example the servo bandwidth is ob .

Fig. 14

Bode diagram for rules of bode.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Calculated Frequency Response The closed-loop response of Figure 5 can also be found from the calculated output/input ratio of Eq. (8.1-12), which combines the motor responses ðom Þ and the amplifier response ðoa Þ as a second-order response with a resonance on from Eq. (8.1-14) and a closed-loop gain from Eq. (8.1-12). The closed-loop gain of Eq. (8.1-12) can be simplified as follows: Closed-loop gain ¼

Ka Km 1 þ Ka Km KTA

(8.2-9)

Dividing Eq. (8.2-10) by Ka Km yields

1 Ka Km

1 þ KTA

(8.2-10)

where 1 << 1 Ka Km

(8.2-11)

Closed-loop gain ¼ 1=KTA

(8.2-12)

Thus

The resonance is found from Eq. (8.1-14). The damping factor is d¼

ta þ tm om 6 1 þ Ka Km KTA 2

(8.2-13)

Therefore the calculated closed-loop response has a gain of 1=KTA and a bandwidth at on as shown in Figure 15.

Position Loop Response The velocity servo drive of Figure 5 can be included in a position loop of Figure 16. The closed-loop mathematical solution is given in Eq. (8.2-14) to (8.2-22). For comparison, the closed position-loop response is shown for both the approximate Bode plot response and the calculated response in Figures 17 and 18, respectively.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 15

Frequency response for Eq. (8.1-12).

Calculated Position Loop Response Gjo ¼ KD 6ðVmjo =Rjo Þ6ð1=joÞ

(8.2-14)

H¼1

(8.2-15)

yA =yD ¼ G=ð1 þ GHÞ

(8.2-16)

Fig. 16

Positioning servo-drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 17

Bode diagram for position loop of Figure 16.

yAjo Vmjo 1 1 ¼ KD 6 6 6 yDjo Rjo jo 1 þ KD Vmjo 6 Rjo

1 jo

(8.2-17)

yAjo KD Ka Km ¼ (8.2-18) yDjo jo½ð1 þ jota Þð1 þ jotm Þ þ Ka Km KTA  þ KD Ka Km yAjo KD Ka Km ¼ (8.2-19) yDjo jo½ðjoÞ2 tm ta þ ðta þ tm Þjo þ ð1 þ Ka Km KTA Þ þ KD Ka Km yAjo KD Ka Km ¼ (8.2-20) yDjo ðjoÞ3 tm ta þ ðjoÞ2 ðta þ tm Þ þ joð1 þ KD Ka Km Þ þ KD Ka Km

Fig. 18

Bode diagram for calculated solution to Figure 16.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The general solution: yAjo K i h 2v ¼ ðjoÞ yDjo w 1 þ j wv o 2 þ o2dm jo þ 1

(8.2-21)

KD Kv ¼ KTA

(8.2-22)

m

The position loop has an open-loop gain which is referred to as the velocity constant. For Figure 16, the open-loop gain is the forward-loop gain since there is no feedback term and the feedback loop is open. Under these conditions, an input position command will cause the servo drive to move at some output velocity. Thus the open-loop gain is Kv ¼ speed=error ðEÞ

(8.2-23)

For a steady-state condition at a constant velocity, the error (called following error) is E ¼ speed=Kv

(8.2-24)

The following error is the amount the output position lags the input command for a steady-state velocity. Therefore, to minimize the following error, the position loop gain should be large and practical as long as servo stability is maintained. For Figure 16, the open-position loop gain is the product of gain KD and the closed-velocity loop gain. From Eq. (8.2-12) the steady-state velocity loop gain is 1/KTA. Therefore the steady-state open-position loop gain (velocity constant) is Kv ¼ KD =KTA

(8.2-25)

Assuming a Kv ¼ 1 ipm/mil or 16.67/second the KD of Figure 16 and Eq. (8.2-22) is KD ¼ 16:676KTA ¼ 16:6760:0268 ¼ 0:48 volts=volt

8.3

(8.2-26)

NICHOLS CHARTS

Determining frequency-response characteristics from the servo block diagram has been shown in Section 8.2 using approximate Bode diagrams or by a mathematical solution with block diagram algebra. There are numerous computer programs that can analyze servo drive block diagrams and provide any desired output (transient or frequency response, Nyquist

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

charts, etc.). Graphical solutions to block diagrams are presented here only to explain the basics of industrial servo drive analysis. One other graphical analysis to obtain the frequency response from the system block diagram is known as the Nichols chart. In the use of a Nichols chart the open-loop phase angle is plotted as phase margin: Phase margin ¼ 180 þ jphase shiftj

(8.3-1)

Note that the phase shift is usually a negative angle. The Nichols charts have not been used extensively, and the purpose of this discussion is to explain their purpose and how to apply them to practical problems. The analysis of feedback control systems requires the knowledge of two important things: 1. What will the system do when a sudden input is applied? This is called the transient response to a unit step. 2. How long will the output follow the input and stay in step as the input is varied at faster rates? This is referred to as the frequency response to a sinusoidal input. It is the second method of analysis that gives the Nichols chart its function. Having discussed approximate frequency response (Bode diagrams) in Section 8.2, it is now possible to use this information to discuss the use of this graphic method to close the servo loop. It is assumed that all the minor loops have been replaced with equivalent blocks and transfer functions to the place where only one block remains with the major loop as shown in Figure 19. The open loop (C/E) is usually known in the form of an attenuation plot and phase margin versus frequency. The problem, then, is what does the frequency plot look like when the loop is closed (C/R)? This is the function of the Nichols chart. Assume some major open loop with the

Fig. 19

Block diagram for closed-loop servo.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

transfer function Gjo ¼ 0:25ð1 þ 6:25joÞ=ðjoÞ2 ð1 þ 0:25joÞ2

(8.3-2)

The Nichols chart enables the designer to close the loop for an open-loop transfer function of Eq. (8.3-2). The mechanics of such a chart are as follows: The chart consists of families of curves for various closed-loop magnitude (M) values and phase shift (a) values. The scales are plotted with the gain of jGj in decibels on the ordinate and the phase margin in degrees on the abscissa. Such a chart is shown in Figure 20. From the open-loop magnitude and phase, values of jGj and phase margin are read for a series of frequencies. At the point where the value of jGj and the phase margin intersect on the Nichols chart, a value of (C/R) or M is given, along with the closed-loop phase angle. When a series of frequency points are chosen and the value of jGj and phase margin are plotted as just described, a closed-loop frequency response results. Considering the transfer function given in Eq. (8.3-2), the Bode plot is shown in Figure 21. Figure 22 shows the magnitude jGj and phase margin

Fig. 20

Nichols chart.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 21

Open-loop plot for Eq. (8.3-2).

plotted on the Nichols chart. Figure 23 shows the closed-loop response of Cjo =Rjo ¼ M ¼ Gjo =ð1 þ Gjo Þ

(8.3-3)

Performance is determined by how close the open-loop response of Figure 21 comes to encircling the origin of the Nichols chart of Figure 22. The bandwidth of the closed-loop response and the height of the closed-loop response peak (M) are related to how close the open-loop response comes to the origin of the Nichols chart. If the open-loop response encircles the origin, the servo system is unstable. The closer the open-loop response is to the origin, the more oscillatory the closed-loop response will be.

8.4

SERVO ANALYSIS TECHNIQUES

Previous sections discussed the components of servo drives as applied to block diagrams. Block diagram algebra was also discussed with some graphical methods to obtain the closed-loop frequency response. Servo analysis techniques are now discussed in more depth. When a servo drive is used as a prime mover on a machine, it is very important to know the technical capabilities of the servo drive, since this is a

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 22

Nichols chart for Eq. (8.3.-2).

measure of the machine’s ability to make rapid changes in velocity during machining. A simple example can be stated to illustrate this point. If it is required to mill a rectangular pocket in a piece of metal, the machine should be capable of milling into a corner at the desired feed, change direction (90 degrees), and have a square corner as a result. This is also about as severe a condition possible for a servo to meet. In some respects this is an indication of the servo performance capabilities.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 23

Closed-loop frequency response for Eq. (8.3-2).

The question then arises, how do we measure this capability, what criterion do we have to compare with? Feedback control system designers have two ways to measure performance. These measuring techniques are called the frequency response and the transient step response.

The Frequency Response What does this frequency response tell us? To understand the relevance of servo loop bandwidth to servo performance it is necessary to define what is meant by servo bandwidth. ‘‘Bandwidth’’ is a term describing the frequencyresponse characteristics of a servo drive. In its simplest definition, the frequency response is a measure of how well an output of a servo follows the input as a sinusoidal input driving frequency increases. At low frequencies the output of a servo faithfully follows the input, so the ratio of output/ input is 1. This is often referred to as the flat part of the response. At high frequencies, the output of a servo no longer follows the input. The higher the servo bandwidth, the greater the capability of the servo to follow rapid changes in the commanded input.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

As the frequency increases beyond the flat part of the response, the servo output begins to lag behind the input signal. There is a phase lag and an amplitude attenuation between the output and input. When the phase lag reaches 180 degrees the servo is unstable. Thus it is a requirement to establish some performance specification to state how much phase lag is acceptable before instability occurs. Most industrial axis servo drives use a velocity servo inside a positioning servo system. It is critical that the position servo drive have an open-loop phase lag less than 180 degrees to avoid instability of the servo. As an index of performance, a maximum of 135 degrees phase lag has been used as a criterion. This means there are 45 degrees phase lag left before instability occurs, and this 45 degrees is referred to as the phase margin before the servo goes unstable. Thus, the total acceptable phase lag of the internal velocity loop and other components cannot be allowed more than 135 degrees phase lag. In any positioning servo system with an internal velocity servo there will be a required integration of the velocity output of the velocity servo to the position output of the position loop. This integration component takes place in the servo motor and is measured by the position transducer (often a resolver). Any integration component has a fixed 90 degrees phase lag at all frequencies. Thus the total allowable phase lag of 135 degrees for the velocity servo drive and integration is now reduced to 45 degrees (1358–908) for the closed velocity loop servo. Therefore the useful velocity servo bandwidth occurs at a frequency where the closed velocity servo loop has a phase lag of 45 degrees (often stated as the 45-degree phase shift frequency).

Defining Position-Loop Gain A term used to define the performance criteria of a position servo is referred to as the ‘‘position-loop gain’’ or ‘‘velocity constant.’’ This gain is actually the open-position loop gain. The open-loop gain is a parameter that represents the product of the individual gains in the position servo loop without the feedback being closed. Without the feedback being closed, a position command will cause the servo drive to have motion. The ratio of output/input will therefore be in units of velocity/position. As an example, if the input command has the units of inches, the open-position loop output will be inches per second. Thus the open-position loop gain, Kn , is in units of (inches per second/inches) or 1/seconds. A position-loop gain, in units of 1/ seconds, does not have a practical meaning to many machine control people

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

working in industry. By making a units conversion of Kv ¼ 1=sec660 sec=min61 in:=1000 mils ¼ 1=sec60:06 in:=min=mil or (inches per minute/mil)616:67 ¼ 1=sec Then the open-position loop gain expressed as some velocity (ipm) divided by distance (mils) has a practical significance.

The Transient Step Function Response It is also possible to measure the performance of a positioning servo by its response to a step change in input. The system block diagram for a positioning system can be represented with a single block with the overall gain Kv, as shown in Figure 24. The gain of the system is then called the velocity constant Kv by definition. A step function response (typically a second-order response) can be plotted for a command change in position versus time as shown in Figure 25. The step function response can vary from one extreme of self-sustained oscillations (curve a) to an overdamped system (curve d). Curve c represents a system with critical damping (damping factor ¼ 1), which is considered an ideal performance for a positioning servo drive, and curve b represents a system with approximately one overshoot. The transient response of a velocity servo drive can also be used to observe the change in velocity versus time for a step change in input. The response will be characteristically like Figure 25 except the controlled variable will be velocity instead of position. Velocity servo drives are characteristically adjusted to have a time response of one overshoot.

Servo Analysis and the Machine Designer Two things, system ‘‘stability’’ and ‘‘accuracy,’’ determine the performance of the positioning system. The accuracy of a positioning system is directly related to the drive resolution (see Section 12.4). Each component in the

Fig. 24

Simplified block diagram for a position loop.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 25

Step function response.

positioning system contributes to the overall gain. Thus the power amplifier, drive motor, gearing, and feedback have individual gains that contribute to the overall system gain when they are cascaded together. The important point as far as the mechanical designer is concerned is the mechanical design of the machine parts, drive screw, and gearing as related to the inertias involved. System performance or stability is partially dependent on keeping the total reflected inertia to the motor shaft as small as practical (see Section 12.2). It is quite possible the resultant total inertia and the drive resonant frequencies will not make it possible to get the required positioning system performance and some mechanical redesign may be necessary.

8.5

SERVO COMPENSATION

Once the components of an industrial servo drive have been selected and the block diagram constructed, it is necessary to make sure the drive will be stable. Thus it will be necessary to compensate the servo drive. This section

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

discusses some of the accepted methods of compensating a servo drive to make it stable. A stability criterion for servo drives is that the open-loop difference in phase shift of the output referred to the input, for a sinusoidal input, should not be more than 135 degrees (for 45 degrees phase margin with the output lagging the input). If a servo drive motor (actuator) and its associated amplifier (without servo compensation) are the main components of the servo drive, closing the servo drive loop through the feedback will most likely result in an unstable servo drive because of excessive phase lag. Thus additional components must be used to make the servo drive stable. The addition of these components is referred to as compensating the servo, equalizing the servo, or synthesizing the servo. Using Bode frequency response plots for an uncompensated electric velocity servo drive shown in Figure 26a, the output lags the input at 600 rad/sec or 95 Hz. This can be ascertained from the open-loop response at 0 dB gain with approximately 170 degrees phase lag. At 170 degrees phase lag, the drive will be oscillatory; as seen in the transient response of Figure 26b. At 180 degrees phase lag the drive will be unstable. Therefore some type of compensation must be added to make the drive stable.

Fig. 26a

Bode diagram for an uncompensated electric drive.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 26b

Transient response of the uncompensated servo of Figure 26a.

Compensation in servo drives can be accomplished with analog circuits or digital algorithms. The compensation is usually put in the forward loop amplifier. These compensating elements are described using frequency-response representations of the actual analog or digital algorithms. The first type of classical compensating element is referred to as lead and lag compensation. A typical lead compensation has the transfer function of K

ðt1 s þ 1Þ ðt2 s þ 1Þ

t1 > t2

(8.5-1)

where a lag term is used to put a limit on the differential effect of the compensation. Without this limit the differential response would excite unwanted higher frequencies (noise). A frequency plot of a lead and lag compensation is shown in Figure 27.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 27

Bode diagram for lead compensation.

The second type of compensation used is the lag and lead compensation, which has a transfer function of K

ðt1 s þ 1Þ ðt2 s þ 1Þ

t2 > t1

(8.5-2)

The lag compensation has the frequency-response characteristic shown in Figure 28. Also note that these compensations often are identified by the lower frequency term. Thus, a lag and lead network would be called a lag and the lead and lag network would be called a lead. These lead and lag types of servo compensation can be used to modify the frequency of a servo to make it stable. It is desired to decrease the phase lag of the servo output relative to the input for a sinusoidal input. A typical example is shown in Figure 29a where the servo of Figure 26a had a lag/lead (shorthand for lag and lead) compensation added to the servo amplifier. It should be noted that the lead term in the numerator of the amplifier compensation is the same as one of the motor frequencies in the denominator and thus they will cancel each other. With the lag term at o ¼ 0.2 rad/sec, the phase lag will be about 90 degrees when the open-loop response magnitude is equal to 0 dB. Since the servo would be unstable for a

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 28

Bode diagram for lag compensation.

180 degrees phase lag, the 90 degrees phase lag of Figure 29a leaves a margin of 90 degrees for a stable drive. The classical use of lead/lag circuits has been with analog operational amplifiers to compensate servo drives. Another servo compensation method is the use of proportional, integral, and differential (PID) algorithms. This type of servo compensation is used widely with digital control techniques. A PID servo block diagram with the associated approximate (Bode) frequency response is shown in Figure 30. The differential part of the compensation should be used with caution since the amplitude response increases with frequency, which may amplify some undesirable noise. In general, many servo drives use proportional plus integral (PI) type compensation. The servo block diagram and approximate (Bode) frequency responses are shown in Figures 31 and 32. The foregoing types of servo compensation have an important application with velocity or type 0 servo drives. For good servo regulation and minimum servo error, it is desirable to have a large gain at low frequencies. With lag/lead compensation (Figure 28), the low-frequency gain should be as high as practical to maintain good servo regulation. It is also important to reduce this gain with increasing frequency. Therefore the lag term o2 in Figure 28 should start to reduce the gain starting at a low

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 29a

Bode diagram for a compensated electric drive.

frequency such as 0.1 rad/sec. Reducing the gain as frequency increases will minimize the excitation of resonant frequencies in the servo drive. In addition, the lead term of Figure 28 will provide some phase lead, which will reduce the overall phase lag of the open-loop response of the servo drive. The application of lag/lead compensation is shown in Figure 29a, with a stable transient response shown in Figure 29b. No servo compensation results in a servo drive that will be oscillatory or unstable (Figure 26b). In addition, the use of high open-loop gain increases the drive stiffness, since the stiffness is proportional to the product of the velocity servo loop gains and the position-loop gain (see Section 12.3). Using PI compensation or PID compensation has the same application as lag/lead compensation. It should be noted in Figures 30, 31, and 32 that the integral or lag component of the compensation has no lowfrequency gain limit. This results in a servo drive with even higher drive stiffness, and improved servo regulation. In commercial servo drives that use digital-type PID servo compensation, the values of Kp (proportional compensation), Ki (integral compensation), and Kd (differential compensation) are usually dimensionless numbers. These numbers differ from one manufacturer to another depending on what manufacturer supplies the digital compensation circuitry.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 29b

Transient response of the compensated servo of Figure 29a.

The actual units of the proportional gain ðKp Þ should be A/rpm. The actual units of the integral gain ðKi Þ should be A/sec/rpm. Since the numerical values assigned to Kp and Ki are different from one manufacturer to another, it is useful to the servo engineer to be able to relate the compensation values of one manufacturer to another in their actual units. To illustrate the use of actual compensation values, three servo drive manufacturers are considered. The first drive to be considered is a Fanuc brushless DC design using a model 20S/3000 motor. The actual proportional compensation can be calculated as follows: Kp ¼ 26pðrad=secÞ6ðmin =60secÞ6260:7626p6206 ðJm þ JL Þ ðA=rpmÞ KT Jm ¼ 0:17 ðkg-cm=sec2 Þ ðmotor inertiaÞ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 30

PID compensation.

JL ¼ 0:17 ðkg-cm=sec2 Þ ðload inertiaÞ KT ¼ 7 ðkg-cm=AÞ ðmotor torque constantÞ

  I Ki Kp s þ Ki Ki Kp ¼ ¼ Kp þ sþ1 ¼ V2 s s s Ki Kp Ki t1 ¼ o¼ Ki Kp "K p #  I Kp 1 Ki jo þ 1 ¼ Ki þ ¼ Kp jo ! ? ¼ Ki V2 Ki jo jo

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 31

PI compensation.

Fig. 32

Varying Kp compensation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Kp ¼

26p ð0:17 þ 0:17Þ 6267626p6206 ¼ 0:894 ðA=rpmÞ 60 7

The actual integral gain can be calculated as follows: Ki ¼ 26pðrad=secÞ6ðmin=60 secÞ6ð26p620Þ2 6

ðJm þ JL Þ KT

ðA=sec=rpmÞ Ki

26p ð0:17 þ 0:17Þ 6ð26p620Þ2 6 ¼ 80:4 ðAmp=sec=rpmÞ 60 7

The second drive to be considered is an Allen-Bradley brushless DC servo. Allen-Bradley uses per unit values for the proportional and integral compensation values. The actual proportional compensation can be calculated as follows: Kp ¼ Kp;pu

Rated current ðAÞ ¼ ðA=rpmÞ 1000 ðrpmÞ

The actual integral compensation can be calculated as follows: Ki ¼ Ki;pu

Rated current ðA=secÞ ¼ ðA=sec=rpmÞ 1000 ðrpmÞ

For example, for the 1326-ABC3E motor: Gain ¼ 30% Kp;pu ¼ 16 Ki;pu ¼ 1482 Rated current ¼ 49 A Kp ¼ 16649=1000 ¼ 0:78 A=rpm Ki ¼ 1482649=1000 ¼ 72:62 A=sec=rpm The third drive to consider is an Indramat vector-controlled induction motor design. The actual proportional compensation can be calculated as follows: P ¼ %Pgain 6Proportional68:85 ¼ ðlb-in.-minÞ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Pðlb-in.-minÞ 26p ðradÞ 6 ¼ ðA=rpmÞ KT ðlb-in:=AÞ ðrevÞ I ¼ %Igain 6Integral68:85 ¼ ðlb-in.-min=secÞ

Ki ¼

Ki ¼

Iðlb-in.-minÞ 26pðradÞ 6 ¼ ðA=sec=rpmÞ KT ðlb-in:=AðsecÞ ðrevÞ

For example, for the 2AD180D motor: Proportional ¼ 5:32 N-m-min Integral ¼ 379:8 N-m=sec %Pgain ¼ 0:8 %Igain ¼ 0:4 KT ¼ 17:5 lb-in:=A ðmotor torque constantÞ P ¼ 0:865:3268:85 ¼ 37:67 lb-in.-min 37:67 626p ¼ 13:52 A=rpm Kp ¼ 17:5 I ¼ 0:46379:868:85 ¼ 1344 lb-in.-min 1344 626p ¼ 482:48 A=sec=rpm Ki ¼ 17:5

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

9 Indexes of Performance

The state of the art of the machine industry has experienced a rapidly accelerating evolution since the outset of the first industrial revolution. Modern machines are required to perform to varying requirements of accuracy, productivity, and surface finish requirements. It is the purpose of this discussion to establish some guidelines of machine system design to attain the desired performance requirements of accuracy, surface finish, etc. Modern industrial machines are considered as a total system comprised of the control, drive, and machine. There are two kinds of performance criteria. One, referred to as indexes of performance (I.P.), relates to specifying servo drive stability in relation to design parameters. The second, performance characteristics, relates to parameters such as drive stiffness, resolution, maximum acceleration, and the effects of friction. To some extent the performance characteristics are a function or the end result of the indexes of performance that are selected.

9.1

DEFINITION OF INDEXES OF PERFORMANCE FOR SERVO DRIVES

During World War II, the growth of feedback control theory was given a stimulus that has continued for five decades. As the state of the art improved, control system performance could be predicted with increased accuracy. However, as the state of the art in feedback control theory

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

increased in reliability, it also increased in complexity. There is a need to specify performance in relation to design parameters. An accepted technique to relate performance, I.P., is an attempt to optimize performance using constraints of design criteria. Some I.P.s in common use are: A.

From the frequency response: 1. 2. 3.

B.

Phase margin Gain margin Gain margin versus control loop bandwidth

From the step response: 1. 2. 3. 4.

Percent overshoot Rise time Delay time Settling time

These I.P.s are given for the frequency response and step response because these two methods of analysis are also related to diagnostic techniques to measure actual performance of a feedback control system. There are numerous types of feedback control applications. The I.P. to be discussed are limited to hydraulic and electric feed drives used with position control systems. The frequency-response characteristics of these drives are considered as the analytical tool in specifying I.P.s. Frequency-response characteristics describe how well the output of a given control system will follow the input variations as a function of frequency. In brief, the frequency response describes the dynamic characteristics of the drive. In addition, required compensation can be determined from the frequency-response characteristics; required compensation is an important part of the synthesis of feedback control drive systems. The following is a description of the various indexes of performance. ‘‘Phase margin’’ is the amount of phase shift remaining between the output controlled variable and the input reference at the crossover frequency (Kv of Figure 1) before 1808 of phase shift occurs. (Note that phase shift is a negative angle.) Feedback control systems with 180 degrees phase shift at the open-loop crossover frequency are unstable. Therefore, phase margin is a measure of how much additional phase shift can be tolerated before instability will occur. As an I.P., the phase margin at the crossover frequency should be 458 or more. Values of phase margin less than 458 will result in an oscillatory condition that will have increasing oscillation with less phase margin until self-sustained oscillation will occur at a phase margin of 08 (see Section 8.4).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1 Open-position loop frequency response for a hydraulic drive with an internal tachometer loop.

‘‘Gain margin’’ is the amount the position-loop gain can be raised to produce a phase margin of zero or a phase shift of 1808. As an I.P. the gain margin should be no less than 2.0 (or 6 dB). In Figure 1 the gain margin is shown as 6 dB or a gain of 2. Gain margin versus control loop bandwidth is the most important of the I.P.s. This I.P. relates the gain margin to the attainable servo bandwidth. The bandwidth of the position loop, velocity loop, and hydraulic or mechanical resonance have a relationship that is the basis for system stability. Hydraulic and electric drives are next considered separately. These I.P.s are summarized in Eq. (9.1-1) to (9.1-6). They are derived in Section 9.2.

Hydraulic Drives 1. Hydraulic drive with position loop only: Minimum allowable hydraulic resonance ¼ oh ¼ 100 rad=sec Kv ¼

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oh 3

(9.1-1)

2. Hydraulic drive with tachometer loop: Minimum allowable hydraulic resonance ¼ oh ¼ 200 rad=sec. 1 oc ¼ velocity loop bandwidth ¼ oh 3 1 1 Kv ¼ oc ¼ oh 2 6

(9.1-2) (9.1-3)

Electric Drives Electric drive with tachometer loop: 1 1 ¼ ¼ o1 mechanical time constant tme o1 ¼ lead compensation oe oc ¼ velocity loop bandwidth ¼ 2 1 Kv ¼ oc 2

ome ¼

9.2

(9.1-4)

(9.1-5) (9.1-6)

INDEXES OF PERFORMANCE FOR ELECTRIC AND HYDRAULIC DRIVES

I.P.s were defined at the beginning of this section. Of the available I.P.s, gain margin versus control loop bandwidth is the most important index. The I.P. for hydraulic drives is considered first.

Hydraulic Drive without a Tachometer Loop The first hydraulic feed drive considered is the simplest case of a position loop (single servo loop). Bandwidth for a stable servo drive is a function of the hydraulic damping factor, illustrated with the block diagram for this servo loop shown in Figure 2. The position feedback uses a resolver for this example. Since the valve response ðov Þ is usually greater than the other system responses, it is neglected in the open-loop response represented by the transfer of Eq. (9.2-1). BðsÞ K K  A fb  ¼ EðsÞ Dm s s22 þ 2dh s þ 1 oh o h

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-1)

Fig. 2 Hydraulic servo-drive block diagram without a tachometer loop.

Let KA Kfb ¼ Kv Dm sffiffiffiffiffiffiffiffiffiffiffiffi 2bD2m oh ¼ JT V c

(9.2-2)

Also, for sinusoidal analysis s ¼ jo. BðjoÞ Kv Kv  ¼  ¼  o2 EðjoÞ jo ðjoÞ2 þ 2dh jo þ 1 jo 1
2 þ 2jd o o2h

oh

oh

(9.2-3)

oh

The critical frequency occurs at o ¼ oh : BðjoÞ Kv Kv  ¼ ¼
2dh oh EðjoÞ jo 1
o2h þ 2jd oh h h oh o2

(9.2-4)

h

This denotes an amplitude of Kv =2dh oh at a phase angle of 1808. For a gain margin of 2 at o ¼ oh , B ðoh Þ <
6 dB E

(9.2-5)

which can be observed from Figure 3, since the height of the attenuation peak at the hydraulic resonance should not be higher than 0 dB for the system to be stable.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 3

Bode diagram for hydraulic positioning servo (no tachometer loop).

Therefore, combining Eq. (9.2-4) and (9.2-5) yields Kv <
6 dB 2dh oh

or

0:5

(9.2-6)

Although the hydraulic resonance is the limiting resonance in most hydraulic drives, it is possible to have a lower frequency mechanical resonance in the servo loop and it will then be the limiting resonance. Since hydraulic damping factors in industrial drives are typically 0:2 < dh < 0:4

(9.2-7)

for stability, Kv ¼ 2dh ð0:5Þ ¼ dh oh

(9.2-8)

From Eq. (9.2-7), the required ratio of Kv =oh will vary as 0:2 <

Kv < 0:4 oh

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-9)

Therefore, an I.P. for hydraulic positioning drives can be established. The velocity constant Kv should be less than the limiting resonance in the servo loop by the following factor: Kv ¼ ð0:2 to 0:4Þoh

(9.2-10)

A specific recommendation is made that the hydraulic resonance (or other limiting resonance) by three times the desired velocity constant Kv . Commercial contouring controls using the soft servo technique have velocity constants ðKv Þ ranging from 0.6 ipm/mil to 2 ipm/mil, where 0:6 ipm 1000 mil min 6 6 ¼ 10=sec mil inch 60 sec 2 ipm 1 6 ¼ 33:4 sec mil 0:06

(9.2-11) (9.2-12)

Using the I.P. Kv ¼

oh 3

(9.2-13)

and using Kv ¼ 33:4=sec as a worst-case condition, the lowest hydraulic resonance should not be less than oh 3 oh ¼ 100 rad=sec

33:4 ¼

(9.2-14)

for a position loop without a tachometer minor loop feedback.

Hydraulic Drive with Tachometer Feedback The major share of commercially available soft servo drives use a tachometer feedback to increase the drive stiffness and limit the overshoot in step inputs. There is a further limitation on the relation of the velocity constant ðKv Þ to the hydraulic resonance in this case. The block diagram for this case is shown in Figure 4. The open-loop transfer characteristics for this case are shown in Figures 5 and 6. Figure 5 shows the minor tachometer loop frequencyresponse characteristics. From the closed-loop response of Figure 5, the bandwidth is oc . The relationship between the crossover frequency oc and the hydraulic resonance should again be related by the type of relationship described by Eq. (9.2-10): oc ¼ ð0:2 to 0:4Þoh ¼ 0:3oh

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-15)

Fig. 4

Hydraulic servo-drive block diagram.

When a position loop is added to the tachometer loop, the bandwidth oc of the minor loop must be considered. The position-loop velocity constant Kv must occur at a lower frequency than oc , the crossover frequency of the velocity loop. The separation of the velocity constant Kv and oc can be considered as another I.P. For adequate stability, the relation of Kv to oc

Fig. 5

Block diagram for a hydraulic velocity servo.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 6 Bode diagram for a hydraulic positioning servo (with a tachometer feedback loop).

should be Kv <

oc 2

(9.2-16)

The I.P. in Eq. (9.2-16) is justified as follows. The frequency-response characteristics of Figure 6 are replotted in Figure 7 with the phase-margin characteristics. Phase margin is shown for several values of hydraulic damping factors. Phase margin was calculated by computer from the open-loop frequency characteristics of Figure 6. The open-position loop block diagram of Figure 4 and the closed-velocity loop frequency response of Figure 5 can be reduced to the block diagram shown in Figure 8 by using the equivalent closed-velocity loop and combining all the loop gains into one gain constant Kv . The open-loop equation for this hydraulic positioning drive with an inner tachometer loop is BðsÞ K  v  ¼  EðsÞ s s þ 1 s22 þ 2dh s þ 1 oc oh o h

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-17)

Fig. 7 Bode diagram for the open-position frequency response for a hydraulic drive (with a tachometer feedback loop).

By definition the phase margin in a position loop is: g ¼ 180
phase shift

(9.2-18)

In this example the phase-shift characteristic for varying frequency with s ¼ jo is 2d o

y ¼
90
tan
1

h o o
Tan
1  h 2  oc 1
o2

oh

Fig. 8 Hydraulic drive block diagram for a position loop.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-19a)

Combining Eq. (9.2-18), (9.2-19a), and (9.2-19b) yields 2d o

g ¼ 180
90
tan
1

h o o
tan
1  h 2  ¼ phase margin oc 1
o o2

(9.2-19b)

h

The closed-loop frequency characteristics of the position loop were determined by computer for various values of velocity constant ðKv Þ and hydraulic damping factors ðdh Þ. The amplitude of the closed-position loop frequency-response characteristic is found from yo 1 ¼ M ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 cos g yi 1
þ 12 jGj

(9.2-20)

jGj

The closed-position loop phase is found from y ¼ tan
1


sin g G
cos g

(9.2-21)

In closed-loop positioning systems the amplitude ðMÞ is related to overshoot in transient step inputs. The larger the peak amplitude of M, the greater the transient overshoot and the tendency for oscillation in response to steps inputs (Figure 9). As in I.P., the amplitude of the closed-loop frequency response ðMÞ should be limited to 1.3. If the maximum values of closed-position loop amplitude ratio ðMÞ are plotted for various hydraulic damping factors and position-loop velocity constants, an additional I.P. can be realized. Such a graphical plot is shown in Figure 10. From Figure 10, which relates the closed-position loop peak amplitude, position loop velocity constant, and hydraulic damping factor (with oc ¼ 0:33oh ), an I.P. can be given that the velocity constant ðKv Þ should be less than the bandwidth of the velocity loop by at least 1.5 for hydraulic damping factors up to 0.8. For a margin of safety and to minimize overshoot, a recommended I.P. is Kv ¼

oc 2

(9.2-22)

This relation states that the position-loop velocity constant ðKv Þ should be one-half the bandwidth ðoc Þ of the tachometer loop. In turn, oc should be

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 9

Frequency response versus transient response.

one-third the hydraulic resonance ðoh Þ. oh 3 oc Kv ¼ 2

oc ¼

(9.2-23) (9.2-24)

Therefore, 1 oh oh ¼ Kv ¼ 6 2 3 6

(9.2-25)

Since the highest Kv used with the soft servo technique is often considered as 2 ipm/mil or 33.4/sec, the hydraulic resonance should not be less than oh ¼ Kv 66

(9.2-26)

oh ¼ 33:466 ¼ 200 rad=sec

(9.2-27)

A hydraulic resonance of 200 rad/sec is used as an I.P. in sizing hydraulic drives. Therefore, oh > 200 rad=sec

(9.2-28)

Electric Drive with Tachometer Feedback Most commercial electric feed servo drives have a tachometer minor servo loop. The I.P., however, are not as complex or limiting in performance as with the hydraulic resonances of the fluid drive. Forward-loop compensation is usually used to increase the steady-state gain, which will greatly increase the drive stiffness. Lag compensation, o2 , is used to provide a high

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Gain versus hydraulic damping factor.

static loop gain for the velocity servo loop. Additionally, the compensation includes a lead term, o1 , in Figure 11. Since the electric motor has two separate time constants, electrical and mechanical, the drive can readily be compensated by letting the lead term of Figure 11 equal the mechanical time constant, tme ¼ 1=ome . As can be observed in Figure 12, the next limiting factor on the frequency response of the velocity loop drive will be the drive motor time constant, te ¼ 1=oe . Summarizing, an I.P. for the electric servo

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 11

Electric drive block diagram.

drive to make the drive-compensation lead-time constant equal the mechanical motor time constant is t1 ¼ tm

Fig. 12

Bode diagram for an electric velocity servo.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-29)

The motor mechanical time constant is

tm ¼

Ra JT Ke KT

(9.2-30)

The bandwidth of the tachometer loop is dependent on the value of the electric time constant. As an I.P., the bandwidth of the tachometer loop ðoc Þ of Figure 12 should be one-half the electrical time constant, the separation required between two single time constants as discussed with Figure 10 and Eq. (9.2-22): oc ¼

oe 2

(9.2-31)

Velocity loop bandwidth ðoc Þ is defined as the frequency at which the closed velocity loop attenuation is
3 dB. As an example, in Eq. (9.3.85) the ratio of output velocity ðVm Þ to input voltage ðEi Þ will have an amplitude of
3 dB at the velocity loop bandwidth, oc (rad/sec), for s ¼ jo. For electric drives using switching type amplifiers (SCRs), there will be an additional limiting factor on the bandwidths. This factor will be the effect of the form factor on the amplifier transport lag, which in turn adds phase shift with increasing frequency. The phase-shift characteristics for various designs of silicon control rectifier (SCR) circuits versus frequency are shown in chapter 7, figure 11. The transport lag is used in the nonlinear analysis of a single-phase, full-wave SCR amplifier discussed in Section 7.5. The position-loop bandwidth of the electric drive has the same I.P. as the hydraulic drive, where the position-loop gain or bandwidth should be one-half the velocity-loop bandwidth of Figure 13. Kv ¼

oc 2

SUMMARY Hydraulic drive with position loop only: Minimum allowable hydraulic resonance ¼ oh ¼ 100 rad=sec

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(9.2-32)

Fig. 13

Open-position loop for an electric drive (with a tachometer feedback loop).

Hydraulic drive with tachometer loop: Minimum allowable hydraulic resonance ¼ oh ¼ 200 rad=sec 1 oc ¼ velocity loop bandwidth ¼ oh 3 1 1 Kv ¼ oc ¼ oh 2 6 Electric drive with tachometer loop: 1 1 ¼ ¼ o1 mechanical time constant tm 1 Kv ¼ oc ðvelocity-loop bandwidthÞ 2 1 oc ¼ velocity-loop bandwidth ¼ oe 2

om ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

ðlead compensationÞ

10 Performance Criteria

10.1 PERCENT REGULATION There are a number of ways to define performance, such as servo bandwidth, accuracy, regulation, etc. Percent regulation, for machine servo drives, can be defined as how much the controlled variable of a servo will reduce as load is applied to the servo. As such, percent regulation is usually applied to type 0 velocity regulators to describe how well the output velocity will be maintained as load is applied to the output of the drive. Percent regulation is often given as the equation Percent regulation ¼

no load speed
full load speed full load speed

A general block diagram for a DC servo drive with a position loop is shown in Figure 1. Using block diagram algebra the servo drive block diagram can be redrawn for speed as the controlled variable with the load torque as an input, shown in Figure 2. Equations (10.1-1) to (10.1-11) show the calculations for the regulation of a DC drive. Equations (10.1-12) to (10.1-14) rearrange Eq. (10.1-11) for velocity-loop stiffness. Eq. (10.1-15) expresses the velocity stiffness. The position-loop equations are shown in Eq. (10.1-16) to (10.1-27) for the position-loop stiffness.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1 Block diagram for an electric servo.

Fig. 2 Block diagram for electric drive stiffness.

Velocity Loop Vm K 6  T ¼ e JsRa RL s þ 1 1þ

1 KT Ke

JsRa



L Rsþ1

¼

JsRa

KT  s þ 1 þ KT Ke R

L

(10.1-1) Vm ¼h JRa e

1=Ke

L 2 KT Ke R s

þ

JRa KT Ke

sþ1

i¼

1=Ke ðTM Te s2 þ TM s þ 1Þ

(10.1-2)

Vm KA K1 ðT1 s þ 1Þ ¼ A K1 ðT1 sþ1ÞKTA 2 A Ke ðT2 s þ 1Þ½TM Te s þ TM s þ 16 Ke ðT2Ksþ1Þ½T 2 M Te s þTM sþ1 (10.1-3)

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Vm KA K1 ðT1 s þ 1Þ ¼ Ke ðT2 s þ 1Þ½TM Te s2 þ TM s þ 1 þ KA K1 ðT1 s þ 1ÞKTA A (10.1-4) s
?0 Vm s KA K1
?0 ¼ A Ke þ KA K1 KTA Vm S Kv ¼ K2
?0 (10.1-6a) A

(10.1-5) Kfb ¼

K2 KA K1 Kfb Ke þ KA K1 KTA

Vm 1 1 ¼ 6 TL Js 1 þ KT ½Ke þKL1 KTA KA  ðT1 sþ1Þ ðT2 sþ1Þ JsRa ðR sþ1Þ Vm 1 h i ¼ TL Js þ KT 6 Ke þ K1 KTA KA ðT1 sþ1Þ ðT2 sþ1Þ Ra ðRL sþ1Þ s
?0 Vm 1 Ra ¼ ¼ TL KRT 6½Ke þ K1 KTA KA  KT ½Ke þ K1 KTA KA KT 

(10.1-6b) (10.1-7)

(10.1-8)

(10.1-9)

a

Velocity-Loop Regulation For these equations, units are: Ra , ohm KT , in.-lb/A Ke , V/rad/sec K1 , V/V KTA , V/rad/sec KA , V/V

Vm Ra ¼ TL KT Ke þ K1 KTA KA KT Vm Ra ¼ Ti KT Ke þ KVO Ke KT Vm Ra ¼ TL KT Ke ð1 þ KVO Þ

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rad=sec in:-lb

 (10.1-10)

(10.1-11)

Velocity-Loop Stiffness TL KT Ke þ K1 KTA KA KT ¼ Vm Ra K1 KTA KA ¼ open-loop gain (v=v) KVO ¼ Ke Ke KVO ¼ K1 KTA KA

(10.1-12) (10.1-13) (10.1-14)

TL KT ½Ke þ KVO Ke  KT Ke in.-lb ¼ ¼ ð1 þ KVO Þ Ra rad/sec Vm Ra

(10.1-15)

Position Loop Vm 1 h i ¼ ðT1 sþ1Þ KT B Js þ Ra ðLs=Rþ1Þ 6 Ke þ K1 KTA KA ðT 2 sþ1Þ

(10.1-16)

Vm Ra ðLs=R þ 1Þ h i ¼ 1 sþ1Þ B JsðLs=R þ 1Þ þ KT Ke þ K1 KTA KA ðT ðT2 sþ1Þ

(10.1-17)

Vm Ra ðLs=R þ 1ÞðT2 s þ 1Þ ¼ JsðLs=R þ 1ÞðT2 s þ 1Þ þ KT Ke ðT2 s þ 1Þ þ KT ½Ke þ K1 KTA KA ðT1 s þ 1Þ B (10.1-18) y Ra ðLs=R þ 1ÞðT2 s þ 1Þ 1 ¼ 6 K K K K K ðT sþ1Þ R ðLs=Rþ1ÞðT sþ1Þ a 2 * TL s½  6 1 2 A T fb 1 1þ s½

*



ðLs=rþ1ÞðT2 sþ1Þ

(10.1-19) y ¼ TL s½ *

Ra ðLs=R þ 1ÞðT2 s þ 1Þ  þ Ra ðLs=R þ 1ÞðT2 s þ 1Þ

K1 K2 KA KT Kfb ðT1 sþ1Þ ðLs=Rþ1ÞðT2 sþ1ÞRa

* ¼ denominator of10.1-18 s?0 y Ra ¼ TL K1 K2 KA KT Kfb KV ¼

K1 K2 KA KT Kfb C þ K1 KTA KA

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(10.1-20)

(10.1-21) (10.1-22)

KVO ¼ K1 KTA KA 6

KT ðLs=R þ 1ÞRa JT s þ Ke KT

(10.1-23)

s?0 K1 KTA KA Ke K1 KTA KA Kfb KVO ¼ Ke þ KVO Ke y Ra Ra ¼ ¼ TL ðK1 K2 KA KT Kfb ÞKT Kv KT Ke ð1 þ KVO Þ

KVO ¼

(10.1-24) (10.1-25) (10.1-26)

Stiffness Variables for this equation have the following units: Ke , V/rad/sec KT , in.-lb/A Kv , 1/sec KVO , v/v Ra , ohms

TL Kv KT Ke ð1 þ KVO Þ ¼ y Ra

(10.1-27)

10.2 SERVO SYSTEM RESPONSES In the analysis of different types of servos they are often modeled in their simplest form, which would include only the frequency characteristics below the servo bandwidth. This can be done for the type 0 and type 1 servos, as discussed in this section. The frequency characteristics (dynamics) above the servo bandwidth could introduce some errors in the analysis, but past experience demonstrates that the major contribution to system performance occurs in the frequency characteristics below the servo bandwidth. Therefore, assuming a type 1 positioning servo drive can be modeled as shown in Figure 3, the output response can be predicted depending on what type of input (velocity step, position step, velocity ramp, or position ramp) will be

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 3 (a) Position response for a velocity step input. (b) Position response for a velocity ramp input. (c) Position response for a position step input.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

applied to the model. For each input in Figure 3 the output response for velocity, position, acceleration, and errors is shown. The models for the type 1 servo of Figure 3 are often referred to as a naked servo, meaning there are no frequency characteristics below the bandwidth velocity constant Kv . A type 0 velocity regulator servo drive can be modeled as shown in Figure 4. The first model is simplified by not including any of the system dynamics above the servo bandwidth. The second model is more characteristic of velocity servo drives using proportional and integral servo composition. These response characteristics are more representative of commercially available velocity servo drives.

Fig. 4 (a) Velocity response for a velocity step input. (b) Velocity response for a velocity step input.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

11 Servo Plant Compensation Techniques

Servo compensation usually implies that some type of filter network such as lead/lag circuits or proportional, integral, or differential (PID) algorithms will be used to stabilize the servo drive. However, there are other types of compensation that can be used external to the servo drive to compensate for other things in the servo plant (machine) that can, for example, be structural resonances or nonlinearities such as lost motion or stiction. These machine compensation techniques are shown in Figure 1 and are valid for either hydraulic or electric servo drives.

11.1 DEAD-ZONE NONLINEARITY Stiction, sometimes referred to as stick-slip, occurring inside a positioning servo, can result in a servo drive that will null hunt. The definition of a null hunt is an unstable position loop that has a very low periodic frequency such as 1 Hz or less with a small (a few thousandths) peak-to-peak amplitude (limit cycle). The most successful way to avoid stiction problems is to use antifriction machine way (rollers or hydrostatics) or use a way linear material that has minimal stiction properties. If stiction-free machine slide ways cannot be provided, the use of a small dead-zone nonlinearity placed inside the position loop, preferably at the input to the velocity servo, has

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1

Servo plant compensation techniques.

had some success in overcoming a null hunt problem. However the dead zone must be very small (e.g., 0.001 in.); otherwise, the servo drive will have an instability from too much lost motion. A simple analog dead-zone nonlinear circuit is shown in Figure 2. The same function can be provided with a digital algorithm in computer control of machines.

11.2 CHANGE-IN-GAIN NONLINEARITY In some industrial servo drives it is a requirement to position to a very low feed rate to obtain a smooth surface finish. This requirement usually occurs

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 2 Dead-zone nonlinearity.

in a turning machine application. At feed rates below 0.01 ipm, the requirement for a smooth surface may not be easily attainable because the servo drive may have a cogging problem at these low federates. Increasing the forward loop gain to the velocity drive can overcome the lowfeed cogging problem but will result in an unstable servo drive. As a compromise, a change in gain nonlinear circuit can be used to improve the low-feed-rate smoothness and still have a stable servo drive. The object is to have a high forward-loop gain in the velocity servo (which is inside a position loop). For normal operation, the high servo loop gain is reduced by the change-in-gain circuit at a low velocity to its normal gain, thus maintaining a stable servo drive. This type of nonlinear circuit has been used successfully for smooth

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 3

Change-in-gain nonlinearity.

feed rates down to 0.001 ipm. The analog version of a change-in-gain nonlinearity is shown in Figure 3. With digital controls a digital algorithm can be used.

11.3 STRUCTURAL RESONANCES Structural resonances or machine dynamics, as it often referred to, is certainly not a new subject. However, on the morning of November 7, 1940, the nation awoke to the destruction of the Tacoma Narrows Bridge. A 42 mile-per-hour gale caused the bridge to oscillate thus exciting the structural resonances of the bridge to a final destruction frequency of about 14 Hz and a peak-to-peak amplitude of 28 ft. The destruction of the bridge was a wake-up call to the importance of dynamic analysis in structural

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

design in addition to static analysis and design. Some sixty years later the technology of dynamic analysis is now well known. To further investigate machine resonances, a typical linear industrial servo drive can be represented as in Figure 4. The mechanical components of this servo drive are referred to as the servo system plant. The servo plant may have a multiplicity of resonant frequencies resulting from a number of degrees of freedom. In actual practice there will be some resonant frequencies that are high in frequency and far enough above the servo drive bandwidth so that they can be ignored. In general there will be a predominant low resonant frequency that could possibly be close enough to the servo drive bandwidth to cause a stability problem. Therefore a single degree of freedom model as shown in Figure 5 can represent the predominant low-resonant frequency, where: BL ¼ viscous friction coefficient (lb-in.-min/rad) T ¼ driving torque, developed by the servo motor (lb-in.) K ¼ mechanical stiffness of the spring mass model (lb-in./rad) JM ¼ inertia of the motor (lb-in.-sec2) JL ¼ inertia of the load (lb-in.-sec2) S ¼ laplace operator From Newton’s second law of motion, the classical equations for this servo plant (industrial machine system) can be written. In most industrial machines it can be assumed that the damping BL is zero. Therefore the

Fig. 4 Block diagram of a machine feed drive.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 5

Machine slide free-body diagram.

motor equation is:

T M ¼ J M s2 yM þ KðyM
yL Þ

(11.3-1)

Also the load equation is:

0 ¼ J L s yL þ KðyL
yM Þ

(11.3-2)

Solving for the motor position yM and the load position yL :

ðJ M s2 þ KÞyM ¼ T M þ yL TM þ KyL yM ¼ JM s 2 þ K

(11.3-3) (11.3-4)

and

ðJ L s2 þ KÞyM ¼ KyM KyM yL ¼ JL s 2 þ K

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(11.3-5) (11.3-6)

Further solving for yM and yL by combining Eq. (11.3-4) and (11.3-6):

TM þ KðKyM =JL s2 þ KÞ JM s 2 þ K 2 TM ðJL s þ KÞ þ K 2 yM yM ¼ ðJM s2 þ KÞðJL s2 þ KÞ TM ðJL s2 þ KÞ K 2 yM þ yM ¼ 2 2 2 ðJM s þ KÞðJL s þ KÞ ðJM s þ KÞðJL s2 þ KÞ K 2 yM TM ðJL s2 þ KÞ ¼ yM
2 2 ðJM s þ KÞðJL s þ KÞ ðJM s2 þ KÞðJL s2 þ KÞ yM ¼

ððJM s2 þ KÞðJL s2 þ KÞ
K 2 ÞyM ¼ T M ðJ L s2 þ KÞ

(11.3-7) (11.3-8) (11.3-9) (11.3-10) (11.3-11)

2

TM ðJL s þ KÞ ðJM s2 þ KÞðJL s2 þ KÞ
K 2 TM ðJL s2 þ KÞ ¼ JM JL s4 þ KðJM þ JL Þs2 TM ðJL s2 þ KÞ ¼ 2 s ðJM JL s2 þ ðJM þ JL ÞK TM ðJL s2 þ KÞ ¼ 2 s ðJM þ JL ÞððJM JL =JM þ JL Þs2 þ KÞ

yM ¼

(11.3-12)

yM

(11.3-13)

yM yM

(11.3-14) (11.3-15)

Let:

J ¼ JM þ JL J P ¼ J M J L =ðJ M þ J L Þ yM ¼

TM ðJL s2 þ KÞ s2 Jðs2 JP þ 1Þ

(11.3-16)

yL ¼

KyM JL s 2 þ K

(11.3-17)

Also:

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K TM ðJL s2 þ KÞ 6 ðJL s2 þ KÞ s2 Jðs2 JP þ KÞ TM K yL ¼ 2 2 s Jðs Jp þ KÞ yL 1 ¼ TM ðJ=KÞs2 ðs2 ðJp =KÞ þ 1Þ yL 1 ¼ 2 TM ðJ=KÞs ððs2 =o2r Þ þ 1Þ sffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K KðJM þ JL Þ or ¼ load resonant frequency ¼ ¼ Jp JM JL yL ¼

(11.3-18) (11.3-19) (11.3-20) (11.3-21) (11.3-22)

From a practical point of view industrial machines and their servo drives (hydraulic and electric) are to this day still subject to resonant frequency stability problems. Most industrial servo drives use an inner velocity servo inside a position servo loop. Hydraulic servo drives have the added variable of hydraulic fluid resonance, which can be a limiting factor of stability. The hydraulic resonance or can be observed as a typical second order response in the Bode frequency response of Figure 6. For hydraulic drives having a low damping factor dh , the resonant peak may be higher than 0 dB gain, which will result in a resonant oscillation. There are a number of methods to compensate for this resonant oscillation. First, a small cross-port damping hole of about 0.002 in. can be used across the motor ports. Secondly, the velocity loop differential compensation can be varied, which quite often eliminates the oscillation. Lastly, the velocity loop gain could be lowered, which can also lower the velocity servo bandwidth. As an index of performance (I.P.) the hydraulic resonance should by proper sizing be above 200 Hz, and the separation between the velocity servo loop bandwidth oc and the hydraulic resonance oh should be three to one or greater. Brushless DC electric drives do not usually have velocity loop resonance problems unless a more compliant coupling is used internally in the motor to couple a position transducer to the motor shaft. Both hydraulic and brushless DC electric drives can have resonance (stability) problems if the machine is included in the position servo loop. This is an ongoing problem with industrial machines, in spite of all the available technology to minimize stability problems. A typical position servo Bode frequency response is shown in Figure 7. As a figure of merit the separation between the velocity loop bandwidth oc and the position-loop velocity constant K v (gain) should be two to one or greater. The machine

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 6 Hydraulic velocity servo.

resonance or should be at least three times greater than the velocity servo bandwidth oc . However in actual practice the machine resonance inside the position loop is often quite low (such as 15 Hz). The resonant peak in this case should be above 0 dB gain, resulting in a resonant oscillation. There are a number of control techniques that can be applied to compensate for machine structural resonances that are both low in frequency and inside the position servo loop. The first control technique is to lower the position-loop gain Kv (velocity constant). Depending on how low the machine resonance is, the position-loop gain may have to be lowered to about 0.5 ipm/mil (8.33/ sec). This solution has been used in numerous industrial positioning servo drives. However, such a solution degrades servo performance. For very large machines this may not be acceptable. The I.P. that the servo loop gain (velocity constant) should be lower than the velocity servo bandwidth by a factor of two, will be compromised in these circumstances. A very useful control technique to compensate for a machine resonance is the use of notch filters. These notch filters are most effective

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 7

Position loop frequency response.

when placed in cascade with the input to the velocity servo drive. These notch filters should have a tunable range from approximately 5 Hz to a couple of decades higher in frequency. The notch filters are effective to compensate for fixed structural resonances. If the resonance varies due to such things as load changes, the notch filter will not be effective. Since most of these unwanted resonant frequencies are analog sinusoidal voltages, a notch filter can effectively remove these frequencies. In digital control the algorithm for the notch filter can be used. A simple analog notch filter is shown in Figure 8 as it appeared in Electronics Magazine, December 7, 1978. This filter is equivalent to a twin T-notch filter but it is an active filter versus passive networks, so there are no signal losses. The frequency of the notch is set by the selection of resistor R. For a 1-microfarad (mF) capacitor (C), the values for R versus the notch frequency are shown in Figure 9. The depth of the notch is adjusted by varying the potentiometer P1 . Frequency responses of the notch filter for values of R ranging from 40 K ohms to 200 ohms are shown in Figure 10. A 40-Hz notch filter frequency response is shown in Figure 11 with a number of potentiometer settings to

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 8 Notch filter circuit. (Reprinted with permission from Penton Publishing.)

show the change in the depth of the notch filter as the potentiometer is varied. A few case histories are of interest. In a hydraulic servo valve feed drive, pump pulsations of 500 Hz traveled through the machine piping to the servo valve, the hydraulic motor, and finally the feedback tachometer of the velocity loop. The high sensitivity of the tachometer (100 V/1000 rpm) sensed the 500-Hz vibration and generated this voltage into the servo drive electronics, where it was amplified through the entire drive, causing an undesirable vibration. A 500-Hz notch filter at the tachometer output feed to the servo amplifier eliminated the vibration problem. In another case, the switching frequency of a numerical control (125 Hz) beat with the single-phase, full-wave DC SCR drive frequency (120 Hz) producing a 5-Hz signal that appeared in the machine servo drive, causing what appeared to be a 5-Hz instability. Using a notch filter frequency of the control switching frequency, the 5-Hz beat frequency signal was eliminated. In another case history a 45-Hz resonance existed in an air bearing of a rotary position feedback transducer. Once this resonance was excited it was amplified through the electronics drive and the machine slide vibrated at 45 Hz. A 45-Hz notch filter applied at the input to the velocity drive eliminated the problem. There are numerous possibilities where unwanted

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 9

Notch filter resistance nomogram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Notch filter characteristics.

Fig. 11

Depth of notch filter characteristics.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

signal frequencies appear in industrial servo drives. This simple notch filter can readily be used to eliminate all kinds of undesirable signals.

11.4 FREQUENCY SELECTIVE FEEDBACK Another technique that has been very successful with industrial machines having low-frequency machine resonances, is known as ‘‘frequency selective feedback’’. In abbreviated form it requires that the position feedback be located at the servo motor eliminating the mechanical resonance from the position servo loop, resulting in a stable servo drive but with significant position errors. These position errors are compensated for by measuring the slide position through a low-pass filter; taking the position difference between the servo motor position and the machine slide position; and making a correction to the position loop, which is primarily closed at the servo motor.

Compensator Operation The complete control circuit has more complexity than comparing two position transducer outputs. Figure 12 can be used to discuss the actual operation of the compensating system. This particular compensating scheme used an instrument servo to perform the compensating function. A software based frequency selective feedback system will also be discussed. For this compensating system, the machine-feed servo drive uses a position measuring feedback resolver (1) connected electrically in series with a correction resolver (5). Any correction required during positioning is introduced into the numerical control feedback circuit with the correction resolver (5). The compensator circuit includes the positioning servo-motor position measuring resolver called a compensator feedback resolver (2), a machine slide linear position measuring transducer (3), and an instrument type correction servo drive. The difference between the feed servo-drive motor position and the machine slide position is measured with the compensator feedback resolver (2) and the linear resolver (3). However, an additional correction resolver (4) is included in the circuit. Therefore, the instrument correction servo error is a function of three resolver positions. This is shown in the block diagram of Figure 13. The resolver positions are shown as angle y. The total correction error is shown as a function of the three resolver positions yc ¼ ym
y c
y s :

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(11.4-1)

Fig. 12

Analog block diagram for frequency selected feedback.

Fig. 13

Simplified block diagram for frequency selective feedback.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

As a position error is developed between the feed servo-drive motor position ym and machine slide position ys , an error is developed at the instrument servo drive (Figure 13). The error is also a function of yc . ye ¼ ym
yc
ys ¼ ðym
ys Þ
yc

(11.4-2)

It is significant to note that the bandwidth of the instrument correction servo must be a low frequency such as 1.5 Hz. This is required to eliminate the machine structural dynamics from the machine slide feedback position ðys Þ and is the key to the frequency selective feedback control function. A correction yc is developed at the output of the instrument servo drive, which is a function of: yc ¼ G1 ½ðym
ys Þ
yc 

(11.4-3)

The correction yc is added to the main-feed servo drive by means of resolver (5). The correction yc causes the feed servo drive to move by the amount of the correction. Therefore both ym (at the motor) and ys (at the machine slide) move by the amount of the correction yc . The correction can be shown mathematically to be approximately a function of ðym
ys Þ as follows: Since:

yc ¼ Gl ½ðym
ys Þ
yc  yc þ yc Gl ¼ ðym
ys ÞGl yc ¼

ðym
ys ÞG1 ð1 þ G1 Þ

1 1=G1 þ 1 1=G1 5 much smaller than 1 Therefore: yc %ðym
ys Þ yc ¼ ðym
yc Þ

from Eq: (11.4-3) (11.4-4) (11.4-5) (11.4-6)

(11.4-7)

The correction can also be shown graphically in Figure 14. For illustration the relation of the feed servo-drive position ym to the machine slide position ys will remain constant during the correction. Figure 14 shows that the machine slide position ys moves by the amount of the correction. The feed servo-drive motor position moves by the same amount. From another point of view the instrument servo-drive error must be reduced to zero after the correction is made. From Eq. (11.4-2) the error is: ye ¼ ðym
ys Þ
yc

(11.4-2)

Assuming the initial condition at the start of the correction is such that the

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 14

Correction diagram for frequency selective feedback.

machine slide position ys is short of the correct position (assume 10 in.) by 0.01 in. and that the location of ym and ys are as follows: ym ¼ a related distance of 10 in. ys ¼ a related distance of 9:99 in: After correction the related positions will be— ym ¼ a related distance of 10:01 in: ys ¼ a related distance of 10 in: The relation between ym and ys is the same but the machine position ys has moved to the correct position.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Meanwhile the correction yc (which was equivalent to 0.01 in.) had to be included in the error of the instrument servo. Since the relation of ðym
ys Þ remains constant, a correction must be made in the instrument servo loop to reduce the error to zero: yc ¼ 0 ¼ ðym
ys Þ
yc

(11.4-8)

The compensator correction resolver (4) serves the purpose to reduce the instrument servo error to zero. As the correction process becomes a continual process, as in normal machine operation, the resolver (4) will continually be in motion to reduce the error to zero in the correction loop of the instrument servo. At machine traverse feeds the correction will not be effective, but this is not important since there will not be any machining operations at the traverse feeds.

Software version of frequency selective feedback The software version for the correction is shown in Figure 15. This version can be added to a machine axis that exhibits a structural resonance problem. The drive is assumed to be a typical commercial electric servo drive with a current loop and a velocity loop inside a position servo loop. The difference between the motor position and the machine slide position is used as an input to a low-pass digital filter. This filter has a very low bandwidth of about 1.5 Hz. The reason for this is to remove structural dynamic frequencies from the correction process.

Fig. 15

Software version of frequency selective feedback.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The machine axis prior to adding frequency selective feedback, is assumed to have a position transducer on the machine slide. To add frequency selective feedback, an instrument gearbox must be added to the rear of the servo motor. A position transducer such as a resolver will be geared to the motor shaft with a ratio determined by the resolution of the machine slide feedback transducer.

11.5 FEEDFORWARD CONTROL The problem of developing an industrial servo drive with high-performance capabilities for accurate positioning is a subject of much importance. On multiaxis industrial machine servos using classical type 1 servo control, it is a requirement that each machine axis have matched position-loop gains to maintain accuracy in positioning. Quite often this means that all machine axis servo drives must have their position-loop gains K v adjusted to the poorest performing axis. Consider the basic approach to the design of a poisoning servo drive illustrated in Figure 16. This is the classical type 1 servo, which exhibits characteristic errors e in position that are well known for various inputs yi . Consider a simplified block diagram of the system where GðsÞ is the servo drive and inner-loop transfer function typically of the following form: GðsÞ ¼

Kv sFðsÞ

(11.5-1)

F ðsÞ is a polynomial that represents the dynamics of the servo drive and servo plant. What is really desired of the servo of Figure 17 is that yi ¼ y0 under all conditions. That is, for any yi ; e ¼ 0. Clearly, this is not possible for a type 1 servo described by Figure 17.

Fig. 16

Type 1 servo block diagram.

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Fig. 17

Simplified position-loop block diagram.

Consider, then, a compensator Gc that could possibly produce a system response as desired, placed cascade with the system as shown in Figure 18. yoðsÞ GðsÞ ¼ GcðsÞ ¼1 yiðsÞ 1 þ GðsÞ 1 þ GðsÞ 1 Thus : GcðsÞ ¼ ¼ þ1 GðsÞ GðsÞ Since

(11.5-2) (11.5-3)

Thus, the desired system may be represented in block diagram form as Figure 19. It is convenient to rearrange the diagram of Figure 19 so that the actual error term e ¼ yi
yo appears, as Figure 20. As pointed out by Eq. (11.5-1) GðsÞ ¼

Kv for a practical position servo: sFðsÞ

(11.5-4)

Thus the block diagram of Figure 20 becomes Figure 21. sFðsÞ FðsÞ ¼ oiðsÞ Kv Kv dyiðtÞ ¼ dt

Note that yiðsÞ

(11.5-5)

Where

(11.5-6)

oiðtÞ

Furthermore, a well-designed, high-performance servo GðsÞ should have dynamics described by F ðsÞ sufficiently negligible with respect to the integration crossover that F ðsÞ need not be synthesized in the forward compensating path. This assumption is important to the simplicity of the concept, but it is possible to check if it is valid in any specific application.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 18

Compensator with position servo.

Fig. 19

Block diagram with feedforward.

Fig. 20

Zero-error feedforward block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 21

Complete feedforward block diagram.

Because the inverse of a type 1 system differentiates rather than integrates the compensating path is simply a velocity signal rather than a position signal. Consider the final design of the system described in Figure 22. From Figure 22   oiðsÞ Kv yoðsÞ ¼ yiðsÞ
yoðsÞ þ Kv sFðsÞ     oiðsÞ Kv Kv yoðsÞ 1 þ ¼ yiðsÞ þ sFðsÞ Kv sFðsÞ yoðsÞ ¼

Fig. 22

ðKv yiðsÞ þ oiðsÞ Þ ðsFðsÞ þ Kv Þ

Zero-error position servo.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(11.5-7) (11.5-8) (11.5-9)

Equation (11.5-9) may be used to illustrate how in the ideal case (where F ðsÞ ¼ 1) the system is zero error in position regardless of the input. yi ¼ Ct therefore yiðsÞ ¼ oi ¼ C

C s2

oiðsÞ ¼

(11.5-10) C s

(11.5-11)

Therefore yoðsÞ ¼

Kv C=s2 þ C=s ðs þ Kv ÞC ¼ 2 sFðsÞ þ Kv s ðsFðsÞ þ Kv Þ

(11.5-12)

Or for F ðsÞ &1, the lag-lead cancel and yoðsÞ ¼

C ¼ yiðsÞ s2

therefore zero error:

(11.5-13)

Consider next the steady-state error for the constant velocity case:   C C s þ Kv eðsÞ ¼ yiðsÞ
yoðsÞ ¼ 2
2 (11.5-14) s s sFðsÞ þ Kv   C s þ Kv (11.5-15) eðsÞ ¼ 2 1
s sFðsÞ þ Kv Applying the final value theorem, the steady-state position error is:   lim C s þ Kv 1
eð?Þ ¼ lim seðsÞ ¼ (11.5-16) sFðsÞ þ Kv s?0 s   lim C s þ Kv 1 1
(11.5-17) ¼ sFðsÞ þ Kv s s?0   lim C sFðsÞ þ Kv
s
Kv 1 (11.5-18) ¼ s?0 s sFðsÞ þ Kv

¼

  lim C FðsÞ
1 s?0 sFðsÞ þ Kv

However, for all possible forms of F ðsÞ lim s?0

F ðsÞ ¼ 1

Thus: eð?Þ ¼ 0

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(11.5-19)

For constant acceleration input: 1 yiðtÞ ¼ at2 2

yiðsÞ ¼

a s3 a ¼ 2 s

(11.5-20)

oiðtÞ ¼ at

oiðsÞ

(11.5-21)

From Eq. 11.5-9 Kv a=s3 þ a=s2 a s þ Kv ¼ 3 sFðsÞ þ Kv s sFðsÞ þ Kv   limseðsÞ a sFðsÞ þ Kv
s
Kv ¼ 2 And eð?Þ ¼ s?0 sFðsÞ þ Kv s

yiðsÞ ¼

eð?Þ ¼

lim a ðFðsÞ
1Þ s?0 s ðsFðsÞ þ Kv Þ

dFðsÞ =ds alim FðsÞ
1 alim ¼ s?0 s2 FðsÞ þ Kv s s?0 ds2 FðsÞ =ds þ Kv lim dFðsÞ ?finite value Then s?0 ds limit ds2 FðsÞ While ?0 s?0 ds 1 Thus eð?Þ ¼ Kv eð?Þ ¼

(11.5-22) (11.5-23) (11.5-24) (11.5-25) (11.5-26) (11.5-27) (11.5-28)

The magnitude of the finite position error will depend in this case on the coefficient of the s term of F(s). If b is this coefficient, then eð?Þ ¼

ab Kv

(11.5-29)

The compensator technique described holds excellent potential for providing an outstanding servo drive for the industrial machine. When properly executed in a well-designed position servo it should virtually eliminate velocity lag errors and reduce acceleration lag errors to low levels. The velocity feedforward approach will eliminate position errors for constant velocity moves on a machine axis if the machine dynamics represented by F ðsÞ in the feedforward term F ðsÞ =K v exactly respresents the G(s) term of the forward position loop. Matching of position-loop gains K v on all axes will not be required. For the case of acceleration and deceleration the velocity feedforward approach will not be effective. For these situations commercial servo-drive manufacturers will use an additional technique of acceleration feedforward.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

12 Machine Considerations

12.1 MACHINE FEEDBACK DRIVE CONSIDERATIONS Machine Feed Drive Considerations for Resolver Feedback The usual configuration for resolver position feedback is to have the resolver geared to the servo drive motor. Computer-aided design programs for hydraulic drive sizing are based on the fact that the hydraulic resonance is the predominant resonance in the servo loop. With electric drives the mechanical time constant is not of prime consideration since it can be compensated for with the drive compensation. These are two important dynamic considerations with electric and hydraulic servo drives using resolver feedback at the servo motor. Since the position feedback resolver is close-coupled to the drive servo motor, all other mechanical resonances and nonlinearities are excluded from the position servo loop. Recommendations for the required indexes of performance (I.P.) are identical with those given in Section 9.2.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Machine Feed Drive Considerations for Direct Machine Slide Position Feedback The main reason that direct feedback could be used was the universal use of the ‘‘soft servo’’ with its low position-loop gain. By definition a soft servo has a low position-loop gain of about 1 ipm/mil with no frequency breaks below the servo bandwidth. The range of gains for the ‘‘soft servo’’ range from 0.5 ipm/mil to 2 ipm/mil. As long as no resonances are within six times the high end of this range (200 rad/sec or 32 Hz, Eq. [9.2-27]), it should be possible to close the position loop and be stable. As previously stated, if the feedback position transducer (resolver) is located at the drive motor, any mechanical resonance in the machine will be outside the servo loop. The effects of these mechanical resonances on the closed-position loop will be reflected load disturbances. However, if direct (e.g., Inductosyn-linear scales) position feedback is to be used, the mechanical resonance in the mechanical drive components will be inside the position loop. For this situation, some guidelines, recommendations, etc. should be put forth to prevent a situation where direct feedback is used on a feed drive that will not be stable.

12.2 BALL SCREW MECHANICAL RESONANCES AND REFLECTED INERTIAS FOR MACHINE DRIVES As a constraint the lower limit on any resonance (hydraulic or mechanical) inside the velocity loop should not be less than 200 rad/sec (Eq. [9.2-27]). Likewise, the bandwith ðoc Þ of the closed velocity loop should not be greater than one-third the lowest resonance in the servo loop, which is usually the hydraulic resonance, oh , in hydraulic drives (Eq. [9.2-23]). With the position loop, there may be a number of different mechanical resonances. These resonances are outside the velocity loop but inside the position loop. Considering the mechanical feed drive components, the one of greatest concern is the drive screw axial resonance. This resonance will probably be the predominant low mechanical resonance in a ball-screw feed drive. Typical values of ball screw stiffness are shown in the graph of Figure 1 for the variables of screw diameter and length. These values are based on constant end bearing and ball nut stiffnesses. In actual practice, these would be varied according to the screw diameter. In Figures 2 to 5, the ball screw resonances are plotted for various screw diameters, lengths, and applied load weights. Resonances occurring outside the velocity loop and inside the position loop will contribute phase shift to the open-position loop frequency-

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Fig. 1 Ball screw stiffness.

Fig. 2

Ball screw resonance.

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Fig. 3

Ball screw resonance.

Fig. 4 Ball screw resonance.

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Fig. 5 Ball screw resonance.

response characteristics. To maintain the desired 458 phase margin for the position loop, the resonances (such as the ball screw) should be sufficiently high relative to the position-loop velocity constant ðKv Þ that they will not reduce the phase margin. It is recommended practice that the lowest resonance inside the position servo loop should be three times higher than the minimum hydraulic resonance (200 rad/sec). If this practice is followed, the reflected phase shift occurring at the velocity constant ðKv Þ from the position loop resonance will not reduce the overall position-loop phase margin. In actual practice this is somewhat academic because in large machinery mechanical resonances do occur above and below 200 rad/sec. As an index of performance (I.P.) it is recommended that the position loop resonance (usually the ball screw) should be at least six times the positionloop velocity constant ðKv Þ or greater than oh . (Refer to Section 9.2, Eq. [9.2-22] to [9.2-28].) A graph displaying this I.P. can be drawn (Figure 6) showing the relation between position-loop velocity constant ðKv Þ and the lowest allowable resonance in the position servo loop. For numerical controls using the ‘‘soft servo’’ technique, it can be determined from the graph in Figure 6 that resonances inside the position loop should, in general, be large

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 6

Lowest recommended resonance inside the position loop.

enough (at least 200 rad/sec) such that the resonance will not have a detrimental effect on the drive. For large machines where the velocity constant ðKv Þ usually cannot be larger than 0.5 ipm/mil for reasons of stability, the lowest allowable resonance (from Figure 6) could be as low as 50 rad/sec. For small machines using the ‘‘soft servo’’ with velocity constants of 2 ipm/mil, the lowest resonance in the position servo loop should not be less than 200 rad/sec. Therefore, it is difficult to make an across-the-board recommendation with changing velocity constants ðKv Þ from machine to machine. One possible way to arrive at a recommendation is to assume that all machines being built use position loop gains up to 2 ipm/mil, and therefore, machines with ball screw resonances under 200 rad/sec cannot use direct feedback. Another possibility, which is more practical, is to list the machines with the recommended maximum allowable position servo loop gains ðKv Þ that can be used with direct feedback. Since all axes on a machine must have the same position loop gain, the machine axis with the lowest resonance will be the determining factor in how low the position loop gain or velocity constant ðKv Þ must be set.

Other Resonance Considerations Additional mechanical resonances inside the position loop that may cause stability problems should also be considered. Reflected structural resonances

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

due to large and sometimes limber machine structures may reflect their resonances into the machine slide as the result of an antinode vibration. Likewise, very large and heavy workpieces placed on an otherwise stable machine slide can cause stability problems when direct position feedback is used. Additionally, mechanical power transmission devices, such as ‘‘wound up’’ gear trains, can be a source of stability problems when they are included inside the position loop with direct slide position feedback. The gearbox can have two types of problems. First, the amount of windup can vary causing a change in spring rate that can cause instability. Second, any nonlinearity such as backlash that occurs as a result of lost windup will result in an unstable servo drive.

Reflected Inertias for Machine Drives An important parameter to consider in sizing a machine servo drive is the total inertia at the servo motor. The definition of inertia is the property of a body, by virtue of which it offers resistance to any change in its motion. For an industrial machine slide the total reflected inertia to the servo motor is made up of the reflected belt and pulley inertia (or gearboxes), the reflected drive screw, and the reflected machine slide with its load weight. Knowing the total inertia at the servo drive motor is a requirement for any form of servo analysis or drive sizing. The reflected inertia at the drive motor can be calculated as

J reflected ¼

Jload ðD2 =D1 Þ2

(lb-in.-sec2 Þ

(12.2-1)

Where D1 ¼ diameter of the motor pulley or gear (in.) D2 ¼ diameter of the load pulley or gear (in.)

A very important consideration in applying a drive to a machine axis is the torque to inertia ratio. A mechanical equation for acceleration torque is: T T ¼ J a (lb-in.) Where J T ¼ total inertia at the motor shaft (lb-in.) a ¼ acceleration at motor shaft (rad=sec2 )

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.2-2)

Equation (12.2-1) can be rearranged as a¼

T (rad=sec2 ) J

(12.2-3)

Equation (12.2-3) is known as the torque to inertia ratio, and has important significance. In a given drive application the machine will have a total inertia (motor inertia plus reflected inertia) that exists by design. For productivity requirements there will be a required acceleration. The question must now be asked: Is there enough acceleration torque to meet these requirements? Quite often this becomes a critical drive-sizing problem providing enough available current from the servo amplifier to accelerate the load required to meet a productivity requirement? For industrial servos using brushless DC drives there are two types of motors to be considered. The brushless DC motor in general industrial applications uses ceramic magnets. For high-performance applications, low inertia motors are used. These motors use Neodymium-iron-boron (NdFeB) magnet material or Samarium cobalt (SmCo) magnet material. A comparison of torque to inertia ratios ðT/JÞ with various total inertia loads is compared in Figure 7. At small inertia loads the torque to inertia ratios for low inertia motors are significantly higher than standard ceramic motors, which is an indication that the low inertia motors are capable of higher performance (and acceleration) with small total inertia loads. These curves also indicate that the low inertia motors have a significant reduction in performance as total load inertia increases. The standard ceramic magnet-type motors have less reduction in performance with added total load inertia. The low inertia drive is usually more expensive than the standard ceramic-magnet brushless DC motor. For larger total inertia loads it is not economically justifiable to use the low inertia type motor since performance will be compromised. If high performance is a requirement, the reflected inertia load should be reduced with a ratio if possible. To further study the performance of brushless DC drives under various inertial loads, a transient step response test will be made for a velocity servo using a lag/lead compensation in the servo amplifier with a gain of 1000 volts/volt. A more detailed analysis is presented in Chapter 14. A 3-volt/1000 rpm tachometer feedback will be used. The block diagram for this servo drive is shown in Figure 8. A simple velocity servo was chosen for this analysis to minimize complexity. A standard ceramic brushless DC motor and a low inertia brushless DC motor will be used for this analysis. The motor parameters are identified as the following:

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

MOTOR

STANDARD MOTOR (Kollmorgan M607B)

LOW INERTIA MOTOR (Kollmorgan EB606B)

KT ¼ Ke ¼ Rmotðl
lÞ ¼ SRmotðl
lÞ SRmotðphaseÞ ¼ Ll
l ¼ Jmot ¼

9.9 lb-in./A 0.3734 v-sec/rad 0.14 ohm 0.189 ohm 0.0945 ohm 0.0038 henry 0.1872 lb-in.-sec2

9.9 lb-in./A 0.3734 v-sec/rad 0.14 ohm 0.189 ohm 0.0945 ohm 0.0038 henry 0.02688 lb-in.-sec2

te ¼

Lðl
lÞ 0:0038 ¼ 0:02 sec ¼ 0:189 SRmotðl
lÞ

0:02 sec

tm ¼

SRmotðphaseÞ Jtot KeðphaseÞ KT

(12.2-4)

(12:2-5)

J total ¼ J mot þ J reflected

(12.2-6)

Transient responses for a multiplicity of load inertias are shown in Figures 9 to 16. In this analysis it is assumed that each drive will be analyzed with identical inertia loads as if a machine axis was being tested with a standard brushless DC motor and then tested again with a low inertia motor. Load inertia (lb-in.-sec2)

Standard motor transient responses

1 X MOTOR 0.2 0.8 2.0

Fig. Fig. Fig. Fig.

9 11 13 15

Low inertia motor transient responses Fig. Fig. Fig. Fig.

10 12 14 16

A ceramic-magnet brushless DC servo motor will have some reduction in servo bandwidth with added load inertia (Figures 9, 11, 13, and 15), but these drives are tolerant of added load inertia without having to be recompensated. Rare earth brushless DC drives are load inertia sensitive (Figures 10, 12, 14, and 16). For brushless DC servo drives with ceramicmagnet armatures, a ratio of approximately four times reflected load inertia to the motor armature can be tolerated. For low-inertia brushless DC motors it is recommended that the reflected load inertia match the motor inertia. Since these motors are used for high-performance applications, it is advisable not to reduce the servo bandwidth with inertia mismatch ratios larger than one. Reflected load inertias can be computed as follows:

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Reflected Machine Slide Inertia Jslide ¼ W6ðL=NÞ2 60:0000656 ðlb-in.-sec2 Þ where: W ¼ total machine slide and load weight (lb) L ¼ ball screw lead (in./rev) N ¼ drive ratio

Fig. 7 Torque to inertia ratio comparisons.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.2-7)

Fig. 8 Velocity servo block diagram.

Fig. 9 Transient step response with a 1 6 inertia load for a standard ceramicmagnet motor.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Transient step response with a 1 6 load inertia for a low inertia motor.

Reflected Drive Screw Inertia The drive screw inertia (without threads) reflected to the servo motor can be calculated from Eq. (12.2-1). Jscrew ¼ X1
½ðX2 þ X3 Þ6X4 6X5  (lb-in.-sec2 ) where: DIA ¼ screw diameter (in.) LGTH ¼ screw length (in.) Ball dia ¼ diameter of the balls in the drive screw nut (in.)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.2-8)

Fig. 11 Transient step response with a 0.2 lb-in.-sec2 inertia load for a standard ceramic motor.

where: X1 ¼ ðDIA4 6LGTH60:000072Þ=N 2 T2 ¼ ATN½L=ð3:141596DIAÞ

(12.2-9) (12.2-10)

X2 ¼ ðball dia=2Þ4 60:10941

(12.2-11)

2

X3 ¼ ½ð3:14159=26ball dia Þ=2 6½DIA=2
ð0:42446ðball dia=2Þ2 Þ X4 ¼ 0:0046046½DIA=2
ð0:42446ball dia=2Þ 6½1= cosðT2 Þ

(12.2-12) (12.2-13)

2

X5 ¼ LGTH=ðL6N Þ

(12.2-14)

The reflected ball screw inertia (including threads) can be calculated as follows:

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Fig. 12 Transient step response with a 0.2 lb-in.-sec2 inertia load for a low inertia motor (equivalent to a 10 6 inertia load).

Jscrew ¼

DIA4 6LGTH67:2610
5 (lb-in.-sec2 ) N2

Pulley Inertias Ratio ¼ screw pulley diameter/motor pulley diameter Woods pulleys, 3-inch pulley inertias (lb-in.-sec2) No.

Pulley no.

Approx. diameter

Inertia

1 2 3

16H300 18H300 19H300

2.39 2.71 2.86

0.0082 0.013579 0.017198

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.2-15)

Fig. 13 Transient step response with a 0.8 lb-in.-sec2 inertia load for a standard ceramic-magnet motor.

No. 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Pulley no.

Approx. diameter

Inertia

20H300 21H300 22H300 24H300 26H300 28H300 30H300 32H300 36H300 40H300 44H300 48H300 60H300 72H300

3.03 3.18 3.35 3.66 3.98 4.30 4.62 4.94 5.58 6.21 6.85 7.48 9.36 11.30

0.02145 0.0263 0.02417 0.03189 0.04275 0.05909 0.0780 0.09990 0.160 0.2249 0.33489 0.47042 0.8005 1.4818

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Fig. 14 motor.

Transient step response with a 0.8 lb-in.-sec2 inertia load for a low inertia

No.

Pulley no.

Approx. diameter

18 19 20 21

84H300 96H300 120H300 156H300

13.21 15.12 18.94 24.67

Inertia 2.5103 4.0924 8.6475 21.798

Woods pulleys, 2-inch pulley inertias (lb-in.-sec2) No.

Pulley no.

Approx. diameter

Inertia

1 2 3 4 5

16H200 18H200 19H200 20H200 21H200

2.39 2.71 2.86 3.03 3.18

0.004066 0.00802 0.01246 0.01527 0.01913

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Fig. 15 Transient step response with a 2 lb-in.-sec2 inertia load for a standard ceramic motor (equivalent to a 10 6 inertia load).

No. 6 7 8 9 10 11 12 13 14 15 16 17

Pulley no.

Approx. diameter

Inertia

22H200 24H200 26H200 28H200 30H200 32H200 36H200 40H200 44H200 48H200 60H200 72H200

3.35 3.66 3.98 4.30 4.62 4.94 5.58 6.21 6.85 7.48 9.36 11.30

0.01869 0.02488 0.03342 0.04555 0.05987 0.07841 0.12445 0.17903 0.26439 0.37604 0.54131 1.1520

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Fig. 16 motor.

Transient step response with a 2 lb-in.-sec2 inertia load for a low inertia

No.

Pulley no.

Approx. diameter

18 19 20 21

84H200 96H200 120H200 156H200

13.21 15.12 18.94 24.67

Inertia 1.9541 3.2206 6.7934 17.3840

In addition to the reflected inertias for a linear machine slide, other configurations are given for a belt drive (Figure 17), a rack drive (Figure 18), a rotary drive (Figure 19), and a rotary drive with a ring gear and pinion (Figure 20) using a ring gear. In selecting the units to express inertias the user should be consistent throughout all calculations. A rotary conversion table is included as Figure 21.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 17

Reflected inertia for a belt drive.

where: N¼

NOUT NIN

L ¼ lead of belt ¼ J REF

N IN < N OUT in rev

 2 WT½lb L (in./rev)6(rev=2p rad) ¼ 6 386½in:=sec2  N

J REF ¼ WTðL=NÞ2 0:0000656 (lb-in.-sec2 )

Fig. 18

Reflected inertia for a rack drive.

where: in rev ¼ WTðL=NÞ2 0:0000656 (lb-in.-sec2 )

L ¼ lead of rack and pinion ¼ J REF

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Fig. 19

Reflected inertia for a rotary drive.

where: N¼

NOUT NIN

J REF TO MOTOR ¼

Fig. 20

N IN < N OUT WT6RADIUS2 (lb-in.-sec2 ) 263866RATIO2

Reflected inertia for a rotary drive using a ring gear.

where: N¼

NOUT NIN

N IN < N OUT

Ratio of table ring gear/pinion ¼ lead ¼ revolutions of table/revolutions of pinion. J REF TO MOTOR ¼ WT ¼ lbs Lead ¼ ln=rev N ¼ ratio Radius ¼ in:

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WT6RADIUS2 263866ðRATIO6LEADÞ

2

(lb-in.-sec2 )

Fig. 21

Rotary inertia conversion table.

12.3 DRIVE STIFFNESS Machine feed drives should have sufficient static stiffness to be insensitive to load disturbances. In addition, a feed drive in a contouring numerical control system must remain stationary or clamped when not in motion. This is called ‘‘servo clamping.’’ Also during the standstill period, the axis at rest must resist the load disturbances caused by the cutter reaction forces. The static stiffness of a feed servo drive is the equivalent spring stiffness of the drive measured at the output. The stiffness can be measured at several different places, such as the drive motor shaft, at the drive input to the ball screw, or at the machine slide. When the stiffness is measured at the rotary components the units of measurement are lb-in./rad. When the stiffness is measured at the machine slide the unit of measurement is lb/in. Drive stiffness, measured as the machine slide displacement in inches caused by a displacement force in pounds, increases as the square of the ratio and with the fineness of the lead of the screw. Therefore, in sizing a servo drive, it is important to consider the benefits of using a drive ratio and small lead (in./rev).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 22

Hydraulic servo-drive stiffness.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 23

Electric servo-drive stiffness.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

It is possible to have a stiffness measurement for each type of feedback servomechanism. For example, a velocity servo drive with tachometer feedback would have a velocity stiffness with the units of lb-in./rad/sec. For the purpose of this discussion the drive stiffness is derived for the case of a positioning servo drive with an internal tachometer loop. To illustrate the engineering analysis of the static drive stiffness, two cases of direct slide positioning are considered: a hydraulic servo drive and an electric servo drive. Servo-drive stiffness equations for both hydraulic and electric drives are summarized in Figures 22 and 23. The derivation for the stiffness of the hydraulic servo drive is followed by the electric servodrive stiffness derivation.

Hydraulic Servo-Drive Stiffness with Direct Feedback A mechanical model of a machine feed slide is shown in Figure 24. It is assumed that the drive screw spring rate is represented by GT . The machine’s slide inertia reflected to the drive screw plus the inertia of the drive screw is represented by JL . The load forces acting on the machine slide are reflected to the drive screw as TL . The friction forces of the machine slide are reflected to the drive screw and represented by BL . The mechanical drive ratio is represented by N.

Fig. 24

Machine slide free-body diagram.

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The equations for the mechanical model of Figure 24 are written in the following steps: y1 ¼

ym N

T1 ¼ NTm

T1 ¼ GT ðy1
yo Þ

For dynamic equilibrium we have: JL s2 yo þ BL syo ¼ TL þ GT ðy1
yo Þ

(12.3-1)

ðJL s2 þ BL s þ GT Þyo ¼ TL þ GT y1 GT ym ðJL s2 þ BL s þ GT Þyo ¼ TL þ N

(12.3-2) (12.3-3)

From the basic torque relationship, T1 GT ¼ ðy1
yo Þ N  N  G T ym
yo Tm ¼ N N GT GT yo Tm ¼ 2 ym
N N Tm ¼

(12.3-4) (12.3-5) (12.3-6)

From Eq. (12.3-3) the output position of the drive screw is yo ðsÞ ¼

TL þ GNT ym ðJL s2 þ BL s þ GT Þ

(12.3-7)

An additional set of equations can be written for the hydraulic drive. The basic flow equations are developed as a first step from Figure 25, using the terms Q1 ; Q2 ¼ hydraulic oil flow in and out (in:3 =sec) P1 ; P2 ¼ hydraulic pressure (psi) KC ¼ hydraulic valve pressure coefficient (in:3 =sec
psi) Kv ¼ hydraulic valve gain (in:3 =sec-in:) Xv ¼ hydraulic valve displacement (in.) qQ DQ KC ¼
¼
qPL DPL

(12.3-8)

Therefore PL ¼ DP1
DP2

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(12.3-9)

Fig. 25

Hydraulic drive pressure.

since DP2 ¼
DP1 Therefore PL ¼ DP1 þ DP1 ¼ 2 DP1 ¼ 2 DP DQ KC ¼
2 DP DQ for one port 2KC ¼
DP

(12.3-10) (12.3-11)

The valve flow is given by Q1 ¼ Kv Xv
2KC P1

(12.3-12)

Q2 ¼ Kv Xv þ 2KC P2

(12.3-13)

Adding Eq. (12.3-12) and (12.3-13) yields Q1 þ Q2
2Kv Xv
2KC ðP1
P2 Þ P1
P2 ¼ PL

(12.3-14) (12.3-15)

Thus, the average load flow in the valve is QL ¼

Q1 þ Q2 ¼ Kv Xv
KC PL 2

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-16)

Motor flow equations are developed as follows for each chamber: Flow in
flow out ¼ theoretical flow þ compressibility V1 Q1
KLcp ðP1
P2 Þ
KLex P1 ¼ sV1 þ sP1 b V2
Q2 þ KLcp ðP1
P2 Þ
KLex P2 ¼ sV2 þ sP2 b

(12.3-17) (12.3-18)

where: V1 ¼ Vc þ fv ðym Þ V2 ¼ Vc
fv ðym Þ sV1 ¼ Dm sym ¼
sV2

(12.3-19) (12.3-20) (12.3-21)

KLcp ¼ cross port leakage (in:3 =sec-psi) KLex ¼ hydraulic external leakage (in:3 =sec-psi) b ¼ bulk modulus (16105 lb=in:2 ) V1 ; V2 ¼ hydraulic volume in and out (in:3 ) fv ðym Þ ¼ variation in each chamber volume (in:3 ) Dm ¼ hydraulic motor displacement (in:3 =rad) Vc ¼ total trapped volume on one side of motor (that oil under compression); valve; and manifold (in:3 ) s ¼ Laplace operator Combining Eq. (12.3-17) to (12.3-21) results in Q1
KLcp P1 þ KLcp P2
KLex P1   Vc fv ym þ ¼ Dm sym þ sP1 b b
Q2 þ KLcp P1
KLcp P2
KLex P2   V c f v ym
¼
Dm sym þ sP2 b b

(12.3-22)

(12.3-23)

Combining Eq. (12.3-22) and (12.3-23) yields the total motor flow equation: Q1 þ Q2 þ ð2KLcp þ KLex ÞðP2
P1 Þ Vc fv ym sðP1 þ P2 Þ ¼ 2Dm sym þ sðP1
P2 Þ þ b b

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-24)

Since Q1 þ Q2 ¼ QL 2 PL ¼ P1
P2

(12.3-25) (12.3-26)

and sP1 þ sP2 ¼ 0

(12.3-27)

since the time rate of change of P1 is equal and opposite to that of P2 . Combining Eq. (12.3-24) to (12.3-27) results in the total motor flow equation   KLex Vc QL ¼ KLcp þ PL þ Dm sym þ sPL 2 2b

(12.3-28)

Rearranging, the total motor flow equation becomes   KLex Vc Dm sym þ KLcp þ PL þ sPL ¼ QL 2 2b

(12.3-29)

ðmotor displacementÞ þ ðleakageÞ þ ðcompressabilityÞ Simplifying the leakage by KLcp þ

KLex ¼ KL 2

(12.3-30)

results in a simplified motor flow equation: Dm sym þ KL PL þ

Vc sPL ¼ QL 2b

(12.3-31)

Applying the torque equation for a hydraulic motor gives Torque generated ¼ Dm PL ¼ Jm s2 ym þ Tm

(12.3-32)

Rearranging terms yields PL ¼

J m s2 Tm ym þ Dm Dm

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(12.3-33)

Differentiating yields

sPL ¼

Jm s 3 Tm ym þ s Dm Dm

(12.3-34)

Combining the motor flow Eq. (12.3-31) with Eq. (12.3-33) and (12.3-34) equals the total flow equation with load torque:

Dm sym þ KL þ

J m s2 Tm KL Vc Jm s3 ym þ þ ym Dm Dm 2b Dm

Tm Vc s ¼ QL Dm 2b

(total flow, in:3 =sec)

(12.3-35)

Letting QL ¼ Q and factoring ym yields 

   Jm Vc 2 KL Jm Tm KL Tm Vc s þ s þ D m ym ¼ Q
þ 2bDm Dm Dm 2bDm   Vc Tm Q
Dm KL þ 2b s  ym ¼  Jm Vc 2 K L Jm s 2bD s þ s þ D m D m m   QDm
Tm KL þ V2bc s  ym ¼  s Jm2bVc s2 þ KL Jm s þ D2m s

(12.3-36)

(12.3-37)

(12.3-38)

Combining with Eq. (12.3-5) yields

ym ¼

QDm


h

i    KL þ V2bc s GNT2 ym
KL þ V2bc s GNT yo   s Jm2bVc s2 þ KL Jm s þ D2m

(12.3-39)

It is desired to combine the mechanical torque equations and the hydraulic flow equations to arrive at one system equation. By rearranging Eq. (12.339) in several steps, Eq. (12.3-43) results. 2 2 A second-order equation can be written as pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ðs =oh Þ þ ð2d s=oh Þ þ 1. Therefore in Eq. (12.3-39) oh ¼ ð2bD2m =Jm Vc Þ is defined as the hydraulic resonance.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

  Jm Vc 2 s þ KL Jm s þ D2m ym 2b     Vc GT Vc GT s yo ¼ QDm
KL þ s y þ K þ m L 2b N 2 2b N    Jm Vc Vc GT þ KL Jm s2 þ D2m s þ KL þ s s3 ym 2b 2b N 2   Vc GT yo ¼ QDm þ KL þ s 2b N    Jm Vc Vc GT KL GT þ s2 KL Jm þ D2m þ s3 s þ ym 2b 2bN 2 N2   Vc GT yo ¼ QDm þ KL þ s 2b N   QDm þ KL þ V2bc s GNT yo   i ym ¼ h Jm Vc 3 Vc GT KL GT 2 2 2b s þ KL Jm s þ Dm þ 2bN 2 s þ N 2 s

(12.3-40)

(12.3-41)

(12.3-42)

(12.3-43)

Drive system Eq. (12.3-7) and (12.3-43) can now be included in a positioning servo drive with an inner tachometer loop as shown in the block diagram of Figure 26. To simplify the solution of the drive system equations, substitutions for some of the blocks are introduced in Figure 26. Rearranging the block diagram of Figure 26 yields the block diagram of Figure 27 for a slide displacement output on a load force input. This block diagram can then be used to solve for the drive system stiffness FL =Xo . Using the substitutions for the block in Figure 26, drive system equations can be written in Eq. (12.3-44) to (12.3-52). From Figure 26 Q ¼
Xo Kfb KD Gv
ym KTA sGv þ Xi

(12.3-44)

But Xi ¼ 0, so ym ¼ QDm Gm þ Gm GF

2p Xo L

(12.3-45)

where: 2pXo ¼ yo L 2p GT L Xo ¼ GL ym þ FL GL L 2p N

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-46)

Fig. 26

Hydraulic drive-servo block diagram.

Fig. 27

Hydraulic drive-servo block diagram for stiffness.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Combining Eq. (12.3-44) to (12.3-46) yields 2p Xo L Gm GF Xo 2p ym ¼
Dm Gm Kfb KD Gv Xo
Dm Gm KTA Gv sym þ L

ym ¼ Dm Gm ð
Kfb KD Gv Xo
KTA sGv ym Þ þ Gm GF

(12.3-47) (12.3-48)

Rearranging Eq. (12.3-48) yields   2pGm GF ym ð1 þ Dm Gm KTA sGv Þ ¼
Dm Gm Kfb KD Gv þ Xo (12.3-49) L  
Dm Gm Kfb KD Gv þ 2p L Gm GF Xo ym ¼ (12.3-50) ð1 þ Dm Gm KTA sGv Þ Using Figure 26, Eq. (12.3-50) can be modified to yield   2p GT
Dm Gm Kfb KD Gv þ 2p L Gm GF Xo ¼ GL Xo L ð1 þ Dm Gm KTA sGv Þ N þ

L GL FL 2p

Rearranging yields    2p GT
Dm Gm Kfb KD Gv þ 2p L L Gm GF
GL GL FL Xo ¼ ð1 þ Dm Gm KTA sGv Þ N L 2p   
Dm Gm Kfb KD Gv þ 2p FL 2p GT 2p L Gm GF 6
GL ¼ ð1 þ Dm Gm KTA sGv Þ Xo N L LGL    2 2p 2pG
D G K K G þ G G FL 2p 1 T m m fb D v L m F ¼
LNð1 þ Dm Gm KTA sGv Þ Xo L GL

(12.3-51)

(12.3-52) (12.3-53) (12.3-54)

for o ¼ 0 at steady state. From Figure 26 the following simplifications are possible. N2 KL GT GT KL GF ¼ N 1 GL ¼ GT Gv ¼ K1 Kv

Gm ¼

(12.3-55) (12.3-56) (12.3-57) (12.3-58)

Equations (12.3-55) to (12.3-58) can then be substituted into Eq. (12.3-54),

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

yielding  2 F 2p 2pGT ¼ GT
LN Xo o¼0 L   N2 2pN 2 GT KL Kfb KD K1 Kv þ 6
Dm KL GT LKL GT N

(12.3-59)

Rearranging Eq. (12.3-59) yields  2  2 2pDm NKfb KD K1 Kv F 2p 2p ¼ GT þ
GT LKL Xo L L FL 2p Dm NKfb KD K1 Kv ¼ stiffness for direct feedback X KL L

(12.3-60) (12.3-61)

It can be shown that the velocity closed-loop equation from Eq. (12.3-35) is ym K1 Kv   ¼ vr Dm 1 þ KD1 Km v KTA

(also see Figure 26)

(12.3-62)

The velocity open-loop gain is Kvo ¼

K1 Kv KTA (V/V) Dm

(12.3-63)

Substituting into Eq. (12.3-62) yields ym K1 Kv ¼ vr Dm ð1 þ Kvo Þ

(12.3-64)

From Figure 26 the position-loop velocity constant is Kv ¼

KD Kfb L K1 Kv 6 2pN Dm ð1 þ Kvo Þ

(12.3-65)

Rewriting Eq. (12.3-65) yields K1 Kv KD Kfb ¼

Kv 2pNDm ð1 þ Kvo Þ L

(12.3-66)

Rewriting Eq. (12.3-61) yields FL 2p Dm N 6 ¼ 6Kfb KD 6K1 Kv L KL Xo

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-67)

Substituting Eq. (12.3-66) into (12.3-67) yields FL 2p Dm N Kv 2pN 6Dm ð1 þ Kvo Þ ¼ Xo L KL L

(12.3-68)

The drive stiffness equation for direct position feedback and measured at the machine slide is FL ¼ Xo



 2p 2 D2m N 2 Kv ðKvo þ 1Þ L KL

(lb/in.)

(12.3-69)

As a dimensional check,   2rad2 rev2 in:6 1 V sec lb lb 6 6 3 2 ¼ 6 26 6 6 2 rev sec V in: in: in in: rad

Linear Electric Servo-Drive Stiffness with Resolver Feedback at the Drive Motor The first step in deriving the drive system equations is to derive the DC electric motor equations, which can be done with the equivalent electrical circuit shown in Figure 28. The equation for the equivalent circuit of Figure 28 is 

 La s þ 1 ia Ra þ Ke Vm ðsÞ Ra eðsÞ
Ke Vm ðsÞ   ¼ R a ia La Ra s þ 1

eðsÞ ¼

Fig. 28

Electric motor diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-70) (12.3-71)

The torque equation for Figure 28 is Torque ¼ KT ia ¼ JT sVm ðsÞ þ TL JT s TL ia ¼ Vm ðsÞ þ KT KT

(12.3-72) (12.3-73)

Solving Eq. (12.3-73) for the motor velocity Vm ðsÞ yields   TL KT ¼ Vm ðsÞ ia
K T JT s

(12.3-74)

Equation (12.3-74) can be reduced as follows: ia KT TL ia KT
TL
¼ Vm ðsÞ ¼ ¼ Vm ðsÞ JT s JT s JT s

(12.3-75)

Equations (12.3-71) and (12.3-75) can be represented by the block diagram of Figure 29: The motor block diagram can now be included in an electric positioning servo drive with an internal tachometer loop as shown in Figure 30. This drive system block diagram can be rearranged as shown in the block diagram of Figure 31 for an output displacement on a torque load input while holding the reference input position yi at zero. This solution will be a linear solution considering all components as linear devices.

Fig. 29

Electric motor block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 30

Electric drive-servo block diagram.

Fig. 31

Electric drive-servo block diagram for stiffness.

The loop equations for drive stiffness of the block diagram of Figure 31 can be written as follows: Vm ðsÞ ¼ AðsÞ

JT sRa



La Ra

Rf Ra



La Ra

 s þ 1 ðt2 s þ 1Þ

 s þ 1 Rf ðt1 s þ 1Þ þ KT Ke Rf ðt2 s þ 1Þ þ    þ K1 KTA R2 ðt1 s þ 1Þ (12.3-76)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The output/input of the block diagram of Figure 31 can then be solved for and is shown in Eq. (12.3-77). yo ðsÞ ¼ TL ðsÞ Rf Ra



La Ra

 s þ 1 ðt2 s þ 1Þ

h   i s JT sRa RLaa s þ 1 Rf ðt1 s þ 1Þ þ KT Ke Rf ðt2 s þ 1Þ þ    þ K1 KTA R2 ðt1 s þ 1Þ    R2 þ Rf ðt1 s þ 1ÞK1 KD KT Kfb (12.3-77) Rin Letting s ¼ jo ! 0 for the steady-state case, Rf Ra yo ¼ TL Rf K1 KD KT ðR2 =Rin ÞKfb

(12.3-78)

TL K1 KD KT R2 Kfb ¼ yo Ra Rin

(12.3-79)

TL KD R2 K1 KT Kfb ¼ 6 yo Rin Ra

(12.3-80)

TL KT K1 R2 KD Kfb ¼ 6 yo Ra Rin

(12.3-81)

It is necessary to present the drive stiffness in terms of readily available parameters. Therefore, the following solutions are presented starting with the motor closed loop followed by the velocity closed loop and finally the position-loop velocity constant. Motor closed-loop equations can be written from Figure 30: Vm ðsÞ K T 6 ¼ eðsÞ 1þ JT sRa La s þ 1 Ra

¼

JT sRa RLaa

T

e

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

KT Ke JT sRa ðRLaa sþ1Þ

KT



Vm ðsÞ ¼ JT Ra eðsÞ K K

1

 s þ 1 þ KT Ke

1=Ke 1=Ke ¼ JT R a 2 þ KT Ke s þ 1 ðtme te s þ tme s þ 1Þ

La 2 Ra s

(12.3-82)

(12.3-83)

where: JT Ra KT Ke La te ¼ Ra

tme ¼

Velocity closed-loop equations can be written from Figure 30: Vm ðsÞ R2 ðt1 s þ 1ÞK1 ¼ Ei ðsÞ Ke ðt2 s þ 1Þðtme te s2 þ tme s þ 1Þ 1 6 R2 ðt1 sþ1ÞK1 KTA 1 þ Ke ðt2 sþ1Þðtme te s2 þtme sþ1ÞRf

(12.3-84)

Equation (12.5-84) reduces to Vm ðsÞ R2 ðt1 s þ 1ÞK1 ¼ Ei ðsÞ Ke ðt2 s þ 1Þðtme te s2 þ tme s þ 1Þ þ R2 ðt1 sþ1ÞK1 KTA

(12.3-85)

Rf

The open position-loop velocity constant from Figure 30 is Kv ¼

KD Vm ðsÞ 6Kfb 6 Rin Ei ðsÞ

with s ¼ 0

(12.3-86)

The position-loop velocity constant including Eq. (12.3-85) yields Kv ¼

KD Kfb R2 ðt1 s þ 1ÞK1 6 (12.3-87) Rin Ke ðt2 s þ 1Þðtme te s2 þ tme s þ 1Þ þ R2 ðt1 sþ1ÞK1 KTA Rf

KD Kfb R2 K1 Kv ¼ Rin Ke þ R2 K1RKfTA Rin

(12.3-88)

Rearranging Eq. (5.5-88) yields Kv ¼

KD Kfb R2 K1 Rf Rin Ke Rf þ R2 K1 KTA Rin

(12.3-89)

Rearranging, Kv Rin Ke þ

Kv R2 K1 KTA Rin ¼ KD K1 R2 Kfb Rf

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-90)

Combining Eq. (12.3-81) and (12.3-90) yields  TL KT Kv R2 K1 KTA Rin ¼ 6 Kv Rin Ke þ yo Ra Rin Rf

(12.3-91)

The velocity open-loop gain from Figure 30 is Kvo ¼

R2 ðt1 s þ 1ÞK1 KTA Ke ðt2 s þ 1ÞRf ðtme te s2 þ tme s þ 1Þ

(12.3-92)

For the steady-state condition s ! 0, the velocity open-loop gain is Kvo ¼

R2 K1 KTA Ke Rf

(12.3-93)

Substituting Eq. (12.3-93) into Eq. (12.3-91) yields TL KT ¼ Kv Rin ðKe þ Kvo Ke Þ yo Ra Rin

(12.3-94)

Therefore, the static stiffness equation for a DC electric position loop with an internal tachometer loop is   TL KT Ke in.-lb ¼ Kv ð1 þ Kvo Þ (12.3-95) yo Ra rad

Drive Stiffness for Three-Phase, Half-Wave SCR Drive Amplifiers Using an Inner Current Loop A linear analysis for the static stiffness at the motor of a DC electric drive using a three-phase, half-wave silicon controlled rectifier (SCR) amplifier and an inner current loop can be made starting with the basic equations for a DC motor, referring to Figures 28 and 29 and Eq. (12.3-70) to (12.3-75). When a current loop is added to the drive amplifier, the motor and current loop will appear as the block diagram shown in Figure 32. To facilitate solving the loop equations, the block diagram of Figure 30 can be modified as shown in Figure 33. The DC electric motor with a current feedback loop can then be incorporated into a position servo loop with an inner velocity servo loop as shown in Figure 34. The system block diagram can be redrawn for a position output and a torque disturbance input in Figure 35.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 32

Electric servo-drive block diagram.

Fig. 33

Electric servo-drive block diagram.

Fig. 34

Electric servo-drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 35

Electric servo-drive block diagram.

The servo drive stiffness can now be solved using the following sequence of equations from the block diagram of Figure 35. Solving for the inner loop first yields Vo ðsÞ 1 ¼ 6 AðsÞ JT s  þ

1 1 þ J1T s

(12.3-98)

KT K1 ðtA sþ1Þ ½ðtB sþ1ÞðRLaa sþ1ÞRa þK1 Kie ðtA sþ1Þ

Ke ðtB s þ 1Þ R2 K2 ðt1 s þ 1ÞKTA þ ðt2 s þ 1ÞRf ðtA s þ 1ÞK1 1

þ 



Vo ðsÞ ¼ (12.3-99) T K1 ðtA sþ1Þ AðsÞ þ    JT s þ ðt sþ1Þ LaKsþ1 ðRa ÞRa þK1 Kie ðtA sþ1Þ B  Ke ðtB s þ 1Þ R2 K2 KTA ðt1 s þ 1Þ þ þ ðtA s þ 1ÞK1 ðt2 s þ 1ÞRf Vo ðsÞ 1 ¼ (12.3-100) T K1 ðtA sþ1Þ AðsÞ JT s þ ðt sþ1Þ LaKsþ1 þ  R þK K ðt sþ1Þ ½ B ðRa Þ a 1 ie A   Ke ðtB s þ 1Þðt2 s þ 1ÞRf þ R2 K2 KTA K1 ðt1 s þ 1ÞðtA s þ 1Þ þ ðtA s þ 1ÞK1 ðt2 s þ 1ÞRf For the steady-state condition, Eq. (12.3-100) yields Vo 1 ¼ K K Ke Rf þR2 K2 KTA K1 T 1 A s!0 Ra þK1 Kie Rf K1 Rf ðRa þ K1 Kie Þ Vo ¼ A s!0 KT ½Ke Rf þ R2 K2 KTA K1 

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-101) (12.3-102)

Solving for the outer loop of Figure 29 yields Vo 1 yðsÞ ¼ 6 TL ðsÞ A s

1 þ Vo 1 A s

Vo yðsÞ ¼ 6 A TL ðsÞ



1 KT K1 ðtA sþ1ÞR2 K2 ðt1 sþ1ÞKD Kfb ½ðtB sþ1ÞðRLaa sþ1ÞRa þK1 Kie ðtA sþ1Þðt2 sþ1ÞRin



(12.3-103) s þ VAo



1 KT K1 KD Kfb ðtA sþ1ÞðT1 sþ1ÞR2 K2 ½ðtB sþ1ÞðRLaa sþ1ÞRa þK1 Kie ðtA sþ1Þðt2 sþ1ÞRin

(12.3-104)

For the steady-state solution, y Vo 1 6V ¼ K K T 1 KD Kfb R2 K2 o A TL 6 A

(12.3-105)

ðRa þK1 Kie ÞRin

as s?0 y ðRa þ K1 Kie ÞRin ¼ TL KT K1 KD Kfb R2 K2 TL K1 K2 KD KT R2 Kfb ¼ y ðRa þ K1 Kie ÞRin

steady-state compliance

(12.3-106)

steady-state stiffness

(12.3-107)

Solving for the closed current loop of Figure 34, ia ðsÞ K1 ðtA s þ 1Þ   ¼ vi ðsÞ ðtB s þ 1Þ La s þ 1 Ra þ K1 ðtA s þ 1ÞKie

(12.3-108)

Ra

For the current closed loop steady state, ia K1 ¼ vi s!0 Ra þ K1 Kie

(12.3-109)

Solving for the motor closed loop from Figure 34 yields the following: Vo ðsÞ K1 ðtA s þ 1ÞKT i   ¼h La vr ðsÞ ðtB s þ 1Þ s þ 1 Ra þ K1 ðtA s þ 1ÞKie JT s Ra

1

6 1þ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

h

K1 ðtA sþ1ÞKT ðtB sþ1ÞKe ½ðtB sþ1ÞðRLaa sþ1ÞRa þK1 ðtA sþ1ÞKie JT s K1 ðtA sþ1Þ

i

(12.3-110)

Simplifying Eq. (13.2-110) yields Vo ðsÞ K1 ðtA s þ 1ÞKT i   ¼h La vr ðsÞ ðtB s þ1Þ s þ1 Ra þ K1 ðtA s þ1ÞKie JT s þ K1 ðtA s þ 1ÞKT Ra

Vo ðsÞ K1 KT K1 ¼ ¼ vr ðsÞ s!0 K1 KT KKe Ke

ðtB sþ1ÞKe K1 ðtA sþ1Þ

(12.3-111) steady state for motor loop

(12.3-112)

1

Solving for the open velocity loop of Figure 34 yields Kvo ðsÞ ¼ 6h

KTA R2 K2 ðt1 s þ 1Þ Rf ðt2 s þ 1Þ

K 1 ðtA s þ 1ÞKT i sþ1ÞKe ðtB s þ 1Þ RLaa s þ 1 Ra þ K1 Kie ðtA s þ 1Þ JT s þ K1 ðtA s þ 1ÞKT ðtðtAB sþ1ÞK 1 



(12.3-113) as s !0 The steady-state open velocity loop gain is Kvo ¼

KTA R2 K2 KTA K1 K2 R2 ¼ Rf Ke Rf KK1e

(12.3-114)

Solving for the closed tachometer loop of Figure 34 yields Vo ðsÞ R2 K2 ðt1 s þ 1Þ Vo ðsÞ 1 ¼ 6 6 ðt 1 v2 ðsÞ ðt2 s þ 1Þ vr ðsÞ 1 þ R2 K2 sþ1Þ

Vo ðsÞ KTA ðt2 sþ1Þ vr ðsÞ Rf

(12.3-115)

Substituting Eq. (12.3-112) into (12.3-115) and reducing Eq. (12.3-115) yields Vo ðsÞ R2 K2 ðt1 s þ 1ÞK1 KT ¼ v2 ðsÞ ðt2 s þ 1ÞK1 KT KKe þ R2 K2 ðt1 s þ 1ÞK1 KT KRTA 1 f Vo R2 K2 K1 KT ¼ v2 s!0 KT Ke þ R2 K2 K1 KT KRTA f

(12.3-116)

For the steady-state closed velocity loop, K1 K2 R2 Rf Vo ¼ v2 Ke Rf þ K1 K2 R2 KTA

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.3-117)

The position-loop velocity constant is the steady state from Figure 34 and Eq. (12.3-117): Kv ¼

KD Kfb Vo KD Kfb K1 K2 R2 Rf 6 ¼ 6 Rin v2 Rin ðKe Rf þ K1 K2 KT R2 KTA Þ

(12.3-118)

Reducing Eq. (12.3-118) yields Kv ¼

K1 K2 R2 KD Kfb   Rin Ke þ K1 K2RRf2 KTA

(12.3-119)

Rearranging Eq. (12.3-114) yields Ke Kvo ¼

K1 K2 R2 KTA Rf

(12.3-120)

Then substituting Eq. (12.3-120) into Eq. (12.3-119) yields Kv ¼

K1 K2 R2 KD Kfb K1 K2 R2 KD Kfb ¼ Rin ðKe þ Ke Kvo Þ Rin Ke ð1 þ Kvo Þ

(12.3-121)

Rearranging yields K1 K2 R2 KD Kfb Rin Ke TL K1 K2 KD KT R2 Kfb ðEq:½12:3
107Þ ¼ yo Rin ðRa þ K1 Kie Þ

Kv ð1 þ Kvo Þ ¼

(12.3-122)

Substituting Eq. (12.3-122) into (12.3-107) yields TL 1 ¼ Kv ð1 þ Kvo ÞKe KT ðRa þ K1 Kie Þ yo TL Kv ð1 þ Kvo ÞKe KT   ¼ yo Ra 1 þ KR1 Ka ie

(12.3-123)

Applying the equation for drive stiffness at the motor of a three-phase, half-wave SCR amplifier with a current and velocity loop to a commercial Gettys drive, Eq. (12.3-123) can be simplified since K1 Kie A=sec ¼ 275 Ra A V=sec Velocity open-loop gain ¼ Kvo ¼ 35,965 V Current open-loop gain ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Therefore, for a specific commercial SCR drive amplifier, Eq. (12.3-123) will reduce to   TL Kv ð1 þ 35,965ÞKe KT in.-lb ¼ (12.3-124) rad yo Ra ð1 þ 275Þ Applying the equation drive stiffness at the motor of a three-phase, half-wave SCR amplifier with a current and velocity loop to a commercial Gettys drive, Eq. (12.3-124) can be used for an example: Motor ¼ 1000 rpm ð30-2Þ Ke ¼ 1:15 V=rad=sec KT ¼ 10:1 in.-lb=A Kie K1 A=sec ¼ open current loop gain ¼ 275 Ra A Rmotor ¼ 0:25 ohm Rinductor ¼ 0:016 ohm Rtransformer ¼ 0:1 ohm Ra ¼ 0:25 þ 0:016 þ 0:1 ¼ 0:36 ohm Kv ¼ 1 ipm=mil ¼ 16:8 sec Kvo ¼ 35,965 V=V TL 16:863596561:15610:1 ¼ 70,888 in.-lb/rad ¼ 0:366275 yo

12.4 DRIVE RESOLUTION One of the most important I.P.s for machine slides that have a very low feed-rate requirement is the drive resolution, which is defined as the difference between the breakaway position error and the run error: Resolution ¼ breakaway error
run error

(12.4-1)

Drive resolution can also be described as the least amount of position error required to cause motion to take place. It is directly related to the quality of surface finish on a contouring controlled milling machine. The units of drive resolution are inches. The drive resolution for hydraulic and electric drives is discussed in this section. A graphical representation of drive resolution is shown in Figure 36. The difference between the breakaway error and the run error is the drive

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

resolution. For machines having a significant static friction above the coulomb friction, the drive resolution can be unacceptable.

Drive Resolution for Hydraulic Drives The drive resolution for a hydraulic drive is a function of the port-to-port leakage at breakaway. Thus Breakaway leakage ðLBA Þ ¼ DPBA 6KL þ

LBA

  3  FBA in: ¼ KL DPBA þ 6100 KTH sec

friction torque of slideBA 6100 KL KTH (12.4-2) (12.4-3)

In Eq. (12.4-1) the breakaway characteristics of the motor and machine slide must be added to obtain a true value for the overall breakaway characteristics. A drive resolution representation is shown in Figure 36. Likewise the run leakage characteristics can be equated as   run friction torque 6100 Run leakage ðLr Þ ¼ KL no-load motor psi þ KTH (12.4-4) Having computed the breakaway and run leakage characteristics, it is possible to compute the breakaway and run velocity from 1 L 1 6 6 Dm N 2p 1 L 1 Run velocity ðVr Þ ¼ Lr 6 6 6 Dm N 2p

Breakaway velocity ðVBA Þ ¼ LBA 6

(12.4-5) (12.4-6)

The contouring control position loop can be modeled as in Figure 37. Kv s EðsÞKv ¼ yo s ¼ Vo ðsÞ yo ðsÞ ¼ EðsÞ

Position following error EðsÞ ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.4-7) (12.4-8) Vo ðsÞ Kv

(12.4-9)

Therefore, the breakaway and run velocities divided by the Kv yield the breakaway and run errors. VBA LBA L ¼ Kv 2pKv Dm N  LKL FBA of slide EBA ¼ DPBA of motor þ 6100 KTH 2pKv Dm N Vr Lr L Er ¼ ¼ Kv 2pKv Dm N   LKL Fr Er ¼ NL motor psi þ 6100 2pKv Dm N KTH EBA ¼

(12.4-10) (12.4-11) (12.4-12) (12.4-13)

The drive resolution can therefore be equated as: Resolution ¼

LKL 2pKv Dm N     FBA Fr 6 DPBA þ 6100
NL psi þ 6100 KTH KTH (12.4-14)

Motor

DPBA

NL psi

Vickers MFA5B Hartman Nutron

600 psi 12 psi 50 psi

300 10 25

For antifriction ways: Friction breakaway ðFBA Þ ¼ run friction ðFr Þ LKL Resolution ¼ ðDPBA
NL psiÞ 2pKv Dm N

(12.4-15)

Drive Resolution for Electric Drives The drive resolution for an electric drive is a function of the static friction breakaway of the drive motor and machine slide. The derivation of the electric drive resolution can be developed similar to the case for a hydraulic

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

drive as follows:

TFBA (12.4-16) KT ¼ motor breakaway þ slide breakaway TFr Run current ðIr Þ ¼ Imr þ (12.4-17) KT Ra L (12.4-18) Breakaway velocity ðVBA Þ ¼ IBA 6 2pKe N Ra L Run velocity ðVr Þ ¼ Ir 6 (12.4-19) 2pKe N

Breakaway current ðIBA Þ ¼ ImBA þ

For a position loop as shown in Figure 36, the position error is velocity/Kv (Eq. [12.4-9]):

VBA (12.4-20) Kv   IBA Ra L Ra L TFBA ¼ ¼ ImBA þ KT Kv 2pKe N Kv 2pKe N

Breakaway error ðEBA Þ ¼

Vr (12.4-21) Kv   Ir Ra L Ra L TFr ¼ Imr þ ¼ KT Kv 2pKe N Kv 2pKe N     Ra L TFBA Ra L TFr ImBA þ Imr þ Resolution ¼
KT KT 2pKv Ke N 2pKv Ke N

Run error ðEr Þ ¼

    Ra L TFBA TFr ImBA þ Resolution ¼
Imr þ KT KT 2pKv Ke N

(12:4-22) (12.4-23)

For antifriction ways: Resolution ¼

Ra L ðImBA
Imr Þ 2pKv Ke N

Motor breakaway currents are given for some examples:

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.4-24)

Fig. 36

Drive resolution representation.

Fig. 37

Positioning servo block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Motor Gettys 10-3 20-1 20-2 30-1 30-2 Inland 4020 4023 5301 5302 4501A 4501B 4503A 4503B Allen-Bradley 1326-ABC4C 1326-ABC3E

ImBA ðAÞ

Imr ðAÞ

1.28 3.0 1.48 3.6 1.3

0.85

0.8 0.8 2.2 1.9 1.8 2.4 1.36 1.81

0.535 0.535 1.46 1.3 0.36 1.5 0.89 1.19

0.832 0.949

0.665 0.712

0.99 0.87

12.5 DRIVE ACCELERATION Feed drive acceleration is a critical factor with large industrial machines. If the acceleration is too high, the resulting excessive forces could be destructive to mechanical components in the drive. Three types of acceleration have been used with industrial feed servo drives: ramp in velocity, exponential increase in velocity, and ‘‘S’’ curve type of acceleration. These are shown graphically in Figure 38. The velocity ramp (a) has constant acceleration as velocity increases. At the time where acceleration ceases and the velocity is a constant a small visible mark will be evident (because of a small overshoot) on a contouring-type machine. This small but objectionable mark can be eliminated by using an exponential increase in velocity (b), which rounds the high point in velocity to become a constant velocity. The objection to this type of acceleration is that the acceleration at time zero is a maximum and is sometimes referred to as ‘‘jerk.’’ To further improve the acceleration characteristics, the ‘‘S’’ type of acceleration can be used. The advantage of the ‘‘S’’ curve acceleration is that the acceleration is zero at time zero and becomes a maximum at a time when the velocity is increasing at its maximum rate.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 38

Graphic representation of servo-drive inputs.

The velocity and acceleration equation can be derived from a simple classic model of a positioning servo. In this model it is assumed that the inner servo loops (velocity, etc.) have bandwidths that are at least twice that of the position loop bandwidth ðKv Þ. The model for this positioning servo is shown in Figure 39. The transfer characteristics for this model are: Kv 1 Kv _ Vo ðsÞ ¼ y_o ðsÞ ¼ y_i ðsÞ 6 yi ðsÞ ¼ 1 þ Kv =s s þ Kv s

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.5-3)

Fig. 39

Positioning servo block diagram.

For a worst-case condition of a step input, VF y_i ðsÞ ¼ s

(12.5-4)

Substituting Eq. (12.5-4) into (12.5-3), Vo ðsÞ ¼

VF Kv sðs þ Kv Þ

(12.5-5)

The inverse transform for Eq. (12.5-5) is Vo ðtÞ ¼ VF
VF e
Kv t ¼ VF ð1
e
Kv t Þ

(12.5-6)

Differentiating Eq. (12.5-6) yields acceleration: V_ o ðtÞ ¼ aðtÞ ¼ VF Kv e
Kv t

(12.5-7)

The greatest acceleration occurs at t ¼ 0. Thus, the motor angular acceleration is aðtÞ max ¼ VF Kv (rad/sec2 Þ where: VF ¼ rad/sec Kv ¼ 1/sec

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.5-8)

It is also possible to calculate the time required to accelerate to a given velocity from Eq. (12.5-6) by rearranging terms:   Vt e ¼ 1
VF   Vt
Kv t ¼ loge 1
VF   1 Vt loge 1
t¼
VF Kv
Kv t

(12.5-9) (12.5-10) (12.5-11)

For example, with a Kv ¼ 25=sec to reach to velocity 98% of final velocity for a step input, the time to reach this velocity is 1 loge ð1
0:98Þ 25 1
1 log ð0:02Þ ¼ 6ð
4:0173Þ ¼ 0:16 sec t¼
25 25 t¼


(12.5-12) (12.5-13)

The acceleration and velocity versus time characteristics are shown in Figure 40. The model of Figure 37 can be used to determine the velocity characteristics and time to reach velocity when there is an initial velocity. Eq. (12.5-6) can be rearranged as  Vo ðsÞ

s þ Kv Kv



¼ y_i ðsÞ

(12.5-14)

with initial conditions 1 Vð0Þ sVo ðsÞ
þ Vo ðsÞ ¼ y_i ðsÞ Kv Kv

(12.5-15)

For a step input, VF y_i ðsÞ ¼ s Vo ðsÞs Vinitial VF
þ Vo ðsÞ ¼ Kv Kv s   s VF Vinitial þ þ 1 Vo ðsÞ ¼ Kv s Kv

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(12.5-16) (12.5-17) (12.5-18)

Fig. 40

Acceleration and velocity characteristics.

Rearranging yields Vo ðsÞ ¼

VF Kv Vinitial þ sðs þ Kv Þ ðs þ Kv Þ

(12.5-19)

The inverse of this transfer function is Vo ðtÞ ¼ VF
VF e
Kv t þ Vinitial e
Kv t Vo ðtÞ ¼ VF þ ðVinitial
VF Þe

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved


Kv t

(12.5-20) (12.5-21)

The time to reach a given velocity from an initial velocity can be solved by rearranging Eq. (12.5-22): VðtÞ
VF ¼ e
Kv t Vinitial
VF   1 VðtÞ
VF t¼
loge Kv Vinitial
VF

(12.5-22)

It is sometimes desirable to know how far the machine axis has moved for a step input in velocity. The distance traveled can be found by integrating Eq. (12.5-22): Z t Z t Z t Z t
Kv t VðtÞ dt ¼ VF dt þ Vinitial e dt
VF e
Kv t dt 0

0

0

0

Vinitial
Kv t Vinitial VF
Kv t VF e þ þ e
XðtÞ ¼ VF t
Kv Kv Kv Kv Vinitial
VF VF
Vinitial
Kv t þ e XðtÞ ¼ VF t þ Kv Kv

(12.5-23) (12.5-24)

If the model of Figure 39 were to be considered for a ramp input, the input of Eq. (12.5-4) would change to AI y_i ðsÞ ¼ 2 s

(12.5-25)

where AI is the initial acceleration. Similar equations for displacement, velocity, and acceleration with the step input can be derived for a ramp input summarized as follows: 2  t t 1 e
Kv t
XðtÞ ¼ AI þ 2
2 (in:) (12.5-26) 2 Kv Kv Kv   1 e
Kv t þ (ips) (12.5-27) Vo ðtÞ ¼ AI t
Kv Kv Ao ðtÞ ¼ AI ð1
e
Kv t Þ

(in:=sec2 )

(12.5-28)

Unfortunately the maximum acceleration at time equals zero produces an objectionable mechanical shock on the drive mechanics (gears, lead screw, etc.). Another problem relates from a desire on the part of the user to call for high acceleration rates to meet productivity requirements. Thus the required drive torque/current is sufficiently high to cause the drive to go into current limit condition. As long as the drive remains in current limit, the velocity increases linearly. This linearly increasing velocity will continue

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

until the required target velocity is reached and exceeded. As the velocity overshoots its target the drive will come out of current limit and the velocity will drop back to its target value with a resulting severe shock on the drive and machine components.

‘‘S’’ curve type of acceleration The predominant advantage in using an ‘‘S’’ curve type of acceleration is that the initial acceleration will be zero at time equals zero. Thus the initial shock or jerk on the drive and machine will be eliminated. However, the maximum acceleration can be larger than the initial acceleration with exponential acceleration depending on how the ‘‘S’’ curve is generated. In commercial servo drive applications the type of ‘‘S’’ curve is a function of the path control generated in the computer control. There are numerous ways to generate ‘‘S’’ curve accelerations. One method is the use of a sinusoid to generate an ‘‘S’’ curve (Figure 41) represented as the following: Velocity ¼ V peak sin wt

(12.5-29)

The velocity profile is a zero-shifted negative cosine wave (Figure 42) with double the peak of the sine wave. Where: y ¼ angle of motor shaft (rad) T ¼ period of the motor shaft rotation (sec) V F ¼ desired final velocity ¼ y=T (rad=sec)

Fig. 41

Sinusoidal curve.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The velocity profile is assumed to accelerate for half the period T, where it will level off at the desired target velocity Vmax. This velocity profile will be the input command to a positioning servo modeled as in Figure 29. The velocity and acceleration are expressed as the following: VF ð1
cos wtÞ (rad=sec) 2 2p 2pT Acceleration ¼ V F sin (rad=sec2 ) T T T VELmax ¼ 2V F for t ¼ 2 2p T for t ¼ ACCELmax ¼ V F T 4 Velocity ¼

(12.5-30) (12.5-31) (12.5-32) (12.5-33)

Where:

V F ¼ traverse (in./min)62p ðrad=revÞ6ratio6 6

Fig. 42

1 ¼ ðrad=secÞ 60ðsec=minÞ

Zero-shifted negative cosine wave.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

1 Lead ðin:=revÞ

o ¼ 2pf ¼

2p ðrad=secÞ T

2p6Ratio (rad=sec2 ) Lead DESMAXCEL ¼ desired acceleration (in:=sec2 ) VF 62p (sec) T ¼ period of the sinusoid ¼ MAXCEL MAXCEL ¼ DESMAXCEL

Using a plotting program, the velocity/acceleration responses for the exponential and sinusoid ‘‘S’’ curve are compared in Figure 43.

12.6 DRIVE SPEED CONSIDERATIONS There are two speed requirements that must be satisfied. First, the speeds must increase to meet the requirements of increased productivity. Feed rates as high as 1000 inches per minute (ipm) are not uncommon. For highvelocity-feed servo drives it is necessary to consider the critical speed at which ‘‘whipping’’ of the ball screw mat occur. Critical speed can be calculated:

Critical speed ¼ 1:56

4:766106 6D1 (rpm) L21

(12.6-1)

Where: D1 ¼ outside diameter
1=2 ball diameter (in:) L1 ¼ longest unsupported length from ball nut to the end bearing (in:)

For safety the maximum speed of the drive screw should be 20% less than the critical screw speed. Additionally two other mechanical ball screw resonances must be considered. If an industrial machine servo uses position feedback from the machine slide, the following two resonances will be inside the position loop. For stability reasons these resonances should be (as an I.P.) three times the elocity servo bandwidth (see Section 9.2).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Ball Screw Resonances Axial Resonance K SCREW ¼ 4 pi ðDIAMETER=2Þ2 6 K NUT ¼ 106106

306106 (lb-in./rad) LENGTH

(12.6-2)

(lb-in./rad)

K BEARING ¼ 116106 (lb-in./rad) Compliance ¼ 1=K NUT þ 0:5K BEARING þ 1=K SCREW

(12.6-3)

K TOTAL ¼ 1=compliance

(12.6-4)

Axial resonance ¼ oa ¼

Fig. 43

KTOTAL 6386 1 6 (Hz) WEIGHT 2p

(12.6-5)

Comparisons for a ramp, exponential, and ‘‘S’’ curve acceleration.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Torsional Resonance Torsional resonance ¼ ot ¼ K TOTAL

ðJM þ JL Þ 1 (Hz) 6 ðJM 6JL Þ 2pi

(12.6-6)

J M ¼ motor inertia (lb-in.-sec2 ) J L ¼ load inertia (lb-in.-sec2 ) While maximum speed is important for increased productivity it is also just as important that the servo-drive motor rotate smoothly at very low feed rates where this is a requirement. For surface finish requirements the motor should provide smooth contouring feed rates to as low as 0.01 ipm. The servo motor must not cog. The worst case occurs with a single-point tool in a machining contouring operation on a turning machine (lathe).

12.7 DRIVE RATIO CONSIDERATIONS Considering a linear drive, the ball screw lead and drive ratio must be selected to match the rated speed of the motor to the maximum feed-rate requirements of the machine axis. In selecting a ratio to use, it should be noted that the available machine thrust is directly related to the ratio used (motor speed being downspeed to the ball screw). Also, the reflected inertia to the motor is reduced by the square of the ratio. The predominant reflected load inertia will be derived from the components closest to the motor shaft. When belt ratios are used it should be noted that the pulley inertias increase as the fourth power of their diameters. Thus, their reflected inertias to the motor may be excessive if large-diameter pulleys are used. A compromise can be made using a doublepulley arrangement with two ratios in series. For cases where the motor is directly connected to a ball screw, it should be noted that the benefit of increasing the machine slide thrust and reducing the reflected load inertia will not be realized without a ratio. Additionally, the drive resolution and stiffness will be reduced. Drive stiffness is an important consideration in selecting a drive ratio. There are two reasons why an industrial servo drive should have high drive stiffness. First, the servo drive should have sufficient static and dynamic stiffness to be insensitive to load disturbances during motion. In addition, the servo drive must remain stationary or clamped when not in motion. The static stiffness of the servo drive is the equivalent spring constant of the drive. The servo drive stiffness increases as the square of the ratio. The drive ratio also reduces the reflected inertia to the drive motor by the square of the ratio,

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

which is directly related to servo stability. Thus the drive ratio is a very significant factor in sizing a servo drive to an industrial machine. There are a number of ways a drive ratio can be obtained. First, the use of spur gears has definite disadvantages since it is very difficult to eliminate backlash or lost motion, which has a direct relation to servo position-loop stability. Wound-up gearboxes have been used but they are very expensive. However, there are commercial gearboxes available (such as planetary gearboxes) that offer minimum backlash. These gearboxes have very high stiffness but the dynamic stiffness must also be considered since the gearbox resonances are often inside the servo position loop and can affect stability. Timing belts and pulleys offer a cheap solution to eliminating backlash. Dynamically they don’t display any resonances below approximately 200 Hz. However, as stated previously, the reflected inertia of a pulley is proportional to the fourth power of its diameter. The smaller pulley, with minimal inertia, is usually at the motor and the larger driven pulley is connected to the machine drive lead screw. The larger pulley inertia reflected to the servo motor limits the practical ratio to about three. Worm and wheel gearboxes can also be used, and are quite often used with rotary servo drives. They have the advantage of providing larger ratios in a small volume. They can wear and introduce backlash but modern commercial worm/wheel gearboxes have adjustments to eliminate the backlash. These worm/wheel gearboxes have a wide range of efficiencies with the higher ratios having lower efficiencies. These efficiencies must be taken into consideration during the drive sizing. The majority of industrial servo applications using worm/wheel gearing are of this type where the losses are frictional. As industrial servo drives are applied to machines, there are two general categories. There is a class of applications where the machine positions from one position to another, often referred to as a transfer servo. These servo drives usually have one load to consider, which is the frictional loss. The other class of industrial servo has some friction along with other loads doing some work as they position. These loads can result from such things as grinding, forming, drilling, milling, etc. Cutter vibration is one form of load disturbance. The worstcase situation can occur with an interrupted turning cut on a lathe where the loads are severe intermittent transient steps in loading. Just about all modern industrial servos have an internal torque (current loop) regulator, which minimizes these load disturbances. However, these load disturbances must be sensed by the servo motor and drive. Therefore, it is important that torque disturbances be reflected back to the servo motor through the gearing to maintain the required stiffness. This is not a problem with spur gearboxes or belt/pulley ratios, but worm/wheel gearboxes have a wide range of efficiencies in back

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

driving the worm/wheel gearbox. In commercial worm/wheel gearboxes the back-driving efficiencies can range from a high of about 95% at high speed using small ratios, down to about 70% in the servo positioning speeds. For higher ratios (the usual case with rotary drive positioning servos) the backdriving efficiencies at high speed range from about 40% down to 0% in the servo positioning range. Typical worm/wheel gearbox efficiencies versus motor speed and ratio are shown in Figure 44. For very low efficiencies a worm/wheel gearbox can greatly diminish the servo drive stiffness. Fortunately, most industrial servo applications are not subject to backdriving load disturbances. Some machine tool applications definitely require that load disturbances be sensed by the servo torque loop to maintain the required stiffness. There are a number of criteria that can be used in selecting a drive ratio. Two of these are discussed next.

Minimum Acceleration Time It is possible to select a gear ratio based on a minimum acceleration time. Provided the constraints for gear, belt, and pulley, etc. based on mechanical stress are met, a guide for selecting a gear ratio can be derived in the following manner (refer to Figure 45). Terms used are:

Fig. 44

Worm/wheel gear ratio back-driving efficiencies.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Jm ¼ motor inertia, lb-in.-sec2 N ¼ gear ratio (overall) Vm ¼ motor velocity, rad/sec Vsc ¼ drive screw velocity, rad/sec a ¼ angular acceleration (at lead screw), rad/sec2 WL ¼ load weight, lbs Tsc ¼ torque at the drive screw, in.-lb JL ¼ loadðWL Þinertia at screw, in.-lb-sec2 Jscrew ¼ inertia of screw at the screw, in.-lb-sec2 Jsc ¼ total inertia reflected to the screw, in.-lb-sec2 The equations are: Vsc ¼ Vð0Þ þ at

(12.7-1)

Vð0Þ ¼ 0 Vsc ¼ at Vsc t¼ a Tsc ¼ Jsc a Tsc a¼ Jsc Jsc t ¼ Vsc Tsc Vm Tsc ¼ N¼ Vsc Tm Jsc ¼ Jscrew þ JL þ N 2 Jm

(12.7-2) (12.7-3)

2

(12.7-4) (12.7-5) (12.7-6) (12.7-7) (12.7-8) 

ðJscrew þ JL þ N Jm Þ Vsc Jscrew JL þ þ NJm Vsc ¼ NTm Tm N N  dt Vsc
ðJscrew þ JL Þ ¼ þ Jm dN Tm N2

t¼



(12.7-9) (12.7-10)

¼ 0 (for minimum acceleration time) (12.7-11) JL Jscrew þ ¼ Jm (12.7-12) 2 N 2 rN ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Jscrew þ JL N¼ (gear ratio for minimum acceleration time) Jm (12:7-13)

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 45

Drive mechanical diagram.

Speed Matching Another method used to select a ratio is to make the rated speed of the drive motor correspond to the traverse feed rate. The required gear ratio can be equated as follows: Motor rated speed6lead of the screw ¼ ratio Maximum ipm feed rate

(12.7-14)

This method is used in the computer-aided drive sizing programs. The ratios are selected by the computer program based on the rated speed of the drive motor and the desired traverse rate. The reason this criterion is selected is to keep the motor speed as high as possible so when the feed rate becomes very low the drive will still have the drive resolution needed for satisfactory surface finish.

12.8 DRIVE THRUST/TORQUE AND FRICTION CONSIDERATIONS One of the most important requirements in sizing a feed servo drive on a machine is to make sure there will be enough slide thrust to manufacture or machine a part. There are three conditions of thrust that must be considered. First, there are the machining thrust requirements, which can range from 5000 lbf to 20,000 lbf. Second, the servo drive must have sufficient forcing capability to accelerate the machine slide rapidly and maximize productivity. Acceleration torque at the servo motor (thrust at the machine slide) is controlled by the position loop in a machine positioning servo drive. The drive motor acceleration torque is the product of the total inertia at the motor and the desired motor acceleration. To keep the acceleration torque within practical limits, the acceleration at the machine

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

slide should not be much above 0:15Gr . Gr ¼ 32:2 ft=sec2 or 23,200 ipm=sec The third thrust condition is the friction losses. There are three types of friction that must be considered: static, coulomb, and viscous. All three exist on machine way systems. The servo-drive motor will have the effects of friction reflected to its shaft. These three types of friction may exist in combination from a multiplicity of sources. Examples of these sources are the machine slide, the wiper configuration on the machine ways, guide clamp arrangements, gibing, and drive screw-bearing arrangements. Static friction, if combined with a mass and spring in the right proportions, will cause a null-hunt instability to occur in a position servo system. Thus it is very important not to use machine slide way-liner materials that exhibit stiction properties. Such materials are, for example, Micarta, Formica, and Synthane. Coulomb friction is usually constant and in opposition to the direction of motion. Practical values could be approximately 20% of a servo motor’s rated torque rating. Coulomb friction is independent of velocity and is referred to as Fc : Fc ¼ cðlbf=lbÞ6 weight ðlbÞ ¼ ðlbfÞ

(12.8-1)

Lastly, viscous friction is velocity dependent and is referred to as Fv : Fv ¼ Bv 6velocityðipmÞ ¼ ðlbfÞ Bv ¼ coefficient of viscous friction ðlbf
min=in:Þ

(12.8-2)

Some machine tool friction way-liner materials have minimal stiction characteristics. Examples of these are Rulon and Turcite. It should be noted that for very heavy weights the friction losses can be excessive. A practical coefficient of coulomb friction for these materials is 0.12 lbf/lb. For very large machines with very heavy load weights, an antifriction way system can be used to minimize friction losses. Two such machine way systems are hydrostatic way systems and roller bearing systems. A practical coefficient of coulomb friction for antifriction ways is 0.01 lbf/lb. Coulomb friction can have nonlinear properties depending on the type of material on the machine slide way surface. This nonlinear property is often referred to as ‘‘stick-slip,’’ which is a function of the static friction Fs (sometimes referred to as the breakaway friction). These types of friction can be represented graphically in Figure 46. Machine slides that have positioning servo drives with the stick-slip characteristics (inside the servo loop) and use position measuring transducers on the machine slide have a very good possibility for a servo

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 46

Friction characteristics.

null-hunt condition. As previously stated, static friction combined with a mass and spring can cause a limit-cycle null hunt when the static and coulomb friction are in the right ratio. Experience shows that a ratio of static friction to coulomb friction equal to two or more can result in a position loop null hunt. Prior to the use of servo drives on machine feed drives, friction-type ways with stick-slip were widely used. With the advent of servo drives on machine feed drives, the driven screws were changed from acme type to ball-bearing type drive screws for increased efficiency. Additionally, the machine slide ways were changed to antifriction-type way surfaces to improve efficiency and eliminate the stick-slip problems where a machine slide was enclosed in the position loop. With machine feed slides using roller bearings or hydrostatic way surfaces, the friction characteristics are minimal. However, machine way surfaces that use Rulon-type way liners are friction-type materials. These materials have minimal stick-slip characteristics, but they do have coulomb type friction. Additionally, other mechanical components of the feed drive contribute friction forces to the total reflected friction characteristics of the servo motor. Typical friction characteristics are compared graphically in Figure 47. A drawing for a typical machine feed drive is shown in Figure 48. The coulomb friction is obtained from a number of sources. In industrial applications most of the coulomb way friction comes from the machine slide weight. However, other major contributors of coulomb friction can come from the way wipers and the nut wiper.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 47

Friction characteristics for different way liner materials.

Fig. 48

Machine slide friction characteristics.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 49

Duty cycle characteristics. (Courtesy of Kollmorgen Corp.)

12.9 DRIVE DUTY CYCLES In sizing a machine servo drive it is important to consider how much thrust (or torque at the servo motor) will be required, how long it can be provided (on time), and how long the drive must rest (off time). These parameters can be calculated when a machine servo drive is sized for a particular application. It is important to the machine designer to know how much overload the drive can stand and what the recovery time will be. Of the three load requirements (load torque, acceleration torque, and friction losses), the servo-motor torque requirement to accelerate the load will usually be the greatest torque. Commercial DC servo motors and brushless DC motors have continuous torque ratings and a duty cycle where torque above rated torque can be supplied for servo forcing (e.g., acceleration) on a limited time basis. As torque requirements exceed the rated value, excessive heating of the motor will occur. These speed/torque ratings are discussed in Section 3.2 and Figures 3.4 and 3.5. These speed torque requirements are for the motor/ amplifier combination. Servo drive requirements calling for repeated

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 50

Duty cycle characteristics.

accelerations and decelerations will result in a rapid temperature rise in the servo motor. For general positioning feed drive requirements, the available (above rated torque) torque for servo forcing is important. When using DC SCR servo drives the servo forcing torques are about 400% of rated torque. For electric servo drives using transistor design amplifiers such as DC pulsewidth modulation (PWM) drives or brushless DC drives, the forcing torques are limited by the transistor designs to about 200% of rated torque. These values will increase as research in solid-state technology advances, with insulated-gate bipolar transistors (IGBT), for example.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

In addition to the limiting of forcing torque by the amplifier, the servo drive motor has thermal limits for heating. Duty cycle characteristics based on the thermal time constant of the servo motor have been provided by some manufacturers in graphical form. A typical duty cycle characteristic for an Inland DC servo drive is shown in Figure 49 for a 200% load requirement. The recommended ‘‘on’’ time for this example is about 4 min. Based on the thermal time constant of a motor, the duty cycle characteristic can be computed from a technical paper by S. Noodleman and B. R. Patel (Duty cycle characteristics for a DC servomotor, IEEE-IAS Trans., IA-9 (5), pp. 563–569, Sept./Oct. 1973). A basic software program from this paper to calculate and print the thermal duty cycle characteristics is shown in Figure 50. A typical print out of this program is shown in Figure 51. This program can be used for DC servo motors or brushless DC servo motors.

Fig. 51

Duty cycle characteristics.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

13 Drive Sizing Considerations

13.1 INTRODUCTION In the application of industrial servo drives to machines it is important that the drive be large enough to meet all the load torque requirements and be stable. It is necessary to make sure that the servo drive torque rating is large enough to meet the load thrust and acceleration requirements plus the machine friction losses; an organized method to accomplish this is referred to as ‘‘sizing the drive.’’ There are many drive-sizing software programs available from commercial servo drive suppliers. It is of critical importance that machine designers size the servo drive in an organized engineering manner. Either manual or computer software sizing must be used to avoid an unacceptable servo-drive performance. It is also a requirement that the software drivesizing programs be documented and interactive with the user. Machine design engineers may not have a feedback control background; thus it is important that the programs have complete documentation (sometimes referred to as remark statements). To be of value to the user, the software programs should be interactive with enough description of what is happening during the drive-sizing process. The criteria of the drive sizing are different for hydraulic and electric drive sizing. Hydraulic servo drives usually have more than ample torque to

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

meet the load requirements. However, the hydraulic fluid has compressibility, which is like a spring inside the servo loop and can cause a minor loop instability. This hydraulic spring is represented as a hydraulic resonance. Any hydraulic servo drive with a large volume of hydraulic oil compressed between the actuator and the servo valve has a potential for an unstable minor loop servo drive. Hydraulic servo valve drives that use long-travel piston actuators have a great potential for servo instability. Likewise, hydraulic servo pump drives with large volumes of hydraulic fluid between the pump and actuator are prone to servo instability. The hydraulic resonance is an indicator of what to expect for stability. An industrial index of performance (I.P.) for the hydraulic resonance is that the resonance should be 200 rad/sec or greater. The hydraulic resonance can be calculated from sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 26b6D2m ðrad=secÞ (13.1-1) oh ¼ Vc 6JT where: Dm ¼ motor displacement (in:3 =rad) Vc ¼ oil under compression (in:3 ) JT ¼ total inertia at the motor (lb-in.-sec2 ) b ¼ bulk modulus of oil ¼ 16105 lb=in:2 In general, hydraulic servo drives also have the added complexity of oil contamination, leakage, and changes in viscosity with temperature. Electric drives do not have the hydraulic medium problems of leakage, compressibility, etc. However, electric drives do have some problems obtaining sufficient torque for load requirements. These torque limitations stem mostly from the amplifier. DC silicon controlled rectifier (SCR) amplifiers have the problem of excessive phase lag due to the circuit transport lag. Transistor design amplifiers have limits on available current. In general, electric servo drives have more than adequate capability for torque requirements and performance for industrial feed drive applications. Section 13.2 consists of a manual hydraulic servo drive-sizing form. This drive sizing will give assurance that the correct size hydraulic servo motor and hydraulic pressure will be used to meet the torque or thrust requirements for a machine feed drive. The hydraulic resonance is of particular importance. To have a stable drive the hydraulic resonance should be at least 200 rad/sec or higher. If this requirement is not met, the drive should be resized with a different ratio, lead, motor, etc.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Section 13.3 consists of a manual electric drive-sizing form. This form can be used for DC or brushless DC servo drives. In addition, most commercial drive suppliers will provide CAD (computer-assisted design) sizing programs on computer disk. As with the hydraulic drive-sizing form, the goal is to provide a servo drive capable of providing enough torque and thrust to meet the load requirements of the machine feed drive. Not only must the drive be large enough to provide the required torque, but it must be stable. Other servo analysis techniques must be used to ascertain if the servo drive is stable. One such technique is the use of simulation software programs. The simulation programs must include the control, drive, and servo plant (machine). Such a machine simulation technique is discussed in Chapter 15.

13.2 HYDRAULIC DRIVE SIZING To illustrate the use of the hydraulic sizing form, it will be presented as applied to a typical machine axis. Blank hydraulic sizing forms are located in the Appendix. The illustrated hydraulic servo drive is a horizontal axis on a machine tool and has a positioning controller. The hydraulic motor is coupled to a ball bearing lead screw with a Thomas coupling. No belt ratio or gear ratio is used. This is called a direct drive. The axis slide uses Turcite way liners having a coefficient of friction of 0.12 lbf/lb. As with all hydraulic servo drives, it is necessary to minimize the effect of the hydraulic spring to have a stable servo drive. Therefore it is a requirement to keep the hydraulic resonance above an I.P. of 200 rad/sec. The machine and servo specifications are: Weight of the load and machine slide Slide friction coefficient Maximum thrust required Length of the drive ball screw Diameter of the ball screw Lead of the ball screw Traverse rate Direct drive (no ratio) Desired servo motor Coupling Position-loop gain Machine slide will be accelerated exponentially Servo valve

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

8000 lb 0.12 lbf/lb 10,000 lbf 90 in. 2 in. 0.375 in./rev 180 ipm Hartman HT-50 Thomas 201DBZ 2 ipm/mil or 34/sec Pegasus 160

HYDRAULIC DRIVE SIZING Machine HBM, Axis Horiz., Traverse rate 180 (ipm) Ball screw: Length 90 (in.), Diameter 2.0 (in.), Lead 0.375 (in./rev) Gearbox # —, Ratio 1 Slide weight 8000 (lbs) Inertia calculations: Slide inertia at motor: Jslide ¼ WT ðlead=ratioÞ2 66:5610
5 (lb-in.-sec2 ) Jslide ¼ 8000 ð0:375=1Þ2 66:5610
5 ¼ 0:07312 lb-in:-sec 2 Ball screw inertia at motor: Jscr ¼ DIA4 6LGTH=ðratio2 Þ67:2610
5 (lb-in.-sec2 ) Jscr ¼ 2 4 690=167 :2610
5 ¼ 0:10368 lb-in:-sec 2 Coupling inertia ¼ 0.144 (lb-in.-sec2) Motor pulley inertia: Pulley number ¼ none Pulley inertia ¼ Screw pulley inertia: Pulley number ¼ none Pulley inertia ¼ Pulley inertias reflected to motor: Jp ¼ Jmot:pul: þ Jscrew pul: =Ratio2 Jp ¼

þ

=

2

(lb-in.-sec2 )

¼

—

—

¼ 0:321 (lb-in.-sec2 )

Total reflected inertia: Jref ¼ Jslide þ Jscr þ Jcoup þ Jp Jref ¼ 0:0731 þ 0:103 þ 0:144 þ Total inertia: Jtot ¼ Jmot þ Jref ¼ 0:0152 þ 0:321 ¼ 0:3362 (lb-in.-sec2 )

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Coefficient of friction (0.12 for Rulon) ¼ 0.12 (lbf/lb) (0.01 for rollers or hydrostatics) 1 Lead 6Gear ratio ¼ VT ðrpmÞ

Maximum motor speed ¼ Traverse (ipm)6

VT ¼ 1806

1 61 ¼ 480 ðrpmÞ 0:375

Drive motor chosen Hartman HT-50 Displacement DM ¼ 5 ðin:3 =revÞ61=6:28 ¼ 0:796 (in:3 =rad) Torque constant KT ¼ 80 ðin:-lbÞ at 100 psi Torque available 1600 (in.-lbs) at 2000 psi Torque available 2400 (in.-lbs) at 3000 psi 1. Establish the maximum thrust required ¼ FM ¼ 10,000 (lbf). 2. Establish the no-load friction force ¼ FF . FF ¼ Coefficient of friction6slide weight FF ¼ 0:1268000 ¼ 960 ðlbfÞ No-load friction torque ¼ TF TF ¼ no-load friction force ðFF Þ6 TF ¼ 9606

lead 1 6 (lb-in.) 6:28 ratio

0:375 1 6 ¼ 57 :32 (lb-in.) 6:28 1

3. Stall thrust ¼ FS ¼ 1:56maximum thrust ðFM Þ (lbf). FS ¼ 1:5610,000 ¼ 15,000 ðlbfÞ thrust ðF Þ (lbf). 4. Stall thrust for efficiency ¼ FST ¼ stall S 0:90 eff FST ¼

15,000 ¼ 16,666 (lbf) 0:9

5. Maximum motor torque ¼ TM ðFM Þ lead 1 ¼ max thrust 6 6:28 6 ratio (lb-in.) 0:9 eff 10,000 0:375 1 6 6 ¼ 663 (lb-in.) (torque needed) 0:9 6:28 1 Lead ¼ 0:375 ðin=revÞ Ratio ¼ 1 Torque available ðat 2000 psiÞ ¼ 1600 (lb-in.) TM ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

6. Drive acceleration. To avoid excessive acceleration forces, the servo drive position loop must control the acceleration. There are three methods that have been used to control acceleration: velocity ramp, velocity exponential, velocity ‘‘S’’ curve. The velocity exponential acceleration has a maximum acceleration (sometimes referred to as ‘‘jerk’’) at time zero from: Acceleration ¼ Kv 6 max velocity6e
Kv t Max velocity ¼ VT ðrpmÞ Kv ¼ position-loop gain ¼ ðipm=milÞ616:8 ¼ ð1=secÞ at t ¼ 0 6:28 6Kv ð1=secÞ6VT ðrpmÞ ¼ ðrad=sec2 Þ 60 6:28 6346480 ¼ 1708 ðrad=sec2 Þ ¼ 60

Accelerationmax ¼ Accelerationmax

This is a ‘‘worst-case’’ condition for acceleration. To avoid an initial maximum acceleration, the ‘‘S’’ curve velocity control can be used. The initial acceleration is zero. The maximum acceleration occurs when the rate of change in velocity is a maximum. The acceleration for the ‘‘S’’ curve is Acceleration ¼

Traverse feed 60 6 ðipmÞ6C 2 6t6e
Ct ðin:=sec2 Þ

where C is a factor determining acceleration rate. Values of C for desired acceleration rates (in./sec2) are: Acceleration (in./sec2)

C

9 18 27 36

5 10 15 20

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The maximum acceleration occurs at t ¼ 1=C. Final velocity ðVT Þ Ratio 6C 2 6t6e
Ct 66:286 (rad=sec2 ) 60 Lead 6 Þ 6eð 66:286 ¼ 6 6 60 ¼ (rad=sec2 )

Accelmax ¼ Accelmax

Ratio ¼ Lead ¼

(in:=rev)

7. Torque at maximum acceleration. TA ¼ JT 6accelerationmax Motor inertia ¼ 0:0152 (lb-in.-sec2 ) Reflected load inertia ¼ 0:32 (lb-in.-sec2 ) JT ¼ total inertia at motor ¼ Jmotor þ Jload

(lb-in.-sec2 )

JT ¼ 0:0152 þ 0:32 ¼ 0:3362 (lb-in.-sec2 ) TA ¼ 0:3362 (lb-in.-sec2 )61708 (rad=sec2 ) ¼ 574 (lb-in.) 8. Torque requirement summary. Friction torque ðTF Þ ¼ 57 :32 (lb-in:) Thrust load ðTM Þ ¼ 663 (lb-in:) Acceleration torque ðTA Þ ¼ 574 (lb-in:) 9. Hydraulic pressure at the motor. torque6100 ðpsiÞ motor torque constant ðKT Þ 100 PFR ¼ Pressure for friction load ðTF Þ ¼ 57 :326 KT ¼ 71:65 ðpsiÞ 100 PTH ¼ Pressure for thrust ðTM Þ ¼ 6636 ¼ 828 ðpsiÞ KT 100 PA ¼ Pressure for max acceleration ðTA Þ ¼ 5746 KT ¼ 717 ðpsiÞ Pressure at motor ¼

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10.

Hydraulic losses. Line loss ¼ 12% of system pressure; approx: 300 psi 300 ðpsiÞ Valve drop ðapprox: 1000 psi at rated flowÞ 1000 ðpsiÞ Pressure for friction load 72 ðpsiÞ No load motor loss at traverse ¼ KPM VT =1000 KPM ¼ 150 psi=1000 rpm ðmotor pressure drop no-loadÞ VT ¼ 480 ðrpmÞ 1506480 No load motor loss ¼ ¼ 72 ðpsiÞ 1000 Total losses 1443 ðpsiÞ

11.

Supply pressure needed. (Note: Load pressure ¼ 23 supply pressure, or 1.5 6 load pressure ¼ supply pressure) Pressure for friction ¼ friction psi ðPFR Þ þ losses ¼ 72 þ 1443 ¼ 1514 psi Pressure for max thrust ¼ 1:5 thrust psi ðPTH Þ þ 300 ¼ 828 þ 300 ¼ 1128 psi Pressure for max accel ¼ accel psi ðPA Þ þ losses ¼ 717 þ 1443 ¼ 2160 psi (Select a pressure greater than the highest of these pressures.) Recommended supply pressure ¼ 2500 ðpsiÞ

12.

Motor size. Effective drive displacement (from thrust requirement) ¼ EDD Total stall thrust ðFST Þ 16,660 lbf ¼ Supply pressure 2500 psi ¼ 6:6 in:3 =in: of drive movement Motor displacement ðfrom torque requirementÞ: Eff drive disp ðEDD Þ 6Lead Ratio ¼ motor displacement required ðin:3 =revÞ 6:6 in:2 60:375 in:=rev ¼ 2:47 ðin:3 =revÞ Displacement of motor chosen DM ¼ 5 ðin:3 =revÞ

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

If displacement of motor chosen is less than required, use the next larger motor. 13.

Flow rating. Theoretical flow at traverse ¼ QT (Select a servo valve larger than required flow at traverse.) 1:3VT DM 1:3648065 ¼ ¼ 13:5 ðgpmÞ 231 231 VT ¼ max motor speed at traverse ¼ 480 ðrpmÞ QT ¼

DM ¼ motor displacement ¼ 5 ðin:3 =revÞ 14.

Maximum accelerating torque TA (lb-in.). PS 6KT
TF 100 PS ¼ hydraulic power supply pressure ¼ 2500 ðpsiÞ KT ¼ motor torque constant ¼ 80 ðlb-in:=100 psiÞ TA ¼

tF ¼ friction torque ¼ 57 :32 ðlb-in:Þ 2500680
57 :32 ¼ 1942 ðlb-in:Þ TA ¼ 100 15.

Maximum decelerating torque TD . ðPs þ PR ÞKT þ TF 100 PS ¼ supply pressure ¼ 2500 ðpsiÞ

TD ¼

PR ¼ relief pressure setting in manifold ¼ 0 ðpsiÞ ð2500 þ 0Þ680 þ 57 :32 ¼ 2057 ðlb-in:Þ TD ¼ 100

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16.

Hydraulic resonance. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 BD2M wH ¼ ðrad=secÞ Vc JT DM ¼ motor displacement ¼ 0:796 ðin:3 =radÞ Vc ¼ oil under compression ¼ Vmotor þ Vvalve þ Vmanifold ¼ ðin:3 Þ ð21 total volumeÞ Vvalve ¼ 0:17 in:3 ðapprox:Þ Vmanifold ¼ 0:3 in:3 ðapprox:Þ 2:91 þ 0:17 þ 0:3 ¼ 3:38 ðin:3 Þ JT ¼ total inertia at the motor ¼ 0:3362 (lb-in.-sec2 ) B ¼ bulk modulus of oil ¼ 16105 ðlb=in:2 Þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 26105 60:796 ¼ 374 ðrad=secÞ oh ¼ 3:3860:3362

17.

Hydraulic damping factor. rffiffiffiffiffiffiffiffiffi Lt BJT dH ¼ DM 2Vc Lt ¼ motor leakage þ valve leakage (in:3 =sec=psi) Valve leakage ¼ 0:0019 in:3 =sec=psi (approx:) Motor leakage ¼ 0:00134 (in:3 =sec=psi) Lt ¼ 0:00134 þ 0:0019 ¼ 0:0032 (in:3 =sec=psi) rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0032 16105 60:3362 dH ¼ ¼ 0:28 0:796 263:38

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Typical motor specifications

Hartman motor HT-5 HT-10 HT-22 HT-50 HT-76

Motor displacement (in.3/rev)

Max. speed (rpm)

Inertia (lb-in. -sec2)

Compressed oil volume (in.3)

Torque constant (in.-lb/ 100 psi)

Leakage (in.3/ sec-psi)

0.5 1.0 2.2 5.0 7.6

2000 2000 1500 1500 1200

0.00174 0.00198 0.006 0.0152 0.0263

0.77 1.31 2.91 4.81

8 16 35 80 121

0.00058 0.00065 0.00134 0.00167

Thomas coupling inertias Coupling

J (lb-in.2)

J (lb-in.-sec2)

101DBZ 126DBZ 163DBZ 201DBZ 226DBZ 263DBZ 301DBZ 351DBZ 401DBZ 451DBZ

4.7 10 22 56 100 210 420 950 1090 3500

0.01216 0.02587 0.0569 0.1449 0.2587 0.5434 1.0869 2.4585 2.8209 9.0579

101DBZA 126DBZA 163DBZA 201DBZA 226DBZA 263DBZA 301DBZA 351DBZA 401DBZA 451DBZA

4.7 10 22 56 100 220 430 970 1900 3500

0.01216 0.02587 0.0569 0.1449 0.2587 0.5434 1.0869 2.4585 2.8209 9.0579

101DBZB 126DBZB 163DBZB 201DBZB 226DBZB 263DBZB 301DBZB 351DBZB 401DBZB 451DBZB

4.7 11 23 57 110 220 440 1000 1900 3600

0.01216 0.02846 0.0592 0.1475 0.2846 0.5693 1.13871 2.5879 4.9171 9.3167

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Used on motor

HT22 HT50

HT10

HT76

13.3 ELECTRIC DRIVE SIZING To illustrate the use of the drive-sizing form it will be presented as applied to a typical machine axis. Blank electric drive-sizing forms are located in the Appendix. The electric servo-drive design problem is for a horizontal axis on a large machine. The machine slide has Rulon way liners with a coefficient of friction of 0.12 lbf/lb. The servo drive will be a brushless DC electric drive. The machine slide is large (50,000 lbs). This weight was selected to illustrate that the mechanical isolation of the ball screw lead and ratio will minimize the reflected inertia of this weight to the servo-motor drive shaft. The servo drive axis will be accelerated exponentially for this drive-sizing problem to illustrate an undersized servo drive. The drive-sizing problem will be followed by two computer drive-sizing program printouts. The first drivesizing printout (see Figure 1) will be for this problem using exponential acceleration. The second drive-sizing printout (see Figure 2) will be for the same problem, but using an ‘‘S’’ curve acceleration to illustrate the advantage of using this type of acceleration. The machine and servo have the following specifications: Machine Axis Traverse rate Total axis weight Slide friction coefficient Required cutting thrust Ball screw diameter Ball screw length Ball screw lead Motor pulley (Woods pulley) Screw pulley (Woods pulley) Motor Position loop gain

Horizontal axis Table 337 ipm 50,000 lbs 0.12 lbf/lb 8000 lbf 3 in. 70 in. 0.375 in./rev 18H300 60H300 Kollmorgan M607-B 1 ipm/mil

Find reflected inertias to the drive motor for the following: Machine slide Ball screw Pulley inertias Motor inertia Total reflected inertia at the drive motor

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1 Printout for drive-sizing problem with exponential acceleration.

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Fig. 2 Printout for drive-sizing problem with ‘‘S’’ curve acceleration.

Size the drive using the manual drive-sizing forms.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

INERTIA CALCULATIONS Slide inertia at the motor:     lead 2 0:375 2 60:0000656 ¼ 50,000 60:0000656 J slide ¼ weight N 3:333 ¼ 0:0415 lb-in.-sec2 Ball screw inertia at the motor: J screw ¼

diameter4 6lead60:000072 34 67060:000072 ¼ N2 3:33332

¼ 0:0367 lb-in.-sec2 Motor pulley inertia Pulley number ¼ 18H300 Pulley inertia ¼ 0.0136 lb-in.-sec2 For Woods pulleys 3- and 2-inch inertias, see Section 12.2 Screw pulley inertia Pulley number ¼ 60H300 Pulley inertia ¼ 0.8005 lb-in.-sec2 Pulley inertias reflected to the motor Jscrew pulley ratio2 0:8005 J pulley ¼ 0:0136 þ ¼ 0:0857 lb-in.-sec2 3:3332 Total reflected inertia ¼ J slide þ J screw þ J pulley ¼ 0:0415 þ 0:0367

J pulley ¼ J motor pulley þ

þ 0:0857 ¼ 0:1639 lb-in.-sec2 0:0196 Total inertia ¼ J total reflected þ J motor ¼ 0:1639 þ 12 2 ¼ 0:3511 lb-in.-sec

ELECTRIC DRIVE SIZING Machine: Horizontal, Axis: Table, Traverse rate: 337 ipm. Ball screw lead: 0.375 in. rev., Length ¼ 70 in., Diameter ¼ 3 in., Gear box: — , Gear ratio ¼ 3.333

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Reflected load inertia ¼ 0.1639 lb-in.-sec2, Slide weight ¼ 50,000 lbs. Slide coefficient of friction: 0.12 lbf/lb. 1. Establish the maximum load or working thrust. Load thrust ¼ 8000 lbf: 2. Establish no-load thrust. Thrustnl ¼ Coefficient of friction6slide weight Thrustnl ¼ 0:12650,000 ¼ 6000 lbf: 3. Torque requirements at the drive motor. (a)

Geared Drives. NOTE: Use speed matching to select the ratio. Motor rated speed ½rpm6lead ½in. / rev Traverse rate½ipm 0:375 ¼ 3:394 Ideal ratio ¼ 30506 337 Ideal ratio ¼

Load thrust torque: lead 1 6 in.-lbs 2p gear ratio 0:375 1 6 ¼ 143 in.-lbs Motor load torque ¼ 80006 2p 3:333

Motor load torque ¼ load thrust6

No-load torque: lead 1 6 2p gear ratio 0:375 1 6 ¼ 107 in.-lbs Motor no-load torque ¼ 60006 2p 3:333

Motor no-load torque ¼ no-load thrust6

Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼ 143 þ 107 ¼ 250 in.-lbs:

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(b)

Direct Drives: Load thrust torque: lead 2p 6 ¼ 2p

Motor load torque ¼ load thrust6 Motor load torque ¼

in.-lbs:

No-load torque: Motor no-load torque ¼ no-load thrust6 Motor no-load torque ¼

6

2p

lead 2p

¼

in.-lbs:

Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼ þ ¼

in.-lbs:

4. Total drive motor torque required must be derated according to the amplifier used. Amplifier a. SCR amplifiers Single-phase full wave Single-phase full wave/inductor Three-phase half wave Three-phase half wave/inductor b. DC generator c. PWM or brushless DC

Derating Form Factor for Motor 1.66 1.2 1.25 1.05 1.0 1.0

Required torque ¼ total torque6form factor ¼ in.-lbs: ¼ 25061 ¼ 250 in.-lbs:

5. Motor traverse speed.

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(a)

Geared Drives: Maximum motor speed ¼ traverse ipm6 Maximum motor speed ¼ 3376

(b)

1 6gear ratio lead

1 63:333 ¼ 2995 rpm 0:375

Direct Drives: 1 lead 61 ¼

Maximum motor speed ¼ traverse ipm6 Maximum motor speed ¼

rpm

6. Drive motor manufacturer selected: Kollmorgan Motor selected: M607B Rated speed of motor selected 3050 rpm Maximum speed of motor selected 3050 rpm Motor rated speed should be equal or larger than the traverse speed. If this requirement is not satisfied, a different gear ratio and/or drive screw lead should be selected to resize the motor for the required torque. 7. Torque for maximum acceleration (based on exponential response to a traverse rate step input of velocity). ipm 1000 mil min 1 6 6 ¼ 16:66 mil 60 sec sec in: 2p 1 rev 6gain a max: ¼ 6 max: motor speed 60 sec min Gain ¼ 1

¼ rad/sec2 ¼ VF 6Kv [
Kv T 16:66 rad 62995 rpm ¼ 5214 a max: ¼ 0:10456 sec sec2 Torquea ¼ J T 6a max : J T ¼ motor inertia + reflected load inertia J T 0:0156612 þ 0:1639 ¼ 0:3511 lb-in.-sec2 T a ¼ 0:3511 lb-in.-sec2 65214 rad/sec2 ¼ 1830 in.-lbs 8. Motor torque requirements summary. Torque for no-load 107 in.-lbs

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Torque for load thrust 143 in.-lbs Torque for maximum acceleration 1830 in.-lbs 9. Motor selected M607-B . Rated torque 33 6 12 ¼ 396 in.-lbs Maximum torque 93 6 12 ¼ 1116 in.-lbs Motor rated torque should be greater than required torque for total thrust. NOTE: Refer to motor curves. Motor maximum torque should be approximately the same or greater than the required torque for maximum acceleration ðTaÞ plus no-load thrust. 10.

Maximum torque ¼ 1116 in.-lb from motor curves at traverse rpm. Imax xKT
Tnoload ¼ torque available for acceleration form factor I max ¼ Motor armature maximum current ¼ 118 amps K t ¼ Motor torque constant ¼ 0:826612 ¼ 9:9 in.-lbs/amp

Ta ¼

T nl ¼ No-load torque ¼ 107 in.-lbs  118 69:9
107 ¼ 1061 in.-lb torque available Ta ¼ 1 for acceleration torque available 1061 ¼a¼ total inertia 0:3511 rad rad ¼ 3022
sec2 sec2

Allowable acceleration ¼

11.

Maximum decelerating torque ¼ Td. I ma 6K T þ T no load form factor I max ¼ Motor armature maximum current ¼ 118 amps  118 69:9 þ 107 ¼ 1275 in.-lb torque available Td ¼ 1 Td ¼

for deceleration

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12.

Motor time constants. Brushless DC Motors 67:7 v 1 volts 6 ¼ 0:646 Ke ¼ 1000 rev=min 2prad/rev6 min =60 sec rad/sec Keðl
LÞ 0:646 ¼ 0:3734 ¼ K eðPHASEÞ ¼ Motor voltage constant ¼ 1:73 1:73 volts-sec/rad (see motor data) K t ¼ Motor torque constant ¼ 9:9 in.-lb./amp (see motor data) RMðL
LÞ ¼ Motor resistance ¼ 0:14 ohms SRMðL
LÞ ¼ Total motor circuit resistance ¼ 1:35RMðL
LÞ ¼ 1:3560:14 ¼ 0:189 ohm SRMðPHASEÞ ¼ SRMðL
LÞ 60:5 ¼ 0:18960:5 ¼ 0:0945 ohm LL
L ¼ Motor inductance ¼ 0:0038 henries J m ¼ Motor armature inertia ¼ 0:0156612 ¼ 0:1872 lb-in.-sec2 J LOAD ¼ Reflected inertia to motor ¼ 0:1639 lb-in.-sec2 J Total ¼ J m þ J LOAD ¼ 0:1872 þ 0:1639 ¼ 0:3511 lb-in.-sec2 LL
L 0:0038 te ¼ Motor electrical time constant ¼ ¼ SRMðL
LÞ 0:189 ¼ 0:0201 sec 1 1 ¼ 49:7 rad/sec oe ¼ ¼ te 0:0201 tm ¼ Motor mechanical time constant ¼

SRMðPHASEÞ J Total K eðPHASEÞ K t

0:094560:3511 ¼ 0:008975 sec 0:373469:9 1 om ¼ ¼ 111 rad/sec tm tm 0:008975 ¼ 0:4465 ¼ te 0:0201 tm ¼

DC Motors K e ¼ Motor voltage constant ¼ data) K t ¼ Motor torque constant ¼ data)

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volts/rad/sec (see motor in.-lb/amp (see motor

Ra ¼ Motor armature resistance ¼ ohms SRa ¼ Total armature circuit resistance ¼ 1.35 Ra ¼ 1.35 6 ¼ ohm henries L ¼ Motor armature circuit inductance ¼ lb-in.-sec2 J m ¼ Motor inertia ¼ 2 J Load ¼ lb-in.-sec J Total ¼ J m þ J L ¼ þ ¼ lb-in.-sec2 L Te ¼ Motor electrical time constant ¼ SR ¼ a ¼ sec 1¼ rad/secP oe ¼ 1=te ¼

T m ¼ Motor mechanical time constant ¼ tm ¼ xx ¼ sec rad/sec om ¼ t1m ¼ tm ¼ ¼ te 13.

Ra J t KeKt

Motor transfer function. (a) If ttme < 4, motor transfer function should be: . 1 Ke oðsÞ ½rad=sec i ¼h 2 s ½volt V ðsÞ þ 1 sþ1 om oe

(b)

If

tm te

oðsÞ V ðsÞ

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om

> 4, motor transfer function should be: . 1 Ke ¼ s ðoe þ 1Þðosm þ 1Þ

14 Adjusting Servo Drive Compensation

14.1 MOTOR AND CURRENT LOOP A drive sizing makes sure that the drive has enough torque to satisfy the load requirements to overcome friction losses, provide sufficient load thrust, and provide enough acceleration torque for the type of acceleration required. Once the drive-sizing requirement is satisfied, the servo drive stability must be addressed. Stabilizing the servo drive is a matter of adjusting the servo compensation. All industrial servo drives require some form of compensation often referred to as proportional, integral, and differential (PID). The process of applying this compensation is known as servo equalization or servo synthesis. In general, commercial industrial servo drives use proportional plus integral compensation (PI). It is the purpose of this discussion to analyze and describe the procedure for implementing PI servo compensation. The block diagram of Figure 1 represents DC and brushless DC motors. All commercial industrial servo drives make use of a current loop for torque regulation requirements. Figure 1 includes the current loop for the servo drive with PI compensation. Since the block diagram of Figure 1 is

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1 Motor and current loop block diagram.

not solvable, block diagram algebra separates the servo loops to an inner and outer servo loop of Figure 2. For this discussion a worst-case condition for a large industrial servo axis will be used. The following parameters are assumed from this industrial machine servo application: Motor: Kollmorgen motor—M607B Machine slide weight: 50,000 lbs Ball screw: Length—70 in. Diameter—3 in. Lead—0.375 in. revolution Pulley ratio: 3.333 JT ¼ total inertia at the motor ¼ 0.3511 lb-in.-sec2 te ¼ electrical time constant ¼ 0.02 second ¼ 50 rad/sec t1 ¼ te Ke ¼ motor voltage constant ¼ 0.646 volt-sec/radian KT ¼ motor torque constant ¼ 9.9 lb-in./amp KG ¼ amplifier gain ¼ 20 volts/volt Kie ¼ current loop feedback constant ¼ 3 volts/40A ¼ 0.075 volt/amp

Fig. 2

Rearranged motor and current loop block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Ra ¼ motor armature circuit resistance ¼ 0.189 ohm Ki ¼ integral current gain ¼ 735 amp/sec/radian/sec The first step in the analysis is to solve the inner loop of Figure 2. The closed-loop response I=e1 ¼ G=1 þ GH where: G ¼ 1=Ra ðte S þ 1Þ ¼ 5:29=½te S þ 1


ð5:29 ¼ 14:4 dBÞ

GH ¼ 0:64669:9=½0:18960:3511½ðte S þ 1ÞS

GH ¼ 96=S½te S þ 1
96 ¼ 39 dB 1=H ¼ J T S=K e K T ¼ 0:3511 S=0:64669:9 1=H ¼ 0:054 S

ð0:054 ¼ 25 dBÞ

Using the rules of Bode, the resulting closed loop Bode plot for I=e1 is shown in Figure 3. Solving the closed loop mathematically: I G 1 ¼ ¼ e1 1 þ GH Ra ðte S þ 1Þ þ Ke KT =JT S JT S ¼ JT Ra Sðte S þ 1Þ þ Ke KT I JT S ¼ 2 e1 J T Ra te S þ J T Ra S þ K e K T J T =K e K T S ¼ ½ðJ T Ra =K e K T Þte S 2 þ ðJ T Ra =K e K T ÞS þ 1
I ð0:3511=0:64669:9ÞS ¼ e1 tm te S 2 þ tm S þ 1 0:054 S ¼ 0:0160:02 S 2 þ 0:01 S þ 1 J T Ra 0:351160:189 ¼ 0:01 sec; ¼ where : tm ¼ 0:64669:9 K eK T om ¼ 1=tm ¼ 100 rad=sec te ¼ 0:02 sec oe ¼ 1=te ¼ 50 rad=sec

(14.1-1)

(14.1-2)

(14.1-3)

(14.1-4)

For a general quadratic: S 2 2 delta þ Sþ1 or o2r or ¼ ½om oe
1=2 ¼ ½100650
1=2 ¼ 70 rad=sec I 0:054S ¼ 2 2 e1 S =70 þ ð2 delta=70ÞS þ 1

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(14.1-5) (14.1-6)

Fig. 3 Current inner loop frequency response.

Fig. 4 Current loop block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Having solved the inner servo loop it is now required to solve the outer current loop. The inner servo loop is shown as part of the current loop in Figure 4. In solving the current loop, the forward loop, open loop, and feedback loop must be identified as follows: The forward servo loop: K i K G 60:054 ð0:02 S þ 1Þ 73562060:054 ð0:02 S þ 1Þ ¼ 0:0002 S 2 þ 0:01 S þ 1 0:0002 S 2 þ 0:01 S þ 1 (14.1-7) 794ð0:02 S þ 1Þ 15:88 S þ 794 ¼ G¼ 2 0:0002 S þ 0:01 S þ 1 0:0002 S 2 þ 0:01 S þ 1 (14.1-8) G¼

Where : K G ¼ 20 volt=volt K ie ¼ 3=40 ¼ 0:075 volt=amp K i K G 60:054794 ð58 dBÞ K i ¼ 794=ð2060:054Þ ¼ 735 79,400 S þ 3,970,000 G¼ S2 þ 50 S þ 5000

(14.1-9)

The open loop: 79,400 S þ 3,970,000 S 2 þ 50 S þ 5000 5955 S þ 297,750 GH ¼ 2 S þ 50 S þ 5000

GH ¼ 0:0756

(14.1-10) (14.1-11)

The feedback current scaling is the following: H ¼ 3 volts=40 amps ¼ 0:075 volts=amp

1=H ¼ 13:33 ¼ 22:4 dB

The Bode plot frequency response is shown in Figure 5. The current loop bandwidth is 6000 rad/sec or about 1000 Hz, which is realistic for commercial industrial servo drives. The current loop as shown in Figure 5 can now be included in the motor servo loop with reference to Figure 2 and reduces to the motor servo loop block diagram of Figure 6.

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Fig. 5 Current loop frequency response.

Fig. 6

Motor servo loop block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The completed motor servo loop has a forward loop only (as shown in Figure 6) where: J T ¼ total inertia at the motor ¼ 0:3511 lb-in.-sec2 K T ¼ motor torque constant ¼ 9:9 lb-in./amp 13:369:9 375 ¼ ð51:5 dBÞ G¼ 0:3511 S ðð jw=6000Þ þ 1Þ S ð0:000166 S þ 1Þ (14.1-12) 375 2,250,090 G¼ ¼ 2 2 0:000166 S þ S þ 0 S þ 6000 S þ 0

(14.1-13)

vo KT jIj 9:9 13:1ð0:02 S þ 1Þ 6 ¼ 6 ¼ (14.1-14) ei J T S jvi j 0:3511 S 0:00000331 S 2 þ 0:02 S þ 1 vo 375ð0:02 S þ 1Þ ¼ ei S 0:00000331 ðS þ 50ÞðS þ 5991Þ

(14.1-15)

vo 375ð0:02 S þ 1Þ ¼ ei S 0:0000033165065991ððS=50Þ þ 1Þððs=5991Þ þ 1Þ

(14.1-16)

vo 375ð0:02 S þ 1Þ ¼ ei Sð0:02 S þ 1Þð0:000166 S þ 1Þ

(14.1-17)

vo 375 ¼ ei Sðð jw=5991Þ þ 1Þ

(14.1-18)

The Bode frequency response for the motor and current loop is shown in Figure 7. The motor and current closed-loop frequency response indicate that the response is an integration that includes the 6000 rad/sec bandwidth of the current loop. This is a realistic bandwidth for commercial industrial servo drives.

14.2 MOTOR/CURRENT LOOP AND POSITION LOOP Usually this response is enclosed in a velocity loop and further enclosed in a position loop. However, there are some applications where the motor and current loop are enclosed in a position loop. Such an arrangement is shown in Figure 8.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 7

Motor and current loop frequency response.

The forward loop transfer function is as follows: G¼

Kv S2 ðð jw=6000Þ þ 1Þ

(14.2-1)

where: K v ¼ K 2 6375. For most large industrial machine servo drives a position loop kv ¼ 1 ipm/mil or 16.6 rad/sec can be assumed. The frequency response for the position loop is shown in Figure 9. This response is obviously unstable

Fig. 8

Position loop block diagram using motor and current loop only.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 9 Position loop frequency response without an inner velocity servo.

with a minus two slope at the zero gain point. It is also obvious that there are two integrators in series, resulting in an oscillator. If a velocity loop is not used around the motor and current loop, some form of differential function is required to obtain stability. By adding a differential term at about 10 rad/sec in Figure 9, the response can be modified to that of a type 2 control, which could have performance advantages. The absence of the minor velocity servo loop bandwidth could make it possible to increase the position-loop velocity constant (position-loop gain) for an increase in the position loop response.

14.3 MOTOR/CURRENT LOOP WITH A VELOCITY LOOP For the purposes of this discussion it will be assumed that the motor and current loop are enclosed in a velocity servo loop. Such an arrangement is shown in Figure 10.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Velocity loop block diagram.

14.4 PI COMPENSATION The servo compensation and amplifier gain are part of the block identified as K 2 . Most industrial servo drives use PI compensation. The amplifier and PI compensation can be represented as in Figure 11. h i Kp
 K s þ 1 i Kp s þ Ki Ki I Ki K2 ½t2 s þ 1
¼ (14.3-1) ¼ ¼ Kp þ ¼ s s s V2 s Kp t2 ¼ o2 (corner frequency) Ki The adjustment of the PI compensation is suggested as follows: 1. For the uncompensated servo Bode plot, set the amplifier gain to a value just below the level of instability. 2. Note the forward loop frequency ðog Þ at 1358 phase shift (458 phase margin) of Figure 14.

Fig. 11

PI compensation block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

3. From the Bode plot for PI compensation of Figure 12, the corner frequency o2 K i =K p should be approximately og =10 or smaller as a figure of merit. The reason for this is that the attenuation characteristic of the PI controller has a phase lag that is detrimental to the servo phase margin. Thus the corner frequency of the PI compensation should be lowered about one decade or more from the 1358 phase shift point ðog Þ of the open loop Bode plot for the servo drive being compensated. For the servo drive being considered, og occurs at 6000 rad/sec.

Fig. 12

PI frequency response.

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Fig. 13

Velocity servo block diagram with PI compensation.

Applying the PI compensation of Figure 11, to the velocity servo drive is shown in Figure 13. In general the accepted rule for setting the servo compensation begins by removing the integral and/or differential compensation. The proportional gain is then adjusted to a level where the velocity servo response is just

Fig. 14

Velocity loop frequency response without PI compensation—K 2 ¼ 1000.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

stable. The proportional gain is then reduced slightly further for a margin of safety. For a gain K 2 ¼ 1000 of the uncompensated servo, the Bode plot is shown in Figure 14. It should be noted that the motor and current loop have a bandwidth of 6000 rad/sec as shown in Figure 7. This is a normal response for industrial servo-drive current loops. The transient response for this servo is shown in Figure 15 as a damped oscillatory response. If the gain K 2 is reduced to a value of 266 for a forward loop gain of 100,000, the Bode plot is shown in Figure 16 with a stable transient response shown in Figure 17. At this point the PI compensation is added as shown in Figure 13. The index of performance (I.P.) for the PI compensation is that the corner frequency o2 ¼ K i =K p , should be a decade or more lower than the 135 degree phase shift (458 phase margin) frequency ðog Þ of the forward loop Bode plot (Figure 16) for the industrial servo drive being considered.

Fig. 15

Velocity servo transient response for frequency response of Figure 14.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

With reference to Figure 16 and Figure 17 of the stable uncompensated servo, the 135 degree phase shift (458 phase margin) occurs at 6000 rad/sec frequency, which is also the bandwidth of the motor/current loop. Using the I.P. of setting the PI compensation corner frequency at one decade or more lower in frequency that the 1358 phase shift frequency point, the corner frequency should be set at 600 rad/sec or lower. With the corner frequency of the PI compensation set at 600 rad/sec (0.001666 sec) the compensated servo is shown in the Bode plot of Figure 18. The transient response is shown in Figure 19 as a highly oscillatory velocity servo drive. Obviously this servo drive needs to have the PI compensation corner frequency much lower than one decade (600 rad/sec) I.P. For a two decade, 60 rad/sec (0.0166 sec) lower setting for the PI corner frequency the Bode response is shown in Figure 20 with a transient response shown in Figure 21 having a single overshoot in the output of the velocity servo drive. By lowering the PI compensation corner frequency ðo2 ¼ KKpi Þ to 20 rad/sec (0.05 sec), a stable velocity servo drive results. The forward loop

Fig. 16

Velocity loop frequency response without PI compensation—K 2 ¼ 266.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

and open loop are defined as follows: H ¼ 0:0286 v=rad=sec

1=H ¼ 34:9 ð30:8 dBÞ

Gain @ o ¼ 1 rad=sec ¼ 100 dB ¼ 100,000 K 2 ¼ 100,000=375 ¼ 266

G¼

Fig. 17

(14.3-2)

K 2 6375ðð jo=20Þ þ 1Þ 100,000ðð jo=20Þ þ 1Þ ¼ ð100 dBÞ S 2 ðð jo=6000Þ þ 1Þ S 2 ðð jo=6000Þ þ 1Þ (14.3-3)

Velocity servo transient response for frequency response of Figure 16.

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G¼

100,000ð0:05 S þ 1Þ 5000 S þ 100,000 ¼ 2 2 S ð0:000166 S þ 1Þ S ð0:000166 S þ 1Þ

(14.3-4)

G¼

30,120,481 S þ 602,409,638 S3 þ 6024 S 2 þ 0 S þ 0

(14.3-5)

GH ¼ 0:02866G ¼

2860ðð jo=20Þ þ 1Þ ð69 dBÞ S 2 ðð jo=6000Þ þ 1Þ

(14.3-6)

The Bode plot for the velocity loop with PI compensation is shown in Figure 22, having a typical industrial velocity servo bandwidth of 30 Hz (188 rad/ sec). The transient response is stable with a slight overshoot in velocity as shown in Figure 23.

Fig. 18 Velocity loop frequency response with PI compensation—K 2 ¼ 266, T 2 ¼ 0:001666 sec.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 19

Velocity servo transient response of Figure 18.

14.5 POSITION SERVO LOOP COMPENSATION Having compensated the velocity servo, it remains to close the position servo around the velocity servo. Commercial industrial positioning servos do not normally use any form of integral compensation in the position loop. This is referred to as a ‘‘naked’’ position servo loop. However, for type 2 positioning drives, PI compensation would be used in the forward position loop. There are also some I.P. rules for the separation of inner servo loops by their respective bandwidths (see Chapter 9). The first I.P. is known as the three to one rule for the separation of a machine resonance from the inner velocity servo. All industrial machines have some dynamic characteristics, which include a multiplicity of machine resonances. It is usually the lowest mechanical resonance that is considered; and the I.P. is that the inner velocity servo bandwidth should be one-third lower than the predominant machine structural resonance.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 20 Velocity loop frequency response with PI compensation—K 2 ¼ 266, T 2 ¼ 0:01666 sec.

A second I.P. is that the position servo velocity constant ðK v Þ or position loop gain, should be one-half the velocity servo bandwidth. These I.P.s are guides for separating servo loop bandwidths to maintain some phase margin and overall system stability. Industrial machine servo drives usually require low position-loop gains to minimize the possibility of exciting machine resonances. In general for large industrial machines the position-loop gain ðK v Þ is set about 1 ipm/mil (16.66/sec). The example being studied in this discussion has a machine slide weight of 50,000 lbs, which can be considered a large machine that could have detrimental machine dynamics. There are numerous small machine applications where the position-loop gain can be increased several orders of magnitude. The technique of using a low position-loop gain is referred to as the ‘‘soft servo.’’ A low position-loop gain can be detrimental to such things as servo drive stiffness and accuracy. The soft servo technique also requires a highperformance inner velocity servo loop. This inner velocity servo loop with its high-gain forward loop, overcomes the problem of low stiffness. For example, as the machine servo drive encounters a load disturbance the

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 21

Velocity servo transient response of Figure 20.

velocity will instantaneously try to reduce, increasing the velocity servo error. However the high velocity servo forward loop gain will cause the machine axis to drive right through the load disturbance. This action is an inherent part of the drive stiffness. For this discussion it will be assumed that the industrial machine servo drive being considered has a structural mechanical spring/mass resonance inside the position loop. The machine as connected to the velocity servo drive is often referred to as the ‘‘servo plant.’’ The total machine/servo system can be simulated quite accurately to include the various force or torque feedback loops for the total system. For expediency in this discussion, a predominant spring/mass resonance will be added to the output of the velocity servo drive. Thus the total servo system is shown in the block diagram of Figure 24. Position feedback is measured at the machine slide to attain the best position accuracy. As stated previously, the I.P. for the separation of the velocity servo bandwidth and the predominant machine resonance should be three to one;

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where the velocity servo bandwidth is lower than the resonance. The machine resonance is shown if Figure 24 as or . Since the velocity servo bandwidth of this example is 30 Hz (188 rad/sec), as in Figure 22, the lowest machine resonance should be three times higher or 90 Hz (565 rad/sec). It is further assumed that this large machine slide has roller bearing ways with a coefficient of friction ¼ 0.01 lbf/lb, and a damping factor ðd ¼ 0:1Þ. Additionally, this industrial servo-driven machine slide (50,000 lbs) will have a characteristic velocity constant ðK v Þ of 1 ipm/mil (16.66/sec). This large machine was used for this discussion as a worst-case scenario since the large weight aggravates the reflected inertia and machine dynamics problems. Most industrial machines are not of this size. The closed-loop frequency response with a mechanical resonance ðor Þ of 90 Hz (565 rad/sec) is shown in Figure 25. A unity step in position transient response is shown in Figure 26. These are acceptable servo responses for the machine axis being analyzed.

Fig. 22 Velocity loop frequency response with PI compensation—K 2 ¼ 266, T 2 ¼ 0:05 sec.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 23

Velocity servo transient response of Figure 22.

Fig. 24

Position loop block diagram with position gain of 1 ipm/mil.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 25

Position loop frequency response with a machine resonance of 90 Hz.

In reality a machine axis weighing 25 tons will have structural resonances much lower than 90 Hz. Machine axes of this magnitude in size will characteristically have structural resonances of about 10 Hz to 20 Hz. Using the same position servo block diagram of Figure 24 with the same position-loop gain of 1 ipm/mil (16.66/sec), and a machine resonance of 10 Hz; the servo frequency response is shown in Figure 27 with the transient response shown in Figure 28. The position servo frequency response shows a 8 dB resonant (62.8 rad/sec) peak over zero dB, which will certainly be unstable as observed in the transient response. Repeating the position servo analysis with a 20 Hz resonance in the machine structure, results in an oscillatory response as shown in Figures 29 and 30. One of the most significant problems with industrial machines is in the area of machine dynamics. Servo motors and their associated amplifiers have very long mean time before failure characteristics. It is quite common to have an industrial velocity servo drive with 20 Hz to 30 Hz bandwidths mounted on a machine axis having structural dynamics (resonances) near or

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

much lower than the internal velocity servo bandwidth. There must be some control concept to compensate for these situations. There are a number of control techniques that can be applied to compensate for machine structural resonances that are both low in frequency and inside the position servo loop. The first control technique is to lower the position-loop gain (velocity constant). Depending on how low the machine resonance is, the positionloop gain may have to be lowered to about 0.5 ipm/mil (8.33/sec). This solution has been used in numerous industrial positioning servo drives. However, such a solution also degrades servo performance. For very large machines this may not be acceptable. The I.P. that the position-loop gain (velocity constant) should be lower than the velocity servo bandwidth by a factor of two, will be compromised in these circumstances.

Fig. 26

Position loop transient response of Figure 25.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

A very useful control technique to compensate for a machine resonance is the use of wien bridge notch filters. These notch filters (see Chapter 11) are most effective when placed in cascade with the position forward servo loop, such as at the input to the velocity servo drive. These notch filters should have a tunable range from approximately 5 Hz to a couple of decades higher in frequency. The notch filters are effective to compensate for fixed machine structural resonances. If the resonance varies due to such things as load changes, the notch filter will not be effective. There are commercial control suppliers that incorporate digital versions of a notch filter in the control; with a future goal to sense a resonant frequency and tune the notch filter to compensate for it. This control technique can be described as an adaptive process. Another technique that has been very successful with industrial machines having low machine resonances, is known as ‘‘frequency selective

Fig. 27

Position loop frequency response with a machine resonance of 10 Hz.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 28

Position servo transient response of Figure 27.

feedback. This control technique is the subject of Section 4 in Chapter 11. In abbreviated form it requires that the position feedback be located at the servo motor eliminating the mechanical resonances from the position servo loop, resulting in a stable servo drive but with significant position errors. These position errors are compensated for by measuring the machine slide position through a low pass filter; taking the position difference between the servo motor position and the machine slide position; and making a correction to the position loop, which is primarily closed at the servo motor.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 29

Position loop frequency response with a machine resonance of 20 Hz.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 30

Position servo transient response of Figure 29.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

15 Machine Simulation

15.1 INTRODUCTION In the world market for industrial machinery there is a requirement that machines must work with a minimum of down time, especially for machine cells with two or more machines. In the manufacture of these machines there are two important requirements. First, the machine drive must be sized correctly to provide the required torques and thrust. Second, the machine feed servo drive must provide the required accuracy and be stable. To meet these requirements a predictive technique should be used before the machine is manufactured to avoid costly retrofits of the servo drive and or the machine itself. These analysis techniques and requirements are shown in Figure 1.

15.2 SIMULATION FUNDAMENTALS Classical control system analysis involves the use of differential equations to describe a physical system. Most physical systems include some combination of mechanical, electrical, and hydraulic components. Once the control system is described by differential equations, it is desired to solve them for some control variable (output) in response to a desired input function such as a step input, ramp input, or other function. In addition, the control

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 1

Trend in world markets for machines.

system may have some initial conditions prior to initiating the input. Including the initial conditions in a differential equation requires an expansion of these equations to include all the required integrals and differentiations of these conditions. Likewise, any nonlinearities such as transport lags, current limits, etc. must be included. Describing functions are often used to represent nonlinear conditions. In summary, systems of linear and nonlinear differential equations are tedious to solve. The use of a shorthand technique to solve control system differential equations is well known as the Laplace transformation in the s plane. These techniques have been used successfully for the solution of analog-type control systems. In the past two decades there has been a transition from analog control systems to all digital controls or some combination of analog and digital controls. These control systems are better described using the Z transform. There are numerous texts on the subject of control systems and digital analysis. With the advent of total computer control, it is possible to use iterative techniques for the solution of control systems described by Laplace transforms. The technique of applying these iterative techniques to a control system has been described as ‘‘simulation.’’ These techniques are discussed to show how this technology can be applied to the control/

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

machine/servo drive system. The result is general simulation programs that represent the individual servo components of a machine system.

Simulation and the Laplace Transform Control systems can be represented by a combination of transformed differential equations in block diagram form. Simulation of a control system then consists of taking the individual blocks, converting the transform into integral form, and letting a computer sum the area under the curve of the equation by rectangular integration. A simple example will suffice to demonstrate the simulation process. Assume the electrical circuit in Figure 2. Applying Kirchoff’s law: S voltage ¼ 0   1 ei ¼ R e þ i Ca s

(15.2-1)

s ¼ Laplace operator Equation (15.2-1) is written directly in Laplace notation. Also: eo ¼ i

1 sCa

(15.2-2)

Equation 14.2-1 can be rewritten as i ¼ ei

sCa Re Ca s þ 1

(15.2-3)

Combining Eq. (15.2-2) and (15.2-3) yields eo ¼

ei Re Ca s þ 1

Fig. 2 Electric R/C network.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(15.2-4)

Let t ¼ Re Ca : Then eo ¼

ei ts þ 1

(15.2-5)

A computer simulation can be made using state variables. The previously mentioned case of simple lag can be represented by the Laplace block diagram in Figure 3, where eo 1 1=t ¼ ¼ V ts þ 1 s½1 þ ð1=tsÞ

(15.2-6)

Then changing Eq. (15.2-6) to integral form eo ½ð1=tÞs
1 X ¼ V ½1 þ ðs
1 =tÞX

(15.2-7)

V ¼ Xð1 þ s
1 =tÞ

(15.2-8)

and


1

eo ¼ Xðs =tÞ

(15.2-9)

Solving for X X ¼ V
Xðs
1 =tÞ

(15.2-10)

Writing the state variable diagram: V

þ1

Q1 X

s
1
1=t

Q1

þ ð1=tÞ

eo

Converting to block diagram form in Figure 4:

Fig. 3

Block diagram.

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Fig. 4

State variable block diagram.

Fig. 5 State variable block diagram.

Fig. 6 State variable block diagram.

Fig. 7 State variable block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Where Q_ 1 ¼ V
ð1=tÞQ1 Q1 ¼ Old Q1 þ Q_ 1 times the computer iteration rate

(15.2-11)

Q1 ¼ Q1 þ Q_ 1 6Delt ðthe sampling rateÞ

(15.2-12)

or

eo ¼ ð1=tÞQ1

(15.2-13)

Combining Eq. (15.2-11), (15.2-12) and (15.2-13) yields eo ¼ eo ð1
Delt=tÞ þ V6Delt=t

(15.2-14)

To use Eq. (15.2-11), (15.2-12) and (15.2-13) in a computer solution, it is necessary to use an interactive loop with a small sample time (Delt) repeated many times. The preceding example is included in the example program:

The preceding example illustrates the simulation procedure for a simple R-C circuit, such as that in Figure 2, represented by a Laplace transform. Servo drives and controls can be represented by block diagrams

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

using Laplace transforms. Each block in the diagram represents a component of the servo. For example, motors, and amplifiers can be represented by individual blocks. The representative transfer functions in these blocks are usually first- or second-order transfer functions. Therefore it is only necessary to know the simulation for these individual ‘‘blocks’’ and then combine the simulations for the total system. As an example, the derivation for the simulation equations representing a single order lag transfer function ½K=ðts þ 1Þ and a second order transfer function ½K=ðS 2 þ b1 s þ b2 Þ are presented, followed by simulation examples for various commonly encountered transfer functions for servo drive components.

Single Order Transfer Function (Figure 5) C K K=t ¼ ¼ R ts þ 1 s þ 1=t

(15.2-15)

In integral form C ððK=tÞs
1 ÞX ¼ R ð1 þ s
1 =tÞX

(15.2-16)

R ¼ Xð1 þ ð1=tÞs
1 Þð1Þ

(15.2-17)

C ¼ XððK=tÞs
1 Þð2Þ

(15.2-18)

Solving for X X ¼ R
ðX=tÞ6s
1

(15.2-19)

Writing the state variable diagram R

þ1

Q_ 1

1=s Q1 1=t

ðK=tÞ

C

Writing the equivalent block diagram (Figure 6) Q_ 1 ¼ R
Q1 =t Q1 ¼ Q1 þ Q_ 1 6Delt C ¼ ðK=tÞ6Q1

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(15.2-20) (15.2-21) (15.2-22)

Second Order Transfer Function K C
? þ b1 s þ b2 C K ¼ R s 2 þ b1 s þ b2 R
?

s2

(15.2-23)

In integral form: C¼

ðK6s
2 ÞX ð1 þ b1 6s
1 þ b2 6s
2 ÞX

R ¼ X þ Xb1 6s
1 þ X6b2 6s
2 C ¼ K6X6s


2

(15.2-24) (15.2-25) (15.2-26)

Solving for X X ¼ R
X6b1 6s
1
X6b2 6s
2

(15.2-27)

Writing the state variable diagram R þ 1 Q2 X s
1 Q2 Q1 s
1 Q1 þ K C
b1
b2

Writing the equivalent block diagram (Figure 7) Q_ 1 ¼ Q2

(15.2-28)

Q1 ¼ Q1 þ Q_ 1 6Delt Q_ 2 ¼
b2 6Q1
b1 6Q2 þ R

(15.2-29)

Q2 ¼ Q2 þ Q_ 2 6Delt C ¼ Q1 6K

(15.2-31) (15.2-32)

(15.2-30)

Summary of Simulation Examples To apply this technique requires a library of the most common types of transfer functions found in control system blocks. A summary of commonly used block transfer functions with the associated simulation equations follows:

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

1. Integration KC
? s Q_ 1 ¼ R

(15.2-33)

Q1 ¼ Q1 þ Q_ 1 6Delt C ¼ K6Q1

(15.2-34) (15.2-35)

R
?

2. Differential s C
? K C ¼ ðR
OldRÞ=ðK6DeltÞ

R
?

(15.2-36)

3. Single Order Lag K C
? ts þ 1 C ¼ C6ð1
Delt=tÞ þ K6R6Delt=t K C R
?
? sþW Q_ 1 ¼ R
W6Q1 Q1 ¼ Q1 þ Q_ 1 6Delt R
?

C ¼ K6Q1 a1 C
?
? b1 s þ b2 Q_ 1 ¼ R
ðb2 =b1 Þ6Q1 Q1 ¼ Q1 þ Q1 6Delt

(15.2-37)

(15.2-38) (15.2-39) (15.2-40)

R

C ¼ ða1 =b1 Þ6Q1

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(15.2-41) (15.2-42) (15.2-43)

4. Second Order Simulations a1 s þ a2 C
? b1 s 2 þ b2 s þ b 3 Q_ 2 ¼ R
ðb2 =b1 Þ6Q2
ðb3 =b1 Þ6Q1 Q2 ¼ Q2 þ Q_ 2 6Delt Q_ 1 ¼ Q2 R
?

(15.2-44) (15.2-45) (15.2-46)

Q1 ¼ Q1 þ Q_ 1 6Delt

(15.2-47)

C ¼ ða2 =b1 Þ6Q1 þ ða1 =b1 Þ6Q2 a1 s þ a2 C R
? 2
? s þ b1 s þ b2 Q_ 1 ¼ Q2 Q1 ¼ Q1 þ Q_ 1 6Delt

(15.2-48)

Q2 ¼
b2 6Q1
b1 6Q2 þ R Q2 ¼ Q2 þ Q_ 2 6Delt

(15.2-49) (15.2-50) (15.2-51) (15.2-52)

C ¼ a2 6Q1 þ a1 6Q2 K R C
? 2
? S þ b1 s þ b2 Q_ 2 ¼ R
b1 6Q2
b2 6Q1 Q2 ¼ Q2 þ Q_ 2 6Delt

(15.2-55)

Q_ 1 ¼ Q2

(15.2-56)

Q1 ¼ Q1 þ Q_ 1 6Delt

(15.2-57)

C ¼ K6Q1 Ks R C
? 2
? s þ b1 s þ b2 Q_ 2 ¼ R
b1 6Q2
b2 6Q1 Q2 ¼ Q2 þ Q_ 2 6Delt

(15.2-58)

(15.2-60)

Q_ 1 ¼ Q2

(15.2-61)

Q1 ¼ Q1 þ Q_ 1 6Delt C ¼ K6Q2

(15.2-62) (15.2-63)

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(15.2-53)

(15.2-54)

(15.2-59)

5. Third Order Simulation a1 s 2 þ a2 s þ a3 C
? þ b1 s 2 þ b2 s þ b3 Q_ 3 ¼ R
b1 6Q3
b2 6Q2
b3 6Q1 Q3 ¼ Q3 þ Q_ 3 6Delt Q_ 2 ¼ Q3 R
?

s3

Q2 ¼ Q2 þ Q_ 2 6Delt Q_ 1 ¼ Q2 Q1 ¼ Q1 þ Q_ 1 6Delt C ¼ a3 6Q1 þ a2 6Q2 þ a1 6Q3

(15.2-64) (15.2-65) (15.2-66) (15.2-67) (15.2-68) (15.2-69) (15.2-70)

This summary of transfer function simulations are useful for application to control system problems.

15.3 MACHINE SIMULATION TECHNIQUES TO PREDICT PERFORMANCE* In the application of feed servo drives to a machine, in addition to sizing the drive to make sure there is enough torque available to satisfy the load requirements, it is also desirable, as a final step in the design of either a hydraulic or electric servo drive, to predict the servo-drive performance (stability, response time, etc.). A useful analytical tool for servo analysis is the ‘‘transient response.’’ This type of analysis is very practical since it closely relates to what an observer would see on a machine if a step input in position or velocity were used. By using servo ‘‘simulation’’ techniques, it is possible to construct a very accurate model of the servo drive and machine. It is possible to observe the effect on performance by changing the drive parameters such as load inertia, backlash, stiction (stick-slip), viscous friction, load thrust, current limit, servo loop gains, machine resonance, and acceleration. Assuming the model is correct, the simulation program allows the machine designer to observe the feed drive performance without having the drive and machine available. This simulation procedure makes it possible to anticipate machine design problems affecting servo performance and correct them before the machine is built.

* Reprinted, with permission, from IEEE-IAS Transactions; Vol. 27, No. 2, pp. 268–274, March/April 1991.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

In the design and synthesis of machine feed servo drives, it is desirable to predict the effects of machine nonlinearities in the machine system on servo performance. The machine system is often referred to as the servo plant. If the effects of these nonlinearities can be minimized or eliminated prior to manufacturing the machine, the cost of redesign and added testing can be avoided. Some machine nonlinearities that are commonly encountered are backlash and stiction. The analytical techniques of classical servo analysis include the use of transformation calculus. To include these machine nonlinearities means some techniques such as ‘‘describing functions’’ must be used. At best, these describing functions approximate a method to linearize a nonlinear function so that they can be used with linear transform analysis. It is possible to use computer iterative techniques in analyzing and synthesizing servo drives. Thus, nonlinearities can be treated much more realistically. Applying these iterative computer techniques to a closed-loop servo drive is referred to as simulating the servo drive/control/machine. It is the purpose of this discussion to show how simulation can be used to accurately represent and predict performance of a machine feed servo drive/ control/machine combination, specifically, performance due to nonlinear effects. Before any analytical techniques can be discussed, it is necessary to establish a servo plant (model) of the machine feed axis. For purpose of this discussion, the servo plant will include one predominant mechanical spring mass system that has a known resonance, a load inertia, viscous damping, backlash, and stiction (stick-slip). The block diagram for this machine tool servo feed axis is shown in Figure 8. The free-body diagram for this machine slide is shown in Figure 9. In Figure 9: Bv ¼ viscous friction coefficient of the machine slide, lbf-min/in. Fa ¼ driving force, developed by the servo motor, applied to the slide, lbf Fb ¼ viscous friction force, lbf Fs ¼ stiction force on the machine slide, lbf Fk ¼ spring force between Mm and Ml , lbf Fl ¼ load force of the machine slide (cutting force), lbf K ¼ mechanical stiffness of the mechanical drive elements, lb/in. Mm ¼ mass of the driving component (motor), lb-in.-sec2 Ml ¼ mass of the load (slide plus any workpiece), lb-in.-sec2 N ¼ ratio s ¼ Laplace operator Vl ¼ velocity of the slide and load, in./sec

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 8

Block diagram of a machine feed drive.

Vm ¼ velocity of the driving component (motor), rad/sec Xm ¼ position of the driving component, in. Xl ¼ position of the slide and load, in. DELX ¼ difference between the load and the driving component positions

Fig. 9 Free body diagram of a machine slide.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The two nonlinearities included in this servo plant are backlash and stiction. The backlash nonlinearity, when combined with the servo plant spring, has a characteristic shown in Figure 10, where: Ks ¼ soft spring constant Kh ¼ hard spring constant The model used for the stiction nonlinearity has the characteristic shown in Figure 11. For a machine tool slide that is not moving, the friction is static ðFs Þ. As soon as the machine slide breaks away (starts to move), the friction drops to a lower Coulomb friction ðFc Þ. These friction forces are not dependent on velocity but are sign dependent on the direction of motion. In Figure 11, Fs ¼ static friction Fc ¼ Coulomb friction From Newton’s second law of motion, the classical equations for this servo plant (machine system) are: Ml s2 Xl ¼
Fb
Fk
Fl Fb ¼ running friction force ¼ BsXl

(15.3-1) (15.3-2)

Fk ¼ spring force ¼ KðXl
Xm Þ ¼ KðVl
Vm Þ=s (without backlash) (15.3-3) Combining Eq. (15.3-1), (15.3-2), (15.3-3) and rearranging: Xl ¼

KXm
Fl ðMl s2 þ Bv s þ KÞ

(15.3-4)

Also from Eq. (15.3-1),
Fb
Fl
Fk Ml s Fa þ Fk Vm ¼ Mm s DELX ¼ ðXl
Xm Þ ¼ ðVl
Vm Þ=s

Vl ¼

(15.3-5) (15.3-6) (15.3-7)

Simulating the Servo Plant Prior to writing simulation equations, it should be noted that all spring constants, inertias (rotary case), etc. are referred to the motor. Since the

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 10

Backlash nonlinearity.

Fig. 11

Stiction nonlinearity.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

stiction force is a nonlinear parameter, it was not included in Eq. (15.3-1) to (15.3-7), but is part of the simulation model. It is assumed that the Coulomb friction will be about 60 lb-in., which is 20% of the rated motor torque, and the static friction is some factor larger than the Coulomb friction. The ratio of static friction to Coulomb friction is an input variable. All scalings of the parameters in the servo simulation are included in a definition of constants as part of the dimensionless simulation program. The equivalent simulation equations from Eq. (15.3-1) to (15.3-7) for this simulation servo plant follow: Vm ¼ Vm þ ðFa þ Fk Þ6DELT=Jm

(15.3-8)

For the nonlinearity of stiction: IF ABS ðV1 Þ < Vo THEN IF ½ABSðFk þ F1 þ Fb Þ
ABSðFs Þ < 0 THEN V1 ¼ V1
½Fk þ F1 þ Fb
SGNðFk þ F1 þ Fb Þ6Fc  6DELT=J1 END IF ELSE V1 ¼ V1
½Fk þ F1 þ Fb þ SGNðV1 Þ6Fc 6DELT=J1

(15.3-9)

(15.3-10)

END IF From the backlash relationship of Figure 10, DELX ¼ DELX þ ðV1
Vm Þ6DELT

(15.3-11)

IF ½ABSðDELXÞ
BLASH > 0 THEN Fk ¼ Kh 6DELX IF ½ABSðDELXÞ
BLASH40 THEN Fk ¼ Ks 6DELX

(15.3-12) (15.3-13)

where DELX is the numerical integration time interval, Vo is close to zero velocity, and BLASH is half the total backlash. A block diagram representing the simulation for the servo plant is shown in Figure 12.

Simulating the Complete Servo Drive Having established a servo plant (machine system), the next step is to model the rest of the machine feed servo drive. It is assumed that this drive will be a commercially available electric pulse-width-modulated (PWM) or a brushless DC servo drive. Further, it is assumed that an outer positioning servo loop will be a commercial numerical controller. The complete machine contouring feed servo drive/machine is shown in Figure 13. It will be

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

assumed that the electric servo drive has a current loop and a velocity loop. The drive also has an adjustable current limit and uses proportional plus integral compensation. The numerical controller (not shown in detail) has an adjustable velocity constant (gain), acceleration control, and feedforward control. The additional definitions of terms in Figure 13 are: ILIM ¼ Kp ¼ Ki ¼ R1 ¼ T1 ¼ W¼

current limit velocity loop proportional gain velocity loop integral gain static regulation of the velocity loop velocity loop error amplifier filter time constant current loop bandwidth

The machine system (servo plant) selected for this simulation is a typical feed axis. The mechanical components and servo parameters of the axis drive have the following values: . Ball screw drive Ball screw length ¼ 95 in. Ball screw diameter ¼ 3 in. Ball screw lead ¼ 0.375 in./rev

Fig. 12

Block diagram of the servo plant.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 13

Machine-feed servo-drive block diagram.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

. Ratio ¼ 3.3 . Axis weight ¼ 18,000 lb (with workpiece) . Inertias Total reflected load inertia to the servo motor ¼ 0.161 lb-in.-sec2 Servo motor inertia ¼ 0.22 lb-in.-sec2 Servo motor rated torque ¼ 310 lb-in. Servo motor rated speed ¼ 3,000 rpm Machine system predominant resonant frequency ¼ 40 [Hz] Approximate machine slide Coulomb friction ¼ 20% of the motor rating (62 lb-in.) . Velocity loop gain set at 30% of maximum . Inductosyn feedback from the machine slide . . . .

Performing the Simulation As previously stated, the transient response for a step input to the servo positioning drive is a practical method of analysis that can be readily used on any machine. Two types of transient response will be used for the purpose of this discussion. First, an input step in velocity is used for the model servo/control/machine system selected. The acceleration is controlled by the position loop. For this model the numerical controller accelerates the velocity with a second-order algorithm. In addition, it is assumed that the cutter reaction forces are negligible. The second type of transient response uses a step input of position to observe the effects of a stiction nonlinearity in the mechanical system of the servo plant. The step in position input is very small in magnitude to observe the conditions necessary to cause the control/servo system to self-sustain a null-hunt (limit-cycle) instability.

Response to a Step Input in Velocity These simulations for the model selected have an input velocity command step to 1000 rpm at the motor or 112 ipm at the machine slide. For these simulations backlash, stiction, and viscous friction will be the variables. Backlash is in inches. Stiction is represented as the ratio of static friction to Coulomb friction. Viscous friction, or running friction as it is sometimes called, is reflected to the servo motor as a torque (lb-in.). The simulations to be made are summarized in the following table:

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Backlash (in.)

Viscous friction (lb-in.)

Ratio Fs =Fc

Figure

0 0.001 0.001 0.001 0.002

0 0 20 30 30

1.0 1.0 1.0 1.0 1.0

15.14 15.15 15.16 15.17 15.18

The first simulation of Figure 14 represents an ideal situation where there is no backlash, stiction, or viscous friction in the feed axis contouring servo drive. As expected, the servo transient response is stable. Positioning system feedforward control following error reaches a maximum of about 0.0015 in. The position loop following error in all the simulations has been amplified 500 times so it can be observed. In simulating a typical machine feed drive, it is important that the resulting performance of the model represent what actually happens on a machine. Machine axis feed drives using antifriction way systems that are

Fig. 14

Velocity simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

underdamped and have a small amount of backlash will be unstable. A typical example is a machine feed drive using roller bearing ways that have minimal damping and a ball screw that has approximately 0.001–0.002 in backlash in the ball nut. Typically this is an unacceptable condition and will result in an unstable axis positioning drive. The simulation for this situation is shown in Figure 15. If the same machine used a non–stick-slip friction way liner material such as Rulon, the viscous friction damping would increase. For this condition the axis would be stable for a backlash of 0.001 in. Figures 16 and 17 show that as damping increases to 30 lb-in. (1100 lbf slide friction) with 0.001 in backlash, the drive becomes stable. However, with 0.002 in. backlash (Figure 18) this same 30 lb-in. viscous friction does not eliminate the instability. Therefore, as known from actual practice, increasing the slide friction adds damping, which can stabilize a machine slide with a small amount of backlash. This fact is verified by the simulation of Figures 16 to 18.

Fig. 15

Velocity simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Response to a Step Input in Position One of the requirements of a machine axis feed positioning servo drive is that it must be stable under all operating conditions. One form of instability is referred to as null hunting (limit cycle), characterized by a low-frequency period of oscillation (approximately 0.5–2 Hz) and low amplitude (up to several thousandths of an inch slide motion). Null hunting is caused by a combination of nonlinear friction (stiction) with a spring and mass. Null hunting in a contouring axis can be minimized by avoiding the use of way liner materials such as Formica, Micarta, brass, or other metals. One of the goals of this section is to illustrate how the nonlinearity of stiction can be simulated in a model of a machine and compared to actual practice. To make this simulation, a disturbance was used as an input to the positioning servo drive to observe under what conditions a self-sustained null-hunt instability would occur. This input disturbance was a small position step input (0.01 in.) to the positioning drive. The simulations made are summarized in the following table:

Fig. 16

Velocity simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Backlash (in.)

Viscous friction (lb-in.)

Stiction ratio Fs =Fc

Coulomb friction (lb-in.)

Figure

0 0.0005 0.0005 0 0 0 0 0

0 0 150 35 35 35 35 35

1.0 1.0 1.0 1.5 1.9 2.0 1.5 2.0

60 60 60 150 150 150 60 120

14.19 14.20 14.21 14.22 14.23 14.24 14.25 14.26

As an ideal situation, the first simulation for a position step input was made for no backlash, stiction, or viscous friction damping. The expected stable transient response is shown in Figure 19.

Fig. 17

Velocity simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 18

Velocity simulation.

Fig. 19

Position simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Machine axis contouring servo drives will be unstable for small amounts of backlash inside the servo loop. As shown in Figure 15, the servo drive was unstable with backlash inside the servo loop for a step input in velocity. As expected, a self-sustained position oscillation will occur (Figure 20) for a 0.01-in. step in position having a small amount (0.0005 in.) of backlash inside the servo loop with no damping due to friction. The resulting oscillation in position has an amplitude equal to the amount of backlash as expected. By adding a large amount of viscous friction (150 lbin. at the motor), there is a noticeable reduction in the velocity oscillation in Figure 21. However, the self-sustained position oscillation is still evident. From actual practice, a small amount of backlash in the position loop will result in an unstable position loop for small position commands or load disturbances. The position loop oscillation may be self-sustained or it may decay, depending on the amount of friction damping. The next simulations were made for various amounts of stiction. There are two parameters to consider for the stiction nonlinearity. The first consideration is the ratio of static friction to the Coulomb friction, and second, the amount of Coulomb friction. Additionally, it is the combination of the stiction nonlinearity with the mass and spring of the machine system or servo plant that results in a null-hunt instability. For the model being used in these simulations having an 18,000-lb load with a mechanical resonance of 40 Hz, a position-loop null hunt did not occur for stiction ratios of 1.5 (Figure 22) or 1.9 (Figure 23). When a null hunt did occur, the stiction ratio was 2.0 and the Coulomb friction was 50% of the motor rating. This null hunt is shown in Figure 24 and has a period of oscillation of about 1.9 Hz. It was possible to modify the machine model and find null-hunt instabilities for other combinations of stiction ratio, amount of Coulomb friction, levels of load inertia, and mechanical system resonances. A different set of conditions that caused a null hunt to take place was for a machine system resonance lowered from 40 to 15 Hz. For a self-sustained null hunt, the Coulomb friction requirement was 20% of the motor rating (62 lb-in.) and a stiction ratio of 1.5. The resulting null hunt of approximately 2.27 Hz is shown in Figure 25. Still another set of conditions to cause a self-sustained null hunt is shown in Figure 26. For this situation, the load inertia reflected to the servo motor was increased from 0.167 to 0.3 lb-in.-sec2, which is larger than the motor inertia (0.22 lb-in.-sec2). Null hunting occurred for a Coulomb friction of 40% of the motor rating (124 lbin.), a machine system resonance of 30 Hz, and a stiction ratio of 2.0. There is an unlimited number of combinations of system parameters that could cause a null hunt to take place. Three sets of conditions were used to demonstrate the interdependence of the machine system resonance, load

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 20

Position simulation.

Fig. 21

Position simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

inertia, Coulomb friction, and the stiction ratio. In actual practice it is very difficult to eliminate the instability of a null hunt. Some solutions to eliminate null hunting include the use of dither in the servo drive. The addition of a vibration (dither) in the drive has been less than satisfactory and could cause a fretting problem on machine way systems. The only reliable solution to prevent null hunting is to avoid the use of stick-slip way liner materials. It has become standard in machine designs to either use antifriction ways on axis slides or use anti–stick-slip way liner materials. It is also desirable to maintain low reflected inertias, and high machine resonances if possible. The foregoing simulation examples serve to illustrate the capabilities of simulating the total system in predicting performance. Some actual case histories reinforce the concept of simulating machine systems in predicting performance before a machine is manufactured. The first example is a machine slide simulation. It has the specifications shown in Figure 27. A position step input of 2 in. was used in the machine simulation and the actual machine slide. In this particular performance study, there was not enough motor torque to accelerate the

Fig. 22

Position simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 23

Position simulation.

Fig. 24

Position simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 25

Position simulation.

inertia load as desired. Under these conditions, the servo drive will go into current limit as the velocity increases. During the period that the drive is in current limit, the velocity will increase at a constant rate. A recording of the velocity and current for the actual slide for a 2-in. step response is shown in Figure 28. The drive is in current limit (200% of rated current) until the rated feed of 400 ipm is reached, at which time the current limit reverses to a limit in the opposite direction. For comparison, a machine simulation is shown in Figure 29. The vertical scales are not the same in Figures 28 and 29. The simulation shows a response the same as the actual case of Figure 28. Both the actual and simulated response are in current limit for about 0.3 sec. This is a common occurrence with large machine slides where the available motor torque is not sufficient to accelerate the machine slide inertia at the desired rate of acceleration. The drive does not have enough torque to inertia ratio for the machine load inertia. For these cases a drive with a larger torque rating should be used or the desired acceleration should be reduced.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 26

Position simulation.

A second example, for a normal machine slide step response of 200 ipm, has the specification shown in Figure 30. The actual step response in velocity is shown in Figure 31. The simulated step response is shown in Figure 32. Both the actual velocity response and the simulated response accelerate smoothly to 200 ipm in about 0.1 sec. The simulated response was commanded to decelerate at 0.1 sec. Therefore the velocity of the simulated response did not quite reach the 200 ipm (3.333 ips) target velocity. The actual current response just happened to be recorded inverted in Figure 31. The actual peak current was 100% rated (45 A) in both the actual and simulated current responses.

15.4 OTHER SIMULATION SOFTWARE The simulation techniques presented in Section 15.3 are narrow in scope since they relate specifically to a servo driven machine slide. There are commercial software simulation programs available such as Simulink (MathWorks, Inc.) and VisSim (Visual Solution, Inc.).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

The VisSim simulation and Simulink software use a Microsoft Windows environment including icons for function blocks in a block diagram. These products can address many applications, some of which are: . . . . . . . . . .

Fig. 27

AC/DC motor design Biochemical processes Process control Motion control Control systems Drive systems Motion control Robotics Nonlinear mechanics And many other applications

Example simulation example.

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Fig. 28

Machine slide transient response.

Fig. 29

Machine simulation.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 30

Example simulation specifications.

Use of these simulation programs does require Windows 3.0þ, at least 2 MB memory, and 1 MB free hard-disk space. The user of these simulation programs should be experienced in the use of Windows software and have an engineering background. As an example to illustrate the use of the VisSim software, a DC motor block diagram simulation is shown in Figure 33. The motor has the following specifications: Motor: Gettys 30-1/M137 DC motor Continuous rated torque ¼ 299 lb-in. Inertia ¼ 0.437 lb-in.-sec2 Ke ¼ 0:56 V-sec/rad

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Fig. 31

Machine slide transient response.

Fig. 32

Machine simulation.

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Fig. 33

DC motor simulation.

KT ¼ 4:99 lb-in./A Ra ¼ 0:08 ohm La ¼ 0:0003 henry The simulation of Figure 33 is for a step input of 100 V. The gain of the motor is 1=Ke or 1.7857 rad-V/sec. Thus a 100-V step input will result in a velocity of 178.57 rad/sec or 1706 rpm.

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16 Conclusion

Part II started in Chapter 7 with a review of the mathematical tools needed to discuss the components of servo drives as in Part I, but from a mathematical point of view. To know how to apply the components of servo drives, it is necessary to understand both the hardware and mathematical description of these building blocks. Continuing in the next logical step, Chapter 8 showed how to put these components (blocks) together to make a servo drive. The tools of block diagram algebra were reviewed. The analysis tools of Bode plots and servo analysis were discussed. This section on the review of control theory concluded with the practical application of servo drive compensation. Having discussed the components of industrial servo drives and how they are put together into a workable drive, the next logical step was to discuss indexes of performance (I.P.s) (Chapter 9) for electric and hydraulic servo drives used in industry. Chapter 10 offered some practical methods for performance criteria. The logic then moved to the area of machinery with a discussion in Chapter 11 of some practical methods to overcome common machine problems with nonlinear circuits. This discussion continued in Chapter 12 with numerous real-world machine considerations. Chapter 13 dealt with the real-world application of commercial drives to machinery. ‘‘Drive sizing’’ was explained with manual forms as an educational tool in understanding what the procedure is. It was stated that many vendors supply software

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

sizing for the asking. Since drive sizing only assures that the drive selected is suitable to provide the required torques, a method must be selected to assure that the drive will have the required performance and stability. Chapter 14 addresses servo compensation for an industrial machine. An example is given for a large-machine servo-driven linear axis. The techniques for using proportional and integral (PI) compensation are addressed for the motor/ current loop, velocity servo loop, and position servo loop. The effect of servo plant dynamics on the position servo loop is discussed. Classic analysis techniques such as phase plane, root locus, etc. have given way to computer simulation techniques. Chapter 15 discussed a simulation modeling technique to analyze the performance capability of the servo drive. The machine simulation was narrow in scope in that it was dedicated to machine slide servo drive analysis. Chapter 15 also mentioned some commercial software programs available in the Windows environment.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Appendix Hydraulic Drive and Electric Drive-Sizing Forms

Hydraulic Drive Sizing , Axis , Traverse rate (ipm) Machine Ball screw: (in.), Diameter (in.), Lead (in./rev) Length , Ratio Gearbox # Slide weight (lbs) Inertia calculations: Slide inertia at motor: Jslide ¼ WTðLead/RatioÞ2 66:5610
5

(lb-in.-sec)2

Ball screw inertia at motor: Jscr ¼ DIA4 6LGTH=ðRatio2 Þ67:2610
5 Coupling inertia ¼ Motor pulley inertia: Pulley number ¼ Pulley inertia ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

(lb-in.-sec2)

(lb-in.-sec2 Þ

Screw pulley inertia: Pulley number ¼ Pulley inertia ¼ Pulley inertias reflected to motor: Jp ¼ Jmot: pul: þ Jscrew pul: =Ratio2 Jp ¼

þ

2

=

(lb-in.-sec2 Þ

¼

Total reflected inertia: Jref ¼ Jslide þ Jscr þ Jcoup þ Jp Jref ¼

þ

þ

þ

¼

(lb-in.-sec2 Þ

Total inertia: Jtot ¼ Jmot þ Jref ¼

þ

¼

Coefficient of friction (0.12 for Rulon) ¼ or hydrostatics)

(lb-in.-sec2 Þ (lbf/lb) (0.01 for rollers

Maximum motor speed ¼ Traverse ðipmÞ6 Gear ratio ¼ VT (rpm) 1 VT ¼ 6 6 ¼

1 6 Lead

(rpm)

Drive motor chosen Displacement DM ¼ (in3 =rev)61=6:28 ¼ (in.-lb) at 100 psi Torque constant KT ¼ Torque available (in.-lbs) at 2000 psi Torque available (in.-lbs) at 3000 psi

(in:3 =rad)

1. Establish the maximum thrust required ¼ FM ¼ 2. Establish the no-load friction force ¼ FF . FF ¼ Coefficient of friction6slide weight FF ¼

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6

¼

(lbf)

(lbf).

No-load friction torque ¼ TF TF ¼ no-load friction force ðFF Þ6 TF ¼

6

6:28

6

1

¼

lead 1 6 (lb-in.) 6:28 ratio

(lb-in.)

3. Stall thrust ¼ FS ¼ 1:56maximum thrust ðFM Þ (lbf). FS ¼ 1:56

¼

(lbf)

4. Stall thrust for efficiency ¼ FST ¼ FST ¼

0:9

¼

stall thrust ðFS Þ (lbf). 0:90 eff

(lbf)

5. Maximum motor torque Max thrust ðFM Þ lead 6 0:9 eff 6:28 1 (lb-in.) 6 ratio

¼ TM ¼

1 6 ¼ (lb-in.) (torque needed) 6:28 (in./revÞ Ratio ¼ Lead ¼ Torque available (at 2000 psi) ¼ (lb-in.) TM ¼

6

6. Drive acceleration. To avoid excessive acceleration forces, the servo drive position loop must control the acceleration. There are three methods that have been used to control acceleration: velocity ramp, velocity exponential, velocity ‘‘S’’ curve. The velocity exponential acceleration has a maximum acceleration (sometimes referred to as ‘‘jerk’’) at time zero from: Acceleration ¼ Kv 6max velocity6e
Kv t Max velocity ¼ VT (rpm)

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Kv ¼ position-loop gain at t ¼ 0 6:28 6Kv (1=sec)6VT (rpm) Accelerationmax ¼ 60 ¼ (rad=sec2 ) 6:28 Accelerationmax ¼ 6 ¼ (rad=sec2 ) 6 60 This is a ‘‘worst-case’’ condition for acceleration. To avoid an initial maximum acceleration, the ‘‘S’’ curve velocity control can be used. The initial acceleration is zero. The maximum acceleration occurs when the rate of change in velocity is a maximum. The acceleration for the ‘‘S’’ curve is Acceleration ¼

Traverse feed (ipm)6C 2 6t6e
Ct (in:=sec2 ) 60

where C is a factor determining acceleration rate. Values of C for desired acceleration rates (in./sec2) are: Acceleration (in./sec2)

C

9 18 27 36

5 10 15 20 The maximum acceleration occurs at t ¼ 1=C. Final velocity ðVT Þ Ratio 6C 2 6t6e
Ct 66:286 (rad=sec2 ) 60 Lead Accelmax ¼ 6 6 6eð
6 Þ 66:286 ¼ (rad=sec2 ) 60 Ratio ¼ Lead ¼ (in:=rev) Accelmax ¼

7. Torque at maximum acceleration. TA ¼ JT 6accelerationmax Motor inertia ¼ (lb-in.-sec2) (lb-in.-sec2) Reflected load inertia ¼ JT ¼ total inertia at motor ¼ Jmotor þ Jload (lb-in.-sec2 ) þ ¼ (lb-in:-sec2 ) JT ¼ TA ¼ (lb-in.-sec2 )6 (rad/sec2 ) ¼ (lb-in.)

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8. Torque requirement summary. Friction torque ðTF Þ ¼ (lb-in.) (lb-in.) Thrust load ðTM Þ ¼ (lb-in.) Acceleration torque ðTA Þ ¼ 9. Hydraulic pressure at the motor. torque6100 (psi) motor torque constant ðKT Þ 100 PFR ¼ Pressure for friction load ðTF Þ ¼ 6 ¼ (psi) KT 100 PTH ¼ Pressure for thrust ðTM Þ ¼ 6 ¼ (psi) KT 100 PA ¼ Pressure for max acceleration ðTA Þ ¼ 6 KT ¼ (psi) Pressure at motor ¼

10.

Hydraulic losses. Line loss ¼ 12% of system pressure, approx. 300 psi (psi) Valve drop (approx. 1000 psi at rated flow) (psi) Pressure for friction load (psi) No load motor loss at traverse ¼ KPM VT =1000 KPM ¼ 150 psi=1000 rpm (motor pressure drop no-load) VT ¼

(rpm)

No load motor loss ¼ Total losses 11.

1506 ¼ 1000

(psi)

(psi)

Supply pressure needed. (Note: Load pressure ¼ 23 supply pressure, or 1.5 6 load pressure ¼ supply pressure) Pressure for friction ¼ friction psi ðPFR Þ þ losses ¼ þ ¼ psi Pressure for max thrust ¼ 1:5 thrust psi ðPTH þ 300 ¼ þ ¼ psi Pressure for max accel ¼ accel psi ðPA Þ þ losses ¼ ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

psi

þ

(Select a pressure greater than the highest of these pressures.) ðpsiÞ Recommended supply pressure ¼ 12.

Motor size. Effective drive displacement (from thrust requirement) ¼ EDD Total stall thrustðFST Þ lbf ¼ Supply pressure psi in:3 =in: of drive movement

¼

Motor displacement (from torque requirement): Eff drive disp ðEDD Þ 6Lead ¼ motor displacement Ratio required ðin:3 =revÞ in:2 6

in./rev ¼

ðin:3 =revÞ ðin:3 =revÞ

Displacement of motor chosen DM ¼

If displacement of motor chosen is less than required, use the next larger motor. 13.

Flow rating. Theoretical flow at traverse ¼ QT (Select a servo valve larger than required flow at traverse.) 1:3VT DM 1:36 6 ¼ ¼ (gpm) 231 231 (rpm) VT ¼ max motor speed at traverse ¼

QT ¼

(in:3 =rev)

DM ¼ motor displacement ¼ 14.

Maximum accelerating torque TA (lb-in.). PS 6KT
TF 100 PS ¼ hydraulic power supply pressure ¼ (psi) (lb-in:=100 psi) KT ¼ motor torque constant ¼ TA ¼

tF ¼ friction torque ¼ 6
TA ¼ ¼ 100

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(lb-in.) (lb-in.)

15.

Maximum decelerating torque TD . ðPS þ PR ÞKT þ TF 100 PS ¼ supply pressure ¼ (psi) PR ¼ relief pressure setting in manifold ¼

TD ¼

TD ¼ 16.

ð þ Þ6 þ 100

¼

(psi)

(lb-in.)

Hydraulic resonance. sffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2BD2M oH ¼ (rad/sec) V c JT DM ¼ motor displacement ¼

(in:3 =rad)

Vc ¼ oil under compression ¼ Vmotor þ Vvalve þ Vmanifold ¼ (in:3 ) ð12 total volume) Vvalve ¼ 0:17 in:3 (approx:) Vmanifold ¼ 0:3 in:3 (approx:) þ

þ

¼

(in:3 )

JT ¼ total inertia at the motor ¼

(lb-in.-sec2 )

B ¼ bulk modulus of oil ¼ 16105 (lb=in:2 ) rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 26105 6 ¼ oh ¼ (rad=sec) 6 17.

Hydraulic damping factor. rffiffiffiffiffiffiffiffiffi Lt BJT dH ¼ DM 2Vc Lt ¼ motor leakage þ valve leakage ðin:3 =sec=psiÞ Valve leakage ¼ 0:0019 in:3 =sec=psiðapprox:Þ Motor leakage ¼ Lt ¼ dH ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

ðin:3 =sec=psiÞ

þ ¼ ðin:3 =sec=psiÞ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 16105 6 ¼ 26

Typical Motor Specifications Max. Compressed Torque Motor Hartman displacement speed Inertia oil volume constant Leakage motor (in.3) (in.-lb/100 psi) (in.3/sec-psi) (in.3/rev) (rpm) (lb-in.-sec2) HT-5 HT-10 HT-22 HT-50 HT-76

0.5 1.0 2.2 5.0 7.6

2000 2000 1500 1500 1200

0.00174 0.00198 0.006 0.0152 0.0263

0.77 1.31 2.91 4.81

8 16 35 80 121

Thomas Coupling Inertias Coupling

J (lb-in.2)

J (lb-in.-sec2)

101DBZ 126DBZ 163DBZ 201DBZ 226DBZ 263DBZ 301DBZ 351DBZ 401DBZ 451DBZ

4.7 10 22 56 100 210 420 950 1090 3500

0.01216 0.02587 0.0569 0.1449 0.2587 0.5434 1.0869 2.4585 2.8209 9.0579

101DBZA 126DBZA 163DBZA 201DBZA 226DBZA 263DBZA 301DBZA 351DBZA 401DBZA 451DBZA

4.7 10 22 56 100 220 430 970 1900 3500

0.01216 0.02587 0.0569 0.1449 0.2587 0.5434 1.0869 2.4585 2.8209 9.0579

101DBZB 126DBZB 163DBZB 201DBZB 226DBZB 263DBZB 301DBZB 351DBZB 401DBZB 451DBZB

4.7 11 23 57 110 220 440 1000 1900 3600

0.01216 0.02846 0.0592 0.1475 0.2846 0.5693 1.13871 2.5879 4.9171 9.3167

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Used on motor

HT22 HT50

HT10

HT76

0.00058 0.00065 0.00134 0.00167

ELECTRIC DRIVE SIZING Machine: Axis: Traverse rate: in./rev Length ¼ in. Ball screw lead: Gearbox: Gear ratio ¼ Diameter ¼ Reflected load inertia ¼ lb-in.-sec2 Slide weight ¼ lbs Slide coefficient of friction: lbf/lb

ipm

1. Establish the maximum load or working thrust. Load thrust ¼

lbf

2. Establish no-load thrust. Thrustnl ¼ Coefficient of friction6slide weight Thrustnl ¼

¼

6

lbf

3. Torque requirements at the drive motor. (a)

Geared drives.

NOTE: Use speed matching to select the ratio. Motor rated speed (rpm)6lead (in./rev) Traverse rate (ipm) Ideal ratio ¼ x ¼

Ideal ratio ¼

Load thrust torque: lead 1 6 2p gear ratio 6 ¼ in.-lbs

Motor load torque ¼ load thrust6 Motor load torque ¼

6

2p

No-load torque: Motor no-load torque ¼ no-load thrust6 Motor no-load torque ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

6

2p

6

lead 1 6 2p gear ratio ¼ in.-lbs

Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼ (b)

þ

¼

in.-lbs

Direct Drives:

Load thrust torque: lead 2p ¼ in.-lbs

Motor load torque ¼ load thrust6 Motor load torque ¼

6

2p

No-load torque: Motor no-load torque ¼ no-load thrust6 Motor no-load torque ¼

6

2p

¼

lead 2p in.-lbs

Total thrust torque: Total thrust torque ¼ load torque þ no-load torque ¼

þ

¼

in.-lbs

4. Total drive motor torque required must be derated according to the amplifier used. Amplifier

Derating Form Factor for Motor

a. SCR amplifiers Single-phase full wave Single-phase full wave/inductor Three-phase half wave Three-phase half wave/inductor b. DC generator c. PWM or brushless DC

1.66 1.2 1.25 1.05 1.0 1.0

Required torque ¼ total torque6form factor ¼ in.-lbs ¼

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

6

¼

in.-lbs

5. Motor traverse speed. (a)

Geared Drives: Maximum motor speed ¼ traverse ipm6 6 1 6

Maximum motor speed ¼ (b)

1 6gear ratio Lead ¼ rpm

Direct Drives: 1 Lead 6 1 ¼ rpm

Maximum motor speed ¼ traverse ipm6 Maximum motor speed ¼

.

6. Drive motor manufacturer selected:

.

Motor selected: rpm. rpm.

Rated speed of motor selected Maximum speed of motor selected

Motor rated speed should be equal or larger than the traverse speed. If this requirement is not satisfied, a different gear ratio and/or drive screw lead should be selected to resize the motor for the required torque. 7. Torque for maximum acceleration (based on exponential response to a traverse rate step input of velocity). . ipm 1000 mil min 6 6 ¼ sec mil  in: 60 sec    2p 1 rev 6gain a max: ¼ 6max: motor speed 60 sec min Gain ¼

¼ rad=sec2 ¼ V F 6K V [
KV T a max: ¼ 0:10456

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

sec

6

rpm ¼

rad sec2

Torquea ¼ J T 6a max: J T ¼ motor inertia þ reflected load inertia þ

JT

lb-in.-sec2

¼

lb-in.-sec2 6

Ta ¼

rad=sec2 ¼

in.-lb

8. Motor torque requirements summary. in.-lbs Torque for no-load in.-lbs Torque for load thrust Torque for maximum acceleration

in.-lbs .

9. Motor selected in.-lbs in.-lbs

Rated torque Maximum torque

Motor rated torque should be greater than required torque for total thrust. NOTE: Refer to motor curves. Motor maximum torque should be approximately the same or greater than the required torque for maximum acceleration (Ta) plus no-load thrust. 10.

Maximum torque ¼ rpm.

in.-lb from motor curves at traverse

I max 6K t
T nl ¼ torque available for acceleration form factor ¼ Motor armature maximum current ¼ amps

Ta ¼ I max

K t ¼ Motor torque constant ¼ in.-lbs/amp in.-lbs T nl ¼ No-load torque ¼ Ta ¼ ½

6




¼

in.-lb

torque available for acceleration torque available T a ¼ ¼a Allowable acceleration ¼ JT total inertia rad ¼ ¼ sec2

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

11.

Maximum decelerating torque ¼ Td. I max 6K t þ T nl form factor I max ¼ Motor armature maximum current ¼ Td ¼ ½ 6 þ

Td ¼

¼ 12.

amps

in.-lb torque available for deceleration

Motor time constants. Brushless DC Motors K eðL
LÞ ¼ 1:73 1:73 volts-sec/rad (see motor data)

K eðPHASEÞ ¼ Motor voltage constant ¼ ¼

K t ¼ Motor torque constant ¼ (see motor data) RMðL
LÞ ¼ Motor resistance ¼

in.-lb/amp

ohms

SRMðL
LÞ ¼ Total motor circuit resistance ¼ 1:35RMðL
LÞ ¼ 1:356 SRMðPHASEÞ ¼ SRMðL
LÞ 60:5 ¼

¼

ohm

60:5 ¼

LL
L ¼ Motor inductance ¼

ohm

henries lb-in.-sec2

J m ¼ Motor armature inertia ¼ J LOAD ¼ Reflected inertia to motor ¼

lb-in.-sec2

lb-in.-sec2 LL
L te ¼ Motor electrical time constant ¼ SRMðL
LÞ

J Total ¼ J m þ JLOAD ¼

þ

¼

¼

¼ sec 1 1 ¼ oe ¼ ¼ rad/sec te tm ¼ Motor mechanical time constant ¼ tm ¼ om ¼ tm ¼ te

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

x ¼ x 1 ¼ tm ¼

sec rad/sec

SRMðPHASEÞ J Total K eðPHASEÞ K t

DC Motors K e ¼ Motor voltage constant ¼

volts/rad/sec (see motor data)

K t ¼ Motor torque constant ¼ in.-lb/amp (see motor data) Ra ¼ Motor armature resistance ¼ ohms Ra ¼ Total armature circuit resistance ¼ 1:35 Ra ¼ 1:356 ¼ ohm L ¼ Motor armature circuit inductance ¼ J m ¼ Motor inertia ¼ J Load ¼

lb-in.-sec

henries

2

lb-in.-sec

2

J Total ¼ J m þ J L ¼ þ ¼ lb-in.-sec2 T e ¼ Motor electrical time constant ¼ L=Ra ¼ 1 ¼ oe ¼ 1=te ¼ rad/sec P Ra J t T m ¼ Motor mechanical time constant ¼ K eK t x ¼ tm ¼ sec x 1 om ¼ ¼ rad/sec tm tm ¼ ¼ te 13.

Motor transfer function. (a) If ttme < 4, motor transfer function should be: . 1 Ke oðsÞ ½rad=sec i ¼h 2 s 1 V ðsÞ ½volt þ sþ1 om oe

(b)

If

tm te

> 4, motor transfer function should be: . 1 Ke oðsÞ   ¼ s s V ðsÞ þ1 þ1 oe

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

om

om

¼

sec

Glossary

Acceleration, maximum from stall The acceleration rate of an unloaded motor, initially at rest, at the instant a current I pk is applied; am ¼

KT Ipk
Tfriction Ja

Ambient temperature The temperature of the cooling medium immediately surrounding the motor. Armature reaction The production of a magnetic field 90 electrical degrees in the direction opposite to the direction of armature rotation. This magnetic field is produced by armature current. Bandwidth The frequency range in which system response is within 3 dB of the specified response. Block diagram A simplified representation of a servo system with each component represented by a block and each block is positioned in the order of signal flow through the system. Bode plot A plot of system gain in decibels versus the log of signal frequency. Break frequency The frequency (lines) at which the asymptotes of the gain versus frequency plot intersect. Characteristic equation The characteristic equation of a servo system is 1 þ GH ¼ 0 where G is the transfer function of the forward signal path and H is the transfer function of the feedback signal path.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Circulating current The current flow in armature conductors that are short-circuited during commutation. Cogging The nonuniform rotation of a motor armature caused by tendency for the armature to prefer certain discrete angular positions. Computer simulation The representation of an actual or proposed system by mathematical models with a computer. Corner frequency See ‘‘break frequency.’’ Coupling ratio A general term to define the relative motion between motor armature and the driven load. Critical damping A critically damped system’s response to a disturbance is a return to its equilibrium state without overshooting its equilibrium state. Crossover frequency The frequency at which the magnitude of the product of the forward path gain and the feedback path gain is unity. Damping The amplitude decay of an oscillatory signal. Damping ratio A measure of system damping expressed as the ratio between actual damping to the value for critical damping. Dead band A range of input signals for which there is no system response. Decibel (dB) A measure of system gain in units of 20 log10(gain). Dielectric test A high-voltage test of the motor insulation system’s ability to withstand an AC voltage. Test criterion limits leakage current to a specified maximum with the test potential at a specified magnitude and frequency applied between the motor case and insulation. Dynamic breaking A method for braking a DC servo motor by controlling armature current during deceleration. Efficiency The ratio of power output to power input: n ¼ Po =Pi . Electrical time constant The electrical time constant of a DC servo motor is the ratio of armature inductance in henries to armature circuit resistance in ohms: te ¼ La =Ra . Fall time The time for the amplitude of system response to decay to 37% of its steady-state amplitude after the removal of the steadystate forcing function. Feedback A function of the directly controlled variable in such form as to be used at the summing point. Field weakening A method of increasing the speed of a wound field by reducing field strength by reducing field current. Figure of merit (FOM) The ratio of the square of peak available motor torque to armature inertia: FOM ¼ Tpk 2 =Ja . ‘‘Maximum power rate’’ is a term used interchangeably with FOM.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Flux biasing A method for controlling the torque constant of a servo motor by varying the field strength of a separate wound field assembly. Form factor The form factor of a current is the ratio of its root mean square (rms) value to its average value. Full load current The armature current of a motor operated at its full load torque and speed with rated voltage applied. Full load speed Armature speed of a motor operated with rated voltage and full load torque. Gain The ratio of system output signal amplitude to system input signal amplitude. Gain margin The magnitude of the system characteristic equation at the frequency for which the phase angle of the product of the forward path and feedback path gain function is
180 degrees. Hard servo A machine servo drive where the position loop gain is higher than 2 inches per minute per mil (33/sec). Hunting Oscillation of the system response about zero error condition due to insufficient damping. Incremental motion system A control system that positions the load in small discrete steps rapidly and repetitively. Inertial match An inertial match between motor and load is obtained by selecting the coupling ratio such that the load inertia referred to the motor shaft is equal to the motor inertia. Lag network An electrical network that increases the delay between system input signal and system output signal. Lead network An electrical network that decreases the delay between system and system output signal. Lead screw A device for translating rotary to linear motion consisting of an externally threaded screw and an internally threaded carriage (nut). Linearity The maximum deviation from a straight line between zero and output at specified speed expressed as a percentage of the output at the specified speed. Loop gain The gain product of the forward path and feedback path gain function. Machine servo compensation The use of an analog filter technique or digital algorithm to stabilize a servo drive. Mechanical time constant The time for a motor to reach 63.3% of its final velocity after applying armature excitation: tm ¼ JT Ra =Ke KT

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

No-load speed Motor speed with no external friction load. Nonlinearities A term used to identify components of a servo drive or servo plant (machine) that have an output characteristic that does not have a continuous linear relation between the output and the input. Phase lock servo A digital control system in which the output of an optical tachometer is compared to a reference square wave to generate a system error signal proportional to both shaft velocity and position. Phase margin The phase angle of the loop gain minus 180 degrees at the crossover frequency. PID An abbreviation for types of servo mechanism compensation (proportional, integral, and differential). Poles Terminology of root locus plotting for the frequency at which system gain goes to unity. Reflected inertia Inertias of machine components reflected to the drive motor shaft. Regulation An expression for the amount of drop in motor speed as load torque is applied to the motor: Regulation = (no-load speed full-load speed)/no load speed Resonance A condition evidenced by a large oscillating amplitude, which results when a small amplitude of a periodic input has a frequency approaching one of the natural frequencies of the drive system. Ripple torque The cyclical variation of generated torque at a frequency proportional to the product of motor speed and the number of commutator segments. Safe operating area curve The boundary on the speed curve inside of which the motor may be operated continuously without exceeding its thermal rating. Servo A learned borrowing from Latin where it meant ‘‘slave.’’ A feedback control system in which the output (controlled variable) follows the input (command) within the bandwidth of the servo. Servo loops Elements of a servo drive, which include a forward loop and summing junction of an amplifier, and a feedback loop. Multiple servo loops can have one servo loop inside another to improve performance and accuracy. Soft servo A machine servo drive where the position loop gain is between 0.5 and 2.0 inches per minute per mil (Kv is between 8 and 33/sec).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Speed regulation For a speed control system, the variance in set speed expressed as a percentage of set speed. Stability A condition where the output of a servo oscillates or is oscillatory for an input that is constant, consistent, and is not oscillating. System order The degree of the system characteristic equation. System stiffness A measure of system accuracy when subjected to disturbance signals. System type The number of poles located at the origin of the loopgain function.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

Bibliography

J. A. Alloca, Transducers: Theory and Application, Reston Publishing (1984). J. L. Bower and P. M. Schultheiss, Introduction to the Design of Servomechanisms, John Wiley and Sons, New York (1958). T. B. Bullock, Servo Basics for the Layman, Industrial Controls Consulting (1991). J. J. D’Azzo and C. H. Houpis, Linear Control System and Design, 3rd ed., McGrawHill, New York (1988). R. C. Dorf, Modern Control Systems, 6th ed., Addison-Wesley, Reading, MA (1992). G. Ellis, Control System Design Guide, Academic Press, San Diego (1991). Eveleigh, Adaptive Control and Optimization Techniques, McGraw-Hill, New York (1967). G. F. Franklin, J. D. Powel, and A. Emami-Naeini, Feedback Control of Dynamic Systems, Addison-Wesley, Reading, MA (1988). H. Gross, Electrical Feed Drives for Machine Tools, John Wiley and Sons, New York (1983). G. H. Jones, U.S. Patent 3,358,201, Apparatus for compensating machine feed drive servomechanisms, December 12, 1967. T. S. Kenjo and Nagamori, Permanent-Magnet and Brushless DC Motors, Clarendon Press, (1985). B. C. Kuo, Automatic Control Systems, 7th ed. Prentice Hall, (1993). E. Lewis and H. Stern, Design of Hydraulic Control Systems, McGraw-Hill, New York (1962). H. E. Merritt, Hydraulic Control Systems, John Wiley and Sons, New York (1967). S. Noodleman and B. R. Patel, Duty cycle characteristics for DC servomotors, IEEE IAS Trans., 1A-9(5) (1973).

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

R. G. Seippel, Transducers, Sensors, and Detectors, Reston Publishing (1984). S. M. Shinners, Minimizing servo load resonance error, Control Engineering, January 15, 1962. G. J. Thaler and M. P. Pastel, Analysis and Design of Nonlinear Feedback Control Systems, McGraw-Hill, New York (1962). G. Young and J. G. Bollinger, A research report on the principles and applications of frequency selective feedback, University of Wisconsin, Department of Mechanical Engineering, Engineering Experimental Station, March 1969. G. W. Younkin, Applying and Sizing Machine Servo Drives, SME paper MS90-300, International Manufacturing Technology Conference, September 6, 1990. G. W. Younkin, Modeling machine tool feed servo drives using simulation techniques to predict performance, IEEE IAS Trans., 27(2):268–274 (1991). G. W. Younkin, Modeling machine tool feed servo drives using simulation techniques to predict performance, IEEE Transactions on Industry Applications, 27(2) (1991). G. W. Younkin, Industrial Servo Control Systems—Fundamentals and Applications, 1st ed., Marcel Dekker, Inc., New York (1996). G. W. Younkin, Brushless DC—Motor and Current Servo Loop Analysis Using PI Compensation. G. W. Younkin and T. B. Bullock, Taking the mystery out of motion-control algorithms, Machine Design, January 12 (1995). G. W. Younkin and T. B. Bullock, Bode diagrams analyze servosystems, Machine Design, February 9, 1995. G. W. Younkin, W. D. McGlasson, and R. D. Lorenz, Considerations for Lowinertia ac drives in machine tool axis servo applications, IEEE IAS Trans., 27(2):262–267 (1991). G. W. Younkin and R. H. Welch, How Temperature Affects a Servo Motor’s Electrical and Mechanical Time Constants, IAS paper 27, IEEE Annual Meeting, Pittsburgh, October 13–18, 2002.

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