linear control theory - the state space approach frederick walker fairman

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LINEAR CONTROL THEORY THE STATE SPACE APPROACH FREDERICK We Faux

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Page 1: Linear Control Theory - The State Space Approach Frederick Walker Fairman

LINEARCONTROLTHEORY

THE STATESPACE APPROACH

FREDERICK We Faux

Page 2: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Linear Control Theory

Page 3: Linear Control Theory - The State Space Approach Frederick Walker Fairman
Page 4: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Linear Control TheoryThe State Space Approach

Frederick Walker FairmanQueen's University,Kingston, Ontario, Canada

John Wiley & SonsChichester New York Weinheim Brisbane Singapore Toronto

Page 5: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Copyright ( 1998 John Wiley & Sons Ltd,Baffins Lane, Chichester.West Sussex P019 IUD. England

National 01243 779777International (rt 44) 1243 779777

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All rights reserved. No part of this publication may be reproduced, storedin a retrieval system, or transmitted, in any form or by any means electronic,mechanical, photocopying, recording, scanning or otherwise, except under the termsof the Copyright, Designs and Patents Act 1988 or under the terms of a licence issuedby the Copyright Licensing Agency, 90 Tottenham Court Road, London W IP 9HE, UKwithout the permission in writing of the Publisher.

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Library of Congress Cataloguing-in-Publication Data

Fairman, Frederick Walker.Linear control theory : The state space approach / Frederick

Walker Fairman.p. cm.

Includes bibliographical references and index.ISBN 0-471-97489-7 (cased : alk. paper)1. Linear systems. 2. Control theory. I. Title.

QA402.3.F3 1998629.8'312-dc2l 97-41830

CIP

British Library Cataloguing in Publication Data

A catalogue record for this book is available from the British Library

ISBN 0 471 97489 7

Typeset in part from the author's disks in 10/12pt Times by the Alden Group, Oxford.Printed and bound from Postscript files in Great Britain by Bookcraft (Bath) Ltd.This book is printed on acid-free paper responsibly manufactured from sustainable forestry, in which at leasttwo trees are planted for each one used for paper production.

Page 6: Linear Control Theory - The State Space Approach Frederick Walker Fairman

To Nancy for her untiring support

Page 7: Linear Control Theory - The State Space Approach Frederick Walker Fairman
Page 8: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Contents

Preface

1 Introduction to State Space

X111

1

1.1 Introduction1.2 Review of Second Order Systems

1.2.1 Patterns of behavior 2

1.2.2 The phase plane 5

1.3 Introduction to State Space Modeling 7

1.4 Solving the State Differential Equation 9

1.4.1 The matrix exponential 9

1.4.2 Calculating the matrix exponential 10

1.4.3 Proper and strictly proper rational functions 12

1.5 Coordinate Transformation 12

1.5.1 Effect on the state model 13

1.5.2 Determination of eAt14

1.6 Diagonalizing Coordinate Transformation 15

1.6.1 Right-eigenvectors 16

1.6.2 Eigenvalue-eigenvector problem 17

1.6.3 Left-eigenvectors 19

1.6.4 Eigenvalue invariance 20

1.7 State Trajectories Revisited 21

1.7.1 Straight line state trajectories: diagonal A 22

1.7.2 Straight line state trajectories: real eigenvalues 23

1.7.3 Straight line trajectories: complex eigenvalues 241.7.4 Null output zero-input response 25

1.8 State Space Models for the Complete Response 261.8.1 Second order process revisited 261.8.2 Some essential features of state models 281.8.3 Zero-state response 29

1.9 Diagonal form State Model 321.9.1 Structure 321.9.2 Properties 331.9.3 Obtaining the diagonal form state model 35

1.10 Computer Calculation of the State and Output 371.11 Notes and References 39

2 State Feedback and Controllability 41

2.1 Introduction 412.2 State Feedback 422.3 Eigenvalue Assignment 44

2.3.1 Eigenvalue assignment via the controller form 45

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viii Contents

2.3.2 Realizing the controller form2.3.3 Controller form state transformation2.3.4 Condition for controller form equivalence2.3.5 Ackermann's formula

2.4 Controllability2.4.1 Controllable subspace2.4.2 Input synthesis for state annihilation

2.5 Controllable Decomposed Form2.5.1 Input control of the controllable subspace2.5.2 Relation to the transfer function2.5.3 Eigenvalues and eigenvectors of A

2.6 Transformation to Controllable Decomposed Form2.7 Notes and References

3 State Estimation and Observability3.1 Introduction3.2 Filtering for Stable Systems3.3 Observers3.4 Observer Design

3.4.1 Observer form3.4.2 Transformation to observer form3.4.3 Ackermann's formula

3.5 Observability3.5.1 A state determination problem3.5.2 Effect of observability on the output

3.6 Observable Decomposed Form3.6.1 Output dependency on observable subspace3.6.2 Observability matrix3.6.3 Transfer function3.6.4 Transformation to observable decomposed form

3.7 Minimal Order Observer3.7.1 The approach3.7.2 Determination of xR(t)3.7.3 A fictitious output3.7.4 Determination of the fictitious output3.7.5 Assignment of observer eigenvalues

3.8 Notes and References

4 Model Approximation via Balanced Realization4.1 Introduction4.2 Controllable-Observable Decomposition4.3 Introduction to the Observability Gramian4.4 Fundamental Properties of Wo

4.4.1 Hermitian matrices4.4.2 Positive definite and non-negative matrices4.4.3 Relating E. to A[W0]

4.5 Introduction to the Controllability Gramian4.6 Balanced Realization4.7 The Lyapunov Equation

4.7.1 Relation to the Gramians4.7.2 Observability, stability, and the observability Gramian

4.8 Controllability Gramian Revisited4.8.1 The least energy input problem4.8.2 Hankel operator

4.9 Notes and References

67

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Contents ix

5 Quadratic Control 115

5.1 Introduction 115

5.2 Observer Based Controllers 116

5.3 Quadratic State Feedback Control 119

5.3.1 Motivating the problem 120

5.3.2 Formulating the problem 121

5.3.3 Developing a solution 122

5.4 Solving the QCARE 127

5.4.1 Stabilizing solutions 127

5.4.2 The Hamiltonian matrix for the QCARE 130

5.4.3 Finding the stabilizing solution 133

5.5 Quadratic State Estimation 137

5.5.1 Problem formulation 137

5.5.2 Problem solution 140

5.6 Solving the QFARE 143

5.7 Summary 145

5.8 Notes and References 145

6 LQG Control 147

6.1 Introduction 147

6.2 LQG State Feedback Control Problem 149

6.2.1 Problem formulation 149

6.2.2 Development of a solution 150

6.3 LQG State Estimation Problem 153

6.3.1 Problem formulation 154

6.3.2 Problem solution 155

6.4 LQG Measured Output Feedback Problem 157

6.5 Stabilizing Solution 158

6.5.1 The Hamiltonian matrix for the GCARE 158

6.5.2 Prohibition of imaginary eigenvalues 159

6.5.3 Invertability of T11 and T21 162

6.5.4 Conditions for solving the GFARE 165

6.6 Summary 166

6.7 Notes and References 166

7 Signal and System Spaces 167

7.1 Introduction 167

7.2 Time Domain Spaces 167

7.2.1 Hilbert spaces for signals 168

7.2.2 The L2 norm of the weighting matrix 170

7.2.3 Anticausal and antistable systems 1727.3 Frequency Domain Hilbert Spaces 173

7.3.1 The Fourier transform 173

7.3.2 Convergence of the Fourier integral 175

7.3.3 The Laplace transform 176

7.3.4 The Hardy spaces: 7d2 and 7{2-L 177

7.3.5 Decomposing L2 space 178

7.3.6 The H2 system norm 1797.4 The H. Norm: SISO Systems 181

7.4.1 Transfer function characterization of the H, norm 181

7.4.2 Transfer function spaces 1837.4.3 The small gain theorem 184

7.5 The H. Norm: MIMO Systems 1857.5.1 Singular value decomposition 185

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x Contents

7.5.2 Induced 2-norm for constant matrices 186

7.5.3 The L,,. Hx norm for transfer function matrices 189

7.6 Summary 190

7.7 Notes and References 191

8 System Algebra 193

8.1 Introduction 193

8.1.1 Parallel connection 193

8.1.2 Series connection 195

8.2 System Inversion 196

8.2.1 Inverse system state model 197

8.2.2 SISO system zeros 198

8.2.3 MIMO system zeros 199

8.2.4 Zeros of invertible systems 200

8.3 Coprime Factorization 201

8.3.1 Why coprime? 202

8.3.2 Coprime factorization of MIMO systems 204

8.3.3 Relating coprime factorizations 205

8.4 State Models for Coprime Factorization 206

8.4.1 Right and left coprime factors 207

8.4.2 Solutions to the Bezout identities 209

8.4.3 Doubly-coprime factorization 212

8.5 Stabilizing Controllers 213

8.5.1 Relating W(s) to G(s),H(s) 214

8.5.2 A criterion for stabilizing controllers 215

8.5.3 Youla parametrization of stabilizing controllers 217

8.6 Lossless Systems and Related Ideas 219

8.6.1 All pass filters 220

8.6.2 Inner transfer functions and adjoint systems 221

8.7 Summary 223

8.8 Notes and References 223

9 H. State Feedback and Estimation9.1 Introduction9.2 H. State Feedback Control Problem

9.2.1 Introduction of P.,9.2.2 Introduction of G1(s)9.2.3 Introduction of J-inner coprime factorization9.2.4 Consequences of J-inner coprime factorization

9.3 H. State Feedback Controller9.3.1 Design equations for K9.3.2 On the stability of A + B2K29.3.3 Determination of 0

9.4 H. State Estimation Problem9.4.1 Determination of T,(s)9.4.2 Duality9.4.3 Design equations for L2

9.5 Sufficient Conditions9.6 Summary9.7 Notes and References

10 Hx Output Feedback Control10.1 Introduction10.2 Development

225

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229

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Contents xi

10.2.1 Reformulation of P. 248

10.2.2 An H, state estimator 251

10.2.3 Introducing estimated state feedback 253

10.3 H, Output Feedback Controllers 254

10.3.1 Central controller 255

10.3.2 Controller parametrization 256

10.3.3 Relation to Youla parametrization 260

10.4 H. Separation Principle 261

10.4.1 A relation between Hamiltonians 262

10.4.2 Relating stabilizing solutions 267

10.4.3 Determination of Lo 269

10.5 Summary 269

10.6 Notes and References 270

A Linear Algebra 271

A.1 Multiple Eigenvalues and Controllability 271

A.2 Block Upper Triangular Matrices 272

A.3 Singular Value Decomposition (SVD) 274

A.4 Different Forms for the SVD 276

A.5 Matrix Inversion Lemma (MIL) 277

B Reduced Order Model Stability 279

C Problems 283

C.1 Problems Relating to Chapter 1 283

C.2 Problems Relating to Chapter 2 285

C.3 Problems Relating to Chapter 3 287

C.4 Problems Relating to Chapter 4 288

C.5 Problems Relating to Chapter 5 290

D MATLAB Experiments 293

D.1 State Models and State Response 293

D.1.1 Controller form 293D.1.2 Second order linear behavior 293D.1.3 Second order nonlinear behavior 295D.1.4 Diagonal form 296

D.2 Feedback and Controllability 297D.2.1 Controllable state models 297D.2.2 Uncontrollable state models 298

D.3 Observer Based Control Systems 299D.3.1 Observer based controllers 301D.3.2 Observer based control system behavior 303

D.4 State Model Reduction 303D.4.1 Decomposition of uncontrollable and/or unobservable systems 304D.4.2 Weak controllability and/or observability 305D.4.3 Energy interpretation of the controllability and observability

Gramians 306D.4.4 Design of reduced order models 307

References 309

Index 313

Page 13: Linear Control Theory - The State Space Approach Frederick Walker Fairman
Page 14: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Preface

This book was written with the intent of providing students and practicing controlengineers with the basic background in control theory needed to use control systemdesign software more productively. The book begins with a detailed treatment of thoseaspects of the state space analysis of linear systems that are needed in the remainder of thetext. The book is organized in the following manner:

The first four chapters develop linear system theory including model reduction viabalanced realization.Chapters 5 and 6 deal with classical optimal control theory.The final four chapters are devoted to the development of suboptimal Hx controltheory.

The mathematical ideas required in the development are introduced as they are neededusing a "just-in-time" approach. This is done to motivate the reader to venture beyondthe usual topics appearing in introductory undergraduate books on "automatic control",to more advanced topics which have so far been restricted to postgraduate level bookshaving the terms "mathematical control theory" and "robust control" in their titles.

This book can be used as the text for either a one or two-semester course at the finalyear undergraduate level or as a one semester course at the beginning postgraduate level.Students are assumed to have taken a basic course in either "signals and systems" or"automatic control". Although not assumed, an introductory knowledge of the statespace analysis of systems together with a good understanding of linear algebra wouldbenefit the reader's progress in acquiring the ideas presented in this book.

Ideas presented in this book which provide the reader with a slightly different view ofcontrol and system theory than would be obtained by reading other textbooks are asfollows:

The so-called PBH test which is usually presented as a test for controllability and/orobservability is used throughout the present book to characterize eigenvalues incontrol problems involving eigenvalue assignment by state feedback and/or outputinjection.An easy to understand matrix variational technique is used to simplify the develop-ment of the design equations for the time invariant, steady-state, quadratic and LQGcontrollers.The relatively simple idea of the L2 gain is used as a basis for the development of theH,,, controller.

Page 15: Linear Control Theory - The State Space Approach Frederick Walker Fairman

xiv Preface

Concerning the style of the book, the beginning section, "Introduction", for eachchapter contains motivational material and an overview of the ideas to be introduced insubsequent sections in that chapter. Each chapter finishes with a section called "Notesand References", which indicates a selection of other sources for the material treated inthe chapter, as well as an indication of recent advances with references.

I would like to thank the following colleagues in the Department of Electrical andComputer Engineering at Queen's University for proof-reading parts of the manuscript:Norm Beaulieu, Steve Blostein, Mingyu Liu, Dan Secrieu and Chris Zarowski. Specialthanks go to my former research student Lacra Pavel for proof-reading and advice onChapters 6, 9 and 10 as well as to Jamie Mingo in the Department of Mathematics andStatistics at Queen's University for his help with some of the ideas in Chapter 7. Thanksgo also to Patty Jordan for doing the figures. Finally, I wish to acknowledge thecontribution to this book made by my having supervised the research of former researchstudents, especially Manu Missaghie, Lacra Pavel and Johannes Sveinsson.

The author would appreciate receiving any corrections, comments, or suggestions forfuture editions should readers wish to do so. This could be done either by post or e-mail:< [email protected] >.

Page 16: Linear Control Theory - The State Space Approach Frederick Walker Fairman

1Introduction to State Space

1.1 IntroductionA well known behavioral phenomenon of dynamic systems is the appearance of an outputin the absence of an input. This effect is explained once it is recognized that the internalstorage of energy in the system at the beginning of the response time will produce anoutput. This kind of behavior is referred to as the system's zero-input response.Alternatively, the production of an output caused solely by an input when there is noenergy storage at the start of the response time is referred to as the zero-state response.These two classes of response are responsible for all possible outputs and in the case oflinear systems we can always decompose any output into the sum of an output drawnfrom each of these classes. In this chapter we will use the example of a second order systemtogether with both the zero-input response and the zero-state response to introduce thereader to the use of the state space in modeling the behavior of linear dynamic systems.

1.2 Review of Second Order SystemsA commonly encountered physical process which we will use in the next two sections tointroduce the state modeling of linear dynamic systems is the electric circuit formed byconnecting an ideal constant resistor Re, inductor Le, and capacitor Ce in series in a closedloop as shown in Figure 1.1

Suppose the switch is closed at t = is < 0 so that there is a current flow i (t), t > 0, and avoltage across the capacitor y(t), t > 0. Then applying Kirchhoff's voltage law yields

Rei(t) + Le ddt) + y(t) = 0

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

i(t) = Cedatt)

Combining these equations gives a second order differential equation in the capacitorvoltage, y(t),

d2Y(t) + a, dy(t) + a2y(t) = 0 (1.1)dt dt

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2 Introduction to State Space

R. L

swit.h

C. y(t)

Figure 1.1 Electric circuit with charged capacitor. Switch closed prior to t = 0

where

Re 1al = Le a2 = LTCL

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

1.2.1 Patterns of behaviorThe differential equation (1.1) is said to govern the evolution of the output, y(t), since itacts as a constraint relating y(t), dy(t) and d2yt)

Tt , to each other at each instant of time. Wewill see now that once the initial conditions, i.e., the values of initial output, y(0), andinitial derivative of the output, y(0), are specified, the differential equation, (1.1),completely determines the output, y(t), for all positive time t c (0, oc). We obtain y(t)as follows.

Suppose we have y(t) such that (1.1) is satisfied. Then denoting the derivatives of y(t)as

dy(t) - g(t) d2y(t) h(t)dt dt2

we see that equation (1.1) becomes

h(t) + aig(t) + a2Y(t) = 0 (1.2)

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

g(t) = sy(t) h(t) = s2y(t)

Otherwise equation (1.2) can only be satisfied at specific instants of time. Therefore withthis assumption assumption (1.2) becomes

P(s)Y(t) = 0 (1.4)

Page 18: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Review of Second Order Systems 3

where p(s) is the second degree polynomial

p(s)=sz +als+az

Finally, equation (1.4) holds for all time, when y(t) is not zero for all time, i.e., the trivialsolution, if and only ifs is any one of the roots, {Ai : i = 1, 2} ofp(s),

z

p(s) = (s - A,) (s - A2) A1,2 = 2 ± (2 - a2 (1.5)

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

es,

0C

EO

(st),

i!!

where i!, (factorial i), is the product

i>0=1 i=0

Notice that a bit of algebra shows us that the derivative of e`t, equation (1.6), has thedesired property of being an eigenfunction for differentiation,

dent

dt

st= se

Now we see from the foregoing that e't satisfies equation (1.1) when s = Al or A2.Therefore any linear combination of es't and e1\2t satisfies equation (1.1) so tha£the outputy(t) is given in general as

y(t) = kles't + k2eA2i ai A2 (1.7)

where the kis are constant scalars chosen so that y(t) satisfies the initial conditions. Wecan be do this by solving the equations which result from setting the given values for theinitial conditions, y(O) and y(0), equal to their values determined from equation (1.7), i.e.,by solving

y(o)

= [Al A2]

[k2](1.8)

for kl, k2. Notice that we can do this only if Al 54 A2. In order to proceed when Al _ A2 wereplace equation (1.7) with

y(t) = (k3 t+k4)eAlt Al = A2 (1.9)

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4 Introduction to State Space

and determine the kis from the appropriate equations to ensure that the initial conditionsare satisfied.

Returning to the behavior of the physical process that is under analysis, notice thatsince Re, Le, and CE, are real, the as are real. As a consequence the roots Ai of p(s).equation (1.5), are both real or are both complex. Moreover when these roots are complexthey are conjugates of each other, i.e., A, = A.

More generally, if all the coefficients of a polynomial of any degree are real, eachcomplex root must be matched by another root which is its complex conjugate. Thisproperty is important in the context of the behavior of linear physical processes since theparameters of these processes, e.g., mass, heat conductivity, electric capacitance, arealways real so that the coefficients of p(s) are always real.

Now a plot of the output, y(t), versus time, t, reveals that there are two basic patternsfor the behavior of the output depending on whether the Ais are real or are complexconjugate pairs.

If the ais are real, we see from equation (1.8) that the kis are also real and the outputy(t) : t e (0, oc) is given as equation (1.7) or (1.9). In this case we see that the outputvoltage y(t) exhibits at most one maximum and decays without oscillation to the time axisas t tends to infinity. Notice from equation (1.5) that the A is are real provided theparameters RP7 Le, Ce have values such that (,)2 > a2.

Alternatively, if the ais are complex, i.e., if (2)2 < a2, then we see from (1.5) thatAl = A and from (1.8) that k, = kz. Thus kleAl t and k2e\2t are complex conjugates ofeach other and their sum which gives y(t), equation (1.7), is real. Incorporating theseconjugate relations for the Ais and the kis in equation (1.7) allows us to write the output asa damped oscillation

y(t) = 2

where

k1 leRepa']` cos(Im[A1]t + 8) (1.10)

k, = Re[kl] +jIm[kl]

B =taneGRM-0

Thus we see from (1.10) that the output voltage across the capacitor, y(t), swings backand forth from its initial value to ever smaller values of alternating polarity. This behavioris analogous to the behavior of the position of a free swinging pendulum. The capacitorvoltage (pendulum position) eventually goes to zero because of the loss of heat energyfrom the system resulting from the presence of Re (friction). In this analogy, voltage andcurrent in the electric circuit are analogous to position and velocity respectively in themechanical process. The inductance Le is analogous to mass since the inductance resistschanges in the current through itself whereas the inertial effect of mass causes the mass toresist change in its velocity.

In addition, notice from equation (1.10) that the frequency of the oscillation, Im[A1], aswell as the time constant associated with the decay in the amplitude of the oscillation,(Re[al])-1, are each independent of the initial conditions and depend on the systemparameters, Ref Le, Ce only, i.e., on al, a2 only.

Page 20: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Review of Second Order Systems 5

The previous discussion leads to the following characterization of the zero-inputresponse of dynamic processes whose behavior can be modeled by second orderdifferential equations with constant coefficients.

(i) The zero-input response, y(t) : t > 0, depends on the set of signals {eA,t : i = 1, 2}referred to as modes of the system where the constants A,, (system eigenvalues), areroots of the polynomial p(s), (characteristic polynomial).

(ii) The steady state zero-input response is zero, i.e., limy(t) = 0, for any initialconditions if and only if all the .his are negative or have negative real part, i.e.,Re[Ai] < 0, i = 1, 2 . In this situation we say that the system is stable.

(iii) We have Re[Ai] < 0, i = 1, 2, if and only if a, > 0 and az > 0. More generally, thecondition ai > 0, i = 1, 2,. - - n for systems whose behavior is governed by differ-ential equations in the order of n. > 2, is necessary but not sufficient for the system tobe stable, i.e., is necessary but not sufficient for all Ais to have negative real part

1.2.2 The phase planeWe have just seen that, when there is no input, a second order system having specified aishas output, y(t), which is specified completely by the initial conditions, y(0) and y(0). Thisimportant observation suggests that the same information concerning the behavior of thesystem is contained in either (a) a plot of y(t) versus t or (b) a plot of y(t) versus y(t).

Thus if we make a plot of y(t) versus y(t), the point representing y(t), y(t) in the y(t)versus y(t) plane traces out a curve or trajectory with increasing time.

The two dimensional space in which this trajectory exists is referred to as the statespace and the two-element vector consisting of y(t) and y(t) is referred to as the state,denoted as x(t) where

x(t)y(t)Ly(t) J

This approach to visualizing the behavior of a dynamic process was used bymathematicians at the end of the last century to investigate the solutions for secondorder nonlinear differential equations, i.e., equations of the form (1.1) but with the aisfunctions of y(t) and/or y(t). The term phase plane plot was used to refer to the statetrajectory in this case. Since, in general, the dimension of the state space equals the orderof the differential equation which governs the output behavior of the process, the statespace cannot be displayed for systems of order greater than two. Even so, the mathema-tical idea of the state space has become of great practical and theoretical importance inthe field of control engineering.

Referring to the previous section, we see that the state trajectory for a dynamic processwhose behavior can be modeled by a second order differential equation with constantcoefficients, can exhibit any one of the following four fundamental shapes.

(i) If the Ais are complex and Re[Ai] < 0 the system is stable and the state trajectoryspirals inwards towards the origin.

(ii) If the .his are complex and Re[Ai] > 0 the system is unstable and the state trajectoryspirals outwards away from the origin.

Page 21: Linear Control Theory - The State Space Approach Frederick Walker Fairman

6 Introduction to State Space

Figure 1.2 Plot of y(t) vs. t and y(t) vs. y(t) when A is complex

(iii) If the .,s are real and both his are negative the system is stable and the statetrajectory moves towards the origin in an arc.

(iv) If the ),s are real and one or both are positive the system is unstable and the statetrajectory moves away from the origin in an arc.

Notice that state trajectories (ii) and (iv) do not occur in the present example of anelectric circuit. This results from the fact that the parameters Re, Lei Ce are positive. Thusthe coefficients, a; : i = 1, 2 of the characteristic polynomial, equation (1.4) are positive sothat the A is are negative or have negative real parts. This implies that we are dealing with astable dynamic process, i.e., state trajectories tend to the origin for all initial states.

So far in this chapter we have used an electric circuit as an example of a system. Weused the character of the behavior of this system in response to initial conditions to

Page 22: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Introduction to State Space Modeling 7

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

1.3 Introduction to State Space ModelingWe saw in the previous section that once a second order system is specified, i.e., once thea;s are given numerical values, the zero-input response is determined completely from thesystem's initial conditions, y(0), y(0). In addition, we noted that the second derivative ofthe output is determined at each instant from y(t) and y(t) through the constraint (1.1).These facts suggest that it should be possible to obtain the zero-input response by solvingtwo first order differential equations involving two signals, X1 (t): x2(t), which are relateduniquely to y(t), y(t). One straightforward way of doing this is to identify y(t) withxl(t)and y(t) with x2(t), i.e.,

y(t) = x1(t) (1.11)

Y (t) = X2(t) (1.12)

An immediate consequence of this identification is that at every instant the derivative ofx2(t) equals x1 (t)

X2(t) = XI (t)

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

d

dt (Y(t)) = -aiy(t) - a2Y(t)

(1.13)

and using equations (1.11-1.13) gives us the differential equation for x1(t) as

z1(t) = -a1x1(t) - a2x2(t) (1.14)

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

z(t) Ax(t) (1.15)

where

-a210

x(t)X2(t) j

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

y(t) = Cx(t) (1.16)

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8 Introduction to State Space

where

C= [O 1]

with C being a row vector referred to as the output matrix.In summary the second order differential equation (1.1) is equivalent to the vector

differential equation (1.15) and the output equation (1.16). These equations, (1.15, 1.16)constitute a state model for the second order system in the absence of an input.Alternatively, the state model can be represented by a block diagram involving theinterconnection of blocks which operate as summers, integrators, and scalar multiplierson the components of the state. The Laplace operator 1 Is is used to indicate integration.

More generally, we can use the foregoing procedure to obtain a state model for thezero-input response of higher order dynamic processes as follows.

Suppose the zero-input response of an nth order process is governed by

y(n) (t) + a, y(n-1) (t) + a2y(n-2) (t) ... + anY(t) = 0

where

YW (t) = d`y(t)dt'

(1.17)

Then we proceed as in the second order case to identify components of the state withderivatives of the output as

x1(t)Y (n

1) (t)x2(t) = Y(n 2)

(t)

xn(t) =Y(t)

(1.18)

Thus using (1.17, 1.18) we obtain a vector differential equation (1.15) having a system

-a,

z,(t) 3,(t) tS

Figure 1.3 Block diagram representation of the state model

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Solving the State Differential Equation 9

matrix A given as

A=

-a1 -a2 -a3 ... -an1 0 0 0

0 1 0 ... 0 (1.19)

L0 0 0 1 01

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

C= [0

... 0 1]

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

p(s) = sn + alsn-1 +a2Sn-2 +

... + an-ls + an

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

p(s) = det[sI - A]

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

1.4 Solving the State Differential EquationRecall that the solution to a scalar differential equation, e.g., (1.1), involves the scalarexponential function, eA'. In this section we will show that the solution to the statedifferential equation, (1.15), involves a square matrix, eA', which is referred to as thematrix exponential.

1.4.1 The matrix exponentialSuppose we are given the initial state x(0) and the system matrix A, either constant or timevarying. Then we obtain a solution to the state differential equation, (1.15), by finding0(t), the square matrix of scalar functions of time, such that

x(t) = O(t)x(0) (1.20)

where 0(t) is referred to as the transition matrix.Since the state at each instant of time must satisfy the state differential equation, (1.15),

the transition matrix is a matrix function of the system matrix A. In this book A isconstant. In this case the dependency of 0(t) on A is captured by the notation

0(t) = eAt (1.21)

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10 Introduction to State Space

where the square matrix eAt is referred to as the "matrix exponential of At" since it can beexpressed as an infinite series reminiscent of the infinite series for the exponential of ascalar- (1.6), i.e.,

eAt l + At +A2t2

+A

3t

3

+Ait'

2! 3!0

! (1.22)

In order to show that the transition matrix given by (1.22) solves the state differentialequation, (1.15), we differentiate the foregoing series expansion for the matrix exponen-tial of At to obtain

deAt

A+ 2A2t+

3A3t2 AeAt

+4A4t3 +...

dt= AeAt

Then using this relation to differentiate the assumed solution

x(t) = eAtx(0)

yields

z(t) = AeAtx(0) = Ax(t)

and we see that (1.23) solves the state differential equation, (1.15).

(1.23)

1.4.2 Calculating the matrix exponentialThere are many ways of determining eAt given A. Some of these approaches are suitablefor hand calculation and others are intended for use with a digital computer. Anapproach of the first kind results from using Laplace transforms to solve the statedifferential equation. We develop this approach as follows.

We begin by taking the Laplace transform of (1.15) to obtain

sX(s) - x(O) = AX(s) (1.24)

where A is 2 x 2 we have

X(s) _

Then rewriting (1.24) as

Xl(s)

X2(s)fxj(t)etdtX, (s) =

0

(sI - A) X(s) = x(0) (1.25)

we see that provided s is such that (sI - A) is invertible, we can solve (1.25) for X(s) as

X(s) _ (sI - A) tx(0) (1.26)

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Solving the State Differential Equation 11

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

adj[sI - A](s1 - A) = det[sI - A]

where when A is an n x n matrix, the adjugate matrix, adj [sI - A], is an n x n matrix ofpolynomials of degree less than n and det[sI - A] is a polynomial of degree n.

Finally, taking the inverse Laplace transform of (1.26) yields

x(t) = G-' [(sI - A)-1]x(0)

and we see, by comparing this equation with (1.23), that

e At _ L-' [(sI - A)-']

Now in the case where A is the 2 x 2 matrix given by (1.15), we have

1

(i_ adj[sI - A] _ s+al

a2sI - A)-

det[sI - A] -1 s1

where

det[sI - A] = s2 + a, s + a2 = (s )Il) (s - A2)

adj[sI - A] =s - a2

1 s+a1

(1.27)

Notice from the previous section that det[sI - A] = p(s), (1.4), is the characteristicpolynomial. In general any n by n system matrix A has a characteristic polynomial withroots {A : i = 1, 2 ... n} which are referred to as the eigenvalues of A. The eigenvalues ofthe system matrix A play an important role in determining a system's behavigr.

Returning to the problem of determining the transition matrix for A, (1.15), we applypartial fraction expansion to the expression for (sI - A)-', (1.27), assuming det[sI - A]has distinct roots, i.e., A A2i to obtain

s + al a2 K1 K2(1.28)

[ -1 s

+] s-'\1 s-1\2

where

K1 = limsa,adj[sI - A]

[(s- A1) det[sI - A]

Al -1\1A2

-A2

Al - A2

A2 -A1A211 -A1 JK2

sera (s - A2)adj[sI - A]

A2 - Aldet[sI - A]]

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12 Introduction to State Space

Finally, taking the inverse Laplace transform of (1.28), we obtain the transitionmatrix a-.

e"' (Al - 1

AI eA,r - AzeA2

eA'' - eA,t

_AiA2(eA,1 - eA't)

. ea'' + A-'z ie(1.29)

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

1.4.3 Proper and strictly proper rational functionsBefore continuing to the next section, notice that when A is an n x n matrix, adj [sI - A] isan n x n matrix of polynomials having degree no larger than n - 1. Thus, since thecharacteristic polynomial for A, det[sI - A], is of degree n, we see from (1.27) that(sI - A)-' is an n x n matrix of strictly proper rational functions.

In general a rational function

r(s) d(s)

is said to be;

(i)

deg[n(s)] < deg[d(s)]

(ii) proper when the degree of its numerator polynomial equals the degree of itsdenominator polynomial, i.e.,

deg[n(s)] = deg[d(s)]

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

1.5 Coordinate TransformationIn Section 1.3 we saw that the zero-input response for a system could be obtained bysolving a state vector differential equation where the components of the state wereidentified with the output and its derivatives. In this section we examine the effect ofchanging this identification.

strictly proper when the degree of its numerator polynomial is less than the degree ofits denominator polynomial, i.e.,

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Coordinate Transformation 13

1.5.1 Effect on the state modelReferring to the second order process used in the previous section, let z(t) denote the stateobtained by setting

y(t)

Y(t)

[xi(t)V

X2(t)(1.30)

where V is any invertible (nonsingular) 2 x 2 matrix of constants. In the previous sectionV was the identity matrix.

Now we see from (1.11, 1.12, 1.30) that the state x(t) used in the previous section isrelated in a one-to-one fashion to the state x(t) as

x(t) = Vx(t) (1.31)

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

(A, C) "'-+ (A, C) (1.32)

where

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

A = V-'AV

C=CV

V x= AVx(t)

which, since V is invertible, can be multiplied throughout by V-I to give

x (t) Ax(t)

where

A = V-'AV

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

y(t) = CX(t)

where

C=CV

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14 Introduction to State Space

Notice that the transition matrix, eA`, which applies in the new coordinates is related tothe transition matrix, eAt, in the original coordinates as

At V-'A V)Y (4.)1e = VV-1e

AtV

(1.33)

1.5.2 Determination of eAt

The flexibility provided by being able to choose the coordinates for the state modelrepresentation of a dynamic process is often of considerable use in the analysis and designof control systems. We can demonstrate this fact by using a change of coordinates tocalculate the transition matrix.

Suppose we are given a two dimensional system matrix A having a characteristicpolynomial, det[sI - A], with distinct roots (eigenvalues), i.e., Al A2. Then we canalways find a coordinate transformation matrix V so that the system matrix A in the newcoordinates is diagonal and

z (t) = Ax(t) (1.34)

where

a 0A = V-l AV = 1

0 A2

with entries along the diagonal of A being the eigenvalues of A.Now when the system matrix is diagonal, the corresponding transition matrix is also

diagonal. We can see this by noting that the state differential equation in thesecoordinates, (1.34), consists of two scalar first order differential equations which areuncoupled from each other

xl(t) = Al -xi(t)

x2(1) = A2x2(t)

so that their solution can be immediately written as

xl (t) = eAl txl (0)(1.35)

x2(t) = eA'`z2(0)

which in matrix form is

xl (t) l f e\tt 0 x1(0)[ A2t 0

(1.36)x2(t)

1

L 0 e )x2(

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Diagonalizing Coordinate Transformation 15

Thus we see that the transition matrix is indeed diagonal

10At

[et0 0e"

J

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

eAt = VeA`V-1

I

with V being the coordinate transformation matrix which makes A diagonal.Now we will see in the next section that, in general, the coordinate transformation

matrix V needed to make A diagonal depends on the eigenvectors of A. However in thespecial case when A is a 2 x 2 companion matrix, (1.15), with \1 # A2i the requiredcoordinate transformation matrix is simply related to the eigenvalues of A as

V=

(1.37)

(1.38)Al A2

1 1

We can see that this coordinate transformation gives rise to a diagonal system matrix byusing

V _(al-az) 1I 11aizI

to obtain

1 1 -a1A1 - 1\1/\2 - a2 -alaz - az - a2A=V AV

(1.39)

Al - A2 a1A1 + aZ + a2 a1A2 + alaz + az]

Then since s 2 + als + a2 = (s - A1) (s - \z) we have a1 = -(Al + A2) and-a2 = A11\2-Therefore the foregoing expression for A reduces to

A=L 10 a] (1.40)

L z

Finally, the expression obtained for eA` using V, (1.38), in (1.37) equals (1.29) whichwas obtained at the end of Section 1.4 through the use of Laplace transforms.

The foregoing approach to the determination of the transition matrix requires thedetermination of a coordinate transformation matrix V which diagonalizes the systemmatrix A. We will see in the next section that the columns of the coordinate transforma-tion matrix required to do this are right-eigenvectors for A.

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

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16 Introduction to State Space

eigenvalues there is at least one right and one left-eigenvector. Moreover we will see thatwhen the eigenvalues of A are distinct, the coordinate transformation V required to makeA diagonal has columns equal to the right-eigenvectors of A. In addition we will see thatV-1 has rows which are the transpose of the left-eigenvectors of A.

1.6.1 Right-Eigen vectorsConsider the special case when A is a two-by-two matrix in companion form, (1.15),having unequal eigenvalues {ai : 1, 2}. Then writing the characteristic polynomial as/\i = -al Ai - a2 we see that

11 021[1ij IIor

i = 1,2 (1.41)

Av'=Aiv' i=1,2

where

(1.42)

v`= 11`J i=1,2

Notice that (1.42) is a general expression relating the ith eigenvalue, right-eigenvectorpair (Ai, v`) for the any square matrix A, where v' is said to be the right-eigenvectorcorresponding to the eigenvalue Ai. These pairs play a major role in the state analysis ofsystems. The particular dependence of v` on A, in the present instance is a result of A beingin companion form.

Continuing with the construction of V to make V- 'AV diagonal, we combine theequations given by (1.42) for i = I and i = 2 to obtain

AV= VA (1.43)

where V, A are given as

V= [v1 v21 A=a1 0

0 A2

Now when A has distinct eigenvalues, i.e., a1 # A2, V is invertible and we can pre-multiply (1.43) by V-1 to obtain

V-'AV =A=A

Thus we see that V is the coordinate transformation matrix required to make A diagonal.More generally, suppose A is any n x n matrix, not necessarily in companion form,

which has distinct eigenvalues, ai Aj : i j. Now it turns out that this condition ofdistinct eigenvalues is sufficient for the eigenvectors of A to be independent, i.e., v` and v1point in different directions. Therefore V has independent columns and is therefore

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Diagonalizing Coordinate Transformation 17

invertible where

V = vI V2 ... v" 1 (1.44)

Av' _ .Aiv' i = 1, 2, , n (1.45)

and V- 'AV is diagonal.In the special case when A is in companion form, (1.19), its eigenvalues,

: i = 1 , 2, . . . , n} are related to its eigenvectors, {v` : i = 1, 2,. , n} as

viT = r)n-I Xn 2 .. a; 1

In order to see that this result holds, set the last entry in v` equal to one. Then taking Ain companion form, (1.19), solve the last scalar equation in (1.45) and use the result tosolve the second to last scalar equation in (1.45). We continue in this way solvingsuccessive scalar equations in (1.45), in reverse order, until we reach the first scalarequation. At this stage we will have all the entries in v'. These entries satisfy the first scalarequation in (1.45) since A, is a root of the characteristic polynomial whose coefficientsappear with negative signs along the first row of A.

In general, when A is not in any special form, there is no special relation between theeigenvalues and the corresponding eigenvectors. Thus in order to determine the eigen-value, right-eigenvector pairs when A is not in any special form we need to determine(A,, v`) : i = 1, 2, n so that the equations

Av` = A1v` i = 1, 2, ... n (1.46)

are satisfied.

1.6.2 Eigenvalue-Eigenvector problemThe problem of determining (a;, v`) pairs which satisfy (1.46) is referred to as theeigenvalue-eigenvector problem. There are well established methods for solving thisproblem using a digital computer. In order to gain additional insight into the nature ofthe eigenvalue-eigenvector problem we consider a theoretical approach to findingeigenvalue-eigenvector pairs to satisfy (1.46).

To begin, suppose we rewrite (1.46) as

(aI-A)v=o (1.47)

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

v = (Al - A)-'o

and the only solution is the trivial solution v = 0. However when (Al - A) is notinvertible, (1.47) can be satisfied by v 0.

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18 Introduction to State Space

Now from Crammer's rule for matrix inversion we have

(Al - A)-1= adj[al - A]det[AI - A]

Therefore Al - A does not have an inverse when det[.\l - A] = 0, i.e., A = A, aneigenvalue of A.

Next recall that singular matrices have dependent columns. Therefore A1I - A hasdependent columns so that we can find scalars {?,,'k : k = 1, 2, - . , n} not all zero such that

n

lE[(AJ - A)]kvk = 0k=1

(1.48)

where [(A11 - A)]k: k = 1, 2. . , n denote columns of a;1 - A. Notice that (1.48) can berewritten as (1.47) with A = A and v = v' where

v`T[v' v' v`

1 2 n]

Since we can always multiply (1.48) through by a nonzero scalar a, the solution, v', to(1.48) or (1.47) is not unique since av' is another solution. More generally, we say that theeigenvectors of a given matrix are determined to within a scalar multiple, i.e., thedirections of the eigenvectors are determined but their lengths are arbitrary.

Assuming that A has a complete set of (n independent) right-eigenvectors, we candecompose any initial state as

n

x(0)a,v`=Va

where

(YT = [a1 az ... an] V = [v1,v2,...vn]

with the a;s being found as

a = V-1x(0)

(1.49)

This decomposition of the state into a linear combination of right-eigenvectors of Aplays an important role in the analysis of system behavior. This is illustrated in the nextsection where we will use the eigenvectors of A to reveal certain fundamental properties ofstate trajectories.

Unfortunately, when an n x n matrix A has multiple eigenvalues, i.e., whendetfAl - AJ does not have n distinct roots, the number of eigenvectors may or may notbe equal to n, i.e., A may or may not have a complete set of eigenvectors. When A does nothave a complete set of eigenvectors, other related vectors, referred to as generalizedeigenvectors, (Appendix), can be used together with the eigenvectors to provide adecomposition of the state space in a manner similar to (1.49). However in this case it

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Diagonalizing Coordinate Transformation 19

is impossible to find a nonsingular matrix V such that V-l AV is diagonal. Then A is saidto be not diagonalizable.

In summary, the condition that A has distinct eigenvalues is sufficient but notnecessary for A to be diagonalizable. Most of the time, little additional insight intocontrol theory is gained by discussing the case where A does not have a complete set ofeigenvectors. Therefore, we are usually able to avoid this complication without loss ofunderstanding of the control theory.

1.6.3 Left-EigenvectorsSuppose A is any n x n matrix having n distinct eigenvalues. Then we have seen that thematrix V having columns which are right-eigenvectors of A is invertible and we can write

V-'AV = A (1.50)

whereAv` =Aiv` : i =

V = [vt v2 ... vn] A=f

Ai 0 ... 0

0 A2 ... 0

L 0 0 ... kNow suppose we post-multiply both sides of (1.50) by V-1 to obtain

V-lA = AV-'

Then if we denote V-1 in terms of its rows as

V-1 =

w1T

w2T

[wnT

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

w1TA

I w2TA

wnTA

Alww2T12

AnwnT

(1.51)

Therefore it follows, by equating corresponding rows on either side of this equation, that

w`TA=a,w`T i= 1 2 ,n (1.52)

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20 Introduction to State Space

which transposing throughout gives

AT wi = A wi i= (1.53)

Thus we see from equation (1.53) that the column vector w` is an eigenvector of ATwith corresponding eigenvalue Ai. However, since the row vector w'T appears on the leftside of A in equation (1.52), w' is referred to as a left-eigenvector of A to distinguish itfrom the corresponding right-eigenvector of A, namely, v'.

Notice that, since V-1 V is an identity matrix, the left and right-eigenvectors justdefined are related to each other as

wiTVj = 0 i j= I i =j

In addition notice that any nonzero scalar multiple of wi

satisfies (1.52), i.e.

(1.54)

ZiTA = A ZiT

Therefore z' is also a left-eigenvector of A and we see from equation (1.54) that in generalthe left and right eigenvectors are related as

ZITvj = 0 i i=,yi 54 0 i =j

This basic characteristic of left and right-eigenvectors, referred to as the orthogonalityproperty, is responsible for a number of fundamental facts relating to the behavior ofstate models.

1.6.4 Eigenvalue invarianceBefore we go to the next section, it is important to note the basic fact that the eigenvaluesof A and of A are the same whenever A is related to A as A = V-1AV for any invertiblematrix V. We can see this by premultiplying both sides of equation (1.45) by V-1 andinserting VV-1 between A and v', viz.,

V-lAVV-lv' _ AiV-iv`

Then setting t` = V-l v` and taking V-1AV = A gives

At' = Ait'

which implies that the eigenvalue, right-eigenvector pairs for A are (At, t'). Thus the

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State Trajectories Revisited 21

eigenvalues of A equal the eigenvalues of A independent of the coordinate transformationmatrix V.

Alternatively, another way we can see this fact is to carry out the followingmanipulations

det[sI - A] = det[sI - V-1 AV] = det[V-'(sI - A)V]

= det V-1 det[sI - A] det V

= det[sI - A]

Thus A and A have the same characteristic polynomial, and since the roots of apolynomial are uniquely dependent on the coefficients of the polynomial, A and Ahave the same eigenvalues.

Finally, since the differential equations modeling the behavior of dynamical processesmust have real coefficients, we can always work in coordinates where the state modelparameters, (A, C), are real. As a consequence, if (A,, v') is a complex eigenvalue-eigenvector pair for A, then (A , v`*) is also an eigenvalue-eigenvector pair for A.

1.7 State Trajectories RevisitedWe saw in Section 1.6.2 that assuming A has a complete set of eigenvectors, any initialstate can be written in terms of the eigenvectors of A, (1.49). In this section this fact is used

'to gain additional insight into the nature of a system's state trajectories and zero-inputresponse.

More specifically, under certain conditions on the matrix pair, (A, C), a system canexhibit a null zero-input response, y(t) = 0 for all t > 0, for some non-null initial state,x(0) q. When this occurs we say that the state trajectory is orthogonal (perpendicular)to CT, denoted CT lx(t), since the output depends on the state as y(t) = Cx(t). Twovectors a, A are said to be orthogonal if

aT,0=0

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

x(t) y(t)v

satisfying

Cv=0

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

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22 Introduction to State Space

any trajectory orthogonal to CT can be decomposed into a sum of straight linetrajectories all lying in the n - 1 dimensional subspace orthogonal to CT. Therefore anunderstanding of straight line trajectories is essential to an understanding of the propertyposed by certain systems of having a null zero-input response to certain initial states.

1.7.1 Straight line state trajectories: diagonal ASuppose A is a 2 x 2, real, diagonal matrix. Then the state trajectory is a straight linewhenever the initial state lies only on one of the two coordinate axes. We can see thisimmediately as follows.

Consider the initial states

X'(0) = L 00)1 X2(0) - [ -t 20(0)

where xl (0) and, -t2(0) are any real nonzero scalars. Then recalling from Section 1.5.2 thatthe transition matrix eA` is diagonal when A is diagonal, we see from (1.36) that the statecorresponding to each of these initial states is

(t) = I Xl (0)eJ

for z(0) = x'(0)

X(t) = [x2(O)0

e1\2tj for x(0) = x2(0)

(1.55)

The foregoing suggests that the trajectory for any initial state in these coordinates

x (0)X1(0)1

Lx2(0)J

can be written as

x(t) _ (x1(0)eA'`)i' + (X2(0)e'2t)i2

where ik : k = 1, 2 are columns from the 2 x 2 identity matrix, viz.,

I= Ii I i2 ]

More generally, when A is a real, diagonal, n x n matrix, the state trajectory

(1.56)

x(t) = Xk(0)e)'k`ik (1.57)

results when the initial state

xT(0) = [X1(0) x2(0)... x(0)1

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State Trajectories Revisited 23

has components which satisfy

zi(0) = 0 for i j4 k

0 fori=k

where i k is the kth column from the n x n identity matrix.The foregoing result implies that the zero-input state response for any initial state can

be written asn

x(t) = E(xk(O)e1kt)ikk-1

whenn

x(0) = E tk(0)ikk=1

In summary, the state trajectory is the vector sum of state trajectories along coordinateaxes where each coordinate axis trajectory depends on one of the system modes in the setof system modes, {eAkt : k = 1, 2 , n}.

Now in order to generalize the foregoing result to the case where A is not diagonal, wesuppose, in the next section, that the foregoing diagonal case resulted from a coordinatetransformation from the original coordinates in which A is given.

1.7.2 Straight line state trajectories: real eigenvaluesRecalling that V, equation (1.44), is the coordinate transformation needed to diagonalize--A and taking z(0) as in (1.57) we have

x(0) = Vx(0) = tk(0)vk (1.58)

where Avk = )\kvk. Then, using the series expansion for the matrix exponential, (1.22), wesee that when x(O) is given by equation (1.58) we have

Ax(t) = eAtx(0) =

(I+At+2)k(o)vk

2 2

- 4(0) 1 + Akt + AZi ... vk =(xk(0)eakt)vk

(1.59)

Now with Ak real we see from equation (1.59) that the point representing the statemoves along the eigenvector, vk, towards the origin when Ak < 0 and away from the originwhen Ak > 0. The case where Ak is complex is taken up in the next subsection.

More generally, assuming A has a complete set of eigenvectors, we can write any initialstate as

n

x(0) _ ryivi (1.60)

i=1

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24 Introduction to State Space

where 'y = V-1x(0) and

V= [ 211 4,2

Then the state trajectory can be written as

7T-[71 72 ' .. 7n]

(1.61)

Now recall, from Section 1.2, that a system is stable if

lim x(t) = 0t 30

for all x(0) (1.62)

Therefore assuming A has real eigenvalues, we see from equation (1.59) that the system isstable if and only if

Ai <0

It should be emphasized that a system is stable if and only if equation (1.62) is satisfied.Therefore if we need to restrict the initial condition in order to ensure that the zero-inputstate trajectory goes to the origin, the system is not stable. For example, referring to theexpansion (1.60), any initial state, x(0), which is restricted so that its expansion satisfies

7i=0 when A, > 0

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

1.7.3 Straight line trajectories: complex eigenvaluesSince the differential equations governing the input-output behavior of the physicalprocesses we are interested in controlling have real coefficients, we can always choose towork in coordinates so that A, C are real matrices. Then since the eigenvalues, if complex,occur in conjugate pairs we see that the corresponding eigenvectors are also conjugates ofeach other, i.e., if (A, v) is a complex, eigenvalue-eigenvector pair, then (A*, v*T) is also aneigenvalue-eigenvector pair of A.

Now if the initial state is any scalar multiple of the real or imaginary part of v, theresulting state trajectory lies in a two dimensional subspace of state space composed fromthe real and imaginary parts of v. More specifically, suppose

x(0) = 2[v + v*T] = 7Re[v]

where 7 is any real scalar and

(1.63)

v = Re[v] +jlm[v]

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State Trajectories Revisited 25

Then from (1.59) we obtain

x(t) = a,-e(t)Re[v] + ai,,,(t)Im[v] (1.64)

where

a, (t) = yeRe[A]t eos(Im[A]t)

aim(t) = -7eRe[A]t sin(Im[A]t)

The foregoing generalizes the observation made at the beginning of the chapter that,for second order systems, spiral shaped trajectories result when the roots of thecharacteristic equation consist of a complex conjugate pair. In the present case wherethe system order n > 2, the spiral shaped trajectory lies in a two-dimensional subspace ofthe n-dimensional state space.

Notice that the initial state was chosen to ensure that the state x(t) is real. However ifwe choose the initial state as

x(0) = a1v +a2v*T

with the real scalars ai satisfying

loll 1a21

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

1.7.14 Null output zero-input responseHaving discussed straight line state trajectories, we return to the problem stated at thebeginning of this section concerning the possibility of having initial states which producenull outputs.

Suppose the output matrix happens to satisfy

Cvk=0 (1.65)

for some eigenvector, vk, of A. Then it follows from (1.59) that if x(O) = 7kvk then

y(t) = Cx(t) ='ykeAktCvk = 0

and we see the output is null for all time when the initial state lies along vk.This effect of the existence of non-null initial states which do not affect the output, is

related to a property of the state model's A and C matrices which is referred to as thesystem's observability (Chapter 3). More is said about this matter in Section 3.5.2.

In order to obtain some appreciation of the importance of this effect consider a statemodel with A having all its eigenvalues, except Ak, in the open left-half plane. Then thisstate model is internally unstable since x(0) = vk produces a trajectory which moves away

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26 Introduction to State Space

from the origin. However if C satisfies equation (1.65) this trajectory has no effect on theoutput and in this case we have

lim y(t) = 0 for all x(0)t x

and the system is externally stable.This demonstrates that while the state model is internally unstable its output behavior

is stable. However, since it is practically impossible to exactly model a physical process,the foregoing stability of the output in response to initial states exists on paper only and isreferred to by saying that the system is not robustly output stable. For this reason, we saythat a system is stable if and only if its A matrix has all its eigenvalues in the open left-halfplane.

Before we leave this section, it is instructive to consider conditions on C whichguarantee that equation (1.65) is satisfied. Recall, from Section 1.6.3, that right and left-eigenvectors corresponding to different eigenvalues are orthogonal,

w`T vj =0

Therefore suppose A has a complete set of eigenvectors so that we can expand C in termsof the left-eigenvectors of A,

C a`WtT (1.66)

Then we see that equation (1.65) is satisfied when C is independent of wk, i.e., when ak = 0in equation (1.66). This structural characterization of C will be used in Chapter 3 todevelop properties of state models which relate to their observability.

1.8 State Space Models for the Complete ResponseSo far we have used the state space to model a system's zero-input response. In this sectionwe take up the use of the state space in connection with the modeling of a system's zero-state response.

1.8.1 Second order process revisitedReturning to the electric circuit used in Section 1.2, suppose we connect a voltage sourceas show in Figure 1.4. Then the differential equation governing the output, y(t) (capacitorvoltage), becomes

ddytZt)+ al dd(tt) + azY(t) = bzu(t) (1.67)

where u(t) is the input (source voltage) and al, a2 are as in (1.1) with b2 = az.

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State Space Models for the Complete Response 27

Figure 1.4 Electric circuit with voltage input

Suppose, as in Section 1.3, that we choose the components of the state as

xl (t) _ y(t)]

(1.68)xz(t) y(t)

so that

zz(t) = x1(t) (1.69)

Then we see from (1.67) that

(t)z

(t) - (t) + b (t)z - (1 70)xl 2ua2x2a,x1dt

.

and from (1.68-1.70) that the state differential equation and output equation are given as

z(t) = Ax(t) + Bu(t) (1.71)

y(t) = Cx(t) (1.72)

where

A-a1

Oz] B=[b

OZ]

C= [ 0 1

with B being referred to as the input matrix. This state model is complete in as much as itcan be used to obtain the output caused by any specified combination of initial state, x(0),and input, u(t). We will see that the matrix product of the initial state and transitionmatrix, which gives the zero-input response, is replaced in the calculation of the zero-stateresponse by an integral, referred to as the convolution integral, of the imput andtransition matrix. Before showing this, consider the following modification of theforegoing state model.

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28 Introduction to State Space

Suppose we rewrite (1.67) as

a, 4y(t) + a,y(t) = u(t)dt dt

where

Y(t) =(t)

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

z,.(t) = A,x,(t) + B,u(t) (1.73)

y(t) = C'x'(t)

where A, = A and

B,= IB= [']b2 0

C, = b2C = [ 0 b2 ]

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

1.8.2 Some essential features of state modelsFirst notice that when we change coordinates by setting x(t) = Vx(t) for some constantinvertible matrix V, the resulting state model in the new coordinates is given by

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

y(t) = Cx(t)

where the parameters for the original and transformed state models are related as

(A, B, C) H (A, B, C)

where

A= V-'AV B= V-'B C= CV

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State Space Models for the Complete Response 29

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

y(t) = Cx(t) + Du(t)

Notice that, unlike the other state model parameters (A, B, C), the D parameter isunaltered by a state coordinate transformation.

Third, some dynamic processes have more than one scalar input and/or more than onescalar output. For instance a dynamic process may have m scalar inputs,{u,(t) : i c [1, m]} and p scalar outputs, {y,(t) : i E [l,p]}. In this situation the systeminput, u(t), and system output, y(t), are column vectors of size m and p respectively

uT (t) = [U1 (t) U2(t) ... Um(t) ]

yT (t) = [yl (t) y2(t) ... yp(t)

Further, in this case, the state model has an input matrix B with m columns and an outputmatrix C with p rows. More specifically, the general form for state models used here isgiven as

z(t) = Ax(t) + Bu(t)

y(t) = Cx(t) + Du(t)(1.74)

where x(t), u(t) and y(t) are column vectors of time functions of length n, m and prespectively and the state model parameters, (A, B, C, D), are matrices of constants of sizen x n, n x in, p x n, and p x m respectively.

Finally, systems having p > 1 and m > 1 are referred to as multiple-input-multiple-output (MIMO) systems and systems having p = 1 and m = 1 are referred to as single-input-single-output (SISO) systems.

1.8.3 Zero-state responseRecall from Section 1.3 that the zero-input response, y,i(t), depends on the transitionmatrix and initial state through multiplication as y,i(t) = CeA`x(0). In this section we willshow that the zero-state response, yzs(t), depends on the transition matrix and the inputthrough integration as

C J eA(`-T)Bu(T)dT + Du(t)t0

where the integral is known as the convolution integral.We begin the development of the foregoing relation by assuming that the initial state

is null, x(O) = 0, and that the state model parameters, (A, B, C, D) are known. Thentaking Laplace transforms throughout the state differential and output equations, (1.74),

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30 Introduction to State Space

gives

sX(s) = AX(s) + BU(s) (1.75)

Yn(s) = CX(s) + DU(s) (1.76)

Next solving (1.75) for X(s) yields

X(s) = (sI - A)-'BU(s)

and substituting this expression for X(s) in (1.76) yields

Y,,, (s) = G(s)U(s)

where

(1.77)

G(s) = C(sI - A)-' B + D (1.78)

Now, recalling from Section 1.4.3 that (sI - A)-1 is a matrix of strictly proper rationalfunctions, we see that Gp(s) is strictly proper where

Gyp(s) = C(sI - A)-1B

so that

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

and

lim G' (S) = 0s-CC

G(s) = Gsp(s) +D (1.79)

lim G(s) = D

Notice that two state models which are related by a coordinate transformation havethe same transfer function. We can see this by calculating the transfer function using thestate model in the transformed coordinates, (A, B, C, D)

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

and substituting A = V-'AV, B = V-1B, C = CV, D = D to obtain

G(s) = CV[V-1(sl -A) V]-'V-1B+D

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

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State Space Models for the Complete Response 31

Next recalling the following general property of Laplace transforms

ffi(t -7)fz(r)drr0

we see that setting

F, (s) = C(sI - A)-1 F2(s) = BU(s)

so that

f, (t) = CeA` f2(t) = Bu(t)

gives the inverse transform of (1.77) fas

yZS(t) = C JeAl`-T1 Bu(T)drr + Du(t) (1.80)

Notice that when D is null and the input is an impulse, u(t) = 6(t), (1.80) gives

YZS (t) = CeA`B

In addition recalling from section 1.4.2 that

G[eA`] = (sI - A)-'

(1.81)

we see that y,, (t), (1.81), has Laplace transform equal to GP(s) as we expect since theLaplace transform of the impulse response equals the transfer function. In addition wesee from (1.81) that the zero-input response equals the impulse response when the initialstate is x(0) = B.

Now we need to define eAt as a null matrix when t < 0 in (1.80, 1.81). This is done tomatch the mathematics to the physical fact that the future input, u(T) : T > t, does notaffect the output at time t, i.e., y,,(t) is independent of future values of the input. Thisproperty of the transition matrix, i.e., q(t) = 0 for all t < 0, is referred to as the causalityconstraint and applies when we use the transition matrix in connection with the zero-stateresponse. Thus the causality constraint forces the integrand in (1.80) to be null for T > tand enables (1.80) to be rewritten as

ty_s(t) = C f eA(t-T)Bu(T)dr + Du(t)0

Notice that when the transition matrix is used in connection with the zero-inputresponse we can interpret 0(-t) for t > 0 as the matrix needed to determine the initialstate from the state at time t, i.e., x(0) = 0(-t)x(t), which implies that 0(-t) is the inverseof 0(t).

At this point we can see, by recalling the principle of superposition, that when a systemis subjected to both a non-null initial state, x(0), and a non-null input, u(t), we can write

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32 Introduction to State Space

the output as

Y(t) =Yzr(t) + y,.,.(t) (1.82)

where

Yzi(t) = CeAtx(O)

/tY_-, (t) = C J eA('-T)Bu(T)dr + Du(t)I

Notice that:

(i) yzi(t), the zero-input response, is caused solely by to x(O)(ii) yZS.(t), the zero-state response, is caused solely by to u(t).

In this section we have developed several essential relations involving the state modelof a dynamic process. These important relations provide the complete response of asystem to any specified input and any specified initial state.

1.9 Diagonal Form State ModelIn subsequent chapters we will encounter a number of fundamental properties of statemodels in connection with their use in the design of feedback control systems. A simpleway of beginning to appreciate these properties is to consider state models in coordinateswhere the system matrix A is diagonal. This type of state model is referred to as a diagonalor normal form state model. We encountered this model earlier in Section 1.5.2 inconnection with the determination of the matrix exponential, and in Section 1.7.1 inconnection with straight line state trajectories.

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

zi(t) = Aix;(t) + b,u(t) i = 1, 2, . . , n (1.83)

with the output being a scalar multiple of these components

y(t) _ c,x;(t) + Du(t) (1.84)

We can visualize these equations as a block diagram of the sort introduced in Section 1.3to model the zero-input response.

Alternatively, when we view this state model in the frequency domain so as to obtainthe plant's transfer function, we see that since the state model's system matrix, A, is

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Diagonal Form State Model 33

ip)

s.(0

a,

Cl

Figure 1.5 Block diagram representation for diagonal form state model

diagonal, we have (sI - A)-' diagonal,

1(s - 'XI)-) 0 ... 0

(sI - A)-10 (s - A2)-' ... 0

0 0 ... (s -

(1.85)

This fact simplifies the dependency of the transfer function on the elements in the B and Cmatrices since we have

G(s) = C(sI - A)-'B + D

= . c`b`+ D1_)s-a,

where

C=[C) C2 ... c,, ] B=

[b)b2

b

(1.86)

1.9.2 PropertiesNotice from (1.83) that the evolution of the components of the state, Xk(t), are dccoupledfrom each other, i.e., xk(t) is independent of {xi(t) : i E [0, n], i 54 k}. This fact plays animportant role in the following discussion.

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34 Introduction to State Space

We begin by noting that this decoupling of components of the state from each otherimplies that any component of the state, say Xk(t), depends on the input through the kthcomponent of the input matrix, bk, only. Therefore, if one of the components of B is zero,the corresponding component of the state is not affected by the input. Alternatively,notice from (1.86) that when bk = 0, the transfer function, G(s), does not have a pole at Akand the zero-state response does not involve the mode eakr

These phenomena are related to a property of state models known as controllability.Roughly speaking, a system is said to be controllable if its state can be forced, by theinput, to equal any point in state space at some finite time . Therefore if bk = 0, theevolution of the kth component of the state, Xk(t), is independent of the input, u(t), andthe state model is not controllable since there is no way of manipulating the input to makexk(tf) equal some specified value after the passage of a finite amount of time, tf. We willsee in the next chapter that controllability is a necessary and sufficient condition for beingable to arbitrarily assign the closed loop system matrix eigenvalues by using statefeedback. Unlike the diagonal form state model, the controllability of a state model inother coordinates, depends on both the A and B matrices.

The foregoing observations were obtained by viewing the system from the input side.We can obtain dual observations by viewing the system from the output side involving theoutput instead of the input. We develop these observations as follows.

First, since equation (1.83) decouples the evolution the components of the state fromeach other, we see that the output, y(t), (1.84), depends on the kth component of the state,xk(t), through Ck only. Therefore, if one of the components of C is zero, the correspondingcomponent of the state does not affect the output. Alternatively, notice from (1.86) thatwhen ck = 0, the transfer function, G(s), does not have a pole at Ak and the zero-stateresponse does not involve the mode e1'kr

The foregoing phenomena are related to the property of state models known asobservability. Recall that we encountered this property in the previous section inconnection with the system's zero-input response. In the present instance we see thatthe observability of a system also relates to the system's zero-state response. Notice thatunlike the diagonal form state model, the observability of a state model in othercoordinates, depends on both the A and C matrices.

In summary, we see from equation (1.86) that if either bk or Ck is zero for some integerk C [0, n], then the system transfer function G(s) does not have a pole at Ak and the zero-state response does not involve the mode e)'kt. Moreover, under these conditions, if wetransformed coordinates and recomputed the transfer function from the parameters ofthe state model in the new coordinates, i.e.,

G(s) = C(sI - A)-I B + D

we would find that the both the numerator polynomial, Cadj(sI -A) B, and denominatorpolynomial, det(sI - A) (or det(sI - A) since the characteristic polynomial is invariantto coordinate transformation), would have a zero at Ak. Since S - Ak would then be acommon factor of Cadj (sI - A)B and det(sI - A), it would cancel from the ratio of thesepolynomials and be absent from the transfer function. Thus the eigenvalue of A at Ak isnot a pole of G(s) and has no effect on the system's input-output.

Given a transfer function, any state model that is both controllable and observable andhas the same input-output behavior as the transfer function, is said to be a minimal

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Diagonal Form State Model 35

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

1.9.3 Obtaining the diagonal form state modelThe relation between the form for the transfer function, equation (1.86) and the diagonalform state model suggests a way of determining a diagonal form state model for a givenSISO transfer function, G(s), having simple poles. This is done as follows.

We begin by using one cycle of division when the given transfer function is proper, toseparate out the strictly proper part of the transfer function. Thus suppose we are given atransfer function which is proper

n

G,(s) =

Y_ 0"-i S'i-0

n-1

Sn + > Qn_i Si-0

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

G,(s) = Gsp(s) + 30

where

n-1

bn-i Sii=0

Gsp(s) = n 1bi =,3i - 00ai i =

n C iS + Rn_i Si=0

7 - 7

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

Gsp(s) _k`

(1.87)i=1 S - Ai

we can obtain the diagonal form state model by factoring the kis as

ki = cibi

and referring to equation (1.86).Notice it is only possible to obtain a diagonal form state model if the transfer function

can be expanded as equation (1.87), i.e., if the system can be decomposed into first orderSystems. When GP(s) has simple poles we can always expand G,sp(s), as equation (1.87).

Notice that if GP (s) has simple poles some of which are complex, the diagonal form

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36 Introduction to State Space

state model will be complex. In some situations we may want the benefits of the diagonalform state model, i.e., decoupled state differential equations, but require the state modelto be real. In this case a reasonable compromise is to combine complex conjugate pairs ofterms in equation (1.87) so that the resulting realization has an A matrix which is blockdiagonal with each pair of complex conjugate poles corresponding to a 2 x 2 real blockon the diagonal of A. More specifically we expand GP(s) as

GSP(s) = Gspr(s) + GSP(.(s) (1.88)

where

G.SPr(S) _ ki Ai reali 1 s Ai

bils + bitGSP, (s) 2

i=1 S + ails +ai2

with the n,, complex pairs of poles of GP(s) being the poles of GS.P,(s), i.e.,

s2 + ails + ai2 = (S - Ai+n,)(S - +n,)

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

A-L J B-

[A3A4 B2

C = [Cl C2]

where (Al, Bl, Cl) constitute a diagonal form state model for GSP,(s), i.e., Al is diagonalwith diagonal elements equal to the real poles and the elements of B1 and Cl satisfyclibil = ki for i = 1, 2, . . n,. Both off diagonal blocks of A, namely A2 and A3 are null.

The remaining blocks constitute a state model for GPc(s) as

A41 0 ... 00 A42 ... 0

A4 = B2 =

B21

B22

L 0 0 ... A4n, J L B2n,. J

C2 =L

C21 C22 ... C2n, J

where {(A4i, B2i, C2f) : i = 1, 2,- n,} are 2 dimensional real parameter state modelswhich we obtain from {G,i : i = 1, 2, nc} with

bits + bi2G`i

_s2+ails +ai2

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Computer Calculation of the State and Output 37

One such real parameter state model is given as

ail ai2 1

A4' 1 0 ]B2 = [0 (1.89)

C21 = I b, I bit I

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

1.10 Computer Calculation of the State and OutputA fundamental problem encountered in using digital computers to analyses controlproblems is created by the fact that the dynamic processes needing to be controlledoperate in continuous time whereas digital computers perform operations in discretetime. In this section we indicate how this fundamental mismatch is overcome.

In order to be able to use a digital computer to calculate the system output y(t) from(1.82) we need to employ a piecewise constant approximation of the input u(t) over thetime interval of interest, [0, tN]. This is done by partitioning the interval into N equal non-overlaping contiguous subintervals each of time duration T. Then the piecewise constantapproximation, u(t), of the input is given by

u(t) = U(tk) for t e [tk, tk+1) (1.90)

where

tk=kT:

k = 0, 1, 2, . . . N} is referred to as the discrete-time equivalent of thecontinuous-time input, u(t), and the values, U(tk), of the discrete-time equivalent inputare referred to as the sampled values of the input. The continuous time "staircase like"function u(t) generated by this approximation is referred to as the sample and holdequivalent of the input. Figure 1.6 shows a plot of a typical input and its sample and holdequivalent.

The output, y(t), at the sample times, tk : k = 0, 1, 2, , N, is computed by repeatingthe calculation indicated in equation (1.82) for each subinterval with kT considered as thestart time for the computation of y([k + 1] T). This is carried out in a cyclical fashion withthe state that is computed at the end of a given subinterval being used as the initial statefor the computation of the state at the end of the next subinterval.

More specifically, the computations are started by specifying the initial state, x(0), andthe input, u(t), over the interval of interest, [0, tN]. The computerprogram then decides onthe sample spacing T. This is done so that the sample and hold version of the input is a"close" approximation to the actual input in the time interval [0, tN]. Then taking the

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38 Introduction to State Space

U(t)

At)

tift2

I

t3

1t5

fU t6 t7

Iis

Figure 1.6 Typical input and its piecewise constant approximation

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

JeA(T_T)Bu(O)dTx(T) = eATx(0) +

0

y(T) = Cx(T) + Du(T)

which can be rewritten as

x(T) = Fx(0) + Gu(0)

y(T) = Cx(T) + Du(T)

where

FeATT

G = f eA(T_T)Bdr

0

t

(1.91)

Then to compute y(2T), we replace x(0) by x(T) and u(0) by u(T) in (1.91). This gives

x(2T) = Fx(T) + Gu(T)

y(2T) = Cx(2T) + Du(2T)

This procedure continues with the output at the (k + 1)th sample time being given as

x[(k + 1)T] = Fx(kT) + Gu(kT)

y[(k+ 1) T] = Cx[(k+ 1)T] +Du[(k+ 1)T]

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Notes and References 39

There are several ways in which the computation of F and G can be carried out. Asimple way of doing this is to truncate the series expansion for the matrix exponential,(1.22), with the upper limit, no, of the sum being chosen large enough to achieve somespecified accuracy. Thus F, Gin (1.91) can be expressed as

no A'T'F=E +E.i!

=o

G=(fTn:AIEi

no AiT i+1E (i+l)!)B+EG

where the error matrices EF and EG are made negligible by choosing no "large enough" sothat we can take F and G as

. no A;7.;F=E

i-o i!

no A;7.;+

i=oG

(i+1)! B

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

1.11 Notes and ReferencesThere have been a number of excellent textbooks expounding on state space analysis ofsystems. Some of these books are [5], [4], [23], [6], [9], [15].Throughout the eighties the book by Kailath [23] was used widely as a text for courses onstate space analysis of linear systems. Although Brockett's book [5], is quite old and waswritten for a more mathematically sophisticated reader, the first half of the book doesgive the reader a nice exposure to the more general nature of the subject. The treatment ofstate modeling in the brief book by Blackman [4], shares the view with the present bookthat understanding basic ideas in the state space can be facilitated by thinking geome-trically.

Concerning books that provide insight into matters pertaining to linear algebra,Strange's book [39] gives an excellent introduction. The book by Golub and Van Loan[17] provides a more advanced treatment which is more orientated towards computercomputation. In addition the book by Brogan [6], plus the appendix to Kailath's bookalso provide useful information on linear algebra.

Page 55: Linear Control Theory - The State Space Approach Frederick Walker Fairman
Page 56: Linear Control Theory - The State Space Approach Frederick Walker Fairman

2State Feedback andControllability

2.1 IntroductionAn, important and well known consequence of feedback is behavioral modification.Feedback is present when the input to a system depends on that system's output. Whenfeedback is used to modify a system's behavior the resulting closed loop system is referredto as a feedback control system. For the purposes of discussion we can separate thecontrol system into a plant and a controller. The plant consists of the physical process tobe controlled plus any transducers, actuators or sensors, needed to interface the physicalprocess output to the controller and the output of the controller to the input of thephysical process. While controllers were implemented in the past by a physical processsuch as an electric filter, their implementation now is often done by a digital computer.Figure 2.1 shows the general setup for a control system.

u2(t)

PLANT

Y2(t)

CONTROLLER

Figure 2.1 Block diagram of the feedback control scheme

ul (t) = external input y, (t) = desired outputu2(t) = controlled input y2(t) = measured output

where u1 (t), u2(t),Yl (t), y2(t) have dimensions ml, m2,p1,P2 respectively.

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42 State Feedback and Controllability

The general form of the state model for the plant is

z(t) = Ax(t) + B, (t)u, (t) + B2u2(t)

Yt (t) = C,x(t) + D>>ut (t) + D12u2(t) (2.1)

Y2(t) = C2x(t) + D21 u, (t) + D22u2(t)

This chapter focuses on the effect of using a feedback signal composed from thecomponents of the plant's state as the controlled input,

u2(t) = Kx(t) (2.2)

where K has m2 rows, one row for each scalar signal making up the controlled input. Thisimportant kind of feedback is referred to as state feedback and is one of two ingredientsneeded to complete the design of the controller.

The need for something in addition to state feedback can be appreciated when it isrecognized that the plant state is usually unknown. Therefore if we are to use statefeedback as a basis for the design of a controller, we must have a means for estimating theplant state. This is done by a state model, referred to as a state estimator, which isdesigned from knowledge of the plant's state model. The information on the plant statecontained in measurements of all or some of the plant's input and output signals issupplied to the state estimator by setting the estimator's input equal to these measure-ments. The state estimator is designed so that its state mimics the state of the plant. Thereare several different types of state estimator depending on the assumptions made aboutthe relation between the actual plant signals and their measurements, e.g., exact,corrupted by random measurement noise, or partly unknown. In subsequent chapterswe will study both the design of these estimators and the effect on the performance of thefeedback control system that is caused by their use in conjunction with state feedback.

In this chapter we assume that the state of the plant's state model is known so that wecan use it to generate the state feedback signal given by equation (2.2). Our goal in usingstate feedback is to produce a closed loop system in which the zero state and zero inputbehavior is changed in some way which is beneficial to the plant's use. The extent to whichthe plant's controlled input can affect the plant state, is referred to as a system'scontrollability. We will see that the controllability of the plant state model determinesthe extent to which the state model of the closed loop system can differ from the plant'sstate model when we use state feedback. The important property of state modelcontrollability will be encountered in several different guises throughout this chapterand the rest of the text.

2.2 State FeedbackApplying state feedback, equation (2.2), to the plant, equation (2.1) gives the state modelfor the feedback system as

i(t) = (A + B2K)x(t) + B, (t)ui (t)

Yi (t) _ (Cl + D12K)x(t) + D u, (t) (2.3)

Y2(t) _ (C2 + D22K)x(t) + D21 U1 (t)

Figure 2.2 gives a block diagram interpretation of state feedback.

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State Feedback 43

uz(t)

CONTROLLER

I

PLANT

K

Figure 2.2 Block diagram representation of state feedback.

Notice that the main effect of using state feedback is the transformation of the systemmatrix from A, when the plant stands alone, to A + B2K, when the plant is embedded in astate feedback loop. Thus we see that the main purpose of state feedback is the assignmentof eigenvalues to the closed loop system matrix A + B2K to achieve a set of specifiedperformance criteria for the behavior of the closed loop system from the external inputu) (t) to the desired output yl (t). This fact leads immediately to the question: can wechoose K so that the eigenvalues of A + B2K are assigned to any specified set of values?The answer to this question is obviously of great importance to the use of state feedbackin the design of feedback control systems. The answer to this question, which is given inthe following theorem, involves the use of the left-eigenvectors of A. Since this propertyapplies in general to any state model (A, B, C, D) we will drop the subscript on B, i.e., B2will be replaced by B.

Theorem 2.1 Whenever A has a left-eigenvector w', i.e., w`TA = A;w'T, such that

W'TB=0

the corresponding eigenvalue of A, A j, is invariant to state feedback, i.e., A, E A(A + BK)for all K.

Proof Suppose the condition in the theorem is satisfied. Then multiplying the closedloop state feedback system matrix, A + BK, on the left by w'T gives

w'T(A+BK) =WITA+w'TBK

However since w'TA = A w'T and w,TBK = 0, the previous equation becomes

w'T(A+BK) =A;w'T

and we see that A is an eigenvalue of the system matrix for the state model of the feedbackcontrol system, since Al E A(A + BK) for all K.

It turns out that when there are no left-eigenvectors of A which satisfy wT B = 0, theeigenvalues of A are all assignable to any specified values by state feedback. This is shownlater in Theorem 2.4. Notice that when we have the condition WT B = 0 so that thecorresponding eigenvalue Al cannot be assigned by state feedback, we refer to A, as an

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44 State Feedback and Controllability

uncontrollable eigenvalue. Conversely, when we have the condition wT B 0 we refer tothe corresponding eigenvalue \i which can be assigned by state feedback, as a con-trollable eigenvalue.

Definition 2.1 An eigenvalue \i of A is said to be a controllable (uncontrollable)eigenvalue of the pair (A, B) if we can (cannot) find K so that Ai 0 )+[A + BK]

Uncontrollable (controllable) eigenvalues for a pair (A, B) are invariant to changingcoordinates. This can be seen as follows.

Suppose \i is an uncontrollable eigenvalue of (A, B). Then we have

w'T B = 0 (2.4)

wiTA = A.wiT(2.5)

Now insert TT-1 between w`T and B in equation (2.4) and between w`T and A, in equation(2.5). In addition, postmultiply both sides of equation (2.5) by T. Then we obtain

gITB=0

where

giT=

wiTT

q`T A = AigIT

B=T-1B A=T_'AT

Thus we see that Ai remains an uncontrollable eigenvalue of the state model in thetransformed coordinates.

Reflection on the foregoing reveals that plant state models which have no eigenvaluesof A which are both uncontrollable and unstable have the property that state feedbackcan be used to make the closed loop system stable. State models having this property arereferred to as stabilizable state models. Conversely, a state model is not stabilizable whenit has uncontrollable eigenvalues which are unstable. This is an important property sincethe stability of a control system is almost always a necessary condition for a controlsystem to be of any use.

Definition 2.2 A plant state model is said to be stabilizable if all its uncontrollableeigenvalues are stable.

Recall that the poles of the transfer function for an SISO output feedback controlsystem are constrained to lie on branches of the root locus. In the present case of statefeedback, the location of the eigenvalues of the closed loop system matrix A + BK areunconstrained provided, of course, that they are controllable. This additional flexibilityof state feedback over output feedback arises from there being one adjustable parameterin K for each component of the state when we use state feedback, whereas there is onlyone adjustable parameter when we use constant output feedback.

2.3 Elgenvalue AssignmentRecall, Section 1.6.4, that a state model's system matrix has the same eigenvaluesindependent of the coordinates used to express the state model. In addition, notice that

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Eigenvalue Assignment 45

if the state, x(t), in the original coordinates is related to the state, x(t), in the newcoordinates as

x(t) = Tx(t)

then the state feedback matrix is transformed as K = KT since

u(t) = Kx(t) = KTx(t)

This fact provides us with the flexibility of being able to carry out the determination of thestate feedback matrix in coordinates which are the most suitable for this computation.

We saw in the previous section, that we can choose K to achieve a specified set of closedloop eigenvalues, {pi : i = 1, 2, .. , n}, provided that each uncontrollable eigenvalue ofthe pair (A, B) equals one of the desired closed loop eigenvalues. Assuming this conditionis satisfied, the state feedback matrix, K, which is needed to make the set of closed loopeigenvalues equal the specified set of eigenvalues can be obtained by equating thecharacteristic polynomial for A + BK, denoted cr(s), to the characteristic polynomial7(s) having roots equal to the specified set of eigenvalues, {µi : i = 1, 2, . . . n}, as

a(s) = 7(s) (2.6)

where

a(s) = det[sI - (A + BK)] = sn + alsi-1 + a2sn-2 +n

7(S)=H(S-µi)=Sn+71sn 1+'Y2sn-2+...+7ni=1

+ an

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

ai = 7i i= 1,2,...n

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

2.3.1 Eigenvalue assignment via the controller formA state model for an nth order SISO system is said to be in controller form when A = A, isa companion matrix and B = B, is the first column of the n x n identity matrix. More

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46 State Feedback and Controllability

specifically the controller form state model has parameters

r -a1 -a2 ... -ai-1 -an

Ac =

1 0 ... 0 0

0 1 ... 0 0 Bc =

1

0

0

LUJ

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

1 -al + kci -a2 + kc2 -a3 + k3 ... -an_1 + kn_1 -an + kn I

1 0 0 0 0

Ac + BcKc = 0 1 0 0 0

L 0 0 0 ... 1 0

with the kris being the elements of the state feedback matrix

Kc = [kc1 kc2 kc3 .. kcn I

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

ai=ai - kci i =1,2,...n (2.7)

Then if we want the closed loop system matrix, Ac + BcKc, to have eigenvalues at{µi : i = 1, 2, , n}, the ais are obtained from

n

fl(s-µi) -Sn+a1Sn-1 +a2sn-2+...+an

i=1

and the elements {kci : i = 1, 2, , n} of the feedback matrix K are easily determinedfrom equation (2.7) as

kci=ai - ai i= 1,2,---n

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

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Elgenvalue Assignment 47

2.3.2 Realizing the controller formHaving seen that the assignment of eigenvalues is easy to do when the system state modelis in controller form, we turn now to the problem of getting a controller form state modelfor a system specified by a transfer function or differential equation.

Theorem 2.2 Suppose we are given the ai, bi parameters in the differential equation ortransfer function model of some plant, i.e.,

n-1 n-1

Y (n) (r) + an-i y") (t) _ bn_i u(" (t)i=0 i=0

or

where

i=0

Y(s) = G3(s)U(s)

with y(i) (0) = u(i) (0) = 0 for all i = 0, 1, 2, , n - 1Then the controller form state model for this plant has parameters (As, B, C,) given by

A, =

n-1

> bn_i s'

GSp = r=0_1 (2.9)n iS + an_i S

-a1 -a2 -a3 ... -an-1 -an1 0 0 ... 0 0

0 1 0 ... 0 0

0 0 0 1 0

C, = f b1 b2 b3

Proof Factor the transfer function as

G,(s) = N(s)D(s)

where

Then we see that

n-1

N(s) = E bn-is` D(s) _i=0

1

B,, =

n-1n [ iS + L. Qn_iS

i=0

II

0

0 (2.10)

0

Y(s) = N(s)Z(s) (2.11)

Z(s) = D(s) U(s) (2.12)

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18 State Feedback and Controllability

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

z(t) ='C-1[Z(S)1

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

and setting

n-1

z(n) (t) + E Qn-iz(') (t) = u(t) (2.13)

i=o

x,(t) =z(n-`)(t) i = (2.14)

in (2.13) gives

z(t) = A,x(t) + B,u(t)

where the matrices A, B, are as stated in the theorem.Finally, since the initial conditions are all assumed zero, we see from (2.11) that

n-1

y(t) = E bn-i z(Z) (t)i=o

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

y(t) = C'x(t)

where the matrix C. is as stated in the theorem.Notice that since GSp (s) is strictly proper, D, = 0. Alternatively when the given transfer

function, GP (s), is proper, we can proceed as in Section 1.9.3 to obtain its controller form(Ar, B, C, Di.) with A, B, C, as in Theorem 2.2 and D = 00 where

G, (s) =

with

n n-l

Nn-i S' E bn-i Si=0 i-0

+00n-1 n-1

Sn + 1: Qn_i S i Sn + YJ ` Qn_i S i

i=0 i=0

bi=0,_Qoai i=1,2,...n

In summary, Theorem 2.2 shows that the controller form state model is easily obtainedwhen we are given either the transfer function or the differential equation governing theprocess behavior.

In some cases we may be given a state model for an SISO plant which is in a form otherthan the controller form. In this situation it may be possible to simplify the determinationof the state feedback matrix K by:

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Eigenvalue Assignment 49

(i) determining the coordinate transformation matrix, T,., to transform the given plantstate model to controller form

(ii) choosing Kc using (2.8) to assign the eigenvalues to the controller form state model(iii) using the coordinate transformation matrix T, to obtain the require state feedback

matrix as K = KKTC'

This approach, while not of practical significance when applied as just described, doesenable the development of a practical computational algorithm for determining K. Thisalgorithm which is referred to as Ackermann's formula will be developed in Section 2.3.4.

At this point we need to develop both a method for determining Tc and conditions onthe given state model which enables its transformation to controller form.

2.3.3 Controller form state transformationSuppose we want to find a coordinate transformation matrix, T,., which transforms agiven state model, (A, B, C), not in controller form to a state model, (A, B, C), which is incontroller form. Then assuming, for the time being, that such a transformation exists, wecan solve this problem by proceeding as follows.

Since, in general A, and A = T-'AT have the same characteristic polynomial, we canuse the coefficients of the characteristic polynomial for the given A matrix, det[sI - A], toconstruct the controller form system matrix A = A, from Theorem 2.2. In addition, sinceB, is the first column of the n x n identity matrix we have B = B, without doing anycalculation. Therefore neither A, nor B, requires the determination of T,. However T, isneeded to determine C = CT, = C,. In what follows we use A,, B, and their relation toT, to develop an algorithm for determining T,, one column at a time.

To begin with, notice that the first column of T, can be obtained immediately frominspection of the relation T,.B = B when T, is written in terms of its columns,

[t't2 ... t"] =t'=B

I

0

0J

where t' is the i th column of T.Having determined the first column of T, as B, we can see how to determine other

columns of T, by inspection of the relation T,A = AT, with A set equal to A, and T,expressed in terms of its columns,

-al -a2 -a3 ... -a"_1 -a"1 0 0 ... 0 0

[t' t2 .. t"]1 0 1 0 ... 0 0 A[t' t2 ... t"]

0 0 0 ... 1 0

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au airare reeadack and Controllability

Thus equating like positioned columns on either side of this equation gives

-alt I + t2 = At'-alt' + t3 = A t2

-ant ' + to = Atn-'

(2.15)

-ant' = At" (2.16)

and we see that the remaining columns of T, can be generated successively from (2.15) bystarting with t' = B and proceeding as

t2 = alt' +At't3 = alt' + A t2

t"=an-1t' +Atn-'

(2.17)

To recap, we can generate the columns of T, by

(i) obtaining the coefficients of the characteristic polynomial for the given A(ii) setting the first column of T, equal to the given B.

(iii) using (2.17) to determine the remaining columns of T, in succession.

Notice that so far we have no assurance that the matrix T, which results from using(2.17) will be invertible. However T, must be invertible if it is to be a coordinatetransformation matrix. We show now that only those state models which have theproperty of controllability will produce a nonsingular T, matrix as a result of applyingthe foregoing algorithm. Before doing this notice that we did not need (2.16) to get (2.17).This extra equation will be used in Section 2.3.5 to derive the Cayley-Hamilton theorem.

2.3.4 Condition for controller form equivalenceIn order to determine conditions on the given state model which guarantees that it can betransformed to a controller form, i.e., to insure that the matrix T, which is generated by(2.17) is invertible, we expand (2.17) by successive substitution to obtain

t' = B

2t = a,B+AB

t3 = a2B + a1AB + A2B

t" = an-,B + an_2AB + + a1An-2B + A"-'B (2.18)

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Eigenvalue Assignment 51

Then writing these equations in matrix form gives

T. = 52R (2.19)

where

1l=[B AB A2B . . An-1B] R=

1 a1 a2 an-1

0 1 a1 ... an-2

0 0 1 ... an-3

0 0 0 1

Now since the product of two square matrices is invertible if and only if each matrix inthe product is invertible, T, is invertible only if both S2 and R are invertible. Howeversince R is an upper-triangular matrix having nonzero elements along its diagonal, R isinvertible for any given state model. Therefore T, is invertible if and only if 52 is invertible.

The matrix 52, which depends on the interaction between the given state model's A andB matrices, is referred to as the given state model's controllability matrix. For systemshaving only one input the controllability matrix is square and the state model is said to becontrollable if its controllability matrix is invertible. However when the input consists ofm scalar inputs, SZ, has nm columns and only n rows. In this case the state model is said tobe controllable if its controllability matrix is full rank, i.e., all its rows are independent orequivalently, n of its nm columns are independent.

More generally, the property of controllability of a given state model is preservedunder coordinate transformation. We can see this by noting that the controllabilitymatrix SZ of the state model in the transformed coordinates is related to the controllabilitymatrix Q of the state model in the original coordinates as

52 = [B A. An 1B]

_ [T-1B T-1ATT-1B ... T-'An-1B]

= T-152 (2.20)

Therefore since SZ has fewer rows than columns we have

aT s2 = ,QT 1 = OT

for some a 54 p only if Q has dependent rows, since /3T = aT T-1 = 0T only if a = 0. Thisimplies that

rank[Q] = rank[T-152] = rank[Q]

where the rank of a matrix equals the maximum number of its columns which areindependent. However since, for any matrix, the number of independent columns equalsthe number of independent rows, we see that, in general

rank[S2] < n

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52 State Feedback and Controllability

since SZ is an n x nm matrix for any n dimension state model having m scalar inputs. Thisfact will be used later in the fourth section of this chapter.

When the state model is single input and controllable, m = 1 and S2, SZ in (2.20) aresquare and invertible. Therefore we can obtain the coordinate transformation matrix, T,from (2.20) as

T = S2SZ-' (2.21)

Then comparing this expression with (2.19) we see that Q-' = QC-1 = R. Moreover thecontrollability matrix for a state model in controller form is readily calculated as

SZ =L

B.. A,. B,. ... An 'Bc.

J

1 -a1 aj - a2

0 1 -a1 *

(2.22)

0 0 0

which, being upper-triangular with fixed nonzero entries along the diagonal, is invertible.Thus all controller form state models are controllable.

To recap, a given single input state model can be transformed through an appropriatecoordinate transformation to a controller form state model if and only if the given systemis controllable. Moreover we have just seen that an uncontrollable state model cannot bemade controllable by a coordinate transformation and that any controller form statemodel is controllable. In addition, we will see that there are several other aspects of systemcontrollability which play important roles in control theory. A more basic view of systemcontrollability is developed in the fourth section of this chapter.

2.3.5 Ackermann's formulaAs mentioned at the end of Section 2.3.1, we could calculate K to assign the eigenvalues ofA + BK, for a given single input controllable system by transforming the given system tocontroller form , finding K, and transforming K, back into the original coordinatesthrough the relation K = KCTc'. We will see now that we can use this idea to develop acomputational algorithm for obtaining K which is known as Ackermann's formula. Inorder to do this we will need the following result.

Theorem 2.3 (Cayley-Hamilton) Any square matrix A satisfies its own characteristicequation.

Before proving this theorem, the following explanation may be needed. Recall that thecharacteristic equation for A,

a(s) = det[sI - A] =s"+als"-' 0 (2.23)

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

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Eigenvalue Assignment 53

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

a(A) = AT +a,A"-i

+ a,A"-2 + + a"I = 0

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

(An + a, An-1 + a2An-2 + ... a"I)B = 0 (2.24)

Now (2.24) holds for any B. Therefore letting B = I in the foregoing equation yields

An+a1A"-I +a2An-2+..anI = 0

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

a(s) = det[sI - (A + BK)]

then the required feedback matrix K is given by

K = -qT [a(A)]

where qT is the last row of the inverse of the controllability matrix, Q, for the given pair(A, B).

Proof For convenience, and without loss of generality, suppose the given plant is ofdimension n = 3. Then the plant state model's system matrix A has characteristicpolynomial

a(s) = det[sI - A] - s3 +a,S2 + a2s + a3

Let the desired characteristic polynomial, a(s), for the state feedback system matrix bedenoted as

a(s) = det[sI - (A + BK)] = s3 + a1s2 + a2s + a3

Then using the coordinate transformation matrix T, developed in Section 2.3.3 wetransform the given state model for the plant to controller form

-a, -a2 -a3 r 11

A,=I B1.=I001 0 0

0 1 0

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

-a, + kc.1 -a2 + kc2 -a3 + kc3 I

A, + BcK, = 1 0 0

0 1 0

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54 State Feedback and Controllability

where

Kc = [ kc1 kc2 kc3 ]

Therefore the desired characteristic polynomial can be written by inspection from thecompanion form for Ac + BcK, as

a(s) = s3 + (al - kc1)s2 + (a2 - kc2)s + (a3 - kc3)

= a(s) - (kcis2 + kc2s + kc.3) (2.25)

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

a(A) = a(A,) _ 0

Therefore we see from (2.25, 2.26) that

a(Ac) = -kc1A2 - kc.2Ac - kcal

where I is the 3 x 3 identity matrix

However we can readily verify that

(2.26)

(2.27)

3TI = 3T i3TA i2T i3TA2 = i1TAc _- c -

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

i3Ta(Ac) = -kc1i1T - kc2i2T - kc3i3T

= -[kcl kc2 kc3 ] = -Kc

(2.28)

Then inserting the identity matrix TC 1Tc between i3T and a(Ac) on the left side of (2.28)and post-multiplying throughout by TC 1 yields

(i3TTc 1)(Tca(Ac)T_ 1) = -KcTc 1 (2.29)

Moreover using the facts that

Tca(Ac)Tc 1 = TcA3 T_ 1 + a1 TcA2 TT 1 + 012TcAcT- 1 + a3I

= A3 + a1A2 + a2A + a3I = a(A) (2.30)

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Controllability 55

and that KCTC-1 = K we see that (2.29) can be rewritten as

i3TTr-la(A) = -K (2.31)

Finally recalling (2.19) we see that

1 * *

TC-1 =R-152-1 = 0 1 * 52-1 .(2.32)

0 0 1

where

1 a1 a2

Q= [B AB A2 B] R= 0 1 a1

0 0 1

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

K =._(i3T1-1)[a(A)]

and the theorem is proved.Notice that since only the last row of 52-1 i.e., qT , is required when we use this result to

compute the feedback matrix, K, we need only compute q which satisfies

in=52Tq

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

2.4 ControllabilitySo far we have encountered the effects of system controllability twice, once in connectionwith the condition for eigenvalue assignment by state feedback, (Theorem 2.1), and againin connection with the existence of a coordinate transformation matrix which transformsa given state model to controller form, (Section 2.3.4). In this section we encounter systemcontrollability in the context of the basic problem of the input's ability to manipulate thestate of a given state model. The following definition of a state model's controllability ismade with this problem in mind.

Definition 2.3 A state model (A, B, C, D), or pair (A, B), is said to be controllable if forevery possible initial state, x(0), we can find at least one input u(t), t E [0, tf] and somefinite final time tf < no so that x(tf) _ 0, i.e., so that the state is annihilated by the input ina finite time.

It is important to notice in the foregoing definition, that tf is required to be finite. Thisis done to prevent all stable systems from being considered to be controllable. Morespecifically, since stable systems have the property that

lim x(t) = 0 for any x(0) when u(t) is nulltcc

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56 State Feedback and Controllability

all stable systems would be controllable if the final time, t f, in Definition 2.3 were allowedto be infinite.

In order to develop the implications of the foregoing definition we need to recall, fromthe first chapter, that the state x(tt) which results from having an initial state, x(0), and aninput, u(t), is given by

x(tf) = eA`t x(0) + J eA(`,-T)Bu(T)dT (2.33)

0

Then if the state is annihilated at time tj, i.e., if .x(tt) = 0, we see from (2.33) that the input,u(t), must be chosen so that

tt

eAtt x(0) = - J eA(t1-T)Bu(T)dT

0

However from the series expansion for eAt, (Section 1.4.1) we see that

(eAtt)-l= e_Atf eA(tf-T) = eAtf e-ATand

so that we can simplify (2.34) as

y

x(0)J

e-ATBu(T)dT

0

(2.34)

(2.35)

This equation is the basic constraint which must be satisfied by any input which drivesthe system state from x(O) to the origin in state space in finite time, tf < no. Notice thatthe following three questions are immediately evident:

1. For each initial state, x(0), is there at least one input, u(t), which satisfies (2.35)?2. If it is not possible to satisfy (2.35) by some input for each initial state, how should the

initial state be restricted to enable (2.35) to be satisfied by some input u(t)?3. If for a specific initial state, x(0), it is possible to satisfy (2.35), what is the specification

of the input u(t) which does so?

Notice that when the answer to the first question is in the affirmative the system iscontrollable in the sense of Definition 2.3.

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

1l = [B AB ... An_1B]

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Controllability 57

Notice that when rank [Q] = n, 1 has n independent columns so that we can always finda constant vector, 'y, for any initial state, x(0), such that

x(0) 5t7 (2.36)

However, if rank[Q] < n, then iZ has fewer than n independent columns, and it is notpossible to find 'y to satisfy equation (2.36) for some x(0). However, those x(O) for whichwe can find 7 to satisfy equation (2.36) are said to lie in the range or image of SZ, denoted as

x(O) E range[Q]

We we will show, later in this section, that those initial states, x(0), satisfyingx(O) E range[Q], are those initial states for which we can find u(t) to satisfy equation(2.35). This fact answers question 2 and leads to the following definition.

Definition 2.4 Initial states, x(0), for which we can (cannot) find u(t) to satisfyequation (2.35) are said to lie in the controllable subspace S,. (uncontrollable subspaceSe) of the state space.

Thus we see from the answer just given to question 2 that

Sc = range[S2]

An important property of the controllable subspace is given in the following theorem.Theorem 2.5 If x(O) Erange[ 1] then Ax(O) Erange[S2].Proof Suppose x(O) E range[1l]. Then we can find a constant vector, 7, to satisfy

x(O) = S27 = B71 + AB72 + ... An-1 B7n (2.37)

where the ryas are constant vectors having the same length as u(t). However, we know fromthe Cayley-Hamilton theorem, (Theorem 2.3), that

n

An = - aiAn-ii=1

where the a;s are coefficients of the characteristic polynomial

det[sI - A] = sn + assn-1 + a2sn-2 + an

Therefore multiplying (2.37) by A and using the Cayley-Hamilton theorem gives

Ax(O) - AB7i + A + .. A B7n-1 + (E_aiA1B)7ni=1

= SZ7 (2.38)

where

T7 = [ -7nan 71 - 7nan-1 7n-2 - 7na2 7,-1 - 7na1]

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58 State Feedback and Controllability

and we see from (2.38) that

Ax(O) E range[Q]

Notice that this theorem implies that if x(O) E range[SZ], then Akx(0) E range[1l] for allintegers k. This fact enables us to show that rank[ 1] = n is a necessary and sufficientcondition for S, to be the entire state space so that there is an input satisfying (2.35) for allinitial states. Thus rank[1lJ = n is a necessary and sufficient condition for a state model tobe controllable in the sense of Definition 2.3. This fact is shown now in the followingtheorem.

Theorem 2.6 If a state model (A, B, C, D) or pair (A, B) is controllable in the sense ofDefinition 2.3 then

rank[ 1] = n

where

SZ = [B AB . . . An-1B]

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

X(0) = - Jtf [I - AT + A2 z - A3 3 ...I Bu(T)dT

3

=Byo+ABy1+A2By2+...

= Q70 + A71 + A272 + .. .

where

7 =[7kn ykn+1 y(k+l)n-1 ] k = 0, 1, 2"'

with

(2.39)

tf T I

u(T)dT 0,1,2...o i!

However from Theorem 2.5 we have

A' Q 7k E range[S2] k = 0, 1, .. (2.40)

Therefore the columns on the right side of (2.39) span the entire state space if and only ifrank[Q] = n. Otherwise when rank[Q] < n there are initial states x(0) 0 range[Q] so that(2.39) is impossible to satisfy by any u(t) and the state model is not controllable in thesense of Definition 2.3. This implies that rank[Q] = n is necessary for the state model to becontrollable in the sense of Definition 2.3.

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Controllability 59

2.4.2 Input synthesis for state annihilationThe following theorem shows that rank[Q] = n is sufficient for a state model (A, B, C. D)or pair (A, B) to be controllable in the sense of Definition 2.3.

Theorem 2.7 If rank[Q] = n then we can satisfy (2.35) for any x(O) by choosing u(t) as

u(t) = -BTe-AT`W-1X(0)

where

W= e-ATBBTe-A r TdTI If

Proof Suppose W is invertible. Then substituting the expression for u(t) given in thetheorem into the right side of (2.35) gives

e-ATBu(T)dr = -Iti

e-ATB [-BTe-AT T] dTW-1x(0)

J = WW-1x(0)

= x(0)

which is the left side of (2.35). Thus we have shown that the state model is controllable inthe sense of Definition 2.3 when W is invertible.

In order to show that W is invertible, suppose rank[Q] = n and W is not invertible.Then we can find a 54 0 such that

aT W = 0T

Now since the integrand in the definition of W is quadratic, i.e.,

MT (T)M(T)drW = f

where

MT(T)=e ATB

we can only satisfy (2.41) if

aTCATB= 0T

(2.41)

for all T E [0, tf] (2.42;

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

e-ATB = 11/3° + A"SZ/31 + A2 1132 + .. .

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60 State Feedback and Controllability

where

okT(_T)kn (_T)kn+l

(kn)! (kn+1)!

(_T)(k+l)n-l

([k+ 1]n - 1)!]k=0,1,2..

Thus (2.42) is satisfied only if

aT S2 = 0T (2.43)

However since rank [Q] = n and I is n x nm, we see that S2 has independent rows and onlya = 0 satisfies (2.43) and (2.41). This proves that W is invertible when rank[Q] = n.

In the next section we will see that when rank[S2] < n we can still use the foregoingapproach to provide a method for determining an input which annihilates an initial statex(0) in a finite time provided x(0) Erange[S2]. This is achieved by using a controllabledecomposed form for the system's state model so that the controllable part of the systemis immediately evident.

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

X(t) =

where x`(t) is of length ni and

x(t) E S, if x2(t) = 0

x(t) E Se if x' (t) = 0

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

x(t) = a,x`(t) + aex`(t) (2.44)

where x`(t) E S, and xc(t) E S, with a, a, being scalars. Notice that SCISCA state model whose state space is orthogonally decomposed in the foregoing manner

is referred to as being in controllable decomposed form. The state model for a system tobe in this form is defined as follows.

Definition 2.5 A state model having m scalar inputs is said to be in controllable

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Controllable Decomposed Form 61

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

A=Al A,

0 A4JB =

B,(2.45)

0

where (A1, B1) is a controllable pair; A1, A2, A4 and B1 have dimensions: nl x n1, n1 x n2,n2 x n2, nl x m respectively with n = nl + n2 being the dimension of the state space.

2.5.1 Input control of the controllable subspaceWe begin the discussion of the effect of an input on the state when the state model is incontrollable decomposed form by noticing that the product of upper-triangular matricesis upper-triangular, and that the coefficients of the series expansion of eAt involve powersof A. Therefore the transition matrix, eAt, for a state model in controllable decomposedform, (2.45), is upper-triangular,

eAtAl

* ti=0 0 A4 11

where the block marked * is not of importance here.Next recall, from the previous section, that if u(t) drives the state from x(O) to the

origin in finite time, tf, then (2.35) is satisfied. Therefore since the eAt is upper-triangularand B has its last n2 rows null when the state model is in controllable decomposed form,we see that (2.35) can be written as

[x2(0)1 _

fo

j E(-1)i AIBI - U(T)dTX (o) ii=o ` /

or

IfX1(0) = - e-AITBIU(T)dY

0

x2 (0) = 0

(2.46)

Thus we see that (2.46) can not be satisfied by any input u(t) if x2 (0) 0. Alternatively,since (A,,BI) is a controllable pair, we can always find u(t) to force xl(tf) _ 0 for anyx1(0) and x2(0).

More specifically, when the state model is in controllable decomposed form, (2.45), wesee that

x2(t) = eA4tx2(0)

so that

xl (tf) = eA,,,X1 (0) +J

t' eAI (tr-T) [A2eA4TX2(0) + Bl u(T)] dT0

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62 State Feedback and Controllability

Therefore, proceeding in the--s ame manner as was done to obtain (2.35) we see that ifx1(tj) is null then we must choose u(t) to satisfy the following equation

J'eq =-A'TBIU(T)dT (2.47)

where

q = x1(0) + J e-A, TA2eA^Tx2(0)dr

0

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

u(t) = -BT aAT `W-lq

where

W =J rf e-A,TBIBT e-AI TdT

To recap, we have just shown that we can always find an input to annihilate theprojection of the initial state onto the controllable subspace. More specifically since anyinitial state can be written as (2.44) with t = 0, we can find u(t) so that x`(tj) _ 0 for anytj<00.

2.5.2 Relation to the transfer functionConcerning the input-output behavior, suppose the state model has A,B as specified by(2.45) and C partitioned as

C=[C1 C2]

with C, being p x n1, i = 1, 2 where p is the number of scalar outputs. Then from the blockupper-triangular form for A, (2.45), we see that

(s1 - A)-'=sI - Al -A2 (sI - A1)-1 (s1 - A1) 'A2(sl - A4)-1

_0 sI - A4 0 (sI - A4)-1

Therefore the system transfer function is given as

G(s) = C(sI - A)-1B

= Cs7-A -1B _Cladj[sI-A1]B1

1( 1) 1 = det[sI-A1](2.48)

Notice that the order of the transfer function is no more than n1. This reduction in theorder of the transfer function results from the eigenvalues of A4 being uncontrollable. We

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conMo

will see in the next chapter that a further reduction in the number of eigenvalues of A thatgo over as poles of the corresponding transfer function may occur as a result of theinteraction of the state model's (A, C) pair. In order to better appreciate the fact that theeigenvalues of A4 are uncontrollable, consider the following.

2.5.3 Eigenvalues and eigenvectors of ASuppose we attempt to determine the left-eigenvectors of A when the state model is givenin controllable decomposed form. Then we need to solve

WT A = \WT

for the pair {A, w}. However from the block structure of A, (2.45), we see that theforegoing equation expands to two equations,

WIT Al = AwIT (2.49)

WITA2 + W 2T

A4 = AW2T

where

(2.50)

wT = [wlT w2T]

with w' being of length n,. Then (2.49) is satisfied when we let wl = 0 and (2.50) becomes

W2TAa = Aw2T (2.51)

Therefore for this choice of w1, we see that A is an eigenvalue of A4, w2 is thecorresponding left eigenvector of A4, and wT = [0 w2 ] is the corresponding left-eigenvector of A. Thus we see that eigenvalues of A4 are eigenvalues of A. Notice thatin this case we have

WTB= [0 WT] ]B1

=0

so that eigenvalues of A which are also eigenvalues of A4 are uncontrollable.Now the remaining eigenvalues of A are eigenvalues of Al. We can see this by noting

from (2.49) that if wl 0 then A is an eigenvalue of AI and wl is the corresponding left-eigenvector of A1. Since in controllable decomposed form the pair (A1,BI) is control-lable, eigenvalues of A which are also eigenvalues of AI are controllable. Moreover sincethe controllable and uncontrollable eigenvalues of any pair (A, B) are disjoint we havethat if A E A[AI] then A 0 A[A4] and Al - A4 is invertible. Therefore when wl is a left-eigenvector of A 1, we can determine the required w2 to make wT = [WIT w2T ] a left-eigenvector of A from (2.50) as

w2T = wITA2(AI - A4)-1

The foregoing property of the eigenvalues of A is immediately evident when oneattempts to use state feedback. Thus if we use state feedback when the state model is in

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64 State Feedback and Controllability

uncontrollable decomposed form, the closed loop system matrix is

A+BKAl+B1K1 A2+B1K2

0 A4

where the state feedback matrix is partitioned as

K = [ K1 K2 ] (2.52)

with K1, K2 being m x n1 and m x n2. Now we just saw that block upper-triangularmatrices have the property that their eigenvalues equal the union of the eigenvalues oftheir diagonal blocks. Thus the eigenvalues of the state feedback system matrix satisfies

A[A + BK] = A[A1 + BI K1 ] U A[A4]

and A + BK has a subset of eigenvalues, A[A4], which is clearly unaffected by statefeedback. However since (AIB1) is a controllable pair, all the eigenvalues of AI + B1K1can be assigned by K1. Notice also, from the definition of stabilizability, Definition 2.2,that a system is stabilizable if, when its state model in controllable decomposed form, theA4 partition of A is stable.

Finally notice that Ackermann's formula, which we developed to assign the eigenva-lues of single input controllable state models, can also be used to assign the controllableeigenvalues of uncontrollable systems. This is done by transforming coordinates to putthe given state model in controllable decomposed form. Ackermann's formula is thenapplied to the controllable pair (A1i B1) to determine the K1 partition of K, (2.52), with K2being chosen arbitrarily. Finally, we can obtain the feedback matrix K0 in the originalcoordinates from Ko = KT-1 where T is the coordinate transformation matrix needed totransform the given state model to controllable decomposed form. In the next section weconsider how this coordinate transformation matrix can be determined.

2.6 Transformation to Controllable Decomposed FormIn this section we will indicate the role played by the controllability matrix in constructinga coordinate transformation matrix to put any state model in controllable decomposedform.

We begin by noting that when a state model is in controllable decomposed form withthe (A, B) pair being given by (2.45), the last n2 rows of the controllability matrix consistof nothing but zeros, i.e.,

11=B1 A1B1 ... AI'-IBI A7'B1 ... An-1B1

0 0 ... 0 0 ... 0

(2.53)

where

QI = [BI A1B1 ... A?'-IBI ] rank[S11] = n1

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Transformation to Controllable Decomposed Form 65

Notice that the columns of SI2 depend on the columns of 1l i i.e., we can always find aconstant matrix O to satisfy

Q2 = S21 -0

Therefore recalling that the subspace spanned by the independent columns of Q is thecontrollable subspace, S, we see that

range[Q] = range(1 0 J /

= Sc

This fact together with computationally robust methods for finding a basis for the rangeof a matrix, e.g., singular value decomposition, (Chapter 7), provides a means forconstructing a coordinate transformation matrix T which takes any state model tocontrollable decomposed form. The following theorem provides some insight into why acoordinate transformation based on the controllability subspace is able to achieve thisresult.

Theorem 2.8 A coordinate transformation matrix T transforms a given state model(A, B, C, D) to controllable decomposed form, (Definition 2.5) if

range[T1] = range[d] = Sc

where Q is the controllability matrix for the given state model and

T = [T1 T2]

with T1, T2 being n x n1, n x n2 and with T invertible where rank(I) = n1.Proof Let QT denote the inverse of T, i.e.,

QT T = I

IQT rI[ T1 T2 ] = I I (2.54)

LQT

L

Then the transformed A, B matrices are given by

TAT TATA - T-1 AT - Q1 1 Q1 2

LQAT, Q2 AT 2TB

B _ T-'B _ Q1

QT B2

Al A2

A3 A4

Recall that for the transformed state model to be in controllable decomposed form,(Definition 2.5), we need to show that A3 and B2 are each null.

To show that B2 = 0, notice from (2.54) that Q2 T1 = 0 so that the columns of T1 areorthogonal to the columns of Q2. However range(T1) = Sc so that we have

Q2 x = to when x E Sc (2.55)

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66 State Feedback and Controllability

Now since B appears as the first m columns of Q,

range[B] c range[f1] = S,. (2.56)

and we see from (2.55) that the columns of B are orthogonal to the columns of Q2 so that

TB2=Q2B=0

To show that A3 = 0, recall from Theorem 2.5 that if x E S, than Ax E S,. Thereforesince

range[T1] = Sc

we have

range[AT1] C S,. (2.57)

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

TA3=Q2AT1=0

Notice, in the foregoing proof, that (2.55-2.57) imply that we can find constantmatrices 01, 02 such that

B= T101

AT1 = T102

and since QTz

Tl = 0 we achieve Q2B = 0 and Q2AT1 = 0 as required for T to transformthe given system to controllable decomposed form.

2.7 Notes and ReferencesThe Ackermann formula was one of the first attempts to develop an algorithm forassigning the eigenvalues to a single input state model by state feedback. More recentwork on this problem has concentrated on developing algorithms which are least sensitiveto errors caused by the need to round off numbers during the execution of the algorithmusing finite precision arithmetic on a digital computer. Further information on thisproblem can be obtained by consulting [43].

The Cayley-Hamilton theorem plays an important role in calculating matrix func-tions and can be used to provide an efficient method for calculating e At once theeigenvalues of A are computed. Modern computer oriented algorithms rely on methodswhich truncate the infinite series expansion for eAt after having put A in a certaincomputationally beneficial form known as real Schur form, [17].

The approach to the problem of transforming a state model to controllable decom-posed form which was discussed in Section 2.6 forms the basis for the command CTRBFin MATLAB.

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3State Estimation andObservability

3.1 IntroductionIn the previous chapter we saw that state feedback could be used to modify basic aspectsof a plant's behavior. However, since the plant state is usually not available, we need ameans for obtaining an ongoing estimate of the present value of the state of the plant'sstate model. In this chapter we will see that this can be done by using measurements of theplant's input and output.

The problem of determining the state of a state model for a plant from knowledge ofthe plant's input and output is referred to in general as the state estimation problem.There are two classical types of state estimation: deterministic state estimation andstochastic state estimation. Deterministic state estimation, which is the subject of thischapter, assumes that the system input and output are known or measured exactly. Thegoal of deterministic state estimation is the determination of an estimate of the statehaving error which tends to decrease with time following the initiation of thr.estimationprocedure. However, in stochastic state estimation the input and output signals are notassumed to be known exactly because of the presence of additive stochastic measurementnoise having known statistics. In this situation the state estimation error is alwayspresent. The goal of stochastic state estimation is the determination of an estimate of thestate so that, in the steady state, the average or expected value of the state estimation erroris null while the expected value of the squared error is as small as possible. Stochastic stateestimation is discussed in Chapter 6.

Both deterministic and stochastic state estimation are further subdivided according towhen the input and output signals are measured relative to when the state estimate isneeded. If the plant's input and output are measured over the time interval [0, T] and if weneed the state estimate at time t, we have

1. a prediction problem if to > T2. a filtering problem if to = T3. a smoothing problem if to < T

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

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68 State Estimation and Observability

plant's state at the present time. Prediction and smoothing problems, which are notdiscussed ii1 this chapter, arise, for example, in hitting a moving target with a projectile byaiming at the target's predicted position and in estimating the true value of data obtainedfrom experiments done in the past when the measured data are corrupted by noise.

The computer implementation of a solution to the filtering problem, either determi-nistic or stochastic, takes the form of a state model which is referred to in general as afilter. The filter's input consists of both the plant's input and output and the filter's outputis the state of the filter, i.e., the filter is designed so that its state is an estimate of the plantmodel's state. In the deterministic case the filter is referred to as an observer. In thestochastic case the filter is referred to as a Kalman filter. The Kalman filter is taken up inChapter 6.

3.2 Filtering for Stable SystemsSuppose we know the parameters (A, B, C, D) of a plant state model as well as the plant'sinput and output, Ju(t), y(t) : t E [0, te]}. Then, from Chapter 1, we can express the plantstate at any time to in terms of the initial state and input as

x(te) = eAtex(O) +I[,

(3.1)

However since the initial plant state is usually unknown we are only able to calculate thezero state response. We use this fact to form an estimate of the plant state at time t, as

x(te) = feABu(r)d(3.2)where from (3.1) we see that the plant state, x(te), and plant state estimate, z(te), differ bythe plant state estimation error, x(te),

z(te) = x(te) - X(te) = eA`ex(O) (3.3)

Now if the plant's state model is stable, we see from (3.3) that the state estimationerror, z(te), approaches the null vector with increasing estimation time, tE1

lim z(te) = 0 for all z(0) = x(0)

In this situation the state estimate, x(te), given by (3.2), can be a reasonably goodapproximation to the plant model's state, x(te), provided to is large enough to ensure thatthe effect of the initial state estimation error 5E(0) is negligible. In this case the estimate issaid to be an asymptotic estimate of the state since the state estimate approaches the statebeing estimated asymptotically with time.

Alternatively, we can view the state estimate obtained in the foregoing approach as theoutput, ye(t), from a system called a state estimator having state model (Ae1 Be, Ce). More

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Observers 69

specifically we have

(t) = A,, B (3.4)

ye(t) C,-i(t) C, I (3.5)

Then since the state of the plant model is governed by

z Ax(t) + Bu(t) (3.6)

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

(t) = Az(t) (3.7)

where

z(t) x(t) - z(t)

Therefore the state estimation error is independent of the plant input and is given by

z(t) = eA`x(0) (3.8)

Now if the initial state of the plant were known we could take z(0) = x(O) as the initialstate for the estimator, equation (3.4). Then z(0) 0 and we see from equation (3.8) thatwe would have the desirable result of exact state estimation, i(t) = x(t), for all time.Usually we don't know the initial plant state. In this case we could set z(0) = 0 in the stateestimator, (3.4), so that z(0) = x(O) and provided A is stable, the state estimate wouldapproach the actual plant state asymptotically for any initial plant state.

However, unstable plants are encountered quite frequently, and must be stabilized byfeedback, e.g., the feedback control of satellite rocket launchers. In these cases theestimate obtained from the state estimator, (3.4), diverges from the plant sate for anyz(0) j4 0 since eA` becomes unbounded with time and the state estimation error given byequation (3.8) grows indefinitely. Notice that it is impossible to know a physicalparameter such as the initial plant state exactly. Therefore in practice it is not possibleto set i(t) = x(O) so as to obtain i(t) = 0 from (3.8). Thus whenever the plant is unstable,we are unable to use (3.4) as an asymptotic state estimator.

Not knowing the initial state and needing to estimate the state to implement stabilizingstate feedback for unstable plants, forces us to seek another approach to state estimation.

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

y(t) = Cx(t) + Du(t) (3.9)

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

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70 State Estimation and Observability

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

y(t) = Cx(t) + Du(t) (3.10)

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

y(t) = y(t) - 33(t) (3.11)

Now since we can form y(t) from u(t), y(t) and z(t) all of which are known, we can usey(t) to indicate the accuracy of the plant state estimate since y(t) 0 is an indication thatthe state estimate differs from the actual state. More importantly, we can use y(t) tocorrect future state estimates by letting y(t) affect z (t) by subtracting Ly(t) to the rightside of (3.4) to obtain

x= Ax(t) + Bu(t) - LL(t) (3.12)

which using (3.10, 3.11) can be rewritten as

x= Fx(t) + Gu(t)

(3.13)y(t)

where

F=A+LC G= [B+LD -L]

When F is stable, the system represented by (3.13) is referred to as an observer. In orderto determine the effectiveness of an observer in obtaining an asymptotic estimate of thestate, consider the differential equation for the state estimation error, z(t) = x(t) - 1(t).We obtain this differential equation by subtracting (3.13) from (3.6) to obtain

z (t) = F5 (t) (3.14)

where

F=A+LC

Then the state estimation error is given by

x(t) = eF`x(0)

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

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Observer Design 71

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

Cv' = 0

the corresponding eigenvalue A, of A is an eigenvalue of A + LC for all L.Proof Suppose the condition of the theorem is satisfied. Then multiplying A + LC on

the right by v' gives

(A + LC)v' = Av' + LCv'

However, since Cv' = 0 and Av' = \;v' we see that the foregoing equation becomes

(A + LC)v' = \iv'

Thus A, is a fixed eigenvalue of the matrix A + LC for all L M

We will see that if v' is an eigenvector of A such that Cv' 54 0 then we can choose L sothat the corresponding eigenvalue, A1, of A can be assigned to any desired value as aneigenvalue of A + LC. This leads to the following definition.

` Definition 3.1 An eigenvalue A. of A is said to be an observable (unobservable)eigenvalue for the pair (A, C) if we can (cannot) find L such that A, is not an eigenvalue ofA+LC.

Reflection on the foregoing reveals that we can design an observer for a given statemodel, i.e., we can find L so that A + LC is stable, provided the plant state model has noeigenvalues which are both unstable and unobservable. Plant state models having thisproperty are referred to as being detectable.

Definition 3.2 A plant state model is said to be detectable if all its unobservableeigenvalues are stable.

Thus we can only determine an observer for a given state model if that state model isdetectable. This is of obvious importance in the implementation of state feedback usingan observer for the stabilization of an unstable plant.

Recall, in the case of state feedback, that the controller form state model facilitates thecalculation of K to achieve a specified set of closed loop eigenvalues. In the presentsituation there is an analogous form for the plant state model which facilitates thecalculation L to assign a specified set of eigenvalues to the observer. This form, which weencounter in the next section, is referred to as an observer form.

3.4 Observer DesignSuppose we are given a state model (A, B, C, D) for the plant and we want to determine Lso that the observer eigenvalues, X[A + LC], equal a specified set {p : i = 1, 2 . n}. Thenprovided any unobservable eigenvalues of the pair (A, C) are in the specified set, we candetermine L so that the observer eigenvalues coincide with the specified set. This could bedone by equating coefficients of like powers on either side of the equation

a(s) = ry(s)

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72 State Estimation and Observability

where

a(s) = det[sI - (A + LC)] = S" + alsn-1 + (x2Sn 2 + - - + anI,

'Y(s) = H(S - µi) = sn + 11s" 1 +'Y2Sn-2 + ... + %ni=1

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

3.4.1 Observer formRecall that any state model (A, B, C, D), for a plant is related to the plant transferfunction G(s) as

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

Then a system having transfer function GT (s) has state model (A, B, C, b) where

A=AT B=CT

C=BT D=DT

State models (A, B, C, D) and (A, B, C, D) are said to be duals of each other when theirparameters are related in this way. When G(s) is symmetric, the dual of any state modelfor G(s) is also a state model for G(s).

In the SISO case G(s) is a scalar so that GT (s) = G(s). Therefore, in this case, we canobtain a canonical state model referred to as an observer form, by forming the dual of thecontroller form state model. Thus if the state model for G(s) is in controller form,(A,.., B,., C,, Dr), the dual state model is in observer form, (A0 = AT , Bo = CT ,Co = BT , Dn = Dr), or more specifically

a 1 0 0 0 b

-a2 0 1 . . . 0 0l 1 b2

Ao =-an_2 0 ... 0 1 0

-an-1 0 .. 0 0 1

-an 0 .. 0 0 0

Bn =

Co = [ 1 0 0 0 0] D,, = D

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Observer Design 73

An indication of the importance of this form is evident from the ease with which we canassign observer eigenvalues through L. Thus when the plant is in observer form,Fo = A + L0Co is in the same form as A

r -a, +1ii1 1 0 0 0 11

-a2 + 4)2 0 1 .

1

0 0 12

Fo

-an_2 + lon-2 0 0 1 0

Lo =In-2

-ai_1 + Inn-1 0 0 0 1 In-1

-an + Ion 0 0 0 0 In

Therefore if we want the eigenvalues of F to satisfy a specified characteristic polynomial,7(s), we have

det[sI - F] = sn + Yisn-rjLJi

and the required elements of Lo, {loj : i = 1, 2, . n} are easily obtained from

Ioi = a, - 'Yi i= 1,2...,n

In order to preserve the eigenvalues of A + LC under a coordinate transformation wesee that if L assigns a desired set of eigenvalues to A + LC in the transformed coordinatesthen we need to replace L by L = TL to assign the same set of eigenvalues to A + LC inthe original coordinates. This is seen by noting that

A + LC = T-1AT + LCT = T-1 (A + TLC)T

3.4.2 Transformation to observer formIn the previous chapter, (Section 2.3.3), we developed an algorithm for generating acoordinate transformation matrix T so that any single input controllable state model istransformed to a controller form state model. This algorithm can be adapted to provide acoordinate transformation matrix so that a single output observable state model istransformed to an observer form state model. This adaptation is made using the dualitybetween the controller and observer forms. Thus by replacing A, B in the algorithm givenin Section 2.3.3 by AT , CT we obtain a matrix T which transforms the pair AT , CT tocontroller form. Then T-T is the coordinate transformation matrix needed to transformthe given state model to observer form. Just as the transformation to controller form isonly possible if (A, B) is a controllable pair so here the pair (AT , CT) must be controllableor equivalently the pair (A, C) must be observable, i.e.,

rank[U] = n (3.15)

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74 State Estimation and Observability

where

C

CA

CA n-1

The matrix U is called the observability matrix. Notice, in general, that U is a pn x nmatrix where p is the number of elements in y(t). Also in general we have rank[U] < nsince the rank of any matrix is never greater than its smallest dimension.

Notice that if a pair (A, C) is observable, i.e., rank[U] = n, we can assign all theeigenvalues of A + LC by choosing L. Otherwise if we are not able to assign all theeigenvalues, one or more of the right-eigenvectors of A satisfies Cv' = 0 (Theorem 3.1)and we can show that

Z3v' = 0 (3.16)

Therefore at least one of the n columns of U is dependent so that rank [ U ] < n and the pair(A, C) not observable.

Alternatively, if all the eigenvalues of A + LC can be assigned, all the right-eigenvec-tors of A, v' satisfy Cv' 0 and (3.16) is not satisfied by any right-eigenvector of A. Thenassuming A has a complete set of eigenvectors, any vector q having n elements can bewritten as

and

Uq 0

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

3.4.3 Ackermann's formulaRecall from the previous chapter that we could determine K to assign the eigenvalues ofA + BK for any controllable single input state model by using Ackermann's formula,(Section 2.3.5). Ackermann's formula can also be used to determine L to assign a specifiedset of eigenvalues to A + LC for any observable single output state model. This is madepossible by the fact that any square matrix and its transpose have the same eigenvalues.Therefore A + LC and AT + CT LT have the same eigenvalues. Thus given an observablestate model for a single output system and a set of desired eigenvalues for A + LC, we candetermine L by replacing A and B in Ackermann's formula by AT and CT , respectively.Ackermann's formula then produces L .T

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Observablllty 75

3.5 ObservabilitySo far we have encountered two effects of system observability. Both the capability ofassigning all the eigenvalues to A + LC and the capability of transforming coordinates sothat the state model is put in observer form requires that the given state model beobservable, i.e.,

rank[U] = n

where Z5 is pn x n observability matrix, (3.15), with p = 1.It is important to notice from the discussion so far, that we can still design an observer

to estimate the state of a given state model when the state model is unobservable, i.e.,when rank[Z3] < n, provided the given state model is detectable, Definition 3.2. Thusobservability is sufficient but not necessary for the existence of an observer for a givenplant state model.

In order to gain further insight into the meaning of observability, we are going toconsider the problem of determining the state at some estimation time from thederivatives of the input and output at the estimation time. Notice that in practice weavoid using signal differentiation since noise acquired in measuring a signal can appeargreatly enlarged in the derivatives of the measured signal. Therefore our intent indiscussing the problem of determining the state from the derivatives of the input andoutput is to provide additional insight into the theoretical nature of observability. We willshow that we cannot solve this problem unless the state model is observable.

3.5.1 A state determination problemSuppose we are given the state model, (A, B, C, D), and the derivatives of the input andoutput at some estimation time, tP7 i.e.,

{y")(te),u(')(te): i =0,1,2,...,n-l}

Then from the output equation

y(te) = Cx(te) + Du(te) (3.17)

we have p equations in the n unknown components of x(te), where p is the number ofelements in y(t). Now if p = n, C is square and if the rows of C are independent, orequivalently if the elements in y(t) are not related by constants, i.e., are independent, thenwe can solve (3.17) for the state at time te as

X(te) = C-' [y(te) - Du(te)]

without requiring derivatives of the input and output signals. However usually p < n andwe need to generate equations in addition to those obtained directly from the outputequation (3.17). We can do this by using the state differential equation

x(te) = Ax(te) + Bu(te) (3.18)

together with the output equation (3.17).

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76 State Estimation and Observability

Suppose p = 1. Then we have one equation, (3.17), which we can use in thedetermination of then components of the state x(te). We can generate a second equationfor this purpose by taking the derivative of (3.17)

y(te)=Cx(te)+Du(te) (3.19)

and using (3.18) to obtain

y(te) = CAx(te) + CBu(te) + Du(te) (3.20)

We can generate a third equation for determining x(te) by differentiating (3.20) andsubstituting (3.18) for the derivative of the state. This gives

y,2)(te) = CA2x(te) + CABu(te) + CBuW'W (te) + Du (2) (te)

Continuing in this way we can generate additional equations in the state until we havethe required n equations. These equations can be collected and given as

z = z3x(te) (3.21)

where

z=Y-rU Y=

u(te)

UM (tU=

u(n-') (te) J

Y(te)

Y(1) (te)

Y(n-1) (te) J

rC I

CA

CAn-' J

D 0 0 0

CB D 0 ... 0

r= CAB CB D ... 0

CAn-2B CAn-3B CAn-4B ... D

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

X(te) = (UTU)-1UTZ

and when p = 1 we have

x(te) = U-Iz

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Observability 77

However when rank[ ZU ] < n, Z5 has dependent columns and we have

0 = UX(te) (3.22)

for some non-null x(te) which we denote by x (te). Solutions to (3.22) when rank[ZU] < nare said to lie in the null space of U denoted

x°(te) E null[U] when Z5x°(te) = 0

Hence if x(te) is any solution to (3.21) then we have x(te) + x°(te) as another solution to(3.21). Therefore we see that if (A, C) is not observable it is impossible to determine thetrue state, x(te), from the derivatives of the system's input and output.

In summary we have shown that the observability of the state model is a necessary andsufficient condition for being able to determine the actual state of the plant state model atany time from derivatives of the input and output at that time.

3.5.2 Effect of observability on the outputContinuing the discussion begun in Section 1.7.4 we see that when (A, C) is anunobservable pair with right-eigenvector satisfying Cvl = 0 then when x(O) = v` weobtain the zero-input response as

y(t) = CeA`x(0) = C[I + At + 2i A2t2 + 3i A3t3 + . . ]x(0)

2 3t2+-3!A (3.23)C[I+At+2A

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

A- I 13 0 B=[1]

0

C= [+1 +1]

Then since A is in companion form, (Section 1.6.1), we can express its right-eigenvectorsas

and any initial state given by

Al=-1, A2=-2)

x(0) = cxv' =

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

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78 State Estimation and Observability

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

G(s)= s+l 1

sZ+3.s+2 s+2

and the unobservable eigenvalue of A at -1 is not a pole of the transfer function. Thiseffect was encountered in Section 1.9.2 in connection with diagonal or normal form statemodels.

We will see, in Chapter 4, that this property of having certain non-null initial stateswhich produce no output when a state model is unobsevable leads to a characterization ofobservability in terms of the rank of a matrix called the observability Gramian. Theobservability Gramian is involved in determining the energy in the output caused by agiven initial state.

3.6 Observable Decomposed FormRecall, from the previous section, that we are unable to determine x°(t) Enull[U ] fromknowledge of a state model's input and output derivatives at time t. We refer to null [ U ] asthe state model's unobservable subspace, denoted S, i.e., So =null [U]. Moreover werefer to range[ UT] as the state model's observable subspace, S°. Now it turns out that anysolution to (3.21) can be written as

x(t) = x°(t) + x°(t)

where

(i) x°(t) Enull[ G ] = Sc, and is arbitrary otherwise;(ii) x°(t) Erange[UT] = S° and depends uniquely on z.

Moreover the observable and unobservable subspaces are orthogonal, SQLS°, i.e.,

X0 T(t)x (t) = 0 (3.24)

for any x° (t) Erange [UT ] and any x° (t) Enull[ U ]. This can be seen by using the fact that

x°(t) E range [Z3T]

if and only if we can find w to satisfy

x°(t) = ?T11,

so that we have

X T (t)x°(t) = (uTw)TX (t) = wTUX (t) = 0

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

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Observable Decomposed Form 79

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

x(t)x2(t)]

where x`(t) is of length n, and

x(t) E So if x2(t) = 0

X(t) E So if x' (t) = 0

(3.25)

Notice that (3.24) is satisfied when the state is given by (3.25). Now the structurerequired for the state model to be in observable decomposed form is given in the followingdefinition. Notice that this structure is dual to the structure of the controllable decom-posed form, Definition 2.5, in the sense described in Section 3.4.1.

Definition 3.3 A state model having p scalar outputs is said to be in observabledecomposed form when its A and C matrices are in the following block forms

Al 0A=

A3 A4

C= [Cl 0]

where (A1, C1) is an observable pair and A1, A3, A4, C1 have dimensionsnl x nl, n2 x n1 i n2 x n2 and p x n1 respectively with n = n1 + n2 being the dimensionof the state space.

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

so that the zero input response is given by

°O t`Y(t) = Ce'4'x(0) = C1 > Ai i! x' (0)

=o

= C1eA'`x'(0) (3.26)

Thus the zero input response is independent of x2(0) and Sc, is as specified inequation (3.25). In addition since (A1,C1) is observable, we could determine x'(t)from derivatives of the output. Thus S. is also as specified in equation (3.25).

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80 State Estimation and Observability

3.6.2 Observability matrixAlternatively, we can also see that the decomposition of the state space given by equation(3.25) is achieved when the state model is in observable decomposed form by inspection ofthe observability matrix.

C, 0CIA 0

C,A1'-1 0C, Ai' 0

where

C,

CIA,

U, _

LcIA;`-'

and we have

rank[U] = rank[s] = n,

x E null[U] if and only if x =

for any x2 of length n - n, .

3.6.3 Transfer functionAgain, notice that the observable decomposed form gives the transfer function as

G(s)_[C, 0]

C,Ai-I 0]

(sI-A,) i 0

I IB71

B,(sI- A4) i

(3.27)

with order equal to or less than n, depending on the controllability of the pair (A,, BI).Notice also that the eigenvalues of A4 are not poles of the transfer function since they areunobservable.

This latter observation is readily verified from the structure of (A + LC) in these

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Observable Decomposed Form 81

coordinates since

A+LC= [Al+L1C1 C

]L2C1 A4

and since A + LC is block lower triangular we have

A[A+LC] = \[A1 +L1C1] U.[A4]

with ) [A4] being eigenvalues of A + LC for all L, i.e., being unobservable eigenvalues of(A, C).

3.6.4 Transformation to observable decomposed formNow we saw in Chapter 2 that the controllable subspace is A-invariant, i.e., if x Erange[1l]then Ax Erange[Q], (Theorem 2.5). The analogous result here is that the unobservablesubspace is A-invariant, i.e., if x Enull[ZS] then Ax Enull[U]. When the state model is inojservable decomposed form we can see this fact directly from (3.27) and the blockstructure of A. Alternatively, we can see that this result holds more generally in anycoordinates by noting that we can always express the last block row of U A using theCayley-Hamilton theorem, Theorem 2.3, as

CA" = -a1 CA" 1 - a2CA"-2 ... a"C

Therefore the last block row of UA depends only on block rows of U and thus if Ux = 0then ZJAx = 0.

Now the A-invariance of the unobservable subspace provides a means for determiningthe coordinate transformation matrix needed to put a given state model in observabledecomposed form.

Theorem 3.2 A coordinate transformation matrix T transforms a given state model(A, B, C, D) having observability matrix ZU with rank[Z3] = n1 to observable decomposedform if

range[T2] = null[U]

where

T=[T1 T2]

with T1, T2 being n x n1, n x n - n1 such that T-1 exists.Proof Let QT denote the inverse of T, i.e.,

(3.28)

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82 State Estimation and Observability

Then the transformed A and C matrices are given by

AT2QT ATE Q TA = T-'AT =

Qz ATl Qz AT2 J A3 A4

C=CT=[CT1 CT2]=[C1 C2]

Now since C appears as the firstp rows of U and T2 is chosen so that UT2 = 0, we haveC2=CT2=0.

Finally since range[T2] = S. and QT is the inverse of T we see from (3.28) that

QTx=0 when xESi,

Then since Ax E So if x c So, we have range [A T2] = S. and thus QTAT2 = 0.

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

3.7.1 The approachSuppose the model for the plant whose state is to be estimated is given as

z(t) = Ax(t) + Bu(t) (3.29)

y(t) = Cx(t) (3.30)

where

YT (t) = [Yi (t) Y2(t) ... yp(t)

Notice that there is no loss of generality in assuming that the state model has D = 0,since when D Owe can replace the left side of (3.30) by q(t) = y(t) - Du(t) and use q(t)in place of y(t) everywhere in the following.

Now we assume throughout that the components {yi(t) : i = 1, 2, p} of the plant

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Minimal Order Observer 83

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

p0 for all t c [0, x) only if [ri = 0 for all i E [1, p]

i-

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

x(t) = C-i y(t)

Since we usually have p < n, C is not invertible. However, even in this case, we can still usey(t) and (3.30) to obtain information on the state. More specifically, we can determinexR (t),the projection of the state on range [C'], from

xR(t) = C#Y(t) (3.31)

where C# = CT (CCT)-1 is referred to as a right inverse of C since CC# = I. Then we canwrite x(t) as

x(t) = XR(t) + xN(t)

where

(3.32)

xR(t) E range[CT] xN(t) E null[C]

Now since we can be obtain xR(t) from (3.31), we see from (3.32) that we can determinethe complete plant state provided we can develop a method for estimating xN(t). Beforedoing this we show that xR(t) can be determined from (3.31).

3.7.2 Determination of xR (t)Notice that the independence of the rows of C ensures that CCT is invertible. We can seethis as follows. Suppose CCT is not invertible. Then we can find y 54 0 such that

YTCCTY=WTW=0

where w = CTy. However this is possible only if w = 0 and therefore CT has dependentcolumns. Since this contradicts the assumption that C has independent rows we musthave CCT is invertible.

Now the expression for C#, (3.31), can be derived as follows. Since xR (t) Erange [CT ] itfollows that

xR(t) = CTw(t) (3.33)

for an appropriate w(t). Then multiplying this equation through by C and using the

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84 State Estimation and Observability

invertability of CCT enables us to determine w(t) as

w(t) _ (CCT) I CxR (t) (3.34)

However from (3.32) we see that

Y(t) = Cx(t) = CxR(t)

and we can rewrite (3.34) as

w(t) = (CCT)-ly(t) (3.35)

Finally, premultiplying this equation by CT and using (3.33) yields (3.31).

3.7.3 A fictitious outputHaving determined the projection, xR(t), of x(t) on range [CT] along null[C], we need todevelop a method for estimating xN(t). This is done by introducing a fictitious output,yF(t), and using it with y(t) to form a composite output, yc(t) as

Yc(t) - LY F(t) j - [ CFC X(t) Ccx(t) (3.36)

where CF is chosen so that Cc is invertible, i.e., CF has n - p independent rows eachindependent of the rows of C. Then we can solve (3.36) for x(t) as

x(t) = Cciyc(t) (3.37)

Before showing how to generate yF(t), it should be pointed out that we can ensure theinvertability of Cc by choosing CF so that

range[CFT-] = null[C] (3.38)

We can see this by using the general result for matrices which is developed at thebeginning of Section 3.6 in terms of the matrix 0, i.e.,

range UT] Lnull [ U]

Therefore we have

range [Cr] Lnull[CF]

and it follows from (3.38, 3.39) that

null[C] f1 null[CF] = {0}

(3.39)

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Minimal Order Observer 85

so that

only if x=0

which implies that Cc is invertible.Now assuming we chose CF to satisfy (3.38) we can readily show that

Ccl C# CF ] (3.40)

where

C# = CT (CCT) t CF = CF(CFCF)-1

This enables (3.37) to be rewritten as

x(t) = C#y(t) + CFyF(t) (3.41)

3.7.4 Determination of the fictitious outputHaving obtained the state of the plant model in terms of the known plant output and thefictitious plant output, (3.41), we need a means for generating the fictitious output. This isdone by using the plant state differential equation to set up a state model which has theplant's input and output, (y(t), u(t)), as input and has state which can be used as anestimate of the fictitious output. The derivation of this state model is given as follows.

We begin by multiplying the plant state differential equation (3.29) by CF to obtain

CFZ(t) = CFAx(t) + CFBu(t) (3.42)

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

CFC#Y(t) + CFCF#.yF(t) = CFAC#y(t) + CFACFyF(t) + CFBu(t) (3.43)

However from (3.38, 3.40) we have CFC# = 0 and CFCF = I. Therefore (3.43) can bewritten as

yF(t) = CFACF#./F(t) + CFAC#y(t) + CFBu(t) (3.44)

Now equation (3.44) suggests that an estimator for yF(t) has differential equation

yF(t) = CFACF#.YF(t) + CFAC#y(t) + CFBu(t) (3.45)

where yF(x) will be an asymptotic estimate of yF(t) provided CFAC#- is stable. Thisfollows from the differential equation for the estimation error, jF(t), which we can

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86 State Estimation and Observability

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

Yr (t) = FRyF(t)

where

Thus we have

so that

FR = CFACF YF(t) = ,F(t) -YF(t)

YF(t) = (eFR`)YF(0)

limy (t) = 0 for all yF(0) (3.46)too

if and only if FR is stable.

3.1.5 Assignment of observer eigenvaluesNow the eigenvalues of FR = CFACF are fixed by the plant state model. Therefore whensome of these eigenvalues are unstable we cannot satisfy (3.46). Recall we encountered thesame sort of difficulty in Section 3.2 in connection with trying to estimate the state of aplant state model without using the plant output. We can overcome the present problemwhen the state model, whose state is to be estimated, is detectable by replacing CFeverywhere in the foregoing development by

T=CF+LRC (3.47)

where LR is used in the same way that L is used in Section 3.3 to assign the eigenvalues ofan observer's system matrix, A + LC. This replacement is effected by replacing yF(t) byz(t) = Tx(t) in (3.36) to obtain

y(t) = CTx(t) (3.48)Z(t)

where

Q=

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

x(t) = My(t) + Nz(t) (3.49)

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Minimal Order Observer 87

where

CT' _ [M N] (3.50)

Now by forming CT-TI = I we can show that M and N must satisfy

CM=I CN=0TM=0 TN=I (3.51)

Therefore recalling from (3.38, 3.40) that

I CCF I [C#

rICF] = I I] (3.52)

we see that the constraints (3.51) on M and N are satisfied when

M = Co - CF LR N = CF

In order to obtain a differential equation for z(t), we follow the approach used to get adifferential equation for yF(t), (3.44). Therefore multiplying equation (3.29) by T andsubstituting for x(t) from (3.49) yields

z(t) = TANz(t) + TAMy(t) + TBu(t) (3.53)

where z(t) = Tx(t) and

TAN = CFACF + LRCACF

TAM = CFAC# + LRCAC# - CFACFLR - LRCACFLR

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

2 (t) = FT2(t) + GTI u(t) ]

Y(t)(3.54)

where

FT = TAN GT = [ TB TAM]

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

2(t) = (eFTt)2(0)

where 2(t) = z(t) - 2(t) and we have an asymptotic estimate of z(t) if and only if FT isstable.

Thus provided FT is stable, we can use z(t) generated by equation (3.54) in place of z(t)

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88 State Estimation and Observability

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

z(t) = My(t) + Nz(t) (3.55)

Notice that the dependency of FT = TAN on LR, (3.53), implies that we are only ableto choose LR so that TAN is stable if the pair (CFACF, CACF) is detectable. We shownow that this detectability condition is satisfied when the plant state model or pair (A, C)is detectable.

Theorem 3.3 The pair (CFACF, CACF) is detectable if and only if the pair (A, C) isdetectable.

Proof Suppose (A, v) is an eigenvalue, right-eigenvector pair for A. Let r be acoordinate transformation matrix given as

r C# CF ] r CF]

(3.56)

Then premultiplying the eigenvalue-eigenvector equation Av = Av by r-1 and insertingrr-1 between A and v gives

Aq = Aq (3.57)

where q = r-1v and

r-1Ar=

[A1 A211 _ CAC# CACFA= _ _

A3 A4 CFAC# CFACF

Then from (3.52) we see that C is transformed as

C=Cr=c[c# cF]=[r 0]Now suppose A is an unobservable eigenvalue of the pair (A, C). Then recalling that

unobservable eigenvalues are invariant to coordinate transformation, we see that q,(3.57), satisfies

Cq=6

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

where q2 has n - p components. Then using the form just obtained for A, we see that theeigenvalue-eigenvector equation, (3.57) becomes

zq2 _ 0

A4g2 [ Aqz ]

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Minimal Order Observer 89

implying that A is an unobservable eigenvalue of the pair (A4, A2) which equals the pair(CFACF, CACF) specified in the statement of the theorem. This shows that theobservability or detectability, (if Re[A] > 0), of (A, C) is necessary for the observabilityor detectability of the pair (CFACF, CACF). Sufficiency can be shown by reversing theforegoing argument.

Notice that when the plant state model's (A, C) pair is observable the coordinatetransformation t, (3.56), just used in the proof of Theorem 3.3, can be used to determineLR, (3.47), to achieve some specified set of eigenvalues for the minimal order observer'ssystem matrix, FT, (3.54). One way of doing this would be to use Ackermann's formula,introduced in the previous chapter, on the pair (A4i A2), (3.57).

The proof of Theorem 3.3 suggests that if we work in coordinates where the statemodel's C matrix is

C = [I 0]

then a minimal order observer to estimate the state in the transformed coordinates can bedetermined by taking T, (3.47), as

T = [LR Q]

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

q(t) = My(t) + Nz(t)

z (t) = TAN2(t) + TAMy(t) + TBu(t)

where

TAN = A4 + LRA2

TAM = LRAI - A4LR - LRA2LR + A3

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

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90 State Estimation and Observability

3.8 Notes and ReferencesThe observer was originally proposed by David G. Luenberger in his Ph.D. thesis in 1964.For this reason the observer is sometimes referred to as a Luenenberger observer.Previously, in 1960, the use of a state model having the same structure as an observerwas proposed by R. E. Kalman for the purposes of least squares signal estimation in thepresence of additive noise. We will encounter the Kalman filter in Chapter 6. For furtherdiscussion of observers the reader is referred to Chapter 4 of [23]

Finally, the term " output injection" is used by some authors to refer to the systemsignal manipulation we use to obtain an observer having system matrix A + LC. Thisterminology should be compared with the use of the term "state feedback" in connectionwith the formation of a system having system matrix A + BK.

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4Model Approximation viaBalanced Realization

4.1 IntroductionIn previous chapters it was noted that whenever a state model (A, B, C) is either notcontrollable or not observable the transfer function resulting from the calculationC(sI - A)-' B has pole-zero cancellations. Thus system uncontrollability and/or unob-servability gives rise to a transfer function with order less than the dimension of the statespace model. Therefore systems that are "almost uncontrollable" or "almost unobser-vable", should be able to be approximated by a transfer function model of order less thanthe dimension of the state model. Reduced order model approximation can be ofimportance in the implementation of feedback control systems. For instance, modelorder reduction may be beneficial when the large dimension of the full state model givesrise to computational problems during the operation of the control system.

This chapter treats the problem of extracting a subsystem of a given full order systemwhich works on that part of the state space which is most involved in the input-outputbehavior of the original system. This is accomplished by introducing two matrices calledthe observability Gramian and the controllability Gramian. It is shown that thesematrices can be used to quantify the energy transfer from the input to the state andfrom the state to the output. In this way it is possible to identify which part of the statespace is controlled most strongly and which part is most strongly observed. Then in orderto construct a reduced order model by extracting a subsystem from the given system, itwill be shown that the strongly controlled subspace must coincide with the stronglyobserved subspace. To achieve this coincidence we change coordinates, if necessary, sothat the Gramians become equal and diagonal. The resulting state model is said to be abalanced realization. The reduced order model is then obtained by extracting a subsystemfrom the balanced realization.

4.2 Controllable-Observable DecompositionThe idea for model order reduction being pursued in this chapter grew out of the basicfact that only the jointly controllable and observable part of a state model is needed tomodel the input-output behavior.

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92 Model Approximation via Balanced Realization

The controllable and observable part of the state space can be viewed as one of fourpossible subspaces for the state space of any given state model. More specifically, considerthe four dimensional state model in diagonal form which is shown in Figure 4.1.

Inspection of the block diagram reveals that the four dimensional state model has atransfer function which is only first order since only the first component of the state isboth affected by the input and affects the output. Thus that part of the state space whichis both controllable and observable forms a one dimensional subspace S, which isspecified as

x(t) E Sao if xi(t) = 0 for i = 2, 3, 4

where

xT (t) = [XI (t) x2(t) x3(t) x4(1) 1

On further inspection of Figure 4.1 we see that there are three other subspaces S(.o, S ,

S,, specified as

x(t) E S'.o if x,(t) = 0 for i = 1, 3, 4

X(t) E S" if xi(t) = 0 for i = 1, 2, 4

X(t) E S16 if xi(t) = 0 for i = 1, 2, 3

where Se.o, Sc.o, Sco, Sao are referred to as follows.

5c.,, is the controllable and observable subspace5co is the controllable and not observable subspaceS,o is the not controllable and observable subspaceS., is the not controllable and not observable subspace

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

u(t)

Figure 4.1 Example of controllable-observable decomposition

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Controllable-Observable Decomposition 93

not in any special form, we can extract a subsystem from the given system which has itsstate space coincident with the controllable and observable subspace. This can be done byusing two successive coordinate transformations as described now.

First we change coordinates so that the state model is in a controllable decomposedform

AA,1 A,2

BC =Bc1

C(_ [ Cc1 C'.2 ] (4.1)L 0 Ar41 0 ]

with (A,.1, B,i) being a controllable pair. The transformation matrix to do this wasdiscussed in Chapter 2.

Notice that, in the new coordinates, we have the full n dimensional state space split intocontrollable and uncontrollable subspaces, S,, S, It remains to split S, into observableand unobservable subspaces, i.e., to split S, into S,,,, and S,c,.

Next we carry out a second coordinate transformation which affects only thecontrollable part of the controllable decomposed state model, (4.1). This is done usinga coordinate transformation matrix, T, which is constructed as

To0

T=0 I

with To being chosen so that the controllable state model (A,,, B,1, C,1) is transformed toobservable decomposed form, i.e., so that

A01 0 Bo1Act = Bc1 = I Ccl = [Col 0] (4.2)

AoA4 J B02

with (A01, C01) being an observable pair. The transformation To to do this was discussedin Chapter 3. Since coordinate transformation preserves controllability, the state model(4.2) is controllable. Therefore the subsystem of the observable part of the state model(4.2), i.e., (A01, Bol, C01) is both controllable and observable. Thus the transfer functionfor the entire system has order equal to the dimension of this subsystem and is given by

G(s) = C01 (sI - A01) 1801

Notice that the foregoing model reduction procedure applies to full order state modelswhich are either not controllable or not observable or both not controllable and notobservable. More important, the reduced order model we obtain has exactly the sameinput-output behavior as the full order model.

In this chapter we are interested in obtaining a reduced order model when the full orderstate model is completely controllable and observable. In this situation any reduced ordermodel will have input-output behavior which differs from the input-output behavior ofthe full order state model. Our goal is to develop a method for achieving a reduced orderstate model having input-output behavior which approximates the input-outputbehavior of the full order state model. We will see that we can develop an approachfor solving this model approximation problem by replacing the concepts of controll-

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94 Model Approximation via Balanced Realization

ability and observability by concepts of energy transfer. We will see that the distributionof energy among components of the state is related to properties of Gramian matriceswhich replace the controllability and observability matrices in this analysis. Just ascoordinate transformations are important by extracting the controllable and observablehart of a state model, so too are coordinate transformations important to being able toseparate out that part of a controllable and observable state model which is mostimportant in the transfer of energy from the system's input to the system's output. Thestate model in coordinates which enable the identification of this part of the system iscalled a balanced realization.

4.3 Introduction to the Observability GramianIn order to define the observability Gramian we need to consider the related idea ofoutput energy. Recall that if the state model (A, B, C) is known, then the zero inputresponse, y(t), caused by any specified initial state, x(0), is given as

y(t) = CeA`x(0) (4.3)

The output energy, E0, for each initial state, x(0), is then defined as the integral of anonnegative, real, scalar function v(t) of y(t), i.e., v(t) > 0 for all t c [0, oc)

Eo =Jac

v(t)dt0

Notice that since v(t) is non-negative real for all t, the output energy E0 for any initialstate x(0) must satisfy

Eo > 0 for all x(0)

Eo=0 only ifv(t)=0 fort>0

A simple way to choose v(t) so that Eo has these properties is as follows.

if y(t) is a real scalar then choose v(t) = y2(t)

if y(t) is a complex scalar then choose v(t) = I y(t)I2

if y(t) is a real vector then choose v(t) = yT (t)y(t)

if y(t) is a complex vector then choose v(t) = y*(t)y(t)

Notice that when y(t) is a scalar, y*(t) denotes the complex conjugate of y(t) andIy(t) I2= y* (t) y(t). Alternatively when y(t) is a complex vector, y* (t) denotes the transposeof y(t) and the complex conjugation of each entry in the resulting row vector. Thus if y(t)is a complex vector of length p then

p

y*(t)y(t) = Elyi(t)I2i=1

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

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Introduction to the Observability Gramian 95

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

E,, _ fy(t)v(t)dt (4.4)

so that after substituting from (4.3) we obtain

E,, = x* (0) (fOC

eA*iC*CeA`dt) x(0) (4.5)0

r = x* (0) Wox(0) (4.6)

where the constant n x n matrix WO is called the observability Gramian.Notice that in order for the foregoing integral to converge, or equivalently for W00 to

exist, the observable eigenvalues of (A, C) must be in the open left half plane. This can beseen by referring to section 7 of Chapter 1. Alternatively, if (A, C) is observable WO existsonly if A is stable. Notice also that W,, must be

1. an Hermitian matrix, i.e., WO = W o,2. a non-negative matrix, denoted W0 > 0.

Requirement number one results from the integral defining WO, (4.5). Hermitianmatrices have entries in mirror image positions about the diagonal which are conjugatesof each other, i.e., (W0)0 = (W0)ji. When W0 is real we require WO = WT and WO isreferred to as a symmetric matrix. Hermitian (symmetric) matrices have eigenvalues andeigenvectors with some rather striking properties which we will develop in the nextsection.

Requirement number two results from the output energy being nonnegative for allinitial states since a non-negative matrix, W, has the property that

x*(0) 0

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

x(0) Wox(0) = x(0)*M*Mx(0)n

= Z*Z = Elzilzi=1

where z = Mx(0). Since E, J1zilz> 0 for any z we have Wo > 0. Moreover sinceJ:i= 1 zi1z= 0 if and only if z = o, we have x(0)* Wox(0) = 0 for some x(0) o if and

only if M and hence Wo is singular, i.e., if and only if W has one or more zeroeigenvalues. More is said about this in the next section.

Finally the observability of a state model is closely tied to the nonsingularity of itsobservability Gramian. This can be seen by noting from (4.4) that Eo = 0 if and only ify(t) = o for all time. However we saw in Chapter 3 that the zero input response is null,y(t) = o, for some x(0) 54 o if and only if (A, C) is unobservable. Therefore it follows that

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96 Model Approximation via Balanced Realization

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

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

4.4.1 Hermitian matricesRecall that a matrix Wo is said to be Hermitian if it equals its conjugate transpose,Wo = W *. This implies the ij`h and ji`h entries are conjugates of each other. Theeigenvalues and eigenvectors of Hermitian matrices have some rather striking propertieswhich are given as follows.

Theorem 4.1 If Wo is Hermitian then:

(i) W. has real eigenvalues(ii) WO has right eigenvectors which satisfy v`*vj = 0, when aoi ,-- Qoj

Proof (i) Let (a0, v) be any eigenvalue-eigenvector pair for W0, i.e.,

Wov = a0v

Then premultiplying (4.7) by v* gives

v*W0,v = aw*v

and taking the conjugate transpose throughout this equation gives

v* W*v = a*v*v

Then subtracting (4.9) from (4.8) we obtain

v*(W, Wo)v = (a0 - (4.10)

Now since WO is Hermitian, the left side of (4.10) is zero. Therefore either a", = a**, orv*v = 0. However v*v = 0 holds only if v = o, and since v is an eigenvector of WO, we havev*v 54 0. Therefore (4.10) can only be satisfied if a,, = a* which shows that ao must bereal.

Proof(ii) Let (ao;, v` : i = 1, 2) be two eigenvalue-eigenvector pairs for W withaol 54 ao2. Then we have

Wool = aol,vl (4.11)

Yi'ov2 = ao2v2 or v2- W* = ao2v2* (4.12)

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Fundamental Properties of W. 97

Next operate on these equations as follows:

(i) premultiply (4.11) by v2*

(ii) postmultiply the second form for (4.12) by v1(iii) subtract the result of (ii) from (i)

This yields

v2*(W - W* v1 = a - a* v2*vl0 0) ( 01 02) (4.13)

However, since W0 is Hermitian, the left side of (4.13) is zero. Moreover we are assumingthat a01 # a 2. Therefore (4.13) is satisfied only if VII V2 = 0.

Notice that when W0 is real, i.e., symmetric, eigenvectors corresponding to differenteigenvalues are orthogonal, i.e., v`T vj = 0 : i # j, a01 Qoj.

Recall that if a matrix has distinct eigenvalues, it has a complete set of eigenvectors andcan be diagonalized by a matrix having columns made up from its eigenvectors. Thuswhen W0 has distinct eigenvalues we can use its eigenvectors {v` : i = 1, 2, . n} to form amatrix Vo which diagonalizes W0

Vo 1 W0 V0 = E0

where

E0 =

laol 0 ... 0 10 a02

... 0

0 0 - a0n J

= diag[aol, a02, ' , aon]

V0 = [v1 v2 ..V

n]

However if the eigenvectors are normalized,

v`*v`=1:i=1,2, n

then we see from Theorem 4.1 that

VOVo=

rvl*

V2*

vn*

[v1 v2 ... vn] = I

(4.14)

(4.15)

Therefore Vo 1 = Vo so that V*WOV = E0. Matrices satisfying the condition Vo 1 = V*are referred to as unitary matrices.

Alternatively, when W0 is real then (4.15) becomes

V;VO=I

and therefore V0 -1 = VT

oso that VT W0 V0 = E0. Matrices satisfying the condition

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98 Model Approximation via Balanced Realization

Vt = VT0 are referred to as orthogonal matrices. It can be shown that any Hermitianmatrix car, be diagonalized by a unitary matrix even if the Hermitian matrix has multipleeigenvalues.

To recap, we have encountered the following matrix types.Definition 4.1

If M is complex then

M is Hermitian if M M*M is unitary if M-1 = M*

If M is real then

M is symmetric if M = MTM is orthogonal if M-I M7

4.4.2 Positive definite and non-negative matricesSuppose that we change coordinates, viz., x(O) Tz(O), with T nonsingular. Then foreach initial state x(O) in the old coordinates, there is a unique initial state z(O) in the newcoordinates and vice versa. Thus we see from the expression for the output energy, (4.6),that the output energy can be rewritten in terms of the initial state in the new coordinatesas

E,, = z*(0) W0z(0) (4.16)

where W,, = T * WOT.Now if we choose T as the unitary matrix V0, (4.14), then

Wo = V;W0Vo = F.,

and (4.16) becomes

n

E = aoi (4.17)

However since there is a unique z(O) for each x(O) we have that E0 > 0 for all x(0) ifand only if E0 > 0 for all z(0). Thus we see from (4.17) that E,, > 0 for all x(0) if and onlyif all eigenvalues of W0 are non-negative, i.e.,

o,>0 i=1,2,...n

Otherwise, if W0 were to have some negative eigenvalues, we could choose z(0) so that

zi(0))4 0 if a0; < 0 and z;(0) = 0 if a0r > 0 i = 1, 27 , n

which would give E0 < 0, for the initial state x(0) = Tz(0) and the requirement for Eo tobe non-negative for all initial states would be violated.

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Fundamental Properties of W, 99

Finally, we can show by a similar argument that Eo > 0 for all x(O) 4 0 if and only ifall eigenvalues of W are positive.

To recap, we have defined the matrix properties of positive definiteness and non-negativeness as follows.

Definition 4.2 W, said to be:

(i) positive definite, denoted as W > 0 when x* W,,x > 0 for all x 54 0

(ii) non-negative, denoted as W > 0 when x* Wax > 0 for all x

Moreover these properties impose certain restrictions on the eigenvalues of Wnamely

(i) if Wn > 0 then o-,,, > 0 for Q,,, E .X[

(ii) if W0 > 0 then a,,, > 0 for Q,,, E

4.4.3 Relating E. to A[W0]In order to begin to appreciate the role played by the eigenvalues of W in achieving alower order state model approximation we consider the problem of finding the initialstate, xmax(0), which produces the most output energy, Eomax, when the initial state isconstrained so that

xmax(0)xmax(0) = 1 (4.18)

When x(0) is constrained in this way we say x(0) is normalized to one. Notice that thisconstraint on the initial state removes the possibility of obtaining an unbounded outputenergy by choosing an initial state with an unbounded component.

We begin the development of a solution to this problem by noticing that if we use acoordinate transformation matrix T which is unitary, the constraint (4.18) is preserved,i.e., if T satisfies T*T = I then we have

x*(0)x(0) = z*(0)T*Tz(0) = z*(0)z(0) (4.19)

Therefore if x*(0)x(0) = 1 so does z*(0)z(0).Now we have seen that using the unitary matrix T = V as a coordinate transforma-

tion matrix makes the observability Gramian diagonal in the new coordinates. Thereforeour problem can be solved by first solving the problem in coordinates in which theobservability Gramian is diagonal and then transforming the solution, Zmax(0) in thesecoordinates back into the original coordinates.

Suppose we are in coordinates where the observability Gramian is a diagonal matrix,E0. Suppose also that we have reorder the components of the state in these coordinates sothat the largest entry on the diagonal of E,, is first, i.e., a01 > Q,,, : i = 2, 3, , n. Then wesee from equation (4.17) that the output energy satisfies

n n

Eo = aotI zt(0)12 < a01 z1(0) 2 (4.20)

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100 Model Approximation via Balanced Realization

However since we are assuming that x(0) is normalized to one, we see from (4.19) thatz(0) is also normalized to one

n

z'(0)z(0) _ zi(0) I2= 1 (4.21)

Therefore under this constraint we see from (4.20) that the output energy, E0, is boundedby the observability Gramian's largest eigenvalue, i.e.,

Eo < (Yoi (4.22)

Notice that if we choose

zT(0) = [1 0 ... 0]

then (4.21) is satisfied and the output energy becomes

n

Eo =0,niIZi(0) 2

= Qoti=0

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

xmax(0) = Voz(0) (4.23)

rl

0= [VI V2 ... vn]

0

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

E. =

when

x(0) = vk

The foregoing analysis indicates that the relative importance of the components of z(0)in producing output energy depends on the size of an eigenvalues of Wo relative othereigenvalues of Wo. We can use this observation to suggest how we are going to be able toobtain a reduced order state model approximation. We do this as follows.

Suppose we choose the coordinates for the state space so that the eigenvalues of Wo arein descending size, i.e., aol > 0o2 > > aon with W0 diagonal. Suppose we consider z(0)

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Introduction to the Controllability Gramian 101

to have all its elements equal, zi(0) = z.(0). Then the output energy is given by

Eo = z'(0)

001

0

0 0

f2

of0

Qon

where

z(0) = Qoi

'Y Izi(0)12

Moreover, we can always choose an integer n1 C [0, n] large enough so that

Eo 1 >> E02

where

n, n

E'ol = 1' poi . E,,2 = 1 jaii=1 i=ni+1

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

(4.24)

Since the last n - n1 components of the initial state are less important to the zero inputresponse than the first n1 components under the foregoing condition of an equal elementinitial state, a reduced order state model in these coordinates could be formed as(AI, B1, C1) where Al is the n1 x n1 partition in the top left corner of A and '01, C, arecorresponding partitions of B, C.

However, our interest in this chapter is in obtaining a reduced dimension state modelwhose input-output behavior approximates the input-output behavior of the fulldimension model. The foregoing argument for neglecting the last n - n1 componentsof z(t) focuses only on the zero input response. Therefore what is missing from theprevious argument, if we are interested in modeling the input-output behavior, is theeffect the input has on the neglected components of z(t). The effect of the input ondifferent components of the state is related to properties of a matrix known as thecontrollability Gramian.

4.5 Introduction to the Controllability GramianThe controllability Gramian, which we introduce in this section, is involved in char-acterizing the energy delivered to the state when the input is a unit impulse and the initialstate is null. Since the Laplace transform of the zero state response to this input is thegiven system's transfer function the distribution of energy among the components of thestate in this situation is appropriate to our goal of obtaining a reduced order transferfunction approximation. We begin the development of the controllability Gramian asfollows.

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102 Model Approximation via Balanced Realization

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

x(t) _ f etT)B6(-r)dy = eB

Then the energy, E,., transferred from this input to the state is given by

E,. = J x x*(t)x(t)dt = J B*e 'eA'Bdto 0

(4.25)

Notice that the integrand obtained here is strikingly different from the one obtained inthe integral for the output energy, (4.5). We proceed now to express E,. (4.25) as anintegral which appears as the dual of the integral expression of E,,. More specifically, weshow now that E(. can be expressed as the trace of the integral obtained earlier for E0, (4.5)with A replaced by A* and C replaced by B*. The trace of a matrix is a scalar which isdefined as follows.

Definition 4.3 The trace of any square matrix M of size n, is defined as the sum of itsdiagonal entries,

trace[M] _ mu

where m;j is the ij`h entry in M.In order to proceed we need the following result involving the trace of the product of

two, not necessarily square, matricesTheorem 4.2 Given two matrices M, N of size p x n, and n x p then

trace[MN] = trace[NM]

Proof Let MN = Q. Then

P

trace [MN] _ q,;t=

where q;; is the result of multiplying the i th row of M by the i th column of N,

n

qi1 _ E m,ni;J=i

Therefore we can rewrite the trace of the matrix product as

p n

trace[MN] _ 1: 1: mjnj;i-i i=1

Next interchanging the order of the summations gives

n P

trace[MN] _ nj;m;i-t =1

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Introduction to the Controllability Gramian 103

and we see that

P

Y'n1,my = S11i=1

where sjj is the jch entry on the diagonal of S = NM, and the result followsAn immediate consequence of this theorem is that the trace of any square matrix is the

sum of its eigenvalues. More specifically if M = V- 'AV, then

n

trace [M] = trace [ V-1 A V] = trace [A VV_ 1 1 = \,J

r=1

where \, E A[M].Returning to the problem of re-expressing the energy E, (4.25), we use Theorem 4.2

with

M=B*eA*1 N=eAtB

to obtain

E, = trace[W,] (4.26)

where W, the controllability Gramian, is given by

W, = eA`BB*eA' `dtf (4.27)

Notice that W, is Hermitian and that We is non-negative since the integrand in (4.27)is quadratic. Notice also that if we replace A by A* and B by C* in the foregoing definitionof the controllability Gramian, We., (4.27), we obtain the observability Gramian, W0,(4.6). Thus all the properties discussed in connection with Wo hold for W,

More specifically, since W, is Hermitian, W, has real eigenvalues and can bediagonalized by a unitary matrix, V, i.e.,

V* W'V' = E, (4.28)

where V-1 = V* and E, is real and diagonal. Notice that V,, has columns which arenormalized right eigenvectors of We and the diagonal entries in the diagonal matrix E,are the eigenvalues of W. In addition recalling the discussion following the proof ofTheorem 4.2 that the trace of a square matrix equals the sum of its eigenvalues, we have

E, = trace [EC]

Now we will see in the next section that we can always transform coordinates so thatthe controllability Gramian is diagonal. Assuming we have done this transformation, wepermute components of the state, if need be, so that the entries along the diagonal of Ecappear in descending size, i.e., o- , > uc(i+,) : i = 1, 2, n - 1. Then we can always

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104 Model Approximation via Balanced Realization

choose nl so that

where

Ec.1 >> E,,

r,,

E, = E,1 + E,.2 with E,1 = E Q,.i and with E,.2 =i=1 i=n,+1

This inequality implies that the first nl components of the state in these coordinatesreceive most of the energy from the input. If in addition, in_ these coordinates, theobservability Gramian happened to be diagonal with entries on the diagonal being indescending size, the first nl components of the state would also be relatively moreimportant than the last n - nl components in the transfer of energy to the output. Wecould then obtain an approximation to the full order transfer function for the system as

Gapx(s) = Cbl (sI Ab1) 1Bbl

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

AblAb =

A,,3

Cb = [ Cb1

Ab2

A,,4Bb =

Bbl l I nt

[Bb2J In - nl

Cb2 ]

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

with Eb having entries in descending size along its diagonal. A state model having thisproperty is referred to as a balanced realization. A reduced order approximation, Gap,, (s),to the full order system is obtained from the balanced realization in the manner justdescribed.

Finally, we are assuming the pair (A, B) is controllable so that every component of thestate receives some of the energy supplied by the input. Therefore, from the foregoing, wehave all eigenvalues of W, bigger than zero or W,. > 0 when (A, B) controllable. Inaddition, it turns out that

=rank[1]

4.6 Balanced RealizationSuppose we are given the controllability and observability Gramians, W,, W0, for acontrollable and observable state model, {A, B, C, D}, of the system to be approximated.Then in this section we will show how to construct a coordinate transformation matrix so

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Balanced Realization 105

that the state model in the new coordinates is a balanced realization. Just as in Section 4.2we saw that a succession of two state coordinate transformations are required to achieve acontrollable-observable decomposed form, so too in this section we will see that we need asuccession of two transformations to achieve a balanced realization. The first coordinatetransformation gives a new state model with controllability Gramian equal to an identitymatrix. The second coordinate transformation keeps the controllability Gramian diag-onal and diagonalizes the observability Gramian obtained from the first transformationso that the resulting Gramians are equal and diagonal.

To develop these transformations we need the fact that, if the coordinates are changedusing coordinate transformation matrix T, then the Gramians (We, W0)for the statemodel in the new coordinates are related to the Gramians (We.., W0) for the state model inthe original coordinates as

W, = T-'W,,T-* W,, = T* W0T (4.29)

where x(t) = Ti(t). These relations can be obtained by replacing (A, B, C) in theintegrals for W(. and W0 by (T-'AT, T-1B, CT). We use (4.29) now to develop a thesequence of three coordinate transformations T; : i = 1, 2 and P so that Tb = T1T2Ptransforms the given controllable and observable state model to a balanced realization,where P is the permutation matrix required to put the diagonal elements of the Gramiansin decreasing size. We will show now that these coordinate transformation matrices aredetermined from the Gramians for the given state model. In the next section, we will showthat the Gramians for the given state model can be determined by solving two linearmatrix equations known as Lyapunov equations.

For the time being suppose we are given the controllability and observabilityGramians, W, W0 for the state model we wish to approximate. Then we form the firstcoordinate transformation as

T1 = V,E'c

where VC* W, V, = E, with V *C = VC-land E > 0 is diagonal with E lE l = E c . Then

using (4.29) we see that the Gramians which result from the first coordinate transforma-tion are given as

Wc =T11W.T1*=I WO =T*W0Tj

Next, we take the second coordinate transformation as

T2= VOr'O

where V o W o V o = Eo and Eo > 0 with ( > 1 4 )4 = YO. Then using (4.29) we see that theGramians which result from the second transformation are given by

Wc=TZ'WcTz'=Eo

W. = T*2 W,T2 = E0

and we have the Gramians equal and diagonal.

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106 Model Approximation via Balanced Realization

It remains to permute the components of the state so that the entries on the diagonal ofthe Gramians are ordered in decreasing size. This corresponds to a third coordinatetransformation, this time using a permutation matrix P as the coordinate transformationmatrix so that

PW(.P = PWW,P = Eh

where

Eb = diag[Ob1, Ub2, ... Qbn] with abl ? .. > (yb

and the obis are referred to as the Hankel singular values for the system. More is saidabout this in Section 4.9

At this stage we can obtain the balanced realization, (Ab, Bb, Ch), from(T-1 AT,

T-1 B, CT) with T = T 1 T,P. Then partitioning the balanced realization as

Abi Ah2 Bbl I n1Ab = Bb = (4.30)

An_ 3 Ab4 Bbl .. n- n1

Cb=[Cb1

with n1 large enough so that

Cb2 I

n n

O'bi >> 9bii=1 i=nj+1

we obtain the reduced order state model as (Ab1 i Bbl, Chl i D) and the reduced ordertransfer function model as

Gapx(s) = Cb1(sI - Ab I) -'Bbl

Recall that we are assuming that the full order state model is stable, i.e., Ab is stable.Therefore we get a bounded output for every bounded input to the full order, unreducedsystem. Thus if Gap,,(s) is to be a meaningfully reduced order approximation to the fullorder transfer function, we expect Gapx(s) to be stable. Otherwise the behavior of thereduced order system would be dramatically different from the behavior of the originalfull order system. The following theorem specifies conditions which guarantee that thereduced order model is stable.

Theorem 4.3 If in (4.30) we have 0bn, 5 ahn,tl then Ab1 is stable.Proof AppendixIn the next section we show that the Gramians for a given state model can be

determined by solving linear matrix equations called Lyapunov equations. In this waythe calculation of the integrals in the definition of the Gramians is avoided. We will alsosee that further relations between the Gramians for a state model and the observability,controllability, and stability of that state model are uncovered by considering theGramians to be solutions to Lyapunov equations.

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The Lyapunov Equation 107

4.7 The Lyapunov EquationWe begin this section by generalizing the output energy as follows. Recall that in theabsence of an input the observability Gramian gives the output energy caused by anyinitial state as

E0 = x*(0)Wox(0)

Now suppose we replace this constant quadratic function of the initial state by ea(t), atime varying quadratic function of the state given as

e,,(t) x*(t)Px(t) (4.31)

where P is a positive definite Hermitian matrix and x(t) = eAtx(0). Notice that thisfunction is always positive

eo(t) > 0 for all t c (0, oo) and all x(0) / 0 (4.32)

Next assume that et,(t) has a derivative which is a negative quadratic function of thestate

den(t) _ _x*(t)Qx(t) < 0 for all t c (0, oo) and all x(0) 0 (4.33)dt

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

lim 0 (4.34)t-x

However since P is invertible we have eo(t) = 0 only if x(t) = 0 and (4.34) implies that

lim x(t) = 0 for all x(0)t oo

(4.35)

Therefore equations (4.31), and (4.33) imply that the state model is internally stable, i.e.,the state model's A matrix has all its eigenvalues in the open left half plane.

Now if we differentiate (4.31) and use z(t) = Ax(t) we can express the time derivativeof eo(t), as

eo(t) = x*(t)(A*P+ PA)x(t)

These observations lead us to formulate the following theorem concerning the stabilityof A.

Theorem 4.4 If for any Hermitian Q > 0 we obtain P > 0 as a solution to the linearmatrix equation

A*P + PA = -Q (4.36)

then A is stable.

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108 Model Approximation via Balanced Realization

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

v*(A*P+ PA)v = -v*Qv

which recalling v*A* = A*v*becomes

(A* + A)v*Pv = -v*Qv (4.37)

However since v* Pv > 0 and -v* Qv < 0 for all v p, we see that (4.37) is satisfied onlyif (A* + A) = 2Re[A] < 0. This shows that all eigenvalues of A lie in the open left halfplane and therefore A is stable. 0

The linear matrix equation (4.36) is called a Lyapunov equation after the Russianmathematician who originated this approach in the study of stability.

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

A* Wo + W,,A = C*C (4.38)

Analogously, the controllability Gramian, W, is the solution the Lyapunov equation

AW, + WcA* = -BB* (4.39)

The foregoing relations between the Gramians and the solutions to Lyapunovequations are shown in the following theorem.

Theorem 4.5 If F is a square stable matrix and if

eFtQeF*tdt = W

then

FW+WF*=-Q

Proof We begin the proof by defining the matrix function

Z(t) = eFtQeF'r

Then we differentiate Z(t) to obtain

dZ(t) = FeFtQeF't + eFtQF*eF tdt

(4.40)

(4.41)

Next recalling the series expansion of e F,t, we see that F * and e F t commute. Therefore

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The Lyapunov Equation 109

the foregoing equation can be written as

dZ(t)dt

= FZ(t) + Z(t)F* (4.42)

Finally, since A is being assumed stable, we have Z(oc) = 0. Therefore integrating(4.42) from 0 to cc gives

fdZ(t) _ f3" [FZ(t) Z(t)F*]dt

Z(oc) - Z(0) = F[10C

eFt QeF t dt1 + [ J eF`QeF `dt] F*0

-Q=FW+WF*

Now there are two important observations we can make when we compare theLyapunov equation (4.36) with the Lyapunov equations for the Gramians (4.38, 4.39).

First, notice that unlike Q on the right side of (4.36), C*C and BB* in (4.38, 4.39) areonly non-negative.

Second, (4.38, 4.39) are linear equations in the elements of the matrices WO, WC.Therefore we may be able to solve (4.38, 4.39) for matrices Wo and We when A is notstable. However since the observability and controllability Gramians (4.5, 4.27) are onlydefined when A is stable, the matrices W0, We obtained by solving (4.38, 4.39) are systemGramians only if A is stable.

The implications of these observations are taken up next.

4.7.2 Observability, stability, and the observability GramianWhat follows are theorems which describe how the concepts named in the title of thissection are interrelated.

Theorem 4.6 If (A, C) is observable and if P > 0 satisfies the Lyapunovequation

A*P+PA = -C*C (4.43)

then A is stable.Proof Suppose, contrary to the theorem, that A is not stable, i.e., Av = Av with

Re[A] > 0 for some A E A[A]. Then proceeding as in the proof of Theorem 4.4 we obtain

2Re[A](v*Pv) _ -v*C*Cv (4.44)

Now since (A, C) is observable we have Cv = z 0 and v * C * Cv > 0. However we areassuming that Re[A] > 0. Therefore (4.44) is satisfied under these conditions only if wehave v*Pv < 0 which contradicts the condition P > 0 specified in the Theorem. Thereforeour assumption that A is unstable is false.

Notice, in connection with the foregoing proof, that when A is unstable and (A, C) isobservable we have

rTW = lim J eA+`C*CeA`dt (4.45)o T-.oc 0

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110 Model Approximation via Balanced Realization

is unbounded and the observability Gramian, W0, is undefined. However this does notmean that we are unable to find P to satisfy (4.43) in this case.

The foregoing theorem has the following corollary.Theorem 4.7 If A is stable and P is positive definite and satisfies (4.43) then (A, C) is

observable.Proof Suppose, contrary to the theorem, that (A, C) is not observable. Then we must

have a (A, v) pair, Av = A v and Cv 0. Therefore proceeding as in the proof of Theorem4.6 we obtain

2Re[A](v*Pv) = -v*C*Cv 0 - (4.46)

Now in order to satisfy (4.46) we must have Re[A] = 0 or v*Pv = 0 or both. Thecondition Re[A] = 0 is impossible since we are assuming that A is stable. The conditionv*Pv = 0 is also impossible since we are assuming that P > 0. Thus (4.46) can not besatisfied and therefore (A, C) must be an observable pair. 0

To recap, the foregoing pair of theorems state that A is stable when (4.43) is satisfied byP > 0 if and only if (A, C) is an observable pair. Since the observability Gramian W0,(4.45) is defined when A is stable and positive definite when (A, C) is an observable pair, ifthe solution P to (4.43) satisfies P > 0 when (A, C) is not an observable pair P is not theobservability Gramian. The following theorem characterizes the effect of having a P > 0as a solution to the Lyapunov equation (4.43) when the pair (A, C) is not observable.

Theorem 4.8 If the pair (A, C) is not observable and P > 0 satisfies the Lyapunovequation (4.43) then A has imaginary axis eigenvalues.

Proof Since the pair (A, C) is not observable, we can choose a (A, v) pair so thatAv = Av and Cv = 0. Then we see from (4.46), that

2Re[A](v*Pv) = 0 (4.47)

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

equation

A*P+ PA = -C*C

is satisfied by P > 0 if and only if the pair (A, C) is observable. Alternatively, if thisLyapunov equation is satisfied by P > 0 when (A, C) is not observable, the unobservableeigenvalues of A lie on the imaginary axis. Notice from the proof of Theorem 4.7 thateigenvalues of A that are not on the imaginary axis are observable and lie in the left halfplane.

Analogously, A is stable when the Lyapunov equation

AP+PA*=-BB*

is satisfied by P > 0 if and only if the pair (A, B) is controllable. Alternatively, if thisLyapunov equation is satisfied by P > 0 when (A, B) is not controllable, the uncontrol-lable eigenvalues lie on the imaginary axis and the controllable eigenvalues are all in theleft half plane.

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Controllability Gramian Revisited 111

4.8 Controllability Gramian RevisitedIn this section we develop another use for the controllability Gramian which provides uswith a different way of viewing the process of model reduction and a new way of thinkingabout a system's input-output behavior.

4.8.1 The least energy input problemAn alternative use for the controllability Gramian, W,., arises in connection with thesolution to the problem of finding the input uo(t) : t c (-oc, 0] which expends the leastenergy in taking the null state at time t = -cc to some specified state at time t 0.

In order to solve this problem we need to recall, from Chapter 1, that the basicequation relating the initial state and input to the final state is

ft

x(tf) = et x(-oc) + ettT)Bu(T)dr (4.48)x

Then, in the present situation, we have x(-oo) = 0, and t= 0 so that (4.48) becomes

0

x(0) = f e T Bu(T)dT (4.49)

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

()u(T)d7- (4.50)

achieves its minimum value, Eu,, determined from

E,,,, = min Ec,U(t)

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

uo(t) = B*eA t W ix(0) (4.51)

where W, is the controllability Gramian, (4.27). Notice that this input satisfies (4.49) andhas energy determined as

fEau = Z(*(T)Z!(T)dT

0

= x*(0)Wc 1 (f e ATBB*e A*TdY) We'x(0)CX_

= x*(0)W, Ix(0) = E 0

Notice that u0(t) exists provided W, is invertible. This requirement that the (A, B) pairbe controllable is a necessary condition for being able to satisfy (4.49) for any x(0), (seeChapter 2 Section 2.4).

Recall that the coordinate transformation matrix, V,, (4.28), that diagonalizes W, isunitary, VC-1 = V*C. Thus we see that this coordinate transformation also diagonalizes

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112 Model Approximation via Balanced Realization

WC I since

(V * W ,.V,)-,= V *W,:-1 V, = E

Therefore we can use this fact to rewrite E,,,,, as

E,,,o = x`(0)VcV*WC V*x(0)

= z«(0)E-z(0) _ zr(bi (4.52)bbl

wherez(0) = V*x(0)

Then if the components of z(O) are equal and have been permuted so the entries alongthe diagonal of E, are ordered, abt % ab(t+1) :

i = 1, 2, n - 1, we see from (4.52) thatmost of the energy in the input is needed to achieve components in the lower partition ofz(0). This implies that the lower partition of z(t) is harder to control than the upperpartition.

Now in the following we make use of the foregoing alternative interpretation of thecontrollability Gramian to provide further insight into the use of a balanced realization toachieve model reduction.

Recall that a balanced realization has both Gramians equal to the same diagonalmatrix with the entries encountered in descending the diagonal forming a nonincreasingsequence. Then we saw, in Section 4.4.3, that most of the energy supplied to the output byan initial state with equal component values comes from components in the upperpartition of the initial state. However, we have just seen that the components in the upperpartition of the state are more easily manipulated by the input than components in thelower partition. Therefore, in balanced coordinates, the subspace of the state spacecorresponding to the upper partition of the full state is more involved in the input-outputbehavior than the subspace corresponding to the lower partition of the state. Thissuggests that we can capture most of the full order system's input-output behavior byusing a reduced order state model formed by truncating the balanced realization in themanner given at the end of Section 4.6.

4.8.2 Hankel operatorWe can obtain further insight into a system's input-output behavior by using the presentinterpretation of the controllability Gramian. This is done by noting that:

(i) the controllability Gramian is used to characterize the system's transformation ofu(t) : t E (-oc, 0] into x(0);

(ii) the observability Gramian is used to characterize the system's transformation ofx(O) into y(t) : t c [0, no) when u(t) = 0 : t E [0, no).

This suggests that, in addition to the concept of a system implied by its transferfunction, G(s), i.e., that the system transforms or maps system inputs, u(t) : t E [0, no), tosystem outputs, y(t) : t c [0, no), we can view the system as a map from past inputs,u(t) : t E (-no, 0] to future outputs, y(t) : t c [0, no). This map is referred to as the

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Controllability Gramian Revisited 113

system's Hankel operator and is denoted as rG. When a system is stable its Hankeloperator has an operator norm which is referred to as the Hankel norm. Operators andoperator norms will be developed in more depth in Chapter 7. For the time being we wantto show that the Hankel norm of a system equals the largest entry on the diagonal of theGramians when the state model for the system is a balanced realization.

Suppose we are given a strictly proper transfer function, G(s), and minimal statemodel, (A, B, C), for some stable system. Then the Hankel operator, FG, for this systemhas output y(t) : t C [0, oc) produced by input u(t) : t c (-oc, 0] when x(-oc) = 0 andu(t) _ 0 : t E [0, oc).

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

JIG(s)IIH = max f0 y*(t)y(t)dtf° u*(t)u(t)dt

'u(

(4.53)

where (.)' indicates the positive square root. There is a close connection between asystem's Hankel norm and that system's balanced realization as shown in the followingtheorem.

Theorem 4.9 The Hankel norm IIG(s)IIH for a stable system having transfer functionG(s) is given as

IIG(S) IH='\max[WcWo] = 0bI (4.54)

where ob1 is the largest entry on the diagonal of the Gramians in balanced coordinates.Proof Recalling (4.29) we see that the eigenvalues of the product of the controllability

and observability Gramians are invariant to a coordinate transformation since we have

T-1 WcT-*T *W0T = T-1 WCWOT (4.55)

Thus when T is the coordinate transformation which puts the system in balanced formwe have

A[Eb] = {Qhi : i = 1, 2, ... n}

where {obi 1, 2, n} are referred to as the Hankel singular values.Next recall from Section 4.4.3 that when the state model is a balanced realization, the

maximum output energy, Eomax, produced by an initial state, z(0), constrained to satisfyz*(0)z(0) = a2, when u(t) = o : t c [0, oc), is given by

Eomax = max [E0] = 1Zma:(O)] *Eb[Zmax(O)]

= a2Qb1z(o)

,*(a) z(0)=.2

where

[Zmax(O)]T= [a 0 ... 0]

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114 Model Approximation via Balanced Realization

Alternatively, recalling the solution to the least energy input problem, we see from(4.52) that using the least energy input uo(t) : t c (-oe, 0] when z(-oo) = o to producez"'a (0) gives the least energy as

Ecmin min [E .u] [Zmax(0)] Eh I Zmax(0)

z(0)=_ma'(0)

Therefore we see that the maximum value of the ratio of energy out to energy in is given as

Eomax = 2

E. min

which from (4.53, 4.55) yields (4.54). 0

4.9 Notes and ReferencesHermitian matrices play an important role in many branches of science and engineering,e.g., quantum mechanics. In general a square matrix having repeated eigenvalues may notbe able to be diagonalized by a similarity transformation, i.e., we may not be able to findV such that V-1 MV is diagonal. However a Hermitian matrix can always be diagonalizedeven if it has repeated eigenvalues, [39].

The term "Gramian" arose originally in connection with the problem of determining ifthe vectors in a set of vectors {vi : i = 1, 2,- . . , n} were linearly independent. TheGramian G was defined as the scalar whose value was determined as G = det[M],where (M)y = vi *vj. The independence of the vis required G to be nonzero, [5]. Asused in system theory, the term Gramian describes a matrix, not a scalar, which is formedin manner reminiscent of the foregoing matrix G.

Balanced realization was first used in the area of digital filters to combat inaccuracies(roundofl) resulting from computation errors brought about by the requirement thatcalculations be done using finite precision arithmetic [29]. The use of a balancedrealization to obtain a reduced order model approximation and the method given herefor calculating a balanced realization was originally presented in [27]. Theorem 4.8 andthe proof of Theorem 4.3 (Appendix) were originally given in [34]. The idea of a balancedrealization was extended to unstable linear systems, [22] and to a class of nonlinear statemodels, [38].

Finally, we need to take note of the following important aspect of model reduction. Wewill see in the remainder of this book that optimal controller design techniques, e.g.,quadratic, Gaussian, and Hx feedback control theories, make use of an observer as partof the feedback controller. Since the observer order matches the plant order, there can beproblems in implementing controllers for use with high order plants. Unfortunately, thestability of the closed loop system may not be preserved when the controller designedusing the optimal control theory is replaced by a reduced order model. A good summaryof approaches to this problem up until 1989 is given in [2]. A more recent approach tosolving this problem, involving H,,, robust control theory is given in [30] for linear plantsand in [32] for a class of nonlinear plants.

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5Quadratic Control

5.1 IntroductionIn the preliminary part of this chapter, we introduce the basic approach to feedbackcontrol known as observer based feedback control. In this scheme the plant state x(t) inthe state feedback control signal u(t) = Kx(t) + v(t) (Chapter 2) is replaced by theestimated plant state z(t) obtained from an observer (Chapter 3). As we will see in theremainder of this book, this scheme has enabled the development of several optimizationbased controller design techniques. In addition to quadratic control treated in thischapter, we will see that observer based feedback control is used in linear quadraticGaussian, LQG, control and in H,,, control.

Following this, we take up the use of quadratic control to determine the K and Lmatrices in an observer based feedback control setup. Quadratic control arises when thereis a need to maintain the control system's steady-state output constant in response to aconstant control input. In this situation the control scheme is said to operate as aregulator. In practice, impulse like disturbance signals, i.e., signals having large ampli-tude and short time duration, enter the control loop and cause the output to jiggle aboutits desired level. A well known example of this is the regulation of the pointing direction ofan outdoor dish antenna. Unfortunately the dish acts like a sail and experiences torquedisturbances because of wind gusts which tend to change its pointing direction. Thereforewe need to choose K and L so as to alter the closed-loop system dynamics, i.e., assign theeigenvalues of the closed-loop system matrix, in a way which reduces the effect on theoutput of these impulse like disturbance signals. In quadratic control this choice of K andL is done so as to minimize the energy in the output signal caused by the disturbance.

We will see that the determination of the optimal K and L matrices needed to achievethis minimum requires the solution to two nonlinear matrix equations known as algebraicRiccati equations. Special solutions to these equations, referred to as stabilizing solutionsare required. We will develop conditions on the plant and optimization criteria which willensure the existence of such solutions. We will encounter the Riccati equation again inconnection with other optimal control schemes developed in this book. More specifically,we will see that the algebraic Riccati equation is involved in the solution of thedisturbance attenuation problem when the disturbances are viewed as randDm signalswith known statistics (Chapter 6) and when the disturbance signals are restricted to theSO-called L2 class of signals (Chapters 9 and 10).

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116 Quadratic Control

5.2 Observer Based ControllersAlthough errors in the estimated plant state in an observer based control system produceeffects at the control system's output, we will see that the dynamics of the observer do notcontribute to the control system's zero-state response. More specifically, we will showthat the control system's transfer function equals the transfer function for the closed loopfeedback system when exact state feedback is used, i.e., when the estimated state equalsthe plant state.

Suppose the plant state model parameters (A, B, C, D) are given. Then recalling thedevelopment of the observer, (Chapter 3), we see that the plant state, x(t), and theobserver state, i(t), are governed by the following vector differential equations

z(t) = Ax(t) + Bu(t) (5.1)

x (t) _ (A + LC)x(t) + (B + LD)u(t) - Ly(t) (5.2)

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

u(t) = Kx(t) + v(t) (5.3)

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

where

z,(t) = A,x,(t) + B,v(t) (5.4)

y(t) = C,x,(t) + D,v(t) (5.5)

A BKA_ -LC A+LC+BKC, C DK]

Ki(t)

At)

D,=D

PLANT yW

STATEFEEDBACK

OBSERVER

X, (t) _ X(t)I

Figure 5.1 Setup for observer based feedback control

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Observer Based Controllers 77/

Notice that since the plant and observer are each of dimension n, the closed-loop

system state model is of dimension 2n. However even if the plant state model is minimal,

i.e., observable and controllable, the closed-loop state model, (5.4, 5.5), is not control-lable. We can see this by changing coordinates so that the resulting state model is incontrollable decomposed form, (Section 2.5). We do this as follows.

Recall that controllable decomposed form requires the transformed input matrix B, to

have a lower partition which is null. Since B, is partitioned into two equal halves, acoordinate transformation matrix, T, which does this is given by

T=

Therefore, after using this coordinate transformation matrix the control system statemodel parameters, (5.4, 5.5), become

Act 14,2Ac =

0 Ac4

Cc = [ C + DK -DK ] Dc = D

where

xc =

Ac 1 =A+ BK Ac2 = -BK Ac4 = A + LC

z(t) = x(t) - i(t)

Notice, from the null block structure of Ac and Bc, that the plant state estimation error,z(t), is uncontrollable and that the control system transfer function is given as

Gc(s) = Cc(sI -Ac)-1

Bc +,6,

K -DK ] (sI - Act) 1

+ D[B]

= [ C + D0 (sI - Ac4) 0

_ (C + DK)[sI - (A+BK)]-1B+D (5.7)

Therefore provided the plant state model is minimal, i.e., (A, B) controllable and(A, C) observable, we have (A + BK, B) controllable and (A + BK, C + DK) observableso that there are no pole-zero cancellations in the foregoing transfer function and thecontrol system and plant have the same order. Notice this control system transferfunction is the same as we obtained in Chapter 2 when we used exact state feedback.Therefore we can conclude that the control system's input-output or zero-state behavioris unaffected by the presence of the observer. However, the control system's zero-inputresponse is affected by the observer. We can show this as follows.

From the control system's state model, (5.6), we see that the differential equations

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118 Quadratic Control

governing the plant state and plant state estimation error are

.k (t) = A,1x(t) + Ace (t) + Bv(t) (5.8)

x (t) = Ac4x(t)

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

z(t) = e(A+LC)tz(0)

Therefore substituting this expression for z(t) in (5.8) yields

i(t) _ (A + BK)x(t) - BKe(A+LC)tx(O) + Bv(t) (5.10)

and we see that the evolution of the plant state depends on the state estimation error.However, this effect which is caused by the presence of the observer in the feedback

loop, disappears with time. More specifically, recall from Chapter 3 that L must always bechosen to make A + LC is stable. Therefore we see from (5.9) that

limi(t) = 0 for any i(0)t Oc

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

±(t) = (A + BK)x(t) + Bv(t) (5.11)

Notice that (5.11) is the differential equation for the plant state when exact state feedbackis used.

Finally, recalling that the eigenvalues of the system matrix A, are preserved undercoordinate transformation, (Section 1.6.4), and that block upper-triangular matriceshave eigenvalues which are the union of the eigenvalues of the diagonal blocks,(Appendix), we see from (5.6) that

A[A,.] = \[A,] = \[A + BK] U A[A + LC]

Thus half the closed loop system eigenvalues depend on K and are independent of L, andthe other half depend on L and are independent of K. This spliting of the closed loopeigenvalues is responsible, in part, for being able to break the optimal control problemstreated in this book into two simpler problems:

1. the state feedback control problem2. the state estimation problem

This fact, referred to as the separation principle, has been a major theme running throughthe development of linear control theory since its discovery in the 1960s.

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Quadratic State Feedback Control 119

In summary, the use of an observer-estimated plant state in place of the true plant statein a state feedback control configuration leads to a feedback control system havingtransfer function (steady-state, input-output behavior) which is the same as we wouldobtain using exact state feedback. However, initially at the start-up of the control system,

any departure of the observer-estimated state from the true plant state will affect thecontrol system output. This effect diminishes with time as the observer-estimated stateapproaches the true plant state.

Recall, from Chapters 2 and 3, that we needed the specification of eigenvalues forA + BK and A + LC in order to determine K for the design of the state feedback controland L for the design of the state estimator, respectively. However, in practice, we are onlygiven some desired goals for the behavior of the control system. Therefore, in thissituation, we need methods for turning specified goals for the behavior of the controlsystem into values for the K and L matrices. In the remainder of this chapter, and insubsequent chapters, we will show in each case that by interpreting the specified goal forthe control system's behavior as a mathematical optimization problem, we can developequations for determining the required K and L.

5.3 Quadratic State Feedback ControlThere are many possible control system design objectives in addition to the mandatoryrequirement that a control system be stable. One of these additional requirements ariseswhen there are large-amplitude, short-duration disturbance signals entering the controlloop and we want to maintain the output constant. In this situation we need to design thecontroller both to stabilize the closed-loop system and to diminish, as much as possible,the effect of the disturbance signals on the plant output. The solution to the optimalquadratic control problem attempts to achieve this goal by determining K and L both tostabilize the closed-loop system and to minimize the energy in the departure of the controlsystem output from a desired steady-state value. As mentioned in the previous section, wecan use the separation principle to split this problem into two simpler subproblems:

1. the quadratic state feedback control problem and2. the quadratic state estimation problem

with the solution to 1. determining K, and the solution to 2. determining L.In this section we develop the design equations for K which solves the quadratic state

feedback control problem. When K satisfies these design equations, the sum of theenergies in the output and feedback signal caused by the disturbance signal, is minimized.One of these design equations is a nonlinear matrix equation known as the quadraticcontrol algebraic Riccati equation (QCARE). The theoretical background needed tosolve this equation is provided in the next section, Section 5.4.

In Section 5.5 we develop the design equations for L to solve the quadratic stateestimation problem. Again we will see that this involves the solution of an algebraicRiccati equation. This equation is closely related to the QCARE and is known as thequadratic filtering algebraic Riccati equation (QFARE). The required solution to theQFARE can be obtained using procedures developed in Section 5.4 for solving theQCARE.

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120 Quadratic Control

5.3.1 Motivating the problemLet the disturbance input to the plant be denoted as ut (t); let the controlled input involvedin implementing the state feedback be denoted by uz(t). Then the plant state model isgiven by the equations

i(t) = Ax(t) + B, ui (t) + Bzuz(t) (5.12)

y(t) = Cx(t) + Di ui (t) + Dzuz(t) (5.13)

and setting U2(t) = Kx(t) + v(t), where v(t) is the command input, we obtain a statefeedback control system having state model given as

z(t) Ax(t) + B, ul (t) + Bzv(t) (5.14)

y(t) = Cx(t) + D1 u1 (t) + Dzv(t) (5.15)

hw ere

A=A+B2K C=C+D2K

ul (t) = disturbance input y(t) = control system output

U2(t) = controlled input v(t) = control system command input

However, if

(i) the disturbance input is null, ui (t) = 0(ii) the control system command input is constant, v(t) = vs,

(iii) the feedback matrix K makes the control system stable so that all eigenvalues of .4 liein the open left half plane,

then the constant steady-state control system state, xs, is obtained from the controlsystem state differential equation, (5.14) by setting i(t) = 0. Then using xs in (5.15) givesthe steady-state control system output y,.as

y, = Cx.,. + Dzv., X, = -A-iBzvs

PLANT

CONTROLLER

Figure 5.2 Setup for disturbance input attenuation

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Quadratic State Feedback Control 121

wherex,. = limx(t)t-x

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

ui (t) 0 for t c [t1, t, + T]

= 0 otherwise

Therefore, shifting the time origin to the instant immediately following the end of thedisturbance, i.e., to time tl + T, we are left with the state immediately following the end ofthe disturbance being given by

x(0) = x, + xd(0)

However since A + BK is assumed stable we have

lira xd(t) = 0t-oo

and the state of the plant returns to its steady-state value with time,

limx(t) = x,too

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

5.3.2 Formulating the problemRecall, from the previous chapter, that we used the observability Gramian and its relationto the output energy to obtain a measure of the zero-input response. We use this idea hereto reformulate the problem just discussed as the problem of choosing K to minimize theeffect of a non-null initial plant state on the sum of the energy in the output and the energyin the feedback signal, i.e.,

min JQC = JQCO

where

JQc= J (Y*(t)Y(t)+Puz(t)u2(t))dt

= f[x*(t)*x(t) +uz(t)[PI]u2(t)]dt (5.16)

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122 Quadratic Control

with p being a positive scalar and JQc being referred to as the cost or the performanceindex for the quadratic control problem.

Notice that the reason for including the feedback signal energy in the performanceindex, i.e., for including u2(t) in the integrand, is to limit the feedback signal u, (t) = Kx(t)to a level that can be physically implemented. A more detailed explanation of this is givenas follows.

When we choose K to minimize JQc, we discover that the smaller p the larger theelements in K. If we were to let p be zero so that there would be no contribution to thecost, JQc, from the feedback control signal, u,(t), we would find that the K required tominimize JQ(. would have unbounded entries. Thus in this case the required feedbackcontrol signal u2(t) would be an impulse. This physically unrealizable requirementprohibits the use of p = 0. However, while increasing the size of p has the beneficialeffect of reducing the size of the required feedback signal, the effect of the disturbance onthe output is increased. Thus a beneficial choice for the size of p is a compromise betweenlimiting the size of the control signal, u2 (t) and limiting the effect of the disturbance on theoutput.

In order to simplify the development to follow, we will assume that there is no directfeedthrough of the feedback controlled input u2(t) to the plant output, i.e.,

we assume D2 = 0

In addition, since it is always possible to choose coordinates for the state space so thatthe parameters of the state model are real, we will assume that these parameters are real.

Finally, for generality in what follows, we will replace C and pI in (5.16) by Q, andR, respectively so that the performance index becomes

JQC = f [x*(t)Qcx(t) + u*(t)R,u2(t)]dt (5.17)

where Q, R, are real, symmetric matrices with Q. > 0 and R, > 0 implying that for allx(t) and u(t) we have

x*(t)Q'x(t) > 0

uz(t)R,u2(t) > 0

These restrictions on Q, and R, are needed to ensure that the contributions to JQc fromboth u2(t) and x(t) are never negative and so that R(. is invertible.

In summary, we have formulated the problem of designing a state feedback controllerto combat the effect of unknown short-duration disturbance inputs as an optimizationproblem. The solution to this optimization problem requires that K be determined so thata quadratic performance index in the disturbed state and the controlled input isminimized.

5.3.3 Developing a solutionFrom the discussion earlier in this section, we suppose that the command input is null,v(t) = 0, and that the time origin is taken at the instant a short-duration disturbance has

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Quadratic State Feedback Control 123

just ended. This leaves the plant state at some non-null value, x(0) = Xd and thereafter wehave the plant state given by

x(1) = e(A -B,K)tx

and the controlled input given by

u2(t) = Kx(t)

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

JQC -xd 0

where

(5.18)

(5.19)

`dt] xd (5.20)

Qc = Q, + K*R,.K

A=A+B2K

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

frx eA tQcea'dt = PQc (5.21)0

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

JQC = xjPQcxd (5.22)

Recall from Chapter 4 that PQC, (5.21), is the solution to the Lyapunov cquation

A*PQc+PQCA+Q,=0 (5.23)

Therefore, since A, Q,., (5.20), depend on K, so does the solution, PQC, to this Lyapunovequation and the performance index, JQc, (5.22). We denote this fact by writing PQc andJQc as PQC(K) and JQC(K, xd) respectively.

From the foregoing we conclude that our goal is to find K such that

(i) K is stabilizing, i.e., A + B2K is stable, and(ii) the solution, PQC(K), to (5.23) minimizes JQC(K, Xd), (5.22), for all initial states, xd.

In the following development we use K(, to denote the value of K which satisfies (i) and(ii). Thus K,, satisfies

JQC(K, xd) JQC(K0, xd) K 54 Ko (5.24)

for all disturbed initial states, xd.

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124 Quadratic Control

Now we are going to use a matrix variational approach in connection with theLyapunov equation, (5.23) to develop an equation for PQCO, where PQCO denotes thematrix PQc which solves (5.23) when K = K0. We begin by noting that any real andstabilizing K can be written as

K = K0 + E(6K) (5.25)

where c is a positive real scalar and 6K is a real matrix having the same dimension as Kwith c(6K) being referred to as a perturbation on K, , the desired value for K.

Since PQc depends on K through the Lyapunov equation, (5.23), the perturbation,c(6K), on Ko produces a series expansion for PQc about its optimal value PQcO,

PQc PQco + E(6I PQc) + HOT (5.26)

where HOT (higher order terms) stands for terms in the expansion involving powers of Ewhich are 2 or greater. The term (61PQc) is a symmetric matrix which depends on (6K) ina manner to be determined.

Next we substitute the expansion of PQc, (5.26), into the expression for the cost JQc,(5.22). This gives an expansion for JQc about its minimum value JQco

JQC = JQco + E(6lJQc) + HOT (5.27)

where

JQco = XdPQc0Xd (61JQc) = Xd(61PQc)Xd

and again HOT stands for terms in the expansion involving powers of E which are 2 orgreater.

Now since the terms indicated by HOT in both (5.26) and (5.27) depend on{E` : i= 2,3,- - .1 we see that

limHOTI _ 0

(5.28)Ego L J

Therefore it follows that the scalar, (6IJQc), in (5.27) and the matrix, (6IPQc), in (5.26)can be expressed as

(6IJQc) = lim JQc - JQcoe 0 E

(61 PQc) = lim I PQc - PQc°JE

(5.29)

(5.30)

Notice that (5.29) is reminiscent of the derivative in calculus. This fact together withideas we encounter in using calculus to obtain minima of a scalar function of a scalar

variable enables us to visualize the present problem of minimizing a scalar function JQc ofa matrix K as follows.

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Quadratic State Feedback Control 125

Suppose, for simplicity we assume u, (t) is a scalar. Then K is a row vector with n scalarentries. Then we can think of JQc together with the components of K as forming an n + 1dimensional space. Imagine making a plot of the scalar JQ(, on the "vertical axis" vs. Kwith K being represented by a point in a "horizontal" hyperplane, or plane when n = 2.The resulting surface in this space would appear as a bowl with the bottom of the bowl asthe "point" {JQco, K,}, i.e., the optimal point. At this point (61JQc), (5.29), is zero bothfor all 6K and for all initial disturbed states Xd. This is analogous to the derivative of ascalar function of a scalar variable being zero at points where the function has a local

minimum.Notice from the dependency of (61J(?c) on (61PQc), (5.27), namely

(61JQ(,-) = Xd*(61PQ(')xd

that (61 JQc) is zero for all Xd if and only if (61 PQc) is a null matrix. Therefore the optimalvalue for K, Ko, makes (61PQc) null for all (6K) which maintain K, (5.25), stabilizing.Therefore in order to determine Ko we need to determine an equation for the matrix(61 PQc) which involves (SK). We can do this by substituting the expansions for K and forPQc (5.25, 5.26), in the Lyapunov equation, (5.23). The results of doing this aredetermined as follows.

We begin by denoting the result of using the expansion of K, (5.25), in A and Q,, (5.20),as

A = Ao + eB2(6K)

Qc = Qco + e[(6K)*RcK0 + KoRc(6K)] + HOT

where

Ao = A + B2K0 Qco = Qc + K*RC.KO

(5.31)

Then substituting these expressions for A, Qc as well as the expansion of PQC, (5.26), inthe Lyapunov equation, (5.23), yields

AoPQCo + PQCOAo + Qco + c(M + M*) + HOT = 0 (5.32)

where

M = A0*(61PQc) + (6K)*M1

M1 = RcK0 + B-PQco

Notice that the first three terms in (5.32) sum to a null matrix since PQco satisfies theLyapunov equation, (5.23), when K = Ko. Thus (5.32) becomes

e(M + M*) + HOT = 0 (5.33)

Then dividing (5.33) through by c, letting c go to zero, and using the limiting property,

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126 Quadratic Control

(5.28), yields

M+M*=0

Finally, substituting for M from (5.32) yields the following Lyapunov equation relating(61PQC) to (6K)

A,*,(61PQC) + (6,PQC)Ao + (SK)*M1 + M16K = 0 (5.34)

Recall, from discussion earlier in this subsection, that K = K00 when 61 PQc is a nullmatrix for all 6K which maintain K, (5.25) stabilizing. Therefore when 61PQC is a nullmatrix, (5.34) becomes

(6K)*M1 + M16K = 0 (5.35)

which'is satisfied for all stabilizing perturbations, 6K, only if M1 is a null matrix.Therefore we see from the definition of M1, (5.32) that

Ko = -RC-- 1 B*PQco (5.36)

where we have used the assumption R, > 0 to insure that Rc is invertible.Notice, at this stage of the development, that the state feedback matrix Ko is readily

determined from (5.36) once we have PQco. It turns out that PQco is a solution to anonlinear matrix equation known as the quadratic control algebraic Riccati equation(QCARE). This equation can be developed by setting K = Ko, (5.36), in the Lyapunovequation (5.23). After carrying this out, we get the QCARE as

A*PQCo + PQCoA - PQCoRQCPQCO + Qc = 0

where

RQc = B2RC.1B*

z

(5.37)

Having determined the QCARE it remains to solve it for PQco such that A + B2Ko isstable when PQCo is used to determine KO from (5.36).

We can recap this section as follows. We began by introducing a matrix variationalapproach and using it to determine an equation for the optimal feedback matrix, Ko,(5.36). This equation involves PQco, the matrix which solves the Lyapunov equation,(5.23), when K = K0. Then we saw that by eliminating Ko from the Lyapunov equationby using (5.36) we obtained a nonlinear matrix equation in PQco referred to as theQCARE, (5.37).

In the next section we will be concerned with conditions on the plant state modelparameters and the parameters of the performance index, R0 and Q,, which ensure theexistences of the so-called stabilizing solution, PQco, to the QCARE, i.e., the solutionwhich makes A - B2RC 1B*PQco stable.

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Solving the QCARE 127

5.4 Solving the QCAREWe have just seen that the determination of the state feedback matrix Ko, (5.36), whichminimizes the performance index JQc, (5.17), requires the determination of a stabilizingsolution to the QCARE, (5.37). The main goal of the present section is to establishconditions on the plant state model parameters and weighting matrices Q, and R, which

ensure the existence of a stabilizing solution to the QCARE. We will be aided in this taskby resorting to a 2n x 2n matrix called the quadratic control Hamiltonian matrix. Beforedoing this we consider aspects of the closed loop stability which are brought out using thefact the PQco is the solution to the Lyapunov equation, (5.23), when K = K0.

5.4.1 Stabilizing solutionsConsider the following theorem and its proof.

Theorem 5.1 Ao is stable if

(i) (A, B2) is a stabilizable pair and(ii) PQco satisfies the QCARE with PQco > 0 and

(iii) (A, Q,) is a detectable pair.

where Ao = A + B2Ko with Ko given by (5.36) and PQco satisfying (5.37).Proof The necessity of (i) is obvious since we have, from Chapter 2, that a pair (A, B) is

stabilizable if we can find K such the A + BK is stable. Therefore assuming (i) holds, wewill show that (ii) and (iii) together are sufficient for Ao to be stable.

Notice from (5.36, 5.37) that

-PQCoRQCPQCo = PQCOB2Ko = KoB2PQco

Therefore adding and subtracting PQCoRQcPQCo on the left side of the QCARE, (5.37),gives the Lyapunov equation, (5.23), with K = KO,

A PQCo + PQCoAo + Qco-O

where

(5.38)

Ao=A+BzKo Qco=K*R,K0+Q,

We proceed now to use the ideas from Chapter 4 concerning stability in connection withthe solution of Lyapunov equations.

First we rewrite (5.38) by expanding Qco, (5.31), using Ko, (5.36). This enables (5.38) tobe rewritten as

AoPQco + PQcoAo = -PQCOB2R- 1BzPQco - QC (5.39)

Then pre and post-multiply this equation by v'* and v' respectively where Aov' = Av'yields

2Re , (v '*PQ v') = -v'* (PQ B R-1B*PQ )v` - v'*Q v` (5.40)i] co co z 2 co

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128 Quadratic Control

Next recalling that R,. > 0 and that Q,. > 0 we have

PQC,,B2R-' B;PQCo > 0

and the right side of (5.40) is either negative or zero, i.e., either

(a) - v'*(PQCoB2Rc'B2PQCo)v' - v *Q,,v< < 0

or

(b) - v'* (PQCOB2R- ' B2PQCn

Suppose (a) holds. Then we see from (5.40) that

B2PQCoVV' _ 0 and

2Re[a,] 0 (5.42)

and neither Re[A,] nor v'*PQCov' can be zero. Therefore we see from condition (ii) in thetheorem that v'*PQCOv' > 0. Thus (5.42) is satisfied only if Re[al] < 0.

Alternatively, suppose (b) holds. Then we see from (5.41) that both

v'* (P QCOB2RC'BzPQCo)v' = 0

This implies that

and we see that

,' = 0

w'=0

(5.41)

(5.43)

Atv' = Aov' _ (A - B2R-'B*PQCO)v' = Av' (5.44)

Thus (a v') is an eigenvalue, right-eigenvector pair for A satisfying Q,v' = o implyingthat A, is an unobservable eigenvalue for the pair (A, Q,.). However (A, Q,.) is detectable,condition (iii). Therefore if (b) holds we have Re[a,] < 0.

To recap, since the right side of (5.40) is either negative or zero and we have shown thatin either case Re[A] < 0 when the conditions in the theorem hold. Thus we have shownthat the conditions in the theorem are sufficient for Ao to be stable.

We will show in the next subsection that condition (iii) in the statement of theforegoing theorem is not necessary for Ao to be stable. Instead, we will show that onlycondition (i) in Theorem 5.1 and the condition that any imaginary axis eigenvalues of Abe observable eigenvalues for the pair (A, Q,) are needed to ensure the existence of astabilizing solution to the QCARE. Some appreciation of this weakening of condition (iii)can be obtained by reconsidering the foregoing proof as follows.

First, notice that when case (b) in the proof holds, (5.40) implies that

and v'* QCv' = 0

2Re[A1] (vi*PQCovr) = 0

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Solving the QCARE 129

and either

or

(bl) : Re[A1] 0

(b2) : PQCov` = 0

However, recall from the proof that when case (b) holds the eigenvalue a, of A,, is also anunobservable eigenvalue for the pair (A, Q,). Therefore if we impose the condition that(A, Q,) has no unobservable imaginary axis eigenvalues, then case (bl) is impossible.

Second, though not apparent at this stage in the discussion, we will see in the nextsubsection that if both condition (i) in Theorem 5.1 and the condition that (A, Q,.) haveno unobservable imaginary axis eigenvalues are satisfied, then eigenvectors of A can onlysatisfy the condition for case (b2) to apply if the corresponding eigenvalue of A is stable.

Finally, notice in the proof of Theorem 5.1, that case (b) is impossible when the pair(A, Q,) is observable and thus PQC,, > 0 when (A, Qc) is observable. Again, though notobvious now, we will show that we can replace this condition for PQcO, > 0 in case (b) bythe condition that (A, Q,) have no stable unobservable eigenvalues. Therefore thiscondition together with condition (i) in Theorem 5.1 and the condition that (A, Qc)have no imaginary axis eigenvalues are the necessary and sufficient conditions forPQC'>0.

The uniqueness of the stabilizing solution is shown in the following theorem.Theorem 5.2 The stabilizing solution to the QCARE is unique, when it exists.Proof Suppose PQcI and PQC2 are two stabilizing solutions to the QCARE. This

enables us to write

A*PQcI + PQC1A - PQC1RQCPQc1 + Q, = 0

A*PQC2 + PQC2A - PQc2RQCPQc2 + Qc = 0

where A; = A - RQcPQci : i = 1, 2 are each stable.Next taking the difference between these two equations yields

A* (APQc) + (APQC)A - PQCI RQCPQCI + PQc2RQCPQc2 = 0

where (OPQc) = PQcI - PQc2Then adding and subtracting PQcI RQcPQc2 enables this equation to be rewritten as

A1(oPQc) + (APQc)A2 = 0 (5.45)

Now proceeding in the same manner as was done to prove Theorem 4.5, we can showthat when A 1, A2 are stable the matrix equation

A1(1X PQc) + (APQc)A2 = M

has solution APQC which can be written as the integral

APQC = jehhutMeA2tdt

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130 Quadratic Control

However since the right side in (5.45) is null, M is null and the integrand of theforegoing integral is null for all time. Therefore we have APQC = 0 which implies thatPQc1 = PQc2 and the stabilizing solution to the QCARE is unique. 0

Having examined certain properties of stabilizing solutions to the QCARE, we turnnext to questions concerning the existence and calculation of such solutions. Animportant tool for answering these questions is the Hamiltonian matrix.

5.4.2 The Hamiltonian matrix for the QCAREIn this section we will show the relation between the solutions of the QCARE andproperties of a 2n x 2n matrix made up from the parameters of the QCARE, (5.37). Thismatrix, called the quadratic control Hamiltonian matrix, is defined as

HQc

A RQCI

Q, A*J(5.46)

where RQC = B2Rc 1 Bz.Notice that the blocks of HQc depend on all the matrices in the QCARE, (5.37), except

its solutions PQc. Recall that we are interested in the stabilizing solution denoted PQCo. Inthe following theorem, we give a property of HQc which is important in establishingconditions for the existence of a stabilizing solution to the QCARE.

Theorem 5.3 It is not possible to find a stabilizing solution to the QCARE, (5.37), ifHQC, (5.46), has imaginary axis eigenvalues.

Proof Define the 2n x 2n matrix T as

T= (5.47)

where PQc is any Hermitian matrix. Notice that T is invertible for all Hermitian matrices,PQc, having inverse given by

T-1 =

Next let HQc be related to HQc as

1 A RQcHQC = T HQCT =

where

Z -A*

A A - RQCPQC

Z = A*PQC + PQCA - PQCRQCPQc + Q,

Notice that

(5.48)

A[-HQC] = A[HQc]

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Solving the QCARE 131

independent of PQc. Therefore if we choose PQc so that Z is null, HQC is block upper-

triangular and

)[HQc] = A [A] U A[-A"] (5.49)

However since complex eigenvalues of A occur in conjugate pairs, we havea[A*] _ A[A]. This together with the fact that A[-A] = -A[A] allow us to rewrite (5.49) as

A[HQC] = A[A] U -A[A] (5.50)

This shows that the eigenvalues of HQc are mirror images of each other across theimaginary axis, i.e., if 77 E A[HQC] then -17 c A[HQc].

Now when we choose PQc so that Z is null, PQc is a solution to the QCARE, (5.37).Moreover, recalling the expression for the feedback matrix K0 (5.36), we see that A,(5.48) can be rewritten as

A=A-RQCPQC=A+B2K0

when PQc = PQco, the stabilizing solution to the QCARE. However, if HQc has animaginary axis eigenvalue, (5.50) shows that this eigenvalue must also be an eigenvaluefor Afor all Hermitian matrices PQc which satisfy the QCARE. Therefore it is impossibleto choose PQC to satisfy the QCARE, i.e., to make Z null, so that A is stable.

To recap, the 2n eigenvalues of HQC split into two sets of n eigenvalues each such thateach eigenvalue in one set has a corresponding eigenvalue in the other set which is itsmirror image across the imaginary axis. Thus if PQC is a stabilizing solution to theQCARE, the eigenvalues of A, (5.48), equal n of the eigenvalues of HQC which are in theopen left half plane. However, if HQC has imaginary axis eigenvalues, HQC does not haven eigenvalues in the open left half plane. Therefore, in this situation, a stabilizing solutionto QCARE does not exist since the eigenvalues of A must include at least one imaginaryaxis eigenvalue of HQc for all solutions, PQc, to the QCARE.

We need now to establish conditions which ensure that HQc does not have imaginaryaxis eigenvalues. This is done in the following theorem.

Theorem 5.4 The quadratic control Hamiltonian matrix HQc, (5.46), is devoid ofimaginary axis eigenvalues if

(i) (A, RQc) has no uncontrollable eigenvalues on the imaginary axis and(ii) (A, Q,) has no unobservable eigenvalues on the imaginary axis

Proof Suppose HQC has an eigenvalue atjw. Then we have

(5.51)HQcv = jwv

r RQC r

1

r 1

or

A*L Q,. L V2 J -lw L V2 J

(jwI - A)vl = RQCV2 (5.52)

(jwI + A*)v2 = Q,vl (5.53)

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132 Quadratic Control

Next, we premultiply (5.52) by v*2 and (5.53) by vi to obtain

v;(jwl - A)v, = v*RQw2 (5.54)

vi(jwI + A*)v2 = ?)IQ ,VI (5.55)

Then taking the conjugate transpose on either side of (5.55) and recalling that Q,. isHermitian enables us to rewrite (5.55) as

vz(jwI - A)vi = -v*Q,vl (5.56)

Now we see, by comparing (5.54) with (5.56), that

vzRQw2 = -?)IQ ,Vi (5.57)

However Q, and RQ(- are each non-negative. Therefore the only way (5.57) can besatisfied is for each side to be zero. Thus we have

RQCV2 = 0 (5.58)

Qcvi = 0 (5.59)

and (5.52, 5.53) become

Avi = jwvl (5.60)

A*v2 = jwv2 (5.61)

This shows that jw is an eigenvalue of A and that vi, v2 are the corresponding right andleft-eigenvectors of A.

However if v2 J 0, (5.58, 5.61) imply that (A, RQc) has an uncontrollable eigenvalueon the imaginary axis and condition (i) is not satisfied. Therefore if (i) is satisfied, we musthave V2 = 0.

Alternatively, if vi 54 0, (5.59, 5.60) imply that (A, Q,.) has an unobservable eigenvalueon the imaginary axis and condition (ii) is not satisfied. Therefore if (ii) is satisfied, wemust have vi = 0.

From the foregoing we see that if conditions (i) and (ii) are satisfied, (5.51) holds only ifv = 0. Thus, contrary to the assumption we made at the beginning of the proof, HQC hasno imaginary axis eigenvalues when the conditions of the theorem are satisfied.

Notice that since Re is non-singular and RQc = B2R-'B*, we have

v2RQC = 0 only if v2B2 = 0

Therefore condition (i) in the foregoing theorem is equivalent to

(i) (A, B2) has no uncontrollable eigenvalues on the imaginary axis.

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

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Solving the QCARE 133

imaginary axis eigenvalues, the conditions in the foregoing theorem are not sufficient to

ensure that the QCARE has a stabilizing solution.

5.4.3 Finding the stabilizing solutionThe proof of Theorem 5.3 suggests a way of determining the stabilizing solution to theQCARE. More specifically, we saw that when the stabilizing solution to the QCAREexists, we can find T, (5.47), such that HQc, (5.48), is block upper-triangular with the topdiagonal block being stable. This suggests that the stabilizing solution to the QCARE canbe determined from the eigenvectors of HQC corresponding to the stable eigenvalues.However, it turns out that we can avoid having to compute eigenvectors. Instead we canuse Schur decomposition on HQC to determine the stabilizing solution to the QCARE. Inorder to see how to do this we need the following basic result form linear algebra.

Theorem 5.5 If

MU = UW (5.62)

where M and W are square with W being invertible and U having independent columns,then(i) each eigenvalue of W is an eigenvalue of M, i.e., A[W] C A[M](ii) range[U] is an M-invariant subspace corresponding to A[W], i.e., range[U] is

spanned by the eigenvectors of M corresponding to the eigenvalues of W.

Proof Let (A,, v') be any eigenvalue-eigenvector pair for W. Then premultiplyingWv' = A1v` by U and using (5.62) we obtain

Mw=A;w

where w = Uv' and we see that A, is also an eigenvalue of M. This completes the proof of(i).

Next assuming the eigenvectors of W are complete, we can express any vector q of thesame dimension as these eigenvectors as

njqa;vr=i

so that post-multiplying (5.62) by q gives

Ms = r

where

n,

r = Up P=L(Aja)v'

S = Uq

*ad we see that M maps range[U] into itself, i.e., range[U] is M-invariant.

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134 Quadratic Control

Finally, notice that if V1 is a matrix whose columns are the eigenvectors of m

corresponding to the eigenvalues of M which are also eigenvalues of W, then

MV1 = V1A1

where Al is diagonal with the eigenvalues of W along its diagonal. Then post-multiplyingthis equation by the non-singular matrix O such that U V 1O and inserting 00-1between V1 and Al gives

MU UWwhere

U= V1O W 0- -1A1U

Thus )[W] = \[A1]. Moreover since O is non-singular, we have

range[U] = range[V10] = range[V1]

This shows that range[U] is spanned by eigenvectors of M corresponding to theeigenvalues of W. 0

In order to develop a relation between the quadratic control Hamiltonian matrix andthe QCARE which we can use as a basis for solving the QCARE for its stabilizingsolution, we proceed in the following manner.

Suppose we have the stabilizing solution, PQC,,. Then the QCARE, (5.37), can bewritten as

A*PQCO + QC_ -PQCO (A - R PQc Qco)

and we see that

IHQc I -PQco ] = [ -PQCO] (A - RQCPQco) (5.63)

where

A RQCHQC _

Q,. -A

Now since A - RQcPQcO is stable, we see from Theorem 5.5 that (5.63) implies

Irange I -

PQCn

is the HQc-invariant subspace corresponding to the stable eigenvalues of HQC.The standard way of computing this subspace, when it exists, is to use Schur

decomposition. This numerically reliable method computes an orthogonal (unitarywhen HQC is complex) matrix T such that the transformed quadratic control Hamilto-nian matrix, HQC, (5.48), is block upper-triangular with first diagonal block, H11, being

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Solving the QCARE 135

stable, i.e.,

T*HQCT = HQC =

Notice from (5.48) that H11 = A - RQCPQC,, = A,,, the stable system matrix for the statefeedback control system.

Now since T is unitary, T* = T I , we see that premultiplying (5.64) by T and equating

the first n columns on either side of the resulting equation gives

HQcTI = T1H11 (5.65)

H11 H12

0 H,2(5.64)

where T is partitioned into n x n blocks, T;1, with T 1 denoting the first n columns of T,i.e.,

TI =T11

T21T = [T1 T2] = I

T11 T12 I

L T21 T22

Then we see from Theorem 5.5 and (5.65) that

(i)7A[H11] C A[HQC](ii) range[T1] is an HQC-invariant subspace corresponding to A[H11].

Therefore since H11 is stable, the columns of T1 span the stable eigenspace of HQC, i.e.,the columns of T1 can be expressed in terms of the eigenvectors of HQC whosecorresponding eigenvalues lie in the open left half plane.

Recall that range[T1] =range[T10] for any non-singular matrix O of appropriatedimension. Hence if T, I is non-singular the columns of T 1 T-11 I also span the stableeigenspace of HQC. Therefore assuming TI, is invertible, we postmultiply (5.65) by Tilland insert TI-11 T11 between T 1 and H11 on the right side of (5.65) to obtain

HQC (T1IH1ITill)QC [T21']T21IT111

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

PQCo = -T21T111 (5.66)

and we have the stabilizing solution to the QCARE.Notice that in order to use the foregoing approach to solve the QCARE for its

stabilizing solution, not only must HQC have no imaginary axis eigenvalues, (Theorem5.3), but in addition TI, must be invertible. Notice also that when TI, is invertible wehave PQCo > 0 only if T21 is also invertible. The following theorem establishes conditionson the parameters of the plant and control cost which ensure that T11 and T21 areinvertible.

Theorem 5.6 Suppose we can find a unitary matrix T so that H11 is stable and

T*HQCT = HQC _ I HII H12 I

L 0 H22 ]

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136 Quadratic Control

where

HQc=A RQ(

A* JQiT

T1T, Ti i T12 I-[] L T.,1 T22

Then we have the following results:

(i) T11 is invertible if (A, B2) is stabilizable;(ii) if T11 is invertible, then T21 is invertible if (A, Q,,) has no stable unobservable

eigenvalues.

Proof (i) From (5.65) and the blocks in HQC and H11 we have

AT11 +RQCT21 = T11H11 (5.67)

Q,711 - A*T21 = T21Hn (5.68)

Then letting let (A, v) be any eigenvalue-right-eigenvector pair for H11 and post-multiplying (5.67, 5.68) by v gives

AT11v + RQcT21v = AT1Iv (5.69)

QcTlly - A*T21v = AT21v (5.70)

Now suppose T, I v = 0. Then T 11 is not invertible and (5.69, 5.70) become

RQCW = 0 (5.71)

A*w = -Aw (5.72)

where w = T21v. Then if w = 0 we would have

v=0

implying that T has dependent columns and is therefore not invertible. This is impossiblesince we are assuming that T is unitary. Therefore it, 54 0.

Now since H11 is stable, (5.71, 5.72) imply that (A, RQC) is not stabilizable, and sinceRQ = B2R 1BZ with RC-1 non-singular, we have (A, B2) is not stabilizable. This contra-dicts the condition given in (i). Therefore no eigenvector v, of H1 I satisfies T11 v = 0 when(A, B2) is stabilizable. Thus assuming the eigenvectors of H1I are complete, the foregoingcondition shows that T11 is invertible when (A, B2) is stabilizable.

Proof (ii) Suppose T11 is invertible and T21v = 0 for an eigenvector v of H11. Then(5.69, 5.70) become

As = As

Q's=0

(5.73)

(5.74)

where T 11 v = s.Therefore since H11 is stable and A is an eigenvalue of H11, we have that (A, Q,) has a

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Quadratic State Estimation 137

stable unobservable eigenvalue. This contradicts the condition given in (ii). Therefore noeigenvector v of H11 satisfies T21v 0 if (A, Q,) has no stable unobservable eigenvalues.Finally assuming H11 has a complete set of eigenvectors, the foregoing condition showsthat T21 is non-singular when (A, Q,.) has no stable unobservable eigenvalues.

Although more involved, it can be shown that the foregoing theorem holds even when

H, j does not have a complete set of eigenvectors. This is done using generalizedeigenvectors.

In summary, in this section, we have considered the problem of solving the QCARE

for its stabilizing solution. The so-called Hamiltonian matrix was introduced for thispurpose. By using this matrix it was shown that the computationally useful technique ofSchur decomposition could be employed to obtain the stabilizing solution to the QCAREwhen it exists. In addition, the Hamiltonian matrixenabled the development of importantconditions on the plant and performance index parameters which are necessary andsufficient for the existence of a stabilizing solution to the QCARE.

5.5 Quadratic State EstimationSo far we have solved the quadratic control problem under the assumption that the stateof the plant is known. In this case the controlled input is

u2s(t) = Kox(t)

However, when the plant state is unknown, we must resort to the use of an observer inan observer based feedback control scheme as discussed at the beginning of this chapter.Doing this and using the optimal quadratic state feedback matrix K0, developed inprevious sections, gives the controlled input as

u2(t) = Koz(t) = U2S(t) - u2E(t) (5.75)

where

U2E(t) = Koz(t) 1(t) = x(t) - 1(t)

Since u2S(t) = Kox(t) is the controlled input which minimizes the quadratic statefeedback control cost, JQC, (5.17), the departure of the controlled input, u2E(t), from itsoptimal value, U2S(t), causes the cost JQC to increase. Our goal in this section is to designan observer to minimize this increase in the quadratic control performance index. Weapproach this goal by solving the optimization problem of choosing the observer's Lmatrix so as to minimize the energy in the state estimation error caused by both animpulse disturbance in the measurement of the plant output and an initial plant stateestimation error.

5.5.1 Problem formulationRecall, from Section 3.3, that ideally the observer estimates the state of the plant based onthe true plant output, y(t) = Cx(t), where for simplicity, we assume, as in the quadraticstate feedback control problem, that the plant is strictly proper. However, in practice,

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138 Quadratic Control

only a measured plant output is available and this output differs from the true plantoutput by an additive measurement noise signal, iv(t). Thus if y,,, (t) denotes the measuredoutput we have

Y "'(t) = y(t) + w(t)

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

v(t) T' u2(t)PLANT

yR)

CONTROLLER

Figure 5.3 Setup for output measurement noise attenuation

w(t) = output measurement noise y(t) = control system output

u2(t) = controlled input v(t) = control system command input

Now applying y,,, (t) to the input of an observer, we obtain an estimate of the plant statefrom

x= Ax(t) + Bu(t) + L[y(t) - y,,,(t)]

y(t) = Cx(t)

and we see that the plant state estimation error, z(t) = x(t) - z(t), is governed by thedifferential equation

(t) = Az(t) + Lw(t) A A + LC (5.76)

As in the state feedback control problem, we assume the unknown disturbance is a large-amplitude, short-duration signal which we model as an impulse, w(t) = S(t - to).

Suppose the control system has been in operation with a constant command input andno measurement noise, for a time which is great enough so that the plant state estimatehas converged to the steady-state value for the plant state, i.e.,

z(t) = x, i(t) = 0

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Quadratic State Estimation 139

Then shifting the time origin to to, the time where the output measurement noise impulse

occurs, we have w(t) = 6(t), z(0) = 0 and we see from (5.76) that the state estimation

error becomes

:x(t) = eAIL (5.77)

Now we want to return the state estimation error to its null value as rapidly as possiblefollowing the occurrence of the measurement noise impulse. Therefore (5.77) implies thatL = 0 is the best choice for this purpose since then z(t) = 0 and the disturbance wouldhave no effect on the state estimate. However, this approach to setting up the quadraticstate estimation problem does not take into account effects on the state estimation errorwhich arise from the initial plant state being unknown. For instance, if the plant isunstable, we have seen in Chapter 3 that choosing L null causes any estimation errorpresent at start-up to grow without bound.

To overcome this problem, we reformulate the foregoing optimization problem byassuming that the initial state estimation error has an additional component not causedby the output measurement noise, w(t), i.e., we take z(0) as

i(O) = Xd

Then we see from (5.76) that the plant state estimation error, z(t), caused by bothz(0) = id and w(t) = 6(t) is given by

z(t) = eAtzd + eAIL (5.78)

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

min JQE = JQEO

where

JQE = fx*(t)x(t)dt

However recalling the use we made of the trace of a matrix in connection with thedevelopment of the controllability Gramian in Chapter 4, we see here that JQE can berewritten as

JQE -,trace[Joe

z(t)5*(t)dt]0

which using (5.78) becomes

JQE = trace [f° e itMetdt] (5.79)

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140 Quadratic Control

where

M = LL* + xdL* + L.z d + xdz d A = A + LC

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

Qe = Qe + LReL*

where Re, Qe are real symmetric matrices with Qe > 0, Re > 0. Then our problem is tochoose L to minimize JQE such that A is stable, where

JQE = trace

with

I 0

eAt eeA `dt]

A = A + LC Qe = Qe + LReL*

(5.80)

The development of equations for doing this proceeds along lines similar to those used todevelop the equations for Ko in Section 5.3.

5.5.2 Problem solutionWe begin the development by noting that the required stability of the observer ensuresthat the integral, (5.80), converges to a real symmetric matrix PQE so that

JQE = trace[PQE] (5.81)

where

PQE = Joe -

eA0

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

APQE + PQEA* + Qe = 0

(5.82)

(5.83)

Notice that since A and 0, (5.80), depend on L so does PQE and JQE. We use thenotation PQE(L) and JQE(L) to denote this fact. Therefore our goal is to find L so that

(i) A(L) is stable, and(ii) the solution, PQE(L), to (5.83) minimizes JQE(L), (5.81).

We denote the value of L that satisfies (i) and (ii) as L,,. Thus Lo, the optimal value of L,satisfies

`QeeA''dt

JQE(L) > JQE(LO) L L.

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Quadratic State Estimation 141

Now we employ the matrix variational technique used in Section 5.3 to develop Lo.

Thus suppose L is perturbed away from Lo as

L = Lo + E(6L) (5.84)

where c is a small positive real scalar and (6L) is any matrix having the same dimensions

as L such that A is stable.Then we can write the expansion of PQE about PQEo as

PQE = PQEo + E(61PQE) + HOT (5.85)

where HOT (higher order terms) stands for terms in the series involving powers of c which

are 2 or greater. Next we obtain the expansion of JQE about its minimum value, JQEO, bysubstituting (5.85) in (5.81). Doing this and using the fact that the trace of sum of matricesis the sum of the trace of each matrix gives

JQE = trace [PQE] = trace[PQEO] + E trace [(61PQE)] + HOT

= JQEO + E(61JQE) + HOT (5.86)

Notice that

(61JQE) = trace[61PQE] (5.87)

Now comparing the present development of equations for Lo with the developmentused to obtain equations for Ko, Section 5.3, we see that the optimal value for L, Lo,makes (61 JQE) = 0 for all 6L. Therefore we see from (5.87) that an equation for (61 PQE) isneeded.

We begin the determination of an equation for (61PQE) by using (5.84) to express Aand Qe , (5.80), as

A=Ao+E(6L)C

Qe = Qo + E[LoRe(SL)* + (SL)ReLO*] + HOT

where

Ao=A+LOC Qo = Qe + LoReLo*

Then substituting these expressions for A, Q and the expansion for PQE, (5.85), in theLyapunov equation, (5.83), yields

AOPQEo + PQEoAo + Qo + E[M + M*] + HOT = 0 (5.88)

where

M = Ao(61PQE) + Ml (SL)* M1 = LoRe + PQEoC*

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142 Quadratic Control

Now the first three terms on the left side of (5.88) sum to a null matrix since PQEosatisfies the Lyapunov equation, (5.83), when L = Lo. Thus using this fact and recallingfrom Section 5.3 that

limrHOTI = 0

f-0 E J

we see that dividing (5.88) by c and letting a go to zero gives

(61 PQE)A, + A0 (bl PQE) + (b L)Ml* + M1(b L)* = 0 (5.89)

Notice that (5.89) is a Lyapunov equation in (61PQE). Therefore, since A0 is requiredto be stable we must have (b1 PQE) > 0. Thus no eigenvalue, Qi, of b1 PQE is negative, i.e.,{Qi > 0 : i = 1, 2 , n}. Moreover, referring to the development in Section 5.3, we seethat JQE is minimized by choosing L so that 61JQE = 0 for all bL. Thus after rewriting(5.87) as

61JQE = trace[(61PQE)] Qii=1

we see that 6]JQE = 0 for all SL only if all the ais are zero for all bL which implies that(61PQE) is null for all iL.

Now referring to (5.89) we see that if (b1 PQE) is null then

(iL)M* + M1(8L)* _ 0

which holds for all SL only if M1 is null. Therefore Lo must satisfy

M1 = L0Re + PQEoC* _ 0

which since Re > 0 can be solved for Lo as

Lo = -PQEoC*Re (5.90)

Notice that in the same way that K00 depends on PQco, the stabilizing solution to theQCARE, so Lo depends on PQEo, the stabilizing solution to an algebraic Riccati equationcommonly referred to as the quadratic filtering algebraic Riccati equation (QFARE).This equation is obtained by substituting Lo, (5.90) in the Lyapunov equation, (5.83).Thus after some algebra we obtain the QFARE as

APQEo + PQEoA* - PQEoRQEPQEo + Qe = 0 (5.91)

where RQE = C * Re 1 C. This completes the development of the equations for the optimalvalue of L.

In summary, the same variational approach which was used in Section 5.3 to determineequations for the optimal state feedback matrix Ko was used to determine equations forthe optimal observer matrix L,,. In the next section we will find PQEo so that the QFARE,

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Solving the QFARE 143

(5.91), is satisfied and A - PQEORQE is stable by resorting to a Hamiltonian matrixapproach similar to the approach used to solve the QCARE for its stabilizing solution.

5.6 Solving the QFARE

In this section we exploit the dual nature of the quadratic state feedback and stateestimation problems. We begin by recalling the QCARE, (5.37) and QFARE, (5.91)

QCARE : A*PQCo + PQCUA - PQCORQCPQCo + Qr = 0

QFARE : APQEo +PQEOA*

- PQEoRQEPQEo + Qe = 0

where

RQC = B2 Re. -'B2 RQE = C * Re 1 C

Then comparing these equations we obtain th correspondences given in Table 5.1.Recall that if (A*, C*) is stabilizable then (A, C) is detectable. Therefore the foregoing

correspondences together with Theorem 5.1 in Section 5.4.1, imply the followingsufficient conditions for the existence of a stabilizing solution to the QFARE.

Theorem 5.7 Ao is stable if

(i) (A, C) is a detectable pair and(ii) PQEo satisfies the QFARE with PQE,, > 0 and

(iii) (A, Qe) is a stabilizable pair.

where Ao = A + L0C with Lo given by (5.90) and PQEo satisfying (5.91).Next recalling the quadratic control Hamiltonian matrix, HQC, (5.46), we see from the

foregoing correspondences that the quadratic estimation Hamiltonian matrix, HQE,needed to solve the QFARE for its stabilizing solution should be defined as

HQE _[ A* R

lQE

Qe -AIL J

(5.92)

where RQE = C * Re CThen substituting PQE in place of PQC in the similarity transformation matrix T, (5.47)

Table 5.1 QCARE-QFARE Symbol Correspondences

Estimation Control

A A*C * Bz

Re Rr

Qe QrPQEo PQCo

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144 Quadratic Control

gives the transformed quadratic estimation Hamiltonian matrix as

T-1H T=H RQEQE

QE - [Y -A

where

A = A - PQERQE

Y = PQEA* + APQE - PQERQEPQE + Qe

Notice that HQE and HQE have the same eigenvalues for all T. Therefore we can see, bychoosing T so that Y is null, that HQE has eigenvalues which are mirror images of eachother across the imaginary axis. This means that any imaginary axis eigenvalues of HQEare also eigenvalues of A for all PQE that make Y null, i.e., that satisfy the QFARE. Thuswhen HQE has imaginary axis eigenvalues the QFARE does not have a stabilizingsolution, cf., Theorem 5.3. Conditions which ensure that HQE has no imaginary axiseigenvalues are determined by analogy with Theorem 5.4.

Theorem 5.8 The quadratic estimation Hamiltonian matrix, HQE, (5.92), is devoid ofimaginary axis eigenvalues if

(i) (A, C) has no unobservable eigenvalues on the imaginary axis and(ii) (A, Qe) has no uncontrollable eigenvalues on the imaginary axis

Assuming a stabilizing solution to the QFARE exists, we can use the Schur decom-position of HQE to determine this solution as was done in Section 5.4.3 to determine thestabilizing solution to the QCARE. Therefore referring to Section 5.4.3, we use the first ncolumns of T, the unitary or orthogonal matrix which transforms HQE to a block uppertriangular matrix with the first diagonal block stable. Then the stabilizing solution to theQFARE is given as

PQEo = -T21T111

where

T T1 T2]7'11 T12=[=T21 T22

Conditions required for the existence of the stabilizing solution to the QFARE can beobtained by analogy with Theorem 5.6.

Theorem 5.9 Suppose we can find a unitary matrix T so that H11 is stable and

T*HQET = HQE =H11 H12

0 H22

where

I1 [T1

T2

rT11 T12

HQE = I

At RQC

T=Qe A T21 1 T22L J

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Summary 145

Then we have the following results:

(i) Tl1 is invertible if (A, C) is detectable;

(u) if T11 is invertible, then T21 is invertible if (A, Q,) has no stable uncontrollableeigenvalues.

Finally we can show that the stabilizing solution to the QFARE is unique, when itexists, by proceeding in the same fashion as was used to prove Theorem 5.2. Thiscompletes the discussion of the stabilizing solution to the QFARE.

5.7 SummaryWe began this chapter with a description of observer based state feedback and itsproperties. This was followed by the derivation of design equations for both an optimalquadratic state feedback controller and for an optimal quadratic state estimator(observer). In each case the design equations were nonlinear matrix equations referredto as algebraic Riccati equations, i.e., the QCARE and QFARE. The requirement for thestate feedback system and observer to be stable restricted the use of the solutions of eachof these equations to their so-called stabilizing solution. Conditions for the existence ofthese stabilizing solutions were developed, in each case, in terms of the parameters of theplant and performance index. This was done through the use of a related four-blockmatrix known as a Hamiltonian matrix. We saw, in Theorems 5.4 and 5.6, that theQCARE has a stabilizing solution if and only if (A, B2) is stabilizable and (A, Q,) has nounobservable imaginary axis eigenvalues. Alternatively we saw in Theorems 5.8 and 5.9that the QFARE has a stabilizing solution if and only if (A, C) is detectable and (A, Qe)has no uncontrollable imaginary axis eigenvalues.

5.8 Notes and ReferencesThe algebraic Riccati equation has played a role of great importance in the evolution ofcontrol theory. An extensive view of this subject is given in [3]. The use of the Schurdecomposition of the Hamiltonian matrix to solve the quadratic state feedback controlproblem, is implemented in the MATLAB Control System Toolbox under the command1qr. More detailed information on the Schur decomposition is given in [17].

The matrix variational technique used here to develop the design equations for theoptimal K and L matrices was used previously to reduce the effect of inaccurate modelingof the plant on the performance of a quadratic control system [28]. For more recent usesof this approach see [13] and the references therein.

Since the stabilizing solutions, P, to either of the algebraic Riccati equationsencountered in this chapter are uniquely determined from the corresponding Hamilto-nian matrix, H, we have is a function. This fact has lead to the use of H Edom(Ric)to denote those H that admit a stabilizing P and P =Ric(H) to denote the stabilizingsolution P determined from If. This notation has become standard in the controlliterature. For a more detailed discussion see Chapter 13 of [47].

When either the plant is time-varying, or when the control cost is evaluated over a finiteinterval of time, the optimal controller is time-varying and is determined by solving twoRiccati matrix differential equations. The development of these conditions is beyond theCope of the present text. The reader is referred to [ 1 ] and [ 10]for more information on thisToblem.

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6LQG Control

6.1 IntroductionThis chapter continues the development, begun in Chapter 5, of methods for choosing theK and L matrices in an observer based feedback control system. As in Chapter 5, our goalhere is to combat the effect that unknown disturbances have on the control system'soutput. Thus, as discussed in Section 5.3.1, we can assume, without loss of generality, thatthe control system's command input is null so that the control system output is causedsolely by the disturbance input. The assumed sporadic, short duration character of thedisturbances used in Chapter 5, is replaced here by disturbances characterized as beingpersistent random signals which when applied as an input to a linear system produce anoutput with bounded average power, where y(t) has bounded average power if

T

lim 1 y*(t)y(t)dt < 00T-.oc T o

Notice that the persistence of the disturbances makes the output energy unbounded.Therefore unlike the quadratic control problem, we can no longer use the output energycaused by the disturbance input as a measure of the control system's ability to attenuatedisturbances. Instead the steady-state average power output is used as a performancemeasure.

We will assume throughout that the disturbances w;(t) are zero mean Gaussianrandom vectors with covariance given as

E[w(ti)w*(Woo Woi W02

W1o W11 "12

W20 W21 W22

S(tl - t2) (6.1)

where

wo(t)w(t) = w, (t)

W2(t)

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148 LQG Control

and E[.] is the expectation operator from probability theory. Moreover we will assumethat the output caused by these random disturbances is ergodic so that the steady-stateaverage power at the output is given as

T

lim T / y' (t)y(t)dt = rlim E[y`(t)y(t)] (6.2)T-oc -0c,

Now we will begin by assuming that the plant to be controlled has state model specifiedas

z(t) = Ax(t) + wo(t) + B2u2(t) (6.3)1

yi(t) = Cjx(t) + W1(t) +D12U2(t) (6.4)

y2(t) = C2x(t) + w2(t) + D22u2(t) (6.5)

where

w.(t) = disturbances Y1 (t) = controlled output

u2(t) = controlled input y2(t) = measured output

and u2(t),y,(t),Y2(t) have dimensions m2ip1,p2 respectively.Our goal here is to choose K, L in an observer based feedback control system so that

JGC is minimized where

JGC = lim E[yi (t)yI (t)]t 00

Since the disturbance signals are assumed to be zero mean Gaussian random vectors,this type of optimal control is known by the acronym LQG (Linear Quadratic Gaussian).More recently this optimal control problem was interpreted as a minimum norm problemand the term 7-12 optimal control was also used. More will be said about this in Chapter 8following the development of ideas involving signal and system spaces.

WI(t)

u2(t)

PLANT

CONTROLLER

Figure 6.1 LQG control configuration.

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LQG State Feedback Control Problem 149

As in the quadratic control problem, the separation principle allows us to obtain thesolution to the LQG control problem by combining the solutions to two subproblems:

(i) the LQG state feedback control problem(ii) the LQG state estimation problem.

The LQG control problem is also referred to as

(iii) the LQG measured output feedback control problem.

In what follows we assume that all matrices are real so that we can write (.)T in placeof (.)-

6.2 LOG State Feedback Control ProblemIn this section we develop the design equations for the optimal state feedback controlmatrix K which minimizes JGC, (6.6). We will do this using the variational approachintroduced in the previous chapter. We will see that the determination of the optimal Krequires the solution of an algebraic Riccati equation.

6.2.1 Problem formulationSuppose the plant state, x(t), is known. Then we want to determine the state feedbackmatrix, K, involved in generating the controlled input, u2 (t) = Kx(t), so that JGC, (6.6), isminimized. Notice from (6.3, 6.4) that under state feedback the system relating thedisturbances to the controlled output is given as

z(t) = Ax(t) + wo(t) (6.7)

Y, (t) = Cx(t) + wi (t) (6.8)

where

A=A+B2K C=C,+D12K

and wo(t) and w1 (t) are unknown zero mean Gaussian random vectors having covar-iances given by (6.1).

Now in order to proceed we must have w1 (t) = 0. This requirement can be seen bynoting from (6.1) that when w1(t) 0 the performance index, JGC, (6.6), is unbounded forall K because of the impulsive nature of the covariance of w1(t). Therefore in order for theLQG state feedback control problem to be properly posed we must have w1 (t) = 0.Therefore under this assumption the statement of the LQG state feedback controlproblem becomes:

min JGCK

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150 LQG Control

given that

JGC iim E[yT (t)Yi(t)]t DU

X(t) = Ax(t) + wo(t)

Y, (t) = Cx(t)

with A stable where (A, C) are given in (6.7, 6.8) and wo(t) is characterized from (6.1) as

E[wo(ti)wo (t2)1 = Woo 6(tl - t2)

The variational method used in the previous chapter is used now to develop equationsfor K to solve this problem.

6.2.2 Development of a solutionSince A is stable and JGC, (6.9), involves the steady-state behavior, the performance isunaffected by the initial state of the plant. Therefore only the zero state controlled outputresponse is needed. Thus referring to (6.10, 6.11) we see that this response is given by

Yi(t) = C J reA(`-T)wo(rr)dr (6.12)

Then using (6.12) we see that the expectation needed in the cost, (6.9), can be written as

t

tE[Yi (t)TYI (t)] = E[10

wo (TI )eAT (t-TI) CT d7-1J

CeA(t-T2)wo(T2)dT21 (6.13)

Now we can express the right side of this equation by employing the trace relation usedin the development of the controllability Gramian in Chapter 4. This relation is restatedhere as

(vTe)v2 = trace[v2VT e] (6.14)

for any compatible vectors v; and matrix e.Therefore identifying wo (T,) with v, and wo (T2) with v2 enables (6.13) to be rewritten as

E[yT(t)y,(t)]o \ o wo(T2)wo (T1 )eAT(t-T,)CTdT1)CeA(t-T2)dT211 (6.15)

Next evaluating the expectation using (6.1) reduces (6.15) to

E [YT (t)Yi (t)] = trace [WO06(T2 -.)eAT (t-TI) CTdrl) CeA(t-T2)dr21 (6.16)

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LQG State Feedback Control Problem 151

Finally, use of the sifting property of the impulse gives

00eATE[yi (t)y1(t)] = trace[it

W

and substituting T = t - T2 and allowing t -> oc enables us to express the performanceindex, JGC, (6.9) as

JGC = trace[WOOPGC]

where

Now recalling Theorem 4.5 in Chapter 4, we see that PGC satisfies the Lyapunovequation

ATPGC+PGCA+CTC=0 (6.19)

and is the ooservability Gramian for the pair (A, Q. Thus since (A, C) depends on K sodoes PGC

Now the equations which determine K so that JGC, (6.9), is minimized can bedeveloped by following the variational approach used in the previous chapter. Thus ifK,, denotes the optimal value for K we have

JGC(K) > JGC(KO) = JGCo

and expanding JGC and PGC about Ko yields

JGC = JGCo + f(6,JGC) + HOT

= trace[ WOOPGCo] + Etrace[ Woo(S1PGC)] + HOT (6.20)

where

PGC = PGCo + f(S1 PGC) + HOT

and a is a small positive real scalar with SK being any real matrix having the samedimension as K such that A is stable. In addition, recall that HOT denotes higher orderterms in a and therefore satisfies

iimHOT -0

E-0 f

GC =

x eATT

CPGC CeAT dT

0

K=Ko+E(6K)

(6.18)

(6.21)

Thus we see from (6.20, 6.21) that

(S1JGC) = lira fJGC - JGCOI = traceWOO Pf J [ (b1 GC)] (6.22)

E-0

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152 LQG Control

Recall from the previous chapter that JGC is minimum when (SIJGC) = 0 for all 6K.Thus we see from (6.22) that we need a relation between (S1 PG(') and SK. This relation isdeveloped by substituting the expansions for K and PGC as given in (6.20) into theLyapunov equation (6.19). This gives

PGC'()A()+A0PGCo+C0Co+e(E1+F1)+HOT =0 (6.23)

where

A = A + B2K

C. = Cl + D12Ko

El = (S1PG(')A()+ Ao (blPGC) + PGC()B2(6k) + (SK)T Bz PGCo

F1 = Co D12(SK) + (SK)T D12Co

Notice that the first three terms on the left side of (6.23) sum to a null matrix since PGC()satisfies (6.19) when K = K(). Therefore dividing (6.23) through by c, letting E go to zero,and using (6.21) yields

Ao (SlPGC) + (61PGC)Ao + (SK)TM1 + M1 SK = 0

where

(6.24)

M1 = B2 PGCo + D1T2C1 + D12D12Ko

and we see that (S1PGC) is governed by a Lyapunov equation.Now in order for (61JQc), (6.22), to be zero for all SK, we must have Woo(SIPGC) null

for all SK. However Woo is independent of SK. Therefore we can have Woo(61PGC) nullfor all 6K only if (S1PGC) is null for all SK.

Thus we see from (6.24) that if (S1PGC) = 0 for all SK then we must have Ml = 0.Therefore assuming that the columns of D12 are independent so that D12D12 invertible, wecan solve M1 = 0 for Ko as

T 1Ko -(Dl2D12) Qo (6.25)

where

Qo=B2PGCo+D1zC1

As in the case of the optimal quadratic control problem, the matrix PGCo is obtained asthe solution to an algebraic Riccati equation. We develop this equation by substituting,K(), (6.25), in the Lyapunov equation, (6.19). Thus after substituting for A, C from (6.7,6.8) in (6.19) we obtain

TQO +KaDT12 D12K0, _ 0 (6.26)

which after substituting for K() from (6.25) becomes

ATPGC()+P A-2 T(DTD ) l T TTD) 1

=0ccoQT

12 12 QO + Cl C] +QT (D12

l2 QO -

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LQG State Estimation Problem 153

Finally substituting for Qo, (6.25), gives

[AT - CTi D12(Dl2Di2) ' B2 ]PGCo + PGC(,[A - B2(Dl2D12) ' Dl2C1]

-PGCOB2(D12D12) 'B2PGCO + C1 C1 C1 D12(D12D12) 1DiC1 =0

which we can write more compactly as

A TI PGCo + PGCaAi - PGCoR1 PGCo + Q1 0

where

T 1 TAl = A - B2(D12Di2) Di2C1

R1 B2(DT12D12)

1B T2

Q1 = CI [I - D12(Dl2D12) 1Di2]C1

Notice that

Al - RIPGCO = A + B2Ko

(6.27)

Now the algebraic Riccati equation (6.27) is referred to as the Gaussian controlalgebraic Riccati equation, (GCARE). Thus we obtain the optimal K, K0, by solving theGCARE for its stabilizing solution, i.e., the solution PGCo which makes A + B2K00 stable.As in the previous chapter, we will show that the GCARE can be solved for its stabilizingsolution by using an appropriate Hamiltonian matrix. This matter is considered furtherafter taking up the state estimation problem.

D12 be invertible isBefore leaving this section, notice that the requirement that D12similar to the requirement we encountered in Section 5.3.2 that R, be invertible. Both R,and DT D12 play the role of introducing a penalty on the use of the controlled input in12

minimizing the control costs JQC and JGC respectively.

6.3 LQG State Estimation ProblemIn this problem we assume that the plant state x(t) is unknown and that we want to designan observer to estimate it from the controlled input, u2(t) and measured output, y2(t),(6.3, 6.5). Unlike the problem just treated, the controlled output y1(t), (6.4) is not used.Our goal now is to choose the observer matrix L so that JGE, the steady-state averagepower in the state estimation error, z(t), is minimized where

JGE = lim E[iT (t)x(t)] = lim trace [E [X(t).XT (t)] ]tax t-.oo(6.28)

Notice that this performance index involves the steady-state, state estimation errorcovariance matrix PGE where

PGE = lim E [x(t)iT (t)]t 00

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154 LQG Control

In order to proceed we need to review the role played by L in the evolution of theestimation error. This is done as follows.

6.3.1 Problem formulationRecall, from the beginning of this chapter, that the state model for the plant having themeasured output, y2(t), as its only output is given as

z(t) = Ax(t) + wo(t) + B2u2(t) (6.29)

Y2(t) = C2x(t) + w2(t) + D22u2(t) (6.30)

where the unknown disturbances wo(t) and wa'2(t) are zero mean Gaussian random vectorswith covariance matrices as specified by (6.1). Then referring to Chapter 3, we see that anobserver for this state model can be cast as

(t) = Ai(t) + B2u2(t) + L[.v2(t) - y2(t)] (6.31)

y2(t) = C2-i(t) + D22u2(t) (6.32)

where we have replaced the unknown disturbances by their expected values.Again as in Chapter 3, we can obtain a differential equation for the state estimation

error by subtracting (6.31) from (6.29) and using (6.30, 6.32). This gives

x (t) = Az(t) +. (t) (6.33)

where

z(t) = x(t) - z(t)

A=A+LC2 B=[I L]

Notice from (6.1) that the composite zero mean Gaussian random disturbance vector,w(t), has covariance

E[w(tt), . T ( 1 ) ] = W 5(t1 - t2) (6.34)

where

W _ f Woo WO21

W20 W221

Recall from Chapter 3, that in the absence of disturbances, we chose L so that A isstable. This ensures that the observer generates a state estimate which approaches theplant state for any initial plant state estimation error. Unfortunately this beneficialasymptotic behavior of the state estimator is not possible in the presence of persistent

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LQG State Estimation Problem 155

disturbances like those being considered here. However the expected value of the stateestimation error will be null in the steady-state when A is stable and the expected values ofthe disturbances are null. Estimators which have this desirable property are said to be

unbiased.

Therefore since we are assuming zero mean disturbances, we obtain unbiasedestimates of the plant state provided our choice of L to minimize JGE, (6.28), isconstrained by the requirement that A be stable. This plus the fact that the performanceindex, JGE, (6.28), concerns the steady-state behavior of the estimator, implies that we canignore the state estimation error response due to any initial state estimation error.Therefore using (6.33) we obtain

f -T)Bw(T)dT (6.35)

Now this relation can be used to develop an expression for the state estimation errorcovariance matrix PGE. We do this by following the steps we used to develop PGC in theprevious section. Doing this yields

E [ft dT2 f teA(t-T')Bi,(Tl)wT

(T2)BT eAT (t-T2)dTa]

_ I t d, I teA(t-T')BWt(Ti - T2)BT eAT (t-T2)dTl

= J'dT2eA('_T2) B WBT eAT (t-T2) (6.36)

Finally, we can obtain the steady-state error covariance matrix, PGE, by setting'r = t - 7-2 and taking the limit as t tends to infinity. This yields

lim E[X(t)XT (t)] =fc

eWBTeATTdr = PGE (6.37)

Notice from Chapter 4, (Theorem 4.5), that PGE, (6.37), is also the controllabilityGramian for (A, h) and satisfies a Lyapunov equation

APGE + PGEAT + BWBT = 0 (6.38)

where PGE depends on the observer matrix L since both A and h depend on L. Thus we seefrom (6.28) that we need to adjust L so that A is stable and so that the solution to (6.38),PGE, has minimum trace. A solution to this problem is obtained now by using the dualitybetween the present problem and the LQG state feedback control problem which weSolved in the previous section.

6.3.2 Problem solutionWe begin by using the dependency of A and C, on K, (6.7, 6.8) to expand the Lyapunov'libation (6.19) and by using the dependency of 4 and h on L, (6.33) to expand the

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156 LQG Control

Lyapunov equation (6.38). This gives

AT PGC + PGCA + KT B2 PGC + PGC,B2K

+CT C1 + CI D12K + KTD12C1 + KT D 2D12K = 0

APGE +PGEAT + LC2PGE + PGEC2 LT

+W00 + LW20 + W02LT + LW22LT = 0

(6.39)

(6.40)

In order to compare the symbols in these Lyapunov equations we make use of the factthat covariance matrices are symmetric. Therefore we can factorize W, (6.34), as

W =B

1

[B T Dz ] =D21

Woo W02

W20 W221(6.41)

so that

Woo = BIBT Woe = Wo = B1D21 W22 = D21D21

Notice that these factorizations allow us to relate wo (t) and w2 (t) to a zero mean Gaussianrandom vector ul (t) as

wo(t) = Blul(t) w2(t) = D21u1(t) (6.42)

where

E[ul (tl)uT (t2)] = I6(tl - t2)

Now we can use the foregoing factorizations, (6.41), to rewrite the Lyapunov equation(6.40) as

APGE + PGEAT + LC2PGE + PGEC2 LT

+BIBT +LD21BT +BID21LT +LD21D2iLT = 0(6.43)

Then comparing terms in (6.43) with terms in (6.39) we obtain the symbol correspon-dences for the LQG state feedback control problem and for the LQG state estimationproblem given in Table 6.1.

Thus we see that the formulas for the optimal L and PGE, i.e., L0, and PGEO, can beobtained by substituting from these correspondences in the formulas for Ko, PGCO givenin (6.25, 6.27)). Doing this yields

L. = - [PGECz + B1Dz21 ] (D21D21 1

A2PGEo + PGEOA2 - PGEOR2PGEo + Q2 = 0

(6.44)

(6.45)

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LQG Measured Output

Table 6.1 GCARE-GFARE Symbol Correspondences

Estimation

PGE

A

L

C2

B1

D21

where

Notice that

Control

PGCAT

KTBT2CT

DTie

T T 1

A2 = A - B1D21 (D21 D21) C2

T T 1

R2 = C2 (D21 D21) C2

Q2 = B1 [I - D22 (D21 Dz1) 1 D21 ] BT

A2 - PGEOR2 = A + LoC2

Now the algebraic Riccati equation (6.45) is referred to as the Gaussian filteringalgebraic Riccati equation, (GFARE). Thus L,, is obtained by substituting PGEO in (6.44)where PGEO is called the stabilizing solution to the GFARE since the resulting L,, makesA + LoC2 stable. As in the previous chapter, the GFARE can be solved for its stabilizingsolution by using an appropriate Hamiltonian matrix. We will show this in Section 6.5.

Before going on, notice that the foregoing solution to the LQG state estimationproblem requires that D21 D21 be invertible. This implies that D21 must have independentrows. Thus when this condition is satisfied the elements in the random vector-]v2(t) aredeterministically independent of each other in the sense that a subset of the elements inw2(t) can not be used to determine another element in w2(t) exactly. When this conditionis not satisfied, the estimation problem is referred to as being singular.

6.4 LQG Measured Output Feedback ProblemHaving solved the LQG state feedback control problem and the LQG state estimationproblem, the separation principle allows us to determine the LQG measured outputfeedback controller by combining these solutions as follows.

First, replacing the plant state by the observer's estimate of the plant state in theexpression for the controlled input gives

u2(t) = K0 c(t)

where K,, is obtained from (6.25, 6.27) and z(t) is the state of the observer. Thensubstituting this expression for u2(t) in the observer's state differential equation, (6.31,

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158 LQG Control

6.32), yields the state model for the controller as

z (t) = A,ez(t) - L0y2(t)

u2(t) = K,,-i(t)

where

(6.46)

(6.47)

Ace = A + L,,(C2 + D22K0,) + B2K0

and Lo is obtained from (6.44, 6.45).This completes the formulation of the controller which solves the LQG control

problem. It remains now to develop conditions on the plant state model parameterswhich ensure the existence of stabilizing solutions to the GCARE and the GFARE.

6.5 Stabilizing SolutionSince stabilizing solutions to both the GCARE and GFARE are needed to implement anLQG controller, sufficient conditions for the existence of these solutions, expressed asconditions on the plant parameters, are of considerable importance to designers of thesecontrollers. As in the quadratic control problem, the stabilizing solutions to the GCARE,(6.27), and to the GFARE, (6.45), can be computed, when they exist, by usingappropriate Hamiltonian matrices. We will use this fact, in the same manner as wasdone in the previous chapter, to develop conditions on the parameters of the plant whichensure the existence of stabilizing solutions for the GCARE and GFARE. The duality ofthe GCARE to the GFARE allows us to obtain these results for both by treating theGCARE in depth and extending the results obtained to the GFARE through the use ofduality.

6.5.1 The Hamiltonian matrix for the GCAREReferring to Section 5.4 we see that the GCARE has a stabilizing solution if and only ifthe associated Hamiltonian matrix HOC has

(i) no eigenvalues on the imaginary axis,(ii) T11 invertible where

T1 =T11 1

T21

is any 2n x n matrix whose columns span the invariant subspace ofHGC corresponding to the stable eigenvalues of HGC.

where HGC is composed from the GCARE, (6.27), in the same manner as HQC iscomposed from the QCARE in Section 5.4.2. Thus we see that

HccrA1, R1rJ

(6.48)=Q1 -A,

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Stabilizing Solution 159

with

Al =A Bz(DT1zD1z) 1D T12

T -'TR1 = B2(D12D12) B2

Q1 = Ci [I - D1z(DI2D1z) 1D1z1C1

Also recall, from the previous chapter, that condition (1) is necessary and sufficient forthe existence of a nonsingular 2n x 2n matrix T, (which can be orthogonal), so that HGCcan be transformed to block upper triangular form such that then x n matrix H11 is stablewhere

T_1HcCT =H11 Hlz

H2z T21 T22(6.49)

Then condition (ii) is required to construct the stabilizing solution to the GCARE from

Pcco =-Tz1T-111

and we see that PGCO > 0 requires that T21 be invertible.In what follows we obtain necessary and sufficient conditions on the plant parameters

so that (i) and (ii) hold. We do this by modifying theorems obtained in the previouschapter taking into account the differences in the dependency of the Hamiltonianmatrices HQC and HGC on the plant parameters. Conditions on the plant parameterswhich ensure that PcCo > 0 will also be developed.

6.5.2 Prohibition of imaginary eigenvaluesRecall that in the quadratic state feedback control problem we obtained conditions on theplant and performance index which ensured that HQC has no imaginary axis eigenvalues.These conditions were given in Theorem 5.4. Therefore using Theorem 5.4 and compar-ing HQC with HGC, we see that HGC, (6.48), has no imaginary axis eigenvalues if:

(a) the pair (AI, R1) has no uncontrollable imaginary axis eigenvalues and(b) the pair (A 1, Q 1) has no unobservable imaginary axis eigenvalues.

Notice that these conditions are not immediately applicable to the given data of a plantstate model and performance index since they do not explicitly state how theseparameters are involved in contributing to the presence of imaginary axis poles forHGC. This deficiency is overcome in the following theorem.

Theorem 6.1 HGC, (6.48), has no imaginary axis eigenvalues if

(i) (A, B2) has no uncontrollable imaginary axis eigenvalues,(ii)

T= [TIl T12 l

11

A-jwl B2rank = n + m2 w E (-oc, oc)

Cl D12

where B2 is n x m2

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160 LQG Control

Proof (i) Referring to the proof of Theorem 5.4, we see that since D12D1z isnonsingular we have

gTRI = 0 if and only if gTB2 = 0

and therefore A is an uncontrollable eigenvalue of (A1, R1) if and only if A is anuncontrollable eigenvalue of (A, B2). Thus condition (i) here is equivalent to condition(i) in Theorem 5.4.

Proof (ii) We begin by considering the pair (A1i Q1) where

Al = A - Bz(D1zD12)-1D C1

T= [I - T01 D12(D12D12)

1

D1z ] C1

Thus if jw is an unobservable eigenvalue of (A1, 01) we have

Alv = jwv 01V=0

which can be rewritten as

(A -jwI)v - Bz(D1iDI2) 1D1zCly = 0

(I - D12(D1zDl2) 1Dlz)C1v = 0

Now these equations can be written as

00

where

A - jwI Bz

C1 DlO z J

I 0T

-1DT

D12D12) 12C1 I

(6.50)

Since z is invertible, OD has independent columns only if O has independent columns.However (6.50) implies that ez has dependent columns. Therefore a must havedependent columns and

rank[8] < n + mz

This shows that condition (ii) is sufficient for (A1, 01) to have no imaginary axiseigenvalues. Now we want to use this fact to show that condition (ii) is sufficient for(A,, Q1) to have no imaginary axis eigenvalues.

In order to proceed we need to be able to factor D. as

C D,T

(6.51)

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Stabilizing Solution 161

where here

T -'TD, = Ip, - D12(D12D12) D12

However, we can only factorize Dc in the foregoing manner if D, is symmetric and non-negative. The symmetry of Dc is obvious. We can show Dc. > 0 as follows.

Since D12 has independent columns, we can solve the matrix equation

D12u =Yr y,. E range[D12]

as

Therefore it follows that

and

u = (D1zD12) 1D12Yr

T I TYr = D12u = [D12(Dl2D12) D12]Yr

T 1 T ]Yr = Dr = 0 (6.52)fh, - D12(D12D12) D12 J

However since any pl dimensional real vectors y can be written uniquely as

Y=Yr+yL (6.53)

where

Yr E range[D12] yL E null [D,Z]

we see from (6.52) that

DcY = YL

and Dc. is said to be a projector.Finally, since in general yT yL = 0, we see from (6.54, 6.53) that

T TY DcY = YLYL

(6.54)

(6.55)

for all real pl dimensional vectors, y. Therefore since the right side of the foregoingequation can not be negative we have shown that Dc > 0 and therefore we can factorizeDc as (6.51) and we can proceed to show that when condition (ii) is satisfied (A1, Q1) hasno imaginary axis eigenvalues.

Suppose condition (ii) is satisfied. Then we showed earlier in the proof that anyeigenvector-imaginary axis eigenvalue pair (v, jw) of Al satisfies

Qlv$0 (6.56)

where

Q1 = DeC1 =(D)

DCI

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162 LQG Control

Thus we must have

so that

or

This shows that

Dc.C1u 0

D'('Clv :) 0T

i T iVT CI (D` DCCIV = vT Q1V 0

Qlv:A o

for any eigenvector v of Al associated with an imaginary axis eigenvalue which impliesthat the pair (AI, Q1) has no unobservable imaginary axis eigenvalues. E

The foregoing proof reveals an important property of Dc which we will make use of inChapter 9 and 10 in connection with the H,,, feedback control problem. Not only is D,.square and non-negative but in addition is a contraction since it has the property that

uTU<yTy

for anyp1 dimensional vectors u, y which are related through Dc as u = Dr.y. This can beseen by noting from (6.53-6.55) that

TT Y. + YIYL U T U = YTY Y = Y YL

A matrix or operator having this property is referred to as a contraction.In addition, notice that matrices which are contractions have eigenvalues which are

bounded above by one, i.e.,

'max [Dr] < 1

This property can be seen by refering back to Chapter 4 and the discussion of thesignificance of the largest eigenvalue of a non-negative matrix (Section 4.6.3).

Finally, notice that this constraint on the eigenvalues of D,, is also a consequence of D,being a projector since D, projects from a higher to a lower dimenional space.

6.5.3 Invertability of T11 and T21Having established conditions on the plant state model which ensure that HGC has noeigenvalues on the imaginary axis, it remains to determine conditions on the plant whichensure that T11, (6.49), is non-singular. We can do this by appropriately modifyingTheorem 5.6 as follows.

Theorem 6.2 If HGC, (6.48), has no imaginary axis eigenvalues so that we can find T,(6.49), so that H11 is stable where

T 1 HccT = [ H11 H12 T = [ T, T21 T1 T11I

(6.57)0 H22 T21

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Stabilizing Solution 163

then we have

(i) T11 invertible if (A, B2) is stabilizable;(ii) if T11 is invertible, then T71 is invertible if (A1. D.C1) has no stable unobservable

eigenvalues where

T 1 TAI = A - B2(D12D12) D12C1

TD, = I - D12(D12D12) 1Dlz = (D`l D'.

Proof (i) We begin by recognizing that (6.57) implies that

HGCTI = T1H11

which we can expand as

A1T11 +R1T21 = T1IH11 (6.58)

Q1T11-ATITz1=Tz1H11 (6.59)

Now suppose H11 has an eigenvalue-eigenvector pair (A, v) such that

T11v=0

Then post-multiplying (6.58, 6.59) by v we see that

R1w=0 (6.60)

(6.61)

where w = T21 v.Now if w = 0 we would have

v=oT21

implying that T1 and hence T has dependent columns. This is impossible since T isinvertible. Therefore w 54 0.

Finally since H11 is stable we have Re[A] < 0 and therefore (6.60, 6.61) imply that(A1, R1) is not stabilizable. This is equivalent to (A,B2) not being stabilizable sinceRI = B2(D1zD12)-1Bz . Therefore no eigenvector, v, of H11 satisfies T11v = 0 when(A, B2) is stabilizable. Thus assuming the eigenvectors of H11 are complete we haveshown that T, I must be invertible when (A, B2) is stabilizable. If H11 lacks a complete setof eigenvectors, the proof is completed by resorting to generalized eigenvectors.

Proof (ii) Suppose T21v = 0 with s = T11v 0 for some eigenvector, v, of H11. Then

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164 LQG Control

post-multiplying (6.58, 6.59) by v we obtain

A1s = As

i iQ1s=MTMs=C1

T(Dr, D(CIs=0

where

(6.62)

(6.63)

M = DIC1

However since, in general, we have null [MT] I range [M], we see that (6.63) is satisfiedonly if s Enull[M]. Therefore if (6.63) is satisfied then

DICCIs = 0 (6.64)

Moreover since H11 is stable, (6.62, 6.64) imply that (A,, D2CC1) has a stableunobservable eigenvalue. Therefore no eigenvector, v, of H11 satisfies T21v = 0 when(A1i DIC C1) has no stable unobservable eigenvalues. Thus assuming the eigenvectors ofH11 are complete we have shown that T21 is invertible when (A1i DI.C1) has no stableunobservable eigenvalue. Again as in the proof of (i), if H11 lacks a complete set ofeigenvectors, the proof is completed by resorting to generalized eigenvectors. 0

Notice that when (A, C1) is not observable neither is (A1i Therefore when theGCARE has a stabilizing solution, a necessary condition for T21 to be non-singular sothat PGCo > 0 is that (A, C1) have no stable unobservable eigenvalues. However thiscondition is not sufficient for PGCO, > 0. This can be seen from the following considera-tions.

First, reviewing (6.51-6.56) we see that

rank[D,] = rank P1 - m2

where null [D"] is of dimension m2 and it may turn out that C1v j 0 but D,2.C1v = 0.Second, since D12D12 = RD is symmetric and invertible it can be factored and its

square root RD is invertible. Hence we can write (6.51) as

T l 1

ri tl-T

(D2 ) D2 =I-D12(R D) (RD D12

which implies that

DC

(R2 1 T DTD) 12

has orthonormal columns. Therefore it is impossible to have a non-null vector q such that

i TD7cq = 0 and (RD) Dlz g = o

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Stabilizing Solution 165

Thus if v is an eigenvector of A which satisfies C, v o but D, C, v = o, then we have(RID)-TD

12C17) 0 so that

A1v= [A B2(RD) (Rv)-T

D7' .w

and v is an eigenvector of A but not of A,. Notice in this case that unlike the eigenvectorsof A, the eigenvectors of A, depend on B2 implying that the positive definiteness of thestabilizing solution depends on B2.

In summary, the GCARE has a stabilizing solution, PGCO if

(1c) (A, B2) is stabilizable;

(2c)

A-jwI B2rank = n + m2 w E (-oo, oc)

C, D12

Finally if a stabilizing solution, PGco, exists then

(3c) PGco > 0 if (A,,D2CC,) has no stable unobservable eigenvalues.

6.5.4 Conditions for solving the GFAREIn order to obtain conditions on the plant parameters which ensure that the GFARE hasa stabilizing solution, we can use the correspondence between matrices in the GCAREand the GFARE given in Table 6.1. Therefore substituting these correspondences in (Ic-3c) we see that the GFARE has a stabilizing solution, PGEU if

(1e) (AT , Cr) is stabilizable;(2e)

AT - IwI Czrank

TT= n+ p2 w E(- oc, oc)

B1 D21

Moreover, if a stabilizing solution, PGEO, exists then

(3e) PGEo > 0 if (A2 , DPBT) has no stable unobservable eigenvalues, where

z T ;(De) D`e = Imi - D21 (D21 D 1) D21

However these conditions can be restated in a more convenient form by using the factsthat the stabilizability of any pair (A, B) is equivalent to the detectability of the pair(AT , BT) , and that the rank of a matrix equals the rank of its transpose. Thus the GFAREhas a stabilizing solution, PGEo, if

(1e) (A, C2) is detectable;

rrA-jwI B, 11ranklLlL

JJ=n+p2 wE (oo,oo)

C2 D21

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166 LQG Control

Moreover, if a stabilizing solution, PGEO exists then

(3e) PGEO, > 0 if A2, B1 De ) has no stable uncontrollable eigenvalues.

6.6 SummaryIn this chapter we have given a derivation of the design equations for the LQG optimalfeedback controller for a linear time-invariant continuous-time plant. The performancecriterion or cost function which is minimized by the optimal LQG controller is the steady-state expected or average value of a quadratic form in the output vector to be controlled(the controlled output) when the disturbance input is a zero mean Gaussian randomvector. Necessary and sufficient conditions were given for being able to design the LQGcontroller.

6.7 Notes and ReferencesAgain, as in the quadratic control problem treated in the previous chapter, we have seenthat the algebraic Riccati equation plays a central role in the design of optimalcontrollers. For other instances where the algebraic Riccati equation arises in controltheory see Chapter 13 of [47]. The LQG optimal control problem was recently referred toas the H2 optimal control problem. The reason for this will be given in Chapter 7.Assuming the appropriate existence conditions are satisfied, the LQG controller can becalculated using the command h2lqg in the MATLAB Robust Control Toolbox.

The observer obtained by solving the LQG state estimation problem was originallygiven by Kalman as an alternative to the Wiener filter for extracting a desired signal froman additive combination of the desired signal and noise. This aspect of the LQG problemhas had a great impact on a wide variety of industrial problems. There are many bookswhich deal with this subject. For example, informative treatments are given in [7, 10, 25].

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7Signal and System Spaces

7.1 IntroductionIn the previous two chapters we were concerned with the problem of designing controllersto attenuate the effects of disturbance inputs on the output of a feedback control system.We assumed that the disturbance input was either an impulse (Chapter 5) or a randomsignal with an impulsive covariance (Chapter 6). In the next two chapters we developideas needed to solve the disturbance attenuation problem for a broader class ofdisturbance input. Signals in this class have finite energy and are denoted by L2[0, oc)where L is used in recognition of the mathematician H. L. Lebesgue, pronounced"Lebeg", the subscript 2 is used in recognition of the quadratic nature of energy, andthe bracketed quantities indicate the time interval over which the signals are not alwayszero. Therefore f (t) E L2 [0, oc) if

(i) f (t) has bounded L2 norm (finite energy),

[100Ift112< oo (7.1)

where [.] denotes the positive square root and f * (t) denotes the conjugate transpose off(t).(ii) f (t) is null for negative time

f (t) = p for all t c (-oc, 0) (7.2)

In this chapter we will use the foregoing idea of the L2 norm for signals to developsystem norms defined in both the time and frequency domains. This is made possible byParseval's theorem which relates the L2 norm in the time domain to the L2 norm in thefrequency domain.

7.2 Time Domain SpacesIn classical control theory there was no apparent need to consider signals defined fornegative time since control problems usually have a well defined start time. However the

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168 Signal and System Spaces

need for more sophisticated mathematics to deal with new approaches to controlproblems requires that we enlarge the domain of definition of signals to include thenegative time axis and to consider the operation of systems in both positive and negativetime.

7.2.1 Hilbert Spaces for SignalsConsider a signal vector, f (t), which unlike signals in L2 [0, cc), is not zero for all negativetime but still has finite energy over the time interval (-cc, oc), i.e.,

[f:f* (t)f(t)dtj

Any f (t) which satisfies (7.3) belongs to the normed space G2(-0c, oc).Alternatively, another related normed space, denoted G2(-0e, 0], consists of finite

energy signals which are null for all positive time. Notice that L2 [0, oc) and L2 (-0e, 0] areeach subspaces of L2( cc, 0e), i.e.,

.C2(-oc, 0] C L2(-0c, Oc) £2[0, oc) C £2(-00, Oc)

Now from the foregoing we see that if f (t) E L2(-00, 00) it is always possible to write

f(t) =f+(t) +f-(t) (7.4)

where f+ (t) E G2[0, oc) and f (t) E G2( oc, 0] with

f+(t) =f (t) t > 0 f+(0) _ icf (0)

f (t) = f (t) t < 0 f_ (0) = (1 - r,) .f (0)

where ,c c [0, 1].Notice that once n is specified, f+(t), f (t) are uniquely dependent on f (t). Moreover

since f+ (t) and f (t) are nonzero on disjoint intervals, we see that

t:ftfdt=0 (7.5)

Now (7.4) together with (7.5) enables the L2 norm of any f (t) E L2(-oc, oc) to bedecomposed in the following fashion

lf(0112 = [f-f (t)f (t)dt]

_ [llf+(t)112+lIf(t)11212 2

which is reminiscent of the Pythagorean theorem for right-angled triangles.

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Time Domain Spaces 169

In the foregoing signal analysis we encountered an integral involving two signals in,C2(-oc, oo). In general the quantity

J'b

(7.7)

is referred to as the inner product of a(t) and /.3(t) for any equal length vectorsa(t), p(t) E L2(a, b) and is denoted as < a(t), /3(t) >, i.e.,

< a(t), 0(t) > _ I a` (t)3(t)dt (7.8)h

Notice that, unlike the norm which involves only one signal and is always a real non-negative scalar, the inner product involves two signals and is not restricted to be positiveor real. Notice that the norm and the inner product are related as

a(t)M2 = [< a a >]2 (7.9)

where again [.]2' denotes the positive square root.A signal space with an inner product is referred to as inner product space. Under

additional technical constraints (completion), an inner product space is referred to as aHilbert space. Hilbert space has found widespread use in applied mathematics, physicsand engineering. In addition to being normed spaces, G2[0, oc), G2(-oo, oc), andG2(-o0, 0] are each Hilbert spaces with inner product defined by (7.7).

An important consequence of inner products is the property of orthogonality. Twosignals are said to be orthogonal or form an orthogonal pair if they have an inner productwhich is zero. Thus we see from (7.5) that f+(t), f (t) E G2(-oc, oo) are orthogonal.

The classical example of signals which are orthogonal arises in connection with theFourier series decomposition of periodic signals. This decomposition exploits theperiodicity and orthogonality of the following signal pairs: {cos(wt), sin(wt)};{cos(nwt), cos(mwt)}; {sin(nwt), sin(mwt)}; where n, m are unequal integeis and theinterval of integration used in the inner product is the period of the periodic signal beingdecomposed.

Two spaces S1, S2 are said to be orthogonal, denoted Sl 1 S2 if

< a(t), 0(t) > = 0 for all a(t) E S1 and all /3(t) E S2

Thus we see that G2(-o0, 0] 1 ,C2 [0, oc). More important, £2(-00, 0] n L2 [0, 00) _ 0 sothat ,C2 (-oo, 0] and ,C2 [0, oc) are orthogonal complements of each other, denoted by

£2(-cc, 0] = ,Cz [0, oo) G2[0, oc) = L2 (-oc, 0]

and the decomposition of any signal f (t) E G2(-o0, cc) given by (7.4) is captured by thefollowing relation between the involved Hilbert spaces

.C2(-oc, oc) = G2(-oc, 0] ®G2[0, oc) (7.10)

where ® is referred to as the direct sum of the spaces involved.

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170 Signal and System Spaces

7.2.2 The L2 Norm of the Weighting MatrixRecall, from Chapter 1, that the zero state response of a single-input, single-outputsystem can be calculated from its state model as

y(t) = J t CeA(t-T)Bu(T)dr (7.11)0

with impulse response denoted yj(t) being given as

yj(t) = CeAtB (7.12)

In addition recall from Section 1.8.3 that the causality constraint requires

yj(t) = 0 for all t < 0

Therefore the square of the L2 norm of yj(t) for stable systems can be determined byrecalling the observability and controllability Gramians, W0, W, from Chapter 4. Thuswe have

llYI(t)112 = f y (t)Y,(t)dt = J x B*eA TC*CeATBdr

= B* WOB (7.13)

Alternatively, since yI (t) is a scalar, we can determine the square of the L2 norm of yI (t) as

11

YI(t)112 = f "0 YI(t)y(t)dt = J x CeA`BB*eA tC*dt

= CW,C* (7.14)

Thus we see from (7.13) or (7.14) that the impulse response of a stable system has finite L2norm and hence yI(t) E £2[0, oc).

The extension of the foregoing result to multi-input, multi-output systems is moreinvolved since, in this case, CeA`B is a matrix referred to as the system's "weightingmatrix". This terminology is used in recognition of the fact that CeAtB applies differingweights to the components in the system's input vector in the calculation of the zero stateresponse

t

y(t) = 1 CeA(t-T)Bu(T)dr

0

Notice that the Laplace transform of a system's weighting matrix is the system'stransfer function. Notice also that the weighting matrix is the impulse response when thesystem is single-input, single-output. In what follows we will use the sum of the energies ineach entry in the weighting matrix to generalize the L2 norm of the impulse response tothe multi-input, multi-output case.

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Time Domain Spaces 171

Suppose u(t) is a vector of length m and y(t) is a vector of length p, i.e., the system hasm scalar inputs and p scalar outputs and the weighting matrix, CeAt B, is p x m. Then if theinput is

u(t) = u,(t) = I;,,6(t)

where I'm is the ith column of the m x m identity matrix, then the output is

y(t) = Yt(t) = CeAtBi

where B` is the it" column of the n x m matrix B.Now we want to calculate the sum of the energy in each of the outputs in the set of

outputs {y,(t) : i = 1, 2, m} resulting from the application of each input in the set{u, : i = 1, 2, m} respectively. Clearly, we can write the desired energy as

IYt(t)112

"= ( ( [Y,(t)]*Y,(t)dt

0 (B`)*eA*tC*CeA`B`dt (7.15)

where y, (t) is the pm dimensional vector

YI(t) =

LYi (t)

Now we can rewrite (7.15) as

IIY,(t)112 = (trace[J(C

eA`B)*CeA`Bdt] I; (7.16)/

= (trace (7.17)

Alternatively, recall (Theorem 4.2) that for any matrix M we have

trace[M*M] = trace[MM*]

Therefore setting M = CeA`B, we see that (7.16) can be rewritten as

ILYi(t)112 = (trace [CWcC*])l' (7.18)

Finally, since 11y, (t) 112, (7.16), is the sum of the energy in each scalar signal in thep x m

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172 Signal and System Spaces

matrix CeA`B, we define the L2 norm of the p x m matrix CeA'B as

ICe4'BI12= (trace[f(ceAtB)*ceAtBdt1)(7.19)

which we see from (7.17, 7.18) can be written in either of the following two forms

CeA`B 2 = (trace [B*WOB])2 (7.20)

= (trace [CWc.C*])Z (7.21)

We will see in Section 7.3.6 that the foregoing time domain L2 norm of the weightingmatrix equals an appropriately defined frequency domain L2 norm of the system'stransfer function which is referred to as the H2 norm.

7.2.3 Anticausal and antistable systemsWe begin by considering the single-input, single-output case. Suppose u(t) E £2(-oo, 0].Then substituting negative t in (7.11) and considering t decreasing, i.e., reversing thedirection of integration in (7.11), we see that the output for negative time is given as

y(-fit]) = - J Ce Bu(T)dTr t E [0, oo)

or

y(t) = - I CeA('-T)Bu(T)dr t E (-oc, 0] (7.22)

Notice that this system operates in negative time, i.e., runs backward in time. Thus animpulse input at the time origin, u(t) = b(t), affects the output at earlier (negative) times.This effect is contrary to nature where dynamic processes have the general behavioralproperty that the output now is independent of the future input and is solely caused bypast initial conditions and the cumulative effect of the input over the past interval of time.Natural systems having this behavior, where the cause precedes the effect, are referred toas causal systems. Artificial systems having the predictive behavior exhibited in (7.22),where the effect precedes the cause, are referred to as anticausal systems.

Now if we let u(t) = 6(t) then we see from (7.22) that the impulse response of ananticausal system is given by

yj(t) = -CeA'B t E (-oc, 0]

y,(t) = 0 for all t > 0(7.23)

Notice that yI(-oo) is zero only if -A is stable, i.e. only if A has all its eigenvalues inthe open right half-plane or equivalently, only if no eigenvalue of A is in the closed lefthalf-plane. Systems that have this property are referred to as being antistable. Recall that

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Frequency Domain Hilbert Spaces 173

a system is unstable if at least one eigenvalue of its A matrix lies in the closed right half-plane. Thus we see that antistable systems are a special class of unstable systems.

The introduction of this class of system allows us to express the transfer function of anunstable system, provided it has no imaginary axis poles, as the sum of two transferfunctions, one for a stable system and one for an antistable system. We can obtain thissum decomposition by expanding the given transfer function in partial fractions.

The foregoing readily extends to the multi-input, multi-output case as follows. If ananticausal system is antistable the system's weighting matrix satisfies

lim W(t) _ 0` x

and the system's A matrix has no eigenvalues in the closed left half-plane. Moreover, thesystem's zero state response is given by

y(t) = - 1 W(t - T)u(T)dT t E (-no, 0]

W(t) = -CeArB t E (-oo, 0]

so that

11 W(t) (trace V 00W t W t dt

Finally, in summary, we have shown that the weighting matrices for causal systemswhich are stable satisfy

W(t) E G2[0, oo)

whereas the weighting matrices for anticausal systems which are antistable satisfy

W(t) E £2(-oo, 0]

7.3 Frequency Domain Hilbert SpacesWe begin this section by relating the time domain Hilbert space £2(-nn, no) to thefrequency domain Hilbert space G2. This relation is made possible by Parseval's theoremin the general theory of the Fourier transform. Following this we introduce two frequencydomain Hilbert spaces, 7-12 and 1I which constitute an orthogonal decomposition of £2and are known as Hardy spaces.

7.3.1 The Fourier transformRecall that the Fourier transform and the inverse Fourier transform are defined as

-F[f(t)] =F(jw) _ f:ttdt (7.24)

.f (t) = 27rf F(jw)ej"`dw (7.25)

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174 Signal and System Spaces

Also recall the following well known sufficient condition for the convergence of theintegral defining the Fourier transform off (t), (7.24)

[f*(t)f(t)]'dt < oc (7.26)

where again is to be interpreted as the positive square root. Notice that when f (t) is ascalar, (7.26) reduces to f (t) being absolute value integrable, i.e.,

j9C I f (t) I dt < oc (7.27)

Signals satisfying (7.26), or (7.27) if appropriate, are said to be in the Lebesgue spaceG1(-oc, oc).

Now we are interested in signals f (t) E L2 (- 00, oc). Therefore we need to be concernedabout the convergence of the Fourier integral, (7.24) when f (t) E L2(-oo, oe), f (t)G1(-oe, oc). It turns out that the Fourier integral of signals in this latter subspaceconverges for allmost all w except for, what is referred to as, a set of measure zero. Thismeans that Parseval's theorem

xf*(t)f(t)dt = 27rf F*(jw)F(jw)dw (7.28)

applies to all f (t) E ,C2 (-oc, oc) independent of the values assigned to F(jw) at points wwhere the Fourier integral does not converge. This fact is discussed further in the nextsubsection.

The important conclusion to be drawn from the foregoing is that the L2 norm of asignal in the time domain, If (t) MM2, equals an appropriately defined L2 norm in thefrequency domain, JIF(jw)112,

If(t)112 = JIF(jw)MM2

where

(7.29)

11.f(t)MM2 =[jf*

x(t) f(t)dtJ2 (7.30)

x

JIF(jw)112 = r lf x F*(jw)F(jw)dwz (7.31)2,

Notice that the definition of the frequency domain L2 norm, (7.31) includes the scalingfactor (27r)-1 which is missing from the definition of the time domain L2 norm, (7.31).

Any F(jw) satisfying

JIF(jw)112 < oc

is said to belong to the L2 frequency domain space denoted G2. Notice that the frequencyand time domain L2 norms are denoted by L2 and C2(-oe, oo) respectively.

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Frequency Domain Hilbert Spaces 175

Now Parseval's theorem, (7.28), allows us to calculate the time domain inner productof fl (t), f2(t) E G2(-oc, oc) in the frequency domain as

<.ft (t),.f2(t) > = < Fi (jw), F2 (j-) > (7.32)

where F, (jw) = F [ f; (t)] and

<f1(t),f2(t) >_f:fl*(t)f2(t)dt (7.33)

1x

< Fi (jw), F2(jw) >= f Fi (jw)F2(jw)dw (7.34)27r x

Thus L2 is an inner product space which can be shown to be complete. Therefore L2 is aHilbert space with inner product defined by (7.34).

The reason which make it possible for us to extend Parseval's theorem fromG1(-oc, oc) to £2(-oo, oo) arises from the theory of Lebesgue integration which isbeyound the scope of this book. However the following subsection is included in anattempt to provide some insight into this matter.

7.3.2 Convergence of the Fourier integralSuppose f (t) is specified as

f(t) _ (t+ l)-it > 0

=0 t<0Then

or more specifically

f(t) E £2(-OC,oc) f(t)OLI(-oc,oc)

12 °° 1fm(

t+l) dt<oc and f ')cHowever we can determine the Fourier transform of (7.35) at all frequencies except

w = 0. To see this we need to show that

fLsin(wt)dt < oc and j1___cos(wt)dt < oc w54 0

We can do this by noting that

f 1 sin(wt)dt = Ck (7.36)1 + t

k=O

where fork=0,1,2... we havetk+1

Ck 1 + tsin(wt)dt

Lk7r

tk = -w

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176 Signal and System Spaces

Then since f (t) is a positive monotonically decreasing function, the foregoing series isan alternating series satisfying

ICkl > lk+1 limek=0 k0,1,2...k-x

which is known to converge. In a similar manner we can show that the integral involvingcoswt converges. Therefore in this example (7.24) defines F(jw) unambiguously for everyw except w = 0.

Now since F(jO) is undefined in this example, suppose we set it to n. Then we wouldfind that the L2 norm IIF(jwl12 is independent of 11. This fact is stated mathematically, inthis case, by saying that the Fourier transform off (t) specified as (7.35) is uniquelydetermined for all w E (-oo, oc) except w = 0 which is a point of measure zero.

The foregoing example demonstrates the fact that, in general, when f (t)OG1 (-oc, oc)and f (t) E £2(-oc, oc) the Fourier transform F(jw) is defined uniquely as an element ofthe Hilbert space L2 but is only determined "almost everywhere" as a point function ofjw, i.e., is uniquely determined except on a set of measure zero. Therefore the Fouriertransform is a unitary operator between Hilbert spaces L2(-oo, oc) and C2 since for anyfi(t),f2(t) E G2(-oc, oo) we have

<-F [fl(t)],-F [f2(t)] >=<fl(t),f2(t) > (7.37)

where the inner product on the left is defined by (7.34) and the inner product on theright is defined by (7.33). Hilbert spaces which are related in this way are said to beisomorphic and the unitary operator between the Hilbert spaces is called an isomorph-ism. In the present case the isomorphism is the Fourier transform and L2 (-00, 00) isisomorphic to C2-

7.3.3 The Laplace transformIn order to characterize the properties of the Fourier transforms of causal and anticausalsignals in L2[0, oc) and G2(-o0, 0], respectively, we need to define two Laplace trans-forms, the usual one denoted L+[f (t)] for causal signals and one denoted G_ [f (t)] foranticausal signals. These transforms are defined as

L+[ f (t)] = lim f f (t)e-srdt : Re[s] > a+ (7.38)a-0 -

G_ [ f (t)] = lim foo

f (t)e rdt: Re[s] < a (7.39)

where a, a+, and are real scalars with a being positive.Recall that the existence of the Laplace transform, L+[ f (t)], requires that f (t) have a

lower bounded exponential order, i.e., there must exist a finite real scalar a+, called theabscissa of convergence, such that

lim f (t)e `r` = 0 a > a+ (7.40)roc=oc a<a+

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Frequency Domain Hilbert Spaces 177

This implies that L [ f (t)] has no poles to the right of a line called the axis ofconvergence which is parallel to the imaginary axis and which cuts the real axis at theabscissa of convergence, s = a+. Now it turns out that f (t) E L2 [0, oo) then f (t) satisfies(7.40) with a+ = 0. Therefore the integral (7.38) converges for Re[s] > 0 whenf (t) E L2 [0, oo). This implies that L [f (t)] is devoid of poles in the open right half-plane when f+(t) E L2[0, oo).

We can use a similar argument to show that L- [f (t)] is devoid of poles in the open lefthalf-plane when f (t) E L2(-oo, 0]

Now suppose we are given a causal signal f+(t) E .C2[0, oo) and an anticausal signalf (t) E L2(-00, 0]. Then we have

f+(t)e "+r c Li (-no, oo) a+ E (0, oc)

f_(t)e-a-t c LI (-oo, oo) a_ E (-oo, 0)

and

.7 t o j:f+t)e e dt=F a +'w

-T [f-(t)er]

= t f (t)e '-re-Jwrdt = 1;1(a- +jw)

for any real scalars a+ > 0, a_ < 0.Recall that the Fourier integral, (7.24), converges for almost all w when

f (t) E L2(-oo, oo). Therefore if we take a+ and a_ zero we have the Fourier transformsforf+(t) and f_(t) as

F+(Jw) = L+[.f+(t)]j, j(7.41)

F-(Jw) = C- [f- (t)] j,

with {F+(jw) : w E S+}, {F_(jw) : w E S-} being sets of arbitrary finite elements whereS+, S_ are sets of measure zero. Then we see that f (t) E L2 [0, oo) (f (t) E L2 (-oo, 0]) hasa Fourier transform almost everywhere on the imaginary axis if and only if the Laplacetransform G+[ f (t)] (G- [ f (t)]) has no poles in the closed right (left) half-plane. Functionshaving these properties are said to belong to the Hardy space 7I2 (7-l2 ).

7.3.4 The Hardy spaces:'HZ and L2

In this section we will introduce frequency domain spaces denoted 7-12 and 7-1 so that wecan write any F(jw) E L2 as

F(Jw) = F, (Jw) + F2(Jw)where

Fi(Jw) E x2 F2(Jw) E xz

These spaces are called Hardy spaces after the mathematician G. H. Hardy whocarried out extensive studies on these and other related spaces, e.g., 7-1 space.

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178 Signal and System Spaces

The definition of the Hardy spaces 7-12 and 7-Lz involve ideas from the theory ofcomplex functions of a complex variable. A function of this sort is said to be analytic at apoint, if it has a derivative at that point. Points where the derivative does not exist arereferred to as singular points. The only singular points a rational function has are itspoles.

7-12 consists of all complex functions of a complex variable which are analytic at everypoint in the open right half-plane and have a finite L2 norm, (7.31). Alternatively, theHardy space 7-t consists of all complex functions which are analytic in the open left half-plane and have a finite L2 norm, (7.31). The analytic requirement in these definitions isneeded to ensure that (7.31) is a well defined norm for these spaces. No function in either7i2 or 71 can have imaginary axis poles.

In the rational case, 712 consists of all strictly proper functions which have no poles inthe closed right half-plane, whereas 7-1 consists of all strictly proper functions whichhave no poles in the closed left half-plane. In summary, if F(s) is real rational then

(i) F(s) E 712 if and only if F(s) is strictly proper and has no poles in the closed righthalf-plane.

(ii) F(s) E 7-1 if and only if F(s) is strictly proper and has no poles in the closed left half-plane.

(iii) F(s) E G2 if and only if F(s) is strictly proper and has no poles on the imaginary axis.

Notice that if F(s) is the transfer function of some system, then in case (i) the system isstable, and in case (ii) the system is antistable.

7.3.5 Decomposing £2 spaceRecall that the L2 time domain Hilbert spaces decompose as

G2(-OC, Oc) = £2(-x, 0] ® £2[0, oo)

Then since the Fourier transform maps G2 [0, oo) onto 7-12 and G2 (-OO, 0] onto 7-12 and thesame inner product is used for G2, 712 and R21, we have

G2 = 7-12 ® 7-1

and the Fourier transform is an isomorphism such that

,C2(-OO, no)

£2[0, no)G2(-oc,0]

is isomorphic tois isomorphic tois isomorphic to

L2

7-112

As an illustration, suppose we are given the following signal

f (t) = e-at t > 0

=ent t<0

(7.42)

where a, b are real and positive.

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Then we can write f (t) as

where

Frequency Domain Hiibert Spaces 179

f(t) =f+(t) +f(t)

f(t)=e-"t f(t)=O fort>0f+ (t)=0 f (t)=eht fort<0

Since the Laplace transforms of these signals are given by

G+[f+(t)] = s + a Re[s] > -a

,C-[f (t)] = s 1 6 Re[s] < b

we see that both G+[,f+(t)] and G_[f (t)] are free of poles on the imaginary axis.Therefore f (t) has a Fourier transform which can be written as

a[f(t)] =l

+jw

bjw+aHowever G+[f+ (t)] and G_ [ f-(t)] are each zero at w = oc with G+[f+ (t)] being

analytic in the closed right half-plane and G- [f (t)] being analytic in the closed left half-plane. Therefore we have

L+[f+(t)] E'H2 £-[f (t)] E R2 a[f(t)] E £2

The foregoing example demonstrate that, in general, systems with transfer functionsatisfying G(jw) E G2 can be written as

G(s) = G1 (s) + G2(s)

where

G1 (s) E'H2 G2 (s) E H2

7.3.6 The H2 system normRecall from Section 7.2.2 that when the state model for a given system is minimal, the timedomain L2 norm of the weighting matrix is finite only if A is stable and D is null. Thereforein this case the system's transfer function

G(s) = G+ [CeA`B] = C(sI - A)-1B

is strictly proper, i.e., G(oc) = 0 and has no poles in the closed right half-plane. Thus we

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180 Signal and System Spaces

have G(s) E N2 and using Parseval's theorem we have

ICeA'B 12 =(trace[f(ceAtB)(ceA1B)dt)

x

/x= t race I 1 J G`(Jw)G(Jw)d

z

w = I27f x G(1w112 (7.43)

Thus the time domain L2 norm of the system's weighting matrix equals the frequencydomain L2 norm of the system's transfer function, G(s). Moreover, since in this case wehave G(s) E N2, this norm is usually referred to as the system's H2 norm.

Notice that if G(s), U(s) E N2 with G(s) rational, then G(s) U(s) E N2 since each termin the product is analytic in the open right half-plane has no poles on the imaginary axiswith the product being zero at infinity. Thus the zero state response from a stable causalsystem having a strictly proper transfer function satisfies Y(s) E 7-12 or y(t) E G2[0, oo)for any input u(t) E L2[0, oo).

In order to extend this result to the case where the transfer function is only proper weneed another Hardy space. In the next section we will see that transfer functions of stablesystems which are only proper, i.e., G(oc) 0, lie in a Hardy space denoted N. Thus ifU(s) E N2 and G(s) E Rx then the product G(s) U(s) is analytic in the open right half-plane and has no poles on the imaginary axis. Clearly, in the case when U(s) is rational wehave G(oc)U(oo) = 0 and Y(s) E N2 or y(t) E £2[0, oc) for any input u(t) E £2[0, oo). Itturns out that this result holds in the more general case when U(s) is irrational. Thus wehave

Y(s) = G(s) U(s) E N2 (7.44)

when

G(s) E N,, U(s) E N2

Just as there is a frequency domain norm for functions in either N2 or HI,, namely theL2 norm, we will show in the next section that there is a norm for functions in 7-Lx, calledthe H norm. Thus H. is a normed space. However, unlike N2 or Nz there is no innerproduct defined on N. Therefore Nx is not a Hilbert space.

Now the space of rational transfer functions that have no poles in the closed left half-plane and are finite at infinity is denoted N. Notice that we do not use N' to denote thisspace since the concept of orthogonality is missing. Therefore in addition to (7.44) wehave the dual result

Y(s) = G(s) U(s) E 7-l12

when

(7.45)

G(s) E Nw- U(s) E H21

We will see in the next three chapters that the Hardy spaces, Nom, NM, 712 and NZ , that

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The H. Norm: SISO Systems 181

we are examining in this chapter are essential to the solution of the 7-L control problem.This solution consists of a feedback controller which stabilizes the plant and constrainsthe H,, norm of the plant's transfer function, from disturbance input to desired output, tobe less than a specified scalar. The H, norm is introduced now as follows.

7.4 The HO. Norm: SISO SystemsAt the end of the previous section we introduced the H2 norm of a system's transferfunction, JIG(s)112, by relating it to the time domain L2 norm of the system's impulseresponse, CeA`B. In this section we introduce another type of frequency domainsystem norm referred to as the "H infinity norm" denoted JIG(s)110,. We do this nowby relating JJG(s)jI,, to an equivalent time domain norm referred to as the "L2 systemgain".

Suppose we are given a stable single-input, single-output system having zero stateresponse y(t) when the input is u(t). Then we define the L2 system gain, yo, as

yo su 11Y(t)112

(,)EL2A-) I1*t)112(7.46)

where "sup" is the abbreviation for supremium or smallest upper bound of a set, which inthis case is the set of positive real numbers generated by the ratio of the L2 norms in (7.46)as u(t) is varied over G2[0, oc). Notice that "sup" is used instead of "max" in case the setover which the search is carried out does not contain a maximal element. In addition,notice that we do not consider the null input to avoid dividing by zero.

Now the use of a signal norm to define a system norm as in (7.46) is referred to bysaying that the signal norm induces the system norm. Therefore the L2 system gain, (7.46),is induced by the L2 signal norm. Notice that, for a given system, this norm is, in general,not equal to the L2 norm of the system's impulse response, CeA`B.

Since we are assuming the system is stable, we saw in the previous section thaty(t) E L2[0, oc) when u(t) E G2[0, 00). In addition we can use the fact that G2[0, oc) isisomorphic to R2 to rewrite the L2 system gain, (7.46) in the frequency domain as

yo = sup11 Y(S) 112

Uz II U(S)112u(,)#a

(7.47)

Therefore comparing (7.46) and (7.47) we see that the L2 system gain in the timedomain is equivalent to the frequency domain system norm induced by the H2 norm. Thisnorm is referred to as the "H infinity norm", denoted as I G(s) 11, Notice that this normand the H2 norm (Section 7.3.6), for a given system are, in general, not equal.

7.4.1 Transfer function characterization of the Hoc normSuppose we are given a stable system and that we can choose the system input so that theratio in the definition of the L2 system gain, (7.46), is close to being maximized. More

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182 Signal and System Spaces

specifically suppose the zero state response, yopt(t), caused by the input uopt(t) satisfies

Yopt(t)112 (7.48)yo= uoptt 112

+ E

where E is a small positive scalar in the sense that c « /0.Then recalling that linear systems satisfy the principle of superposition, we have

ayop,(t) when auopt(t) for any constant a. Therefore (7.48) can be written as

-YO = I(Ynor(t)1I2+E

where

Ynor(t) = cYopt(t) a= 1

Iuopt(t)112

This shows that we can obtain the L2 system gain, 7o, (7.46), or Hx, system norm,JIG(s)II., (7.47), by restricting u(t) or U(s) to satisfy the following conditions:

(i) u(t) E G2[0, oc) or U(s) E 7-12

(ii) Iu(t)112 = 1 or IIU(s)MM2 = 1

Thus (7.46, 7.47) become

-YO= sUP IIY(t)112 = supU(s)EW2 U(s)EH2u(r)II2-1 IIU(.,)112=I

Y(S)112

x

SuP2

(1

't J G(jw) U(jw)12dwI t"(0112-1

Notice that

2

27rfIG(Jw)U(iw) 12dw < 2

JQmax

I

U(jw)12dw

where

Amax = sup IG(Jw)lWE(-oo,ac)

and amax is referred to as the Lx norm of the function G(jw).However, since U(s) is restricted in (7.49) as

1

U(S)HH2= (f:Ujw2dw)= 1

we see that (7.50) becomes

I G(jw)U(jw)12dw <1 f'

27r2Qmax

(7.49)

(7.50)

(7.51)

(7.52)

(7.53)

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The H. Norm: SISO Systems 183

Now it turns out that we can choose U(jw) in (7.53) subject to (7.52) so that the left sideof (7.53) is arbitrarily close to Amax. This implies that the supremium in (7.49) is (T,,ax andwe have

70 = 0max

sup G(jw)( = IIG(jw)Iloo (7.54)we(-ao,x)

Notice that

sup IG(jw)I < oowE (-oo,oo)

provided G(s) has no imaginary axis poles. However we have

sup1Y(t) 112 < o0

-0)CE20.m) Iu(t)112

only if G(s) has no poles in the closed right half-plane, since y(t) is unbounded for someu(t) E G2[0, no) otherwise. Therefore IIG(jw)IIc, is referred to as:

(i) the L,,. norm of G(s) when the system is unstable and G(s) has no poles on theimaginary axis including s = oo;

(ii) the H,,,, norm of G(s) when the system is stable and G(cc) < no.

In addition, transfer functions that have bounded H,,, (L) norm form a normed spacedenoted 7-l (G,). Notice that 7-( C Gam. More is said about this in the next subsection.

Now there are two different ways of interpreting I I G(jw) I oo . One arises from definingIG(jw)I1. as

IG(jw)Ilo=suPY 'w

Uz 11v(>w)112

Then in this case we are considering G(jw) to be an operator which maps U(jw) E G2 toY(jw) E G2 and 11 G(jw) I I oo to be the operator norm induced by the frequency domain L2norm or, if the system is stable, by the time domain L2 norm.

Alternatively, if we take (7.54) to be the definition of IG(iw)IIoc then we areconsidering G(jw) to be a function of w and 11 G(jw) 11,,c is a function norm.

7.4.2 Transfer function spacesSince any proper rational transfer function with no imaginary axis poles has a finite L',norm, we can define three normed transfer function spaces depending on the location ofthe transfer function's poles. Therefore assuming the rational function G(s) is proper wehave

(i) G(s) E H,,, when all poles of G(s) are in the open left half-plane(ii) G(s) E 'H- when all poles of G(s) are in the open right half-plane

(iii) G(s) E G,,c, when G(s) has no poles on the imaginary axis.

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184 Signal and System Spaces

Notice that when G(s) E L we can use the partial fraction expansion (after one cycleof division) to write G(s) as

G(s) = G`(s) + G_(s)

where

G'_ (s) E Hx G-(s) E R-

Then assuming some scheme for making G+(s), G_(s) unique, e.g., we can insist thatG - (s) be strictly proper, we can capture this spliting of a transfer function in G, into thesum of stable and antistable transfer functions as a direct sum decomposition of thespaces involved

G,, = 7-c E6 7-lx (7.55)

Alternatively, suppose two stable systems having transfer functions G, (s), G2 (s) E 7-form the components of a cascade connection. Then the transfer function of thecomposite system, GI(s)G2(s), is stable since its poles are contained in the poles of thecomponent transfer functions. In addition, the transfer function of the composite systemis proper or strictly proper since the product of proper rational functions is properindependent of any pole-zero cancellations that might occur. Therefore GI (s)G2(s) Eif GI(s), Gz(s) E 7-L .

Now we can readily show, from the definition of the H,,, norm, (7.47), that the H,Cnorm of the transfer function of the cascade connected system is related to the H normsof the transfer functions of the component systems as

G1(s)jj.jjG2(s)11.? JIG1(s)G2(s)II. (7.56)

This result plays an important role in control theory. An example of its use is given inthe following subsection.

7.4.3 The small gain theoremPrior to the use of the H,o norm in control theory, the H,o norm appeared in a number ofdifferent guises in various engineering applications. For example, the resonant gain usedto characterize the performance of an electric filter is the H,, norm of the transferfunction relating the filter's input and output voltages. Another instance of this occurswhen the Nyquist stability criterion is applied to determine the stability of a feedbacksystem having transfer function T(s)

T(s)G(s)

1 + G(s)H(s)

Then when G(s), H(s) E 7-L the distance from the origin to the point on the polar plot ofG(jw)H(jw) which is furthest from the origin is jG(s)H(s)jj,,. This geometricalobservation leads to the following result which is known as the small gain theorem.

Recall that when G(s)H(s) E 7-L,,,, the Nyquist criterion states that the feedbacksystem is stable if and only if the polar plot of G(jw)H(jw) does not encircle the point

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The H. Norm: MIMO Systems 185

-1 +j0. Therefore since the foregoing geometrical interpretation of JJG(s)H(s)JJ,,.implies that the polar plot of G(jw)H(jw) cannot have any encirclements of -1 +j0 if

G(s)H(s)JJx< I (7.57)

we have (7.57) as a sufficient condition for the feedback system to he stable whenG(s)H(s) E H,,. This result, referred to as the small gain theorem, is important inconnection with robust stabilization where a single fixed controller is required to stabilizeany one plant in a set of plants. A simplified example of the application of the small gaintheorem in this context is given as follows.

Suppose we are given a fixed controller having transfer function H(s) E H,, whichsatisfies

JH(s) J x< a (7.58)

Then using (7.56) we can show that (7.57) is satisfied if

J G(s) Ja

(7.59)

Therefore any stable plant satisfying (7.59) is stabilized by the given fixed controller.Notice that by interchanging the role of the controller and plant, the foregoingdevelopment leads to a characterization of a set of stable controllers any one of whichcan be used to stabilize a given fixed stable plant.

This completes the development of the H,o (L,.) norm for single-input, single-outputsystems. We need now to proceed to develop this norm for multi-input, multi-outputsystems.

7.5 The H., Norm: MIMO SystemsWe have just seen that the H,0 norm of a single-input, single-output systefn is the H2induced norm of the system's transfer function, G(s). In order to extend this idea to themulti-input, multi-output case, where G(s) is a matrix, we need to introduce the induced2-norm for constant matrices, denoted as JIG(jw) 11 (for fixed w). Then we can obtain theH,o norm of the matrix G(j w) as the largest induced 2-norm as w varies over (-oc, oc),i.e.,

JG(jw)JJoo= sup (WE(-oo,oo)

G(jw)MM)

7.5.1 Singular value decompositionBefore we can develop the induced 2-norm for a constant matrix we need to introduce thesingular value decomposition (SVD) of a constant matrix. This is done as follows.

Recall, from Chapter 4, that Hermitian matrices have real eigenvalues and orthogonaleigenvectors. Moreover notice that the product matrices M*M and MM* formed fromany complex matrix M, are Hermitian and nonnegative. Therefore these product

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186 Signal and System Spaces

matrices have orthogonal eigenvectors and real nonnegative eigenvalues. This generalobservation leads to the matrix analysis technique known as SVD. The followingtheorem, which is proved in the appendix, defines the SVD of a matrix.

Theorem 7.1 Any p x m matrix of complex constants, M, which has rank r can bedecomposed as

M = UE V* (7.60)

where U, and V are p x p and m x m unitary matrices with

U=[U1 U2][E0 OO 0 V=[V1 V2]

Ulispxr U2ispxp-rV1 ismxr V2 ismxm-r

and Eo is an r x r, real, positive definite, diagonal matrix denoted as

ai 0

0 Q2

Eo = = diag[a1, a2, ... , Qr]

with diagonal entries referred to as singular values and ordered so that

0-i > 5i+1 i = 1

Proof See AppendixThe SVD of a matrix has become ever more useful since the establishment of practical

methods for its computation in the early 1970s. The SVD is of practical use wherever werequire the solution of problems involving matrices having small rank. One problem ofthis sort, which we encountered in Chapter 4, involves the determination of a lower orderstate model approximation of a state model having controllability and/or observabilitymatrices which are almost rank deficient, i.e., the state model is almost uncontrollableand/or unobservable. We were able to obtain a solution to this problem without requiringthe SVD by using the state model's Gramians. Since the Gramians are nonnegativeHermitian matrices, they enjoy the nice properties which make the SVD so useful,namely, real and nonnegative eigenvalues and mutually orthogonal eigenvectors.

7.5.2 Induced 2-norm for constant matricesWe now proceed with the development of the constant matrix norm induced by the 2-norm for constant vectors. This is done in the proof of the following theorem byemploying the SVD and its properties.

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The H. Norm: MIMO Systems 187

Theorem 7.2 Any constant p x m matrix M has a norm induced by the 2-norm forconstant vectors which equals or,, the largest singular value of M.

Proof The induced 2-norm for a matrix M, I(M)1, is defined as

IMMM = sup 11M X11 (7.61)

where Cm is the vector space of all vectors of length m having constant complex entriesand Ilxjj equals the positive square root of the sum of the squared absolute value of eachentry in x, i.e.,

m 3

i xi l -x*J

(7.62)

Notice that unlike the L2 norm for a time varying vector which we used earlier, the 2-norm for a constant vector has no subscript, i.e., 1l'MM2 denotes the L2 norm whereas 11.11denotes the 2-norm.

Next since the transformation Mx is linear, multiplying x by a constant scalar changesboth the numerator and the denominator in (7.61) by the same amount so that the ratio in(7.61) remains unchanged. Therefore we have the alternative definition of the induced 2-norm of M given by

IM11 = SUPI

u=u=

MX 11 (7.63)

Since them x m matrix V in the SVD of M, (7.60), is invertible, we can use its columns{v' : i = 1, 2, m} as a basis for Cm. Thus we can express any x E Cm as

mxaiv`=Vai=o

where

T[al a2 ... amI

with a c Cm.Now, since V is unitary we have

IIx112 = a*V*Va

=am=s' ail2=

110112

(7.64)

(7.65)

andand the 2-norms of x and of a are equal. In addition since V is invertible there is a uniquea for each x and vice versa so that

supllMxjj = supjjMVaJj (7.66)-C' EC^11

11=i 11.11=7

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188 Signal and System Spaces

Then using the SVD of M, (7.60) we can write

I MVa = (a*V*M*MVa)-

r_ Eaz

where is the m x m diagonal matrix given by

Ez

Ez = o

M 00

a2

)

z 222]Eo = diag[a1, az, ... , ar

Therefore from (7.63), (7.66) and (7.67) we have

M =Sz aIz

(7.67)

(7.68)

Now the summation on right side of the foregoing equation satisfies the followinginequality

r r

a2 ail2 alEcx;z=1 r=i

(7.69)

where a, is the largest singular value of M. However since a is restricted in (7.68) to haveunit norm, we see from (7.69) that

IIMII < al (7.70)

Notice that if the components of a satisfy

cti1=1 a;=0 i>2thenllal=land

a?Iailz= a 2

which determines the supremium in (7.68) as

IIMII al

Notice, in the foregoing proof, that the supremium is achieved for a = I, the firstcolumn of the m x m identity matrix. Therefore, letting c = I' in (7.64) gives the resultthat the unit 2-norm x that gives the supremium in (7.63) is the right singular vector of M,

'v , corresponding to the largest singular value of M, a1 . Thus

IlMxll_IIaII

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The H. Norm: MIMO Systems 189

when x = Qv1 for any complex scalar, /3. Further consideration of this fact leads us to theimportant conclusion that

MxMI <-011

with equality only when x is a scalar multiple of v1. Finally, notice that 1IMII = IMM is a scalar.

for all x E C' (7.71)xI1

when

7.5.3 The L, H,,, norm for transfer function matricesHaving developed the idea of the induced 2-norm for constant matrices, we are now in aposition to give the definition of the L,,, norm of a transfer function matrix, G(jw). Thisnorm is defined as the supremium of the induced 2-norm of G(jw) over all w c (-oc, oc)

IG(jw)IIx= sup al(jw)Lie (-0070c)

(7.72)

where a1(jw) is the largest singular value of G(jw).As in the SISO case, when an MIMO system is stable its transfer function matrix G(s)

has finite L,, norm, which is referred to as the Hx norm and G(s) is lies in the normedspace denoted by 7-lx. Alternatively, when the system is antistable, its transfer functionmatrix G(s) has finite Lx norm, (7.72), and G(s) lies in the normed space denoted 7-c.Finally notice that, as in the SISO case, all G(s) with finite L', norm are denoted byG(s) E L. Thus 7-L,,, and 7-i; are each subspaces of G,,, which are disjoint and completethe G,, space so that (7.55) holds for MIMO systems.

In subsequent development we will need the following result which is an immediateconsequence of the foregoing.

Theorem 7.3 A real rational transfer function G(s) E G,,,, has an L , norm less than afinite positive scalar -y, i.e.,

IG(jw)II.<y

if and only if F(jw) is positive definite for all w, i.e.,

F(jw)>0 dwE(-oo,oc)where

F(jw) = -y2I - G*(jw)G(jw)

Proof Let the SVD (Theorem 7.1) of the p x m matrix G(jw) be given as

G(jw) = U(jw)E(jw)V*(jw)

where U(jw), V (j w) are p x p and m x m unitary matrices respectively and

[Eo(jw) O1E(Jca)

ILJ Eo(jc.)=diag[al(jw),a2(jw),...,a.(jw)J

0

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190 Signal and System Spaces

with r = rank [G(jw] and

a1 (Jw) > ai(Jw) i > I (7.73)

Then using the SVD of G(jw) we can write F(jw) as

F(jw) = 722 - V * (7.74)

where is the m dimensional diagonal matrix

E,,(Jw) = diag 1011 (Jw), a2(Jw), ... , a2 (Jw), 0, ... 0]

Next pre and post multiplying (7.74) by V* (jw) and V(jw) respectively gives

V * (Jw)F(Jw) V (Jw) = EF

where

(7.75)

EF(jw) = diag[(72 - ai (Jw)),(,Y2 - a2(Iw)), ... (ryz - a; (1w), Yz, ... y2]

Now since V (jw) is unitary and therefore invertible, we have

F(jw) > 0 if and only if V* (jw)F(jw) V(jw) > 0

Thus we see from (7.75) that

F(jw) > 0 if and only if [ryz - 0,1(jw)] > 0 (7.76)

However, since the L,,, norm for G(s) is defined as

G(jw)ll.= sup at(Jw)WE(-oo,oo)

we see from (7.76) that

F(jw) > 0 for all w c (-oo, oo) if and only if G(jw) 1x<7

and the theorem is proved.In the next chapter we will use the foregoing theorem to obtain necessary and sufficient

conditions on the state model parameters, {A, B, C, D}, of a system which ensures thatthe system's transfer function has L,,. norm less than a given finite real scalar.

7.6 SummaryWe began by formalizing the idea, used earlier in Chapters 5 and 6, of relating the size of asignal to the energy in the signal. This led to the time domain Hilbert space G2(-oc, oo)and its orthogonal subspaces L2(-oo, 0] and L2[0, oo). Following this we used the Hilbertspace isomorphism from L2 (-oo, oo) in the time domain to £s in the frequency domain tointroduce the frequency domain spaces 7-L2, NZ known as Hardy spaces.

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Notes and References 191

Next we introduced the notion of the size of a system. This was done by using theinduced operator norm to characterize a system's ability to transfer energy from its inputto its output. As with signal norms we saw that there is an equivalence between a timedomain system norm known as the system's L2 gain and a frequency domain normknown as the H, norm. Unlike the Hardy spaces, 7-12, 7-l; for signals which are equippedwith an inner product, the Hardy spaces, 7-t 71 for systems are normed spaces only.

7.7 Notes and ReferencesThere are many texts on Hilbert space. A readily accessible treatment of this subjectwhich includes material related to the control problems being considered here can befound in [46]. Unfortunately books on Hardy spaces require a considerable backgroundin functional analysis and complex analysis. Some of these references are given in [14, p.13]. An interesting treatment of complex analysis as it applies to Laplace and Fouriertransforms is given in [26].

The example following (7.27) in Section 7.3.2 which is used to discuss the Fouriertransform of signals in L2 that are not in LI is taken from [45, p. 272]. The convergence ofthe series (7.36) is discussed in [36, p. 71]. The applicability of Parseval's theorem forfunctions in £2(-oc, no) is discussed in [37, p. 185]. An excellent reference for the SVD is[17].

The small gain theorem is used to solve a wide variety of robust control problemsincluding both linear, [47] [18] and nonlinear, [42] [21] [33], systems. Finally a good basicintroduction to the ideas in this and the next chapter can be found in [11].

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8System Algebra

8.1 IntroductionIn the previous chapter we were interested in characterizing signals and systems in termsof normed spaces in both the time and frequency domains. This led to the introduction ofthe frequency domain spaces 712, H21 and G2 for signals and 7L, , R.-, and G. for systems.

In the present chapter we consider a number of operations involving systems in thesespaces. We will be especially interested in obtaining state models for the systems whichresult from these operations. To facilitate this endeavour, we use the compact equivalencerelation

G(s) Sr A B 1C D

to denote that

G(s) = C(sI - A)-' B + D

We begin the discussion of these operations by considering the effect of connectingsystems in parallel and in series.

8.1.1 Parallel connectionOne of the simplest examples of operations with systems consists of connecting severalsystems in parallel so that all systems have the same input and the output from theconnection is the sum of the outputs from each system. The connection of the first ordercomponent systems resulting from a partial fraction expansion of the system's transferfunction is an example of the parallel connection of systems.

The input-output constraints which ensure that r component systems are connected inparallel are

u(t)=u,(t) i= 1,2,.--rr

y(t) = Eyi(t)i=1

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194 System Algebra

G,(s)

G,(s)

Ga(s)

Figure 8.1 Parallel Connection

so that

Y(s) = G(s) U(s)

wherer

G(s) _ Gi(s)i=1

with Gi(s) : i 1, 2 . r being the transfer functions of the component systems.Thus assuming we know state models for each of the component systems as

A BGi(s)

s

Cj Di

we want to determine state model parameters {A, B, C, D} for the composite system, i.e.,

A BG(s)

C D

We do this as follows.First we write the state equations for each component system as

xi(t) Aix'(t) + Biui(t)(8.5)

yi(t) = Clxi(t) +Dlui(t)

where i = 1, 2, r. Next we use (8.5) together with the constraints imposed by (8.2, 8.3)to give a state model for the composite system as

z(t) = Ax(t) + Bu(t) (8.6)

y(t) = Cx(t) + Du(t)

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Introduction 195

where

A=

A, 0 ... 00 A2 ... 0

B=B2B`1

L Br L xr(t)r

C=[C( C2 Cr] D=ED;r=(

Notice that the dimension of the state model for the composite system is the sum of thedimensions of each component system. Moreover, MIMO systems can be connected inparallel if and only if each of the systems has: (i) same number of inputs, and (ii) the samenumber of outputs. Notice in the case of the partial fraction expansion of G(s), the A,s arethe poles of G(s).

Finally notice that if we are given G(s) E ,C,,, then we can use partial fractionexpansion to decompose G(s) into the parallel connection of two systems such thatG(s) = G( (s) + G2(s) where G( (s) E 7-L and G2(s) E 7-lx with the eigenvalues of A( (A2)being in the open left (right) half plane. This fact was given in the previous chapter as thedirect sum decomposition of the spaces involved, i.e., Gx = 7-L 7-lx.

8.1.2 Series connectionNext consider a series connection of two systems, often referred to as system composition.The constraints imposed by this connection are

so that

where

u(t) = u2(t) y2(t) = ui(t)

Y(s) = G(s) U(s)

G(s) = GI(s)G2(s)

Y, (t) = y(t)

Now we can develop the state model for the composed system, by imposing theconstraints (8.7) on the state models for the component systems, (8.4), with r = 2. Theresult is as follows.

zl (t) = A, x1(t) + Bl C2x2(t) + B(D2u(t)

$22(t) = A2x2(t) + B2u(t)

y(t) = C, xl(t) +D,C2x2(t) +D,D2u(t)

u(t) Gz(s) GI(s) y(t)

Figure 8.2 Series Connection

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196 System Algebra

which can be rewritten as

x (t) Al BiC2(t) B1Dz-Y1

il(t) 0 A2 I Lx2(t) + B,

]U(t)

y(t) _ [Cl D1C21 X2(t)I +DiD2u(t)

and we see that the composed system has state model

G1(s)G2(s)[A B]C D

where

A= [A1 B1C2I B= [B1D2]

O A2 B2

C=[C1 D1C21 D=D1D2

Notice that the constraint ul (t) = y2(t) implies that the number of inputs to the systemlabeled 1 must equal the number of outputs from the system labeled 2. Alternatively, wecould connect the systems in reverse order if the number of inputs to the system labeled 2and the number of outputs from the system labeled 1 are equal. In this case the transferfunction for the composed system would be G2(s)G1(s). Notice that we get the samecomposed system independent of the order in which the component systems areconnected in series only if the matrices G1(s) and G2(s) commute so thatG1(s)G2(s) = G2(s)G1(s). Thus in the SISO case the result of composing systems isindependent of the order in which the component systems are connected.

In the remainder of this chapter we will take up the problem of determining the statemodel for several other types of system operation. One of these problems concerns thedetermination of the state models for the systems needed in a series connection so that theresulting composed system has a specified transfer function. This problem is known as thesystem factorization problem. The requirement that the factors have certain propertiesgives rise to a number of different classes of system factorizations, several of which are ofgreat importance to the development of control theory.

8.2 System InversionBefore we begin considering system factorization we need to consider the related problemof finding a system inverse.

Suppose we are given two SISO systems having transfer functions

G1(s) bpsm + b1 sm-1 + ... + b",

(8.10)

G2(S)

s" + alsn-1 + ... + a"

s" + als"-1 + ... + a"(8.11)

bpsm+blsin-1+...+b

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ay stem Inversion 197

Then the series connection of these systems gives a transfer function which is unity, i.e.,the series connected system has its output equal to its input. Thus it would appear that thesystem labeled 1 has the system labeled 2 as its inverse.

However, physical processes have the property that their sinusoidal steady state gaincannot be unbounded at infinite. Therefore we require G1(oc) and G2(oc) to be finite.Now Gi (oc) (G2(oc)) is finite if and only if m < n (n < m). Therefore we can only haveboth G1(oc) and G2(oc) finite if in = n. This means that a given physical process havingtransfer function G1 (s) has an inverse if and only if G1 (s) is proper (not strictly proper) orequivalently only if the state model for G1 (s) has a nonzero D matrix.

In the MIMO case, a system having transfer function G1 (s) has an inverse if and only ifits state models have a D matrix which is invertible. Thus only systems having the samenumber of inputs as outputs can have an inverse. The state model for the inverse system ofa given invertible system can be developed as follows.

8.2.1 Inverse system state modelSuppose we are given the transfer function and state model for an MIMO system

G(s)A BC D D-1 exists

so that state equations for the system are

z(t) = Ax(t) + Bu(t) (8.12)

y(t) = Cx(t) + Du(t) (8.13)

Then solving (8.13) for u(t) gives

u(t) = -D-1Cx(t) +D-1y(t) (8.14)

which when substituted in (8.12) gives

z(t) = (A - BD-1C)x(t) + BD-1y(t) (8.15)

Thus we see from (8.14, 8.15) that the inverse system has output u(t), input y(t), andstate model given as

L C" D"G!-'(s)-'

AX

BX

where

A" =A - BD-1C B" = BD-1

CX = -D-1 C D" = D-i

(8.16)

As we will see, the state model for the system inverse is useful in developing results incontrol theory.

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198 System Algebra

Notice that the transfer functions for ani-nvertible SISO system and its inverse (8.10,8.11) are reciprocals of each other. Therefore the poles of the given system equal the zerosof the inverse system and vice versa. A similar statement can be made for invertibleMIMO systems. To see this we need interpret the idea of a system zero in terms of asystem's state model. This will be done by first considering system zeros for SISO systems.

8.2.2 SISO system zerosRecall that an SISO system has a system zero at s = so if its transfer function, G(s), is zerofor s = so or

Y(so) = 0 U(so) 0

where

(8.17)

Y(s) G(s) U(s)

Notice that (8.17) implies that G(so) = 0.Now in order to arrive at an interpretation of system zeros which we can use in the

MIMO case, we consider a system's zeros in terms of a state model for the system.Suppose we are given a controllable and observable state model for the system,

G(s) [A ' ] (8.18)

Then taking x(0) = o since system zeros are defined in terms of the system's zero stateresponse, the Laplace transform of the state equations for the system yields

rsI-A -B1 [X(s)1 _ r0

C D U(s) Y(s)(8.19)

Therefore if so is a system zero we have

cI - A] X(so)+ U(so) _ 0 U(so)

54

0 (8.20)

which implies that the column

[-B]D

is dependent on the n columns in

Therefore s = so is a system zero if and only if the (n + 1)side of (8.19) is not invertible or

soI - A -Brank C < n + 1

x (n + 1) matrix on the left

(8.21)

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System inversion 199

As illustration, suppose we are given a second order system in controller form with Dzero. Then (8.19) is

a [X2(s)] _ [0' ] U(s)_S

[c c2]

[ 001

= Y(s)

which after eliminating Xl (s) becomes

(s2 + ais + a2)X2(s) - U(s) = 0 (8.22)

(cps+c2)X2(s) = Y(s) (8.23)

Notice from (8.22) that if U(so) 0 then X(so) 0. Consequently (8.23) implies thatY(so) = 0 for U(so) 0 only if so is a zero of the polynomial

cis + C2

However

cls + C2

s2 + als + a2G(s) =

which shows that so is indeed a zero of G(s).By proceeding in the same fashion, we can show, when D 0, that the foregoing state

model has a system zero at so which is a root of the polynomial

D(s2+als+a2)+cls+c2

which is the numerator of G(s).The foregoing ideas are used now to extend the definition of a system zero to MIMO

systems.

8.2.3 MIMO system zerosSuppose the system denoted by (8.18) has m inputs and p outputs with m < p and with allm columns of B independent. Then a complex number .s0 is a system zero if

G(so)U(so) = o for some U(so) 54 o (8.24)

Notice that unlike the SISO case where (8.17) implies that G(s0) = 0, (8.24) does notimply that G(so) = O.

Now with appropriate modification, (8.20) implies that .s = so is a system zero if at leastone of the m columns of

[-B]D

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200 System Algebra

is dependent on the columns of

Therefore, recalling that the rank of a matrix cannot exceed its smallest dimension, wesee from (8.21) that so is a system zero if

soI - A -Brank C

D< n + m (8.25)

Alternatively, when m > p the matrix in the foregoing inequality cannot have rankgreater than n + p so that (8.25) is satisfied for all s0 and (8.25) is meaningless.

In order to overcome this problem we define so to be a system zero if

ranksoI - A B <

[C D

where

p = max I rankSCC C D

sI - A B

We want now to apply the foregoing ideas to invertible systems. Since invertiblesystems are square, i.e., m = p, we can use (8.25) to define the system zeros for this class ofsystem.

8.2.4 Zeros of invertible systemsSuppose G(s) is invertible. Then we see, in this case, that the condition for s0 to be asystem zero, (8.25), is equivalent to

dets0I-A -B -0

[C D(8.26)

Notice that if the state model is not controllable the foregoing condition is alsosatisfied when so is an uncontrollable eigenvalue of the system since we have aneigenvalue-left-eigenvector pair (A, w) of A which satisfies

wT(AI-A)=0 and wTB=o

Therefore

aI A B[wT 0]

C D =[0

0] (8.27)

and (8.26) holds for A = so. Thus, in this case, s0 is both a system zero and a system pole.Alternatively, if (A, C) is an unobservable pair a similar argument can be used to showthat at least one of the eigenvalues of A is both a system zero and a system pole.

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Coprime Factorization 201

However, if the state model is both controllable and observable with D is nonsingular,then there is no system zero which equals a system pole. In addition, each system zero ofG(s) is a system pole of G-1 (s) and vice versa. To see this suppose s0 is a system zero. Then(8.26) is satisfied and is equivalent to

det i oI - A D MI = 0 (8.28)

where M is any nonsingular matrix of appropriate dimension.Now suppose we choose M as

1 0 1M=

-D_1C I]Then the left side of (8.28) becomes

dets0I-A

IL C

B1

Ms0I -A+BD-1C

D ]] = det

0-B1

DJ

= det [soI - A + BD-1 C] det[D] (8.29)

However since G(s) is invertible, we have det[D] # 0 and (8.29, 8.28) imply

det[s0I - A"] = 0

where

A" =A-BD-1C

and A" is the system matrix for G-1(s), (8.16). Therefore so is a pole of G- '"(s) as well asbeing a zero of G(s).

Finally, we will see that invertible systems which are stable with stable inverses play animportant role in control theory. These systems, which are referred to as "units", form asubspace U of H,,c and are characterized in terms of their transfer functions G(s) as

G(s) E U if and only if G(s), G-1(s) E 7-L

In the case of an SISO system, this definition implies that a system is a unit if it has atransfer function which is proper with no poles or zeros in the closed right-half plane.

8.3 Coprime FactorizationProblems in system factorization are concerned with the determination of two systems,called factors, whose series connection has the same input-output behavior as the systembeing factored. Additional constraints on the factors determine different classes offactorization.

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202 System Algebra

Coprime factorization can be characterized as a series connection of a stable systemand the inverse of a stable system with no unstable pole-zero cancellations betweenfactors. As an example consider the system with transfer function G(s) where

(s + 1)(s - 2)G(s) _

(s + 3)(s - 4)(s + 5) (8.30)

Then a coprime factorization of this system is indicated as follows:

G(s) =Ni(s)MI '(s)

where

Ni(s) _(s- 2)(s+ 1)

Mi(s) =(s - 4)(s + 5)

(s + 3)a(s) a (s)

with a(s) being any polynomial of degree 2 having no zeros in the closed right half plane.Notice that this choice of degree for a(s) ensures that M, (s) is invertible whereas the

restriction on the location of the zeros of a(s) ensures that a(s) is not involved in anyunstable pole-zero cancellations between N, (s) and M-1' (s). Thus even if a(s) has zeros ats = -1 and/or -5 the resulting pole-zero cancellations in N, (s) and M-11 (s) are allowedsince they are stable. More important, notice that the closed right-half plane zero of N1 (s)at s = 2 and of M, (s) at s = 4 are different so that there are no unstable pole-zerocancellation between N, (s) and Mi 1(s). Stable systems which do not share any systemzeros in the closed right-half plane are said to be coprime.

Notice, from the foregoing example, that the form of coprime factorization,N(s)M-1(s), bears a striking resemblance to the expression for a transfer function interms of its numerator polynomial N(s) and denominator polynomial M(s). Of coursethe "numerator" and "denominator" in the present context are not polynomials but areeach rational and can therefore be thought of as transfer functions for "numerator" and"denominator" systems.

8.3.1 Why coprime?In the foregoing example the numerator N1 (s) and denominator M1 (s) are coprime. Inorder to better appreciate this fact we give a second factorization of G(s) in which thefactors are not coprime

Suppose we re-express G(s), (8.30), as

G(s) = N2(s)M21(s) (8.31)

where

N2(s) =(s - 6)(s - 2)(s + 1)

MZ(s) _(s - 4)(s + 5)(s - 6)

(s + 3)J3(s) 0(s)

with 13(s) restricted to be any degree 3 polynomial with no zeros in the closed right halfplane. Although this factorization has the desired property that N2 (s), M2 (s) E R. with

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Coprlma Factorltatlon 203

M2(s) being invertible, it does not satisfy the requirement that N2(s) and M2(s) becoprime since they share a right-half plane zero at s = 6. Thus (8.31) is not a coprimefactorization of G(s). We can demonstrate the importance of not having unstable pole-zero cancellations between factors as follows.

Suppose that we have a strictly proper scalar transfer function G(s) having one of its npoles, n > 2, at s = so on the positive real axis with the remaining n - I poles being in theopen left-half plane. These specifications imply that G(s) can be written as

G(s) = q(s) [s - (so + E)](s - sO) P(s)

0<E<oc, so>0

where the degree of q(s) is less than p(s) with q(so) j 0 and p(s) having no zeros in theclosed right half plane. Now the system having this transfer function is stable only if f = 0so that there is an unstable pole-zero cancellation. However, closer inspection of theconsequences of such a pole-zero cancellation reveals that we cannot rely, in practice, onunstable pole-zero cancellations to make an unstable system stable. This becomesimmediately evident when we consider the system's impulse response, ,C+1 [G(s)] = g(t)

g(t) = Koeso( + r(t)

where

Ko = p(so)(so)

with r(t) bounded for 0 < E < oc.Notice that if c = 0, Koes0t is missing from g(t). Then the system's impulse response is

bounded and the system is stable. However, if c 0, then Koes0` is present in g(t) and thesystem's impulse response tends to infinity with time and the system is unstable. Thus thesystem's stability is catastrophically sensitive to E.

Conversely, if so is in the open left-half plane then Koes°` tends to zero witlT'time and thesystem is stable for all E.

Comparing the effect on the system behavior of the two types of pole-zero cancellation,stable and unstable, we see that the lack of robustness of a system's stability to unstablepole-zero cancellation makes this type of pole-zero cancellation a detriment to reliablecontrol system design. We will see that the coprimeness and the stability of the factors in acoprime factorization can be exploited to provide a means for determining controllerswhich not only make a feedback control system input-output stable but in additionprevent unstable pole-zero cancellations between the controller and plant so that theclosed loop system is internally stable.

Returning to coprime factorization, we saw earlier in this chapter that system zeros ofan invertible system are poles of that system's inverse. Therefore, we see now, thatunstable pole-zero cancellations between N(s) and M-1 (s) are prevented if and only ifM(s) and N(s) have no common system zeros in the closed right-half plane, i.e., if andonly if M(s) and N(s), are coprime.

In addition, since N(s) is always stable, M-1(s) is unstable if G(s) is unstable.Moreover, since M(s) and N(s) are coprime, the unstable poles of G(s) are the only

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204 System Algebra

unstable poles of M-1 (s). Finally, since M(s) is stable, M-1 (s) has no zeros in the closedright-half plane. Therefore the closed right-half plane zeros of G(s) are the only closedright-half plane zeros of N(s).

8.3.2 Coprime factorization of MIMO systemsMoving on to MIMO systems, we need to distinguish between two types of coprimefactorization depending on the ordering of the factors. Thus a p x m real rational transferfunction matrix G(s) can be coprime factored as

G(s) = N(s)M-1(s) (8.32)

or as

G(s) = M-1(s)N(s) (8.33)

where N(s), N(s) E 7-h are both p x m and M(s), M(s) E 7-L,,, are m x m and p x prespectively. In addition both pairs {N(s), M(s)}, and {N(s), M(s)} must be coprime.Notice that the two varieties of factorization (8.32, 8.33) are referred to as right and leftcoprime factorizations since the denominator is on the right in (8.32) and on the left in(8.33). In the SISO case these two varieties coincide since products of scalars commute.

We give now, in the theorem to follow, an alternative characterization of coprimeness.We will see later that this characterization is also useful in the problem of determiningcontrollers which make the closed loop control system internally stable.

Theorem 8.1 The factors M(s), N(s) E R. {M(s), N(s) E rh} are right {left}coprime if and only if there exists X(s), Y(s) E 7-Lx {X(s) Y(s) E 7I, }which satisfythe following Bezout, (pronounced "bzoo") identity

X(s)M(s) + Y(s)N(s) = I (8.34)

{M(s)X (s) + N(s) ?(s) = 11 (8.35)

Proof (if) Consider the SISO case. Suppose M(s), N(s) are not coprime, i.e.,M(so) =N(so) = 0 for some so in the closed right-half plane. Then since X(s) and Y(s)are stable, X(so) < no, Y(so) < no. Therefore we have X(so)M(so) = 0, Y(so)N(so) = 0and (8.34) cannot be satisfied. This shows, in the SISO case, that satisfaction of theBezout identity is sufficient for M(s), N(s) to be coprime. The MIMO case is treated asfollows.

Rewrite the Bezout identity, (8.34), as

L(s)R(s) = I (8.36)

where

L(s) = [X(s) Y(s)] R(s) =M(s)

[ N(s) I

and L(s), R(s) E H,,. are m x (m + p) and (m +p) x m respectively.

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Coprime Factorization 205

Next suppose that M(s), N(s) are not coprime with so being a zero of M(s) and of N(s)which lies in the closed right-half plane. Then so is a system zero of R(s) and we see from(8.24) that there is a complex vector U(so) :/ o such that

R(so)U(so) = 0

However since L(s) E 7-1,x, all elements of L(so) are finite. Therefore we have

L(so)R(so)U(so) = 0

and the Bezout identity (8.36) is not satisfied. Therefore we have shown that, in theMIMO case, if the Bezout identity, (8.36) or (8.34), is satisfied then M(s),N(s) arecoprime.

We can show, in a similar way, that (8.35) is sufficient for M(s), N(s) to be coprime.Proof (only if) This will be done in the next section by using certain state models

we will determine for the transfer functions 'M(s), N(s), M(s), N(s), X(s), Y(s), X(s),Y(s). 0

Before going on to the development of state models for the various systems involved incoprime factorization, we need to consider the possibility of there being more than onecoprime factorization for a given system. This is done now as follows.

8.3.3 Relating coprime factorizationsThe coprime factors in a coprime factorization of a given system are not unique.Moreover the factors in two different coprime factorizations of the same transfer functionare related by a unit, (Section 8.2.4). This fact is shown in the following theorem.

Theorem 8.2 Suppose G(s) has right and left coprime factorizations

G = N1(s)MI ' (s)

= M1' (s)Nj (s)

Then G(s) has right and left coprime factorizations given by

G = N2(s)M2 1 (S)

= M2 '(s)N2(s)

if and only if there exists units R(s), L(s) E U such that

N2(s) = N1(s)R(s) N2(s) = L(s)Ni(s)

M2(s) = M1(s)R(s) M2(s) = L(s)MI(s)(8.37)

Proof (if) If R(s) E 7-I then we see from (8.37) that N2(s), M2(s) E H. since thecomposition (series connection) of stable systems is stable. In addition we see from (8.37)that

N2(s)Mz' (s) = Ni (s)R(s)R-' (s)Mi' (s) = G(s) (8.38)

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206 System Algebra

However since N1(s), M1(s) are coprime the Bezout identity (8.34) is satisfied. There-fore pre and post-multiplying the Bezout identity by R-1 (s) and R(s), and using (8.37) weobtain

X2(s)M2(s) + Yi(s)N,(s) = I

where

(8.39)

X2(s) = R '(s)X1(s) Y2(s) = R-1(s)Y2(s)

Finally if R-1 (s) E 7-1, then we have X2 (s), Y2 (s) E Nx and M2 (s) and N2 (s) arecoprime from Theorem 8.1. Therefore we have shown that N2(s)M2_1(s) is an additionalright coprime factorization of G(s) if R(s) E U,,.

We can show, in a similar fashion, that if L(s) E Ux then M21(s)N2(s) is a second leftcoprime factorization of G(s).

Proof (only if) Consideration of (8.38) shows that the invertability of R(s) is necessary.Therefore suppose R(s) is invertible but R(s) U. Then R(s) 7-L and/or R-1 (s) $ 7 Lx.

Suppose R(s) 7(x. Then since N1(s)Mi 1(s) is a coprime factorization of G(s), wesee from (8.37) that either N2(s), M2(s) or N2(s), M2(s) E 7-"'. Suppose N2(s),M2 (S) Then N2(s)M21(s) is not a coprime factorization of G(s). Alternatively,suppose N2(s), M2(s) E 7-1 . Then we see from (8.37) that all unstable poles of R(s) arecancelled by zeros of N1 (s) and M, (s). This implies that N1 (s) and M1 (s) share closedright-half plane zeros and therefore, contrary to assumption, N1(s)Mi (s) is not acoprime factorization of G(s). Therefore R(s) E 7-x

Alternatively, if R-1(s) 0 71(,,, then R(s) has closed right-half plane zeros. Thus we seefrom (8.37) that N2(s) and WS) share closed right-half plane zeros and are therefore notcoprime. Thus N2(s)M21(s) is not a coprime factorization of G(s).

The foregoing shows that N2(s)M21(s) is an additional right coprime factorizations ofG(s) only if R(s) in (8.37) is a unit.

In the same way we can show that MZ1(s)N2(s) is a second left coprime factorizationof G(s) only if L(s) in (8.37) is a unit.

In summary, we have seen in this section that M(s) is always proper since it is alwaysinvertible. Alternatively, N(s) is proper when G(s) is proper and strictly proper when G(s)is strictly proper. We will see in the next section that Y(s) is always strictly proper so thatY(oo) _ 0. Therefore the product Y(s)N(s) is always strictly proper so thatY(oc)N(oc) _ 0 no matter whether G(s) is proper or strictly proper. Thus we see thata necessary condition for satisfying the Bezout identity, (8.34), is X(cc) = M-1 (0c) whichimplies that the state models for X (s) and M(s) must have D matrices which are inversesof each other. Similarly, state models fork(s) and M(s) must have D matrices which areinverses of each other.

8.4 State Models for Coprime FactorizationIn what follows we develop, from a given minimal state model for the system beingfactored, minimal state models for its coprime factors, M(s), N(s), M(s), N(s) as well asthe factors X(s), Y(s), X(s), k(s) needed to satisfy the Bezout identities, (8.34, 8.35).

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2 'State Models for Coprlme`Fi "brI ff ff

8.4.1 Right and left coprime factorsSuppose we are given a controllable and observable (minimal) state model for the systemto be factored

G(s)s A B

(8 40)C D .

Then applying state feedback

u(t) = Kx(t) + v(t) (8.41)

with K chosen to make A + BK is stable, yields the following state equations for theclosed loop system

±(t) = (A + BK)x(t) + Bv(t) (8.42)

y(t) _ (C + DK)x(t) + Dv(t) (8.43)

Now we can interpret (8.41-8.43) as two systems having transfer functions N(s), M(s)with each system having v(t) as input

Y(s) = N(s)V(s) (8.44)

U(s) = M(s)V(s) (8.45)

herew

5N(s) _ A + BK BM(s)

A+BK B(8.46)

C+DK D K I

Notice that N(s), M(s) E 7-1 since A + BK is stable. In addition notice that M(s) isinvertible since its state model has a D matrix which is an identity matrix. This allows us tosolve (8.45) for V(s) and substitute the result in (8.44) to obtain

Y(s) = N(s)M-1(s) U(s)

The following simple example provides an illustration of the relation of the foregoingstate models, (8.46) to coprime factorization.

Suppose we want to obtain a coprime factorization for

G(s) =s-1

s(s - 2)

We begin by obtaining a minimal state model for G(s), say in controller form, so that

s [A B1

G(s) = L C D J

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208 System Algebra

where

A [1 0] B= [011

C=[1 -1] D=0

Next we need to choose K so that A + BK, is stable. Suppose we do this so that A + BKhas a multiple eigenvalue at -1. Then K is

K=[-4 -11and A + BK, and C + DK in the state models for the factors, (8.46) are

A+BK= [ 1201] C+DK=C=[1 -1]

Finally, we get the transfer functions for the coprime factors from the state models,(8.46), as

N(s) = (C+DK)[sI - (A+BK)]-1B=s-1

(s+ 1)2

M(s) = K[sI - (A+BK)] 1B+I =sz - 2s(s+ 1)z

and we see that

N(s)M-1(s) = G(s)

The required coprimeness of N(s), M(s) in this example is seen by inspection.We can show that the factors, M(s), N(s), (8.46), are coprime in general by using the

definition of a system zero given in Section 8.2.3 of this chapter. This is done as follows.We begin by noting that the state models for the factors, (8.46), have the same system

matrix, A + BK, and the same input matrix, B. Therefore the states of these systems areequal for all time if their inputs are equal and if their initial states are equal.

Recall from Section 8.2.3 that the system zeros are defined in terms of the zero stateresponse. Therefore in order to investigate the coprimeness of the factors specified by(8.46), we take the initial states of these state models null, i.e., equal. Moreover, we notefrom, (8.44, 8.45) that the inputs to these state models are the same being equal to v(t).Therefore the states of the state models for the factors M(s) and N(s), (8.46), are assumedequal in the following investigation of the system zeros of the factors.

Suppose so is a common system zero of the factors M(s) and N(s) given by (8.46). Thenwe see from (8.20) that the following conditions must be satisfied

[so, - A - BK]X(so) - BU(so) = 0 (8.47)

[C + DK]X(so) + DU(so) = 0 (8.49)

KX(so) + U(so) = 0 (8.49)

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State Models for Coprime Factorization 209

Thus substituting for U(s0) from (8.49) in (8.47, 8.48) yields

[soIA]X(so)=oCX(so) - Q

which implies that the pair (A, C) is unobservable. This contradicts our assumption thatthe state model for G(s) is minimal. Thus so cannot be both a zero of N(s) and a zero ofM(s) which is sufficient for N(s), M(s) to be coprime.

Notice, from the foregoing demonstration of the coprimeness of M(s) and N(s),(8.46), that the detectability of the state model used for G(s), (8.40), is both necessary andsufficient for the coprimeness of these factors, i.e., for there to be no common closed right-half plane system zeros for M(s), N(s).

In the next section we will provide additional evidence for the coprimeness of the pairM(s), N(s), (8.46), by finding state models for X(s), Y(s) to satisfy the Bezout identity,(8.34). Before doing this we determine state models for some left coprime factors of G(s)by utilizing the fact that the right coprime factorization of GT (s), i.e.,

GT (s) = 1VT(s)M-T (s) (8.50)

becomes the desired left coprime factorization of G(s) after transposition, i.e.,

G(s) = M-' (s)N(s)

We do this as follows.From (8.40) we have

GT(s)[41T CT 1

BT DT J

Then from (8.46) we see that the right coprime factors of GT (s) are given as

BT + DT LTAT + CT LT DTCTJMT (s)

s AT + CT LT

LT

CT

I

where LT replaces K in (8.46). Notice that as K was chosen to make A + BK stable in(8.46), so must LT be chosen now to make AT + CT LT stable. Therefore after doing therequired transposition we obtain the left coprime factors for G(s) as

N(s) S [A+LC B+LDC D J

M(s)[A+LC L

C I

Notice that we have N(s), M(s) E ?-L) with M(s) being invertible.

(8.51)

8.4.2 Solutions to the Bezout identitiesIn this section we will see that there is an important connection between the coprimefactors for a controller in an observer based feedback control system (Section 5.2), and

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c I u sysrem Algebra

the pairs {X(s), Y(s)} and {X(s), Y(s)} in the Bezout identities, (8.34, 8.35). Thisconnection is used here to establish state models for these pairs. We will see later that thisconnection can be used to characterize all controllers which impart stability to a closedloop control system.

In order to obtain a coprime factorization for the controller, we need to recall that therelevant equations defining an observer based controller for the plant, (8.40) are

z (t) = Ai(t) + Bu(t) + y(t)]

y(t) = Ci(t) + Du(t)

u(t) = Ki(t)

Then we see from these equations that the state equations for the controller are

k (t) = Ai(t) + By(t)

u(t) = ci(t) (8.52)

where

A=A+BK+LC+LDK B= -L C =K

so that the controller transfer function H(s) is given by

A BH(s) 0

Notice that the controller is strictly proper. Notice also that since we are assuming thatthe plant has m inputs and p outputs, the controller hasp inputs and m outputs. ThereforeH(s) is an m x p matrix.

Next using (8.46, 8.51) we see that the right and left coprime factorizations of thecontroller are

H(s) = NH (s)MH' (s) = MHl (s)NH(s)

where

NH (s)

NH (s)

rA+BK B S A+BK BlLC O MH(s) - K I J

I

A+LC B s A+LC LC O1

MH(s)C

with k and L being chosen so that the factors are stable. One way of doing this is tochoose

K=C+DK L=-(B+LD)

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State Models for Coprime Factorization 211

where K, L were used in the previous section to obtain state models for the left and rightcoprime factorization of G(s), (8.46, 8.51). Notice that this choice provides the requiredstability since the A matrices for the state models of the right and left coprime factors ofH(s) are given as

A+BK=A+BK A+LC=A+LC

Therefore using this choice for K and L we see that the coprime factor state models forthe controller become

NH(S) , f K+LC -0L1 MH(s) .`. [K+LC I (B+LD)1(8.54)

We show now that the Bezout identities, (8.34, 8.35), can be satisfied by making use ofthe foregoing factorizations of the observer based controller for G(s).

Theorem 8.3 If a system has transfer function G(s) with minimal state model specifiedas (8.40) and coprime factorizations specified as (8.46, 8.51), then the Bezout identities,(8.34, 8.35), are satisfied by

X(s) = MH(s) Y(s) = -NH(S) (8.55)

X(s) = MH(S) Y(s) = -NH(s) (8.56)

where MH(S), NH(S), MH(s), and NH(s) are given by (8.53, 8.54)Proof We need to establish that

MH(s)M(s) - NH(s)N(s) = I (8.57)

We can do this by recalling the state model for systems connected in series, (8.9) andusing the state models for MH(s), M(s), NH(s), N(s) (8.54, 8.46) to obtain

rA+LC -(B+LD)K (B+LD)MH(s)M(s) S IL 0 A+ BK J IL B (8.58)

[ K K ] I

A + LC -L(C + DK) -LDNH(s)N(S) S [ 0 A + BK ] [ B ] (8.59)

[ K 0 ] 0

Then changing coordinates for the state representation of NH(s)N(s) by using the

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212 System Algebra

coordinate transformation matrix T

yields

[A+LC -(B+LD)K] [-(B+LD)11NH(s)N(s) 0 A + BK B

L[ K K ] 0

Finally comparing the state models given by (8.60) and by (8.58) we see that

MH(s)M(s) = NH(s)N(s) + I

(8.60)

which shows that (8.57) is satisfied.We can show that (8.56, 8.51, 8.53) satisfy the Bezout identity, (8.35) by proceeding in

a similar fashion. M

8.4.3 Doubly-coprime factorizationThe foregoing right and left coprime factorizations of G(s), (8.46, 8.51) constitute what isreferred to as a doubly-coprime factorization. This type of factorization is defined fromthe following observations.

First recall, in general, that any right and left coprime factorizations of G(s) and ofH(s) satisfy

G(s) = N(s)M-1(s) = M-1(s)1V(s)

H(s) = NH(S)MH1(S) = M11 1(S)NH(S)

Therefore it follows that

M(s)N(s) - N(s)M(s) = 0 (8.61)

MH(s)NH(s) - NH(S)MH(s) = 0 (8.62)

Next recall, from Theorem 8.3, that if we use the particular coprime factorizations forG(s) and H(s) given by (8.46, 8.51, 8.53, 8.54) then we have the following Bezout identities

MH(s)M(s) - NH(s)N(s) = 1 (8.63)

M(s)MH(s) - N(s)NH(s) = I (8.64)

Finally, notice that (8.61-8.64) can be written as the single matrix equation

MH(s) -NH(S) M(s) NH(S) I 0]I

(8.65)]-N(s) M(s) N (s) MH(s) 0 I

Now we use this relation to define a doubly-coprime factorization as follows.

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Stabilizing Controllers 213

Suppose a system with transfer function G(s) has left and right coprime factorizations

G(s) = M-' (s)N(s)

= N(s)M-1 (s)

Then these factorizations taken together constitute a doubly-coprime factorization ofG(s) if there exists a set of transfer functions, {MH(s), NH(s), MH(s), NH(s)} E 7-1,x,which satisfies (8.65).

In the next section, we use the foregoing relation between the doubly-coprimefactorization of a plant and the stabilizing controller for the plant to provide a meansof characterizing all controllers which stabilize a given plant, i.e., all controllers whichrender the feedback control system internally stable.

8.5 Stabilizing ControllersRecall, from the discussion in Section 8.3.1, that achieving stability through unstablepole-zero cancellations is not robust. Therefore we cannot rely, in practice, on unstablepole-zero cancellations between the mathematical models for the plant and controller tolead to the implementation of a stable feedback control system. To avoid this pitfall weneed a controller with model H(s) which when used with a plant having model G(s)makes the state model for the closed loop system internally stable. Controller models,H(s), having this property are said to stabilize the plant model, G(s), or, alternatively, aresaid to be stabilizing.

Alternatively, the internal stability of the state model for the closed loop system isequivalent to the input-output stability from {ui (t), u2(t)} as input to {y, (t), y2(t)} asoutput for the interconnection shown in Figure 8.3. Satisfaction of this condition impliesthat O(s) E 7-1 where

Y1(s) ll - p(s) U, (S)

l (8.66)[ Y2 (S) J

E)[ U2 (S) J

We will show that O(s) E 7-L is equivalent to W(s) E H. where

El (s) Ui (s)

[ E2(S) 1W(S) [ U2(s) I

(8.67)

In order to establish conditions on G(s) and H(s) which insure that W(s) E 7-L. weneed to determine relations between the blocks of W(s) and G(s), H(s). Before doing thisconsider the following fundamental aspect of feedback control systems.

Recall that transfer functions of physical processes must be bounded at infinity.Therefore to be physically meaningful G(s), H(s) and W(s) must be either proper orstrictly proper. Assuming that G(s) and H(s) are proper or strictly proper, the feedbackcontrol system is said to be well-posed if W (s) is proper or strictly proper. We will see thatW(s) is proper or strictly proper if and only if

[I - G(oc)H(oc)]-' exists

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214 System Algebra

G(s)

H(s)

Ydt)

K

Ti

Figure 8.3 Setup for Plant Stabilization Criterion

U2(t)

Notice that if either G(oc) or H(oc) is null, i.e., strictly proper, this condition issatisfied and the feedback control system is well-posed. Recall that H(s) is strictly proper,independent of whether G(s) is proper or strictly proper, in the case of an observer basedcontroller, (8.52). Therefore observer based control systems are always well-posed.

We proceed now to establish expressions for the partitions of W(s) in terms of G(s)and H(s).

8.5.1 Relating W(s) to G(s), H(s)We begin by noting from Figure 8.3 that

or

where

El (s) = U1 (s) + H(s)E2 (s) (8.68)

E2(s) = U2 (s) + G(s)E1(s) (8.69)

U(s) = W-1(s)E(s)

W_ (S)I -H(s)

-G(s) I

Recalling that G(s), H(s) are p x m and m x p respectively we see that W-1(s) andW (s) are each (m + p) x (m + p). Now the determination of W (s), (8.67), in terms ofG(s), H(s) proceeds as follows.

First substituting (8.69) in (8.68) and solving for El (s) as well as substituting (8.68) in(8.69) and solving for E2 (s) gives

El (s) = [I - H(s)G(s)]-1 U, (s) + [I - H(s)G(s)]-1 H(s) U2 (s) (8.70)

E2(s) = [I - G(s)H(s)]-1G(s)U1(s) + [I -G(s)H(s)]-1

U2(s) (8.71)

Alternatively, substituting for El (s) from (8.70) in (8.69) gives

E2(s) = G(s)[I - H(s)G(s)]-1

U1 (s) + [I + G(s)[I - H(s)G(s)]-1H(s)] U2(s) (8.72)

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Stabilizing Controllers 215

Then comparing (8.71, 8.72) yields the following equalities

[I - G(s)H(s)]-'G(s) = G(s)[I - H(s)G(s)]--' (8.73)

[I - G(s)H(s)]-1 = I + G(s)[I - H(s)G(s)]-'H(s) (8.74)

Alternatively, by substituting for E2(s) from (8.71) in (8.68) and comparing theresulting expression for Ei (s) with the expression for E1(s) given by (8.70) leads toequations (8.73, 8.74) with G(s) and H(s) interchanged.

The foregoing results, which can also be obtained using the matrix inversion lemma,(Appendix), enable W(s), (8.67), to be expressed in terms of G(s), H(s) in the followingways

W(s) =W1 (s) W2 (s)

W3 (S) W4 (S)

[I - H(s)G(s)]-1 [I - H(s)G(s)]-'H(s)G(s)[I - H(s)G(s)]-' 1 + G(s)[I - H(s)G(s)]-1 H(s) j

I + H(s)[I - G(s)H(s)]-1G(s) H(s)[I - G(s)H(s)]-1

(8.75)

[I - G(s)H(s)]-'G(s) [I - G(s)H(s)]-1

J

(8.76)

Thus we see from the dependency of the blocks of W (s) on [I - G(s)H(s)] -1 that W (s)is proper or strictly proper if and only if [I - G(s)H(s)]-' is proper or strictly properwhich requires that I - DGDH be invertible where G(oc) = DG and H(oc) = DH.

Finally, notice from (8.67, 8.75, 8.76) that we can write O(s), (8.66), as

O(s)W3(S) W4(S) - 1 ]

W1 (S) - I W2(s)

Therefore we have O(s) E 7-l, is equivalent to W(s) E 7-l

8.5.2 A criterion for stabilizing controllersWe showed, in the previous section, that W(s) E Na is equivalent to the correspondingclosed loop system being internally stable. In this section we give a more useful criterionfor deciding if a controller is stabilizing. We will use this criterion in the next section todevelop a characterization of all controllers which are capable of stabilizing a given plant.The criterion to be developed here is stated in the following theorem.

Theorem 8.4 A controller having transfer function H(s) with right coprime factoriza-tion H(s) = NH(s)MH' (s) stabilizes a plant having transfer function G(s) with leftcoprime factorization G(s) = (s)NG(s) if and only if

A(s) E U (8.77)

where

O(s) = MG(s)MH(s) - NG(s)NH(S)

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216 System Algebra

Proof Recall that .(s) E U, is equivalent to A(s), 0-' (s) E 7{x. Since 0(s) isdefined as the sum and product of transfer functions in R., we have 0(s) E 7-{x,irrespective of the internal stability of the closed loop system. Therefore we need toshow that 0-' (s) E 7t is necessary and sufficient for W(s) E 7-k.

In order to show that 0-' (s) E N . is sufficient for W(s) E 7Ix we assume0-' (s) E 71 and proceed as follows.

From the given coprime factorization of the plant and controller we have

G(s)H(s) = (s)NG(s)NH(s)MH' (s)

or

MG(s)G(s)H(s)MH(s) = NG(s)NH(s)

Therefore 0(s), (8.77), can be written as

L(s) = MG(s)[I - G(s)H(s)]MH(s)

This allows us to express [I - G(s)H(s)]-'as

[I - G(s)H(s)]-i = MH(s)0-'

(s)MG(s)

Then we can use this expression to rewrite W(s), (8.75, 8.76) as follows

_ I + NH(s)0_' (s)NG(s)W (S)

MH(s)0 i(s)NG(s)

NH(s)0-'

(s)MG(s)

MH (s)A-'

(s) MG (s)(8.78)

Therefore, assuming that 0-' (s) E 7-I we see that W (s) E 7-L since each element ofW (s) consists of products and sums of transfer functions in 7-(0. This shows that (8.77) issufficient for the controller to be stabilizing. In order to show that A-1(s) E 'H,,, isnecessary for W(s) E 'H,,,, we assume W(s) E 7-L and proceed as follows.

Since H(s) = NH(s)MH' (s) is a coprime factorization, the following Bezout identity issatisfied

XH(s)MH(s) + YH(s)NH(s) = I

for some XH(s), YH(S) E 7-L.. Then post-multiplying this Bezout identity by 0-' (s)MG(s) and by 0-' (s)NG(s) gives the following two equations

XH(s)MH(s)0-1(s)MG(s) + YH(s)NH(s)0-'(s)MG(s) = QI(s)

XH(s)MH(s)0-'(s)NG(s) + YH(s)NH(s)0 '(s)NG(s) = Q2(s)

where

Q1 (S) = 0 (s)MG(s)

Q2(S) = 0-'(s)9G(s)

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Stabilizing Controllers 217

However we can rewrite these equations using W(s), (8.78), as

XH(s)W4(s) + YH(s)W2(s) = Q(s) (8.79)

XH(s)W3(s) + YH(s)[W1(s) - I] = Q2 (S) (8.80)

Therefore, since all terms on the left sides of (8.79, 8.80) belong to 7-lx, we have

Q1(s), Q2 (S) E R. (8.81)

However the factors of an 7i c, function need not be in H,,,. Therefore we are not able touse (8.81) to conclude that 0 1(s) E

Since G(s) = MG- 1(s)NG (s) is a coprime factorization, the following Bezout identity issatisfied

MG(s)XG(s) + NG(s) YG(s) = I

for some XG(s), YG(s) E 7-x. Then pre-multiplying this equation by 0-1 (s) yields

Q1(s)XG(s) + Q2(s)YG(s) = 0 (s)

and we see that 0-1 (s) E R., since all terms on the left side of this equation belong toThis shows that (8.77) is necessary for the controller to be stabilizing.

By interchanging the role of the plant and controller in the foregoing theorem, we seethat a left coprime factored controller, H(s) = MH1(s)NH(s), stabilizes a right coprimefactored plant, G(s) = NG(s)MG1(s), if and only if

0(s) E U", (8.82)

where

0(s) = MH(s)MG(s) - NH(s)NG(s)

In summary, we see that we can use either (8.77) or (8.82) to determine if a givencontroller stabilizes a given plant. We can use this fact together with a doubly-coprimefactorization of the plant to provide a parametrization of all the plant's stabilizingcontrollers.

8.5.3 Youla parametrization of stabilizing controllersWe have just shown that if there are coprime factorizations of the transfer functions forthe plant and controller which makes 0(s), (8.77), or 0(s), (8.82), a unit, i.e., 0(s) E U,,,;,or 0(s) E U,,,,, then the corresponding closed loop system is internally stable and thecontroller is said to be stabilizing. However, if the coprime factorizations of the plant andcontroller happen to also make both 0(s) and 0(s) identity matrices, then not only is thecontroller stabilizing but, in addition, the left and right coprime factorizations of theplant constitute a doubly-coprime factorization, (Section 8.4.3), of the plant.

The foregoing observation is used now to show that we can construct a set ofstabilizing controllers by using a doubly-coprime factorization of the plant. The elementsof this set are generated by varying an m x p matrix, Q(s), over 7-I .

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21 8 System Algebra

Theorem 8.5 Given a doubly-coprime factorization of a p x m plant transfer functionmatrix

G(s) = MG'(s)Nc(s) = NG(s)MG1(s) (8.83)

where

MH(s) -NH(s)f

MG (Y) NH(S)l f

I d

L J- L ]-NG(s) MG (s) NG(s) MH(s) 0 I

then either of the following forms for the m x p controller transfer function matrixstabilize the given plant

H(s) = N,(s)M- 1(s) (8.85)

H(s) = MC' (s)Nc(s) (8.86)

for all m x p matrices Q(s) satisfying Q(s) E 7-L00where

Nc(s) = NH (S) + MG(s)Q(s)

M, (s) = MH(s) + NG(s)Q(s) (8.87)

NJ(s) = NH(S) + Q(s)MG(s)

Me.(s) = MH(s) + Q(s)NG(s)

Proof Recall that for (8.85) to be a coprime factorization of H(s), we must have

(i) N.(s), Me(s) E H.(ii) N(.(s), Me(s) coprime

Since Nc(s), Mc(s), (8.87), are each dependent on sums and products of matrices in7-lx, condition (i) is satisfied. In order to show that condition (ii) is satisfied, recall thatNe(s), M,.(s) are coprime if X(s), Y(s) E l,, satisfy the Bezout identity

X(s)M,(s) + Y(s)N,(s) = I (8.88)

Now substituting for M,. (s), Ne.(s) from (8.87) and choosing

X(s) = MG(s) Y(s) = -NG(s) (8.89)

yields the following equation

X(s)M,(s) + Y(s)N,(s) = e1 + ®2

where

E)I = "1 G(s)MH(S) - NG(s)NH(s)

02 = `MG(s)NG(s) - NG(s)MG(s)I Q(s)

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Lossless Systems and Related Ideas 219

Then we see that e1 = I since the plant transfer function factorization is doubly-coprime, (8.84). Moreover we have e2 = 0 from (8.83). Therefore (8.88) is satisfied andcondition (ii) is satisfied. Thus (8.85) is indeed a coprime factorization of H(s).

Finally, in order to show that the controllers, (8.85), are stabilizing, notice that we havejust shown that

MG (s)(s)Me(s) - NG(s)N,.(s) = I

Therefore, referring to Theorem 8.4, we have 0(s) = I E Ux so that the controllers givenby (8.85) are stabilizing.

This completes the proof that the controllers given by. (8.85). stabilize the given plant.The proof that the controllers given by (8.86) are stabilizing can be done in a similarfashion.

8.6 Lossless Systems and Related IdeasA linear time-invariant system is said to be lossless if it is stable and if it has zero stateresponse y(t) corresponding to any u(t) E G,[0. CO such that

Ily(t)112= 11146 2 (8.90)

Notice, from Sections 7.4 and 7.5 of the previous chapter. that a lossless system has L2gain equal to one. However it is important to note that this condition is necessary but notsufficient for a system to be lossless, i.e., for (8.90) to be satisfied for all u(t) E G2[0, oe).

In contrast, recall from the previous chapter that G, [0. x) is isomorphic to G2 so thatthe L2 norms in the time and frequency domains are equivalent. Therefore a system islossless if its transfer function G(s) E 7-l, maps U(s) E 7-l, to Y(s) E 7L2 such that

I Y(jW 2= 1 UU";)11, . (8.91)

However since

x11 Y(jW 2 -f U,(jw)G.(j.,)G(

27r , U(.1 )d.c

we see that (8.91) is satisfied for all U(s) E R2 if and only if

G*(jw)G(jw) = Im for all E (-x. )c) (8.92)

where G(jw) is p x m. Notice that p > m is necessary for (8.92) to be satisfied.Mathematically, we can consider lossless systems to be unitary operators. Recall, from

Chapter 7, that the Fourier transform is an isometry or unitary operator between the timedomain Hilbert space G2(-x, oc) and the frequency domain Hilbert space £2. In thepresent instance (8.90) implies that an m-input p-output tossless system is an isometry orunitary operator between the in-dimensional input Hilbert space, G2[0. cc), and p-dimensional output Hilbert space, G2[0, oc). Alternati,, ek. (8.91) implies that a losslesssystem is a unitary operator between the m-dimensional input Hardy space, 71,, and thep-dimensional output Hardy space, 7-12.

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zzv 3 ih-xwfaAn important property of lossless systems can be seen as follows. Notice that if we

connect two systems with transfer function GI (s), G2(s) E ?-I is series with GI (s) beingthe transfer function of a lossless system, then we have JIG, (s) III= 1 and from Section7.4.2 we see that

IIGI(s)G2(s)II.< JIG, (s)II,IIG2(s)IIx= IIG2(s)II. (8.93)

However (8.92) implies that the L2 norm of the output from the composite system isindependent of GI (s). Therefore the inequality (8.93) becomes an equality

IIGI(s)G2(S)11,= IIG2(S)II.

In the next chapter we will see that this type of invariance of the H,,. norm plays animportant role in the development of a solution to the H,,. control problem.

Now since the linear models of the time-invariant physical processes being consideredhere, e.g., plants, have transfer function matrices whose elements are ratios of poly-nomials having real coefficients, we have

G' (jw) = GT ( jw) for all w E (-oo, oo)

and (8.92) is equivalent to

GT (-s)G(s) = Im for all complex s (8.94)

We can gain additional insight into these matters by reviewing the earlier developmentof these ideas in connection with electric filters.

8.6.1 All pass filtersRecall that a stable time-invariant linear SISO system having transfer function G(s) has asteady-state output given as

when its input is given by

where

y(t) = A. sin(wt + 60)

u(t) = A; sin(wt)

A,, = A, G- (jw)G(jw) =1G(jwI A,

e0=tanI [Im[G(jw)] 1

Re[G(jw)]L J

This system is referred to as an all pass filter if

I G(jwl = 1 for all w c (-oo, oc) (8.95)

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Lossless Systems and Related Ideas 221

or ifG(-s)G(s) = 1 for all complex s (8.96)

so that the amplitude A. of the steady state output equals the amplitude Ai of the input forall frequencies. Thus comparing (8.95, 8.96) with (8.92. 8.94) we see that all pass filters arelossless systems. Moreover if G(s) is the transfer function of an all pass filter, then we seefrom (8.96) that G(s) is proper, with G(O) = 1, and has poles which are mirror images ofits zeros across the imaginary axis, i.e., ifpi is a pole of G(s) then pi must be a zero of G(s).For example, the first order system having transfer function

G(s) = {..,

- 1)

1)

is an all pass filterAn all pass transfer function is a generalization of the concept of an all pass filter. This

generalization is done by relaxing the requirement of stability. Thus G(s) is said to be anall pass transfer function if (8.96) is satisfied. Notice that the mirror imaging of poles andzeros implied by (8.96) means that when G(s) is an all pass transfer function it has noimaginary axis poles, i.e., G(s) E G

8.6.2 Inner transfer functions and adjoint systemsIn order to distinguish between stable and unstable all pass transfer functions, stable allpass transfer functions are referred to as being inner. Thus a transfer function is innerwhen it is the model for a lossless system.

In the case of a transfer function matrix the following generalization is made. IfG(s) E 7-i is p x m, then G(s) is said to be inner if (8.97) is satisfied or co-inner if (8.98) issatisfied where

G (s)G(s) =I.. if p > in (8.97)

G(s)G (s) = I,, if p < in (8.98)

for all complex s with

G -(s) = G z(-s)

Notice that p > m (p < m) is necessary for G(s) to be inner (co-inner). Also notice thatif the transfer function is square and inner then G-1(s) = G (s) E H-. Moreover, sincedimensionally square systems have the property that the poles of G-1 (s) are the zeros ofG(s), (Section 8.2.4), we see that in this case that all system zeros of G(s) lie "in" the openright half plane, i.e., lie in the region of the complex plane where G(s) is analytic. This isthe origin of the terminology "inner". However in the nonsquare case this concentrationof zeros in the open right half plane is not necessary for a transfer function to be inner,e.g.,

rs± 11

G(s)=

1

ILs-t- r2 1 J

is inner without having any zeros in the right half plane.

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zzz System Algebra

The system whose transfer function is G (s) is referred to as the adjoint system(relative to a system having transfer function G(s)). Notice that the state model for G (s)is related to the state model for G(s) as

G(s) s

since

I

A BlC D J I

G _ ( s ) BTA T DTT (8.99)

GT (-S) = BT {- [SI - (-AT)] 1 }CT + DT

Properties which the parameters of a minimal state model must satisfy in order for asystem to have an inner transfer function are given in the following theorem.

Theorem 8.6 G(s) E Nx is inner if it has a minimal state model

G(s)s A B

C D

with parameters that satisfy the following conditions:

D TD = I

BTWo+DTC=0

ATWO+W0A+CTC=0

Proof From the state model for G (s), (8.99), and the rule for composing two systems,(8.9), we obtain

[-A T -CTCI r_CTDG (s)G(s) 0 A L B

[BT DTC] DTD

Next transforming coordinates using the coordinate transformation matrix

T=

gives

G (s) G(s) s

II0

XI

-ATA

A2

0 J

L [ C] C2 ]

LB2J

DTDI

where A, = -AT X - XA- CTC and

I] DB-XB]B11 _ [-CCT [_CT

Ji

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Summary 223

Finally, applying the conditions given in the theorem yields

X= W, A2=0 Bi=0 C,=0 DTD=I

and the state model for G (s)G(s) becomes

G (s)G(s)s

[ C D ]

where

Thus we have

-AT t r «]0 A B2

C= [ C1 0] D I

G (s)G(s) = C(sI - A)-1B - D

=I

and the theorem is shown.In developing a solution to the H. state feedback control problem in the next chapter,

we will extend the foregoing characterization of state models so that the resulting systemis J-lossless with J-inner transfer function.

8.7 SummaryIn this chapter we have established state models for certain interconnections andoperations with systems. This included the ideas of system inversion, system zeros,system coprimeness and lossless systems. We will see in the next two chapters that theoperation of coprime factorization coupled with the requirements on a system's statemodel for that system to be lossless, plays an important role in developing a solution tothe H, feedback control problem.

8.8 Notes and ReferencesThe treatise by Vidyasagar [44], on the consequent ramifications of the coprimefactorization approach to feedback control places coprime factorization in the mathe-matical setting of algebra known as ring theory. The term "unit" refers to elements in aring that have an inverse in the ring. The development of state models for the coprimefactors used here is treated in [14].

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9HO, State Feedback andEstimation

9.1 IntroductionIn the previous two chapters we developed ideas for characterizing signals and systems. Inthe next two chapters we use these ideas to develop the design equations for a controllerwhich is required to

(i) stabilize the closed-loop control system and(ii) constrain the L norm of the closed-loop transfer function from disturbance input,

uI (t), to desired output, y, (t), to be no greater than -,. a specified positive scalar.

Figure 9.1 gives the setup for this problem. Before going on to consider this setup inmore detail, we make the following more general observations on this control problem.

First, we should recall that to avoid having a system's input-output stabilitydependent on the physically unrealistic requirement of pole-zero cancellation, we haveused the term "stability" to refer to a system's internal stability. Thus when requirement(i) is satisfied so that the closed-loop system is internally stable, the L , norm mentionedin requirement (ii) is an H norm. Therefore controllers satisfying (i, ii) are said to solvethe Hx control problem.

Alternatively, recall from Chapter 7, that the H,, system norm is induced by the H2

signal norm in the frequency domain or the L, signal norm in the time domain. Thereforewhen requirement (i) is satisfied, (ii) can be interpreted as the requirement that the closed-loop control system's L2 gain from ui (t) to yi (t) be no greater than -y. Thus the Hxcontrol problem is sometimes referred to as the L- gain control problem.

In addition, notice that satisfaction of requirement (ii) does not imply that requirement(i) is satisfied since the L,,. norm is defined for both stable and unstable systems providedonly that the system's transfer function have no imaginary axis poles. This differs withboth the quadratic and LQG control problems since the performance indices, JQc andJcc; are infinite when the closed-loop system is unstable.

Concerning the setup for the Hx control problem. it has become customary to replacethe actual plant in the feedback control configuration, Figure 9.1, by a composite systemreferred to as a generalized plant. The generalized plant consists of the plant to becontrolled plus any other systems which the control system designer may want to connect

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226 H State Feedback and Estimation

v,(t)

GENERALIZEDPLANT

CONTROLLER

y2(t)

Figure 9.1 Setup for H. Control Problem

ul (t) = disturbance input y1 (t) = desired outputu2(t) = controlled input y2(t) = measured output

where u1 (t), U2 (t), yi (t),Y2 (t) have dimensions m1 i m2,p1 ,p2 respectively.

in cascade with the plant (referred to as weights or filters). For instance, the filters can beused to take into account any known spectral characteristics of the disturbance signals,e.g., narrow band noise. Then when the feedback control system is implemented, thecontroller determined by applying the control algorithm to the generalized plant is usedwith the actual plant.

In what follows the signals {ui(t), yi(t) : i = 1, 2} shown in Figure 9.1 are assumed tobe the various inputs and outputs associated with the generalized plant which has real-rational transfer function G(s) with given real-parameter state model

A

G(s)

ICL CZ]J

[B1 B2]

[D11 D121

LD21_ D22

As in the LQG and quadratic control problems, a solution to the H. output feedbackcontrol problem takes the form of an observer-based controller and involves thestabilizing solutions to two algebraic Riccati equations. Thus in order to develop asolution to the H,,,, output feedback control problem we need first to consider two simplerproblems namely, the H, state feedback control problem, and the Hx state estimationproblem. These basic Hx problems are taken up in this chapter with the H,, outputfeedback control problem being addressed in the next chapter.

As we progress through the development of solutions to these problems, we will seethat the generalized plant's D matrix plays a pivotal role. However, only certain parts ofthe D matrix play a role in each problem. For instance, the existence of a solution to theHx state feedback control problem requires D12 to have independent columns and onlythe first pl rows of D, denoted by D1, are involved in the solution to this problem where

D1 = [D11 D12]

However, the existence of a solution to the H,,, state estimation problem requires D21to have independent rows and only the first m1 columns of D, denoted by D2, are involved

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H,. State Feedback Control Problem 227

in the solution to this problem where

D11D2 =

D,,

In the next chapter we will see that the last p, rows of D. denoted by D3, are involved inthe solution to the Hx output feedback control problem where

D3=[D21 D2,

9.2 HO.) State Feedback Control ProblemAssuming that the state of the generalized plant, x(t), is known we want to choose thestate feedback matrix, K2, so that setting the controlled input as

u2(t) = K2x(t) (9.2)

makes the closed-loop system internally stable and the transfer function from ul (t) toyi (t) have H,,, norm no greater than a specified positive constant ry.

Since the measured output, y2 (t), is not needed in this problem, we take the plant as thegeneralized plant, (9.1), with y2(t) missing. Thus we assume the parameters of G, (s) areknown where

Yi (s) = G1 (s)Ul (s)

'2 (s)

with

G1(s)= 1A

B IC, D1

where

B = [B, B2] D, _ `Dii Dt,

The relation between the signals and systems in the H state feedback control problemare shown in Figure 9.2. where the additional signals ul (t) and K, x(t) are going to be usedin the development of a solution, K2, to the H,;, state feedback control problem.

Now after implementing state feedback we have

Y, (s) = TT.(s)C, (s) (9.4)

where

T (s) s A + B2 K2 B,1Cl+D,2K

and we want to find K2 such that

(i) : T, (S) E R. and ii) y (9.5)

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228 H. State Feedback and Estimation

U90

K3 X(t)

K, x(t)

STATECONTROLLER

PUtNr

N

Figure 9.2 Setup for the H,,. State Feedback Control Problem

Next recall from Chapter 7 that the H , norm is defined by

1ITc(S)IIoo= Sup amax[Tc(jw)]WE(-oo,oo)

where amaxtTc(jw)] is the largest singular value of TT(jw). Now since

Jim TT(jw) = D11W-00

we see that it is not possible to satisfy condition (ii), (9.5), if

amax[Dll] > 7where

amax[D11] = max{6 } 6i E A[DllD11J

In subsequent sections we will see that the more restrictive condition

tmax[Dii] < 7

must be satisfied in order to solve the H. state feedback control problem. In addition, wewill see that this condition is also needed to solve the H. state estimation problem.

We can summarize the strategy we will employ now to develop the design equations forK2 as follows.

1. Start by assuming stabilizing feedback, i.e., assume K2 achieves condition (i), (9.5).2. Define a scalar quadratic performance index, Pry, in terms ofy, (t) and ul (t) such that

P. G 0 implies condition (ii), (9.5), is satisfied.3. Use G, (s), (9.3), to introduce a transfer function G, (s) relating Jul (t), y1 (t)} as output

to {u1(t),u2(t)} as input under the constraint of stabilizing state feedback, i.e.,u2(t) = K2x(t) such that G, (s) E 7-l 0. This enables P..1 to be rewritten as a quadraticin u, (t) and u2(t) with U1 (s) E 712 and U2(s) restricted to stabilize G1(s).

4. Convert P-, to a quadratic in {V1(s), V,(s)} E 712 where {V1(s), V2(s)} arise as aconsequence of using J-inner coprime factorization on G, (s).

We will see that the design equations for K2 result from carrying out the last step.

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H., State Feedback Control Problem 229

9.2.1 Introduction of P.Suppose K2 is chosen so that condition (i), (9.5), is satisfied. Then we see from Chapter 7that

y1(t) E .C2[0, oo) if u1(t) E G2[0, Or)

and we can interpret condition (ii), (9.5), in the time domain as a constraint on the L2 gain

I1Y1(t)II2<- 7I1u1(t)(2 (9.7)

Next define P', as

P'Y _ -7211 u1 (t) 112+I11Y1 (t) lit

Then condition given in (9.7) is equivalent to

P7 <0 ul(t)EL'10.x) (9.9)

Now since L2 [0, no) is an inner product space we can rewrite P. as

Pyl t)IIJ-pnp 11t)I>

where

-72Im CJ.mp = 0 Ip

(9.10)

However, recalling from Chapter 7 that L, [0, oc) is isomorphic to 71,, we see that P.y,(9.10), can be rewritten in the frequency domain as

U1(S)

-7 IIU1(s)II+I1Y1(S)!2

and condition (ii), (9.5), is satisfied if

P< 0 U, (s) E 7-L,

(9.11)

(9.12)

It is important to remember that the foregoing result assumes stabilizing statefeedback.

9.2.2 Introduction of G, (s)Recalling from (9.3) that YI (s) depends on U, (s) and L 2 (s ), we see that we can relate

Page 245: Linear Control Theory - The State Space Approach Frederick Walker Fairman

cav N . W M Feedbaek ai) i ftdll'/on

{ UI (s) , Yl (s) } to { Ui (s), Uz (s) } through the use of a transfer function 61(s) as

[ U, (s)1 = Gi (s) [U1(s)1

L Yi (s) JJ Uz (s) J

where

(s)=A B

C, = [0

IC, D] C,

D, _Dig D1z[Im

Now under the constraint that the state feedback is stabilizing we have

H,X (s) Y (s) E H for (i): Ui (s) E xz

(9.13)

(ii): U2(s) = K2X(s)

where A[A + B2K2] are all in the open left half plane, and X(s) is the Laplace transform ofthe closed-loop system state.

Therefore we can use (9.13) to rewrite the inner product defining P.,, (9.11), in terms ofU, (s), U,_ (s) I as

Ui (s) Ui (s)P-, = CGt (s)

Uz(s) ,

Jrymp6, (s)Uz(s)

U, (s) Ui (s)(s)

Uz(s)(9.14)[U2(S)jJ7m

whereU2(s) = K2X(s)

In what follows, it is important to remember that even though Gl (s) may possibly beunstable, the restriction that the controlled input, U2(s) = K2X(s), must be stabilizingmaintains P,., given as either (9.8) or (9.14), finite.

9.2,3 Introduction of J-inner coprime factorizationRecall, from Chapter 8, that G, (s) E G, has a right coprime factorization given by

G, (s) = N(s)M-1(s) (9.15)

where N(s) , M (s) E 7-l,, with N(s), M(s) being coprime. Now we are going to make use ofa special type of right coprime factorization in which N(s) in (9.15) satisfies

NT (-s)JJn:nN(s) = 1 cmn, (9.16)

J.mp =-lzlmi -72lnm, 00 !pi n

!m

A system having transfer function N(s) E 7-l, which satisfies (9.16) is said to be J-innerand a right coprime factorization (9.15) which has IV(s) J-inner is referred to here as a J-

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H,,. State Feedback Control Problem 231

inner coprime factorization. The utility of using this type of coprime factorization herecan be seen immediately as follows.

Let G1 (s) have a J-inner coprime factorization as denoted by (9.15, 9.16). Then usingthe fact that N(s) is J-inner we see that Pry, (9.14), can be rewritten as

\[Ul(S)] 1T 1

[U25] (S)

/= 2([M (-s)] JmnM (s))

)

where

(9.17)

U2(s) = K7X(s)

Next recall, from Chapter 8, that since M(s). N(s) E R, we can obtain {Ui(s), U2(s)}and { U1(s), Y1(s)} from V1(s), V2(s) E 7-12 as

U1(s) Y 1 s)

U2 (S)M(S) IV2(s)I

[U1(s) V1(s)

Y1(S)IN(s)IV2(s)I

Then since N(s) E R. we have

Ui(s)E R2

Y1(s) Jfor all

V, (s)

Finally, by using (9.18) in (9.17) we can rewrite P. as

V1 b'1 (s)

Pry

(s)

= , Jrymrn

V2(S) T"-(s)

x211 V1(S)112+(V2(s)

(9.21)

Notice that unlike U2(s), which is restricted to a subspace of 7-12 since it must providestabilizing state feedback, V1(s), V2 (s) are not restricted to any subspace of 7-12. We willuse this freedom in the choice of the V;s to ensure that P_ < 0.

9.2.4 Consequences of J-inner coprime factorizationSuppose we have a J-inner coprime factorized for G1 (s). (9.15, 9.16). Then to proceed weneed to consider state models for these coprime factors. Therefore, recalling (8.46) andSection 8.3.3 we see that state models for the coprime factors, M(s). N(s), (9.15), can bedetermined from the state model for 61 (s), (9.13), as

rA+BK B,- 11M(s)

I1

N(s)

K JI

A+BK B.I 1

C1 +D1K D1,

(9.18)

(9.19)

E 712 (9.20)

(9.22)

(9.23)

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232 H, State Feedback and Estimation

where K, 0 are chosen so that M(s), N(s) E l,. with N(s) satisfying (9.16); 0 is aconstant nonsingular (m1 + mz) x (m1 + mz) matrix which along with K is partitioned soas to conform with the partitioning of B and D1, viz.,

0=LA103

Oz

04K=

LK2J

We show now that the H,,, state feedback control problem is solved when K and z arechosen so that the foregoing coprime factorization is J-inner. The equations needed todetermine K and 0 are developed in subsequent sections.

We begin by using (9.18) to express V, (s), V2(s) in terms of U, (s), U2(s) as

f V, (S) l = M 1(s) U1( s)(9.24)

LVz(s)] LU2(s)I

Then using the state model for the inverse system developed in Chapter 8, we see that thestate model for the inverse of M(s), (9.22), is

A BM (s) -`

-OK 0 (9.25)

Notice that the system and input matrices for the generalized plant, G, (s), (9.3), andfor ft-1 (s), (9.24), are the same. Therefore the states of these systems are equal for alltime if their initial states are equal. The initial state of each system is the same since each isnull. This is because we are basing our development of the H. state feedback controlleron transfer functions which relate the system output to the system input under theassumption the initial state is null.

The foregoing observation together with the state model for X-1-1 (s), (9.25), allows usto write the Visas

VI (S)

) J(9.26)

Vz (s)-OKX (s) + 0 L

U2 (ss

where X(s) is the Laplace transform of the state of the state model for G, (s), (9.3).Now suppose we choose U, (s) and U2(s) as

r U (s)11U, (s) _ r K1 ]X(s)+IUz(s) LK20 J

(9.27 )

where U, (s) E Nz is an external disturbance input whose effect will be seen to beequivalent to the actual disturbance input U1 (s) E 7-I2. Then substituting (9.27) in(9.26) yields

f Vi (s)1 = f 1 1 Ui(s) 01 (S) E xz (9.28)

L V-(S')J Lo3]

We show now that P-; < 0 for U, (s) E Nz, (9.12). This is done by first showing that (9.28)

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H State Faadback Coni rl°Pr rem zra

implies that P7 < 0 for U1 (S) E N2 and then showing that this implies P. < 0 forU1(s) E N2.

Recall that

IlVt(s)112 -1 V; (jw)V;(jw')dw27r x

so that using (9.28) in P7, (9.21) can be written as

x<

Ui (jw) [-72Oi . + J A31 U1(jw)dw (9.29)P7 =27r

Therefore we see from (9.29) that

if [-72AIA1 + A3*A3j < 0

then P7 < 0 for all non-null Ul (s) E H2 (9.30)

P7 = 0 for Ul (s) = c1

However recall, from the beginning of this section, that the H, state feedback controlproblem has a solution only if D11 satisfies (9.6). We will show in Section 9.3.3 that

if Qmax[Dll} < 7

then -7201A1+A1A3 <0

Therefore (9.31, 9.30) imply that

(9.31)

if Qmax [Dl l } < 7

then P7 < 0 for all non - null Ul (s) E N2 (9.32)

P7=0 for Ul (s) = o

We need next to show that (9.32) implies (9.12). To see this notice from (9.27) that

U1(s) = Ui(s) - K1X(s) (9.33)

Now we will show in Section 9.3.3 that the state feedback. u,(t) = K2x(t) stabilizes thegeneralized plant by itself without requiring ul (t) = K1 x (t). Thus X(s) E 7-12 for anyinput Ul (s) E N, provided u2(t) = K2x(t). Therefore we see from (9.33) that U11 (s) E N2for all Ul (s) E N2. This fact together with (9.32) implies that (9.12) is satisfied, i.e.,

P7 < 0 for Ui (s) E N2 and U1 (s) K1 X (s)

P7 = 0 for Ul (s) = Kl X (s)(9.34)

as is required to satisfy requirement (ii), (9.5), for the solution of the Hx state feedbackcontrol problem.

Page 249: Linear Control Theory - The State Space Approach Frederick Walker Fairman

AIM try ek aim` RIiia0on

Notice that the disturbance input Ul (s) = K1X(s) maximizes Pry. Therefore U1 (s) _K1 X (s) is referred to as the worst disturbance input for the H,, state feedback controlsystem.

In summary, we have shown that a solution to the H,, state feedback problem isobtained when (9.6) is satisfied and when we choose K and A so that the right coprimefactorization of 61 (s), (9.13), has numerator N(s), (9.23), which is J-inner. The equationsfor K which make N(s) J-inner are the design equations for a solution to the Hx statefeedback problem. These equations are developed in the next section.

9.3 H., State Feedback ControllerIn order to obtain a J-inner coprime factorization of G1(s), (9.13), we need conditions onthe state model parameters of a given system which ensure that the system is J-inner.These conditions are given in Theorem 9.1. A similar set of conditions was given in theprevious chapter in connection with inner systems, (Theorem 8.6). Once we haveTheorem 9.1 we can apply it to the state model parameters for N(s), (9.23). In this waywe determine an equation relating A to D1 and an expression for K involving thestabilizing, solution Xx to an algebraic Riccati equation.

9.3.1 Design equations for KNotice from (9.13, 9.15) that N (s) is an (m1 + pi) x (m1 + m2) matrix. Now for N(s) to beJ-inner, the conditions given in the following theorem must be satisfied.

Theorem 9.1 A system having an (m1 + p1) x (ml +M2) transfer function N(s) is D-inner if N(s) E 7-1 and the parameters of a minimal state model for the system

N(s) AN BNCN DNJ

satisfy

(1) J,,,,, = DNTNJy pDN

(ii) 0 = BTVX + DATrJ7mpCN

(iii) 0 = ANX + XAN + CNJ7mpCN

where J?m,, Jn,n are defined as

J-"=

_^L J7mm = _ IPi mz

Proof The proof is similar to the proof of Theorem 8.7. Here we need to obtain astate model for NT (-s)J,mpN(s) and change coordinates with coordinate transformationmatrix T given by

T=L X0 1

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H. State Feedback Controller 235

The resulting state model then reveals that in order for NT (-s)J_;,,,PN(s) = Jry,,,,,, we needto impose the conditions in the theorem.

Now we want to choose 0 and K in the state model for V(s), (9.23), so that N(s) is D-inner. We do this by equating N(s) to N(s) and applying Theorem 9.1.

To begin, recall that the D matrix in a state model is unchanged by a change ofcoordinates. Therefore from N(s) = N(s) we have

DN = D10-I

Then using condition (i) in Theorem 9.1, we obtain

DJc = OTJ7mmA

where

T -y2lm, +D11D11D D I -Jc =D1JrymP

DT12D1 l

TD1A7D17D1,

(9.35)

(9.36)

Notice that Djc is square. We will show later in this section that Djc is invertible if D12 hasindependent columns and D11 satisfies (9.6). This inverse is needed to determine K.

Continuing with the development, we see from Theorem 9.1 and (9.23) that furtherconsequences of setting N(s) = N(s) are

AN = A + BK (9.37)

BN=Bz 1 (9.38)

CN=C1+DlK (9.39)

Then using (9.35, 9.38, 9.39) in condition (ii) of Theorem 9.1 we obtain

0 TBTX+O TDiJ,,»P(Cl 1-D1K) =0

which after multiplying on the left by OT becomes

(9.40)

Now we see from the definitions of Cl and D1, (9.13), that

C CT D1 (9.41)

Ci JympCI = Ci Cl (9.42)

as

Thus using (9.41, 9.36) and assuming Dj. is invertible enables us to solve (9.40) for K

K = -D,'[BT X Di C1' (9.43)

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236 H,o State Feedback and Estimation

Notice that K depends on the state model parameters for the given generalized plant,(9.3), and on X. We will show now that X is a solution to an algebraic Riccati equation.

To begin, we substitute from (9.37, 9.39) in condition (iii) of Theorem 9.1 and use(9.41, 9.42) to obtain

ATX + XA + KT BT X + XBK + KT D Cl + CI Dl K + KT D,,K + Cl C1 = 0 (9.44)

Next grouping terms involving K and KT and using (9.43) to substitute -D_K forBT X + Di C1 enables (9.44) to be rewritten as

ATX + XA - KT C1 Cl = 0 (9.45)

Finally substituting for K from (9.43) gives the equation

AT.1X + XR,o1X + Q., = 0 (9.46)

where

A.1 = A - Cl

BDJC'BT

Q.] =C (1-D1Dj!Di)C1

with

721., +D 1D11D1=[D11 D12] D!,- T1)12111

Notice that (9.46) is an algebraic Riccati equation like the GCARE (Section 6.2). Werefer here to (9.46) as the HCARE. Since A.1 - Ro1X = A + BK we have K as astabilizing state feedback matrix if the solution X to (9.46) makes A,,1 - Ra1X stable.As in the quadratic and LQG state feedback control problems, we reserve a specialnotation for these solutions which in this case is X,,. Notice that X > 0.

In addition, notice that unlike D12D12, Dj, is not sign definite, i.e., neither non-negative nor non-positive. Thus unlike R1 in the GCARE, (section 6.2.2), R1 in theHCARE is not sign definite. This makes the determination of conditions on the statemodel for the generalized plant which ensure the existence of a stabilizing solution to theHCARE more involved than in the case of the QCARE.

9.3.2 On the stability of A + B2K2We have just developed equations for the determination of K to effect a J-inner coprimefactorization of G, (s), a system which is made up from the given generalized plant. ThusA + BK is a stability matrix. However the H,,, state feedback controller, (9.2), affects onlyu2(t) through the state feedback equation u2(t) = K2x(t). In addition, recall from Section9.2.4, that condition (ii), (9.5), is satisfied provided K2 is the last m2 rows of the matrix K

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H. State Feedback Controller 237

which is needed in the J-inner coprime factorization of G1(s). Therefore it remains toestablish that condition (i), (9.5), is satisfied when K is determined in the foregoingmanner.

More specifically, assuming that the HCARE, (10.96), has a stabilizing solution XX,and that K, (9.43), is determined using X = X,,,, we need to show that A + B2K2 has all itseigenvalues in the open left half plane when K2 is the last n7, rows of K. This is done in theproof of the following theorem.

Theorem 9.2 If the pair (A, B2) is stabilizable and X, > 0 is a stabilizing solution tothe HCARE, (10.96), then A[A + ,K2] are call in the open left half plane where

K= IK1JK,

with K given by (9.43).Proof Suppose X. is a stabilizing solution to (9.46). Then A,,1 - R,,.1X,,. is stable

where

A.1 - R.1X. = A. - B1K1 (9.47)

with A2 = A + B2K2.Notice, from Chapter 2, that the stabilizabilty of (A. B,) is necessary for A2 to be

stable. In addition, recall from Chapter 3, that a pair (A. C) is detectable if we can find Lsuch that A + LC is stable. Alternatively, when (A, C) is detectable all unstableeigenvalues of A are observable, i.e., if Re[A] > 0 when At = A v. then Cv j4 0. Thereforesince A2 + B1 K1, (9.47), is stable, we have (A,, K1) detectable. Thus if K1 v = 0 whereA,v = Av then Re[A] < 0. We will use this fact later in the proof.

Next we add and subtract both XB2K2 and its transpose to and from (9.45) to obtain

AT X. + XXA2 = O (9.48)

where

O=XXB2K2+KTBZXx+KTDjK-CTC1

Now we are going to show that O < 0. Then we will see that this fact plus the fact that(A2, K1) is detectable implies that A2 is stable.

We begin by expanding KT Dj,K

T T T M1 M2, KiK [K1 KZ ] MT M K,2 4

= KT M1 K1 + KT M, K2 + KT M ; K1 -- K. M4K2

where we see from (9.46) that the M;s are given as

M1 = 21., + DT11 D11 M, = DT1D1, M4 = DT 12 D12

Then 0, (9.48), becomes

O = (X,B2 + KT M2)K2 + K. (BZXX, + M2 K1) Kr_111K1 - KT M4K2 CTCI

(9.49)

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238 H. State Feedback and Estimation

Next after rearranging (9.43) as

M1K1 +M2K2 _DJ`K LMzK1+M4K2)

we see that

B2 Xx + M2 K1 = - (M4K2 + D 12 C1) (9.50)

Therefore using (9.50) in 8, (9.49), gives

8=-(M4K2+DzCi)TK2-K2 (M4K2+D12C1)+K1M1KI+KKM4K2-CTC1

=-KZM4K2-C1D12K2-KZD12C1+K1M1K1-C1C1

_ -(Cl + D12K2)T (C1 + D12K2) + KT M1K1 (9.51)

Finally, recall from (9.6) that we require all the eigenvalues ofD11ID11 to be less that ryesince this is a necessary condition for the solution of the H,,. state feedback controlproblem. We can readily show that under this condition we have M1 < 0. Thus we seefrom (9.51) that 8 < 0. We can now proceed to show that A2 is stable.

Pre and post multiplying (9.48) by v* and v where A2v = Av gives

v*8v (9.52)

Now since we have just shown that 8 < 0 we have either v*8v < 0 or v*8v = 0.If v*8v < 0 it follows from the condition X,,,, > 0 that v*Xxv > 0 and therefore

Re[A] < 0

If v* 8v = 0 we see from (9.51) that

-V* (C1 +D12K2)T(C1 +D12K2)v+v*(K[M1K1)v=0 (9.53)

and since M1 < 0 we see that (9.53) requires

K1v = 0 and (Cl + D12K2)v = 0 (9.54)

Finally, recalling from the beginning of the proof that the pair (A2, K1) is detectable, wesee that if K1v = 0 then Re[A] < 0. M

To recap, we have shown that the state feedback matrix K2 which solves the H, statefeedback control problem can be determined as the last m2 rows of the matrix K, (9.43),when X = Xx the stabilizing solution to the HCARE, (9.46). A remarkable feature ofthis solution, shown in Theorem 9.2, is that while Xx makes A + BK stable we also haveA + BK2 stable. Before discussing conditions on the generalized plant which ensure theexistence of a stabilizing solution to the HCARE, we pose and solve the H... stateestimation problem.

Page 254: Linear Control Theory - The State Space Approach Frederick Walker Fairman

H,, 'Nt eedbacYD 7f9.3.3 Determination of 0Recall that in Section 9.2.4 we made use of the condition

-'Y 1 O1 + 0303 <

We will show now that this condition is satisfied when

amax[D11] < y and DT D1, is invertible

We do this by first developing an expression for 0 in terms of D11 , D12 so that (9.36) issatisfied. This is done in the following theorem.

Theorem 9.3 If

Qmax[D11] < 7 and DT D1, is invertible

then 0 can be taken as

O=

where

O = D1T2D12

with

© `/2D1zD1101;2

1 0

LJ_ _ D4

E) = eT/201/2 r = rr ,r1; 2

r = Y2Im1 - D11(Ir. - D120-1D2 )D11

M1 M2 TDJc = T = J_ JFM2 M4

where JJ,,,,,,, is given in (9.16) and

MI = -'Y2I,, + DIID11 M2 = DI1D1 M4 = D 2D12

Proof Consider the constant symmetric matrix factorization given by

M1 M21 _ I M2M4 1 M1 - M2Ma I MT

MT M4J 0 I [ 0 :'t14

IM4 1 M2

0l

I

Then identifying the blocks of DJc with the M;s, (9.55). we see from (9.56) that

(9.55)

(9.56)

--D j,.

I DI,D120 1I kr °1 1 1 0J1

(9.57)DI1 I0 I 0 al, O-1D 12

where r, O are defined in the theorem.

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240 Hc State Feedback and Estimation

Next assume for the time being that r, 6 can be factored in the manner specified in thetheorem. Then we can write

r:]=wT J7mm

where J.y,,,,,, was given in (9.16) and

Finally refering to (9.57) we have

AF F 00 011

4I

T0 =L

® D,2D11 I

so that D j can be factored as (9.55) provided ®, r have square root factors.Recall from the proof of Theorem 6.1 that a matrix has a square root factor provided it

is symmetric and non-negative. Now ® and IF are seen to be symmetric by inspection.However unlike ® it is not evident that .F is non-negative.

To show that F is non-negative notice that it can be expressed as

F = -Y2 I., - D i D,D11 (9.58)

where D, was used in Section 6.5.2

T 1 TD, = In, - D12(D12D12) D12

Next recall from Chapter 7 that the singular value ai and corresponding singularvectors v', u' for D11 are related as

D1iD11v' = QZV'

where

Therefore noting that

we see from (9.58) that

D DT ui = 2ui11 11

ti T yi = uiT u' = 1

D11v'=Qiu' D,Iu'=Qiv

v'Trv' = 72 - o u`TD.u' (9.59)

However we saw, following the proof of Theorem 6.1, that D, is a contraction so that

UiT DCu' < UiTU' = 1 (9.60)

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Hx State Feedback Controller 241

Therefore from (9.59, 9.60) we have

v'TI'v' > 72 - Q',

and

v`TFv' > 0

for all eigenvectors, v' : i = 1, 2 . . p1 ofD111 D11 provided the singular values of D11 satisfy

a1<7 i=1,2...p1

Finally, since D11D11 is symmetric, its eigenvectors, v', are complete and it follows thatif

amax[Du} < 1

then

vT l'v > 0

for any p1 dimensional vector v. Therefore F > 0 when the necessary condition (9.6) issatisfied, and we can factor F in the manner stated in the theorem. 0

The foregoing expression for 0 can now be used to prove (9.31). Thus we see fromTheorem 9.3 that

-1'201 +0303=-}Ylni+D11D11

and (9.31) follows.Finally, recall that K, (9.43), requires DJ, to be invertible. We can show that DJ, is

indeed invertible, by using the factorization of DJ, given in Theorem 9.3 to obtain anexpression DEC' as

1 1 1 -rDJc = 0 J} n

Then we see, from the lower triangular structure of A. (9.55), that

Q-1 0-041Q3Q' 1 041

1r ' - 1 D -1

E)-1D T Djjr-112 O 1-O-1Di

Dur-1DTD120 1

(9.61)

(9.62)

where F, O are given in (9.55).In the next chapter we will use the form for A given in Theorem 9.3 to develop the

solution to the H,,. output feedback control problem. Before we can do this we need todevelop a solution to the H,,. state estimation problem.

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242 H,,,, State Feedback and Estimation

9.4 HO. State Estimation ProblemRecall from the introduction to this chapter that our ultimate goal is the determination ofan H,, controller which uses the measured output to determine the controlled input to thegeneralized plant so that the resulting closed loop system is internally stable and hastransfer function from the disturbance input to the desired output, T0(s), with H,,. normno greater than 7.

In this section we treat the problem of determining an estimator or observer for thegeneralized plant state, i(t), from knowledge of the measured output, y2(t), andcontrolled input, u2(t). We need to do this so that we can use this state estimator inconjunction with the H,, state feedback controller determined in the previous twosections to obtain the desired H, controller. Notice that since the state of the generalizedplant depends on the disturbance input u1 (t), which is unknown, it is not possible todesign an observer to give an asymptotic estimate of the generalized plant state. Insteadwe design the observer so that the L2 gain from the disturbance input to desired outputestimation error, Y, (t), is no greater than 7, i.e., so that

IIY1(t)II2< 7IIu1(t)112 (9.63)

where

Y1 (t) =Yi(t) - Yi(t) (9.64)

and j1 (t) is the estimate of the desired output,

91(t) = C1X(t) +D12u2(t) (9.65)

Alternatively, we must determine the observer so that the transfer function TT(s) fromu1 (t) to Y, (t) satisfies

(i) Te(s) E ?-lx and (ii) : IITe(s)II,c< 7 (9.66)

As in the quadratic and LQG state estimation problems where the solution wasobtained by dualizing the solution to the corresponding state feedback control problem,we will see now that the solution to the H,, state estimation problem can be obtained bydualizing the solution we obtained in previous sections for the H , state feedback controlproblem.

9.4.1 Determination of Te (s)Recalling the development of observers given in Chapter 3, we see that an observer whichestimates the state of the generalized plant (9.1) using u2(t) and y2(t) is constructed as

X (t) = AX(t) + L2Y2(t) + B2u2(t) (9.67)

where

Y2(t) =Y2(t) -Y2(t)

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H... State Estimation Problem 243

However, from the state model for the generalized plant, (9.1), we see that

y2(t) = C2x(t) +D21u1(t) +D22u2(t) y, (t) = C1x(t) +D,2u2(t) (9.68)

Thus using (9.67, 9.68) we see that the generalized plant state estimation error,z(t) = x(t) - i(t), is the solution to the differential equation

x (t) = (A+L2C2)X(t) + (B1 + L1D:1)u1(t) (9.69)

In addition, we see from (9.1, 9.64, 9.65) that the desired output estimation error,y, (t) = yl (t) - yl (t), depends on the state estimation error and disturbance input as

5 (t) = -C1. (t) - D11u, (t) (9.70)

Therefore we see from (9.69, 9.70) that the transfer function T,(s) from ul (t) to 91(t)has state model given as

A + L2C2 (B1 + L,D,1)

-C1 -D1, I(9.71)

Recalling the discussion following (9.5), we see from this state model for Te(s) thatcondition (ii), (9.66), cannot be satisfied if

Qmax Dlll >

Thus the same necessary condition, (9.6), which we required to solve the H" statefeedback control problem, is needed now to solve the H, state estimation problemnamely

imax[Dlll < (9.72)

We will see now that the design equations for the determination of L2 so thatconditions (i, ii), (9.66), are satisfied can be obtained by using the fact that the presentHx state estimation problem is the dual of the H,, state feedback problem solvedpreviously in this chapter.

9.4.2 DualityRecall, from Chapter 7, that the H. norm of T,(s) is defined as

sup ntav(jw1

where rr,,,ax(jw) is the largest singular value of TT,(i, ).More specifically, recall that amax(jw) is the positive square root of the largest

eigenvalue of either of the Hermitian matrices T,7 (. j i) T /w 1 or T j jw:) T, (jw). Thereforethe largest singular value of TT(jw) equals the largest singular value of TT (jw). Asdiscussed in Chapter 8, the adjoint system corresponding to a system having real-rational

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244 H. State Feedback and Estimation

transfer function Te(s) has transfer function Te'(-s) and satisfies

T e (S) Is jw= Te (-S) I s jw

Therefore condition (ii), (9.66), is equivalent to the condition

IITe(-S)II.<--r (9.73)

where we see from (9.71) that

Te (-s) s - AT + C? L2) -CI- (BT + D21 LZ) -D11

(9.74)

Notice that condition (i), (9.66), is satisfied if and only if Te (-s) is antistable sinceTa(s) E 7-l(s) is satisfied if and only if TT(-s) E 7{-(s).

Thus we see from the foregoing that the H. state estimation problem and the H,,. statefeedback control problem are dual problems in the sense that a solution to the H. stateestimation problem requires that

(i) : TT(-s) E 71 and (ii) : IITe (-s)II.< ry (9.75)00 -whereas a solution to the H. state feedback control problem requires that

(i) : T, (s) E 7i and (ii) : IITT(s)IIx<'r (9.76)

9.4.3 Design Equations for L2In order to exploit the foregoing dual nature of the H,, state feedback control problemand H,,. state estimation problem, (9.75, 9.76), we need to establish a symbol correspon-dence between these problems. This can be done by comparing the state models for Ta(s)and TT (-s), (9.4, 9.74). This gives the following correspondences between their para-meters.

Recall that we determine K2 in the Hx; state feedback problem by first solving theHCARE, (10.96), for its stabilizing solution, X,,.. Then K2 is the last m2 rows of the matrixK which we determine by setting X X,,,, in (9.43). Therefore, proceeding in ananalogous manner we determine L2 as the last p2 columns of the matrix L obtained byusing the stabilizing solution Y,, to an algebraic Riccati equation. More specifically, thedesign equations for determining L are obtained by replacing parameters in (9.46) and(9.43) using the parameter equivalences in Table 9.1. Thus doing this and also replacing-X with Y gives

A Txe Y + YA31_2 - YRx2 Y + Q.2 = 0

and

(9.77)

L=-(YCT+B1Dz)D.,, (9.78)

Page 260: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Sufficient CohdRIorre c3a

where

with

Table 9.1 State model Parameter Correspondences

Control Estimation

AB2

K2

B1

Cl

D12

D11

D2 =DitD21

-AT-CTLT

-CT-BIT

-D21

-D Til

Axe =A-Bt D T2D

tCoRx2 = CT Duo C

Qx2 = B1 [I - Dz Dr4)1 D2_j B

I IP ;-Dj1Di DlID21 1="JODztDlTt

D2tDT21

Notice that (9.77) is an algebraic Riccati equation like the GFARE, (Section 6.3.2)which is referred to here as the HFARE. The solution Y which makesA2 - YR2 = A + LC stable is referred to as the stabilizing solution to the HFARE andis denoted Yx and satisfies Yx > 0. Just as A + B1K2 is stable when K is determined from(9.43) using Xx so too is A + L2C2 stable when L is determined from (9.78) using Yxwhere

L=[L1 L,l

Just as the independence of the columns of D1, and condition (9.72) ensure that Dj, isinvertible so too does the independence of the rows of D,1 and condition (9.72) ensurethat Dje. is invertible.

Finally, recall that in the Hx state feedback problem we required (A, B2) to bestabilizable and the HCARE to have a solution Xx > 0 in order for A + B2K2 to bestable. We see by analogy that in the H,x state estimation problem we need the HFARE tohave a solution Y,,, > 0 and (A, C2) to be detectable in order for A + L2C2 to be stable.

9.5 Sufficient ConditionsRecall that a given state model for the generalized plant must satisfy certain conditions inorder for the GCARE and GFARE to have stabilizing solutions (Section 5 of Chapter 6).A similar set of conditions which ensure the existence of stabilizing solutions to the

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246 H. State Feedback and Estimation

HCARE and HFARE are

(Al) (A, B2) and (A, C2) are respectively stabilizable and detectable pairs.(A2)

(A - jwI) B2

C1 D12and

(A - ju)I)B1

C2 D21

have respectively independent columns and independent rows for all real W.(A3) D12 and D21 have respectively independent columns and independent rows.

Taken together, (Al-A3) constitute a sufficient condition for the solution of both the H.state feedback control problem and the H. state estimation problem.

Notice that (A1,A2) ensure that the Hamiltonian matrices corresponding to theHCARE and HFARE have no imaginary axis eigenvalues. This can be seen by notingthe similarity of the HCARE with the GCARE in Chapter 6 and referring to Theorem 6.1in Section 6.5.2.

9.6 SummaryIn this chapter we have given a derivation of the design equations for solving the H... statefeedback control and state estimation problems for a given linear time-invariant,continuous-time plant. Unlike the solutions to the corresponding quadratic or LQGproblems obtained in Chapters 5 and 6, solutions obtained here for the H,,. problems donot minimize any performance criterion or cost. Rather the solutions to the H,,. problemsachieve a specified upper bound on the L2 gain from disturbance input to desired outputwith the resulting closed loop systems being stable. As in the quadratic and LQG controlproblems, algebraic Riccati equations still play a central role in the solution. In the nextchapter we will see how the solutions obtained in this chapter can be used to providecontrollers which use the measured plant output instead of the plant state to achieve thejust mentioned H,, norm bounding and closed loop stability.

9.7 Notes and ReferencesA great deal of work has been done on the H,, control problem since its inception over adecade ago. One of the first approaches solved this problem by beginning with a Youlaparametrization of all stabilizing controllers for the given plant and finished by having tosolve a model matching problem. This work is the subject of [14]. This and other earlyapproaches which are reviewed in [8] and [20] are quite complicated and have controllerswith unnecessarily high dimension. Indications for a simpler approach came fromresearch on stability robustness [35], which lead to the two-Riccati-equation solutiongiven in this chapter [12, 16]. Following this similar results were obtained using J typefactorizations, e.g., J spectral, J lossless coprime, [19], [41], [20], [40]. Discussion of thisapproach in connection with the chain-scattering formulation of the generalized plant isgiven in [24].

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10HO, Output Feedback andControl

10.1 IntroductionHaving seen how to solve the basic H, problems of state feedback control and stateestimation for the given generalized plant, we come now to the problem of determining acontroller which generates the controlled input from the measured output such that:

(i) the closed loop system is internally stable and(ii) the H,, norm from the disturbance input to desired output is no greater than a given

positive scalar -y.

This problem is referred to as the H,, output feedback control problem. The setup forthe output feedback control problem was given previously at the start of Chapter 9.Again, as in that chapter, we are assuming that the generalized plant has known transferfunction G(s) and state model, i.e.,

Y2((ss) )

G(s)IU2(s) G(s)

[Dc (lo.l)

Y

where

U, (s) = disturbance input Yj (s) = desired outputU2(s) = controlled input Y2(s) = measured output

In the development to follow, we will see that there are many controllers which solvethe H, output feedback control problem. Moreover. we will show that these controllersconstitute a constrained subset of Youla parameterized controllers for the system havingU2(s) as input and Y2(s) as output. Recall, from Theorem 8.5, that Youla parameterizedcontrollers are stabilizing. Therefore condition (i) is satisfied. We will show how toconstrain this set so as to achieve condition (ii).

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248 H. Output Feedback and Control

10.2 DevelopmentRecall, from Section 9.3.2, that the partition K2 of K which effects a J-inner coprimefactorization of G1 (s) stabilizes the generalized plant. In addition, recall from Section9.2.4 that P. < 0 for all U1 (s) E 7-12 and that this implies that P..1 < 0 for all U1 (s) E 7-L2where

U1(s) = K1X(s) + U, (s) (10.2)

This fact is used now to develop a solution to the H output feedback control problem.The plan for doing this is given as follows.

1. We begin by assuming that the disturbance input U1 (s) is given by (10.2) with K1determined in Section 9.3.1.

2. Then we show that the signals V, (s), V2 (s) can be replaced in the expression for thequadratic performance index P7, introduced in Section 9.2.1, by the "external"disturbance input U1 (s) and an output-like signal, Y,, (s). This gives P7 = P7o whereP7o is usgd to denote the new expression of P.Y. The output-like signal, Y0(s), dependson U, (s), U2 (s) and the state of the given state model for the generalized plant.

3. Next we determine an estimator for the generalized plant state which uses yo(t) andachieves P7o < 0 for all U1 (s) E 7{2. This is accomplished by adapting the H. stateestimator obtained in the previous chapter.

4. Following this we replace the actual generalized plant state in the He state feedbackcontroller determined in Chapter 9, by the estimate of this state which is generated bythe state estimator developed in the previous step. This leads to a controller whichachieves P7o < 0 for all U1 (s) E 7I2 but requires as an input in addition to themeasured output y2 (t) .

5. Since y0(t) is not available for use as an input to the controller, the remainder of thedevelopment concentrates on eliminating the need for y,,(t) as an input to thecontroller.

10.2.1 Reformulation of P,(Recall from the development of the solution to the H,,, state feedback control problem(Section 9.2.1) that P7 < 0 for all u1 (t) E £2[0, no) implies that condition (ii) is satisfied.We are going to use this fact to solve the Hx output feedback control problem. Alsorecall, (Section 9.2.3), that P7 was expressed in terms of the signals V, (s), V2(s), whichresulted from a J-inner coprime factorization, i.e.,

P7 = -'Y211 V1(s)112+II V2 (S) 112

and recalling (9.25) we have

V, (S) _ j1T1(S)

U, (S)1

VAS) I LU2(S)J

(s)= A B

_AK A

(10.3)

Page 264: Linear Control Theory - The State Space Approach Frederick Walker Fairman

beveropinem Z"

Now in order to incorporate the solution to the Hx state estimation problem we needto re-express P,y in terms of the generalized plant state. We do this by assuming the statefeedback matrix K is determined in the manner specified in Section 9.3.1 and that A canbe taken, from Theorem 9.3, as

I

A,

A3 A04 - Le T12D1 D11(10.4)

where

e = OT/201/2 = D12D12

T = pT/2Vt12 = 2Im, - Dit (I 1 - D120 1 D )D11

Then substituting for Ut(s) from (10.2) we see that the state models for the generalizedplant G(s), (10.1), and M -1(s) , (10.3), become

G(s)s A + B1 K1 B

(10.5)

111 1(s)

C

`

+ D2K1

A+BIK1

0

D

B

L1 0 (10.6)

-04K2 J3 4

where

Y1(s)

Y2(s)

V, (S)

V2(S)

I G(s)

_ M_1(

U

U

s)

1(s)

2(s)

Ut(s

[U-1(s)]

Notice that the state differential equations and inputs to the foregoing state models forG(s) and M-1(.s) are identical. In addition notice that the initial states of each of thesemodels are identical because we are using transfer functions to relate inputs to outputswhich requires that the initial state of each model be null. Therefore the states of thesemodels are identical for all time and equal to the state of the generalized plant when thedisturbance input is given by (10.2). This observation together with the output equationof the state model for M-'(s), (10.6), reveals that V, (,s). V,(s) are dependent on thegeneralized plant state X(s) and are given as

V1(s) = OtUt(s) (10.7)

V2(s) = -04K2X(s) + X13 C1(s ) - A4L"'(s) (10.8)

We can use these expressions for the V,(s)s together with the relations for the s,(10.4) to re-express P7, (10.3), in the manner given in the following theorem.

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250 H Output Feedback and Control

Theorem 10.1 If the state feedback matrix K is determined as in the previous chapterso that its partition K2 solves the H. state feedback control problem, then P7, (10.3), canbe re-expressed as P,y = P. 0 where

P7o = -72 I I U1(S) 112+I I Yo (5)112

with

Y,,(s) _ -D12K2X(s)+D11U1(s)+D12U2(s)

and X(s) being the state of the generalized plant.Proof From (10.7) we see that

(10.9)

-72111'1 (S) 11`2= -721101 U] (S) 112=-72 (Al U1 (S), Al Ul (S)) (10.10)

where

(A1 UI (S), 01 U1 (s)) =27f f. U1(IW)O1 01 U1 (w)d

Similarly from (10.8) we see that

111'2(5)112 =11A4K2X(S)II2+I A4 U2 (S)112+11A3 U1(s)112-2(L4K2X(s), A 4 U2 (s))

- 2 (A4K2X(s), A3 UI (s)) + 2(A4 U2 (s), A3 U1(s)) (10.11)

However from the dependency of the DEs on D11 and D12, (10.4), we have

AT01 = Im, - 7 2DIi D11 + 7 2DIIID12 (D12D12) -'DT

T T 1 T03 03 = DIID12(D12D12) D12D11

T T T T0403 =D12D11 0404=D12D12

Therefore using these relations in (10.10, 10.11) gives

y2

and

11'1(s)II2= -7211 U1 (5)112+IIDI1 U1 (S)II2-IIA3U1(S)112 (10.12)

II1'z(S}112 = IID12K2X(s)II2+IID12U2(S)112+IIA3U1(S)II2

- 2(D12K,X (s), D12 U2(s)) - 2(D12K2X(s), D11 Ul (s))

+ 2(D12 U2(s), D11 U1(s)) (10.13)

Finally substituting (10.12) and (10.13) in P. ,(10.3), gives P7 - P.y , (10.9).

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Development z13-1

Notice that, in the ideal case when the generalized plant state is available for statefeedback, we can set U2(s) = K2X(s) so that Y,,(s) = D,1 Ul (s) and P,,,, (10.9), becomes

Pyo = (U1, +D D11)U1)

which satisfies P.70 < 0 for all U1 (s) E 7-12 since we are assuming that o',,,ax [D11 ] < ry.Alternatively, since X(s) is unknown in the present case, we use its estimate, X(s), to

generate the controlled input. Therefore replacing U_(s) by K2X(s) in Y,(s) andsubstituting the result in P_,0, (10.9), yield

P,O _ -1r211U1(s)II2+l

where

Yo (s) = Y. (S) - Y"(S)

= D12K2X(s) - D11 C71 (s)

X (S) = X (S) - i(s)

(10.14)

Thus we see from (10.14) that in order for P, < 0 for all U1(s) E 7-12 we need todevelop an estimator for x(t) so that

(i) : T0'-(s) E7h (ii) : (10.15)

where

Yo(s) = To'(s) U1(s) (10.16)

which is an H,, state estimation problem. We obtain a solution to this problem byadapting the solution to the H state estimation problem obtain in Section 4 ofChapter 9.

10.2.2 An H., state estimatorA comparison of the H,, state estimation problem just posed with the H, stateestimation problem solved in Chapter 9, reveals that the role played by the desiredoutput estimation error, 5 (t), is now being played by i (r). the error in estimating j,, (1)

-

In addition, in the present situation, we are using it, (1) in place of u] (t).Therefore in order to use the solution to the HY, state estimation problem given in

Chapter 9 we need to develop a state model for a system which relates ul (t), u2(t) as inputto yo(t), Y2(t) as output and has the same state as the generalized plant. We can obtainsuch a system by replacing the desired output, Y1 (s) from G( s). (10.5), by Y,,(s), (10.9).Thus doing this and denoting the resulting transfer function by G0(s), we obtain a systemwhich is specified as

Ga(s)s A B] [ Y,, (s) ] 1(s) G,.,(s! [ L i (s) (10.17)

CO D Y, (S) (s) G,,4 Ls) I U, (S)

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252 H, Output Feedback and Control

where

[

-D12K2A° = A + B1K1 C. _

C2 + D21 K1

Now we could develop a solution to the present H. state estimation problem byobtaining T, (s), (10. 15), from the state model for G. (s), (10.17) and proceeding as we didwith T° in the previous chapter in the development of an H , state estimator for thegeneralized plant. However, since our goal here is to determine a stable estimator for thestate of G° (s) subject to the requirement that

II Y°(S) MM2< vlI U1(S) 112 (10.18)

we can obtain the design equations for the desired estimator by replacing the parametersin the design equations for the H,,, state estimator for the generalized plant state model,G(s), (10.1), by the corresponding parameters in the state model for G°, (10.17).Inspection of the state models for G(s) and G°(s) shows correspondences between thestate model parameters as given in Table 10.1.

Therefore the H,,. state estimator for 6,, (s), (10.17), has differential equation

z (t) = A°z(t) + B2u2(t) + L°y°(l) - y°(t)

l (10.19)[Y2(t) - y2(t) J

Notice that we are assuming temporarily that y°(t) is known and can be used in thestate estimator, (10.19). This assumption will be removed later since the final controllerwe are seeking has only y2 (t) as input and u2 (t) = K2-i(t) as output. Notice also that L° in(10.19) corresponds to L in the Hx state estimator for the generalized plant given inSection 9.4.1.

Now the design equations for L° are obtained by replacing the parameters in theHFARE and the equation for L which we obtained in Section 9.4.3 by the correspondingparameters as indicated in Table 10.1. Thus L° is obtained (when it exists) by finding thesolution Yx3 to the following ARE such that A. + L°C° is stable

Ax3 Yoo3 + Yoo3A0o3 - Y°o3Roo3 Y°o3 + Qoc3 = 0

L. = -[Yoc3Co +BID2JDJO'

where

Ax3=A° B1D T 2Dj°1C°

Ra3 = Cr ° D1J°

Q.x3 = Qx2 = Bl (I - D2 DJo D2)Bi

(10.20)

with

D- _ [D11 A°=A+B1K1D21

-72Ip, + D11 D1 i D11D2i -D12KzDJ0 = r r C0 - [D21 D I I D21 D21 C2 + D21 K1

Page 268: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Table 10.1 Parameter correspondences for the general-ized plant, G(s), and G0(s).

G(s) Ga (s)

A AD

B B

C C"

C1 -D1,K2C2 C2 - D,1 K1

D D

As in the H,,, state estimation problem treated in Chapter 9, the solution, Y3, to theARE, (10.20), which makes A.3 - Yo,3Rx3 = Ao + L0C,, stable, satisfies YY3 > 0 and isreferred to as the stabilizing solution to equation (10.20). Notice that the parameters ofthe ARE, (10.20), involve the solution to the H, state feedback control problem.Therefore the existence of a stabilizing solution to equation (10.20) depends on thesolution to the H,,. state feedback control problem. This observation suggests that theexistence of solutions to both the H,,, state feedback control problem and the H. stateestimation problem for a given generalized plant, does not guarantee the existence of asolution to the H,,, output feedback control problem for that plant. This contrasts withthe LQG output feedback control problem, which has a solution whenever both the LQGstate feedback control and LQG state estimation problems have solutions.

We will see in Section 10.4 that the existence of a solution to the Hx output feedbackcontrol problem not only requires that both the HCARE and HFARE have stabilizingsolutions, X,,, and Yom, but in addition the product of these solutions satisfies thefollowing inequality

max Y. X. <

10.2.3 Introducing estimated state feedbackSo far we have developed an H. state estimator for the generalized plant under theassumption that the state of the generalized plant is being used to generate the statefeedback signal, u2(t) = K2x(t). In this case the estimate of and Of Y2 (t),y2 (t) 112(,) based on the estimate of the generalized plant state, s(i), are seen from (10.17) tobe

-D12K2z(t) - D12t1;(t) (10.21)

y2(01.,(t) = (C2 + D21K1)i t) + D221t2(t) (10.22)

However, if the controlled input is generated as

u2(t) = K2z(t) (10.23)

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254 H Output Feedback and Control

we see from (10.21, 10.22) that

Y" (t) =Y0(t) -Y.(t) -Y.(t)

Y2 (t) =52(t) -Y2(t) = (C2+D21K, +D22K2)k(t) -Y2(t)

(10.24)

(10.25)

where we have y,(t), (10.21), null when the controlled input satisfies (10.23). This factplays an important role in the development of the controller.

Continuing, we see that use of (10.24, 10.25) in (10.19) yields

z (t) _ (A + BK + Lo2C2 + Lo2D3K)z(t) - L. y°(r) (10.26)Y2O]

where

D3 = [ D21 D221 L. = [ Lo, Lot

and the controlled input is generated by a controller with transfer function r(s) as

Y,(s) Lo

UZ(S)r(s) [

Y2(S) Ir(s)

[A+BK+L02(C2+D3K)K2 0 , (10.27)

However, since y,(t) is unknown we need to eliminate it as part of the input to thiscontroller in a manner which preserves Pry, < 0 for all U, (s) E 712. We will do this in thenext section.

Before we go on to develop these y,(t)-free controllers, we summarize what has beenachieved so far in the development of a solution to the H. output feedback controlproblem.

1. We began by incorporating the solution to the H. state feedback control problem inthe performance index, P, used in the previous chapter. The expression we obtainedwas denoted P.,,. It was shown that Pry, depends on y,(t), (10.9), an output from asystem, (10.17), whose state equals the state of the generalized plant.

2. Next, in Section 10.2.2, we used the solution to the H,, state estimation problemobtained in Chapter 9 to obtain an H,. state estimator based on the output y,(t) sothat the error in estimating y,(t), i.e., y,(t), satisfies 1Iy,(t)112< ryIIut(t)112.

3. Finally, in Section 10.2.3, we showed that by using the estimated state to generate thestate feedback signal as u2(t) = K2k(t) and the H,. state estimator based on y,(t) gavea controller which satisfied P.,, < 0 for all U1 (s) E 7-t2 or equivalently all U1(s) E 712 asrequired to solve the H, output feedback control problem.

10.3 Hx Output Feedback ControllersAs mentioned before, the controller given by equation (10.27) cannot be used as it has theunknown output-like signal, as an input. In this section we will show how thisunknown signal can be removed from the input to the controller, (10.27).

I

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One way of doing this which leads to a single controller referred to as the centralcontroller, is to assume that the disturbance input is the worst disturbance, (Section9.2.4), so that U1 (s) is null.

Another way of doing this is to relate y,(t) to #1 (t) through a system. This approachyields a family of controllers any one of which solves the H. output feedback controlproblem. We will see that this family is a subset of the Youla parametrized stabilizingcontrollers for the given generalized plant which we encountered in Chapter 8.

10.3.1 Central ControllerRecall, from the previous section, that the error in estimating y,(t) satisfiesv0(t) = -yo(t), (10.24). Therefore we can write the performance index P.yo, (10.14) as

P70 _ --y (10.29)

and the relation (10.16) as

Y0(s) = -To'(s)U1(s) (10.29)

However recall that we determined the estimator, (10.19), so that

'Y U,(5) (10.30)

Thus we see from (10.28, 10.29, 10.30) that

P0O <0 U1(s) EN,

as is required to satisfy condition (ii) in the statement of the Hx output feedback controlproblem.

Now U1(s) is unknown. However suppose we generate the controlled input, u2(t),from the controller, (10.27), under the assumption that the disturbance input is the worstdisturbance, (Section 9.2.4), i.e., under the assumption that u1 (t) is null. Then in thissituation we see that Y(,(s), (10.29), is null and therefore the controlled output, (10.27), isgiven as

U2(s)=12(s)Y2(s) r2(s)IIA+BK-L tC,±D3K) -021 (10.31)

where

D3='[D21 D

This controller is referred to as the central controller.In essence the central controller is an H, state estimator of the generalized plant under

the restriction that the disturbance input is the worst disturbance and the controlled inputdepends on the state estimate as u2(t) = K2z(t). where K, solves the H, state feedbackcontrol problem. This is one way of obtaining a controller which is free from y(t).

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256 H,. Output Feedback and Control

10.3.2 Controller parametrizationSuppose the unknown output which is present in the controller, (10.27), is replacedby Y, (s) where

Y,,(s) ='D (s)U1(s) (10.32)

with

(i) : E floc (ii) : II'D(s)II.<_ 'Y

Notice that when UI (s) E 7-t2 we have Y,, (s) E 7-12 and II YID(S) I I2 7I I U1(s) I I2. Thuswhen Y,,(s) is replaced by Y,,(s) in (10.28) we have

P.y0 = (U1 (S), (-727+4*(s)4?(s))U1(s))

which implies that P.yo G 0 for U1 (s) E 7{2 as required to satisfy condition (ii) in thestatement of the Hx output feedback control problem.

Next suppose we have W (s) E 7-L such that

(S) Wl ( S) W2(s) l[U2

j - 1

fY.(s)

(10.33)/

/L Y2 (S) W3 (s) W4 (8) J L U1(s)

Then replacing Y0(s) in (10.33) by

Y0(s) = 4'(s)Ul(S)

and solving (10.33) for U2(s) gives the desired controller as

U2(s) = H(s)Y2(s) (10.34)

where

H(s) = [WI(s)-D(s)+ W2(s)][W3(s)4?(s)+ W4(s)]-1

with I(s) constrained as specified in (10.32). The specification of W(s) is given in thefollowing theorem

Theorem 10.2 The state model for the transfer function W(s), (10.33), satisfiesW(s) E 71, and has state model

W1(S) W2 (s) s

W3 (S) W4 (S)

A+BK [B2D12 -L02]

K2

C2+D3K

where

T2Die = (DD12) 1Di

01 (10.35)

I

Page 272: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Proof Since K was determined in the solution of the H. state feedback controlproblem so that A + BK is stable, (Section 9.3.1), we have W (s) E 7I .

In order to determine W2(s) and W4(s) we choose 4)(s) null so that (10.34) becomes

U2 (s) = W2 (s) W4 ' (s) Y2 (s)

and comparing this equation with (10.31) yields

F2(s) = W2(s)W41(s)

(10.36)

(10.37)

Now since W2(s), W4(s) E H,,,, we can determine W2 (s) and W4(s) by coprimefactorizing I'2(s). More specifically, we choose PI/j,(s) = Nc(s) and W4(s) = Mr(s)where r2(s) = Nr(s)Mr 1(s) with Nr(s), Mr(s) E 7-{,, and : 'r(s),:L1r(s) coprime.

Recall from Section 8.4.1 that if the state model for I _ (s) is denoted as

rZ (s),. Ar Br I

[Cr DF]

then state models for the right coprime factors of F,(s) are given as

s Ar +BrKr BrNr (s)

Cr + DrKr Dr

where Kr is chosen so that Nr(s), Mr(s) E 71,,.Now from (10.31) we have

Mr- (S) S

Ar

Br = -L,,2

Kr I JIAr+BrKr Br

Cr=K, Dr=O

so that choosing KI = (C2 + D3K) gives the coprime factors as

Nr(s) = W2(s)

Mr (s) = W4 (s)

A + BK -Lo2K, O

A + BK - L,,,

C,+D;K I

(10.38)

where Nr(s), Nlr(s) E 71,E since K was determined in the solution of the Hx statefeedback control problem, (Section 9.3.1) so that A + BK is stable. This completes thedetermination of state models for W2(s) and W4 (s).

Next in order to determine W, (s) we recall from (10. 17) and (10.33) that

Yo(s) = Go,(s)U,(s) (10.39)

U2(s) = W,(s)Yo(s) + 14",(s)L,(,s) (10.40)

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258 H,. Output Feedback and Control

Now if the system inverse, G-Z (s), were to exist we could solve (10.39) for U2(s) as

U2(s) = Go' (s) Y. (s) - Go-2 (s) Gol (s) U, (s) (10.41)

and we would have Wl (s) = (s) from a comparison of (10.41) with (10.40). Howeverwe see from (10.17) that Go2(s) is not invertible in general since its state model,

G02(s)5 LA+B1K1 B2

JL -D12Kz D12

has a D matrix, D12, which is not invertible in general, (Section 8.2). This difficulty can beovercome by recalling that we are assuming D12D12 is invertible since this is a necessarycondition for the existence of a solution to the H. state feedback control problem. ThusG02(s) = D12G02(s) is invertible where

A + B1 K1 B2G02 (S) T T

-D12D12K2 D12 12

Therefore we rewrite (10.39, 10.40) as

(10.42)

Yo (s) = Gol (s) Ul (s) + Got (s) U2 (s) (10.43)

U2(s) = W1 (s)Yo(s)+ W2(s)U1(s) (10.44)

where

Yo(s) = u12Y0(s)

Goi(s) = D12Goi(s) : i = 1,2

W1(s) = Wl (s)D12

Then solving (10.43) for U2(s)

O(s) = Goz (s) Y,, (s) - Gnz (s)Go1(s) U, (s) (10.45)

and comparing (10.44) with (10.45) yields

Wl (s) = Go2 (s) (10.46)

Finally, using the form for the state model of a system inverse, (Section 8.2), we seefrom (10.42) that

1 A+BK B2(D12D12)-1Got (s) = T 1

K2 (Dl2D12)

and (10.47) yields

W s W Drf A+ BK B2D12

1( ) 1(.s) lzK2 Dt12

(10.47)

(10.48)

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AIT

where Die is a left inverse of D12, i.e., D12D12 = I, given as

Dig = (D12D12) 1D1 (10.49)

This completes the determination of W, (s).In order to determine W3(s), rewrite the second block equations in (10.33) and in

(10.17) as

Y2(s) = W3(s)Y,,(s) + W4(s)G1(s) (10.50)

Y2(s) = G0 (s)U1(s) +Ga4(s)L'2(s) (10.51)

Then substituting for U2(s) from (10.40) in (10.51)

Y2(s) = G04(s)W1(s)Y0(s) + (Go3(s) + 6,,4(S) 1f 2(S)) UI (s) (10.52)

and comparing (10.50) with (10.52) reveals that

W3 (S) = GA(S) WI (S)

Then using the state models for Go4(s), (10.17), and WI (s), (10.48) together with therule for combining systems in series, (Section 8.1), we obtain

s f Au, Brv 1=W3(s)

LCw D14,1

rA + B1K1 B2K2

IL 0 A -1- BK

B,D1L

[C2 +D21K1 D2,K, l D,2Di2

Next this state model can reduced without affecting W3 (,%). We can do this through theuse of an appropriate change of coordinates. Therefore if we let the coordinatetransformation matrix be

T=

the state model for W3 (s) in the new coordinates is

W3 (s)

where

T-1AwT T-1BIr

CwT Div L CII. DII,

A+BK B,K, I B2D',[ 0 A± B1 K1 IL

0JI

[C2+D3K D,-K2; D,2Di,

D3 = [ D21 D-

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260 H.. Output Feedback and Control

Finally, notice that W3(s) is given by

W3(S) = Cw(sI - Aw) 'Bw +Dµ,

= (C2 + D3K)[sI - A - BK]-'B2Diz + D22Di2

which implies that W3(s) has a state model given as

W3 (S) S

C

A + BK B2D12

C2 + D3K D22D12D12 = (D12D12) 'Diz

In summary, in this section we have determined a family of controllers, H(s), (10.34,10.35), any one of which solves the H,,) output feedback control problem. In the nextsection we relate this family of controllers to a subset of the Youla parametrizedcontrollers.

10.3.3 Relation to Youla parametrizationRecall, from Theorem 8.5, that the Youla parametrization of stabilizing controllers,H(s), for a plant, G(s), involves the doubly coprime factorization of G(s), (Section 8.4.3).These controllers are given by

H(s) = NN(s)M,' (s) (10.53)

where

N,(s) = MG(s)QG(s) +NH(s)QG(S) E 71.

Me(s) = NG(s)QG(s) + MH(s) 1

with

G(s) Mc1(s)NG(s) =NG(s)MG'(s)

MH(s) -NH(S) MG (S) NH(S) I 01-NG(s) MG(S) L NG(s) MH(s) I I0 I

Now in the present situation, the H,o output feedback controller is connected betweenthe generalized plant's measured output, y2(t) and controlled input, u2(t). Moreover, thiscontroller must stabilize the generalized plant for all disturbance inputs satisfyingU1 (S) E 7-12 including the worst disturbance, ul (t) = K,x(t) (Section 9.2.4). Therefore,assuming the disturbance input is the worst disturbance, we see that the controller muststabilize a plant G(s) = G22(s), where from (10.5) we have

G(s) - Gzz (s) AG BG _ A + B1 Ki B2- l f

CG DG J = L C2 + D21 Ki D22

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l $ep8rstm c rrnrcrpP5 =jn

Thus recalling the state models for the systems involved in a doubly coprimefactorization, (8.46, 8.53), we see that G22(s) has a doubly coprime factorization given by

NH (s) s r AG + BGKG

L KG

LGl0

BG J

DG

MH (s)AG + BGKG

CG - DGKG

LAG + BG KGMG (S) = KG

(10.55)

NG(s) s [ AG + BGKGCG + DGKG

Then substituting the state model parameters as specified in (10.54) and setting KG = K2and LG = -Lo2 in (10.55) gives

A+B1K1 +B,K, 1B, -L07 ]MG (S) NH (S)

K2 1 0 (10.56)NG(S) MH(s) C2 + D21 K1 + D22K2 D,, I JI

However, since D12QG(S) E H. if QG(S) E 7-1,,,, we can replace QG(S) in the generalexpression for the parametrized stabilizing controller, (10.53)), by D124)(s) so that

H(s) = [MG(S)'D(s) + NH(S)][NG(S)'b(S) + MH(S); ID(s) E 7I (10.57)

where

MG (s) NH (s)

NG(S) MH(s)

A+B1K1 +B2K I B,D1, -L.,

K, Di, 01C2+D21K1 +D,,K, ILD22Di2

I

Thus controllers satisfying (10.57) stabilize the generalized plant and have the sameform as the H,, output feedback controllers determined in the previous section, (10.34,10.35). Notice that in addition to the restriction D (s) E 7-t, which ensures that the controlsystem is internally stable, in the present problem we also require Ij(D(s)jjx<'y in order

- 1f1(1112.that the control system satisfies the condition !1y'1(t)i12<

10.4 Hx Separation PrincipleRecall, from Chapter 6, that we obtained the controller which solves LQG outputfeedback control problem directly from the corresponding solutions to the basic LQGplant state estimation and plant state feedback control problems. We referred to this factas the separation principle. However in the H,;; case, being able to solve the correspond-ing basic H, state feedback control and state estimation problems is only a necessarycondition for being able to solve the Hx output feedback control problem.

In this section we will show that the additional condition needed to ensure the existenceof the solution to the H, output feedback control problem is

j < ^,'2 (10.58)Amax[YDXc

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262 H,. Output Feedback and Control

where X,,,., YY are stabilizing solutions to the HCARE and HFARE, (9.46, 9.77). Beforedoing this we show now that Y.3 and Lo, (10.20), needed to construct the controllers,(10.34, 10.35) can be calculated directly from the stabilizing solutions to the HCARE andHFARE as follows:

Yx3=ZY,

L,, Z(L1 72 YxCT1K) ZL2

I

where

Z = (j 7-2 y XX)-1

CJK=D11K1+D12K2+C1

(10.59)

(10.60)

with L being the solution to the H,,. state estimation problem for the given generalizedplant (9.78).

The approach used to show these results relies on a similarity relation between theHamiltonian matrices associated with the HFARE, (9.77), and the ARE, (10.20).

10.4.1 A relation between HamiltoniansWe are going to make use of the following ideas from Chapter 5.

The Hamiltonian matrix HP associated with the ARE

is given as

AP+PAT PRP+ Q = 0 (10.61)

HP =[ AT R l

Q AJ(10.62)

The eigenvalues of HP exhibit mirror image symmetry across the imaginary axis.

A[HP] = A[(A PR)"I U A[-(A - PR)]

This fact is seen from the following similarity transformation of Hp

Hp=T-1HPT= I

where

(A PR)T RAP + PA* - PRP + Q -(A - PR)

T= [ P I

(10.63)

with HP being block-upper triangular when P satisfies (10.61).

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Hoc Separation Principle 263

Notice that equating the first n columns on either side of (10.63) yields

HpI pl = [ p](A-PR)T

so that if P is a stabilizing solution to (10.61) then

range -PI

(10.64)

is the stable subspace of Hp.Now the relation between the Hamiltonian matrices associated with the HFARE,

(9.77), and the ARE, (10.20) which is given in the following theorem will be used to show(10.59, 10.60).

Theorem 10.3 The Hamiltonian matrix H2 corresponding to the HFARE (9.77) andthe Hamiltonian matrix H3 corresponding to the ARE (10.20) are similar and related as

H2 = --IH341

where

THi = Axi Roci

i 3 (D _Qxi -Axi

(10.65)

Proof Suppose (10.65) is valid. Then carrying out the matrix multiplication which isindicated on the right side of (10.65) enables (10.65) to be rewritten as

(Ax3 - 72Qoo3Xoo)T

7-2(AxT 3Xx - X -,-4',3) - _! 4XxQx3Xx + Rx3

H2Qoo3 -(Ax3 - Qx3XxJ

Then equating like positioned blocks on either side of (10.66) yields

(10.66)

Qxz = Qx3 (10.67)

Ax2 =Axi - _Y-2Qx3Xx (10.68)

Rx2 = Y2 XxAx3) - +

aX-QY3X +Rx3 (10.69)

Now we show that (10.65) is satisfied by showing that (10.67-10.69) are satisfied.Recall from (10.20) and the HFARE, (9.77), that

Axi = A, - B1 D? D.J1 C,, A,, _ -4 - B, D D.J] C

Rx3 = CrDJo Cn Rs2 = C D IC (10.70)

Qx3 Bi (I - Dz D2)BI Qa., = Bi (J - DAD.,,, D2) B

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264 H. Output Feedback and Control

where

D111D2

IAo = A+B1K1

D21 1

Dly +D11Dli D11D21

C D12K2r0 T T

o

D21D11 D21D21 C2+D21K1

and (10.67) is satisfied.Next in order to show (10.68) we use (10.70) so that

A.3 - A.2 = B1K1 + B1DZ DuoC1 + D12K2

-Dz1 K1(10.71)

Now we are going to re-express the right side of this equation using results from the H.state feedback control problem.

Recall from the solution to the H. state feedback control problem given in Chapter 9,(9.43), that

DlcK = -(BT X. + DTI C1) (10.72)

Now writing Dj, as

T `Y2Im, 0Dj,, D1 D1-

P) 0

where

D1 = [D11 D121

enables (10.72) to be rewritten as

T 2 TD11C1K = Y K1 - B1 Xoc

T TD12C1K = -B2 X.

where

CIK=D1K+C1

Then K1 is obtained from (10.74) as

K1 y 2 (Bi X:X: + Dli CIK)

Next, we can readily show that

(10.73)

(10.74)

(10.75)

(10.76)

(10.77)

C, D12K2 J L IK-D2K1 =C - Co

0

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Hx Separation Principle 265

where

-D21K1 0

which after substituting for K1 from (10.77) becomes

K1 1C1 +D12Kz

-rCI

However writing (10.70), as

gives

Dz _ D11

D21 J

- 7 2D2 (BjXti + DT CI K)

f Ins -Yz r

D11D11L

'Y-2Dz1Dl

_ 2 [1 - Y 2D11Di-ryL 2 T

-'Y D21D11

Ip, - 1 2D11Di1_'[2D21D1i

and (10.78) can be rewritten as

CDK1j

7-2 Dj° 11,1

JCIK

-Dz1 1 J

2Bi X. (10.78)

T11D1

21D 1

zD,Bi Xx (10.79)

Returning to (10.71) we see that use of (10.79, 10.77) on the right side of (10.71) gives

Ax3-Ax2 =1 `B1DT f JC

and we see from the definition of Qx3, (10.70), that (10.68) is satisfied.Finally it remains to show that (10.69) is satisfied. i.e.. that

where

Rx2 = R,3 c2 (10.80)

(Z = %2(Ax3Xx + XxAx3) -

We begin by using (10.67, 10.68) to express C) as

-4X, Q13Xx

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266 H.. Output Feedback and Control

Then using the definitions for Axe and Q.2, (10.70), enables S2 to be written as

SZ = -Y-2 (ATX. + XXA - CT MT Xx X"cMC) + ry-4 (XOOBI Bi XOC - X0MD2B1 X. )

(10.81)

where

M = BI DT Dol

Next we are going to re-express ATX" + XXA for use in (10.81).Recall from the development of the HCARE given in the previous chapter that

ATX,Q+X,.A-KTD,,K+CI CI =0

which we can rewrite using the expression for Dj,, (10.73), as

(10.82)

However, from the definition of CIK, (10.76), we see that expanding C KCIK and Ci C1Kgives

KT DI DIK = CIKCIK - CCIKC1KCl +

CTC1

which when substituted in (10.82) yields

ATX". + X,oA = CIKClK - CTiCIK - C1KC1 -'YZKI K1

Finally we can expand this relation by substituting for KI from (10.77) to obtain

(10.83)AT Xx + XXA = FCK Y 2FX - 'Y ZX0B1BiX.

where

T T TFCK = C1KC1K - C1 CIK -

CT

F,- = XX B! DIII ClK + CI KDI1 B1 X. + C1xD11D1i ClK

Returning to (10.81), we can use (10.83) to rewrite (10.81) as

S2 =72 (FCK - CT MT X,, - X, MC) - 7-4(FX + XXMD2Bj TX')

This completes the expansion of S2 for use in showing (10.80)Finally using (10.79) yields

(10.84)

Co = C + ^[-2Dn,[

] CIK + 7-2D2Bi X. (10.85)

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Hx Separation Principle 267

and from R,,,3, (10.70), we have

2 T TR.3 = R.2 - Y [FCx C M Xx - XMCjR

+7 Xx] (10.86)

where M, FCK and FX are given in (10.81, 10.83)Finally we see from (10.86) and (10.84) that (10.80) is satisfied so that (10.69) is

valid. M

10.4.2 Relating stabilizing solutionsThe relation between Hamiltonian matrices, (10.65), is used now to show that (10.59) issatisfied and that condition (10.58) is required in order to solve the H,,, output feedbackcontrol problem.

Recall from (10.64) that the Hamiltonian matrix, H,, corresponding to the HFAREsatisfies

H2[_y00] = [I] (Ax, - YxRx2)T (10.87)

so that

Irange

is the stable subspace of H2.Therefore premultiplying (10.87) by 4), inserting 4)-14) following H2 and using (10.65)

yields

H I - -Y-,X. Y.

=

I - -,-`X,3 -Yx -Y.

which indicates that

range- Yx

(A 2 - YoRco2)T

is the stable subspace of H3, since A,c2 - YxR,, is stable.Now since the range of a matrix is unchanged by post-multiplication by a nonsingular

matrix, we have

I I-,-2Xx Yx Irange = range i

l

Yx (I Xx Yx )Y.

provided I y-2Xx Y, is invertible. Therefore we see that

1range _ I-Yx(I-7 `x, Y')

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268 H.0 Output Feedback and Control

is the stable subspace of H3. However we saw in the previous subsection that

Irange- Yoo3

is the stable subspace of H3. Thus we see that

Yx3=Yx(I-7-2X.Y.) 1=(I-y 2YYXX) 1YY=ZYY (10.88)

and (10.59) is established. Notice in (10.88) that the rightmost equality is a consequence ofthe matrix inversion lemma, (Appendix). We show next that (10.58) needed to ensure theexistence of the solution to the H. output feedback control problem follows from(10.88).

Recall that Ya.3 and Y are each stabilizing solutions to AREs so that we haveYo.3, Y. > 0. Thus we see from (10.88) that Z > 0. However in the derivation of (10.88)we required Z to be invertible. Therefore we can conclude that Z > 0 is required for theexistence of a solution to the H,,, output feedback control problem. Now the condition onYom, X,,. which ensure that Z > 0 is given in the following theorem.

Theorem 10.4 Z > 0 if and only if

Amax[YooXoo] <y2

where

(10.89)

Z_ (I-y_2y XX)-i

Proof Recall from Chapter 4, that Z > 0 if and only if Z-1 > 0. In addition recall thatZ-1 > 0 if and only if Az > 0 for all aZ E A[Z-11

Let AZ-i be any eigenvalue of Z-1. Then

det(AZ-.I-Z-1) =0

and from the dependency of Z-1 on YxX,,. we have

det(AyxI - YYX.) = 0

where Ayx E A[Y.XQ.] and

A YX = y2 (l - AZ_,)

Then solving for aZ gives

(10.90)

(10.91)

AZ-1 = 1 y ZAYX (10.92)

and we see that Az-) > 0 if and only if Ayx < y'.Finally we see from (10.92) that aZ > 0 for all aZ C A[Z] if and only if the largest

eigenvalue of Y., Xc, Amax[Y,.X,,o], satisfies the condition given in the theorem.

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Summary Zb

10.4.3 Determination of L.The relation between L,, and (L, Z), (10.60), can now be established directly form (10.88).As mentioned earlier, this relation enables Y,3 to be calculated directly from thestabilizing solutions to the HCARE and the HFARE since L depends on Yx and Zdepends on both X. and Y.

In order to establish (10.60), recall that Lo, was determined in (10.20) to be

Lo = -[Y,-3C0 + B1D2 ]DJo' (10.93)

Therefore substituting for Yx3 and C0 from (10.59) and (10.85) yields

L0 _ -[ZYxCT + (I + y ZZYxX,,)B,D2 DJo - ; 'ZYxC KLIn 01(10.94)

where

CIK = C1 + D11K1 + D1-K2

However from the definition of Z, (10.88), we have

I+7-2ZYOGXOC = Z(Z-1 +7-'Y, XX) = Z

so that (10.94) becomes

L, = +B1D2)D- - l 2ZYCl'I 0]=ZL -ry 2ZY"'C, [I 01.

= [Z(L1 +7-2Y".C ) ZL,]K

where L was obtained in Section 9.4.3.

(10.95)

10.5 SummaryIn this chapter we have shown that if the HCARE and HFARE developed in the previouschapter have stabilizing solutions, Xx, Y,, and provided these solutions satisfy

max[YxXxl <

then we can solve the Hx output feedback control problem. In addition we have shownthat when these conditions are satisfied the controllers, H(s), which solve the H,,, outputfeedback control problem are given by

U2 (S) = H(s) Y2 (S) H(s) = V,.(s)M- 1(s)

where

N,, (s) = W1(s)- (s) + W2(s)(s) xY Y

M, (S) = W3(s)c(s) + W4(s) }

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270 H. Output Feedback and Control

with W(s) having state model

W1(s)

W3 (S)

WAS)

W4 (S)

A+BK [B2D12 ZL2]

K2

D12[C2+D3KI D22D12 I01

Die =(Dl2D12)-1D12

and with K, L being determined from

Z= (I-ry-2Y.XX)-1

K=-Dj, (BTX.+DiC1)

L = - (Y,oCT + BID T )D-1

where X., Y. are stabilizing solution of the HCARE and HFARE, viz.,

ATa01 X + XA.1 - XR,1X + Q., = 0

A.2Y+YA002-YR.2Y+Q.2=0

with

A.1 C1 =A-B1DZDuoC=BDicI BT R,.2= CT DJ01C

Qoo1 =CTi (I - D1Dr,1 Di )C, B1 [I - Dz D-o D2]BT

[Di.]D1 = [D11 D12] D2 = D21

and

1_721m1 + D? D11 DT D12-T r TT n

T DDT

D21D11 2121

Conditions on the generalized plant state model parameters which ensure that theHCARE and the HFARE have stabilizing solutions were given at the end of Chapter 9.

10.6 Notes and ReferencesH, control can be used in conjunction with the small gain theorem (Section 7.4.3) toprovide a means for designing controllers to stabilize plants having inexactly specifiedmodels [47, pp. 221-229]. These ideas were extended recently to plants having a class ofnonlinear state models [31]. Additional material on the ideas presented in this chapter canbe found in [18].

-72In, +D11D1i DI D21 1Dj0_[

Page 286: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Appendix A:Linear Algebra

Most of the current "first" books on control systems with terms like "automatic control"or "feedback control" in their titles have an appendix containing a summary of matrixtheory. These can be consulted for this purpose. The topics treated in this appendix areselected to give the reader additional insight into issues alluded to in the text.

A.1 Multiple Eigenvalues and ControllabilityWhen some of the roots of the characteristic polynomial. det[AI -

A]for an n x n matrix

A are repeated, A may not have a complete set of n independent eigenvectors. Notice thatif

det[AI - A] = (A Al ' (A - A2)"' (A - ) (A.1)

where the Aks are unique a, i Al i 54j i, j E [1. m], then we say that Ak has multiplicitynk or is repeated nk times. Since the degree of the characteristic polynomial, (Al),.is n, thedimension of the A matrix, and since an nth degree polynomial always has n roots,counting multiplicities, we have

nk =Yik=1

Now there is at least one eigenvector, vkl. corresponding to each eigenvalueAk : k c [l, m]. However since Ak is repeated nk times. we obtain a complete set ofeigenvectors only if there are nk independent eigenvectors. t, k' : i = 1, 2, nk, corre-sponding to each Ak, k c [l, m], i.e., only if

At, - Ak ?;

has nk independent solutions v = vk', i = 1,27 ... nk for each k E I, fn].Any square matrix of size n having a complete set of n eigenvectors gives rise to a

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272 Appendix A: Linear Algebra

diagonal matrix under the transformation

V-1AV = A

where the n columns of V are eigenvectors of A, i.e.,

lV= [v11 v12 ... vln, V21 v22 ... V2n2 ... vm1 vm2 ... ,vmn,,,1

and A is a block diagonal matrix having n, x ni diagonal blocks, Ai : i = 1, 2, m, withAi at each diagonal position, i.e.,

A=

A 1 0 ... 0 ai 0 ... 01

0 A2 ... 0 0 ai ... 0

0 0 An I 1 0 0 ... Ai

Matrices having a complete set of n eigenvectors are referred to as diagonalizablematrices. Any matrix whose eigenvalues are simple, i.e., do not repeat, is a diagonalizablematrix. Alternatively, a matrix A that does not have a complete set of eigenvectors cannotbe diagonalized by any invertible matrix V in the transformation V-1A V. In this case A issaid to be defective, i.e., defective matrices are not diagonalizable matrices. For moredetail the reader is referred to [17, pp. 338-339].

Concerning the controllability of a state model having an A matrix with multipleeigenvalues, it turns out that when A has more than one eigenvector corresponding to anyeigenvalue it is not possible to control the state from one input, i.e., the pair (A, B) isuncontrollable for all B having only one column.

The foregoing fact can be shown as follows. Whenever A has two independent right-eigenvectors corresponding an eigenvalue A, A has two independent left-eigenvectors,w 1, w2 corresponding to this eigenvalue. Then any linear combination of these left-eigenvectors is also a left-eigenvector of A, i.e.,

w TA = .\wTwhere

1v=a1w1+a2w-

with the ais being arbitrary, constant, non-zero scalars. Therefore we can always choosethe ais for any vector B of length n so that

wTB=O

showing that A is an uncontrollable eigenvalue for all vectors B.

A.2 Block Upper Triangular MatricesBlock upper triangular matrices play an important role in many different situations incontrol theory. They are directly related to invariant subspaces. A subspace W of V issaid to be an A-invariant subspace if for all x E W we have Ax E W.

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Block Upper Triangular Matrices 273

Suppose we are given a square matrix of size n which has a block upper triangularstructure, e.g.,

A=At A2

A3 A4JA3 = 0 (A.2)

where At, A2, A3, and A4 are nt x n1, n1 x n2, n2 x n1, and n, x n, with n = n1 +n2. Thenany x of the form

x=t

X

0

where x1 is any vector of length n1, lies in an A-invariant subspace, since x c W

r 1

Ax = [AtxE W

0

In this case W is called an eigenspace corresponding to the eigenvalues of A which are alsoeigenvalues of A t . To see this consider the eigenvalue-eigenvector equation for A. We canrewrite this equation in the following way

Atv1+A1v' =Aivl (A.3)

A4v`, = A;r2 (A.4)

where

v!t

v =r

v,

is the partitioned form for the eigenvector v` corresponding to eigenvalue A,.Suppose that vZ is null. Then we see from (A.3) that t '1 must be an eigenvector of At

corresponding to eigenvalue A,. Thus the eigenvalues of At are eigenvalues of A i.e.A(A1) ci A(A) with the corresponding eigenvectors of A being

I0I

with A1v' = a;vlReal Schur decomposition [17, p. 362], provides a computationally robust method for

obtaining an orthogonal matrix Q such that

QTAQ

has the form (A.2). In addition there are ways of doing this so that the eigenvalues of A 1are a specified subset of the eigenvalues of A.

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274 Appendix A: Linear Algebra

A.3 Singular Value Decomposition (SVD)In this section we give a proof of Theorem 7.1, viz.,

Theorem 7.1 Any p x m matrix of complex constants, M, which has rank r can bedecomposed as

M = UEV' (A.5)

where U, and V are p x p and m x m unitary matrices with

U=[U1 U2] E= V=[V1 V2]

Ui ispxr U2ispxp-rVl ismxr V2ismxm-r

and Eo is an r x r, real, positive definite, diagonal matrix denoted as

Eo =

L 0 0 ... ar]

diag[a,, a2, ... , ar]

with diagonal entries referred to as singular values and ordered so that

ai > ai+i i = 1

Proof Let (a?, vi) denote an eigenvalue-eigenvector pair for M*M an m x m Hermitian non-negatve matrix. Thus we have

M*Mvi = a2vi (A.6)

with ai real and positive so that we can always order the eigenvalues as

a1>a2>...>ar>02 2 2

ar+I =ar+2 = =am=0

Notice that since rank[M] = r, only r of the eigenvalues of M'M are non-zero.In addition, since M*M is Hermitian, we can always choose the eigenvectors so that

they are orthonormal

I i=1,2,. mv;vi=0 fori j

(A.7)

(A.8)

Next we form a matrix V from the foregoing orthonormal eigenvectors of M"M so

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Singular Value Decomposition (SVV) 275

that (A.6) can be written as the matrix equations

M*MV1 = V1E, (A.9)

M*MV2 = 0 (A.10)

where

V = [V1 V2] Eo2 = diag[a1,U2,...,Qr]

V1 = [vl V2 ... Vr] V2 = ?r+1 1'r+2... vm

Notice that the orthonormal nature of the eigenvectors, (A.7, A.8), makes the m x rmatrix V1 satisfy V i V 1 = I, and the m x m - r matrix V2 satisfy V i V2 = I. Thus wehave V * V = VV * = I, i.e., the m x m matrix V is unitary.

Next, pre-multiplying (A.9) by Vi and then pre- and post-multiplying the resultingequation by E01 gives

which we can interpret as

E0-1V*M*MViE0-1 =I

U*U1 =I (A.11)

where

U1=MV1EO1

Notice from (A.11) that the p x r matrix U1 has orthonormal columns. Therefore wecan always choose a p x p - r matrix U2 having orthonormal columns so that the p x pmatrix U

U= [U1 U2j

is unitary, i.e., U* U = UU* = I.We show next that the foregoing unitary matrices, U. V satisfy (A.5) in the statement

of Theorem 7.1.We begin by expanding U*MV in terms of its blocks

U*MV=

U*JMV1 U*MV2

U*MV1 U;MV,(A.12)

Then we see that the 1,1 and 1,2 blocks can be reduced by substituting for U1 from(A.11) and using (A.9, A.10). Doing this yields

U'MV1 = EO 1 ViM*MV1 E(-) 1 Eo = Eo

UIMV2=Eo1V*M*MV2 =E(1u l1 =0

Next the reduction of the 2,1 block in (A. 12) relies on using the expression for U1,

Page 291: Linear Control Theory - The State Space Approach Frederick Walker Fairman

(A.11) to enable MV, to be replaced by U1 E0 so that

U*MV1 = UZ U1 Eo = 0

Finally in order to reduce the 2,2 block in (A. 12) we use the fact that MV2 = 0. This isseen by noting, from the construction of V2, (A.10), that

M*Mvi = 0

where v, is any column of V2. This implies that

v; M*Mvi = 0

and letting zi = Mvi we see that

which is only possible if zi is null. Thus Mvi must be null and since vi is any column of V2we have M V2 = 0. Therefore we see that the 2,2 block of (A.12) becomes

U*MV2 = U20 = 0

Thus from the foregoing we see that (A.12) is reduced to

U*MV =

and we have the form for the SVD of M which is given in the theorem.

Eo 0[o 01

A.4 Different Forms for the SVDThere are occasions where constraints imposed on the SVD by the dimensions of thematrix being decomposed can be taken into account so as to provide a more explicit formfor the SVD. This possibility arises from the fact that the rank of a rectangular matrixcannot exceed the smaller of its dimensions. Therefore since M is p x m, we have

rank[M] = r < min(p,m)

and the SVD of M, (A.5), can be expressed in one of the following forms depending onwhich of the dimensions of M is larger.

If p < m, i.e., M has more columns than rows, then (A.5) can be written as

M= UU[EE 01,VP*

Vp2

= L" E, Vpl

where U, V in (A.5) equal Up and [ Vp, V.21 respectively. In addition Vpl, Vp2 are

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Matrix Inversion Lemma (MIL) 277

m x p and m x (m - p) with Ep being a p x p diagonal matrix

EP = diag[a1, a2, ... , a,. 0..... 0]

In this case it is customary to refer to thep diagonal entries in EP as the singular values ofM and the p columns of Up and of Vpl as the left and right singular vectors of Mrespectively.

Similarly, if p > m, i.e., M has more rows than columns, then (A.5) can be written as

EmM = [ Uml U. . .2 11, 0

= Um1Em Vn,

where U, V in (A.5) equal [ Uml Um2 ] and V,,, respectively. In addition Uml, Um2 arep x m and p x (p - m) with Em being the m x m diagonal matrix

Em = diag[al, a2, ... a,.. 0, ... 0]

Again it is customary to refer to the m diagonal entries in E,,, as the singular values of Mand the m columns of Uml and of Vm as the left and right singular vectors of Mrespectively.

A.5 Matrix Inversion Lemma (MIL)The following relation has been of considerable importance in the development of controltheory.

Lemma If 1 and E are nonsingular matrices then

L-' = R

where

L=SI+WED

R=Q-' -E- -Do-

Proof Expanding the product RL yields

RL=I+Q-'wE-D -r1 -F,

where

(A.13)

r,1 c 1)-i

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278 Appendix A: Linear Algebra

Then we see that

F1+I2=S2 +E 'YE10

= S2 I E ) E + 4iSZ Y]E4i

(A.14)

Finally substituting (A.14) in (A.13) gives RL = I. 0

Page 294: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Appendix B: ReducedOrder Model Stability

As mentioned at the end of Section 4.6, since the full order model is stable, we require thereduced order model to also be stable. Otherwise the input-output behavior of thereduced order model cannot approximate the full order model behavior as the differencebetween the outputs of the full and reduced order models diverges in response to acommon bounded input. Since the reduced order model has a system matrix A 1 which is apartition of the full order system matrix Ab, in balanced coordinates,

Ab =A3 A4 mn-n1

Al Az Inl

we need a criterion which can be applied to ensure that Al has all its eigenvalues in theopen left half plane. A practical criterion which ensures this property was given, withoutproof, as Theorem 4.3 in Section 4.6. In this appendix we are going to prove this theorem.

Theorem 4.3 If in balanced coordinates, no diagonal entry in Ebl is the same as anydiagonal entry in i h2 then A 1 will be stable where

EhlWc=Wo=F'6=

0

Proof Suppose Al is not stable. Then we have A E A(A1, such that

A1v = Xv w*Ai = Aw* Re,A > 0

Now we are going to show that under the conditions of the theorem, Ab is unstable.However this is impossible since a balanced realization is only possible if the given systemis stable.

To begin, consider expanding the Lyapunov equations for the controllability and

0

r;,7

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280 Appendix B. Reduced Order Model Stability

observability Gramians as

AI A2 Eb1

A3 A4] 0

I

A*

I A3

11

EbI

A2 A4 0

bbl

Ill 0

0 1 [Ai

Eb2 A2 A4JA3

JBBJ[B*B2* ] (B.2)

EbI 011

Al A2 [C;l[ C1 C2] (B.3)

0 Ebz A3 A4 C2*

Then we see that EbI satisfies the following Lyapunov equations

AIE61 + Eb1Aj = -BIB*

AlEbl + Eb1A1 = -C*CI

(B.4)

(B.5)

Now since EbI is diagonal with all diagonal entries being positive and real we haveEbI > 0. Therefore we see from the proofs of Theorems 4.7 and 4.8 that the assumedunstable eigenvalue, A, of A 1 must lie on the imaginary axis and be an unobservable anduncontrollable eigenvalue of the pairs (A,, C1) and (A1, B1) respectively. Therefore thecorresponding right and left eigenvectors, (B. 1), satisfy

w*BI = 0 (B.6)

CIV = 0 (B.7)

Next we show that there are many eigenvectors corresponding to this eigenvalue. Inwhat follows we show how these eigenvectors are related.

To begin, postmultiply (B.5) by the eigenvector v, and use (B.7) to obtain

AT EbI V = -EbI AV or (Ebl v)*A l = -A* (Eblv)* (B.8)

However, A is on the imaginary axis. Therefore -A* = A and we see from (B.8) thatEbIV is an additional left eigenvector of Al corresponding to the uncontrollableeigenvalue A and therefore we have

v*EbIBI = 0 (B.9)

Next premultiply (B.4) by v*EbI, and use (B.8, B.9) to obtain

av*Ebl = -v*EbIAI or A,Eb1v = aEbIV

Thus we see that both v and E2h1 v are (independent) right eigenvectors of A 1 correspond-ing to the eigenvalue A.

Continuing in this fashion we can generate sets of left and right eigenvectorscorresponding to A as

S,,. = {E;,lv: 1 1,3,5;---,2k+ 1} (B.10)

(B.11)

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Appendix B: Reduced Order Model Stability 281

where k is an integer which is chosen large enough so that we can find a matrix Q to satisfy

Eb1M = MQ (B.12)

where M is the n1 x (k + 1) matrix

M= Iv 2 2k1v ... Ehlz'b

Notice from (B. 12) that under the transformation Ebl, the range of M is unchanged. Thisfact plays an important role in what is to follow.

Now we saw in Theorem 5.5, that (B12) implies that every eigenvalue of Q is aneigenvalue of E2b1

A[Q] c a[Ebl] (B.13)

This important fact will be used to show that A3M = 0. Notice that since the elements inS S, are left and right eigenvectors, respectively, of A I corresponding to the uncontrol-lable and unobservable eigenvalue A, we have

v*Eb1B1 = 0 i = 1 3. . 2k + 1 (B.14)

C1Eb1v=0 i=0,2, 2k (B.15)

Returning to (B2, B.3) we have

A2Eb2 + EbIA3 = -BIB; (B.16)

A2E61 + Eb2A3 = -C;CI (B.17)

Then premultiplying (B.16) by v*Ebl, and postmultiplying (B.17) by E',1v and using(B.14, B.15) gives

Eb2A2Ebly = -A3E HIV i = 1.3... 2k+ 1

A*Eh 1 v = -Eb2A3Ehly i = 0.2..... 2k

Next we can rewrite (B. 18) as

Eb2A2 IEe1V Eb1 V

and (1319) as

2k IEhl _ -A'Eb1M

Eb2A2[EblV Ebly ... Ebi l -Eh2A3M

(B. 18)

(B.19)

(B.20)

(B.21)

where Eb2 was used to premultiply (B.19) and M was defined previously in (B.12).Now since the left sides of (B.20, B.21) are equal we have

A3Eb1M = Eb2A3M (B.22)

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282 Appendix B: Reduced Order Model Stability

Moreover we can use (B.12) to rewrite the expression on the left side of this equation.Doing this gives rise to the following Lyapunov equation

A Q - EbzO = 0 (B.23)

whereA=A3M

Now it turns out if Q and E62 do not have any eigenvalues in common then 0 = 0 isthe only solution to (B.23). Indeed Q and Eb2 do not share any eigenvalues since theeigenvalues of Q were shown to be a subset of the eigenvalues of Eb1, (B.13), and from thestatement of the theorem we have that no eigenvalue of Eb is an eigenvalue of E62. Thuswe have A3M = 0.

Since the columns of M consist of all the right-eigenvectors of A 1 corresponding to theimaginary axis eigenvalue A of A1, we have that (A, q) is an eigenvalue right-eigenvectorpair for the full system matrix Ab where

q- [0v]

since

A,

Abq _ IA3 A4] Lo] =

A[] = Aq

However, since A lies on the imaginary axis, Ab is unstable. This contradicts the factthat A,, is stable. Therefore our assumption at the start of the proof that A 1 is unstable isfalse and the proof is complete.

Page 298: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Appendix C:Problems

The following problems illustrate the basic ideas introduced in the earlier chapters of thebook. Problems dealing with more advanced principles are under development.

C.1 Problems Relating to Chapter 11. Use the fact that any matrices M2 and M3 have the same eigenvalues as the matrix

Ml ifM=MT

2 1 M3 = S-1M1S

for any nonsingular matrix S, to determine relations between the unspecifiedelements in the matrices Ak : k = 1, 2, 3, 4, so that each matrix has the same set ofeigenvalues. If this set is {1, -2, 3} specify the values for the akjs.

all a12 a13

Al = 1 0 0

1

A2 = a,l 0 1

0 1 0 La31 0 0

0 0 a13 ro 1 0

A3 = 1 0 a23 A4 = 0 0 1

0 1 a33 L a31 (132 a33

2. Find the eigenvalues {ak; : i = 1, 2} for each of the following system matricesAk : k c [1, 6]. Then calculate the corresponding right eigenvectors, vk forAk : k = 1, 2 letting the first component of each eigenvector be one.

Ol 1 A2 = I -l 3-)

J

, { 41J

-5rA4 = [ 12 131

AS -6 2 1A

2 2

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284 Appendix C. Problems

3. Using Laplace transforms, and without calculating numerical values, determinewhich of the transition matrices corresponding to the Aks, k E [1, 6], given in theprevious question have entries dependent on only one mode, e

4. A certain system has a zero-input state response of

where

x, (t) = 6e8' + e-2t x2(t) = 8e81 - 2e-2t

Find, without using Laplace transforms, the eigenvalues and eigenvectors of the Amatrix for the system's state model.

5. Using the trajectories obtained when the initial state is an eigenvector, sketch thestate trajectory for the given initial states x`(0) : i = 1, 2 and system matrix A. Useyour result to determine a relation between the elements of C, namely, c,, and c2i sothat the system output is bounded for all initial states.

A= [48

2] x1(0)=

[_l]x2(0)-

[431

6. Calculate the eigenvalue left-eigenvector pairs, {a;, w` : i c [1, 2]}, for A given in theprevious question. From this result determine an output matrix C such that for anyinitial state the zero-input response y(t) depends only on the mode a 2'. What is therelation between the C matrix you obtain here and the C matrix you obtained in theprevious question.

7. Suppose a state model has system matrix, A, which is partially specified as

A=a 2

1 0

If the last component of each right-eigenvector is 1, find:

(i) the initial state, x(0), which gives a zero input response of

y(t) = 4e-t when C = [ -1 1]

(ii) the output matrix C which gives a zero input response of

y(t) = 4e-twhen x(0)

21

8. Recall that if q5(t) is a transition matrix then

(i) 0(0) =I (ii) q(t) = AO(t)

If M(t) is a transition matrix

- 3 2e-2` ate' 2 e - 2M(t) [aie'+03e` - e 21 a4el + e 2t

find the ai.s and the corresponding system matrix A.

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Problems AilAtn4 I CROW I M5

9. Determine diagonal form state models for systems having transfer functions

G1(s) _3s-1 s2-3s+2

G (s)2s2+3s+2 s2+3s+2

10. If the transfer function for a certain system is

G(s) =s+4

(s+1)(s+1+j)(s+1-j)

find the real parameters a1, a2, a3, c1, c2, c3 in the state model for this system where

a1 0 0 1

A = 0 a2 a3 B= 1

0 1 0 0

C=[C1 C2 C3] D=0

11. Use the eigenvectors of A to find B (other than the null vector) so that the state goes tothe origin with time when the input is a unit impulse. u(t) = b(t) and the initial state isnull. It is known that one of the eigenvalues of A is at +l and

+2 +1 -2

A= 1 0 0

0 1 0

12. If the steady-state output is zero when the input is a unit impulse, u(t) = S(t), find(without using Laplace transforms) the values for h2 and b3 when

A=2 -3/2 4/3 1

0 -1 8/3 B = b,

0 0 3 b;

C=[1 1 1] D=0

C.2 Problems Relating to Chapter 21. A certain plant is known to have a controller form state model with a system matrix,

A, which has eigenvalues at {-1, 1,2, 3}. Further, this state model for the plant isknown to have direct feedthrough matrix, D. which is zero. If the plant's steady stateoutput is y(oc) 5 when the plant input and initial state are u(t) = I and x(0) = 0,find the controller form state model for the plant.

2. Suppose we are given the transfer function G(s) for some plant. If the state model forthe plant is in controller form, determine the initial state, x'(0), in each case, so that

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286 Appendix C. Problems

the corresponding output, y;(t), is as specified. The input is zero.

y, (t) = 12e-' y2(t) = 6e-'+ 12e-2'

G(s) - (s + 4) (s + 5)(s + 1)(s + 2)(s + 3)

3. Given that A[A] = { 1, 2, -3} determine left-eigenvectors of A, w', assuming the firstentry in each left-eigenvector is one. Use your result to find a and /3 in B so that x(oc)is null when u(t) = 6(t) and x(0) = o

0101

1

A= 7 B= a-6 0 0 13

4. Suppose there is a set of three state models each having the same A matrix butdifferent B and C matrices as given. Determine which models if any can betransformed to controller form. Do this without using Laplace transforms.

case(i) Bl =L 2 J

C, = [1 1]

4 3A=

[

case(ii) B2 = 14 C2=[2 1]8 2 J

11case(iii) B3 = 1 C3 = [-4 3]

5. Specify for each state model, { (A, B,, C;) : i = 1, 2, 3}, given in the previous problem:(i) the controllable eigenvalues, (ii) the uncontrollable eigenvalues. In addition,specify which of these three state models is stabilizable.

6. Given the state models

AI =

1 0 -1 0 0

B1= 2 A2= 1 0 1 B2= 0

0 0 0 0 1

C1=[1 0 0] D1=0 C2=[-1 1 1] D2=0

find, without using Laplace transforms, a coordinate transformation matrix T ineach case so that the state models in the new coordinates are each in controller form.Specify T and the controller form state model, (A, B, C) in each case.

7. Use your knowledge of the controllable decomposed form and the eigenvalues oflower triangular matrices to rapidly determine the polynomials

p, (s) = det[sI - A] p2(s) = Cadj[sI - A]B

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Problems Aelefln# to Chapter 3 zaz

if the state model parameters are

0 7 -6 0 0 0

1 0 0 0 0 0

A= 0 1 0 0 0 B= 03 5 -2 -3 -2 1

-4 6 4 1 0 0

C=[3 -1 2 1 1] D=0

Determine the transfer function which is free from coincident poles and zeros, i.e.,whose numerator and denominator are coprime.

8. Determine B so that the steady state output is 1(x) = 10 when u(t) = 5 for allpositive t and

+2 1 0

A= +1 0 1 C=[1 0 0]

-2 0 0

C.3 Problems Relating to Chapter 31. Find the numerator polynomial, Num(s), and denominator polynomial, Den(s), of

the transfer function G(s) corresponding to the following state model

-5 1 0 0- ro

A=-10 0 1 0

B =0

-10 0 0 1 I

4- 0 0 0- 4

C=[1 0 0 0] D=1

where

G(s) =Num(sDen(s)

2. Suppose there is a set of three state models each having the same A matrix butdifferent B and C matrices as given. Determine which models if any can he

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288 Appendix Ct Problems

transformed to observer form. Do this without rusing Laplace transforms.

case(i) B, =L

21J

Cl =[I 1]

A =[S

2]case(ii) B2 = [4J C2=[2 1]

case(iii) B3 = [ 1J

C3 = [-4 3]

3. Specify for each state model, J (A, B;, C1) : i = 1, 2, 3J}, given in the previous problem:(i) the observable eigenvalues, (ii) the unobservable eigenvalues. Specify which ofthese state models allow the design of an observer to estimate their state .

4. Suppose a plant state model is in controller form and is 3 dimensional with A havingeigenvalues at 1,2,-3. If the C matrix in this model is partially specified as

C=[1 -2 q]

determine values for q which would make it impossible to design an observer whichcould be used to obtain an asymptotic estimate of the plant state.

C.4 Problems Relating to Chapter 41. If every entry in a square matrix is either 1/2 or -1/2 determine the size of the matrix

and the sign pattern of its entries if the matrix is orthogonal.2. Find real numbers a, b, c1, c2, and c3 such that M is orthogonal where

a b clM.a 0 c2

a -b c3

3. M, and M2 are square matrices having eigenvalue - right eigenvector pairs as follows:

for M,: f0,11 and {[1,2]T,[2 - 1]T}

forM2: {0,1} and {[1,2]T,[2 1]T}

respectively. Is either of these matrices symmetric? If so which one. Justify youranswer.

4. For each matrix, specify which of the following properties holds: positive definite,non-negative, indefinite, i.e., not positive definite, not nonnegative, not negativedefinite. (Hint: use the distinguishing properties of the eigenvalues of each type ofmatrix)

M,=[ 12 521 M2=[2it

1 0 0 1 0 2

M3= 0 2 0 M4= 0 0 0

0 0 3 2 0 1

Page 304: Linear Control Theory - The State Space Approach Frederick Walker Fairman

PiOSNMF *MMV r GM "Mr IF zes

5. Which of the following matrices could be either a controllability or an observabilityGramian. Give reasons for your answer in each case.

M1 =

6

M2 = -'l13 =

The observability Gramian for a system is determined to be

Is there an initial state vector, x(0) other than the null vector, which when the input,u(t), is always zero, produces an output, y(t), which is always zero? If "no" justify. If"yes", give an example, i.e., give an x(0), other than the null vector, which has thisproperty.

7. If, in the previous problem, the initial state vector is restricted such thatxT(0)x(0") = 1, (and u(t) is always zero), find the largest value possible for theintegral from 0 to oo of the square of the output signal v(t) under this restriction.Specify x(0) (subject to the restriction) which would produce this value.

8. Determine, in each case, if the pairs (A, C) are observable. Calculate the observabilityGramian manually by solving the appropriate Lyapunov equation in each case.Determine, in each case, if the corresponding Gramian is positive definite. Relateyour findings on the positive definiteness to your findings on the observability in eachcase

case(i): A =

case(ii): A =

C= '1 1]

C=`0 1]

9. Determine the observability Gramian for the given system assuming that the r;s arereal.

0 1

oil

0

A= r1 0 B= 0

0 r2 r3 1

C= [0 0 1] D= 0

What values of the r;s make the Gramian singular" What values of the r,s make thesystem unobservable? Are these values related? if so explain.

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290 Appendix C: Problems

10. A certain plant has state model matrices (A, B) and observability Gramian Wo givenas

Given that D = 0 and recalling that the integral defining W° does not exist when A isunstable, find the transfer function of the plant using the Lyapunov equation relatingA, C, WO to find a and C.

11. Suppose that a certain system is known to have maximum output energy equal to 2for all possible initial states satisfying x (0) + x2(0) = 1, when u(t) - 0. Suppose theinitial state that achieves this is xT(0) _ [ f]. In addition suppose that this systemhas minimum output energy equal to 1 for some initial state which also satisfiesx2 (0) + x2(0) = I when u(t) -- 0. Find the system's observability Gramian and theinitial state that gives this minimum output energy.

C.5 Problems Relating to Chapter 51. An observer based controller (Fig. 5.1) generates a feedback signal, f (t), from

f (t) = Hz(t)

z(t) = Fz(t) + Gov(t) + G2y(t)

where

-6 1 O1

0

0 12

F= -11 0 1 G, = 0 1 Gz = 0 HT = 0

-18 0 0- 1 12 0

If the plant output equals the first component of the plant state vector, find: (i) theclosed loop transfer function from v(t) to y(t), (ii) the characteristic polynomial forthe observer.

2. If in the previous problem the plant state model is in controller form and the plantoutput is the sum of the first three components of the plant state, determine the closedloop transfer function from v to y for the observer based feedback control systemwhere

F=

0 4 0 0

-1 -4 0 0

5 7 -13 -80 0 13 0

G1 =

0 0 -1

0 0 T 1

1 GZ 10H

6

0 -12 0

3. If in the observer based feedback control system shown in the Figure 5.1 we are giventhe transfer functions from v(t) to y(t), Gy(s), and from y(t) to f (t), Gyf(s), where

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Problems for Cnap[ar3 xVF

f(t) is the feedback signal, determine the controller state model matrices (Q, R, S, T)

when

1 -2G,,y.(s)

s3 + 3s2 +2s+ 1

with plant model parameters given as

-3 -2 0

A= 1 0 0

0 1 0

G, f(s) _ s't3.s-+2s+3

CT =0

0

1

and controller equations

z(t) = Qz(t) + Rv(t) + Sy(t) f (t) = T7( t)

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Page 308: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Appendix D:MATLAB Experiments

D.1 State Models and State ResponseThe intention of this lab is to introduce some of the , avs MATLAB can be used inconnection with state models and to provide further experience in thinking in terms of thestate space.

D.1.1 Controller formGiven the transfer function model for some system, the command

[a, b, c, d] = tf 2ss(nurn. den)

can be invoked to find an equivalent state model, (use the help facility), where num andden are entered as row vectors containing coefficients of the numerator and denominatorof the transfer function. Try the following examples

1

) =G ( .s, -s`+3s+?

G (s)s2 2s + 3

z3s2 2s+ 1

Check your result by using the command "ss2tf '. (use the help facility), to convertback to the transfer function models.

D.1.2 Second order linear behaviorRecall that the state at any time is represented by a point in an n-dimensional space andthe path traced out by this point is referred to as a state trajectory. Unlike cases where thesystem order is greater that 2 , we can readily display the trajectory for a second ordersystem as a graph with the two components of the state forming the axes of the graph.

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294 Appendix D: MATLAB Experiments

Consider the system whose behavior is governed by the following second order lineardifferential equation

yl2l(t) +2Cwnyl'l (t) w.2u(t)

where

yU> (t) = dal t)

Then the controller state model is

2-2(Wn -Wn Br lA

1 0 0

C= [0 w2 D=0

Now we are going to investigate the zero-input state response by obtaining the statetrajectories for one value of natural frequency, say w = 1 and different values of thedamping ratio, say

(=0.3,1,10,-0.3

each for some initial state, say xT(0) = [2 0]One way to do this is to invoke the command, (use the help facility),

initial(A, B, C, D, A)

Then from the resulting graph of the output vs. time that is obtained on the screen, youcan decide over what interval of time you will need the state.

Suppose you decide you need to know the state in the time interval [0, 5] seconds every0.5 seconds. We can create a vector, i, of these times by incorporating this data in a vectoras start: increment: end. This is done here by typing

t=0:0.5:5

and

[ y, x, t] = initial(A, B, C, D, x0)

This produces a column y, followed by two columns, one for each component of thestate. The entries in these columns give the values of the output, and components of thestate at 0.5 s intervals from 0 to 5 s. Now you can obtain the graph of the state trajectoryby entering

plot(x(:, 1), x(:, 2))

where x(:, i) is all rows in the ith column of x. Your graph can be labelled using the

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State Models and State Response 295

procedure (on line help) where you enter each line (there are four) on the same lineseparated by commas.

Explain why these plots are markedly different for different signs on (.

D.1.3 Second order nonlinear behaviorConsider the system whose behavior is governed by the second order nonlineardifferential equation, referred to as a van der Pol equation,

Y121 (t)- (1

-Y2(t))YI'i(t) i,(t) = 0

Then if we assign componets of the state to derivates of the output in the same manneras done to get the controller form in the linear case, the behavior of the components of thestate are seen to be governed by the following first order nonlinear differential equations

x,(t) _ (1 -xz(t))xl(t) x,(t)

z2(t) =xI(t)

Now the state trajectory obtained for a specific initial state can be determined usingMATLAB by creating an m-file called vdpol.m. To create this file open an editor and type

functionyp = vdpol(t, q);

yp = [(1 - q(2) * q(2)) * q(1) - q(2); q(l)];

Save the file and invoke one of the ordinary (vector) differential equation solvers, sayODE23

[t, q] = ode23(`vdpol`, [t0, tf". q0)

where tO is the start time, tf is the finish time and qO is the initial state, entered as a columnvector by typing

qO = [xi(0);x2(0)-

The sample values of the components of the state are returned as the columns of q.Use t0 = 0, tf = 25 and try getting the state for each of the initial states

10

0

21and

0

Obtain a graph of the state trajectory in each case by typing

plot(q(:, 1). q(:. 2)

Notice that q(:, 1) is all the elements in the first column of q produced by the simulation asthe values of the first component of the state.

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296 Appendix D: MATLAB Experiments

In the second case where xT (0) = [2 0] it is instructive to concentrate on the latter partof the trajectory, say from time step 40 to the end of the simulation. To find out what thisend value is type

size(q)

Then if this end value is 200 type

plot(q(40 : 200, 1),q(40 :200,2))

to obtain the trajectory from time 40 to time 200.Generalize the nature of the trajectories in terms of their dependency on the initial

state. Compare the state trajectories you obtain for the present nonlinear system withthose you obtained for the linear system in the previous section.

D.1.4 Diagonal formSo far we have only considered the controller form state model. There are several otherimportant canonical forms for the state model of a linear dynamic system. One of these isthe diagonal or normal form. This form is characterized by having its system matrix Awhich is diagonal. This simplifies solving the state differential equation since thederivative of each component of the state is dependent on itself only.

Recall that if [A, B, C, D] is a state model for a given system so is [Q-1AQ, Q-1B, CQ,D] for any nonsingular matrix Q of appropriate size, where Q is a coordinate transforma-tion matrix. Also recall that if the eigenvalues of A are distinct, we can change coordinatesso that A in the new coordinates is diagonal. This is done by letting Q have columns equalto the eigenvectors of A. We are going to use these facts.

Use MATLAB to determine the parameters, (A, B, C, D), of the controller form statemodel for the system having transfer function

G(s)s2 - s+20

s4+5s3+5s2-5s-6

Use the command "roots" to decide if all the roots of the denominator are poles ofG(s). Calculate the eigenvalues and eigenvectors of A by typing

[Q, E] = eig(A)

and the characteristic polynomial of A by typing

poly(A)

How do the poles of G(s) compare with the eigenvalues of A, i.e., diagonal entries in E?Should each of these sets be the same and if not why not?

Use Q to transform the controller form state model to a diagonal form state model.

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Feedback and Controllability 297

This can be done by typing

aa = inv(Q) * A * Q

bb = inv(Q) * B

cc = CA * Q

dd = D

How do the diagonal entries of as compare with the poles of G(s)?Finally, invoke the command "residue" to determine the residues, poles and any

constant term D (nonzero only if G(s) is proper). This is done by typing

[res, poles, d] = residue(nuni, den)

How do the residues of G(s) relate to the entries in bb and cc?

D.2 Feedback and ControllabilityIn order for you to proceed as rapidly as possible with the tasks described in what follows,you should plan a subscripting scheme for the various matrices you will be entering in thecomputer.

D.2.1 Controllable state modelsEnter the state model

2 1 5 9 3 1

0 -3 -6 3 2 bA= 0 0 1 0 0 B= b3

0 0 4 a44 0 b4

0 0 5 7 -6 b5

C=(1 1 1 1 1] D=0

with the following assignment to the variable entries

b; = 1 : i = 2,3,4.5 and a44 = -2

Obtain the poles and the zeros of the transfer function using the command "ss2zp".Are there any pole-zero cancellations indicating that part of the state space is superfluousto the input-output behavior?

Next obtain the controllability matrix using the command ..ctrb" and check its rankusing the command "rank".

Using the command "eig" find the eigenvalues of A. Then find the state feedback gainmatrix K which shifts the unstable eigenvalues to -1. Try doing this with both thecommand "place" and the command "acker". Check using the command "eig" to seethat the eigenvalues of the state feedback system matrix .4 - BK are those you specified.

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298 Appendix D. MATLAB Experiments

Obtain the poles and zeros of the state feedback system using the command "ss2zp". Arethere any pole-zero cancellations? What effect does the state feedback have on thetransfer function zeros? Do other experiments if need be so that you can give a generalreasons for your answers to the foregoing questions.

Obtain the feedback matrix which produces a closed loop transfer function which hasno finite zeros (they are all cancelled) and poles which are all in the open left half-plane.Does the rank of the controllability matrix change. Explain your answer by usingeigenvectors to identify uncontrollable eigenvalues.

D.2.2 Uncontrollable state modelsIn this section we are going to examine the use of state feedback when the given statemodel for the plant is not controllable. One way to obtain an uncontrollable system toenable this study, is to form the input matrix B so that it is linearly dependent on a subsetof the eigenvectors of A. We do this as follows.

Invoke "eig" to get the eigenvectors of A. These will be displayed in a square matrixyou specify in the argument for "eig". Suppose this matrix is V and D is used as thediagonal4natrix displaying the eigenvalues. Create a new input matrix which we call BIby combining a subset of the columns of V, i.e., some of the eigenvectors of A. Supposewe do this by combining all the eigenvectors corresponding to the unstable eigenvalues ofA with all the weights used to combine these eigenvectors equal to 1. This can be doneusing the MATLAB subscripting rules (note the use of :).

Obtain the controllability matrix ("ctrb") for (A, BI) as well as the rank of thiscontrollability matrix. Obtain the transfer function for the state model (A, Bl, C, D) andcomment, with an explanation, on any pole-zero cancellations. What is the order of thesystem indicated by the transfer function and does the transfer function indicate that thissystem is stable?

To gain more insight into the nature of uncontrollable state models, invoke "ctrbf" totransform coordinates so that the state model. is in controllable decomposed form,(Section 2.5). In this case we see that the block matrix consisting of the last two rows andcolumns of A and the last two elements of B in these coordinates has special significance.If (Abu are the system and input matrices in the controllable decomposed form thenwe can extract the subsystem corresponding to these rows and columns by typing

AC = A ,, , , , . ( 4 :5 , 4 :5 )

BC = Bbar (4 : 51 1)

CC = Cbcr (1, 4: 5)

Then calculate the transfer function using the command"ss2zp" for the system(AC, BC, CC) and compare it with the transfer function you obtained earlier for thesystem in the original coordinates. Explain what you see.

Next can you find a state feedback matrix K in these coordinates so that the resultingsystem is internally stable?

Repeat the foregoing by making up a new B matrix from a linear combination of theeigenvectors corresponding to the stable eigenvalues of A. Notice that the new AC matrixwill be a 3 x 3.

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Observer Based Control Systems 299

Controllability and Repeated EigenvaluesRecall that we showed in Appendix A that single-input systems are uncontrollable for anyinput matrix B when the system matrix A has more than one eigenvector associated withany repeated eigenvalue. To investigate this fact we proceed as follows.

Assume A is as given at the beginning of this lab. We can recall A by typing A. Then usethe command "eig" to determine the eigenvalues and eigenvectors of A. Are theeigenvalues distinct? Next change the 4,4 element in A by typing

A(4,4) = A(4,4) + 3

Find the eigenvectors and eigenvalues of the new A matrix. Are the eigenvaluesdistinct? Is there a complete set of eigenvectors? Construct an input matrix B so that thenew (A, B) pair is controllable. Explain.

D.3 Observer Based Control SystemsRecall that an observer is used to provide an estimate of the plant state from measure-ments of the plant's input and output. In this lab you will examine the design andbehavior of observers and the effect of using an observer's estimate of the plant state inplace of the actual plant state in a state feedback control system.

In order for you to proceed as rapidly as possible with the tasks described in whatfollows, read the lab through and plan the subscripting scheme you will use in connectionwith the various matrices you will need to enter in the computer. This is especiallyimportant in connection with the second section where you will need to make a blockdiagram of the control system with each input and output numbered for use with thecommands "blkbuild" and "connect".

Observer Estimates of the Plant State[To Be Done Prior to Using Computer]Suppose you are given the unstable second order plant

dd2 d t)2+a

t+a71'(t) = u(t)i dt

with ai = 0 and a2 = - 1. Write the observer form state model for the plant and design anobserver (Section 3.3)

z(t) = Fz(t) + Gu(r)

r(t)

where

F=A+LC G=1B-LD -L

with F having eigenvalues at -1, -2.Now you will attempt to verify experimentally using MATLAB that the observer state

and plant state approach each other with time. Use the eigenvalues of the observer and

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300 Appendix D: MATLAB Experiments

plant state model to determine a suitable length of time over which to let the simulationoperate.

[To Be Done Using Computer]Begin by entering the observer form plant state model matrices: A, B, C, D. Then use

the command "acker" or "place" to determine the observer matrix L so that the observereigenvalues are at - I ,-2. This is done by typing

L = acker(AT, CT, p)T

where p is a column vector of the desired observer eigenvalue locations. Construct F andG and check the eigenvalues of F using the command "eig". Now we are going to checkthis observer's ability to track the components of the state of the plant by generating boththe plant and observer states when the plant input is a unit step.

Invoke the command "step" to obtain the plant output, y and state x. Try using a timeinterval of ten seconds in steps of 0.1 s. In order to do this you will need to create a rowvector t having elements equal to the computation times. This can be done by typing

t=0:0.1:10

You will need tin the last section of this lab.Next use the command "Isim" to obtain the state of the observer when the plant input

is a step and y is the plant output in response to a step. This can be done by typing

[y 1, z 1 ] = Isim (F, G, CI , DB, UB, t)

where

UB = [uu, y] DB =[0; 0]

and uu is a column vector of l's of the same length as y with y being a column vector ofsamples of the plant output you just computed. The input sequence, uu, could begenerated by typing

uu = ones(101,1)

Once you have zt, the state of the observer, (a matrix having two columns of samplesof the components of the observer's state) you can compare the observer state, zl, justobtained, with the plant state x by generating the error

err = (zl - x)

However recall that since the plant is in observer form, the plant output is the firstcomponent of the plant state. Therefore we need only compare the second components ofthe plant and observer states in order to characterize the performance of the observer as astate estimator. In addition in attempting to comment on this performance, you mightfind it helpful to examine the relative error, i.e.,

zl (:, 2) - x(:, 2)/x(:, 2)

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Observer Based Control Systems 301

For instance if we want to compute this for time 10 we would type

[zl(101, 2) - x(101, 2)]/x(101.2)

Now the initial state of the plant in the foregoing generation of the plant output is null.However in practice we do not know the initial state of the plant. Therefore to test theobserver's ability to estimate the plant state when the initial state of the plant andobserver are not equal, suppose you thought the initial plant state was xT(0) _ [1, 1].Then setting the initial state of the observer equal to your assumption, use the command"lsim" with the observer, i.e., type

z0 = [1, 1]

[ y2, z2] = lsim(F, G, C, DB. UB. t. z0)

Next obtain the history of the estimation error over the simulation time and obtain aplot of this history. This can be done using element by element division by typing

rele = (z2 - x). /x

Notice that since the actual initial plant state is null. the elements in the first row of thematrix xs are zero, so that the first listing in "rele" has NaN (not a number) or no as itsfirst entry. Try other guesses for the "unknown" initial plant state, e.g., set the initialobserver state to zO = [100; 100] and obtain a plot of the resulting relative error in theestimate of the plant state.

D.3.1 Observer based controllers[To Be Done Prior to Using Computer]Suppose you want to use the observer studied in the previous section to implement a

state feedback controller. In this section you will begin to study the interaction whichtakes place between the plant and the observer when they are connected via a feedbackmatrix K in a closed loop configuration.

Choose the feedback matrix K so that if exact state feedback were used the closed loopsystem would have eigenvalues at -3, -4. Do this for the plant state model in observer formwhich you used in previous sections.

Next referring to Section 5.2, obtain the system equations for the feedback system, i.e.,let the plant input u be given as u = Kz + v. The state vector q for the closed loop systemhas four entries, viz.,

7 T'Tq=[x ,

where x is the state for the plant (in observer form) and = is the state for the observer youdesigned in the previous section.

[To Be Done Using Computer]Use the command "acker" to find the feedback matrix K so that the plant state model

(in observer form) has its eigenvalues shifted to -3, -4. Then denote the observer based

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302 Appendix D: MATLAB Experiments

controller state model as

[a3, b3, c3, d3] = [F, G, K, d3] d3 = [0, 0]

where F,G, were determined in the previous section and K was just determined.Now you are going to use the commands "blkbuild" and "connect" to form the

observer based feedback control system by interconnecting the various subsystemsinvolved. This is done in two steps. First enter each subsystem, either as transferfunctions, (ni.di) or as state models (ai, bi, ci, di). In each case i should identify a differentsubsystem and should be the integers starting with one. In the present case there are threesubsystems: the input-controller (i = 1), the plant (i = 2) and the observer-controller(i = 3). The input-controller is trivial and is done by entering its transfer function asn 1 = I and d I = 1. Then the plant and observer-controller can be entered in terms oftheir state models as [a2,b2,c2,d2] and [a3,b3,c3,d3] respectively.

Once you have entered all the subsystems, type

nblocks = 3

blkbuild

A message appears "state model [a, b, c, d] of the block diagram has 4 inputs and 3outputs". You are now ready to proceed to the second step which involves specifying howto interconnect the subsystems you have just identified. This is done through the use of aconnection matrix, Q.

The connection matrix, Q, specifies how the subsystems inputs and outputs are to beconnected to form the observer based control system. Notice that the outputs yI, y2, andy3 are the outputs from the input-controller, the plant and the observer-controllerrespectively. The inputs ul, u2, u3 and u4 are the input to the input-controller, the plantand the two inputs to the observer. It is important to keep the ordering of the inputs to theobserver consistent with the ordering used in the previous section in forming the statedifferential equation for the observer.

Draw a block diagram showing how all inputs and outputs relate to one another.The connection matrix, Q, has rows corresponding to subsystem inputs with the integer inthe first position of each row identifying the subsystem input. Subsequent entries in agiven row are signed integers which identify how the subsystem outputs are to becombined to make up the subsystem input. Zero entries are used when a subsystemoutput does not connect to a subsystem input. The sign associated with each integer entryin Q indicates the sign on the corresponding output in the linear combination ofsubsystem outputs making up a given input. Since there are three inputs which dependon the subsystem outputs, Q will have three rows.

Using your block diagram, enter the matrix Q.Once Q is entered, the inputs and outputs of the feedback control system must be

declared by typing

inputs = [1]

outputs = [2]

Page 318: Linear Control Theory - The State Space Approach Frederick Walker Fairman

State Modei i educilbf"i 3U

where here the input to the input-controller, i = 1, is the observer based feedback controlsystem input and the output from the plant, i = 2, is the output from the feedback controlsystem.

At this stage you are ready to use the connection information in the Q matrix to formthe observer based feedback control system from the multi-input multi-output system[a, b, c, d] you formed using "blkbuild". The connected system [ac, be, cc, dc], i.e., theobserver based feedback control system, is obtained by typing

[ac, be, cc, dc] = connect(a, b, c, d, Q. inputs. outputs)

Check your result with the state model you obtained in the pencil-and-paper work youdid before starting this computing session.

Once you have the correct closed-loop state model, calculate the poles and zeros of thetransfer function for the closed-loop system by using "ss2zp" (don't use k on the left sideif you have already used it for the feedback matrix). Are there any common poles andzeros indicated? If so, why? Compare this transfer function with the transfer function ofthe closed loop system assuming exact state feedback (without the observer). Discussyour observations by relating them to the theory of observer based control systems.

Finally, check the controllability and observability of the closed loop state model byusing the commands"etrb" and "obsv" and computing ranks by using "rank".

D.3.2 Observer based control system behavior[To Be Done Prior To Using Computer]For the plant introduced in the first section, and assuming the states are available for

feedback, calculate the output for a step input for the closed loop system with statefeedback such that the closed loop system has poles at -3, -4. Do this for the initialcondition xO = [0, 0] and xO = [10, 10]. Make a rough sketch of the output in each case forcomparison with the results you will obtain using MATLAB.

[To Be Done Using Computer]Using "Isim" obtain the state and output of the state model of the observer based

control system that you obtained in the previous section as a result of using "blkbuild"followed by "connect", i.e., [ac, be, cc, dc]. Do this over a ten second interval in steps of 0.1s (use the t vector and step input vector, uu, which you created for use in the first part ofthis lab). Obtain a plot of the output when the plant state is .v(0) = [0, 0] and [10, 10] whenthe observer state is 70 = [0, 0] in each case.

D,4 State Model ReductionIn previous labs on state feedback and state estimation we saw that controllability andobservability play a key role in being able to (i) assign feedback system poles (ii) assignobserver poles. Recall that the transfer function model of a system has order less than thedimension of the state model when the state model is unobservable and/or uncontrol-lable. Therefore the input-output behavior of a given state model which is "almost"uncontrollable and/or unobservable should be able to be approximated using a statemodel whose dimension is less than the dimension of the given state model. Thispossibility was introduced theoretically in Chapter 4. In this lab we are going to do

Page 319: Linear Control Theory - The State Space Approach Frederick Walker Fairman

301 Appendix D. MATLAB Experiments

some simulation studies to further investigate the effectiveness of this model orderreduction technique.

D.4.1 Decomposition of uncontrollable and/orunobservable systems

Recall that we can always transform the coordinates so that a given state model that is notcontrollable is transformed to controllable decomposed form

Ac. -_1Ac1'] Bc =Ac3 Bcz

Cc _ [ Cpl Cr2 ]

where (Acz, Bci2) is controllable (done using "ctrbf '). Alternatively if the given model isunobservable we can change coordinates so the state model becomes

A[Ac! A02l B_

[B.21

B.,L0 A,4 J

C = [ 0 C02 ]

where (A,,4, C02) is observable (done using "obsvf ")Therefore when a given state model is both uncontrollable and unobservable we can

use the foregoing decompositions to obtain a state model which has a state space which isdivided into four subspaces. In the following list of these subspaces C, C indicatescontrollable and uncontrollable, respectively with a similar notation for observable andunobservable. Thus the state space splits .up into

I a subspace CO of dimension nl2 a subspace CO of dimension n23 a subspace CO of dimension n;4 a subspace CO of dimension n4

Notice that if the dimension of the given state model is n then

4

n = E n;

An example demonstrating the use of MATLAB to do this decomposition can becarried out as follows. Enter the system state model matrices

Al = Bl = I

-1 5 7 9

0 -2 0 8

0 0 -3 6

0 0 0 -4

1

0

0

Cl=[0 1 0 1] D1=0

Page 320: Linear Control Theory - The State Space Approach Frederick Walker Fairman

State Model Reduction 305

and determine the corresponding transfer function by entering

[zl,pl,kl] =ss2zp(A1,B1,Cl,D1)

What is the order of the system? What can be said about the controllability andobservability of the state model? We can use the eigenvector tests for controllability andobservability to answer this question by interpreting the result of making the followingentries

[V1,DR1] = eig(A1)

C1*V1

and

[W1, DL1] = eig(A1' )

B1'* W1

As a further check on your results, obtain the rank of the controllability andobservability matrices using the commands "ctrb". "obsv" and "rank". Use your resultsto specify basis vectors for each subspace in the decomposition of the state space.

D.4.2 Weak controllability and/or observabilityEnter the following system state model matrices

-1 5 7 9 1

0 -2 0 8 1

A2 = B2 =0 0 -3 6 0

0 0 0 -4 1

C2=[0 1 .01 11 D2=0

and repeat the previous section for this state model. Contrast the results you obtain nowwith those you obtained in the previous section.

Next obtain the controllability and observability Gramians. (We, W(,), for the systemyou just entered. Do this by typing

Q0=C2'*C2QC=B2* B2'

WO = lyap(A2'.QO)

WC = lyap(A2. QC)

(Use "help" for information on the command "Iyap".) Obtain the eigenvalues of theproduct of the Gramians. Record these values for future reference.

Next use the command "balreal" to obtain a balanced realization by typing

[A2b, B2b, C2b,g2, t2] = balreal(A2. B2. C2)

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306 Appendix D: MATLAB Experiments

Determine the eigenvalues of the product of the Gramians you just entered andcompare them with the entries in g2. Comment on any correspondences.

Next obtain a reduced order model by discarding parts of the balanced realizationcorresponding to elements of g2 that are relatively small. For example supposeg2(3) << g2(2). Then since the elements of g2 are arranged in descending size we canobtain a reduced order model, (A2r, B2r, C2r) by typing

A2r = A2b(1 : 2, 1 :2)B2r = B2b(1 : 2)

C2r = C2b(1 : 2)

Use use the command "ss2zp" to determine the poles and zeros of the transfer functionfor the reduced order system. Is there any obvious relation between the original systemtransfer function and the reduced order system transfer function?

DA.3 Energy interpretation of the controllability andobservability Gramians

Recall, from Chapter 4, that the sum of the diagonal entries in the controllabilityGramian equals the energy transferred into the state over all positive time, from theinput when the input is a unit impulse at time zero. We can check to see if this is the casehere.

Recall from Chapter 1 that the zero state response to a unit impulse is the same as thezero input response to an initial state x(0) = B. Therefore we can use "initial" with theinitial state set to B. As before we need to decide on the time interval and sample times fordoing this. Suppose we use an interval of 10 s with samples taken every 0.1 s. Then wecreate a vector of sample times by typing .

ts=0:0.1 : 10

Notice that is has 101 elements.Therefore using the command "initial" with B2 in the position reserved for the initial

state, we can obtain the resulting state by typing

[Y2, X2, ts] = initial(A2, B2, C2, D2, ts)

where X2 has four columns with each column corresponding to a different component ofthe state and each row corresponding to a sample time.

Recall that the energy transferred to the state is given by

/x 4

EE = J xT (t)x(t)dt = > Ec

where

rxJ x2(t)dt

0

Page 322: Linear Control Theory - The State Space Approach Frederick Walker Fairman

State Model Reduction 307

is the energy transferred to the i`h component of the state. Using this fact we can computean approximation, EAc, to E, by entering

eei = X2(1 : 101, i)' * X2(1 : 101. i)

with i = 1, 2, 3,4 in succession and then enter

14

EAc 10Eeei1-1

Compare your result with the trace of the controllability Gramian by typing

ETc = trace(WC)

Recall, from Chapter 4, that the energy, E0, which is transferred to the output from aninitial state can be determined using the observability Grarnian as

Eo = fy2(t)dt xT 0) Wx(0)

Now we can use the procedure just described for computing E,. to compute E0 when theinitial condition x(0) = BI by using the data stored previously in Y2. Compare yourresult with the result obtained by using the observability Gramian

(B2) * WO * B2

D.4.4 Design of reduced order modelsIn this part of the lab you will examine the balanced realization approach to model orderreduction. The system to be reduced is a degree eight Butter,,worth analog filter. Thissystem has a transfer function with constant numerator and a denominator of degreeeight. The poles are located in the open left half-plane at unit distance from the origin. Alist of the poles can be obtained by entering

[zbw, pbw, kbw] = huttap(8 )

We can check that the poles are equidistant from the origin by examining the diagonalelements of the matrix obtained by entering

pbw * pbir'

A balanced realization (state model) for the Buttersa orth filter is obtained by firstcalculating a controller form state model and then transforming it to balanced form.Therefore enter

[A3, B3, C3. D31 _ zp2ss(,-bu. pbu. kbw )

and

[A3b, B3b, Cb3,g3, t3] = balreal(A3. B3, C3)

Page 323: Linear Control Theory - The State Space Approach Frederick Walker Fairman

308 Appendix D: MATLAB Experiments

Recall that the reduced order model is obtained by discarding the latter part of thestate vector in balanced coordinates so that the corresponding discarded part of g3 isnegligible in the sum of the elements in g3. Notice that there is a significant decrease in thesize of the elements in g3 in going from the fifth to the sixth element. Therefore a reducedorder model obtained by discarding the last three elements in the state should give a goodapproximation to the full order system. The state model for the reduced order approx-imation can therefore be obtained by entering

A3b5 = A3b(l 5, 1 : 5)

B3b5 = B3b(l : 5)

C3b5 = C3b(1 : 5)

Notice that this reduced order state model approximation is also balance. This can beseen by generating the controllability and observability Gramians by typing

QOb5 = C3b5' * C3b5

QCb5 = B3b5 * B3b5'

Wob5 = lyap(A3b5', QObS)

WcbS = lyap(A3b5, QCb5)

The final part of this lab concerns the performance of the reduced order model. Onepossible test would be to compare the step responses of the reduced and full order models.

`y3b, x3b] = step(A3b, B3b, C3b, D3, 1, ts)

[y3b5, x3b5] step(A3b5, B3b5, C3b5, D3, 1, ts)

Note any marked differences in the plots on the screen and proceed to compare theerror between the reduced order model output and the full order model output. This canbe done using element by element division, i.e.,

rele =()-3b5 - y3h)./y3b

plot(ts, rele)

Repeat the foregoing by deleting more than the last three elements of the state toobtain a reduced order model and use the step response to compare the performance ofthis approximation with the performance of the fifth order approximation just studied.

Page 324: Linear Control Theory - The State Space Approach Frederick Walker Fairman

References

[1] Anderson, B. D. O. and Moore, J. B. Optimal Control. Linear Quadratic Methods. Prentice-Hall, Englewood Cliffs, New Jersey, 1990.

[2] Anderson, B. D. O. and Liu, Y. "Controller reduction: Concepts and Approaches", IEEETrans. Automatic Control, TAC-34, pp. 802-812, 1989.

[3] Bittanti, S., Laub, A. J. and Willems, J. C. (Eds.) The Rice ati Equation. Springer-Verlag,Berlin, 1991.

[4] Blackman, P. F. Introduction to State-variableAnal-

ysis. The Macmillan Press, London, 1977.[5] Brockett, R. W. Finite Dimensional Linear Systems, John Wiley and Sons. New York, 1970.[6] Brogan, W. L. Modern Control Theory. 3rd ed., Prentice-Hall. Englewood Cliffs, N. J., 1991.[7] Brown, R. G. Introduction to Random Signal Analysis and Kalman Filtering. John Wiley and

Sons, Chichester, 1983.[8] Chandrasekharan, P. C. Robust Control of'Dynamic Srstenu. Academic Press, London,

1996.

[9] Chen, C. T. Introduction to Linear Systems Theor. Holt Rinhart and Winston, New York,1970.

[10] Dorato, P., Abdallah, C. and Cerone, V. Linear Quadratic Control, An Introduction.Prentice-Hall, Englewood Cliffs, N. J., 1995.

[11] Doyle, J. C., Francis, B. A. and Tannenbaum. A. R. Feedback Control Theory. MaxwellMacmillan Canada, Toronto, 1992.

[12] Doyle, J. C., Glover, K., Khargonekar, P. P. and Francis, B. A. "State space solutions tostandard H2 and H, control problems", IEEE Trans. Automatic Control, AC-34, pp. 831-847, 1989.

[13] Fairman, F. W., Danylchuk, G. J., Louie, J. and Zaroww ski. C. J. "A state-space approach todiscrete-time spectral factorization", IEEE Transactions on Circuits and Systems-II Analogand Digital Signal Processing, CAS-39, pp. 161-170. 1992.

[14] Francis, B. A. A Course in H, Control Theory, Springer-Verlag, Berlin, 1987.[15] Furuta, K., Sano, S. and Atherton, D. State Variable .Methods in Automatic Control, John

Wiley and Sons, Chichester, 1988.[16] Glover, K. and Doyle, J. C. "A state space approach to H, optimal control", Lecture Notes

in Control and Information Science. 135, pp. 179-2 18. 1989.[17] Golub, G. H. and Van Loan, C. F. Matrix Computation.,, 2nd ed., The Johns Hopkins

University Press, Baltimore, Maryland, 1989,[18] Green, M. and Limebeer, D. J. N. Linear Robust Control. Prentice-Hall, Englewood Cliffs,

New Jersey, 1995.

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[19] Green, M., Glover, K., Limebeer, D. J. N. and Doyle, J. "A J spectral factorization approachto H. control", SIAM Journal on Control and Optimization, 28, pp. 1350-1371, 1990.

[20] Green, M. "H. controller synthesis by J lossless coprime factorization", SIAM Journal onControl and Optimization, 30, pp. 522-547, 1992.

[21] Isidori, A. "H. control via measurement feedback for affine nonlinear systems", Int. J.Robust and Nonlinear Control, 4, pp. 553-574, 1994.

[22] Jonckheere, E. A. and Silverman, L. M. "A new set of invariants for linear systems-applications to reduced order compensator design", IEEE Trans. Automatic Control, AC-28, pp. 953-964, 1983.

[23] Kailath, T. Linear Systems, Prentice-Hall, Englewood Cliffs, N. J., 1980.[24] Kimura, H. Chain-Scattering Approach to H. Control, Birkhausser Boston, 1997.[25] Kwakernaak, H. and Sivan, R. Linear Optimal Control Systems, Wiley Interscience, New

York, N. Y., 1972.[26] LePage, W. R. Complex Variables and the Laplace Transform for Engineers, Dover, 1980.[27] Moore, B. C. "Principal component analysis in linear systems: controllability, observability,

and model reduction", IEEE Trans. Automatic Control, AC-26, pp. 17-32, 1981.[28] Missaghie, M. M. and Fairman, F. W. "Sensitivity reducing observers for optimal feedback

cont&ol", IEEE Transactions on Automatic Control, AC-22, pp. 952-957, 1977.[29] Mullis, C. T. and Roberts, R. A. "Synthesis of minimum roundoff noise in fixed point digital

filters", IEEE Trans. Circuits and Systems, CAS-23, 551-562, 1976.[30] Mustafa, D. and Glover, K. "Controller reduction by H. balanced truncation", IEEE

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Int. J. Robust and Nonlinear Control, 6, pp. 691-726, 1996.[32] Pavel, L. and Fairman, F. W. "Controller reduction for nonlinear plants-an L2 approach",

Int. J. Robust and Nonlinear Control, 7, pp. 475-505, 1997.[33] Pavel, L. and Fairman, F. W. (1998). "Nonlinear H. control: a J-dissipative approach",

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1987.[38] Scherpen, J. M. A. "H, balancing for nonlinear systems", Int. J. Robust and Nonlinear

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control", IEEE Trans. Automatic Control, AC-37, pp. 770-784, 1992.[43] Varga, A. "A multishift Hessenberg method for pole assignment of single-input systems",

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Page 327: Linear Control Theory - The State Space Approach Frederick Walker Fairman
Page 328: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Index

Ackermann formula 52, 64, 74adjoint system 221

adjugate matrix 11, 18

algebraic Riccati equation (ARE)145, 166, 224, 246

GCARE 153GFARE 157HCARE 236HFARE 244QCARE 126QFARE 142

all pass system 221

anticausal, antistable systems 172

asymptotic state estimation 69

average power 147

balanced realization 104Bezout identity 204, 209

controllerform 47

7-(,,-central. parametrized 255

115, 119, LQG 157quadratic 120

convolution integral 29

coordination transformation 12, 28coprime factorization 201, 204,

doubly 212

J-inner 230

state models 206covariance 147

causality constraint 31causal system, signal 172

Cayley-Hamilton theorem 52

characteristic polynomial 9

coinner function 221

companion matrix 9

computer determination of state 37, 66contraction 162controllability 34, 55

Gramian 101, 112matrix 51, 57, 63

controllabledecomposed form 60, 93, 118

eigenvalue 44subspace 56

decomposition of a space 169, 178detectable system 72

direct sum 169

disturbance 115. 120, 138. 147, 154, 225, 232,242. 248

dom(Ric) 145

doubly coprime factorization 212

dual system 72

eigenvalue 17, 96assignment 43. 64, 74, 82, 86controllable. observable 44, 72invariance 20

eioenvector 17, 96left-eigenvector 19right-eigenvector 16

tests (controllability, observability) 43, 71energy 94. 102. 110. 167

feedback 41

filtering 68

Page 329: Linear Control Theory - The State Space Approach Frederick Walker Fairman

314 Index

filtering (contd.)-H 242LQG (Kalman) 90, 155, 166quadratic 138

Fourier transform 173

Gaussian random vector 147

Gramiancontrollability 101, 112

observability 94, 109

Hamiltonian matrices 130, 158H. relation 262

Hankel norm 1 I I

Hardy spaces72,12 177

7-tM 183

Hermitian matrix 95Hilbert space 167

induced norm 181

initial condition response(see zero-input response)

inner function 221

inner product space 169

input-output response(see zero-state response)

invariant subspace 133

inverse

left matrix 259right matrix 83

system 197

isomorphic, isomorphism 178, 229

Laplace transform 176

Lebesgue spacestime domain: G'(-x, x

167, 174frequency domain: LE, G_ 174

L2 gain 181

linear combination 3, 18lossless system 219LQG control

requirement for solution 165

state estimation 153

state feedback 149

Lyapunov equation inbalanced realization 107

H_ control 234

LQG control 151, 155quadratic control 123-129, 140

matrixHermitian 95nonnegative 96positive definite 98

matrix exponential 9matrix fraction description

(see coprime factorization)matrix square root 160

measurement noise 138

minimal realization 34reduction to 91

nonnegative matrix 96

norms2-norm for vectors 94H2 signal norm 167

H2 system norm 172, 179H. system norm 181, 191Hankel norm 112

induced 2-norm for matrices 186

L2 induced system norm 181

L2 time domain signal norm 95, 167L. function norm 182

null space of a matrix 78

null vector, matrix 17, 36

observability 34, 76Gramian 94, 109matrix 73

observabledecomposed form 82, 94eigenvalue 72

subspace 78

observer 70

form 72

minimal order 82observer-based controllers 116

orthogonal 21

compliment 169, 178matrix 98

spaces, signals 169

output feedback 149, 157, 254output injection 90

Parseval's theorem 174PBH (Popov-Belevich-Hautus) tests

Page 330: Linear Control Theory - The State Space Approach Frederick Walker Fairman

Index 315

(see eigenvector tests)performance index

Gaussian 148, 153H 224, 233, 242-252quadratic 121, 139

phase plane 5

pole placement(see eigenvalue assignment)

pole-zero cancellation 34

positive definite matrix 98

projector 161

proper 12

quadratic control 115

requirement for solution 147

state estimation 137

state feedback 119

range of a matrix (operator) 57, 77, 163rank of a matrix 51, 59, 76rational function 12

realizationbalanced 104

minimal 34, 35Ric(H) 145

robustness(see stability)

separation principle 118, 149, 157, 261singular value decomposition 185, 274small gain theorem 184, 191spaces

Hardy 179, 185Hilbert 168, 173Lebesgue 167

stabilityinternal, external 25

reduced order model 279

robustness 26, 69, 203stabilizable system 43

stabilizing controllers 213, 215

stabilizing solution 128, 135, 158, 227, 267stable system 6. 109state

computation 37, 66estimation 67, 153, 242, 251feedback 41. 120, 151, 227

state model formscontrollable decomposed 60controller 48diagonal (normal) 32observable decomposed 78

observer 72state trajectory 5. 21strictly proper 12

symmetric matrix(see Hermitian)

system factorization 201

system interconnections 193

system inverse 196

system zero 198

trace of a matrix 102, 150transfer function 34

proper. strictly proper 12

transformation tocontrollable decomposed form 64

controller form 49observable decomposed form 81

observer form 73

transition matrix 9. 31

unitary matrix 98

weighting matrix 170

well posed feedback control 214

worst disturbance 234, 259

Youla parametrization 217, 260

zero-input (initial condition) response 1, 25,32

zero-state (input output) response 1, 32

Page 331: Linear Control Theory - The State Space Approach Frederick Walker Fairman

LINEAR CONTROL THEORYTHE STATE SPACE APPROACH

Frederick W. FairmanQueen's University, Kingston, Ontario, Canada

Incorporating recent developments in control and systems research,Linear Control Theory provides the fundamental theoretical backgroundneeded to fully exploit control system design software. This logically-structured text opens with a detailed treatment of the relevant aspectsof the state space analysis of linear systems. End-of-chapter problemsfacilitate the learning process by encouraging the student to put his orher skills into practice.

The use of an easy to understand matrix variational technique todevelop the time-invariant quadratic and LQG controllers

a A step-by-step introduction to essential mathematical ideas as theyare needed, motivating the reader to venture beyond basic conceptsThe examination of linear system theory as it relates to control theoryThe use of the PBH test to characterize eigenvalues in the statefeedback and observer problems rather than its usual role as a testfor controllability or observabilityThe development of model reduction via balanced realizationThe employment of the L2 gain as a basis for the development ofthe H,, controller for the design of controllers in the presence ofplant model uncertainty

Senior undergraduate and postgraduate control engineering studentsand practicing control engineers will appreciate the insight this self-contained book offers into the intelligent use of today's control systemsoftware tools.

JOHN WILEY & SONSChichester - New York Weinheim Brisbane Singapore . Toronto

i