Schaum's Outline of Theory and Problems of State Space and Linear Systems

SCHAUM'S OUTLINE OF THEORY AND PROBLEMS OF STATE SPACE and LINEAR SYSTEMS .. BY DONALD M.WIBERG., Ph.D. Associate ...

Author: Donald M. Wiberg

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SCHAUM'S OUTLINE OF

THEORY AND PROBLEMS OF

STATE SPACE and

LINEAR SYSTEMS

.. BY

DONALD M.WIBERG., Ph.D. Associate Professor of Engineering University of California, Los Angeles

S~BAUM'S OUTLINE SERIES McGRAW·HILL BOOK COMPANY New York, St. Louis, San Francisco, Dusseldorf, Johannesburg, Kuala Lumpur, London, Mexico, Montreal, New Delhi, Panama, Rio de Janeiro, Singapore, Sydney, and Toronto

Copyright © 1971 by McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,. recordiIig, or otherwise, without the prior written permission of the publisher.

07-070096-6 1234567890 SHSH 754321

Preface The importance of state space analysis is recognized in fields where the time behavior of any physical process is of interest. The concept of state is comparatively recent, but the methods used have been known to mathematicians for many years. As engineering, physics, medicine, economics, and business become more cognizant of the insight that the state space approach offers, its popularity increases. This book was written not only for upper division and graduate students, but for practicing professionals as well. It is an attempt to bridge the gap between theory and practical use of the state space approach to the analysis and design of dynamical. systems. The book is meant to encourage the use of state space as a tool for analysis and design, in proper relation with other such tools. The state space approach is more general than the "classical" Laplace and Fourier transform theory. Consequently, state space theory IS applicable to all systems that can be analyzed by integral transforms in time, and is applicable to many systems for which transform theory breaks down. Furthermore, state space theory gives a somewhat different insight into the time behavior of linear systems, and is worth studying for this aspect alone. In particular, the state space approach is useful because: (1) linear systems with timevarying parameters can be analyzed in essentially the same manner as time-invariant linear systems, (2) problems formulated by state space methods can easily be programmed on a computer, (3) high-orde-r linear systems can be analyzed, (4) multiple input-multiple output systems can be treated almost as easily as single input-single output linear systems, and (5) state space theory is the foundation for further studies in such areas as nonlinear systems, stochastic systems, and optimal control. These are five of the most important advantages obtained from the generalization and rigorousness that state space brings to the classical transform theory. Because state space theory describes the time behavior of physical systems in a mathematical manner, the reader is assumed to have some knowledge of differential equations and of Laplace transform theory. Some classical control theory is needed for Chapter 8 only_ No knowledge of m:atrices or complex variables is prerequisite. The book may appear to contain too many theorems to be comprehensible and/or useful to the nonmathematician. But the theorems have been stated and proven in a manner suited to show the range of application of the ideas and their logical interdependence. Space that might otherwise have been devoted to solved problems has been used instead to present the physical motivation of the proofs. Consequently I give my strongest recommendation that the reader seek to understand the physical ideas underlying the proofs rather than to merely memorize the theorenls. Since the emphasis is on applications, the book might not be rigorous enough for the pure mathematician, but I feel that enough information has been provided so that he can tidy up the statements and proofs himself._ The book has a number of novel features. Chapter 1 gives the fundamental ideas of state from an informal, physical viewpoint, and also gives a correct statement of linearity. Chapter 2 shows how to write transfer functions and ordinary differential equations in

matrix notation, thus motivating the material on matrices to follow. Chapter 3 develops the important concepts of range space and null space in detail, for later application. Also exterior products (Grassmann algebra) are developed, which give insight into determinants, and which considerably shorten a number of later proofs. Chapter 4 shows how to actually solve for the Jordan form, rather than just proving its existence. Also a detailed treatment of pseudoinverses is given. Chapter 5 gives techniques for computation of transition matrices for high-order time-invariant systems, and contrasts this with a detailed development of transition matrices for time-varying systems. Chapter 6 starts with giving physical insight into controllability and observability of simple systems, and progresses to the point of giving algebraic criteria for time-varying systems. Chapter 7 shows how to reduce a system to its essential parameters. Chapter 8 is perhaps the most novel. Techniques from classical control theory are extended to time-varying, multiple input-multiple output linear systems using state space formulation. This gives practical methods for control system design, as well as analysis. Furthermore, the pole placement and observer theory developed can serve as an introduction to linear optimal control and to Kalman filtering. Chapter 9 considers asymptotic stability of linear systems, and the usual restriction of uniformity is dispensed with. Chapter 10 gives motivation for the quadratic optimal control problem, with special emphasis on the practical time-invariant problem and its associated computational techniques. Since Chapters 6, 8, and 9 precede, relations with controllability, pole placement, and stability properties can be explored. The book has come from a set of notes developed for engineering course 122B at UCLA, originally dating from 1966. It was given to the publisher in June 1969. Unfortunately, the pUblication delay has dated some of the material. Fortunately, it also enabled a number of errors to be weeded out. Now I would like to apologize because I have not included references, historical development, and credit to the originators of each idea. This was simply impossible to do because of the outline nature of the book. I would like to express my appreciation to those who helped me write this book. Chapter 1 was written with a great deal of help from A. V. Balakrishnan. L. M. Silverman helped with Chapter 7 and P.K.C. Wang with Chapter 8. Interspersed throughout the book is material from a course given by R. E. Kalman during the spring of 1962 at Caltech. J. J. DiStefano, R. C. Erdmann, N. Levan, and K. Yao have used the notes as a text in UCLA course 122B and have given me suggestions. I have had discussions with R. E. Mortensen, M. M. Sholar, A. R. Stubberud, D. R. Vaughan, and many other colleagues. Improvements in the final draft were made through the help of the control group under the direction of J. Ackermann at the DFVLR in Oberpfaffenhofen, West Germany, especially by G. GrUbel and R. Sharma. Also, I want to thank those UCLA students, too numerous to mention, that have served as guinea pigs and have caught many errors of mine. Ruthie Alperin was very efficient as usual while typing the text. David Beckwith, Henry Hayden, and Daniel Schau~ helped publish the book in its present form. Finally, I want to express my appreciation of my wife Merideth and my children Erik and Kristin for their understanding during the long hours of involvement with the book. DONALD M. WIBERG University of California, Los Angeles June 1971

CONTENTS Chapter

1

Page

MEANING OF STATE

•••••

t

.......................

,.

..................

.

1

Introduction to State. State of an Abstract Object. Trajectories in State Space. Dynamical Systems. Linearity and Time Invariance. S~Tstems Considered. Linearization of Nonlinear Systems.

Chapter

2

METHODS FOR OBTAINING THE STATE EQUATIONS ......

16

Flow Diagrams. Properties of Flow Diagrams. Canonical Flow Diagrams for Time-Invariant Systems. Jordan Flow Diagram. Time-Varying Systems. General State Equations~

Chapter

3

ELEMENTARY MATRIX THEORy...........................

38

Introduction. Basic Definitions. Basic Operations. Special Matrices. Determinants and Inverse Matrices. Vector Spaces. Bases. Solution of Sets of Linear Algebraic Equations. Generalization of a Vector. Distance in a Vector Space. Reciprocal Basis. Matrix Representation of a Linear Operator. Exterior Products.

Chapter

4

MATRIX ANALySIS...........................................

69

Eigenvalues and Eigenvectors. Introduction to the Similarity Transformation. Properties of Similarity Transformations. Jordan Form. Quadratic Forms. Matrix Norms. Functions of a Matrix. Pseudo inverse.

Chapter

5

SOLUTIONS TO THE LINEAR STATE EQUATION ...........

99

Transition Matrix. Calculation of the Transition Matrix for Time-Invariant Systems. Transition Matrix for Time-Varying Differential Systems. Closed Forms for Special Cases of Time-Varying Linear Differential Systems. Periodically-Varying Linear Differential Systems. Solution of the Linear State Equations with Input. Transition Matrix for Time-VarYing nifference Equations. Impulse Response Matrices. The Adjoint System.

Chapter

6

CONTROLLABILITY AND OBSERV ABILITY ................. 128 Introduction to Controllability and Observability. Controllability in TimeInvariant Linear Systems. Observability in Time-Invariant Linear Systems. Direct Criteria from A, B, and C. Controllability and Observability of TimeVarying Systems. Duality.

CONTENTS Page Chapter

7

CANONICAL FORMS OF THE STATE EQUATION ........... 147 Introduction to Canonical Forms. Jordan Form for Time-Invariant Systems. Real Jordan Form. Controllable and Observable Forms for Time-Varying Systems. Canonical Forms for Time-Varying Systems.

Chapter

8

RELATIONS WITH CLASSICAL TECHNIQUES .............. 164 Introduction. Matrix Flow Diagrams. Steady State Errors. _ Root Locus. Nyquist Diagrams. State Feedback Pole Placement. Observer Systems. Algebraic Separation. Sensitivity,Noise Rejection, and Nonlinear Effects.

Chapter

9

STABILITY OF LINEAR SySTEMS .......................... 191 Introduction. Definitions of Stability for Zero-Input Linear Systems. De~ finitions of Stability for Nonzero Inputs. Liapunov Techniques. Liapunov Functions for Linear Systems. Equations for the Construction of Liapunov Functions.

Chapter

10

INTRODUCTION TO OPTIMAL CONTROL ................... 208 Introduction. The Criterion Functional. Derivation of the Optimal Control Law. The Matrix Riccati Equation. Time-Invariant Optimal Systems. Out~ put Feedback. The Servomechanism Problem. Conclusion.

INDEX ......................................................... 233 /

Chapter 1 Meaning of State 1.1 INTRODUCTION TO STATE To introduce the subject, let's take an informal, physical approach to the idea of state. (An exact mathematical approach is taken in more advanced texts.) First, we make a distinction between physical and abstract objects. A physical object is an object perceived by our senses whose time behavior we wish to describe, and its abstraction is the mathematical relationships that give some expression for its behavior. This distinction is made because, in making an abstraction, it is possible to lose some of the relationships that make the abstraction behave similar to the physical object. Also, not all mathematical relationships can be realized by a physical object. The concept of state relates to those physical objects whose behavior can change with time, and to which a stimulus can be applied and the response observed. To predict the future behavior of the physical object under any input, a series of experiments could be performed by applying stimuli, or inputs, and observing the responses, or outputs. From these experiments we could obtain a listing of these inputs and their corresponding observed outputs, i.e. a list of input-output pairs. An input-output pair is an ordered pair of real time functions defined for all t == to, where to is the time the input is first applied. Of course segments of these input time functions must be consistent and we must agree upon what kind of functions to consider, but in this introduction we shall not go into these mathematical details. Definition 1.1:

The state of a physical obiect is any property of the object which relates input to output such that knowledge of the input time function for t == to and state at time t = to completely determines a unique output for t == to.

Example 1.1. Consider a black box, Fig. 1-1, containing a switch to one of two voltage dividers. Intuitively, the state of the box is the position of the switch, which agrees with Definition 1.1. This can be ascertained by the experiment of applying a voltage V to the input terminal. Natural laws (Ohm's law) dictate that if the switch is in the lower position A, the output voltage is V/2, and if the switch is in the upper position B, the output voltage is V/4. Then the state A determines the ihput-output pair to be (V, V /2), and the state B corresponds to (V, V/4).

Fig. 1-1

1.2 STATE OF AN ABSTRACT OBJECT The basic ideas contained in the above example can be extended to many physical objects and to the abstract relationships describing their time behavior. This will be done after abstracting the properties of physical objects such as the black box. For example, the color of the box has no effect on the experiment of applying a voltage. More subtly, the value of resistance R is immaterial if it is greater than zero. All that is needed is a listing of every input-output pair over all segments of time t ~ to, and the corresponding states

at time to. 1

2

MEANING OF STATE

Definition 1.2:

[CHAP.!

An abstract object is the totality of input-output pairs that describe the behavior of a physical object.

Instead of a specific list of input time functions and their correspondIng output time functions, the abstract object is usually characterized as a class of all time functions that obey a set of mathematical equations. This is in accord with the scientific method of hypothesizing an equation and then checking to see that the physical object behaves in a manner similar to that predicted by the equation. Hence we can often summarize the abstract object by using the mathematical equations representing physical laws. The mathematical relations which summarize an abstract object must be oriented, in that m of the time functions that obey the relations must be designated inputs (denoted by the vector u, having m elements 'Ui) and k of the time functions must be designated outputs (denoted by the vector y, having k elements Yi). This need has nothing to do with causality, in that the outputs are not "caused" by the inputs. Definition 1.3:

The state of an abstract object is a collection of numbers which together with the input u(t) for all t ~ to uniquely determines the output y(t) for all t ~ to.

In essence the state parametrizes the listing of input-output pairs. The state is the answer to the question "Given u(t) for t ~ to and the mathematical relationsh~ps of the abstract object, what additional information is needed to completely specify y(t) for t ~ to?" Example 1.2. A physical object is the resistor-capacitor network shown in Fig. 1-2. An experiment is performed by applying a voltage u(t), the input, and measuring a voltage yet), the output. Note that another experiment could be to apply yet) and measure u(t), so that these choices are determined by the experiment. The list of all input-output pairs for this example is the class of all functions u(t). yet) which satisfy the mathematical relationship (1.1) RCdy/dt + y = u

R

I

=

y(to)

e(to-O/RC

+

ic

it

e(T-O/RCU(T)

dT

c

u(t)

1

This summarizes the abstract object. The solution of (1.1) is

yet)

ww

I

r

yet)

T1

Fig. 1-2

(1.2)

to

This relationship explicitly exhibits the list of input-output pairs. For any input time function u(r) for ~ to. the output time function yet) is uniquely determined by y(to), a number at time to· Note the distinction between time functions and numbers. Thus the set of numbers y(to) parametrizes all inputoutput pairs, and therefore is the state of the abstract object described by (1.1). Correspondingly, a choice of state of the RC network is the output voltage at time to· T

Example 1.3. The physical object shown in Fig. 1-3 is two RC networks in series. The pertinent equation is R2C2

WN

d2 y/dt 2

+ 2.5RC dy/dt + y 'V'NI/'

(1.3)

= u

I r 1_~~~l R

u(t)

Ic

2R

li I

C

T

Fig. 1-3

yet)

CHAP.!]

MEANING OF STATE

3

with a solution y(t)

+

_2_ dy (t )[eCto~t)I2RC _ 3RC dt 0

eCto-02/RC]

(1.4)

Here the set of numbers y(to) and

~~ (to)

parametrizes the input-output pairs, and may be chosen as state.

Physically, the voltage and its derivative across the smaller capacitor at time to correspond to the state.

Definition 1.4:

A state variable, denoted by the vector x(t), is the time function whose value at any specified time is the state of the abstract object at that time.

Note this difference in going from a set of numbers to a time function. The state can be a set consisting of an infinity of numbers (e.g. Problems 1.1 and 1.2), in which case the state variable is an infinite collection of time functions. However, in most cases considered in this book, the state is a set of n numbers and correspondingly x(t) is an n-vector function of time. Definition 1.5:

The state space, denoted by

~,

is the set of all x(t).

Example 1.4. The state variable in Example 1.2 is x(t) = y(t}, whereas in Example 1.1 the state variable remains either A or B for all time. Example 1.5. The state variable in Example 1.3 is

x(t) =

y(t») dy . ( dt (t)

The state representation is not unique. There can be many different ways of expressing the relationship of input to output. Example 1.6. In Example 1.3, instead of the voltage and its derivative across the smaller capacitor, the state could be the voltage and its derivative across the larger capacitor, or the state could be the voltages across both capacitors.

There can exist inputs that do not influence the state, and, conversely, there can exist outputs that are not influenced by the state. These cases are called uncontrollable and unobservable, respectively, about which much more will be said in Chapter 6. Example 1.7. In Example 1.1, the physical object is state uncontrollable. No input can make the switch change positions. However, the switch position is observable. If the wire to the output were broken, it would be unobservable. A state that is both unobservable and uncontrollable makes no physical sense, since it cannot be detected by experiment. Examples l.2 and 1.3 are both controllable and observable.

One more point to note is that we consider here only deterministic abstract objects. The problem of obtaining the state of an abstract object in which random processes are inputs, etc., is beyond the scope of this book. Consequently, all statements in the whole book are intended only for deterministic processes.

4

MEANING· OF STATE

[CHAP. 1

1.3 TRAJECTORIES IN STATE SPACE The state variable x(t) is an explicit function of time, but also depends implicitly on the starting time to, the initial state x(to) = Xo, and the input U(T). This functional dependency can be written as x(t) = ~(t; to,xo, U(T)), called a trajectory. The trajectory can be plotted in n-dimensional state space as t increases from to, with t an implicit parameter. Often this plot can be made by eliminating t from the solutions to the state equation. Example 1.8. Given Xl(t) = sin t and X2(t) = cos t, squaring each equation and adding gives is a circle in the XIX2 plane with t an implicit parameter.

xi + x~

== 1.

This

Example L9. In Example 1.3, note that equation (1-4) depends on t, u(r), x(to) and to, where x(t o) is the vector with components y(to) and dy/dt(to). Therefore the trajectories + depend on these quantities. Suppose now u(t) = 0 and Re = 1. Let Xl = yet) and Xz = dy/dt. Then dx/dt = X2 and d 2y/dt 2 = dX2/dt. Therefore dt = dXl/x2 and so d2y/dt 2 = x2dxZ/dxl' Substituting these relationships into (1.3) gives

which is independent of t.

This has a solution

+ 2X2 = C(2Xl + x2)4 [Xl (to) + 2X2(tO)}/[2xl(tO) + xZ(tO)]4. Typical Xl

where the constant C = trajectories in state space are shown in Fig. 1-4. The one passing through points xl(t O) = 0 and X2(t O) = 1 is drawn in bold. The arrows point in the direction of increasing time, and all trajectories eventually reach the origin for this particular stable system.

--------~----_+~--~~~--~----+_-------------Xl

1.4 DYNAMICAL SYSTEMS In the foregoing we have assumed that an abstract object exists, and that sometimes we can find a set of oriented mathematical relationships that summarizes this listing of input and output pairs. Now suppose we are given a set of oriented mathematical relationships, do we have an abstract object? The answer to this question is not always affirmative, because there exist mathematical equations whose solutions do not result in abstract objects. Example 1.10. The oriented mathematical equation yet) = ju(t) cannot give an abstract object, because either the input or the output must be imaginary.

If a mathematical relationship always determines a real output y(t) existing for all

t ~ to given any real input u(t) for all time t, then we can form an abstract object. Note that by supposing an input u(t) for all past times as well as future times, we can form an abstract object from the equation for a delayor y(t) = u(t - T). [See Problem 1.1.J

CHAP. 1]

MEANING OF STATE

5

However, we can also form an abstract object from the equation for a predictor If we are to restrict ourselves to mathematical relations that can be mechanized, we must specifically rule out such relations whose present oufputs depend on future values of the input. y(t) = u(t + T).

Definition 1.6:

A dynamical syste'tn is an oriented mathematical relationship in which: (1) A real output y(t) exists fOl' all t ~ to given a real input u(t) for all t. (2) Outputs y(t) do not depend on inputs u(,) for. > t.

Given that we have a dynamical system relating y(t) to u(t), we would like to construct a set of mathematical relations defining a state x(t). We shall assume that a state space description can be found for the dynamical system of interest ,satisfying the following conditions (although such a construction may take considerable th~ught): Condition 1:

A real, unique output y(t) = fI(t, +(t; to, Xo, U{T)), u(t)) exists for all t> to given the state Xo at time to and a real input U(T) for 7' ~ to.

Condition 2:

A unique trajectory q,{t; to, Xo, u(.)) exists for all t > to given the state at time to and a real input for all t ~ to.

Condition 3:

A unique trajectory starts from each state, i.e. (1.5)

Condition 4:

Trajectories satisfy the transition property q,(t; to, x(to), U(T))

= 'P(t; t l , X(tl)' u(.))

for to < tl < t

(1.6)

where Condition 5:

(1.7) \

Trajectories +(t; to, Xo, u(,)) do not depend on inputs u(.) for • > t.

Condition 1 gives the functional relationship y(t) = 7j(t, x(t), u(t)) between initial state and future input such that a unique output is determined. Therefore, with a proper state space description, it is not necessary to know inputs prior to to, but only the state at time to. The state at the initial time completely summarizes all the past history of the input. Example 1.11. In Example 1.2., it does not matter how the voltage across the capacitor was obtained in the past. All that is needed to determine the unique future output is the state and the future input.

ConditiDn 2 insures that the state at a future time is uniquely determined. Therefore knowledge of the state at any time, not necessarily to, uniquely determines the output. For a given u(t), one and only one trajectory passes through each point in state space and exists for all finite t ~ to. As can be verified in Fig. 1-4, one consequence of this is that the state trajectories do not cross one another. Also, notice that condition 2 does not require the state to be real, even though the input and output must be real. Example 1.12.

The relation dy/dt = u(t) is obviously a dynamical system. A state space description dx/dt = ju(t) with output y(t) = -jx(t) can be constructed satisfying conditions 1-5, yet the state is imaginary.

6

MEANING OF STATE

[CHAP. 1

Condition 3 merely requires the state space description to be consistent, in that the starting point of the trajectory should correspond to the initial state.. Condition 4 says that the input uCr) takes the system from a state x(to) to a state x(t),' and if X(tl) is on that trajectory, then the corresponding segment of the input will take the system from X(tl) to x(t). Finally, condition 5 has been added to assure causality of the input-output relationship resulting from the state space description to correspond with the causality of the original dynamical system. Example 1.13. We can construct a state space description of equation (1.1) of Example 1.2 by defining a state x(t) = yet). Then condition 1 is satisfied as seen by examination of the solution, equation (1.2). Clearly the trajectory ¢(t; to, xo, U(T)) exists and is unique given a specified to. Xo and U(T), so condition 2 is satisfied. Also, conditions 3 and 5 are satisfied. To check condition 4, given x(to) = y(t o) and U(T) over to:::: T:::: t, then x(t)

where x(t l )

(1.8)

=:

=

+

x(to)eCto-tl)/RC

RlC

f

tl eCTo-tl)/RC U(TO) dTo

(1.9)

to

Substitution of (1.9) into (1.8) gives the previously obtained (1.2). Therefore the dynamical system (1.1) has a state space description satisfying conditions 1-5.

Henceforth, instead of "dynamical system with a state space description" we will simply say "system" and the rest will be understood. 1.5 LINEARITY AND TIME INVARIANCE Definition 1.7: Given any two numbers a, {3; two states Xl(tO), X2(tO); two inputs UI('), uz{-r); and two corresponding outputs YI(T), Y2(T) for T ~ to. Then a system is linear if (1) the state X3(tO) = aXl(to) + {3X2(tO), the output Y3er) = aYl(T) + {3Y2(T), and the input U3(T) = aUI(T) + j1U2(T) can appear in the oriented abstract object and (2) both Y3(T) ~nd X3(T) correspond to the state X3(tO) and input U3(T). The operators pet; to, Xo, u(r») = x(t) and n(t; pet; to,xo, U(T)) = Yet) are linear on {U(T)} EB {x(to)} is an equivalent statement. Example 1.14. In Example 1.2, XI(t)

=

xI(tO)eCto-O/RC

+

RIC

ft

eCT-tJ/RCUl(T) dT

to

yz(t)

=

x2(t)

=

x2(tO)eCto-t)/RC

+

ic ft

eCT-tJ/RCuZ(T) dT

to

are the corresponding outputs YI(t) and Y2(t) to the states Xl(t) and X2(t) with inputs Ul(T) and UZ(T). Since any magnitude of voltage is permitted in this idealized system, any state X3(t) = aXI(t) + (3X2(t), any input U3(t) = aUl(t) + /3uz(t), and any output Y3(t) = aYI(t) + j3Y2(t) will appear in the list of inputoutput pairs that form the abstract object. Therefore part (1) of Definition 1.7 is satisfied. Furthermore, let's look at the response generated by X3(tO) and U3(T). yet)

=

xa(to)e{t; ~v(t), Yl(t O), ••• , Yn(t O), to). Suppose now that the initial conditions are changed: y(to) = Yl(tO) + Xl(tO),

dy dt (to)

=

yz(to)

+ X2{tO),

... ,

d n - 1y dt n- 1 (to) = Yn(t O)

+ Xn(tO)

where X1(tO), X2(tO) and Xn{t o) are small. Furthermore, suppose the input is changed slightly to u(t) = w(t) + v{t) where v(t) is small. To satisfy the differential equations, d(cf>l

+ x 1)/dt

=

11 (4)1 + Xl' cf>2 + X2 ,

••• ,

cf>~ + X n , W

+ v, t)

CHAP. 1]

MEANING OF STATE

9

If 11' [2' ... , In can be expanded about 0 (positive velocity) and decreases for x2 < 0, gIvmg the motion of the system in the direction of the arrows as t increases. For instance, starting at the initial conditions xl(t O) and xz(to) corresponding to the point numbered 1, the system moves along the outer trajectory to the point 6. Sim'ilarly point 2 moves to point 5. However, starting at either point 3 or point 4, the system goes to point 7 where the system motion in the next instant is not determined. At point 7 the output y(t) does not exist for future times, so that this is not a dynamical system.

1.5.

Given the electronic device diagrammed in Fig. 1-6, with a voltage input u(t) and a voltage output y(t). The resistors R have constant values. For to ~ t < t l , the switch S is open; and for t === t l , S is closed. Is this system linear? R

s u(t)

R

y(t)

R

Fig. 1-6

Referring to Definition 1.7, it becomes apparent the first thing to do is find the state. No other information is necessary to determine the output given the input, so there is no state, i.e. the dimension of the state space is zero. This problem is somewhat analogous to Example 1.1, except that the position of the switch is specified at each instant of time. To see if the system is linear, sincex(t) = 0 for all time, we only need assume two inputs Ul(t) and U2(t). Then Yl(t) = uI(t)/2 for to ~ t < tl and Yl(t) = uI(t)/3 for t:=, t l • Similarly Y2(t);::: u2(t)/2 or uz(t)/3. Now assume an input o.:ul(t) + {3U2(t). The output is [o.:ul(t) + {3u2(t)]J2 for to === t < tl and Ia::Ul(t) + {3u2(t)]/3 for t> tl' Substituting Yl(t) and Y2(t), the output is aYl(t) + {1Y2(t), shOWing that superposition does hold and that the system is linear. The switch S can be considered a time-varying resistor whose resistance is infinite for t < tl and zero for t::=: t l . Therefore Fig. 1-6 depicts a time-varying linear device.

12 1.6.

MEANING OF STATE

[CHAP.!

Given the electronic device of Problem 1.5 (Fig. 1-6), with a voltage input u(t) and a voltage output y(t). The resistors R have constant values. However, now the position of the switch S depends on y(t). Whenever y(t) is positive, the switch S is open; and whenever y(t) is negative, the switch S is closed. Is this system linear? Again there is no state, and only superposition for zero state but nonzero input need be investigated. The input-output relationship is now yet) =

[5u(t)

+ u(t) sgn u(t)]l12

where sgn u = +1 if u is positive and -1 if u is negative. Given two inputs Ul and U2 with resultant outputs Yl and Yz respectively, an output y with an input u3 = aul + {3~ is expressible as

= [5( aul

y

+ (3U2) + (aul + fJ u 2)

sgn (aul + !3u z)]l12

To be linear, ay + j3Y2 must be equal y which would be true only if aUl

+ j3u2 sgn U2

sgn ul

=

(aUl

+ (3u2) sgn (aul + !3U2)

This equality holds only in special cases~ such as sgn Ul = sgn u2 = sgn (au1 + j3u2), so that the system is not linear.

1.7.

Given the abstract object characterized by =

y(t)

xoe to - t

+

,t

J to

e",-tu(T) dT

Is this time-varying? This abstract object is that of Example 1.2, with RC = 1 in equation (1.1). By the same procedure used in Example 1.16, it can also be shown timeinvariant. However, it can also be shown time-invariant by the test given after Definition 1.8. The input time function u{t) is shifted by T, to become (t). Then as can be seen in Fig. 1-7,

u

~ (t) = u(t - T)

Starting from the same initial state at time to + T,

Y(u)

xoe to + T-u

=

f

+

e t - u u(.,- - T) dT

17

to+T

Let g =

T -

Y(CT)

T:

=

xoeto+T-U

+

f

u-T

~ to

~

Yat

y (t + T)

CT

=

= t +T xoe to -

+

I~(t)

~

to

Iy(t)

ft ee-tuW dg to

which is identical with the output y(t).

t

t

Original System

e~+T-uuW dg

gives t

t

to

to

Evaluating

t

+T

t+ T

~ to+T

Shifted System Fig. 1-7

t+T

t

t

MEANING OF STATE

CHAP. I]

13

Supplementary Problems 1.8.

Given the spring-ruass system shown in Fig. 1-8. What is the physical object, the abstract object, and the state variable? "

1.9.

Given the hereditary system yet) =

it""

K(t, T) U(r) dT

where

K(t, T) is some single-valued continuously differentiable function

of t and T. What is the state variable? Is the system time-varying?

Is the system linear? Fig. 1-8

1.10.

Given the discrete time system x(n + 1) = x(n) + u(n), the series of inputs u(O), u(I), ... , u(k), and the state at step 3, xeS). Find the state variable x(m) at any step m =: O.

1.11.

An abstract object is characterized by yet) =u(t) for to ~ t < t l , and by dy/dt = du/dt for t =: t l . lt is given that this abstract'object will permit discontinuities in yet) at t l . What is the dimension of the state space for to === t < tl and for t =: tl?

1.12.

Verify the solution (1.2) of equation (1.1), and then verify the solution (1.3) of equation (1.4). Finally, verify the solution Xl + 2xz = C(2xl + xZ)4 of Example 1.9.

1.13.

Draw the trajectories in two dimensional state space of the system

1.14.

Given the circuit diagram of Fig. 1-9(a), where the nonlinear device NL has the voltage-current relation shown in Fig.- 1-9(b). A mathematical equation is formed using i = Cv and v = let) = f(Ov):

v=

y+ y

= O.

(1IC)f-l(v)

where 1- 1 is the inverse function. Also, veto) is taken to be the initial voltage on the capacitors. Is this mathematical relation a dynamical system with a state space description?

C

~ ~

NL

Ca)

vet)

i

(b) Fig. 1-9

1.15.

Is the mathematical equation y2 + 1 = u a dynamical system?

1.16.

Is the system dy/dt = tZy time-varying?

Is it linear?

1.17.

Is the system dy/dt = l/y tim'e-varying?

Is it linear:

1.18.

Verify that the system of Example 1.16 is nonlinear.

1.19.

~how

1.20.

Show equation (1.10) is time-invariant if the coefficients j = 0, 1, ... , n, are not functions of time.

1.21.

equation (1.10) is linear.

Given dXl/dt = Xl + 2, dX2/dt = Xz + u, y = Xl + xz· Does this system have a state space description:

(a)

(b) What is the input-output relation?

(c)

Is the system linear?

£ri

and Pj for

i = 1,2,

"'J

nand

14

MEANING OF STATE

[CHAP; 1

= u(t + T)

1.22.

What is the state space description for the anticipatory system yet) tion 5 for dynamical systems is violated?

1.23.

Is the system dx/dt = etu, y = e-tx time-varying?

1.24.

What is the state space description for the diiferentiator y = du/dt?

1.25.

Is the equation y = j(t)n a dynamical system given values of jet) for to

===

in which only con-

t

=== tl

only?

Answers to Supplementary Problems

x

1.8.

The physical object is the spring-mass system, the abstract object is all x obeying M + kx = 0, and the state variable is the vector having elements x(t} and dx/dt. This system has a zero input.

1.9.

It is not possible to represent

=

yet)

y(t o)

+

ft K(t, r) u(r) d-r,

unless K(t, r)

= K(t o, r)

to

(This is true for to ~ T for the delay line.) For a general K(t, r), the state at time t must be taken as U(T) for - 0 0 < T < t. The system is linear and time varying, unless K(t, r) = K(t - r), in which case it is time-invariant. k-1

1.10.

x(k)

= x(3) + i=3 ~ u(i)

for

k

= 4,5, . ..

3-k

and x(k)

= x(3) -

~ u(3 - i) i=l

for k

= 1,2,3.

Note we

need not "tie" ourselves to an "initial" condition because anyone of the values xCi) will be the state for i = 0, 1, ... n. 1.11.

The dimension of the state space is zero for to::::: t < tb and one-dimensional for t:=: t 1• Because the state space is time-dependent in general, it must be a family of sets for each time t. Usually it is possible to consider a single set of input-output pairs over all t, i.e. the state space is timeillVariant. Abstract obj ects possessing this property are called uniform abstract objects. This problem illustrates a nonuniform abstract object.

1.12.

Plugging the solutions into the equations will verify them.

1.13.

The trajectories are circles. It is a dynamical system.

1.14.

It is a dynamical system, because vet) is real and defined for all t ~ to. However, care must be taken in giving a state space description, because j-l(V) is not single valued. The state space description must include a means of determining· which· of the lines 1-2, 2-3 or 3-4 a particular voltage corresponds to.

1.15.

No, because the input u

1.16.

It is linear and time-varying.

1.17.

It is nonlinear and time-invariant.

1.21.

(a)

Yes

(b) yet)

(c) No

t-

e

to

[x 1(t O)

u(t)

1---...----- y(t)

---+--f + Fig. 2-6

This interchange could not be done if 0:. were a general function of time. In certain special cases it is possible to use integration by parts to accomplish this interchange. If o:.(t) = t, then (2.1) can be integrated to y(t}

=

y(t o)

+

ft 'T[Y('T) + u('T)] d-r to

18

METHODS FOR OBTAINING THE STATE EQUATIONS

Using integration by parts, yet)

::::

y(to) -

It f'l'" [y(~) + ~

uW] dg dT

~

+ t

it

[yeT)

[CHAP. 2

+ U(T)] dT

4

which gives the alternate flow diagram shown in Fig. 2-7.

u(t)

--+-1

r------+---.-. y( t)

+ Fig. 2·7

Integrators are used in continuous time systems, delayors in discrete time (sampled data) systems. Discrete time diagrams can be drawn by considering the analogous continuous time system, and vice versa. For time-invariant systems, the diagrams are almost identical, but the situation is not so easy for time-varying systems. Example 2.2. Given the discrete time system y(k

+ l + 1)

:::: ay(k + 1)

+ au(k + l)

(2.2)

with initial condition y(l). The analogous continuous time systems is equation (2.1), where d/dt takes the place of a unit advance in time. This is more evident by taking the Laplace transform of (2.1),

and the

£

aY(S)

+ aU(S)

zY(z) - zy(l) :::: aY(z)

+ aU(z)

transform of (2.2),

Hence from Fig. 2-5 the diagram for (2.2) can be drawn immediately as in Fig. 2-8.

u(k

+ 1)

> - - - + - - -....... y(k + 1)

---+~

Fig. 2-8

If the initial condition of the integrator or delayor is arbitrary, the output of that integrator or delayor can be taken to be a· state variable. Example 2.3. The state variable for (2.1) is yet), the output of the integrator. To verify this, the solution to equation (2.1) is yet) = y(to) e(Y.Ct-to) + 0'. f t e(Y.Ct-T) U(T) dT to

Note y(t o) is the state at to, so the state variable is yet). Example 2.4. The state variable for equation (2.2) is y(k + l), the output of the delayor. manner similar to the previous example.

This can be verified in a


CHAP.2J

19

Example 2.5.

From Fig. 2-7, the state is the output of the second integrator only, because the initial condition of the first integrator is specified to be zero. This is true because Fig. 2-7 and Fig. 2-5 are equivalent systems. Example 2.6.

A summer or a scalor has no state associated with it, because the output is completely determined by the input.

2.3 CANONICAL FLOW DIAGRAMS FOR TIME ..INVARIANT SYSTEMS Consider a general time-invariant linear differential equation with one input and one output, with the letter p denoting the time derivative d/dt. Only the differential equations need be considered, because by Section 2.2 discrete time systems follow analogously. p1l.y

+ (Xlpn-Iy + ... + O'.n-IPY + anY

= fJopnu

+ f3 1 pn- 1u + ... + fJn-IPu + fJnu

(2.3)

This can be rewritten as p""(Y - f3 ou )

+ pn-l(aIY - f3 IU) + ... + p(an-Iy - f3n - Iu) + anY -

because aip1O-iy = pn-iaiy , which is not true if and rearranging gives y

=

f3 0u

a

i

f3 nu

=0

depends on time. Dividing through by.pn

1 1 1 aly) + ... + pn-l (f3 1O - Iu - <X1o-IY) + pn (fJ nu - anY)

+ P(f3 1u -

(2.4)

from which the flow diagram shown in Fig. 2-9 can be drawn starting with the output y at the right and working to the left. u(t) -

-------+----.....,...-

.......

I---~y(t)

Fig. 2-9.

Flow Diagram of the First Canonical Form

The output of each integrator is labeled as a state variable. The summer equations for the state variables have the form Y

Xl = X2 =

Xn -

1

X10

+ f3 0u -a1Y + X,2 + f3 I U -a2Y + X3 + f32U Xl

-an-IY -anY

+ Xn + f3 n -

+ f3nU

(2.5) Iu

20


[CHAP. 2

Using the first equation in (2.5) to eliminate y, the differential equations for the state variables can be written in the canonical matrix form

d dt

Xl

-a

X2

-a

1

1

0

2

0

1

..........

,.

Xn

...........

0 0

- an - 1 -an

0 0 "

0 0

/3 1 /3 2 -

Xl X2

a 1/3 o a 2/3 0

+

III

1

X n -1

0

Xn

u f3 n -

1 -

/3 n

-

(2.6)

f3 o

an- 1 a rJ30

We will call this the first canonical form. Note the Is above the diagonal and the a'S down the first column of the n x n matrix. Also, the output can be written in terms of the state vector

y

=

(I 0 '"

0 0)

(2.7)

Note this form. can be written down directly from the original equation (2.3). Another useful form can be obtained by turning the first canonical flow diagram "backwards." This change is accomplished by reversing all arrows and integrators, interchanging summers and connection points, and interchanging input and output. This is a heuristic method of deriving a specific form that will be developed further in Chapter 7. y(t)

u(t)

Fig. 2-10.

Flow Diagram of the Second Canonical (Phase-variable) Form

Here the output of each integrator has been relabeled. variables are now

....................

Xn

y

-a1X n flnX1

,.

....

a ZX n - 1 -

l1li

•••••••••

••• -

,.

••

O!n-1 X 2 -

0

The equations for the state

.............

O!n X 1

+ f3n - 1x 2 + ... + f3 1xn + f3 o[U -

II

II

•••••••••

II

....

.

+u

Q1X n -

••• -

O!n_1 X 2 -

O!n X 1]


CHAP.2J

21

In matrix form, (2.8) may be written as

X2

d dt

1 0

0 0

Xl

=

•

•

•

•

..

•

•

..

•

•

.. ~,.

..

0

0

-an -

Xn

and

•

0 1 •

0 0 •

•

•

•

•

•

..

..

ill

•

X2

+

•

1

0 1

•

0 0

Xl

-an -

-a 1

2

u

(2.9)

0 1

Xn

(2.10)

y

This will be called the second canonical form, or phase-variable canonical form. Here the Is are above the diagonal but the a'S go across the bottom row of the n x n matrix. By eliminating the state variables x, the general input-output relation (2.8) can be verifie~d. The phase-variable canonical form can also be written down upon inspection of the original differential equation (2.3).

2.4

JORDAN FLOW DIAGRAM

The general time-invariant linear differential equation (2.3) for one input and one output can be written as y

(2.11)

By dividing once by the denominator, this becomes y

(2.12)

Consider first the case where the denominator polynomial factors into distinct poles Ai, i = 1,2, ... , n. Distinct means Ai 01= Aj for i 01= j, that is, no repeated roots. Because most practical systems are stable, the Ai usually have negative real parts. pn

+ a 1pn-1 + ... + O::n-1P + an

=

(p - A 1 )(P - A2) '

••

(p -

(2.13)

An)

A partial fraction expansion can now be made having the form

y

f30u

+

P1 P - A1

---\u

Pz Pn +- u + ... + --\-u P - A2 P ---:: .ll.n

(2.14)

Here the residue Pi can be calculated as (f3 i

-

a1f3o)Af-'1 + (f3 2 -

0::2f3 o)Ar-

Z

+ ... + ({3n-l -

(Ai - Al)(Ai - "-2)' .. (Ai -

\-1)('\ -

f3

O::n- 1 0)Ai

+ (f3n -

Ai+l)' .. (\ - An)

0::)30)

(2.15)

The partial fraction expansion (2.14) gives a very simple flow diagram, shown in Fig. 2-11 follo,ving.

22


u(t) --~P---------l

[CHAP. 2

.r---y(t)

+

Fig. 2-11.

Jordan Flow Diagram for Distinct Roots

Note that because p. and A. can be complex numbers, the statesx.t are complex;'valued functions of time. The state equations assume the simple form ~

~

(2.16) Xn = AnXn-+U

Consider now the general case. For simplicity, only one multiple root (actually one Jordan block, see Section 4.4, page 73) will be considered, because the results are easily extended to the general case. Then the denominator in (2.12) factors to (2.17)

instead of (2.13). Here there are sion for this case gives y

=

PIU

P2U

v

identical roots. Performing the partial fraction expan-

f3 oU + (P-Al)V + (P-Al)V-l + ... +

Pvu P-AI

+

PiJ+I U

P-Av+I

+ ... +

Pnu

(218)

P-An'

The residues at the multiple roots can be evaluated as =

k

= 1,2, ... , v

(2.19)

where f(p) is the polynomial fraction in P from (2.12). This gives the flow diagram shown in Fig. 2-12 following.

CHAP. 2]


u(t}------"""'"--------~------------..,...f

23

y(t)

+

+

Fig. 2~12.

Jordan Flow Diagram with One Multiple Root

The state equations are then

A1X v - l

XV-1

Xv = XV+1

Xn y

).IX

+ Xv

v+ U

Av+lXv +l

(2.20)

+U

+U = (3ou + PIX! + P2X2 + ... + PnX n AnXn

24

METHODS FOR OBTAINING THE 'STATE EQUATIONS

(CHAp. 2

The matrix differential equations associated with this Jordan· form are Al 1 0 0 Al 1

Xl X2

..... W'.,. ....

d

X v -1

dt

Xv

0 0 0

=

Xv+l ill

Xn

0 0

and

II

..

lit

...........

0 0 0

0

•••••

0

0 0

ill

0

0 0

W'

.....

0 0

0#'""

..... "

Al 1 0 0 Al 0

............

0

(PI P2

...

"

ill

0 0

Xl X2

..

0 0 0

o AII+1

0

0

y

0 0

XI/-1 Xv

+

0 1

XII + 1

1

Xn

1

u

(2.21)

.....................

0

0

n

P )

An

(j:)

+

f3 0u

(2.22)

In the n X n matrix, there is a diagonal row of ones above each Al on the diagonal, and then the other A's follow on the diagonal. Example 2.7. Derive the Jordan form of the differential system

jj Equation (2.23) can be written as

y ::::

u+ u

+ 2y + y ==

(~: 1~)2 u

whose partial fraction expansion gives

o

y

(p

(2.23)

+ 1)2 U +

1

P

+ 1u

Figure 2-13 is then the flow diagram.

yet)

+ u(t)--+40-/

+

Fig. 2-13 Because the scalor following Jordan form are

Xl

is zero, this state is unobservable.

The matrix state equations in


25

- 2.5 TIME-VARYING SYSTEMS Finding the state equations of a time-varying system is not as easy as for time-invariant systems. However, the procedure is somewhat analogous, and so only one method will be given here. The general time-varying differential equation of order n with one input and one output is shown again for convenience. (1.10)

=

Differentiability of the coefficients a suitable number of times is assumed. Proceeding in a manner somewhat similar to the second canonical form, we shall try defining states as in (2.8). However, an amount Ylt) [to be determined] of the input u(t) enters into all the states.

Xl =

x 2 + Yl(t)U

x2

X3

+ Y2(t)U (2.24)

Xn- l

xn

xn

+ Yn-l(t)u

-al(t)X

Y

Xl

n- l¥2(t)X

n_ 1 -

••• -

an(t)xl

+ Yn(t)u

+ Yo(t)u

By differentiating y n times and using the relations for each state, each of the unknown Yi can be found. In the general case,

= (2.25)

= Example 2.8. Consider the second order equation d 2y dt 2

+

dy Il'l(t) dt

+

=

1l'2(t)y

Then by (2.24) Y

==

Xl

d 2u f3o(t) dt2

+

du f31(t) dt

+

(2.26)

f32(t)U

+ yo(t)u

(2.27)

and differentiating,

(2.28)

Substituting the first relation of (2.24) into (2.28) gives

y :;:

x2

+

[Yl(t)

+ yo(t)]u + Yo(t)u

(2.29)

Differentiating again,

(2.90)

From (2.24) we have

X2 = -al(t)X2 - 1l'2(t)Xl + Y2(t)U

(2.31)

Now substituting (2.27), (2.29) and (2.30) into (2.31) yields

y-

[Yl(t)

+ yo(t)]u =

+ 2Yo(t)]u - Yo(t) u [Yl(t) + Yo(t)]u - Yo(t)u}

- [Yl(t)

-ll'l(t){y -

-

t'X2(t)[y - Yo(t)u]

+ Y2(t)u

(2.32)

26


[CHAP. 2

Equating coefficients in (2.26) and (2.32), Y2

+ Yl + Yo +(Y1 + Yo)a1 + Yoa2 Yl + 210 + alYO = 131

=

132

(2.33) (2.34)

Yo = 130

(2.35)

Substituting (2.35) into (2.34), (2.36) and putting (2.35) and (2.36) into (2.33),

Y2 = Po -

PI + /32 + (alf3o + 2Po -

+ f3oa1 -

f3l)a1

(2.37)

f30az

Using equation (2.24), the matrix state equations become J\ d dt

0 0

X2

•••••

0

xn

-an(t)

II

41

1

0

0

1

.......

ill,.

0 -an_ 1 (t)

y

II

ill

..

'II."

0 0 ill

.......

,.

......

Xi Xz ,.

+

..

0

1

-an_2 (t)

-a 1(t)

xn

u

(2.38)

Yn-l(t) Yn(t)

(1 0 ... 0)

2.6 GENERAL STATE EQUATIONS Multiple input-multiple output systems can be put in the same canonical forms as single input-single output systems. Due to complexity of notation, they will not be considered here. The input becomes a vector u(t) and the output a vector y(t). The components are the inputs and outputs, respectively. Inspection of matrix equations (2.6), (2.9), (2.21) and (2.88) indicates a similarity of form. Accordingly a general form for the state equations of a linear differential system of order n with m inputs and k outputs is

+ B(t)u = C(t)x + D(t)u

dx/dt = A(t)x y

where

(2.39)

x(t) is an n-vector,

u(t) is an m-vector, y(t) is a k-vector,

A(t) is an n X n matrix,

B(t) is an n

X

m matrix,

C(t) is a k x n matrix, D(t) is a k x m matrix. In a similar manner a general form for discrete time systems is x(n + 1)

y(n)

= =

A(n) x(n)

C(n) x(n)

+ B(n) u(n) + D(n) u(n)

where the dimensions are also similar to the continuous time case.

(2.40)

CHAP.2J


27

Specifically, if the system has only one input u and one output y, the differential equations for the system are dx/dt = A(t)x + b(t)u

=

Y

ct(t)x

+ d(t)u

and similarly for discrete time systems. Here c(t) is taken to be a column vector, and ct(t) denotes the complex conjugate transpose of the column vector. Hence ct(t) is a row vector, and ct(t)x is a scalar. Also, since u, y and d(t) are not boldface, they are scalars. Since these state equations are matrix equations, to analyze their properties a knowledge of matrix analysis is needed before progressing further.

Solved Problems 2.1.

Find the matrix state equations in the first canonical form for the linear timeinvariant differential equation y

with initial conditions y(O) state variables.

= Yo,

+ 5y + 6y y(O)

=

= yo.

it + u

(2.41)

Also find the initial conditions on the

Using p = dldt, equation (2 .."-1) can be written as p2y + 5py + 6y = pu + u, and rearranging, 1 1 Y = -(u-5y) + ~(u-6y) p p2_ The flow diagram of Fig.

2~14

Dividing by p2

can be drawn starting from the output at the right.

u----------~--~------~~--~

.>--t---.... Y

Fig. 2..14

Next, the outputs of the integrators are labeled the state variables an equation can be formed using the summer on the left:

X2 =

-6y

Xl

and x2 as shown.

Now

+u

Similarly, an equation can be formed using the summer on the right:

Xl = Also, the output equation is y =

Xl'

X2 -

5y

+u

Substitution of this back into the previous equations gives

Xl

-5X1

+ X2 + u (2.42)

28


[CHAP. 2

The state equations can then be written in matrix notation as

with the output equation y

The initial conditions on the state variables must be related to Yo and Yo, the given output initial conditions. The output equation is xl(t) = yet), so that Xl(O) = yeO) = Yo. Also; substituting yet) = Xl (t) into (2.42) and setting t == 0 gives

yeO)

= -5y(D)

+ X2(O) + u(O)

Use of the given initial conditions determines X2(O) =

Yo + 5yo -

u(O)

These relationships for the initial conditions can also be obtained by referring to the flow diagram at time t = O.

2.2.

Find the matrix state equations in the second canonical form for the equation (2.41) of Problem 2.1, and the initial conditions on the state variables. The flow diagram (Fig. 2-14) of the previous problem is turned "backwards" to get the flow diagram of Fig. 2-15. y......----I

.....+----u

Fig. 2-15

The outputs of the integrators are labeled Xl and x2 as shown. These state variables are different from those in Problem 2.1, but are also denoted Xl and Xz to keep the state vector x(t) notation, as is conventional. Then looking at the summers gives the equations y

X2 =

Xl

+ X2

-6Xl -

5x2

=

(2.43)

+u

(2.44)

Furthermore, the input to the left integrator is

This gives the state equations d

dt and

(Xl) X2

=

(-~ -~)(::)

y

The initial conditions are found using (2.43),

and its derivative

(1

1)(:~)

+

G)

u

/"""-'

CHAP. 2]


29

Use of (2 ..M) and (2.45) then gives

Yo = X2(0) - 6x 1 (0) - 5x z(0)

+ u(O)

(2.47)

Equations (2.46) and (2.47) can be solved for the initial conditions

2.3.

Xl(O)

-2yo - lyo

X2(O)

SYo

+ iYo -

+ lu(O) lu(O)

Find the matrix state equations in Jordan canonical form for equation (2.41) of Problem 2.1, and the initial conditions on the state variables. The transfer function is y

A partial fraction expansion gives

-1

Y

=

p+2 u

+

2

p+8 u

From this the flow diagram can be drawn:

+ +

u---..

J----~y

+

Fig. 2-16

The state equations can then be written from the equalities at each summer:

(-~ _~)(::) Y

=

+

(~) u

(-1 2) ( :: )

From the output equation and its derivative at time to,

Yo = 2xz(O) - Xl (0) Yo = 2X2(0) - 0: 1(0) The state equation is used to eliminate Xl(O) and X2(0):

Solving these equations for Xl(O) and x2(O) gives

3yo - Yo

Xl(O)

u(O) -

xz(O)

l[u(O) - 2yo - Yo]

30

2.4.


[CHAP. 2

Given the state equations

! (~:)

(-~ -!)(~:)

=

(1

y

+

G)

u

1)(:~)

Find the differential equation relating the input to the output. In operator notation, the state equations are

Eliminating

Xl

and

X2

PX l

X2

PX2

- 6X l -

then gives p2y

+ 5py + 6y

5x2

=

+U

pu,

+u

This is equation (2.41) of Example 2.1 and the given state equations were derived from this equa,.. tion in Example 2.2 .. Therefore this is a way to check the results.

2.5.

Given the feedback system of Fig. 2-17 find a state space representation of this closed loop system. K G(s) = - -

R(s)---+.-t

8+1

H(s) =

1-------4~-_

C(s)

1

8

+3

Fig. 2-17

The transfer function diagram is almost in flow diagram form already. U sing the Jordan canonical form for the plant G(s) and the feedback H(s) separately gives the flow diagram of Fig. 2-18. r----- -----------1

I

I

1 - + - - 9 - - - _ c(t)

r ( t ) - -........

+

I I I IL __________________ ! ~

+

Fig. 2-18

Note G(s) in Jordan form is enclosed by the dashed lines. Similarly the part for H(s) was drawn, and then the transfer function diagram is used to connect the parts. From the flow diagram, d

dt

(Xl) X2

=

(-1 -1) (Xl) + (1) K -3

X2

0

ret),

c(t)

=

(K

O)(~~)

, 2.6.

31


CHAP. 2]

Given the linear, time-invariant, multiple input-multiple output, discrete-time system y/n + 2)

+ (l:lyl(n + 1) + a2Yl(n) + Y2YZ(n + 1) + YSY2(n) == y 2(n + 1) + Y Y2(n) + (Xsyl(n + 1) + YI(n) == (X4

1

(31u1(n) (32u/n)

+ 8Iu 2(n) + 82u2(n)

Put this in the form

+ Bu(n) Cx(n) + Du(n)

x(n+ 1)

Ax(n)

y(n)

where

y(n) =

(~;~:?),

u(n) =

(=;~:?).

The first canonical form will be used. Putting the given input-output equations into z operations (analogous to p operations of continuous time system),

Z2Y1 Dividing by

Z2

and

z

+ alZYI + (X2Yl + Y2 z Y2 + YSY2 == zY2 + YIY2 + (lSZYI + ()!4VI ==

respectively and solving for YI and

[hul f32 u I

+ 0lu2 + 02U 2

Y2'

Starting from the right, the flow diagram can be drawn as shown in Fig. 2-19.

+

Fig. 2-19

Any more than three delayors with arbitrary initial conditions are not needed because a fourth such delayor would result in an unobservable or uncontrollable state. From this diagram the state equations are found to be

32

2.7.


[CHAP. 2

Write the matrix state equations for a general time-varying second order discrete time equation, i.e. find matrices A(n), B(n), C(n), D(n) such that x(n + 1)

y(n)

= =

A(n) x(n) C(n) x(n)

+ B(n) u(n) + D(n) u(n)

(2.48)

given the discrete time equation y(n + 2)

+ a 1(n) y(n + 1) + a 2 (n) y(n)

f3 o(n) u(n + 2)

+ f3 1(n) u(n + 1) + {32(n) u(n)

(2.49)

Analogously with the continuous time equations. try xl(n

+ 1) =

x2(n)

x2(n + 1) = -al(n) x2(n) yen) = xl(n)

+ "h(n) u(n) a2(n) Xl (n)

(2.50)

+ Y2(n) u(n)

(2.51)

+ Yo(n) u(n)

(2.52)

Stepping (2.52) up one and substituting (2.50) gives yen

+ 1) =

X2(n)

+ Yl(n) u(n) +

yo(n + 1) u(n + 1)

(2.53)

Stepping (2.53) up one and substituting (2.51) yields yen

+ 2) =

+ Y2(n) u(n) + YI(n + 1) u(n + 1) + Yo(n + 2) u(n + 2) (2.49) and equating coefficients of u(n), u(n + 1) and

-al(n) X2(n) - a2(n) XI(n)

Substituting (2.52). (2.53). (2.54) into gives Yo(n)

J3o(n - 2)

YI(n)

J31(n-1) - al(n)J3o(n-2)

Y2(n)

J32(n) - al(n)J31(n -1)

+

(2.54) u(n + 2)

[al(n)al(n -1) - a2(n)]J30(n - 2)

In matrix form this is

(1 0)

y(n)

2.8.

(::~:D +

yo(n) u(n)

Given the time-varying second order continuous time equation with zero input, y

+ O!I(t)y + a 2(t)y =

0

(2.55)

write this in a form corresponding to the first canonical· form

(~:) .Y

- (=:;m =

~)(~:)

(y,(t) y.(t» (~:)

(2.56) (2.57)

2]


33

To do this, differentiate (2.57) and substitute for Xl and Xz from (2.56),

if

= (rl - Yial - Yz a 2)Xl

Differentiating again and substituting for

y

=

C"il -

Xl

+

(Yl + yz)xz

(2.58)

and X2 as before,

2a{Yt - 0:111 - tl2Y2 - 2azY2 - Cl!2Yl + a1C1!zYz + a1Y1)x1

+

(2ft - a111 - Cl!zYz + YZ)X2

Substituting (2.57), (2.58) and (2.59) into (2.55) and equating coefficients of equations Yl - alYl

+ (0:2 -

+ (a1 -

2C1!z)yz

Y2

tl1hl

+ (CI!~ -

=

Xl

(2.59)

and Xz gives the

0

a2 a 1 + Cl!2 --'- a2)YZ

+ 2·h -

a211 = 0

In this case, Yl(t) may be taken to be zero, and any non-trivial yz(t) satisfying 12

+ (a1 -

2a2)YZ

+ (a~ -

Cl!Z a l

+ IX2 -

0:2h2 = 0

(2.60)

will give the desired canonical form. This problem illustrates the utility of the given time~varying form (2.38). It may always be found by differentiating known functions of time. Other forms usually involve the. solution of equations such as (2.60), which may be quite difficult, or require differentiation of the ai(t). Addition of an input contributes even more difficulty. However, in a later chapter forms analogous to the first canonical form will be given.


Given the discrete time equation yen + 2) + 3y(n + 1) + 2y(n) = u(n + 1) + 3u(n), find the matrix state equations in (i) the first canonical form, (ii) the second canonical form, (iii) the Jordan canonical form.

2.10.

Find a matrix state equation for the multiple input-multiple output system

YI + atih + a2YI + Y3112 + Y4YZ yz + 11112 + 1zY2 "+ a3111 + Cl!4Y1 2.11.

(h u 1 + SlUZ

:::::

f32 u 1

+ S2 U Z

Write the matrix state equations for the system of Fig. 2-20, using Fig. 2-20 directly.

u

2.12.

:::::

Xz p

+

Consider the plant dynamics and feedback compensation illustrated in Fig. 2-21. Assuming the initial conditions are specified in the form v(O), v(O), ii (0), ·v·(O), w(O) and w(O), write a state space equation for the plant plus compensation in the form x = Ax + Bu and show the relation between the specified initial conditions and the components of xeD).

+

Xl

p

u---~

~----~y

500 p(p + 5)(1'2 + P + 100)

v w _ _ _ _....,

+ p + 100 1'2 + 2p + 100 1'2

Fig. 2-21

34


[CHAP. 2

2.13.

The procedure of turning "backwards" the flow diagram of the first canonical form to obtain the second canonical form was never· justified. Do this by verifying that equations (2.8) satisfy the original input-output relationship (2.3). Why can't the time-varying flow diagram corresponding to equations (2.24) be turned backwards to get another canonical form?

2.14.

Given the linear time-varying differential equation

y + al(thi + a2(t)y

= f3o(t) U + fll(t)U

Yo.

with the initial conditions y(O) = Yo, y(O) = (i)

+ f32(t)U

Draw the flow diagram and indicate the state variables.

(ii) Indicate on the flow diagram the values of the scalors at all times. (iii) Write down the initial conditions of the state variables. 2.15.

Verify equation (2.37) using equation (2.25).

2.16.

The simplified equations of a d-c motor are Motor Armature:

Ri

+

L

~; =

V - Kf :

Inertial Load: Obtain a matrix state equation relating the input voltage V to the output shaft angle state vector

8

using a

x

2.17.

The equations describing the time behavior of the neutrons in a nuclear reactor are 6

In

Prompt Neutrons:

(p(t) - {l)n

+ i=l :I ""-iCi

Delayed Neutrons: 6

where f3 =

:I f3i

and p(t) is the time-varying reactivity, perhaps induced by control rod motion.

i=1

Write the matrix state equations. 2.18.

Assume the simplified equations of motion for a missile are given by

z + K ¢ + K a + Kaf3

Lateral Translation: Rotation:

if; + K 4a + K5f3 =

Angle of Attack:

2

1

0

Ka

a::: ¢ -

= 0

z M(l) = Ka(l)a

Rigid Body Bending Moment:

+ K B (l)f3

where z::;: lateral translation, ¢ = attitude, a == angle of attack, f3 == engine deflection, M(l) bending moment at station 1. Obtain a state space equation in the form dx

dt where

u = f3

and

y

(i)

x

==

(M~l)) ,

=

(D

Ax

+ Bu,

y::::: - ex

+ Du

taking as the state vector x

(ii)

x =

G)

(iii)

x


CHAP. 2] 2.19.

'Consider the time-varying electrical network of Fig. 2-22. The voltages across the inductors and the current in the capacitors can be expressed by the relations

ea -

el

:t

il -

i2

d dt (Glel)

(L1i l )

di l

=

del Glert

-il

. dL I '/,1&

+

L ldt

dGI

+

cI&

o el -

eb

:t (L 2i 2 )

d~

=

L2dt

35

o

. dL2

+ t2dt

It is more convenient to take as states the inductor fluxes PI

=

it

(e a - el) dt

+ PI(tO)

to

ft (el -

P2

eb) dt

+

P2(t O)

(i1 - i 2) dt

+

q1(tO}

to

and the capacitor charge ql

=

ft to

Obtain a state space equation for the network in the form

d:x dt

A(t)x

+

B(t)u

yet)

C(t)x

+

D(t)u

where the state vector x, input vector u, and output vector yare

x=GJ x=G)

(i)

(ii)

2.20.

u

(::)

u

(::)

y

y

=G) =G)

Given the quadratic Hamiltonian H = -iqTVq + ipTTp where q is a vector of n generalized coordinates, p is a vector of corresponding conjugate momenta, V and Tare n X n matrices corresponding to the kinetic and potential energy, and the superscript T on a vector denotes transpose. Write a set of matrix state equations to describe the state.

Answers to Supplementary Problems 2.9.

(-3 ~) x(n) + (~) u(n); -2

yen)

x(n+ 1)

(_~ _~) x(n)

+

(~) u(n);

yen)

(3

x(n + 1) =

(-~ _~) x(n)

+

(~) u(n);

yen)

(-1 2) x(n)

(i)

x(n + 1)

(ii)

(iii)

=

::::

(1 0) x(n)

1) x(n)

36 2.10.


[CHAP. 2

Similar to first canonical form

. x

C"

-a3 -a4

(~

y

~)

1 -Y3 0 -Y4 0 -[1 0 -Y2

-0:2

0 0

0 1

1

x

+

0

~l)U

(~'

°2

f32

~) x

or similar to second canonical form

(-;

.

X

1 -0:1

0

0 -0:4

~.)x (~1 ~)U +

0

-Y4 -Y3 -Y2 -Y1

(~

Y

0 0

0

1

{12

°2

~)x

2.11.

:t G;) Y

2.12.

For

Xl

= v)

X2

= V,

::li3

x

= V,

X4

= ·v:

C~' (1

x5

+"'1

-"'2 0

+:2) (::) +(~)

-"'3

X3

U

"'3

oo)G:)

=w, X6=W

the initial conditions result immediately and

1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 -500 -105 -6 0 0 00 0 0 0 1 1 100 1 0 -100 -2 0

0

0

0

x

+

0 500 0 0

u

2.13.

The time-varying flow diagram cannot be turned backwards to obtain another canonical form because the order of multiplication by a time-varying coefficient and an integration cannot be interchanged.

2.14.

(i)

u--------~--------------------~~----------------__,

I - - - -. .

Fig. 2-23

y

The values of Yo, Y1 and Y2 are given by equations (2.33), (2.34) and (2.35).

(ii)

(iii) X1(0) = Yo - Yo(O) u(O)

2.16.

d

dt

37


CHAP. 2]

( i) fJ

x2(0) = Yo - (fo(O)

+ Y1(0»

u(O) - Yo(O) U(O)

fJ

::::

(0 1 0) (

dfJ/dt

2.17.

!£ dt

(~1). :

Os

o 2.18.

(i)

c

(~

=

1

~)

(ii)

-Ka

1

K2 K S

(iii)

o -K4

c == 2.19.

(~

1 'Kex

0 0

(i)

(ii)

o1 ) -0.. 1 1 -LL2 2

A

c

2.20.

;

de/dt

D

::::

The equations of motion are Using the state vector x :::: (q1

qn PI

...

X :::: (

Pn)T, the matrix state equations are

0

-V

T)o

x

)

Chapter

3

Elementary Matrix Theory 3.1 INTRODUCTION This and the following chapter present the minimum amount of matrix theory needed to comprehend the material in the rest of the book. It is recommended that even those well versed in matrix theory glance over the text of these next two chapters, if for no other reason than to become familiar with the notation and philosophy. For those not so well versed, the presentation is terse and oriented towards later use. For a more comprehensive presentation of matrix theory, we suggest study of textbooks solely concerned with the subject. 3.2 BASIC DEFINITIONS Definition 3.1: A matrix, denoted by a capital boldfaced letter, such as A or so that the row numbers are in natural 1, 2, 3 order instead of the column numbers. From this, it can be concluded in general that it takes exactly the same number of permutations to undo PI' P2;' . " Pn to 1,2, ... , n as it does to permute 1,2, ... , n to obtain Pl' P2 • .. " Pn • Therefore p must be the same for each product term in the series, and so the determinants are equal.

=

=

=

60 3.4.

ELEMENTARY MATRIX THEORY

[CHAP. 3

A Vandermonde matrix V has the form

Prove V is nonsingular if and only if Bi "1= OJ for i "1= j. This will be proven by showing det V =

(0 2 -

° )(° °

01)(03 -

1)' •• (On -

3-

2

0n-l)(On -

On-2)' •• (On -

01)

II

=

(OJ -

0i)

1 :=:i< ;===n

For n = 2, det V = O2 - 01' which agrees with the hypothesis. By induction if the hypothesis can be shown true for n given it is true for n -1, then the hypothesis holds for n :=: 2. Note each term of det V will contain one and only one element from the nth column, so that det V = "10 If

On

°

= 0i,

01' 2 , •

for

00' 8 n -

1

i = 1,2, .. 0, n -1, then det V = 0 because two columns are equal. are the roots of the polynomial, and Yo

But

"In-l

+ "lIOn + ... + "In_l 0: - 1

+ Yl(Jn + ... + "In_18~-1

is the cofactor of

0:-

1

=

"In-l (8 n -

01)(On -

O2 )' •• {On -

Therefore

(In-I)

by the Laplace expansion; hence

"I n - l

det

(~~ ~~ ....

(J~-2

~.".~~)

.............

(J;-2

. ••

O~=~

By assumption that the hypothesis holds for n - 1,

Combining these relations gives det V

3.5.

Show' det(!

~)

=

(on - 01)«(Jn - 82 )'

•• (8 n -

IT

On-I)

(OJ -

(Ji)

l===i O.

Definition 4.9:

An n x n Hermitian matrix Q is nonnegative definite if its associated quadratic form ~ is never negative. (It may be zero at times when x is not zero.) Then Q is nonnegative if and only if all its eigenvalues are ~ O.

Example 4.10. ~ ~i 2~1~2 + ~i

= -

= (~1 -

~2)2 can be zero when ~1

=:

~2' and so is nonnegative definite.

The geometric solution of constant Q. when Q is positive definite is an ellipse in n-space.

Theorem 4.9:

Au

A unique positive definite Hermitian matrix R exists such that RR = Q, where Q is a Hermitian positive definite matrix. R is called the square root ofQ.

P1'oof: Let U be the unitary matrix that diagonalizes Q. Then Q = UAU t . Since is a positive diagonal element of A, defineAl/2 as the diagonal matrix of positive A1/ 2. Q = UAlIZAl/2Ut

=

UAl/2UtU.A 1 / 2 U t

Now let R = UA1!2Ut and it is symmetric, real and positive definite because its eigenvalues are positive. Uniqueness is proved in Problem 4.5. One way to check if a Hermitian matrix is positive definite (or nonnegative definite) is to see if its eigenvalues are all positive (or nonnegative). Another way to check is to use Sylvester's criterion. Definition 4.10: The mth leading principal minor, denoted detQm, of the n x n Hermitian matrix Q is the determinant of the matrix Qm formed by deleting the last n - m rows and columns of Q.

Theorem 4.10:

A Hermitian matrix Q is positive definite (or nonnegative definite) if and only if all the leading principal minors of Q are positive (or nonnegative) .

....

A proof is given in Problem 4.6. Example 4.11. Given Q = {qij}'

Then Q is positive definite if and only if

o < det Ql = If

O. But Q. can be written another way:

2 -_ (;2;1) (q22 .

+

q22;2

q12

q12) (~2) qu ~1

which is positive definite if q22> 0 and qllq22 - qi2 > O. The conclusion is that Q is positive definite if det Q > 0 and either q22 or qu can be shown greater than zero.

4.6

MATRIX NORMS

Definition 4.11: A nOTm of a 'matrix A, denoted [IAxii ~ K!ix!1 for all x.

IIAII,

is the mininlum value of

K.

such that

Geometrically, multiplication by a matrix A changes the length of a vector. Choose the vector Xo whose length is increased the most. Then IIAII is the ratio of the length of Axo to the length of Xo. The matrix norm is understood to be the same kind of norm as the vector norm ·in its defining relationship. Vector norms have been covered in Section 3.10. Hence the matrix norm, is to be taken in the sense of corresponding vector norm. Example 4.13. To find jjU112, where

ut = U-l, consider any nonzero vector x. IIUxll~ = xtutUx ::::: xtx = Ilxil~ and so

Theorem 4.11:

= 1

Properties ,of any matrix norm are: (1) (2) (3) (4) (5)

(6)

Proof:

IIUl1 2

Since

IIAII =

IIAxl1 ~ IIAlllixl1 IIAII = max IIAull, whel~e the maximum value of IIAul1 is to be taken over those u such that Ilull = 1IIA+Blj ~ IjAl1 + IIBII IIABjj ~ IIAIIIIBII IAI ~ IIAII for any eigenvalue A of A. IIAII = 0 if and only if A = O. K

min ,

substitution into Definition 4.11 gives (1).

To show (2), consider the vector

II Ax II

u=

X/a, where a =

= IlaAul1 = ialllAul1

~

Kllxll

Ilxll. Then Klailiull

=

Division by Ja) gives IIAulj ~ Kilull, so that only unity length vectors need be considered instead of all X in Definition 4.11. Geometrically, since A is a linear operator its effect on the length of aX is in direct proportion to its effect on x, so that the selection of Xo should depend only on its direction. To show (3),

II(A+B)xil

=

IIAx+Bx)1

~

IIAxl1 + IIBx!1

~

(IIAII + IIBII) Ilxli

for any x. The first inequality results from the triangle inequality, Definition 3.46(4), for vector norms, as defined previously.

CHAP.4J

MATRIX ANALYSIS

79

Bx and x: IIAllllBxl1 ~ IIAllllBllllxl1

To show (4), use Definition 4.11 on the vectors

[[ABxll = IIA(Bx)11

for any x.

~

To show (5), consider the eigenvalue problem Ax of vectors Definition 3.46(5),

Since x is an eigenvector,

IAlllxl1 llxll ~ 0

=

IIAxl1 = IIAxI[

= Ax.

Then using the norm property

I]Allllxll

~

and (5) follows.

To show (6), IJAII = 0 implies IIAxI] = 0 from Definition 4.11. Then Ox for any x. Therefore A = O. The converse 1[011 == 0 is obvious.

Ax =

]jAII 2 --

Theorem 4.12:

where pZmax is the maximum eigenvalue of AtA, and further-

pmax'

more

o To calculate Proof:

IIA1Jz'

Ax= 0 so that

~

Pmin

~

IIAxllz Ilxllz

~

Pmax

find the maximum eigenvalue of AtA.

Consider the eigenvalue problem

AtAgi

= P;gi

Since (x, AtAx) = (Ax, Ax) = [[Ax]!; ~ 0, then AtA is nonnegative definite and pf ~ O. Since At A is Hermitian, PT is real and the gj can be chosen orthonormal. Express any x n

in CV n as x = ~

i=1

eigi .

Then n

~ ]1~iAgill: i=1

I]AxI]~

n

Ilxll;

~ II~igill~

i=1 n

n

~

=

gfp;

i=1 n

~ ~f

i=1

n

n

Since (pnmin ~ ~f ~ ~ ~TP7 ~ (pT)max ~ ~L taking square roots gives the inequality of i=l

i=1

1=1

the theorem. Note that IIAxliz = Pmaxllxl12 when x is the eigenvector gi belonging to (P~)lnaK'

4.7 FUNCTIONS OF A MATRIX Given an analytic scalar function f(a) of a scalar a, it can be uniquely expressed in a convergent Maclaurin series, f(a)

where Ik

= dkf(a)/da k

evaluated at

a = O.

Definition 4.12: Given an analytic function f(a) of a scalar a, the function of an n

i

A is f(A) =

tkAk/k!.

k=O

Example 4.14. Some functions of a matrix A are cosA

=

(cosO)I eAt

=

+

+ (-cosOjA2/2 + ... + (-1)mA2m/(2'm)! + + {eO)At + (e O)A2 t 2/2 + ... + Aktk/k! + ...

(-sinO)A

(eO)I

X

n 1natrix

80

[CHAP~4

MATRIX ANALYSIS

Theorem 4.13: If T-IAT = J, Proof:

then

Note Ak = AA· .. A =

f(A)

= Tf(J) T-l.

TJT-ITJT~l

... TJT-l

00

f(A)

"'1:, fkAk/k! = = k=O

t

= TJkT-l.

Then

fk,TJkT- 1/k!

k=O

Theorem 4.14: If A = TAT-l, where.A. is the diagonal matrix of eigenvalues f(A)

Proof:

Ai, then

= f(A1)xlr! + f(A )x r t + ... + f(An)xnr~ 2

2

From Theorem 4.13, f(A) = Tf(A)T-l

(4.13)

But

o f(A)

= o

Therefore /(A')

f(A)

I(A2)

== (

o

0)

(4.14)

f(An}

Also, let T = (Xl Ix21 ... IXn) and T-l = (rl Ir21 .- .. Irn)t, where ri is the reciprocal basis vector. Then from (4.13) and (4.14),

f{A)

The theorem follows upon partitioned multiplication of these last two matrices. The square root function /(0:) == a 1l2 is not analytic upon substitution of any Ai. Therefore- the square root R of the positive definite matrix Q had to be adapted by -always taking the positive square root of Xi for uniqueness in Theorem 4.9. Definition 4.13: Let f(a) == a in Theorem 4.14 to get the spectral representation of A

==

TAT~l:

A

=

CHAP. A]

MATRIX ANALYSIS

81

Note that this is valid only for those A that can be diagonalized by a similarity transformation. To calculate I(A) Jk

= T/(J)T-I, =

we must find I(J). From equation (4.7),

(Ll1(Al~..

0

o

Lmn(An)

)k

= _

(

L~l(Al~..

k

0

)

Lmn(An)

0

where the last equality follows from partitioned matrix multiplication. Then

0)

/(Lll)

~

I(J)

=

IkJk/k!

k=O

(

o '. I(Lmn)

(4.15)

,

Hence, it is only necessary to find I(L) and use (4.15). By calculation it can be found that for an l x l matrix L, k ) ( l-l

Ak-U-IJ

k ) Ak -CZ- 2 J ( l- 2

o

o where (mn)

= (n-m)!m!' n!

Then from f(L) =

t

(4.16)

the number of combinations of n elements taken m at a time.

fkLk/k!, the upper right hand terms

all

are

n=O all

=

t

k,.=l-l

Ik(

l

~-)Ak-(t-l}/k! = 1

but

1:

k=l-l

f

I k Ak -CI-l)/[(l-1)!(k-l+1)lJ

k=l-l

I k Ak -

a - 1 )/(k-l+1)!

(4.17)

(4.18)

The series converge since I(A} is analytic, so that by comparing (4.17) and (.4-.18), 1 dl-I/(A} (l- I) ! dA l - 1

Therefore /()")

I(L)

=

(

[(l-l)!J_ldl_ll/dAI_l)

dl/dA

0.... !~~)........... .[~l.~ .2} .!].~1.~1~~~/.~>".1~~

..

°

0

...

(4.19)

I(A}

From (4.19) and (4.15) and Theorem 4.13, I(A) can be computed. Another almosLequivalent ,method that can be used comes from the Cayley-Hamilton theorem.

Theorem 4.15 (Cayley-Hamilton): Given an arbitrary n x n matrix A with a characteristic polynomial ¢(>..) = det (A - AI). Then ¢(A) = O. The proof is given in Problem 5.4.

82

MATRIX ANALYSIS

[CHAP. 4

Example 4.15.

Given A

Then det (A - }'I) = ¢(}.) =

By the Cayley-Hamilton theorem,

=

¢(A)

A2 -

6A

+

G!

=

51

~~)

6}.

}.2 -

+5

(4.20)

- G~) + G~) 6

5

= (~ ~)

The Cayley-Hamilton theorem gives a means of expressing any power of a matrix in terms of a linear combination of Am for -m = 0,1, ... , n-l. Example 4.16.

From Example 4.15, the given A matrix satisfies 0 = A2 - 6A + 51. terms of A and I by A2 = SA - 51

Then A2 can be expressed in (4.21)

Also A3 can be found by multiplying (4.21) by A and then using (4.21) again: A3 = 6A2 - 5A =

6(6A -:- 51) - 5A = 31A - 301

Similarly any power of A can be found by this method, including A-I if it exists, because (4.21) can be multiplied by A-I to obtain A-I = (61-A)/5

Theorem 4.16: For an n

X

n matrix A, I(A)

=

YIAn-1

+ Y2 An - 2 + ... + Yn-I A + Yn I

where the scalars Yi can be found from f(J) = Yl In-l

+ Y2Jn-2 + ... + Yn- l J + Yn I

Here 1(1) is found from (4.15) and (4.19), and very simply from (4.14) if A can be diagonalized. This method avoids the calculation of T and T-l at the expense of solving n

I(J) = ~

Yi Jn - i

for the

Yi'

i=1

Pro 01; Since theorem, then

I(A)

'11.-1

= k=O 1: fkAk/k!

and

l: <XkmA.m

Ale =

by the Cayley-Hamilton

m=O

The quantity in brackets is Yn-m • Also, from Theorem 4.13, I(J)

=

T-lf(A)T

=

n

T-1 ~ YiAn-iT i=l

=

n

~

"'liT-IAn-iT

i=1

Example 4.17.

For the A given in Example 4.15, cos A = riA + Y21. Here A has eigenvalues cos A = YtA + Y21 we obtain cos }.1 Y1}.1 + "'12 COS}.2

Yl}.2

+ "'12

}.1

= 1,

}.2

= 5. From

CHAP. 4]

MATRIX ANALYSIS

Solving for Yl and Y2 gives cos

cosA

cos 2

1- 5(32

cos 1-5

2) 3

+

83

(1

5 cos 1 - cos 5 5- 1 \0

1( 1-1) + co~ (1 11) -1

1

2

1

Use of complex variable theory gives a very neat representation of f(A) and leads to other computational procedures.

Theorem 4.17: If f(a} is analytic in a region containing the eigenvalues >"i of A, then

=

f(A)

21. £: /(s) (sI - A)-1 ds 7iJ J'

where the contour integration is around the boundary of the region. Proof:

Si~ce

that f(L) = 27ij

f

.f

f(A) = Tf(J)T-l = T [2~j /(s) (s I

~ L)-l ds.

[(s) (sl - J)-l as]T-l, it suffices to show

Since

sI - L

s

(

~~ s.~~~ .. .

o

~

.. ::: .... ..)

0

s-)..,

...

1)

then (S - >..)1-1 (81 - L)-l

(s

The upper right hand term

all

2y)' (

(8 - >..)1-2

~

(~~. ~).'~~ .. : : : .... ~ .~ ~..

... ......

o

o . ..

(8 - >..)1-1

of 21. £: /(8) (sI - L)-l ds is 7iJ J' 1

27ij

all

f

f(s}ds (8 + >..}1

Because all the eigenvalues are within the contour, use of the Cauchy integral formula then gives 1 d1- 1f(>..) -(Z---l)-! d>..l-l

which is identical with equation (4.19). Example 4.18. Using Theorem 4.17, cos A

2~i

=

For A as given in Example 4.15, (81 - A)-l

=

s- 3 ( -2

-2 ) 8 -

Then cos A

_ 2- 1:

-

3

f -1

cos s(81 - A)-l ds

1

=

82 -

68

+5

(8 -

8 (8 - 3 2)

cos 21rj:r (s - 1)(8 - 5)

2

s - 3

3

2

ds

84

MATRIX ANALYSIS

[CHAP. 4

Performing a matrix partial fraction expansion gives cosA

_1 f cos1 (-2 2) ds + lf~ (2 2)d

=

2rri

-4(8 -1)

2 -2

271"i

4(8 - 5)

2

2

8

4.8 PSEUDOINVERSE When the determinant of an n X n matrix is zero, or even when the matrix is not square, there exists a way to obtain a solution "as close as possible" to the equation' Ax = y. In this section we let A be an m X n real matrix and first examine the properties of the real symmetric square matrices ATA and AAT, which are n X nand m x m respectively.

Theorem 4.18: The matrix BTB is nonnegative definite. Proof: Consider the quadratic form ~ = yTy, which is never negative. Let y where B can be m x n. Then xTBTBx:::"" 0, so that BTH is nonnegative definite.

= Bx,

Fromthistheorem, Theorem 4.6 and Definition 4.9, the eigenvalues of ATA are either zero or positive and real. This is also true of the eigenvalues of AAT, in which case we let B = AT in Theorem 4.18.

Theorem 4.19: Let A be an m x n matrix of rank r, where r ~ m and r ~ n; let gf be an orthonormal eigenvector of AT A; let fi be an orthonormal eigenvector of AAT, and let p~ be the nonzero eigenvalues of ATA. Then ~

(1)

pf

(2)

Ag i

(3)

Agi = 0

for i = r+1, .. . ,n

(4)

ATf.t

Pigi

for i = 1,2, .. . ,r

(5)

AT£.t

0

for i = r+1, .. . ,m

are also the nonzero eigenvalues of AAT. Pifi

for i = 1,2, ... , r

Proof: From Problem 3.12, there are exactly r nonzero eigenvalues of ATA and AAT. Then ATAg.-t = p~g, for i = 1,2, -. .. , rand ATAg;• = 0 for i = r + 1, ... , n. Define an t t m-vector hi as hi == Ag/Pi for i 1,2, ... , r. Then

=

AA'I'bi = AATAgJpi = A(pfg)/Pi = pfhi

Furthermore, hih; = gfATAg/ PiPj = Pjgfg/Pi = 8ij

Since for each i there is one normalized eigenvector, hi can be taken equal to fi and (2) is these must be the eigenvalues of AAT and proven. Furthermore, since there are r of the p?, t (1) is proven. Also, we can find fi for i = r+ 1, ... , m such that AATfi= 0 and are orthonormal. Hence the fi are an orthonormal b~sis for 'Vm and the gi are an orthonormal basis for CV n • To prove (3), ATAgi = 0 for i= r+l, ... ,m. Then IIAgill~ = gfATAgi = 0, so that Agi = O. Similarly, since AATfi = 0 for i == r+ 1, ... , m, then ATfi = 0 and (5) is proven. Finally, to prove (4), (ATAg)/p~

P1?g./p·1 ~

'-CHAP. 4]

MATRIX ANALYSIS

85

Example 4.19. Let

A

--

6 (6

0 4 0 1 0 6 ).

=

AAT

= r = 2,

Then m

(52 86

86) 73

n

= 4.

ATA

=

72 6 24 6 1 0 0 16 24 ( 36 6 0

36) 6

0 36

pi

The eigenvalue of AAT are = 100 and P~ = 25. The eigenvalues of ATA are 0, O. The eigenvectors of ATA and AAT are

p; = p! =

0.84

gl g2

=

0.24 -0.12 (-0.49

ga

g4 f1

f2

0.08

= = =

0.00

0.24

pi = 100, pi = 25,

O.48)T

0.64 -O.72)T 0;78

O.49)T

0.04 -0.98 -0.06 -0.13)T 0.6

0.8}T

0.8 -O.6)T

From the above, propositions (1), (2), (3), (4) and (5) of Theorem 4.19 can be verified directly. Computationally it is easiest to find the eigenvalues p2 and p2 and eigenvectors fl and f2 of AAT, and then obtain the gj from propositions (4) and (3). 1 2 l'

Theorem 4.20: Under the conditions of Theorem 4.19; A = ~

pJigT.

i=l

Proof; The m x' n matrix A is a mapping from CVn to CVm , Also, gl' g2' ... , gn and £1' £2' .•. ,fm form orthonormal bases for CV n and CV m respectively. For arbitrary x in G() n'X

n

=

~ ~,g.

i=l

1

1

where ~i = gfx

(4.22)

Use of properties (2) and (3) of Theorem 4.19 gives r

n

~ ~iAgi

=

Ax

i=l

+ i==r+l ~

~iAgi

r

From (4.22), gj = gix and so Ax = ~ PifigTx. theorem is proven. i=l

Since this holds for arbitrary x, the

Note that the representation of Theorem 4.20 holds even when A is rectangular, and has no spectral representation. Example 4.20. Using the A matrix and the results of Example 4.19, 6 (6

°

= 10 (0.6) ~0.84 0.08 0.24 0.48) + 5 ( 0.8) (0.24 -0.12 0.64 -0.72)

4 0) 1 0 6

-0.6

0.8

Definition 4.14: The pseudoinverse, denoted A -I, of the m x n real matrix A is the n x rn r

real matrix A -1 = ~ p:-lg j fJ. ~

i=l

Example 4.21. Again, for the A matrix of Example 4.19,

=

0.1

(

o,84) 0.08 (0.6 0.8) 0.24 0.48

+

0.2

(

0.24) -0.12 (0.8 -a.6) 0.64 -0.72

=

(-~~~~!! ~:~:~:) 0.1168 -0.0576 -0.0864 0.1248

86

MATRIX ANALYSIS

[CHAP. 4

Theorem 4.21: Given an m

X n real matrix A and an arbitrary m-vector y, consider the equation Ax = y. Define Xo = A - Iy . Then IIAx - Yl12 ~ IIAxo - YI12 and for those z # Xo such that jlAz - Yl12 : : : II Ax o- y112, then Ilzi 12 > Ilxo 112.

In other words, if no solution to Ax::::: y exists, Xo gives the closest possible solution. If the solution to Ax = y is not unique, Xo gives the solution with the minimum norm. P1~oof; Using the notation of Theorems 4.19 and 4.20, an arbitrary m-vector y and an arbitrary n-vector x can be written as m

n

y ::::: ~ 'lJifp

= i=l ~ ~igi

X

i=l

where

rh:::::

fry and ~i

= gJx.

Then use of properties (2) and (3) of Theorem 4.19 gives

n

Ax - y

=

(4.23)

m

r

~ ~iAgi

~ (~iPi - 'I])fi

i=l

i=l

+

~ 'l]ifi

(4.24)

i=r+l

Since the fi are orthonormal, r

m

~ (~iPi -

IIAx-YII~

i=l

+

7]i)2

~

'1];

i= r+l

To minimize IIAx - Yl12 the best we can do is choose gi = 7]/Pi for i::::: 1,2, ... , r. those vectors z in G()n that minimize IIAx - ylk can be expressed as

Then

r

z

:::::

~

i=l

where

~i

for i

= r+l, ... , n

giTJ/Pi

is arbitrary.

+

But

The z with minimum norm must have ~i::::: 0 for i::::: r + 1, ... , n. Then using T

(4.23) gives z with a minimum norm ::::: ~

i=l

Example 4.22. Solve the equations

6x!

+ 4X3 =

1

7]i :::::

fT y

from

T

g = ~ p:-lgif[y = A - I y ::::: Xo.

p:-l7]i i 1

i=l

1

6xl + x 2 + 6X4 = 10.

and

This can be written Ax = y, where y = (1 10)T and A is the matrix of Example 4.19. Since the rank of A is 2 and these are four columns, the solution is not unique. The solution Xo with the minimum norm is 0.4728) 0.1936 -0.4592 ( 1.1616

= y can be expressed as (I - A -IA)z, where z is any arbitrary n-vector.

Theorem 4.22: If it exists, any solution to Ax

~i

= A -Iy +

Proof; For a solution to exist, 'lJi = 0 for i = r + 1, ... , m in equation (4.24), and for i = 1,2, ... , r. Then any solution x can be written as

::::: 7]/Pi

r

X

where ti for i::::: r

+ 1, ... , n

n

Z

x

= "" .tt::..; g.'., i=l

I].

where f:. - I

=

= g!'z. ~

T

=

~ gi"fJ/Pi

i=l

n

+ i=r+l ~

are arbitrary scalars.

(4.25)

gi'i

Denote in

G()n

an arbitrary vector

Note from Definition 4.14 and Theorem 4.20,

T

~ ~

~.tt::..; k=l i=l

p-lg.f!'f gTp i t1kkk

(4.26)

CHAP. 4]

MATRIX ANALYSIS

87 n

Furthermore, since the gi are orthonormal basis vectors for 'V n , 1 = n

~

(1- A -IA)z

~ gig;.

Hence

i=l

7l.

gigTz

~ gi~i

=

(4.27)

i=r+l

i=r+1

From equation (4.23), TJi = fry so that substitution of (4.27) into (4.25) and use of Definition 4.14 for the pseudo inverse gives x = A-Iy + (I-A-1A)z Some further properties of the pseudoin verse are: 1.

If A is nonsingular, A - I = A -1.

2.

A -IA =1= AA -1 in general.

3.

AA-IA=A

5.

(AA -I)T = AA-I

6.

(A-IA)T=A-IA

7.

A -lAw = w for all w in the range space of AT.

8.

A

9.

A -1(y + z) = A -1y + A - IZ space of AT.

-IX

=0

for all x in the null space of AT, for all y in the range space of A and all z in the null

10.

Properties 3-6 and also properties 7-9 completely define a unique A - I and are sometimes used as definitions of A -I.

11.

Given a diagonal matrix A = diag (AI, A2, •.. , An) where some Ai may be zero. Then A -I = diag (A;\ .\.2\ ... , .\.;;1) where 0- 1 is taken to be O.

12.

Given a Hermitian matrix H Ht. Let H UAUt where ut = V-I. H-I = UA -Jut where A - I can be found from 11.

13.

Given an arbitrary m x n matrix A. Let- H where H- 1 can be computed from 12.

14.

15.

=A (AT)-I = (A-1)T

16.

The rank of A, AtA, AAt, A -I, A -1A and AA -1 equals tr (AA -1)Q = r.

17.

If A is square, there exists a unique polar decomposition A = UH, where H2 = At A, U = AA -I, ut = U-I. If and only if A is nonsingular, H is positive definite real symmetric and U is nonsingular. When A is singular, H becomes nonnegative definite and U becomes singular.

18.

If A(t) is a general matrix of continuous time functions, A -1(t) may have discontinuities in the time functions of its elements.

=

=

= AtA.

Then A-I

Then

= H- At = (AH-I)t 1

(A -1)-1

Proofs of these properties are left to the solved and supplementary problems.

88

MATRIX ANALYSIS

[CHAP. 4


Show that all similar matrices have the same determinants and traces. To show this, we show that the determinant of a matrix equals the product of its eigenvalues and that the trace of a matrix equals the sum of its eigenvalues, and then use Theorem 4.2. Factoring the characteristic polynomial gives det (A - AI) =

Setting A = 0 gives det A det (A - AI)

=

(AI - },,)(A2 - A)' .. (An - A)

=

AIA2" 'A n

+ ... + (Al + A2 + ... + An )(-A)n-l + (-A)n

Furthermore,

Al A2' .. An.

(al - Aed /\ (a2 -l\e2) /\ ... /\ (an - Ae n )

+ (-A) [el /\ a2 /\ ... /\ an. + al /\ ez /\ ... /\ an + ... + a1/\ a 2 /\ .•. + '" + (- A)n-l [al /\ e2 /\ ... /\ en + el /\ a2 /\ ... /\ en + ... + el /\ e2 /\ ... /\ an] + (-A) ne l/\ e2 /\ ... /\ en

al /\ a2 /\ ... /\ an

/\ en]

Comparing coefficients of -A again gives AIA2'" An = a 1 /\ a2/\ ..• /\ an' and also Al

+ A2 + ... + An

= al /\ ez /\

.•. /\ en

+

el /\ 32

/\ ••• /\

en

+ '" + el /\ e2 /\

... /\ an

However, al /\ e2 /\ ... /\ en :;:: (anel

and similarly e 1 /\

32/\ ••• /\

Al

4.2.

+ a21e 2 + ... + an1en) /\ ~2 /\

en =

etc.

a22'

+ A2 + ... + An

••• /\ en

Therefore

= all

+ a22 + ... + ann

tr A

Reduce the matrix A to Jordan form, where (a)

A

(! =: =~)

(c)

( -~o

(d)

3 -4

(b)

(a)

A

Calculation of det (8 : X vector

(

Xi

0-1

1

~ =~)

( -~o -~

A

0-1

1-3

-~~

~)

=

A 1 =!X)

0 gives Al = 1, 1.2 = 2, A, = 3.

The eigen-

is solved for from the equations

8-Ai -8 -2 ) 4 3

( ~~o -~ -~)

A

-3 - Ai -4

-2 1 - Ai

Xi

(0)

=

0 0

giving

Xl

where the third element has been normalized to one in

= X2

G)' G)' G) x,

and

Xs

=

and to two in

X3

Xl'

-2)( D G DG D G -8 -3 -2 -4 1

(b) Calculation of

det

(-l-A 0

0

3 2

3 2 1

1

=

3

0

2

2 0

1

-1

)

1-A

-4

1

-3 - A

=:

0

gives Al = A2 = AS = -l.

=

Then

MATRIX ANALYSIS

CHAP.4J

89

Solution of the eigenvalue problem (A - (-l)l)x = 0 gives

G~ =DGJ G) =

and only one vector, (a 0 O)T where a is arbitrary. one Jordan block Lll(-l), so

Therefore it can be concluded there is only

C~ -D 1

J

==

Lu(-I)

Solving,

1-1)

G

2 -4

-1 0

G)

::::

tl

1 -2

gives

tl

==

2 -4

xl

_Finally, from

(f3 2a a)T where f3 is arbitrary.

1-1)

==

:::::

t2

1 -2 we find t2 == (y 2f3 - a {3 - a)T where y is arbitrary. respectively gives a nonsingular (Xl I tl l,t2 ), so that

(~

o

(c)

Choosing a, {3 and y to be 1,0 and 0

~ -~)(-~ ~ =~)(~ ~ -~)

1 -2

0

1 -3

0

(-~ -: ~)

==

1 -1

0

0 -1.

=

Since A is triangular, it exhibits its eigenvalues on the diagonal, so Al == A2 = A3 -l. The solution of iicAx = (-1)x is x = (a P 2f3)T, so there are two linearly independent eigenvectors a(1 0 O)T and p(O 1 2)T. Therefore there are two Jordan blocks L 1(-I) and L 2 (-1). These can form two different Jordan matrices J 1 or J 2 : (

0) (-1~ -~1_~0)

L (-1) I 0 L 2 (-1)

=

It makes no difference whether we choose J 1 or J 2 because we merely reorder the eigenvectors and generalized eigenvector in the T matrix, i.e., A(X21 Xl I t l ) = (X21 Xl I t 1)J2

How do we choose a and f3 to get the correct

Xl

to solve for

2o -1) 0

o

tl?

From (4.19)

tl

0

from which f3 = 0 80 that Xl = (a 0 O)T and tl = (y 8 28 - a)T where y and {} are also arbitrary. From Problem 4.41 we can always take a = 1 and y, 8, etc. = 0, but in general

C~ -~ ~DG 20~a 2;)

G28~" 2;)C~ -~ -D

Any choice of a,/3, y and 0 such that the inverse of the T matrix exists will give a similarity transformation to a Jordan form. (d) The A matrix is already in Jordan form.

A

==

-I and T-l(-I}T

==

-I.

Any nonsingular matrix T will transform it, since This can also be seen from the eigenvalue problem

(A - AI)x

=

(-I - (-1)I)x = Ox = 0

so that any 3-vector X is an eigenvector. The space of eigenvectors belonging to -1 is three dimensional, so there are three Jordan blocks LU(A) = -1, L 1Z (A) = -1 and LlS(A) = -1 on the diagonal for A == -1.

90 4.3.

[CHAP. 4

MATRIX ANALYSIS

Show that a general normal matrix N (Le. NNt = NtN), not necessarily with distinct eigenvalues, can be diagonalized by a similarity transformation U such that ut = U-l. The proof is by induction. First, it is true for a 1 X 1 matrix, because it is already a diagonal matrix and U = I. Now assume it is true for a k -1 X k -1 matrix and prove it is true for a t l k Xk matrix; i.e. assume that for U k-l -- Uk-l

Let

Form T with the first column vector equal to the eigenvector Xl belonging to AI' an eigenvalue of Nk;. Then form k - 1 other orthonormal vectors x2' X3,' •• , Xk from 'V k using the Gram~Schmit process, and make T = (xl I x21 ... I Xk)' Note TtT = I. Then, AIXll

( ;; }Xl IX, I ... I

xkl (

n1 x2

nr Xk)

ntx2

ntxk

~~~2.'... ~~~~ ........ ~.;.~k. AIXkl

and

where the

aij

are some numbers.

But TtNkT is normal, because

Therefore

AIo (

a12 a22 ill

"

•••

oil

o

ak2

..• ............

•.•

a1k) a2k oil

..

oil

akk

(A'at2t ~

....

:

aIk

0

a;z

..............

a;k

0)

a~k

'"

•••

"

•

..

==

•

a~k

(Aiaf2 •

"

0

a;2

....

II

....

a~k a;k

••. "

......

•••

0a~2 ) (At at2 0

ill

..

..

akk

•

0

..

"

...

ill

ak2

a

lk

a12

•.• ••

"

..

•.•

a2k "

..

"

..

)

.

akk

Equating the first element of the matrix product on the left with the first element of the product on the right gives

Therefore a12, ala, ..• , alk must all be zero so that

where

and Ak~l is normal. Since A k - l is k -1 X k - 1 and normal, by the inductive hypothesis there t t exists a U k-l = U-l k-l such that Uk - 1 Ak - l Uk - 1 = D, where D is a diagonal matrix.

MATRIX ANALYSIS

CHAP. 4]

91

Then

Define Sk such that

st Sk = I

and

Therefore the matrix TS k diagonalizes N k , and by Theorem 4.2, D has the other eigenvalues of N k on the diagonaJ.

A2, A3, ••• , Ak

Finally, to show TS k is unitary, I = StSk = StISk

4.4.

Prove that two n x n Hermitian matrices A = At and B = Bt can be simultaneously diagonalized by an orthonormal matrix U (i.e. utu = I) if and only if AB = BA. If A = UAut and B = UDUt, then AB = UAUtUDUt

==

uAnut =UDAUt = UDUtuAut = BA

Therefore aU matrices that can be simultaneously diagonalized by an orthonormal U commute. To show the converse, start with AB = BA. Assume A has distinct eigenvalues. Then Axi = AiXi, so that AB~ = BAxi = AiB~. Hence if Xi is an eigenvector of A, so is B~. For distinct eigenvalues, the eivenvectors are proportional, so that BXi = Pi~ where Pi is a constant of proportionality. But then Pi is also an eigenvalue of H, and ~ is an eigenvector of B. By normal~ izing the ~ so that Xi tXi = 1, U = (Xl I ... Ixn) simultaneously diagonalizes A and B.

If neither A nor B have distinct eigenvalues, the proof is slightly more complicated. Let Abe an eigenvalue of A having multiplicity m. For nondistinct eigenvalues, all eigenvectors of A belong in the m dimensional null space of A - A-I spanned by orthonormal xi' X2'" •• , "m' Therem

fore B~ = .~ cijXj' t=l

where the constants cij can be determined by Cij =

"T

BXi'

Then for

C = {cij} and X = (Xl Ix21 ... Ix n ), C = XtBX = XtBtX = ct, so C is an m X m Hermitian matrix. Then C = UmDmU! where Dm and Um' are m X m diagonal and unitary matrices respectively, Now A(XUm ) = ;\(XUm ) since linear combinations of eigenvectors are still eigenvectors, and Dm = Ut,XtBXUm . Therefore the set of m column vectors of XU m together with all other normalized eigenvectors of A can diagonalize both A and B. Finally, (XUm)t(XUm ) = (UlxtxUm ) = (UkImUm ) = 17/l> so that the column vectors XUm are orthonormal.

4.5.

Show the positive definite Hermitian square root R, such that R2 = Q, is unique. Since R is Hermitian, UA i ut = R where U is orthonormal. Also, R2 and R commute, so that both Rand Q can be simultaneously reduced to diagonal form by Problem 4.4, and Q = UDUt. Therefore D = .Ai. Suppose another matrix 8 2 = Q such that S = V A2 vt. By similar reasoning, D = Since a number> 0 has a unique positive square root, .A2 =.A i and V and U are matrices of orthonormal eigenvectors. The normalized eigenvectors corresponding to distinct eigenvalues are unique. For any nondistinct eigenvalue with orthonormal eigenvectors Xl' x2" .. , x n ,

Ai.

AXXt

92

MATRIX ANALYSIS

[CHAP. 4

and for any other linear combination of orthonormal eigenvectors, Yl' Y2' ... ; Ym ,

1 where Tt = Tm m'

Then

Hence Rand S are equal even though U and V may differ slightly when Q has nondistinct eigenvalues.

4.6.

Prove Sylvester's theorem: A Hermitian matrix Q is positive definite if and only if all principal minors det Qm > O. If Q is positive definite, ~ == (x, Qx) ==== 0 elements of x. For those XO whose last m Therefore Qm is positive definite, and all its determinant of any matrix equals the product If det'lm

> 0 for

for any x. Let Xm be the vector of the first m - n elements are zero, (x m• Qmxm) = (xO, QXO) ==== O. eigenvalues are positive. From Problem 4.1, the of its eigenvalues, so det'!m > O.

m = 1,2, ... , n, we proceed by induction.

Assume now that if det Q 1 > 0, ... , det Qn-l possess an inverse. Partition Qn as

0, then

Qn-l

For n = 1, det Q = Al

is positive definite and must

+-q-nn---q-:~----\q-)(: I Q;:l

: )(_Q_:-...,...-1

=

> O.

q )

(4.28)

We are also given det Qn > O. Then use of Problem 3.5 gives det'tn = (qnn - qtQ;~lq) detQn-l' so that qnn - qtQ;~lq > O. Hence

> for any vector (xt-l 1 x~).

Then for any vectors y defined by

substitution into (x, Qx) and use of (4.28) will give (y, QnY)

4.7.

Show that if

[IAII < 1,

then (I - A)-l =

Sk =

~ An

Ti=O

O.

iAn.

n=O

k

Let

>

eo

and

S =

~ An. n=O 11

Then

~ Akll n=k+l

00

:::

~ 11

=k+l

11Allk

o

CHAP. 4]

MATRIX ANALYSIS

93

by properties (3) and (4) of Theorem 4.11. Using property (6) and IIAII < 1 gives S = lim Sk' k--+«! Note JIAk-t 1 11 ~ IIAll k + 1 so that lim Ak+l = O. Since (I-A)Sk == !-Ak+l, taking limits as k_eo

k - co gives {I - A)S == I. Since S exists, it is (1 - A)-I. because lI(I - A)xlj :=: (1-IIAli)llxll :=: Ilxll·

4.8.

Find the spectral representation of

This is called a contraction mapping,

1 1 2)

(o

A

-1

1 -.2 0 -1

.

3

The spectral representation of

A

~ Ai~rit.

==

i=1

=

11.2 ::::: 1 - j and A3 1 + j, and eigenvectors The reciprocal basis ri can be found as

1

1

( -0.5 0

0

Then

4.9.

(1)

(-~.5 )(0 0

-2)

+

(1-

= (1

( :::::

j)(f)

AI::::: 1,

1 -0.5)T, x2::::: (j 1 O)T and xs::::: (-i 1 O)T.

j _j)-1

1 1

A

Xl

The matrix A has eigenvalues

(-0.5j

0.5

0.5

0

-i i

1- j)

0 1

-4 ) 2 -2i

1

2+2j

+

(1

+

j)(-f)

(0.5j

0.5

1 + j)

Show that the relations AA -lA = A, A -lAA- 1 = A -I, (AA -1)T = AA - I and (A -IA)T = A -IA define a unique matrix A -I, that can also be expressed as in Definition 4.15. r

Represent

r

~ Pifig[

A

~ p-lgkf~.

A-I =

and

k=1

i=1

Then

k

AA-1A r

r

r

~ ~

~ PiP~lpkfig'[ gjfJfkg~

i=l j=1 k=1

,

r

AA -IA

~ Pifig'[

=

A

i=l

Similarly A -IAA - I

== A -I, and from equation

(4.~6),

r

A-IA

=

~ gigf

(A-IA)T

==

i=l

r

Similarly

(AA -1)T

== (~ fifT) T == AA -I. i=1

To show uniqueness, assume two solutions X and Y satisfy the four relations.

Then

1. AXA==A

3. (AX)T

== AX

5. AYA==A

7. (Ay)T == AY

2. XAX=X

4. (XA)T == XA

6. ¥A¥=Y

8. (¥A)T = YA

and transposing 1 and 5 gives

9. ATXTAT = AT

10. A1'yTAT = AT

MATRIX ANALYSIS

94

[CHAP. 4

The following chain of equalities can be established by using the equation number above the equals sign as the justification for that step. X ~ XAX ~ ATXTX ~ ATyTATXTX ~ ATyTXAX ~ AT¥TX ~ YAX

~ YAYAX ~ YAyxTAT

yyTATXTAT ~ yyTAT

b

b

YAY ~ Y

Therefore the four relations given form a definition for the pseudoinverse that is equivalent to Definition 4.15.

4.10. The outcome of y of a certain experiment is thought to depend linearly upon a parameter x, such that y= aX + (3. The experiment is repeated three times, during which x assumes values Xl = 1, X2 = -1 and X3 = 0, and the corresponding outcomes Y1 = 2, and Y2 == -2 and Y3 = 3. If the linear relation is true,

+ (3 a(-l) + P

2

a(l)

-2 3

a(O)

+ f3

However, experimental uncertainties are such that the relations are not quite satisfied in each case, so a and f3 are to be chosen such that 3

L

(Yi -

i=l

f3)2

aXi -

is minimum. Explain why the pseudoinverse can be used to select the best and then calculate the best a and f3 using the pseudoinverse.

a

and f3,

is minimized.

Since

g2 = (0 1).

Since

The equations can be written in the form y = Ax as

3

Defining Xo = A-I y , by Theorem 4.21, ATA =

(~

:),

then

p,

=

V2

f i = Ag/Pi' then f1 = (1 -1 0)/"/2 Definition 4.15 to be

and

Ily - Axolli = P2 =

va.

~ (Yi - aOxi - f30)2

an;' g, = (1 0)

£2 = (1 1 1)/V3.

and

and

Now A-I can be calculated from

0)

t -i (i 1 ! so that the best ao

=2

and f30

= 1. n

Note this procedure can be applied to ~ (Yi - axf - (3xi - y)2, etc. i=1

4.11.

Show that if an n X n matrix C is nonsinguIar, then a matrix B exists such that C == eB" or B = In C. Reduce C to Jordan form, so that C = TJT-I. Then B = In C = Tin JT-1, problem is to find In L(A) where L(X) is an l x l Jordan block, because

so that the

CHAP. 4]

MATRIX ANALYSIS

95

U sing the Maclaurin series for the logarithm of L(A.) - AI,

=

In L(A.)

00

In [A.J + L(A.) - A.I]

I In A. -

~ (-iA.) - i [L(A.) - A.I]i

i=l

Note all the eigenvalues ~ of L(A.) - A.I are zero, so that the characteristic equation is by the Cayley-Hamilton theorem, [L(A.) - A.I]l = 0, so that

~l

= O.

Then

[-1

In L(A.)

=

~ (-iA.)-i [L(A.) - A.I]i

I In A. -

i=l

Since A. ¥= 0 because C-l exists, In L(A.) exists and can be calculated, so that In J and hence In C can be found. Note B may be complex, because in the 1 X 1 case, In (-1) = hr. Also, in the converse case where B is given, C is always nonsingular for arbitrary B because C-l = e- B •


Why does at least one nonzero eigenvector belong to each distinct eigenvalue?

4.13.

Find the eigenvalues and eigenvectors of A where

(~

A

~ -~).

-2

4.14. 4.15.

Suppose all the eigenvalues of A are zero.

0-1

Can we conclude that A = O?

Prove by induction that the generalized eigenvector tt of equation (4.9) lies in the null space of (A - AiI)l+l.

4.16.

Let x be an eigenvector of both A and B.

Is x also an eigenvector of (A - 2B)?

4.17.

Let Xl and x2 be eigenvectors of a matrix A corresponding to the eigenvalues of A.! and A.2' where A.l # A.2' Show that ax! + j3x2 is not an eigenvector of A if a # 0 and j3 #- O.

4.18.

U sing a similarity transformation to a diagonal matrix, solve the set of difference equations

(~)

where

utu =

4.19.

Show that all the eigenvalues of the unitary matrix U, where of one.

4.20.

Find the unitary matrix U and the diagonal matrix .A. such that

utC{ ~ -t)u

=

A

Check your work.

4.21.

Reduce to the matrix

A

1~ 3/~5 -4~/5) (

to Jordan form.

I,

have an absolute value

96

MATRIX ANALYSIS

[CHAP. 4

4.22.

Given a 3 X 3 real matrix P oft 0 such that

4.23.

Given the matrix

• Find the transformation matrix T that reduces A 101 to Jordan form, and identify the Jordan blocks Lij(}\i)'

4.24.

Find the eigenvalues and eigenvectors of

4.25.

Find the square root of

4.26.

Given the quadratic form ~

4.27.

Show that ~ A n/n! converges for any A. n=O

4.28.

Show that the coefficient an of I in the Cayley-Hamilton theorem An + alAn-l + is zero if and only if A is singular.

4.29.

Does

4.30.

Let the matrix A have distinct eigenvalues A1 and A2'

4.31.

Given a real vector

A

:;:::

(-

p2:;:::

O.

Find its Jordan form.

~ ~ ~)

A -_ (1 +od2

What happens as e ~ O?

(! :). ==

+ 2~1~2 + 4~~ + 4~2~3 + 2~~.

~i

Is it positive definite?

I;Q

... + anI:;::: 0

A2f(A):;::: [f(A)]A2?

x:::::: (Xl

X2

•••

grad x

0::

and a scalar

xn)T :;:::

Does A3(A - A1I){A - A2I) a.

== O?

Define the vector

(aa/aX1 aa/fJX2 .•• fJa/fJxn)T

Show grad x xTQx = 2Qx if Q is symmetric, and evaluate grad x x TAx for a nonsymmetric A. n t Show that I:;::: ~ Xjri'

4.32.

i=l

4.33.

Suppose A2:;::: O.

4.34.

Find eat, where A is the matrix of Problem 4.2(a).-

4.35.

Find eAt for

4.36.

Find the pseudoinverse of

4.37.

Find the pseudoinverse of A -- (13

4.38.

Prove that the listed properties 1-18 of the pseudoinverse are true.

4.39.

Given an m

X

Can A be nonsingular?

A == (

0'

-w

"'). 0'

(4-03). 0

n real matrix A and scalars Agi

Show that only n 4.40.

26

== 1 gives fif j

:;:::

aij

such that

Of

:;::: O~fi

Afi =

if gTgj =

aij'

Given a real m X n matrix A. Starting with the eigenvalues and eigenvectors of the real symmetric (n + m) X (n + m) matrix ( : \

!T),

derive the conclusions of Theorems 4.19 and 4.20.

CHAP. 4]

4.41.

MATRIX ANALYSIS

97

Show that the T matrix of J = T-IAT is arbitrary to within n constants if A is an n X n matrix in which each Jordan block has distinct eigenvalues. Specifically, show T = ToK where To is fixed and Kl

=

K

0

.0."

(

...

0)

.~~ .. :::....~.

° ° ".

Km

where K j is an l X l matrix corresponding to the jth Jordan block L j of the form

~1

(: : :' ::: ::~:

:: :: ... :;_1)

... .... ...

where

O::i

...

is an arbitrary constant.

(!-I) .

4.42.

Show that for any partitioned matrix (A I 0),

4.43.

Show that

4.44.

Show that another definition of the pseudoinverse is A-I = lim (ATA + d')-lAT, where P is any positive definite symmetric matrix that commutes with ATA. €--+O

4.45.

Show

Ixt Axl

IIA -III

===

(A 10)-1

IIAI1211xlii .

= 0 if and only if A = O.


Because det (A - Ail) := 0, the column vectors of A - Ail are linearly dependent, gIvmg a null space of at least one dimension. The eigenvector of Ai lies in the null space of A - Ail.

4.13.

A

4.14.

No

4.16.

Yes

4.18.

Xl(n) = 2 and x2(n) = 1 for all n.

4.20.

A

= 3,3, -3,

Xs

0

u

1

0

+ (3(0

= a(-2 0 1)

~}

G ~.8). GD

~

U

0

4.21.

T

J

0.6

-0.8

;:::::

0.6

1 0);

x-s::::: y(1 0 1)

COl) V2 0 1

0 0-1 1 1 0

G D

1

4.22.

J =

0 0

0

or

J

=

G

0

0

0

98

4.23.

MATRIX ANALYSIS

T

G: T: 0)

=

for

a,T, e arbitrary, and

J

=

G

~),

1 1

o

4.24.

Al = 1 + el2, A2 = 1- e/2, xl = (1 0), x2 = (1 e); as tions on the eigenvalues break the multiplicity.

4.25.

2 (1 21)

4.26.

Yes

4.27.

Use matrix norms.

4.28.

If A-I exists,

A-I

[CHAP. 4

= a;l[An-1 + a 1An-2 + ... ].

Ii

~ 0,

If an

At

=

= AIA2"

value is zero and A is singular. 4.29.

Yes

4.30.

Yes

4.31.

(AT + A)x

4.33.

No

4.34.

-4e t + 6e 2t - est Set - ge 2t + eSt ( -4e t + 8e 2t + eat

eAt

= 0,

Slight perturba-

then at least one eigen-

eSt)

-e t + 2e2t 2et - 3e2t + eSt -et + e2t + eSt

wt)

1(4

25

:-3

=

00)

5~ GD

4.37.

A-I

4.40.

There are r nonzero positive eigenvalues

Pi

to the eigenvalues Pi are the eigenvectors

(! I~T) 4.48.

.An

Xl = X20

sin cos CJJt

4.35.

4.36.

-3e t + 4e2t - eSt 6e t - 6e2t + eat -8e t + 2e2t + eSt

A2'

all one big Jordan block.

IxtAxI

===

then gives the desired result.

and r nonzero negative eigenvalues

(g~) f,

IIxl1211Axl12 by Schwartz' inequality.

and to -Pi are (_g~). f,

-Pi'

Corresponding

Spectral representation of

Chapter 5 Solutions to the Linear State Equation 5.1 TRANSITION MATRIX From Section 1.3, a solution to a nonlinear state equation with an input u(t) and an initial condition xo can be written in terms of its trajectory in state space as x(t) = t/J(t; U(T), Xo, to). Since the state of a zero-input system does not depend on u(,), it can be written x(t) = +(t; Xo, to). Furthermore, if the system is linear, then it is linear in the initial condition so that from Theorem 3.20 we obtain the Definition 5.1:

The transition matrix, denoted x(t) = ,pet; Xo, to) = (t(t, to)xo.

~(t,

to), is the n x n matrix such that

This is true for any to, i.e. x(t) = cI»(t, T) XCi) for 7' > t as well as ,=== t. Substitution of = 4t(t, to)xo for arbitrary Xo in the zero-input linear state equation dx/dt = A(t)x gives the matrix equation for (t(t, to),

x(t)

Bq,(t, to)/Bt = A(t) ",(t, to)

Since for any Xo, Xo

(5.1)

= x(t o) = (t(to, to)xo,

the initial condition on q,(t, to) is (5.2) q,(to, to) = I Notice that if the transition matrix can be found, we have the solution to a time-varying linear differential equation. Also, analogous to the continuous time case, the discrete time transition matrix obeys cJt(k+l,m)

A(k)4»(k,m)

(5.3)

(5.4)

q,(m,m) = I

with so that x(k)

=

= ~(k, m) x(m).

Theorem 5.1:

Properties of the continuous time transition matrix for a linear, timevarying system are (1) transition property 4t(t2, to) = 4t(t2, t 1) CP(tl, to)

(5.5)

(2) inversion property (5.6)

(3) separation property CP(tl, to) :::: B(tl) 8- l (to)

(5.7)

(4) determinant property det cI»(tl, to)

(5.8)

= e'f1

to [tr A(T)] d-r

and properties of the discrete time transition matrix are (5) transition property cp(k, m) :::: 4t(k, l) IP(l, m) 99

(5.9)

100

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

(6) inversion property (5.10)

(7) separation property 4l(m, k) = 8(m)9- 1 (k)

(5.11)

(8) determinant property

det4l(k,m) = [detA(k-I)][detA(k-2)]···[detA(m)]

for k>m

(5.12)

In the continuous time case, cp-l(t, to) always exists. However, in rather unusual circumstances, A( k) may be singular for some k, so there is no guarantee. that the inverses in equations (5.10) and (5.11) exist. Proof of Theorem 5.1: Because we have a linear zero-input dynamical system, the transition relations (1.6) and (1.7) become lfJ(t, to) x(to) = 4l(t, tl) X(tl) and X(tl) = q;(tl, to) x(to). Combining these relations gives 4t(t, to) x(to) = cp(t, t 1 ) 4t(tl, to) x(to). Since x(to) is an arbitrary initial condition, the transition property is proven. Setting t2 = to in equation (5.5) and using (5.2) gives cIJ(to, t 1) cIJ(tl, to) = I, so that if det cp(to, t 1 ) #- 0 the inversion property is proven. Furthermore let 8(t) = cp(t, 0) and set tl = in equation (5.5) so that cIJ(tz, to) = 8(t2) 4t(O, to). Use of (5.6) gives cp(O,to) = cp-l(to, 0) = g-l(tO) so that the separation property is proven.

°

To prove the determinant property, partition det cIJ

into its row vectors 4»1' 4»21 •.. , +n' Then

rp

= +1 /\ +2 /\ ... /\ +n

and d"t. / dt /\ '1"2 A. /\ ""1

d(det cp)/dt

• • • /\..1.. "t"n

+

..I.. /\

d..l.. / dt /\ ...

"t"I"1'2

/\..1.. 't'n

+ ' .. + +1/\ +2 /\ ... /\d+Jdt

(5.13)

From the differential equation (5.1) for 4l, the row vectors are related by n

d+/dt

~ aik(t) +k

=

for i = 1,2, .. . ,n

k=l

Because this is a linear, time-varying dynamical system, each element aik(t) is continuous and single-valued, so that this uniquely represents d+/dt for each t.

+1/\ ... /\ d+/dt /\ ... /\ +n

n

+1/\ ... /\ L

k=l

+

aik

k /\ ••• /\

'Pn

=

aii

+

1 /\'"

/\

+i /\ ... /\ 9>n

Then from equation (5.13), d( det q,)/ dt

==:

an

+

1 /\(1)2 /\ ••• /\

+n +

a22+1 /\ +2 /\

•.• /\

+n + ... + ann+!

/\+2/\ ••• /\

+n

[tr A(t)] det cp Separating variables gives d(det cIJ)/det cp

= tr A(t) dt

Integrating and taking antilogarithms results in det cp(t, to)

=

ye

ftt

[tr A(T)] dr

0

where y is the constant of integration. Setting t = to gives det 4t(to, to) = det I == 1 = y, so that the determinant property is proven. Since efW = 0 if and only if j(t) = -00, the inverse of q,(t, to) always exists because the elements of A(t) are bounded. The proof of the properties for the discrete time -transition matrix is quite similar, and the reader is referred to the supplementary problems.

CHAP. 5]

TH~

SOLUTIONS TO

101

LINEAR STATE EQUATION

5.2 CALCULATION OF THE TRANSITION MATRIX FOR TIME ..INV ARIANT SYSTEMS

Theorem 5.2:

The transition matrix for a time-invariant linear differential system is

=

7l', the rows of .;.(to, t) B(t) are linearly independent time functions, and from Problem 6.30, page 145, the system is completely controllable for it - to> 7T. The system is not totally controllable because for every to. if t2 - to < 11, either 11(T) or 12(T) is zero for to::: T::: i 2· If t - to

However, for systems with analytic A(t) and B(t), it can be shown that complete controllability implies total controllability. Therefore rank Q = n is necessary and sufficient for complete controllability also. (Note fl(t) and f2(t) are not analytic in Example 6.4.) For complete controllability in a nonanalytic system with rank Q(t) < n, the rank of q,(t, 1') B(1') must be_ found.

6.6 DUALITY In this chapter we have repeatedly used the same kind of proof for observabiIity as was used for controllability. Kalman first remarked on this duality between observing a dynamic system and controlling it. He notes the determinant of the W matrix of Problem 6.8, page 142, is analogous to Shannon's definition of information. The dual of the optimal control problem of Chapter 10 is the Kalman filter. This duality is manifested by the following two systems:

CHAP. 6]

CONTROLLABILITY AND OBSERVABILITY

System #1:

dx/dt y

System #2:-

dwldt z

= =

139

A(t)x + B(t)u C(t)x

+ D(t)u

= -At(t)w + Ct(t)V

= Bt(t)w + Dt(t)v

Then system #1 is totally controllable (observable) if and only if system #2 is totally observable (controllable), which can be shown immediately from Theorems 6.10 and 6.11.


Given the system dx

dt

=

1 o 2-1) + (0) (

0 u, 1

lOx 1 -4 3

y

(1 -1 l)x

find the controllable and uncontrollable states and then find the observable and unobservable states. Following the standard procedure for transforming a matrix to Jordan form gives A = TJT-l as

G-~ -D

C~ ~

=

Then f == T- 1b == (0 1 O)T and gT shown in Fig. 6-8.

D( ~ : DC~ -~ 0

== cTT = (0 1 1).

The flow diagram of the Jordan system is Zl

1

o

+

8-2

+ Z2 U

o---+---+{ 1

+

1

Y

8-2

+

o

za

1

Fig. 6-8

z2

The element zl of the state vector z is controllable (through Z2) but unobservable. The element is both controllable and observable. The element za is uncontrollable but observable. Note Theorem 6.1 is inapplicable.

140 6.2.


[CHAP. 6

Find the elements of the state vector z that are both controllable and observable for the system of Problem 6.1. Taking the Laplace transform of the system with zero initial conditions gives the transfer function. Using the same symbols for original and transformed variables, we have SXl

Xl

+ 2X2

~ Xs

(6.13) (6.14)

SX2

(6.15) (6.16)

From (6.15), Xl = 4x2 + (s - 3)xs - u. Putting this in (6.13) gives (48 - 6)x2 = Substituting this in (6.14), (s - 1)(8 - 2)2XS = (8 -1)2U. Then from (6.16),

-- [(s(8 -_1)2 _ 0 + ~Ju 2)3 2)2

y

(s -

-

(8-1)(8-2) (8 -1)(8 - 2)2

(8 -

l)u -

(8 -

2)2 XS •

u

Thus the transfer function k(s) = (8 - 2)-1, and from Theorem 6.7 the only observable and controllable element of the state vector is z2 as defined in Problem 6.1.

6.3.

Given the system of Problem 6.1. Is it totally observable and totally controllable? Forming the P and Q matrices of Theorems 6.9 and 6.8 gives

P

Q

Then rank P = 2, the dimension of the observable state space; and rank Q = 2, the dimension of the controllable state space. Hence the system is neither controllable nor observable.

6.4.

Given the time-invariant system

and that u(t) = e- t and y(t) What happens when = O?

=2 -

(de- t • Find X1(t) and X2(t). Find X1(O) and X2(O).

Q'

Since y = xl' then XI(t) = == -e- t + te-t. Then xl(D) because dXl/dt = o. There is no Theorem 6.5. (For a = 0, the

x2(t)

6.5.

2 - cde-t. Also, dX1/dt = aX2, so differentiating the output gives = 2 and x 2(O) = -1. When a = 0, this procedure does not work way to find X2(t), because x2 is unobservable as can be verified from system is in Jordan form.)

The normalized equations of vertical motion y(r, 0, t) for a circular drumhead being struck by a force u(t) at a point r = ro, 0 = 00 are iJ2y

V 2y

at2

+

27ir 8(r - ro) 8(0 - 00) u(t)

where y(rl, 0, t) = 0 at r = rl, the edge of the drum. modes of vibration of the drum?

(6.17)

Can this force excite all the

The solution for the mode shapes is 00

y(r, 0, t)

:::::

~

oc

~ I n (K m r/rl)[X2n,m (t) cos 2n1T09

m=l n=O

+ X2n +l,m (t)

sin 2n'lTO]

CHAP. 6]


where ICm is the mth zero of the nth order Bessel function In(r). harmonicm == 1, n == 1 into equation (6.17) gives

141

Substituting the motion of the first

+ y cos 2;;8ou ;\x31 + y sin 2;;oou

d2 x 2 ldt 2

;\X21

d2x3ldt2

where ;\ == K~ + (2;;)2 and y == ToJ1(lClrO!rl). Using the controllability criteria, it can be found that one particular state that is not influenced by u(t) is the first harmonic rotated so that its node line is at angle 8 0 , This is illustrated in Fig. 6-9. A noncolinear point of application of another force is needed to excite, or damp out, this particular uncontrollable mode.

r

+ Y

r--------~----------~r------

Fig. 6-9

6.6.

Consider the system

1

dx

1

dt

o

where Ul and U2 are two separate scalar controls. is totally controllable if (1)

b

(2)

b

(3)

b

Determine whether the system

(0 0 O)T

=

(0 0 1)T (1 0

oy

For each case, we investigate the controllability matrix Q == (B I AB I A2B) for A

G~D

and

For b = 0 it is equivalent to scalar control, and by uncontrollable. For case (2), 1 0 0 1 Q 1 1

G

G b)

B

condition (I) of Theorem 6.2 the system is 0 0 1

2 1 1

D

The first three columns are linearly independent, so Q has rank 3 and the system is controllable. For case (3), 1 1 1 2 :::: Q o 1 o 1 o 1 o 1 The bottom two rows are identical and so the system is uncontrollable.

142


6.7.

Investigate total controllability of the time-varying system

Gg)x + (~)u

~~ The Q(t) matrix of Theorem 6.10 is

et(l-

Q(t)

Then det Q(t) = et(e t

6.8.

+ 2t -

2).

[CHAP. 6

t»)

-e t

Since et + 2t = 2 only at one instant of time, rank Q(t) = 2(e.d.).

Show that the time-varying system dx/dt = A(t)x + B(t)u is totally controllable if and only if the matrix W(t, ,) is postive definite for every, and every t> T, where

it

W(t, ,)

cp(T, 'rJ) B{7]) Bt('rJ) 4>t('1, 'rJ) d'rJ

Note that this criterion depends on cJt(t, T) and is not as useful as Theorems 6.10 and 6.11. Also note positive definite W is equivalent to linear independence of the rows of 4>(7,7]) Bh) for T ~ 1) ~ t. If W(t,7") is positive definite, W-I exists.

Then choose

u(7") = -Bt(r) 4>t(t 1, r) W-I(t o, t l ) x(t I ) Substitution will verify that

so that the system is totally controllable if W(t,7") is positive definite. Now suppose the system is totally controllable and showW(t, '7") is positive definite. First note for any constant vector k,

=

(k, Wk)

f·,.

t

ktcJlo(r, 'TJ) B(7]) Bt(7J) 4Jt(r, 17)k dn -

ft IiBth) 4Jt(r, 7])kll~

dn

==

0

'T

Therefore the problem is to show W is nonsingular if the system is totally controllable. Suppose W is singular, to obtain a contradiction. Then there exists a constant n-vector z # 0 such that (z, Wz) = O. Define a continuous, m-vector function of time f(t);:::: -Bt(t) q;t(to, t)z. But

stl

ftlllf(-r)II~ dT to

ztcJlo(t o' t) B(t) Bt(t) cJlot(to, t)z dt

to

(z, W(t l , to)z)

Hence f(t) = 0 for all t, so that 0

Itt

£t(t) u(t) dt

=

0

for any u(t).

Substituting for f(t) gives

to

t!

o

-

S

ztcJlo(to, t) B(t) u(t) dt

(6.18)

to

In particular, since the system is assumed totally controllable, take u(t) to be the control that transfers 0 to cJlo(tl' to)z # O. Then t!

z

J

4J(to, t) B(t) u(t) dt

to

Substituting this into equation (6.18) gives 0 = ztz which is impossible for any nonzero z. Therefore no nonzero z exists for which (z, Wz) ~ 0, so- that W must be positive definite.

CHAP. 6]


143


Consider the bilinear scalar system d~/dt = u(t) ~(t). It is linear in the initial state and in the control, but not both, so that it is not a linear system and the theorems of this chapter do not apply. The flow diagram is shown in Fig. 6-10. Is this system completely controllable according to Definition 6.1? ~(t)

u(t)---~

Fig. 6-1.0

6.10.

Given the system dx dt

(

~ ~ ~)

-2

0 -2

x

+

(-~) u,

y

= (-2 1 O)x

-1

Determine which states are observable and which are controllable, and check your work by deriving the transfer function. 6.11.

Given the system

dx

y

dt

= (1 1)x

Classify the states according to their observability and controllability, compute the P and Q matrices, and find the transfer function. 6.12.

Six identical frictionless gears with inertia I are mounted on shafts as shown in Fig. 6-11, with a center crossbar keeping the outer two pairs diametrically opposite each other. A torque u(t) is the input and a torque yet) is the output. Using the angular position of the two outer gearshafts as two of the elements in a state vector, show that the system is state uncontrollable but totally output controllable.

+

1-----I-----k:~>L.-.4--_ Y (t) u(t)

Fig. 6-11

Fig. 6-12

6.13.

Given the electronic circuit of Fig. 6-12 where u(t) can be any voltage (function of time). Under what conditions on R, Ll and L2 can both i 1(t 1 ) and i 2(t 1 ) be arbitrarily prescribed for tl > to. given that i 1(to) and i 2 (tO) can be any numbers?

6.14.

Consider the simplified model of a rocket· vehicle

Under what conditions is the vehicle state controllable?

144

CONTROLLABILITY AND· OBSERVABILITY

[CHAP. 6

6.15.

Find some other construction than equation (6.2) that will transfer a zero initial condition to an arbitrary z(t 1).

6.16.

Prove Theorem 6.5.

6.17.

Prove Theorem 6.6.

6.18.

What are the conditions similar to Theorem 6'.2 for which a two-input system is totally controllable?

6.19.

Given the controllable sampled data system ~(n

u(n + 1) - u(n)

+ 2) + 3Hn + 1) + 2~(n)

Write the state equation, find th~ transition matrix in closed form, and find the control that will force an arbitrary initial state to zero in the smallest number of steps. (This control depends' upon these arbitrary initial conditions.) 6.20.

Given the system with nondistinct eigenvalues dx

dt

y = (0 1 -l)x

=

Classify the elements of the state vector z corresponding to the Jordan form into observable/not observable, controllable/not controllable. 6.21.

Using the criterion Q:::: (b I Ab

6.22.

Consider the discrete system

I ... I An-1b), x(k

where x is a 2-vector, u is a scalar,

+ 1)

develop the result of Theorem 6.1. :::: Ax(k)

= (~

A

+ bu(k)

~),

b

(~).

=

(b) If the initial condition is

(a) Is the system controllable?

x(O)

=

(~),

find the control

sequence u(O), u(l) required to drive the state to the origin in two sample periods (i.e., x(2) = O}. 6.23.

Consider the discrete system

x(k

+ 1)

where x is a 2-vector, y is a scalar,

A

(a) Is the system observable? initial state x(O).

= Ax(k), =

(~

ctx(k) ct

=

(1 2).

(b) Given the observation sequence y(1) = 8,

y(2) = 14, find the

Hint: See Problem 6.8.

6.24.

Prove Theorem 6.8, constructing a bounded U(T) instead of equation (6.7).

6.25.

Given the multiple input-multiple output time-invariant system dx/dt = Ax + Bu, y = Cx + Du, where y is a k-vector and u is an m-vector. Find a criterion matrix somewhat similar to the Q matrix of Theorem 6.8 that assures complete output controllability.

6.26.

Consider three linear time-invariant systems of the form i = 1,2,3 (a) Derive the transfer function matrix for the interconnected system of Fig. 6-13 in terms of

A (i), Bm and C m , i = 1, 2, 3.

y(3)

u '---_----,

, Fig. 6-13

---J

CHAP.6J

CONTROLLABILITY AND OBSERV ABILITY

145

(b) If -the overall interconnected system in part (a) is observable, show that Sa is observable. (Note that u(i) and y(i) are vectors.) 6.27.

Given the time-varying system dx

dt

=

(~

y = (e- t e- 2t)x

Is this system totally controllable and observable? 6.28.

Prove Theorem 6.9 for the continuous-time case.

6.29.

Prove a controllability theorem similar to Theorem 6.10 for the discrete time-varying case.

6.30.

Similar to Problem 6.8, show that the time-varying system dx/dt = A(t)x + B(t)u is completely controllable for t1 > t if and only if the matrix Wet, T) is positive definite for every T and some finite t > T. Also show this is equivalent to linear independence of the rows of 4t(T, '1]) BC?]} for some finite 7J > T.

6.31.

Prove that the linear time-varying system dxldt = A(t)x, y = C(t)x is totally observable if and only if M(tll to) is positive definite for all t1 > to, where M(t 1, to)

Itl

=

cltt(t, to) ct(t) C(t) 4t(t. to) dt

to

Answers to Supplementary Problems ~(t1)

can be reached from

~(to)

= 0, so the system is uncontrollable.

6.9.

No nonzero

6.10.

The states belonging to the eigenvalue 2 are unobservable and those belonging to -1 are uncontrollable. The transfer function is 3(8 + 3)-1, showing only the states belonging to -3 are both controllable and observable.

6.11.

One state is observable and controllable, the other is neither observable nor controllable.

Q"

6.14.

MB ¥= 0 and Za ¥= 0

6.15.

Many choices are possible, such as

=

(1-1) 1 -1

and

h(s)

=

2

8

+1

n

u(t)

~

Ilk{U[t- to - (t1 - toHk -l)/n] -

U[t - to - (tl - to)k/n])

k=l

where U(t - T) is a unit step at t = T; another choice is u(t)

n

=:::

~ Ilke-A~t.

In both cases the

k=1

expression for 6.18.

Ilk

is different from equation (6.4) and must be shown to have an inverse.

Two Jordan blocks with the same eigenvalue can be controlled if 111/22 - 112/21 ¥= 0, the!'s being the coefficients of Ul and Uz in the last row of the Jordan blocks.

146 6.19.


:::::

(-3-2

.p(n, k)

==

(_1)n-k

(U(O) )

=

!

x(n +

1)

u(1)

6.20.

~) x(n) + (_~) u(n);

(-1

~)

-2

(-2)n-k

+

6(1310 -2-5)C~I(0»)

y(n)

==

(1 0) x(n)

(22-1-1)

X2(O)

The flow diagram is shown in Fig. 6-14 where

zl

and

z2

are controllable,

u------------_

~

::::: 1 and

p

::= 0 in T

--

(~~ ;~ 2p~ ~) ~

p -

.•

~

=2.

6.22.

Yes; u(O)

= -4,

6.23.

Yes; x(O)

= (!).

6.25.

rankR::::: k, where R = (CD I CAB , ... I CAn-IB I D)

6.26.

D(s)::= C(3)(Is - A(3»-IB(3) [C(l) (Is - A(D)-IB(1)

6.27.

It is controllable but not observable.

u(l)

Z2

and

Za

are observable.

~------------------y

Fig. 6-14

for

[CHAP. 6

+ C(2)(Is -

A(2»-IB(2)]

Chapter 7 Canonical Forms of the State Equation 7.1 INTRODUCTION TO CANONICAL FORMS The general state equation dx/dt = A(t)x + B(t)u appears to have all n2 elements of the A(t) matrix determine the time behavior of x(t). The object of this chapter is to reduce the number of states to m observable and controllable states, and then to transform the m2 elements of the corresponding A(t) matrix to only m elements that determine the inputoutput time behavior of the system. First we look at time-invariant systems, and then at time-varying systems. 7.2 JORDAN FORM FOR TIME ..INVARIANT SYSTEMS Section 2.4 showed how equation (2.21) in Jordan form can be found from the transfer function of a time-invariant system. For single-input systems, to go directly from the form dx/dt = Ax + bu, let x = Tz so that dz/dt = Jz + T-1bu - where T-IAT = J. The matrix T is arbitrary to within n constants so that T = ToK as defined in Problem 4.41, page 97. For distinct eigenvalues, dz/dt = Az + K-ITol bu, where K is a diagonal matrix with elements k ii on the- diagonal. Defining g = TOlb, the equation for each state is dzJdt = AiZi + (giu/k ii ). If gi = 0, the state Zi is uncontrollable (Theorem 6.1) and does not enter into the transfer function (Theorem 6.7). For controllable states, choose k ii == gi. Then the canonical form of equation (2.16) is attained. For the case of nondistinct eigenvalues, look at the l x l system of one Jordan block with one input, dz/dt = Lz + T-1bu. If the system is controllable, it is desired that T-1b = el, as in equation (2.21). Then using T = ToK, we require TOI b = Kel = (at al-l ... al)T, where the ai are the l arbitrary constants in the T matrix as given in Problem 4.41. In this manner the canonical form of equation (2.21) can be obtained. Therefore by transformation to Jordan canonical form, the uncontrollable and unobservable states can be found and perhaps omitted from further input-output considerations. Also, the n 2 elements in the A matrix are transformed· to the n eigenvalues that characterize the time behavior of the system. 7.3 REAL JORDAN FORM Sometimes it is easier to program a computer if all the variables are real. A slight drawback of the Jordan form is that the canonical states z(t) are complex if A has any complex eigenvalues. This drawback is easily overcome by a change of variables. We keep the same Zi as in Section 7.2 when Ai is real, but when M is complex we use the following procedure. Since A is a real matrix, if A is an eigenvalue then its complex conjugate A* is also an eigenvalue and if t is an eigenvector then its complex conjugate t* is also. Without loss generality we can look at two Jordan blocks for the case of complex eigenvalues.

of

147

148

CANONICAL FORMS OF THE STATE EQUATION

[CHAP. 7

If Re means "real part of" and 1m means "imaginary part of", this is d (Rez + jImz) dt Rez - jImz

o

ReL - jImL

)(Rez + jlmZ) Rez - jlmz

ReL Rez - ImL Imz + jReL Imz + jlmL Rez) ( ReL Rez - ImL Imz - jReL Imz - jlmL Rez By equating real and imaginary parts, the system can be rewritten in the "real" Jordan form as ReL -ImL)(Rez) ( ImL ReL Imz 7.4 CONTROLLABLE AND OBSERVABLE FORMS FOR TIME-VARYING SYSTEMS

We can easily transform a linear time-invariant system into a controllable or observable subsystem by transformation to Jordan form. However, this cannot be done for .timevarying systems because they cannot be transformed to Jordan form, in general. In this section we shall discuss a method of transformation to controllable and/or observable subsystems without solution of the transition matrix. Of course this method is applicable to time-invariant systems as a subset of time-varying systems. We consider the transformation of the time-varying system

(7.1) into controllable and observable subsystems. The procedure for transformation can be extended to the case y == CX(t)x + DX(t)u, b.ut for simplicity we take Dx(t) = O. We adopt the notation of placing a superscript on the matrices A, Band C to refer to the state variable because we shall make many transformations of the state variable. In this chapter it will always be assumed that A(t), B(t) and C(t) are differentiable n - 2, n - 1 and n - 1 times, respectively. The transformations found in the following sections lead to the first and second ("phase-variable") canonical forms (2.6) and (2.9) when applied to time-invariant systems as a special case. Before proceeding, we need two preliminary theorems.

Theorem 7.1:

If the system (7.1) has a controllability matrix QX(t), and an equivalence transformation x(t) = T(t) z(t) is made, where T(t) is nonsingular and differentiable, then the controllability matrix of the transformed system QZ(t) = T-l(t) QX(t) _ and rank Qz(t) == rank QZ(t).

Proof: The transformed system is dz/dt = Az(t)z + B%(t)u, where AZ = T-l(AxT - dT/dt) and BZ = T-IBx. Since QX = (Qi IQ~ I ... I Q~) and QZ is similarly partitioned, we need to show Q~:::: T-l Q~ for k = 1,2, ... , n using induction. First Q~ = HZ = T-IBx = T-IQ:. Then assuming Q~-l = T-IQ~_l'

Q% =

=

-AzQ~_l

+ dQ~_/dt

-T-l(AXT -

dT/dt)(T-IQ~_l)

T-l(~AxQ%_l

+ dQ%_tldt) =

+ dT-l/dt Q~-l + T-ldQ~_l/dt T-1Q%

for k = 2,3, ... , n. Now QZ(t) = T-l(t) QX(t) and since T(t) is nonsingular for any t, rank QZ(t) = rank QX(t) for all t.

CHAP.7J

CANONICAL FORMS OF THE STATE EQUATION

149

It is reassuring to know that the controllability of a system cannot be altered merely by a change of state variable. As we should expect, the same holds for observability.

Theorem 7.2:

If the system (7.1) has an observability matrix PX(t), then an equivalence transformation x(t) = T(t) z(t) gives PZ(t) = PX(t) T(t) and rank PZ(t) = . rank PX(t).

The proof is similar to that of Theorem 7.1. Use of Theorem 7.1 permits construction of a Tc(t) that separates (7.1) into its controllable and uncontrollable states. Using the equivalence transformation ,,(t) = Tc(t) z(t), (7.1) becomes d (Zl) dt Z2

(7.2)

=

where the subsystem

= A~l (t)Zl + B:(t)u

dzddt

(7.3)

is of order nl === n. and has a controllability matrix QZ1(t) = (Q~ll Q~ll ... I Q~J ranknl(e.d.). This shows Zl is controllable and Z2 is uncontrollable in (7.2).

with

I

The main problem is to keep Tc(t) differentiable and nonsingular everywhere, i.e. for all values of t. Also, we will find Q: (Qo

Z1

.

... 0

n1

= Tc(t) z(t)

and if Q:

=

Cf)

!~~)(~:')

+

fft (
is obtained.

1

First assume (7.1) can be transformed by x(t)

Using induction,

Therefore

= V (S IR)

Z

l 1 Qn1+

o

•••

to the form of (7.2).

then for i

= 1,2, ... ,

(Q~~ 1 )

(7.4)

0:i

1---"'" y(t)

+

Fig. 8-1

Also, flow diagrams of transfer functions (block diagrams) can be drawn in a similar manner for time-invariant systems. We denote the Laplace transform of x(t) as ..c{x}, etc. 164

CHAP. 8]

165

RELATIONS WITH CLASSICAL TECHNIQUES

Example 8.2. The block diagram of the system considered in Example 8.1 is shown in Fig. 8-2.

.e{u}---...J

l---oC{Y}

Fig. 8-2

Using equation (5.56) or proceeding analogously from block diagram manipulation, this can be reduced to the diagram of Fig. 8-3 where H(s)

=

(6/8 (28 (3

.e{U}--:1

+ 3)/8 2)

2>[8(~ ~) - (~ ~)r (~ ~)

H(s)

I~-""'"

.e{y}

Fig.8-S

Vector block diagram manipulations are similar to the scalar case, and are as useful to the system designer. Keeping the system representation in matrix form is often helpful, especially when analyzing multiple input-multiple output devices. 8.3 STEADY STATE ERRORS Knowledge of the type of feedback system that will follow an input with zero steady state error is useful for designers. In this section we shall investigate steady state errors of systems in which only the output is fed back (unity feedback). The development can be extended to nonunity feedback systems, but involves comparing the plant output with a desired output which greatly complicates the notation (see Problem 8.22). Here we extend the classical steady state error theory for systems with scalar unity feedback to timevarying multiple input-multiple output systems. By steady state we mean the asymptotic behavior of a function for large t. The system considered is diagrammed in Fig. 8-4. The plant equation is dx/dt = A(t)x + B(t)e, the output is y = C(t)x and the reference input is d(t) = y(t) + e(t), where y, d and e are all m-vectors. For this system it will always be assumed that the zero output is asymptotically stable, i.e. lim C(t)cI»A_BC(t, T) t~oo

=

0

where alP A-BC(t, T)/at = [A(t) - B(t)C(t)]cI» A-BC (t, T) and lP A-BC(t, t) = I. Further, we shall be concerned only with inputs d(t) that do not drive Ily(t)11 to infinity before t = co, so that we obtain a steady state as t tends to infinity.

1:==-:===> y( t)

d(t)==~

+

Zero output is asymptotically stable

Fig. 8-4.

Unity Feedback System with Asymptotically Stable Zero Output

166


Theorem 8.1:

Proof:

tends to

co.

For the system of Fig. 8-4, lim e(t) = 0 if and only if d = C(t)w + g t-oQ where dw/dt == A(t)w + B(t)g for all t:::=:, to in which g(t) is any function such that lim g(t) = 0 and A, B, C are unique up to a transformation t-::¢ onw.

Consider two arbitrary functions f(t) and h(t) whose limits may not exist as t If lim [f(t) - h(t)] = 0, then f(t) - h(t) = r(t) for all t, where r(t) is an t_'"'

arbitrary function such that lim r(t) then for all t, t_oo d(t)

=

[CHAP. 8

y(t)

+

r(t)

=

=

O.

From this, if 0

=

lim e(t) = lim [d(t) - y(t)], t-oQ

C(t)q,A_BC(t,tO)x(to)

+

t-oo

rt C(t)cI»A_BC(t,T)B(T)d(i)dT + Jto

r(t)

Jrtot C(t) q, A-BC(t, or) B(T) d(T) dT + g(t) + C(t).p A-BC (t, to) w(to) where the change of variables g(t) = r(t) + C(t)q,A-BC(t, to)[x(to) - w(to)] is one-to-one for =

arbitrary constant w(to) because lim C(t)cI»A-BC (t, to) = O.

This Volterra integral equation

t-oo

for d(t) is equivalent to the differential equations dw/dt = [A(t) - B(t) C(t)]w + B(t)d and d = C(t)w + g. Substituting the latter equation into the former gives the set of equations that generate any d(t) such that lim e(t) = O. t_oo

Conversely, from Fig. 8-4, dx/dt = A(t)x + B(t)e = [A(t) - B(t) C(t)]x + B(t)d. Assuming d = C(t)w + g and subtracting dw/dt = A(t)w + B(t)g gives d(x-w)/dt = [A(t) - B(t) C(t)](x-w)

Then

lim e

t-co

lim (d - y) t-oo

=

lim [g - C(t)(x - w») t-co

From the last part of the proof we see that e(t) = g(t) - C(t) CPA-Be(t, to)[x(to) - w(to)] regardless of what the function g(t) is. Therefore, the system dw/dt = A(t)w + B(t)g with d = C(t)w + g and the system dx/dt = [A(t) - B(t) C(t)]x + B(t)d with e = d - C(t)x are inverse systems. Another way to see this is that in the time-invariant case we have the transfer function matrix of the open loop system H(s) = C(sI - A)-lB relating e to y. Then for zero initial conditions, ..e{d} = [H(s) + I]..c{g} and .,e{e} = [H(s) + I]-l..c{d} so that .,e{g} = .,e{e}. Consequently the case where g(t) is a constant vector forms a sort of boundary between functions that grow with time and those that decay. Of course this neglects those functions (like sin t) that oscillate, for which we can also use Theorem 8.1. Furthermore, the effect of nonzero initial conditions w(to) can be incorporated into g(t). Since we are interested in only the output characteristics of the plant, we need concern ourselves only with observable states. Also, because uncontrollable but observable states of the plant must tend to zero by the assumed asymptotic stability of the closed loop system, we need concern ourselves only with states that are both observable and controllable. Use of equation (6.9) shows that the response due to any Wi(tO) is identical to the response due to an input made up of delta functions and derivatives of delta functions. These are certainly included in the class of all g(t) such that lim g(t) O. t-"" Since the case g(t) = constant forms a sort of boundary between increasing and decreasing functions, and since we can incorporate initial conditions into this class, we may take g(t) as the unit vectors to give an indication of the kind of input the system can follow with zero error. In other words, consider inputs

=

for i = 1,2, .. . ,m

CHAP. 8]


167

which can be combined into the matrix function C(t)

rt 4tA(t,T)B(T)(e Ie 1... lem)dT

J to

1

=

2

_ Inputs of this form give unity error, and probably inputs that go to infinity any little bit slower than this will give zero error. Example 8.3. Consider the system of Example 8.1 in which e(t) = U2(t) and there is no input u 1(t).

~~:::::

unity feedback system is then

y

=

e- t {[3X,(O)

(3 2)x

tends to zero asymptotically. equations

[( ~

2)]

~) - (~) (3 + 2x,(O)]

cos V2 t

x

whose output

+ ~ [x.(O) + x,(O)]

lim get) = 0,

t-oo

t}

(3 2) (:;)

+

g

generate the class of inputs d(t) that the system can follow with zero error.

Solving this system of equations gives dU)

sin V2

Consequently Theorem 8.1 applies to the unity feedback system, so that the d

where

The zero input,

=

3W1(O)

+

2w2(0)

+

3tW2(0)

+

it

[3(t - '1")

o

+ 2Jg(T}

dT

+

gCt)

For get) = 0, w~ see that the system can follow arbitrary steps and ramps with zero error, which is in agreement with the classical conclusion that the system is of type 2. Also, evaluating

it

[3(t - T)

+ 2J dT

=

1.5t2

+

2t

o

shows the system will follow t2 with constant error and will probably follow with zero error any function t 2 -r: for any e > O. Thisis in fact the case, as can be found by taking get) = t-e. Now if we consider the system of Example 8.1 in which

2)]

and there is no input u2(t),

then the closed loop system -is

~~:::::

y ::::: O.5x2(O) + [3Xl(O) be used.

which does not tend to zero asymptotically so that Theorem 8.1 cannot

Definition 8.1:

+ 1.5x2 (O)]e- 6t

[(

~ ~) - (~) (3

e(t) ~ Ul(t) X.

The output of this system is

The system of Fig. 8-4 is called a type-l system (l = 1,2, ... ) when lim e(t) = 0 for the inputs dt = (t - to)l-lU(t - to)ei for all i = 1,2, ... , m. t- 00

In the definition, U(t - to) is the unit step function starting at t = to and ei is the ith unit vector. All systems that do not satisfy Definition 8.1 will be called type-O systems. Use of Theorem 8.1 involves calculation of the transition matrix and integration of the superposition integral. For classical scalar type-l systems the utility of Definition 8.1 is that the designer can simply observe the power of s in the denominator of the plant transfer function and know exactly what kind of input the closed loop system will follow. The following theorem is the extension of this, but is applicable only to time-invariant systems with the plant transfer function matrix H(s) = C(sI - A)-lB. Theorem 8.2:

The time-invariant system of Fig. 8-4 is of type l ~ 1 if and only if

H(s) = s-IR(s) + P(s) where R(s) and P(s) are any matrices such that lim SR-1(S) = 0 and lilim 81- 1P(s)11 < 00. s-o

8-0


168

[CHAP. 8

Proof: From Theorem 8.1, the system is of type 1 if and only if ..e{(d1 1 d 2 ! ... I dm)} = (l-l) Is-II = [H(s) +1]G(s) where .,e{gi}, the columns of G(s), are the Laplace transforms of any functions gi(t) such that 0 = lim gi(t) = lim S~{gi} where S.,e{gi} is analytic for

Re s ~ O.

t-eo

8-0

First, assume H(s) = s-IR(s) + P(s) where lim SR-l(S) neighborhood of s = O. Choose 8-0 G(s)

= (1-1) I S-I [H(s) + I]

=

0 so that n-l(s) exists in a

= (l-l)! [R(s) + SIP(S)

-1

+ sll]-1

Since [H(s) + I] -1 is the asymptotically stable closed loop transfer function matrix, it is analytic for Re s === O. Then sG(s) has at most a pole of order 1-1 at s= 0 in the region Re s ~ O. In some neighborhood of s = 0 where R-l(S) exists we can expand

= (l-l)! s[R(s) + SlP(S) + sIIJ-l = (l-l)! SR-l(S)[1 + Z(s) + Z2(S) + ... ] Z(s) = SR-l(S)[SI-lP(S) +SI-II). Since IimZ(s) = 0, this expansion is valid s-o

sG(s)

where small

lsi,

and lim sG(s) = O.

for

Consequently sG(s) has no pole at s = 0 and must be

8-0

analytic in Re s:::'" 0 which satisfies Theorem 8.1. Conversely, assume lim sG(s) == ·0 where sG(s) is analytic in Re s === O. Write H(8) = s-o s-IR(s) +P(s) where P(s) is any matrix such that lilim sl-IP(s)ll < 00 and R(s) = 8 1[H(8) - P(s)] is still arbitrary. Then 8-0 (1-1) ! s-Il = [s-IR{s)

+ P(s) + I]G(s)

can be solved for SR-l(S) as (l-l)! SR-l(S) = sG(s)(1 + W(S))-1 = 8G(s}[1 + W(s)

+ W2{S) + ... ]

where (l-l) ! W(s) = [SI-IP(S) + SI-11]sG(s). This expansion is valid for IlsG(s)!l small enough, so that R-l(S) exists in some neighborhood of s = O. Taking limits then gives lim SR-l(S) = O. 8-0

We should be careful in the application of Theorem 8.2, however, in light of Theorem 8.1. The classification of systems into type 1 is not as clear cut as it appears. A system with H(s) = (s + f)-1 can follow inputs of the form e- Et • As f tends to zero this tends to a step function, so that we need only take E- 1 on the order of the time of operation of the system. Unfortunately, for time~varying systems there is no guarantee that if a system is of type N, then it is of type N - k for all k === O. However, this is true for time~invariant systems. (See Problem 8.24.). Example 8.4.

(~+~ sa

... JI

..e{d}

+

1

~

-1-~) 8 2

+ 128 + 382

----7> ~{y}

0

82

-

Fig. 8-5

The system sho~ in Fig. 8-5 has a plant transfer function matrix H(s) that can be written in the form H(s)

in which

=

_1 8

2

(-68-1 + 1

9

II;~ sP(s) I

-1) 0

-

+

( 0 1281

I ( lim

S-foO

12

+3

0

+ 38

-1 ) 0

=

s-2R(s)

00

+

pes)

CHAP. 8]


and where

l' (0

=

1m s

s-o

169

0

#

-1

Since lim SR-l(S) has a nonzero element, the system is not of type 2 as appears to be the case upon 8-0

first inspection.

Rewrite H(s) in the form (where R(s) and P(s) are different)

!

H(s)

Again,

s

lilim pes) II
O. Then an error constant matrix table can be formed for time-invariant systems of Fig. 8-4. Steady State Error Constant Matrices

In the table

System Type

Step Input

Ramp Input

Parabolic Input

0

lim [I + H(S)]-l s-+o

*

*

1

0

lim R-l(8)

*

2

0

S-I-O

lim &-1(8)

0

s-O

* means the system cannot follow all such inputs.

Example 8.5. The type-1 system of Example 8.4 has an error constant matrix

;~ R-l(s)

=

the input were (t - to)U(t - t o)e2J the steady state output would be [(t - to)U(t - to) can follow with zero steady state error an input of the form (t - to)U(t - t o)el'

(~ _~). + 6]e2'

Thus if

The system

8.4 ROOT LOCUS Because root locus is useful mainly for time-invariant systems, we shall consider only time-invariant systems in this section. Both single and multiple inputs and outputs can be considered using vector notation, i.e. we consider dx/dt = Ax

+ Bu

y = Cx+Du

(8.1)

Then the transfer function from y to u is H(s) = C(sI - A)-IB + D, with poles determined by det (sI - A) = O. Note for the multiple input and output case these are the poles of the whole system. The eigenvalues of A determine the time behavior of all the outputs.

170


[CHAP; 8

We shall consider the case where equations (8.1) represent the closed loop system. Suppose that the characteristic equation det (sI - A) = 0 is linear in some parameter K so that it can be written as SU

+ (8 1 + o/lK)sn-l + (8 2 + o/2K)sn-2 + ... + (8 n- 1+ o/n_1K)s +

+ o/nK =

8n

0

This can be rearranged to standard root locus form under K variation, o/l sn -l + o/2sn - 2 + ... + o/n-l s + o/n -1

K.

sn

+ 0lsn-1 + ... + ()n-Is + Bn

The roots of the characteristic equation can be found as K varies using standard root locus techniques. The assumed form of the characteristic equation results from both loop gain variation and parameter variation of the open loop system. Example 8.6. Given the system of Fig. 8-6 with variable feedback gain

K.

d(t) ==::::::::::\

+

Fig. 8-6

The closed loop system can be written as dx

dt The characteristic equation is root locus shown in Fig. 8-7.

82

1)

-K ( -K

+ (3 + K)S + 41( =

-3 x O.

+

(1

1

Putting it into standard root locus form leads to the

Fig. 8-7

Example 8.7. The feedback system of Fig. 8-8 has an unknown parameter a. Find the effect of variations in a upon the closed loop roots.' Let sinh a: = K. The usual procedure is to set the open loop transfer function equal to -1 to find the closed loop poles of a unity feedback system.

(8 + 3) sinha 82

+ 38 + sinh a

Fig. 8-8

y

171


CHAP. 8]

-1

=

(8 + 8)" 82

+ 3s +

Ie

This can be rearranged to form the characteristic equation of the closed loop system, (8 + 3)IC ::.: O. Further rearrangement gives the standard root locus form under K variation. -1

K

82

+ 38 + + Ie

8+4 8(8+ 3)

This happens to give the same root locus as in the previous example for sinh IX

8.5 NYQUIST DIAGRAMS First we consider the time-invariant single inputsingle output system whose block diagram is shown in Fig. 8-9. The standard Nyquist procedure is to plot G(s) H(s) where s varies along the Nyquist path enclosing the right half s-plane. To do this, we need polar plots of GUw) H(jw) where w varies from -00 to

===

O.

.('{e}

+

G(s)

+ .e{v}

+00.

Using standard procedures, we break the closed loop between e and v. Then setting this up in state space form gives v = ctx+de dx/dt = Ax + be

H(s)

Fig. 8~9

(8.2)

Then GUw)H(jw) = ctUwI - A)-lb + d. Usually a choice of state variable x can be found such that the gain or parameter variation K of interest can be incorporated into the c vector only. Digital computer computation of {iwI - A)-lb as w varies can be most easily done by iterative techniques, such as Gauss-Seidel. Each succeeding evaluation of (jwi+lI - A)-lb can be started with the initial condition (jwl- A)-lb, which usually gives fast convergence. Example 8.8. Given the system shown in Fig. 8-10. The state space form of this is, in phase-variable canonical form for the transfer function from e to 'I), ' dx

dt Then

Ie

ct(j6lI - A)-lb

jw(jw + 1) •

'U

= (K O)x

giving the polar plot of Fig. 8-11. ImGH

.e{e}

ReGH

+

.c{v}

1 8+1

Fig. 8~lO

About the only advantage of this over standard techniques is that it is easily mechanized for a computer program. For multiple·loop or multiple-input systems, matrix block diagram manipulations give such a computer routine even more flexibility.

RELATIONS WITH CLASSICAL TECHNIQUES Example 8.9. Given the 2 input - 2 output system with block diagram shown in Fig. 8-12. and V2, == . Then dx/dt -:::: Ax + hIe! +-b2 e2 and VI == ctx. The loop connecting VI and el can be closed, so that 61 ::: VI x. Then

+

elx

+

:::ct

.c{V2} = 50

el (81 -

A - bi

(CHAP. 8

..e{v}

et) -lb2.,e{ e2}

that we can ask -the computer to give us the polar plot of A - bIer )-lb2 •

cd (jwI -

Fig. 8.. 12

8.6 STATE FEEDBACK POLE PLACEMENT Here we discuss a way to feed back the state vector to shape the closed loop transition matrix to correspond to that of any desired nth-order scalar linear differential equation. For time~invariant systems in particular, the closed loop poles can be placed where desired. This is why the method is called pole placement, though applicable to general time-varying systems. To "place the poles," the totally controllable part of the state equation is transformed via x(t) == T(t)z(t) to phase-variable canonical form (7.12) as repeated here:

dz

=

dt

o

1

0

0

0

0

0

1

0

0

••

It

•

"

••••••••••

0 a2(t)

0

aI(t)

II'

ill

Z

............

+

1 an(t)

0 a3(t)

u

(7.12)

0 1

Now the scalar control u is constructed by feeding back a linear combination· of the z state variables· as u = kt(t)z where each ki(t);:;:;;: -alt) - ai(t). [-al(t) - a 1(t)]zl + [-a2(t) - a2(t)Jz2 + ... + [-an(t) - an(t)]zn Each ai(t) is a time function to be chosen. This gives the closed loop system U

;:;:;;:

dz

dt

o

i

0

0

o

0

1

0 z

==

o

o

o

1 -a 1 (t) -a2(t) -ocs(t) -an(t) n V Then Zl(t) obeys z~n) + an(t)zi + '" + a2(t)zl + a1(t)Zl = 0 and each Zi+l(t) for i = 1,2, ... , n - 1 is the ith derivative of Zl(t). Since the a/t) are to be chosen, the corresponding closed loop transition matrix q,z(t, to) can be shaped accordingly. Note, however, thatx(t) =T(t)4tAt, to)zo so that shaping of the transition matrix iPz(t, to) must be done keeping in mind, the effect of T(t). This minor complication disappears when dealing with time-invariant systems. Then T(t) is constant, and furthermore each ai(t) is constant. In this case the time behavior of x(t) and z(t) is essentially the same, in that both AX and AZ have the same characteristic equation An + an ",n-1 + ... + a 2 A+ a l = O.. For the closed loop system to have poles at the desired values YI' 12' •.. , Yn, comparison of coefficients of the A.'s in (A-Yl)(A.-a2)· •• (A.-an) = 0 determines the desired value of each Q t•• Example 8.10; Given the system

~~

==

(~ ~ ~)" + \1(~) 2 -3

u

t

It is desired to have a time-invariant closed loop system with poles at 0, ~1 and -2. Then the desired system will have a characteristic equation A3 + 3A2 + 2A := O. Therefore we choose u:= (-2 1 -(t + 3»)x, so kT

==

(-2

1

-(t + 3»).

CHAP. 8]


173

For multiple-input systems, the system is transformed to the form of equation (7.13), except that thesul?system dw/dt = A~Wi + bf'ui must be in phase-variable canonical form (7.12) and for i ¥= i, the A{j(t) must be all zeros except for the bottom row. Procedures similar to those used in Chapter 7 can usually attain this form, although general conditions are not presently available. If this form can be attained, each control is chosen as ui = kt (t)Wi - e!Aijwj for j ¥= i to "place the poles" of Aii(t) and to subtract off the coupling terms~

Why bother to transform to canonical form when trial and error can determine k? Example 8.11. Place the poles of the system of Problem 7.8 at PI' pz and Pa.

We calculate

A-1 det

[(

_~_

This is A3

-

(k 13 + k21 + k22 + kZ3

- ku -

k13 -

+ 5)A2 + [ku + 2k 13 + 3kz1 + 4k22 + 5k23 + k 13(k 21 + k 2Z ) -

2k21 -

4k22 -

6k 23 -

k n (k 22

+ k 23 ) + k 12 (k 21 + k Z3 ) + k 13 (k 21 -

kZ3(kn

k 2Z )

+ k 1Z ) + 8]A

-

4

It would take much trial and error to choose the k's to match (A - Pl)(A - PZ)(A - Pa) = A3

-

(PI + pz + Pa)A2

+

(PIP2

+ PzPa + PIPa)A -

PIPzPa

Trial and error is usually no good, because the algebra is nonlinear and increases greatly with the order of the system. Also, Theorem 7.7 tells us when it is possible to "place the, poles", namely when Q(t) has rank n everywhere. Transfor!llation to canonical form seems the best method, as it can be programmed on a computer. State feedback pole placement has a number of possible defects: (1) The solution appears after transformation to canonical form, with no opportunity for obtaining an engineering feeling for the system. (2) The compensation is in the feedback loop, and experience has shown that cascade compensation is usually better. (3) All the state variables must be available for measurement. (4) The closed loop system may be quite sensitive to small variation in plant parameters. Despite these defects state feedback pole placement may lead to a very good system. Furthermore, it can be used for very high-order and/or timevarying systems for which any compensation may be quite difficult to find. Perhaps the best approach is to try it and then test the system, especially for sensitivity. Example 8.12. Suppose that the system of Example 8.10 had t - e instead of t in the lower right hand corner of the A(t) matrix, where e is a small positive constant. Then the closed loop system has a characteristic equation AS + 3AZ + 2A - e = 0, which has an unstable root. Therefore this system is extremely sensitive.

8.7 OBSERVER SYSTEMS Often we need to know the state of a system, and we can measure only the output of the system. There are many practical situations in which knowledge of the state vector is required, but only a linear combination of its elements is known. Knowledge of the state, not the output, determines the future output if the future input is known. Conversely knowledge of the present state and its derivative can be used in conjunction with the state equation to determine the present input. Furthermore, if the state can be reconstructed from the output, state feedback pole placement could be used in a system in which only the output is available for measurement. In a noise-free environment, n observable states can be reconstructed by differentiating a single output n-l times (see Section 10.6). Ina noisy environment, the optimal reconstruction of the state from the output of a linear system is given by the Kalman-Bucy filter. A discussion of this is beyond the scope of this book. In this section, we discuss an observer system that can be used in a noisy environment because it does not contain differentiators. However, in general it does not reconstruct the state in an optimal manner.

174

RELATIONS WITH· CLASSICAL TECHNIQUES

[CHAP. 8

To reconstruct all the states at all times, we assume the physical system to be observed is totally observable. For simplicity, at first only single-output systems will be considered. We wish to estimate the state of dx/dt = A(t)x + B(t)u, where the output y = et(t)x. The state, as usual, is denoted x(t), and here we denote the estimate of the state as x(t). First, consider an observer system of dimension n. The observer system is constructed as

+ k(t)[et(t)x -

dx/dt == A(t)x

y]

+ B(t)u

(8.3)

where k(t) is an n-vector to be chosen. Then the observer system can be incorporated into the flow diagram as shown in Fig. 8-13.

~

Physical system

------+-0\.-----------

Observer system

-------..1.\

Fig. 8-13

Since the initial state x(to.), "where to is the time the observer system is started, is not known, we choose x(to) == O. Then we can investigate the conditions under which x(t) tends to x(t). Define the error e(t) == x(t) - (t). Then

x

de/dt

=

dx/dt - di/dt = [A(t)

+ k(t)ct(t)]e

(8.4)

Similar to the method of the previous Section 8.6, k(t) can be chosen to "place the poles" of the error equation (8 ..4-). By duality, the closed loop transition matrix 1j"(t, to) of the adjoint equation dp/dt = -At(t)p - c(t)v is shaped using v == kt(t)p. Then. the transition matrix fIJ(t, to) of equation (8 ..4.) is found as fIJ(t, to) = iJt(to' t), from equation (5.64). For time-invariant systems, it is simpler to consider dw/dt = Atw + cv rather than the adjoint. This is because the matrix At + ckt and the matrix A + ke t have the same eigenvalues. This is easily proved by noting that if A is an eigenvalue of At + ck t , its complex conj ugate A* is also. Then A* satisfies the characteristic equation det (A *1 - At - ckt) = O. Taking the complex conjugate of this equation and realizing the determinant is invariant under matrix transposition completes the proof. Hence the poles of equations (8.3) and (8.4) can be placed where desired. Consequently the error e(t) can be made to decay "as quickly as desired, and the state of the observer system tends to the state of the physical system. However, as is indicated in Problem 8.3, we do not want to make the error tend to zero too quickly in a practical system. Example 8.13. Given the physical system dx

y = (1 l)x

dt

Construct an observer system such that the error decays with poles at -:-:-2 and -3. First we transform the hypothetical sy~em

dw (it

=

CHAP. 8]

175


to the phase variable canonical form : where

~) z,

w = (:

(_~ _~)Z

=

+

G)v

obtained by Theorem 7.7. We desire the closed loop system to have the

teristic equation 0 = (x. + 2) (X. + 3) = x.2 + 5X. + 6. Therefore choose k = (-1 O)t and the observer system is constructed as

(-3 0)

A

2 -2 x +

(1)

v = (-4 -l)z = (-1 O)w.

charac~ Then

(-1)

0 Y +0 u

N ow we consider an observer system of dimension less than n. In the case of a single~ output system we only need to estimate n - 1 elements of the state vector because the known output and the n -1 estimated elements will usually give an estimate of the nth element of the state vector. In general for a system having k independent outputs we shall construct an observer system of dimension n - k. We choose P(t) to be certain n - k differentiable rows such that the n

)-1 = C(t)

P(t) (

(H(t) I G(t)) exists at all times where H has n - k columns.

x n matrix

The estimate

x

is constructed as x(t) = H(t)w

+ G(t)y

or, equivalently, (P(t) ) C(t)

x=(

w)

(8.5)

y

Analogous to equation (8.3), we require P d~/dt = P[Ax

+ L(Cx -

y)

+ Bu}

where L(t) is an n x k matrix to be found. (It turns out we only need to find PL, not L.) This is equivalent to constructing the following system to generate w, from equation (8.5), dwldt = (dPldt)

x+ Pdx/dt =

Fw - PLy

+ PBu

where F is determined from FP = dP/dt+PA+ PLC. Then (F1-PLl(

(8.6)

~)

so that F and PL are determined from (F j-PL) = (dP/dt + PA)(H IG). (8.6) the error e = P(x -,x) = Px - w obeys the equation de/dt = Fe.

= dP/dt +PA

From (8.5) and

The flow diagram is then as shown in Fig. 8-14.

~

Physical system

- - - - - - - - ...... 1. . ......-------

Observer system

Fig.8~14

Example 8.14. Given the system of Example 8.13, construct a first-order observer system. Since

c = (1 1), choose P =

(:r

(PI pz)

with

PI # P2"

Then

1 (1

PI - P2

-1

-P2) PI

(H I G)

-~-------fOol·1

176

RELATIONS WITH CLASSICAL TECHNIQUES Therefore A

=

X

1 (1)

Pi - P2

and F

PL

+ PA)H =

==

(dP/dt

==

-(dP/dt

-1

1 (0)

+

w

Pi - P2

1

Y

P.l (-2 1)( 1) = ~(2 1)(-P2) =

-3PI

(P, PI - P2

+ PA)G =

[CHAP. 8

2 -2

-1

2 -2

PI - P2

+ 4P2

PI - P2

PI

pi -

2p~

Pi - P2

so that (PI - P2)dw/dt = (-BPI + 4p2)W - (pi - 2p~)y - PI(Pl- P2)U is the first~order observer. A bad choice of PI/P2 with 1 < PI/P2 < 4/3 gives an unstable observer and makes the error blow up.

The question is, can we place the poles of F by proper selection of P in a manner similar to that of the n-.dimensional observer? One method is to use trial and error, which is sometimes more rapid for low-order, time-invariant systems. However, to show that the poles of F can be placed arbitrarily, we use the transformation x = Tz to obtain the canonical form

:t (~:)

=

(~~: ..~~............~)(~:) All

Zt

A!2

(~~ ~~

y

..

o

where the subsystem dzi/dt ical form

= Aiizi + Biu

An

• • •

and Yi

."

= ct Zi

~

)

Z

ct is in the dual phase variable canon-

i

(0

in which B, is defined from T-I B =

(~:)

T-lBu

Zl

...........

0

+

= 1,2, ... , l

o l)Zi

(8.7)

and n, is the dimension of the ith subsystem.

As per the remarks following Example 8.10, the conditions under which this form can always be obtained are not known at present for the time-varying case, and. an algorithm is not available for the time-invariant multiple-output case. However, assuming the subsystem (8.7) can be obtained, we construct" the observer equation (8.6) for the subsystem (8.7) by the choice of Pi = (I Ik i ) where k(t) is an (14 -1)vector that will set the poles of the observer. We assume ki(t) is differentiable. Then

(~)

=

(*)

and

(::r

=

(-H-T-)

(8.8)

CHAP; 8]

We find

RELATIONS WITH ·CLASSICAL TECHNIQUES

F,

=

=

(dP'/ dt + p,AiilH.

[(0 I dkd dt l 0

0 0

1 0

=

Fi

+ (I Ik'lAii1(

from which

ki1(t) ki2(t) ki3(t)

0 0 0

1

+)

177

......................... o 0 1 k i• ni - (t) 1

By matching ,coefficients of the "characteristic equation" with "desired pole positions," we make- the error decay as quickly as desired. Also, we find PiLi as PiLi = -(dPJdt + PiAii)Gi . '_

Then

X

- '(A) T.z = T i, Zl

A

A

=

•

A

where

Zi=

Example 8.15. Again, consider the system of Example 8.13.

~~

(-~ _~) x

-

+ (- ~ )

y = (1 l)x

u

To construct an observer system with a pole at -2, use the transformation x = Tz where (:

~).

(Tt)-l =

Then equations (8.7) are

(0-2)

dz dt

1 -4

z

+

(-4) u

y

-1

(0 1)z

-

The estimate ~ according to equation (8.8) is now obtained as

(-H+)(;)

A

Z

where kl = -2 sets the pole of the observer at -2. system is dw/dt = -2w + 2y - 2u. Therefore A

X

=

and the error PT-1x - w = of Fig. 8-15.

u

T~ 2Xl

+ x2 -

8+4

82

Then F = -2 and PL = -2 so that the observer

+ 48 + 2

= w decays with a time constant of 1/2.

Y

t----~

1 8+2

Fig. 8-15

This gives the block diagram

178

8.8


[CHAP. 8

ALGEBRAIC SEPARATION

In this section we use the observer system of Section 8.7 to generate a feedback control to place the closed loop poles where desired, as discussed in Section 8.6. Specifically, we consider the physical open loop system dx/dt

=

A{t)x

+ B(t)u + J(t)d

y = C(t)x

(8.9)

with an observer system (see equation (8.6)),

= x=

dw/dt

F(t)w -P(t)L(t)y

H(t)w

+ P(t)B(t)u + P(t)J(t)d

+ G(t)y

(8.10)

and a feedback control u(t) that has been formed to place the poles of the closed loop system as u = W(t)x (8.11) Then the closed loop system block diagram is as in Fig. 8-16. u

d

G

~ Ph:iCa!

u

·1-.--~--

system - - -.....

J=:==:.J

. . Feedback Observer ----------+-1..-

~

Fig. 8-16

Theorem 8.3:

(Algebraic Separation). For the system (8.9) with observer (8.10) and feedback control (8.11), the characteristic equation of the closed loop system can be factored as det (AI - A - BW) det (AI - F).

This means we can set the poles of the closed loop system by choosing W using the pole placement techniques of Section 8.6 .and by choosing P using the techniques of Section 8.7.

Proof: The equations governing the closed loop system are obtained by substituting equation (8.11) into equations (8.9) and (8.10):

:t(~) Changing variables to e gives

(p:W~~~~ic F::B~)(~) + (~J)d = Px -

wand using HP + GC = I and FP

= dP / dt + PA + PLC

CHAP. 8]


179

Note that the bottom equation de/dt = Fe generates an input -WHe to the closed loop of observer system dx/dt = (A + BW)x. Use of Problem 3.5 then shows the characteristic equation factors as hypothesized. Furthermore, the observer dynamics are in general observable at the output (through coupling with x) but are uncontrollable by d and hence cancel out of the closed loop transfer function. Example 8.16. For the system of Example 8.13, construct a one-dimensional observer system with a pole at -2 to generate a feedback that places both the system poles at -1. We employ the algebraic separation theorem to separately consider the system pole placement and the observer pole placement. To place the pole of

dx

y

dt

== (1 l)x

using the techniques of Section 8.6 we would like

u == (-2 3/2)x which gives closed loop poles at -1. However, we cannot use x to form iL, but must use from the observer system with a pole at -2, which was constructed in Example 8.15.

dw/dt = -2w

+ 2y -

l::

as found

2(u + d)

We then form the control as u

=

(-2 3/2)~

Thus the closed loop system is as in Fig. 8-17.

8 +4 82 +48+2

1 8+2

y

I - - -- _ -_ _

+

Fig.S~17

Note that the control is still essentiaIIy in the feedback loop and that no reasons were given as to why plant poles at -1 and observer pole at -2 were selected. However, the procedure works for highorder, multiple input- multiple output, time-varying systems.

8.9 SENSITIVITY, NOISE REJECTION, AND NONLINEAR EFFECTS Three major reasons for the use of feedback control, as opposed to open loop control, are (1) to reduce the sensitivity of the response to parameter variations, (2) to reduce the effect of noise disturbances, and (3) to make the response of nonlinear elements more linear. A proposed design of a feedback system should be evaluated for sensitivity, noi_f?e tejection, and the effect of nonlinearities. Certainly any system designed using the pole placement techniques of Sections 8.6, 8.7 and 8.8 must be evaluated in these respects because of the cookbook nature of pole placement.

180


[CHAP; 8

In this section we consider these topics in a very cursory manner" mainly to show the relationship with controllability and observability. Consequently we consider only: small percentage changes in parameter variations, small noise compared, withthe'signal, -and nonlinearities that are almost linear. Under these assumptions we will show how each effect produces an unwanted input into a linear system and then how to minimize this unwanted input. First we consider the effect of parameter variations. Let the subscriptN refer to the nominal values and the subscript a refer to actual values. Then the nominal system',(the system with zero parameter variations) can be represented by

=

dXN/dt

AN(t)xN + B(t)u

YN = C(t)XN

+ D(t)u

(8.12)

These equations determine XN(t) and YN(t), so these quantities are assumed known. If some of the elements of AN drift to some actual Aa (keeping B, C and D fixed only for simplicity), then Aa(t)Xa + B(t)u dXa/dt Ya

= C(t)Xa + D(t)u 8y = Ya - YN, subtract

(8.13)

Then let ax = Xa - XN, SA = Aa - AN, equations (8.12) from (8.13), and neglect the product of small quantities BA Bx. Warning: That SA 8x is truly small at all times must be verified by simulation. If this is so, then d(8x)/dt 8y

== ==

AN(t) 8x

+ BA(t) XN

C(t) 8x

(8.14)

In these equations AN(t), C(t) and XN(t) are known and 8A(t), theyariation of the parameters of the A(t) matrix, is the input to drive the unwanted signal 8x. For the case of noise disturbances d(t), the nominal system remains equations (8.i2) but the actual system is dKa/dt = AN(t)Xa + B(t)u + J(t)d Ya

=

C(t)Xa + D(t)u + K(t)d

(8.15)

Then we subtract equations (8.12) from (8.15) to obtain _d(8x)/dt

=

By =

AN(t) 8x

+ J(t)d

eft) 8x+ K(t)d

Here the noise d(t) drives the unwanted signal

_(8.16)

ax.

Finally, to show how a nonlinearity produces an unwanted signal, consider a scalar input into a nonlinearity as shown in Fig. 8. .18. This can be redrawn into a linear system with a large output and a nonlinear system with a small output (Fig. 8-19). XN

XN

d

Fig. 8-18

* +

Fig. 8. . 19

dN

d

+ +

3d

CHAP. 8J-

RELATIONS WITH -CLASSICAL TECHNIQUES

181

Here the unwanted signal is 8d which is generated by the nominal XN. This can be incorporated into a block diagram containing linear elements, and the effect of the nonlinearity can be evaluated in a manner similar to that used in deriving equations (8.16). d(8x)/dt = AN(t)

ax + j(t)8d

8y = C(t) 8x + k(t) Sd

(8.17)

Now observability and controllability theory can be applied to equations (8.14), (8.16) and (8.17). We conclude that, if possible, we will choose C(t), AN(t) and the corresponding input matrices B(t), D(t) or J(t), K(t) such that the unwanted signal is unobservable with respect to the output 8y(t), or at least the elements of the state vector associated with the dominant poles are uncontrollable with respect to the unwanted signal. If this is impossible, the system gain with respect to the unwanted signal should be made as low as possible. Example 8.17. Consider the system

(-~ _ ~ ) x +

d:x/dt =

G)

u.

The nominal value of the parameter

~ is zero

and the nominal input u{t) is a· unit step function.

=

dxNldt

If XN{O) = 0,

then

xN{t)

=

(~=

equation (8.14),

:=:t).

d(ax)ldt

=

(-~ -~)

XN

G)

The effect of small variations in a can be evaluated from

(-~ _~) ax

+

:=:')

(~ ~)G =

Simplifying, d(8x)/dt

+

(-~ _~) Sx +

G)

0(1- e-')

We ean eliminate the effects of a variation upon the output if c is chosen such that the output observability matrix (ctb ctAb) == O. This results in a choice ct = (0 y) where y is any number.

Furthermore, all the analysis and synthesis techniques developed in this chapter can be -used to analyze and design systems to reduce sensitivity and nonlinear effects and to reject noise. This may be done by using the error constant matrix table, root locus, Nyquist, and/or pole placement techniques on the equations (8.14), (8.16) and (8.17).

182


[CHAP. 8


~~

For the system of Fig. 8-4, let with y = (1 l)x.

= (~

(t

_~)x + ~ 2)e

t+1

t+1

for t

t+1

~ 0,

Find the class of inputs that give a constant steady state error.

The closed loop system with zero input (d(t)

dt

is

_~) -. (t 12)(1 l)]X

[Ci1

dx

= 0)

t+1

t+1

which has a double pole at -1. Therefore the zero output of the closed loop system is asymptotically stable and the remarks following Theorem 8.1 are valid. The class of all d(t) that the system can follow with lim e(t) = 0 is given by t--+

OQ

Cl1

dw dt

with

+ g.

(1 l)w

d

1t

) w

t+1

(t~.2).

+

g

t+1

The transition matrix for this system is

1

+1

T

(t +

eT - t

1 - eT -

t(1- eT - t

t

1

»)

+ te'T-t

Then d(t)

(Notice this system is unobservable.) K(t+1)

f

t

For constant error let get)

'7"+2 (r+1)2dT

K(t

+ 1HIn (t + 1) -

== K,

In (to + 1)

an arbitrary constant.

+ (to + 1)-1 -

(t

Then

+ 1)-1]

to

and the system follows with error K all functions that are asymptotic to this. Since the system is reasonably well behaved, we can assume that the system will follow all functions going to infinity slower than Ie(t + 1) In (t + 1).

8.2.

Given the

multiple~input

system

(-~ -~ ~)(::) + (~ -~)(~:)

:t (::)

0

Z3

0 -3

1

Za

0

Place the poles of the closed loop system at -4, -5 and -6. Transform the system to phase variable canonical form using the results of Problem 7.21.

(-~ 1)("' 1 -2 -8

x

4

9

Then (

z

leI -1

0 0

0 -1

K2

0

~

le - 1 3

0

0

le2

0

0

~)z le3

)(-! 1

5/2 :--4

3/2

1/2) -1

1/2

x

CHAP. 8]


and

( 0 1 0) (1 1 1)(Q'/(l

dx

o

0 -6 -11

dt

1 -6

x

+

-1 -2 -3 1 4 9

183

Kl)( )

1C2

-K2

lCa

0

::

To obtain phase variable canonical form for ul' we set

which gives

aKl

= 1/2,

"'2

= -1,

lI:a:::::

1/2.

For a # 0, we have the phase variable canonical system dx

dt

~ ~)

=

-11

x

-6

+

(~ -~ +~ 1/2Q' ~;::)(::) 1

4

To have the closed loop poles at -4, -5 and -6 we desire a charateristic polynomial AS + 15A.2 + 74A + 120. Therefore we choose ul -114xl - 63x 2 - 9xa and U2 OXI + OX2 + OXa.

=

=

In the case a = 0, the state Zt is uncontrollable with respect to Ul and, from Theorem 7.7, cannot be put into phase variable canonical form with respect to u 1 alone. Hence we must use U2 to control Zl. and ean assure a pole at -4 by choosing u2 = -3z 1 • Then we have the single-input system

can be transformed to phase variable canonical form, and ul:;:::: -12z2 + 6zs' The above procedure can be generalized to give a means of obtaining multiple-input pole placement.

8.3.

Given the system d 2 y/dt2 = O. Construct the observer system such that it has a double pole at -yo Then find the error e'= x - x as a function of y, if the output really is y(t) + 'J](t), where 'Y)(t) is noise. The observer system has the form d~ dt

=

(~

1)A -+ (kl)z

o

X

k

A [(1 O)x - y - 1]]

The characteristic equation of the closed loop system is A(A - k 1) - k2 = O. A double pole at -y has the characteristic equation A2 + 2YA + y2 = O. Hence set kl = -2y and k 2 :;:::: _y2. Then the equation for the error is

Note the noise drives the error and prevents it from reaching zero. The transfer function is found from

.e {(::)}

.e{1]}

82

+ 2ys + y2

(2YS+ y2) y 2s

As y ~ 0::>, then el ~ 1] and ez ~ drJldt. If 1](t) = 170 cos IUt, then 1U110. the amplitude of d11/dt, may be large even though '110 is small, because the noise may be of very high frequency. We conclude that it is not a good idea to set the observer system gains too high, so that the observer system can filter out some of the noise.

184 8.4.


[CHAP. 8

Given the discrete-time system x(n+ 1)

!) x(n) + (n urn),

= (_~

y(n)

=

(1 O)x(n)

Design a feedback controller given dominant closed loop poles in the z plane at (1 ± 1)/2. We shall construct an observer to generate the estimate of the state from which we can construct the desired control. The desired closed loop characteristic· equation is ",2 - X+ 1/2 = o. Hence we choose u = 3~/2 - 2~2. To generate ~l and ~2 we choose a first-order observer with a pole at 0.05 so that it will hardly affect the response due to the dominant poles and yet will be large enough to filter high-frequency noise. The transformation of variables

x=

(~ ~)z

z(n+1)

gives

(~-~)z(n) + (~)u(n),

=

yen}

(0 l)z(n)

Use of equation (8.8) gives P = (1 -0.05)

Then the observer is ""x

and

(-13 o1)""z

-_

where

and wen) is obtained from wen + 1) = O.05w(n) - 2.1525y(n) + u(n}.

8.5.

Find the sensitivity of the poles of the system

(-1; a' _~)x

~~ to changes in the parameter

a

where

lal

is small.

We denote the actual system as dXa/dt = Aaxa and the nominal system as dXN/dt = ANxN, where AN == Aa when a = O. In general, AN == Aa - SA where !lSAIl- is small since Ja\ is small. We assume AN has distinct eigenvalues xf so that we can always find a corresponding eigenvector Wi from ANWi == Wi. Denote the eigenvectors of A~ as vt, so that A~ vi = vt . Note the eigenvalues Xf are the same for AN and Taking the transpose gives

xf

xf

Air.

--

.I\i . Nvti

which we shall need later. Next we let the actual eigenvalues Xf = Xf into the eigenvalue equation for the actual Aa gives

ANWi

==

Xf Wi

and multiplying by

vt SA Wi + vT AN SWi

=

I~n\i

+ SXL.

Substituting this

(;\.f + SXi ) (Wi + 8Wi)

(AN + 8A)(wi + OWi) =

Subtracting

(8.18)

vr

(8.19)

gives

v1 Wi + xf v! OWi + v1 (OXi I -

SA) 8wi

Neglecting the last qua~tity on the right since it is of second order and using equation (8.18) then leads to

v! 8Awi VTWi

Therefore for the particular system in question, Ad.

=

-1 (

+ a2 2

ct.) -2

-1 0) ( 2 -2

+

(ai30 0ct.)

(8.20)

CHAP. 8]


185

Then for AN we find

(~)

-1

-2

(~)

W2

Using equation (8.20),

=

V1

2

-'1

+

(1 0) ( a0

-2

+

(-2 1)

aO·

V2

=

(~) (-~)

).(21)

(~2 ~)(~)

-2 -,-. 2a::

For larger values of a note we can use root locus under parameter variation to obtain exact values. However, the root locus is difficult computationally for very high~order systems, whereas the procedure just described has been applied to a 51st-order system.

8.6.

Given the scalar nonlinear system of Fig. 8-20 with input a sin t. Should K> 1 be increased or decreased to minimize the effect of the nonlinearity? y

a sin t

a sin t

+

+

Fig. 8·21

Fig. 8·20

The nominal linear system is shown in Fig. 8~21. The steady state value of eN = (a:: sin t)/(K - 1). We approximate this as the input to the unwanted signal d 2 (Sy)/dt2 == e~, which gives the steady state value of ay::;::: a3 (27 sin t - sin 3t)/[36(K -1)3]. This approximation that d 2 (oy)/dt2 = e~ instead of must be verified by simulation. It turns out this a good approximation for lal ~ 1, and we can conclude that for lal

Fig. 8~27

Reduce the block diagram to obtain HI isolated in a

Fig. 8-28

188 8.11.


[CHAP. 8

Determine whether the scalar feedback system of Fig. 8-29 in which yet) is related to e(t) as dx

dt

:::::

(~ _4~-1)X

+

y

d---+I

+

(t~2) e

t>O

y = (1 O)x (i) can follow a step input 0: with zero error, (ii) can follow a ramp input at with zero error.

Fig. 8-29

8.12.

Given the time-varying system d2y/dt 2 + aCt) dy/dt + f3(t)y ::::: u. Find a feedback control u such that the closed loop system behaves like d 2z/dt 2 + 8(t) dz/dt + ¢(t)z ::::: O.

8.13.

Given the system

-2) 1 1 o (

dx

-1

dt

1

1

x

::::: (0 1 -1)x

y

0-1

Construct a third-order observer system with poles at 0, -1 and -1. 8.14.


==

(1 1-2) o

1

-1

1

x

(~

Y

0-1

Construct a first-order observer system with a pole at -1. 8.15.


y = (1 O)x

Construct a first-order observer system with a pole at -3 and then find a feedback control u =: kl~l + k 2 '052 that places both the closed loop system poles at -2. 8.16.

Given the system dx

_1. 4

dt

(3 1) x 1

y

3

(1 1)x

Construct a first-order observer system with a pole at -4. 8.17.

Given the system dx/dt = A(t)x + B(t)u where y = C(t)x + D(t)u. What is the form of the observer system when D(t) ~ O? What is the algebraic separation theorem when D(t) ¥= O?

8.18.


=

_!(3

and

4 1

y = (0 1)x

where x(O) = o. As a measure of the sensitivity of the system, assuming jo:(t) 1 ~ 1, find ay(t) as an integral involving aCt) and u(t). Hint: It is easiest to use Laplace transforms. 8.19.

Given the nominal system dXN/dt = AN(t)xN + BN(t)UN and YN == CN(t)XN + DN(t)UN' What is the equation for 8x corresponding to equation (8.1.0 when the parameters of the system become Aa(t). Ba(t), Ca(t), Da(t) and ua(t)? .

8.20.

Given the system

dx/dt

== (

1

+ t + fet) f(t)

U(t») t2

x.

Choose u(t) such that at least one state

will be insensitive to small variations in f(t), given the nominal solution XN(t). 8.21.

Given that the input d(t) is generated by the scalar system dw/dt = a(t)W and d(t) = [yet) + p(t)Jw, under what conditions on pet) can the system dUJ/dt = c.:(t)x + f3(t)e with y = y(t)x follow any such d(t) with zero error?

CHAP. 8] 8.22.

189


Given the general nonunity feedback time-invariant system of Fig. 8-30. Under what conditions can lim e(t) = O? Set F = I and H;::; I and derive Theorem 8.2 from these conditions. t-co

+ .({y}

.e{d}--------.."l

J...----.,/'

.e{e}

Fig. 8-30 8.23.

In the proof of Theorem 8.1, why is d(t)

=

ft C(t)

IPA-BC Ct, '7") B(r) d(r) dT

+

get)

to

equivalent to

dw/dt

= [A(t) -

B(t) C(t)]w + B(t)d

and

d

+

C(t) q.A-BC(t, to) w(to)

= C(t)w + g?

8.24.

Show that if a time-invariant system is of type N, then it is of type N - k for all integers k such that 0 ~ k ~ N.

8.25.

Given the system of Fig. 8-31 where the constant matrix K has been introduced as compensation. Show that (a)

The type number of the system cannot change if K is nonsingular.

(b)

The system is of type zero if K is singular. d(t) )t

+

-

-

00

(3K -1 K

O

===

st

st

111I'(t, T)lh

~

=:

II~(T, to)lh dT

IlcJI(t, t o)I]1 dT

ft 111I'(t, T) cJI(T, t o)lh d1"

::::

~

111I'(t, to)lll (t-

to)

to

Define the time T =: {3ICB-1KE so that at time t =: to + T, {3K o - I K === 111II(t o + T, t o)111{3K B -lKE. Hence 114t(to + T, to)lb :::::: e- 1 , so that using the argument found in the last part of the proof just given for Theorem 9.3 shows that j]4a(t, t o)I]1 :::::: Kle-Ct-to)/T, which proves Theorem 9.5.

9.3.

Is the scalar linear system dx/dt = 2t(2 sin t -l)x uniformly asymptotically stable? We should expect something unusual because IACt)1 increases as t increases, and A(t) alternates in sign. We find the transition matrix as 0, then there is a T in to::= T::= tl such that "'fT) = 0 and w(t) > 0 for T ~ t == tp But using the mean value theorem gives ",(tl) :=:",(t 1) - weT) :=: (t l - T) dw/dt ~ 0, which is a contradiction. Assertion (ii):

Proof:

If d",/dt - a(t)",

~

0 and ",(to) "",. 0, then w(t):=: 0 for t::::: to.

Multiplying the given inequality by e-I} 0 if and only if the real parts of the eigenvalues of A are in the left half plane. Furthermore, v == xt(M-l)tM-1x decays as fast as is possible for a quadratic in the state. This is because the state vector decays with a time constant '1- 1 equal to one over the real part of the maximum eigenvalue of A, and hence the square of the norm of the state vector decays in the worst case with a time constant equal to 1/2'1. To investigate the time behavior of v, we find dv/dt = xtQx := -(M-1x)t(.A. + .4. t) (M-1x)

=== -2'l/(M- 1x)t(M- 1x)

=

-2'l]v

Fot a choice of M-1x equal to the unit vector that picks out the eigenvalue with real part becomes an equality and then v decays with time constant 1/2'1.

9.8.

'l},

this

In the study of passive circuits and unforced passive vibrating mechanical systems we often encounter the time-invariant real matrix equation FdZyldt 2 + Gdy/dt + Hy = 0 where F and H are positive definite and G is at least nonnegative definite. Prove that the system is stable Ls.L. Choose as a Liapunov. function v

dyT (F + FT) dy dt dt

+

r

Jo

t dyT (G

dT

This is positive definite since· F and H are, and v

TdYT)(H ( Y dt 0

+ GT) dy dT + dT

yT(H + HT)y

[CHAP. 9

STABILITY OF LINEAR SYSTEMS

206

which is a positive definite quadratic function of the state. dv dt

2Y

dyT [ d dt F dt2 +

dy G dt

+

By

]

+

[ d2Y F dt'},

Also,

+

dy G dt

+

J T

By

dy dt

==

o

By Theorem 9.6, the system is stable i.s.L. The reason this particular Liapunov function was chosen is because it is the energy of the system.


Given the scalar system dx/dt (a)

= tx + u

and y == x.

Find the impulse response h(t, to) and verify

ft [k(t, 'T)[

t2

dT :::::

yrI;; e

/

2

•

tlJ

(b) Show the system is not stable i.s.L. (c)

Explain why Theorem 9.5 does not hold.

9.10.

Given the system dx/dt = -x/t + u and y = x. (a) Show the response to a unit step function input at time to > 0 gives an unbounded output. (b) Explain why Theorem 9.5 does not hold even though the system is asymptotically stable for to > O.

9.11.

By altering only one condition in Definition 9.6, show how Liapunov techniques can be used to give sufficient conditions to show the zero state is not asymptotically stable.

9.12.

Prove that the real parts of the eigenvalues of a constant matrix A are < u if and only if given any symmetric, positive definite matrix Q there exists a symmetric, positive definite matrix P which is the unique solution of the set of n(n + 1)/2 linear equations -2uP + ATp + PA = -Q.

9.13.

Show that the scalar system dx/dt = -(1 + t)x is asymptotically stable for t theory.

9.14.

Consider the time~varying network of Fig. 9-4. Let x 1(t) == charge on the capacitor and X2{t) = flux in the inductor, with initial values XIO and X20 respectively. Then L{t) dXl/dt = x2 and L(t) C(t) dX2/dt + L(t)Xl + R(t) C(t)x2 = O. Starting with R + (2 L/RC) 1 ) pet) ( 2/R 1

===

0 using Liapunov

CCt)

L(t)

find conditions on R, L, and C that guarantee asymptotic stability.

Fig. 9-4

9.15.

Using the results of Example 9.16, show that if a > 0, 0 < fJ < 1 and a 2 > fJ2(a 2 + p.-2), then the Mathieu equation d2 y/dt2 + ady/dt + (1 + fJ cos 2t/p.)y = 0 is uniformly asymptotically stable.

9.16.

Given the time-varying linear system dx/dt = A(t)x with initial condition Xo where the elements of A(t) are continuous in t. Let H(t) be a symmetric matrix defined by B(t) = (A + AT)/2. Let Amin(t) and Amax(t) be, for each t, the smallest and the largest eigenvalue of H(t). Using the Liapunov function vex) =

XTX,

[[XO!!2 B 9.17.

show

it

Amin(T) d-r

-

to:::::

Ilx(t)112 :::::

IIxol12B

it

Jlmax(T)dT

to

What is the construction similar to that of Problem 9.7 for P in the discrete-time case ATPA-P = -Q?

CHAP. 9]

9.18.

STABILITY OF LINEAR SYSTEMS

207

Show that if the system of Example 9.22, page 200, is changed slightly to dxl/dt = X2 - €X1 + + x~ )/2 and dX2/dt = Xl - €x 2 + (xi + x~ )/2, where E> 0, then the system is uniformly asymptotically stable (but not globally).

(xi

9.19.

d

Given the system

-x

dt

Construct a Liapunov function for the system in the case o(t) = O. This Liapunov function must give the necessary and sufficient condition for stability, i.e. give the exact stability boundaries on a. Next, use this Liapunov function on the system where oCt) is not identically zero to find a condition on oCt) under which the system will always be stable. 9.20.

Given the system dx/dt = (A(t) + B(t»x where dx/dt = A(t)x has a transition matrix with norm 114I(t"r)11 :::::: e-K2Ct-r) • Using the Gronwall-Bellman inequality with p. = 0, show Ilxll:o::: Ilxolle{K3-K2)(t-to) if IIB(t)II:o::: IC3e-K2t.

9.21.

Show that Ilx(t)11 = IIxolle to

Jt IIACT)lIdr + ft eoT(t IlA(ll)lld1J B(r) U(T) dT

B(t)u with initial condition x(t o) = Xo.

for the system dx/dt = A(t)x +

to


9.10.

ft

(a)

it~~

(b)

¢(t, to) =

(e)

The system is not externally stable since (3 depends on t in Theorem 9.4.

(a)

yet)

(b)

The system is not uniformly asymptotically stable.

t2/2

Ih(t, T)I dT = e

==

2

2

et 12e-tol

e-~/2 dr ~ ~ et2/Z

-00

Z

(t - t~ /t)/2 for

t:=:: to

>

O.

9.11.

Change condition (4) to read dv/dt> 0 in a neighborhood of the origin.

9.12.

Replace A by A - uI in equation (9.5).

9.13.

Use v

9.14.

0 < Kl :0::: R(t) === IC2 < o < ICg == L(t) === /C4 < o < /Cs == C(t) === ICe

UT(T) R(T) U(T) dT

where R(T) is an m x m symmetric time-varying positive definite matrix to be fixed at the outset by the designer. R is positive definite to assure each element of u is bounded. This is the generalized control "energy", and will be added to p2(X, 0). The relative magnitudes of IIQ(-r)11 and IIR(T)II are in proportion to the relative values of the response and control energy. The larger IIQ(T)II is relative to IIR(T)II, the quicker the response and the higher the gain of the system.

10.3 DERIVATION OF THE OPTIMAL CONTROL LAW The mathematical problem statement is that we are given dx/dt = A(t)x + B(t)u and want to minimize v[x, uJ

(10.2)

Here we shall give a heuristic derivation of the optimal control and defer the rigorous derivation to Problem 10.1. Consider a real n-vector function of time p(t), called the costate or Lagrange multiplier, which will obey a differential equation that we shall determine. Note for any x and u obeying the state equation, pT(t)[A(t)x + B(t)u - dx/dt] = O. Adding this null quantity to the criterion changes nothing.

v[x, u] Integrating the term pTdx/dT by parts gives v [x,

where V1[X(t 1)]

=

=

u]

VI[X(t1)]

txT(t1)SX(t1) - XT(tl)P(tl)

St

1

v

2

[x] = v

3

+ v2 [x] + v3[U]

[U]:

(i xTQx

(10.3)

+ xTAp + x.T dp/dT) d.,

(10.4)

+ uTBTp) dT

(10.5)

rt'(!uTRU

J to

+ XT(to)p(to)

Introduction of the costate p has permitted v[x, u] to be broken into vi'v 2 and VS' and heuristically we suspect that v[x, u] will be minimum when VI'V 2 and Vs are each independently minimized. Recall from calculus that when a smooth function attains a local minimum, its derivative is zero. Analogously, we suspect that if VI is a minimum, then the gradient of (10.3) with respect to x(t 1) is zero: (10.6)

and that if x is zero:

v

2

is a minimum, then the gradient of the integrand of (10.4) with respect to dp/dt

=

_AT(t)p - Q(t)xop

(10.7)

CHAP. 10]

INTRODUCTION TO OPTIMAL CONTROL

211

Here XOP(t) is the response x(t) using the optimal control law UOP(x(t), t) as the input to the system (10.1). Consequently we define p(t) as the vector which obeys equations (10.6) and (10.7). It must be realized that these steps are heuristic because the minimum of v might not occur at the combined minima of VI' v 2 and v 3 ; a differentiable function need not result from taking the gradient at each instant of time; and also taking the gradient with respect to-x does not always lead to a minimum. This is why the final step of setting the gradient of the integrand of (10.5) with respect to u equal to zero (equation (10.8)) does not give a rigorous proof of the following theorem. Theorem 10.1: Given the feedback system (10.1) with the criterion (10.2), having the costate defined by (10.6) and (10.7). Then if a minimum exists, it is obtained by the optimal control (10.8)

The proof is given in Problem 10.1, page 220. To calculate the optimal control, we must find p(t). This is done by solving the system equation (10.1) using the optimal control (10.8) together with the costate equation (10.7), !i (XOP) ( A(t) -B(t)R- 1(t)B T(t»)(X OP ) dt

\ -Q(t)

p

_AT(t)

P

(10.9)

with xop(to) = Xo and p(tl) = SXOP(tl)' Example lOA. Consider the scalar time-invariant system dx/dt = 2x + u with criterion X 2 (t 1 )

v

Then A = 2, B = 1, R = 1/4, Q = 3, S = 2.

+~

f

1

(3x 2

+ u 2/4) dt

o Hence we solve

-4)(XOP)

2 ( -3 -2

with xop(O) = x o, p(l) = 2x(1).

(X;;g»)

p

U sing the methods of Chapter 5,

=

[e~4t (!

P(O») :) + e~4t (_! -4)](X 2 p(O) O

(10.10)

Evaluating this at t = 1 and using p(l) = 2x op (1) gives p(O)

15e+ 8 + 1 10e+8 - 2

Then from Theorem 10.1, 2e+ 4t

(10.11)

Xo

+ 30e8 - 4t

1 - 5e+8

Xo

(10.12)

Note this procedure has generated u OP as a function of t and Xo. If Xo is known, storage of UOP(xo, t) as a function of time only, in the memory of a computer controller, permits open loop control; i.e. starting at to the computer generates an input time-function for dxop/dt = Ax°P + BuOP(xo, t) and no feedback is needed. However, in most regulator systems the initial conditions Xo are not known. By the introduction of feedback we can find a control law such that the system is optimum for arbitrary Xo. For purposes of illustration we will now give one method for finding the feedback, although this is not the most efficient way to proceed for the problem studied in this chapter. We can eliminate Xo from dxop/dt = Axop + BuoP(xQ, t) by solving for XOP(t) in terms of Xo and t, i.e. XOP(t) = .,,(t; Xo, to) where q, is the trajectory of the optimal system. Then XOP(t) = ,p(t; Xo, to) can be solved for Xo in terms of XOP(t) and t, Xo = xo(XOP(t), t). Substituting for Xo in the system equation then gives the feedback control system dxop/dt =

AxoP + BuOP(xOP(t), t).

212


[CHAP. 10

Example 10.5. To find the feedback control for Example 10.4, from (10.10) and (10.11) (5eS-4t _ e + 4t)

Xo

1

5e+ 8 -

Solving this for

Xo

and substituting into (10.12) gives the feedback control 2

+ 30e 8 (1-t)

1 - 5e8(1-t) XOp(t) Hence what goes in the box marked

* in Fig. 10-1 is the time-varying 2 + BOeS(l-t) K(t)

=:!

and the overall closed loop system is

1-

gain element K(t) where

5880 - 0

4 + 20e80 - O -1------::5,...-,88=(::-l----;-t~) ::cop

10.4 THE MATRIX RICCATI EQUATION To find the time-varying gain matrix K(t) directly, let p(t) = P(t) XOP(t), Here P(t) is an n X n Hermitian matrix to be found. Then UOP(x, t) = -R-1BTpxop so that K = -R-IBTP. The closed loop system then becomes dxop/dt = (A - BR-IBTP)XOP , and call its transition matrix qJcl(t, i). Substituting p = PXOP into the bottom equation of (10.9) gives (dP/dt)xOP

+ P dxoP/dt =

-QxoP -

ATPXOP

Using the top equation of (10.9) for dxoP/dt then gives

o=

(dP/dt

+ Q + ATp + PA -

PBR-IBTp)xOP

But XOP(t) = ~el(t, to)xo, and since Xo is arbitrary and lite! is nonsingular, we find the n x n matrix P must satisfy the matrix Riccati equation (10,13)

This has the "final" condition P(tl) = S since P(tl) = P(tl) X(tl) = SX(tl). Changing independent variables by T = tl - t, the matrix Riccati equation -becomes dP/dT = Q + ATp + PA - PBR-lBTp where the arguments of the matrices are tl - T instead of t. The equation can then be solved numerically on a computer from T = 0 tOT = tl - to, starting at the initial condition P(O) = S. OccaSionally the matrix Riccati equation can also be solved analytically. Example 10,6. For the system of Example 10.4, the 1 X 1 matrix Riccati equation is

-dP/dt == 3 with P(l) = 2.

+ 4P -- 4P2

Since this is separable for this example,

f

t tl

dP 4(P - 3/2)(P

1

P(t) -- 3/2

+ 1/2) == "8 In P(tl) -- 3/2

Taking antilogarithms and rearranging, after setting tl

pet)

==

=:!

1 P(t) + 1/2 - SIn P(t1) + 1/2

1, gives

15 + e8(t-U 10 -- 2e8(t-l)

Then K(t) = -R-1BTP = -4P(t), which checks with the answer obtained in Example 10.5.

CHAP. 10]


213

Another method of solution of the matrix Riccati equation is to use the transition matrix of (10.9) directly. Partition the ttansition matrix as ~ (~l1(t, i) at ~21(t, T)

~12(t, T)) ~22(t, T)

(A

;: :

-Q

-BR-IBT)(~l1(t, i) -AT IP 21 (t, T)

T))

~12(t, 0

BR-IBTnx)TIIx = -xTQx - xTnBR-lBTnx

0 because Q is positive definite and nBR-tBTn is nonnegative definite.

< 0

215


CHAP. 10]

Since there are in general n(n + 1) solutions of (10.16), it helps to know that TI is the only positive definite solution.

Theorem 10.6: The unique positive definite solution of (10.16) is n. Proof:

Suppose there exist two positive definite solutions TIl and II2.

Then

o

Q-Q

(10.17)

Recall from Problem 9.6, that the equation XF + GX = K has a unique solution X whenever Ai(F) + Aj(G) =F 0 for any i and j, where Ai(F) is an eigenvalue of F and "AG) is an eigenvalue of G. But A - BR~lBTTIi for i = 1 and 2 are stability matrices, from Theorem 10.5. Therefore the sum of the real parts of any combination of their eigenvalues must be less than zero. Hence (10.17) has the unique solution III - ll2 = O. Generally the equation (10.16) can be solved for very high-order systems, n ~ 50. This gives another advantage over classical frequency domain techniques, which are not so easily adapted to computer solution. Algebraic solution of (10.16) for large n is difficult, because there are n(n + 1) solutions that must be checked. However, if a good initial guess P(t2) is available, i.e. P(t2) = 8P(t2) + II where 8P(t2) is small, numerical solution of (10.13) backwards from any t2 to a steady state gives II. In other words, 8P(t) = P(t) - n tends to zero as t -? -00, which is true because:

Theorem 10.7: If A, B, Rand Qare constant matrices, and R, Q and the initial state P(t2) are positive definite, then equation (10.13) is globally asymptotically stable as t -00 relative to the equilibrium state II. --,)0

Proof:

BP obeys the equation (subtracting (10.16) from (10.13)) -d8P/dt = F T8P

+ 8P F

-8P BR-IBT 8P

where F:;:::: A - BR-IBTn. Choose as a Liapunov function 2V = tr[8PT (Iln T)-18P]. For brevity, we investigate here only the real scalar case. Then 2F = -Qrr- 1 - B2R- 1rr and 2V = rr- 2 SP2, so that dV /dt = n~2 sP d8P/dt = n- 2 Sp2 (Qrr- l + B2R~IP) > 0 for all nonzero sP since Q, R, P and II are all > O. Hence V 0 as t -00, and SP O. It can be shown in the vector case that dV/dt > 0 also. --,)0

--,)0

--,)0

Example 10.10. For the system of Example 10.8, consider the deviation due to an incorrect initial condition, i.e. suppose P(t l ) = € instead of the correct value P(t l ) = O. Then P(to) = tanh (tl - to + tanh- l €). For any finite €, lim P(t o) = 1 = II. to- -00

Experience has shown that for very high-order systems, the approximations inherent in numerical solution (truncation and roundoff) often lead to an unstable solution for bad initial guesses, in spite of Theorem 10.7. Another method of computation of n will now be investigated.

Theorem 10.8: If Ai is an eigenvalue of H, the constant 2n x 2n matrix corresponding to (10.9), then -xf is also an eigenvalue of H. Denote the top n elements of the ith eigenvector of H as fi and the bottom n elements as gi. Then Proof:

Ai

(!:)

(10.18)

=

or \gi

or

-AI

Ct)

=

-Qfi

-

ATgi

(-B::'B :::~)(~~,) T

216

[CHAP~'


10

This shows -Ai is an eigenvalue of HT. Also, since detM = (detMT)* for any matrix M, then for any ~ such that det (H - ~I) = 0 we find that det (HT - ~I) = O. This shows that if ~* is an eigenvalue of HT, then ~ is an eigenvalue of H. Since -Ai is an eigenvalue of HT, it is concluded that -At is an eigenvalue of H. This theorem shows that the eigenvalues of H are placed symmetrically with regard to the jw axis in the complex plane. Then H has at most n eigenvalues with real parts < 0, and will have exactly n unless some are purely imaginary. Example 10.11. The H matrix corresponding to Example 10.8 is H = which are symmetric with respect to the imaginary axis.

(-10-1). 0

This has eigenvalues +1 and -1,

Example 10.12. A fourth-order system might give rise to an 8 X 8 H having eigenvalues placed as shown in Fig. 10-2. 1m].. x

x

---~--------~--~--~~-----~~-ReA

x

x

Fig. 10-2

The factorization into eigenvalues with real parts < 0 and real parts> 0 is another way of looking at the Wiener factorization of a polynomial p{(2) into p+(w) p-(w). Theorem 10.9: If AI, A2, ••. , An are n distinct eigenvalues of H having real parts < 0, then n = GF-1 where F = (£1 j £21 ... j in) and G = (gtl g2j ... Ign) as defined in equation (10.18). Proof: Since n is the unique Hermitian positive definite solution of (10.16), we will show first that GF-1 is a solution to (10.16), then that it is Hermitian, and finally that it is positive definite.

Define an n X n diagonal matrix A with AI, A2, •.• , An its diagonal elements so that from the eigenvalue equation (10.18), , A ( -Q

from which FA GA

=

-BR-IBT)(F) _AT G

AF - BR-IB TG

(10.19)

-QF - ATG

(10.20)

Premultiplying (10.19) by F-l and substituting for A in (10.20) gives GF-1AF - GF-IBR-IBTG = -QF - ATG

Postmultiplying by F-l shows GF-1 satisfies (10.16). Next, we show GF-l is Hermitian. It suffices to show M = FtG is Hermitian, since GF-l = Ft- 1MF-l is then Hermitian. Let the elements of M be mjk = gk; for j =F k,

ff

CHAP. 10]

217


~)(!:) + (~)t (_~ ~) AkG:) } ~) Since the term in braces equals 0, we have m jk

= mk.j

+

(_~ ~)H }(!:)

and thusGF-I is Hermitian.

Finally, to show GF-I is positive definite, define two n x n matrix functions of time as Then 8(co) = 0 = ,,(co). Using. (10.19) and (10.20), d8/dt (AF - BR-1BTG)eAtF-I

8(t) = FeAtF-I and ,,(t) = GeAtF-I.

d,,/dt = -(QF + ATG)eAtF-l

Then GF-I so that

i«)

9 t (0) ,,(0) (eAtF-l)t(FtQF

+ GtBR-IBTG)(eAtF-l) dt

Since the integrand is positive definite, GF-l is positive definite, and GF-l = II. Corollary 10.10: The closed loop response is x(t) = FeACt-to)F-lxo with costate p(t) GeACt-to) F-l Xo. The proof is similar to the proof that GF-l is positive definite. Corollary 10.11: The eigenvalues of the closed loop matrix A - BR-l BTU are AI, A2, ••• , An and the eigenvectors of A - BR-l BTn are fl' f2, ... , f n • The proof follows immediately from equation (10.19). Furtherlnore, since AI, A2, •.• , An are assumed distinct, then from Theorem 4.1 we know f1, f2, ... , fn are linearly independent, so that F-l always exists. Furthermore, Theorem 10.5 assures Re Ai < 0 so no "-i can be imaginary. Example 10.13. The H matrix corresponding to the system of Example 10.8 has eigenvalues -1 Corresponding to these eigenvalues,

= Al

and +1 = A2'

and where a and f3 are any nonzero constants. Usually, we would merely set a and {3 equal to one, but for purposes of instruction we do not here. We discard A2 and its eigenvector since it has a real part > 0, and form F;:::: It = a and G = 01 = a. Then II = GF-l = 1 because the a's cancel. From Problem 4.41, in the vector case F = FoK and G = GoK where K is the diagonal matrix of arbitrary constants associated with each eigenvector, but still D = GF-I = G oK(F oK)-l = GoFo.

Use of an eigenvalue and eigenvector routine to calculate n from Theorem 10.9 has given results for systems of order n ~ 50. Perhaps the best procedure is to calculate an approximate 110 = GF-l using eigenvectors, and next use Re (no + n~)/2 as an initial guess to the Riccati equation (10.13). Then the Riccati equation stability properties (Theorem 10.7) will reduce any errors in 110, as well as provide a check on the eigenvector calculation. 10.6 OUTPUT FEEDBACK Until now we have considered only systems in which the output was the state, y = Ix. For y = Ix, all the states are available for measurement. In the general case y = C(t)x, the states must be reconstructed from the output. Therefore we must assume the· observahility of the closed loop system dx/dt = F(t)x where F(t) = A(t) - B(t) R-I(t) BT(t) P(t; tl).

218


[CHAP. 10

To reconstruct the state from the output, the output is differentiated n -1 times. y

=

C(t)x

=

Nl(t)X

dy/dt = Nl(t)dx/dt

+ dN /dtx 1

d n- 1y/dtn- 1 = (Nn - 1F

= (NIF+dNddt)x = N2x

+ dNn-ddt)x

= Nnx

where NT = (Ni I ..• IN~) is the observability matrix defined in Theorem 6.11. Define a nk x k matrix of differentiation operators H(d/dt) by H(d/dt) = (I Ild/dt I ..• \Idn-1/dtn - 1). Then x = N-I(t) H(d/dt)y. Since the closed loop system is observable, N has rank n. From Property 16 of Section 4.8, page 87, the generalized inverse N-r has rank n. Using the results of Problem 3.11, page 63, we conclude the n-vector x exists and is uniquely determined. The optimal control is then u = R-1BTPN-IHy. Example 10.14. Given the system y = (1 O)x

with optimal control u = k1xl output is u = k1y + k2 dy/dt.

+ k2X2'

Since y =

Xl

and dx/dt =

x

System (10.1)

I I

C

X2,

the optimal control in terms of the

I

y

I

u

State estimator

-R-IBTp

Fhr.l0-3

Since u involves derivatives of y, it may appear that this is not very practical. This mathematical result arises because we are controlling a deterministic system in which there is no noise, i.e. in which differentiation is feasible. However, in most cases the noise is such that the probabilistic nature of the system must be taken into account. A result of stochastic optimization theory is that under certain circumstances the best estimate of the state can be used in place of the state in the optimal control and still an optimum is obtained (the "separation theorem"). An estimate of each state can be obtained from the output, so that structure of the optimal controller is as shown in Fig. 10-3. 10.7 THE SERVOMECHANISM PROBLEM Here we shall discuss only servomechanism problems that can be reduced to regulator problems. We wish to find the optimum compensation to go into the box marked * * in Fig. 10-4.

o

-e

** L

>1

System (10.1)

x; I c

y

d---+:~~~~~~!~1~.~~~~1~·~~~11 Fig.l0~4.

U

The Servomechanism Problem

;>

219


CHAP. 10]

The criterion to be lllinimized is

r 2J 1

+ -

leT(tl) Se(tl)

1'1

[eT(T) Q(T) e(T)

+

UT(T) R(T) U(T)] dT

(10.21)

to

Note when e(t) = y(t) - d(t) is minimized, y(t) will follow d(t) closely. To reduce this problem to a regulator, we consider only those d(t) that can be generated by arbitrary z(to) in the equation dz/dt A(t)z, d C(t)z. The coefficients A(t) and C(t) are identical to those of the open loop system (10.1).

=

Example 10.15. GiYen the open loop system

=

~o ~)x + (:~)u

dx dt

y = (1 1 l)x

b3

-2

Then we consider only those inputs d(t) = ZlO + Z20(t + 1) + zsoe- 2t • In other words, we can consider as inputs only ramps, steps and e- 2t functions and arbitrary linear combinations thereof.

Restricting d(t) in this manner permits defining new state variables w dw/dt

=

A(t)w

+ B(t)u

e = C(t)w

=x -

z.

Then (10.22)

Now we have the regulator problem (10.22) subject to the criterion (10.21), and solution of the matrix Riccati equation gives the optimal control 1:1S u = -R-IBTPW. The states w can be found from e and its n -1 derivatives as in Section 10.6, so the content of the box marked * * in Fig. 10-4 is R-IBTPN-IH. The requirement that the input d(t) be generated by the zero-input equation dz/dt = A(t)z has significance in relation to the error constants discussed in Section 8.3. Theorem 8.1 states that for unity feedback systems, such as that of Fig. 10-4, if the zero output is asymptotically stable then the class of inputs d(t) that the system can follow (such that lim e(t) = 0) is generated by the equations dw/dt = A(t)w + B(t)g and d(t) = C(t)w + g t-CQ where g(t) is any function such that lim g(t) = O. Unfortunately, such an input is not t ...... «>

reducible to the regulator problem in general. However, taking g == 0 assures us the system can follow the inputs that are reducible to the regulator problem if the closed loop zero output is asymptotically stable. Restricting the discussion to time-invariant systems gives us the assurance that the closed loop zero output is asymptotically stable, from Theorem 10.5. If we further restrict the discussion to inputs of the type d i = (t - to)lU(t - to)ei as in Definition 8.1, then Theorem 8.2 applies. Then we must introduce integral compensation when the open loop transfer function matrix H(s) is not of the correct type l to follow the desired input. Example 10.16. Consider- a system with transfer function G(s) in which G(O) -# i.e. it contains no pure integrations. We must introduce integral compensation, lis. Then the optimal servomechanism to follow a step input is as shown in Fig. 10-5 where the box marked ** contains R-1bTIIN-IH(s). This is a linear combination of 1, s, ... , sn since the overall system G (s) lsi s of order n + 1. Thus we can write the contents of ** as kl + k2 s + ... + kns n• The compensation for G(s) is then k1/s + k2 + kas + ... + kns n- 1. In a noisy environment the differentiations cannot be realized and are approximated by a filter, so that the compensation takes the form of integral plus proportional plus a filter. OX) ,

9Yo

d(t)

~ ~L. -_e_(t_)_a_'_K_:~~~~_U~(~t)~~.~'_!__'''''_-_-_-Y:r-) -step

___

Fig. 10-5

220

[CHAP. 10


10.8 CONCLUSION In this chapter we have studied the linear optimal control problem. For time-invariant systems, we note a similarity to the pole placement technique of Section 8.6. We can take our choice as to how to approach the feedback control design problem. Either we select the pole positions or we select the weighting matrices Q and R in the criterion. The equivalence of the two methods is manifested by Corollary 10.11, although the equivalence is not one-to-one because analysis similar to Problem 10.8 shows that a control u = kTx is optimal if and only if Idet {jwI - A - bkT ) I ::::""Idet (jwI - A)I for all w. The dual of this equivalence is that Section 8.7 on observers is similar to the Kalman-Eucy filter and the algebraic separation theorem of Section 8.8 is similar to the separation theorem of stochastic optimal control.

Solved Problems 10.1. Prove Theorem 10.1, page 211, and Theorem 10.2; page 213. The heuristic proof given previously was not rigorous for the following six reasons. (1) By minimizing under the integral sign at each instant of time, i.e. with t fixed, the resulting minimum is not guaranteed continuous in time. Therefore the resulting optimal functions XOp(t) and pet) may not have a derivative and equation (10.7) would make no sense. (2)

The open loop time function u(t) was found, and only later related to the feedback control u(x, t).

(3)

We wish to take piecewise continuous controls into account.

(4)

Taking derivatives of smooth functions gives local maxima, minima, or inflection points. We wish to guarantee a global minimum.

(5)

We supposed the minimum of each of the three terms

(6)

We said in the heuristic proof that if the function were a minimum, then equations (10.6), (10.7) and (10.8) held, i.e. were necessary conditions for a minimum. We wish to give sufficient conditions for a minimum: we will start with the assumption that a certain quantity vex, t) obeys a partial differential equation, and then show that a minimum is attained.

PI' 1'2,

and

P3

gave the minimum of v.

To start the proof, call the trajectory x(t} = +(t; x(to), to) corresponding to the optimal system = Ax + Buop(x, t) starting from to and initial condition x(t o). Note ~ does not depend on u since u has been chosen as a specified function of x and t. Then p[x, uOP(x, t)] can be evaluated if cf. is known, simply by integrating out t and leaving the parameters t l , to and x(to). Symbolically,

dx/dt

p[x, UOP]

;:;:::

icf.T(tl; x(t o), to)SI/>(t 1 ; x(to), to)

+ ~ ft!

[4>T(T; x(t o), to) Q(T) 4>(T; x(to), to)

+

(UOP)T(I/>, T)RuoP(~, T)] dT

to

Since we can start from any initial conditions x(to) and initial time to, we can consider v[x, u Op] + 1 variables x(to) and to, depending on the fixed parameter t l • Suppose we can find a solution vex, t) to the (Hamilton-Jacobi) partial differential equation

v(x(to), to; t 1 ) where v is an explicit function of n

:~ + txTQx - t(grad x v) TBR -lBT gradx v + (grad x v ) Tax = 0

(10.23)

with the boundary condition V(X(tl)' t l ) ::::::: txT(tl) Sx(t 1). Note that for any control u(x, t), !uTRu + l(grad x v)TBR-IBT grad x v + (gradx v)TBu ;:;::: t(u + R-IBT gradx'I.')TR(u + R-IBT gradxv) :::: 0 where the equality is attained only for u(x, t) ;:;::: -R-'-lBT gradx v

(10.24)

Rearranging the ineqUality and adding lxTQx to both sides, }XTQx + iuTRu ~ }XTQX - !(gradxv)TBR-IBT gradxv -

(gradxv)TBu

Using (10.23) and dx/dt = Ax + Bu, we get iXTQx

+ iuTRu ~

-

[~~ + (gradx v)T(Ax + Bu) ]

= -

fft v(x(t), t}


CHAP. 10]

221

Integration preserves the inequality, except for a set of measure zero:

Given the boundary condition on the Hamilton-Jacobi equation (10.23), the criterion v[x, u] for any control is

so that if v(x(t o), to) is nonnegative definite, then it is the cost due to the optimal control, (10.2 .0: uOP(x, t) = -R-l(t) BT(t) gradxv(x, t) To solve the Hamilton-Jacobi equation (10.23), set vex, t) = txTP(t)x where pet) is a timevarying Hermitian matrix to be found. Then grad x v = P(t)x and the Hamilton-Jacobi equation reduces to the matrix Riccati equation (10 . 13). Therefore the optimal control problem has been reduced to the existence of positive definite solutions for t < tl of the matrix Riccati equation. Using more advanced techniques, it can be shown that a unique positive definite solution to the matrix Riccati equation always exists if I!A(t)!! < 00 for t < t l • We say positive definite because

and all nonzero x(to), which proves Theorem 10.2. Since we know from Section 10.4 that the matrix Ricca-ti equation is equivalent to (10.9), this proves Theorem 10.1.

10.2. Find the optimal feedback control for the scalar time-varying system dx/dt = - x/2t + U

'" It t 0 mInImIze

v

= "21

it!

(x 2 + u 2 )dt.

to

Note the open loop system has a transition matrix 0 for 0 < t < tl and for t < tl ~ 0.. However, for -t1 < t < 0 < t l , the interval in which the open loop system escapes, PCt) < O. Hence we do not have a nonnegative solution of the Riccati equation, and the control is not the optimal one in this interval. This can only happen when IIACt) II is not bounded in the interval considered.

= y3e. An open loop control d(t) = -t has been found (perhaps by some other optimization scheme) such that the nonlinear system dynldt = y~d with initial condition yn(O) = 1 will follow the nominal path Yn(t) = (1 + t2)-1I2. Unfortunately the initial condition is not exactly one, but y(O) = 1 + € where € is some very small, unknown number. Find a feedback control that is stable and will minimize the error y(tl) - Yn(t l ) at some time tl.

10.3. Given the nonlinear scalar system dy/dt

Call the error y(t) - YnCt) = xCt) at any time. The equations corresponding to this system are

dy dt

Assume the error

dYn dt

Ixl

~

+

IYI

dx dt

=

(Yn+ x)3

Consider the feedback system of Fig. 10-6 below.

Cd +u)

lui ~ Idl so that !3Ynx2d + x 3 d + 3y;xu + 3Ynx2 u + x 3 ul

and corrections

13y;xd + y!u! ?

(10.25)

222


(CHAP. 10

y(t)

d(t)

+ x(t)

Fig. 10-6

Since dy"Jdt == y!d, we have the approximate relation 3t

dx

at We choose to minimize

=

- 1

+~

i

+ t2 x +

1

(1

+ t 2)3/2 U

tl R(t) u 2 (t) dt o where R(t) is some appropriately chosen weighting function such that neither x(t) nor u(t) at any time t gets so large that the inequality (10.25) is violated. Then K(t) = -P(t)/[(1 + t 2 )3J2R(t)1 where pet) is the solution to the Riccati equation

v

!X 2 (t l )

=

dP

6tP

-ill

=

-1

p2

+ t2

(1

-

+ t2)3R(t)

with P(t 1) = 1. This Riccati equation is solved numerically on a computer from t1 to O. the time-varying gain function K(t) can be found and used in the proposed feedback system.

10.4.

Then

Given the open loop system with a transfer function (8 + £1')-1. Find a compensation to minimize

= ~ fo"> [(y -

d O)2

+ (du/dt)2] dt

where do is the value of an arbitrary step input. Since the closed loop system must follow a step, and has no pure integrations, we introduce integral compensation. The open loop system in series with a pure integration is described by dX2/dt = - ax2 + u, and defining Xl = u and du/dt = p. gives dXl/dt = du/dt = IL so that dx

y = (0 1)x

dt

Since an arbitrary step can be generated by this, the equations for the error become dw dt

subject to minimization of

v

1

reo

2" In

o

(~ ~ )

o

e (e

2

+ (~

+ p.2) dt.

_~) II

~

(0 1)w

The corresponding matrix Riccati equation is

+

II

G_:) -

II

(~ ~) II

Using the method of Theorem 10.9 gives H

y

A (

+ 0: 1

~)

'Where V2 Y = ~ a + V0: - 4 + V0:4 - 4 and A-I = 0:2 + ya + 1. The optimal control is then JL = 4(wl + yw 2 ). Since e = w 2 and WI = o:e + de/dt, the closed loop system is as shown in Fig. 10-7 below. 2

4

-Va2 -

223


CHAP. 10]

+'z

-e

S t e p - - -......

·1

~(y

t

+ a + 8)

fJ-

·1

~

1

8(8 + a)

y

I

Fig. 10-7

Figure 10-8 is equivalent to that of Fig. 10-7 so that the compensation introduced is integral plus proportional.

Step

+

~(l+a;y)

1 8+a:

-

Fig. 10-8

10.5. Given the single..,input time-invariant system in partitioned Jordan canonical form d (

dt

zc) z:;-

=

where Zc are the j controllable states, Zd are the n - j uncontrollable states, J e and J d are j x j and (n-j) x (n- i) matrices corresponding to real Jordan form, Section 7.3, bisa j-vector that contains no zero elements, U is ascalar control, and the straight lines indicate the partitioning of the vectors and matrices. Suppose it is desired to minimize the quadratic criterion, v where

v

= ~ foo [z~Qczc + p-1U2JdT.

Here

p>

0

to

is a scalar, and Qc is a positive definite symmetric real matrix. Show that the optimal feedback control for this regulator problem is of the form u(t) = -kTzc(t) where k is a constant j-vector, and no uncontrollable states Zd are fed back. From the results of this, indicate briefly the conditions under which a general time-invariant singleinput system that is in J014dan form will have only controllable states fed back; i.e. the conditions under which the optimal closed loop system can be separated into controllable and uncontrollable parts. The matrix II satisfies equation (10.16). Q

(

··Qoc

I 0°),

R-l --

p

and partition

For the II

Now if ne can be shown to be constant and (10.16),

==

=

ncd

~se g:en,

(~). II~d

I lId

A =

C' I;)

B = (:),

Then

can be shown 0, then kT == pbTnc'

But from

ICHAP.IO


224 or

Within the upper left-hand partition,

o

+ DeJe -

= JeDe

pDebbTDe

+ Qc

the matrix Riccati equation for the controllable system, which has a constant positive definite solution Dc. Within the upper right-hand corner,

o=

JeDed

+ DedJd -

pDcbbTDed

+0

This is a linear equation in Ded and has a solution Ded = 0 by inspection, and thus u

= -pbTUczc•

x

N ow a general system = Ax + Bu can be transformed to real Jordan form by x = Tz, where T-1AT = J. Then the criterion v

;:;:::

If~ (xTQx + p-1u Z) dr -2

-If~ (zTTTQTz + p- 1U 2 ) dr 2 t

=

to Zd

(* ' Q

0

gives the matrix Riccati equation shown for z, and

if TTQT =

0)

i.e. only the con-

trollable states are weighted in the criterion. Otherwise uncontrollable states must be fed back. If the uncontrollable states are included, v diverges if any are unstable, but if they are all stable the action of the controllable states is influenced by them and v remains finite.

10.6. Given the controllable and observable system

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

=

C(t)x (10.26) It is desired that the closed loop response of this system approach that of an ideal or model system dw/dt = L(t)w. In other words, we have a hypothetical system dw/dt = L(t)w and wish to adjust u such that the real system dx/dt = A(t)x + B(t)u behaves in a manner similar to the hypothetical system. Find a control u such that the error between the model and plant output vector derivatives becomes small, i.e. minimize y y v (10.27) 2 to (it - Ly Q dt - Ly ) + u1'Ru] dt dx/dt

1It! [(d

)T

y

(d

Nate that in the case A, L = constants and B = C = I, R == 0, this is equivalent to asking that the closed loop system dx/dt = (A + BK)x be identical to dx/dt = Lx and hence have the same poles. Therefore in this case it reduces to a pole placement scheme (see Section 8.6). Substituting the plant equation (10.26) into the performance index (10.27), V

=

2"1 jtt {[(C· + CA -

to

LC)x

+ CBu]TQ[(C. + CA -

LC)x

+- CBu] + uTRu} dt

(10.28)

This performance index is not of the same form as criterion (10.2) because cross products of x and u appear. However, let 11 = u + (R + BTQB) -1 BTCTQ( dCI dt + CA - LC)x.Since R is positive definite, zTRz > 0 for a~y nonzero z. Since BTQB is nonnegative definite, 0 < zTRz + zTBTQBz = zT(R + BTQB)z so that R = R + BTQB is positive definite and hence its inverse always exists. Therefore the control 11 can always be found in terms of u and x. Then the system (10.26) becomes dx/dt =

and the performance index (10.28) becomes V

=

2"1

It!

Ax + BI1 A

(xTQx

(10.29)

+ uTRu) dt A

AA

(10.30)

to

in which, defining M(t) = dC/dt + CA -LC and K(t) = (BTQB + R)-lBTCTQM,

A

A

=

A-BK,

Q = CT[(M - BK)TQ(M - BK)

+

KTRK]C

A

and

R = R

+ BTQB

CHAP. 10]

INTRODUCTION TO OPTIMAL CONTROL A

225

A

Since R has been shown positive definite and Q is nonnegative definite by a similar argument, the regulator problem (10.29) and (10.30) is in standard form. Then 11 = -R-lB'IJ'x is the optimal solution to the regulator problem (10.29) and (10.30), where P is the positive definite solution to A

A

-dP/dt = Q + ATP

with boundary condition P(tl ) = O.

A

+ PA -

A

PBR-1B'IJ'

The control to minimize (10.27) is then u = -R-lBT(P

+ BTCTQM)x

A

Note that cases in which Q is not positive definite or the system (10.29) is not controllable may give no solution P to the matrix Riccati equation. Even though the conditions under which this procedure works have not been clearly defined, the engineering approach would be to try it on a particular problem and see if a satisfactory answer could be obtained.

10.7. In Problem 10.6, feedback was placed around the plant dx/dt = A(t)x + B(t)u, y = C(t)x so that it would behave similar to a model that might be hypothetical. In this problem we consider actually building a model dw/dt = L(t)w with output v = J(t)w and using it as a prefilter ahead of the plant. Find a control u to the plant such that the error between the plant and model output becomes small, Le. minimize (10.31) Again, we reduce this to an equivalent regulator problem. be written as

The plant and model equations can (10.32)

and the criterion is, in terms of the (w x) vector, v

~

s,:T(:Y(-J

C)TQ(-J C)

G) + Jat uTRu

Thus the solution proceeds as usual and u is a linear combination of wand x. However, note that the system (10.32) is uncontrollable with respect to the respect to the w variables. In fact, merely by setting L = 0 and using J(t) to generate the input, a form of general servomechanism problem results. The conditions under which the general solution to the servomechanism problem exists are not known. Hence we should expect the solution to this problem to exist only under a more restrictive set of circumstances than Problem 10.6. But again, if a positive definite solution of the corresponding matrix Riccati equation can be found, this procedure can be useful in the design of a particular engineering system. The resulting system can be diagrammed as shown in Fig. 10-9.

v(t)

x(t)

Feedback Fig. 10·9

y(t)

226


[CHAP.lO

10.8. Given the time-invariant controllable and observable single input-single output system dx/dt = Ax + bu, y = cTx. Assume that u = kTx, where k has been chosen such that 2v

=

i

oo

(qy2 +u2) dt has been minimized.

Find the asymptotic behavior of

the closed loop system eigenvalues as q --? 0 and q --?

00.

Since the system is optimal, k = -bTn where n satisfies (10.16), which is written here as

o=

qccT

+ ATD + nA -

nbbTn

(10.33)

If q = 0, then n = 0 is the unique nonnegative definite solution of the matrix Riccati equation. Then u = 0, which corresponds to the intuitive feeling that if the criterion is independent of response, make this control zero. Therefore as q ~ 0, the closed loop eigenvalues tend to the open loop eigenvalues.

To examine the case q""" 00, add and subtract sn from (10.33), multiply on the right by (sl - A)-l and on the left by (-81 - AT)-l to obtain

o =

q(-sl - AT)-lccT(sl - A)-l -

nCsI - A)-l -

(-sl - AT)-ln -

(-81 - AT)-lnbbTn(sI - A)-l (10.34)

Multiply on the left by b T and the right by h, and call the scalar G(8);:= cT(sl - A)-lb and the scalar H(s) = +bTlI(sI - A)-lb. The reason for this notation is that, given an input with Laplace transform pCs) to the open loop system, x(t) = .,c-l{(sl - A)-lbp (s)} so that ..e{y(t)} = cT~{x(t)} =

and

cT(sI - A) -lbp(s) = G(s) pes)

-.,c{u(t)} = bTlI~{xCt)} ==

bTlI(sl - A)-lbp(s) = H(s) pes)

In other words, G(s) is the open loop transfer function and H(s) is the transfer function from the input to the control. Then (10.34) becomes

o =

qG(-s) G(s) - H(s) - H(-s) - H(s) H(-s)

Adomg 1 to each side and rearranging gives 11 + H(s)12 = 1

+ qjG(S)\2

(10.35)

It has been shown that only optimal systems obey this relationship. Denote the numerator of H(s) as nCB) and of G(s) as neG), and the denominator of H(s) as d(H) and of G(s) as d(G). But d(G)::::;: det (sl - A) = d(H). Multiplying (10.35) by Id(G)]2 gives ]d(G)

+ n(H»)2

;:=

]d(G»)2

+ qjn(G)12

(10.36)

As q -') 00, if there are m zeros of neG), the open loop system, then 2m zeros of (10.96) tend to the zeros of jn(G)]2. The remaining 2(n - m) zeros tend to 00 and are asymptotic to the zeros of the equation 8 2(n-m)::::;: q. Since the closed loop eigenvalues are the left half plane zeros of jd(G) + n(H)12, we conclude that as q ~ 00 m closed loop poles tend to the m open loop zeros and the remaining n - m closed loop poles tend to the left half plane zeros of the equation s2(n-m) ::::: q. In other words, they tend to a stable Butterworth configuration of order n - m and radius q'l where 2(n - m) ;:= y-l. If the system has 3 open loop poles and one open loop zero, then 2 open loop poles are asymptotic as shown in Fig. 10-10. This is independent of the open loop pole-zero configuration. Also, note equation (10.36) requires the closed loop poles to tend to the open loop poles as q ~ o.

1

00

2 ( qy 2 + u ) dt is quite o general for scalar controls since c can be chosen by the designer (also see Problem 10.19).

Furthermore, the criterion

Of course we should remember that the results of this analysis are valid only for this particular criterion involving the output and are not valid for a general quadratic in the state.

Fig. to-tO

CHAP. 10]


227

S upplementaryProblems 10.9.

Given the scalar system dx/dt = 4x + u with initial condition x(o) = xo'

i

u to minimize 21' =

00

o

10.10.

(9x 2 + u 2 ) dt.

We desire to find u to minimize

(b) (e)

G-fs)x+G)u

dx dt

Consider the plant

(a)

Find a feedback control

fooo (4xi + u

21'

2)

dt.

Write the canonical equations (10.9) and their boundary conditions. Solve the canonical equations for xCt) and pet). Find the open loop control u(xo. t).

10.11.

For the system and criterion of Problem 10.10, (a) write three coupled nonlinear scalar equations for the elements ¢ij of the pet) matrix, (b) find pet) using the relationship pet) = p(t) x(t) and the results of part (b) of Problem 10.10, (e) verify the results of (b) satisfy the equations found in Cal, and that PCt) is constant and positive definite, (d) draw a flow diagram of the closed loop system.

10.12.

Consider the plant

and performance index v

~

=

f

T

(qxi

+ u 2 ) dt

o

Assuming the initial conditions are Xl(O) = 5, x 2 (O) = -3, find the equations (10.9) whose solutions will give the control (10.8). What are the boundary conditions for these equations (10.9)? 10.13.

Consider the scalar plant

dx/dt = x

and the performance index

v

+U

-

Ca) Using the state-costate transition matrix .p(t, T),find PCt) such that pet) lim PCt)?

= PCt) xCt).

(b) What is

t~-""

10.14.

Given the system of Fig. 10-11, find K(t) to minimize 21' =

it1

(X2

+ u 2/p) dt

and find lim K(t).

~

10.15.

~~""


(-2o 1) 1

x

+

_1_ l---4x.--_ 8+a

Zero input

(1)° u

Find the control that minimizes 21' =

10.16.

f""o CxTx + u

2)

Fig.l0~11

dt

v

Consider the motion of a missile in a straight line. Let;· = v and = u, where r is the position and v is the velocity. Also, u is the acceleration due to the control force. Hence the state equation is

l)(V + (°1) O r

)

2v

where qv and qr are scalars> 0.

=

It is desired to minimize

u.

2

qvv (T)

+

2

qr r (T)

+

fT u2(T) dT o

[CHAP. 10


228

Find the feedback control matrix M(e) relating the control u(t) to the states ret) and vet) in the form M(e) u(t)

(r(t») vet)

where

e = T - t. Here

8

is known as the "time to go".

(0)

10.17.

0 Given the system dx = ( 1) x + u with criterion 21' = dt 0 0 1 root locus of the closed loop roots as q varies from zero to CIJ.

10.18.

Consider the matrix

2

fOl (qx~ + u 2) dt.

Draw the

o

/rJ

-b -a

H

associated with the optimization problem

dx/dt = ax

= 1.: (T

v

(qx2

Jo

2

+ btl, + r-u.2) dt

xeD) = xo;

peT) = 0

Show that the roots of H are symmetrically placed with respect to the jl.) axis and show that the location of the roots of H depends only upon the ratio q/r for fixed a and b. 10.19.

Given the system (10.1) and let S = 0, Q(t) = 17Qo(t) and R(t) = pRo(t) in the criterion (10.2). Prove that the optimal control law u op depends only on the ratio 7J/ p if Qo(t) and Ro(t) are fixed.

10.20.

Show that the transition matrix (10.14) is symplectic, i.e.

( ~rl (t, 7")

4J22 (t,7")

de--t (4111(t, 7')

and also show

0 I) (~l1(t,

q,~1 (t, 7"») (

~12(t,7")

4J 21 (t, 7')

~12(t, 4J 22 ( t,

-I

0

T»)

~)

7")

~21(t,7")

1.

7")

10.21.

Using the results of Problem 10.20, show that for the case S = 0, P{t} = .p;}i (t, 7') .p21(t, T). provides a check on numerical computations.

10.22.

For the scalar time-varying system u(x, t) to minimize

P= ~ itl (3x o

10.23.

2

dx/dt

-x tan t

+ u,

This

find the optimal feedback control

+ 4u2)dt.

Find the solution pet) to the scalar Riccati equation corresponding to the open loop system with finite escape time dx/dt = -(x + u)/t with criterion v

=~

Itl

(6x 2 + u 2 ) dt.

Consider the behavior

to

for 0 10.24.

(t, T)BBT«>T(t" T) dTJ-l x(t)

and

P(t)

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