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

a[z alz I Z (
................

d
••• ,

l1li

••••••••••

"

,.

"

•

•

•

•

..

•

•

••

"

..

II

II

ah

+ aYn Xn "

+ a/l v au

a/z + al2 . + aYn Xn au v

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

a/n a/n I n(
a/n

II.

•••••

al

n + -a v Yn Xn + -a U

where each a//aYi is the partial derivative of li(Y l ' Yz' ... , Yn' u, t) with respect to Yp evaluated at Y1 =
:t (~:) = (;~~;~~ ..~~~~~:.......... ~~~~~) (~:) (:~1:) +

Xn

aln/aYl a/n/aY2

.••

aln/aYn

Xn

v

aln/au

which is, in general, a time-varying linear differential equation, so that the nonlinear equation has been linearized. Note this procedure is valid only for Xl, X2, ••• , Xn and v small enough so that the higher order terms in the Taylor's series can be neglected. The matrix of al/aYj evaluated at Yj =
This linear equation gives the solution yet) == Xl(t) + t- 1 and dy/dt:= equation, and is valid as long as Xl' X2 and 'V are small.

X2 - t- 2

for the original nonlinear

Example 1.22. Given the system dy/dt == ky - y2 + u. Taking u(t) := 0, we can find two constant solutions ¢(t):::: 0 and 1f;(t) := k. The equation for small motions x(t) about ¢(t) := 0 is dx/dt == It:x + u so that y(t) ~ x(t), and the equation for small motions x(t) about 1f;(t) :::: k is dx/dt:= -kx + u so that y(t) = k + x(t).

10

MEANING OF STATE

[CHAP. 1

Solved Problems 1.1.

Given a delay line whose output is a voltage input delayed T seconds. What is the physical object, the abstract object, the state variable and the state space? Also, is it controllable, observable and a dynamical system with a state space description? The physical object is the delay line for an input u(t) and an output y(t) = u(t ~ T). This equation is the abstract object. Given an input time function u(t) for all t, the output yet) is defined for t ~ to, so it is a dynamical system. To completely specify the output given only u( t) for t. ~ to. the voltages already inside the delay line must be known. Therefore, the state at timetIJ is x(to) = u[t o- T , to) ~ where the notation u[t2 , tl) means the time function u(r) for 7 in the interval t2 === 7 < t 1 • For E> 0 as small as we please,· U[to-T, to) can be considered as the uncountably infinite set of numbers {u(to - T), 'u(to - T

+ e),

•.. , u(t o - e)}

In this sense we can consider the state as consisting of an infinity of numbers. variable is the infinite set of time functions x(t)

=

U[t-T, t)

==

{u(t - T), u(t - T

+ e),

... , u(t -

Then the state

e)}

The state space is the space of all time functions T seconds long, perhaps limited by the breakdown voltage of the delay line. An input u(t) for to === t < to + T will completely determine the state T seconds later, and~ any state will be observed in the output after T seconds, so the system is both observable and controllable. Finally, x(t) is uniquely made up of x(t - 7) shifted r seconds plus the input over T seconds, so that the mathematical relation yet) == u(t - T) gives a system with a state space deseri ption.

1.2.

Given the uncontrollable partial differential equation (diffusion equation) 2

=

ay(r, t)

at

a y(r, t) ar2

with boundary conditions y(O, t) = y(l, t) = O.

0.(

+ u r,

t)

What is the state variable?

The solution to this zero input equation for t:=:: to is 00

y(r, t)

where en

==

i

==

~ cne-n2172Ct-to)

sin nrrr

n=l

I y(r, to) sin n"r dr. All that is needed to determine the output is y(r, to), so that o . y(r, t) is a choice for the state at any time t. Since y(r, t) must be known for almost all r III the interval 0,::::: r == 1, the state can be considered as an infinity of numbers similar to the case of Problem 1.1.

1.3.

Given the mathematical equation (dy/dt)2 = y2 + Qu. with a state space description?

Is this a dynamical system

A real output exists for all t === to for any u(t), so it is a dynamical system. The equation can be written as dy/dt == s(t)y, where set) is a member of a set of time functions that take on the value +1 or -1 at any fixed time t. Hence knowledge of y(t o) and set) for t ~ to uniquely specify yet) and they are the state.

1.4.

I)} +

Plot the trajectories of d 2y dt 2 Changing variables to y =

= Xl

if (dy/dt-y~ ,if (dy/dt-y < 1) and dy/dt

= X2' or

Ou

the mathematical relation becomes

CHAP. 1]

MEANING OF STATE

11

The former equation can be integrated immediately to xz(t) = X2(tO), a straight line in the phase plane. The latter equation is solved by multiplying by dx/2 to obtain x2 dx2i2 + Xl dXl/2 ::= O. ' can b . x 2 (t) + Xl2 (t) = Xz2 (to) + Xl2 (to). The result is an equation of circles ThIS e 'm t egrat e d to 0 btam 2 in the phase plane. The straight lines lie to the left of the line X2 - Xl ::= 1 and the circles to the right, as plotted in Fig. 1-5.

2.-----~------~------+_--_+~, 3.-----~

__----~____

- - + __ _~

------~4~====~====~~====~~~~--_t--~-------------Xl 6.-------~----_.

____~

Note Xl increases for x2 > 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)

<

1 results in an imaginary output.

+ X2(t O)] + ft to

et -7"Iu(T)

+ 2] dT

MEANING OF STATE

CHAP.!]

15

1.22.

No additional knowledge is needed other than the input, so that state is zero dimensional and the state space description is yet) = u(t + T). It is not a dynamical system because it is not realizable physically if u(t) is unknown in advance for all t::=" to. However, its state space description violates only condition 5, so that other equations besides dynamical systems can have a state space description if the requirement of causality is waived.

1.23.

Yes.

yet) =

e-txo

+ ft

e(T-t)

u(r) dT,

and the contribution of the initial condition Xo depends

to

on when the system is started. If Xo = 0, the system is equivalent to dx/dt = -x + u = x, which is time-invariant.

where

y

1.24.

If we define du/dt

=

lim [u(t+£)-u(t)]I€ £-1-0

so that y(to) is defined, then the state space is zero dimensional and knowledge of u(t) determines yet) for all t ::=" to. Other definitions of du/dt may require knowledge of u(t) for to - € ~ t,.o:: to, which would be the state in that case. 1.25.

Obviously yet) is not defined for t > tl> so that as stated the equation is not a dynamical system. However, if the behavior of engineering interest lies between to and t l , merely append y = Ou for t ::=" tl to the equation and a dynamical system results.

Chapter 2 Methods for Obtaining the State Equations 2.1

FLOW DIAGRAMS

Flow diagrams are a simple diagrammatical means of obtaining the state equations. Beca use only linear differential or difference equations are considered here, only four basic objects are needed. The utility of flow diagrams results from the fact that no differentiating devices are permitted. Definition 2.1:

A summer is a diagrammatical abstract object having n inputs Ul(t),U2(t), ... , un(t) and one outputy(t) that obey the relationship y(t) = ±Ul(t) ± U2(t) ± ... ± un(t)

where the sign is positive or negative as indicated in Fig. 2-1, for example. Ul(t) - - - - - - - _

) - - - - - - - _ y(t)

Fig. 2~1.

Definition 2.2:

A scalar is a diagrammatical abstract object having one input u(t) and one output y(t) such that the input is scaled up or down by the time function a(t) as indicated in Fig. 2-2. The output obeys the relationship y(t) = a(t) u(t). U(t)

. .------..

-------~.8 Fig. 2-2.

Definition 2.3:

Summer

y(t)

Scalor

An integrator is a diagrammatical abstract object having one input u(t), one output y(t), and perhaps an initial condition y(to) which may be shown or not, as in Fig. 2-3. The output obeys the relationship y(t)

y(t o)

+

ft u(,) d, to

~>-O)----_-,

U(t) _ _ _ _ _ _ _+-1'

Fig. 2-3.

16

Integrator at Time t

y(t)

CHAP. 2]

METHODS FOR OBTAINING THE STATE EQUATIONS

Definition 2.4:

17

A delayor is a diagrammatical abstract object having one input u(k), one output y(k), and perhaps an initial condition y(l) which may be shown or not, as in Fig. 2-4. The output obeys the relationship y(i + l + 1)

u(k)

= u(i + l)

for j

= 0,1,2, ...

~>-)

--------.!. Fig. 2~4.

-----_ .. y(k)

Delayor at Time k

2.2 PROPERTIES OF FLOW DIAGRAMS Any set of time-varying or time-invariant linear differential or difference equations of the form (1.10) or (1.11) can be represented diagrammatically by an interconnection of the foregoing elements. Also, any given transfer function can also be represented merely by rewriting it in terms of (1.10) or (1.11). Furthermore, multiple input and/or multiple output systems -can be represented in an analogous manner. Equivalent interconnections can be made to represent the same system. Example 2.1. Given the system dyldt = o:.y

+ o:.U

(2.1)

with initial condition y(to). An interconnection for this system is shown in Fig. 2-5.

>--.....----+- y(t)

u(t)--+--l

+ Fig. 2-5 Since 0:. is a constant function of time, the integrator and scalo"r can be interchanged if the initial condition is adjusted accordingly, as shown in Fig. 2-6.

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

METHODS FOR OBTAINING THE STATE EQUATIONS

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

METHODS FOR OBTAINING THE STATE EQUATIONS

[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]

METHODS FOR OBTAINING THE STATE EQUATIONS

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

METHODS FOR OBTAINING THE STATE EQUATIONS

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]

METHODS FOR OBTAINING THE STATE EQUATIONS

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

METHODS FOR OBTAINING THE STATE EQUATIONS

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

METHODS FOR OBTAINING THE STATE EQUATIONS

[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

METHODS FOR OBTAINING THE STATE EQUATIONS

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

METHODS FOR OBTAINING THE STATE EQUATIONS

[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]

METHODS FOR OBTAINING THE STATE EQUATIONS

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.

METHODS FOR OBTAINING THE STATE EQUATIONS

[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

METHODS FOR OBTAINING THE STATE EQUATIONS

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.

METHODS FOR OBTAINING THE STATE EQUATIONS

[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]

METHODS FOR OBTAINING THE STATE EQUATIONS

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.

Supplementary Problems 2.9.

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

METHODS FOR OBTAINING THE STATE EQUATIONS

[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

METHODS FOR OBTAINING THE STATE EQUATIONS

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.

METHODS FOR OBTAINING THE STATE EQUATIONS

[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

METHODS FOR OBTAINING THE STATE EQUATIONS

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
A

=

(

0

2

1J

x2

j)

sin t

Definition 3.2:

A row of a matrix is the set of all elements through which one horizontal line can be drawn.

Definition 3.3:

A column of a matrix is the set of all elements through which one vertical line can be drawn.

Example 3.2. The rows of the matrix of Example 3.1 are (0 2 j) and (1i x 2 sin t).

(~).

(:,) and

The columns of this matrix are

(Si~ t) .

Definition 3.4:

A square matrix has the same number of rows and columns.

Definition 3.5:

The order of a matrix is m x n, read m by n, if it has m rows and n columns.

Definition 3.6:

A scalar, denoted by a letter that is not in boldface type, is a 1 x 1 matrix. In other words, it is one element. When part of a matrix A, the notation atj means the particular element in the ith row and jth column.

Definition 3.7:

A vector, denoted by a lower case boldfaced letter, such as a, or with its contents displayed in braces, such as {at}, is a matrix with only one row or only one column. Usually a denotes a column vector, and aT a row vector. 38

CHAP. 3] Definition 3.8:

ELEMENTARY MATRIX THEORY

39

The diagonal of a square matrix is the set of all elements aij of the matrix in which i = j. In other words, it is the set of all elements of a square matrix through which can pass a diagonal line drawn from the upper left hand corner to the lower right hand corner.

Example 3.3. Given the matrix B

The diagonal is the set of elements through which the solid line is drawn, b 11 , b22 , b33 , and not those sets determined by the dashed lines.

Definition 3.9:

The trace of a square matrix A, denoted tr A, is the sum of all elements on the diagonal of A. n

trA

~ au i=l

3.3

BASIC OPERATIONS

Definition 3.10:

Two matrices are equal if their corresponding elements are equal. A = B means aij = bij for all i and j. The matrices must be of the same order.

Definition 3.11: Matrix addition is performed by adding corresponding elements. A + B = C means aij + bij = Cji, for all i and j. Matrix subtraction is analogously defined. The matrices must be of the same order. Definition 3.12: Mat1'ix ditfe1'entiation or integration means differentiate or integrate each element, since differentiation and integration are merely limiting operations on sums, which have been defined by Definition 3.1l. Addition of matrices is associative and commutative, i.e. A + (B + C) = (A + B) + C and A + B = B + A. In this and all the foregoing respects, the operations on matrices have been natural extensions of the corresponding operations on scalars and vectors. However, recall there is no multiplication of two vectors, and the closest approximations are the dot (scalar) product and the cross product. Matrix multiplication is an extension of the dot product a· b = a 1b 1 + a2 b2 + ... + anb n, a scalar. Definition 3:13: To perform matTix multiplication, the element Cij of the product matrix C is found by taking the dot product of the ith row of the left matrix A and the jth column of the right matrix B, where C = AB, so that n

Cij

= ~ aikb kj • k=l

Note that this definition requires the left matrix (A) to have the same number of columns as the right matrix (B) has rows. In this case the matrices A and B are said to be compatible. It is undefined in other cases, excepting when one of the matrices is 1 x 1, i.e. a scalar. In this case each of the elements is multiplied by the scalar, e.g. aB = {ab ii } for all i and j. Example 3.4. The vector equation y = Ax, when y and x are 2 X 1 matrices, i.e. column vectors, is

40

ELEMENTARY MATRIX THEORY

[CHAP. 3

2

where Yi = ~

k=l

aikxk

for i = 1 and 2. Yi

=

But suppose x = Bz, so that 2 ( 2 ) ~ aik ~ bk-zk=l 3= 1 J 1

=

Then

2

~ c--z-

j=1

~J

j

so that y = A(Bz) = Cz, where AB = C.

Example 3.4 can be extended to show (AB)C = A(BC), i.e. matrix multiplication is asso.;. ciative. But, in general, matrix multiplication is not commutative, AB =F BA. Also, there is no matrix division. Example 3.5.

To show AB # BA, consider

and note D oF C.

AB

(~ ~)(~ ~)

D

BA

(~. ~)(~ ~)

C

Furthermore, to show there is no division, cQnsider

(~ ~)(~ ~)

BF

C

Since BA = BF = C, "division" by B would imply A = F, which is certainly not true.

Suppose we have the vector equation Ax = Bx, where A and Bare n X n matrices. It can be concluded that A = B only if x is an arbitrary n-vector. For, if x is arbitrary, we may choose x successively as el, e2, ... , en and find that the column vectors al = h1, a2 = h2, ... , an = b n • Here ej are unit vectors, defined after Definition 3.17, page 41. Definition 3.14: To partition a matrix, draw a vertical and/or horizontal line between two rows or columns and consider the subset of elements formed as individual matrices, called submatrices. As long as the submatrices are compatible, i.e. have the correct order so that addition and multiplication are possible, the submatrices can be treated as elements in the basic operations. Example 3.6.

A 3 X 3 matrix A can be partitioned into a 2 X 2 matrix All, a 1 X 2 matrix A21 , a 2 X 1 matrix A 12 , and a 1 X 1 matrix A22 •

( -~-::--+-~-:: )

A

A similarly partitioned 3 X 3 matrix B adds as A

+

B

and multiplies as AB

Facility with partitioned matrix operations will often save time and give insight.

ELEMENTARY MATRIX THEORY

CHAP.3J

41

3.4 SPECIAL MATRICES Definition 3.15: The zero matrix, denoted 0, is a matrix whose elements are all zeros. Definition 3.16: A diagonal matrix is a square matrix whose off-diagonal elements are all zeros. Definition 3.17: The unit matrix, denoted I, is a diagonal matrix whose diagonal elements are all ones. Sometimes the order of I is indicated by a subscript, e.g. In is an n X n matrix. Unit vectors, denoted ei, have a one as the ith element and all other elements zero, so that 1m = (el J e2 J . . . J em). Note IA = AI = A, where A is any compatible matrix. Definition 3.18: An upper triangular matrix has all zeros below the diagonal, and a lower triangular matrix has all zeros above the diagonal. The diagonal elements need not be zero. An upper triangular matrix added to or multiplied by an upper triangular matrix results in an upper triangular matrix, and similarly for lower triangular matrices. Definition 3.19: A transpose matrix, denoted AT, is the matrix resulting from an interchange of rows and columns of a given matrix A. If A = {au}, then AT:= {ajd, so that the element in the ith row and jth column of A becomes the element in the jth row and ith column of AT. Definition 3.20: The complex conjugate tranBpose matrix, denoted At, is the matrix whose elements are complex conjugates of the elements of AT. Note (AB)T

= BTAT

and (AB)t = BtAt.

Definition 3.21: A matrix A is sym'metric if A = AT. Definition 3.22: A matrix A is Hermitian if A = At. Definition 3.23: A matrix A is normal if AtA = AAt. Definition 3.24: A matrix A is skew-symmetric if A = -AT. Note that for the different cases: Hermitian A = At, skew Hermitian A = -At, real symmetric AT = A, real skew-symmetric A = -AT, unitary AtA == I, diagonal D, or orthogonal AtA = D, the matrix is normal. 3.5 DETERMINANTS AND INVERSE MATRICES Definition 3.25: The determinant of an n X n (square) matrix {au} is the sum of the signed products of all possible combinations of n elements, where each element is taken from a different row and column. n!

det A

~ (-1)palPla2P2"

·anpn

(3.1)

Here Pl, P2, ... , pn is a permutation of 1,2, ... , n, and the sum is taken over all possible permutations. A permutation is a rearrangement of 1,2, ... , n into some other order, such as 2, n, ... ,1, that is obtained by successive transpositions. A transposition is the interchange of places of two (and only two) numbers· in the list 1, 2, ... , n. The exponent p of -1 in (3.1) is the number of transpositions it takes to go from the natural order of 1,2, ... , n to Pl, P2, ... ,pn. There are n! possible permutations of n numbers, so each determinant is the sum of n! products.

[CHAP. 3

ELEMENTARY MATRIX THEORY

42

Example 3.7. To find the determinant of a 3 X 3 matrix, all possible permntations of 1,2,3 mllst be found. forming one transposition at a time, the following table can be formed. P

PI' P2' Pa

0

1, 2,

3

1

3,

2,

1

2

2,

3,

1

3

4

2, 1, 3, 1,

2

5

1,

2

3,

Per-

3

This table is not unique in that for P = 1, possible entries are also 1,3,2 and 2,1,3. However, these entries can result only from an odd p, so that the sign of each product in a determinant is unique. Since there are 8! = 6 terms, all possible permutations are given in the table. Notice at each step only two numbers are interchanged. Using the table and (3.1) gives

+ (-1)1a12a21a33 + (-1)2a12u23aSI + (-1)3u13a22a31 +

det A = (-I)Oallu22u33

Theorem 3wl:

(-1)4a18u21a32

Let A be an n X n matrix. Then det (AT)

+ (-1)5ullu23a32

= det (A).

Proof is given in Problem 3.3, page 59.

Theorem 3.2:

Given two n

X

n matrices A and B. Then det (AB) = (det A) (det B).

Proof of this theorem is most easily given using exterior products, defined in Section 3.13! page 56. The proof is presented in Problem 3.15, page 65. Definition 3.26: Elementary row (or column) operations are: (i)

Interchange of two rows (or columns).

(ii) Multiplication of a row (or column) by a scalar.

(iii) Adding a scalar times a row (or column) to another row (column). To perform an elementary row (column) operation on an n X m matrix A, calculate the product EA (AE for columns), where E is an n x n(m X m) matrix found by performing the elementary operation on the n x n(m X m) unit matrix I. The matrix E is called an elementary matrix. Example 3.8. Consider the 2 X 2 matrix A = {au}. obtain E.

To interchange two rows, interchange the two rows in I to

EA

==

To add 6 times the second column to the first, multiply AE

~)

==

Using Theorem 3.2 on the product AE or EA, it can be found that (i) interchange of two rows or columns changes the sign of a determinant, i.e. det (AE) = -detA, (ii) multiplication of a row or column by a scalar a multiplies the determinant by a, i.e. det (AE) = a det A, and (iii) adding a scalar times a row to another row does not change the value of the determinant, i.e. det (AE) = det A.

CHAP.3J

ELEMENTARY MATRIX THEORY

43

Taking the value of a in (ii) to be zero, it can be seen that a matrix containing a row or column of zeros has a zero determinant. Furthermore, if two rows or columns are identical or multiples of one another, then use of (iii) will give a zero row or column, so that the determinant is zero. Each elementary matrix E always has _an inverse E-l, found by undoing the row or column operation of I. Of course an exception is a = 0 in (ii). Example 3.9. The inverse of E

=

G~)

is E-l

=

G~).

The inverse of E

= (~ ~)

is E-l

= (-~

n.

Definition 3.27: The determinant of the matrix formed by deleting the ith row and the jth column of the matrix A is the minor of the element aij, denoted detMij. The cofactor Co = (-l)Hi detMij. Example 3.10. The minor of detM22 •

Theorem 3.3:

a22

of

a

3 X 3 matrix A

is

det M22 =

alla33 - a13aSl'

The cofactor

The Laplace expansion of the determinant of an n n

~

for any column j or det A

aijCij

i=l

=

n

~

aijCij

X

c22

= (-1)4 det M22 =

n matrix A is detA

for any row i.

j=l

Proof of this theorem is presented as part of the proof of Theorem 3.21, page 57. Example 3.11. The Laplace expansion of a 3 X 3 matrix A about the second column is det A =

Corollary 3.4:

+ a22cZ2 + aS2c32 - a I 2(a21 a 33 - a23 a 31) + a22(a ll aaa a12c12

a13 a 31) -

a32(alla23 - a13 a 21)

The determinant of a triangular n x n matrix equals the product of the diagonal elements.

Proof is by induction. The corollary is obviously true for n = 1. For arbitrary n, the Laplace expansion about the nth row (or column) of an n x n upper (or lower) triangular matrix gives detA = annCnn • By assumption, Cnn = ana22' • ·an-l,n-l, proving the corollary. Explanation of the induction method: First the hypothesis is shown true for n = no, where no is a fixed number. (no = 1 in the foregoing proof.) Then assume the hypothesis is true for an arbitrary n and show it is true for n + 1. Let n = no, for which it is known true, so that it is true for no + 1. Then let n = no + 1, so that it is true for no + 2, etc. In this manner the hypothesis is shown true for all n ~ no. Corollary 3.5:

The determinant of a diagonal matrix equals the product of the diagonal elements.

This is true because a diagonal matrix is also a triangular matrix. n

The Laplace expansion for a matrix whose kth row equals the ith row is det A = for k # i, and for a matrix whose kth column equals the jth column the Laplace

1: akjCij

j=1

n

expansion is det A = ~ ~kCij. But these determinants are zero since A has two identical i=l rows or columns.

[CHAP. 3

ELEMENTARY MATRIX THEORY

44

Definition 3.28: The Kronecker delta

8ij

== 1 if

i

=j

and

8 ij

= 0 if i =1= j.

Using the Kronecker notation, n

:l: i=1

Ski

aikCij

detA

= {cu}

Definition 3.29: The adjugate matrix of A is adj A of cofactors ofA.

T,

(3.2)

the transpose of the matrix

The adjugate is sometimes called "adjoint", but this term is saved for Definition 5.2. Then (3.2) can be written in matrix notation as A adj A

=

I det A

(3.3)

Definition 3.30: An m x n matrix B is a left inverse of the n x m matrix A if BA = 1m , and an m x n matrix C is a right inverse if AC = In. A is said to be nonsingular if it has both a left and a right inverse. If A has both a left inverse B and a right inverse C, C == IC = BAC = BI = B. Since BA = 1m and AC = In, and if C = B, then A must be square. Furthermore suppose a non .. singular matrix A has two inverses G and H. Then G = GI = GAB = IH == H so that a nonsingular matrix A must be square and have a unique inverse denoted A -1 such that A-1A== AA-1 = I. Finally, use of Theorem 3.2 gives detA detA-l:;:::: detI:;:::: 1, so that if detA::: 0, A can have no inverse.

Theorem 3.6:

Cramer's rule. Then

Given an n

X

A-I

n (square) matrix A such that detA =1= O. de! A adj A

The proof follows immediately from equation (3.3). Example 3.12. The inverse of a 2 X 2 matrix A. is

Another and usually faster means of obtaining the inverse of nonsingular matrix A is to use elementary row operations to reduce A in the partitioned matrix A II to the unit matrix. To reduce A to a unit matrix, interchange rows until an =1= O. Denote the interchange by E 1 • Divide the first row by an, denoting this row operation by E 2 • Then E2E1A has a one in the upper left hand corner. Multiply the first row by Ui1 and subtract it from the ith row for i = 2,3, ... , n, denoting this operation by Ea. The first column of EaE2EIA is then the unit vector el. Next, interchange rows E3E2EIA until the element in the second row and column is nonzero~ Then divide the second row by this element and subtract from all other rows until the unit vector e2 is obtained. Continue in this manner until Em' .. E1A = I. Then Em'" El = A-I, and operating on I by the same row operations will prod Uc€) A -1. Furthermore, det A -1 = det E1 det E 2 • •• det Em from Theorem 3.2. Example 3.13. To find the inverse of

(_~ ~),

adjoin the unit matrix to obtain

(-~ ~ I ~ ~)

ELEMENTARY MATRIX THEORY

CHAP. 3]

45

Interchange rows (det El = -1):

o

1 1

~)

1

Divide the first row by -1 (det E2 = -1): (

1 -1 1

o

I

0 -1) 1 0

It turns out the first column is already ell and all that is necessary to reduce this matrix to I is to add the second roW to the first (det Es = 1).

(~

o 1

I

1-1) 1

0

The matrix to the right of the partition line is the inverse of the original matrix, which has a determinant equal to [(-l)(-l)(l)} ~l = 1.

3.6 VECTOR SPACES Not all matrix equations have solutions, and some have more than one. Example 3.14. (a)

The matrix equation

has no solution because no

~

exists that satisfies the two equations written out as ~

0

2~

1

(b) The matrix equation

(~) is satisfied for any

~.

To find the necessary and sufficient conditions for existence and uniqueness of solutions of matrix equations, it is necessary to extend some geometrical ideas. These ideas are apparent for vectors of 2 and 3 dimensions, and can be extended to include vectors having an arbitrary number of elements. Consider the vector (2 3). Since the elements are real, they can be represented as points in a plane. Let (t 1 t 2 ) = (2 3). Then this vector can be represented as the point in the t l , t2 plane shown in Fig. 3-1.

• (2 3)

~~-+~~~~~~.---~.-~---.----~--~

Fig.3-1.

Representation of (2 3) in the Plane

46

ELEMENTARY MATRIX THEORY

[CHAP. 3

If the real vector were (1 2 3), it could be represented as a point in (~1 ~2 g3) space by drawing the ~3 axis out of the page. Higher dimensions, such as (1 2 3 4), are harder to draw but can be imagined. In fact some vectors have an infinite number of elements. This can be included in the discussion, as can the case where the elements of the vector are other than real numbers. Definition 3.31: Let CUn be the set of all vect01'S with ncomponents. Let al and a2 be vectors having n components, i.e~ al and a2 are in CUn. This is denoted al E CUn, a2 E CUn. Given arbitrary scalars (31 and !32, it is seen that ((31al + !32a2) E
is an infinite plane. To represent diagrammatically these and, in general, rvn for any n, one uses the area enclosed by a closed curve. Let 1l be a set of vectors in G[}n. This can be represented as shown in Fig. 3-2. G[}2

Fig.-3-2.

A Set of Vectors 11 in "Un

Definition 3.32: A set of vectors 11 in rvn is closed under addition if, given any al E 11 and any a2 E 11, then (at + a2) E 11. Example 3.15. (a)

Given 11 is the set of all 3-vectors whose elements are integers. This subset of 'V3 is closed under addition because the sum of any two 3-vectors whose elements are integers is also a 3-vector whose elements are integers.

(b)

Given 'U is the set of all 2-vectors whose first element is unity. This set is not closed under addition because the sum of two vectors in 1l must give a vector whose first element is two.

Definition 3.33: A set of vectors U in G[}n is closed under scalar multiplication if, given any vector a E'U and any arbitrary scalar /3, then (3a E 'U. The scalar f3 can be a real or complex number. Example 3.16. Given 11 is the set of all 3-vectors whose second and third elements are zero. Any scalar times any vector in 'U gives another vector in 1l, so U is closed under scalar multiplication.

Definition 3.34: A set of n-vectors 'U in ti()n that contains at least one vector is called a vector space if it is (1) closed under addition and (2) closed under scalar multiplication. If a E 1{, where 'U is a vector space, then Oa::::: 0 E'U because multiplication. Hence the zero vector is in every vector space.

'U

is closed under scalar

Given 81, a2, ... , an, then the set of all linear combinations of the (linear manifold).

3i

is a vector space

1l

3.7 BASES Definition 3.35: A vector space 'U in G(Jn is spanned by the vectors 31, a2, ... , ak (k need not equal n) if (1) al E 11,- a2 E 11, ... , ak E 11 and (2) every vector in 'U is a linear combination of the ai, 32, . . . , ak.

· CHAP. 3]

ELEMENTARY MATRIX THEORY

47

Example 3.17. Given a vector space 'U in 'Va to be the set· of all 3-vectors whose third element is zero. Then (1 2 0), (1 1 0) and (0 1 0) span 'U because any vector in 'U can be represented as (a (1 0), and

=

(ex (1 - 0)

(,8 - a)(l 2 0)

+

(2a - (1)(1 1 0)

+

0(0 1 0)

Definition 3.36: Vectors

a1, a2, .•. , ak E G()n are linearly dependent if there exist scalars !31, !32, .•• , /3k not all zero such that /31a1 + (32a2 + ~ .. + j3kak O.

=

Example 3.18. The three vectors of Example 3.17 are linearly dependent because

+1(1

2 0) -

1(1 1 0) -

1(0 1 0)

=

(0 0 0)

Note that any set of vectors that contains the zero vector is a linearly dependent set. Definition 3.37: A set of vectors are linearly independent if they are not linearly dependent.

Theorem 3.7:

If and only if the column vectors a1, a2, •.• , an of a square matrix A are linearly dependent, then det A = O.

Proof: If the column vectors of A are linearly dependent, from Definition 3.36 for some 13 1, i32 , ••• , f3 n not all zero we get 0 = f3 1a 1 + {32a2 + ... + f3nan, Denote one nonzero f3 as f3 i , Then 1 0 o

o

1

o

o

0

o

o 0

1

Since a matrix with a zero column has a zero determinant, use of the product rule of Theorem 3.2 gives det A det E = O. Because det E = f3 i ¥- 0, then det A = O. Next, suppose the column vectors of A are linearly independent. Then so are the column vectors of EA, for any elementary row operation E. Proceeding stepwise as on page 44, we find E 1 , ••• , Em such that Em'" ElA = I. (Each step can be carried out since the column under consideration is not a linear combination of the preceding columns.) Hence, (detEm)" . (detEI)(detA) = 1, so that detA ¥- O. Using this theorem it is possible to determine if al, a2, ... , ak, k ~ n, are linearly dependent. Calculate all the k x k determinants formed by deleting all combinations of n- k rows. If and only if all determinants are zero, the set of vectors is linearly dependent. Example 3.19.

Consider

gives det

(!

(~) and ( ; ) . ~)

= -12.

Deleting the bottom row gives det

G:)

Hence the vectors are linearly independent.

= O.

Deleting the top row

There is no need to check the

determinant formed by deleting the middle row.

Definition

3~38:

A set of n-vectors ai, a2, ... , ak form a basis for (2) they are linearly independent.

'11

if (1) they span ,'11 and

[CHAP. 3

ELEMENTARY MATRIX THEORY

48 Example 3.2U.

Any two of the three vectors given in Examples 3.17 and 3.18 form a basis of the given vector space, since (1) they span 1..( as shown and (2) any two are linearly independent. To verify this for (1 2 0) and (1· 1 0), set

=

=

This gives the equations f3t + {J2 0, 2{Jl + {J2 O. The only solution is that {J1 and {J2 are both zero, which violates the conditions of the definition of linear dependence. Example 3.21. Any three non coplanar vectors in three-dimensional Euclidean space form a basis of C0 3 (not necessarily the orthogonal vectors). However, note that this definition has been abstracted to include vector spaces that can be subspaces of Euclidean space. Since conditions on the solution of algebraic equations are the goal of this section, it is best to avoid strictly geometric concepts and remain with the more abstract ideas represented by the definitions. Consider 'U to be any infinite plane in three-dimensional Euclidean space 'V s. vectors in this plane form a basis for 11.

Theorem 3.8:

Any two noncolinear

If al, a2, ... ,ak are a basis of 'U, then every vector in uniquely as a linear combination of a1, a2, ... , ak.

'U

is expressible

The key word here is uniquely. The proof is given in Problem 3.6. To express any vector in 'U uniquely, a basis is needed. Suppose we are given a set of vectors that span 'U. The next theorem is used in constructing a basis from this set.

Theorem 3.9:

Given nonzero vectors a1, a2, ... , am E "Un. The set is linearly dependent if and only if some ak, for 1 < k ~ m, is a linear combination of al, a2, ... , ak-l.

Proof of this is given in Problem 3.7. This theorem states the given vectors need only be considered in order for the determination of linear dependency. We need not check and see that each ak is linearly independent from the remaining vectors. Example 3.22.

=

Given the vectors (1 -1 0), (-2 2 0) and (1 0 0). They are linearly dependent because (-2 2 0) -2(1 -1 0). We need not check whether (1 0 0) can be formed as a linear combination of the first two vectors.

To construct a basis from a given set of vectors al, a2, ... , am that span a vector space test to see if a2 is a linear combination of al. If it is, delete it from the set. Then test if 83 is a linear combination of al and a2, or only a1 if a2 has been deleted. N ext test 84, etc., and in this manner delete all linearly dependent vectors from the set in order. The remaining vectors in the set form a basis of 'U.

11,

Theorem 3.10: Given a vector space Ii with a basis a1, a2, ... , am and with another basis bl, b2 , ••• , hI. Then m· = l.

Proof: Note al, hi, h2, ... , hI are a linearly dependent set of vectors. Using Theorem 3.9 delete the vector hk that is linearly dependent on al, hI, ... , h k -1. Then a1, hi, ... , b k -'-l, hk+ 1, ••• , bl still span 'U. Next note a2, a1, hI, ... , hk~l, b k +h .•• , hI are a linearly dependent set. Delete another b-vector such that the set still spans 'U. Continuing in this manner gives aI, •.. , 82, a1 span 'U. If l < m, there is an 81 + 1 that is a linear combination of aI, ... , 82, al. But the hypothesis states the a-vectors are a basis, so they an must be linearly independent, hence l ~ m. Interchanging the b- and a-vectors in the argument gives m ~ l, proving the theorem.

CHAP. 3]

ELEMENTARY MATRIX THEORY

49

Since all bases in a vector space 11 contain the same number of vectors, we can give the following definition. Definition 3.39: A vector space 11 has dimension n if and only if a basis of 11 consists of n vectors. Note that this extends the intuitive definition of dimension to a subspace of 'V m • 3.8 SOLUTION OF SETS OF LINEAR ALGEBRAIC EQUATIONS Now the rneans are at hand to examine the solution of matrix equations. Consider the matrix equation

o

Ax

If all ~i are zero, x = 0 is the trivial solution, which can be obtained in all cases. To obtain nontrivial solution, some of the ~. must be nonzero, which means the ai must be linearly dependent, by definition. Consider the set of all solutions of Ax = O. Is this a vector space? a

~

(1) Does the set contain at least one element? '

Yes, because x

=0

is always one solution.

(2) Are solutions closed under addition?

Yes, because ifAz = 0 and Ay = 0, then the sum x = z + y is a solution of Ax=O.

(3) Are solutions closed under scalar multiplication?

Yes, because if Ax = 0, then j3x is a solution of A({3x) = O. So the set of all solutions of Ax = 0 is a vector space. Definition 3.40: The vector space of all solutions of Ax = 0 is called the null space of A.

Theorem 3.11: If an m

n matrix A has n columns with r linearly independent columns, then the null space of A has dimension n - r. X

The proof is given in Problem 3.8. Definition 3.41: The dimension of the null space of A is called the nullity of A. Corollary 3.12:

If A is an n x n matrix with n linearly independent columns, the null space has dimension zero. Hence the solution x = 0 is unique.

Theorem 3.13: The dimension of the vector space spanned by the row vectors of a matrix is equal to the dimension of the vector space spanned by the column vectors. See Problem 3.9 for the proof. Example 3.23. 1 Given the matrix (

2 3).

246

It has one independent row vector and t,herefore must have only one

'

independent column vector.

Definition 3.42: The vector space of all y such that Ax = y for some x is called the range space of A. It is left to the reader to verify that the range space is a vector space.

50

ELEMENTARY MATRIX THEORY

[CHAP. 3

Example 3.24. The range space and the null space may have other vectors in common in addition to the zero vector. Consider

=

A

(~ ~)

b

=

(~)

c

=

(;)

Then Ab = 0, so b is in the null space; and Ac = b, so b is also in the range space.

Definition 3.43: The rank of the m

x n matrix A is the dimension of the range space of A.

Theorem 3.14: The rank of A equals the maximum number of linearly independent column vectors of A, i.e. the range space has dimension r. The proof is given in Problem 3.10. Note the dimension of the range space plus the dimension of the null space = n for an m x n matrix. This provides a means of determining the rank of A. Determinants can be used to check the linear dependency of the row or column vectors.

Theorem 3.15: Given an m

X

n matrix A. Then rank A = rank AT

= rankATA = rankAAT.

See Problem 3.12 for the proof.

Theorem 3.16: Given an m x n matrix A and an n x k matrix B, then rankAB ===-rank A,

rank AB ~ rank B. Also, if B is nonsingular, rank AB = rank A; and if A is nonsingular, rank AB = rank B.

See Problem 3.13 for the proof. 3.9 GENERALIZATION OF A VECTOR From Definition 3.7, a vector is defined as a matrix with only one row or column. Here we generalize the concept of a vector space and of a vector. Definition 3.44: A set 'U of objects x, y, z, ... is called a linear vector space and the objects x, y, z, ... are called vectors if and only if for all complex numbers a and (3 and all objects x, y and z in 'U: (1)

x + y is in

(2)

x

'U,

(3)

+ y = y + x, (x + y) + z = x + (y + z),

(4)

for each x and y in 'U there is a unique z in 1-l such that x + z

(5)

aX

(6)

a {(3x) =

(7)

Ix = x,

(8)

a(x + y)

(9)

(a + (3)x

is in

= y,

'U,

(a(3)x,

+ ay, aX + (3x.

aX

The vectors of Definition 3.7 (n-vectors) and the vector space of Definition 3.34 satisfy this definition. Sometimes a and (3 are restricted to be real (linear vector spaces -over the field of real numbers) but for generality they are taken to be complex here.

ELEMENTARY MATRIX THEORY

CHAP. 3]

51

Example 3.25. The set 11 of time functions that are linear combinations of sin t, sin 2t, sin St, ... is.a linear vector space. Example 3.26. The set of all solutions to dx/dt = A(t, x is a linear vector space, but the set of all solutions to dx/dt = A(t) x + B(t) u for fixed u(t) does not satisfy (1) or (5) of Definition 3.44, and is not a linear vector space. Example 3.27. The set of all complex valued discrete time functions x(nT) for n = 0,1, . .. is a linear vector space, as is the set of all complex valued continuous time functions x(t).

All the concepts of linear independence, basis, null space, range space, etc., extend immediately to a general linear vector space. Example 3.28. The functions sin t, sin 2t, sin St, .. ' form a basis for the linear vector space 11 of Example 3.25, and so the dimension of 1l is countably infinite.

3.10 DISTANCE IN A VECTOR SPACE The concept of distance can be extended to a general vector space; To compare the distance from one point to another, no notion of an origin need be introduced. Furthermore, the ideas of distance involve two points (vectors), and the "yardstick" measuring distance may very well change in length as it is moved from point to point. To abstract this concept of distance, we have Definition 3.45: A metric, denoted p(a, b), is any scalar function of two vectors a E 14 and b E 14 with the properties (1)

p(a, b) ~ 0

(2)

p(a, b) = 0 if and only if a points coincide),

(3)

p(a, b)

=

(distance is-always positive),

p(b, a)

=b

(zero distance if and only if the

(distance from a to b is the same as distance

from b to a), (4)

p(a, b)

+ p(b, c)

~

p(a, c)

(triangle inequality).

Example 3.29. (a) An example of a metric for n-vectors a and b is

p(a, b)

=

[1

(a-b)t(a-b) (a - b)t (a - b)

+

J1I2

(b) For two real continuous scalar time functions x(t) and yet) for to:::: t === tv one metric is p(x, y)

=

[5,:'

[x(t) - yet)], dt

J12

Further requirements are sometimes needed. By introducing the idea of an origin and by making a metric have constant length, we have the following definition.

Definition 3.46: The norm, denoted Iiall, of a vector a is a metric from the origin to the vector a E 11, with the additional property that the "yardstick" is not a function of a. In other words a norm satisfies requirements (1)-(4) of a metric (Definition 3.45), with b understood to be zero, and has the additional requirement (5)

Ilaall = lailiali.

52

ELEMENTARY MATRIX THEORY

[CHAP. 3'

The other four properties of a metric read, for a norm, (1)

]]all

(2)

Ila[[

(3)

[Iall = Ila]1 (trivial), Ila[ I + Ileil ~ Iia + ell·

(4)

~ 0,

=0

if and only if a = 0,

Example 3.30. A norm

in

'V2 is

II(al a2)112

~

Ilall2 :::: v'lall2 + la212.

Example 3.31. A norm in 'V 2 is

II(al a2)lll = Ilall l = lal! + la21.

Example 3.32. A norm in 'V n is the Euclidean norm

Ila112::::.ya:Fa

In the above examples the subscripts 1 and 2 distinguish different norms. Any positive monotone increasing concave function of a norm is a metric from 0 to a. Definition 3.47: The inner product, denoted (x, y), of any two vectors a and b in 11 is- a complex scalar function of a and b such that given any complex numbers a and {3, (1) (a, a) ;:".. 0, (3)

(a, a) = 0 if and only if a = 0, (a, b)* = (h, a),

(4)

(aa + ph, c)

(2)

=

a*(a, c)

+ ,8*(b, c).

An inner product is sometimes known as a scalar or dot product. Note (a, a) is real, (a, ab) = a(a, b) and (a, 0) = Example 3.33. An inner product in 'V n is

o.

n

(a, b) = atb = ~

i=l

at bi •

Example 3.34. An inner product in the vedor space 'U of time functions of Example 3.25 is (x, y)

=

i7To x*(t)

y(t) dt

Definition 3.48: Two vectors a andb are orthogonal if (a, b) = O. Example 3.35. . In three-dimensional space, two vectors are perpendicular if they are orthogonal for the inner product of Example 8.83.

Theorem 3.17: For any inner product (a, b) the Schwarz inequality [(a, b)]2::::::: (a, a)(b, b) holds, and furthermore the equality holds if and only if a = ab or a or b . . is the zero vector. Proof: scalar j3,

If a or b is the zero vector, the equality holds, so take b 0::::::: (a + ph, a + fih) = (a, a)

-=F

O. Then for any

+ ,8*(b, a) + j3(a, h) + 1J312(b, b) where the equality holds· if and only if a + Ph = O.. Setting j3 = -(b, a)/(b, b) and rearranging gives the Schwarz inequality.

ELEMENTARY MATRIX THEORY

CHAP. 3]

53

Example 3.36.

Using the inner product of Example 3.34,

f

1 Theorem 3.18: Proof:

x*(t) y(t)

(f

dt I' '""

For any inner product,

Ix*(t)I'

dt) (f

ly(t)12

dt)

V(a + b, a + b) ..,,::; V(a, a) + V(b, b).

(a+b,a+b) = l(a+b,a+b)1

=

I(a, a)

+ (h, a) +

(a, h) + (h, b)] ~ I(a, a)1 + I(b, a)] + ](a, h)1 + I(h, b)1

Use of the Schwarz inequality then gives (a + b, a + b) ~ (a, a) + 2v'(a, a)(b, b) + (b, b)

and taking the square root of both sides proves the theorem. This theorem shows that the square root of the inner product (a, a) satisfies the triangle inequality. Together with the requirements of the Definition 3.47 of an inner product, it can be seen that y(a, a) satisfies the Definition 3.46 of a norm. Definition 3.49: The natural norm, denoted

Ila112'

of a vector a is

IIal12 =

V(a, a).

Definition 3.50: An orthonormal set of vectors aI, a2, ... ,ale is a set for which the inner product if i # if i = j

{~

j}

where 8ij is the Kronecker delta. Example 3.37.

An orthonormal set of basis vectors in 'V n is the set of unit vectors

ell e21 ••.

I

en, where

ei

is defined as

o

o =

1

_

ith position

o o

Given any set of k vectors, how can an orthonormal set of basis vectors be formed from them? To illustrate the procedure, suppose there are only two vectors, al and a2. First choose either one and make its length unity:

becomes Because hI must have length unity, Yl = into its components:

IIa11121,

Now the second vector must be broken up

54

ELEMENTARY_ MATRIX THEORY

[CHAP. 3

Here YZb 1 is the component of a 2 along b p and a 2 - 1 2b 1 is the component perpendicular to hi" To find 12' note that this is the dot, or inner, product 12 = b 1t . a 2 • Finally, the second orthonormal vector is constructed by letting a z -1Zhl have unit length: az - YZh l

IIa2 ~ 1 2b 1 11z This process is called the Gra1n-Schmit orthonormalization procedure. The orthonormal hj is constructed from aj and the preceding hi for i = 1, ... , j -1 according to j-1

aj

i~ (bi , aj)bi

-

j-1

II aj

i~ (hi' aj)hi 112

-

Example 3.38.

ai

a~ = (1 -2 1)/,;2.

Given the vectors = (1 1 0). By the formula, the numerator is 82 -

(b l ,az)b1

=

Then

1181112 =,;2,

~(-~) - ~(i O)~(-~)~(~) Y2 Y2 V2 1

1

Making this have length unity gives

-{2

1

so

_1 2V2

0

hi = (1 1 0)//2.

(-!) 2

b~ = (3 -3 2)/-122.

Theorem 3.19: Any finite dimensional linear vector space in which an inner product exists has an orthonormal basis. Proof: Use the Gram-Schmit orthonormalization procedure on the set of all vectors in 'U, or any set that spans 'U, to construct the orthonormal basis.

3.11 RECIPROCAL BASIS Given a set of basis vectors hit h2' ... , b n for the vector space x E 'V n can be expressed uniquely as

~n,

an arbitrary vector

(3.4) Definition 3.51: Given a set of basis vectors h1, h2, ... , hn for G()n, a reciprocal basis is a set of vectors r1, r2, ... , rn such that the inner product i, j

=

1, 2, ... , n

(3.5)

Because of this relationship, defining R as the matrix made up of the ri as column vectors and B as the matrix with hi as column vectors, we have RtB

=

I

which is obtained by partitioned multiplication and use of (3.5). Since B has n linearly independent column vectors, B isnonsingular so that R is uniquely determined as R = (B-1)t. This demonstrates the set of ri exists, and the set is a basis for 'Vn. because R-1 = Bt exists so that all n of the rr are linearly independent. Having a reciprocal basis, the coefficients- f3 in (3.4) can conveniently be expressed. Taking the inner product with an arbitrary ri on both sides of (3.4) gives (ri' x)

= f3 t (ri' hi) + f3 2(ri' h z) + ... + f3 n (ri , h n )

Use of the property (3.5) then·gives (:3i = (ri'x),

~,CHAP. 3]

ELEMENTARY MATRIX THEORY

55

Note that an orthonormal basis is its own reciprocal basis. This is what makes "breaking a vector into its components" easy for an orthonormal basis, and indicates how to go about it for a non-orthonormal basis in CVn.

'3.12 MATRIX REPRESENTATION OF A LINEAR OPERATOR Definition 3.52: A linear operator L is a function whose domain is the whole of a vector space 'U 1 and whose range is contained in a vector space 112 , such that for a and b in 'U1 and scalars ll' and /3, L(ll'a + ,8b) = ll'L(a)

+ ,BL(b)

Example 3.39.

An m X n matrix A is a linear operator whose domain is the whole of CU n and whose range is in CUm, i.e. it transforms n-vectors into m-vectors. , Example 3.40.

Consider the set 'U of time functions that are linear combinations of sin nt, for n = 1,2, . ...

Then

00

any x(t) in 'U can be expressed as x(t) = time is a linear operator, because

~ ~n sin nt.

The operation of integrating with respect to

n=l

Example 3.41.

The operation of rotating a vector in CU 2 by an angle ¢ is a linear operator of 'V 2 onto CU 2 • Rotation of a vector aa + f3b is the same as rotation of both a and b first and then adding a times the rotated a plus fi times the rotated b.

Theorem 3.20: Given a vector space

1,11 with a basis b 1 , b 2 , ••• , bn, ... , and a vector space with a basis C1, C2, ••• , Cm, • • •• Then the linear operator L whose domain is 'U 1 and whose range is in 112 can be represented by a matrix {YjJ whose ith column consists of the components of L(b i ) relative to the basis Cl, C2, ••• , Cm, • • • •

112

Proof will be given for n dimensional Then x

=

:t

pib i •

'U 1

and nt dimensional

Furthermore since L(b i ) is a vector in

1l2 •

'U 2 ,

Consider any x in

'U 1•

determine Yji such that

i=1

t

L(x)

PiL(b)

i=l

n

Hence the jth component of L(x) relative to the basis {c j } equals ~ {Yji} times a vector whose components are the Pi. i=l

Yji,811

i.e. the matrix

Example 3.42.

The matrix representation of the rotation by an angle ¢ of Example 3.41 can be found as follows. Consider the basis e 1 = (1 0) and e2 = (0 1) of CU 2• Then any x = (fi1 f32) = file1 + f32 e2' Rotating e 1 clockwise by an angle ¢ gives L(e1) (cos 4»e1 - (sin ¢)e2. and similarly L(e2) (cos ¢)e2 + (sin 1»ev so that 1'11 = cos 1>, 1'21 = -sin ¢, Y12 = sin 1>. Y22 = cos ¢. Therefore rotation by an angle 1> can be represented by COS 1> sin L ( -sin 1> cos ¢

=

=

¢)

[CHAP. 3

ELEMENTARY MATRIX THEORY

56 Example 3.43.

An elementary row or column operation on a matrix A is represented by the elementary matrix E as given in the remarks following Definition 3.26.

The null space, range space, rank, etc., of linear transformation are obvious extensions of the definitions for its matrix representation.

3.13 EXTERIOR PRODUCTS The advantages of exterior products have become widely recognized only recently, but the concept was introduced in 1844 by H, G. Grassmann. In essence it is a generalization of the cross product. We shall only consider its application to determinants here, but it is part of the general framework of tensor analysis, in which there are many applications. First realize that a function of an m-vector can itself be a vector. For example, z = Ax: + By illustrates a vector function z of the vectors x and y. An exterior product,pP is a vector function of pm-vectors, 31, a2, ••• , 3 p • However,,pP is an element in a generalized vector space, in that it satisfies Definition 3.44, but has no components that can be arranged as in Definition 3.7 except in special cases. The functional dependency of P = 3i /\ 3j /\ ... /\ 3k for p = 0,1,2, ... , n is a vector in an abstract vector space, denoted /\1YU, such that for any complex numbers a and (3, (aai

+ (3aj) /\ ak /\

ai/\'"

3i /\

1\3j/\"

' •. /\ at

'/\ak/\'"

aj /\ ... /\ 3k #- 0 if

Equation (8.6) says

~p

=

a(ai /\ 3k /\ ... /\ al) /\ar

=

ai, 3j, ••• , 3k

-3i/\"

+ (3(aj /\ ak /\

' / \ ak/\"

••• /\ aL)

(8.6)

/\31

(8.7)

'/\3j/\'"

are linearly independent.

is multilinear, and (3.7) says pP is alternating.

Example 3.44. The case p= 0 and p = 1 are degenerate, in that 1\ o'U is the space of all complex numbers and 1\ 111 = 1J, the original vector space of m-vectors having dimension n. The first nontrivial example is then 1\2'U. Then equations (3.6) and (3.7) become the biliriearity property (3.8)

(3.9)

and also Interchanging the vectors in (3.8) according to (3.9) and multiplying by -1 gives ak /\ (a3t

By (3.10) we see that (aai) /\ (a3) # a(3i /\ 3j).

Hi /\

Note that setting 3) = bina tion of ai' ~ /\ 3j = O. Let b 1• b 2 ,

••• ,

+ f3 a j)

=

aj is linear in either 3i

a(3k /\ 3i)

3i

3j

(bilinear) but not both because in general

n

Then

(3.10)

Furthermore if and only if

in (3.9) gives

b n be a basis of 'U.

or

+ /3(ak /\ aj)

3 i :::::::

~

/-:=1

3j

n

a'kbk

and

3j

=

:I 1=1

I'tbl> so that

is a linear com-

ELEMENTARY MATRIX .THEORY

':CHAP.3] ,'Since hk /\ hk ::::: 0 and hk /\ b t

:::::

57

for k> l, this sum can be rearranged to

-b[ /\ b k

1.,..1

11.

=

ai /\ aj

~

~ (llk'Yl- lll'Yk)b k /\h l

(3.11)

1=1 k=l

Because ai /\ aj is an arbitrary vector in /\2'11, and (3.11) is a linear combination of hie /\ hb then the vectors hk /\ h z for 1"";;: k < l ".,;;: n form a basis for /\2'11. Summing over all possible k and l satisfying this relation shows that the dimension of /\211 is n(n - 1)/2

::::: (;).

Similar to the case /\ 2U, if any ai is a linear combination of the other a's, the exterior product is zero and otherwise not. Furthermore the exterior products bi /\ h j /\ ••• /\ bk for 1 ~ i < j < ... < k ~ n form a basis for/\ P'U, so that the dimension of /\ Pl.{ is n! (n-p)ip!

(8.12)

In particular the only basis vector for 1\ nu is hi /\ b2 /\ ... /\ bn, so 1\ nu is one-dimensional. Thus the exterior product containing n vectors of an n-dimensional space U is a scalar. Definition 3.54: The determinant of a linear operator L whose domain and range is the vector space U with a basis hi, h2' ... , bn is defined as L(h~) /\ L(h2) /\ ... /\

detL

hi /\ b 2 /\

••• /\

L(bn) bn

(3.13)

The definiiton is completely independent of the matrix representation of L. If L is an 11, x

n matrix A with column vectors aI, a2, ... , an, and since el, e2, ... , en are a basis detA

=

Ae1 /\ Ae2 /\ ... '/\ Aen el /\ e2 /\ . . . /\ en

Without loss of generality, we may define ,and det I

= el /\ ez /\ ... /\ en = 1.

det A =

a1 /\ a2 /\ • . • /\

an

e1 /\ e2 /\ ... /\ en

e1 /\ e2 /\ ...

Aen

= 1,

so that (3.14)

a1 /\ a2 /\ • • . /\ an

N ow note that the Grassmann notation has a built-in multiplication process. Definition 3.55: For an exterior product ~p = a1/\ . . . /\ ap in /\ PU and an exterior product ",q = C1 /\ ••• /\ Cq in /\ qu, define exterior rnultiplication as cpP /\ ",q

=

a1 /\ ••• /\

ap

/\

C1 /\ ••• /\ Cq

(3.15)

This definition is consistent with the Definition 3,53 for exterior products, and so cpP /\ ",q is itself an exterior product.

Also, if 1n

=n

then

a n -1 is an n-vector since ~n-l has dimension n from since u and /\ n-'-lU must coincide.

a1/\ ••• /\ an-2 /\

(3.12), and must equal some vector in

'U

:Theorem 3.21. (Laplace expansion). Given an 11, x n matrix A with column vectors Then det A = a1 /\ 82 /\ ... /\ an = a[(a2 /\ ... /\ an).

ai.

C':,y Proof: Let ei be the ith unit vector in 'Vnand Ej be the ith unit vector in 'Vn-l, i.e. ei has ',''1,'b.components and Ej has n -1 components. Then ai = alie1 + a2ie2 + ... + aniCn so that

58

ELEMENTARY MATRIX THEORY

[CHAP. 3

The first exterior product in the sum on the right can be written el /\ a2 /\ ... /\ an = el /\ (al2e1

+ a22e2 + ... + an2en) /\

••• /\ (alnel

+ ... + annen )

Using the multilinearity property gives el /\ a2 /\ . • . /\ an

0) ( a22

el /\

:

/\ . . . /\

anz

Since detIn = 1 = detln -

(0 ) aZn :

ann

then

1,

Performing exactly the same multilinear operations on the left hand side as on the right hand side gives e,

1\

[a22 (:J

an,(:'J} ... [a2n (:.) + ... + ann(:JJ

+ ... +

1\

(~22) /\ ... /\ (7")

=

an2

=

detMn

Cn

ann

where Cll is the cofactor of all. Similarly, n

=

ej /\ 32 /\ ••• /\ 3n

and

Cjl

a1 /\ 32/\

~

••• /\ 3 n

aj1Cjl

== (Cn

C21 ••• Cn 1)a1

j=1

so that a2/\ .•. /\ an = (Cll C21 ••• Cnl)T and Theorem 3.21 is nothing more than a statement of the Laplace expansion of the determinant about the first column. The use of column interchanges generalizes the proof of the Laplace expansion about any column, and use of det A = det AT provides the proof of expansion about any row.

Solved Problems 3.L

Multiply the following matrices, where elements.

(!)

(i)

(1, 2)

(ii)

(~) (3

4)

aI, 32, hI

(iii)

(:D

(iv)

(a,la2)

and b 2 are column vectors with n

(b'lb2 )

(v)

(~ ~J(j)

~D

(vi)

(a, Ia2)

(~

~)

U sing the rule for multiplication from Definition 3.13, and realizing that multiplication of a k X n matrix times an n X m matrix results in a k X m matrix, we have

+ 2 X 4)

(i)

(1 X 3

(ii)

1 X 3) ( (2 X 3)

= (11)

(1 X4») (2 X 4)

=

(iii)

(v)

(iv)

(vi)

3J

ELEMENTARY MATRIX THEORY

59

Find the determinant of A (a) by direct computation, (b) by using elementary row and column operations, and (c) by Laplace's expansion, where

A

(1

::=

0 2 3 0

0 0 1

D

0

(a) To facilitate direct computation, form the table Ph Pz, Pa, P4

P

0 1 2

2

3

1

2

1

3

1

P

PI' P21 Ps, P4

4

12

3

1

2

4

4

3

13

3

1

4

2

4

2

14

3

2

4

1

3

1

3

2

4

15

3

2

1

4

4

1

4

2

3

16

3

4

1

2

5

1

4

3

2

17

3

4

2

1

6

2

4

3

1

18

4

3

2

1

7

2

4

1

3

19

4

3

1

2

8

2

3

1

4

20

4

2

1

3

9

2

3

4

1

21

4

2

3

1

10

2

1

4

3

22

4

1

3

2

11

2

1

3

4

23

4

1 , 2

3

= 24 terms in this table. Then detA = +1· 2 ·1 • 2 - 1 • 2 • 8 • 0 + 1· 0 • 8 • 0 -

There are 4!

= 4

l ' 0 • 3' 2

+ 1'6·3'0 -1'6·1'0 + 0'6'1'0 - 0'6'1'0 + 0'0'1'2 - 0·0·8'0 + 0'1'8'0 - 0'1'1·2 + 0'1'3'2 - 0·1·g'0 + 0·2·g·0 - 0,2,1-2 + 0'6'1'0 - 0'6'3'0 + 2·0·3'0 - 2'0·1·0 + 2-2-1-0 - 2·2'1'0 + 2'1·1'0 -2-1'3-0

(b) Using elementary row and column operations, subtract I, 3 and 4 times the bottom row from

the first, second and third rows respectively. This reduces A to a triangUlar matrix, whose determinant is the product of its diagonal elements l ' 2 ·1· 2 = 4. (c)

Using the Laplace expansion about the bottom row,

det A

=

2 det

(~ ~ ~ )

=

2•2

4

131

3.3.

Prove that det (AT) = det A, where A is an n X n matrix. If A

= {Ctij},

then AT

= {Ct

n!

ji }

so that

det (~.T)

= ~ (-1)PCtP1 l a

p2 2' •.

uPnn •

Since a determinant is all possible combinations of products of elements where only one element is taken from each row and colUmn, the individual terms in the sum are the same. Therefore the only question is the sign of each product. Consider a typical term from a 3 X 3 matrix: aala12u23, i.e, P1 3, P2 1, Ps 2. Permute the elements through a12a31a23 to Ct12a2Sa3i> 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<.i===n-I

det A det C, where A and Care n x nand m x m matrices

respecti vely. Either det A=::O or det A :/= O. If det A = 0, then the column vectors of A are linearly dependent.

Hence the column vectors of (:) are linearly dependent, so that det(!

and the hypothesis holds.

~)

= 0

If det A oF 0, then A-I exists and

(!

~)

(!

:)(~ ~)(!

A~IB)

The rightmost matrix is an upper triangular matrix, so its determinant is the product of the diagonal elements which is unity. Furthermore, repeated use of the Laplace expansion about the diagonal elements of I gives det (

! ~)

=

det A

det

(! ~ )

Use of the product rule of Theorem 3.2 then gives the proof.

=

det C

3.6.

61

ELEMENTARY MATRIX THEORY

CHAP.8J

Show that if at, a2, ... , ak are a basis of 'U, then every vector in 'U is expressible uniquely as a linear combination of aI, a2, ... , a". Let x be an arbitrary vector in 'U. Because x is in 'U, and 'U is spanned by the basis vectors aI' a2' ... , ak by definition, the question is one of uniqueness. If there are two or more linear combinations of a l1 a2' ... , ale that represent x, they can be written as k

x

=

~ f3iai

i=1

and

k

~

x

aiai

i=l

Subtracting one expression from the other gives

Because the basis consists of linearly independent vectors, the only way this can be an equality is for f3i = ai' Therefore all representations are the same, and the theorem is proved. Note that both properties of a basis were used here. If a set of vectors did not span 'U, all vectors would not be expressible as a linear combination of the set. If the set did span 'U hut were linearly dependent, a representation of other vectors would not be unique.

3.7.

Given the set of nonzero vectors aI, a2, ... , am in G()n. Show that the set is linearly dependent if and only if some ak, for 1 < k ,.:::: m, is a linear combination of aI, a2, ... , ak-l.

If part:

If ak is a linear combination of aI' a2' •.. , ak -1> then k-t ak

where the f3i are not all zero since

o

= f3lal

ak

=

is nonzero.

~ f3i ai

i=1

Then

+ f32a2 + ... + f3k-lak-l +

(-l)ak

+ Oak+l + ... + Oam

which satisfies the definition of linear dependence. Only

if part:

If the set is linearly dependent, then

where not all the f3i are zero. linear combination is

3.8.

Find that nonzero 13k such that all

f3i = 0 for

i> k.

Then the

Show that if an In. x n n1atrix A has n columns with at most r linearly independent columns, then the null space of A has dimension n - r. Because A is m X n, then ai E CUm' X E "Un- Renumber the ai so that the first r are the independent ones. The rest of the column vectors can be written as

(3.16)

beca use a r + I' .•. , an are linearly dependent and can therefore be expressed as a linear combination of the linearly independent column vectors. Construct the n - r vectors Xl' X2 • ••. ,x n - r such that

62

ELEMENTARY MATRIX THEORY

Xl

/311

/321

/3n- r.1

/312

/322

/3n-r,-2

f31r

=

[CHAP. 3

/32r

X2

-1 0

0 -1

0

0

/3n-r,r

Xn - r

0 0 -1

Note that Ax1 = 0 by the first equation of (3.16), Ax2 = 0 by the second equation of (3.16), etc., so these are n - r solutions. Now it will be shown that these vectors are a basis for the null space of A, First, they must be linearly independent because of the different positions of the -1 in the bottom part of the vectors. To show they are a basis, then, it must be shown that all solutions can be expressed as a linear combination of the "4. i.e. it must be shown the Xj span the null space of A. Consider an arbitrary solution x of Ax = O. Then ~1

/311

~r

f31r -~r+l

~r+ 1

X

/3n-r, 1

/321

/32r

-

-1

~r+2

-

0

.,

.

0"1

ar

/3n-r,r

-

~n

+

0

0

~r+2

0

-1

0

0

~n

0

0

-1

0

n-r

Or, in vector notation,

x

=

~

i=l

+

-~r+ixt

S

where s is a remainder if the xi do not span the null space of A. If s = 0, then the "4 do span the null space of A. Check that the last n - r elements of s are zero by writing the vector equality as a set of scalar equations. n-r

Multiply both sides of the equation by A.

~ -~r+iAxi

Ax

i=l

Since Ax = 0 and r

gives

n- r

3.9.

:£

O"i~ = O.

i=l Xi are

~

= 0, As = O.

+

As

Writing this out in terms of the column vectors of A

But these column vectors are linearly independent, so

ai

= O.

Hence the

a basis, so the null space of A has dimension n - r.

Show that the dimension of the vector space spanned by the row vectors of a matrix is equal to the dimension of the vector space spanned by the column vectors. Without loss of generality let the first r-column vectors 3 i be linearly independent and let s of the row vectors ai be linearly independent. Partition A as follows: 1

aLr+l

atn

arn

. , .... , .......... , . , ... . ~

A

arl

ar 2

a r +l,l

U r +1,2

•.•

arr

I

ar. r + 1

a r +1,r

1

a r +Lr+l

...

U r +1,n

---------L------

.......................... ··1···'·······,·······,··· I am • r +l

I

aLr+l

1

ar. r + 1

..... ·1·'·'·········,·······,· =

Xr

Xr + l i a r +1, r + 1

• • .

am

ar + L

n

..... 'I······················· Xm

1 a-m,r-t 1

63

ELEMENTARY MATRIX THEORY

3]

Yl

' .,

Yr+

Yr

I

' • •

Yn)

-----~-~---"""---

( '~~~ , , .,' •• ,' , • ~~r' • , ~~:.~~ ~' , , ',' .. ,' • , ;~~

that Xi = (ail ai2 for some nonzero bi ,

r+l

yr

and = (ali a2j a r + l,j)' Since the ~ Let the vector b T = (b l bz .,' br + 1) so that

,"

SO

air)

Xi

are T-vectors, ~

i=l

bixi

= 0

1+1

o

~ biXi

i=l

Therefore bTYj = 0 for j = 1,2, ' .. ,1", T

Since the last n -

l'

r

Then

Yi

o=

ai

r

= ~ a··y·

so that

~J J

3=1

column vectors a i are linearly dependent,

(b TYl b TY2

bTYi

'"

= ~

]=1

aijbTYj

= 0 for i =

bTYt bTYr+ 1 . . .

bTYrJ

=

for i

Cl'ijaj

= 1'+ 1, ' , "n.

3=1

r+

blal

= ~

1, ' , "n,

Hence

+ b2a 2 + ... + br + 1 arT 1

Therefore l' + 1 of the row vectors ai are linearly dependent, so that s::::: '/", The same argument leads to r:::::: 8, so that r = 8,

Now consider AT.

Show that the rank of the m X n matrix A equals the maximum number of linearly independent column vectors of A. Let there be r linearly independent column vectors, and without loss of generality let them be r

aI'

a2 •

The

' , ., aT'

ai

= ~ j=1

aijaj

for i =

r

+ 1, . , "

Any Y in the range space of A can be ex-

n,

pressed in terms of an arbitrary x as n

y = Ax

=

~

aixi

i=l

This shows the ai for i = 1, ...• r span the range space of A, and since they are linearly independent they are a basis, so the rank of A = r.

For an m x n matrix A, give necessary and sufficient conditions for the existence and uniqueness of the solutions of the matrix equations Ax = 0, Ax = b and AX = B in terms of the column vectors of A. For Ax = 0, the solution x = 0 always exists, A necessary and sufficient condition for uniqueness of the solution x = 0 is that the column vectors a 1.,." an are linearly independent. To'show the necessity: If a i are dependent, by definition of linearly dependent some nonzero ~i n

exist such that

~ ai~[ =

O.

Then there exists another solution x = (~l , .. ~n) # O.

i=l

To show

n

sufficiency: If the a[ are independent, only zero ~i exist such that

~ ai~i

= Ax = 0,

i=l

For Ax = b, rewrite as

b

=

n

~ aixi'

Then from Problem 3.10 a necessary and sufficient

i=l

condition for existence of solution is· that b lie in the range space of A, i.e. the space spanned by the column vectors. To find conditions on the uniqueness of solutions, write one solution as n (1']1 '1]2 ' • , 7J n )

and another as (~1 g2 ' ., ~?l)'

n

Then b= ~ a i 7Ji = ~ ai~i so that 0 = i=1

i=1

The solution is unique if and only if aI' ... , an are linearly independent,

n

~

i=l

(1,11 -

~i)ai'

64

ELEMENTARY ·MATRIX THEORY

[CHAP. 3

Whether or not b = 0, necessary and sufficient conditions for existence and uniqueness of solution to Ax = b are that b lie in the vector space spanned by the column vectors of A· and that the column vectors are linearly independent. Since AX = B can be written as Axj = b j for each column vector Xj of X and hi of B, by the preceding it is required that each hi lie in the range space of A and that all the column vectors form a basis for the range space of A.

3.12. Given an m

X

n matrix A.

Show rank A

= rank AT = rankATA = rankAAT.

By Theorem 3.14, the rank of A equals the number r of linearly -independent column vectors of A. Hence the dimension of the vector space spanned by the column vectors equals r. By Theorem 3.13, then the dimension of the vector space spanned by the row vectors equals r. But the row vectors of A are the column vectors of AT, so AT has rank r. To show rank A = rankATA, note both A and ATA have n columns, Then consider any vector y in the null space of A, i.e. Ay O. Then ATAy 0, so that y is also in the null space of ATA. Now consider any vector z in the null space of ATA, i.e. ATAz = O. Then zTATAz = llAzll~ = 0, so that Az = 0, i.e. z is also in the nun spaca of A. Therefore the null space of A is equal to the null space of ATA, and has some dimension k. Use of Theorem 3.11 gives rank A = n-k = rank ATA. Substitution of AT for A in this expression gives rank AT = rank AAT.

=

=

3.13. Given an m x n matrix A and an n X k matrix B, show that rankAB ==== rank A, rankAB ==== rankB. Also show that if B is nonsingular, rank AB = rank A, and that if A is nonsingular, rankAB = rankB. Let rank A = r, so that A has r linearly indepedendent column vectors r

ai =

n

~ (lijaj _ for i = r + 1, ... , n.

Therefore AB

j=l

i=l

of B, using partitioned multiplication. AB

~

=

i=l

+.

aib'[

bi

= ~ aib: where

a 1, ••. , a"..

Then

are the row vectors

Hence

i

~=r+l

~

]=1

a'ijajb;

=

~

i=1

ai (bi +

:i akib~)

k=r+l

so that all the column· vectors of AB are made up of a linear combination of the r independent column vectors of A, and therefore rank AB ~ r. Furthermore, use of Theorem 3.15 gives rank B = rank BT. Then use of the first part of this problem with BT substituted for A and AT for B gives rankBTAT === rank B. Again, Theorem 3.15 gives rank AB = rank BTAT, so that rank AB === rank B. If A is nonsingular, rank B = rank A -l(AB) ~ rank AB, using A-I for A and AB for B in the first result of this problem. But since rank AB === rank B, then rank AB = rank B if A-1 exists. Similar reasoning can be used to prove the relnaining statement.

3.14. Given n vectors Xl, X2, ••• , Xn in a generalized vector space possessing a scalar product. Define the Gram matrix G as the matrix whose elements gij = (Xi, Xj). Prove that detG = 0 if and only if Xl,X2, •• • ,Xn are linearly dependent. Note that G is a matrix whose elements are scalars, and that Xi might be a function of time. Then from Theorem 3.7, /31g1 + fJ2g2

Suppose det G = O.

n

vector of G.

Then 0 =

~ fJiU'lj i=1

Multiplying by a constant

o =

+ ... + [:3ngn =

n

=

~ [:3i(Xi, x).

i=1

137 and summing still gives zero, so that n

n

j=l

i=l

~ /3; ~ fJ-i(xi, Xj)

0, where g is a column

ELEMENTARY MATRIX THEORY

'CHAP.8J

n

Use of property (2) of Definition 3.47 then gives dependence.

:.I

f3~xi

=

65

0, which is the definition of linear

i:::: 1

n

Now suppose the

xi

are linearly dependent.

Taking the inner product with any n ~ Yigj

xi

0 =

gives

Then there exist Yi such that n

n

~ Yj(xt. Xj)

j=1

~ YjUij

=

j=1

.~ YjXj

for any i.

=

O.

Therefore

j=1

= 0 and the column vectors of G are linearly dependent so that detG = O.

j=l

3.15. Given two linear transformations Ll and L2J both of whose domain and range are in 11. Show det(LIL2) = (detLl)(detL2) so that as a particular case detAB = det BA = det A det B. Let U have a basis hi. b 2 ,

••• ,

Using exterior products, from (3.13),

bn'

L 1L 2 (b 1)

L 1L 2 (b 2 ) /\ ••• /\ L 1L 2 (h n ) hI /\ b2 /\ • • • /\ b n

/\

If L2 is singular, the vectors L 2 (b i ) = Ci are linearly dependent, and so are the vectors L 1L 2 (b i ) = Then det (L 1L 2 ) = 0 = det L 2 • If L2 is nonsingular, the vectors Ci are linearly independent and form a basis of 'U. Then Cl/\ C2 /\ ••• /\ Cn ¥= 0, so that

£1 CCi)'

Cl /\

c2

/\ • • • /\

en

Using exterior products, prove that det (I + ab T) det (I + ab T )

=

= 1 + bTa.

(el + b 1a) /\ (e2 + b 2 a) /\ ••• /\ (en + bna) el

/\

e2 /\ ... /\ en

+

b 1a /\ e2/\ ... /\ en

+

+

b2e l /\ a /\ e3 /\ ... /\ en

+

bnel/\ ... /\ e n - l /\ a

Use of Theorem 3.21 gives det (I + ab T)

but

el

=

1

+

b1aT (e2/\'"

/\

en) -

b2a T(el/\ ea /\ ... /\ en)

+ '" +

(-l)n-l bnaT(e l

+

+ ... +

/\ . . . /\

en-I)

= (e2/\ ... /\ en), etc., so that det (I + ab T)

=

1

+

aTblel

a T b2e Z

aTbne n

=

1

+

aTb

Note use of this permits the definition of a projection matrix P = 1 - abT(aTb}-l such that det P = 0, Pa = 0, PTb = 0, p2 = P, and the transformation y = Px leave:;; only the hyperplane bTx pointwise invariant, i.e. bTy = bTpx = bT[I - abT(aTb)-l]x :::; (b T - bT)x == O.

66

ELEMENTARY MATRIX THEORY

[CHAP. 3

Supplementary Problems 3.17.

Prove that an upper triangular matrix added to or multiplied by an upper triangular matrix results in an upper triangulal' matrix. -

3.18.

Using the formula given in Definition 3.13 for operations with elements, multiply the following matrices

~)(: ~ s~ t)

(:

Next, partition in any compatible manner and verify the validity of partitioned multiplication.

~

2

3.19.

Transpose (

3.20.

Prove IA = AI = A for any compatible matrix A.

3.21.

Prove all skew-symmetric matrices have all their diagonal elements equal to zero.

3.22.

Prove (AB)t = BtAt.

3.23.

Prove that matrix addition and multiplication are associative and distributive, and that matrix addition is commutative.

3.24.

Find a nonzero matrix which, when multiplied by the matrix B of Example 3.5, page 40, results in the zero matrix. Hence conclude AB = 0 does not necessarily mean A = 0 or B = O.

3.25.

How many times does one particular element appear in the sum for the determinant of an n X n matrix'!

3.26.

Prove (AB)-1 = B-IA-l if the indicated inverses exist.

3.27.

Prove (A-l)T = (AT)-I.

3.28.

Prove det A-I = (det A)-I.

3.29.

Verify since

j

sin t

)

and then take the complex conjugate.

'

d et

(~ :)(-~ ~)

(21 2) (-1 (~ 2) 3

det 3

(~

1

.

~) -I

~)

=

det

(~

!)

Also verify

(-~ ~) (~ -1

~) -1

4

A

=

G D· 1 5

Find A-I.

3.30.

Given

3.31.

If both A-I and B-1 exist, does (A + B) -1 exist in general'!

3.32.

Given a matrix A(t) whose elements are functions of time.

3.33.

Let a nonsingular matrix A be partitioned into Alh A 12 , A21 and A22 such that Au and A22 - A21 Ail Al2 have inverses. Show that

dA

Show dA-l/clt = -A-laTA-I. 1

A-I

and if

A21

=

= 0, then

ELEMENTARY MATRIX THEORY

CHAP. 3]

67

3.34.

Are the vectors (2 0 -1 8), (1 -3 4 0) and (1 1 -2 2) linearly independent?

3.35.

Given the matrix equations

0'0'2122) (~1) ~2

(b)

(a)

(0)

=

:~:) G:)

(0)

0'32

0

(~)

=

0

Using algebraic manipulations on the scalar equations ai1~1 + ~2~2 = 0, find the conditions under which no solutions exist and the conditions under which many solutions exist, and thus verify Theorem 3.11 and the results of Problem 3.11. 3.36.

Let x be in the null space of A and y be in the range space of AT,

3.37.

Given

3.38.

For A =

matrix A = (

(_~

2 3 4)

~

-!).

1 -1 -1

Show xTy

== O.

Find a basis for the null space of A.

.

show that (a) an arbitrary vector z = (::) can be expressed as the

sum of two vectors, z = x + y. where x is in the range space of A and y is in the null space of the transpose of A, and (b) this is true for any matrix A. 3.39.

Given n X k matrices A and B and an m X n matrix X such that XA = XB. can we conclude A = B?

3.40.

Given x, yin '"'(..(, where b 1, b 2 ,

•• "

b n are an orthonormal basis of 'U.

Under what conditions

Show that

n

~ (x, bi)(bi • y)

(x, y)

i=l

3.41.

Given real vectors x and y such that IIxl12 = Ily112' Show ex + y) is orthogonal to

3.42.

Show that rank (A + B)

3.43.

Define T as the operator that multiplies every vector in 'Va by a constant a and then adds on a translation vector to. Is T a linear operator?

3.44.

Given the three vectors 81 = ev'IT9 -4 3), a 2 = (v'li9 -1 7) and a3 = eVi19 -10 -5), use the Gram-Schmit procedure on al' a 2 and as in the order given to find a set of orthonormal basis vectors.

3.45.

Show that the exterior product 4-P = 31/\ ... /\ a p satisfies Definition 3.44, i.e. that is an element in a generalized vector space /\P'ti, the space of all linear combinations of p-fold exterior products.

3.46.

Show (aIel + a2e2 + aSe3) /\ ({he1 + f32e2 + f3a e s) = illustrating that 3/\ b is the cross product in 'Va.

3.47.

Given vectors Xl. X2' •• "Xn and an n X n matrix A such that Yl' Y2' ... , Yn are linearly independent, where Yi = Axi . Prove that XII x2' ... ,xn are linearly independent.

3.48.

Prove that the dimension of

3.49.

Show that the remainder for the Schwartz inequality

~

rank A

(x - y).

+ rank B.

/\P'U =

(a2f33 - O!sf32)e1

0!3f31)e2

+ (a1f32 -

a2f31)ea.

n! (n - p)! p!

1

(a, a)(b, b) -

+ (alf33 -

I(a, b)12

n

n

~=1

,=1

= -2.~ .~

JUibj - u j b i l

2

for n-vectors. What is the remainder for the inner product defined as (a, b) =

f

a*(t) b(t) dt?

68

ELEMENTARY MATRIX THEORY

[CHAP. 3

Answers to Supplementary Problems 2a

3.18.

+ x2

aj

bx 2

(

3.24.

t

t)

'l7")

0 2

3.19.

+ sin

b sin

x2

sin t

-j

I" ( -_22:

I" ) :

for any a and fJ

3.25.

(n -1)! as is most easily seen by the Laplace expansion.

3.30.

A-I

=

(~:;~ ~ -~) -1

3.S1.

No

3.34.

No

3.37.

(5 -4 1 O)T

3.38.

(a) x

=

-1

1

and (6 -5 0 l)T are a basis.

(_~) a

ent .they span

and y

=::

(~) ()

where a and () are arbitrary, and since x and yare independ-

G[J2'

3.39.

The n column vectors of X must be linearly independent.

3.43.

No, this is an affine transformation.

3.44.

b 1 == (-v'i19 -4 3)/12, are required.

3.48.

One method of proof uses induction.

3.49.

~

ff

b2

== (0 3 4)/5

1a(t) beT) - aCT) bet) 12 dT dt

and

a3

is coplanar with

al

and a 2 so that only b 1 and b 2

Chapter 4 Matrix Analysis 4.1 EIGENVALUES AND EIGENVECTORS Definition 4.1:

An eigenvalue of the n x n (square) matrix A is one of those scalars A that permit a nontrivial (x # 0) solution to the equation

= AX AI)x = O.

(4.1)

Ax

Note this equation can be rewritten as (A exist only if det (A - AI) = O. Example 4.1. Find the eigenvalues of

(~

Then

{G :) -

A

(~

The characteristic equation is

:).

The eigenvalue equation is

n}(::) det

Nontrivial solution vectors x

C~

A 3

(~) ~ A)

or

=

O.

Then (3 - A)(3 - A) - 4 = 0 or 1.2 - 6A + 5 = 0,

a second-order polynomial equation whose roots are the eigenvalues

Al:= 1,

A2 = 5.

Definition 4.2:

The characteristic polynomial of A is det (A - AI). Note the characteristic polynOll1ial is an nth order polynomial. Then there are n eigenvalues Al, A21 ••. , An that are the roots of this polynomial, although some might be repeated roots.

Definition 4.3:

An eigenvalue of the square matrix A is said to be distinct if it is not a repeated root.

Definition 4.4:

Associated with each eigenvalue Ai of the n x n matTix A there is a nonzero solution vector Xi of the eigenvalue equation AXi = AiXi. This solution vector is called an eigenvector.

Example 4.2 •• In the previous example, the eigenvector assocated with the eigenvalue 1 is found as follows.

(~)

or Then 2xl + 4X2 = 0 and Xl + 2X2 = 0, from which where the scalar x2 can be any number.

Xl

=

---:2X2'

Thus the eigenvector

Xl

is

Xl

=

(-~) X2

Note that eigenvectors have arbitrary length. This is true because for any scalar the equation Ax = Ax has a solution vector aX since A(ax) = aAx = aAx = A(ll'x).

69

a',

70

MATRIX ANALYSIS

Definition 4.5:

[CHAP. 4

An eigenvector is normalized to unity if its length is unity, i.e. Ilxll = l. Sometimes it is easier to normalize x such that one of the elements is unity.

Example 4.3. The eigenvector normalized to unit length belonging to the eigenvalue 1 in the previous example is Xl

1(-2) h

= _r;;:

v5

1

I

..

, w ereas norma 1'" Izmg Its fi rst e ement to umty gIves

Xl

=

( 1) -1/2

.

4.2 INTRODUCTION TO THE SIMILARITY TRANSFORMATION Consider the general time-invariant continuous-time state equation with no inputs, dx(t)/dt = Ax(t)

(4.2)

where A is a constant coefficient n x n matrix of real numbers. given as x(O) = Xo.

The initial condition is

Example 4.4. Written out, the state equations (4.2) are dXI(t)/dt

= allxI(t)

dX2(t)/dt

=

a2l x l(t)

+ al2x 2(t) + ... + alnXn(t) + a22x 2(t) + ... + a2nx n(t)

and the initial conditions are given, such as

Now define a new variable, an n-vector y(t), by the one to one relationship y(t) = M-l.x(t)

(4.3)

It is required that M be an n X n nonsingular constant coefficient matrix so that the solution x can be determined from the solution for the new variable y(t). Putting x(t) = My(t) into the system equation gives M dy(t)/dt = AMy(t)

Multiplying on the left by M-l gives dy(t)/dt

Definition 4.6:

= M-1AMy(t)

(4.4)

The transformation T-IAT, where T is an arbitrary matrix, is called a similarity transformation on A. It is called similarity because the problem is similar to the original one but with a change of variables from x to y.

Suppose M was chosen very cleverly to make M-1AM a diagonal matrix.A.. Then dy(t)

----a;r--- =

!!:.dt

(

Yl(t) ) Y2(t) : Yn(t)

=

( ~ ~~ ..

o

0

..

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

An

Yn(t)

.A.y( t)

CHAP. 4]

MATRIX ANALYSIS

71

Writing this equation _out gives dyJdt = AiYi for i = 1,2, ... , n. The solution can be expressed simply as Yi(t) = Yi(O)e Ait • Therefore if an M such that M-1AM =.A. can be found, solution of dx/dt = Ax becomes easy. Although not always, such an M can usually be found. In cases where it cannot, a T can always be found where T-1AT is aln10st diagonal. Physically it must be the case that not all differential equations can be reduced to this simple form. Some differential equations have as solutions teA;t, and there is no way to get this solution from the simple form. The transformation M is constructed upon solution of the eigenvalue problem for all the eigenvectors Xi, i = 1,2, ... , n. Because AXi = A1Xi for i = 1,2, ... , n, the equations can be "stacked up" using the rules of multiplication of partitioned matrices: (AXI I AX21 ... I AXn) (AlXl I A2 x 21 ... IAnXn)

(Xl

!

XQ

I . . . IXn)

(~~ ~2 ••

o

•••••••••••

0

...

~ ~) •

An

M = (Xl IX2! . . . IXu)

Therefore

(4.5)

When M is singular, A cannot be found. Under a number of different conditions, it can be shown M is nonsingular. One of these conditions is stated as the next theorem, and other conditions will be found later.

Theorem 4.1:

If the eigenvalues of an n linearly independent.

X

n matrix are distinct, then the eigenvectors are

Note that if the eigenvectors are linearly independent, M is nonsingular.

Proof: The proof is by contradiction. Let A have distinct eigenvalues. Let Xl, XQ, ••• , Xu be the eigenvectors of A, with Xl, XQ, ••• ,XI,; independent and Xk+ 1, ••• , Xu dependent. Then k

~ {3 l..J x.! ..::;.;

i=1

for j = k+1, k+2, .. . ,n where not all (3 tJ..

= O.

Since x.J is an eigenvector,

k

\Xj

= Aj ~ f3ijXi

for j

i=1

= k+1, .. . ,n

k

Also,

~f3ijAxi =

Axj --

i=l

Ii:

~ f3 ij \Xi

i=l

Subtracting this equation from the previous one gives

o

Ii:

=

~ f3ij(\ - ,\)xi

i=1

But the Xi, i = 1,2, ... , k, were assumed to be linearly independent. Because not all f3 ij are zero, some Ai = Aj. This contradicts the assumption that A had distinct eigenvalues, and so all the eigenvectors of A must be linearly independent.

4.3 PROPERTIES OF SIMILARITY TRANSFORMATIONS To determine when a T can be found such that T-IAT gives a diagonal matrix, the properties of a similarity transformation must be examined. Define S = T-1AT (4. 6)

72

MATRIX ANALYSIS

[CHAP. 4

Then the eigenvalues of 8 are found as the roots of det (8 - AI) det (S - AI)

= O.

But

det (T-IAT - AI) det (T-IAT - AT-lIT) det [T-l(A - AI)T]

Using the product rule for determinants, det (S - AI) Since detT-l have proved

Theorem 4.2:

= (det T)-l

=

det T-l det (A - AI) det T

from Problem 3.12, det (S - AI)

= det (A -

AI).

Therefore we

All similar matrices have the same eigenvalues.

Corollary 4.3: All similar matrices have the same traces and determinants. Proof of this corollary is given in Probleln 4.1. A useful fact to note here "~lso is that all triangular matrices B display eigenvalues on the diagonal, because the detetminant of the triangular matrix (B - AI) is th~ _product of its diagonal elements.

Theorem 4.4:

A matrix A can be reduced to a diagonal matrix A by a similarity transformation if and only if a set of n linearly independent eigenvectors can be found.

Proof: By Theorem 4.2, the diagonal matrix A must have the eigenvalues of A appearing on the diagonal. If AT = TA, by partitioned matrix multiplication it is required that Ati = Ait, where t are the column vectors of T. Therefore it is required that T have the eigenvectors of A as its column vectors, and T-l exists if and only if its column vectors are linearly independent. It has already been shown that when the eigenvalues are distinct, T is nonsingular. So consider what happens when the eigenvalues are not distinct. Theorem 4.4 says that the only way we can obtain a diagonal matrix is to find n linearly independent eigenvectors. Then there are two cases: Case 1. For each root that is repeated k times, the space of eigenvectors belonging to that root is k-dirnensional. In this case the matrix can still be reduced to a diagonal form. Example 4.5. Given the matrix

A = (

~ ~ ~).

-1

0

Then

det (A - AI) = -i\.(1 - i\.)2

0

0, 1 and 1. For the zero eigenvalue, solution of Ax = 0 gives x = (0 1 -1). the eigenvalue problem is

(

This gives the set of equations

~ ~ ~) (:~)

-1

and the eigenvalues are

0

0

X3

(1) ( : ; ) X3

For the unity eigenvalue,

CHAP. 4]

MATRIX ANALYSIS

73

Therefore all eigenvectors belonging to the eigenvalue 1 have the form

where

xl

and

X2

are arbitrary. Hence' any two linearly independent vectors in the space spanned by The transformation matrix is then

(0 1 0) and (1 0 -1) will do.

T

M

=

C~ ~-D

and T-IAT = A., where A has 0, 1 and 1 on the diagonal in that order.

Note that the occurrence of distinct eigenvalues falls into Case 1. Every distinct eigenvalue must have at least one eigenvector associated with it, and since there are n distinct eigenvalues there are n eigenvectors. By Theorem 4.1 these are linearly independent. Case 2. The conditions of Case 1 do not hold. to a diagonal form by a similarity transformation. Example 4.6. Given the matrix

A

=

(~ ~).

Then the matrix cannot be reduced

Since A is triangular, the eigenvalues are displayed as 1 and 1.

Then the eigenvalue problem is

which gives the set of equations X2 = 0, 0 = O. All eigenvectors belonging to 1 have the form Two linearly independent eigenvectors are simply not available to form M.

(Xl

O)T.

Because in Case 2 a diagonal matrix cannot be formed by a similarity transformation, there arises the question of what is the simplest matrix that is almost diagonal that can be formed by a similarity transformation. This is answered in the next section. 4.4 JORDAN FORM The form closest to diagonal to which an arbitrary n x n matrix can be transformed by a similarity transformation is the Jordan form, denoted J. Proof of its existence in all cases can be found in standard texts. In the interest of brevity we omit the lengthy development needed to show this form can always be obtained, and merely show how to obtain it. The Jordan form J is an upper triangular matrix and, as per the remarks of the preceding section, the eigenvalues of the A matrix must be displayed on the diagonal. If the A matrix has r linearly independent eigenvectors, the Jordan form has n - r ones above the diagonal, and all other elements are zero. The general form is

J

=

(4.7)

74

MATRIX ANALYSIS

[CHAP. 4

Each Lji(Ai) is an upper triangular square matrix, called a Jordan block, on the diagonal of the Jordan form J. Several Lji(Ai) can be associated with each value of Ai, and may differ in dimension from one another. A general Lji(Ai) looks like

Lji(Ai)

Ai 1 0 0 Ai 1 0 0 Ai

0

0

Ai

0 0

(4.8)

................. 0

0

where Ai are on the diagonal and ones occur in all places just above the diagonal.

Example 4.7. Consider the Jordan form

J

--

(Ao; :01':1 000). 1\

o

0 0

Because all ones must occur above the

A2

diagonal in a Jordan block, wherever a zero above the diagonal occurs in J there must occur a boundary between two Jordan blocks. Therefore this J contains three Jordan blocks,

There is one and only one linearly independent eigenvector associated with each Jordan block and vice versa. This leads to the calculation procedure for the other column vectors tt of T called generalized eigenvectors associated wit~ each Jordan block Lji(Ai): AXi

AiXi

Atl

Ait1

+ Xi (4.9)

Note the number of

tl

equals the number of ones in the associated Lji(Ai).

A(Xi I t1 It21 ... Itt I ... )

(AiXi I Ait1 + Xi I Ait2 + t1

=

(Xi

Ih It21

Then

I ... I Ait! + tl-1 I •.. )

... Itl I ... )Lji(Ai)

This procedure for calculating the tl works very well as long as Xi is determined to within a multiplicative constant, because then each tz is determined to within a mUltiplicative constant. However, difficulty is encountered whenever there is more than one Jordan block associated with a single value of an eigenvalue. Considerable background in linear algebra is required to find a construction procedure for the t! in this case, which arises so seldom in practice that the general case will not be pursued here. If this case arises, a trial and error procedure along the lines of the next example can be used. Example 4.8. Find the transformation matrix T that reduces the matrix A to Jordan form, where

A

G-~ D.

CHAP. 4]

MATRIX ANALYSIS

75

The characteristic equation is (2 - ;\)(3 - ;\)(1 -;\) + (2 -;\) = O. A factor 2 -;\ can be removed, and the remaining equation can be arranged so that the characteristic equation becomes (2- ;\)3 = O. Solving for the eigenvectors belonging to the eigenvalue 2 results in

Therefore any eigenvector can be expressed in a linear combination as

What combination should be tried to start the procedure described by equations (4,.9)? expression gives

Trying the general

Then 7"2

+ 73 =

-7"2 -

73

f3 -[3

=

=

These equations are satisfied if a f3 This gives the correct x a(1 1 -l)T. Normalizing x by setting = 1 gives t = ('7'1 '7'2 1 - 72)T. The transformation matl'ix is completed by any other linearly independent choice of x, say (0 1- _1)T, and any choice of '7'1 and 7'2 such that t is linearly independent of the choices of all x, say '7'1 = 0 and '7'2 = 1. This gives AT = TJ, or

a

'0

(~ ! ~)( ~ ~~) o -1

1

-1

0 -1

(~~ ~)(~ ~ ~) -1

0 -1

0

0

2

4.5 QUADRATIC FORMS Definition 4.7: A quadratic form £?( is a real polynomial in the real variables

~1' ~2' n

••• ,

~n

n

containing only terms of the form a:ij~i~j' such that ~ = i~ j~ aij~i~j' where aij is real for all i and i. Example 4.9. Some typical

quadr~tic

forms are 7~i

.w.l

.w.2 = 3~i .w.3 .w.4

Theorem 4.5: Proof:

=

-

2~1~2

+ ~~ + 5~1~3 -

7~2~1

all~i

+ a12~i~2 + a21~2~1 + a22~~ t~i + (1- t2)~1~2 - et~;

All quadratic forms E<. can be expressed as the inner product (x, Qx) and vice versa, where Q is an n X n Hermitian matrix, i.e. Qt = Q.

First

~

to (x, Qx): (4.10)

76

MATRIX ANALYSIS

Let Q = {qij}

= i{aij+ajJ.

Next, (x, Qx) to

~

Then qij = qw so Q is real and symmetric, and ~=XTQX.

(the problem is to prove the coefficients are real): n

(x, Qx)

n

= i=lj=l LL

qi"~i~' J

and

J

Then

n

(x, Qx) = n

So

[CHAP. 4

(x, Qx) =

Theorem 406:

lex, Qx) + t(x, Qtx) =

n

LL

i=1 j=1

Re (qi')~i~' J

J

=~

n

! i=lj=l L ~ (qt·J + q:)~i~' J J

and the coefficients are real.

The eigenvalues of an n X n Hermitian matrix Q = Qt are real, and the eigenvectors belonging to distinct eigenvalues are orthogonal.

The most important case of real symmetric Q is included in Theorem 4.6 because the set of real symmetric matrices is included in the set of Hermitian matrices. Proof:

The eigenvalue problems for specific Ai and Aj are (4.11)

Since Q is Hermitian,

=

QtXj

A.jXj

Taking the complex conjugate transpose gives xfQ 3

>..~J xfJ

=

(4.12)

Multiplying (4.12) on the right by Xi and (4.11) on the left by xT gives XjQXi - xjQXi

=

0

=

(>..~ - >..)xTxi

If j = i, then yxtxi is a norm on Xi and cannot be zero, so that \ = >"i, meaning each eigenvalue is real. Then if j -:F i, >..*J. - A.t = A.J - A..t But for distinct eigenvalues, >...J - A.t =/= 0, so xj Xi = 0 and the eigenvectors are orthogonal. Theorem 4.7:

Even if the eigenvalues are not distinct, a set of n orthonormal eigenvectors can be found for an n X n normal matrix N.

The proof is left to the solved problems. Note both Hermitian and real symmetric matrices are normal so that Theorem 4.6 is a special case of this theorem.

Corollary 4.8:

A Hermitian (or real symmetric) matrix Q can always be reduced to a diagonal matrix by a unitary transformation, where U-IQV =.A and V-I = vt.

Proof: Since Q is Hermitian, it is also normal. Then by Theorem 4.7 there are n orthonormal eigenvectors and they are all independent. By Theorem 4.4 this is a necessary and sufficient condition for diagonaIization. To show a transformation matrix is unitary, construct V with the orthonormal eigenvectors as column vectors. Then

xt1 xt2

CHAP. 4]

-MATRIX ANALYSIS

77

xTx

But because they are orthonormal. Then utu = I. Since the column j = (Xi'X) = 8ij vectors of U are linearly independent, U- 1 exists, so multiplying on the right by U-l gives ut = U-l, which was to be proven. x

Therefore if a quadratic form fL = xtQx is given, rotating coordinates by defining gives ~ == ytutQUy = ytAy. In other words, ~ can be expressed as

= Uy

£t =

Al1Ylj2

+ A21Y21 2 + ... + An IYnl 2

where the Ai are the real eigenvalues of Q. Note ~ is always positive if the eigenvalues of Q are positive, unless y, and hence x, is identically the zero vector. Then the square root of ~ is a norm of the x vector because an inner product can be defined as (x, Y)Q = xtQy. Definition 4.8:

An 11, X n Hermitian matrix Q is positive definite if its associated quadratic form ~ is always positive except when x is identically the zero vector. Then Q is positive definite if and only if all its eigenvalues are> 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

<

is replaced by

~.

q11:

0

<

det Q 2 = det

(q~l q12

q12); QZ2

.•. ,

0

< det Q n = det Q

Q is nonnegative definite.

Rearrangement of the elements of Q sometimes leads to simpler algebraic inequalities.

MATRIX ANALYSIS

78

[CHAP. 4

Example 4.12. The quadratic form Q.

=

is positive definite if ql1

>

0 and qnq22 - qi2

GI --

2+

~

qll;1

2 q12;1;2

>

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

Solved Problems 4.1.

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 •

Supplementary Problems 4.12.

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.

Answers to Supplementary Problems 4.12.

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

=


eA(t-T)

(5.14)

and for a time-invariant linear difference system is cp(k, m)

=

Ak-m

(5.15)

00

Proof:

The Maclaurin series for

eAt

=L

Aktk/k!,

which is uniformly convergent

k=O

as shown in Problem 4.27. Differentiating with respect to t gives deAt/dt

00

=

~ Ak+ltk/k!, k=O

so substitution into equation (5.1) verifies that eA(t-T) is a solution. ~urthermore, for t =T, eA(t~T) = I, so this is the unique solution starting from cf{T, T) = I. Also, substitution of cp(k, m) = AIe-m into equation (5.3) verifies that it is a solution, and for k = m, Ak-m = I. Note that eAeB:oF eBe A in general, but eAt0eAtl = eAtleAto = eA(tO+t1 ) and AeAt = eAtA, as is easily shown using the Maclaurin series. Since time-invariant linear systems are the most important, numerical calculation of eAt is often necessary. However, sometimes only x(t) for t == to is needed. Then x(t) can be found by some standard differential equation routine such as Runge-Kutta or Adams, etc., on the digital computer or by simulation on the analog computer. When eAt must be found, a number of methods for numerical calculation are available. No one method has yet been found that is the easiest in all cases. Here we present four of the most useful, based on the methods of Section 4.7. 1. Series method: (5.16)

2. Eigenvalue method: (5.17)

and, if the eigenvalues are distinct,

3. Cayley-Hamilton:

n-l

L

eAt

yJt)Ai

(5.18)

i=O n-l

eJt

where the Yi(t) are evaluated from eLnO'l)t

0

o

e L21 (Al)t

ill

=

~ Yi (t)Jl.

Note that from (4.15),

i=O

o o

o o

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

o o

o o

eL;r(Al)t

0

o

eL12{A2)t

o

o

o

o

o o ill

ill

ill

....

ill

ill. ill.

o o

(5.19)

102

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

(5.20)

then

e"it (

t~-1 e"?it/(l- 1) ')

teAit

.~ ••• ~~it• ........... ~~~~ ~".it~~l.~ ~: ~ o

0

...

(5.21) ,

el\it

4. Resolvent matrix: where R(s)

= (sI -

1:.- 1 {R(s)}

(5.22)

A)-I.

The hard part of this method is computing the inverse of (sI - A), since it is a polynomial in s. For matrices with many zero elements, substitution and elimination is about the quickest method. For the general case up to about third order, Cramer's rule can be used. Somewhat higher order systems can be handled from the flow diagram of the Laplace transformed system. The elements rij (s) of R(s) are the response of the ith state (integrator) to a unit impulse input of the jth state (integrator). For higher order systems, Leverrier's algorithm might be faster.

Theorem 5.3:

Leverrier's algorithm. Define the n x n real matrices F 1, F 2 , scalars Bl, B2 , ••• , Bn, as follows:

••• ,

Fn and

-trAFdl -trAF2/2 en = -tr AFnln

Then (sI -A)-1 Also, AFn + enI

= 0,

=

sn-l F1 + sn-2F2 + ... + sFn- 1 + Fn sn + (hs n- 1 + ... + en-1S + en

(5.23)

to check the method. Proof is given in Problem 5.4.

Having R(s), a matrix partial fraction expansion can be performed. First, factor detR(s) as (5.24) detR(s) = sn + fh8 n- 1 + ... + On-1S + en = (s - 1.1)(8 - 1.2)' .. (s - An) where the Ai are the eigenvalues of A and the poles of the system. Next, expand R(s) in matrix partial fractions. If the eigenvalues are distinct, this has the form R(8)

1

--R1 A1

S -

+

1

--R" S - 1.2 ~

+ '" +

1

--\-Rn S -

(5.25)

I\n

where Rk is the matrix-valued residue (5.26)

SOLUTIONS TO THE LINEAR STATE EQUATION

CHAP.5J

103

For mth order roots A, the residue of (3 - A) - i is m

(n~ - i)! ds

I

i

d - -1 '-- m i

-

[(

s-AmRs )

Then eAt is easily found by taking the inverse Laplace transform. roots, equation (5.25) becomes

=

eAt

d'ltRl

(5.27)

()] S=X

In the case of distinct

+ eA2tR2 + ... + eAntRn

(5.28)

Note, from the spectral representation equation (5.17),

=

Ri

so that the eigenvectors

Xi

t

Xiri

and their reciprocal rj can easily be found from the R i .

In the case of repeated roots,

=

.,c-l{(S - Ai)-m}

tm- 1eAitj(m -1)!

To find Ak, methods similar to those discussed to find

(5.29)

are available.

eAt

1. Series method:

(5.30)

2. Eigenvalue method:

(5.31)

and for distinct eigenvalues

=

Ak

3. Cayley-Han1ilton:

n

~ A~xir7

i=1

11-1

~ Yj(k)Ai

(5.32)

i=O n-l

where the Yi(k) are evaluated from Jk = ~ Yr(1c)Ji where from equation (4.15), i=l

L~l(Al)

0

o

o

o

L;'l(Al)

o

o

o o

o o

L~l (AI)

0

o

L~C2(A2)

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

o

o

o

"

.......

o o o o I

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

(5.33)

ill".

o

and if Lji(Ai) is l x l as in equation {5.20},

1-l)[(l-1) 1(1c - l + 1) 1] _1) ~.....A.~ .... : : : ...(~.! ~: ~:~'! ~ (~~ ~~!. (.k. ~ .l.~~: :J.-:

A~

(

o

kA~-I

0

(k 1Ar+

...

(5.34)

Ai

4. Resolvent matrix: (5.35)

where

R(z) = (zI - A)-I.

Since R(z) is exactly the same form as R(8) except with z for s, the inversion procedures given previously are exactly the same.

104

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

The series method is useful if Ak = 0 for some k = k o• Then the series truncates at ko - 1. Because the eigenvalue problem Ax = Ax can be multiplied by A k-l to obtain 0= AkX = AA k-1X = Akx, then A = O. Therefore the series method is useful only for systems with ko poles only at the origin. Otherwise it suffers from slow convergence, roundoff, and difficulties in recognizing the resulting infinite series. The eigenvalue method is not very fast because each eigenvector must be computed. However, at the 1968 Joint Automatic Control Conference it was the general consensus that this was the only method that anyone had any experience with that could compute eAt up to twentieth order. The Cayley-Hamilton method is very similar to the eigenvalue method, and usually involves a few more multiplications. The resolvent matrix method is usually simplest for systems of less than tenth order. This is the extension to matrix form of the usual Laplace transform techniques for single input-single output that has worked so successfully in the past. For very high order systems, Leverrier's algorithm involves very high powers of A, which makes the spread of the eigenvalues very large unless A is scaled properly. However, it involves no matrix inversions, and gives a means of checking the amount of roundoff in thatAFn + onI should equal o. In the case of distinct roots, R. = xir:t so that the eigenvectors can easily be obtained. Perhaps a combination of both Leverrier's algorithm and the eigenvalue method might be useful for very high order systems.

5.3 TRANSITION MATRIX FOR TIME-VARYING DIFFERENTIAL SYSTEMS There is NO general solution for the transition matrix of a time-varying linear system such as there is for the time-invariant case. Example 5.1. We found that the transformation A = TJT-l gave a general solution cp(t, to)

for the time-invariant case.

=

cp(t - to)

=

eACt-to)

=

T

eJCt-to)

T-l

For the time-varying case, dx/dt = A( t)x

Then A(t) = T(t) J(t)T-l(t),where the elements of T and J must be functions of t. change of variable x = T(t)y results in dy/dt

=

Attempting a

J(t)y - T-1(t) (dT(t)/dt)y

which does not simplify unless dT(t}/dt = 0 or some very fortunate combination of elements.

We may conclude that knowledge of the time-varying eigenvalues of a time-varying system usually does not help. The behavior of a time-varying system depends on the behavior of the coefficients of the A(t) matrix. Example 5.2. Given the time-varying scalar system dE/dt = ~ sgn (t - t 1) where sgn is the signum function, so that sgn (t - t 1) = -1 for t < tl and sgn (t - t 1 ) = +1 for t > t 1• This has a solution ~(t) = ~(to)e-(t-to) for t < tl and ~(t) = ~(tl)e(t-tl)for t> t l . For times t < t 1, the system appears stable,J~~t actually the solution grows without bound as t ~ co. We shall see in Chapter 9 that the concept of staBility must be carefully defined for a time-varying system.

Also, the phenomenon of finite escape time can arise in a time-varying linear system, whereas this is impossible in a time-invariant linear system.

105

SOLUTIONS TO THE LINEAR STATE EQUATION

CHAP. 5]

Example 5.3. Consider the time-varying scalar system Then and the solution is represented in Fig. 5-1. infinity in a finite time.

to

The solution goes to

tl

Fig. 5-1

These and other peculiarities make the analysis of time-varying linear systems relatively more difficult than the analysis of time-invariant linear systems. However, the analysis of time-varying systems is of considerable practical importance. For instance, a timevarying linear system usually results fronl the linearization of a nonlinear system about a nominal trajectory (see Section 1.6). Since a control system is usually designed to keep the variations from the nominal small, the time-varying linear system is a good approximation. Since there is no general solution for the transition matrix, what can be done? In certain special cases a closed-form solution is available. A computer can almost always find a numerical solution, and with the use of the properties of the transition matrix (Theorem 5.1) this makes a powerful tool for analysis. Finally and perhaps most importantly, solutions for systems with an input can be expressed in terms of the transition matrix. 5.4 CLOSED FORMS FOR SPECIAL CASES OF TIME-VARYING LINEAR DIFFERENTIAL SYSTEMS Theorem 5.4: A general scalar time-varying linear differential system dUdt the scalar transition matrix

= a(t)~

has

rta:("I/)dll

cp (t) ,T = eh Proof:

Separating variables in the original equation,

taking antilogarithms gives

~(t) = foef:u(71) d71 •

Theorem 5.5:

= A(T) A(t)

If A(t) A(T)

d~/~

= a(t)dt.

Integrating and

for all t, T, the time-varying linear differential sys-

.-

tem dx/dt == A(t)x has the transition matrIx 4'(t, T)

= est

A(7)

d'TI

T

•

This is a severe requirement on A(t), and is usually met only on final examinations. Proof:

Use of the series form for the exponential gives I

+

i

t l'

A("l) d",

+

1 2!

r rt J. A("l) dr; J. A(e) de t

Taking derivatives, a ftr A(l1) d1'l

+ .,.

(5.36)

(5.37)

-e at

But from equation (5.36), A(t)e f;A(71) d1) This equation and (5.37) are equal if and only if A(t)

it

A(1]) d1]

ft A(1]) d"l A(t)

106

SOLUTIONS TO THE LINEAR STATE EQUATION

Differentiating with respect to

T

[CHAP. 5

and multiplying by -1 gives the requirement A(t) A(1-)

= A(T) A(t)

Only A(t) A(T) = G(t, T) need be multiplied in the application of this test. of T for t and t for T will then indicate if G(t, ,) = G(T, t). Example 5.4. Given A(t)

Theorem 5.6:

==: ( :

~).

Then from

A(t) A(T) = G(t, T);;:::

(

t2,,2 0

fl..,. + 1

t) ,we

Substitution

see immediately

A piecewise time-invariant system, in which A(t) = Ai for ti ~ t~ tt+l for i = 0,1,2, . .. where each Ai is a constant matrix, has the transition matrix

Proof;

Use of the continuity property of dynamical systems, the transition property (equation (5.5)) and the transition matrix for time-invariant systems gives this proof. Successive application of this theorem gives ~(t,

to)

q,(t, to)

eAoCt-to)

for to === t ::::: tl

eAICt-tl) eAoCtl-tO)

for tl::::: t ::::: t2

etc. Example 5.5. Given the flow diagram of Fig. 5-2 with a switch S that switches from the lower position to the upper position at time t l • Then dxidt = Xl for to === t < tl and dxidt ==: 2xl for tt == t. The solutions during each time interval are Xl(t) ==: xlOe t - to for to === t < tl and XI(t);;::: X'1(t 1)e 2(t-t1 ) for tl === t, where Xl(tt) = xlOetl-to by continuity.

2 Fig.5~2

It is common practice to approximate slowly varying coefficients by piecewise constants. This can be dangerous because errors tend to accumulate, but often suggests means of system design that can be checked by simulation with the original system. Another special case is that the nth order time-varying equation dny d n- 1 y dy tn- + a t n- 1 _n_1 + ... + a t - + a y dtn dt n-l dt n

0

can be solved by assuming y ==: tAo Then a scalar polynomial results for A, analogous to the' characteristic equation. If there are multiplicities of order m in the solution of this polynomial, y = (1n t)m-lt"" is a solution for i = 0, 1,2, ... , m-l.

CHAP. 5]

SOLUTIONS TO THE LINEAR STATE EQUATION

107

A number of "classical" second order linear equations have closed form solution in the sense that the properties of the solutions have been investigated. Bessel's equation: (5.38)

Associated Legendre equation:

=

(1- t2) Y - 2ty + [n(n + 1) - m2/(1- t 2 )]y Hermite equation:

Laguerre equation: with solution Ln(t), or

y-

2ty

+ 2ay =

0

ty + (1- t)y + ay = ty + (k + 1- t)y

+ (a -

0

0

k)y

=

0

with solution dkLn(t)/dt k • Hypergeometric equation: Ordinary:

t(1- t)ii + [y - (a + fi

Confluent:

tii + (y-t)iJ -

Mathieu equation: or, with

T

+ l)t]y ay

ii + (a + fi cos t)y =

o

afiy

=0 0

(5.89)

= cos t, 2

The solutions and details on their behavior are available in standard texts on engineering mathematics and mathematical physics. Also available in the linear time-varying case are a number of methods to give q.(t, T) as an infinite series. Picard iteration, Peano-Baker integration, perturbation techniques, etc., can be used, and sometimes give quite rapid convergence. However, even only three or four terms in a series representation greatly complicate any sort of design procedure, so discussion of these series techniques is left to standard texts. Use of a digital or analog computer is recommended for those cases in which a closed form solution is not readily found.

5.5 ,PERIODICALLY..VARYING LINEAR DIFFERENTIAL SYSTEMS Floquet theory is applicable to time-varying linear systems whose coefficients are constant or vary periodically. Floquet theory does not help find the solution, but instead gives insight into the general behavior of periodically-varying systems. Theorem 5.7:

(Floquet). Given the dynamical linear time-varying system dx/dt = A(t)x, where A(t) = A(t + 1Il). Then q.(t, T)

==

P(t, T)e RCt -

T

)

where P(t, T) = P(t+lIl, T) and R is a constant matrix.

108

SOLUTIONS TO THE LINEAR STATE EQUATION

Proof:

The transition matrix satisfies aq,(t, -r)/at = A(t) q,(t, -r)

Setting t

[CHAP. 5

= t + w,

and using A(t)

= A(t + w)

q,(-r, -r) = I

with

gives

aq,(t +w, -r)/at = A(t + w) q,(t+w, -r)

=

A(t) ~(t +w,-r)

(5.40)

It was shown in Example 3.25 that the solutions to dx/dt = A(t)x for any initial condition form a generalized vector space. The column vectors 4>i(t, T) of 4t(t, -r) span this vector space, and since det q,(t, T) 01= 0, the +Jt, T) are a basis. But equation (5.40) states that the Pi(t + w, 'f) are solutions to dx/dt = A(t)x, so that 1'1

.pi (t +w, T) =

~. Cji 4>j(t, -r)

for i

= 1,2, ... , n

j=l

Rewriting this in matrix form, where C = {Cji}, 4J(t+w, -r)

=

q,(t, -r)C

(5.41)

Then

c

= q,(-r, t) q,(t+w, T)

Note that C-l exists, since it is the product of two nonsingular matrices. Problem 4.11 the logarithm of C exists and will be written in the form C = eWR

Therefore by (5.42)

If pet, -r) can be any matrix, it is merely a change of variables to write q,(t, T) = pet, -r)eR(t-'I")

(5.43)

But from equations (5.43), (5.41) and (5.42), P(t+w, -r) = 4t(t+w, -r)e-R(t+~-T)

== q,(t, -r)eWRe-R(t+IiJ-'I")

q,(t, T)e-RCt-'I") = pet, T)

From (5.41)-(5.43), R = w- 1 ln [q,(" t) q,(t+W,T)] and P(t,-r) = c)(t,T)e- RCt - T ), so that to find Rand pet, -r) the solution q,(t, T) must already be known. It may be concluded that Floquet's theorem does not give the solution, but rather shows the form of the solution. The matrix pet, -r) gives the periodic part of the solution, and eRCt - T ) gives the envelope of the solution. Since eRCt-'I") is the transition matrix to dz/dt = Rz, this is the constant coefficient equation for the envelope z(t). If the system dz/dt = Rz has all poles in the left half plane, the original time-varying system x(t) is stable. If R has all eigenvalues in the left half plane except for some on the imaginary axis, the steady state of z(t) is periodic with the frequency of its imaginary eigenvalues. To have a periodic envelope in the sense that no element of z(t) behaves exponentially, all the eigenvalues of R must be on the imaginary axis. If any eigenvalues of R are in the right half plane, then z(t) and x(t) are unstable. In particular, if the coefficients of the A( t) matrix are continuous functions of some parameter a, the eigenvalues of R are also continuous functions of (1:', so that periodic solutions form the stability:boundaries of the system. Example 5.6._ Consider the Mathieu equation d2 x/dt 2 + (0: + f3 cos t)x = 0 (5.39). Its periodic solutions are called Mathieu functions, which exist only for certain combinations of - IX . and f3. The values of IX and 13 for which these periodic solutions exist are given by the curves in Fig. 5.3 below. These curves then form the boundary for regions of stability.

CHAP. 5]

SOLUTIONS TO THE LINEAR STATE EQUATION

109

Fig. 5-3 Whether the regions are stable or unstable can be determined by considering the point f3 = 0 and Since the curves are stability boundaries, regions 2 and 6 are stable. Similarly all the odd numbered regions are unstable and all the even numbered regions are stable. The line f3 =0, a:::::: 0 represents a degenerate case, which agrees with physical intuition. a:

< 0 in region 1. This is known to be unstable, so the whole region 1 is unstable.

It is interesting to note from the example above that an originally unstable system might be stabilized by the introduction of a periodically-varying parameter, and vice versa. Another use of Floquet theory is in simulation of cJt(t, T). Only c)(t,7") for one period w need be calculated numerically and then Floquet's theorem can be used to generate the solution over the whole time span.

5.6 SOLUTION OF THE.LINEAR,STATE EQUATIONS WITH INPUT

Knowledge of the transition matrix gives the solution to the linear state equation with input, even in time-varying systems.

Theorem 5.8:

Given the linear differential system with input dx/dt = A(t)x y

=

+ B(t)u

C(t)x + D(t)u

_with transition matrix q,(t:. T) obeying ac)(t, T)/at (5.1)]. Then 4l(t,to)x(to)

x(t)

y(t)

=

(2.39)

= A(t) 4l(t, T)

[equation

rt cJt(t,T)B(T)U(T)dT

+ Jto

C(t) 4l(t, to) x(t o)

+

it to

C(t) 4t(t, T) B(T) U(T) dT

+

D(t)u

(5.44)

The integral is the superposition integral, and in the time-invariant case it becomes a convolution integral. Proof: Since the equation dx/dt = A(t)x has a solution x(t) = lIt(t, to), in accordance with the method of yariation of parameters, we change variables to k(t) where x(t) = cIJ(t, to) k(t)

(5.45)

110

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

Substituting into equation (2.39),

=

dx/dt

(acp/at)k

+ ipdk/dt ==

Use of equation (5.1) and multiplication by

~(to,

dk/dt =

Integrating from to to t, k(t)

=

Since equation (5.45) evaluated at t

t) gives

t) B(t) u(t) _-

rt +(to, 1') B(T) U(T) dT

k(to)

+

= to

gives x(to) = k(t o), use of (5.45) in (5.46) yields

=

4J(to, t) x(t)

~(to,

A(t)+k + B(t)u

x(to)

(5.46)

Jto +

rt ~(to, 1') B(T) U(T) dT Jto

Multiplying by +(t, to) completes the proof for x(t). D(t) u(t) gives y(t).

Substituting into y(t) == eft) x(t) +

In the constant coefficient case, use of equation (5.14) gives t

x(t)

and y(t)

=

=

+ Jr eA
eACt-to)x(to)

CeA
(5.47)

r CeA
+ _

t

(5.48)

to

This is the vector convolution integral.

Theorem 5.9:

Given the linear difference equation

=

+ B(k) u(k) y(k) = C(k) x(k) + D(k) u(k) (2.40) with transition matrix ~(k, m) obeying iII(k + 1, m) = A(k) .(k, m) [equax(k + 1)

A(k)x(k)

tion (5.3)]. Then k---:-l

x(k)

k-2

k-l

;=m

i=j+l

= II A(i) x(m) + L TI i=m

-

A(i) B(j) u(j)

+ B(k -1) u(k -1)

(5.49)

where the order of multiplication starts with the largest integer, i.e. A(k-1) A(k - 2)· ... Proof:

Stepping equation (2.40) up one gives x(m +2) = A(m + 1) x(m+ 1)

+ B(m + 1) u(m+ 1)

Substituting in equation (2.40) for x(m + 1), x(m + 2)

=

A(m + 1) A(m) x(m)

+ A(m + 1) B(m) u(m) + B(m + 1) u(m + 1)

Repetitive stepping and substituting gives x(k) = A(k -1)· .. A(m) x(m)

+ A(k -

1)· .. A(m + 1) B(m) u(m)

+ A(k -1)·· ·A(m +2) B(m + 1) u(m + 1) + '" + A(k-1)B(k- 2) u(k- 2) + B(k - 1) u(k - 1) This is equation (5.49) with the sums and products written out.

CHAP. 5]

SOLUTIONS TO THE LINEAR S'rATE EQUATION

5.7 TRANSITION MATRIX FOR

TIME~V ARYING

111

DIFFERENCE EQUATIONS

Setting B = 0 in equation (5.49) gives the transition matrix for time-varying difference equations as k-l

iIJ(k, m)

= II

for k > m

A(i)

i=m

(5.50)

For k = m, iIJ(m, m) = I, and if A -l(i) exists for all i, 4»(k, m) Then equation (5.49) can be written as

x(k)

4»(k, m) x(m)

+

k-l

~ iIJ(k, j=m

m-l

= II A -l(i)

for k < m.

i=k

i + 1) B(j) uU)

(5.51)

This is very similar to the corresponding equation (5.44) for differential equations except the integral is replaced by a sum. Often difference equations result from periodic sampling and holding inputs to differential systems.

net)

o

X

'T J

u(tk )

dx/dt = A(t)x + B(t)u -y = C(t)x D(t)u

Hold

yet)

+

T

Fig. 5-4

In Fig. 5-4 the output of the hold element is u(k) = U(tk) for tk ~ t < tk+ 1, where = T for all k. Use of (5.44) at time t = tk+1 and to = tk gives

~k+l - tk

x(k + 1)

(5.52)

Comparison with the difference equations (2.40) results in

where the subscript

8

refers to the difference equations of the sampled system. For time-

invariant differential systems, As = eAT, Bs = iT eACT-r)B dT, C

S

::::::

C and Ds

= D.

Since

in this case As is a matrix exponential, it is nonsingular no matter what A is (see the comment after Problem 4.11). Although equation (5.50) is always a representation for iIJ(k, m), its behavior is not usually displayed. Techniques corresponding to the differential case can be used to show this behavior. For instance, Floquet's theorem becomes cp(k, m) = P(k, m)Rk-m, where Also P(k, m) = P(k + 00, m) if A(k) = (k + (0).

(k1!n)!y(k+n)

+ a l (k+~~l)l

y(k+n-1)

+ ... +

lX _ n

1

(k+l)y(k+1) + fXny(k)

=

0

has solutions of the form )..k/k!. Piecewise time-invariant, classical second order linear, and series solutions also have a corresponding discrete time form.

112

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

5.8 IMPULSE RESPONSE MATRICES With zero initial condition x(to)

= 0,

from (5.44) the output y(t) is

= .( C(t) 4o(t, T) B(T) U(T) dT +

y(t)

(5.53)

D(t) u(t)

This suggests that y(t) can be written as a matrix generalization of the superposition integral, y(t)

=

(5.54)

where H(t, T) is the impulse response matrix, i.e. hij(t,T) is the response of the ith output at time t due to an impulse at the ith input at time T. Comparison of equations (5.53) and (5.54) gives t===T

=

H(t, T)

(5.55)

t
In the time-invariant case the Laplace transformation of H(t, 0) gives the transfer function matrix ~{H(t,O)} = C(sI-A)-lB +D (5.56) Similarly, for discrete-time systems y(k) can be expressed as k

y(k)

~

==

H(k, m) u(m)

(5.57)

m=-<X)

where H(k,m)

=

C(k) 4J(k, m. + 1) B(m) D(k)

k>m k= m

o

k<m

{

(5.58)

Also, the Z, transfer function matrix in the time-invariant case is Z{H(k,O)}

=

(5.59)

C(zl-A)-lB+D

5.9 THE ADJOINT SYSTEM The concept of the adjoint occurs quite frequently, especially in optimization problems. DeJinition 5.2:

The adioint, denoted La, of a linear operator L is defined by the relation

=

(p, Lx)

We are concerned with the system dx/dt = A(t)x. Defining L Using the scalar product (p,x) = equation (5.60) using integration by parts.

Lx = O.

(p,Lx)

=

i

tl

ptA(t)x dt -

to

= p(t~),

= A(t) -

Id/dt, this· becomes

ptxdt, the adjoint system is found from

to

dx pt dt dt

to

. ( ' [xtA tp + xt

For the case p(to) :::: 0

it1 it!

(5.60)

for all x and p

(LaP, x)

~~J dt

+

pt(to) x(t:) -

we find La = At(t) + Id/dt.

pt(t,) x(t o)

CHi\.P.5}

SOLUTIONS TO THE LINEAR ,STATE EQUATION

113

Since Lx = 0 for all x, then (p,Lx) = 0 for all xand p. Using (5.60) it can be concluded LaP = 0, so that the adjoint 8ystem is defined by the relation dp/dt = -At(t)p

(5.61)

Denote the transition matrix of this adjoint system as "M(t, to), i.e., o~(t,

to)/ot

=

-At(t) ~(t, to)

Theorem 5.10: Given the system dx/dt -At(t)p. Then

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

to) = I

(5.62)

and its adjoint system dp/dt = tl

=

pt(tl) X(tl)

~(to,

with

"itt( t, to)

and

r pt(t) B(t) u(t) dt

+ J to

pt(to) x(to)

= • -I( t, to)

(5.63)

(5.64)

The column vectors I/Ii(t, to) Ofik(t, to) are the reciprocal basis to the column vectors tpi(t, to) of +(t, to). Also, if u(t) = 0, then pt(t) x(t) = scalar constant for any t.

Proof;

Differentiate pt(t) x(t) to obtain d(ptx)/dt

=

(dpt/dt)x

+ pt dx/dt

Using the system equation dx/dt = A(t)x + B(t)u and equation (5.61) gives d(ptx)/dt = ptBu

Integration from to to tl then yields (5.63). Furthermore if u(t) = 0, pt(to) x(to)

=

pt(t) x(t)

From the transition relation, x(t) = .(t, to) x(to) and pet) pt(to) Ix(to)

= 'i'(t, to) p(to)

so that

= pt(to) 'i't(t, to) eIt(t, to) x(to)

for any p(to) and x(to). Therefore (5.64) must hold. The adjoint system transition matrix gives another way to express the forced solution of (5.44): x(t)

The variable of integration T is the second argument of (J(t, 7"), which sometimes poses simulation difficulties. Since eIt(to, T) = .-l(T, to) = 'i't(T, to), this become~ x(t)

q,(t, to) [x(to)

+

J:

,.t(T, to) OtT) U(T) dTJ

(5.65)

in which the variable of integration 7" is the first argument of "!V(T, to). The adjoint often can be used conveniently when a final value is given and the system motion backwards in time must be found.

Exa~:~:n5.7~x/dt

= (3:

~:)x.

Use the adjoint system to find the set of states (",(1) "2(1)) that

permit the system to pass through the point x1(2) = 1.

(1 3)

The adjoint system is dp/dt = -t 2 'k(t, T)

=

2 p.

..

.

O.2e2r-2t2

(~ ,~)

This has a transItIon matrIx

O.2e(~/2)",-(rJ2) (_~ -~) +

114

SOLUTIONS TO THE LINEAR STATE EQUATION

Since pt(2)x(2) = pt(l}x{l). if we choose pt(2) = (10), Pl(l) xl(l) + P2(!) x 2(1). But pel) = .(1,2) p(2), so that

[CHAP; 5

(10)x(2) = xl(2) = 1 = pt(l)x(l) =

then

= 0.2(3e-1.5 + 2e6 2e6 - 2e-1.5)

pt(l)

The set of states xC!) that gives xl(2) = 1 is determined by 1

=

+ (OAe 6 -

(O.6e-1.5 + OA( 6)xl(1)

0.4e-1.5)x2(1)

Solved Problems 5.1.

Given 4»(t, to), find A(t). Using (5.2) in (5.1) at time check on any solution.

5.2.

to = t, A(t) =

a~{t,

to)/at evaluated at- to = t.

This is a quick

Find the transition matrix for the system

o

dx CIt

-4 -1

by (a) series method, (b) eigenvalue method, (c) Cayley-Hamilton method, (d) resolvent matrix method. In the resolvent matrix method, find (81 - A)-l by (1) substitution and elimination, (2) Cramer's rule, (3) flow diagram, (4) Leverrier's algorithm. (a) Series method.

eAt

=

From (5.16),

G D C~ 0 1 0

+

C- +!'2+ ...

eAt

= 1+ At/!! + A 2 t 2 /2!

0

-4t -t

o) 4t

(./2 0

o

t

=

+

0

+ . ".

Substituting for A,

0)

o

6t 2 -8t2 2t2 _2t2

0

+ ...

4t(1- 2t +o 4t2/2 + ... ) 1 - 2t + 4t2/2 + 2t(1 - 2t) +

+ 4t2/2 - 2t(1 ~ 2t) + ... -tel - 2t + 4t2/2 + ... )

1- 2t

) ...

Recognizing the series expression for e- t and e- 2t gives

o

-eAt

(1- 2t)e- 2t

o 2t 4te-

-te- 2t

(1 + 2t)e- 2t

)

(b) The eigenvalues of A are -1, -2 and -2, with corresponding eigenvectors (1 0 O)T and (0 2 l)T. The generalized eigenvector corresponding to -2 is (0 1 l)T, so that

(-~ -~ o -1

~)

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

=

0

0

1

1

0

0 -2

0 -1

2

Using equation (5.17),

eAt

=

(~ ~ ~)(e~t e~2t te~2t)(~ ~ _~) 2t o

1

1

0

0

e-

Mult.iplying out the matrices gives the answer obtained in (a).

0 -1

2

SOLUTIONS TO THE LINEAR STATE EQUATION

CHAP. 5] (c)

Again, the eigenvalues of A are calculated to be -1, -2 and -2. (5.18), eJt

=

C~'

0 e- 2t

t'~2)

roO

==

e-2t

0

0) (1

0 1 0

~

+

~

Y1

115

To find the Yi (t) in equation

-2o

o

0) +

1 -2

which gives the equations

e- t

= Yo -

e- 2t

Yo -

+ Y2 2Y1 + 4Y2

Y1 -

4Y2

te- 2t

=

Y1

Solving for the Yi' 'Yo = 4e- t - 3e- 2t

-

2te- 2t

'Y1 = 4e- t - 4e- 2t

-

3te- 2t

e- t - e- 2t

Y2

-

te- 2t

Using (5.18) then gives eA '

=

(4e-' -

3e- 2' -

2te- 2')

o

(~

1

o

0)

o

12 -16

4

-4

Summing these ll'latrices again gives the answer obtained in (a). (d1) Taking the Laplace transform of the original equation,

s.,e(X 1) - XlO

-"e(X 1)

S.,e(X2) -

-4.,e(x2)

X20

+ 4.,e(xs)

-"e(X2)

s.,e(xa) - Xao

Solving these equations by substitution and elimination,

"e(Xg)

== - (8

;~)2 + [s ~ 2 + (8'; 2)2J xao

Putting this in matrix form .,e(x) = R(s)xo,

1

Res)

o o

o

o 1

8

2

+2-

(s + 2)2

1

(8 + 2)2

Inverse Laplace transformation gives eat as found in (a). (d2) From (5.22),

4

(s + 2)2

_1_+_2_

s+2

(8+2)2

Y2

C 0) 0 0

o4

o

-4 4

116

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

U sing Cramer'S rule, 8

R(s)

(.+1)~.+2)2

:::::

+ 2)2

o + 1) -(8 + 1)

:

(

4(s o + 1) ) (8 + 1)(8 + 4)

8(8

Performing a partial fraction expansion, R(s)

=

.! C D .!2G D (.:2 G D 1

0 0 0

:

0

0 1 0

+

-2

P

+

-1

Addition will give R(s) as in (d1).

(dS) The flow diagram of the Laplace transformed system is shown in Fig. 5·5.

Fig. 5-5

For x10 ::::: 1, .(x1)::::: l/(s + 1) and .e(X2)::::: .e(xs) ::::: O. For X20

== 1,

.(X1):::::

0,

For XgO ::::: 1, .e(X1)::::: 0,

+ 2)2 and .,e(xs)::::: 4/(s + 2)2. .(X2):::: -l/(s + 2)2 and .e(xs);;;:: (s + 4)/(8 + 2)2.

.(X2)::::: s/(s

Therefore, 1

R(s)

0

0

s+l

4

8

0 0

(s + 2)2

(8 + 2)2

-1 (s + 2)2

8+4 (8 + 2)2

Again a partial fraction expansion can be performed to obtain the previous result. (d4) Using Theorem 5.3 for Leverrier's algorithm, F1

:::::

F2 :::::

I

(1 0;)+51 0-4 -1

o

Fa:::::

(4

0

~ -8 -1

~) +81

01

:::::

5

82

:::::

8

4

83

-4

Using equation (5.23),

~G RCs)

=

0 1 0

D+·G S3

+

582

0

0 1 -1

+

D+ G

0 -1

88

+

4

A partial fraction expansion of this gives the previous result.

D

SOLUTIONS TO THE LINEAR STATE EQUATION

CHAP.5J

5.3.

117

Using (a) the eigenvalue method and then (b) the resolvent matrix, find the transition matrix for

(a) The eigenvalues a.re -1 and -2. with eigenvectors (1 O)T and (1 -1)T respectively. The reciprocal basis is (1 1) and (0 -1). Using the spectral representation equation (5.31),

(b) From equation (5.35), -1

R(z)

z+2

)-1

==

so that z 1 z+

,2"-1 (

5.4.

1

+ 1)(z + 2)

(Z. +0 2 z +1) 1

z)

(-2».:)

+ 2.

(-l)k (-2)k

z

o

z+2

Prove. Leverrier's algorithm, (sI-A)-1

z 1- z z+

(z

sn-1F1 + sn-2F2

=

sn + 8lS n -

l

+ ... + 8Fn - 1 + Fn + ... + 8n-18 + fin

(5.23)

and the Cayley-Hamilton theorem (Theorem 4.15). Let det (sl - A) ::::: ¢(s) == sn + 8 l S n - 1 + . " + 8 n - l S + On' Use of Cramer's rule gives ¢(8)(sl - A)-I::::: F(s) where F(s) is the adjugate matrix, i.e. the matrix of signed cofactors trans'" posed of (sI - A).

An intermediate result must be proven before proceeding, namely that tr F(s) == d¢/ds. In the proof of Theorem 3.21, it was shown the cofactor cli of a general matrix B can be represented as elj ::::: ej /\ b2 /\ ••• /\ bn and similarly it can be shown cij == b l /\ ... /\ hi-l A ej /\ bi+ 1/\ ••• /\ b n • so that letting B == sl - A and using tr F(s) == c11(s) + c22(8) + ... + enn(s) gives tr F(s)

== el /\ (se2 - a2) /\ ...

+ ... +

1\

(sen - an)

(sel - al) /\ ...

+ 1\

(se l - al) /\ e2/\ ... /\ (sel - an)

(8en -l - an-I) /\ en

But

and drp/ds

==

el/\ (se2 - a2) /\ ... /\ (sen - an)

+ .. , +

(sel - al) /\ ... /\ en

==

tr F(s)

and so the intermediate result is established. Substituting the definitions for ¢(s) and F(s) into the intermediate result,

== tr (sn-lF l + sn-2F2 + ... +sFn - 1 +Fn) == sn-l trFl + sn-2 trF2 + .. , + s trFn- 1 + trFn

trF(s)

drp/ds :::::

nsn-l

+ (n -1)8 1 S n - 2 + ... + 8n -

l

Equating like powers of s, tr Fk + 1 for k

== 1,2, ... , n-1. (sn

and tr Fl

==

n.

(5.66)

(n - k)ok-

Rearranging (5.23) as

+ 8 18 n - 1 + ... + 8 n)I ==

and equating the powers of s gives I

==

== F l , 8k-I

(sl - A)(sn-lFl

8n l :=

==

+ sn-2F2 + ... + Fn)

-AFnl and

-AFk

+ F k +l

(5.67)

118

SOLUTIONS TO THE ::i:..lNEARS'TATE EQtTATION for k = 1,2, .. . ,n-I. by taking _their trace

[CHAP. 5

These are half of the relationships required.- The other half are obtained

and substituting into (5.66) to get kBk

= -trAFk

which are the other half of the relationships needed for the proof. To prove the Cayley-Hamilton theorem, successively substitute for F k + 1, i.e.,

Using the last relation onI = -AFn then gives

which is the Cayley-Hamilton theorem.

5.5.

Given the time-varying system dx

at Find the transition matrix and verify the transition properties [equations (5.5)-{5.8)]. Note that A(t) commutes with A(T), i.e., A(t) A(T)

=

It can be shown similar to Problem 4.4 that two n X n matrices Band C with n independent eigenvectors can be simultaneously diagonalized by a nonsingular matrix T if and only if B commutes with C. Identifying A(t) and A(T) for fixed t and T with Band C, then A(t) = T(t)A(t)T-l(t) and A(T) = T(T)A(T)T-l(T) means that T(t) = T(T) for all t and- T. This implies the matrix of eigenvectors T is constant, so dT/dt = O. Referring to Example 5.1, when dT(t)/dt = 0 then a general solution exists for this special case of A(t)A(T) = A(T)A(t).

For the given time-varying system, A(t) has the eigenvalues Al = a + je- t and i\.2 = a - je-t. Since A(t) commutes with A(r), the eigenvectors are constant, (j -l)T and (j 1)T. Consequently T

=

(

j -1

~)

A(t)

=

a -01e-t (

T'-l

From Theorem 5.5,

Substituting the numerical values, integrating and multiplying out gives

::;::

-1

2

(-1 -1) -j

1

CHAP.5J

SOLUTIONS TO THE LINEAR STATE EQUATION

119

To check this, use Problem 5.1:

Setting 7" = t in this gives A(t). To verify ~(t2' to) ::;:: ~(t2' t l ) ~(tl' to), note cos (e-to - e- t 2)

=

cos (e-to - e- t l + e-tl - e- t2 )

::;::

cos (e-to - e-tl) cos (e-tl - e- t2 ) -

sin (e-to - e-t1) sin (e-t1 - e- t2)

and so that sin (e-tl- e- t2») cos (e-t1 - e- t2 ) COS (e-to - e- t1 ) -sin (e-to - e- t1 )

X e-a(ti-to) (

»)

sin (e-to - e- t1 cos (e-to - e- t1 )

To verify 4t- l (t l • to) ::;:: Iilo(to, til, calculate Iilo-I(t I ; to) by Cramer's rule: 4t- I (t t) 1, 0

ea(to-t1)

::;::

cos2 (e-to - e:-t1)

Since cos2 (e-to - e-tl) equals 4t(to, til.

+ sin2 (e-to -

8(t)

+

.

8m2

e-t1)

(e-to - e tI)

1

::;::

and

(COS. (e-to -

e-t1) -sin (e-to - e-tl») cos (e-to - e-t1)

sm (e-to - e-t1)

sin (e-to - e-t1) ::::: -sin (e-t1- eto), this

»)

sin (1- e- t cos (1- e- t )

::;::

Then 8- I (t)

::::

e- at

COS

.(1- e- t ) -sin (1- e- t

sin (1- e- t )

(

»)

cos (1- e- t )

Since we have cos (e-to - e- t1 )

=

cos (e-to -1) cos (1- e- t1)

-

sin (e- to - 1) sin (1 - e-tl)

and a similar formula for the sine, multiplication of B(tI ) and 8- I (to) will verify this. Finally, det !ft(t, 7") ::;:: e2a (t-'t) and trA(t)::;:: 2a, so that integration shows that the determinant property holds.

5.6.

Find the transition matrix for the system

=

dx/dt

(0

(~+

-1 -

a)/t

1)x

-2a - lit

Writing out the matrix equations, we find dxl/dt::;:: x2 and so d 2x l ldt2

+

(2a

+ lit) dxlldt +

[1

+ (0:2 + a}/tJxl

:::::

0

Multiplying by the integrating factor teat, which was found by much trial and error with the equation, gives t(e at d 2x l /dt 2

+

2/Xe at dx1/dt

+

a2eatxl)

which can be rewritten as td2(eatXl)/dt2

+

+

(eat dx1/dt

d(eatxI)ldt

+

+

teatxI

o:eatxl) ::;::

+

teatXI

0

This has the same form as Bessel's equation (5.38), so the solution is xI(t) x2(t)

=

cle-atJO(t)

+

dx1ldt

-o:Xt(t)

::;::

c 2e- at Y O(t)

+

e-at[-cIJI(t)

+ c2dYo/dt]

0

120

SOLUTIONS TO THE LINEAR STATE EQUATION To solve for the constants

CI

eaT

and

[CHAP. 5

C2,

C~2(T)X~T~Xl(T»)

80 that x(t) = FG(t) G-l(T) F-lx (T), where

F = (1

-a

~)

Yo(t) ) dYo/dt

and so ~(t, T)= FG(t) G-l(T)F-I. This is true only for t and- T> 0 or t and T < 0', because at the point t = 0, the elements of the original A(t) matrix blow up. This accounts for Y 0(0) = 00. Admittedly this problem was contrived, and in practice a man~made system would only accidentally have this form. However, Bessel's equation often occurs in nature, and knowledge that as t ~ co, J 0 (t) ,..., V2lrrt cos (t - '/T/4) and Yo (t) ,.., V2/'/Tt - sin (t - 7r/4) gives great insight into the behavior of the system.

5.7.

Given the time-varying difference equation x(n + 1) = A(n) x(n), where A(n) = Ao if n is even and A(n) = Al if n is odd. Find the fundamental matrix, analyze by Floquet theory, and give the conditions for stability if Ao and Al are nonsingular. From equation (5.50), AIAOA1' .. AIAo AoAIAO' .. AIAO

.z,(k,m)

1

=

if k is even and m is even if k is odd and m is even

Al AoAI ... AoAl

if k is odd and m is odd

AoAIAo' .. AOA!

if k is even and

m is odd

=

For m even, .z,(k, m) P(k, m) (A 1A o)(k-m)/2, where P(k, m) I if k is even and P(k, m) = (AoAl-I)1I2 if k is odd. For m odd, (J(k, m) = P(k, m)(AoAI)Ck-m) 12, where P(k; m) = I if k is odd and P(k,m) = (AIAO-I)1I2 if k is even. For instability, the eigenvalues of R = (AIAo)l/2 must be outside the unit circle. Since the eigenvalues of B2 are the squares of the eigenvalues of B, it is enough to find the eigenvalues of A1Ao. This agrees with the stability ~nalysis of the equation x(n + 2) = A1AoX(n).

5.8.

Find the impulse response of the system d2 y/dt2

+ (1- 20:) dy/dt +

(a 2 - 0: + e- 2t )y

~

u

Choose Xl = Y and X2 = et (dxl/dt - aXI) to find a state representation in which A(T) commutes with A(t). Then in matrix form the system is

y

=(1 O)x

From equation (5.55) and .z,(t, T) obtained from Problem 5.5, B(t, T)

=

(37'

+ a(t-7') sin (e- T - e- t )

This is the response yet) to an input u(t) = S(t - T).

5.9.

In the system of Problem 5.8, let u(t) Find the complete response.

= eCa - 2)t,

y(to)

= Yo

and (dy/dt)(to)

From equations (5.44) and Problem 5.5, yet)

=

eaCt-to)

cos (e-to - e-t)yo

+

eat st e-7" sin (e- r - e- t ) dT to

= ayo.

CHAP. 5]

SOLUTIONS TO THE LINEAR STATE EQUATION

Changing variables from

T

to

11

in the integral, where e- T

e- t =

-

11,

121

gives

Notice this problem cannot be solved by Laplace transformation in one variable.

5~10.

Given a step input U(8)

= 6/8

into a system with a transfer function

8+1

=

H(8)

+ 58 + 6

82

Find the output y(t) assuming zero initial conditions. The easiest way to do this is by using classical techniques.

=

..e{y(t)}

6(8 + 1) = 1: + _3_ _ _ 4_ 8 8+2 8+3 + 582 + 68 determines y = 1 + 3e- 2t - 4e- 3t•

=

U(8) H(8)

Taking the inverse Laplace transform

s3

Doing this by state space techniques shows· how it corresponds with the classical techniques. From Problem 2.3 the state equations are

(-~ _~) (:~) y

+

(~) u

2)(:;)

(-1

The transition matrix is obviously 4)(t, 7)

The response can be expres.sed directly in terms of (5..M). yet)

==

4)(t, to)O

==

1

+

f

t

(-1 2)

(e-2Ct-T) 0

to

+

3e- 2t

-

(5.68)

4e- 3t

This integral is usually very complicated to solve analytically, although it is easy for a computer. Instead, we shall use the transfer function matrix of equation (5.56).

(-1 (8 +02)-1 2)

-<:{H(t,O)}

=

2 8+3

(8

0) (1)

+ 3)-1

1

8+1

1 8+2

82

+ 58 + 6

==

R(s)

This is indeed our original transfer function, and the integral (5.68) is a convolution and its Laplace transform is ..e{y(t)}

whose inverse Laplace transform gives yet) as before.

5.11. Using the adjoint matrix, synthesize a form of control law for use in guidance. We desire to guide a vehicle to a final state x(tt), which is known. x(tf )

=

'fl(tt, t) x(t)

+

jtt t

*(tt, 1') B(T) U(T) dT

From (5.44),

122

SOLUTIONS TO THE LINEAR STATE EQUATION

[CHAP. 5

Choose u(t) = U(t)c where U(t) is a pre specified matrix of time functions that are easily mechanized, such as polynomials in t. The vector c is constant, except that at intervals of time it is recomputed as knowledge of x(t) becomes better. Then c can be computed as c

[ft' ~(tf'

=

T) B(.) U(T) dTJ-'

[~(tf' t) x(t)

-

x(tf )]

However, this involves finding C)(tb t) as the transition matrix of dxldt == A(t)x with x(t) as the initial condition going to x (tt). Therefore C)(tj, t) would have to be computed at each recomputation of c, starting with the best estimates of x(t). To avoid this, the adjoint transition matrix 'i'(r, tf ) can be found starting with the final time tf' and be stored and used for all recomputations of c because, from equation (5.64), 'l't(r, tf) = .p(tf , r) and c is found from c

=

[f: ~t(T.

tf) B(.) U(.) dTJ-'

[~t(t. 1,) x(t)

-

x(tf )]

Supplementary Problems 5.12.

Prove equations (5.9), (5.10), (5.11), and (5.12).

5.1S.

Given cp(k, m), how can A(k) be found?

5.14.

Prove that AeAt == eAtA and then find the conditions on A and B such that

5.15.

Verify that 4t(t,7") = eACt - r ) and cp(k, m) == Ak-m satisfy the properties of a transition matrix given in equations (5.5)-(5.12).

5.16.

Given the fundamental matrix (J(t, r)

=

~

_12 (e-e-

4Ct

-

r)

+

4ct - r ) -

11 e-e-

4Ct

-

r

==

eBeA.

1)

) -

4Ct -1')

eAe B

+1

What is the state equation corresponding to this fundamental matrix?

5.17.

Find the transition matrix to the system

dxldt

5.18.

Calculate the transition matrix .p(t,O) for

dxldt

form,

_!)

==

(~

==

(~ ~)x

x.

using (a) reduction to Jordan

(b) the Maclaurin series, (e) the resolvent matrix.

5.19.

Find eAt by the series method, where series method is the easiest.

5.20.

Find

eAt

A

G~ D·

This shows a case where the

using the resolvent matrix and Leverrier's algorithm, where

A ==

(-~ -~ ~). o

5.21.

0-3

Find eAt using the Cayley-Hamilton method, where A is the matrix given in Problem 5.20.

CHAP. 5] 5.22.

SOLUTIONS TO THE LINEAR STATE EQUATION

Use the eigenvalue method to find

eAt

123

for

(-~ -~ -~)

A

-1 -1

2

5.23.

Use the resolvent matrix and Cramer's rule to find eAt for A as given in Problem 5.22.

5.24.

Use the resolvent matrix and Cramer's rule to find Ak for A as given in Problem 5.22.

5.25.

Find eAt by using the Maclaurin series, Cayley-Hamilton and resolvent matrix methods when A

5.26.

~ (~-~).

Find the fundamental, or transition, matrix for the system

(-~

dx

at

u

o

using the matrix Laplace transform method. 5.27.

Given the continuous time system

(=i =D + G)

dxldt = y

x

::::;

(1 O)x

+

(~)

==

x(O)

u

4u

Compute yet) using the transition matrix if u is a unit step function. Compare this with the solution obtained by finding the 1 X 1 transfer function matrix for the input to the output. 5.28.

Given the discrete time system x(n+1)

(=! =!)

=

yen)

::::;

x(n)

+ (~) u(n)

(1 O)x(n)

+

x(O)

::::;

(~)

4u(n)

Compute yen) using the transition matrix if u is the series of ones 1,1,1, ... ,1, .... 5.29.

==

(a) Calculate q:,(t, to) for the system

dx/dt

(b) Calculate q:,(k, m) for the system

x(k + 1)::::;

(

-1 0

(

-1 0

21)X

using Laplace transforms.

~) x(k)

using

5.30.

How does the spectral representation for not distinct?

5.31.

In the Cayley-Hamilton method of finding eAt, show that the equation eJt

eAt

2: transforms.

extend to the case where the eigenvalues of A are n-i

~ Yi(t)Ai can always i=O

be solved for the Yi(t).

For simplicity, consider only the case of distinct eigenvalues.

5.32.

Show that the column vectors of *'(t, r) span the vector space of solutions to dx/dt == A(t)x.

5.33.

Show A(t) A(T) == A(T) A(t) when A(t) == a:(t)C, where C is a constant n X n matrix and a(t) is a scalar function of t. Also, find the conditions on ai/t) such that A(t) A(r) == A(r) A(t) for a 2 X 2 A(t) matrix.

5.34.

Given the time-varying system dx

Cit Find the transition matrix.

Hint: Find an integrating factor.

124 5.35.

5.36.

SOLUTIONS TO THE LINEAR STATE EQUATION

Given the time-varying periodic system R is constant.

= P(k, m)Rk-m where

t)

sin t sin dx/dt == ( . t . t sm sm

q,(t, to) and verify it satisfies Floquet's result

5.37.

~(k, m)

Prove Floquet's theorem for discrete time systems, if A(k) = A(k + w).

P(k + w, m)

[CHAP. 5

x.

P(k, m)

=

Find the transition matrix

q,(t, to) = pet, to)eRCt-to)

where P is periodic and

Also find the fundamental matrix of the adjoint system.

The linear system shown in Fig. 5-6 is excited by a square wave set) with period 2 and amplitude = 1. The system equation is ii + [,8 + a sgn (sin 7Tt)]y = O.

Is(t)1

8(t)o----f

' - -_ _ _- - 0

Multiplier

yet)

+

Fig. 5-6 It is found experimentally that the relationship between a and ,8 that permits a periodic solution can be plotted as shown in Fig. 5-7.

----------------~~--~--~--~--~--_r----r_----fi o 4 6 5 2 3

Fig. 5-7 Find the equation involving a and ,8 so that these lines could be obtained analytically. (Do not attempt to solve the equations.) Also give the general form of solution for all a and ,8 and mark the regions of stability and instability on the diagram. 5.38.

Given the sampled data system of Fig. 5-8 where S is a sampler that transmits the value of e(t) once a second to the hold circuit. Find the state space representation at the sampling instants of the closed loop system. Use Problem 5.10.

ret)

0

+

.(t)

j-j

Hold

H

8+1

82

oy(t)

+ 5s + 6

Fig. 5-8

5.39.

Find tt( t, T) and the forced response for the system ~(to)

=

~o.

t2;j + t~

+ n :::: p(t)

with

'I1(t o) =

no

and

CHAP. 5] 5.40.

SOLUTIONS TO THE LINEAR STATE EQUATION

Consider the system d

dt

(Xl) X

(:

2

-(~:)

!)(::) + (;) (~ ~)(::)

=

.

or

u

125

Ax+Bu

x

y = Cx + Du

or

Find the transfer functions .,e{Yl}/.,e{ul} and .,e{Y2}/.e{u2} using the relation "c{y}/.,e{u} = [C(ls - A)-IB

5.41.

+ DJ

The steady state response xss(t) of an asymptotically stable linear differential system satisfies the equation dxs/dt = A(t)xss + B(t) u(t) but does not satisfy the initial condition xss(to) = Xo and has no reference to the initial time to. T) is known.

~(t,

(a)

Verify, by substitution into the given equation, that xss(t)

where f{t, T),

f

t

ft~(t'T)B(T)U(T)dT

=

is the indefinite integral evaluated at d

dt

fset)

f(t, T) dT

=

let, {J(t»

dj3

dt -

T

=

t.

Hint: For an arbitrary vector function

do:. f(t, o:.(t» Cit

+

f~(t)

a(t)

a

aff(t, T) dt"

a(t)

= ACt + T), B(t) = B(t + T), u(t) = u(t + T), and the system is stable. expression for a periodic xss(t) = xss(t + T) in the form

(b) Suppose ACt)

xssCt)

=

K(T)

ft+T

41(t, T)

Find an

~(T) UCT) dT

t

where K(T) is an n X n matrix to be found depending on T and independent of t. 5.42.

Check that h(t) = = .sCt - 7').

et+a(t-T)

sin (e- T -

e- t )

satisfies the system equation of Problem 5.8 with

u(t)

2

(1 _ t 2) d y _

!

dy

to the input u(t) =

5.43.

Find in closed form the response of the system tyt 2 ..... 1 with zero initial conditions.

5.44.

Consider the scalar system dy/dt = -(1 + t)y + (1 + t)u. If the initial condition is yeO) = 10.0, find the sign and magnitude of the impulse in u(t) required at t 1.0 to make y(2) 1.0.

5.45.

Given the system

dt 2

t dt

=

dx/dt

=

0 -3/t2) 2/t x. ( -1

u

=

Find the relationship between Xl(t) and X2(t) such

that x 1(t j ) = 1, using the adjoint. 5.46.

Show that if an n X n nonsingular matrix solution T(t) to the equation dT/dt = A(t)T - TD(t} is known explicitly, where D(t) is an n X n diagonal matrix, then an explicit solution to dx/dt = A(t)x is also known.

SOLUTIONS TO THE LINEAR STATE EQUATION

126

[CHAP. 5

Answers to Supplementary Problems 5.13.

A(k) =
5.14.

If and only if AB = BA does

5.16.

A

(-2 -2)

5.17.

!p(t, r)

=

5.20.

(

e

o

"11

l[e

t -,.

(~) (1

2)

= 9, "12 = 26, 'Y3 = 24 0.5.- 2t

=

e4(T-t)

G

(6e-2t - 8e -3t + 3e -4t)I

5.21.

eAt

5.22.

eAt

5.23.

(sI - A)-l

(-~) (-1

2)]

0) +

o

o

1

D/s

k

-1

o

+ B/(s -

1)

+ e/(s _1)2

+

( 1-1 -1

1

0

0

O.5( e - 2t - 2e -3t + e -4t)A2

where

C~ -~ -D

e

O.5e- 4t

-

12e- 3t + 5e- 4t )A

1-1

-k -1

-k

k-1

(~t

o o

(~ ~ =~)

=

1

(

1 1

+ 0.5(7 e - 2t -

F(s)/s(s - 1)2

D

Ak

+

01)

eAt

5.24.

eDe A •

-2 -2

t

5.18.

=

eAeD

B

=

(-~ -~ ~) -1 -1

1)

2k + -2k;1

-1

1 ~ €2t)

5.25.

eAt

5.27.

yet)

![ll- 2e- (t-t o)/2 - e- (t-t o)] U(t - to)

5.28.

y(k)

4

+ [7 -

5.29.

eIt(t, to)

=

(

5.30.

Let the generalized eigenvector tt of A have a reciprocal basis vector s1' etc.

3(-l)k - 4(-1/2)k]/12

et~-t

et-to - eto- t ) , et-to

eAt

5.31.

=

cJt(k,m)

eAtxlrt

=

(-l~k-m

1 - (~1)k-m)

Then

+ eAttlst + teAtxlsT + ...

A Vandermonde matrix in the eigenvalues results, which is then always invertible.

0

5.34.

=

4t(t, to)

=

5.36.

where 4»(to, t). 5.37.

127

SOLUTIONS TO THE LINEAR STATE EQUATION

CHAP. 5]

eT (COSh sinh

T T

= cos to - cos t,

T

Let 1'2 = f3

+ a,

82

41'(2,1)

T)

sinh cosh

T

= f3 - a. =

4»(t, to) = pet, to)

so

Then sin IS

Also

~t(t, to)

e2R = 4»(2,0) = cp(2, 1) cp(l, 0) where (sin 0)/0) cos 8

COSO

-s

(

O.

R

and

cosy ( -I' sin I'

=

4»(1,0)

Y)/Y)

(sin cosy

For periodicity of the envelope zCt + 471/8) = zCt), eigenvalues '}-. of e 2R = e:!:.:Jo.

=

det ('}-./ - e2R)

'}-.2 ~

'}-.2 cos (J

+

1

=

'}-.2 -

'}-.

tr e2R

+

det e 2R

det cp(2, 1) det 41(1,0) = 1

=

2 cos 8

2 cos I' cos e; -

(1'/8

+ S/y)

sin y sin S

The stability boundaries are then determined by

=

±2

2 cos y cos 0 -

(1'/8

+ Sly)

sin I' sin 0

The solution is of the form .pCt, T) = PCt, T)eR(t-T) and the given curves form the stability boundaries between unstable regions and periodic regions. Reference: B. Van Der Pol and M. J. O. Strutt, On the Stability of the Solutions of Mathieu's Equation, Philosophical Magazine, 7th series, vol. V, January-June 1928, pp. 18-38.

5.38.

x(k + 1)

=

5.39.

'I](t)

'1]0

1)

1 + e- 2 )/2 ( (1- e- 3 )/3

cos (In t/to)

e- 2 (5e-3 - 2)/3

+ ~oto sin (In tlto)

x(k)

-

+

(3 -

3e- 2 ) r(k) 2 - 2e- 3 6

ft sin (1n rlt) p(r)/T dT to

5.41.

K(T) = (e- RT - 1)-1

5.42.

Since h(t) is an element of the transition matrix, d2h/dt2

+

(1- 2) dh/dt

+

(0;2 -

0;

+ e-Zt)h

Also, since dh/dt and h are continuous in t, lim 1:'--+0

=

5.43.

yet)

5.44.

u(t)

5.46.

Let x

=

(1/4 ~

Vt Z -1 )(t -

1 - 10e- 4 2e- 5/2 oCt - 1.0)

= T(t)z.

i7"+1:'

d 2h/dt2 dt

T-(,

to)

+

(sin -1 t - sin- 1 t o)/2

o

for t #'

r

6

Chapter Controllability and Observability 6.1 INTRODUCTION TO CONTROLLABILITY AND OBSERV ABILITY

Can all the states of a system be controlled and/or observed? This fundamental question arises surprisingly often in both practical and theoretical investigations and is most easily investigated using state space techniques. Definition 6.1:

A state Xl of a system is controllable if all initial conditions Xo at any previous time to can be transferred to Xl in a finite time by some control function u(t, xo).

If all states Xl are controllable, the system is called completely controllable or simply controllable. If controllability is restricted to depend on to, the state is said to be controllable at time to. If the state can be transferred from Xo to Xl as quickly as desired independent of to, instead of in some finite time, that state is totally controllable. The system is totally controllable if all states are totally controllable. Finally, we may talk about the output y instead of the state X and give similar definitions for output controllable, e.g. an output controllable at time to means that a particular output y can be attained starting from any arbitrary Xu at to. To determine complete controllability at time to for linear systems, it is necessary and sufficient to investigate whether the zero state instead of all initial states can be transferred to all final states. Writing the complete solution for the linear case, fIJ(tl, to) x(to)

+

i

h

CP(tl, i) B(i) u(or) di to

x

which is equivalent to starting from the zero state and going to a final state (tt) = x(t1) - tIt(tt, to) x(to). Therefore if we can show the linear system can go from 0 to any x(t 1), then it can go from any x(to) to any X(tl)' The concept of observability will turn out to be the dual of controllability. Definition 6.2:

A state x(t) at some given t of a system is observable if knowledge of the input u(-r) and output y(.) over a finite time segment to < or === t completely determines x(t).

If all states x(t) are observable, the system is called completely observable. If observability depends on to, the state is said to be observable at to. If the state can be determined for 'T in any arbitrarily small time segment independent of to, it is totally observable. Finally, we may talk about observability when U(i) = 0, and give similar definitions for zero-input observable. To determine complete observability for linear systems, it is necessary and sufficient to see if the initial state x(to) of the zero-input system can be completely determined from y(-r), because knowledge of x(to) and U(i) permits x(t) to be calculated from the complete solution equation (5.44).

128

CHAP. 6]

CONTROLLABILITY AND OBSERVABILITY

129

We have already encountered uncontrollable and unobservable states in Example 1.7, page 3. These states were physically disconnected from the input or the output. By physically disconnected we mean that for all time the flow diagram shows no connection, i.e. the control passes through a scalor with zero gain. Then it follows that any state vectors having elements that are disconnected from the input or output will be uncontrollable or unobservable. However, there exist uncontrollable and unobservable systems in which the flow diagram is not always disconnected. Example 6.1. Consider the time-varying system

%t (::)

( ao f3O)(Xl) X2

+

(eat) eSt

U

From the flow diagram of Fig. 6-1 it can be seen that u(t) passes through scalors that are never zero.

uo-------+

Fig. 6-1 For zero initial conditions, xl(tl)e-atl

=

st!

U(T) dT

=

x2(tl)e-f3tl

to

Only those xz(t 1) = x 1(t 1)e tt C,6-O:) can be reached at tlo so that X2{t1) is fixed after Xl(t 1) is chosen. fore the system is not controllable.

There-

6.2 CONTROLLABILITY IN TIME-INVARIANT LINEAR SYSTEMS For time-invariant systems dxldt = Ax + Bu in the case where A has distinct eigenvalues, connectedness between the input and all elements of the state vector becomes equivalent to the strongest form of controllability, totally controllable. We shall first consider the case of a scalar input to avoid complexity of notation, and consider distinct eigenvalues before going on to the general case.

Theorem 6.1:

Given a scalar input u(t) to the time-invariant system dx/dt = Ax + bu, where A has distinct eigenvalues Ai. Then the system is totally controllable if and only if the vector f = M-1b has no zero elements. M is the modal matrix with eigenvectors of A as its column vectors.

Proof; Only if part: Change variables by x = Mz. Then the system becomes dz/ dt = Az + fu, where.A. is the diagonal matrix of distinct eigenvalues Ai_ A flow diagram

of this system is shown in Fig. 6-2 below. If any element ii of the f vector is zero, the element of the state vector Zi is disconnected from the control. Consequently any x made up of a linear combination of the z's involving Zt will be uncontrollable. Therefore if the system is totally controllable, then all elements it must be nonzero.

130

CONTROLLABILITY AND OBSERVABILITY

U

[CHAP. 6

o-------to

Fig. 6-2

If part: Now we assume all Ii are nonzero, and from the remarks following Definition 6.1 we need investigate only whether the transformed system can be transferred to an arbitrary Z(t1) from z(to) = 0, where t1 can be arbitrarily close to to. To do this, note i = 1,2, .. . ,n It is true, but yet unproven, that if the

o to Z(tl).

Ii are nonzero, many different

(6.1)

u(.-) can transfer

Now we construct a particular U{-r) that will always do the job. Prescribe U(T) as

==

U(T)

n

~ I-'-ke-A~(1'-tl)

(6.2)

k=l

where the JLk are constants to be chosen. (6.1) gives

==

n

~

Substituting the construction (6.2) into equation

f. JLk( e Ak (tl-1'),

k=l

e At (tC 1'J)

for all i

(6.3)

t

where the inner product is defined as

Equation (6.3) can be written in matrix notation as ZI(tl)/h Z2(t1 )112 Zn(tl)/In

=

U11 g21

...........

gln

gIn g2n

g12 U22 g2n

.,

••

41

1-'-1

P-2

(6.4)

......

Unn

f1-n

where gik = (e Ak (tl- T ), eAtCtl-1'J). Note that because Ii =F 0 by assumption, division by Ii is permitted. Since the time functions eAiCtC1') are obviously linearly independent, the Gram matrix {Yik} is nonsingular (see Problem 3.14). Hence we can always solve equation (6.4) for }-thjJ.2, ••• , p.n, which means the control (6.2) will always work. Now we can consider what happens when A is not restricted to have distinct eigenvalues.

CHAP. 6]

Theorem 6.2:

un

CONTROLLABILITY AND OBSERVABILITY

Given a scalar input u(t) to the time-invariant system dx/dt = Ax + bu, where A is arbitrary. Then the system is totally controllable if and only if: (1) each eigenvalue Ai associated with a Jordan block Lji(Ai) is distinct from an eigenvalue associated with another Jordan block, and (2) each element fi of f = T- 1b, where T-1AT = J, associated with the bottom row of each Jordan block is nonzero.

Note that Theorem 6.1 is a special case of Theorem 6.2. Proof: Only if part: The system is assumed controllable. The flow diagram for one l x l Jordan block Lji(Ai) of the transformed system dzldt = Jz + fu is shown in Fig. 6-3. The control u is connected to Zl, Z2, ••• , Z!-l and Zl only if it is nonzero. It does not matter if 11,/2, ... and fl-1 are zero or not, so that the controllable system requires condition (2) to hold. Furthermore suppose condition (1) did not hold. Then the bottom rows of two different Jordan blocks with the same eigenvalues [Lvi(Ai) and L1ji(Ai)] could be written as

dzv/dt

AiZv

+ /,,11-

u

~-----------------------------~.~~---------------Fig. 6-3

Consider the particular state having one element equal to fTJzv - fvzn. d(fT/zv- f"zTJ)ldt

= fT/(AiZv + fvu) = Ai (fTJ Zv - !"Z1j)

/,,(AiZT/

Then

+ /TJu)

Therefore /1/Z,,(t) - /vZTJ(t) = [fTJzv(O) - /vzTJ(O)]d,!t and is independent of the control. We have found a particular state that is not controllable, so if the system is controllable, condition (1) must hold.

If part: Again, a control can be constructed in similar manner to equation (6.2) to show the system is totally controllable if conditions (1) and (2) of Theorem 6.2 hold. Example 6.2. To illustrate why condition (1) of Theorem 6.2 is important, consider the system

(~ ~)(:~) y

=

(1

+

G)

u

-3)(:~)

=

=

Then .,e{X1} = (8.,e{u} + xlO)/(s - 2) and .e{X2} (.e{u} + x2o)/(s - 2) so that y{t) (XIO - 3x20)e2t regardless of the action of the control. The input is physically connected to the state and the state is physically connected to the output, but the output cannot be controlled.

For discrete-time systems, an analogous theorem holds.

132 Theorem 6.3:

CONTROLLABILITY AND OBSERVABILITY

[CHAP. 6

Given a scalar inputu(m) to the time-invariant system x(m + 1) = Ax + bu(m), where A is arbitrary. Then the system is completely controllable if and only if conditions (1) and (2) of Theorem 6.2 hold.

Proof: Only if part: This is analogous to the only if part of Theorem 6.2, in that the flow diagram shows the control is disconnected from at least one element of the state vector if condition (2) does not hold, and a particular state vector with an element equal to f7J zv - fv z7J is uncontrollable if condition (1) does not hold.

If part: Consider the transformed systems z(m + 1) = Jz(m) + fu(m), and for simplicity assume distinct roots so that J = 4. Then for zero initial condition,

For an nth order system, the desired state can be reached on the nth step because Zl(n)//I

,.\n-l 1

Z2(n)//2

,.\n-l 2

••••

Zn(n)/fn

.\n-2 1 ,.\n-2 2

.\n-l n

ill

.......

,.\n-2 n

1 1

u(O) u(1)

(6.5) ill

••

ill

....

,.

..

1

u(n-l)

Note that a Vandermonde matrix with distinct elements results, and so it is nonsingular. Therefore we can solve (6.5) for a control sequence u(O),u(1), .. • ,u(n-l) to bring the system to a desired state in n steps if the conditions of the theorem hold. For discrete-time systems with a scalar input, it takes at least n steps to transfer to an arbitrary desired state. The corresponding control can be found from equation (6.5), called dead beat control. Since it takes n steps, only complete controllability was stated in the theorem. We could (but will not) change the -definition of total controllability to say that in the case of discrete-time systems, transfer in n steps is total control. The phenomenon of hidden oscillations in sampled data systems deserves some mention here. Given a periodic function, such as sin rot, if we sample it at a multiple of its period it will be undetectable. Referring to Fig. 6-4, it is impossible to tell from the sample points whether the dashed straight line or the sine wave is being sampled. This has nothing to do with controllability or observability, because it represents a failure of the abstract object (the difference equation) to represent a physical object. In this case, a differential-difference equation can be used to represent behavior between sampling instants.

Fig. 6-4

6.3 OBSERVABILITY IN TIME-INVARIANT LINEAR SYSTEMS Analogous to Theorem 6.1, connectedness between state and output becomes equivalent to total observability for dx/dt = Ax + Bu, Y = ctx + dTu, when the system is stationary and A has distinct eigenvalues. To avoid complexity, first we -consider scalar outputs and distinct eigenvalues.

· CHAP. 6]

CONTROLLABILITY AND OBSERVABILITY

Theorem 664:

133

Given a scalar output y(t) to the time-invariant system dx/dt = Ax + Bu, y = ctx + dTu, where A has distinct eigenvalues Ai. Then the system is totally observable if and only if the vector gt = ctM has no zero elements. M is the modal matrix with eigenvectors of A as its column vectors.

Proof: From the remarks following Definition 6.2 we need to see if x(to) can be reconstructed from measurement of y(T) over to < T === t in the case where U(T) = O. We do this by changing variables as - x = Mz. Then the system becomes dz/dt = Az and y = ctMz = gtz. The flow diagram for this system is given in Fig. 6.5.

Each Zi(t) = zi(to)ehiCt-tO) can be determined by taking in measurements of y(t) at times Tk = to + (tl - to)k/n for k = 1,2, ... , n and solving the set of equations

y

11

Y(i k)

= i=l L g,zi(tO)e"-ikCtl-to)/n t

for giZi(tO)' When written in matrix form this set of equations gives a Vandermonde matrix which is always nonsingular if the Ai are distinct. If all gi oF 0, then all Zi(t O) can be found. To find x(t), use x(to) = Mz(to) and dx/dt = Ax + Bu. Only if gi oF 0 is each state connected to y.

Fig. 6-5

The extension to a general A matrix is similar to the controllability Theorem 6.2.

Theorem 6.5:

Given a scalar output y(t) from the time-invariant system dx/dt = Ax + Bu, = ctx + dTu, where A is arbitrary. Then the system is totally observable if and only if:

y

(1) each eigenvalue Ai associated with a Jordan block Lji(Ai) is distinct from an eigenvalue associated with another Jordan block, and (2) each element gi of gt = etT, where T-IAT top row of each Jordan block is nonzero.

= J,

associated with the

The proof is similar to that of Theorem 6.2.

Theorem 6.6:

Given a scalar output y(m) from the time-invariant system x(m + 1) = Ax(m) + Bu(m), Y(11~) = etx(m) + dTu(m), where A is arbitrary. Then the system is completely observable if and only if conditions (1) and (2) of Theorem 6.5 hold.

The proof is similar to that of Theorem 6.3. Now we can classify the elements of the state vectors of dx/dt = Ax + Bu, y = Cx + Du and of x(m + 1) = Ax(m) + Bu(m), y = Cx + Du according to whether they are controllable or not, and whether they are observable or not. In particular, in the single input-single output case when A has distinct eigenvalues, those elements of the state Zi that have nonzero Ii and gi are both controllable and observable, those elements Zj that have zero !J but nonzero gj are uncontrollable but observable, etc. When A has repeated eigenvalues, a glance., at Fig. 6-3 shows that Zk is controllable if and only if not all of /k, h+l' ... , f~ are zero, and the eigenvalues associated with individual Jordan blocks are distinct. Unobservable and uncontrollable elements of a state vector cancel out of the transfer function of a single input-single output system.

134

CONTROLLABILITY AND OBSERVABILITY

Theorem 6.7:

[CHAP. 6

For the single input-single output system dx/dt = Ax + bu, Y = ctx + du, the transfer function ct(sI ~ A)-lb has poles that are canceled by zeros if and only if some states are uncontrollable and/or unobservable. A similar statement holds for discrete-time systems.

Prool; First, note that the Jordan flow diagram (see Section 2.4) to represent the transfer function cannot be drawn with repeated eigenvalues associated with different Jordan blocks. (Try it! The elements belonging to a particular eigenvalue must be combined.) Furthermore, if the J matrix has repeated eigenvalues associated with different Jordan blocks, an immediate cancellation occurs. This can be shown by considering the bottom rows of two Jordan blocks with identical eigenvalues A. d dt

(Zl) Z2 y

Then for zero intial conditions ..({y}

..({Zl} = (s~ A)-lb1..({u} [Cl(S - A)-lb 1

+ C2(S -

and ..({Z2} = (s - A)-lb 2..({u}, so that

A)-lb 2

+ d]..c{u}

Combining terms gives ..c{y}

This is a first-order transfer function representing a second-order system, so a cancellation has occurred. Starting from the transfer function gives a system representation of dz/dt = Az + u, Y = (c1b l + c2 b2 )z + du, which illustrates why the Jordan flow diagram cannot be drawn for systems with repeated eigenvalues. Now consider condition (2) of Theorems 6.2 and 6.5. Combining the flow diagrams of Figs. 6-2 and 6-5 to represent the element Zi of the bottom row of a Jordan block gives Fig. 6-6. I

I

I UO-O---~"I

I

·G--+Qf---~

·

y

I I

Fig. 6-6

Comparing this figure with Fig. 2-12 shows that fiu i = Pi' the residue of Ai' If and only if P'i = 0, a cancellation occurs, and Pi = 0 if and only if Ii and/or Ui = 0, which occurs when the system is uncontrollable and/or unobservable. Note that it is the uncontrollable and/or unobservable element of the state vector that is canceled from the transfer function.

6.4 DIRECT CRITERIA FROM A, B, AND C If we need not determine which elements of the state vector are controllable and observable, but merely need to investigate the controllability and/or observability of the whole system, calculation of the Jordan form is not necessary. Criteria llsing only A, Band C are available and provide an easy and general way to determine if the system is completely controllable and observable.

CHAP. 6]

CONTROLLABILITY AND OBSERV ABILITY

Theorem 6.8:

135

The time-invariant system dx/dt = Ax + Bu is totally controllable if and only if the n x nn1, matrix Q has rank n, where

Q

=

(BIABI ... I An-IB)

Note that this criterion is valid for vector u, and so is more general than Theorem 6.2. One method of determining if rank Q =, n is to see if det (QQT) =1= 0, since rank Q = rank QQT from property 16, page 87. However, there exist faster machine methods for determining the rank. Proof; To reach an arbitrary x(tt} from the zero initial state, we must find a control u(.) such that x(tI)

(6.6) n

Use of Theorem 4.16 gives

L

eACtl- r )

·'/i(.)An-i so that substitution into (6.6) gives

i=l

(BWI + ABw2 + ...

X(tl)

Jrto

Q ( W~'nl)

+ An-1Bwn)

t1

where w k =

'Yn+l-k(')U(.) dT.

Hence X(tl) lies in the range space of Q, so that Q must

have rank n to reach an arbitrary vector in "Un' trollable, Q has rank n.

Therefore if the system is totally con-

Now we assume Q has rank n and show that the system is totally controllable. time we construct u(,.) as U(T) =

/l/3(T -

to)

+ /l20(1)(. -

to)

+ ... +

/ln8(n-D(. -

to)

This (6.7)

where /lIe are constant 'Tn-vectors to be found and 8 Ck )(-.) is the kth derivative of the Dirac delta function. Substituting this construction into equation (6.6) gives

since

stl to

x(t 1 ) = eACt1-T)B8(,. -

eACtl-to)B/ll

to)

dT

=

+

eACtl-tO) AB/l2

eACtl-to)B

f_:

+ ... + eACtcto) An-IB/l n

(6.8)

and the defining relation for 8Ck ) is

OCk)(t -~) g(~)

d~

dkg/dtk

Using the inversion property of transition nlatrices and the definition of Q, equation (6.8) can be rewritten

From Problem 3.11, page 63, a solution for the ILi always exists if rankQ = n. Hence SODle (perhaps not unique) always exists such that the control (6.7) will drive the system to

ILi

X(t1).

The construction for the control (6.7) gives some insight as to why completely controllable stationary linear systems can be transferred to any desired state as quickly as possible. No restrictions are put on the magnitude or shape of u(.). If the magnitude of the control is bounded, the set of states to which the system can· be transferred by t1 are called the reachable states at t l , which has the dual concept in observabiIity as recoverable states at t l . Any further discussion of this point is beyond the scope of this text.

CONTROLLABILITY AND OBSERVABILITY

136

[CHAP. 6

A proof can be given involving a construction for a bounded u(t) similar to equation (6.2), instead of the unbounded u(t) of (6.7). However, as tl ...,.. to, any control must become unbounded to introduce a jump from x(t o) to x(t l ). The dual theorem to the one just proven is Theorem 6.9:

The time-invariant system dx/ dt = Ax + Bu, y = Cx + Du is totally observable if and only if the kn X n matrix P has rank n where-

P

Theorems 6.8 and 6.9 are also true (replacing totally by completely) for the discrete-time system x(m + 1) = Ax(m) + Bu(m), y(m) = Cx(m) + Du(m). Since the proof of Theorem 6.9 is quite similar to that of Theorem 6.8 in the continuous-time case, we give a proof for the discrete-time case. It is sufficient to see if the initial state x(l) can be reconstructed from knowledge of y(m) for l ~ m < 00, in the case where u(m) = O. From the state equation, y(l)

=

Cx(l)

y(l + 1) ::;: Cx(l + 1)

=

CAx(l)

y(l + n -1) = CAu-1x(l)

This can be rewritten as y(l)

)

.

(

Y(I+~-l)

=

Px(l)

and a unique solution exists if and only if P has rank n, as shown in Problem 3.11, page 63. 6.5

CONTROLLABILITY AND OBSERVABILITY OF TIME-VARYING SYSTEMS

In time-varying systems, the difference between totally and completely controllable becomes important. Example 6.3. Consider the time~varying scalar system or unity as shown in Fig. 6-7.

dx/dt = -x + b(t)u

and y = x, where b(t) is either zero

Y

u

L...--_---f

-1 '--_.....

Fig. 6~7 If b(t) = 0 for to == t < to + at, the system is uncontrollable in the time inter~al [to, to + at). If b(t) = 1 for to + at:::;:: t :::;:: t l , the system is totally controllable in the time interval to + a. t == t ~ t I •

However, the system is not totally controllable, but is completely controllable over the time interval == t """ tl to reach a desired final state x(t l )·

to

CHAP. 6]

CONTROLLABILITY AND OBSERVABILITY

137

Now suppose b(t) = 0 for t1 < t === t 2 • and we wish to reach a final state x(t2 ). The state x(t2 ) can be reached by controlling the system such that the state at time t1 is x(t1) = e(t2 -h)x(t2 ). Then with zero input the free system will coast to the desired statex(t2 ) = (t2 • t 1 ) x(t l ). Therefore if the system is totally controllable for any time interval te === t === te + fl.t, it is completely controllable for all t:=, te'

For the time-varying system, a criterion analogous to Theorem 6.8 can be formed. Theorem 6.10: The time-varying system dx/dt = A(t)x + B(t)u is totally controllable if and only if the nlatrix Q(t) has rank n for times t everywhere dense in [to, t l ], where Q(t) = (Q1 I Q21 ... [Qn), in which Ql = B(t) and Qk+l = -A(t)Qk + dQk/dt for k = 1,2, ... , n-l. Here A(t) and B(t) are assumed piecewise differentiable at least n - 2 and n -1 tinles, respectively.

The phrase "for times t everywhere dense in [to, t l ]" essentially means that there can exist only isolated points in t in which rank Q < n. Because this concept occurs frequently, we shall abbreviate it to "Q(t) has rank n(e.d.)". Proof: First we assume rank Q = n in an interval containing time 'fJ such that to < 'fJ < tl and show the system is totally controllable.

Construct u( T) as

n

U(T)

~ ILk8Ck-1)( T

-

"l)

(6.9)

k=l

To attain an arbitrary x(tt} starting from x(to) = 0 we must have

Jrto

t1

n

k~l

q.(tl' ,) B(,) U(T) dT

k 1 d d,k-l

[q,(tl, ,) B(T)]

I '/"=7)

(6.10)

ILk

But (6.11)

Note 0 = dl/dt = d(q.q,-l)/dt (6.11) becomes

= Acpq,-l +cpdcp-l/dt

Similarly,

dk -

so that

d~-l/dt

= -cp-lA

and equation

1

(6.12)

dT k - 1 [CP(tl' T) B(T)]

Therefore (6.10) becomes

A solution always exists for the

because rank Q = nand

ILk

cp

is nonsingular.

Now assume the system is totally controllable and show rank Q = n. From Problem 6.8, page 142, there exists no constant n-vector z =1= 0 such that, for times t everywhere dense in to";:::: t ..<::: t l , ztq.(to, t) B(t) = 0 By differentiating k times with respect to t and using equation (6.12), this becomes ztq.(to, t) Qk(t) = O. Since q.(to, t) is always nonsingular, there is no n-vector yt = ztq.(to, t) # 0 such that

o=

(ytQl J ytQ2 J

•..

J

ytQn)

=

ytQ

= Y1ql + Y2q2 + . . . + Ynqn

where qi are the row vectors of Q. Since the n row vectors are then linearly independent, the rank of Q(t) is n(e.d.).

138

CONTROLLABILITY AND OBSERV ABILITY

[CHAP. 6

= A(t)x + B(t)u, y;:::: C(t)x + D(t)u is totally observable if and only if the matrix P(t) has rank n(e.d.), where PT(t) = (Pi I I ... I PJ) in which PI = C(t) and P k + I = PkA(t) + dPk/dt for k = 1,2, ... , n-l. Again A(t) and B(t) are assumed piecewise differentiable at least n - 2 and n -1 times, respectively.

Theorem 6.11: The time-varying system dx/dt

pr

Again, the proof is somewhat similar to Theorem 6.10 and will not be given. The situation is somewhat different for the discrete-time case, because generalizing the proof following Theorem 6.9 leads to the criterion rank P = n, where for this case PI

= C(l),

P2

= C(l + 1) A(l), ... ,

Pk

=

C(l + k - 1) A(l + k - 2)· .. A(l)

The situation changes somewhat when only complete controllability is required. Since any system that is totally controllable must be completely controllable, if rank Q(t) has rank n for some t> to [not rank n(e.d)] then x(t o) can be transferred to X(tl) for tl ~ t. On the other hand, systems for which rank Q(t) < n for all t might be completely controllable (but not totally controllable). Example 6.4.

Given the system

_(ao

where (2k

+

X2

< (2k + 1)rr + l)rr ::: t < 2(k + 1)7T

2krr ::: t

=

O)(Xl)

(J

(fl(t»)u h(t) k = 0, ±1, ±2, ...

Q(t)

At each instant of time, one row of Q(t) is zero, so rank Q(t) :::: 1 for all t.

However,

> 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.

Solved Problems 6.1.

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.

CONTROLLABILITY AND OBSERVABILITY

[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]

CONTROLLABILITY AND OBSERVABILITY

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

CONTROLLABILITY AND OBSERVABILITY

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]

CONTROLLABILITY AND OBSERVABILITY

143

Supplementary Problems 6.9.

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.

CONTROLLABILITY AND OBSERVABILITY

:::::

(-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:<: such that it hasn - nl zero rows.

Theorem 7.3:

The system (7.1) is reducible to (7.2) by an equivalence transformation if and only if Qx(t) has rankn~(e.d.) and can be factored as QX = Vl(S 1R) where VI is an 11. X nl differentiable matrix with ranknl everywhere, S is an nl X mnl matrix with rank nl(e.d.), andR is any nl X m(n - nl) matrix.

Note here we do not say how the factorization QX

·Proof:

Q~ = (~:) = Q:+l

=

(
(Qo

Z1

.

... 0

n1

= Tc(t) z(t)

and if Q:

=

Cf)

!~~)(~:')

+

fft (
A~, Q:' ; dQ:'1dt) QZI

1 .' '.'

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:<:1). ~

... 0

where F(t) is the nl X m(n-nl) matrix manufactured by the iteration process (7.4) for i = nl, nl + 1, ... , n-l. Since QZl has rank nl(e.d.), QZ must also. Use of Theorem 7.1 and the nonsingularity of Tc shows QX has ranknt(e.d.). Furthermore, let Tc(t) (Tl(t) IT2(t»),

=

so that

. Tc(t) Qz(t) (T1QZl TtF)

Since Tl(t) is an n X nl diiferentiablematrix with ranknl everywhere and QZl is an nl x 1nnl matrix with rank nl(e.d.), the only if part of ,the theorem has been proven:

150

CANONICAL FORMS OF THE STATE EQUATION

[CHAP. 7

For the proof in the other direction, since QX factors, then

where Vz(t) is any set of n - nl differentiable columns making Vet) = (VI Vz) nonsingular. But what is the system corresponding to the controllability matrix on the right? From Theorem 6.10,

(~,)

C~+,)

= BZ. =

-

Also,

(!i:

!~:)(~i) +

:t (~i)

i = 1,2, ... , n,-l

and

Therefore 0 = -A~l(t)S(t), and since Set) has ranknl(e.d.) _and A~l(t) is continuous, by Problem 3.11, A~l(t) = O. Therefore the transformationV(t) = (VI(t) Vz(t» is the required equivalence transformation. The dual relationship for observability is x(t) =TO(t) z(t) that transforms (7.1) into

=

y

where the subsystem dza/dt

=

A~a(t)Z3

Y

=

(CHt) 0) (:)

(7.6)

C~(t)Z3

is of order n3 ~n and has an observability matrix pza(t) with rankn3(e.d.). has n - n3 zero columns.

Theorem 7.4:

(7.5)

Then PZ(t)

The system (7.1) is reducible to (7.5) by an equivalence transformation if and only if Px has rank n3(e.d.) and factors as px

= (~) Ba where Ba is an

n3 X n differentiable matrix with rankna everywhere and S is a kn3 x na matrix with rank ns(e.d.).

7.3.

(Rs)

Here TO = -1 where R4 is any , R4 set of n - na differentiable rows making TO nonsingular and S is the observability matrix of the totally observable subsystem. The proof is similar to that of Theorem

We can extend this procedure to find states WI, Wz, Ws and W4 that are controllable and observable, uncontrollable and observable, controllable and unobservable, and uncontrollable and unobservable respectively. The system (7.1) is transformed into

y

in which

Wi

=

(C~ C~

0 O)w

is an ni-vector for i = 1,2,3,4 and where the subsystem

(7.7)

CHAP.7J

CANONICAL FORMS OF THE STATE EQUATION

( A~ A~

d (WI) dt Wa

(~:: ~::)

has a controllability matrix d dt

(Wl)

o

A~

)(WI) + (B~) B~ Ws

has an observability matrix

y

(~:: ~:: )

u

of rank n, + n3 ~ n( e.d.) and where the subsystem

=

W2

151

(7.8)

=

of rank n, + n2 "" n(e.d. ). Hence these subsystems

are totally controllable and totally observable, respectively. Clearly if such a system as (7.7) can be found, the states WI, W2, Ws, W4 will be as desired because the flow diagram of (7.7) shows W2 and W4 disconnected from the control and Ws and W4 disconnected from the output.

Theorem 7.5:

The system (7.1) is reducible to (7.7) by an equivalence transformation if and only if px has ranknl+n2(e.d.) and factors as PX = H I UI +R2U Z and QX has rank ni + ns (e.d) and factors as QX = VlSl + V3Sa•

Here Hi( t) is a kn x ~ matrix with rank ni( e.d.), Si( t) is an ~ X mn matrix with rank ni( e.d.), Ui(t) is an ni x n differentiable matrix with rank ~ everywhere, Vi(t) is an n x ~ differentiable matrix with rank ni everywhere, and Ui(t) Vj(t) = 8ijI~. Furthermore, the rank of Hi and Si must be such that the controllability and observability matrices of (7.8) have the correct rank.

Proof:

First assume (7.1) can be transformed to (7.7). By reasoning similar to (7.4),

(

~l

QI3

Q3I

Q33

o

o

o

P I2 P 22 Ga2 G42

FI2

0

pz

F32 0

0 0 0 0

so that QX and px have rank ni +na (e.d.) and ni + n2 (e.d.), respectively. Let

Then and

QX

px

VI (Qll QI3 F lZ F I4)

+

Va(Q31 Q33 Fa2 F 34)

(Pi1 piz Gi3 Gi4)TUl

+

(Pil pJ2 Gi3 G~)TU2

which is the required form. Now suppose px and QX factor in the required manner. Then, by reasoning similar to the proof of Theorem 7.3, Hi = (P~P~ G~ GL)T for i = 1,2 and Si = (Qil Qi3 Fi2 Fi4) for i = 1, 3, and A~, A~ , A~, A;, A~, A:1 and A~ have all zero elements. Theorem 7.5 leads to the following construction procedure: 1. Factor pXQx = HISI to obtain HI and SI. 2. Solve pXVl

= HI

3. Check that UIVl

and UIQx

=I

n1 ,

= 81

for VI and U I.

15.2

CANONICAL FORMS OF THE STATE EQUATION

4. Factor px - RIUl = R2U2 and QX - VlSl

= Vasa

[CHAP. 7

to obtain R2, U 2 , Va and S3.

5. Find the reciprocal basis vectors of U 2 to form V2.

6. Find V4 as any set of n4 differentiable columns making T(t) 7. Check that

UiVj = SUInt.

nonsingular~

8. Form T(t):::: (VI V 2 Va V4 ).

Unfortunately, factorization and finding sets of differentiable columns to make T(t) nonsingular is not easy in general. Exam~]e

7.1·

GIven P x =

1.)

(Sin t sin t 1 .

Obviously rank PX = 1 and it is factored by inspection, but suppose we

try to mechanize the procedure for the general case by attempting elementary column operations analogous to Example 3.13, page 44. Then

t)-2)

sin t ( sin t

(11

l)((Sin t)-i -(sin 1 0 (sin t)-l

00)

The matrix on the right is perfect for pz, but the transformation matrix is not differentiable at t = i1r for i = ... - 1, 0, 1, ' , ,

However, if a(t) and f3(t) are analytic functions with no common zeros, let E(t)

=

a(t) ( f3(t)

-f3(t») a(t)

Then (a(t) P(t»E(t) = + f32(t) 0) and E(t) is always nonsingularand differentiable. This gives us a means of attempting a factorization even if not all the elements are analytic. (a 2(t)

If a(t) and P(t) are analytic but have common zeros Cl, tz, cali be fixed up as

(;~Z -~l~D

E(t)

Jl

... , Ck,

••• , the matrix E(t)

(1- tll:k)-"Yk(l:k)

where Pk is the order of their common zero Ck and Yk(C k) is a convergence factor. Example 7.2. Let a:(t) Then

= t2

and /3(t) = t S ,

Their common zero is tl

=0

with order 2

= Pl'

Choose

Yt(fl}

= rio

E(t)

Note E(O) is nonsingular.

Using repeated applications of this form of elementary row or column operation, it is obvious that we can find T(t) = En(t)En-l(t) ... El(t) such that pxT = (pzll 0), and similarly for QX if px or QX is analytic and of fixed rank r then px factors as PZIU l •

===

n.

Also, denoting T-l -=

The many difficulties encountered if PX or QX changes rank should be noted. filters or controllers in this case can become quite complicated.

... (UU~).,

Design of

7.5 CANONICAL FORMS FOR TIME-VARYING SYSTEMS As discussed in Section 5.3, no form analogous to the Jordan form exists for a general time-varying linear system. Therefore we shall study transformations to forms analogous to the first and second canonical forms of Section 2.3. -

CHAP. 7]

CANONICAL FORMS OF THE STATE EQUATION

153

Consider first the transformation of a single-input, time-varying system

= A(t)x + b(t)u

dX/dt

(7.9)

to a form analogous to (2.6),

dz

0

o

o

o

1

o

o z

==

dt

-an_let) -an(t)

Theorem 7.6:

1

0 0

0 0

+

u

1

o

o

1

(7.10)

The system (7.9) can be transformed to (7.10) by an equivalence transformation if and only if Qx(t), the controllability matrix of (7.9), is differentiable and has rank n everywhere.

Note this implies (7.9) must be more than totally controllable to be put in the form (7.10) in that rank Q = n everywhere, not n(e.d.). However, using the methods of the previous section, the totally controllable states can be found to form a subsystem than can be put in canonical form.

Proof:

This is proven by showing the needed equivalence transformation T(t) = First let x = Qx(t)w, where the n x n

QX(t)K, where K is a nonsingular constant matrix. matrix QX(t) is partitioned as (qi j ••• j q~). Then

0

dw ([t

=

0 0

-1 0 -1 ..

oil

ill

•

•

•

o 0 b x = Qxb w = qf

This is true because qf+l for i = 1,2, ... , n-l.

(-l)nan (-l)n-lan-l (-1)n- 2an-2

0 0 0 ..

•

..

•

..

,.

•

•

ill

ill

ill

~

•

•

..

..

..

•

•

..

W

+

1 0 0

u

(7.11)

•

o

... -1

and QX-l(AxQx-dQx/dt) =AW so that -AXq:+dqf/dt = Also AWen = QX-l(AXq~ - dq~/dt). Setting w = Kz, where

=

K

(~~:~~; ~.: D .. : : : .

.

will give the desired form (7.10). If the system can be transformed to (7.10) by an equivalence transformation, from Theorem 7.1 QX = TQz. Using Theorem 6.10 on (7.10), QZ = K-l which has rank n everywhere and T by definition has rank n everywhere, so QX must have rank n everywhere. Now we consider the transformation to the second canonical form of Section 2.3, which is also known as phase-variable canonical form. 0 0

dz dt

=

... II.

0 al(t)

ill

1

0

0

0

1

0

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

0 a2(t)

ill.,. .. 1'''''

0 as(t)

1'

.....

1 an(t)

0 0

z

+

u 0 1

(7.12)

154

CANONICAL FORMS OF THE STATE EQUATION

[CHAP. 7

The controllability matrix QZ of this system is

o o

0 0

o

(_1)n-1

(_1)n-2

Qn-l,n-l

and

o 1

-1

qll

where qii = (-l)i an for 1 === i Qik

qn-2,2

qn-1,2

qn-2,1

Qn-l, 1

=== n;

(-l)ian-i+ k

+

and for 1 === k < i

=== n,

(_1)k i-f1 Un-jQi-k,j+l j=O

Theorem 7.7:

+

±

(-1)H1

dqi-~;-i+1

j=1

The system (7.9) can be transformed to (7.12) by an equivalence transformation if and only if QX(t), the controllability matrix of (7.9), is differentiable and has rank n everywhere.

Proof: The transformation matrix T = QXQZ-l, and QZ has rank n everywhere. Therefore the proof proceeds similarly to that of Theorem 7.6, transforming to (7.11) and then setting z = QZw. To determine the ai(t) from the ai(t) obtained from (7.11), note -AzQz + dQz/dt = _QzAw when written out in its columns gives (q~ Iq~

1... I q~ I q~ + 1) =

(q~ I q3

I ... I q! I -QZAwen)

Therefore q!+l = -QzAwen , which gives a set of relations that can be solved recursively for the ai(t) in terms of the lXi(t} of (7.11). Example 7.3. Suppose we have a second-order system.

Then

and

By equating the two expressions it is possible to find, recursively, a2 =

-a1

and

at:= -0:2 -

dal/dt.

It appears that the conditions of Theorems 7.6 and 7.7 can be relaxed if the hz(t) of equation (7.10) or (7.12) is left a general vector function of time instead of en. No results are available at present for this case. Note that if we are given (7.9), defining y = Zl in (7.12) permits us to find a corresponding scalar nth order equation dnyldtn = alY + a2 dy/dt + ... + andn-lyldtn-1 + u. For the case where u is a vector instead of a scalar in (7.9) a possible approach is to set all elements in u except one equal to zero, and if the resulting QX has rank n everywhere then the methods developed previously are applicable. If this is not possible or desirable, the form dw (ft

==

(o~~ ~~...........~. ..

0

...

hi

may be obtained, where the fi are in general nonzero n-vector functions of time, and

(7.13)

CHAP. 7]

CANONICAL FORMS OF THE STATE EQUATION

155

is of the form (7.11); and for i =F j,

A~ (~ ~ =

:.~~.')

.. .. ::: ...

o

0

a~

...

This form is obtained by calculatingm QX matrices for each of the m columns of the B matrix, Le. treating the systems as m single-input systems. Then from any l of these single-input QX matrices, choose the first nl columns of the first QX matrix, the first ?t2 columns of the second QX matrix, ... , and the first nz columns of the lth QX matrix such that these columns are independent and differentiable and nl + n2 + ... + nz = n. In this order these columns form the T matrix. The proof is similar to that of Theorem 7.6. To transform from this form to a form analogous to the first canonical form (7.10), use a constant matrix similar to the K matrix used in the proof of Theorem 7.6. However, as yet a general theory of reduction of time-varying linear systems has not been developed.

Solved Problems 7.1.

Transform 1 3 -1

x(m+1)

to Jordan form. From Example 4.8, page 74, and Problem 4.41, page 97,

J

=

G

1 2

To

o

~

~

(

-1

~),

o 1

o

Then

from which

a2

= 2,

C~ ~ -D(-O al:::::

-1,

PI:::::

-3 are obtained.

(~ -1

=

0'

The proper transformation is then

~ ~)(-~0 -~0 -3~)

0 -1

I)

-1

::

(=~1 -2~ -~) 8

Substitution of x::::; Tz in the system equation then gives the canonical form 1

z(m+ 1)

::::

2

o

156 7.2.

CANONICAL FORMS OF THE STATE EQUATION

[CHAP. 7

Transform

to real Jordan form. From Problem 4.8, page 93, and Problem 4.41, page 97,

x

1

=

(

-~.5

~

-j)(a ~ ~

0

o 13

o

~ )z

{3*

Substitution into the system equation gives

0) + (-2/a)

o o Then a = -2 and 13 = 1 - j.

7.3.

o

1-j

dz/dt

1

z

+j

(1 - j)/f3

(1

u

+ j)/f3*

Putting this in real form. give$

Using the methods of Section 7.4, reduce

~o ~)x+( ~)u,

dx/dt

-1

(i)

to a controllable system, and observable system.

(i)

Form the controllability matrix

y

-2

(ii) to an observable system,

Q

(1 1 2)x

=

(iii) to a controllable

(~-~ ~) -2

0-2

Observe that Q = (b I-Ab I A 2b) in accord with using the form of Theorem 6.10 in the case, and not Q = (b 1Ab 1A 2b) which is the form of Theorem 6.8. Performing elementary row operations to make the bottom row of QZ = 0 gives time~invariant

Then the required transformation is x = Tz where

T-l

=

(-~ ~

D

and

CHAP.7}

CANONICAL FORMS OF THE STATE EQUATION Using this change of variables, dzldt = T-1(AT - dTldt)z

(ii)

+ T- 1bu.

(-~ -~ Dz + G)u

=

dz/dt

157

y = (-1 -1 2)z

Forming the observability matrix gives

= (

px

-~

:

DG -~ D (-1 -3 0) (-1) 0=

(

-~

;

where the factorization is made using elementary column transformations. formation x = TOz,

=

dzldt

0 2 -2 -2

0 1

+

z

0 -2

(1 2 O)z

y

u

Using the trans-

(iii) Using the construction procedure,

(-~ -~ -~)

=

PXQZ

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

-1 -1 -1

= Then pZV1

=

(-~

1

-0

(

1

(-1 -1 -1)

2)('') (-0

2

1

4

5

=

1'21

0

0

0

0

0

RlSl

=

gl

gives

VI .

=

C~V31) 1'31

1'31

1'31

and

U1Qa:

=

(un U12

ud (

~ -2

-1

0 0

-D

=

8 1 gives U1

(-1 -1 -1)

(1

U13 -

1

U13)

Note UI V1 = 1 for any values of UtB and 1'31' The usual procedure would be to pick a value for each, but for instructional purposes we retain U13 and val arbitrary. Then pa: - RIU 1

~ ~ :~: ~ ~ :~:)

=

5-

ula

5-

u1a

(~ ~ :~:) 5-

(0 1 1)

u13

and

Choosing

1'31

= 0 and

Uta

== 1 gives

o

G::-0

T

1

Using UiVj = 8 ij we can determine

T

=

~ ~)-1

G~ -D G

o -1

CANONICAL FORMS OF THE STATE EQUATION

158

[CHAP; 7

The reduced system is then y = (1 1 O)z

where Zl is controllable and observable. Z2 is observable and uncontrollable, and Z3 is controllable and unobservable. Setting VSI = U 13 == 1 leads to the Jordan form, and setting vSl = U 13 == 0 leads to the system found in (H), which coincidentally happened to be in the correct controllability form in addition to the observable form.

7.4.

Reduce the following system to controllable-observable form.

(-j

dx/dt

3+t

1-:~+t2)

-t

-1 0 0

0

t2 - 2t - 1 t

2 0

x

+

(~) 0

u

(0 t2 0 -l)x

y

0

Using the construction procedure following Theorem 7.5 1

px

=

(~ (~

-5 -16-2t 2 0 0

2 0 0

t2 -t2

-6t[40) 0 -1- t ) -St2 0 S -St + 2t2 - 12t - 1 0 0 -3t4 + 2tS - 24t2 + 15t - 18

+ 2t

t2 - 4t+ 2 -t2+ 6t-6

-t2t2+ 2t ) (

t2 - 4t

-t2

+2

+ 6t -

81

(2 2 2 2)

6

(0 1 0 0)

Factoring px - RIU 1 and QX - VI 8 1 gives U2 = (0 0 0 1)

8 3 = (t -5 -16 - 2t -6t - 40)

=

V3 =

(D

Vr =

-1

Then vi (0 0 0 1) and (0 0 1 0) and T:::::: (VI I V2 1Vsl that the equivalence transformation

T=O~~D

will also put the system in the form (7.7).

+-::: =~2t+

-3t3 ( -3t4 + 2t3

v4).

-

24t2

)

- 1 15t - 18

It is interesting to note

CHAP. 7]

7.5.

CANONICAL FORMS OF THE STATE EQUATION

159

Reduce the following system to controllable form:

(

=

dx/dt

-t)

~

sin t

x

--t- -cost

+ ( cost t ) u

First we calculate QX and use elementary row operations to obtain E(t)QX

t ( -cos t

::::

cos t

t)(

t cos t) cos2 t

t

cos t

=

The required transformation is

=

T

7.6.

Put the time-invariant system

x(m+l)

1

E-l(t)

=

t2

(j

+ cos2 t

(t

-ctos

cos t

t)

~ j)x(m) + G)u(m)

(i) into first canonical form (7.10) and (ii) into phase-variable canonical form (7.12). Note this is of the same form as the system of Problem 7.3, except that this is a discrete-time system. Since the controllability matrbc: has the same form in the time-invariant case, the procedures developed there can be used directly. (See Problem 7.19 for time-varying discrete-time systems.) From part (i) of Problem 7.3, the system can be put in the form 1

z(m+ 1)

(

0

-2 -1

o

0

Therefore the best we can do is put the controllable sUbsystem

into the desired form.

The required transformation is

T

to the first canonical form

To obtain the phase-variable canonical form, T

160

CANONICAL FORMS OF THE STATE EQUATION

[CHAP. 7

1 0)(-1)2 (1o -1)-1( 2 -2 -1 Then

( 1 10)

T

from which we obtain the phase-variable canonical form

-2

Zp(m+ 1)

By chance, this also happens to be in the form of the first canonical form.

7.7.

Put the

time~varying

system

dx/dt

=

Gs~lt)X + (-Du

for t

~

0

(i) into first canonical form, (ii) into second canonical form, and second-order equation with the given state space representation.

(iii) find a scalar

To obtain the first canonical form,

T

-2t -3+ t)(-1 sin

==

~)

0

from which

where 6

+

1 2t -

(

sin t

6 -

8 - 4t - 2t2 + (t -1) sin t - cos t ) 6t - 2t2 - (t + 6) sin t - 3 cos t + sin2 t

To obtain the phase-variable form,

-2t

==

T

+ sin -3

t)

(0

1

-1

-a1(t)

)-1

2t - 2a1 - sin t (

al

+3

from which

The second-order scalar equation corresponding to this is

where y

7.8.

==

Zpl.

Transform to canonical form the multiple-input system

dx/dt

1 2-1) (0 1) (1-4 3 1 1 010

x+

01

u

To obtain the form (7.13), we calculate the two QX matrices resulting from each of the columns of the Bx matrix separately:

CHAP.7J

CANONICAL FORMS OF THE STATE EQUATION

161

QX (u1 only)

Note both of these matrices are singular. However, choosing T as the first two columns of (U1 only) and the first column of Q% (U2 only) gives

QX

so that

(-~

dw/dt

4 4

o

Also, T could have been chosen as the first column of Q% (u1 only) and the first two columns of Q:t (u2 only).

To transform to a form analogous to the first canonical form, let w ==Kz where

K

0 -1

G D

;:::

0

Then dz/dt

(-~

=

Dz + G ~)u

0

4 -4

Supplementary Problems 7;9.

Transform

dx/dt

G-~ -D CD

7.10.

Transform

dx/dt

(=~ ~) x + (_:) u

7.11.

U sing the methods of Section 7.4, reduce

dxldt

x

=

+

u

to Jordan form.

to real- Jordan form.

G_~ -Dx G)u +

y = (1 -1 1)x

(i) to a controllable system, (ii) to an observable system, (iii) to an observable and controllable

system. 7.12.

Prove Theorem 7.2, page 149.

7.13.

Prove Theorem 7.4, page 150.

7.14.

Show that the factorization requirement on Theorem 7.3 can be dropped if T(t) can be nonsingular and differentiable for times t everywhere dense in [to, tiJ.

162 7.15.

7.16.

CANONICAL FORMS OF THE STATE EQUATION

Consider the system

!£

(Xl) =

dt X2 elsewhere, and f 2 (t) = 0 for 0 form of equation (7.2)1

===

[CHAP. 7

(fl(t») u

where fl(t) = 1 - cos t for 0 === t ::::: 2 and zero h(t) t === 2 and 1- cos t elsewhere. Can this system be put in the

Find the observable states of the system

(~

dx/dt

6 2 8 4

2 8 4

!)

6

2

6

(1 1 1 l)x

y

x

7.17.

Check that the transformation of Problem 7.5, page 159, puts the system in the form of equation (7.2), page 149, by calculating AZ and B2.

7.18.

Develop Theorem 7.3 for time-varying discrete-time systems.

7.19.

Develop the transformation to a form similar to equation (7.11) for time-varying discrete-time systems.

7.20.

Reduce the system

-t+2) t+2 o -t+

o

dx/dt

1

x

+

(1) 1

1

u

0

to a system of the form of equation (7.2). 7.21.

Given the time-invariant system dx/dt = Ax + enu where the system is in phase-variable canonical form as given by equation (7.12). Let z = Tx where z is in the Jordan canonical form dz/dt = .A.z + bu and.A. is a diagonal matrix. Show that T is the Vandermonde matrix of eigenvalues.

7.22.

Verify the relationship for the system.

7.23.

Solve for the at in terms of the Example 7.3, page 154.

7.24.

Transform the system

qik

dx dt

in terms of the

Cl'i

for i

= 1,2,:3 -11

=

a"

following equation (7.12) for a third-order

(third-order system) in a manner analogous to

4

-1/2 ( -27/2

1 6

to phase-variable canonical form.

7.25.

Transform the system

dx/dt

-~ _~ _~) x + (e~t) u

( -e t

e- t -2

0

to phase-variable canonical form. 7.26.

Using the results of Section 7.5, find the transformation x = Tw that puts the system -1

dx/dt

-2 3

-3 into the form

1.27.

dw/dt

-1 1

Obtain explicit formulas to go to phase-variable canonical form directly in the case of time-invariant systems.

CANONICAL FORMS OF THE STATE EQUATION

CHAP.7J 7.28.

Use the duality principle to :find a transformation that puts the system dx/dt into the form

dz

Prove that

y

:::::

dt

7.29.

163

IjTl1 <

00

(0 0 '"

== A(t)x

and y::::: C(t)x

0 l)z

for the transformation to phase-variable canonical form.

Answers to Supplementary Problems

C~ 2D -1

0 0

where f3 is any number # O.

7.9.

T

7.10.

T

7.11.

There is one controllable and observable state, one controllable and unobservable state, and one uncontrollable and observable state.

7.15.

No.

7.16.

The transformation

:::::

(-: -8) -12

QZ::::: V1(1 0) but VI does not have rank one everywhere.

T

( 4~~

:::::

r41

r42

r ..)

8000

puts the system into the form. of equation (7.6), for any form can be used hut is more difficult algebraically.

7.24.

r4i

that make T nonsingular.

Also, Jordan

-3/2 1 1) (

T-l

5/2 1-2 -1 -1 1

A

7.25.

=

G-~ D

7.26.

T

7.28.

This form is obtained using the same transformation that puts the system dw/dt::::: At(t)w + ct(t)u into phase-variable canonical form.

7.29.

The elements of Qz-l are a linear combination of the elements of QZ, which are always finite as determined by the recursion relation.

Chapter 8 Relations with Classical Techniques 8.1 INTRODUCTION

Classical techniques such as block diagrams, root locus, error constants, etc., have been used for many years in the analysis and design of time-invariant single input-single output systems. Since this type of system is a subclass of the systems that can be analyzed by state space methods, we should expect that these classical techniques can be formulated within the framework already developed in this book. This formulation is the purpose of the chapter. 8.2 MATRIX FLOW DIAGRAMS

We have already studied flow diagrams in Chapter 2 as a graphical aid to obtaining the state equations. The flow diagrams studied in Chapter 2 used only four basic objects (summer, scalor, integrator, delayor) whose inputs and outputs were scalar functions of time. Here we consider vector inputs and outputs to these basic objects. In this chapter these basic objects will have the same symbols, and Definitions 2.1-2.4, pages 16-17, hold with the following exceptions. A summer has n m-vector inputs Ul(t), U2(t), ... , un(t) and one output m-vector y(t) = :±:Ul(t) ± U2(t) ± ... ± un(t). A scalor has one m':'vector input u(t) and one output k-vector y(t) = A(t)u(t), where·A(t) is a k x m matrix. An integrator has one m-vector input u(t) and one output m-vector y(t)

= y(to) +

it

u(,.) d,.. _ To denote

to

vector (instead of purely scalar) time function flow from one basic object to another, thick arrows will be used. Example 8.1. Consider the two input - one output system y

-

(3 2)x

This can be diagrammed as in Fig. 8-1.

x(t)

u(t)====>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

RELATIONS WITH CLASSICAL TECHNIQUES

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]

RELATIONS WITH CLASSICAL TECHNIQUES

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

RELATIONS WITH CLASSICAL TECHNIQUES

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]

RELATIONS WITH CLASSICAL TECHNIQUES

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 <

(-6S-8-1++ 1 _~-1) + (03 0 2

9s-

-1)

12

00

8-0

s-IR(s)

+

P(s)

but now

lim sR-l(s) s-o

lim

=

s-o

8(

0 -8

1

+\28)

(~ ~)

;;:::;

9s - 6

1

+ 128

Since the closed loop system has poles at -1, -1, -0.5, and -0.5, the zero output is asymptotically stable, Therefore the system is of type 1.

To find the error constant matrix of a type-l system, we use block diagram manipulations on Fig. 8-4 to get ~{e} = [1+H(s)]-l.,e{d}. If it exists, then lim e(t) = t--+oo

lim s[1 + S-l R(s)

+ P(S)]-l~{ d}

s--+O

lim SI+l[SII + R(s)

+ sZP(s)]-I.,e{d}

=

lim SI+IR-l(S)~{d}

5--+0

8--+0

for any l> 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

RELATIONS WITH CLASSICAL TECHNIQUES

[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

RELATIONS WITH CLASSICAL TECHNIQUES

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]

RELATIONS WITH CLASSICAL TECHNIQUES

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

RELATIONS WITH CLASSICAL TECHNIQUES

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

RELATIONS WITH CLASSICAL TECHNIQUES

[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]

RELATIONS WITH CLASSICAL TECHNIQUES

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

RELATIONS WITH CLASSICAL TECHNIQUES

[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

RELATIONS WITH CLASSICAL TECHNIQUES

[CHAP. 8

Solved Problems 8.1.

~~

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]

RELATIONS WITH CLASSICAL TECHNIQUES

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.

RELATIONS WITH CLASSICAL TECHNIQUES

[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]

RELATIONS WITH CLASSICAL TECHNIQUES

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 <" 1 we increase K to make ay/y become smaller.

e!

8.7.

A simplified, normalized representation of the control system of a solid-core nuclear rocket engine is shown in Fig. 8-22 where Bn, ST, 8P, 8Pcr and BV are the changes from nominal in neutron density, core temperature, core hydrogen preSSUFe, control rod setting, and turbine power control valve setting, respectively. Also, G1(s), G2 (s) and G3 (s) are scalar transfer functions and in the compensation K1, K2 , KS and K4 are scalar constants so that the control is proportional plus integral. Find a simple means of improving response. liP

+

+

8V

Fig. 8·22

186

RELATIONS WITH CLASSICAL TECHNIQUES

[CHAP. 8

The main point is to realize this "multiple loop" system is really a multiple input-multiple output system. The control system shown has constrained SPcr to be an inner loop. A means of improving response is to rewrite the system in matrix form as shown in Fig. 8-23. This opens up the possibility of "cross coupling" the feedback, such as having BPcr depend on 8P as well as ST. Furthermore it is evident that the system is of type 1 and can follow a step input with zero steady state error.

>

Fig. 8-23

8.8.

Compensate the system (S+Pl)-I(S+P Z)-1 to have poles at -"'1 and -"'2' with an observer pole at -7ro, by using the algebraic separation theorem, and discuss the effect of noise 7J at the output. The state space representation of the plant is

(-:tP. _p,l_ p.)(:~) + G}' + G)d

! (::)

y = (1 O)x

The feedback compensation can be found immediately as

=

u

(PIPZ - ?Tl'ITz

P1

+ pz -?T1 -

To construct the observer system, let P = (al (lz). (F!-PL)(

~)

(-?To j-PL)

?Tz) ~

Then

(~1 ~z)

from which Also PB

= PJ = £lZ'

Therefore the estimator dynamics are

!(:)

=

-?To ( : )

=

(~r(:)

+

[?TO(P1 + pz - ?To) - P1PZ]Y

+

u

+

To construct the estimate,

(~:)

=

(

~1

£lZ

Vo - : , -

The flow diagram of the closed loop system is shown in Fig. 8-24.

+ A

Xt

=Y Fig. 8-24

p.)(:)

d

CHAP. 8]

187

RELATIONS WITH CLASSICAL TECHNIQUES

Note the noise 'tJ is fed through a first-order system and gain elements to form ?.t. If the noise level is high, it is better to use a second-order observer or a Kalman filter because then the noise goes through no gain elements directly to the control, but instead is processed through first- and second-order systems. If there is no noise whatsoever, flow diagram manipulations can be used to show the closed loop system is equivalent to one compensated by a lead network, i.e. the above flow diagram with 'I] = 0 can be rearranged as in Fig. 8-25.

d

y

1

+

Fig. 8-25

Note the observer dynamics have cancelled out, and the closed loop system remains secondorder. This corresponds to the conclusion the observer dynamics are uncontrollable by d. However any initial condition in the observer will produce an effect on y, as can be seen from the first flow diagram.

Supplementary Problems 8.9.

Given the matrix block diagram of Fig. 8-26. indicated inverse exists.

Show that this reduces to Fig. 8-27 when the

y

d

+

=d====::;:j1 Fig. 8-26 8.10.

Given the matrix block diagram of Fig. 8-28. single feedback loop.

y

G(I+HG)-l

;>

Fig. 8~27

Reduce the block diagram to obtain HI isolated in a

Fig. 8-28

188 8.11.

RELATIONS WITH CLASSICAL TECHNIQUES

[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.

Given the system dx dt

==

(1 1-2) o

1

-1

1

x

(~

Y

0-1

Construct a first-order observer system with a pole at -1. 8.15.

Given the system dx dt

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.

Given the system dx dt

=

_!(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

RELATIONS WITH CLASSICAL TECHNIQUES

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

+

-

-<

K

...,..

H(s)

">

Fig. 8-31 8.26.

Design a system that will follow d(t) = sin t with zero steady state error.

Answers to Supplementary Problems 8.10.

This cannot be done unless the indicated inverse exists (Fig. 8-32). The matrices must be in the order given. y

d

+

Fig. 8-32

190 8.11.

RELATIONS WITH CLASSICAL TECHNIQUES

The closed loop system output Theorem 8.1 applies. Then d(t)

=

+

w,(t O)

C:)']

~ [1 -

(~)2 [(2 f - 1) Xl(t O) + (t -

=

y

t

+

w2(t O)

to)

[CHAP. 8

X 2 (t O) ]

tends to zero, so

0

Jot L[1 - (f)'] y(.)

d,-

'""

In t

for t > some t,

so that the system can follow steps but not ramps with zero error. 8.12.

For the state equation dx dt

corresponding to the given input-output relation, 'U,

=

a - fJ)x

(fJ - ¢

which shows the basic idea behind "pole placement," d2y/dt 2 +

Q'

8.13.

d~ dt

G-~ -D~ G)

8.14.

dw/dt

-w

+

+ (0

dy/dt

[(0 1

+ fJy

= (a - fJ) dy/dt

+ (fJ -

¢}y

-l)~-YI

.A-

1)y,

X

8.15. 1

d(t)--+~

82 + 38

+2

1 - - -__- - _

yet)

Fig. 8-33 8.16.

This cannot be done because the system is unobservable.

8.17.

Subtract D(t)u from 11 entering the observer and this reduces to the given formulation .

8.18.

~{oy(t)}

8.20.

d(8x)/dt

=

.e{a:(t) XN(t)}

(s

+ 1){s + t) 1

.....

tt(t») Sx

+ t.+ t(t)

(

p{x (t)} -

where

t2

j{t)

=

Q(t)

N

_

-

.e{u(t)}

4(s

+ 1)(8 + t)

+ (XN1) ot XN2

(:;~

dXNl/dt - (1 + t + t(t)XNl - U(t)xN2 ) dXN2Idt - t(t)XN1 - t2XN2

Choose u(t) such that det Q(t) = 0 for all t. 8.21.

- itto

pet) = e

a(1))

d1) [

that lim e(t) = 0 and t-+-

8.22.

+

e(t) K

ft y(t)e I to

tCl (1)) d1)

T

fJ(T) fJ(T) dT

]

+ y(t)/C

where e(t) is any function such

is an arbitrary constant.

00

lim [F(s) - G(s)(I + G{s) H(8»-1]8~{d} = 0

s-+-o

8.24.

Use Theorem 8.2.

8.25.

U 5e Theorem 8.2.

8.26.

One answer is R(s) == 8(82 + 1)-1.

and is analytic for s =::: O.

Chapter 9 Stability of Linear Systems 9.1 INTRODUCTION Historically, the concept of stability has been of great importance to the system designer. The concept of stability seems simple enough for linear time-invariant systems. However, we shall find that its extension to nonlinear and/or time-varying systems is quite complicated. For the unforced linear time-invariant system dx/dt = Ax, we are assured that the solution x(t) = TeJCt-to)T-lxo does not blow up if all the eigenvalues of A are in the left half of the complex plane. Other than transformation to Jordan form, there are many direct tests for stability. The Routh-Hurwitz criterion can be applied to the characteristic polynomial of A as a yesor-no test for the existence of poles in the right half plane. More useful techniques are those of root locus, Nyquist, Bode, etc., which indicate the degree of stability in some sense. These are still the best techniques for the analysis of low-order time-invariant systems, and we have seen how to apply thenl to multiple input-multiple output systems in the previous chapter. Now we wish to extend the idea of stability to time-varying systems, and to do this we must first examine carefully what is meant by stability for this type of system. 9.2 DEFINITIONS OF STABILITY FOR ZERO-INPUT LINEAR SYSTEMS A type of "stability" results if we can say the response is bounded. Definition 9.1:

For every Xo and every to, if there exists a constant K depending on Xo and to such that Ilx(t)11 ~ K for all t ~ to, then the response x(t) is bounded.

Even this simple definition has difficulties. The troubie is that we must specify the response to what. The trajectory of x(t) =
For all negative values of xo, this is well behaved. For values of time tl = In Xo - In (xo -1), so that lim x(t) = 00,

Xo

> 1, the denominator vanishes at a

t-t 1

It can be concluded that boundedness depends upon the initial conditions for nonlinear equations in general.

Theorem 9.1:

The boundedness of the response x(t) of the linear system dx/dt = A(t)x is independent of the initial condition Xo.

191

STABILITY OF LINEAR SYSTEMS

192 Proof:

Since

Ilxoll

is a

Ilx(t)11 = 11q,(t; 0, Xo, to)11 = 114J(t, to)xoll :-:=: 114t(t, to)llllxoll constant, if Ilx(t)11 becomes unbounded as t ~ it is solely 00,

[CHAP. 9

due to

cp(t, to).

Now we shall consider other, different types of stability. First, note that x = 0 is a steady state solution (an equilibrium state) to the zero-input linear system dx/dt = A(t)x. We shall define a region of state space by Ilxll < 10, and see if there exists a small region of nonzero perturbations surrounding the equilibrium state x = 0 that give rise to a trajectory which remains within Ilx!1 < E. If this is true for all E> 0, no matter how small, then we have Definition 9.2:

The equilibrium state x = 0 of dx/dt = A(t)x is stable in the sense of Liapunov (for short, stable Ls.L.) if for any to and every real number E> 0, there is some 0 > 0, as small as we please, depending on to and E such that if Ilxoll < 0 then Ilx(t)11 < E for all t > to.

This definition is also valid for nonlinear s!ystems with an equilibrium state x = O. It is the most common definition of stability, and in the literature "stable Ls.L." is often shortened to "stable". States that are not stable Ls.L. will be called unstable. Note stability Ls.L. is a local condition, in that 0 can be as small as we please. Finally, since x = 0 is an obvious choice of equilibrium state for a linear system, when speaking about linear systems we shall not be precise but instead will say the system is stable when we mean the zero state is stable. Example 9.2. Consider the nonlinear system of Example 9.1. Ix(t)1

If

::::

Xo ===

1, then

11 + (e t -1)(1- xo)1

In Definition 9.2 we can set 0 = € > 0 if e:::: 1, and if € > 1 we set 0:= 1 to show the zero state of Example 9.1 is stable i.s.L. Hence the zero state is stable Ls.L. even though the response can become unbounded for some Xo' (This situation corresponds to Fig. 9-2(b) of Problem 9.1.) Of course if the response became unbounded for all Xo # 0, the zero state would be considered unstable. Another point to note is that in the application of Definition 9.2, in the range where e is small there results the choice of a correspondingly small 8. Example 9.3. Given the Van der Pol equation

with initial condition x(O}:::; xo' The trajectories in state space can be plotted as shown in Fig. 9-1. We will call the trajectory in bold the limit cycle. Trajectories originating outside the limit cycle spiral in towards it and trajectories originating inside the limit cycle spiral out towards it. Consider a small circle of any radius, centered at the origin but such that it lies completely within the limit cycle. Call its radius € and note only xo::::: 0 will result in a trajectory that stays within Jixii2 < E. Therefore the zero state of the Van der Pol equation is unstable but any trajectory is bounded. (This situation corresponds to Fig. 9-2(e) of Problem 9.1.)

Theorem 9.2:

Note

Ilx(t)11

The transition matrix of a linear system is bounded as jl4J(t, to)jj < K(tO) for all t === to if and only if the equilibrium state x = 0 of dx/dt = A(t)x is stable Ls.L.

is bounded if

Ilcp(t, to)11

is bounded.

CHAP. 9]

STABILITY OF LINEAR SYSTEMS

193

Proof: First assun1e 1)I'Jl(t, to))1 < K(tO) where K is a constant depending only on to. If we are given any € > 0, then we can always find 8 = dK(tO) such that if Ilxoll < 8 then E = K(to)8 > 111'Jl(t, to)llllxoll == 114I(t, to)xoll = Ilx(t)jj. From Definition 9.2 we conclude stability i.s.L. Next we assume stability Ls.L. Let us suppose cp(t, to) is not bounded, so that there is at least one element ipij(t, to) that becomes large as t tends to co. If Ilxoll < 8 for a nonzero 8, then the element Xj of Xo can be nonzero, which results in a trajectory that eventually leaves any region in state space defined by Ilxll < E. This results in an unstable system, so that we have reached a contradiction and conclude that cp(t, to) must be bounded. Taken together, Theorems 9.1 and 9.2 show that boundedness of Ilx(t)1I is equivalent to stability i.s.L. for linear systems, and is independent of Xo. When any form of stability is independent of the size of the initial perturbation Xo, we say the stability is global, or speak of stability in the large. Therefore another way of stating Theorem 9.1 is to say (local) stability i.s.L. implies global stability for linear systems. The nonlinear system of Example 9.1 is stable i.s.L. but not globally stable i.s.L. In practical applications we often desire the response to return eventually to the equilibrium position x = 0 after a small displacement. This is a stronger requirement than stability Ls.L. which only demands that the response stay within a region jlxll < E.

Definition 9.3:

The equilibrium state x = 0 is asymptotically stable if (1) it is stable Ls.L. and (2) for any "to and any Xo sufficiently close to 0, x(t) -7 0 as t -7 co.

This definition is also valid for nonlinear systems. It turns out that (1) must be assumed besides (2), because there exist pathological systems where x(t) -7 0 but are not stable Ls.L. Example 9.4. Consider the linear harmonic oscillator

~~

4I(t,O)

= -

=

(_~ ~) x

t

(cos -sin t

with transition matrix

t)

sin cos t

The A matrix has eigenvalues at ±j. To apply Definition 9.2, Ilx(t)112::= 11~(t, t o)I[21[xoI12 = I[xolb < E = 0 since 11~(tJ t o)112 ::::: 1. Therefore the harmonic oscillator is stable Ls.L. However, x(t) never damps out to 0, so the harmonic oscillator is not asymptotically stable. Example 9.5. The equilibrium state :Ii = 0 is asymptotically stable for the system of Example 9.1, since any small perturbation (xo < 1) gives rise to a trajectory that eventually returns to O.

In all cases except one, if the conditions of the type of stability are independent of to, then the adjective uniform is added to the descriptive phrase. Example 9.6. If S does not depend on to in Definition 9.2, we have uniform stability i.s.L. Example 9.7. The stability of any time-invariant system is uniform.

The exception to the usual rule of adding "uniform" to the descriptive phrase results because here we only consider linear systems. In the general framework of stability definitions for time-varying nonlinear systems, there is no inconsistency. To avoid the complexities of general nonlinear systems, here we give Definition 9.4:

If the linear system dx/dt = A(t)x is uniformly stable i.s.L. and if for all to and for any fixed p however large Ilxoll < p gives rise to a response x(t) -7 0 as t -7 co, then the system is unifo1'mly asymptotically stable.

The difference between Definitions 9.3 and 9.4 is that the conditions of 9.4 do not depend on to, and additionally must hold for all p. If p could be as small as we please, this would be analogous to Definition 9.8. This complication arises only because of the linearity, which in turn implies that Definition 9.4 is also global.

194

STABILITY OF LINEAR SYSTEMS

Theorem 9.3:

[CHAP. 9

The linear system dx/dt = A(t)x is uniformly asymptotically stable if and only if there exist two positive constants 1<1 and 1<2 such that ]]q,(t, to)11 ~ I
The proof is given in Problem 9.2. Example 9.8. Given the linear time-varying scalar system dx/dt = -x/to This has a transition matrix .p(t, to) = to/to For initial times to > 0 the system is asymptotically stable. However, the response does not tend to 0 as fast as an exponential. This is because for to < 0 the system is unstable, and the asymptotic stability is not uniform. Example 9.9. Any time-invariant linear system dx/dt = Ax is uniformly asymptotically stable if and only if all the eigenvalues of A have negative real parts.

However for the time-varying system dx/dt = A(t)x, if A(t) has all its eigenvalues with negative real parts for each fixed t, this in general does not mean the system is asymptotically stable or even stable. Example 9.10. Given the system

with initial conditions x(O) = Xo. The eigenvalues are Al = /C and A2 eigenvalues have real parts less than zero. However, the exact solution is 2X1(t)

=:

3(XlO

For any nonzero real

+ x20)e 5Kt /C

(XIO

+ 3x20)e7Kt

and

=:

31(.

Then if

2X2(t):::: (XlO + 3x20)e-Kt - (x lO

/C

< 0,

both

+ x20)e- SKt

the system is unstable.

There are many other types of stability to consider in the general case of nonlinear time-varying systems. For brevity, we shall not discuss any more than Definitions 9.1-9.4 for zero-input systems. Furthermore, these definitions of stability, and those of the next section, carry over in an obvious manner to discrete-time systems. The only difference is that t takes on discrete values. 9.3 DEFINITIONS OF STABILITY FOR NONZERO INPUTS In some cases we are more interested in the input-output relationships of a system than its zero-input response. Consider the system

dx/dt = A(t)x + B(t)u

with initial condition x(to) = xo and where ]]A(t)ll < KA , a constant, and other constant, for all t:::-: to, and A(t) and B(t) are continuous. Definition 9.5:

(9.1)

y = C(t)x

11B(t)11 < KB'

an-

The system (9.1) is externally stable if for any to, any xo, and any u such that ]]u(t)]] ~ 8 for all t:::-: to, there exists a constant E which depends only on to, Xo and 8 such that ]Iy(t)]]-:::. E for all t ~ to.

In other words, if every bounded input produces a bounded output we have external stability.

Theoretn 9.4:

The system (9.1), with Xo = 0 and single input and output, is uniformly externally stable if and only if there exists a number f3 < 00 such that

rt ]h(t, 7")1 d7" ~ f3

Jto for all t

~

to where h(t, 7") is the impulse response.

CHAP. 9]

STABILITY OF LINEAR SYSTEMS

195

If f3 depends on to the system is only externally stable. If Xo oF 0, we additionally require the zero-input system to be stable i.s.L. and C(t) bounded for t> to so that the output does not become unbounded. If the system has multiple inputs and/or multiple outputs, the

it IIH(t, ,)111

criterion turns out to be

IIvl1 1 = i=l L IVil y(t) =

f3 where the l1 norm of a vector v in

===

'V n is

(see Sections 3.10 and 4.6).

First we show that if

Proof:

dT

to

n

Jtrto h(t,.) u(.) dT, Jy(t)1

it

~

Ih(t, ')ldT

f3, then we get external stability. Since

we can tak:onorms on both sides and use norm properties to obtain

i

:=::;

t1

I

~

h (t, .)1 JU(T)) dT

8

===

i

t1

~

~

)h(t, .)1 dT

8/3

From Definition 9.5 we then have external stability. Next, if the system is externally stable we shall prove

Jr

t

Ih(t, .)ldT

====

f3 by contra-

to

diction. We set U1(T) = sgn h(t,.) for to:::::=. === t, where sgn is the signum function which has bound 1 for any t. By the hypothesis of external stability,

=

Iy(t))

rt )h(t, T)Jd• Jto

Now suppose

. : : :. f3 is not true.

Then by taking suitable values of t and to

we can always make this larger than any pI"eassigned number and

to

= (jo in such a way that i91h(8, .)1 dT 8

<

a. Again we set U2(T) = sgn h(O, T) so that

Since a is any preassigned number, we can set a =

h(B, T)U2(r)dT = y(B).

eo

Suppose we choose t = B

80

1 arrive at a contradiction. a

>

a.

€

and

Example 9.11.

Consider the system dy/dt = u. at time •.

Then

f

t

IU (t -

This has an impulse response h(t,.) = U(t - .), a unit step starting

.) I dT = t - to which becomes unbounded as t tends to

00.

Therefore this

to

system is not externally stable although the zero-input system is stable i.s.L.

In general, external stability has no relation whatsoever with zero-input stability concepts because external stability has to do with the: time behavior of the output and zeroinput stability is concel"ned with the time behavior of the state. Example 9.12.

Consider the scalar time-varying system

where

8(t, to)

==

ft 0'(.)

d..

Then the transition matrix is
Therefore

to

so that

ft

to

Ih(t,

.)1 d. <

ft

Ih(t,

.)1 d. =

1.

Thus the system is externally stable.

-00

can be almost anything, any form of zero-input stability is open to question.

However, since a(t)

196

STABILITY OF LINEAR SYSTEMS

[CHAP. 9

However, if we make C(t) a constant matrix, then we can identify the time behavior of the output with that of the state. In fact, with a few more restrictions on (9.1) we have Theorem 9.5:

For the system (9.1) with l1 norms, C(t) = I, and nonsingular B(t) such that lIB-1(t)111 < K n - 1, the system is externally stable if and only if it is uniformly asymptotically stable.

To have B(t) nonsingular requires u to be an n-vector. If B is constant and nonsingular, it satisfies the requirements stated in Theorem 9.5. Theorem 9.5 is proved in Problem 9.2. 9.4 LIAPUNOV TECHNIQUES The Routh-Hurwitz, Nyquist, root locus, etc., techniques are valid for linear, timeinvariant systems. The method of Liapunov has achieved some popularity for dealing with nonlinear and/or time-varying systems. Unfortunately, in most cases its practical utility is severely limited because response characteristics other than stability are desired. Consider some metric p(x(t), 0). This is the "distance" between the state vector and the 0 vector. If some metric, any metric at all, can be found such that the metric tends to zero as t ~ et:J, it can be concluded that the system is asymptotically stable. Actually, Liapunov realized we do not need a metric to show this, because the triangle inequality (Property 4 of Definition 3.45) can be dispensed with. Definition 9.6:

A time-invariant Liapunov junction, denoted v(x), is any scalar function of the state variable x that satisfies the following conditions for all t ~ to and all x in the neighborhood of the origin: (1)

v(x) and its partial derivatives exist and are continuous;

(2)

v(O) = 0;

(3)

v{x) > 0 for x ¥: 0;

(4)

dv/dt = (grad x v)T dx/dt < 0 for x

~

o.

The problem is to find a Liapunov function for a particular system, and there is no general method to do this. Example 9.13.

Given the scalar system dxldt = -x. We shall consider the particular function 1](x) = x 2 • Applying the tests in Definition 9.6, (1) 1](x) = x 2 and B7JIBx = 2x are continuous, (2) 1](0) = 0, (3) 7J(x) = x 2 > 0 for all x # 0, and (4) d7Jldt = 2x dxldt = -2X2 < 0 for all x # o. Therefore 7J(x) is a Liapunov function.

Theorem 9.6:

Suppose a time-invariant Liapunov function can be found for the state variable x of the system dx/dt = £(x, t) where £(0, t) = O. Then the state x = 0 is asymptotically stable.

Here we shall only prove x ~ 0 as t ~ co. Definition 9.6 assures existence and continuity of v and dv/dt. Now consider v(p{t; to, xo)), i.e. as a function of t. Since v > 0 and dv/dt < 0 for p ~ 0, integration with respect to t shows V{tp(tl; to, xo)> V(p(t2; to, xo)) for to < t1 < t2 • Although v is thus positive and monotone decreasing, its limit may not be zero. (Consider 1 + e- t .) Assume v has a constant limit K> O. Then dv/dt = 0 when v = K. But dv/dt < 0 for p ~ 0, and when p = 0 then dv/dt = (gradxv)Tf(O, t) = O. So dv/dt= 0 implies p = 0 which implies v = 0, a contradiction. Thus v ~ 0, assuring x ~ O. If Definition 9.6 holds for all to, then we have uniforn1 asymptotic stability. Additionally if the system is linear or if we substitute "everywhere" for "in a neighborhood of the origin" in Definition 9.6, we have uniform global asymptotic stability. If condition (4) is weakened to dv/dt ~ 0, we have only stability i.s.L.

Proof:

197

STABILITY OF LINEAR SYSTEMS

CHAP. 9]

Example 9.14. For the system of Example 9.13, since we have found a Liapunov function v(x) = x2 , we conclude the system dx/dt = -x is uniformly asymptotically stable. Notice that we did not need to solve the state equation to do this, which is the advantage of the technique of Liapunov.

Definition 9.7:

A time-varrying Liapunov junction, denoted v(x, t), is any scalar function of the state variable x and time t that satisfies for all t ~ to and all x in a neighborhood of the origin: (1)

v(x, t) and its first partial derivatives in x and t exist and are continuous;

(2)

v(O, t) = 0;

(3)

v(x, t) ~ a(l]xlD > 0 for x ¥ 0 and t ~ to, where a(O) = 0 and is a continuous nondecreasing scalar function of ~;

(4)

dv/dt = (grad x v)T dx/dt + av/at

< 0 for x

¥

a(~)

O.

Note for all t ~ to, v(x, t) must be == a continuous nondecreasing, time-invariant function of the norm Ilxll.

Theorem 9.7:

Suppose a time-varying Liapunov function can be found for the state variable x(t) of the system dx/dt = f(x, t). Then the state x = 0 is asymptotically stable.

Proof: Since dv/dt < 0 and v is positive, integration with respect to t shows that v(x(to), to) > v(x(t), t) for t > to. Now the proof must be altered from that of Theorem 9.6 because the time dependence of the Liapunov function could permit v to tend to zero even though x remains nonzero. (Consider v = x2e~t when x = t). Therefore we require v(x, t) == a(llxll), and a = 0 implies Ilxll = O. Hence if v tends to zero with this additional assumption, we have asymptotic stability for some to. If the conditions of Definition 9.7 hold for all to and if v(x, t) ...:::: P(I!xID where P(~) is a continuous nondecreasing scalar function of ~ with (3(0) = 0, then we have uniform asymptotic stability. Additionally, if the system is linear or if we substitute "everywhere" for "in a neighborhood of the origin" in Definition 9.7 and require a(jjxjD -? co with Ilxll-? 00, then we have uniform global asymptotic stability. Example 9.15. Given the scalar system dx/dt = x. The function x 2 e- 4t satisfies all the requirements of Definition 9.6 at any fixed t and yet the system is not stable. This is because there is no a(llxlD meeting the requirement of Definition 9.7.

It is often of use to weaken condition (4) of Definition 9.6 or 9.7 in the "following manner. We need v(x, t) to eventually decrease to zero. However, it is permissible for v(x, t) to be constant in a region of state space if we are assured the system trajectories will move to a region of state space in which dv/dt is strictly less than zero. In other words, instead of requiring dv/dt < 0 for all x =1= 0 and all t == to, we could require dv/dt === 0 for all x =I" 0 and dv/dt does not vanish identically in t == to for any to and any trajectory arising from a nonzero initial condition x(to). Example 9.16. Consider the Sturm-Liouville equation d 2 y/dt2 + pet) dy/dt + q(t)y = O. We shall impose conditions on the scalars pet) and q(t} such that uniform asymptotic stability is guaranteed. First, call y = Xl and dy/dt = X2' and consider the function vex, t) = + x~/q(t). Clearly conditions (1) and (2) of Definition 9.7 hold if q(t) is continuously differentiable, and for (3) to hold suppose l/q(t):=: iCl > 0 for all t so that a(llxll) = IlxJl~ min {1, iCl}' Here min {1, iCl} = 1 if iCl === 1, and min {l, iCt} = iCl if iCl < 1. For (4) to hold, we calculate dv dXI X2 dX 2 x~ dq dt 2Xl at + 2 q(t) Cit - q2(t) at

-xi

198

STABILITY OF LINEAR SYSTEMS

[CHAP. 9

Since dXl/dt = Xz and dX2/dt = -P(t)X2 - q(t)Xl' then dv dt

2p(t) q(t) -

+

dq/dt

q2(t)

2

x2

Ki.

Hence we require 2p(t) q(t) + dq/dt > 0 for all t ==- to, since q-2 > But even if this requirement is satisfied, dv/dt 0 for x 2 0 and any Xl' Condition (4) cannot hold. However, if we can show that when X2 = 0 and Xl ~ 0, the system moves to a region where X2 ~ 0, then vex, t) will still be a Liapunov function. Hence when X2 = 0, we have dX2/dt = -q(t)xl' Therefore if q(t) # 0, then X2 must become nonzero when Xl ~ O. Thus vex, t) is a Liapunov function whenever, for any t ==- to, q(t) #- 0 and lIq(t) ==- Kl > 0, and 2p(t) q(t) + dq/dt > 0. Under these conditions the zero state of the given SturmLiouville equation is asymptotically stable. Furthermore we can show it is uniformly globally asymptotically stable if these conditions are independent of to and if q(t) === K2' The linearity of the system implies global stability, and since the previous calculations did not depend on to we can show uniformity if we can find a P([lxiD ==- vex, t). If q(t) === /C2, then [lxll~ max {1, /CZl} = ,8([lx[[) ==- vex, t).

=

=

For the discrete-time case, we have the analog for Definition 9.7 as A discrete-time Liapunov function, denoted v(x, k), is any scalar function of the state variable x and integer k that satisfies, for all k> ko and all x in a neighborhood of the origin:

Definition 9.8:

(1)

v(x, k) is continuous;

(2)

v(O, k) = 0;

(3)

for x =1= 0; ~v(x, k) = v(x(k + 1), k + 1) - v(x(k), k)

(4)

v(x, k)

~

a(llxlD > 0

< 0 for x =1= O.

Suppose a discrete-time Liapunov function can be found for the state variable x(k) of the system x(k + 1) = f(x, k). Then the state x = 0 is asymptotically stable.

Theorem 9.8:

The proof is similar to that of Theorem 9.7. If the conditions of Definition 9.7 hold for all ko and if v(x, k) ~ {3(llxID where (3a) is a continuous nondecreasing function of ~ with (3(0) = 0, then we have uniform asymptotic stability. Additionally, if the system is linear or if we substitute "everywhere" for "in a neighborhood of the origin" in Definition 9.8 and require a(llxlD -? 00 with Ilxll-? 00, then we have uniform global asymptotic stability. Again, we can weaken condition (4) in Definition 9.8 to achieve the same results. Condition (4) can be replaced by ~v(x, k) ~ 0 if we have assurance that the system trajectories will move from regions of state space in which .6.v = 0 to regions in which L1v < O. Example 9.17. Given the linear system x(k + 1) = A(k) x(k), where IIA(k)11 as a discrete-time Liapunov function vex) = Ilxll. Th,en ill'

= v(x(k + 1»

- v(x(k»

<1

for all k and for some norm. Choose

= Ilx(k + 1)11 - Ilx(k)[1 :::: [[A(k) x(k)11 - Ilx(k)[1

Since IIA(k)11 < 1, then IIA(k} x(k)11 === IIA(k)llllx(k)11 < Ilx(k)11 so that Av < O. The other pertinent properties of v can be verified, so this system is uniformly globally asymptotically stable.

9.5 LIAPUNOV FUNCTIONS FOR LINEAR SYSTEMS The major difficulty with Liapunov techniques is finding a Liapunov function. Clearly, if the system is unstable no Liapunov function exists. There is indeed cause to worry that no Liapunov function exists even if the system is stable, so that the search for such a function would be in vain. However, for nonlinear systems it has been shown under quite general conditions that if the equilibrium state x = 0 is uniformly globally asymptotically stable, then a time-varying Liapunov function exists. The purpose of this section is to manufacture a Liapunov function. We shall consider only linear systems, and to be specific we shall be concerned with the asymptotically stable real linear system

dx/dt = A(t)x

(9.2)

199

STABILITY OF LINEAR SYSTEMS

CHAP. 9]

IIA(t)112:::=; K3(tO)

in :,vhich there exists a constant K3 depending on to such that

f

11cf>(r, t)11 2 dT

exists for some norm.

and where

Using Theorem 9.3 we can show that uniformly

11cf>(t, r)11 : :=; Kle-K2Ct~T)

asymptotically stable systems satisfy this last requirement because so that

However, there do exist asymptotically stable systems that do not satisfy this requirement. Example 9.18. Consider the time-varying scalar system dx/dt = -x/2t. This has a transition matrix (t, to) = (to/t)1/2. The system is asymptotically stable for to > O. However, even for t === to > 0, we find

f'" II
Note carefully the position of the arguments of <1>.

Theorem 9.9:

For system (9.2),

=

where pet)

i'"

vex, t)

= XTP(t)X is a time-varying Liapunov function,

q"T(T, t) Q(T) q"(,, t) dT in which Q(t) is any continuous

positive definite symmetric matrix such that IIQ(t)ll:::=; K5(tO) and Q(t) is positive definite for all t ~ to if £ > 0 is small enough.

a

This theorem is of little direct help in the determination of stability because q,,(t, ,) must be known in advance to compute vex, t). Note pet) is positive definite, real, and symmetric.

Proof:

Since q" and Q are continuous, if pet)

00

for all t

IIPll ~

to then condition (1) of

~

i""jj4J(r, t)W dr.

Since the in-

tegral exists for the systems (9.2) under discussion, then condition (1) holds. obviously holds. To show condition (3),

Condition (2)

Definition 9.7 is satisfied. Use of

vex, t)

=

i

IIQII : :=;

<

KS

gives

KS

CO

(q,,(T, t) X(t))TQ(,) eIl(r, t) x(t) dT

=

i

oo

XT(T) Q(T) X(T) dT

Since Q(t) - EI is positive definite, then xT(Q - cI)x ~ 0 so that XTQX ~ €xTx x( T) # O. Therefore

> 0 for any

Since IIA(t)112:::=; Ka for the system (9.2), then use of Problem 4.43 gives

vex, t) ~ =

€K3 1

-€K- 1

a

ieJ> IIA(T)1121Ix(T)II~ dT

~

Jtr""xT(dx/dT) d, = €K3lllx(t)II~/2

for all x # 0 . Finally, to satisfy condition (4), since dv/dt = -xT(t) Q(t) x(t) < 0 since Q is positive definite.

f.co xTAxdr

-€K;l

a(llxll) > 0

=

vex, t)

Example 9.19. Consider the scalar system of Example 9.8, dx/dt = -x/to Then 0, when it is asymptotically stable, because

IIA(t)112

=t- 1

<

tol

= K3(tO)

and

foo

114>(r, t) 112 dr =

t

For simplicity, we choose Q(r) = 1 so that pet) = time-varying Liapunov function for all t === to > O.

fOO (t/T)2d'T= t

t.

f

00

A Liapunov function

(t/r)2 dr = t

t

Then

vex, t)

= tx 2 ,

which is a

STABILITY OF LINEAR SYSTEMS

200 Consider the system

dx/dt

=

A(t)x

[CHAP. 9

+ f(x, u(t), t)

(9.3)

where for t ~ to, f(x, u, t) is real, continuous for small Ilxjl and Ilull, and 11£11-'1> 0 as ljxll--')o 0 and as Ilull--')o O. As discussed in Section 8.9 this equation often results when considering the linearization of a nonlinear system or considering parameter variations. (Let ax from Section 8.9 equal x in equation (9.3).) In fact, most nonswitching control systems satisfy this type of equation, and Ilfll is usually small because it is the job of the controller to keep deviations from the nominal small. Theorem 9.10: If the system (9.3) reduces to the asymptotically stable system (9.2) when f = 0, then the equilibrium state is asymptotically stable for some small Ilx[[ and j[u[l·

In other words, if the linearized system is asymptotically stable for t:::".. to, then the corresponding equilibrium state of the nonlinear system is asymptotically stable to small disturbances. Use the Liapunov futiction v(~~-t)

= XTP(t)X

given by Theorem 9.9, in which Then conditions (1) and (2) of Definition 9.7 hold for this as a Liapunov function for system (9.3) in the same manner as in the proof of Theorem 9.9. For small Ilx\j it makes little difference whether we consider equation (9.3) or (9.2), and so the lower bound a(llxlD in condition (3) can be fixed up to hold for trajectories of (9.3). For condition (4) we investigate dv/dt = d(xTpx)/dt = 2xT P dx/dt + x T (dP/dt)x = 2x TPAx + 2x TPf + x T(dP/dt)x. But from the proof of Theorem 9.9 we know dv/dt along motions of the system dx/dt = A(t)x satisfies dv/dt = -xTQx = 2xTPAx + x T(dP/dt)x so that -Q = 2PA + dP/dt. Hence for the system (9.3), dv/dt = -xTQx + 2x T Pf. But Ilfll-? 0 as Ilxll--')o 0 and [[ull--')o 0, so that for small enough Ilxll and Ilull, the term -xTQx dominates and dv/dt is negative. Proof;

q,(t, T) is the transition matrix of the asymptotically stable system (9.2).

Example 9.20. Consider the system of Example 9.1. The linearized system dx/dt = -x is uniformly asymptotically stable, so Theorem 9.10 says there exist some small motions in the neighborhood of x::::: 0 which result in the asymptotic stability of the nonlinear system. In fact, we know from the solution to the nonlinear equation that initial conditions Xo < 1 cause trajectories that return to x::::: O.

Example 9.21. Consider the system dx/dt = -x + u(l + x - u), whereu(t) = e, a small constant. Then I(x, e, t) ::::: 1:(1 + X - e) does not tend to zero as x ~ O. Consequently Theorem 9.10 cannot be used directly. However, note the steady state value of x = 1:. Therefore redefine variables as z = X-e. Then dz/dt = -z + eZ, which is stable for all I: ~ 1. Example 9.22. Consider the system

dXl/dt =

X2

+ xl(xi + ~)/2

and

The linearized system dx/dt = X2 and dX2/dt = -Xl is stable i.s.L. because it is the harmonic oscillator of Example 9.4. But the solution to the nonlinear equation is xi + xi ::::: [(xio + X~o)-l- t]-l which is unbounded for any xlO or X20' The trouble is that the linearized system is only stable i.s.L., not asymptotically stable.

9.6 EQUATIONS FOR THE CONSTRUCTION OF LIAPUNOV FUNCTIONS From Theorem 9.9, if we take the derivative of P(t) we obtain dP/dt

_q,T(t, t) Q(t) q,(t, t)

+

i«l

[8f1l(T, t)/8t]T Q(T)cfJ(,-, t) dT

+ s,~ .pT(T, t)Q(T)[a.p(T, t)/atJdT Since

cfJ'(t, t)

== I and

8lJJ(T, t)/8t == 8cfJ- 1(t, T)/8t = -q,(T, t) A(t), dP/dt

+ AT(t) P(t) + P(t) A(t) =

we obtain

-Q(t)

(9.4)

STABILITY OF LINEAR SYSTEMS

CHAP.9J

201

Theorem 9.11: If and only if the solution P(t) to equation (9.4) is positive definite, where A(t) and Q(t) satisfy the conditions of Theorem 9.9, then the system (9.2), dx/dt = A(t)x, is uniformly asymptotically stable. Proof: If P(t) is positive definite, then v = XTP(t)X is a Liapunov function. system (9.2) is uniforn1Iy asymptotically stable, we obtain equation (9.4) for P.

If the

Theorem 9.11 represents a rare occurrence in Liapunov theory, namely a necessary and sufficient condition for asymptotic stability. Usually it is a difficult procedure to find a Liapunov function for a nonlinear system, but equation (9.4) gives a means to generate a Liapunov function for a linear equation. Unfortunately, finding a solution to equation (9.4) means solving n(n + 1)/2 equations (since P is symmetric), whereas it is usually easier to solve then equations associated with dx/dt = A(t)x. However, in the constant coefficient case we can take Q to be a constant positive definite matrix, so that P is the solution to (9.5)

This equation gives a set of n(n + 1)/2 linear algebraic equations that can be solved for P after any positive definite Q has been selected. It turns out that the solution always exists and is unique if the system is asymptotically stable. (See Problem 9.6.) Then the solution P can be checked by Sylvester's criterion (Theorem 4.10), and if and only if P is positive definite the system is asymptotically stable. This procedure is equivalent to determining if all the eigenvalues of A have negative real parts. (See Problem 9.7.) Experience has shown that it is usually easier to use the Lienard-Chipart (Routh-Hurwitz) test than the Liapunov method, however. Example 9.23.

Given the system d2y/dt 2 + 2dy/dt + y up the state equations dx =

=::

O. To determine the stability by Liapunov's method we set y

dt

=::

(1 O)x

We arbitrarily choose Q = I and solve

PI2)(-1-2 0 1)

(~ =~)(:~~ The solution is P

---c ~

::::

-21- (31 1) l'

(-1o 0) -1

P22

Using Sylvester's criterion we find Pn = 3

-

and so P is positive definite and the system is stable. immediately says the system is stable.

>0

and detP = 1

> 0,

But it is much easier to use Routh's test which almost

We may ask at this point what practical use can be made of Theorem 9.9, other than the somewhat comforting conclusion of Theorem 9.10. The rather unsatisfactory answer is that we might be able to construct a Liapunov function for systems of the form dx/dt::::: A(t)x + f(x,t) where A(t) has been chosen such that we can use Theorem 9.9 (or equation (9.5) when the system is time-invariant). A hint on how to chooseQ is given in Problem 9.7. Example 9.24. Given the system dx/dt = Ax + f:

d2y/dt2 + (2

+ e~t) dy/dt + y

::::: O.

We set up the state equation in the form

dx dt

As found in Example 9.23,

y

2v

dv/dt

= 3xf

=

+ 2XIX2 + x~. + 2x TPf

(1 O)x

Then

d(x'tex)/dt = 2xTP dx./dt =

:':::: -xTQx

-

2x TP(Ax + f)

= -xi - x~ - (Xl + x2)e~tx2

STABILITY OF LINEAR SYSTEMS

202

dv -x

dt

T (

[CHAP. 9

1 e- t /2

Using Sylvester's criterion, 1 > 0 and 1 + e- t > e- 2t/4, so the given system is uniformly asymptotically stable.

For discrete-time systems, the analog of the construction procedure given in Theorem 9.9 is "" v(x, k) = ~ xT(k) cpT(m, k) Q(m) q,(m, k) x(k) m=k

and the analog of equation (9.5) is ATPA - P = -Q.

Solved Problems 9.L

A ball bearing with rolling friction rests on a deformable surface in a gravity field. Deform the surface to obtain examples of the various kinds of stability.

Global Asymptotic Stability

Asymptotic Stability but Unbounded

Global Stability i.s.L.

(a)

(b)

(c)

Stability i.s.L. but Unbounded

Unstable but Bounded

Unstable and Unbounded

Cd)

(e)

(f)

Fig. 9-2

We can see that if the equilibrium state is globally asymptotically stable as in Fig. 9-2(a), then it is also stable i.s.L., as are (b), (c) and (d), etc. If the shape of the surface does not vary with time, then the adjective "uniform" can be affixed to the description under each diagram. If the shape of the surface varies with time, the adjective "uniform" mayor may not be affixed, depending on how the shape of the surface varies with time.

9.2.

Prove Theorems 9.3 and 9.5. We wish to show dx/dt = A(t)x is uniformly asymptotically stable if and only if JI4J(t, to)!! ..:= for all t ~ to and all to. Also we wish to show dx/dt = A(t)x + B(t)u, where IIB-l(t)lh ..:= K'B-l' is externally stable if and only if Ijofl(t, to)11 ..:= K'le- K2 (t-tO), and hence if and only if the unforced system is uniformly asymptotically stable. K'le-KzCt-to)

Since IJx(t)l! ~ IIIJJ(t, to)l) !!~oll, if !!!fJ(t, to)11 :::: K'le-KzCt-to) and ))xol!":= 8, then Ilx(t)!! ~ IelO and tends to zero as t ~ which proves uniform asymptotic stability. Next, if dx/dt = A(t)x is uniformly asymptotically stable, then. for any p > 0 and any 'I'J > 0 there is a tl (independent of to by uniformity) such that if Ilxojl":= p then I1x(t)II":= 'I'J for all t> to + t l • In fact, for p = 1 and 7J = e- 1, there is a tl = T such that 'I'J = e- 1 ~ Ilx(to + T)II = 11~(to + T, to)xoll. Since I)xoll = 1, from Theorem 4.11, property 2, 11q:,(to + T, to)xoll = 114J(to + T, to)ll. Therefore for all to, 11q:,(to + T, to)11 ..:= e- 1 and for any positive integer k, (1;),

114t(to + kT, to)ll : : : : llq:,(to + kT, to + (k - 1)T) II' .. II<JI(to + 2T, to + T)llllofI(to + T, to) II ..:= e- k Choose k such that to + kT === t < to + (k + l)T and define K'1 = 11q:,(t, to + kT)lle < from (I;)

Theorem 9.2.

Then

IlofI(t, to)11 ..:= 114>(t, to + kT)llll4>(t o + kT, to)11 Defining

K2

= 1fT proves Theorem 9.3.

203

STABILITY OF LINEAR SYSTEMS

CHAP. 9] If

lllJ1(t, t o)lll :::::

ft lIB(t, -dill d1"

K1e-K2Ct-to),

ft II

III (t,

:0:::

~

then

1")11 !lB(1") II dT

iCDiCl (1

11
iCO s t

:0:::

~

~

- e-

K2 Ct -

to) )!K2

{3

iC iCl/ KZ D

:0:::

so that use of Theorem 9.4 proves the system is uniformly externally stable. If the system is uniformly externally stable, then from Theorem 9.4, {3

=

ftljH(t,'T)lhdT

===

ftlllll(t,T)B(T)111dT

to

Since IIB-l(T)lh:::::

/CD-I'

to

then

ft 11~(t, T)111 dT

:0:::

to

for all to and all t. Also, for the system (9.1) to be uniformly externally stable with arbitrary Xo, it must be uniformly stable i.s.L. so that II~(T, t o)111 < K from Theorem 9.2. Therefore

>

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 (t, to) = 0 and the system is asymptotit-oo

t-co

cally stable. This is true globally, i.e. for any initial condition xo, because of the linearity of the system (Theorem 9.1). Now we investigate uniformity by plotting (t, to) starting from three different initial times: to = 0, to = 2" and to = 417" (Fig. 9-3). The peaks occur at «2n + 1)1T, 2nn) = e1TC4 -'oT)C4n+1) so that the vertical scale is compressed to give a reasonable plot. We can pick an initial time to large enough that some initial condition in IXol < 8 will give rise to a trajectory that will leave Ixi < E for any E > 0 and (3 > O. Therefore although the system is stable i.s.L., it is not uniformly stable Ls.L. and so cannot be uniformly asymptotically stable.

Fig. 9-3

9.4.

If

~(t)

...::::

I<

+ Jrt [pCr) ~('T) + {J.(1")] dT

where p(t)

~o

and

to

~(t) == [K +

f

t -

e

fT p('I)d'l)

to

JL

()

T

d ] T

K

is a constant, show that

eitto p(T)dT

to

This is the Gronwall-Bellman lemma and is often useful in establishing bounds for the response of linear systems. We shall establish the following chain of assertions to prove the Gronwall-Bellman lemma.

STABILITY OF LINEAR SYSTEMS

204

If dw/dt == 0 and w(to)::= 0, then wet)

Assertion (i):

~

[CHAP. 9

0 for t

~ to.

Proof: If ",(t 1) > 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}
oCt, to)

fta(r) dr,

:=:

gives

to

o

~ e- 1H t,to)

dw/dt - a(t)e- OCt• to)

Applying assertion (i) gives e-o(t.to) '" ~ 0, and since Assertion (iii):

If

d¢/dt - a(t)¢

Let '" =

Proof:

Assertion (iv):

Proof:

¢ - y

Set '"

:=:

and y =

¢

Assertion (v):

0"

K,

,(t) and "(to) '"

(K +

K.

f

9.5.

===

> 0 then '"

¢(to):=: y(to),

then

t

(K +

then _(t) '"

e- OCT, to) ,u(r) dr) eOCt , to)

~ O.

¢(t) === yet).

s,:

e-8(T,t,) pH
.8(t, t,).

in assertion (iii), because dy/dt-

to

II;

ft [per) HT) + ,u(T)] dT,

+

so that the given inequality says

to

Since pet) (iv) gives

",)/dt

The Gronwall-Bellman lemma is true.

Let ",(t) =

Proof:

e-OCt.to)

and

dy/dt - a(t)y

dCe-OCt. to)

in assertion (ii).

If dw/dt - a(t)"

a( t) :::::; ,u( t) and y( to) :::::

==

:::::;

0 we have dw/dt = pet) ~(t) + p.(t) :=; pet) wet)

Show that if A{t) remains bounded for all t have finite escape time.

Also show

===

+ ,u(t),

to (lJA{t)Jl

and w(to) =

L:

Ilx(t)l) :::: II Xo

+

ft A(T) X(T) dT II .~ to

II xol!

+

Ilf

t A(T)

Applying assertion

K), then the system cannot

IIx(t)11 ~ Ilxolle~:IIA(T}lIdT

Integrating dx/ dt :::::: A( t)x between to and t gives x( t) = xo norm of both sides gives

K.

~(t):=: ",(t).

+

for any A{t).

it

A(T) x(T) dT.

Taking the

to

X(T) dT II :=; l)xoll

+

to

f

t

J]A(T)llllx(T)l1
Since j)A(r))I ~ 0, we use the Gronwall-Bellman inequality of Problem 9.4 with pet) = 0 to obtain I)x(t)!) ~ !)xo)le~:IIACT)lldT. If ))A(t))1 ~ K, then Ilx(t)I):=; IlxolleKCt-to), which shows x(t) is bounded for finite t and so the system cannot have finite escape time. However, if I)A(t)11 becomes unbounded, we cannot conclude the system has finite escape time, as shown by Problem 9.3.

9.6.

Under what circumstances does a unique solution P exist for the equation ATp + PA

= -Q?

We consider the general equation BP + PA = C where A, Band C are arbitrary real n X n matrices. Define the column vectors of P and C as Pi and ci respectively, and the row vectors of A as n

aJ for

i:=:

1,2, .. . ,n.

Then the equation BP+PA

= C can be written as BP+i=l ~ Pia[ = C,

in turn can be written as

[( ~ ~ ::: ~) ... ... ... o 0 ... B

=

(f)

which

205

STABILITY OF LINEAR SYSTEMS

CHAP•. 9]

B

A,

Call the first matrix in brackets and the second and the vectors p and c. Then this equation can be written (13 + A)p = c. Note ~ has eigenvalues equal to the eigenvalues of B, call them f3i' repeated n times. Also, the rows of A - Aln are linearly dependent if and only if .the rows of AT - ~.In are linearly dependent. This happens if and only if det (AT -Aln) = O. So A has eigenvalues equal to the eigenvalues of A, call them ai' repeated n times. Call T the matrix that reduces B to Jordan form J, and V = {vij} the matrix that reduces A to Jordan form. Then

A

A

+ A are ai + f3 j for i, j = 1, 2, ... , n. A unique solution to (B + A)p = c exists if and only if det (:8 + A):= IT (ai + f3) #- o. alIi.j Therefore if and only if ai + f3j # 0 for all i and i, does a unique solution to BP + PA = C exist. If B =_AT, then f3j == aj, so we require ai + aj # O. If the real parts of all aj are in the left half plane (for asymptotic stability), then it is impossible that ai + fX.j := O. Hence a unique solution for Hence the eigenvalues of B

P always exists if the system is asymptotically stable.

9.7.

Show that PA + ATp = -Q, where Q is a positive definite matrix, has a positive definite solution P if and only if the eigenvalues of A are in the left half plane. This is obvious by Theorem 9.11, but we shall give a direct proof assuming A can be diagonaIized as M-IAM =.4.. Construct P as (M-l)tM-l. Obviously P is Hermitian, but not necessarily real. Note P is positive definite because for any nonzero vector x we have xtPx = xt(M-l)tM-1x = (M-1x)t(M-1x) > O. Also, xtQx:= -(M- 1x)t(.4. + .A. t)(M-1x) > 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.

Supplementary Problems 9.9.

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

Answers to Supplementary Problems 9.9.

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 <

==

x2 OJ;)

CQ

CQ

o < /C7 === 1 + R(L/RZ o < "8 === 1 + RL/RZ

.

C/2)

+ CL/RC -

L/R

9.17.

P == (Mt)-lM-l which is always positive definite, and Q == M-l(I - .At A)M-l definite if and only if each eigenvalue of A has absolute value less than one.

9.18.

Use any Liapunov function constructed for the linearized system.

9.19.

If Q

==

21,

require 4 + 9.20.

Ilx))

===

then P

2aO -

!Jxolle-

K2

==21 (2+a a

02(1


2

The system is stable for all a if

(J

==

O.

If

e oF 0, we

+ a2/2)2 > o. +

ft e-

K2

~

9.21.

which is positive

Use Problem 9.4 with Ilx(t)11


IIB(r)11 Ilx)) dT so apply the Gronwall-Bellman inequality to

Ilxl)e K2t

==

~(t),

:0:::

JJxolleK2to

+

/Cg

f

t

Ilxll dT to

lixoll = ", jjA(t)11 = pet), and IIB(..-) U(T»)J = P.(T).

Chapter

10

Introduction to Optimal Control 10.1

INTRODUCTION

In this chapter we shall study a particular type of optimal control system as an introduction to the subject. In keeping with the spirit of the book, the problem is stated in such a manner as to lead to a linear closed loop system. A general optimization problem usually does not lead to a linear closed loop system. Suppose it is a control system designer's task to find a feedback control for an open loop system in which all the states are the real output variables, so that

==

dx/dt

A(t)x

+ B(t)u

y

=

Ix

+ Ou

(10.1)

In other words, we are given the system shown in Fig. 10-1 and -wish to design what goes into the box marked *. Later we will consider what happens when y(t) == C(t) x(t) where C(t) is not restricted to be the unit matrix.

d( t)

~

=====+:;;;'+

I

u(t)

I

I

x(t) = y(t)

~======:=>=S::qystem (10.1) 1=======:::;-1 =====;» ;::::1

~

Fig. 10-1.

*

r:::::

========::::J.

Vector Feedback Control System

A further restriction must be made on the type of system ,to be controlled. Definition 10.1: A regulator is a feedback control system in which the input d(t) == O. For a regulator the only forcing terms are due to the nonzero initial conditions of the state variables, x(to) Xo ¥ 0 in general. We shall only study regulators first because (1) later the extension to servomechanisms (which follow a specified d(t)) will become easier, (2) it turns out that the solution is the same if d(t) is white noise (the proof of this result is beyond the scope of the book), and (3) many systems can be reduced to regulators.

=

Example 10.1. Given a step input of height a, u(t)::::: aU(t), where the unit step U(t - to) = 0 -for t < to and U(t - to) = 1 for t === to, into a scalar system with transfer function 1/(8 + f3) and initial state xo. But this is equivalent to a system with a transfer function 1/[8(8 + f3)], with initial states a and Xo and with no input. In other words, we can add an extra integrator with initial condition a at the input to the system flow diagram to generate the step, and the resultant system is a regulator. Note, however, the input becomes a state that must be measured and fed back if the system is to be of the form (10.1).

Under the restriction that we are designing a regulator, we require d(t) == o. Then the input u(t) is the output of the box marked * in Fig. 10-1 and is the feedback control. We shall assume u is then some function to be formed of the present state x(t) and time. This is no restriction, because from condition 1 of Section 1.4 upon the representation of the state of this deterministic system, the present state completely summarizes all the past history of the abstract object. Therefore we need not consider controls u = u(x(7'), t) for to 6 7' ~ t, but can consider merely u = u(x(t), t) at the start of our development.

208

CHAP. 10]

INTRODUCTION TO OPTIMAL CONTROL

209

10.2 THE CRITERION FUNCTIONAL We desire the system to be optimal, but must be very exact in the sense in which the system is optimal. We must find a mathematical expression to measure how the system must be optin1al in comparison with other systems. A great many factors influence the ~ngineering utility of a system: cost, reliability, consumer acceptance, etc. The factors mentioned are very difficult to measure and put a single number on for purposes of comparison. Consequently in this introductory chapter we shall simply avoid the question by considering optin1ality only in terms of system dynamic performance. It is still left to the art of engineering, rather than the science, to incorporate unmeasurable quantities into the criterion of optimality. In terms of performance, the system response is usually of most interest. Response is the time behavior of the output, as the system tries to follow the input. Since the input to a regulator is zero, we wish to have the time behavior of the output, which in this case is the state, go to zero from the initial condition Xo. In fact, for the purposes of this chapter there exists a convenient means to assign a number to the distance that the response [X(7) for to ~ T ~ t 1] is from O. Although the criterion need not be a metric, a metric on the space of all time functions between to and tt will accomplish this. Also, if we are interested only in how close we are to zero at time t l , a metric only on all X(tl) is desired. To obtain a linear closed loop system, here we shall only consider the particular quadratic metric l(x,O)

=

txT(tl) SX(tl)

+ ~

rtlxT(T) Q(T) X(T) dT

J to

where S is an n x n symmetric constant matrix and Q(T) is a n x n symmetric time-varying matrix. If either S or Q(T), or both, are positive definite and the other one at least nonnegative definite then f(x,O) is a norm on the product space {x(t l), X(T)}. It can be shown this requirement can be weakened to S, Q nonnegative definite if the system dxldt = A(t)x with y = yQ(t) x is observable, but for simplicity we assume one is positive definite. The exact form of Sand Q( T) is to be fixed by the designer at the outset. Thus a number is assigned to the response obtained by each control law u(x(t), t), and the optimum system is that whose control law gives the minimum p{x, 0). The choice of Q(T) is dictated by the relative importance of each state over the time interval to ~ t < t 1• Example 10.2. Consider the system of Problem 2.18, page 34, with the choice of state variables (i). If the angle of attack 0: = rp - K6Z is to be kept small, we can minimize the integral of 0:2 = xTHTHx where H = (0 -Ka 1 0). Then Q = HTH is nonnegative definite and we must choose S to be positive definite. Furthermore, if 0: is of importance only during the first part of the missile's flight, Q might be chosen as a function of time, such as Q(t) = HTHe- t •

The choice of S is dictated by the relative importance of each state at the final time, t 1• Example 10.3. Consider the missile of Problem 2.18, page 34, with -the choice of state variables (i), whose target is stationed at z = O. Then it makes no difference what path the missile flies to arrive near z = 0 at t = t l • Therefore choose Q = O. What matters is how small .z2(t1} is. Also, we do not want z(t l }, ¢(tl ) or .if> (t 1) to be too large, so choose

s

=

(~

where El. E2' E3 are small fixed positive numbers to be -chosen by trial and error after finding the closed loop system for each fixed Ei' If any Ei = 0, an unstable system could result, but might not.

210

INTRODUCTION TO OPTIMAL CONTROL

[CHAP. 10

For the system (10.1), the choice of control u(x:(t), t) to minimize p(x, 0) is u = -B-l(t) Xo 8(t - to) in the case where B(t) has an inverse. Then x(t) = 0 for all t> to and p(x, 0) = 0, its minimum value. If B(t) does not have an inverse, the optimal control is a sum of delta functions and their derivatives, such as equation (6.7), page 135, and can drive x(t) to 0 almost instantaneously. Because it is very hard to mechanize a delta function, which is essentially achieving infinite closed loop gain, this solution is unacceptable. We must place a bound on the control. Again, to obtain the linear closed loop system, here we shall only consider the particular quadratic

It!

1

"2 J to

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

INTRODUCTION TO OPTIMAL CONTROL

[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]

INTRODUCTION TO OPTIMAL CONTROL

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, <)22(t, i)

(10.14)

Then

X(t)) ( p(t)

;::::

Eliminating x(t 1) from this gives p(t) ::::: P(t) x(t) ::::: [<)21(t, t 1)

+ ~22(t, tl)S][IPU(t, t 1) +CP12(t, t1)S]-lX(t)

(10.15)

so that P is the product of the two bracketed matrices. A sufficient condition (not necessary) for the existence of the solution P(t) of the Riccati equation is that the open loop transition matrix does not become unbounded in to ~ t ~ t 1• Therefore the inverse in equation (10.15) can always be found under this condition. Example 10.7. For Example 10.4, use of equation (lO.10) gives pet} from (10.15) as

pet)

[3e 4 (t-lJ [2e 4 ( t - l )

-

3e- 4 (t-l)

+ 6e- 4

(t-l)

+ 2(6e4 (t-l) + 2e- 4 (t-1)}]l8 + 2(4e4 (t-l) - 4e- 4 (t-l)}]l8

in Example 10.6. We have a useful check on the solution of the matrix Riccati equation.

which reduces to the answer obtained

Theorem 10.2: If S is positive definite and Q(t) at least nonnegative definite, or vice versa, and R(t) is positive definite, then an optimum v[x, UOP] exists if and only if the solution P(t) to the matrix Riccati equation (10.13) exists and is bounded and positive definite for all t < t l • Under these conditions v[x, nOP] ::::: txT(tO) P(to) x(to).

Proof is given in Problem 10.1, page 220. It is evident from the proof that if both Sand Q(t) are nonnegative definite, P(t) can be nonnegative definite if v[x, u OP] exists.

Theorem 10.3: For S(t), Q(t) and R(t) symmetric, the solution P(t) to the matrix Riccati equation (10.13) is symmetric. Proof: Take the transpose of (10.13), note it is identical to (10.13), and recall that there is only one solution P(t) that is equal to S at time t l •

Note this means for an n X n P(t) that only n(n + 1)/2 equations need be solved on the computer, because S, Q and R can always be taken symmetric. Further aids in obtaining a solution for the time-varying case are given in Problems 10.20 and 10.21. 10.5 TIME ..INVARIANT OPTIMAL SYSTEMS SO far we have obtained only time-varying feedback gain elements. The most important engineering application is for time-invariant closed loop systems. Consequently in this section we shall consider A, B, Rand Q to be constant, S = 0 and t 1 ...,. 00 to obtain a constant feedback element K. Because the existence of the solution of the Riccati equation is guaranteed if the open loop transition matrix does not become unbounded in to ~ t ~ tIl in the limit as tl ~ 00 we should expect no trouble from asymptotically stable open loop systems. However, we wish to incorporate unstable systems in the following development, and need the following existence theorem for the limit as tl 00 of the solution to the Riccati equation, denoted by II. -)0

214

INTRODUCTION TO OPTIMAL CONTROL

[CHAP. 10

Theorem 10.4: If the states of the system (10-.1) that are not asymptotically stable are controllable, then lim P(to; t 1 ) = n(to) exists and n(to) is constant and positl -+ "" tive definite. Proof: Define a control U1(T) = _BT(T) cpT(to, T) W-l(to, t 2) x(to) for to ,.:=:: T ~ t21 similar to that used in Problem 6.8, page 142. Note W drives all the controllable states to' zero at the time t2 < 00. Let U1(T) = 0 for T> t 2. Defining the response of the asymptotically

stable states as Xas(t), then

=

v[x, nl]

12"" xIsQXas dt

<

00.

Therefore

r (xTQx + U1TRu1) dt + J~(r.<J xIsQas dt J~ t2

<

co

Note v[x, u 1] ~ a(to, t 2) XT(tO) x(to) after carrying out the integration, since both x(t) and ul(t) are linear in x(to). Here a(to, t 2 ) is some bounded scalar function of to and t 2 • Then from Theorem 10.2, txT(tO) n(to) x(to) = v[x, nOP] ~ a(to, t 2) xT(tO) x(toY < co Therefore Iln(to)112 ~ 2a(to, t 2 ) < co so n(to) is bounded. It can be shown that for S Ps=o (to; t) ~ Ps=o (to; t l ) for to ~ t ~ t l , and also that when S > 0, then

= 0,

lim IIPs>o (to; t) - Ps=o (to; t)11 = 0 -+

t

00

Therefore, lim P(to; tl) must be a constant because for tl large any change in P(to; tl) must tl-+C() be to increase P(to; t l ) until it hits the bound n(to). Since we are now dealing with a time-invariant closed loop system, by Definition 1.8 the time axis can be translated and an equivalent system results. Hence we can send to -7 -co, start the Riccati equation at P(t1) = S =0 and integrate numerically backwards in time until the steady state constant n is reached. Example 10.8. Consider the system dx/dt = u with criterion -dP/dt = 1- p2

with P(t 1) = O. II

=

~f·l(X2 + u 2 ) dt.

J to

= lim P(to; t l ) = lim tanh (t 1 - to) = U op

Then the Riccati equation is'

This has a solution P(to) = tanh (t1 - to).

~-~

The optimal control is

p

~-~

= -R-lBTIIx = -x.

lim tanh (tl

~--~

-

Therefore to)

= 1

Since II is a positive definite constant solution of the matrix Riccati equation, it satisfies the quadratic algebraic equation (10.16) Example 10.9. For the system of Example 10.8, the quadratic algebraic equation satisfied by II is 0 has solutions ±1, so that II = 1 which is the only positive definite solution.

= 1- II2.

This

A very useful engineering property of the quadratic optimal system is that it is always stable if it is controllable. Since a linear, constant coefficient, stable closed loop system always results, often the quadratic criterion is chosen solely to attain this desirable feature.

Theorem 10.5: If II exists for the constant coefficient system, the closed loop system is asymptotically stable. Proof: Choose a Liapunov function V for all nonzero x. Then dV/dt = xTII(Ax- BR-IBTnx)

for all x

¥=

+ (Ax -

= xTrrx. Since

n is positive definite, V> 0

BR-IBTnx)TIIx = -xTQx - xTnBR-lBTnx

0 because Q is positive definite and nBR-tBTn is nonnegative definite.

< 0

215

INTRODUCTION TO OPTIMAL CONTROL

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~'

INTRODUCTION TO OPTIMAL CONTROL

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

INTRODUCTION TO OPTIMAL CONTROL

~)(!:) + (~)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

INTRODUCTION TO OPTIMAL CONTROL

[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

INTRODUCTION TO OPTIMAL CONTROL

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

INTRODUCTION TO OPTIMAL CONTROL

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}

INTRODUCTION TO OPTIMAL CONTROL

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
pet)

=

o.

t 2_ t2 - _ t_l

ti + t2

Then P(t) > 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

INTRODUCTION TO OPTIMAL CONTROL

(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

INTRODUCTION TO OPTIMAL CONTROL

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

INTRODUCTION TO OPTIMAL CONTROL

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

INTRODUCTION TO OPTIMAL CONTROL

[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]

INTRODUCTION TO OPTIMAL CONTROL

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.

~~""

Given the system dx dt

(-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

INTRODUCTION TO OPTIMAL CONTROL

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.

<

to

< ti,

for to

< tl <

0 and for -It1

Given the system shown in Fig. 10-12.

1

00

2v =

[(y - d)2 +

u 2]

<

to

<0<

tl

where

1'5

= 3/2.

Find the compensation K to optimize the criterion

dt.

o

dlt)

~ "'p ~____-_e_(t_)_.:I_K---,~

Fig. 10-12

__

-.-4'1

U_(t_)

9Yo

!

. I

If---Y,..(t_)--

CHAP. 10]

229

INTRODUCTION TO OPTIMAL CONTROL

10,25.

In Problem 10.6, show that if C = B :::: I in equation (10.26) and if R :::: 0 and Q:::: I in equation (10.27), then the system becomes identical to the model.

10.26.

Use the results of Problem 10.4 to generate a root locus for the closed loop system as a varies. What happens when a:::: O?

10.27.

Given the linear time-invariant system dx/dt:::: Ax + Bu with criterion

J

t

_

2v::::

!

(xTQx + uTRu) dt

tu

where Q and R are both time-invariant positive definite symmetric matrices, but Q is n X n and R is m X m. Find the feedback control u(x, t; t l ) that minimizes v in terms of the constant n X n matrices K n , K 12 , K 21 , K22 where

in which the 2n-vector (::) satisfies the eigenvalue problem given in equation (10.18), where AI' A2' .• " An have negative real parts and An+ 1. An+2' .•• , t..2n have positive real parts and are

distinct. 10.28.

Given only the matrices G and F as defined in Theorem 10.9, note that G and F are in general complex since they contain partitions of eigenvectors. Find a method for calculating II:::: GF-l using only the real number field, not complex numbers.

10.29.

(a)

Given the nominal scalar system I (''''

"2 J

d~/dt

=

~

+ (1-

with

~o::::

HO), and the performance criterion

1)~2 + ,u2] dt for constant 71 > 1. ConstTuct two Liapunov functions for the o optimal closed loop system that are valid for all 'TJ > 1. Hint: One method of construction is given in Chapter 10 and another in Chapter 9.

J ::::

n

[(-17 2 -

(b) Now suppose the actual system is d~/dt:::: €~3 + ~ + (1-. Using both Liapunov functions obtained in (a), give estimates upon how large € > 0 can be such that the closed loop system remains

asymptotically stable. In other words, find a function asymptotic stability of the closed loop system.

10.30.

f('I].~)

such that

€

< I('I],~) implies

Given the system dx/dt = A(t)x + B(t)u with x(to):::: Xo and with criterion function J :::: xT(tf ) x(tf )!2

+

Itl

uTu dt/2

to

(a)

What are the canonical equations and their boundary conditions?

(b)

Given the transition matrix 'P(t, T) where aCJJ(t, 'r}/at = A(t)CJJ(t, T) and CJJ(7,7):::: I. canonical equations in terms of 'P(t, '1") and find a feedback control u(t).

(e)

Write and solve the matrix Riccati equation.

Solve the

INTRODUCTION TO OPTIMAL CONTROL

230

Answers to Supplementary Problems 10.9.

u = -9x

10.10.

(a) dx/dt = x2 dX2/dt =

-v'5 X2 -

dPI/ dt

- 4X l

dp2/ dt (b) Xl(t)

X2(t) pz(t)

P2

10.11.

tt(t)

(a) 0

o

=

XI0

X2(O)

X20

PI (00)

0

0 P2( 00) + V5P2 (2xlO + x20)e- t - (xlO + x 20 )e- 2t -(2XlO + x 20 )e- t + 2(XI0 + x20)e- 2t (V5 - 1)(2xlO + x20)e- t + (4 - 2V5 )(XlO + x20)e- 2t -Pl

Pl(t) = 4(2xlO + x20)e- t (c)

Xl(O)

(1-

2(XIO + x20)e- 2t

V5 )(2X lO + x20)e- t + (2Y5 -

4)(XI0 + x20)e- 2t

4 - ¢~2


v'5 ¢12 -

¢12¢22

(d)

u

+

Fig.l0~13

10.13.

(a)

(b)

pet) =

lim pet)

t--c<:>

10.15.

[8(1 + V2) + 1] e2 \1'2 (T-t)

-

1 -8(1- V2)

[V2 -1 + S]e2V2
Any control results in v = 00

8

since there is an unstable uncontrollable state.

[CHAP. 10

INTRODUCTION TO OPTIMAL CONTROL

CHAP. 101 10.17.

This is the diagram associated with Problem 10.8.

10.18.

f... =

10.19.

u op = Ro1bTPo1J/p where Po depends only on the ratio 1J/p as

±ya2 + b 2 q/r

-dPo/dt

10.20.

Let

= q,T(t, to)Eq,(t, to)

tP(t, to)

q,T(t, to)(HTE

of time in

10.22.

u

10.23.

P(t) -

(

Il:J

- E

+ ATpo + PoA -

where

E

::::

tt)

x

For 0

<

It -

2" tan - 2 -

tan t -

•

t

0

< t 1 and

t

0

00

Then tP(t o, to)

10.26.

At ()':::: 0, the closed loop system poles are at

10.27.

X(t») ( p(t)

< t 1 < 0,

ei77r/8

Note p(t1 ) = 0 and solve for x(t 1) in terms of x(t).

P(t)

F

=

(Re fl

and

Gsimilarly..

(a)

From Chapter 10, VI:::: ('I]

and

I

and

e j97r / 8 •

Then p(t):::: P(t) x(t).

< 7J~-2

Then

F and Gare real and + 1)~2

n == GF-1.

and from Chapter 9, V 2 =

~2/27J.

for both VI and V 2 •

(b) u = -BT«>T(t"

t)[

«>(t, t , ) -

obvious from (b) above.

f2i

I 1m f1 I Re f3 I 1m fa I ... I 1m f 2m - 1 I f2m I ... I fn)

(a) dx/dt = Ax - BBTp and dp/dt:::: -ATp with x(to):::: Xo and P(tl).= x(t I ).

(c)

0

> O. For -'Vt t < to < 0 < tt,

Let the first 2m:=: n eigenvectors be complex conjugates) so that £21-1::: for i = I, 2, ... , m. Define

(b) e

==

as t tends to -yt 1 from the right.

K:::: 1

10.30.

(_~ ~).

1

10.24.

10.29.

PObTRolbPo7]/p

dtP/ dt

=

= 0, so tP(t, to) = O. This shows that for every increasing function there is a corresponding decreasing function of time.

6t~t - 6t 1 t 6 3t 5 + 2t 5

-

= Qo

+ EH)iP(t, to)

and tends to -

10.28.

231

s.:

«>(t, T)BBT«>T(t" T) dTJ-l x(t)

and

P(t)

<0

INDEX Abstract object, 2-6 uniform, 14 Adjoint of a linear operator, 112 Adjoint system, 112-114 Adjugate matrix, 44 Aerospace vehicle, state equations of, 34 Algebraic equation for time-invariant optimal control, 214 separation, 178 Alternating property, 56 Analytic function, 79 Anticipatory system. See State variable of anticipatory system; Predictor Associated Legendre equation, 107 Asymptotic behavior of optimal closed loop poles, 226 stability, 193 state estimator (see Observer systems; Steady state errors)

Component. See Element Contour integral representation of f(A), 83 Contraction mapping, 92-93 Control energy, 210 law, 208-211 Controllable form of time-varying systems, 148-152 Controllability, 128-146 of discrete-time systems, 132, 136 of the output, 144 for time-invariant systems with distinct eigenvalues, 129-130 for time-invariant systems with nondistinct eigenvalues, 131, 135 of time-varying systems, 136-138, 142 Convergence of the matrix exponential, 96 Convolution integral, 109 ,Costate, 210 Cramer's rule, 44 Criterion functional, 209

Basis, 47 Bessel's equation, 107 BIBO stability. See External stability Block diagrams, 164-169 Bounded function, 191 input-bounded output stability (see External stability) Boundedness of transition matrix, 192 Butterworth configuration, 226

Dead beat control, 132 Decoupling. See Pole placement of multiple-input systems Delay line, state variables of. See State variable of a delay line Delayor, 4, 17, 164 Delta function derivatives of Dirac, 135 Dirac, 135 Kronecker, 44 Derivative formula for integrals, 125 of a matrix, 39 Determinant, 41 by exterior products, 57 of the product of two matrices, 42 of similar matrices, 72 of a transition matrix, 99-100 of a transpose matrix, 42 of a triangular matrix, 43 Diagonal matrix, 41 Diagonal of a matrix, 39 Difference equations, 7 from sampling, 111 Differential equations, 7 Differentiator, state of. See State variable of a diiferentiator Diffusion equation, 10 Dimension, 49 of null space, 49 of l'ange space, 50 Disconnected flow diagram, 129 Discrete-time systems, 7 from sampling, 111

Cancellations in a transfer function, 133-134 Canonical equations of Hamilton, 211 flow diagrams (see Flow diagrams, first canonical form; Flow diagrams, second canonical form) Cauchy integral representation, 83 Causality, 2-6 Cayley-Hamilton theorem, 81 Characteristic polynomial, 69 Check on solution for transition matrices, 114 Closed forms for time-varying systems, 105 loop eigenvalues, 217 loop optimal response, 217 under addition, 46 under scalar multiplication, 46 Cofactor of a matrix element, 43 Column of a matrix, 38 Commutativity of matrices, 40 Compatible, 39-40 Compensator pole placement, 172 integral plus proportional, 219-223 Complex conjugate transpose matrix, 41

233

234

INDEX

Distinct eigenvalues, 69-71 Dot product. See Inner product Duality, 138-139 Dynamical systems, 4-6

Gram-Schmidt process, 53-54 Grassmann algebra. See Exterior product Gronwall-Bellman lemma, 203 Guidance control law, 121-122

Eigenvalue, 69 of a nonnegative definite matrix, 77 of a positive definite matrix, 77 of similar matrices, 72 of a time-varying system, 194 of a triangular matrix, 72 of a unitary matrix, 95 Eigenvector, 69 generalized, 74 expression for II = GF-l, 216,229 of a normal matrix, 76 Element, 38 Elementary operations, 42 Ellipse in n-space, 77 Equilibrium state, 192 Equivalence of pole placement and optimal control, 220 Equivalence transformation, 148 Estimation of state. See Observer systems Existence of solution to matrix equations, 63 to Riccati equation, 213 to time-invariant optimal control problem, 214 Exp At, 101-111 Exponential stability. See Stability, uniform asymptotic Exterior product, 56 External stability, 194

Hamilton-J acobi equation, 220 Hamiltonian matrix, 211 Harmonic oscillator, 193 Hereditary system. See State variable of hereditary system Hermite equation, 107 Hermitian matrix, 41 nonnegative definite, 77 positive definite, 77 Hidden oscillations in sampled data systems, 132 Hypergeometric equation, 107

Factorization of time-varying matrices, 152 Feedback optimal control, 211-231 Finite escape time, 104-105,204 First canonical form, 19,20 for time-varying systems, 153 Floquet theory, 107-109 Flow diagrams, 16-24 of first canonical form, 19 interchange of elements in, 17 of Jordan form, 21-24 of second canonical (phase variable) form, 20 for time-varying systems, 18 Frequency domain characterization of optimality, 220, 226 Function of a matrix, 79-84 cosine, 82 exponential, 96-104 logarithm, 94 square root, 80, 91 Fundamental matrix. See Transition matrix Gauss elimination (Example 3.13), 44 Gauss-Seidel method, 171 Global minimum of criterion functional, 220 stability, 193 Gradient operator, 96 Gram matrix, 64

Impulse response matrices, 112 Induced norm. See Matrix, norm of Induction, 43 Inner product, 52 for quadratic forms, 75 Input-output pair, 1-6 Input solution to the state equation, 109 Instantaneous controllability and observability. See Totally controllable and observable Integral of a matrix, 39 representation of teA), 83 Integral plus proportional control, 219-223 Integrator, 16,164 Inverse matrix, 44 of (81 - A), computation of, 102 system, 166 Jacobian matrix, 9 Jordan block, 74 Jordan form flow diagram of, 21-24 of a matrix, 73~75 of state equations, 147 Kronecker delta, 44 Lagrange multiplier, 210 Laguerre equation, 107 La place expansion of a determinant, 43, 57 Least squares fit, 86 Left inverse, 44 Legendre equation, 107 Leverrier's algorithm, 102 Liapunov function construction of, for linear systems, 199-202 discrete-time, 198 time-invariant, 196 time-varying, 197 Linear dependence, 47 independence, 47 manifold, 46 network with a switch, 11, 12

INDEX Linear (cont.) system, 6-7 vedor space, 50 Linear operator, 55 matrix representation of, 55 null space of, 49 range of, 49 Linearization, 8, 9 Logarithm of a matrix, 94 Lo.ss matrix. See Criterion functional Lower triangular matrix, 41 Mathieu equation, 107-109 Matrix, 38. See also Linear operator addition and subtraction, 39 characteristic polynomial of, 69 compatibility, 39 differentiation of, 39 eigenvalue of, 69 eigenvector of, 69 equality, 39 function of, 79-84 fundamental, 99-111 generalized eigenvector of, 74 Hermitian, 107 Hermitian nonnegative definite, 77 Hermitian positive definite 77 integration of, 39 ' Jordan form of, 73-75 logarithm of, 94 multiplication, 39 norm of, 78 principal minor of, 77 rank of, 50 representation of a linear operator, 55 Riccati equation, 211-231 square, 38 state equations, 26, 27 transition, 99-111 unitary, 41 Matrizant. See Transition matrix Mean square estimate, 94 Metric, 51 Minimal order observer, 175-177 Minimum energy control (Problem 10.30), 229 norm solution of a matrix equation, 86 Minor, 43 principal, 77 Mi.ssile equations of state, 34 Model-following systems, 224-225 Motor equations of state, 34 Multilinear, 56 Multiple loop system, 185-186 Natural norm, 53 of a matrix, 79 Neighboring optimal control, 221 Noise rejection, 179-181 Nominal optimal control, 221 Nonanticipative system. See Causality

235 Nonexistence of solution to the matrix Riccati equation, 221, 228 of trajectories, 10 Nonlinear effects, 179-181 Nonnegative definite, 77 N on singular , 44 Norm, 51 of a matrix, 78 natural, 53 Normal matrix, 41 Normalized eigenvector, 70 Null space, 49 Nullity, 49 n-vectors, 38 Nyquist diagrams, 171 Observable form of time-varying systems, 148-152 Observability, 128-146 of discrete-time systems, 133-138 of time-invariant systems, 133-136 of time-varying systems, 136-138, 145 Observer systems, 173-177 Open loop optimal control, 211 Optimal control law, 210 Order of a matrix, 38 Oriented mathematical relations, 2-4 Orthogonal vectors, 52 Orthonormal vectors, 53 Output feedback, 217-218 Parameter variations, 170, 171 Partial differential equation of state diffusion, 10 wave, 141 Partial fraction expansion, 21,22 of a matrix, 102 Partition of a matrix, 40 Passive system stability, 205 Penalty function. See Criterion functional Periodically varying system, 107-109 Permutation, 41 Phase plane, 11 Phase variable canonical form, 20, 21 for time-varying systems, 153-4 Physical object, 1 Piecewise time-invariant system, 106 Polar decomposition of a matrix, 87 Pole placement, 172-173, 220, 224 Poles of a system, 169 Pole-zero cancellation, 134 Positive definite, 77 Predictor, 5 Principal minor, 77 Pseudoinverse, 84-87 Quadratic form, 75 Random processes, 3 Rank, 50 n(e,d.), 137 Range space, 49

236 Reachable states, 135 . Reciprocal basis, 54 Recoverable states, 135 Regulator, 208 Representation of a linear operator, 55 spectral, 80 for rectangUlar matrices, 85 Residue of poles, 21, 22 matrix, 102-103 Response function. See Trajectory Riccati equation, 212-213 negative solution to, 221 Right inverse, 44 RLC circuit, 2, 6, 11, 35 stability of, 205-207 Root locus, 169 Row of matrix, 38 Sampled systems, 111 Scalar, 38 equation representation for time-varying systems, 153-154, 160 product (see Inner product) transition matrix, 105 Scalar, 16,164 Schmidt orthonormalization. See Gram-Schmidt process Schwarz inequality, 52 Second canonical form, 20-21 for time-varying systems, 153-154 Sensitivity, 179-181 Servomechanism, 208 problem, 218-220 Similarity transformation, 70-72 Simultaneous diagonalization of two Hermitian matrices, 91 of two arbitrary matrices, 118 Skew-symmetric matrix, 41 Span, 46 Spectral representation, 80 Spring-mass system. See State variable of spring-mass system Square root of a matrix, 77 Stability asymptotic, 194 BIBO, 194 external, 194 i.s.L., 192 in the large, 193 of optimal control, 214 of solution of time-invariant Riccati equation, 215 of a system, 192 uniform, 193 uniform asymptotic, 193 State of an abstract object, 1-3 controllable, 128 observable, 128 of a physical object, 1

INDEX State (cant.) reachable, 135 recoverable, 135 State equations in matrix form, 26,27 solutions of, with input, 109-110 solutions of, with nonzero input, 99-109 State feedback pole placement, 172-173 State space, 3 conditions on description by, 5 State transition matrix, 99-111 State variable, 3 of anticipatory system, 14 of delay line, 10 of differentiator, 14 of diffusion equation, 10 of hereditary system, 10 of spring-mass system, 10 Steady state errors, 165 response, 125 Sturm-Liouville equation, 197 stability properties of, 198 Submatrices, 40 Sufficient conditions for optimal control, 220 Summer, 16,164 Superposition, 7 integral, 109 Switch. See Linear network with a switch Sylvester's criterion, 77 Symmetric matrix, 41 Symmetry of eigenvalues of H, 215 of solution to matrix Riccati equation, 213 Symplectic, 228 Systems, 7, 8 dynamical, 5 non anticipative, 5 physical, 1 Time to go, 228 Time-invariant, 7 optimal systems, 213-217 Time-varying systems flow diagram of, 18 matrix state equations of, 25-26 transition matrices for, 104-111 Totally controllable and observable, 128 Trace, 39 of similar matrices, 72 Trajectory, 4, 5 Transfer function, 112 Transition matrix, 99-111 property, 5 Transpose matrix, 41 Transposition, 41 Triangle inequality, 51 Triangular matrix, 41 Type-l systems, 167

INDEX Undetermined coefficients, method of. See Variation of parameters, method of Uniform abstract object. See Abstract object, uniform Uniform stability, 193 asymptotic, 193 Uncontrollable, 3 states in optimal control, 223 Uniqueness of solution to ATp + PA = -Q, 204 to matrix equations, 63 to the time-invariant optimal control, 215 for the square root matrix, 91 Unit matrix. 41 vector, 41 Unitary matrix, 41 for similarity transformation of a Hermitian matrix, 76 Unit-time delayor. See Delayor

237 Unity feedback systems, 165 Unobservable, 3 Unstable, 192 Upper triangular matrix, 41 Van der Pol equation, 192 Vandermonde matrix, 60 Variation of parameters, method of, 109 Variational equations. See Linearization Vector, 38, 46, 50 Vector flow diagrams, 164 Voltage divider, 1, 3 Wave equation, 141 White noise input, 208 Wiener factorization, 216 Zero matrix, 41 Zero-input stability, 191 z transfer function, 18