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Linear Control Systems

Topics : 1. Controllability 2. Observability 3. Linear Feedback 4. Realization Theory

c Claudiu C. Remsing, 2006. Copyright All rights reserved.

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C.C. Remsing

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Intuitively, a control system should be designed so that the input u(·) “controls” all the states; and also so that all states can be “observed” from the output y(·). The concepts of (complete) controllability and observability formalize these ideas. Another two fundamental concepts of control theory – feedback and realization – are introduced. Using (linear) feedback it is possible to exert a considerable influence on the behaviour of a (linear) control system.

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AM3.2 - Linear Control

3.1

Controllability

An essential first step in dealing with many control problems is to determine whether a desired objective can be achieved by manipulating the chosen control variables. If not, then either the objective will have to be modified or control will have to be applied in some different fashion. We shall discuss the general property of being able to transfer (or steer) a control system from any given state to any other by means of a suitable choice of control functions. 3.1.1 Definition.

The linear control system Σ defined by x˙ = A(t)x + B(t)u(t)

(3.1)

where A(t) ∈ Rm×m and B(t) ∈ Rm×` , is said to be completely controllable (c.c.) if for any t0 , any initial state x(t0 ) = x0 , and any given final state xf , there exist a finite time t1 > t0 and a control u : [t0 , t1 ] → R` such that x(t1 ) = xf . Note :

(1)

The qualifying term “completely” implies that the definition holds for

all x0 and xf , and several other types of controllability can be defined. (2)

The control u(·) is assumed piecewise-continuous in the interval [t0 , t1 ].

3.1.2 Example.

Consider the control system described by x˙ 1 = a1 x1 + a2 x2 + u(t)

x˙ 2 = x2 .

Clearly, by inspection, this is not completely controllable (c.c.) since u(·) has no influence on x2 , which is entirely determined by the second equation and x2 (t0 ). We have

Z xf = Φ(t1 , t0 ) x0 +

t1

t0

Φ(t0 , τ )B(τ )u(τ ) dτ

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C.C. Remsing

or

Z 0 = Φ(t1 , t0 ) x0 − Φ(t0 , t1 )xf +

t1

Φ(t0 , τ )B(τ )u(τ ) dτ . t0

Since Φ(t1 , t0 ) is nonsingular it follows that if u(·) transfers x0 to xf , it also transfers x0 −Φ(t0 , t1 )xf to the origin in the same time interval. Since x0 and xf are arbitrary, it therefore follows that – in the controllability definition – the given final state can be taken to be the zero vector without loss of generality. Note : For time-invariant control systems – in the controllability definition – the initial time t0 can be set equal to zero .

The Kalman rank condition For linear time-invariant control systems a general algebraic criterion (for complete controllability) can be derived. 3.1.3 Theorem.

The linear time-invariant control system x˙ = Ax + Bu(t)

(3.2)

(or the pair (A, B) ) is c.c. if and only if the (Kalman) controllability matrix h C = C(A, B) : = B AB A2 B . . .

has rank m. Proof :

i Am−1 B ∈ Rm×m`

( ⇒ ) We suppose the system is c.c. and wish to prove that

rank (C) = m. This is done by assuming rank (C) < m, which leads to a contradiction. Then there exists a constant row m-vector q 6= 0 such that qB = 0,

qAB = 0,

. . .,

qAm−1 B = 0.

In the expression Z t x(t) = exp(tA) x0 + exp(−τ A)Bu(τ ) dτ 0

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AM3.2 - Linear Control

for the solution of (3.2) subject to x(0) = x0 , set t = t1 , x(t1 ) = 0 to obtain (since exp(t1 A) is nonsingular) Z −x0 =

t1

exp(−τ A)Bu(τ ) dτ.

0

Now, exp(−τ A) can be expressed as some polynomial r(A) in A having degree at most m − 1, so we get Z t1 −x0 = (r0 Im + r1 A + · · · + rm−1 Am−1 )Bu(τ ) dτ. 0

Multiplying this relation on the left by q gives qx0 = 0. Since the system is c.c., this must hold for any vector x0 , which implies q = 0, contradiction. (⇐) We asume rank (C) = m, and wish to show that for any x0 there is a function u : [0, t1] → R` , which when substituted into Z t x(t) = exp(tA) x0 + exp(−τ A)Bu(τ ) dτ

(3.3)

0

produces x(t1 ) = 0. Consider the symmetric matrix Z t1 Wc : = exp(−τ A)BB T exp(−τ AT ) dτ. 0

One can show that Wc is nonsingular. Indeed, consider the quadratic form associated to Wc T

α Wc α =

Z

t1

ψ(τ )ψ T (τ ) dτ

0

=

Z

0

t1

kψ(τ )||2e dτ ≥ 0

where α ∈ Rm×1 is an arbitrary column vector and ψ(τ ) : = αT exp (−τ A)B. It is clear that Wc is positive semi-definite, and will be singular only if there

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C.C. Remsing

exists an α ¯ 6= 0 such that α ¯ T Wc α ¯ = 0. However, in this case, it follows (using the properties of the norm) that ψ(τ ) ≡ 0, 0 ≤ τ ≤ t1 . Hence, we have τ2 2 τ3 3 T α ¯ Im − τ A + A − A + · · · B = 0 , 0 ≤ τ ≤ t1 2! 3! from which it follows that α ¯T B = 0 ,

α ¯ T A2 B = 0 ,

α ¯T AB = 0 ,

···

This implies that α ¯T C = 0. Since by assumption C has rank m, it follows that such a nonzero vector α ¯ cannot exist, so Wc is nonsingular. Now, if we choose as the control vector u(t) = −B T exp(−tAT )Wc−1 x0 ,

t ∈ [0, t1 ]

then substitution into (3.3) gives Z t1 T T −1 x(t1 ) = exp(t1 A) x0 − exp(−τ A)BB exp(−τ A ) dτ · (Wc x0 ) 0 = exp(t1 A) x0 − Wc Wc−1 x0 = 0

as required.

3.1.4 Corollary. reduces to

Proof :

2

If rank (B) = r, then the condition in Theorem 3.1.3 h rank B AB . . .

Define the matrix

h Ck : = B AB · · ·

i Am−r B = m.

i Ak B ,

k = 0, 1, 2 . . .

If rank (Cj ) = rank (Cj+1 ) it follows that all the columns of Aj+1 B must be linearly dependent on those of Cj . This then implies that all the columns of Aj+2 B, Aj+3 B, . . . must also be linearly dependent on those of Cj , so that rank (Cj ) = rank (Cj+1 ) = rank (Cj+2 ) = · · ·

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AM3.2 - Linear Control

Hence the rank of Ck increases by at least one when the index k is increased by one, until the maximum value of rank (Ck ) is attained when k = j. Since rank (C0 ) = rank (B) = r and rank (Ck ) ≤ m it follows that r + j ≤ m, giving j ≤ m − r as required. 3.1.5 Example.

2

Consider the linear control system Σ described by " # " # −2 2 1 x˙ = x+ u(t) . 1 −1 0

The (Kalman) controllability matrix is Σ

C=C =

"

1 −2 0

1

#

which has rank 2, so the control system Σ is c.c. Note :

When ` = 1, B reduces to a column vector b and Theorem 2.4.3 can be

restated as : A linear control system in the form x˙ = Ax + bu(t) can be transformed into the canonical form w˙ = Cw + du(t) if and only if it is c.c.

Controllability criterion We now give a general criterion for (complete) controllability of control systems (time-invariant or time-varying) as well as an explicit expression for a control vector which carry out a required alteration of states. 3.1.6 Theorem.

The linear control system Σ defined by x˙ = A(t)x + B(t)u(t)

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C.C. Remsing

is c.c. if and only if the symmetric matrix, called the controllability Gramian, Wc (t0 , t1 ) : =

Z

t1

Φ(t0 , τ )B(τ )B T (τ )ΦT (t0 , τ ) dτ ∈ Rm×m

(3.4)

t0

is nonsingular. In this case the control u∗ (t) = −B T (t)ΦT (t0 , t)Wc (t0 , t1 )−1 [x0 − Φ(t0 , t1 )xf ] ,

t ∈ [t0 , t1 ]

transfers x(t0 ) = x0 to x(t1 ) = xf . Proof :

(⇐) Sufficiency. If Wc (t0 , t1 ) is assumed nonsingular, then the

control defined by u∗ (t) = −B T (t)ΦT (t0 , t)Wc (t0 , t1 )−1 [x0 − Φ(t0 , t1 )xf ] ,

t ∈ [t0 , t1 ]

exists. Now, substitution of the above expression into the solution Z t x(t) = Φ(t, t0 ) x0 + Φ(t0 , τ )B(τ )u(τ ) dτ t0

of x˙ = A(t)x + B(t)u(t) gives Z x(t1 ) = Φ(t1 , t0 ) x0 +

t1

Φ(t0 , τ )B(τ )(−B T (τ )ΦT (t0 , τ )Wc(t0 , t1 )−1 ·

t0

[x0 − Φ(t0 , t1 )xf ]) dτ ] = Φ(t1 , t0 ) x0 − Wc (t0 , t1 )Wc (t0 , t1 )−1 [x0 − Φ(t0 , t1 )xf ] = Φ(t1 , t0 ) [x0 − x0 + Φ(t0 , t1 )xf ] = Φ(t1 , t0 )Φ(t0 , t1 )xf = xf .

(⇒) Necessity. We need to show that if Σ is c.c., then Wc (t0 , t1 ) is nonsingular. First, notice that if α ∈ Rm×1 is an arbitrary column vector, then from

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AM3.2 - Linear Control

(3.4) since W = Wc (t0 , t1 ) is symmetric we can construct the quadratic form Z t1 T α Wα = θT (τ, t0)θ(τ, t0 ) dτ Z

=

t0 t1

t0

kθk2e dτ ≥ 0

where θ(τ, t0 ) : = B T (τ )ΦT (t0 , τ )α, so that Wc (t0 , t1 ) is positive semi-definite. Suppose that there exists some α ¯ = 6 0 such that α ¯T W α ¯ = 0. Then we get ¯ (for θ = θ when α = α ¯) Z t1 ¯ 2 dτ = 0 kθk e t0

¯ t0) ≡ which in turn implies (using the properties of the norm) that θ(τ, 0,

t0 ≤ τ ≤ t1 . However, by assumption Σ is c.c. so there exists a control

v(·) making x(t1 ) = 0 if x(t0 ) = α. ¯ Hence Z t1 α ¯=− Φ(t0 , τ )B(τ )v(τ ) dτ. t0

Therefore kαk ¯ 2e = α ¯T α ¯ Z t1 v T (τ )B T (τ )ΦT (t0 , τ )α ¯ dτ = − t0 t1

= −

Z

¯ t0 ) dτ = 0 v T (τ )θ(τ,

t0

which contradicts the assumption that α ¯ 6= 0. Hence Wc (t0 , t1 ) is positive definite and is therefore nonsingular. 3.1.7 Example.

2

The control system is " # " # −2 2 1 x˙ = x+ u(t). 1 −1 1

Observe that λ = 0 is an eigenvalue of A, and b =

"

1

#

is a corresponding 1 eigenvector, so the controllability rank condition does not hold. However, A is

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C.C. Remsing

similar to its companion matrix. Using the matrix T =

"

0

1

1 −1 before (see Example 2.5.1) and w = T x we have the system w˙ =

"

0

#

1

0 −3

w+

"

1 0

#

#

computed

u.

Differentiation of the w1 equation and substitution produces a second-order ODE for w1 : w ¨1 + 3w˙ 1 = 3u + u. ˙ One integration produces a first-order ODE Z w˙ 1 + 3w1 = 3 u(τ ) dτ + u which shows that the action of arbitrary inputs u(·) affects the dynamics in only a one-dimensional space. The original x equations might lead us to think that u(·) can fully affect x1 and x2 , but notice that the w2 equation says that u(·) has no effect on the dynamics of the difference x1 − x2 = w2 . Only when the initial condition for w involves w2 (0) = 0 can u(·) be used to control a trajectory. That is, the inputs completely control only the states that lie in the subspace span [ b Ab ] = span {b} = span

"

1 1

#

.

Solutions starting with x1 (0) = x2 (0) satisfy x1 (t) = x2 (t) =

Z

t

u(τ ) dτ + x1 (0).

0

One can steer along the line x1 = x2 from any initial point to any final point x1 (t1 ) = x2 (t1 ) at any finite time t1 by appropriate choice of u(·). On the other hand, if the initial condition lies off the line x1 = x2 , then the difference w2 = x1 − x2 decays exponentially so there is no chance of steering to an arbitrarily given final state in finite time.

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AM3.2 - Linear Control

Note :

The control function u∗ (·) which transfers the system from x0 = x(t0 )

to xf = x(t1 ) requires calculation of the state transition matrix Φ(·, t0) and the controllability Gramian Wc (·, τ0 ). However, this is not too dificult for linear timeinvariant control systems, although rather tedious. Of course, there will in general be many other suitable control vectors which achieve the same result .

If u(·) is any other control taking x0 = x(t0 ) to

3.1.8 Proposition. xf = x(t1 ), then

Z

t1

t0

Proof :

ku(τ )k2e dτ

Z

>

t1 t0

ku∗ (τ )k2e dτ.

Since both u∗ and u satisfy Z t1 xf = Φ(t1 , t0 ) x0 + Φ(t0 , τ )B(τ )u(τ ) dτ t0

we obtain after subtraction Z t1 0= Φ(t0 , τ )B(τ ) [u(τ ) − u∗ (τ )] dτ. t0

Multiplication of this equation on the left by T [x0 − Φ(t0 , t1 )xf ]T Wc (t0 , t1 )−1

gives

Z

t1

(u∗ )T (τ ) [u∗ (τ ) − u(τ )] dτ = 0

t0

and thus

Z

t1

t0

ku∗ (τ )k2e dτ =

Therefore Z t1 0< ku∗ (τ ) − u(τ )k2e dτ

=

t0

= =

Z

Z

Z

t1 t0 t1 t0 t1 t0

Z

t1

(u∗ )T (τ )u(τ ) dτ.

t0

[u∗ (τ ) − u(τ )]T [u∗ (τ ) − u(τ )] dτ ku(τ )k2e + ku∗ (τ )k2e − 2(u∗ )T (τ )u(τ ) dτ

ku(τ )k2e − ku∗ (τ )k2e dτ

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C.C. Remsing

and so Z t1 Z 2 ku(τ )ke dτ = t0

t1

t0

ku

∗

(τ )k2e

∗

+ ku (τ ) −

u(τ )k2e

as required. Note :

dτ >

Z

t1

t0

ku∗ (τ )k2e dτ 2

This result can be interpreted as showing that the control u∗ (t) = −B T (t)ΦT (t0 , t)Wc (t0 , t1 )−1 [x0 − Φ(t0 , t1 )xf ]

is “optimal”, in the sense that it minimizes the integral Z t1 Z t1 u21 (τ ) + u22 (τ ) + · · · + u2` (τ ) dτ ku(τ )k2e dτ = t0

t0

over the set of all (admissible) controls which transfer x0 = x(t0 ) to xf = x(t1 ), and

this integral can be thought of as a measure of control “energy” involved.

Algebraic equivalence and decomposition of control systems We now indicate a further aspect of controllability. Let P (·) be a matrixvalued mapping which is continuous and such that P (t) is nonsingular for all t ≥ t0 . (The continuous maping P : [t0 , ∞) → GL (m, R) is a path in the e obtained from Σ by the general linear group GL (m, R).) Then the system Σ

transformation

x e = P (t)x

is said to be algebraically equivalent to Σ. 3.1.9 Proposition.

If Φ(t, t0 ) is the state transition matrix for Σ, then e t0 ) P (t)Φ(t, t0 )P −1 (t0 ) = Φ(t,

e is the state transition matrix for Σ.

Proof :

We recall that Φ(t, t0 ) is the unique matrix-valued mapping sat-

isfying ˙ t0 ) = A(t)Φ(t, t0 ), Φ(t,

Φ(t0 , t0 ) = Im

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AM3.2 - Linear Control

and is nonsingular. Clearly, e 0 , t0 ) = Im ; Φ(t

differentiation of

x e = P (t)x

gives

x e˙ = P˙ x + P x˙

= (P˙ + P A)x + P Bu

= (P˙ + P A)P −1 x e + P Bu.

e is the state transition matrix for We need to show that Φ x e˙ = (P˙ + P A)P −1 x e + P Bu .

We have

d P (t)Φ(t, t0 )P −1 (t0 ) dt ˙ t0 )P −1 (t0 ) = P˙ (t)Φ(t, t0 )P −1 (t0 ) + P (t)Φ(t, h i = P˙ (t) + P (t)A(t) P −1 (t) P (t)Φ(t, t0 )P −1 (t0 ) h i e t0 ). = P˙ (t) + P (t)A(t) P −1 (t) Φ(t,

e˙ t0 ) = Φ(t,

3.1.10 Proposition.

e If Σ is c.c., then so is Σ.

e are The system matrices for Σ

Proof :

e = (P˙ + P A)P −1 A

and

e is so the controllability matrix for Σ Z t1 f = e 0 , τ )B(τ e )B e T (τ )Φ e T (t0 , τ ) dτ W Φ(t t0 t1

=

Z

2

e = PB B

P (t0 )Φ(t0 , τ )P −1 (τ )P (τ )B(τ )B T (τ )P T (τ ) P −1 (τ )

t0

= P (t0 )Wc (t0 , t1 )P T (t0 ).

T

ΦT (t0 , τ )P T (t0 ) dτ

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C.C. Remsing

f=W fc (t0 , t1 ) is nonsingular since the matrices Wc (t0 , t1 ) Thus the matrix W

and P (t0 ) each have rank m.

2

The following important result on system decomposition then holds : 3.1.11 Theorem.

When the linear control system Σ is time-invariant

then if the controllability matrix C Σ has rank m1 < m there exists a control system, algebraically equivalent to Σ, having the form " # " #" # " # x˙ (1) A1 A2 x(1) B1 = + u(t) x˙ (2) 0 A3 x(2) 0 h i y = C1 C2 x

where x(1) and x(2) have orders m1 and m − m1 , respectively, and (A1 , B1 ) is c.c. We shall postpone the proof of this until a later section (see the proof of Theorem 3.4.5) where an explicit formula for the transformation matrix will also be given. Note :

It is clear that the vector x(2) is completely unaffected by the control u(·).

Thus the state space has been divided into two parts, one being c.c. and the other uncontrollable.

3.2

Observability

Closely linked to the idea of controllability is that of observability, which in general terms means that it is possible to determine the state of a system by measuring only the output. 3.2.1 Definition. by

The linear control system (with outputs) Σ described x˙ = A(t)x + B(t)u(t) y = C(t)x

(3.5)

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is said to be completely observable (c.o.) if for any t0 and any initial state x(t0 ) = x0 , there exists a finite time t1 > t0 such that knowledge of u(·) and y(·) for t ∈ [t0 , t1 ] suffices to determine x0 uniquely. Note :

There is in fact no loss of generality in assuming u(·) is identically zero

throughout the interval. Indeed, for any input u : [t0 , t1 ] → R` and initial state x0 , we have y(t) −

Z

t

C(t)Φ(t, τ )B(τ )u(τ ) dτ = C(t)Φ(t, t0 )x0 .

t0

Defining yb(t) : = y(t) −

we get

Z

t

C(t)Φ(t, τ )B(τ )u(τ ) dτ

t0

yb(t) = C(t)Φ(t, t0 )x0 .

Thus a linear control system is c.o. if and only if knowledge of the output yb(·) with zero input on the interval [t0 , t1 ] allows the initial state x0 to be determined.

3.2.2 Example.

Consider the linear control system described by x˙ 1 = a1 x1 + b1 u(t) x˙ 2 = a2 x2 + b2 u(t) y = x1 .

The first equation shows that x1 (·) (= y(·)) is completely determined by u(·) and x1 (t0 ). Thus it is impossible to determine x2 (t0 ) by measuring the output, so the system is not completely observable (c.o.). 3.2.3 Theorem.

The linear control system Σ is c.o. if and only if the

symmetric matrix, called the observability Gramian, Wo (t0 , t1 ) : =

Z

t1

t0

is nonsingular.

ΦT (τ, t0)C T (τ )C(τ )Φ(τ, t0) dτ ∈ Rm×m

(3.6)

C.C. Remsing

85

(⇐) Sufficiency. Assuming u(t) ≡ 0, t ∈ [t0 , t1 ], we have

Proof :

y(t) = C(t)Φ(t, t0 )x0 . Multiplying this relation on the left by ΦT (t, t0 )C T (t) and integrating produces

Z

t1

ΦT (τ, t0 )C T (τ )y(τ ) dτ = Wo (t0 , t1 )x0

t0

so that if Wo (t0 , t1 ) is nonsingular, the initial state is Z t1 x0 = Wo (t0 , t1 )−1 ΦT (τ, t0)C T (τ )y(τ ) dτ t0

so Σ is c.o. (⇒) Necessity. We now assume that Σ is c.o. and prove that W = Wo (t0 , t1 ) is nonsingular. First, if α ∈ Rm×1 is an arbitrary column vector, Z t1 T α Wα = (C(τ )Φ(τ, t0)α)T C(τ )Φ(τ, t0)α dτ ≥ 0 t0

so Wo (t0 , t1 ) is positive semi-definite. Next, suppose there exists an α ¯ 6= 0 such that α ¯T W α ¯ = 0. It then follows that C(τ )Φ(τ, t0)α ¯ ≡ 0,

t0 ≤ τ ≤ t1 .

This implies that when x0 = α ¯ the output is identically zero throughout the time interval, so that x0 cannot be determined in this case from the knowledge of y(·). This contradicts the assumption that Σ is c.o., hence Wo (t0 , t1 ) is positive definite, and therefore nonsingular. Note :

2

Since the observability of Σ is independent of B, we may refer to the

observability of the pair (A, C).

Duality 3.2.4 Theorem.

The linear control system (with outputs) Σ defined by x˙ = A(t)x + B(t)u(t) y = C(t)x

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AM3.2 - Linear Control

is c.c. if and only if the dual system Σ◦ defined by T T x˙ = −A (t)x + C (t)u(t) is c.o.; and conversely.

y = B T (t)x

We can see that if Φ(t, t0 ) is the state transition matrix for the

Proof :

system Σ, then ΦT (t0 , t) is the state transition matrix for the dual system Σ◦ . Indeed, differentiate Im = Φ(t, t0 )Φ(t, t0 )−1 to get 0=

d ˙ t0 )Φ(t, t0 )−1 + Φ(t, t0 )Φ(t ˙ 0 , t) Im = Φ(t, dt ˙ 0 , t) = A(t)Φ(t, t0 )Φ(t, t0 )−1 + Φ(t, t0 )Φ(t ˙ 0 , t). = A(t) + Φ(t, t0 )Φ(t

This implies ˙ 0 , t) = −Φ(t0 , t)A(t) Φ(t or Φ˙ T (t0 , t) = −AT (t)ΦT (t0 , t). Furthermore, the controllability matrix WcΣ (t0 , t1 )

=

Z

t1

Φ(t0 , τ )B(τ )B T (τ )ΦT (t0 , τ ) dτ

t0

(associated with Σ ) is identical to the observability matrix WoΣ (t0 , t1 ) (associated with Σ◦ ). Conversely, the observability matrix WoΣ (t0 , t1 ) =

Z

t1

ΦT (τ, t0)C T (τ )C(τ )Φ(τ, t0) dτ

t0 ◦

(associated with Σ ) is identical to the controllability matrix WcΣ (t0 , t1 ) (associated with Σ◦ ).

2

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C.C. Remsing

Note :

This duality theorem is extremely useful, since it enables us to deduce

immediately from a controllability result the corresponding one on observability (and conversely). For example, to obtain the observability criterion for the time-invariant case, we simply apply Theorem 3.1.3 to Σ◦ to obtain the following result.

3.2.5 Theorem.

The linear time-invariant control system x˙ = Ax + Bu(t) y = Cx

(3.7)

(or the pair (A, C)) is c.o. if and only if the (Kalman) observability matrix

has rank m. 3.2.6 Example.

O = O(A, C) : =

C CA CA2 .. . CAm−1

∈ Rmn×m

Consider the linear control system Σ described by " # " # −2 2 1 x+ u(t) x˙ = 1 −1 0 y=x . 1

The (Kalman) observability matrix is Σ

O=O =

"

1 0 −2 2

#

which has rank 2. Thus the control system Σ is c.o.

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AM3.2 - Linear Control

In the single-output case (i.e. n = 1), if u(·) = 0 and y(·) is known in the form γ1 eλ1 t + γ2 eλ2 t + · · · + γm eλm t assuming that all the eigenvalues λi of A are distinct, then x0 can be obtained more easily than by using x0 = Wo (t0 , t1 )

−1

Z

t1

ΦT (τ, t0)C T (τ )y(τ ) dτ.

t0

For suppose that t0 = 0 and consider the solution of x˙ = Ax in the spectral form, namely x(t) = (v1 x(0)) eλ1 t w1 + (v2 x(0)) eλ2 t w2 + · · · + (vm x(0)) eλm t wm. We have y(t) = (v1 x(0)) (cw1)eλ1 t + (v2 x(0)) (cw2 )eλ2t + · · · + (vm x(0)) (cwm)eλmt and equating coefficients of the exponential terms gives γi (i = 1, 2, . . ., m). vi x(0) = cwi This represents m linear equations for the m unknown components of x(0) in terms of γi , vi and wi (i = 1, 2, . . ., m). Again, in the single-output case, C reduces to a row matrix c and Theorem 3.2.5 can be restated as : A linear system (with outputs) in the form x˙ = Ax y = cx

can be transformed into the canonical form v˙ = Ev if and only if it is c.o.

y = fv

89

C.C. Remsing

Decomposition of control systems By duality, the result corresponding to Theorem 3.1.11 is : 3.2.7 Theorem.

When the linear control system Σ is time-invariant then

if the observability matrix OΣ has rank m1 < m there exists a control system, algebraically equivalent to Σ, having the form " # " #" # " # x˙ (1) A1 0 x(1) B1 = + u(t) x˙ (2) A2 A3 x(2) B2 y = C1 x(1) where x(1) and x(2) have orders m1 and m − m1 , respectively and (A1 , C1 ) is c.o. We close this section with a decomposition result which effectively combines together Theorems 3.1.11 and 3.2.7 to show that a linear time-invariant control system can split up into four mutually exclusive parts, respectively • c.c. but unobservable • c.c. and c.o. • uncontrollable and unobservable • c.o. but uncontrollable. 3.2.8 Theorem.

When the linear control system Σ is time-invariant it is

algebraically equivalent to x˙ (1) A A12 A13 11 x˙ 0 A 0 22 (2) = x˙ (3) 0 0 A33 x˙ (4) 0 0 0 y = C2 x(2) + C4 x(4)

B 1 A24 x(2) B2 + u(t) A34 x(3) 0 A44 x(4) 0 A14

x(1)

where the subscripts refer to the stated classification.

90

AM3.2 - Linear Control

3.3

Linear Feedback

Consider a linear control system Σ defined by x˙ = Ax + Bu(t)

(3.8)

where A ∈ Rm×m and B ∈ Rm×` . Suppose that we apply a (linear) feedback, that is each control variable is a linear combination of the state variables, so that u(t) = Kx(t) where K ∈ R`×m is a feedback matrix. The resulting closed loop system is x˙ = (A + BK)x.

(3.9)

The pole-shifting theorem We ask the question whether it is possible to exert some influence on the behaviour of the closed loop system and, if so, to what extent. A somewhat surprising result, called the Spectrum Assignment Theorem, says in essence that for almost any linear control system Σ it is possible to obtain arbitrary eigenvalues for the matrix A + BK (and hence arbitrary asymptotic behaviour) using suitable feedback laws (matrices) K, subject only to the obvious constraint that complex eigenvalues must appear in pairs. “Almost any” means that this will be true for (completely) controllable systems. Note :

This theorem is most often referred to as the Pole-Shifting Theorem, a

terminology that is due to the fact that the eigenvalues of A + BK are also the poles of the (complex) function z 7→

1 · det (zIn − A − BK)

This function appears often in classical control design. The Pole-Shifting Theorem is central to linear control systems theory and is itself the starting point for more interesting analysis. Once we know that arbitrary sets of eigenvalues can be assigned, it becomes of interest to compare the performance

91

C.C. Remsing

of different such sets. Also, one may ask what happens when certain entries of K are restricted to vanish, which corresponds to constraints on what can be implemented.

Let Λ = {θ1 , θ2 , . . . , θm } be an arbitrary set of m com-

3.3.1 Theorem.

plex numbers (appearing in conjugate pairs). If the linear control system Σ is c.c., then there exists a matrix K ∈ R`×m such that the eigenvalues of A + BK are the set Λ. Proof (when ` = 1) :

Since x˙ = Ax + Bu(t) is c.c., it follows that there

exists a (linear) transformation w = T x such that the given system is transformed into w˙ = Cw + du(t) where

C =

0

1

0

...

0

0 .. .

0 .. .

1 .. .

...

0 .. .

0

0

0

...

1

−km −km−1 −km−2 . . .

−k1

The feedback control u = kw, where k :=

h

km km−1 . . .

,

k1

d=

0

. 0 1 0 .. .

i

produces the closed loop matrix C + dk, which has the same companion form h i as C but with last row − γm γm−1 · · · γ1 , where k i = ki − γ i ,

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

Since C + dk = T (A + bkT )T −1 it follows that the desired matrix is K = kT

(3.10)

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AM3.2 - Linear Control

the entries ki (i = 1, 2, . . ., m) being given by (3.10). In this equation ki (i = 1, 2, . . ., m) are the coefficients in the characteristic polynomial of A ; that is, det (λIm − A) = λm + k1 λm−1 + · · · + km and γi (i = 1, 2, . . ., m) are obtained by equating coefficients of λ in λm + γ1 λm−1 + · · · + γm ≡ (λ − θ1 )(λ − θ2 ) · · · (λ − θm ). 2 Note :

The solution of (the closed loop system) x˙ = (A + BK)x

depends on the eigenvalues of A + BK, so provided the control system Σ is c.c., the theorem tells us that using linear feedback it is possible to exert a considerable influence on the time behaviour of the closed loop system by suitably choosing the numbers θ1 , θ2 , . . . , θm .

If the linear time-invariant control system x˙ = Ax + Bu(t)

3.3.2 Corollary.

y = cx

is c.o., then there exists a matrix L ∈ Rm×1 such that the eigenvalues of A + Lc are the set Λ. This result can be deduced from Theorem 3.3.1 using the Duality Theorem. 3.3.3 Example.

Consider the linear control system " # " # 1 −3 1 x˙ = x+ u(t) . 4 2 1

93

C.C. Remsing

The characteristic equation of A is charA (λ) ≡ λ2 − 3λ + 14 = 0 which has roots

√ 3±i 47 . 2

Suppose we wish the eigenvalues of the closed loop system to be −1 and −2, so that the characteristic polynomial is λ2 + 3λ + 2 . We have k1 = k1 − γ1 = −3 − 3 = −6 k2 = k2 − γ2 = 14 − 2 = 12. Hence

i 1h K = kT = 12 −6 8

"

−1 1

"

−11 −21

3 5

#

=−

h

15 4

9 4

i

.

It is easy to verify that 1 A + bK = 4

1

−1

#

does have the desired eigenvalues. 3.3.4 Lemma.

If the linear control system Σ defined by

h is c.c. and B = b1 b2 · · ·

x˙ = Ax + Bu(t) i b` with bi 6= 0, i = 1, 2, . . ., `, then there

exist matrices Ki ∈ R`×m , i = 1, 2, . . ., ` such that the systems x˙ = (A + BKi )x + bi u(t)

are c.c. Proof :

For convenience consider the case i = 1. Since the matrix h i C = B AB A2 B . . . Am−1 B

94

AM3.2 - Linear Control

has full rank, it is possible to select from its columns at least one set of m vectors which are linearly independent. Define an m × m matrix M by choosing such a set as follows : h M = b1 Ab1 . . .

Ar1 −1 b1 b2 Ab2 . . .

i Ar2 −1 b2 . . .

where ri is the smallest integer such that Ari bi is linearly dependent on all the preceding vectors, the process continuing until m columns of U are taken. Define an ` × m matrix N having its r1th column equal to e2 , the second column of I` , its (r1 + r2 )th column equal to e3 , its (r1 + r2 + r3 )th column equal to e4 and so on, all its other columns being zero. It is then not difficult to show that the desired matrix in the statement of the Lemma is K1 = N M −1 . 2 Proof of Theorem 3.3.1 when ` > 1 : Let K1 be the matrix in the proof of Lemma 3.3.4 and define an ` × m matrix K 0 having as its first row some vector k, and all its other rows zero. Then the control u = (K1 + K 0 )x leads to the closed loop system x˙ = (A + BK1 )x + BK 0 x = (A + BK1 )x + b1 kx where b1 is the first column of B. Since the system x˙ = (A + BK1 )x + b1 u is c.c., it now follows from the proof of the theorem when ` = 1, that k can be chosen so that the eigenvalues of A + BK1 + b1 k are the set Λ, so the desired feedback control is indeed u = (K1 + K 0 )x.

2

95

C.C. Remsing

If y = Cx is the output vector, then again by duality we can immediately deduce 3.3.5 Corollary.

If the linear control system x˙ = Ax + Bu(t) y = Cx

is c.o., then there exists a matrix L ∈ Rm×n such that the eigenvalues of A + LC are the set Λ. Algorithm for constructing a feedback matrix The following method gives a practical way of constructing the feedback matrix K. Let all the eigenvalues λ1 , λ2, . . . , λm of A be distinct and let h i W = w1 w2 . . . wm

where wi is an eigenvector corresponding to the eigenvalue λi. With linear feedback u = −Kx, suppose that the eigenvalues of A and A − BK are ordered so that those of A − BK are to be µ1 , µ2 , . . . , µr , λr+1 , . . . , λm

(r ≤ m).

Then provided the linear system Σ is c.c., a suitable matrix is f K = fg W

f consists of the first r rows of W −1 , and where W αr α1 α2 ··· g = β1 β2 βr r Y (λi − µj ) j=1 r if r > 1 Y αi = (λi − λj ) j=1 j6=i λ1 − µ1 if r = 1 h iT f Bf β = β1 β2 . . . βr = W

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AM3.2 - Linear Control

f being any column `-vector such that all βi 6= 0. 3.3.6 Example.

Consider the linear system " # " # 0 1 2 x˙ = x+ u(t) . −2 −3 1

We have

W =

"

λ1 = −1, # 1 1

−1 −2

λ2 = −2 W −1 =

,

and " 2

1

−1 −1

#

.

Suppose that µ1 = −3,

µ2 = −4,

We have α1 = 6, f Bf gives and β = W " # β1 β2

=

"

2

1

−1 −1

f = W −1 . so W

α2 = −2 #"

2 1

#

f=

"

5f1 −3f1

#

Hence we can take f1 = 1, which results in i h g = 56 23 .

Finally, the desired feedback matrix is i h h K = 1 · 56 23 W −1 = 3.3.7 Example.

We now obtain

26 15

8 15

i

.

Consider now the linear control system " # " # 0 1 2 1 x˙ = x+ u(t) . −2 −3 1 0 "

β1 β2

#

=

"

5f1 + 2f2 −3f1 − f2

#

.

97

C.C. Remsing

so that f1 = 1,

f2 = 0 gives

However f1 = 1,

K=

26 15

8 15

0

0

f2 = −1 gives β1 = 3,

β2 = −2

.

so that g =

h

2 1

i

f we now have and from K = f g W " # " # i 1 h 3 1 K= . 2 1 W −1 = −1 −3 −1

3.4

Realization Theory

The realization problem may be viewed as “guessing the equations of motion (i.e. state equations) of a control system from its input/output behaviour” or, if one prefers, “setting up a physical model which explains the experimental data”. Consider the linear control system (with outputs) Σ described by x˙ = Ax + Bu(t)

(3.11)

y = Cx

where A ∈ Rm×m , B ∈ Rm×` and C ∈ Rn×m . Taking Laplace transforms of (3.11) and assuming zero initial conditions gives sx(s) = Ax(s) + Bu(s) and after rearrangement x(s) = (sIm − A)−1 Bu(s).

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AM3.2 - Linear Control

The Laplace transform of the output is y(s) = Cx(s) and thus y(s) = C (sIm − A)−1 Bu(s) = G(s)u(s) where the n × ` matrix G(s) : = C (sIm − A)−1 B

(3.12)

is called the transfer function matrix since it relates the Laplace transform of the output vector to that of the input vector. Exercise 41 Evaluate (the Laplace transform of the exponential) L eat (s) : =

Z

∞

e−st eat dt

0

and then show that (for A ∈ Rm×m ) :

−1

L [exp(tA)] (s) = (sIm − A)

.

Using relation (sIm − A)−1 =

sm−1 Im + sm−2 B1 + sm−3 B2 + · · · + Bm−1 charA (s)

where the ki and Bi are determined successively by B1 = A + k1 Im , k1 = −tr (A),

Bi = ABi−1 + ki Im ; ki = − 1i tr (ABi−1 ) ;

i = 2, 3, . . ., m − 1

i = 2, 3, . . . , m

the expression (3.12) becomes G(s) =

sm−1 G0 + sm−2 G1 + · · · + Gm−1 H(s) = χ(s) χ(s)

(3.13)

99

C.C. Remsing

h i (k) where χ(s) = charA (s) and Gk = gij ∈ Rn×` ,

k = 0, 1, 2, . . ., m−1. The

n × ` matrix H(s) is called a polynomial matrix, since each of its entries is itself a polynomial; that is, (0)

(1)

(m−1)

hij = sm−1 gij + sm−2 gij + · · · + gij Note :

.

The formulas above, used mainly for theoretical rather than computational

purposes, constitute Leverrier’s algorithm.

3.4.1 Example.

Consider the electrically-heated oven described in section

1.3, and suppose that the values of the constants are such that the state equations are x˙ =

"

−2

2

1 −1

#

x+

"

1 0

#

u(t).

Suppose that the output is provided by a thermocouple in the jacket measuring the jacket (excess) temperature, i.e. h i y = 1 0 x. The expression (3.12) gives G(s) =

h

1 0

i

"

s+2

−2

−1 s + 1

using (sI2 − A)−1 =

#−1 "

1 0

#

=

s+1 + 3s

s2

1 adj (sI2 − A). charA (s)

Realizations In practice it often happens that the mathematical description of a (linear time-invariant) control system – in terms of differential equations – is not known, but G(s) can be determined from experimental measurements or other considerations. It is then useful to find a system – in our usual state space form – to which G(·) corresponds.

100

AM3.2 - Linear Control

In formal terms, given an n × ` matrix G(s), whose elements are rational functions of s, we wish to find (constant) matrices A, B, C having dimensions m × m, m × ` and n × m, respectively, such that G(s) = C(sIm − A)−1 B and the system equations will then be x˙ = Ax + Bu(t) y = Cx.

The triple (A, B, C) is termed a realization of G(·) of order m, and is not, of course, unique. Amongst all such realizations some will include matrices A having least dimensions – these are called minimal realizations, since the corresponding systems involve the smallest possible number of state variables. Note :

Since each element in (sIm − A)−1 =

adj (sIm − A) det (sIm − A)

has the degree of the numerator less than that of the denominator, it follows that lim C(sIm − A)−1 B = 0

s→∞

and we shall assume that any given G(s) also has this property, G(·) then being termed strictly proper.

3.4.2 Example.

Consider the scalar transfer function g(s) =

s2

2s + 7 · − 5s + 6

It is easy to verify that one realization of g(·) is " # " # h i 0 1 0 A= , b= , c= 7 2 . −6 5 1 It is also easy to verify that a quite different triple is " # " # h i 2 0 1 , b= , c = −11 13 . A= 0 3 1

101

C.C. Remsing

Note :

Both these realizations are minimal, and there is in consequence a simple

relationship between them, as we shall see later.

Algebraic equivalence and realizations It is now appropriate to return to the idea of algebraic equivalence of linear control systems (defined in section 3.1), and discuss its implications for the realization problem. The (linear) transformation x e = Px

produces a linear control system with matrices e = P AP −1 , A

e = P B, B

e = CP −1 . C

(3.14)

Exercise 42 Show that if (A, B, C) represents a c.c. (or c.o.) linear control system, e B, e C). e then so does (A,

Exercise 43 Show that if two linear control systems are algebraically equivalent, then their transfer function matrices are identical (i.e. e m − A) e −1 B). e C(sIm − A)−1 B = C(sI

Characterization of minimal realizations We can now state and prove the central result of this section, which links together the three basic concepts of controllability, observability, and realization. 3.4.3 Theorem.

A realization (A, B, C) of a given transfer function ma-

trix G(·) is minimal if and only if the pair (A, B) is c.c. and the pair (A, C) is c.o.

102

Proof :

AM3.2 - Linear Control

(⇐) Sufficiency. Let C and O be the controllability and ob-

servability matrices, respectively; that is,

h C = C(A, B) = B AB A2 B . . .

Am−1 B

i

and

C

CA 2 O = O(A, C) = CA . . .. m−1 CA

We wish to show that if these both have rank m, then (the realization of) e B, e C) e of G(·) has least order m. Suppose that there exists a realization (A,

e having order m. G(·) with A e Since

e m − A) e −1 B e C(sIm − A)−1 B = C(sI it follows that

e exp(tA) eB e C exp(tA)B = C which implies that

eA ei B e, CAi B = C

i = 0, 1, 2, . . ..

103

C.C. Remsing

Consider the product

C

CA h OC = B AB .. . CAm−1 CB CAB CAB CA2 B = .. .. . . m−1 CA B CAm B e C C eA e h e ee = B AB .. . eA em−1 C

...

An−1 B

i

...

CAm−1 B

...

CAm B .. .

...

CA2m−2 B

...

em−1 B e A

i

eC. e = O

Exercise 44 Given A ∈ R`×m and B ∈ Rm×n , show that if rank (A) = rank (B) = m, then rank (AB) = m. [Hint : Use the results of Exercise 3.]

eCe also has rank m. However, The matrix OC has rank m, so the matrix O eCe cannot be greater than m. the rank of O e That is, m ≤ m, e so there can be

no realization of G(·) having order less than m.

(⇒) Necessity. We show that if the pair (A, B) is not completely controllable, then there exists a realization of G(·) having order less than m. The corresponding part of the proof involving observability follows from duality. Let the rank of C be m1 < m and let u1 , u2 , . . . , um1 be any set of m1 linearly independent columns of C. Consider the (linear) transformation x e = Px

with the m × m matrix P defined by h P −1 = u1 u2 . . . um1

um1 +1 . . .

um

i

(3.15)

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AM3.2 - Linear Control

where the columns um1 +1 , . . . , um are any vectors which make the matrix P −1 nonsingular. Since C has rank m1 it follows that all its columns can be expressed as a linear combination of the basis u1 , u2 , . . . , um1 . The matrix h i AC = AB A2 B . . . Am B contains all but the first ` columns of C, so in particular it follows that the vectors Aui , i = 1, 2, . . ., m1 can be expressed in terms of the same basis. Multiplying both sides of (3.15) on the left by P shows that P ui is equal to the ith column of Im . Combining these facts together we obtain e = P AP −1 A h = P Au1 . . . " # A1 A2 = 0 A3

Aum1

...

Aum

i

where A1 is m1 × m1 . Similarly, since u1 , u2 , . . . , um1 also forms a basis for the columns of B we have from (3.14) and (3.15) " # e = P B = B1 B 0

where B1 is m1 × `. Writing

h i e = CP −1 = C1 C2 C

we have (see Exercise 43 and also Exercise 60)

e m − A) e −1 B e G(s) = C(sI " #−1 " # h i sI − A −A2 B1 m1 1 = C1 C2 0 sIm−m1 − A3 0 " #" # h i (sI − A )−1 (sI − A )−1 A (sI −1 − A ) B m1 1 m1 1 2 m−m1 3 1 = C1 C2 −1 0 (sIm−m1 − A3 ) 0 = C1 (sIm1 − A1 )−1 B1

105

C.C. Remsing

showing that (A1 , B1 , C1 ) is a realization of G(·) having order m1 < m. This contradicts the assumption that (A, B, C) is minimal, hence the pair (A, B) must be c.c.

2

We apply the procedure introduced in the second part of

3.4.4 Example.

the proof of Theorem 3.4.5 to split up the linear control system

4

3

5

2

x˙ = 1 −2 −3 x + 1 u(t) 2 1 8 −1

(3.16)

into its controllable and uncontrollable parts, as displayed below : "

x˙ (1) x˙ (2)

#

=

"

A1 A2 0 A3

#"

x(1) x(2)

#

+

"

B1 0

#

u(t)

where x(1) , x(2) have orders m1 and m − m1 , respectively, and (A1 , B1 ) is c.c. The controllability matrix for (3.16) is h

B AB A2 B

i

=

2

6

1

3

18

9 −1 −3 −9

which clearly has rank m1 = 1. For the transformation x e = P x we follow

(3.15) and set

P −1 =

2 1 0

1 0 1 −1 0 0

where the column in (3.16) has been selected, and the remaining columns are simply arbitrary choices to produce a nonsingular matrix. It is then easy to

106

AM3.2 - Linear Control

compute the inverse of the matrix above, and from (3.14) " # 3 −2 −1 A A 1 2 −1 e = P AP = 0 A 8 5 = 0 A3 0 3 −1 " # 1 B 1 e = PB = 0 = B . 0 0

Notice that the transformation matrix is not unique. However, all possible e will be similar to 3×3 matrix above. In particular, the eigenvalues matrices A of the uncontrollable part are those of A3 , namely the roots of λ − 8 −5 0 = det (λI2 − A3 ) = = λ2 − 7λ − 23 −3 λ + 1

and these roots cannot be altered by applying linear feedback to (3.16). Note :

For any given transfer function matrix G(·) there are an infinite number

of minimal realizations satisfying the conditions of Theorem 3.4.3. However, one can show that the relationship between any two minimal realizations is just that of algebraic equivalence : If R = (A, B, C) is a minimal realization of G(·), then e = (A, e B, e C) e is also a minimal realization if and only if the following holds : R

e = P AP −1 , A

e B, BP

e = CP −1 . C

Algorithm for constructing a minimal realization We do not have room to discuss the general problem of efficient construction of minimal realizations. We will give here one simple but nevertheless useful result. 3.4.5 Proposition. Rn×`

Let the denominators of the elements gij (s) of G(s) ∈

have simple roots s1 , s2 , . . . , sq . Define Ki : = lim (s − si )G(s), s→si

i = 1, 2, . . ., q

107

C.C. Remsing

and let ri : = rank (Ki ),

i = 1, 2, . . . , q.

If Li and Mi are n × ri and ri × ` matrices, respectively, each having rank ri such that Ki = Li Mi then a minimal realization of G(·) is s1 Ir1 O M1 M2 s2 Ir2 , B = A= . , .. .. . O sq Irq Mq

h C = L1 L2 . . .

i Lq .

(To verify that (A, B, C) is a realization of G(·) is straightforward. Indeed,

C(sIm − A)−1 B =

=

h

L1 · · ·

i Lq

1 s−s1 Ir1

··· .. .

0

···

M1 . .. 1 Mq s−sq Irq 0

L1 M1 Lq Mq K1 Kq +···+ = +···+ = G(s). s − s1 s − sq s − s1 s − sq

Since rank C(A, B) = rank O(A, C) = m, this realization is minimal.) Consider the scalar transfer function

3.4.6 Example.

g(s) =

s2

2s + 7 · − 5s + 6

We have (s − 2)(2s + 7) = −11 , r1 = 1 s→2 (s − 2)(s − 3) (s − 3)(2s + 7) = lim = 13 , r2 = 1. (s − 2)(s − 3)

K1 = lim K2 Taking L1 = K 1 ,

M1 = 1,

L2 = K 2 ,

M2 = 1

108

AM3.2 - Linear Control

produces a minimal realization A=

"

2 0 0 3

However, b=

"

#

b=

,

#

m1 m2

"

#

1 1

and

c=

h

c=

,

h

11 −m 1

i

−11 13

13 m2

.

i

can be used instead, still giving a minimal realization for arbitrary nonzero values of m1 and m2 .

3.5

Exercises

Exercise 45 Verify that the control system described by "

x˙ 1 x˙ 2

#

=

"

0

1

0

0

#"

x1 x2

#

+

"

0

0

1

−1

#"

u1 (t) u2 (t)

#

is c.c.

Exercise 46 Given the control system described by x˙ =

"

#

−1 −1 2 −4

x + bu(t)

find for what vector b the system is not c.c.

Exercise 47 For the (initialized) control system x˙ =

"

−4

2

4 −6

#

x+

"

1 2

#

u(t) ,

x(0) =

"

2 3

#

apply a control in the form u(t) = c1 + c2 e−2t so as to bring the system to the origin at time t = 1. Obtain, but do not solve, the equations which determine the constants c1 and c2 .

109

C.C. Remsing

Exercise 48 For each of the following cases, determine for what values of the real parameter α the control system is not c.c. −1 1 −1 0 (1) x˙ = α 0 −1 x + 2 u(t) ; 0 1 3 1 (2) x˙ =

"

#

2 α−3 0

2

x+

"

1

1

0

α2 − α

#

u(t).

In part (2), if the first control variable u1 (·) ceases to operate, for what additional values (if any) of α the system is not c.c. under the remaining scalar control u2 (·) ?

Exercise 49 Consider the control system defined by "

x˙ 1 x˙ 2

#

=

"

−1

0

0 −2

#"

x1 x2

#

+

"

1 3

#

u(t).

Such a system could be thought of as a simple representation of a vehicle suspension system: in this interpretation, x1 and x2 are the displacements of the end points of the platform from equilibrium. Verify that the system is c.c. If the ends of the platform are each given an initial displacement of 10 units, find using u∗ (t) = −B T (t)ΦT (t0 , t)Wc−1 (t0 , t1 )[x0 − Φ(t0 , t1 )xf ] a control function which returns the system to equilibrium at t = 1.

Exercise 50 Prove that U (t0 , t1 ) defined by U (t0 , t1 ) =

Z

t1

Φ(t0 , τ )B(τ )B T (τ )ΦT (t0 , τ )dτ

t0

satisfies the matrix differential equation ˙ t1 ) = A(t)U (t, t1 ) + U (t, t1 )AT (t) − B(t)B T (t), U(t,

U (t1 , t1 ) = 0.

Exercise 51 In the preceding exercise let A and B be time-invariant, and put W (t, t1 ) = U (t, t1 ) − U0 where the constant matrix U0 satisfies AU0 + U0 AT = BB T .

110

AM3.2 - Linear Control

Write down the solution of the resulting differential equation for W using the result in Exercise 32 and hence show that U (t, t1 ) = U0 − exp ((t − t1 )A) U0 exp (t − t1 )AT .

Exercise 52 Consider again the rabbit-fox environment problem described in section 1.3 (see also, Exercise 29). If it is possible to count only the total number of animals, can the individual numbers of rabbits and foxes be determined ?

Exercise 53 For the system (with outputs) x˙ =

"

−1

−1

2

−4

#

x,

y=

h

1 2

i

x

find x(0) if y(t) = −20e−3t + 21e−2t .

Exercise 54 Show that the control system described by x˙ 1 = x2 ,

x˙ 2 = −2x1 − 3x2 + u,

y = x1 + x2

is not c.o. Determine initial states x(0) such that if u(t) = 0 for t ≥ 0, then the output y(t) is identically zero for t ≥ 0.

Exercise 55 Prove that V (t0 , t1 ) defined by V (t0 , t1 ) =

Z

t1

ΦT (τ, t0 )C T (τ )C(τ )Φ(τ, t0 )dτ

t0

satisfies the matrix differential equation V˙ (t, t1 ) = −AT (t)V (t, t1 ) − V (t, t1 )A(t) − C T (t)C(t),

V (t1 , t1 ) = 0.

Exercise 56 Consider the time-invariant control system x˙ =

"

−1 −1 2 −4

#

x+

"

1 3

#

u(t).

Find a 1 × 2 matrix K such that the closed loop system has eigenvalues −4 and −5.

111

C.C. Remsing

Exercise 57 For the time-invariant control system

1 0

−1

x˙ = 1 2

1

1 x + 0 u(t) 3 1

2 2

find a suitable matrix K so as to make the closed loop eigenvalues −1, −1 ± 2i.

Exercise 58 Given the time-invariant control system x˙ = Ax + Bu(t) where

0

1 0

0 1 , −11 6

A= 0 6

1 0

B= 0 1 1 1

find a suitable matrix K which makes the eigenvalues of A − BK equal to 1, 1, 3.

Exercise 59 Determine whether the system described by

x˙ =

−1

0

3

1

0 x + 1 u(t) 0 −3 −1

0 −3 1

is c.c. Show that under (linear) feedback of the form u = αx1 + βx3 , the closed loop system has two fixed eigenvalues, one of which is equal to −3. Determine the second fixed eigenvalue, and also values of α and β such that the third closed loop eigenvalue is equal to −4.

Exercise 60 Show that if X =

"

A B 0

C

#

is a block matrix with A and C invert-

ible, then X is invertible and X

−1

=

" A−1

−A−1 BC −1

0

C −1

#

.

Exercise 61 Use Proposition 3.4.5 to obtain a minimal realization of 1 G(s) = g(s) where g(s) = s3 + 2s2 − s − 2.

"

(s2 + 6)

(s2 + s + 4)

(2s2 − 7s − 2)

(s2 − 5s − 2)

#

112

AM3.2 - Linear Control

Exercise 62 Show that the order of a minimal realization of 1 G(s) = 2 s + 3s + 2

"

(s + 2)

2(s + 2)

−1

(s + 1)

#

is three. (Notice the fallacy of assuming that the order is equal to the degree of the common denominator.)

Exercise 63 If (A1 , B1 , C1 ) and (A2 , B2 , C2 ) are realizations of G1 (·) and G2 (·), respectively, show that " A1 A= 0

B1 C2 A2

#

B=

,

"

#

0 B2

C=

,

h

C1

0

i

is a realization of G1 (·)G2 (·), assuming that this product exists.

Exercise 64 Verify that algebraic equivalence e = P AP −1, A

e = P B, B

e = CP −1 C

can be written as the transformation " #" #" P 0 sIn − A B P −1 0 In

−C

0

#

0

0 In

=

"

e B e sIn − A −Ce 0

#

.

Exercise 65 Determine values of b1 , b2 , c1 and c2 such that R= and e= R

"

−2

"

0

1

−6

−5

0

0 −3

# #

,

"

b1

,

"

0

b2

1

#

#

h

,

,

h

1

c1

!

1

i

c2

i

!

are realizations of the transfer function g(s) =

s+4 · s2 + 5s + 6

Determine a matrix P such that the algebraic equivalence relationship holds between the two realizations.

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