Modern Control Engineering

Example Problems and Solutions 715

Since the solution of Equation (9–114) is

att=T, we have

(9–117)

Substituting Equation (9–117) into Equation (9–116), we obtain

(9–118)

On the other hand,y(0)=Cx(0). Notice that the complete output controllability means that the
vectorCx(0)spans the m-dimensional output space. Since eATis nonsingular, if Cx(0)spans the
m-dimensional output space, so does CeATx(0), and vice versa. From Equation (9–118) we obtain

Note that can be expressed as the sum of AiBj; that is,

where

andai(t)satisfies

(p: degree of the minimal polynomial of A)

andBjis the jth column of B. Therefore, we can write CeATx(0)as

From this last equation, we see that is a linear combination of CAiBj (i=0, 1, 2,p,
p-1; j=1, 2,p,r). Note that if the rank of Q, where

ism, then so is the rank of P, and vice versa. [This is obvious if p=n. If p<n, then the CAhBj
(wherephn-1)are linearly dependent on CBj,CABj,p,CAp-1Bj. Hence, the rank of

Q=CCB  CAB  CA^2 B  p  CAp-^1 BD (pn)

CeAT x(0)

CeAT x(0)=-a

p- 1

i= 0 a

r

j= 1

gij CAi Bj

eAt= a

p- 1

i= 0

ai(t) Ai

gij=

3

T

0

ai(t)uj(T-t)dt=scalar

3

T

0

eAt Bu(T-t)dt= a

p- 1

i= 0 a

r

j= 1

gij Ai Bj

1

T

0 e

At Bu(T-t)dt

=- C

3

T

0

eAt Bu(T-t)dt

CeAT x(0)=- CeAT

3

T

0

e-At Bu(t)dt

=CeATcx(0)+

3

T

0

e-^ At Bu(t)dtd= 0

y(T)=Cx(T)

x(T)=eATcx(0)+

3

T

0

e-^ At Bu(t)dtd

x(t)=eAtcx( 0 )+

3

t

0

e-^ At Bu(t)dtd