Modern Control Engineering

846 Chapter 10 / Control Systems Design in State Space

Noting that Ais a stable matrix and, therefore, we obtain

Let us put

Note that the elements of are finite sums of terms like where the liare

the eigenvalues of Aandmiis the multiplicity of li.Since the lipossess negative real parts,

exists. Note that

Thus Pis Hermitian (or symmetric if Pis a real matrix). We have thus shown that for a stable A

and for a positive-definite Hermitian matrix Q, there exists a Hermitian matrix Psuch that

We now need to prove that Pis positive definite. Consider the following Her-

mitian form:

Hence,Pis positive definite. This completes the proof.

A–10–16. Consider the control system described by

(10–173)

where

Assuming the linear control law

(10–174)

determine the constants k 1 andk 2 so that the following performance index is minimized:

J=

3

q

0

xT xdt

u=-Kx=-k 1 x 1 - k 2 x 2

A= B

0

0

1

0

R, B=B

0

1

R

x# =Ax+Bu

=0, for x= 0

=

3

q

0

AeAt xB* QAeAt xBdt 7 0, for xZ 0

x* Px=x*

3

q

0

eA*^ t QeAtdtx

A* P+PA=-Q.

P*=

3

q

0

eA*^ t QeAtdt=P

3

q

0

eA*^ t QeAtdt

eAt elit^ ,teli^ tp,tmi-^1 elit^ ,

P=

3

q

0

Xdt=

3

q

0

eA*^ t QeAtdt

• X( 0 )=-Q=A*a
  3

q

0

Xdtb+ a

3

q

0

Xdtb A

X(q)= 0 ,

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