Modern Control Engineering

Example Problems and Solutions 845

The observer controller is clearly a lead compensator.

The Bode diagrams of System 1 (closed-loop system with full-order observer) and of Sys-

tem 2 (closed-loop system with minimum-order observer) are shown in Figure 10–54. Clearly, the

bandwidth of System 2 is wider than that of System 1. System 1 has a better high-frequency noise-

rejection characteristic than System 2.

A–10–15. Consider the system

wherexis a state vector (n-vector) and Ais an n*nconstant matrix. We assume that Ais non-

singular. Prove that if the equilibrium state x= 0 of the system is asymptotically stable (that is, if

Ais a stable matrix), then there exists a positive-definite Hermitian matrix Psuch that

whereQis a positive-definite Hermitian matrix.

Solution.The matrix differential equation.

has the solution

Integrating both sides of this matrix differential equation from t=0tot=q, we obtain

X(q)-X( 0 )=A*a

3

q

0

Xdtb+ a

3

q

0

Xdtb A

X=eA*^ t QeAt

X

=A* X+XA, X( 0 )=Q

A* P+PA=-Q

x# =Ax

Frequency (rad/sec)

Bode Diagrams of Systems

− 300

− 100

− 200

− 250

− 50

− 150

0

− 100

− 50

Phase (deg); Magnitude (dB)

50

0

10 −^1100101102

System 1

System 2

System 1

System 2

Figure 10–54
Bode diagrams of System 1
(system with full-order
observer) and System 2
(system with minimum-
order observer).
System1=
(296s+1024)
(s^4 +20s^3 +144s^2
+512s+1024);
System2= (28s+128)
(s^3 +12s^2 +48s+128).