Example Problems and Solutions 839MATLAB Program 10–27
% Response to initial condition
A = [0 0 1 0;0 0 0 1;-36 36 -0.6 0.6;18 -18 0.3 -0.3];
B = [0;0;1;0];
K = [130.4444 -41.5556 23.1000 15.4185];
Ke = [14.4 0.6;0.3 15.7];
F = [0 0;0 0;1 0;0 1];
Aab = [1 0;0 1];
Abb = [-0.6 0.6;0.3 -0.3];
AA = [A-BK BKF; zeros(2,4) Abb-KeAab];
sys = ss(AA,eye(6),eye(6),eye(6));
t = 0:0.01:4;
y = initial(sys,[0.1;0;0;0;0.1;0.05],t);
x1 = [1 0 0 0 0 0]*y';
x2 = [0 1 0 0 0 0]*y';
x3 = [0 0 1 0 0 0]*y';
x4 = [0 0 0 1 0 0]*y';
e1 = [0 0 0 0 1 0]*y';
e2 = [0 0 0 0 0 1]*y';
subplot(3,2,1); plot(t,x1); grid; title('Response to initial condition'),
xlabel('t (sec)'); ylabel('x1')
subplot(3,2,2); plot(t,x2); grid; title('Response to initial condition'),
xlabel('t (sec)'); ylabel('x2')
subplot(3,2,3); plot(t,x3); grid; xlabel('t (sec)'); ylabel('x3')
subplot(3,2,4); plot(t,x4); grid; xlabel('t (sec)'); ylabel('x4')
subplot(3,2,5); plot(t,e1); grid; xlabel('t (sec)');ylabel('e1')
subplot(3,2,6); plot(t,e2); grid; xlabel('t (sec)'); ylabel('e2')
r=0 y–y Observer u
+– controller
4
s(s+ 2)Figure 10–51 Plant
Regulator system.
A–10–14. Consider the system shown in Figure 10–51. Design both the full-order and minimum-order observers
for the plant. Assume that the desired closed-loop poles for the pole-placement part are located atAssume also that the desired observer poles are located at
(a)s=–8, s=–8for the full-order observer
(b)s=–8for the minimum-order observer
Compare the responses to the initial conditions specified below:
(a) for the full-order observer:
x 1 (0)=1, x 2 (0)=0, e 1 (0)=1, e 2 (0)= 0s=- 2 +j2 13 , s=- 2 - j2 13