Modern Control Engineering

Example Problems and Solutions 619

Clearly, for a unit-step disturbance input, the steady-state output is zero, since

The response to a unit-step disturbance input can be obtained easily with MATLAB. MATLAB
Program 8–9 produces a response curve as shown in Figure 8–48(a). From the response curve, we
see that the settling time is approximately 2.7 sec. The response damps out quickly. Therefore, the
system designed here is acceptable.

tlimSqcd(t)=limsS 0 sCd(s)=slimS 0

s^2

(s+10)As^2 +4s+ 16 B

1

s

= 0

MATLAB Program 8–9

% Response to unit-step disturbance input

numd = [1 0];

dend = [1 14 56 160];

t = 0:0.01:5;

[c1,x1,t] = step(numd,dend,t);

plot(t,c1)

grid

title('Response to Unit-Step Disturbance Input')

xlabel('t Sec')

ylabel('Output to Disturbance Input')

% Response to unit-step reference input

numr = [10.4 47 160];

denr = [1 14 56 160];

[c2,x2,t] = step(numr,denr,t);

plot(t,c2)

grid

title('Response to Unit-Step Reference Input')

xlabel('t Sec')

ylabel('Output to Reference Input')

For the reference inputr(t),the closed-loop transfer function is

The response to a unit-step reference input can also be obtained by use of MATLAB Program 8–9.
The resulting response curve is shown in Figure 8–48(b). The response curve shows that the max-
imum overshoot is 7.3%and the settling time is 1.2 sec. The system has quite acceptable response
characteristics.

=

10.4s^2 +47s+ 160

s^3 +14s^2 +56s+ 160

Cr(s)

R(s)

=

10.4As^2 +4.5192s+15.385B

s^3 +14s^2 +56s+ 160