Modern Control Engineering

Example Problems and Solutions 625

The open-loop transfer function of the designed system is

Open-loop transfer function

We shall next check the unit-step response and the unit-ramp response of the designed system.

The closed-loop transfer function is

The closed-loop poles are located at

s=3.0032+j5.6573

s=3.0032-j5.6573

s=0.0975

MATLAB Program 8–12 will produce the unit-step response curve of the designed system. The re-

sulting unit-step response curve is shown in Figure 8–55. Notice that the closed-loop pole at

s=−0.0975 and the plant zero at s=−0.1 produce a long tail of small amplitude.

C(s)

R(s)

=

6.104s^2 +40.6104s+ 4

s^3 +6.104s^2 +41.6104s+ 4

=

6.104s^2 +40.6104s+ 4

s(s^2 + 1 )

=

40 (0.1526s+ 1 )

s

s+0.1

s^2 + 1

MATLAB Program 8–12

% Unit-Step Response

num = [6.104 40.6104 4];

den = [1 6.104 41.6104 4];

t = 0:0.01:10;

step(num,den,t)

grid

Figure 8–55
Unit-step response of
C(s)/R(s)=(6.104s^2 +
40.6104s+4)/(s^3 +
6.104s^2 +41.6104s+4). Time (sec)

0 1 2 3 4 5 6 798 10

Amplitude

1.4

0.8

0

1.2

0.4

0.2

1

0.6

Step Response