Modern Control Engineering

622 Chapter 8 / PID Controllers and Modified PID Controllers

The closed-loop transfer function is given by

The closed-loop poles are located at and s=–0.3333.A unit-step response curve

is shown in Figure 8–51. The closed-loop pole ats=–0.3333and a zero ats=–0.5714produce

a long tail of small amplitude.

A–8–7. Consider the system shown in Figure 8–52. Design a compensator such that the static velocity

error constant is 4 sec−^1 , phase margin is 50°, and gain margin is 10 dB or more. Plot unit-step and

unit-ramp response curves of the compensated system with MATLAB. Also, draw a Nyquist plot

of the compensated system with MATLAB. Using the Nyquist stability criterion, verify that the

designed system is stable.

Solution.Since the plant does not have an integrator, it is necessary to have an integrator in the

compensator. Let us choose the compensator to be

where is to be determined later. Since the static velocity error constant is specified as

4 sec−^1 , we have

Kv=limsS 0 sGc(s)

s+0.1

s^2 + 1

=limsS 0 s

K

s

Gˆc(s)

s+0.1

s^2 + 1

=0.1K= 4

Gˆc(s)

Gc(s)=

K

s

Gˆc(s), limsS 0 Gˆc(s)= 1

s=- 1 ;j 13

C(s)

R(s)

=

2.3333(s+1)(s+0.5714)

s^3 +s+2.3333(s+1)(s+0.5714)

Time (sec)

082 4 6 10 12

Amplitude

0.4

0.8

1.2

0.6

1

0.2

0

Unit-Step Response of Compensated System

Figure 8–51

Unit-step response of

the compensated

system.

Gc(s) s+ 0.1

s^2 + 1

+−

Figure 8–52

Control system.

Openmirrors.com