Modern Control Engineering

Example Problems and Solutions 559

Since the phase margin of the compensated system is 50°, the gain margin is 12 dB, and the

static velocity error constant is 10 sec–1, all the requirements are met.

We shall next investigate the transient-response characteristics of the designed system.

Unit-Step Response: Noting that

we have

To determine the denominator polynomial with MATLAB, we may proceed as follows:

Define

c(s)=40(s+0.4)(s+0.2)=40s^2 +24s+3.2

b(s)=s(s+1)(s+4)=s^3 +5s^2 +4s

a(s)=(s+4)(s+0.02)=s^2 +4.02s+0.08

=

40(s+0.4)(s+0.2)

(s+4)(s+0.02)s(s+1)(s+4)+40(s+0.4)(s+0.2)

C(s)

R(s)

=

Gc(s)G(s)

1 +Gc(s)G(s)

Gc(s)G(s)=

40(s+0.4)(s+0.2)

(s+4)(s+0.02)s(s+1)(s+4)

MATLAB Program 7–32

num1 = [40 24 3.2];

den1 = [1 9.02 24.18 16.48 0.32 0];

bode(num1,den1)

title('Bode Diagram of Gc(s)G(s)')

Frequency (rad/sec)

Bode Diagram of Gc(s)G(s)

− 300

− 250

− 200

− 150

− 100

− 50

0

− 100

Phase (deg); Magnitude (dB)

− 50

0

50

100

10 −^410 −^310 −^210 −^1100101102

Figure 7–153
Bode diagram of the
open-loop transfer
functionGc(s)G(s)
of the compensated
system.