Modern Control Engineering

Example Problems and Solutions 551

The lead compensator thus determined is

whereKcis determined as

Thus, the transfer function of the compensator becomes

MATLAB Program 7–26 produces the Bode diagram of this lead compensator, which is shown

in Figure 7–147.

Gc(s)=9.5238

s+3.0101

s+14.3339

= 2

0.3322s+ 1

0.06976s+ 1

Kc=

K

a

=

2

0.21

=9.5238

Gc(s)=Kc

s+3.0101

s+14.3339

=Kca

0.3322s+ 1

0.06976s+ 1

Frequency (rad/sec)

Bode Diagram of G 1 (s) = 20/[s(s + 1)]

− 180

− 140

− 130

− 150

− 160

− 170

− 120

− 20

− 10

0

Phase (deg); Magnitude (dB)

40

30

20

10

100 101

Figure 7–146
Bode diagram of
G 1 (s).

MATLAB Program 7–26

numc = [9.5238 28.6676];

denc = [1 14.3339];

w = logspace(-1,3,100);

bode(numc,denc,w)

title('Bode Diagram of Gc(s) = 9.5238(s + 3.0101)/(s + 14.3339')