Modern Control Engineering

Solution.We shall design a lag–lead compensator of the form

Then the open-loop transfer function of the compensated system is Gc(s)G(s). Since the gain K

of the plant is adjustable, let us assume that Kc=1. Then From the requirement

on the static velocity error constant, we obtain

Hence,

K=40

We shall first plot a Bode diagram of the uncompensated system with K=40. MATLAB Pro-

gram 7–30 may be used to plot this Bode diagram. The diagram obtained is shown in Figure 7–151.

=

K

4

= 10

Kv=limsS 0 sGc(s)G(s)=limsS 0 sGc(s)

K

s(s+1)(s+4)

limsS 0 Gc(s)=1.

Gc(s)=Kc

as+

1

T 1

bas+

1

T 2

b

as+

b

T 1

bas+

1

bT 2

b

556 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

MATLAB Program 7–30

num = [40];

den = [1 5 4 0];

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

bode(num,den,w)

title('Bode Diagram of G(s) = 40/[s(s + 1)(s + 4)]')

Frequency (rad/sec)

Bode Diagram of G(s) = 40/[s(s + 1)(s + 4)]

− 250

− 100

− 150

− 200

− 50

− 40

− 20

0

Phase (deg); Magnitude (dB)

40

20

10 −^1100101

Figure 7–151

Bode diagram of

G(s)= 40 Cs(s+1)(s+4)D.

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