Modern Control Engineering

Example Problems and Solutions 387

We now choose the compensator in the following form:

Then the open-loop transfer function of the compensated system becomes

To determine by the root-locus method, we need to find the angle deficiency at the desired

closed-loop pole The angle deficiency can be found as follows:

Hence, the lead compensator must provide 132.73°. Since the angle deficiency is –132.73°,

we need two lead compensators, each providing 66.365°. Thus will have the following form:

Suppose that we choose two zeros at s=–2.Then the two poles of the lead compensators can be

obtained from

or

(See Figure 6–95.) Hence,

Gˆc(s)=Kca

s+ 2

s+9.9158

b

2

=9.9158

sp= 2 +

3.4641

0.4376169

3.4641

sp- 2

=tan(90°-66.365°)=0.4376169

=Kca

s+sz

s+sp

b

2

Gˆc(s)

Gˆc(s)

Gˆc(s)

=-132.73°

Angle deficiency=-143.088°- 120 °-109.642°+ 60 °+ 180 °

s=- 2 +j2 13.

Gˆc(s)

=Gˆc(s)

s+ 4

sAs^2 +0.1s+ 4 B

Gc(s)G(s)=Gˆc(s)

s+ 4

2s+0.1

1

s

2s+0.1

s^2 +0.1s+ 4

Gc(s)=Gˆc(s)

s+ 4

2s+0.1

− 12 − 10 − 8 − 6 − 4 − 2 0 2 4

j 4

j 2

• j 4


• j 2

66.365°

sp

s = −2 + j 23

jv

s

Figure 6–95
Pole and zero of

Gˆc(s).