Modern Control Engineering

Example Problems and Solutions 377

A–6–16. Consider a unity-feedback control system whose feedforward transfer function is given by

Design a compensator such that the dominant closed-loop poles are located at

and the static velocity error constant Kvis equal to 80 sec–1.

Solution.The static velocity error constant of the uncompensated system is

SinceKv=80is required, we need to increase the open-loop gain by 128. (This implies that we

need a lag compensator.) The root-locus plot of the uncompensated system reveals that it is not

possible to bring the dominant closed-loop poles to by just a gain adjustment alone.

See Figure 6–86. (This means that we also need a lead compensator.) Therefore, we shall employ

a lag–lead compensator.

Let us assume the transfer function of the lag–lead compensator to be

whereKc=128.This is because

and we obtain Kc=128.The angle deficiency at the desired closed-loop pole is

The lead portion of the lag–lead compensator must contribute. To choose we may use the

graphical method presented in Section 6–8.

60 ° T 1

Angle deficiency=- 120 °- 90 °- 30 °+ 180 °=- 60 °

s=- 2 +j2 13

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

10

16

= 80

Gc(s)=Kc±

s+

1

T 1

s+

b

T 1

≤±

s+

1

T 2

s+

1

bT 2

≤

• 2 ;j2 13

Kv=^1016 =0.625.

s=- 2 ;j2 13

G(s)=

10

s(s+2)(s+8)

Real Axis

− 10 − 5 0 5 10

Imag Axis

10

− 10

6

− 6

− 8

4

0

2

− 2

8

− 4

Root-Locus Plot of G(s) = 10/[s(s+2)(s+8)]

Desired closed-loop

pole

Complex conjugate
closed-loop pole
Figure 6–86
Root-locus plot of

Cs(s+2)(s+8)D.

G(s)= 10