Modern Control Engineering

Example Problems and Solutions 375

SinceKvis specified as 50 sec–1, we have

Thus

Now choose Then

ChooseT=10.Then the lag compensator can be given by

The angle contribution of the lag compensator at the closed-loop pole is

which is small. The magnitude of Gc(s)at is 0.981. Hence the change in the location
of the dominant closed-loop poles is very small.
The open-loop transfer function of the system becomes

The closed-loop transfer function is

To compare the transient-response characteristics before and after the compensation, the unit-step
and unit-ramp responses of the compensated and uncompensated systems are shown in Figures
6–85(a) and (b), respectively. The steady-state error in the unit-ramp response is shown in Figure
6–85(c). The designed lag compensator is acceptable.

C(s)

R(s)

=

10s+ 1

s^3 +4.005s^2 +10.02s+ 1

Gc(s)G(s)=

s+0.1

s+0.005

10

s(s+4)

s=- 2 +j 6

=-1.3616°

(^) /Gc(s) 2
s=- 2 +j 16
=tan-^1

16

- 1.9

• tan-^1

16

- 1.995

s=- 2 +j 16

Gc(s)=

s+0.1

s+0.005

b= 20

Kˆc=1.

Kˆc b= 20

Kv=limsS 0 sGc(s)

10

s(s+4)

=Kˆc b2.5= 50