Modern Control Engineering

396 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method

B–6–19.Referring to the system shown in Figure 6–109, de-

sign a compensator such that the static velocity error con-

stant is 20 sec–1without appreciably changing the original

location of a pair of the complex-conjugate

closed-loop poles.

As=- 2 ;j 213 B

Kv

B–6–20.Consider the angular-positional system shown in

Figure 6–110. The dominant closed-loop poles are located

at The damping ratio zof the dominant

closed-loop poles is 0.6. The static velocity error constant

is 4.1 sec–1, which means that for a ramp input of 360°sec

the steady-state error in following the ramp input is

It is desired to decrease evto one-tenth of the present

value, or to increase the value of the static velocity error con-

stant to 41 sec–1. It is also desired to keep the damping ratio

zof the dominant closed-loop poles at 0.6. A small change in

the undamped natural frequency vnof the dominant closed-

loop poles is permissible. Design a suitable lag compensator to

increase the static velocity error constant as desired.

Kv

ev=

ui

Kv

=

360 °sec

4.1 sec-^1

=87.8°

Kv

s=-3.60;j4.80.

B–6–21.Consider the control system shown in Figure 6–111.

Design a compensator such that the dominant closed-loop

poles are located at and the static velocity

error constant Kvis 50 sec–1.

s=- 2 ;j2 13

Gc(s)^16

+– s(s+ 4)

Figure 6–109

Control system.

+ Gc(s) s(s+ 10) (^820 s+ 20)

Figure 6–110

Angular-positional system.

+– Gc(s) s(s+ 2) (^10 s+ 5)

Figure 6–111

Control system.

B–6–24.Consider the system shown in Figure 6–114, which

involves velocity feedback. Determine the values of the am-

plifier gain Kand the velocity feedback gain so that the

following specifications are satisfied:

1.Damping ratio of the closed-loop poles is 0.5

2.Settling time2 sec

3.Static velocity error constant

4.0<Kh<1

Kv 50 sec-^1

Kh

R(s) 1 C(s)

s

Kh

K

+– +– 2 s+ 1

Figure 6–114

Control system.

B–6–23.Consider the control system shown in Figure 6–113.

Design a compensator such that the unit-step response curve

will exhibit maximum overshoot of 25%or less and settling

time of 5 sec or less.

+ Gc(s) s(s+^2 1) (s+^1 s+ 2)

Figure 6–112

Control system.

Gc(s)^1

s^2 (s+ 4)

+–

Figure 6–113

Control system.

B–6–22.Consider the control system shown in Figure 6–112.

Design a compensator such that the unit-step response curve

will exhibit maximum overshoot of 30%or less and settling

time of 3 sec or less.

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