0195136047.pdf

16.2 FEEDBACK CONTROL SYSTEMS 801

for the constant field current, withKma constant, and

Td=Kmia=Bmωo+Jm

dωo

dt

(3)

neglecting load torque. Transforming Equations (1), (2), and (3) and rearranging, we get

o

Eg

=

Km

(Ra+sLa)(Bm+sJm)+Km^2

Accounting for the amplifier gainKa, we have

o

Ee

=

KaKgKm

(Rf+sLf)

[

(Ra+sLa)(Bm+sJm)+Km^2

]

The block diagram, including the feedback loop due to the tachometer, is shown in Figure

E16.2.3(b).

EXAMPLE 16.2.4

Consider an elementary feedback control system, as shown in Figure 16.2.3, withH=1. The
output variablecand the inputeto the direct transmission path are related by

d^2 c

dt^2

+ 8

dc

dt

+ 12 c= 68 e

(a) Assuming the system to be initially at rest, fore=u(t), find the complete solution for

c(t) when the system is operated in open-loop fashion without any feedback.

(b) With the feedback loop connected and with a forcing function of a unit step, i.e.,R(s)=

1/s, describe the nature of the dynamic response of the controlled variable by working

with the differential equation for the closed-loop system, without obtaining a formal

solution forc(t).

(c) By working solely in terms of transfer functions, discuss the closed-loop behavior.

Solution

(a) Laplace transforming the given equation with zero initial conditions, we have

s^2 C(s)+ 8 sC(s)+ 12 C(s)=

68

s

(1)

or

C(s)=

68

s(s^2 + 8 s+ 12 )

=

68

s(s+ 2 )(s+ 6 )

=

K 0

s

+

K 1

s+ 2

+

K 2

s+ 6

(2)

Evaluating the coefficients of the partial-fraction expansion, we get

C(s)=

17

3

(

1

s

)

−

17

2

(

1

s+ 2

)

+

17

6

(

1

s+ 6

)

(3)

The corresponding time solution is then given by