Modern Control Engineering

Example Problems and Solutions 639

(a)

t (sec)

Response to Unit-Step Reference Input

0.6

0.8

1

1.2

1.4

0.4

0.2

0

0 0.5 1 1.5 2 2.5 3

y(r

t)

Figure 8–68
(a) Response to unit-
step reference input;
(b) response to unit-
ramp reference
input; (c) response to
unit-acceleration
reference input.

Next, we consider the responses to reference inputs. The closed-loop transfer function

Y(s)/R(s)is

Let us define

Then

To satisfy the requirements on the responses to the ramp reference input and acceleration

reference input, we use the zero-placement approach. That is, we choose the numerator of

Y(s)/R(s)to be the sum of the last three terms of the denominator, or

from which we get

(8–18)

Hence, the closed-loop transfer functionY(s)/R(s)becomes as

The response curves to the unit-step reference input, unit-ramp reference input, and unit-

acceleration reference input are shown in Figures 8–68(a), (b), and (c), respectively. The maximum

Y(s)

R(s)

=

12.403s^2 +74.028s+120.148

s^3 +12.403s^2 +74.028s+120.148

=0.74028+

1.20148

s

+0.12403s

Gc(s)=

0.12403s^2 +0.74028s+1.20148

s

100 sGc(s)=12.403s^2 +74.028s+120.148

=

100 sGc(s)

s^3 +12.403s^2 +74.028s+120.148

Y(s)

R(s)

=

Gc(s)Gp(s)

1 +Gc 1 (s)Gp(s)

Gc1(s)+Gc2(s)=Gc(s)

Y(s)

R(s)

=

CGc1(s)+Gc2(s)DGp(s)

1 +Gc1(s)Gp(s)