Modern Control Engineering

Example Problems and Solutions 347

From Figure 6–63 we notice that the response of the system with k=0.4490is oscillatory.

(The effect of the closed-loop pole at s=–2.9021on the unit-step response is small.) For the

system with k=1.4130,the oscillations due to the closed-loop poles at

damp out much faster than purely exponential response due to the closed-loop pole at s=–0.6823.

The system with k=0.4490(which exhibits a faster response with relatively small overshoot)

has a much better response characteristic than the system with k=1.4130(which exhibits a slow

overdamped response). Therefore, we should choose k=0.4490for the present system.

Example Problems and Solutions

A–6–1. Sketch the root loci for the system shown in Figure 6–64(a). (The gain Kis assumed to be posi-

tive.) Observe that for small or large values of Kthe system is overdamped and for medium val-

ues of Kit is underdamped.

Solution.The procedure for plotting the root loci is as follows:

1.Locate the open-loop poles and zeros on the complex plane. Root loci exist on the negative

real axis between 0 and –1and between –2and–3.

2.The number of open-loop poles and that of finite zeros are the same. This means that there

are no asymptotes in the complex region of the splane.

3.Determine the breakaway and break-in points. The characteristic equation for the system is

or

K=-

s(s+1)

(s+2)(s+3)

1 +

K(s+2)(s+3)

s(s+1)

= 0

s=-2.1589;j4.9652

t Sec

012 3 4 5 6 7 8 9 10

Outputs

c1 and

c^2

1.2

0.4

0

0.6

0.2

1

0.8

Unit-Step Responses of Two Systems

k= 1.4130

k= 0.4490

Figure 6–63
Unit-step response
curves for the system
shown in Figure 6–61
when the damping
ratiozof the
dominant closed-
loop poles is set
equal to 0.4. (Two
possible values of k
give the damping
ratiozequal to 0.4.)