Modern Control Engineering

Example Problems and Solutions 351

The breakaway and break-in points are found from

from which we get

or

Thus, the breakaway or break-in points are at s=0ands=–1.2.Note that s=–1.2is a double

root. When a double root occurs in at point s=–1.2, at this point. The

value of gain Kat point s=–1.2is

This means that with K=4.32the characteristic equation has a triple root at point s=–1.2.This

can be easily verified as follows:

s^3 +3.6s^2 +4.32s+1.728=(s+1.2)^3 = 0

K=-

s^3 +3.6s^2

s+ 4

2

s=-1.2

=4.32

dKds= 0 d^2 KAds^2 B= 0

s(s+1.2)^2 = 0

s^3 +2.4s^2 +1.44s= 0

dK

ds

=-

A3s^2 +7.2sB(s+0.4)-As^3 +3.6s^2 B

(s+0.4)^2

= 0

(a) (b)

jv

• 4 – 3 – 201 s

j 3

j 1

• j 1


• j 3


• j 2

j 2

K(s+ 0.4)

s^2 (s+ 3.6)

• 1

–60°

60 °

+

Figure 6–66
(a) Control system; (b) root-locus plot.