Modern Control Engineering

Example Problems and Solutions 353

or

which can be further simplified to

For sZ–1.6,we may write this last equation as

which gives the equations for the root locus as follows:

The equation v=0represents the real axis. The root locus for 0Kqis between points

s=–0.4ands=–3.6.(The real axis other than this line segment and the origin s=0corre-

sponds to the root locus for –qK<0.)

The equations

(6–29)

represent the complex branches for 0Kq. These two branches lie between s=–1.6and

s=0. [See Figure 6–66(b).] The slopes of the complex root-locus branches at the breakaway

point (s=–1.2)can be found by evaluating of Equation (6–29) at point s=–1.2.

Since the root-locus branches intersect the real axis with angles

A–6–5. Consider the system shown in Figure 6–67(a). Sketch the root loci for the system. Observe that
for small or large values of Kthe system is underdamped and for medium values of Kit is
overdamped.
Solution.A root locus exists on the real axis between the origin and –q. The angles of asymp-
totes of the root-locus branches are obtained as

The intersection of the asymptotes and the real axis is located on the real axis at

The breakaway and break-in points are found from Since the characteristic equation is

s^3 +4s^2 +5s+K= 0

dKds=0.

s=-

0 + 2 + 2

3

=-1.3333

Angles of asymptotes=

; 180 °(2k+1)

3

= 60 °, - 60 °, - 180 °

tan-^113 = 60 °, ; 60 °.

dv

ds

2

s=-1.2

=;

A

• s
  s+1.6

2

s=-1.2

=;

A

1.2

0.4

=; 13

dvds

v=;(s+1.2)

A

• s
  s+1.6

v=-(s+1.2)

A

• s
  s+1.6

v=(s+1.2)

A

• s
  s+1.6

v= 0

vcv-(s+1.2)

A

• s
  s+1.6

dcv+(s+1.2)

A

• s
  s+1.6

d = 0

vCs(s+1.2)^2 +(s+1.6)v^2 D= 0

vAs^3 +2.4s^2 +1.44s+1.6v^2 +sv^2 B= 0