Modern Control Engineering

344 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method

Notice that the adjustable variable kdoes not appear as a multiplying factor. The characteristic

equation for the system is

(6–26)

Define

Then Equation (6–26) becomes

(6–27)

Dividing both sides of Equation (6–27) by the sum of the terms that do not contain K, we get

or

(6–28)

Equation (6–28) is of the form of Equation (6–11).

We shall now sketch the root loci of the system given by Equation (6–28). Notice that the

open-loop poles are located at s=j2, s=–j2, s=–5,and the open-loop zero is located at s=0.

The root locus exists on the real axis between 0 and –5.Since

we have

The intersection of the asymptotes with the real axis can be found from

as

The angle of departure (angle u) from the pole at s=j2is obtained as follows:

Thus, the angle of departure from the pole s=j2is 158.2°. Figure 6–62 shows a root-locus plot

for the system. Notice that two branches of the root locus originate from the poles at and

terminate on the zeros at infinity. The remaining one branch originates from the pole at s=–5

and terminates on the zero at s=0.

Note that the closed-loop poles with z=0.4 must lie on straight lines passing through the

origin and making the angles with the negative real axis. In the present case, there are two

intersections of the root-locus branch in the upper half splane and the straight line of angle 66.42°.

Thus, two values of Kwill give the damping ratio zof the closed-loop poles equal to 0.4. At pointP,

the value of Kis

Hence

k=

K

20

=0.4490 at point P

K=^2

(s+j2)(s-j2)(s+5)

s

2

s=-1.0490+j2.4065

=8.9801

;66.42°

s=;j2

u= 180 °- 90 °-21.8°+ 90 °=158.2°

s=-2.5

slimSq

Ks

s^3 +5s^2 +4s+ 20

=slimSq

K

s^2 +5s+p

=slimSq

K

(s+2.5)^2

Angles of asymptote=

; 180 °(2k+1)

2

=; 90 °

slimSq

Ks

(s+j2)(s-j2)(s+5)

=slimSq

K

s^2

1 +

Ks

(s+j2)(s-j2)(s+5)

= 0

1 +

Ks

s^3 +5s^2 +4s+ 20

= 0

s^3 +5s^2 +4s+Ks+ 20 = 0

20k=K

s^3 + 5 s^2 + 4 s+ 20 ks+ 20 = 0

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