Modern Control Engineering

Section 6–9 / Parallel Compensation 345

At point Q, the value of Kis

Hence

Thus, we have two solutions for this problem. For k=0.4490,the three closed-loop poles are

located at

For k=1.4130,the three closed-loop poles are located at

It is important to point out that the zero at the origin is the open-loop zero, but not the

closed-loop zero. This is evident, because the original system shown in Figure 6–61 does not

have a closed-loop zero, since

G(s)

R(s)

=

20

s(s+1)(s+4)+20(1+ks)

s=-2.1589+j4.9652, s=-2.1589-j4.9652, s=-0.6823

s=-1.0490+j2.4065, s=-1.0490-j2.4065, s=-2.9021

k=

K

20

=1.4130 at point Q

K=^2

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

s

2

s=-2.1589+j4.9652

=28.260

jv

j 6

j 5

j 4

j 3

j 2

j 1

• j 6


• j 5


• j 4


• j 3


• j 2


• j 1


• 7 – 6 – 5 – 4 – 3 – 2 – 110 s

s=–2.1589+j4.9652

Q

P

s = –1.0490 +j2.4065

s=–2.9021 66.42°

Figure 6–62
Root-locus plot for
the system shown in
Figure 6–61.