Modern Control Engineering

Example Problems and Solutions 529

Solution.Figure 7–123(b) shows a complete Nyquist plot in the Gplane. The answers to the three

questions are as follows:

(a) The closed-loop system is stable, because the critical point (–1+j0)is not encircled by the

Nyquist plot. That is, since P=0andN=0, we have Z=N+P=0.

(b) The open-loop transfer function has one pole in the right-half splane. Hence,P=1.(The

open-loop system is unstable.) For the closed-loop system to be stable, the Nyquist plot must

encircle the critical point (–1+j0)once counterclockwise. However, the Nyquist plot does

not encircle the critical point. Hence,N=0. Therefore,Z=N+P=1. The closed-loop sys-

tem is unstable.

(c) Since the open-loop transfer function has one zero, but no poles, in the right-half splane, we

haveZ=N+P=0. Thus, the closed-loop system is stable. (Note that the zeros of the

open-loop transfer function do not affect the stability of the closed-loop system.)

A–7–8. Is a closed-loop system with the following open-loop transfer function and with K=2stable?

Find the critical value of the gain Kfor stability.

Solution.The open-loop transfer function is

This open-loop transfer function has no poles in the right-half splane. Thus, for stability, the

–1+j0point should not be encircled by the Nyquist plot. Let us find the point where the Nyquist

plot crosses the negative real axis. Let the imaginary part of G(jv)H(jv)be zero, or

from which

Substituting into G(jv)H(jv), we obtain

The critical value of the gain Kis obtained by equating –2K/3to–1,or

Hence,

The system is stable if 06 K 632 .Hence, the system with K=2is unstable.

K=

3

2

-

2

3

K=- 1

Gaj

1

12

bHaj

1

12

b=-

2K

3

v= 1  12

v=;

1

12

1 - 2 v^2 = 0

=

K

• 3 v^2 +jvA 1 - 2 v^2 B

G(jv)H(jv)=

K

jv(jv+1)(2jv+1)

G(s)H(s)=

K

s(s+1)(2s+1)