Modern Control Engineering

Example Problems and Solutions 531

Re

Im

4

3

v= 2.45

v= 2

v= 1.5

–1

• 2


• 1

1

123

v= 1

6 v=^0

(^98)
10
v= 0.5
2.65
1 +jv
2.65e– 0.8jv
1 +jv
Figure 7–126
Polar plots of
and2.65/(1+jv).
2.65e-0.8jv(1+jv)
A–7–10. Consider a unity-feedback system whose open-loop transfer function is
Using the Nyquist plot, determine the critical value of Kfor stability.
Solution.For this system,
The imaginary part of G(jv)is equal to zero if
Hence,
Solving this equation for the smallest positive value of v, we obtain
Substitutingv=2.4482intoG(jv), we obtain
The critical value of Kfor stability is obtained by letting G(j2.4482)equal–1. Hence,
or
Figure 7–126 shows the Nyquist or polar plots of 2.65e–0.8jv/(1+jv)and2.65/(1+jv). The first-
order system without transport lag is stable for all values of K, but the one with a transport lag of
0.8 sec becomes unstable for K>2.65.

K=2.65

0.378K= 1

G(j2.4482)=

K

1 +2.4482^2

(cos1.9586-2.4482sin1.9586)=-0.378K

v=2.4482

v=-tan0.8v

sin0.8v+vcos0.8v= 0

=

K

1 +v^2

C(cos0.8v-vsin0.8v)-j(sin0.8v+vcos0.8v)D

=

K(cos0.8v-jsin0.8v)(1-jv)

1 +v^2

G(jv)=

Ke-0.8jv

jv+ 1

G(s)=

Ke-0.8s

s+ 1