Modern Control Engineering

circle, and the phase angle of this point is –120°. Hence, the phase margin is 60°. The fact that

pointG(j1.15)is on the unit circle verifies that at v=1.15radsec the magnitude is equal to 1

or 0 dB. (Thus,v=1.15is the gain crossover frequency.) Thus,K=0.188gives the desired phase

margin of 60°.

Note that in writing ‘text’ in the polar diagram we enter the textcommand as follows:

text(x,y,' ')

For example, to write ‘Nyquist plot’ starting at point (0.3, –1.7), enter the command

text(0.3,–1.7,'Nyquist plot')

The text is written horizontally on the screen.

A–7–23. If the open-loop transfer function G(s)involves lightly damped complex-conjugant poles, then

more than one Mlocus may be tangent to the G(jv)locus.

Consider the unity-feedback system whose open-loop transfer function is

(7–32)

Draw the Bode diagram for this open-loop transfer function. Draw also the log-magnitude-versus-

phase plot, and show that two Mloci are tangent to the G(jv)locus. Finally, plot the Bode diagram

for the closed-loop transfer function.

Solution.Figure 7–141 shows the Bode diagram of G(jv). Figure 7–142 shows the log-magni-

tude-versus-phase plot of G(jv). It is seen that the G(jv)locus is tangent to the M=8-dB locus

atv=0.97radsec, and it is tangent to the M=–4-dB locus at v=2.8radsec.

G(s)=

9

s(s+0.5)As^2 +0.6s+ 10 B

546 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

40

20

0

dB

• 20


• 40

0 

• 90 


• 180 


• 270 


• 360 
  0.1 0.2 0.4 1 2 4 10
  v in rad/sec

Figure 7–141

Bode diagram of

G(s)given by

Equation (7–32).

Openmirrors.com