Modern Control Engineering

Recall that for the lead compensator the maximum phase-lead angle fmis given by Equation

(7–25), where ais1/bin the present case. By substituting a=1/bin Equation (7–25), we have

Notice that b=10corresponds to fm=54.9°. Since we need a 50° phase margin, we may choose

b=10. (Note that we will be using several degrees less than the maximum angle, 54.9°.) Thus,

b=10

Then the corner frequency (which corresponds to the pole of the phase-lag portion of

the compensator) becomes v=0.015radsec. The transfer function of the phase-lag portion of

the lag–lead compensator then becomes

The phase-lead portion can be determined as follows: Since the new gain crossover frequen-

cy is v=1.5radsec, from Figure 7–111,G(j1.5)is found to be 13 dB. Hence, if the lag–lead com-

pensator contributes –13dB at v=1.5radsec, then the new gain crossover frequency is as

desired. From this requirement, it is possible to draw a straight line of slope 20 dBdecade, pass-

ing through the point (1.5 radsec,–13dB). The intersections of this line and the 0-dB line and

–20-dB line determine the corner frequencies. Thus, the corner frequencies for the lead portion

s+0.15

s+0.015

= 10 a

6.67s+ 1

66.7s+ 1

b

v= 1 bT 2

sinfm=

1 -

1

b

1 +

1

b

=

b- 1

b+ 1

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

v in rad/sec

dB

60

40

20

0

• 40
  
  
  • 20
  
  
  
  

90 °

0

• 90 °


• 180 °


• 270 °
  0.01 0.02 0.04 0.1 0.2 0.4 0.6

G

GcG

G

GcG

Gc

Gc

16 dB

124610

• 32 °

50 °

Figure 7–111

Bode diagrams for G

(gain-adjusted but

uncompensated

open-loop transfer

function),Gc

(compensator), and

GcG(compensated

open-loop transfer

function).

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