Modern Control Engineering

Clearly, the denominator of the closed-loop transfer function is a product of two lightly damped

second-order terms (the damping ratios are 0.243 and 0.102), and the two resonant frequencies are

sufficiently separated.

A–7–24. Consider the system shown in Figure 7–144(a). Design a compensator such that the closed-loop

system will satisfy the requirements that the static velocity error constant=20sec–1,phase

margin=50°, and gain marginG10 dB.

Solution.To satisfy the requirements, we shall try a lead compensator Gc(s)of the form

(If the lead compensator does not work, then we need to employ a compensator of different

form.) The compensated system is shown in Figure 7–144(b).

Define

whereK=Kca. The first step in the design is to adjust the gain Kto meet the steady-state per-

formance specification or to provide the required static velocity error constant. Since the static ve-

locity error constant Kvis given as 20 sec–1, we have

or

K=2

With K=2,the compensated system will satisfy the steady-state requirement.

We shall next plot the Bode diagram of

G 1 (s)=

20

s(s+1)

=10K= 20

=limsS 0

s10K

s(s+1)

=limsS 0 s

Ts+ 1

aTs+ 1

G 1 (s)

Kv=limsS 0 sGc(s)G(s)

G 1 (s)=KG(s)=

10K

s(s+1)

=Kc

s+

1

T

s+

1

aT

Gc(s)=Kc a

Ts+ 1

aTs+ 1

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

Gc(s)

G(s) G(s)

10

s(s+ 1)

(b)

10

s(s+ 1)

(a)

+– +–

Figure 7–144

(a) Control system;

(b) compensated

system.

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