Modern Control Engineering

Section 8–7 / Zero-Placement Approach to Improve Response Characteristics 611

which can be simplified to

24+25K-30Ka+5Ka^2 +j(–16-60K+20Ka)=0

By equating the real part and imaginary part to zero, respectively, we obtain

24+25K-30Ka+5Ka^2 =0 (8–9)

–16-60K+20Ka=0 (8–10)

From Equation (8–10), we have

(8–11)

Substituting Equation (8–11) into Equation (8–9), we get

a^2 =13

ora=3.6056or–3.6056.Notice that the values of Kbecome

K=1.3210 fora=3.6056

K=–0.1211 fora=–3.6056

Since is in the feedforward path, the gainKshould be positive. Hence, we choose

K=1.3210, a=3.6056

Then can be given by

To determine and we proceed as follows:

(8–12)

Thus,

To check the response to a unit-step disturbance input, we obtain the closed-loop transfer
functionY(s)/D(s).

=

5s

s^3 +12.605s^2 +52.63s+85.8673

=

5s

s(s+1)(s+5)+5K(s+a)^2

Y(s)

D(s)

=

Gp

1 +Gc1 Gp

Kp=9.5260, Ti=0.5547, Td=0.1387

=9.5260a 1 +

1

0.5547s

+0.1387sb

Gc1(s)=

1.3210As^2 +7.2112s+ 13 B

s

Kp ,Ti, Td,

=

1.3210s^2 +9.5260s+17.1735

s

=1.3210

(s+3.6056)^2

s

Gc1(s)=K

(s+a)^2

s

Gc1(s)

Gc1(s)

K=

4

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