Modern Control Engineering

600 Chapter 8 / PID Controllers and Modified PID Controllers

Second, note that

Notice that the characteristic equation for Y(s)/D(s)and the one for Y(s)/R(s)are identical.

We may be tempted to choose a zero of at s=–1to cancel a pole at s=–1of the

plant However, the canceled pole s=–1becomes a closed-loop pole of the entire system,

as seen below. If we define as a PID controller such that

(8–7)

then

The closed-loop pole at s=–1is a slow-response pole, and if this closed-loop pole is included in

the system, the settling time will not be less than 1 sec. Therefore, we should not choose as

given by Equation (8–7).

The design of controllers and consists of two steps.

Design Step 1: We design to satisfy the requirements on the response to the step-

disturbance input D(s).In this design stage, we assume that the reference input is zero.

Suppose that we assume that is a PID controller of the form

Then the closed-loop transfer function Y(s)/D(s)becomes

Note that the presence of “s” in the numerator of Y(s)/D(s)assures that the steady-state response

to the step disturbance input is zero.

Let us assume that the desired dominant closed-loop poles are complex conjugates and are

given by

s=–a_jb

=

10s

s^2 (s+1)+10K(s+a)(s+b)

=

10

s(s+1)+

10K(s+a)(s+b)

s

Y(s)

D(s)

=

10

s(s+1)+10Gc

Gc(s)=

K(s+a)(s+b)

s

Gc(s)

Gc(s)

Gc1(s) Gc2(s)

Gc(s)

=

10s

(s+1)Cs^2 +10K(s+b)D

Y(s)

D(s)

=

10

s(s+1)+

10K(s+1)(s+b)

s

Gc(s)=

K(s+1)(s+b)

s

Gc(s)

Gp(s).

Gc(s)

Y(s)

R(s)

=

Gp Gc1

1 +Gp Gc

=

10Gc1

s(s+1)+10Gc

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