Modern Control Engineering

(Chris Devlin) #1
Example Problems and Solutions 145

From the equations just derived, a block diagram of the controller can be drawn, as shown in
Figure 4–31(a). Simplification of this block diagram results in Figure 4–31(b).
The transfer function between Pc(s)andE(s)is

For a practical controller, under normal operation is
very much greater than unity and Therefore, the transfer function can be simplified as
follows:

where

Thus the controller shown in Figure 4–30 is a proportional-plus-integral-plus-derivative one.
If the resistance Rdis removed, or Rd=0, the action becomes that of a proportional-plus-
integral controller.

Kp=

bks
aA

Kpa 1 +

1

Ti s

+Td sb

=

bks
aA

a

Ti+Td
Ti

+

1

Ti s

+Td sb

Pc(s)
E(s)



bksATi s+ 1 BATd s+ 1 B
aATi s

TiTd.

@KaATi sC(a+b)ksATi s+ 1 BATd s+ 1 BD@

Pc(s)
E(s)

=

b
a+b

K

1 +K

a
a+b

A

ks

a

Ti s
Ti s+ 1

ba

1

Td s+ 1

b + – + – + –

E(s) X(s) Pc(s)
K

a
a+b

b
a+b

A
ks

PI(s)

PII(s)

1
Td s + 1

1
Ti s + 1

(a)

(b)

ab+b K

E(s) X(s) Pc(s)

aATis
(a+b)ks(Tis+ 1) (Tds+ 1)

Figure 4–31
(a) Block diagram of
the pneumatic
controller shown in
Figure 4–30;
(b) simplified block
diagram.

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