Modern Control Engineering

Example Problems and Solutions 617

R(s) 1 C(s)

g

1

Tds

+

• 
  

Figure 8–46
Approximate
differentiator.

PID controller Plant G(s)

R(s) C(s)

D(s)

K(as+ 1) (bs+ 1)

s

1

s^2 + 3.6s+ 9

+

• 
  

+

+

Figure 8–47
PID-controlled
system.

Consequently,

Notice that R 1 C 1 andR 2 C 2 determine the locations of the zeros of the controller, whileR 1 ,R 3 ,and

C 1 affect the location of the pole on the negative real axis.R 5 /R 4 adjusts the gain of the controller.

A–8–4. In practice, it is impossible to realize the true differentiator. Hence, we always have to approxi-

mate the true differentiator by something like

One way to realize such an approximate differentiator is to utilize an integrator in the feedback path.

Show that the closed-loop transfer function of the system shown in Figure 8–46 is given by the pre-

ceding expression. (In the commercially available differentiator, the value ofgmay be set as 0.1.)

Solution.The closed-loop transfer function of the system shown in Figure 8–46 is

Note that such a differentiator with first-order delay reduces the bandwidth of the closed-loop

control system and reduces the detrimental effect of noise signals.

A–8–5. Consider the system shown in Figure 8–47. This is a PID control of a second-order plantG(s).As-

sume that disturbancesD(s)enter the system as shown in the diagram. It is assumed that the ref-

erence inputR(s)is normally held constant, and the response characteristics to disturbances are

a very important consideration in this system.

C(s)

R(s)

=

1

g

1 +

1

gTd s

=

Td s

1 +gTd s

Td s

1 +gTd s

Td s

=

R 5 R 2

R 4 R 3

as+

1

R 1 C 1

bas+

1

R 2 C 2

b

sas+

R 1 +R 3

R 1 R 3 C 1

b

Eo(s)

Ei(s)

=

Eo(s)

E(s)

E(s)

Ei(s)

=

R 5

R 4 AR 1 +R 3 BC 2

AR 1 C 1 s+ 1 BAR 2 C 2 s+ 1 B

sa

R 1 R 3

R 1 +R 3

C 1 s+ 1 b