Modern Control Engineering

Example Problems and Solutions 637

Gc 1 (s) Gp(s)

Gc 2 (s)

Y(s)

N(s)

U(s)

D(s)

R(s)

B(s)

+– ++ ++

++

Figure 8–66
Two-degrees-of-
freedom control
system.

2.The response to the unit-step reference input has a maximum overshoot of 25%or less, and

the settling time is 1 sec or less.

3.The steady-state errors in following ramp reference input and acceleration reference input

are zero.

Solution.The closed-loop transfer functions for the disturbance input and reference input are

given, respectively, by

Let us assume that is a PID controller and has the following form:

The characteristic equation for the system is

Notice that the open-loop poles are located ats=0(a double pole) ands=–1.The zeros are

located ats=–a(a double zero).

In what follows, we shall use the root-locus approach to determine the values of aandK. Let

us choose the dominant closed-loop poles at s=–5_j5.Then, the angle deficiency at the desired

closed-loop pole at s=–5+j5is

–135°-135°-128.66°+180°=–218.66°

The double zero at s=–amust contribute 218.66°. (Each zero must contribute 109.33°.) By a

simple calculation, we find

a=–3.2460

The controller is then determined as

The constantKmust be determined by use of the magnitude condition. This condition is

@Gc1(s)Gp(s)@s=- 5 +j5= 1

Gc1(s)=

K(s+3.2460)^2

s

Gc1(s)

1 +Gc1(s)Gp(s)= 1 +

K(s+a)^2

s

100

s(s+1)

Gc1(s)=

K(s+a)^2

s

Gc1(s)

Y(s)

R(s)

=

CGc1(s)+Gc2(s)DGp(s)

1 +Gc1(s)Gp(s)

Y(s)

D(s)

=

Gp(s)

1 +Gc1(s)Gp(s)