Signals and Systems - Electrical Engineering

366 C H A P T E R 6: Application to Control and Communications

Remarks

A control system includes two very important components:

n Transducer:Since it is possible that the output signal y(t)and the reference signal x(t)might not be of

the same type, a transducer is used to change the output so as to be compatible with the reference input.

Simple examples of a transducer are: lightbulbs, which convert voltage into light; a thermocouple, which

converts temperature into voltage.

n Actuator:A device that makes possible the execution of the control action on the plant, so that the output

of the plant follows the reference input.

nExample 6.1: Controlling an unstable plant

Consider a dc motor modeled as an LTI system with a transfer function

G(s)=

1

s(s+ 1 )

The motor is not BIBO stable given that its impulse responseg(t)=( 1 −e−t)u(t)is not absolutely

integrable. We wish the output of the motory(t)to track a given reference inputx(t), and pro-

pose using a so-calledproportional controllerwith transferHc(s)=K>0 to control the motor (see

Figure 6.6). The transfer function of the overall negative-feedback system is

H(s)=

Y(s)

X(s)

=

KG(s)

1 +KG(s)

Suppose thatX(s)= 1 /s, or the reference signal isx(t)=u(t). The question is: What should be

the value ofKso that in the steady state the output of the systemy(t)coincides withx(t)? Or,

equivalently, is the error signal in the steady state zero? We have that the Laplace transform of the

error signale(t)=x(t)−y(t)is

E(s)=X(s)[ 1 −H(s)]=

1

s( 1 +G(s)K)

=

s+ 1

s(s+ 1 )+K

The poles ofE(s)are the roots of the polynomials(s+ 1 )+K=s^2 +s+K, or

s1,2=−0.5±0.5

√

1 − 4 K

For 0<K≤0.25 the roots are real, and complex forK>0.25, and in either case in the left-hand

s-plane. The partial fraction expansion corresponding toE(s)would be

E(s)=

B 1

s−s 1

+

B 2

s−s 2

FIGURE 6.6

Proportional control of a motor.

K

e(t)

x(t)

y(t)

+

−

Proportional

controller Motor

G(s)=s(s^1 +1)