Signals and Systems - Electrical Engineering

6.3 Application to Classic Control 367

for some valuesB 1 andB 2. Given that the real parts ofs 1 ands 2 are negative, their corresponding

inverse Laplace terms will have a zero steady-state response. Thus,

lim

t→∞

e(t)→ 0

This can be found also with the final value theorem,e( 0 )is

sE(s)|s= 0 = 0

So for anyK>0,y(t)→x(t)in steady state.

Suppose then thatX(s)= 1 /s^2 or thatx(t)=tu(t), a ramp signal. Intuitively, this is a much harder

situation to control, as the output needs to be continuously growing to try to follow the input. In

this case, the Laplace transform of the error signal is

E(s)=

1

s^2 ( 1 +G(s)K)

=

s+ 1

s(s(s+ 1 )+K)

In this case, even if we chooseKto make the roots ofs(s+ 1 )+Kbe in the left-hands-plane, we

have a pole ats=0. Thus, in the steady state, the partial fraction expansion terms corresponding

to poless 1 ands 2 will give a zero steady-state response, but the poles=0 will give a constant

steady-state responseAwhere

A=E(s)s|s= 0 = 1 /K

In the case of a ramp as input, it is not possible to make the output follow exactly the input

command, although by choosing a very large gainKwe can get them to be very close. n

Choosing the values of the gainKof the open-loop transfer function

G(s)Hc(s)=

KN(s)

D(s)

to be such that the roots of

1 +G(s)Hc(s)= 0

are in the open left-hands-plane, is theroot-locusmethod, which is of great interest in control theory.

nExample 6.2: A cruise control

Suppose we are interested in controlling the speed of a car or in obtaining a cruise control. How to

choose the appropriate controller is not clear. We consider initially a proportional plus integral (PI)

controllerHc(s)= 1 + 1 /sand will ask you to consider the proportional controller as an exercise.

See Figure 6.7.

Suppose we want to keep the speed of the car atV 0 miles/hour fort≥0 (i.e.,x(t)=V 0 u(t)), and

that the model for a car in motion is a system with transfer function

Hp(s)=β/(s+α)