Signals and Systems - Electrical Engineering

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

FIGURE 6.7

Cruise control system: reference speed

x(t)=V 0 u(t)and output speed of carv(t).

Hp(s)

x(t) e(t) c(t) v(t)

+

−

Pl Controller Plant

Hc(s)= 1 +^1 s

with bothβandαpositive values related to the mass of the car and the friction coefficient. For

simplicity, letα=β=1. The question is: Can this be achieved with the PI controller? The Laplace

transform of the output speedv(t)of the car is

V(s)=

Hc(s)Hp(s)

1 +Hc(s)Hp(s)

X(s)

=

V 0 (s+ 1 )

s(s^2 + 2 s+ 1 )

=

V 0

s(s+ 1 )

The poles ofV(s)ares=0 ands=−1 on the left-hands-plane. We can then writeV(s)as

V(s)=

B

s+ 1

+

A

s

whereA=V 0. The steady-state response is

lim

t→∞

v(t)=V 0

since the inverse Laplace transform of the first term goes to zero due to its poles being in the left-

hands-plane. The error signale(t)=x(t)−v(t)in the steady state is zero. The controlling signal

c(t)(see Figure 6.7) that changes the speed of the car is

c(t)=e(t)+

∫t

0

e(τ)dτ

so that even if the error signal becomes zero at some point—indicating the desired speed had been

reached—the value ofc(t)is not necessarily zero. The values ofe(t)att=0 and at steady–state can

be obtained using the initial- and the final-value theorems of the Laplace transform applied to

E(s)=X(s)−V(s)=

V 0

s

[

1 −

1

s+ 1

]

The final-value theorem gives that the steady-state error is

lim

t→∞

e(t)=lim

s→ 0

sE(s)= 0

coinciding with our previous result. The initial value is found as

e( 0 )=lim

s→∞

sE(s)

=lim

s→∞

V 0

[

1 −

1 /s

1 + 1 /s

]

=V 0