Signals and Systems - Electrical Engineering

6.3 Application to Classic Control 365

The Laplace transform of the overall system output is

Y(s)=Hc(s)G(s)X(s)+η(s)

whereη(s)=L[η(t)]. In this case, E(s)is given by

E(s)=[Hc(s)G(s)−1]X(s)+η(s)

Although we can minimize this error by choosing Hc(s)= 1 /G(s)as above, in this case e(t)cannot be

made zero—it remains equal to the disturbanceη(t)and we have no control over this.

Closed-Loop Control
Assumingy(t)andx(t)in the open-loop control are the same type of signals, (e.g., both are voltages,
or temperatures), if we feed backy(t)and compare it with the inputx(t)we obtain a closed-loop
control. Considering the case of negative-feedback system (see Figure 6.5(a)), and assuming no
disturbance (η(t)=0), we have that

E(s)=X(s)−Y(s)

Y(s)=Hc(s)G(s)E(s)

and replacingY(s)gives

E(s)=

X(s)

1 +G(s)Hc(s)

If we wish the error to go to zero in the steady state, so thaty(t)tracks the input, the poles ofE(s)
should be in the open left-hands-plane.

If a disturbance signalη(t)(consider it for simplicity deterministic and with Laplace transformη(s))
is present (See Figure 6.5(a)), the above analysis becomes

E(s)=X(s)−Y(s)

Y(s)=Hc(s)G(s)E(s)+η(s)

so that

E(s)=X(s)−Hc(s)G(s)E(s)−η(s)

or solving forE(s),

E(s)=

X(s)

1 +G(s)Hc(s)

−

η(s)

1 +G(s)Hc(s)

=E 1 (s)+E 2 (s)

If we wishe(t)to go to zero in the steady state, then poles ofE 1 (s)andE 2 (s)should be in the open left-
hands-plane. Different from the open-loop control, the closed-loop control offers more flexibility in
achieving this by minimizing the effects of the disturbance.