Signals and Systems - Electrical Engineering

6.3 Application to Classic Control 373

− 1 −0.8 −0.6 −0.4 −0.2 0

(a) (b)

−0.5

0

0.5

σ

jΩ

RC= 1

RC= 10

0 1020304050

0

0.2

0.4

0.6

0.8

1

t

vc

(t

)

RC= 10

RC= 1

FIGURE 6.10
(a) Clustering of poles and (b) time responses of a first-order feedback system for 1 ≤RC≤ 10.

Forvi(t)=u(t), so thatVi(s)= 1 /s, then the Laplace transform of the output is

Vc(s)=

1

s(sRC+ 1 )

=

1 /RC

s(s+ 1 /RC)

=

1

s

−

1

s+ 1 /RC

so that

vc(t)=( 1 −e−t/RC)u(t)

The following MATLAB script plots the polesVc(s)/Vi(s)and simulates the transients ofvc(t)for
1 ≤RC≤10, shown in Figure 6.10. Thus, if we wish the system to respond fast to the unit-step input
we locate the system pole far from the origin.

%%%%%%%%%%%%%%%%%%%%%

% Transient analysis

%%%%%%%%%%%%%%%%%%%%%

clf; clear all

syms s t

num = [0 1];

for RC = 1:2:10,

den = [ RC 1];

figure(1)

splane(num, den) % plotting of poles and zeros

hold on

vc = ilaplace(1/(RC∗s ˆ 2 + s)) % inverse Laplace

figure(2)

ezplot(vc, [0, 50]); axis([0 50 0 1.2]); grid

hold on

end

hold off