Signals and Systems - Electrical Engineering

4.9 Response of LTI Systems to Periodic Signals 275

M = length(x);

figure(1)

x1 = [zeros(1, 5) x(1:M)];

z = y(1); y1 = [zeros(1, 5) z zeros(1, M−1)];

t0 =− 5 ∗Ts:Ts:Tend;

for k = 0:M−6,

pause(0.05)

h0 = fliplr(h);

h1 = [h0(M - k - 5:M) zeros(1, M - k - 1)];

subplot(211)

plot(t0, h1, ’r’)

hold on

plot(t0, x1, ’k’)

title(’Convolution of x(t) and h(t)’)

ylabel(’x(τ), h(t-τ)’); grid; axis([min(t0) max(t0) 1.1*min(x) 1.1*max(x)])

hold off

subplot(212)

plot(t0, y1, ’b’)

ylabel(’y(t) = (x∗h)(t)’); grid; axis([min(t0) max(t0) 0.1∗min(x) 0.1∗max(x)])

z = [z y(k + 2)];

y1 = [zeros(1, 5) z zeros(1, M - length(z))];

end

Figure 4.14 displays the last step of the convolution integral simulation. Notice that the steady state

is attained in a very short time (aroundt=0.5 sec). The transient changes every time that the script

is executed due to the random phase.

FIGURE 4.14
Convolution simulation: (a) inputx(t)(solid line) and
h(t−τ)(dashed line), and (b) outputy(t): transient
and steady-state response.

0 0.5 1

(a)

(b)

1.5 2

− 10

− 5

0

5

10

x(

τ),

h

(t

−τ

)

τ

0 0.5 1 1.5 2

−0.5

0

0.5

y(

t)

=

(x

*h

)(t

)

t