Impulse Signal Generation - MATLAB

Impulse signal generation—MATLAB

When defining the impulse or $\delta (t)$ signal, the shape of the signal used to do so is not important. Whether we use the rectangular pulse we considered in this chapter or another pulse, or even a signal that is not a pulse, in the limit we obtain the same impulse signal. Consider the following cases:
(a) The triangular pulse,
\[\Lambda_{\Delta}(t) = \frac{1}{\Delta}\left(1-\left\vert \frac{t}{\Delta}\right\vert\right)\left(u(t+\Delta) - u(t-\Delta)\right). \]
Carefully plot it, compute its area, and find its limit as $\Delta\rightarrow 0$. What do you obtain in the limit? Explain.
(b) Consider the signal
\[S_\Delta (t) = \frac{\sin (\pi t/\Delta)}{\pi t}. \]
Use the properties of the sinc signal $S(t) = \sin (\pi t)/(\pi t)$ to express $S_\Delta (t)$ in terms of $S(t)$. Then find its area, and the limit as $\Delta\rightarrow 0$. Use symbolic MATLAB to show that for decreasing values of $\Delta$ the $S_\Delta (t)$ becomes like the impulse signal.

Solution: MATLAB Script
(a)
% Pr. 1.7
clear all; clf
% part (a)
delta=0.1;
t=[-delta:0.05:delta];N=length(t);
lambda=zeros(1,N);
figure(5)
for k=1:6,
lambda=(1-abs(t/delta))/delta;
delta=delta/2;
plot(t,lambda);xlabel(’t’)
axis([-0.1 0.1 0 330]);grid
hold on
pause(0.5)
end
grid
hold off

(b)
% part (b)
syms S t
delta=1;
figure(6)
for k=1:4,
delta=delta/k;
S=(1/delta)*sinc(t/delta);
ezplot(S,[-2 2])
axis([-2 2 -8 30])
hold on
I=subs(int(S,t,-100*delta, 100*delta)) % area under sinc
pause(0.5)
end
grid;xlabel(’t’)
hold off