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problem_3.m
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problem_3.m
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%% Final Problem 3
%{
let x(t) = sinc^2(pi*t)
then by Lecture 1 triangular pulse example and the symmetry property H(t) <=> h(-f), X(f) = 1-abs(f), abs(f)<1; and 0 elsewhere. Therefore, the Nyquist frequency fc=1.
%}
%%
clear all;
close all;
syms t
x = (sin(pi*t)/(pi*t))^2
figure()
ezplot(x, [-2 2 -.5 1])
legend('x(t)')
N=12;
%% part(a)
F=3;
sum_pos=0;
for k =1:N
sum_pos = sum_pos+subs(x,t,k/F)*sinc(F*(t-k/F));
end
DC=limit(x,t,0)*sinc(F*t); % verify sinc(0) = 1; k =0 value;
sum_neg = 0;
for k=1:N
sum_neg = sum_neg+subs(x,t,-k/F)*sinc(F*(t+k/F));
end
y_a=sum_neg+DC+sum_pos;
%% part(b)
F=2;
sum_pos=0;
for k =1:N
sum_pos = sum_pos+subs(x,t,k/F)*sinc(F*(t-k/F));
end
DC=limit(x,t,0)*sinc(F*t); % verify sinc(0) = 1; k =0 value;
sum_neg = 0;
for k=1:N
sum_neg = sum_neg+subs(x,t,-k/F)*sinc(F*(t+k/F));
end
y_b=sum_neg+DC+sum_pos;
%% part(c)
F=1;
sum_pos=0;
for k =1:N
sum_pos = sum_pos+subs(x,t,k/F)*sinc(F*(t-k/F));
end
DC=limit(x,t,0)*sinc(F*t); % verify sinc(0) = 1; k =0 value;
sum_neg = 0;
for k=1:N
sum_neg = sum_neg+subs(x,t,-k/F)*sinc(F*(t+k/F));
end
y_c=sum_neg+DC+sum_pos;
%% plot part(a)
figure()
ezplot(x, [-2 2 -.5 1])
legend("x(t)")
hold on
ezplot(y_a, [-2 2 -.5 1])
legend("y_a(t)")
title("x(t) and y_a(t)")
hold off
legend('x(t)', 'y_a(t)')
figure()
ezplot(x-y_a, [-2 2 -.5 1])
title("x(t)-y_a(t)")
%% plot part (b)
figure()
ezplot(x, [-2 2 -.5 1])
hold on
ezplot(y_b, [-2 2 -.5 1])
title("x(t) and y_b(t)")
hold off
legend('x(t)', 'y_b(t)')
figure()
ezplot(x-y_b, [-2 2 -.5 1])
title("x(t)-y_b(t)")
%% plot part(c)
figure()
ezplot(x, [-2 2 -.5 1])
hold on
ezplot(y_c, [-2 2 -.5 1])
title("x(t) and y_c(t)")
hold off
legend('x(t)', 'y_c(t)')
%{
For F values F=2 & F=3, y(y) is indistinguishable from x(n);
%}
figure()
ezplot(x-y_c, [-2 2 -.5 1])
title("x(t)-y_c(t)")