Suppose x_c(t) = cos(pi t) +2 cos(3 pi t) is sampled with period T = 1/4 to form

Question

Suppose x_c(t) = cos(pi t) +2 cos(3 pi t) is sampled with period T = 1/4 to form a discrete-time signal x_d[n] = x_c(nT). This signal x_d[n] is then the input to an LTI system with impulse response h_d[n] =delta[n] - sin(pi n/2)/pi n and the output of this system is denoted y_d[n]. We then form a continuous-time signal y_s(t) = sigma_n=-infinity^infinity y_d[n] delta(t - nT) which is the input to an LTI system with impulse response h_c(t) = T sin(pi t/T)/pi t and the output of this system is denoted y_c(t). Sketch X_c(j omega) Sketch X_d(e^j omega) Sketch Y_d(e^j omega) Determine y_c(t). For inputs band-limited such that X_c(j omega) = 0 for |omega| > pi/T, determine the frequency response H(j omega) of the equivalent overall system with input x_c(t)and output y_c(t).

Explanation / Answer

clc;
clear all;
close all;

t = 0:0.01:10;

xc_t = cos(pi*t) + 2*cos(3*pi*t);

figure,
plot(t,xc_t);
title('Plot of x(t)');

n = 0:1/4:10; % 1/4 sampling rate

xd_n = cos(pi*n) + 2*cos(3*pi*n);

figure,
stem(n,xd_n);
title('Plot of x[n]');

Xc_f = fft(xc_t);

figure,
subplot(2,1,1)
plot(abs(Xc_f));
title('Amplitude of Xc');

subplot(2,1,2)
plot(angle(Xc_f));
title('Angle of Xc');

Xd_f = fft(xd_n);

figure,
subplot(2,1,1)
plot(abs(Xd_f));
title('Amplitude of Xd');

subplot(2,1,2)
plot(angle(Xd_f));
title('Angle of Xd');

d_n = [1 zeros(1, 40)];

hd_n = d_n - ((sin(pi*n/2))/(pi*n));

Hd_f = fft(hd_n);

Yd_f = Hd_f.*Xd_f;

yd_n = ifft(Yd_f);

figure,
plot(n,yd_n);
title('plot y(n)');

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