Question

6. Let Xc(r) = cos 2π450t. This signal is sampled with f, = 1 .5kHz: x[n] = xe (nT), n = 0,±1, . . . where T, = 1/f, we have 512 samples of x[n]. (a) Plot X (el") of x[n], n = 0, ± 1 , ±2, . . . (b) Approximately plot the DFT magnitude |x(kl. The DFT size is N = 512.

Answer 1

MATLAB code is given below in bold letters.

clc;
close all;
clear all;

L = 512;

fs = 1.5e3;
Ts = 1/fs;
t = 0:1/fs:1/fs*(L-1);
x = cos(2*pi*450*t);

% DTFT of x[n]
w = -pi:0.01:pi;
n = 0:511;
for k = 1:length(w)
X_c(k) = sum(x.*exp(-1j*w(k)*n));
end

figure;
subplot(211);plot(w,abs(X_c));grid;title('Magnitude DTFT');
xlabel('w');ylabel('Amplitude');
subplot(212);plot(w,angle(X_c)*57.3);grid;title('Phase DTFT');
xlabel('w');ylabel('degrees');

% DFT of x[n]
X = fftshift(fft(x,512));
L = length(X);
figure;
subplot(211);plot(-L/2:L/2-1,abs(X));grid;title('Magnitude DFT');
xlabel('k');ylabel('Amplitude');
subplot(212);plot(-L/2:L/2-1,angle(X)*57.3);grid;title('Phase DFT');
xlabel('k');ylabel('degrees');