Question

Perform an FFT on the function : f(t) = cos(2712t) + sin (2124t) a. Use a At of 0.01 s and 64 data points and plot the power spectrum b. Use a At of 0.05 s and 64 data points and plot the power spectrum c. Discuss why your results vary (if they do)

Answer 1

MATLAB code is given below in bold letters.

clc;
close all;
clear all;

% Question a
% define delta_t
ts = 0.01;

% define fs
fs = 1/ts;

% define time vector
t = 0:ts:63*ts;
x = cos(2*pi*12*t) + cos(2*pi*24*t);

% power of the signal computation
N = length(x);
P = 1/N*sum(x.^2)

% PSD computation
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:fs/length(x):fs/2;

figure;
plot(freq,psdx);grid on;
title('PSD of x(t)');
xlabel('Frequency (Hz)');
ylabel('Power (dB)');

PSD of x(t) Power (dB) O 5 10 15 20 25 30 Frequency (Hz) 35 40 45 50

From the above figure it is observed that the power spectrum has peak at frequencies 12 and 24 Hz which are the frequencies constituting the signal x(t).

% Question b
% define delta_t
ts = 0.05;

% define fs
fs = 1/ts;

% define time vector
t = 0:ts:63*ts;
x = cos(2*pi*12*t) + cos(2*pi*24*t);

% power of the signal computation
N = length(x);
P = 1/N*sum(x.^2)

% PSD computation
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:fs/length(x):fs/2;

figure;
plot(freq,psdx);grid on;
title('PSD of x(t)');
xlabel('Frequency (Hz)');
ylabel('Power (dB)');

PSD of x(t) Power (dB) 0 1 2 3 7 8 9 10 4 5 6 Frequency (Hz)

From the above figure it is observed that the power spectrum has peak at frequencies 4 Hz and 8 Hz which are not the frequencies constituting the signal x(t). This is because of the fact that the sampling frequency does not meet the required Nyquist sampling rate.