DFT Exploration by harrde

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Problem Statement

Sample $sin(2\pi nt)$ at 3Hz, take the DFT, and explain the results.

Solution

Here is the MATLAB code and resulting figures:

f = 3;              % Sampling freq.
T = 1/f;            % Sampling period
t = 0:.01:5;
N2 = 500;          % Number of sampling points
N3 = 30;
t2 = 0:T:N2*T;
t3 = 0:T:N3*T;

x = sin(2*pi*t);    % Signal that is sampled
x2 = sin(2*pi*t2);
x3 = sin(2*pi*t3);

X2 = fft(x2);         % DFT of long signal
X3 = fft(x3);         % DFT of short signal

figure(1)                    %Original signal
plot(t(1:500),x(1:500))
xlabel('Time (s)')
ylabel('x(t)')
title('Original Input Signal')

figure(2)
plot(t2(1:15),x2(1:15))      % Sampled signal
xlabel('Time (s)')
ylabel('x(t)')
title('Sampled Input Signal')

figure(3)                    %DFT of long signal
plot(t2/(N2*T*T),abs(X2))    
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 500 Samples')

figure(4)                    % DFT of short signal
plot(t3/(N3*T*T),abs(X3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 30 Samples')

figure(5)                   % Shifted DFT of long signal
XS2=fftshift(X2);
f2=-1/(2*T):1/(N2*T):1/(2*T);
plot(f2,abs(XS2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 500 Samples')

figure(6)                   % Shifted DFT of short signal
XS3=fftshift(X3);
f3=-1/(2*T):1/(N3*T):1/(2*T);
plot(f3,abs(XS3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 30 Samples')

Explanation

The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points. But as we increase the number of sample points the DFT more closely resembles the expected transform.