Improving the Resolution in FFT

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Kutlu Yigitturk
Kutlu Yigitturk on 2 Jun 2021
Commented: Paul on 3 Jun 2021
I need to get an output like below.
But this is the output I get as a result of the code I wrote.
clc
clear all
close all
N = 5;
n = -N-2: 1 : N+2;
u = unit(n+N/2)-unit(n-N/2);
subplot(2,1,1);
stem(n,u);
grid
title('x(t)');
X = fftshift(fft(u));
subplot(2,1,2)
plot(n,abs(X))
grid
title('|X(w)|')
How can I do this, thank you very much for your help.

Accepted Answer

dpb
dpb on 2 Jun 2021
Look at documentation for nextpow2
  4 Comments
dpb
dpb on 2 Jun 2021
Edited: dpb on 3 Jun 2021
subplot(2,1,1)
stem(-7:7,u)
subplot(2,1,2)
Y=abs(fft(u));
hL=plot(fftshift((Y)));
xlim([1 numel(Y)])
hold on
Y=abs(fft(u,2^nextpow2(u)));
hL(2)=plot(linspace(0,15,numel(Y)),fftshift((Y)));
Y=abs(fft(u,32));
hL(3)=plot(linspace(0.5,15,numel(Y)),fftshift((Y)));
ylim([0 5])
legend('N=15','N=16','N=32','location','northeast')
gives
You'll want to fix up legends to make the points match your above; I just normalized to the range of the first set of points; whether is exactly symmetric in indices in output vector depends on whether is even/odd number of points, of course.
Paul
Paul on 3 Jun 2021
Keep in mind that the output of fft() will need to be adjusted if desired to have the phase of the end result approximate the phase of the DTFT of u.

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