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Improving the Resolution in FFT

조회 수: 2 (최근 30일)
Kutlu Yigitturk
Kutlu Yigitturk 2021년 6월 2일
댓글: Paul 2021년 6월 3일
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.

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dpb
dpb 2021년 6월 2일
Look at documentation for nextpow2
  댓글 수: 4
dpb
dpb 2021년 6월 2일
편집: dpb 2021년 6월 3일
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 2021년 6월 3일
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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