How can I do the FFT of a rectpuls?

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Riccardo Barbero 2016-3-14

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https://ww2.mathworks.cn/matlabcentral/answers/273430-how-can-i-do-the-fft-of-a-rectpuls

评论： Star Strider 2016-3-14

I know that the Fourier transform of a rectpuls is a sinc function. I'm able to plot a rectpuls using the command A*rectpuls(t,B), in witch "t" represent the time on x axis, "B" represent the a length of the base of the rectangle and "A" represent the amplitude. So, a possible MATLAB code to plot it could be:

t=-5:.01:5;

x=2*rectpuls(t,2);

plot(t,x)

Now, i would like to compute the FFT, obtaining a sinc function... How can I do this, using the command fft?

Thank you!

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Answer 1

Star Strider 2016-3-14

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https://ww2.mathworks.cn/matlabcentral/answers/273430-how-can-i-do-the-fft-of-a-rectpuls#answer_213562

在 MATLAB Online 中打开

This is how I would do it:

t=-5:.01:5;
x=2*rectpuls(t,2);
figure(1)
plot(t,x)
grid
Ts = mean(diff(t));                                     % Sampling Interval
Fs = 1/Ts;                                              % Sampling Frequency
Fn = Fs/2;                                              % Nyuquist Frequency
L = length(x);
ftx = fft(x)/L;                                         % Fourier Transform
ftxs = fftshift(ftx);                                   % Shift To Centre
Fv = linspace(-Fn, Fn, L);                              % Frequency Vector
figure(2)
subplot(2,1,1)
plot(Fv, abs(ftxs))                                     % Plot Amplitude
grid
subplot(2,1,2)
plot(Fv, angle(ftxs))                                   % Plot Phase
grid

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Riccardo Barbero 2016-3-14

Ok, I understand! ;)

I don't know if you read my P.S...

Why, using your code, I obtain that the maximum amplitude of sinc is 0.4 (in f=0), instead of 4 (as the area of the rectangle)?

Thank you for your patience

Star Strider 2016-3-14

I overlooked your P.S. before, probably because you posted it while I was writing my previous Comment.

The amplitude of the sinc function calculated with the Fourier transform is correct. The power of the Fourier transformed signal is equal to the power of the time-domain signal, as stated in Parseval’s theorem. I invite you to explore that on your own.

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