How can I do the FFT of a rectpuls?

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!

Answer 1

[Star Strider]

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

[Riccardo Barbero]

And this is the standard procedure to follow everytime I have to compute an FFT? The sampling interval (Ts), the sampling frequency (Fs), the Nyuquist frequency (Fn)...and all the following procedure, are standard?

In other words, if I wanted to compute the FFT of another function, could I use this procedure?

Many thanks

[Riccardo Barbero]

P.S.

I used this code to see your result, but I note that the maximum amplitude of the sinc function, and so, corresponding to the frequency f=0, it should be 4...as the area of the rectangle, but it's value is...0.4.

So, I think that there is a little error... Are you agree with me?

[Star Strider]

My pleasure.

I would definitely use that procedure. Defining everything explicitly (and commenting it) makes your code easier to read and easier to remember if you want to go back to it again. I adapted my code with a few modifications from this documentation for fft.

Note the code between the first two plot images. I modified it a bit here to give you a two-sided fft. Use my code here to give a two-sided fft or the code in the documentation to give you a one-sided fft, depending on what you need to do for any specific application. Both are correct.

[Riccardo Barbero]

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]

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.