[Scilab-users] discrete Fourier transform

[Serge Steer]

Le 21/03/2013 16:41, haasejos a écrit :
> hello,
> for signalanalysis I would like to use discrete Fourier transform (dft). To
> see, how it works, I use the simple example below. Why is *XfA =
> abs(Xf)*2/n* respectively why is  XfA = abs(Xf) wrong?
Why do you say that  XfA = abs(Xf) is wrong
Note however. It is much more efficient using fft instead of dft.

Serge Steer
>
> clear; clc; xdel;
>
> function y = f(x);
>      y = sin(x);
> endfunction;
> n = 200;
> x=linspace(0,2*%pi,n);
> y=f(x);
>
> mat = [x',y'];
> //disp(mat);
> //plot2d(x , y);
>
> //xtitle('DATA','n''','y''');
>
> Xf=dft(y,-1);
> XfA = abs(Xf)*2/n;
> plot2d3([1:n/2]',XfA(1:n/2));
>
>
>
>
> --
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