[Scilab-users] discrete Fourier transform

[Stefan Du Rietz]

On 2013-03-21 17:13, Serge Steer wrote:
--------------------
> 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

Maybe haasejos meant and should have used XfA == abs(Xf)

Stefan

>>
>> 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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