[Scilab-users] discrete Fourier transform

[Mike Page]

Hi,

I think this is just a question of scaling.  There is no "correct" scaling
for the FFT - just some different conventions.  This is explained here
(http://www.mathworks.co.uk/matlabcentral/answers/15770-scaling-the-fft-and-
the-ifft).

The "correct" answer with Scilab would be exactly 100 (N/2) in your case,
but you have a small error because your waveform is not exactly periodic -
the first and last sample are the same, but you should have the last sample
being the one before the first sample for a complete cycle.

The following code does that:

-->n=201;

-->x=linspace(0,2*%pi,n);

-->x=x(1:200);

-->y=sin(x);

-->Xf=dft(y,-1);

-->XfA = abs(Xf);

-->XfA(2)
 ans  =

    100.

HTH,
Mike.


-----Original Message-----
From: users-bounces at lists.scilab.org
[mailto:users-bounces at lists.scilab.org]On Behalf Of haasejos
Sent: 21 March 2013 21:14
To: users at lists.scilab.org
Subject: Re: [Scilab-users] discrete Fourier transform


-->n = 200;

-->x=linspace(0,2*%pi,n);

-->y=sin(x);

-->Xf=dft(y,-1);

-->XfA = abs(Xf); *please look at the difference! *
-->XfA(2)
 ans  =

    99.745189



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