[Scilab-users] discrete Fourier transform

[haasejos]

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?

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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[Serge Steer]

Post by haasejos
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

Post by haasejos
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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[Stefan Du Rietz]

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

Post by Serge Steer

Post by haasejos
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

Post by Serge Steer

Post by haasejos
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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[haasejos]

good evening,
what I meant is, that XfA(2) should be "1". Because this value can be
understood as the amplitude of sin(x). But XfA(2) beeing calculated with
XfA=abs(Xf) (see example) returns "100". This is wrong, isn't it?
Josef



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[Stefan Du Rietz]

On 2013-03-21 20:34, haasejos wrote:
--------------------

Post by haasejos
good evening,
what I meant is, that XfA(2) should be "1". Because this value can be
understood as the amplitude of sin(x). But XfA(2) beeing calculated with
XfA=abs(Xf) (see example) returns "100". This is wrong, isn't it?
Josef

-->n = 200;
-->x=linspace(0,2*%pi,n);
-->y=sin(x);
-->Xf=dft(y,-1);
-->XfA = abs(Xf)*2/n;
-->XfA(2)
ans =
0.9974519

/Stefan

[haasejos]

-->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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[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-***@lists.scilab.org
[mailto:users-***@lists.scilab.org]On Behalf Of haasejos
Sent: 21 March 2013 21:14
To: ***@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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[haasejos]

thanks a lot for your answers!



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