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Topic: correlation function
Replies: 3   Last Post: Nov 19, 2012 5:04 PM

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Dana DeLouis

Posts: 24
Registered: 11/18/12
Re: correlation function
Posted: Nov 18, 2012 5:16 PM
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> I get different results but have no idea why.

Hi. If I'm not mistaken, you are loosing data with the default settings of Fourier.
As a personal technique, since I never remember these, I use these variables.

{
fftDefault={0,1},
fftData={-1,1},
fftSignal={1,-1}
};

What you want to do is use Signal {1,-1}

SetOptions[{Fourier,InverseFourier},FourierParameters->fftSignal];

v=RandomInteger[{1,9},8]
{6,2,7,7,9,4,5,4}

If we assume your Fourier equation is correct, then what you are doing is really Convolution:

InverseFourier[Fourier[v]^2]
{259.,222.,250.,224.,242.,222.,267.,250.}

ListConvolve[v,v,{1,1}]
{259,222,250,224,242,222,267,250}

%%==%
True

If we assume your Correlation equation is correct, then one way might be:

ListCorrelate[v,v,{1,1}]
{276,238,246,221,250,221,246,238}

InverseFourier[Fourier[v]*Fourier[v//Reverse]] //RotateRight
{276.,238.,246.,221.,250.,221.,246.,238.}

%%==%
True


= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Mathematica 8
= = = = = = = = = =



On Sunday, November 18, 2012 4:08:45 AM UTC-5, jure lapajne wrote:
> Hello,
>
> I'm having hard time calculating correlation (autocorrelation) function of
>
> two lists (list). I'm trying two different ways of calculating it. One way
>
> is to use fourier transform and second way is to use Mathematica's function
>
> ListCorrelate. I get different results but have no idea why. Here's my code:
>
>
>
> korelacija1 = ListCorrelate[data, data, {1, 1}];
>
> korelacija11 = Abs[InverseFourier[Abs[Fourier[data]]^2]];
>
>
>
> All elements of "data" are real. I have two Abs in second line because for some reason InverseFourier returns small imaginary parts - I know it shouldn't. It's probably only numerical error.
>
>
>
> Thanks for help.









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