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MathGroup Archive 2000

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Re: image 2-d Fouriertransform ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21922] Re: image 2-d Fouriertransform ?
  • From: "Mariusz Jankowski" <mjkcc at usm.maine.edu>
  • Date: Fri, 4 Feb 2000 02:54:49 -0500 (EST)
  • Organization: University of Southern Maine
  • References: <87av1h$nun@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Maurus, a real signal cannot be uniquely represented by 1 quarter of the 2D
discrete Fourier transform (DFT) coefficients. I will use the following
non-symmetric 4 by 4 array:

img={{1,2,1,0},{1,1,0,0},{0,0,0,0},{0,0,0,0}};

MatrixForm[Partition[InverseFourier[img],{2,2}]]

(* I use InverseFourier as my definition of the DFT, as is common in
engineering, signal processing *)

The coefficients have complex conjugate symmetry - they are all related, not
random.

For a short explanation on the ordering of the coefficients in 1D (think of
a single row or column of your 2D transform) see MathGroup archives from
approx 2 weeks ago.

Also take a look at my notebooks on Fourier series and transform in
Wolfram's courseware catalog http://library.wolfram.com/courseware/. I will
be adding a comprehensive set of notebooks on digital image processing to
the courseware catalog in the next couple of month - they may be helpful, if
you can wait that long.

Hope this helps,

Mariusz


======================================================
Mariusz Jankowski
University of Southern Maine
mjkcc at usm.maine.edu




"maurus" <maurus at gmx.de> wrote in message news:87av1h$nun at smc.vnet.net...
> Hello,
>
> sorry for this question but ..
>
> i have a 2-d image with 8 bit grey level.
> Now i make a 2-d-Fouriertransform of the image (in  Mathematica)
> The result is a Picture with 4 quadrants, were the signals are
> significant.
>
> Now my questions:
> I think, only the first quadrant ist important ? Rest ist random ?
> What does the axis, the results  means ?
>
>
>
> If someone canl help me, i can send an example for better understanding.
>
> In this case, the Helpfunktion in Mathematica ist not to good.
>
> Thanks in advance
>
> Maurus
>
>




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