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- From: ms@specklec.mpifr-bonn.mpg.de (Markus Schoeller)
- Subject: Re: Fourier Transform`s What?
- Message-ID: <1993Jan21.173317.28247@mpifr-bonn.mpg.de>
- Sender: news@mpifr-bonn.mpg.de
- Nntp-Posting-Host: specklec
- Organization: Max-Planck-Institut f"ur Radioastronomie
- References: <1993Jan19.145159.19327@gserv1.dl.ac.uk>
- Date: Thu, 21 Jan 1993 17:33:17 GMT
- Lines: 40
-
- In <1993Jan19.145159.19327@gserv1.dl.ac.uk> g.coulter@daresbury.ac.uk (G. Coulter) writes:
-
- >I wounder if someone could help explain to me Fourier Transforms
- >(FFT's) I don't want the formulae involved just an overview of
- >the technique. ie How it works? what is it used for? What are
- >the results of using the transform & how do you get back to your
- >initial say 2d data, any other general information would also be
- >very welcome.
-
- After there was a response about what a FFT looks like I just want
- to give an example what it is used for.
- If you are collecting images -in my case these are astronomical
- ones- you often have a convolution between the object you are
- looking at and a so called point-spread-function.
- The convolution is defined by a very nasty mathematical operation.
- It looks like this:
-
- i(x) = \int( o(y) p(x-y) dy )
-
- where i(x) is the image, o(x) the object and p(x) the psf.
- \int stands for an integral sign.
- If you do a FT, the convolution results in a multiplication:
-
- I(u) = O(u) P(u)
-
- where the uppercase letters are the FT's corresponding to the
- lowercase letters (letter==function).
- Of course everybody can get O(u) from this formula, if P(u) doesn't
- equal 0 on the whole field. A simple devision yields:
-
- O(u) = I(u) / P(u)
-
- An inverse transformation yields the object o(x).
- Unfortunately you don't know the psf in every case. But that's a much
- more complicated topic.
-
- Hope that helps and everything was in an understandable English.
-
- Markus
-
-