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Step Validity with the Fourier Transform of Convolution

  1. Jul 21, 2015 #1
    A convolution can be expressed in terms of Fourier Transform as thus,

    ##\mathcal{F}\left\{f \ast g\right\} = \mathcal{F}\left\{f\right\} \cdot \mathcal{F}\left\{g\right\}##.

    Considering this equation:

    ##g\left(x, y\right) = h\left(x, y\right) \ast f\left(x, y\right)##

    Are these steps valid if I were to compute for ##f\left(x, y\right)##?

    ##\mathcal{F}\left\{g\left(x, y\right)\right\} = \mathcal{F}\left\{h\left(x, y\right) \ast f\left(x, y\right)\right\} \\

    \mathcal{F}\left\{g\left(x, y\right)\right\} = \mathcal{F}\left\{h\left(x, y\right)\right\} \cdot \mathcal{F}\left\{f\left(x, y\right)\right\} \\

    \frac{\mathcal{F}\left\{g\left(x, y\right)\right\}}{\mathcal{F}\left\{h\left(x, y\right)\right\}} = \mathcal{F}\left\{f\left(x, y\right)\right\} \\

    \mathcal{F}^{-1}\left\{\frac{\mathcal{F}\left\{g\left(x, y\right)\right\}}{\mathcal{F}\left\{h\left(x, y\right)\right\}}\right\} = f\left(x, y\right)##

    Thank you in advance.
     
  2. jcsd
  3. Jul 21, 2015 #2
    One way to quickly check for invalidity is to try it with a few simple exemplar functions. Working does not guarantee validity, but not working guarantees invalidity.
     
  4. Jul 21, 2015 #3

    jasonRF

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    In principle it can be fine, but you will run into trouble at points where
    [tex]
    \mathcal{F}\left\{h\left(x, y\right)\right\} = 0
    [/tex]
    This happens more often than you might think.

    jason
     
  5. Jul 21, 2015 #4
    Indeed. However, isn't solving for ##f\left(x, y\right)## the same as de-convolution?
     
  6. Jul 22, 2015 #5

    jasonRF

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    Yes, this is deconvolution. As long as the denominator has no zeros and your known quantities (g and h) have NO noise, then the direct inversion you are proposing might be okay (EDIT: but also might not be okay!). In the real world, we usually have noise, and there can be zeros of functions, so deconvolution is more complicated. There are many many approaches ( I am not an expert in this)- you can find books, PhD dissertations, etc. on this topic. Google may help you. Note that deconvolution is an exmaple of an inverse problem, so google "inverse problems" and "deconvolution".

    One short hit:
    http://mathworld.wolfram.com/Deconvolution.html

    jason
     
  7. Jul 23, 2015 #6

    jasonRF

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    ecastro,

    I realize that I have been thinking about this in the framework of deconvolving images, or other numerical problems. If you are doing this to solve an analytical integral equation then your approach is certainly one that is used.

    jason
     
  8. Jul 23, 2015 #7
    I am actually deconvolving images. So, is the approach still valid?
     
  9. Jul 24, 2015 #8

    jasonRF

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    You can try it and see how it goes - what have you got to lose? However, it is often the case that the result is very noisy - how noisy depends on the initial image, the kernel that you are dividing by, etc. There are a large number of approaches to this - when I google I see a huge amount of material. I have personally used CLEAN for a case were teh image was sparse, and have worked with people that have used other approaches (maximum entropy based). There are a host of regularization techniques - the field of inverse problems deals with this kind of stuff. Astronomers and geophysicists work a lot in this field. good luck

    jason
     
  10. Aug 2, 2015 #9
    Thank you for your information and references!
     
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