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Fourier Transforms - The Convolution Theorem.

  1. Feb 28, 2014 #1
    Ok so I've seen the convolution theorem written as:

    F(h(x)[itex]\otimes[/itex]g(x))=H(k)G(k)

    (And this is how it appears when I have a quick google).

    My book then does a problem in which is uses:

    F(h(x)g(x))=H(k)[itex]\otimes[/itex]G(k)

    Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
    and F represents a fourier transform


    My question
    - I can't see how these are equivalent at all?

    Many Thanks to anyone who can help shed some light !
     
  2. jcsd
  3. Feb 28, 2014 #2

    SteamKing

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    binbagsss:
    Please post your math questions in one of the Math HW forums. This helps to organize the questions so that posters can receive appropriate responses to questions.
     
  4. Feb 28, 2014 #3

    LCKurtz

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    Take a look at:
    http://ugastro.berkeley.edu/infrared09/PDF-2009/convolution2.pdf

    It gives an argument for the formula on the last page.
     
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