Fourier Transforms - The Convolution Theorem.

  • Thread starter binbagsss
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  • #1
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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 !
 

Answers and Replies

  • #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.
 
  • #3
LCKurtz
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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 !
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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