# Fourier Transforms - The Convolution Theorem.

Ok so I've seen the convolution theorem written as:

F(h(x)$\otimes$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)$\otimes$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 !

SteamKing
Staff Emeritus
Homework Helper
binbagsss:
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LCKurtz
Homework Helper
Gold Member
Ok so I've seen the convolution theorem written as:

F(h(x)$\otimes$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)$\otimes$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.