Fourier Transforms - The Convolution Theorem.

  • Thread starter binbagsss
  • Start date
  • #1
1,233
10
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
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
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
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
773
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.
 

Related Threads on Fourier Transforms - The Convolution Theorem.

  • Last Post
Replies
25
Views
3K
Replies
2
Views
5K
  • Last Post
Replies
12
Views
866
  • Last Post
Replies
1
Views
2K
Replies
1
Views
176
Replies
1
Views
2K
Replies
2
Views
1K
Replies
1
Views
762
Replies
5
Views
1K
Top