Talk:Convolution theorem

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Where do these 2pi's come from? As far as I know, it's just F(f * g) = F(f)F(g), F(fg) = F(f) * F(g)

I believe the 2pi's are applicable depending on which school of thought you come from. Mathematicians use fourier transforms with a 'frequency' term k. Engineers prefer to use the symbol omega for natural frequency, but call it k anyways. I'm not entirerly sure about this (which is why it's in discussion and not on the page) and I don't feel like looking it up. But check out the wikipedia page on Fourier transforms.

The 2pis probably come from what definition of the Fourier transform you choose to use. Wikipedia likes to put a 1/sqrt(2pi) in front of both the transform and the inverse transform whereas others choose to put a 1/2pi in front of the inverse transform alone. This is probably the same thing as the person above mentioned but I'm not sure. I believe that the latter is the form that leads to a convolution theorem with no 2pis.

Is there any chance of getting a derivation of the theorem on here so that we can see what's happening?

--Zapateria 14:43, 7 May 2006 (UTC)

Most authors try do be in how they choose their constants. So if the you chose to use a 1 / sqrt2π in the definitions of Fourier transform and inverse transform then often people like also to put the same factor infront of the definition of convolution to avoid the constant croping up in this theorem. 128.135.197.2 18:21, 17 July 2006 (UTC)