Norman Albers Posted July 5, 2008 Posted July 5, 2008 I need to take the Fourier transform of a product of three terms. The statements I have of the convolution theorem are thus: [math]\frac 1 {2\pi} \int dk F(k)G(k)e^{ik x} = \int du f(u)g(x-u) . [/math] How do we express the inverse, namely: [math] \int dx f(x)g(x)e^{-ik x} [/math]? Can I just follow my nose and write a similar convolution integral in k-space?
Dave Posted July 6, 2008 Posted July 6, 2008 I've found Wikipedia is a good reference for many general equations of the Fourier transform (although you have to be a little careful of the normalization that you're using). Anyway, yes, the convolution identity you've got up there will work in reverse and I believe the proof will work in exactly the same way.
ajb Posted July 6, 2008 Posted July 6, 2008 I don't think there is any real difference between x and k, so the formula will work "both ways". Just be careful of normalisation and any minus sign conventions.
Norman Albers Posted July 6, 2008 Author Posted July 6, 2008 Thanks, Dave and ajb, I hope to lose my fear of convolution.
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