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Calculate the Fourier transform of a product of three functions

  1. Oct 18, 2009 #1
    I have a problem understanding the following:

    I should calculate the Fourier transform of a product of three functions:

    [tex] \mathcal{F} \left[ f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) \right] = \int dx_{1} dx_{2} f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) e^{i p x_{1} + i q x_{2}} [/tex]

    okay, and this goes over into a convolution:

    [tex] = \dfrac{1}{2 \pi} \hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q) = \dfrac{1}{2 \pi} \int dk \hat{f}(p-k) \hat{g}(q-k) \hat{h}(k) [/tex].

    I know how to calculate the convolution between two functions, but here we have three and I don't undestand how to get to the last line here. Could somebody explain that to me, please?
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    HallsofIvy

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    Re: Convolution

    Pretty much the same way you extend the product of two numbers to three or the sum of two numbers to three. Writing f #g for the convolution of two functions, the convolution of three can be written (f#g)#h= f#(g#h).
     
  4. Oct 18, 2009 #3
    Re: Convolution

    ok, I start with:

    [tex] \hat{g}(q) \ast \hat{h}(p+q) = \int dk ~ \hat{g}(p+q-k) \hat{h}(k) [/tex]

    But if I go further and calculate the second convolution, i.e. [tex] \hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q) [/tex], I would get a second integral. But in the equation above there is just one. I don't understand that. Maybe it is trivial, but I just don't see what I have to do?
     
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