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Question about convolutions

  1. Aug 15, 2012 #1
    I know that the convolution of two functions [itex]f(x)[/itex] and [itex]g(x)[/itex] is given by

    [tex]
    (f * g)(y) = \int_{\mathbb R} f(x)g(y-x) dx.
    [/tex]

    But what if I'm trying to convolve a function [itex]f(x)[/itex] with a function [itex]g(x + az)[/itex], where [itex]a[/itex] is some constant? Is it just

    [tex]
    (f*g)(y) = \int_{\mathbb R} f(x)g(y - x + az) dx.
    [/tex]

    If so, why? I can't seem to find a definition of the convolution that makes this obvious.
     
  2. jcsd
  3. Aug 15, 2012 #2

    Mute

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    Homework Helper

    Convolutions are symmetric in f and g:

    $$(f \star g)(y) = (g \star f)(y);$$

    you could thus write the convolution of f(x) and g(x + az) as

    $$\int_{\mathbb{R}} dx~f(y-x)g(x+az),$$

    which is perhaps easier to see.
     
  4. Aug 15, 2012 #3

    mathman

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    Gold Member

    Your question is a little puzzling. How does z relate to x in g(x+az)?
     
  5. Aug 16, 2012 #4
    Okay, let me see if I can be more specific. What I'm really trying to do is to take the Fourier transform...in the [itex]x[/itex] variable...of the product [itex]f(x) g(ax + y)[/itex], where [itex]x,y\in \mathbb R[/itex] are independent (hence unrelated) variables. So I'm interested in

    [tex]
    \int_{-\infty}^{\infty} e^{-i p x} f(x) \cdot g(ax + y) \ dx.
    [/tex]

    I know that Fourier transforms turn products into convolutions...but is it that straightforward in this case?
     
  6. Aug 16, 2012 #5

    mathman

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    Gold Member

    What you need is the opposite. Fourier transform of a convolution is product of Fourier transforms of the individual items. Once you've done that, take the back transform to get what you are looking for.
     
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