1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of the convolution theorem

  1. Dec 9, 2014 #1
    1. The problem statement, all variables and given/known data
    With the Fourier transform of f(x) defined as F(k)=1/√(2π)∫-∞dxf(x)e-ikx and a convolution of g(x) and h(x) defined by f(x)=[g*h](x)=∫-∞h(x-y)g(y)dy show that the Fourier transform of f(x) equals √(2π)H(k)G(k).

    2. Relevant equations
    In problem

    3. The attempt at a solution
    So I understand the general idea, but I'm stuck on a few little issues. I'm going to drop the integral limits from now on.

    F(k)=1/√(2π)∫dx{∫h(x-y)g(y)dy}e-ikx (we can put e-ikx into the integral over y, allowing the next step)

    Now the idea is to let z=x-y (we want to get h(z) which will lead us to the form of a Fourier transform for h). However all proofs seems to say this means dx=dz to give

    F(k)=1/√(2π)∫dz{∫h(z)g(y)e-ikye-ikzdy}

    The question is, why isn't dz=d(x-y)=dx-dy used (given no sources actually discuss this I gather I am being very silly). I also seem

    Then
    F(k)=√(2π)[1/√(2π)∫h(z)e-ikzdz][1/√(2π)∫g(y)e-ikydy]
    =√(2π)H(k)G(k)

    If anybody could explain that it would be great, thanks :)
     
  2. jcsd
  3. Dec 9, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use the standard two-dimensional change-of-variable formula from ##(x,y)## to ##(u,v)##, where ##u = y, v = x-y##:
    [tex] dx \, dy = \frac{\partial(x,y)}{\partial(u,v)} du \, dv [/tex]
    Here, ##\partial(x,y)/\partial(u,v)## denotes the Jacobian of the "transformation" ##x(u,v), \, y(u,v)##.
     
    Last edited: Dec 9, 2014
  4. Dec 9, 2014 #3
    Ahh, silly me, never thought about it as a 2D variable change!

    So we have
    F(k)=1/√(2π)∫{∫h(x-y)g(y)e-ikxdy}dx

    (x,y)->(w,z) where w=y, z=x-y (so that x=w+z, y=w)
    Then the Jacobian |∂(x,y)/∂(w,z)|=|(∂x/∂w)(∂y/∂z)-(∂x/∂z)(∂y/∂w)|=|(1)(0)-(1)(1)|=1
    Then dxdy=dwdz=dydz and so we're sorted - thanks :)
     
  5. Dec 10, 2014 #4
    Sorry but I am encountering some trouble with this again:

    My notes say that we can show that a convolution
    f(x)=∫h(x-y)g(y)dy
    is symmetric by writing z=x-y so then (dz=-dy, limits swap)
    f(x)=∫h(z)g(x-z)dz
    and noting that z is just a dummy variable so we could call it y again if we really wanted so
    f(x)=∫h(y)g(x-y)dy.

    However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused...

    Edit: I gather f(x) is f evaluated at some fixed point x, so that x is fixed and dx=0 here, unlike in the double integral... Correct?
     
    Last edited: Dec 10, 2014
  6. Dec 10, 2014 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In your first integral, ##y## goes from ##-\infty## to ##+\infty##, while in the second, ##z## goes from ##+\infty## to ##-\infty##. When you re-write that as ##z## going from ##-\infty## to ##+\infty##, your ##-dz## will become a ##+dz##.
     
  7. Dec 10, 2014 #6
    The question was, why is dx zero? But I think I may have answered it myself...
     
  8. Dec 10, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Here was your question: "However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused..."
     
  9. Dec 10, 2014 #8
    Sorry I guess it wasn't that clear - but z=x-y, dz=dx-dy, so how does dy=-dz (i.e how is dx=0)...
     
  10. Dec 10, 2014 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Have you forgotten that you are trying to equate a product of FTs to the FT of the convolution? That means that you will be integrating over ##x## as well (times ##e^{\pm i kx}##, of course), but that ##x## lies outside the ##y## or ##z## integral, so is like a constant when you are looking at the convolution for fixed ##x##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proof of the convolution theorem
  1. Convolution theorem (Replies: 0)

Loading...