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!

Fourier transform convolution proof

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Let FT(f) = Fourier transform of f, (f*g)(x) = convolution of f and g.
    Given FT(f*g) = FT(f)FT(g), the first part of the convolution theorem, show that FT[fg] = [FT(f)*FT(g)]/2pi.


    2. Relevant equations

    Duality: FT2f(x) = (2pi)f(-x)

    Convolution: (f*g)(x) = ∫f(u)g(x-u)du


    3. The attempt at a solution

    FT(f*g) = FT(f)FT(g)
    FT2(f*g) = FT[FT(f)FT(g)]
    LHS = FT2(f*g)(x) = (2pi)(f*g)(-x) (by duality) =(2pi)∫f(u)g(-x-u)du (by convolution) = (2pi)∫{[FT2f(-u)][FT2g(x+u)]}/(2pi)2 du
    implying that RHS = FT[FT(f)FT(g)] = (1/2pi)∫FT[FT(f(-u))]FT[FT(g(x+u))] du ≠ FT[fg] = [FT(f)*FT(g)]/2pi

    Where did I go wrong?

    Thanks!
     
  2. jcsd
  3. May 5, 2014 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Try letting f = FT(F) and g=FT(G). Then evaluate FT2(f*g) and FT[FT(f)FT(g)] separately and show they are equal.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted