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2D delta function fourier transform

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Given f(x,y) = DeltaFunction(y - x*tan(theta))

    a) Plot function.
    b) Take fourier transform.
    c) Plot resulting transform.


    2. Relevant equations
    Delta function condition non-zero condition DeltaFunction(0) = Infinity
    Sifting property of delta functions


    3. The attempt at a solution
    I am using Mathematica and can plot a 1D version, DeltaFunction(x), however, I am having trouble extending it to 2D for part a. Part b, I tried using the sifting property integral(deltafunction(y-x*tan(theta))*f(x,y)=f(y-tan(theta)) but could not progress after that point.

    Any help or suggestions would be greatly appreciated.
     
  2. jcsd
  3. Nov 3, 2009 #2
    Remember that one of the defining properties of the delta function is that
    [tex]\int_{-\infty}^\infty f(x)\delta(x)\,dx=f(0)[/tex]​
    for sufficiently chosen functions f. Using http://mathworld.wolfram.com/FubiniTheorem.html" [Broken], you can show that this is also true for double integrals.

    Can you think of a way you might use this to find your fourier transform?
     
    Last edited by a moderator: May 4, 2017
  4. Nov 3, 2009 #3
    I now tried to incorporate that property and Fubini's theorem
    where

    DoubleIntegral[ f(x,y)*Delta(y - x*tan(theta)), x,y,-Inf,Inf] = f(0,0) and plugging f(0,0) into Delta(0-0) gives Delta(0,0) which is Infinity. The Fourier Transform of that is 1?
     
  5. Nov 3, 2009 #4
    No,
    [tex]\iint_{\mathbb{R}^2} f(x,y)\delta(y-x\tan\theta)\,dA[/tex]​
    equals the sum of all the f(x,y)'s with [tex]y-x\tan\theta=0[/tex]. Try expressing that as a (single) integral.
     
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