# 2D delta function fourier transform

1. Nov 3, 2009

### quasartek

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. Nov 3, 2009

### foxjwill

Remember that one of the defining properties of the delta function is that
$$\int_{-\infty}^\infty f(x)\delta(x)\,dx=f(0)$$​
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
3. Nov 3, 2009

### quasartek

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?

4. Nov 3, 2009

### foxjwill

No,
$$\iint_{\mathbb{R}^2} f(x,y)\delta(y-x\tan\theta)\,dA$$​
equals the sum of all the f(x,y)'s with $$y-x\tan\theta=0$$. Try expressing that as a (single) integral.