2D delta function fourier transform

[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.

[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 Fubini's theorem (http://mathworld.wolfram.com/FubiniTheorem.html), 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?

[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?

[foxjwill]

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)

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.