How to Approach the Fourier Transform of an Annulus?

[tx213]

Hi guys,

I've been using this site for a while now, but this is going to be my first post. I want to pick your brains to get some insight on this problem I'm tackling.

I'm trying to take a Fourier Transform of a function. My function is a function of (r,phi) and it is a piecewise function where:

f(r,θ) = 0 , r < r_inner
f(r,θ) = cos(θ)^2 + (-0.5)*sin(θ)^2 , r_inner ≤r ≤ r_outter
f(r,θ) = 0 , r > r_outter

I've attached a figure here.

Can I take the FT of the pieces individually and then sum? My knowledge so far tells me this is OK. Since my function is in polar coordinates, I should take the FT in polar coordinates; is there an efficient (clever) way to go about this given the the nature of the function, perhaps that it is symmetric every n*pi ?

Any suggestions/insight would be really appreciated. Thanks in advance!
T

 

[AlephZero]

The "best" way to do this depends on how you want to use the transform.

A good way to start is probably to write f(r,θ) = R(r)T(θ), where R(r) is 0 or 1, and T(θ} = cos(θ)^2 + (-0.5)*sin(θ)^2.

Using elementary trig formulas, T(θ) = a + b cos(2θ) (work out the constants a and b for yourself!) so its Fourier transform is simple.

You could take the Fourier transform of R(r) for all non-negative values of r, or just restrict the way that you use the function, to the region where it is non-zero.

 

[tx213]

Ah awesome thanks for getting back! I made some more progress.

The Fourier transform in polar coordinates is defined as this (I will just list one of them).
F(ρ,ø) = FT[f(r,θ)] = ∫∫ ƒ(r,θ) * exp( i2$\pi$*ρ*r*cos(ø-θ) ) r dr dθ , r from 0→∞ , θ from 0→ 2 $\pi$.

This works great because f(r,θ) is 0 everywhere except between when r_inner ≤ r ≤ r_routter.
So I only need to take one integral, which is

∫∫ ƒ(r,θ) * exp( i2$\pi$*ρ*r*cos(ø-θ) ) r dr dθ , r from r_inner → r_outter , θ from 0→ 2 $\pi$ , with ƒ(r,θ) = ( (3/4)*cos(2θ) ) + 1/4

Here I am stuck again. How should I think about taking/approaching this integral? Again, thanks in advance for any insight!