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Fourier transform of an annulus

  1. Feb 7, 2014 #1
    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. forupload.jpg 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!
  2. jcsd
  3. Feb 7, 2014 #2


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    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.
  4. Feb 8, 2014 #3
    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[itex]\pi[/itex]*ρ*r*cos(ø-θ) ) r dr dθ , r from 0→∞ , θ from 0→ 2 [itex]\pi[/itex].

    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[itex]\pi[/itex]*ρ*r*cos(ø-θ) ) r dr dθ , r from r_inner → r_outter , θ from 0→ 2 [itex]\pi[/itex] , 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!
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