How to Approach the Fourier Transform of an Annulus?

In summary, the individual pieces of the function can be taken separately and then summed. It is recommended to write the function as a product of two simpler functions in order to simplify the Fourier transform. The transform in polar coordinates is defined as an integral, but since the function is only non-zero in a specific region, the integral can be simplified to only include that region. Further assistance is needed when approaching the integral.
  • #1
tx213
7
0
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!
T
 
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  • #2
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.
 
  • #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!
 

FAQ: How to Approach the Fourier Transform of an Annulus?

1. What is the Fourier transform of an annulus?

The Fourier transform of an annulus is a mathematical operation that decomposes a function defined on a circular region into its constituent frequency components. It expresses the function as a sum of complex exponential functions with different frequencies.

2. How is the Fourier transform of an annulus calculated?

The Fourier transform of an annulus can be calculated using the formula: F(k) = ∫f(θ)e^(-ikθ)dθ, where f(θ) is the function defined on the annulus and k is the frequency variable. This integral can be evaluated using techniques such as integration by parts or the residue theorem.

3. What is the importance of the Fourier transform of an annulus in science?

The Fourier transform of an annulus has many applications in science, particularly in signal processing and image analysis. It allows us to analyze the frequency content of a signal or image and filter out unwanted components. It is also used in solving differential equations and in quantum mechanics.

4. Can the Fourier transform of an annulus be extended to higher dimensions?

Yes, the Fourier transform of an annulus can be extended to higher dimensions, such as a sphere or a cylinder. However, the formula and techniques used may differ slightly from the 2D case. In higher dimensions, the Fourier transform is also used in fields like medical imaging and materials science.

5. Are there any limitations or challenges associated with the Fourier transform of an annulus?

One limitation of the Fourier transform of an annulus is that it assumes the function to be continuous and periodic. In reality, many signals and images are not perfectly continuous or periodic, which can lead to inaccuracies in the transformation. Additionally, computing the Fourier transform can be computationally expensive for large datasets, making it challenging to use in real-time applications.

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