Fourier Transform of a Square Annulus

[DrFurious]

Hello all. I'm trying to compute the Fourier transform of a square annulus analytically. A "square annulus" would be the square analog of an infinitely thing ring (circular annulus). Here's what I know:

The Fourier transform of a circular annulus is a Bessel function. In polar coordinates,

\int_0^\infty\int_0^{2 \pi}\!\delta(r-a)e^{i k r \cos{\theta}}r\,\mathrm{d}r \,\mathrm{d}\theta=2 \pi J_0(ka)

The Fourier transform of a square aperture is a sinc function:
\int_{-1/2}^{1/2}\int_{-1/2}^{1/2}e^{i(k_x + k_y y)}\mathrm{d}x\,\mathrm{d}y=\frac{4\sin\left(\frac{k_x}{2}\right)\sin\left(\frac{k_y}{2}\right)}{k_x k_y}

I was thinking of maybe convolving the functions to get the correct output, but I'm not so sure. Any ideas on how to solve the square annulus problem?

 

[marcusl]

Try writing your annulus as a large square box minus a smaller square box. The FT will be the difference of two sincs.

 

[DrFurious]

I don't know if that quite does it. In that case, the annulus still has some thickness... I want it to be a true delta function.

I think that if one could take the derivative of the two dimensional box function and use that, then maybe something tractable might pop out...

 

[marcusl]

Ah, I missed that you wanted a square line of infinitesimal thickness. In that case use delta functions. The problem is separable--that is, you can transform x and y separately because your function has the form
f(x,y)=g(x)h(y)
where g and h are each the sum of two delta functions. Here's g
g(x)=a[\delta(x+x0)+\delta(x-x0)].
The transform will be the product F(kx,ky)=G(kx)H(ky), where G and H each are cosines.

 

[DrFurious]

So this is getting trickier. I have attached two files: One is the Fourier transform of just a "square", and the other a "square annulus". I did these two numerically.

If you make something like

\frac{4 \sin\left(\frac{k_x}{2}\right) \sin\left(\frac{k_y}{2}\right)}{k_x k_y}-\frac{4 \sin\left(0.999\frac{k_x}{2}\right) \sin\left(0.999\frac{k_y}{2}\right)}{k_x k_y}

You get the correct answer as you get closer to 1. However, there doesn't seem to be a way to do the limit analytically... (just get zero)

Separating the Fourier transform as suggested seems to produce a periodic sinusoidal "lattice", which is also not quite the correct answer.

There must be some technique I'm missing... maybe a way to do Fourier transforms piecewise, or some strange coordinate. I tried making this delta function box in polar coordinates but the integrals were too nasty.

 

[marcusl]

Sorry, you are right, I flubbed the function f. It is a sum of four line segments, each of finite length.
f(x,y)=a[\delta(x+x_0)+\delta(x-x_0)]rect\left(\frac{y}{2y_0}\right)+a[\delta(y+y_0)+\delta(y-y_0)]rect\left(\frac{x}{2x_0}\right).
Now do the x transform of f(x,y), then do the y transform of the result. You will get the sum of two terms, each of which is a product of a sinc times a cosine.

Sorry again to lead you astray.