- #1
DrFurious
- 23
- 0
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,
[tex]\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)[/tex]
The Fourier transform of a square aperture is a sinc function:
[tex]\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}[/tex]
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?
The Fourier transform of a circular annulus is a Bessel function. In polar coordinates,
[tex]\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)[/tex]
The Fourier transform of a square aperture is a sinc function:
[tex]\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}[/tex]
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?