What is the Fourier transform of an Airy Disc?

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elham sa
elham sa 2015-8-7
编辑： David Goodmanson 2017-5-18
I know that an airy disc is:
Jinc(r).^2
and Jinc is the Fourier transform of a Disc, am I right?
If so, then Fourier transform of an AiryDisc should be two discs convolved together??
This sounds incorrect. What is the Fourier transform of an Airy Disc?
Given that AiryDiscs are symmetric, their Fourier transform should be real, Does it have a special shape?

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回答（2 个）

Image Analyst
Image Analyst 2015-8-7
The airy disc is the square of the "Sombrero function" https://en.wikipedia.org/wiki/Sombrero_function which is the 2D Fourier Transform of a circular aperture ( a cylinder or disc). It involves the Bessel function 2*J1(rho)/rho. It is the diffraction pattern you'd get at infinity from a plane wave hitting a wall with a circular hole in it. The airy pattern/disc is the square because it's the intensity of the wavefront rather than the wavefront itself.
See my attached demo, underneath the image.
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Ash128
Ash128 2017-5-17
编辑：Ash128 2017-5-17
Its a scaled autocorrelation of the aperture. For a circular aperture, it is the same as taking the convolution (due to the symmetry and not being a complex function). The best way to visualize this is to consider two disks separated by a distance r. The value of the autocorrelation is the overlap between the two disks. The autocorrelation of the disk would be a cone-like function with a soft edge. Analytically, H =(2/pi) (acos(r/2)-(r/2)*(sqrt(1-(r/2)^2))) ; r<=2 and H=0 ; r>2 where r is the normalized aperture radial coordinate. H is the OTF of the system in this case, which is the FT of the PSF (Airy pattern).

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David Goodmanson
David Goodmanson 2017-5-17
编辑：David Goodmanson 2017-5-18
[modified answer]
Hi elham, The Fourier transform of the Airy intensity pattern is the autocorrelation of the aperture disks, same as convoltion in this case I believe. The aperture (simplest case) is equal to 1 inside a circle of radius a, 0 outside. The Fourier transform of that is the sombrero function J_1(rho)/rho as StarStrider has mentioned, where rho = k*a*sin(theta), theta is the angle of observation from the normal and k is the incoming wave number. Taking the abs value squared of that function and transforming back gives the disk autocorrelation, per Ash128's comment. (It's a strange looking function). Here is an example.
N = 1001;
x = linspace(-50,50,N); delx = x(2)-x(1);
y = linspace(-50,50,N); dely = y(2)-y(1);
a = 5; % aperture radius
[xx yy] = ndgrid(x,y);
z = zeros(size(xx));
z((xx.^2 + yy.^2)<a^2)=1; % construct aperture
figure(1)
imagesc(z); colorbar
% transform of aperture
w = fft2(ifftshift(z))*delx*dely;
figure(2)
imagesc(fftshift(real(w))); colorbar % imaginary part is tiny
% transform airy squared function back
w2 = real(w.*w);
zz = fftshift(ifft2(w2))/(delx*dely);
figure(3)
imagesc(real(zz)); colorbar % imaginary part is tiny
% compare fft result to geometrical overlap of two disks.
% for a slice through the center, abs(x) equals the radius
censlice = zz((N+1)/2,:);
ra = abs(x/a);
olap = (pi*a.^2)*(2/pi)*(acos(ra/2)-(ra/2).*sqrt(1-(ra/2).^2));
figure(4)
plot(x,olap,x,censlice,'o')
The aperture function here is a bit primitive but it works all right.
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Ash128
Ash128 2017-5-17
Hey David, after reading your comment I just figured there should be a factor of two dividing r in the equation above. Because, there should be a nonzero overlap between the two disks when the separation between the centers is less than the diameter. I fixed it in my previous comment.
David Goodmanson
David Goodmanson 2017-5-18
编辑：David Goodmanson 2017-5-18
Hi elham & Ash,
Per Ash's first comment I modified the answer to address the question that was asked.

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