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Fourier transformation of J0(x)

  1. Feb 15, 2010 #1

    KFC

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    In wiki (http://en.wikipedia.org/wiki/Fourier_transform), there said the fourier transform of the Bessel function (zeroth order J0) is a rect function (window). But I also saw a text (about optics) that the Fourier transform on a ring slit (or ring disk) is zeroth-order Bessel function, so which one is correct? If wiki is correct, what is the fourier transform on a ring disk?
     
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  3. Feb 15, 2010 #2

    nicksauce

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    Well presumably it is the same thing, but the first one is in x,y space, and the second in is radial space.
     
  4. Feb 16, 2010 #3
    It is a rectangular window divided by sqrt(1-omega^2). In case of the 2 dimensional Fourier transform, you are considering the function J0(sqrt(x^2 + y^2)) as nicksauce said.
     
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