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Fraunhofer diffraction and convolution of two aperture functions

  1. Apr 21, 2013 #1

    I'm having some trouble understanding the concept of two function convolution in Fraunhofer diffraction.

    Let's say I have an aperture function in the shape of some cosine function (which is always above zero), and I want to calculate the transmission function if I only illuminate 3 such "slits" (so I capture 3 peaks of cosine aperture function). In order to do that, I was told to take the convolution of two Fourier transforms: the transmission function of cosine aperture illuminated over infinite slits and the transmission function of the Rect. function (some kind of box, which captures the 3 peaks of cosine function), which is a Sinc. function.

    Is this correct? I don't quite get the physical meaning of convolution in this particular case.

    Thanks for any explanations.
  2. jcsd
  3. Apr 22, 2013 #2

    Andy Resnick

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    If I understand you correctly, the aperture function could be written as something like cos^2(x)*rect(x/3). Assuming I kept track of the scale factors correctly, this means that 3 cycles of cos^2 (which is always positive) are illuminated.

    Then, the far-field diffraction pattern is the FT of the aperture function, which would be FT(cos^2) convolved with FT(rect(x/3)), or:

    (1/2 Sqrt[\[Pi]/2] DiracDelta[-2 + u] + Sqrt[\[Pi]/2] DiracDelta +
    1/2 Sqrt[\[Pi]/2] DiracDelta[2 + u])**Sinc[3u],

    where DiracDelta[a+u] is a delta function located at u = -a.
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