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Fourier Transform of a Triangular Aperture: Manipulate Exponentials, Sinc functions

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I am attempting to manipulate this equation into a form that, presumably, has 3 Sinc terms. I am attempting to do this because my professor has written "reduce the solution to terms having sinc functions", and I am assuming there are 3, because plotting this equation in mathematica gives 3 "streaks" (this is the fourier transform of a triangular aperture), which, intuitively, corresponds to 3 terms.

    variables are:
    [itex]\text{fy}[/itex]
    [itex]\text{fx}[/itex]

    Constants are:
    [itex]\text{h}[/itex]
    [itex]\text{d}[/itex]


    2. Relevant equations

    This is the equation I have derived. It is correct, but not necessarily of the correct form.

    [itex]\text{G}[\text{fx},\text{fy}]\text{=}\frac{-i d}{2\pi \text{fy}}\left(e^{i \pi (\text{fx} d+\text{fy} h)} \text{Sinc}[\pi (-h \text{fy} + \text{fx} d)]-e^{i \pi (\text{fx} d-\text{fy} h)} \text{Sinc}[\pi (h \text{fy} + \text{fx} d)]\right)[/itex]

    Also, to be clear, [itex]\text{Sinc}[x] = \frac{\text{Sin}[\pi x]}{\pi x}[/itex]

    3. The attempt at a solution

    By putting the Sinc functions in terms of exponentials, I was able to change the form of the equation to this. But I am not sure if this is better or worse, or closer or further from the intended result. Does anyone see a way to get either equation into 3 terms containing Sinc functions?

    [itex]\text{G}[\text{fx},\text{fy}]\text{=}\frac{d (\text{fy} \text{Cos}[2 d \text{fx} \pi ]-\text{fy} h \text{Cos}[2 \text{fy} h \pi ]+i (\text{fy} \text{Sin}[2 d \text{fx} \pi ]-d \text{fx} \text{Sin}[2 \text{fy} h \pi ]))}{2 \text{fy} (-d \text{fx}+\text{fy} h) (d \text{fx}+\text{fy} h) \pi ^2}[/itex]
     
  2. jcsd
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