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2d dft

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate, by hand, the DFT of cos(2*pi(fx*m + fy*n)). It should simplify to something simple. It should NOT be left as a summation.

    2. Relevant equations

    2D DFT Formula:
    [tex]
    \tilde{s(k,l)} = \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} s(m,n) e^{-j 2 \pi (\frac{1}{M}mk + \frac{1}{N}ln)}
    [/tex]

    3. The attempt at a solution


    [tex]
    \tilde{s(k,l)} = \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} cos(2 \pi (f_x m + f_y n)) e^{-j 2 \pi (\frac{1}{M}mk + \frac{1}{N}ln)}
    [/tex]


    One of Euler's Formulas:

    [tex]
    cos(u+v) = \frac{1}{2}(e^{ju + jv} + e^{-ju -jv})
    [/tex]


    After much algebra crunching, I wound up with this:

    [tex]
    \tilde{s(k,l)} = \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} \left[ \frac{1}{2}e^{-j2 \pi
    \frac
    {-f_x m M N + m k N + l n M - f_y n M N}
    {MN}
    } + \frac{1}{2} e^{-j2 \pi
    \frac
    {f_x m M N + f_y n M N + m k N + l n M}
    {MN}} \right]
    [/tex]

    I am really not sure how to simplify this double geometric sum. Does it look recognizable? Was there an earlier simplification I could take?
     
  2. jcsd
  3. Jun 5, 2013 #2

    haruspex

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    Consider just one of the two added terms. As a function of n, it's of the form exp(An+B) = exp(A)nexp(B). So isn't it just the sum of a geometric series?
     
  4. Jun 5, 2013 #3
    So if e^{An+B} = e^{An} e^{B}, then e^{B} simply acts as a constant for the internal geometric sum?

    [(1-A^M)/(1-A)] e^{B} ???

    And then I can take the geometric sum again?

    The original cosine function is not separable...is it?
     
  5. Jun 5, 2013 #4

    haruspex

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    Looks ok to me.
     
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