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In below image, we have scaling property of DFT, how the final equation is obtained from the above equation. That is how we are getting the scaling factor , $ \frac{1}{|ab|} $ in the final equation ? Please explain how we are getting $ \frac{1}{|ab|} $. Thank you

enter image description here

Equation 4.12 is given as enter image description here

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  • $\begingroup$ You'll have to read Eq. (4.12) to find out why! $\endgroup$
    – Atul Ingle
    Feb 18, 2014 at 19:59
  • $\begingroup$ the equation 4.12 is general DFT equation. See I have updated the question $\endgroup$
    – user7834
    Feb 19, 2014 at 0:49

2 Answers 2

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$\sum\limits_{m=0}^{M-1} \sum\limits_{n=0}^{N-1} f(am,bn)e^{-j \cdot 2\pi(\frac {k}{M} \cdot m + \frac {l}{N} \cdot n)} $ (let $m'=am, n'=bn$)

$=\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$

($m'$, $n'$ with the incremental interval of $a$ and $b$ respectively)

Note that $\sum\limits_{m'=0}^{M-1} \sum\limits_{n'=0}^{N-1} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$ $=F(k/a,l/b)$

Here $m'$, $n'$ with the incremental interval of $1$, and the points $m$ and $n$ are evenly distributed on the circle, according to the symmetry, you have

$\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')} = \frac {1}{|ab|}F(k/a,l/b)$

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  • $\begingroup$ Thank you, one more doubt.. why the limits are becoming (M-1)/a and (N-1)b ? need some explanation. I am not good at maths. sorry if this question is very silly.. $\endgroup$
    – user7834
    Feb 19, 2014 at 16:26
  • $\begingroup$ It should be (M-1)a... I mis-typed the formula when using latex. Thanks for pointing out. $\endgroup$
    – lennon310
    Feb 19, 2014 at 16:30
  • $\begingroup$ we get (M-1)a because we are scaling in spatial domain and cancelling that effect by taking 1/|ab| in the final equation. Am I correct? explain me this in text please. $\endgroup$
    – user7834
    Feb 19, 2014 at 16:36
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    $\begingroup$ You are correct. m=m'/a, if m ranges from 0 to M-1, m' will be 0 to a(M-1) $\endgroup$
    – lennon310
    Feb 19, 2014 at 16:40
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    $\begingroup$ you are very welcome! Good luck on your study! $\endgroup$
    – lennon310
    Feb 19, 2014 at 17:32
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But 1/ab is scaling the magnitude. Here limits are for the indices.

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