# Scaling Property in DFT

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

Equation 4.12 is given as

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

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

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

• 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.. Feb 19, 2014 at 16:26
• It should be (M-1)a... I mis-typed the formula when using latex. Thanks for pointing out. Feb 19, 2014 at 16:30
• 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. Feb 19, 2014 at 16:36
• You are correct. m=m'/a, if m ranges from 0 to M-1, m' will be 0 to a(M-1) Feb 19, 2014 at 16:40
• you are very welcome! Good luck on your study! Feb 19, 2014 at 17:32

But 1/ab is scaling the magnitude. Here limits are for the indices.