I really need your help here:
Think of $f(x,y)$ as an image where the dimensions are: $NxM$.
I can't prove this property of 2D transform fourier:
if the transform of $f(x,y)$ is $F(u,v)$ so the transform of $f(x-x_0,y-y_0)$ is $F(u,v)e^{-2{\pi}i(ux_0/{N}+ vy_0/M)}$
Any help will be appreciated. thanks.
after changing $x$ by $x-x_0$ and $y$ by $y-y_0$ we will get: $F'(u,v)=\frac{1}{NM}\Sigma^{N-1}{x=0}\Sigma^{M-1}{y=0}f(x-x_0,y-y_0)e^{-2\pi{i}(\frac{u(x-x_0)}{N}+\frac{v(y-y_0)}{M})}$
and after few trival steps: $F'(u,v)=\frac{1}{NM}\Sigma^{N-1}{x=0}\Sigma^{M-1}{y=0}f(x-x_0,y-y_0)e^{-2\pi{i}(\frac{ux}{N}+\frac{vy}{M})}e^{2\pi{i}(\frac{ux_0}{N}+\frac{vy_0}{M})}$
what can I do now? I searched this proof for a week and I was failed..
– daniel Jan 22 '16 at 13:29$$DFTf(n) = \sum_{n=0}^{N-1} f(n) e^{-2 i \pi n k/N}$$
with $f(n+N) = f(n)$ and $a \in \mathbb{Z}$ :
$$DFTf(n-a) = \sum_{n=0}^{N-1} f(n-a) e^{-2 i \pi n k/N} = \sum_{m=-a}^{N-1-a} f(m) e^{-2 i \pi (m+a) k/N} = \sum_{m=0}^{N-1} f(m) e^{-2 i \pi (m+a) k/N} = e^{-2 i \pi a k/N} DFTf(n)$$
– reuns Jan 22 '16 at 14:16