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Hi someone please explain me the translation property of $\textrm{DFT}$. I am not able to understand it neither from Gonzalez nor from internet. I have done extensive study on this but not able to get it. Really Frustrated!!

I don't understand how shifting is done? What is the difference between $f(x-x_0, y-y_0)$ and $f(x+x_0, y+y_0)$ and what are their corresponding $\textrm{DFT}$'s?

enter image description here

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3 Answers 3

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The idea is the same as the 1D case. Start with the regular old continuous time fourier transform definition (the integral) and calculate the Fourier transform of $e^{j \omega_0 t} f(t)$. Similarly, calculate the inverse Fourier transform (using the integral) for $e^{j \alpha \omega} F(\omega)$.

Alternatively, do this with the discrete time fourier transform (so the transform is a sum and the inverse transform is an integral).

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  • $\begingroup$ can you please share some link or upload a snapshot $\endgroup$
    – Navdeep
    Commented Sep 25, 2014 at 15:14
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    $\begingroup$ Just try calculating it yourself. $\endgroup$
    – Batman
    Commented Sep 25, 2014 at 15:15
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I think all the question is trying to show is that a rotation or phase shift in one domain causes a shift or translation in the other domain. Multiplying by exp(jx) or exp(-jx) is nothing but a phase shift by angle x. I know these are correct properties but I don't know if that answers your question.

-K

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  • $\begingroup$ Please check the link $\endgroup$
    – Navdeep
    Commented Sep 28, 2014 at 16:40
  • $\begingroup$ x[n−D]↔e−j2πkDNX[k] That is, if you delay your input signal by D samples, then each complex value in the FFT of the signal is multiplied by the constant e−j2πkDN. $\endgroup$
    – Navdeep
    Commented Sep 28, 2014 at 16:42
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Multiplying by purely complex exponential is basically a phase shift. If you shift original image, the amplitude of its frequency components remains the same, the only thing that changes is the phase of those components. That's the shifting property of an FFT.

Shifting is done by moving the origin point of an image. If $f(x,y)$ is your original image, then $f(x+x_0,y+y_0)$ is your spatially shifted image where point of image originally located at $(0,0)$ is now located at $(x_0,y_0)$.

The first formula is describing something else — the multiplication of an image by sinusoid (complex exponential). It will translate image's components in the frequency domain.

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