To perform the filtering infrequency domain we perform multiplication of $(-1)^{x+y}$ why??

  • $\begingroup$ Do you mean why does that center the transform? Or do you mean why do we want to center the transform? $\endgroup$ Feb 29, 2020 at 0:12
  • $\begingroup$ I mean 1. How that center the transform and 2. Why do we want to center the transform ?? $\endgroup$ Feb 29, 2020 at 13:44

1 Answer 1


Below 1D argumentation also explains the 2D case.

First consider the DTFT property for the pair $x[n] \longleftrightarrow X(e^{j\omega})$

$$ e^{j\omega_0 n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \omega_0)}) $$

Then recognise that $(-1)^n = e^{j \pi n} $ which yields:

$$ e^{j\pi n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \pi)}) $$

The effect is such that the spectrum $X(e^{j \omega})$ is centralized (shifted) into the $\omega = \pi$ frequency.

In 2D, the DTFT of the image is shifted into the central zone.

  • $\begingroup$ Can you answer why we center the transform in frequency domain filtering?? $\endgroup$ Feb 29, 2020 at 13:47
  • $\begingroup$ You don't have to center anything for spatial-domain filtering. But sometimes filter impulse responses are specified in a shifted way, then you need to shift to image transform as well to align the phases, especially if frequency domain filtering is applied. $\endgroup$
    – Fat32
    Feb 29, 2020 at 19:10

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