# Why we have to multiply the images with $(-1)^{x+y}$ to center the transform in frequency domain filtering?

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

• Do you mean why does that center the transform? Or do you mean why do we want to center the transform? – AnonSubmitter85 Feb 29 at 0:12
• I mean 1. How that center the transform and 2. Why do we want to center the transform ?? – vignesh babu Feb 29 at 13:44

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.

• Can you answer why we center the transform in frequency domain filtering?? – vignesh babu Feb 29 at 13:47
• 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. – Fat32 Feb 29 at 19:10