# Zero padding effect on a FFT of gaussian noise

I have a gaussian noise $\nu(t)$ with variance $\sigma^2$. After a FFT I get $X(\omega)$. If now I do the IFFT on the $X(\omega)$ can I say that the result is still a gaussian noise of variance $\sigma^2$? What is the effect of zero padding on FFT? How the zero padding affects the statistics of the noise after FFT and IFFT? Thanks.

## 1 Answer

Think about both questions separately. First of all, the (I)FFT is just an implementation of the (I)DFT, so I'm going to generalize all this to the DFT.

# Does the zero-padded IDFT retain variance?

Parseval's theorem says power out = constant factor · power in, and the power of the zero-padded sequence is the energy of that sequence divided by it's length – and that length is larger than the original length, whereas the energy stayed the same.

# Does the zero-padded IDFT retain gaussianness?

Long story short: yes. This is a result from the fact that the DFT is effectively a large sum of sufficiently identically distributed random variables. What you need is independence (which would imply whiteness), but as mentioned below, most people mean "WGN" when they say "GN".

# Other effects?

When people say "Gaussian noise" they often mean "white Gaussian noise", but since that would have a constant PSD, and you explicitly made it so that your noise realization's Fourier transform is anything but constant, but comes in a boxcar shape, you obviously lose the whiteness.