I am struggling with the following problem for a while: If I have signal of duration N and I apply window (assume rectangular window) of size M < N how will this influence PSD of that signal after doing fft?

PSD of signal should be scaled (I assume) but by which amount and what is the mathematical explanation behind it?

Thank you for your answers in advance.


The shortest answer to that question is to divide your PSD by the squared sum of window values $w_i$:

$$PSD(n) = \dfrac{2|X(n)|^2}{S^2} $$


$n=0\ldots N/2 $

$S = \sum_{i=0}^{N-1}w_i$

You can read more on that topic in the following article:

Heinzel G., et al - Spectrum and spectral density estimation by the DFT, including a comprehensive list of window functions and some new flat-top windows.

  • $\begingroup$ It's an index of frequency bins: $m=0\ldots N/2 $ $\endgroup$ – jojek Sep 9 '15 at 12:12
  • $\begingroup$ Oh sh... Thank you Peter! Time to get some sleep. Made it $n$. Should be fine now. $\endgroup$ – jojek Sep 9 '15 at 12:15
  • 1
    $\begingroup$ Thank you @jojek. Just one clarification - shouldn't the summation for S go from 0 to M-1 since M is the size of the window? N was in my question length of the signal and window with length M had shorter duration, i.e., M < N. $\endgroup$ – Cali Sep 9 '15 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.