# PSD of a windowed signal

Suppose PSD of a signal $x(n)$ is $PSD_{xx}(\omega)=1$. Now if a window function $w(n)$ of length $0\le n \le N-1$ is applied to $x(n)$, what will the new PSD be?

• What do you mean by "difference between the original and windowed signal?" You've stated the mechanism used to apply the window, so it's not clear what you're asking. Also, stepping back a bit, if you know the PSD of $x[n]$, why are you windowing it and then using the result to estimate a PSD? – Jason R Dec 8 '12 at 20:45
• @JasonR the question is how does windowing a signal affect the value of the estimate of true power spectrum? according to my understanding there will be PSDs one obtained simply from original signal $x(n)$ and other one obtained from $\{w(n)x(n)\}$ so how does windowing the signal affected the PSD? – heavenhitman Dec 8 '12 at 21:18

## 2 Answers

When you "window" a signal in the time domain you are multiplying a block of signal data, element by element, with the window's elements. This corresponds to convolution in the frequency domain.

Windows are usually "pseudo-gaussian", in the sense that they look more-or-less like a bell curve (since they have finite length, of curse, the tails only go so far). Like true gaussians, their Fourier transforms also look like gaussians, so in the frequency domain you will get some smearing. The smearing effect is reduced as the window gets larger.

EDIT: As Dilip and Hilmar pointed out, the correct formulation is $S_{out}(\omega) = S_{in}(\omega)\mid H(\omega)\mid^2$.

• in mathematical form what will be the affect on true PSD? how to related true PSD and PSD estimation? – heavenhitman Dec 14 '12 at 18:08
• The mathematical effect will be to convolve the true PSD with the Fourier transform of the window. – Jim Clay Dec 14 '12 at 20:30
• It would be nice if the downvoter would say why they downvoted. Otherwise it can't be fixed. – Jim Clay Dec 21 '12 at 13:31
• I was the downvote! sorry for the delay. It just needs some clarifications: 1st paragraph, I guess you meant "This corresponds to convolution in the frequency domain", didn't you? The explanation you give seems to follow that track... except the formula corresponds to the PSD of the input signal filtered by a (deterministic) filter. The answer by @Hilmar, with this respect, makes more sense, IMHO. Note that the FTs are convolved, not the PSDs. Note also that the mathematics are true for continuous time and frequency signals, but for discrete ones, one might need to check closer... – Jean-louis Durrieu May 7 '13 at 21:39
• @Jean-louisDurrieu You are correct, I meant convolution in the frequency domain. Good catch. – Jim Clay May 8 '13 at 2:53

Windowing is multiplication in the time domain which corresponds to convolution in the frequency domain, so the spectrum of the windowed signal is the spectrum of the original signal convolved with the spectrum of the window. The PSD is simply the magnitude squared of the spectrum of the windowed signal, i.e. $$P(\omega)=\left \| X(\omega)\ast W(\omega) \right \|^{^{2}}$$ where $\ast$ is the convolution operator. Hence the desire to have the spectrum of the window look like an impulse as that has the least impact on the PSD.