0
$\begingroup$

Welch's method splits a time signal, $x(n)$ into $M$ periodograms $P_m$,

$P_{x_m,M }(k) = \frac{1}{M}|F_k(x_m)|^2$

and averages them to give the Power Spectral Density (PSD),

$S_{x}(k) = \frac{1}{K}\sum_{m=0}^{K-1} P_{x_m,M }(k)$

But if I am not interested in the power can I use something like an averaged Fourier transform to represent the Fourier transform for the whole signal $x(n)$?

I.e. The exact same as above but instead of averaging the periodograms, $|F_k(x_m)|^2$, I just average the fourier transforms $F_k(x_m)$ within each window?

Kind of like a STFT except averaging the windows, and is there any merit to using something like this?

$\endgroup$
2
$\begingroup$

Since an FFT is a linear operator, adding up the complex results of a sequence of FFTs of short windows is the same as doing a single short FFT on the the vector addition of all those short windows.

Note that for signals that are exactly integer periodic in the FFT width (sequential 0% overlapped windows), the vector addition will constructively interfere. But for signals that are not exact integer periodic in the FFT width, there can be significant destructive interference. So the result will be highly scalloped, with a narrow high gain bandpass around each FFT bin center, and potentially a lot of signal loss between bin centers. Sometimes this is useful, sometimes not, but a very different result from a single longer FFT over the entire window set.

$\endgroup$
1
$\begingroup$

You can average complex cross spectra or cross correlation like Welch.

For something like coherence, you do both, complex averages and power averages.

It can tell you a lot about a system as opposed to a single signal.

$\endgroup$
0
$\begingroup$

is there any merit to using something like this?

I don't think so. Unless your signal is somehow phase locked with your analysis window, the individual complex Fourier coefficients will just cancel each other simply because of phase variations.

$\endgroup$

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.