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I want to compare time and frequency marginals of Wigner-Ville distribution and STFT. I can calculate WVD's marginals by simple sum operation since WVD is real. But I could not figure out how to deal with STFT.

For WVD, sum(WVD(x)) and sum((WVD(x))') give the marginals where ' denotes the matrix transpose operation.

Similarly, for STFT, I think I should do two summations, but there are two problems. STFT is complex, and length of STFT is based on the window length and overlap length. In order to compare, length of STFT and WVD should be equal.

I will compare these to the real frequency and time densities, which are $|x(t)|^2$ and $|X(\omega)|^2$, where $X(\omega)$ is the Fourier spectrum of the signal.

Any ideas? Thanks...

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  • $\begingroup$ Usually the STFT con-notates the absolute magnitude of the complex-result from the Fourier transform. So the STFT is $|X(f)|$ to begin with, making it $\in \mathbb{R}^{m \text{x} n}$ $\endgroup$ – Tarin Ziyaee Jun 17 '14 at 13:37
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    $\begingroup$ If you've access to IEEE the following lecture notes might be useful: ieeexplore.ieee.org/xpl/… $\endgroup$ – learner Jun 18 '14 at 5:55

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