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If my signal is

$$ f(t)=\exp[i\phi(t)], $$ how do I show that

$$ \int_{-\infty}^{\infty} {\xi}\left|S_f(u,\xi)\right|^2 d\xi = 2\pi\int_{-\infty}^{\infty} {\phi'(t)}\left|g(t-u)\right|^2 dt, $$

where $$ S_f(u,\xi)=\int_{-\infty}^{\infty}f(t)g(t-u)\exp{(-i\xi t)} dt $$ is the short-time Fourier transform of $f(t)$?

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    $\begingroup$ You might get a better response if you explain a little more where you're getting these equations from and what you hope to get out of it. As it is, the question is a little sparse. $\endgroup$ – Peter K. Sep 10 '13 at 16:19
  • $\begingroup$ Is the greek chsi supposed to be frequency here? $\endgroup$ – Tom Kealy Sep 10 '13 at 17:01
  • $\begingroup$ yes, you are right $\endgroup$ – meta_warrior Sep 11 '13 at 1:12
  • $\begingroup$ I believe what the question is asking is: How do I show that the first moment in frequency of the short-time Fourier transform magnitude squared is equal to a smoothed version of the instantaneous frequency ($\phi'$) ? $\endgroup$ – Peter K. Sep 11 '13 at 8:02
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    $\begingroup$ Given your signal, it's relatively easy to show that the spectrogram is just the absolute value of the window function squared. Then you just make the substitution that chsi = dpsi/dt and change the range of integration to be over the unit circle rather than R. $\endgroup$ – Tom Kealy Sep 12 '13 at 10:58

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