# instantaneous frequency

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 $f(t)$?

• 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. – Peter K. Sep 10 '13 at 16:19
• Is the greek chsi supposed to be frequency here? – Tom Kealy Sep 10 '13 at 17:01
• yes, you are right – meta_warrior Sep 11 '13 at 1:12
• 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 ($\phi'$) ? – Peter K. Sep 11 '13 at 8:02
• 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. – Tom Kealy Sep 12 '13 at 10:58