0
$\begingroup$

For an input signal $x(t)$ and output signal $y(t)$ through an LTI system I $H(t)$ I have found and interesting property that concerns signals' power spectral densities:

$${\lvert H(\omega)\rvert}^2 = \frac{\phi_y(\omega)}{\phi_x(\omega)}$$

This relation gives me information on the magnitude of $H(\omega)$ but no phase information. Is there a way to obtain phase information of $H(\omega)$ from this relation?

|improve this question
$\endgroup$
  • $\begingroup$ hey, what is $\phi_\cdot$ in your question? And where's the conjugate product of transfer functions from your title (maybe I'm just confused)? $\endgroup$ – Marcus Müller Mar 10 '18 at 14:07
  • $\begingroup$ $\phi$ is power spectral density function. $H(\omega)\cdot H^*(\omega) = {\lvert H(\omega)\rvert}^2$ $\endgroup$ – MarkoP Mar 10 '18 at 15:37
1
$\begingroup$

The answer to your question is no.

The relationship you show is well known and used to relate the input to output of Gaussian stochastic processes through a linear filter. Any filter that has the same magnitude response $\mid H(\omega ) \mid$ will produce the same result.

One can often infer a phase by assumptions on the filter such as linear phase or minimum phase but this isn’t usually the case.

If you want the actual phase, you need to compute the cross spectrum.

|improve this answer
$\endgroup$
  • $\begingroup$ Thank you! I already calculated cross spectrum and I know that I can get the information from that but I am looking into different applications of power spectral densities and wanted to check out the possibility of using this relation as well. $\endgroup$ – MarkoP Mar 12 '18 at 8:55
0
$\begingroup$

The easiest way would be $$H(\omega)=\frac{Y(\omega)}{X(\omega)}$$ which gives you both magnitude and phase.

|improve this answer
$\endgroup$
  • $\begingroup$ Yes, exactly but I do not want to use this. I would specifically like to use the ratio of power spectral densities. $\endgroup$ – MarkoP Mar 10 '18 at 15:38
  • $\begingroup$ Why not? There are multiple ways to get the phase spectrum, but without understanding why the easiest way doesn't work for you, it's hard to give you alternatives. You could use cross power spectrum, but that still involves making Fourier transforms. You can't get any phase information from the power spectrum itself. $\endgroup$ – Hilmar Mar 10 '18 at 15:48

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.