I have 4 signals ($w_1(t),w_2(t),u_1(t),u_2(t)$ of which I know the power spectral densities (PSDs) $S_{w_1,w_1}(\omega),S_{w_2,w_2}(\omega),S_{u_1,u_1}(\omega),S_{u_2,u_2}(\omega)$ and the cross PSD $S_{w_1,w_2}(\omega),S_{w_1,u_1}(\omega),S_{w_1,u_2}$,... etc for all.

Is it possible to obtain the PSD of $S_{\Gamma,\Gamma}(\omega)$:


using the PSDs that I know?

  • $\begingroup$ It is usually appreciated if you add your own thoughts and effort for homework type questions. Where are you stuck? What have you tried? $\endgroup$ – Matt L. Jun 28 '16 at 11:40

Yes, it is possible. Note that the PSD $S_{\Gamma,\Gamma}(\omega)$ is the Fourier transform of the auto-correlation function $r_{\Gamma,\Gamma}(\tau)=E\{\Gamma(t)\Gamma(t+\tau)\}$, so we can proceed by finding $r_{\Gamma,\Gamma}(\tau)$:

$$\begin{align}r_{\Gamma,\Gamma}(\tau)=&E\{[w_1(t)-w_2(t)-u_1(t)+u_2(t)]\\ &[w_1(t+\tau)-w_2(t+\tau)-u_1(t+\tau)+u_2(t+\tau)]\}\end{align}$$

Writing this out and exchanging the expectation operator with the summations gives you an expression for $r_{\Gamma,\Gamma}(\tau)$ as the sum of the auto-correlation functions of $w_1(t)$, $w_2(t)$, $u_1(t)$, and $u_2(t)$, and all possible cross-correlations between the different signals (some with positive and some with negative signs). Each cross-correlation shows up twice, e.g., you get a term $r_{u_2,w_1}(\tau)+r_{w_1,u_2}(\tau)$. Since


you have


and the corresponding cross-PSD term becomes


With this result you finally obtain for the PSD of $\Gamma(t)$


| improve this answer | |

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.