# PSD of a sum of two stationary real signals

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 $S_{\Gamma,\Gamma}(\omega)$:

$\Gamma(t)=w_1(t)-w_2(t)-u_1(t)+u_2(t)$

using the PSDs that I know?

• It is usually appreciated if you add your own thoughts and effort for homework type questions. Where are you stuck? What have you tried? – 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

$$r_{y,x}(\tau)=r^*_{x,y}(-\tau)$$

you have

$$r_{x,y}(\tau)+r_{y,x}(\tau)=r_{x,y}(\tau)+r^*_{x,y}(-\tau)$$

and the corresponding cross-PSD term becomes

$$S_{x,y}(\omega)+S_{y,x}(\omega)=S_{x,y}(\omega)+S^*_{x,y}(\omega)=2\text{Re}\{S_{x,y}(\omega)\}$$

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

$$S_{\Gamma,\Gamma}(\omega)=S_{u_1,u_1}(\omega)+S_{u_2,u_2}(\omega)+S_{w_1,w_1}(\omega)+S_{w_2,w_2}(\omega)\\-2\text{Re}\{S_{u_1,u_2}(\omega)+S_{u_1,w_1}(\omega)+S_{u_2,w_2}(\omega)(\omega)+S_{w_1,w_2}(\omega)\}\\+2\text{Re}\{S_{u_1,w_2}(\omega)+S_{u_2,w_1}(\omega)\}$$