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Autocorrelation Of Sum Of Random Processes Homework

The random process $x(t)$ doesn't have a power spectral density (PSD) in the conventional sense because it is non-stationary, due to the presence of the deterministic time-dependent function $d(t)$.

You can compute the auto-correlation function of $x(t)$ as follows:

$$\begin{align}R_x(t_1,t_2)&=E\{x(t_1)x^*(t_2)\}\\&=E\{[d(t_1)+s(t_1)][d^*(t_2)+s^*(t_2)]\}\\ &=d(t_1)d^*(t_2)+d(t_1)E\{s^*(t_2)\}+d^*(t_2)E\{s(t_1)\}+E\{s(t_1)s^*(t_2)\}\\&=d(t_1)d^*(t_2)+d(t_1)\mu_s^*(t_2)+d^*(t_2)\mu_s(t_1)+R_s(t_1,t_2) \end{align}\tag{1}$$

where $\mu_s(t)$ is the (generally time-dependent) mean of the random process $s(t)$, and $R_s(t_1,t_2)$ is its auto-correlation function.

If you assume that $s(t)$ is wide-sense stationary (WSS), and if you further assume (for ease of notation) that $d(t)$ and $s(t)$ are real-valued, then Eq. $(1)$ simplifies to


where I've used the usual substitution $t=t_1$ and $\tau=t_2-t_1$. Note that $(2)$ generally still depends on the absolute time $t$, and, consequently, $x(t)$ is non-stationary and has no PSD.

As already pointed out in a comment, only if $d(t)=d$ is constant (and $s(t)$ is WSS) does the random process $x(t)$ become WSS with auto-correlation function


and its PSD becomes (by taking the Fourier transform of $(3)$)

$$P_x(\omega)=2\pi d(d+2\mu_s)\delta(\omega)+P_s(\omega)\tag{4}$$

where $\delta(\omega)$ is the Dirac delta impulse.

EDIT: As mentioned in a comment by Dilip Sarwate, $P_s(\omega)$ also generally includes a Dirac impulse if the mean $\mu_s$ is non-zero. If you define a zero-mean random process $\tilde{s}(t)$ by


then $P_s(\omega)$ can be written as


Using $(6)$, Eq. $(4)$ becomes

$$P_x(\omega)=2\pi (d+\mu_s)^2\delta(\omega)+P_{\tilde{s}}(\omega)\tag{7}$$

which explicitly shows all Dirac impulses in the PSD of $x(t)$.

Он нас надул. Это кольцо - обман. - Червь удвоил скорость! - крикнула Соши.  - Штрафная санкция. На центральном экране прямо под извещением об ошибке ВР представила зрителям ужасающую картину.