so I'm working on a problem where I need to find a PSD for the sum of a deterministic and stochastic function. So assume $x(t)$ being this sum:

$x(t) = d(t)+s(t)$

where $d(t)$ is the deterministic function and $s(t)$ is the stochastic function. How do I write down the autocorrelation $R(\tau )$ and the PSD for $x(t)$ that we calculate from Fourier transforming $R(\tau )$?

  • $\begingroup$ Does the problem provide the mean and variance of the stochastic function? $\endgroup$ Aug 18, 2015 at 23:27
  • $\begingroup$ No, the solution needs to be a general expression for the autocorrelation of the sum of a deterministic function and stochastic signal.. $\endgroup$ Aug 19, 2015 at 0:23
  • $\begingroup$ If $d(t)$ and $s(t)$ are independent (they must be I guess), then you can write the autocorrelation of $x(t)$ as $R_x(\tau)=R_d(\tau)+R_s(\tau)$. $\endgroup$
    – Oliver
    Aug 19, 2015 at 0:36
  • $\begingroup$ Is s(t) wide sense stationary? $\endgroup$
    – bean
    Aug 19, 2015 at 4:18
  • 2
    $\begingroup$ Note that a non-constant deterministic signal is not a stationary random process, whereas the concept of PSD really only makes sense for stationary random processes. Note that the key difference between x and s is that E[x(t)] = E[ d(t) +s(t) ] = d(t) + E[s(t)]. For x and s to both be stationary processes, d would have to be a constant, i.e., d(t) = d for all t. $\endgroup$
    – bean
    Aug 19, 2015 at 4:35

1 Answer 1


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 $x(t)$.

  • $\begingroup$ +1 I agree with what you have written, but I think (4) needs re-writing a little to help beginners. If $d(t) = d$ is a constant and $s(t)$ is a WSS process with nonzero mean $\mu_s$, then $P_s(\omega)$ includes an impulse $2\pi\mu_s^2\delta(\omega)$ which can be profitably be combined with the one that you have to get $2\pi(d+\mu_s)^2\delta(\omega)$. After all, $d+s(t)$ is just a WSS process with mean $d+\mu_s$, and so its PSD should have an impulse at $0$ to represent the "DC" power of $(d+\mu_s)^2$. $\endgroup$ Aug 19, 2015 at 14:35
  • $\begingroup$ @DilipSarwate: I see what you mean; I think the advantage of $(4)$ is that it explicitly includes the PSD of $s(t)$ (which of course has a delta impulse, as you mentioned). The advantage of your way of writing the equation is that it explicitly shows the sum of all deltas in the total PSD. So both forms are valuable for understanding what's going on. I'll leave $(4)$ and add the other equation that you suggested. $\endgroup$
    – Matt L.
    Aug 20, 2015 at 6:21
  • $\begingroup$ I understand the mathematical derivation done in the answer, but I have the following doubt: if d(t) has a Fourier transform, we know it has a "power spectrum", and if s(t) has a PSD it also has a "power spectrum", since x(t) is the sum of both signals I intuitively think it also has a power spectrum. I can see it's not a PSD, but what is it? I'm sure it exists... $\endgroup$
    – MPA95
    Jun 1, 2021 at 14:18

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