I have white Gaussian noise $F[n]$ with zero mean and autocorrelation $R_F[n_1,n_2]=\delta[n_1-n_2]$.

If now I consider the random process defined as

$$X[n]=u[n]e^{-kn}F[n]$$ Is $X[n]$ a wide-ense stationary random process for all values of $k$?

I tried to solve the autocorrelation function to check if it depended only on the lag but my result was that $X$ is WSS only for $k=0$, but I think that since it's a sort of a linear filtering it should be that it is WSS for all $k$.

  • 1
    $\begingroup$ An even simpler test than the one pointed out by Tendero (+1 to him) is to check whether $\operatorname{var}X[n]$ is the same for all $n$. If not, the process is not WSS. A trivial test is to note that $X[-1] = 0$ has zero variance whereas $\operatorname{var}X[0]=1$; but even restricted to nonnegative $n$, $X[1]=e^{-1}F[1]$ has variance $e^{-2} \neq 1 = \operatorname{var}X[0]$ and we have proved that the $X$ process is not WSS. $\endgroup$ Commented Jan 30, 2018 at 17:10

1 Answer 1


The mean of $X(n)$ is always $\mu_X =0$, because the noise has zero mean. Thus we should check whether the autocorrelation corresponds to a WSS process.

$$\begin{align} R_X(n_1,n_2) &=\mathbb{E}[X(n_1)X(n_2)]\\ &=\mathbb{E}[u(n_1)e^{-kn_1}F(n_1)u(n_2)e^{-kn_2}F(n_2)]\\ &=u(n_1)u(n_2)e^{-k(n_1+n_2)}\mathbb{E}[F(n_1)F(n_2)]\\ &=u(n_1)u(n_2)e^{-k(n_1+n_2)}\delta(n_1-n_2) \end{align}$$

We can evaluate the autocorrelation function in two different cases:

  • If $n_1-n_2\neq0$, then $$R_X(n_1,n_2)=0$$
  • If $n_1-n_2=0$, then $$R_X(n_1,n_2)=u(n_1)e^{-2kn_1}=u(n_2)e^{-2kn_2}$$

The second case shows that $R_X$ depends on the specific values of $n_1$ or $n_2$ and not just their difference, thus the process is not WSS.

  • $\begingroup$ tendero: your answer was nice but, at the end, when you say if $n1-n2 =0$, your really computing the ( scaled) variance of the process so it's okay, that the value doesn't depend on the difference. essentially, there is no difference because it's a variance. so, my point is that it is a WSS. process. $\endgroup$
    – mark leeds
    Commented Jan 30, 2018 at 18:25
  • $\begingroup$ tendero: one thing I see a lot ( I'm not EE ) on this list is that the step function, $u[n]$ is also included when defining a discrete process. Do you know the reason for this ? I don't get why it needs to be included ? I don't see it all the time but sometimes. thanks and if I should start a new question, I can do that. all the best. $\endgroup$
    – mark leeds
    Commented Jan 30, 2018 at 18:28
  • $\begingroup$ @markleeds The variance of the process is not constant, it depends on time. If $n<0$ the variance is $0$; if $n \geq 0$, the variance is $\exp(-2kn)$. So I don't see why you say the process is WSS. $\endgroup$
    – Tendero
    Commented Jan 30, 2018 at 19:05
  • $\begingroup$ @markleeds Regarding the another question, I'm not sure if the step function is so often included when defining processes. For deterministic signals, it may seem familiar because many 'popular' signals wouldn't have a DTFT if they weren't multiplied by $u(n)$, such as $a^n$ or $(n+1)a^n$. The same happens with the $z$-transform. I'm not sure if this is what you are referring to. $\endgroup$
    – Tendero
    Commented Jan 30, 2018 at 19:08
  • $\begingroup$ gotcha. tendero. you are correctt about the variance. it's not constant. I didn't think of $n < 0$. my bad for noise. $\endgroup$
    – mark leeds
    Commented Jan 30, 2018 at 20:01

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