6
$\begingroup$

I was wondering how to calculate the autocorrelation of a deterministic signal $x(t)$ multiplied by a stochastic process $M(t)$, whose autocorellation $R_M(\tau)$ is known a priori. In my case, $x(t)$ is a truncated monolateral exponentially decaying function.

I suppose that the result of such multiplication $y(t) = x(t) \cdot M(t)$ is again a stochastic process, but when approaching the calculation of the autocorrelation of $y(t)$ I obtain something that is not even, therefore I suppose I am making some mistakes. I know that the definition of autocorrelation for deterministic signals is different from the one of stochastic processes, but I do not know how to connect the two of them.

$\endgroup$
5
  • 1
    $\begingroup$ The product is not a wide-sense-stationary process even if $M(t)$ is, so be careful with your calculations. $\endgroup$ Jun 1, 2014 at 2:30
  • $\begingroup$ Just to add to what Dilip said, since the random process that results from taking that product is not WSS, there is no such autocorrelation function $R_y(\tau)$. Instead, the autocorrelation is of the form $R_y(t_1, t_2)$; it will be a function of two variables that correspond to the two time indices in the process that the expectation is taken over. $\endgroup$
    – Jason R
    Jun 1, 2014 at 3:28
  • $\begingroup$ Therefore, in order to take the Fourier transform of $R_y(t_1,t_2)$ I need firstly to take the time average over $t_1$ and $t_2$ of the autocorrelation, right? $\endgroup$
    – Aglar
    Jun 1, 2014 at 9:20
  • $\begingroup$ If $x(t)$ were periodic then things would become a bit easier. $\endgroup$
    – Matt L.
    Jun 1, 2014 at 10:02
  • $\begingroup$ $x(t)$ is truncated. So I think there is room for reaching a closed result ;) $\endgroup$
    – Aglar
    Jun 1, 2014 at 11:04

1 Answer 1

5
$\begingroup$

As correctly pointed out in the comments, in general the process $Y(t)=x(t)M(t)$ is not wide-sense stationary (WSS), i.e. its autocorrelation function depends not only on the time difference parameter $\tau$, but also on the absolute time $t$:

$$R_Y(\tau,t)=E[Y(t+\tau)Y^*(t)]=E[x(t+\tau)M(t+\tau)x^*(t)M^*(t)]=\\ =x(t+\tau)x^*(t)E[M(t+\tau)M^*(t)]=x(t+\tau)x^*(t)R_M(\tau)\tag{1}$$

In your case you just need to evaluate (1) with the given function $x(t)$. However, as expected you will end up with a function of two variables because $Y(t)$ is not WSS.

There are a few special cases in which the resulting process is indeed WSS or can be easily made WSS. The first case is modulation by a complex exponential:

$$x(t)=e^{j\omega_ct}$$

in which case

$$x(t+\tau)x^*(t)=e^{j\omega_c(t+\tau)}e^{-j\omega_ct}=e^{j\omega_c\tau}$$

only depends on $\tau$ and not on $t$. Another case of interest is the case where $x(t)$ is $T$-periodic. In this case the process $x(t)M(t)$ can be made WSS by introducing a random phase epoch $\Theta$ with uniform distribution in the interval $[0,T]$ which is independent of $M(t)$:

$$Y(t)=x(t+\Theta)M(t)$$

The autocorrelation function of $Y(t)$ is then given by

$$R_Y(\tau)=R_M(\tau)\frac{1}{T}\int_0^Tx(\alpha+\tau)x^*(\alpha)d\alpha$$

$\endgroup$
4
  • $\begingroup$ Thanks for your reply. If I understand well, writing $R_Y(t_1,t_2)=R_x(t_1,t_2)R_M(t_1,t_2)$ is wrong. Am I right? $\endgroup$
    – Aglar
    Jun 5, 2014 at 16:36
  • $\begingroup$ @Aglar: Yes, you're right. $\endgroup$
    – Matt L.
    Jun 5, 2014 at 19:45
  • $\begingroup$ Thank you for the great answer. Can you provide more discussions on how to make $Y(t)$ WSS (especially for real $x(t)$)? Or do you recommend any book chapters on this kind of discussions? Thanks. $\endgroup$
    – WDC
    May 3, 2019 at 14:37
  • $\begingroup$ @WDC: For real $x(t)$ you'll have to use a periodic signal and a random phase, as explained in the answer. $\endgroup$
    – Matt L.
    May 3, 2019 at 15:27

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.