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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.

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    $\begingroup$ The product is not a wide-sense-stationary process even if $M(t)$ is, so be careful with your calculations. $\endgroup$ – Dilip Sarwate Jun 1 '14 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 '14 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 '14 at 9:20
  • $\begingroup$ If $x(t)$ were periodic then things would become a bit easier. $\endgroup$ – Matt L. Jun 1 '14 at 10:02
  • $\begingroup$ $x(t)$ is truncated. So I think there is room for reaching a closed result ;) $\endgroup$ – Aglar Jun 1 '14 at 11:04
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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

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

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  • $\begingroup$ 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 '14 at 16:36
  • $\begingroup$ @Aglar: Yes, you're right. $\endgroup$ – Matt L. Jun 5 '14 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 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 at 15:27

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