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Defining:

$X(t)$ WSS random process with autocorrelation function $R_{X}(\tau) = \mathbb{E}[X(t)X(t+\tau)]$.

$Y[n] = X(nT)$ (sampling of $X$ at a rate $\frac1T$) with autocorrelation function $R_Y(\tau) = \mathbb{E}[Y[n]Y[n+\tau]]$.

Which is the relationship between $R_X(\tau)$ and $R_Y(\tau)$?

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  • $\begingroup$ when correlating two different signals, and $X(t)$ and $X(nt)$ are two different signals, we call that cross-correlation not auto-correlation. $\endgroup$ – robert bristow-johnson Feb 26 at 23:26
  • $\begingroup$ What I'm interested about is in the relationship between the autocorrelation of both signals, not in the correlation between both... perhaps I expressed myself wrongly $\endgroup$ – MPA95 Feb 26 at 23:27
  • $\begingroup$ well, it seems that one is a continuous-lag autocorrelation and the other is a discrete-lag autocorrelation, right? only the latter can be computed for all discrete-time lags. the continuous-time autocorrelation is a mathematical or theoretical thing. but computers can compute the discrete-time autocorrelation. $\endgroup$ – robert bristow-johnson Feb 26 at 23:30
  • $\begingroup$ In terms of the question: I'm intersted in the functional relationship between 𝑅x(t) and Ry(t) $\endgroup$ – MPA95 Feb 26 at 23:30
  • $\begingroup$ the difference is the difference between an exact integral and a Riemann-sum approximation to the integral. $\endgroup$ – robert bristow-johnson Feb 26 at 23:30
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The answer to the OP's question is more straightforward than rb-j's comments make it out to be.

$\{X(t)\colon -\infty < t < \infty\}$ is a continuous-time WSS random process with autocorrelation function $$R_X(\tau) = E[X[t)X(t+\tau)], -\infty < \tau < \infty.$$ On the other hand, $\{Y(n)\colon n \in \mathbb Z\}$ where $Y[n] = X(nT)$ is a discrete-time process whose mean function \begin{align} \mu_Y[n] &= E\left[Y[n]\right]\\ &= E[X(nT)]\\ &= \mu_X(nT)\tag{1}\\ &= \mu_X \end{align} is a constant since $\{X(t)\}$ is assumed to be WSS and so its mean function $\mu_X(t)$ has fixed value $\mu_X$. Similarly, the autocorrelation function of $\{Y[n]\}$ is \begin{align} R_{Y,Y}[m,n] &= E\left[Y[m]Y[n]\right]\\ &= E[X(mT)X(nT)]\\ &= R_{X,X}(mT, nT),\tag{2}\\ &= R_X((n-m)T)\end{align} that is, the value of $R_{Y,Y}[m,n]$ depends only on the difference $n-m$ of the arguments $m$ and $n$, and not upon their individual values. We conclude that

$\{Y(n)\colon n \in \mathbb Z\}$ is a discrete-time WSS random process with mean $\mu_X$ and autocorrelation function $$R_Y[m] = E\left[Y[n]Y[n+m]\right] = R_X(mT).$$

Note that just as the random variables $Y[n]$ are regularly spaced samples ($T$ seconds apart) of the $\{X(t)\}$ process, the autocorrelation function $R_Y[m]$ is just regularly spaced samples ($T$ seconds apart) of the autocorrelation function $R_X(\tau)$ of the $\{X(t)\}$ process. Indeed, Eqs. $(1)$ and $(2)$ show that if $\{X(t)\}$ were not WSS, then the mean function $\mu_Y[n]$ would be equal to $E[X(nT)] = \mu_X(nT)$, that is, the mean function of $\{Y[n]\}$ is just regularly spaced samples of the mean function $\mu_X(t)$ of the $\{X(t)\}$ process while $$R_{Y,Y}[m,n] = E\left[Y[m]Y[n]\right] = E[X(mT)X(nT)] = R_{X,X}\left(mT, nT\right),$$ that is, the autocorrelation function $R_{Y,Y}[m,n]$ is just regularly spaced sample values of $R_{X,X}(t_1, t_2) = E[X(t_1)X(t_2)]$ where the samples are on a square grid.


Since the OP's original question and my answer have resulted in a flood of comments and queries from rb-j that are far too many to be answered in comments, I will add something here to further explain my view of the issues, and then ignore any follow-up queries or comments. If you don't like my answer, down-vote it or write your own answer explaining your view of the matter. Both actions will be ignored by me.

The OP's question concerns sampling a wide-sense-stationary (WSS) continuous-time process $\{X(t)\colon t \in \mathbb R\}$ to create a discrete-time process $\{Y[n]\colon n \in \mathbb Z\}$ where $Y[n]$ is defined to be $X(nT)$ where $T$ is a known positive constant. The OP then asks for the relationship between the autocorrelation functions $R_X(t)$ (continuous-time) and $R_Y[n]$ (discrete-time) of the two processes.

My answer, which I have explained in some detail, to this question, is that $R_Y[n]$ is just the sampled version of $R_X(t)$: $$R_Y[n] = R_X(nT)~\text{for all} ~n \in \mathbb Z.$$ This relationship is not a two-way street; given the values of $R_Y[n]$, it is not possible to reconstruct $R_X(t)$. Nor have I claimed anywhere that $R_Y[n]$ and $R_X(t)$ are the same function. "But, but, but,$\ldots$ what about the sampling theorem? What about ergodicity? What about $\ldots$," you splutter. Please re-read the sampling theorem, especially the hypotheses. If $R_X(t)$ is a strictly band limited function (and nowhere has it been asserted that $R_X(t)$ enjoys this property), then it can be reconstructed from its sample values provided the sampling rate is adequately high (exactly how high is not relevant). Given only sample values, one can construct infinitely many possible functions that have the specified sample values. If the sample values are of a function is indeed strictly band limited, then one of these reconstructions will coincide with the function that was sampled. But simply to use the Shannon-Whittaker interpolation formula on $R_Y[n]$ and insist that the result must be $R_X(t)$ is something that boggles my mind, though ymmv. As for ergodicity, the less said the better.

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  • $\begingroup$ now if the thing is WSS, it's also ergodic, no? aren't all time averages the same as probabilistic averages? $$\begin{align} R_X(\tau) &= E[X(t)X(t+\tau)], -\infty < \tau < \infty.\\ &= \lim_{L\to\infty} \frac{1}{2L} \int_{-L}^{L} X(t)X(t+\tau) \, \mathrm{d}t \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm{pdf}\{X(t)=\alpha,X(t+τ)=\beta\}(\alpha,\beta)\, \alpha \beta \, \mathrm{d}\alpha \mathrm{d}\beta \\ \end{align}$$ $\endgroup$ – robert bristow-johnson Feb 27 at 10:56
  • $\begingroup$ now for discrete time, $$\begin{align} R_X(\tau) &= E[X[n]X[n+\tau]], -\infty < \tau < \infty.\\ &= \lim_{L\to\infty} \frac{1}{2L+1} \sum\limits_{-L}^{L} X[n]X[n+\tau] \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm{pdf}\{X[n]=\alpha,X[n+τ]=\beta\}(\alpha,\beta)\, \alpha \beta \, \mathrm{d}\alpha \mathrm{d}\beta \\ \end{align}$$ $\endgroup$ – robert bristow-johnson Feb 27 at 11:01
  • $\begingroup$ it's like one is a Riemann sum of the other. but the $\Delta t$ is fixed at 1. so then the lower bandwidth the WSS signal $x(t)$ is, the better the Riemann sum will approximate the integral. That's like saying the more oversampled $X(t)$ is, the closer the Riemann sum is to the integral. $\endgroup$ – robert bristow-johnson Feb 27 at 11:06
  • $\begingroup$ i think i forgot to restrict $\tau\in\mathbb{Z}$ in the discrete-time case. and it's too late to edit the comment. $\endgroup$ – robert bristow-johnson Feb 27 at 11:08
  • $\begingroup$ i think that if $$ X(t) = \sum\limits_{n=-\infty}^{\infty} X[n] \operatorname{sinc}(t-n) $$ then the summation and integral will be equal. $\endgroup$ – robert bristow-johnson Feb 27 at 11:12

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