I'm given the autocorrelation of a WSS random process enter image description here

and the question asks to find the sampling rate that yields uncorrelated samples.

As far as I understand where looking for the $\tau$'s where $R_x(\tau)=0$ but these never occur.

The solution claims that $n/f_o$ is such sample rate. I can see that this will make the autocorrelation constant but not zero.

  • $\begingroup$ "to find the autocorellation" of what? $\endgroup$ Commented Mar 18, 2022 at 23:53
  • $\begingroup$ That was an error. Fixed it. $\endgroup$
    – Essam
    Commented Mar 19, 2022 at 5:16

1 Answer 1


In short, it seems you're right. Since the whole plot never crosses zero, the autocorrelation is non-zero for any shift, and thus, no periodic sampling would produce uncorrelated samples.

We can also take a more physical look at this: the PSD is the Fourier transform of the ACF. Your ACF has a constant offset, so the PSD will have Dirac at $f=0$; signals with a DC component are correlated, no matter how you sample them, by the sheer fact that there's this DC component.

However, one thing we can do: we can split $R_X$:

$$\newcommand{\K}{{K_X}}\newcommand{\X}{{\tilde X}} R_X(\tau) = 1 + \K(\tau) $$

and consider $1$ and $\K$ separately. $\K$ is the autocovariance of $X$. For WSS $X$, this implies the mean of $X$ absolute squared needs to be $1$, so $\mu_X=e^{j\phi}$; with $j$ being the imaginary unit, and $\phi$ some real constant.

Introducing $\X$,

$$ \tilde X = X -\mu_X $$

we could get a signal with zero mean and thus zero autocorrelation every multiple of $\frac1{f_0}$. But: you need to subtract the mean for that!

A high-pass filter with a true zero at DC will do that.

  • $\begingroup$ I almost thought that the problem was that I was setting the autocorrelation to zero instead of the autocovariance. Two random variables are uncorrelated if their covariance is zero which lead me to think that the autocovariance is what should be set to zero which solved the problem. That is, I did not expect this answer. $\endgroup$
    – Essam
    Commented Mar 19, 2022 at 12:01

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