I've seen that the output of a decimator when a WSS process is passed through remains WSS. I am not able to immediately see why this is. What is a good explanation of why the signal maintains stationarity?

  • $\begingroup$ can you define "decimator" more precisely? I can think of two conflicting definitions; the idea behind the answer is the same for both (WSS implies only dependence on $\Delta t$), but the complexity of answer depends on whether you define a decimator as having an FIR filter or something else. $\endgroup$ – Marcus Müller Dec 1 '18 at 23:24
  • $\begingroup$ It's just a WSS signal x[n] passed through an M-fold decimator (a multirate signal processing block where M=2, used in fractional rate conversion), no filter follows the decimator. So y[n] = x[Mn] $\endgroup$ – J Dolan Dec 1 '18 at 23:31
  • $\begingroup$ That's the easiest case. $\endgroup$ – Marcus Müller Dec 1 '18 at 23:39

The decimator by integer $M$ can be shown to be the following block:

$$ x[n] \longrightarrow \boxed{ \downarrow M } \longrightarrow y[n] = x[Mn] $$

Assuming that the input $x[n]$ is WSS it has the ACS as $$ r_x[k] = E\{ x[n] x^*[n+k] \} $$

Then the output autocorrelation can be defined as: $$ r_y[n,n+k] = E\{ y[n] y^*[n+k] \} = E\{ x[Mn] x^*[Mn+Mk] \} = r_x[Mk]$$

as can be seen, the output auto-correlation $r_y[n,n+k] = r_y[k] = r_x[Mk]$ also depends on the lag $k$ and therefore we can conclude that $y[n]$ is also WSS.

(of course the mean of $y[n]$ is also independent of time $n$.)

  • 1
    $\begingroup$ much nicer notation. More than worthy of my upvote! $\endgroup$ – Marcus Müller Dec 2 '18 at 0:06

just a WSS signal $x[n]$ passed through an $M$-fold decimator (a multirate signal processing block where $M=2$, used in fractional rate conversion), no filter follows the decimator. So $y[n] = x[Mn]$

WSS means that the autocorrelation function only depends on the distance between the points (and that the mean and variance are time-independent); i.e. the general

$$r_{xx}(t_1, t_2) = E\left\{(x(t_1)-\mu_{t_1})(x(t_2)-\mu_{t_2})^*\right\}$$

collapses to $$r_{xx}(t_1, t_1-\tau) = r_{xx}(\tau)\text.$$

For your discrete-time case:

$$r_{xx}[n_1, t_2] = E\left\{(x[n_1]-\mu_{n_1})(x[n_2]-\mu_{n_2})^*\right\}$$

collapses to $$r_{xx}[n_1, n_1-l] = r_{xx}[l]\text.$$

Now, your $M$ is just a specific $l$; after decimation, you just have scaled that $\tilde l = \frac{l}M$. But that doesn't change the autocorrelation property.


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