Note I also asked this in math stackexchange I was reading a wiki page about signal averaging specifically in the section Signal power for sampled signals. At least at first glance the information seems to be quite wrong.

Passage from wikipedia follows:

Considering $n$vectors $V_i,\,i \in\{1,\ldots,n\}$ of signal samples of length $T$ $V_i = \left[s_{i,1}, \ldots, s_{i,T}\right],\quad s_{i,k} \in \mathbb K^T$,

the power $P_i$ of such a vector simply is

$P_i = \sum_{k=1}^T {s_{i,k}^2} = \left| V_i \right|^2$.

Again, averaging the $n$ vectors $V_{i},\,i=1,\ldots,n$, yields the following averaged vector

$V_\text{avg} = \frac 1n \sum_{k=1}^T\sum_{i=1}^n s_{i,k}= \frac 1n \sum_{i=1}^n \sum_{k=1}^T s_{i,k}$.

In the case where $V_n \equiv V_m \forall m,n \in \{ 1,\ldots, n\}$, we see that $V_\text{avg}$ reaches a maximum of

$V_\text{avg, identical signals} = P_i$.

In this case, the ratio of signal to noise also reaches a maximum,

$ \text{SNR}_\text{avg, identical signals} = \frac{V_\text{avg, identical signals}}{N_\text{avg}} = n \text{SNR}$.

This is the oversampling case, where the observed signal is correlated (because oversampling implies that the signal observations are strongly correlated).

Firstly, doesn't $$V_i = \left[s_{i,1}, \ldots, s_{i,T}\right],\quad s_{i,k} \in \mathbb K^T$$

Imply that $V_i$ is actually a vector of functions, $f:T\rightarrow K$ which is not what I think the intended meaning is. Shouldn't $V_i \in K^T$ and $s_{i,k} \in K$? Secondly, for $$V_\text{avg} = \frac 1n \sum_{k=1}^T\sum_{i=1}^n s_{i,k}= \frac 1n \sum_{i=1}^n \sum_{k=1}^T s_{i,k}$$ Isn't this just a scalar since they summed all the vectors and over the vectors? How is this the average vector? It may be that I don't understand something here but I feel like this section is full of mistakes and poorly explains whats going on. Am I correct that this is a mistake in the page or is there something I don't understand?

  • $\begingroup$ additionally, if this is indeed incorrect and someone wants to take what I have above and fix it to explain signal averaging for signal noise ratio that would be greatly appreciated $\endgroup$ – Colin Hicks Oct 2 '19 at 20:42
  • $\begingroup$ I asked this question in math stackexchange as well since a bit time sensitive. I have a lab due tommorow and from what it seems either this page has problems or my lab does. So a better understanding or confirmation would be appreciated $\endgroup$ – Colin Hicks Oct 3 '19 at 0:17
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    $\begingroup$ The section seems to me full of wrongly-written stuff, from indics to non-consistent equations, like $V_\textrm{avg}=P_i$ (unit and square of the unit on each side) $\endgroup$ – Laurent Duval Oct 3 '19 at 8:24

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