Sampling the signal$ $N times increases the signal energy by a factor of $N^2$ and the noise energy by a factor of $N$. Why? This explanation is written for SNR(dB) = signal peak(dB) – noise floor(dB)- $10\log N$
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
1
In the additive model $y=s+n$, when the signal is deterministic, it adds coherently over the "realizations". Hence, its variance $V(\sum s_n) = V(N s) = N^2 V( s)$. And when the noise $w$ is independent identically distributed (IID), then $V(\sum w_n) = NV( n) $. This is a classical result on the Variance of Uncorrelated Variables.
You can find this classical result detailed in Deriving the SNR for averaged signals. Warning, this does reach a limit when data is quantized, in other words the noise floor does not follow $1\sqrt{N}$, but plateaues.
answered Jun 9 '19 at 10:42
-
$\begingroup$ I think the question is about that $-10 \log N$ term in the expression for SNR and not about why the noise behaves as $N$ while signal behaves as $N^2$. $\endgroup$ – Dilip Sarwate Jun 19 '19 at 2:50
