0
$\begingroup$

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)- $10\log N$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.