# SNR contains $\log N$ term where $N$ stands for number of samples

Sampling the signalN 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)- $$10\log N$$

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.
• 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 $N^2$. – Dilip Sarwate Jun 19 '19 at 2:50