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Given a signal $x(t) = s(t) + n(t)$ where $s(t)$ is the desired signal voltage and $n(t)$ is the noise, should the signal to noise ratio of this signal be 20log(xrms/nrms) or 20log(srms/nrms)? i.e., should I use the desired signal or the total signal for the signal to noise ratio?

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  • $\begingroup$ The latter is the SNR $\endgroup$ – BlackMath Oct 16 at 21:57
  • $\begingroup$ For high SNR, it won't make much of a difference as the energy of the signal is much higher than the energy of the noise. $\endgroup$ – Ben Oct 16 at 23:19
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The SNR is the ratio of $S$, the power in the signal $s(t)$, and $N$, the power in the noise $n(t)$. So, $$\text{SNR}_\text{dB} = 10\log_{10}\frac{S}{N}.$$

When using RMS values, we have $S = s_\text{RMS}^2/R$ and $N = n_\text{RMS}^2/R$. The resistances cancel out and then $$\text{SNR}_\text{dB} = 20\log_{10}\frac{s_\text{RMS}}{n_\text{RMS}}.$$

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