# When calculating SNR, is noise included in signal?

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?

• The latter is the SNR – BlackMath Oct 16 '19 at 21:57
• 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. – Ben Oct 16 '19 at 23:19

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 $$\text{SNR}_\text{dB} = 20\log_{10}\frac{s_\text{RMS}}{n_\text{RMS}}.$$