# Question about SNR in continuous and discrete domain

How to make sure the discrete domain's SNR equals to the corresponding SNR of continuous domain?

If the definition of SNR in continuous domain is:

$$SNR=\frac{\int_T|s(t)|^2dt}{noiseVariance}$$

if we sampled $s(t)$ ($T_s$ is the sample period) then what is the corresponding SNR of the discrete domain? Is

$$SNR=\frac{\sum_{n=1}^{N}|s(nT_s)|^2dt}{noiseVariance}$$

?

• Their shouldn't be a $dt$ in the second forumula - do you mean $T_s$? Also, if you sample the signal, you are also sampling the noise - so some adjustments are needed for how you calculate the noise variance. – David Dec 17 '15 at 14:16

In your $snr$ expression, if you put a $1/N$ ratio in front of the numerator summation you'll have the standard expression for SNR with regard to sampled data (digital data).
• Considering the fact that average power of the continuous time signal $x_c(t)$ is numerically equal to the average power of its sampled discrete time version $x[n] = x_c(nT)$ (i.e. avegare power is preserved through sampling) we can easily deduce that SNR in both domains should be numerically same. Hence given that $noiseVariance$ will be same in both domains, all you have to do is to compute average signal power which requires a 1/N factor for the discrete time version and 1/T (T = time length) for the continuous time verison of those SNR equations, as described by @Richard Lyons – Fat32 Apr 15 '16 at 23:47