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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

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

?

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    $\begingroup$ 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. $\endgroup$ – David Dec 17 '15 at 14:16
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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).

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  • $\begingroup$ Thank you for your answer! so if I put a 1/N ratio in front of the numerator summation then the snr I have is corresponding to the continuous domain's snr? $\endgroup$ – Tj L Nov 17 '15 at 12:32
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    $\begingroup$ I don't think the answer is as simple as it is made out to be here. $\endgroup$ – Dilip Sarwate Nov 17 '15 at 15:51
  • $\begingroup$ 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 $\endgroup$ – Fat32 Apr 15 '16 at 23:47

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