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This is probably a dumb question about the signal-to-noise ratio (SNR). I know the SNR of a signal is defined as the signal power divided by the noise power. Now I have a noisy transient voltage decay V=V(t), and the voltage decays toward zero as time increases. How do I calculate the SNR? Someone told me it was simply

$$ SNR=\left[\frac{V(0)}{V(\infty)}\right]^2 $$

But I doubted it since I couldn't link this formula to the original definition of SNR. Can anyone explain?

placeholder
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  • Is the transient the signal, or the noise? And what constitutes the other part? – Chris Stratton Apr 09 '13 at 20:46
  • The transient is noisy. It contains both signal and noise. – shva Apr 09 '13 at 20:48
  • What noise do you have and what signal do you have? Is it fair to assume the signal is totally and exactly the transient and decay or is it just the transient and the decay represents unwanted "effects" that may be regarded as noise? Difficult to say but maybe somebody has an answer. – Andy aka Apr 09 '13 at 20:50
  • If the transient is noisy, is that noise present before the transient kicks-in? If it isn't then you have to think a bit harder about what the noise actually is and define it – Andy aka Apr 09 '13 at 20:51
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    The S/N ratio is \${\left(\frac{V_s(t)}{V_n(t)}\right)}^2\$. If \$V_n\$ is constant (independent of \$V_s\$), and if \$V_s\$ is transient, it means that the ratio is good at the height of the transient when the signal is far above the noise floor. Then as the signal fades to the noise floor, the ratio drops to 1, and from there down to zero. – Kaz Apr 09 '13 at 21:41
  • @Kaz should write that up as an answer! basically as t-> infinity V-> Vnoise so the expression is correct! for THAT signal. – placeholder Apr 09 '13 at 22:38
  • The transient V(t)=Vs(t)+Vn(t), where Vs(t) is the signal and Vn(t) is the noise. According to Kaz's answer, does that mean SNR for transient V(t) is a function of time...? – shva Apr 10 '13 at 00:31

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