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SNR is defined as:

$$SNR=10 \log_{10}\left(\frac{\sigma_s^2}{\sigma_n^2}\right)$$

For instance, when we say the SNR of 10 dB is 10 times higher than that of 20 dB, it is higher in what unit? Amplitude, power or what?

Also, could you relate as to why it's 10 times higher for the above example.

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2 Answers 2

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SNR stands for Signal to Noise Ratio. It is a ratio and as such does not have any units, it describes the proportion of signal to undesired noise. There is no single correct measure of SNR, it differs depending on the application.

In the equation you have given, the SNR is broken down in the following way:

1) Calculate the power ratio

$$\frac{\sigma_s^2}{\sigma_n^2}$$

Your equation requires us to know the variance of the signal and the noise, dividing one by the other gives the power ratio. To answer one of your questions, this is the ratio of power, not amplitude, we could achieve the same result by taking a ratio of the root mean square of the signal and noise - another measure of power.

2) Express the ratio in decibels

$$10\log_{10}$$

This is simply a convenience, the SNR may be a very large number so expressing the value on a logarithmic scale can make the information more manageable. Converting a unit to decibels simply involves taking the log10 of the value and multiplying by 10.

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  • $\begingroup$ Thanks for the answer. so we can say : SNR of 10db is 10 times higher in power that the 20db SNR. right? so , SNR is unit-less on itself? right? How can I express a new relationship for SNR that compares the amplitude? Thanks @pak-9 $\endgroup$
    – SAH
    Commented Aug 27, 2014 at 18:22
  • $\begingroup$ @Electricman: You have it backwards; 20 dB refers to a ratio of 100, while 10 dB refers to a ratio of 10. $\endgroup$
    – Jason R
    Commented Aug 27, 2014 at 18:43
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    $\begingroup$ @JasonR, a 10 dB increase in S/N (my preferred label) is an increase of power by a factor of 10. in $$S/N = 10 \log_{10}\left(\frac{\sigma_s^2}{\sigma_n^2}\right)$$ the two sigma's are standard deviation. the $\sigma^2$ is variance, which is the AC power (the DC power is the square of the mean). $\endgroup$ Commented Aug 28, 2014 at 19:27
  • $\begingroup$ As I learned so far, SNR is signal power to noise power ratio, so its a ratio of power. but Its unit-less. so $SNR = 10 \log_{10}\left(\frac{Pw}{Pn}\right)$ so now my quastion is why $\sigma^2$ has been used instead of power. is it still power ratio? @robertbristow-johnson $\endgroup$
    – SAH
    Commented Aug 28, 2014 at 21:41
  • $\begingroup$ please check the comment above , tnx @JasonR $\endgroup$
    – SAH
    Commented Aug 28, 2014 at 21:41
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SNR is unitless, because when you divide the signal by the noise they have the same units, thus canceling the units out.

That being said, SNR is a power ratio, but mathematically there is no reason you can't do a dB measurement of amplitudes, or anything else.

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  • $\begingroup$ That being said, SNR is a power ratio... do they mean that only power signal(not voltage and ...) can be polluted by noise? $\endgroup$
    – SAH
    Commented Aug 27, 2014 at 18:29
  • $\begingroup$ When expressing an amplitude ratio in dB, the typical convention is to use $20 \log_{10}$, so the result is expressed as a power ratio. $\endgroup$
    – Jason R
    Commented Aug 27, 2014 at 18:44
  • $\begingroup$ @Electricman No, voltage can and is polluted by noise too. It's just that SNR is typically measured on power, not voltage. Mathematically you could do it on voltage though. $\endgroup$
    – Jim Clay
    Commented Aug 27, 2014 at 18:45
  • $\begingroup$ @JasonR, it's not $20 \log_{10}$ that's used for power ratio. that's for (r.m.s.) voltage ratio or current ratio or acoustic pressure or particle velocity. for power ratio it's $10 \log_{10}$. $\endgroup$ Commented Aug 28, 2014 at 19:22
  • $\begingroup$ @robertbristow-johnson: Right. Maybe I didn't word it well, but I was trying to indicate that you would use the $20 \log_{10}$ form when you had amplitudes, so that the resulting "dB quantity" refers to the ratio of the power levels commensurate with the amplitudes. $\endgroup$
    – Jason R
    Commented Aug 28, 2014 at 19:56

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