2
$\begingroup$

There is alternative definition for SNR that mention in Signal-to-noise ratio

It is: $$SNR = \frac{\mu}{\sigma}$$

i.e. mean per standard deviation,

Is it possible to convert this SNR value into dB unit, without knowing the power of the noise?

$\endgroup$

1 Answer 1

4
$\begingroup$

The decibel (dB) specifically is 1/10 (deci) of a bel, Where a bel is the log of the power ratio and is therefore itself a unitless quantity. To express a ratio of powers in dB we use:

$$K = 10 \log_{10} (P_2/P_1) \text{ dB}$$

Where $K$ is the power ratio expressed in dB and $P_1$ and $P_2$ are power quantities.

We see the relationship between deci and the product by 10 above by expressing the same formula in bels (which is never done in practice, but here to show its origin):

$$K = \log_{10} (P_2/P_1) \text{ bel}$$ $$ = 10 \log_{10} (P_2/P_1) \space\space\space\frac{1}{10}\text{bel}$$

In the OP's case $\mu$ and $\sigma$ are typically magnitude quantities, not power, as representing the mean and standard deviation. Therefore these would be squared first to convert to a power quantity in converting to dB as follows:

$$\text{SNR}_{\text{dB}} = 10\log_{10}\bigg(\frac{\mu}{\sigma}\bigg)^2=20\log_{10}\bigg(\frac{\mu}{\sigma}\bigg)$$

This is consistent with the more general cases when the ratio is not a power quantity but a magnitude quantity (such as ratio of voltages, currents or complex phasors), where we must convert it to a power quantity first by squaring as done above, which simply means multiplying the log of the ratio by 20 instead of 10 for those cases.

This is In contrast to any ratio of direct power measurements where we would use $10\log_{10}()$, such as a ratio of variances for example.

$\endgroup$
4
  • 1
    $\begingroup$ Except that $\mu$ and $\sigma$ are generally in "amplitude units" (i.e. volts, current, drams, cubits, whatever). So the power ratio is probably $\left ( \mu / \sigma \right)^2$, and the relationship is probably $\mathrm{SNR}_{dB} = 20 \log_{10} \left ( \mu / \sigma \right)$. Note "generally", "probably", etc., because it depends on the problem at hand -- but I'm having a hard time thinking of a "normal" problem where it wouldn't be $20 \log_{10}$ $\endgroup$
    – TimWescott
    Jan 3, 2022 at 0:24
  • 1
    $\begingroup$ @TimWescott yes true - it should be 20Log here $\endgroup$ Jan 3, 2022 at 0:26
  • $\begingroup$ "deci" stands for one tenth (as in decimeter which is a tenth of a meter). What you mean is "deka". $\endgroup$
    – Max
    Jan 3, 2022 at 10:18
  • $\begingroup$ @Max Yes I agree -- I can show how the 1/10th appears better and will update that- thank you! $\endgroup$ Jan 3, 2022 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.