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I have a voltage signal, on top of which there is some noise. I am interested in the noise components so the way I am trying to do this is to fit the known signal to the data and look at the residuals. This quite obviously gives me some positive and some negative values.

For various reasons, it's useful for me to look at this in units of $\rm{dBV_{pk}}$ so to do the conversion one would normally just go $20\log_{10}(\rm{V_{pk}})$. Now in my case $\rm{V_{pk}}$ is the 'voltage-residual' (data minus the fit).

Obviously the negative data cannot be logged -- you get a real and imaginary component in the form $z = x + iy$ -- but I'd really like to take a look at the noise distribution in units of $\rm{dBV_{pk}}$. Are there any ways I can look at the negative noise residuals in $\rm{dBV_{pk}}$.

One dirty approach would to simply be to take the negative volatge-residuals, take the absolute value, covert into a log scale and then multiply the result by $-1$, which mathematically dubious, this would give me the kind of distribution I am interested in -- as whatever form it is in it should be roughly centered around 0.

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    $\begingroup$ dB is a ratio of power or energy. the use of 20 instead of 10, is implicit squaring. $\endgroup$ – Stanley Pawlukiewicz Aug 11 at 16:40
  • $\begingroup$ yes, this is just basic rules of logarithms: $\rm{dBm} = 10\log_{10}(V^{2}/1\rm{mW} R)$ so $\rm{dBV} = 20\log_{10}(V) - 10\log_{10}(1\rm{mW} R))$. Maybe I am being dumb but I don't see how that helps be look at my negative voltages in decibel units. $\endgroup$ – Q.P. Aug 11 at 16:50
  • $\begingroup$ square of a negative? is ___ $\endgroup$ – Stanley Pawlukiewicz Aug 11 at 18:00
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As hinted to by @StanleyPawlukiewicz in comments, the correct definition to use is, using your style of notation:

$$\rm dBV_{pk} = 10\log_{10}\left({V_{pk}}^2\right) = 20\log_{10}\left(\sqrt{{V_{pk}}^2}\right) = 20\log_{10}(|V_{pk}|).$$

Clearly, $\rm dBV_{pk}$ will not contain information about the sign of $\rm V_{pk}$.

You cannot safely fiddle with the sign of $\rm dBV_{pk}$ based on the sign of $\rm V_{pk}$, as then there will be an ambiguity: Does the sign tell whether $\rm |V_{pk}| > 1$ (from the decibel definition) or whether $\rm V_{pk} > 0$ (based on the sign of $\rm V_{pk}$)?

Perhaps you can convey the sign of $\rm V_{pk}$ as auxiliary information.

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