Does anybody know the math required to convert a logarithmic decibel scale to a linear value between 0 and 1, for an automation lane?
In this particular case - the linear value 1 = +15dB, the linear value 0 = -∞dB.
asked Apr 8, 2016 at 16:54
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'Decibel' is a ratio, not an absolute value. To answer this we'd need to know the range of decibel values and the scale reference.– Jim MackCommented Apr 8, 2016 at 19:47
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Ok so the unit delivering these values actually gives -32768 for the fader right down, but 0 when it is right up. So when I have this value (call it N) it aligns exactly to the physical fader positions with this function (N / 10) + 10 - so that effectively gives the same value as shown on the fader which runs from +10dB down to - infinity with -60 the last number. The scale seems logarithmic to me on the fader.– James HarcourtCommented Apr 8, 2016 at 20:37
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What is the automation lane to be used for?– n00dles ♦Commented Apr 9, 2016 at 13:28
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@MarcW Volume on an audio track in a multi-track editor. The scale runs from +15db to -15dB in side the multitrack editor. In the session file which the multitrack editor loads, the value is stored as a linear point from 0 to 1 where 0.65 = 0dB, 0=-infinity dB and 1 = +15dB.– James HarcourtCommented Apr 9, 2016 at 15:18
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3 Answers
Practical version
dB → gain-multiplier:
g = 2d / 6
gain-multiplier → dB:
d = 6 · log2(g)
I find these definitions far more handy than the ones below: changing the amplitude by a factor of two is quite an intuitively relevant change. But, alas, in pre-computer times people couldn't seem to like logarithms of bases other than ten, so...
Official definition
dB → gain-multiplier:
g = 10d / 20
gain-multiplier → dB:
d = 20 · log10(g)
IMO base-10 is silly, but if you need to do exact calibrations, better use the official version. (Alternatively, use the base-2 version, but replace 6 with the factor 20 · log102 ≈ 6.020599913; this is then exactly equivalent to the base-10 definition.)
So if we set that 0 dB gain is 1.0 factor, and -∞ db gain is 0.0 factor, it means that (if we are considering voltage gain as in a mixing desk fader) :
gain = 20.0*log10(factor)
therefore :
factor = 10^(gain/20.0)
If, as described in a comment, the 0 dB gain is at 0.65 factor, it means the ref is 0.65.
gain in db = 20*log10(factor/0.65)
factor = 0.65*10^(gain/20.0)
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How does this change if +15 dB gain is 1.0 factor and -∞ db gain is 0.0 factor? Commented Apr 8, 2016 at 21:04
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1@JamesHarcourt: just add / remove 15 from the dB scale. Commented Apr 8, 2016 at 21:47
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@leftaroundabout that definitely doesn't work, the values that come out aren't anywhere near the linear values produced, i.e. 0dB = 0.65. Commented Apr 9, 2016 at 15:18
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@JamesHarcourt er, no! 10⁽⁰⁻¹⁵⁾’²⁰ ≈ 0.1778. What did you calculate? Commented Apr 9, 2016 at 15:24
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The application in question is Adobe Audition. Their volume automation is stored in XML, so when there is an automation point of 0dB, it is shown thus: <parameterKeyframe sampleOffset="95176" type="linear" value="0.65" />. So I actually want 0.65! Commented Apr 9, 2016 at 15:32
To readers:
According to the questioner's comment:
The application in question is Adobe Audition. Their volume automation is stored in XML, so when there is an automation point of 0dB, it is shown thus: . So I actually want 0.65!
All inferences are based on the SESX file of Adobe Audition.
Try to get more values:
| value | dB |
|---|---|
| 1 | +15 |
| 0.95 | +12.0095 |
| 0.9 | +9.4189 |
| 0.85 | +7.1339 |
| 0.8 | +5.0899 |
| 0.75 | +3.2409 |
| 0.7 | +1.5528 |
| 0.65 | +0 |
| 0.6 | -1.5528 |
| 0.55 | -3.2409 |
| 0.5 | -5.0899 |
| 0.45 | -7.1339 |
| 0.4 | -9.4189 |
| 0.35 | -12.0095 |
| 0.3 | -15 |
| 0.25 | -18.5371 |
| 0.2 | -22.8661 |
| 0.15 | -28.4472 |
| 0.1 | -36.3132 |
| 0.05 | -49.7604 |
| 0 | -∞ |
It is easy to notice that data with non-positive dB presents in the form of dB=A+Blog(value). Using the FindFit function of Mathematica would be very helpful.
So we can easily know that dB=-15sgn(value-13/20)log(13/6,1-|20value/13-1|) and value=13(1+sgn(dB)(1-(13/6)^(-|dB|/15)))/20.
