Much confusion revolves around linear and non-linear numbers. The following outlines the mathematical process to convert from a number expressed in dB to a linear quantity. How do we convert to decibels and back again?
The formula used in the following example is,
X = 20log (Y/R)
Where X is the sensor value in dB, which is to be converted
And Y is the linear value of the sensor
And where R is the reference value of the dB conversion, is the case of microphones it is usually 0.00002 Pa and the case of accelerometers it is usually 0.000001 m/s2
Taking a microphone signal, if we obtain the reading of 17dB what linear value is being referred to?
17 dB = 20log (Y/0.00002)
Y = 10(17/20) * 0.00002
Y = 0.00014 Pa
If it was necessary to go one step further and convert Y into a raw voltage value, Y would simply have to be multiplied by the sensitivity of the sensor, in the case of a microphone, usually 50mV/Pa.
Thus the raw voltage value would be,
0.00014 Pa * 0.05 = 7µV
But what about the case of a squared value, like power?
An auto power spectrum is a squared quantity, so how would this be converted? A different formula has to be used to take account of the fact the quantity is already squared.
The formula would be,
X dB = 10log (Y2/R2)
And so,
17 dB = 10log (Y2/10) * 0.000022)
Y2 = 10(17/10) * 0.0000000004
Y2 = 101.7 * 0.0000000004
Y = ?0.00000002
Y = 0.00014 Pa
This is because 20log X = 10log X2
James Wren
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