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/s^{2}

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 (Y^{2}/R^{2})

And so,

17 dB = 10log (Y^{2}/10) * 0.00002^{2})

Y^{2} = 10^{(17/10)} * 0.0000000004

Y^{2} = 10^{1.7} * 0.0000000004

Y = ?0.00000002

Y = 0.00014 Pa

This is because 20log X = 10log X^{2}

#### James Wren

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