A little math helps to distinguish between relative levels and absolute levels or to calculate them. For the following calculation of twice the sound pressure level of p0, we use one of the logarithm laws to separate the factor 2 from the sound pressure p0.
We must not forget that the multiplication by 20 applies to both logarithms. After removing the brackets, it becomes clear that there is no reference value and therefore no unit left in the first logarithm. The result of the "logarithmic factor multiplied by 20" receives the pseudo-unit dB without additional designation. So it is a relative level. Similar to amplification by a factor of 2. However, the second logarithm remains an absolute level with units of dBSPL. Likewise the result of the entire equation.
L_{p_{0}}=20*log(2)+20*log(\frac{p_{0}}{p_{0}})
L_{p_{0}}=6\,dB + 0\,dB_{SPL} = +6\,dB_{SPL}
In the “dB” world, doubling the sound pressure means an addition of 6 dB. So when you move into the world of “dB”, not only the size of the numbers change, but the arithmetic operation is also simplified. A multiplication turns into an addition. Conversely, division in the world of factors corresponds to subtraction in the “dB” world.
a) A microphone delivers an electrical signal with a voltage level of -40 dBV. It is connected to a preamplifier which amplifies the signal by a factor of 1000, i.e. 60 dB.
A voltage level of +20dBV is measured at the output of the preamplifier.
b) In absolute silence, a certain microphone delivers a very small noise voltage (quiescent noise or thermal resistance noise, “Johnson noise”) of -120 dBV. Such noise would be amplified by an amplifier in the same way like an audio signal. So:
