C: Summing of Levels

If all signals to be added have the same level and phase, we can use the formula we have already developed in a previous section:

L_{sum}=20*log(\frac{value}{reference value}) +20*log(number of sources)

When it comes to power signals, the prefactor 10 is used before both logarithms!

L_{sum}=10*log(\frac{value}{reference value}) +10*log(number of sources)

a) Three identical audio signals with an identical level of -10 dBV each are connected to a mixing console. After summing these signals with no further amplification, the following measurements are taken at the mixer output:

L_{sum}=-10\,dB_{V} +20*log(3)=-10\,dB_{V} +9,54\,dB=-0,46\,dB_{V}

The addition of levels with different values cannot be done directly. To do this, you have to take a detour back via the respective physical quantity. So first the logarithm is “undone”, then the linear values are added or subtracted and then logarithmized again. Here, it is assumed that all signals are in phase.

L_{sum}=20*log(10^{(\frac{level\,1}{20})}+10^{(\frac{level\,2}{20})}+...10^{(\frac{level\,X}{20})})

For power signals again accordingly with the prefactor or divisor in the exponent 10:

L_{sum}=10*log(10^{(\frac{level\,1}{10})}+10^{(\frac{level\,2}{10})}+...10^{(\frac{level\,X}{10})})

b) The scenario corresponds to that in example a). However, the level of the three signals is now -15 dBV, -10dBV and -5 dBV. After the summation, the following measurements are taken at the mixer output:

L_{sum}=20*log(10^{(\frac{-15\,dB_{V}}{20})}+10^{(\frac{-10\,dB_{V}}{20})}+10^{(\frac{-5\,dB_{V}}{20})})+20*log(0,178\,V+0,316\,V+0,562\,V)=+0,472\,dB_{V}