The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $\frac{1}{10}$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $\frac{1}{100}$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $\frac{1}{1000}$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 \left( \frac{1}{10} \right)^{L[cm]}$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 \left(\log(P_0[mW])-L[cm]\right)$$
Whenever you have a process that is multiplicative like:
$$X \propto e^Y $$
with the parameter $Y$ normal distributed: $$Y \sim N(\mu,\sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_{i} = e^{x_i+\epsilon_i}$$
mathematical-statistics
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