Last time we talked about why a full-scale digital sine wave has a power measurement of -3.01 dB FS (Spoiler: because it's not a square wave.)

This time we'll discuss why an atmospheric sound which generates a root-mean-square pressure of 1 Pascal has a power measurement 94 dB SPL.

As before, dB is defined as 10 log_{10}(*P*_{A}^{2} / *P*_{B}^{2}) where *P*_{B} is a reference level.

Before, we had a digital measurement with an obvious ceiling: sample values of -1 and 1. So the reference point 0 dB FS was defined in terms of the signal with the greatest possible energy.

In the analog domain, there isn't an obvious ceiling. We instead consider the floor - the quietest possible signal that is still audible by human ears.

This is a rather wishy-washy definition, but the convention is to take *P*_{B} = 20 μPa = 0.00002 Pa *exactly*.

So our 0 dB SPL reference point is when *P*_{A} = *P*_{B}: 0 dB SPL = 10 log_{10}(0.00002^{2} / 0.00002^{2}) = 10 log_{10}(1) = 10 (0) = 0.

What if the pressure level is 1 Pascal? This is a quite loud sound, somewhere between heavy traffic and a jackhammer.

1 Pa in dB SPL =

10 log_{10}(1^{2} / *P*_{B}^{2}) =

20 log_{10}(1 / *P*_{B}) =

-20 log_{10}(*P*_{B}) =

-20 log_{10}(2(10^{-5})) =

-20 (log_{10 }2 + log_{10 }10^{-5}) =

-20 ((log_{10 }2) - 5) =

100 - 20 log_{10 }2 ≈ 93.9794 dB SPL

So 1 Pa is actually a tiny bit less than 94 dB SPL; it's closer to 93.98 = (100 - 6.02) dB SPL.

the 94dB SPL are peak or rms??

RMS.