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 log10(PA2 / PB2) where PB 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 PB = 20 μPa = 0.00002 Pa exactly.

So our 0 dB SPL reference point is when PA = PB: 0 dB SPL = 10 log10(0.000022 / 0.000022) = 10 log10(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 log10(12 / PB2) =

    20 log10(1 / PB) =

    -20 log10(PB) =

    -20 log10(2(10-5)) =

    -20 (log10 2 + log10 10-5) =

    -20 ((log10 2) - 5) =

    100 - 20 log10 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.