Why is there a factor 2 in dBFS formula?

Question

The conversion formula from a linear ratio r into dBFS units is the following: dBFS = 20 * log10(r). The value of r can be different things, but in my case, this is a RMS value divided by the maximum value.

However, the linear to dB conversion formula is dB = 10 * log10(v), so I do not understand why the same formula is not used for dBFS conversion. Why does the extra factor 2 is required in the dBFS formula? Is it some kind of convention we just need to accept or is there a more logical explanation?

Answer 1

However, the linear to dB conversion formula is dB = 10 * log10(v)

It is – for power-related units.

For amplitude units, it's $\text{dB} = 20\log_{10}(v)$. Simple as that!

(this serves the purpose that a modification of x dB of the amplitude changes the power by the same amount. Would both use 10·…, things would break. "I amplified my signal by 10 dB" means I scaled the amplitude by $10^{0.5}$, and it means I scaled the power by $10^1$. That's the same thing, because power goes quadratically with amplitude.)