How is this possible, a resolution above 24 bit? If it is correct that the 32-bit-float Audio format uses 24 bit for the resolution in it’s significand (in German Mantisse) and 8 bit for the exponent, which I assume is of value 0 for normalized datasets that contain no value greater than 0dB. Where do I have a misunderstanding?
Or … How are the – so it seems for me in my wrong assumption ?? – not used precious 8 exponent-bits used? (not used because I assume they are all 0 for datasets with 0dB max-value)
How is the normalization in calculations performed to always use the best and finest resolution in calculations because 10 exp 1 is the same as 1 exp 2 (base 10), but the latter is more imprecise in cases other than exactly 0. (11 exp 1 is more precise then 1 exp 2)
As this is quite some interesting technical topic, I looked a bit into the general floating point format, using Wikipedia and a very nice Online Converter and did some binary checks with a few short Mono RAW files and a HEX editor.
Interesting, how the binary string is actually an encoding of 3 segments. And even the usage of some special rules, whether the exponent segment has all 0’s or 1’s. I did not expect that.
WaveLab says in its Global Analysis Apparent Bit Depth (in Deutsch Wahrscheinliche Bit-Auflösung), so to me it seems like an approximation, at which integer bit depth the data would need to be saved to contain all information. This value also seems to correspond with a negative exponent used for the floating point calculations, so in WaveLab it will probably only matter for the very low levels, but not for levels exceeding 0dBFS (or +1/-1 normalized).
