OK, I've done some homework and also followed the links provided (thanks
). I'm slowly getting the drift where I went wrong with my observations - though it's not
totally clear to me yet...
First off, I was confusing value range with dynamic range. Obviously the range of possible sample values (signal level) is much larger for floats than for fixed-point with the same resolution. However, the quantisation noise and thus dynamic range is determined by the
precision only - and that's 24bit for a 32bit float. The precision is the number of digits of the significand, including the implicit bit, but without the sign bit (according to
Goldberg).
When summing two floats, the accuracy of the result is determined by the
difference of the values. For each 6dB difference, there will be 1 bit from the smaller value lost - provided the result isn't held for further calculations in the FPU registers. The absolute value doesn't matter because floats provide the full precision over the whole range. So this means that summing two signals with a difference of ~48dB will merely yield the resolution of an audio CD - 16bit - with regard to the smaller signal. For 64bit summing this would be ~222dB - which provides either much more room to play with or much higher accuracy than 16bit.
A difference of 48dB between sample values (not gain!) doesn't sound very much to me. Taking track and bus gains into account, much higher differences can occur all the time. In my book that makes 64bit mixing almost mandatory. OTOH, multiple tracks/buses going to one destination bus are probably summed in one swoop in the FPU (80bit) anyway, and only the result will be rounded to fit into either 32 or 64 bit. For any further processing the lost bits may or may not make an audible difference.
Once again: a mere 6dB difference and the least significant bit of the smaller value will be lost. This is because each 6dB the exponent increments/decrements, and so the significand of the smaller value needs to be shifted right before summing (yeah, it works when shifting
with the implicit bit - which yields a non-normalised number for the moment) to align the exponents. Hence the LSB falls off the grid when the result is rounded (dithered?) and normalised again after the operation.
My apologies for boring you folks
. I mostly needed to write it down for myself - so I figured, why not littering the forum with it, hehe!
werner