how many bits do you really get with 16 24 32 64 floating point etc?

Another post confirmed what I supspected ( I think ).

It said that if you had 64 bit floating point that 12 bits
got used for exponent and 52 bits for the mantissa.

1 bit of the 12 gets used for the sign so you can go
from +/- 2**11 = +/- 2048 for the exponent.
And 1 bit of the 52 is needed for the sign of the mantissa
so the data value precision equals 15 digits decimal.

So wouldnt regular binary at 16 24 32 bits also use one for the sign (as you have to encode both + and - values of the original voltage values of the signal).

That means you only really have 15 24 or 31 bits relating to the dB range back in the analog domain. There may be 16 24 32 bits of digital dB in the digital domain but not when translated back to analog.

I'd warmly recommend some digital signal processing theory for the OP. I have read couple of other threads you started, and your questions would be answered if you'd just do a few hours of old fashioned reading. Grab a good book on the subject and read it through. A bit of reading will make these things a lot less shady and worrying.

As for the question. No. The amount of bits determine the number of different levels achieveable. There is no need to store those values as negative and positive values. Instead we can just determine a number that serves as a zero point. Everything under that number is negative values and above it positive.

Bottom line is, we are dealing with binary numbers here. If the MSB (Most Significant Bit) is 0, we have negative value. If it's 1, we have a positive value. There would be equal amount of both numbers available, no matter how many bits we use. So, in a sense the + / - sign is built in to the value.

Which is why we use 32 bit FP to work with 24 bit audio. We're freed from the tyranny of an absolute value range while retaining the precision implicit in the initial audio.

Which is why we use 32 bit FP to work with 24 bit audio. We're freed from the tyranny of an absolute value range while retaining the precision implicit in the initial audio.

I'd warmly recommend some digital signal processing theory for the OP. I have read couple of other threads you started, and your questions would be answered if you'd just do a few hours of old fashioned reading. Grab a good book on the subject and read it through. A bit of reading will make these things a lot less shady and worrying.

As for the question. No. The amount of bits determine the number of different levels achieveable. There is no need to store those values as negative and positive values. Instead we can just determine a number that serves as a zero point. Everything under that number is negative values and above it positive.

Bottom line is, we are dealing with binary numbers here. If the MSB (Most Significant Bit) is 0, we have negative value. If it's 1, we have a positive value. There would be equal amount of both numbers available, no matter how many bits we use.
So, in a sense the + / - sign is built in to the value.

thanks
been there done that
proved the theorems
did the exercises

know it quite well

thanks for confirming what i said
you only have 15 bits of magnitude to do d/a with
and one bit of sign to tell whether it is + or -

Each bit doubles the amount of available values. And the most significant bit on the 16 bit system raises the amount of available values from 32 thousand to 64 thousand. The first 32 thousand (when the value of the MSB is 0) are negative values and the other 32 thousand (MSB = 1) are positive.

For a person who claims to have done his homework your understanding of such fundamental as binary system appears to be quite vague.

As for the question. No. The amount of bits determine the number of different levels achieveable. There is no need to store those values as negative and positive values. Instead we can just determine a number that serves as a zero point. Everything under that number is negative values and above it positive.

Basically the same as dedicating a bit to the sign value