Triggaaar;164989 Wrote: 
> 
> "Yes, you only maintain all the resolution of the original 16 bits by
> using all 24 bits - so truncating to 20 will loose you some
> information."

a 24-bit input value some of the lower attentuation values will not end
up truncating any bits, in other words.

Thats part of why the input value is 24 bits even though the data is
only 16 bits.

> 
> reducing the digital volume will NOT take out information from the
> original signal, providing the input signal is 16 bit audio and you
> don't go lower than (IIRC) -35dB
> 

The same thing as above: the original value on a CD is only 16 bits.

So if you take a 16 bit value, and left-shift it (effectively) 8 times,
rotating 0's into the low order bits, you have more wiggle room.

Or if binary isnt your cup of tea: think of it in decimal.  You have a
number, say, between 00 and 99, if you make it between 000 and 990, by
multiplying by 10, you have a bit extra resolution when you do division
without resorting to fractions.  In this case, it actually gets you
through the first N steps of volume attenuation without any truncation.


-- 
snarlydwarf
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