On 27/05/2007, at 7:25 PM, Phil Tyler wrote:
> Giving you 8 bits, theoretically, more resolution of
> low level sounds with 24 bits.
>
> Phil
I suspect it's far more than 8 bits as that would imply 2^8 =3D 256
additional "steps", whereas the actual difference is between 2^16 =3D
65,536 and 2^24 =3D 16,777,216.
The page I posted a link to earlier in the week has this rather
pertinent example:
http://www.24bitfaq.org/#Q0_1_1
"To elaborate further, each bit gives us the ability to represent
about 6dB of dynamic range. A passage that is 6dB louder than another
passage is said to be twice as loud as the other passage. In the 4-
bit example, we theoretically have 24dB of dynamic range that can be
used. But what if recording doesn=92t take advantage of all that
dynamic range? What if the recording never peaks beyond 6dB of its
maximum possible limit? In this case, the recording would only take
advantage of 3 of what we call the least significant (or left-most)
bits, meaning 18dB of dynamic range. 16-bit recordings are capable of
a theoretical maximum limit of 96dB of dynamic range. This means that
a single wave could have up to 65536 discrete values that can be used
to represent it. But if the same wave recorded at 16-bit peaks at
48dB below its maximum possible limit, then there would only be 256
discrete values that can be used to represent it, taking advantage of
only 8 of the least significant bits. The 8 most significant bits
would contain no information whatsoever, and would remain unused. In
the case of 24-bit recording, you=92d have a maximum of 16,777,216
values to choose from, and in the case of a wave peaking at 48dB
below its maximum possible limit, the wave would still have 65536
possible discrete amplitude values that could be used to represent it."
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