Re: Audio recording bitdepth
Re: Audio recording bitdepth
- Subject: Re: Audio recording bitdepth
- From: Bjorn Roche <email@hidden>
- Date: Wed, 2 Dec 2009 22:51:56 -0500
On Dec 2, 2009, at 9:33 PM, Brian Willoughby wrote:
On Dec 2, 2009, at 18:24, Bjorn Roche wrote:
On Dec 2, 2009, at 8:33 PM, Brian Willoughby wrote:
Note that (2^n)-1 is not bit-transparent, except for limited cases
where rounding just happens to restore the original value.
Actually, I meant (2^n)-1. I tested (2^n)-1 for n = 16 and 24 with
single precision floating point numbers pretty extensively and
carefully and they are transparent as well. There is presumably
some reason for this beyond coincidence (not enough of a change to
snap it to the next quanta?), but I don't think I want to think
about it anymore.
Well, your tests are not complete. For n==16, (2^n)-1 creates 14
additional bits of quantization noise. Your tests do not reveal
this because the rounding step covers up the noise. Since 24-bit is
only 8 bits more than 16-bit, there are still enough extraneous bits
to cause the rounding to make it appear transparent, even though it
isn't.
Any processing of the float values would show the added quantization
noise when using (2^n)-1, but by using 2^n you're safe from
additional quantization noise.
You are right that I am not measuring quantization noise converting
from int to float. Usually when audio engineers speak of bit
transparency in this context, they are not referring to quantization
noise; rather they are referring to the ability to recover the
original int value after a conversion to float and back to the
original int value. In this case, you can recover the original int, so
the transformation is bit transparent.
bjorn
-----------------------------
Bjorn Roche
XO Wave
Digital Audio Production and Post-Production Software
http://www.xowave.com
http://blog.bjornroche.com
http://myspace.com/xowave
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