Re: Audio recording bitdepth
Re: Audio recording bitdepth
- Subject: Re: Audio recording bitdepth
- From: Brian Willoughby <email@hidden>
- Date: Wed, 2 Dec 2009 18:33:23 -0800
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.
Brian Willoughby
Sound Consulting
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