Hi Austin,
Todd's recent introduction of this audio paper has been a brilliant stroke
of luck.
You've read the paper and correctly found the "6db per bit" rule as you
call it. You even correctly argue why in audio you get 6db because the
encoding is by voltage and what we really want is power. Since we want
voltage squared there's an extra factor of 2, hence 20*log() instead of
10*log() -- as you've noted.
A few days ago you and I were discussing scanner output and you noted
3db per bit. You said 24db for the 8-bit file and 33db for the 2000 values
(i.e. 11-bit file).
Note that ALL the above numbers and info are from YOUR posts.
And finally you've made the following pair of remarks:
Dynamic range is DIRECTLY based on the number of discernable steps.
Dynamic range is not based on the actual values the data represents.
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So here you have two systems where "the actual values the (digital)
data represents" are different. For the same dynamic range the
number of levels is drastically different. For audio a 16-bit system
using 65536 levels is capable of 96db of dynamic range. To get that
same dynamic range in imaging you'd need 32 bits using 4 BILLION
levels. The difference is entirely because of the encoding of the
data.
Looked at another way, the full 48db dynamic range of a
16-bit scanner output can easily be fit into an 8-bit file by
changing the encoding (i.e. using PS>Adjust>Levels or >Curves) before
chopping off the bottom 8 bits. The dynamic range of the two files
(16-bit and 8-bit) will be the same.
So, Austin, you made all the statements above the dotted line and they
are self-contradictory.
Roy
Roy Harrington
roy@harrington.com
Black & White Photography Gallery
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