Digital BW, The Print

--- In DigitalBlackandWhiteThePrint@yahoogroups.com, "Paul D. DeRocco" 
<pderocco@i...> wrote:
> > From: Roy Harrington
> >
> > Ok.  I think it's interesting to note that it is specifically the
> > mathematics of
> > the linear A/D data representing the ratio of the dynamic range
> > that gives rise to the need for more bits to get more dynamic range.
> > E.G. with a 12 bit integer value the biggest ratio you can
> > represent is 4096/1.
> > If you want a larger dynamic range ratio than 4096 you have to
> > have more bits.
> 
> That's not entirely true. It's certainly true for a single pixel, but the
> eye doesn't see only individual pixels, it sees areas of similar pixels as
> solid colors. Since there's always noise, you can see in-between values by
> averaging across many pixels. That's basically what dithering is.
> 
> --
> 
> Ciao,               Paul D. DeRocco
> Paul                mailto:pderocco@i...

True, but I'm not sure if that ought to play a part in the dynamic range of
the sensor and the bits needed in the A/D to represent that range.  But in
any case you will need to add 1 bit each time you increase the dynamic
range by a factor of 2.

Here's another way to look at it.  At the low end say the A/D is distinguishing between
a value of 1 and 2.  No matter how many samples you average together you'll
get either 1 or 2 since it's integer math.  This is quite different than in the upper
range.  E.g. say you are distinguishing between 100100 and 100200 where the
bottom 2 digits are in the noise.  If you average 10 samples you'll be close to being
able to distinguish 100140 and 100150.  The point being that if its in the bottom
bit the multiple-sample noise reduction doesn't get represented.

Roy