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Digital BW, The Print

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Re: Artifacts with Digital images

2005-07-04 by Roy Harrington

--- In DigitalBlackandWhiteThePrint@yahoogroups.com, "Paul D. DeRocco" 
<pderocco@i...> wrote:
> > From: Roy Harrington
..
> > I think you are confusing dynamic range with smoothness or
> > resolution of gray
> > values thoughout the scale.  Dynamic range is only a measure of
> > how far apart
> > the darks shadows can be from the brightest highlights.  It's a
> > matter of seeing
> > detail in shadows without blowing out the highlights.  Dynamic
> > range is just
> > a ratio -- no more.
> 
> Sure, but the dark end isn't limited by the numeric range--that goes to
> zero, and you can't divide by zero. It's limited by the noise level, which
> means that photographing the inside of your lens cap will give you nonzero
> output. The average of that noise defines your black level.
> 
...
> --
> 
> Ciao,               Paul D. DeRocco
> Paul                mailto:pderocco@i...

But the "min" in the denominator isn't the smallest integer number that may
come out of the A/D.  The min is the minimum signal that produces an output
larger than the output you see from just noise.  

Maybe this example will illustrate what I mean: 
Let's assume we calibrate the A/D such that it reachs beyond the voltages 
output by the sensor.  Say its a 10bit A/D so the possible outputs are 0 to 1023
but the noise level with no signal input reads out 20,  and the max read out you
get is say 1000 -- the clipping point.  What is the dynamic range?   It's the
ratio of the signal input values (i.e. light levels) that produce the clip point =1000
over the input that produces the smallest signal value = 21.  Because of the
linearity of input signal with the digital output the ratio is:
  dynamic range =  ( 1000-20 )/ (21-20) = 980/1 or 980-to-1.

The dark end is very much the limiting area.   Note that 21 is the first signal
and 22 is the second signal.  But the signals involved here are:  (22-20) = 2
and (21-20) = 1.  The second detectable signal must be twice the size -- a full
stop brighter.  To get to just 36 output you need 4 stops brighter.  You need to
be much larger than the noise level to produce decent separation of grays.
That's you need a much larger DynRg in say a scanner that it seems you ever use.

Roy

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