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[filmscanners] RE: Density vs Dynamic range




> > It's (my initial statement you disagree with) hardly an
> over-simplification,
> > in fact, it is about as complete and accurate as you can get.
> It is a plain
> > and simple fact that a dynamic range of 5000:1 requires 13 bits
> to represent
> > that full dynamic range.  Again, 1 is the first discernable
> signal (as in
> > the noise floor), and since you can only measure in increments
> OF the noise
> > floor...a dynamic range of 5000:1 simply means you can have
> every integer
> > value from 1 to 5000...and to represent every integer value
> from 1 to 5000,
> > you need 13 bits.
>
> Perhaps this can be made clearer by bringing back a concrete example
> (similar to one which I think Austin used earlier in this thread).
>
> Suppose that the output of the CCD analog stage swings from -2.5 to +2.5
> volts, and that there is 0.001 volts of noise.  If you install an
> analog-to-digital convertor to sample signals from this system, there is a
> lower limit to the step size that you should choose.  There is no
> advantage
> to choosing a step size smaller than the noise level of 0.001 volt.  Why?
> Because when you measure the voltage and get a result x, any value in the
> range, you really don't know the actual value.  You measured the
> signal plus
> noise.  The signal could have been anything between (x + 0.0005) and (x +
> 0.0005).  So the optimum "granularity" is a sampling system with (2.5 -
> (-2.5))/0.001 = 5000 steps.  You could use a system with fewer
> steps, but it
> would not be optimal.  It would not encode all the information.
> If you used
> a system with more than 5000 steps you would not be measuring with more
> accuracy, you would just be building in excess complexity and expense.
>
> The dynamic range of this system is 5000:1.  You would need a 13 bit
> converter to represent 5000 discrete steps, since 2^13 = 8192 (more than
> enough) but 2^12 = 4096 (not enough).

Hi Julian,

Aside from your known sign switch in the tolerance (+- .0005v), perfectly
described and perfectly correct.

Regards,

Austin

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