[filmscanners] Re: Density vs Dynamic range

["" <peter@softwell.co.uk>]

>
> 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).
>

Julian,

I couldn't have put it better myself, as I tried just a few moments ago!
However my point is that if you can reduce the noise level then you can
increase the number of steps (by halving the step size) with real
benefit, but without altering the range.

My principle argument was that a 5000:1 ratio does not specifically define
that 5000 steps are requires

Just a quick question - do CCDs really use a +/- voltage swing? I'd have
thought that would have introduce noise problems around 0.

Peter, Nr Clonakilty, Co Cork, Ireland

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