It appears to me the word 'sensitivity' was meant as 'range'. Sensitivity
of an analog system is the rate of change. A higher bit A/D could give
higher sensitivity, but would not give a better range (which I believe is
what the paragraph was trying to say), since the range is fixed for a given
system. Also, the analog front end filter for the A/D does not 'resolve'
anything. The CCD doesn't 'resolve' either. The only thing doing any
'resolving' in the described system is the A/D.
Also, it is not a given that a higher bit A/D, in a particular system, will
improve accuracy, though it MAY, depending on the system. Where noise in
the system is greater than the resolution of the A/D, it will not.
-----Original Message-----
Paragraph is clear enough for me to understand. And is perfectly
correct to my judgement.
Slava
--- Austin Franklin <[EMAIL PROTECTED]> wrote:
> > for
> > a given sensitivity from the analog circuitry, changing the
> > A/D won't make any difference to the density ranges
> > that the analog circuitry resolves. It only increases the
> > accuracy with which we read the range of analog values
> > that the CCD *does* resolve.
>
> May be I'm slow today...but that paragraph is really unclear to me,
> and I
> know this stuff quite well. What exactly do you mean by 'for a given
> sensitivity from the analog circuitry'? Sensitivity can describe
> one of
> many characteristics, so this seems ambiguous.
>
> Also, what do you mean by 'to the density ranges that the analog
> circuitry
> resolves'. The above paragraph seems to intermix (confuse) different
> concepts/terms, and really comes across, at least to me, as not very
> comprehdable. I don't think 'resolves' is the right word there.
>
> Also, 'analog values that the CCD *does* resolve'? Again, resolve
> doesn't
> really sound right here... The only thing in the described system
> that is
> 'resolved' is the A/D.
>
