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[filmscanners] Re: Density vs Dynamic range>AUSTIN (2a)
> Why is this so difficult for you to grasp?
It's not. I learned information theory a long time ago.
> The information carrying capacity of something
> depends upon the number of different states
> it _could_ have, not the number of states it
> actually attains.
Yes, but a DC signal never leaves its one, single state, and thus carries no
information, no matter what they theoretical capacity of the channel might
be. Only a changing signal carries information.
A DC signal is equivalent to a channel with a bandwidth of zero. It thus
carries no information.
> A page of text may have a few thousand
> characters on it ...
Not if it is DC. DC means no change. A page equivalent of DC is one solid
color (black, white, or whatever); it cannot contain text, which requires
changes in luminosity that are spatially separated.
> Not true. It carries information about what
> the voltage is, as distinguished from what
> it might be.
If it never changes from that voltage, then it serves no purpose, as the
voltage is fixed and doesn't need to be communicated in the first place. It
is a communications channel with a bandwidth of zero.
> But just because it doesn't change doesn't
> mean that it couldn't change.
If it is DC, it cannot change, by definition. DC does not change; AC does.
Since DC cannot change, it will not change, nor will it ever communicate any
> You'd still need a voltmeter to know that
> it hadn't changed.
If it is DC, you don't have to look at the voltmeter to see that it has not
changed, because DC never changes.
> The distance from Los Angeles to San Francisco
> is a fixed number, but it's still a number
> that has meaning; it conveys information.
It does not convey information, it _is_ information. Not the same thing.
> Look, this is getting stupid.
I'm trying to fix that.
> This began with a discussion of whether a
> scanner captures AC or DC information.
Information is not AC or DC; only communications channels are AC or DC ...
and in practice, a DC channel has a bandwidth of zero, so it really isn't a
communications channel at all.
> If DC doesn't represent information,
> then removing the DC component should have
> no effect on the functioning of the scanner.
Correct. But scanners don't have DC components.
> To make a scanner, you need a frequency
> response down to DC.
But to convey information, you need a non-zero frequency, and thus you need
something _other_ than DC.
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