[filmscanners] RE: Dynamic range

["Austin Franklin" <darkroom@ix.netcom.com>]

Roy,

<sigh>

> > The scanned image data file had a dynamic range of 33dB, and
> the processed
> > file is now 24dB.  The processed image file has less dynamic range, of
> > course.
>
> This is where we differ.  Visually we have two files that "look" alike.

Whether something "looks" alike or not is irrelevant as to whether the image
data file is actually alike or not.  If you are talking about viewing the
file by a human on a computer monitor, you are being limited by the monitor
as well as your eyes, which, of course, has no bearing on whether the two
files are the same or not, much less whether they have the same dynamic
range or not.

> There's a 16-bit file (with only 2000 distinct levels) and an 8-bit file
> with 256 levels.  My contention is that they have identical dynamic range
> as well as identical density range.

And, your contention shows you don't understand what dynamic range is.  The
DENSITY range that the two files represent is the same, but the dynamic
range simply is not. There is far less data, and the ratio, 2000:1 vs 256:1
are obviously different, and since that is the dynamic range, how can they
be the same?  They are obviously not.

Let's use YOUR example:

For simplification's sake:

DR (dB) = overall maximum signal level / minimum discernable signal...

simply showing the two variables that make up dynamic range, and their
interdependency.  If either change, the dynamic range changes.

1) So, let's say, for arguments sake, you have a DENSITY range of X (not
relevant what it actually is) scanned into your 2048 distinct levels (11
bits).  That's a dynamic range of your image file of 2048 to 1.  2048 is the
maximum overall, and 1 is the minimum discernable signal.

2) You now convert that to 256 levels (8 bits)...but let's say we encode it
such that 2048 is STILL the maximum overall signal level, but we simply
don't use the bottom three bits (it's the exact same thing...but makes for a
good example to demonstrate your misunderstanding)...your minimum
discernable signal is now 8....

2048 / 1 = 2048

vs

2048 / 8 = 256

Therefore, in your given example, the minimum discernable signal of the
image file has changed, therefore the dynamic range of the image file has
changed.  Plain and simple.

> Now if we actually print these two files (assuming you have a driver that
> can handle 16-bit data) we'll have two prints that are extremely close
> if not indistinguishable.  Do these two prints also differ in dynamic
> range by a factor of 8?

The prints MAY very well differ in dynamic range, but we don't know how they
were printed, and...this is irrelevant.  Once you print it you can change
the dynamic range, but not necessarily the density range...sure that changes
too, hopefully relatively minimally, but this simply isn't relevant to this
discussion.  The dynamic range of THE IMAGE file is still the same, whether
you print the file or not.  The dynamic range of the PRINT and he IMAGE FILE
MAY (will most likely) be different.

> If you answer "yes" they differ by a factor of 8, it seems
> dynamic range is
> very poor characterization of what we see.

For some people, who understand it and know how to use, it has a LOT of
value.  That does not have to be you, and whether you understand it or not
doesn't change that others do correctly understand it and do believe it has
a LOT of value.

> I.e. a factor of 8 change in
> dynamic range makes a negligible difference in what we perceive.

The issue here is you simply don't get it.  The dynamic range of the prints
is entirely irrelevant to the dynamic range the film was scanned at, for
reasons I've explained in hosts of other posts.

> On the
> other hand if you answer "no", the prints have the same or at
> least similar
> dynamic range, then the dynamic range of the files doesn't come close to
> translating into a dynamic range of the final print.

No one ever said the dynamic range of the print was the same as that of the
image file.  You're the one adding this irrelevance to the issue, and it IS
irrelevant.

> Either way you get a
> strange paradox of what happened to the dynamic range.

Only in your mind is there a "strange paradox", as it is neither strange,
nor a paradox, that is, if you really understand what is going on here.
When you change either the minimum discernable signal (in our case with a
digital image file, this is now 1), or the overall largest signal (as in
number of values, when the MDS is 1, and talking about a digital image
file), you change the dynamic range.  Text book...as I said, neither
strange, nor paradox.

DR (dB) = 10log2**N  Go look it up.

> With scanners and other digital imagining systems we have CCD's which
> respond to intensity of light producing a voltage output.  That voltage
> output is fed into a A/D converter that digitizes the voltage and
> outputs a number representing the voltage.  For sake of example, say
> its a 10-bit A/D and it measures millivolts from 1mv to 1024mv in 1mv
> steps. It's linear in voltage with possible outputs of
> 1,2,3,4...1022, 1023,
> 1024.  So it's got 1024:1 or 30db dynamic range and 1024 potential levels.

That is correct.  FYI, typical A/Ds are +- 3V input, and the analog front
end will expand/contract and shift the voltage as such between the CCD and
the A/D.

> In this case number of levels and dynamic range go hand in hand,
> 1024 levels <==> 1024:1 dynamic range.  This translates exactly to
> the RAW output of a scanner  -- the RAW levels and the dynamic range
> will track together.

OK.

> The KEY, KEY thing here is that this equivalency
> is directly due to the linearity of the A/D tracking with the linearity
> of the voltage output of the CCD.

Well, you're key doesn't work...you're still locked out...as it's simply
wrong.  It does not really track linearly, for one, and two, any "somewhat
linearity" is by mere happenstance and does not mean that density range and
dynamic range are always the same.  And, not all scanners are designed such
that the output of the CCD is simply linearly encoded.  In SOME scanners,
they CAN perhaps be close to the same, but overall that is a mistaken
conclusion.

Now, they MAY be "similar" but they are not the same, and even if they
happened to be, so what?  What does that have to do with any of this
discussion?  Nothing.  It's simply that a scanner designer chose to encode
the data from the CCD that way.  It's simply an encoding, it isn't dictated
by any "rule" anywhere.  You could simply have an analog front end that
thresholded the data, and gave you two bits.  01b for less than .01 density
value, 11b for between .01 density value and 3.6 density value, and 10b for
density value greater than 3.6.  You've now encoded the entire DENSITY range
of the scanner into two bits, but the dynamic range is very small... 10log3
(three values, not four) or 4.5dB.

But...if you scanned that SAME DENSITY range at, say, 10 bits...you would
still have encoded the EXACT SAME DENSITY range, but with a completely
different dynamic range... 10log2**14 or 30dB.

> ...so
> a possible set of raw voltages that would represent evenly spaced
> human perceptions could be 1,2,4,8,16,32,..., 256, 512, 1024.
> There are 10 evenly spaces intervals each being one stop apart.
>
> We could represent these 11 values with just 4 bits, the dynamic
> range that
> they represent is still the same as the original 10-bits i.e. its still
> 1024:1 or 30db.

No, that's wrong, you are merely REPRESENTING DENSITY RANGE values, not
dynamic range.  You are changing the dynamic range by only using 4 bits
instead of 10.  It is simply commonly known, except by you apparently, that
the dynamic range that can be represented by a number of bits is simply =
10log2**N (or 20x, depending on the type of system being described), so 4
bits has a dynamic range of 12dB and 10 bits has a dynamic range of 30dB.
The dynamic range is NOT the same as the density range.  Dynamic range is
DIRECTLY based on the number of discernable steps...no matter WHAT the
density range is.

> The important numbers are still 1024mv and 1mv
> so the ratio
> is still the same. This is analogous to reducing the 2000 levels
> picture to
> the 256 level picture.  The dynamic range is not changed because the top
> level and the bottom level still represent the same real-world
> values so the
> real-world ratios are still the same.

Well, that's simply a complete misunderstanding.  The dynamic range HAS
changed, because the RATIO OF THE MAXIMUM SIGNAL TO THE MINIMUM DISCERNABLE
SIGNAL has changed.

For dynamic range, it simply is not relevant what the bits represent (which
is confusing you), for density range it, of course, the bits do have
meaning, but more so from a relative standpoint than an absolute standpoint
at this point.  For 2000 levels, your maximum signal is 2000, and the
minimum discernable signal is 1, and for 256 levels, your maximum signal is
256, and the minimum discernable signal is 1...so the ratios are 2000:1 and
256:1, clearly different dynamic ranges.

Now, for your try at re-encoding the data, as I've said, for dynamic range,
what the bits represent is not relevant...so for 4 bits, that's 16:1,
whether the value 0xF represent a gazillion volts, or .000001 volts...that
doesn't change the dynamic range that can be represented by 4 bits.  Plain
and simple.

That's what you simply aren't understanding about dynamic range, it is based
on the ability to discern, not the value the discerned/encoded data
represent.  Read that again.

> So, in conclusion, Austin, I'm really not trying to give you an
> excessively
> hard time.

I don't consider you "giving [me] a hard time", I consider you determined to
defend your (IMO mis-) understanding of this concept, for some reason I
simply don't understand.  Your examples don't show anything along the lines
of what you believe, and I've shown why they are simply mistaken or simply
irrelevant.  You've added nothing new to the discussion and keep moving your
questioning of the issue.  It's gone from terminology now it's dynamic range
of prints vs image data...

> I hope you can spend some considering what I have written, I've spent
> a fair amount of time figuring out how to explain my case.

I have considered what you've written, and have drawn the same conclusion,
it's simply a misunderstanding on your part, and you are confusing the value
of the data with the number of discernable data points, it's plain and
simple.  Dynamic range is not based on the actual values the data
represents, simply the discernability of the data.

Austin

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