Digital BW, The Print

Hi Martin,

> Dynamic Range, Some Definitions:

The definition/concept of dynamic range is taught in any good EE program,
and is very basic to most electrical engineers (NOT computer engineers, they
are not the same).  I have probably measured the dynamic range, and written
the specs for, over 100 different systems.  I am an EE, and, as you know,
have been designing both audio and video equipment for over 25 years.  I
don't mean to sound smug, but it is unquestionable that my
definition/understanding of dynamic range is correct.  If it was not, since
I do this for a living, and am subject to direct peer review by hundreds of
different engineers over my career, as well as thousands of customers and
people who read my specs...and if I had it wrong, someone would have pointed
it out by now...and no one has.

Most of the definitions you cited are "pedestrian" definitions written by
non-technical people who are, for the most part, mis-explaining a concept
that they apparently do not understand.  What you are calling "dynamic
range" is merely static range.  Measuring the max and min value of something
merely gives you the STATIC range, NOT the DYNAMIC range.  Comparing them by
division or subtraction still give you a STATIC result, not a DYNAMIC
result.

> So what we have is a variety of definitions that seem to vary
> depending upon what field you are in

They vary simply because the people who wrote the incorrect definitions did
not understand what it was they were saying.  Dynamic range IS dynamic
range, no matter what the application.  The X-Rite definition you supplied
is the best one as far as I am concerned.

> but the common element is
> that they are looking for a meaningful way to describe a
> relationship between the minimum and maximum as a ratio.

That's the issue, they do not describe what they mean by minimum and
maximum...  They are ambiguous terms.  These terms can mean either amplitude
or static values.  Per the diagram I posted in another post, they clearly
mean amplitude.

> Some people say the dynamic range is the difference between the
> min and max and some say it is the ratio.

It is always a ratio, as defined by any dynamic range equation.  Forget web
and dictionary definitions, it's the equation that matters here, and only
the equation.  The equation if finite and definitive.  Your understanding
does not follow the equation.

The dynamic range equation from "Digital Signal Processing in VLSI" by
Richard J. Higgins is:

Dynamic Range (dB) = log10 (largest signal/smallest discernable signal)

The diagram I posted previously is from this very book/equation.  It clearly
shows that largest and smallest are amplitude.  Also note DISCERNABLE.  To
discern something means it must change with respect to what it is you are
trying to discern it from, which in this case is it self.  You need two data
points for "discernability".  You can not discern something that is static
(shows no change).

> Some things such as
> sound volume, CCD response and image density are not direct
> measurements of a physical property but are calculated values
> from actual properties given in logarithmic form.

Absolutely not true.  You absolutely can measure the dynamic range of audio
systems and CCDs and image density.  You CAN run an FFT on the data to
arrive at the dynamic range, but you still have to do direct measurements to
arrive at the values you give to the FFT...you don't just make them up!

> So if a print has a white paper base reflectance of 95% (95% of
> the light shining on this spot bounces off toward your eye) and
> the deepest black has a reflectance of 4% (only 4% of the light
> falling on this dark spot bounces off and the rest is absorbed or
> scattered away from your eye.) The dynamic range would be the
> maximum value of 95% divided by the minimum value of 4%, which is
> 23.75.

Because density is represented in a log form, and dynamic range is
represented in a log form does not make them both the same.  Dynamic range
and density range are NOT the same.

How do you know you really measured 95% and not 95.876593254?

> Usually in photography we do not talk about reflectance but
> instead use density. So for this case the paper base would have a
> density of 0.0223 (the base 10 log of 4%)and the darkest black
> 1.3979 (the base 10 log of 95%). Since these are log values
> calculated from the reflectance you find the dynamic range by
> subtracting them. 1.3979 minus 0.0223 is 1.3757 the dynamic range
> and the range of the Density.

You have NOT measured any dynamic range, as you have NOT measured the noise
in the system.  What you have is the DENSITY range NOT the DYNAMIC range.
Again, you can NOT relate dynamic range to density range simply because they
are expressed in log form.

> So if you are talking about image density (a log value) the
> dynamic range and the range are the same.

No, absolutely wrong.  Because two things use the same numeric
representation it does not mean they ARE the same.  Dynamic range is
measured in DECIBELS, density is measured NOT in DB, but on a calibrated
relative density scale.  NO density measurement equation will contain DB.
That fact alone negates your entire premise that they are the same.

OK, here is the Density measurement equation from "Introduction to
Densitometry" published by the Graphics Communication Association:

Density = log10 1/R (where R is the reflectance)

and to quote the book:

"This is a pure definition, which means that it defines the relationship
between light reflectance and density.  This definition numerically
interprets density, and, more importantly, describes density in a manner
that approximates the way in which the human eye sees objects."

No where in this entire book do they talk about dynamic range.  Also,
density measurements REQUIRE they be compared to a known calibrated sample,
which is why you have to calibrate densitometers.  Measuring dynamic range
does NOT require it be calibrated to any known sample, it is simply relative
in and of it self.  Density measurements aren't, they are relative to an
absolute measurement...hence, they are STATIC.

> Noise generally does not seem to be used in the calculation of
> the dynamic range

Absolutely wrong.  The "smallest discernable signal" in any analog systems
IS the noise in the system.  In a digital system, as in number of bits let's
say, the minimum discernable signal is 1, so 4 bits has a dynamic range of
log10(16/1) or 1.2dB.  The key is "smallest DISCERNABLE" signal, what ever
that might be in that particular system.

> The best example would be in audio
> systems where there is a background hum or hiss. This is what is
> known as the noise floor and in audio the weakest or minimum
> value is this noise level since you cannot hear any sounds that
> have a volume lower than the noise.

Again, you are using ambiguous terms here (weakest/minimum).  The weakest
value is NOT necessarily the noise floor.  You can turn the volume down, and
decrease the noise to almost nothing, and that also doesn't take into
consideration the noise before the volume control.  Also, the noise will be
different depending on the volume, as when you add gain to a system, you
also amplify the noise.

> There is noise in the process of making a print since information
> is lost or degraded as you move from scene to camera to print,
> but once the print is finished the concept of noise does not
> appear to apply

Of course noise applies.  Anything that as distorted the actual image
information has caused noise, anywhere in the system.  It is not relevant if
the print is "finished" or not.

> Finally the "number of tones" present in a print does not depend
> upon the dynamic range or the range

Absolutely wrong...as I've said time and time again, and for some reasons
you don't understand my examples.  The number of tones is SOLELY based on
the dynamic range.  It's the definition OF dynamic range (as defined by any
competent source).

How many tones does 8 bits represent?  256.  How many tones does 9 bits
represent? 512.  Now, you can represent ANY density range with only two
bits, as I have explained in another post.  You can represent any density
range with ANY number of bits.  It is merely a representation.  BUT the
dynamic range will be entirely different, but the density range will be the
same.

> A print
> is an analog image and by definition has continuous tones that
> flow from one to another without any step change or gaps.

Absolutely wrong, again.  NO analog system is completely continuous.  There
is NO such thing.  ALL systems have a resolution analog or not!  It's a
fact.  Because you don't understand that it doesn't doesn't make it not
true, it just means you don't understand/get that.

> Just as
> the sun sets in a smooth motion and not in increments, so the
> shades of gray flow in a continuous tone print.

Your ability to SEE the sun setting absolutely is in increments.  It is well
knows that your vision has a resolution, and you can only discern things
with the accuracy of so much of an arc.  It's a fact.

Quoted from "Modern Optical Engineering" by Warren Smith:

"The resolution of the eye was AT BEST about one minute of arc."

Given that you can NOT discern any movement of the sun/earth smaller than
one minute of arc.

One thing you are missing that everything has resolution...everything, even
all analog systems.

> This then comes back to my original assertion that all continuous
> tone photographic mediums have an infinite number of tones
> whether they are ink jet or silver based.

And that assertion is flawed.  ALL analog systems have resolution, and
therefore have a number of discernable tones, and are NOT continuous.

> Thank you for your patience if you got this far. <G>

I have tried, and don't believe I can not explain anything further to you.
It is not a matter of disagreeing, it is a matter of what you are saying is
just wrong, and is based in ambiguous definitions and flawed assertions, and
you don't want to understand that.

I don't know what you do for a living, and what your background is.  I am
guessing software?

If you would like to ask another expert in the field, I am very happy to
suggest someone to you.  Perhaps they can explain it better to you than I
have been able to.

Regards,

Austin