# To RAW or not to RAW?

Discussion in 'Digital SLR' started by M, Oct 4, 2005.

1. ### Roger N. Clark (change username to rnclark)Guest

Gee, I know we went over this. read noise is insignificant at higher
intensities, and by the numbers above a few stops *darker* than
18% gray, the read noise becomes insignificant. Go back up
Yes, that is not the case here.
I believe you are computing the db wrong. This is voltage, not
power. The correct equations would be

12 bit limited: 10 log(4096) = 36.1 dB
versus
14 bit limited: 10 log(16384)= 42.1 dB

Cameras with 50,000 full well and read noise of 3 (Canon 20Da)
would be 14-bit digitization limited. 6 dB is significant.

You don't think seeing details in the shadows on the above page
is significant? And you are viewing it on a monitor with limited
dynamic range. The whole point of characteristic curves is to
compress dynamic range for viewing. The human eye sees much
greater than the dynamic range of even the best sensor built
by humans. But each little improvement is significant,
and so is the present improvement in electronic sensors.

Yes, but that means these manufacturers are behind.

The Fuji S3 pro cameras are 14-bit now:
http://www.dpreview.com/reviews/fujifilms3pro/page2.asp
So is the Minolta Dimage A1:

Roger

Roger N. Clark (change username to rnclark), Oct 20, 2005

2. ### David J TaylorGuest

Roger N. Clark (change username to rnclark) wrote:
[]
Voltage or current dB are 20 log (ratio), so 72.2dB and 84.3dB.

David

David J Taylor, Oct 20, 2005

3. ### Roger N. Clark (change username to rnclark)Guest

http://www.phys.unsw.edu.au/~jw/dB.html

the factor is 20 when working with power. Power is V*V/R (V=voltage,
R-resistance). The factor of 2 comes from the squared term.
When not working in power, it is 10 log(A/B). But then maybe
discussing dB isn't appropriate as we are not talking power anyway?

In either case, regarding the discussion about signal-to-noise ratios
and dynamic range in a digital camera, the difference between
12-bit and 14-bot encoding is much larger than the 2 dB that was
originally stated by Floyd Davidson.

Roger

Roger N. Clark (change username to rnclark), Oct 20, 2005
4. ### David J TaylorGuest

Roger N. Clark (change username to rnclark) wrote:
[]
As I read that page, they use capital P for power, and their first
equation amounts to:

dB = 10 log (P2/P1)

(where P2 and P1 actually are subscripted).

Later, they use sound pressure (with a lower-case "p"), noting that:

"Sound is usually measured with microphones and they respond
(approximately) proportionally to the sound pressure, p"

and then effectively define:

dB = 20 log (p2/p1)

This is entirely consistent, as twice the voltage (into a constant
resistance) gives twice the current, and therefore four times the power.

20 log (v2/v1) = 20 log (i2/i1) = 10 log ((v2 *i2) / (v1 * i1)) = 6dB

The difference in quantising noise between 12-bit and 14-bit is 12dB (a
factor of four), although the difference in the system noise will depend
on the RMS value of the noise in the source. And that noise is not
constant with signal level as it will include photon-limited noise. Where
the maths becomes a little more complex!

Cheers,
David

David J Taylor, Oct 20, 2005
5. ### Floyd DavidsonGuest

Of course the dynamic range is *not* a voltage, hence your
example above is incorrect. As to dB and "not talking power", I
am astounded at such gross misunderstanding! The point is that
it is a ratio, and dB is a logarithmic representation which gives
a more useful scale.

I realize this is a broader technical discussion than you
fellows are comfortable with, but everything discussed in this
particular article, while not specific to photography, is *very*
*very* basic. If you can't get this right without a knock 'em down
drag 'em out fight, there isn't much point in even trying to
deal with anything complex.
Nobody has said that the difference between 12 and 14 bit
*encoding* is only 2 dB.

I've said that the change from 12 to 14 bit encoding with a
sensor that has at most 74 dB of dynamic range, is only going to
be perhaps 2 dB *for the system as a whole*.

As noted, that is very very basic systems engineering...

Floyd Davidson, Oct 20, 2005
6. ### Floyd DavidsonGuest

Sigh... It *is* significant. Read noise is the predominate
noise at lower brightness levels, and is the parameter which
determines the minimum quantum step used by the ADC (and thus
the dynamic range of the quantizer). That selection then
affects quantization at higher brightness levels, even where
read noise is not the predominate noise.

Claiming that read noise is "insignificant at higher
intensities" is just ignoring how the process actually works.

Not to mention that noise is additive, and given the values
mentioned, the read noise *is* still significant.
See above.
How do you know that?
Hence you are saying that all of the references, such as the
Nikon web page that I cited, are wrong?

The dynamic range of the sensor is:

Dynamic_Range_dB = 20 log * (full_well_capacity / read_noise)

(If you need to review that again, I can provide half a dozen
references...)

(Please note that it is *not* a voltage.)
That *is* exactly my point. 6 dB is significant, and 2 dB is
not.

That describes a sensor with an 80 dB dynamic range. Roughly
8 dB more than a 12 bit ADC can provide (and 4 less than a 14 bit

In *that* case, it is not a matter of 2 or 3 dB.

However, *something* is suspect about those numbers... because
Canon did not use a 14 or even a 13 bit ADC. Why? (I don't
know, but obviously there is more to the story than we are aware
of.)
*All* monitors have a "limited dynamic range". Yours included.
Look up the figures for dynamic range of the human eye when
looking at as *single* scene, as opposed to what can be adjusted
for over time between two different scenes.
Providing data with added dynamic range to a device that cannot
process it isn't productive.
Sure they are. Now, If only /you/ had been on the design teams
for Nikon and Canon! ;-)
Which demonstrates my point that when the sensor actually does
have significantly greater dynamic range, manufacturers *will*
be using 14-bit ADC's in DSLRs. But they won't until there
actually is some benefit to doing so. (I note also that this
particular design has been around for more that 2 years, and
isn't exactly making big news...)

And most of all, as I've said before, the reason any given
camera doesn't have a 14-bit ADC to day is *not* the cost of a
14-bit ADC as some have claimed, and is *not* some lack of
availability, or lack of known engineering concepts. 14-bit
ADCs are available, are not expensive, and *are* used in other
types of cameras where the sensor provides sufficient dynamic
range to make it useful.

Your claim that the manufacturers are somehow "behind", when
they *do* produce other types of cameras with 14-bit ADCs
suggests you don't have the right clues. They obviously do know
exactly what the advantages and disadvantages of such
implementations are, and have chosen their designs based on
experience and good engineering. (And they are not describing
the reasons, so we are left to guess. You've clearly guessed
incorrectly.)
An interesting review... given what it does *not* say. And that
of course is anything at all about the great advantage provided
by the 14-bit ADC the camera uses! Maybe in addition to having
you on every camera design team, we should insist on your input
for these reviews too? ;-)

Floyd Davidson, Oct 20, 2005
7. ### Roger N. Clark (change username to rnclark)Guest

OK, let's work, through examples. Noise is computed as the
square root of the sum of the squared individual noise components.
For full well = 50,000 electrons, read noise = 7.5 electrons
(the Canon 20D) (numbers in electrons):

Signal _______Noise________
level Read Poisson Total
10 7.5 3.3 8.2
100 7.5 10.0 12.5
500 7.5 22.4 23.6
1000 7.5 31.6 32.4
5000 7.5 70.7 71.1
10000 7.5 100.0 100.3
20000 7.5 141.4 141.6
50000 7.5 223.6 223.7

12 bit ADC increment: 50000/4096 = 12.2 electrons/DN

14 bit ADC increment: 50000/16384= 3.05 electrons/DN

(DN = data number = integer value out of ADC)

Conclusion: at signal levels brighter than about 1% of
full well the camera is photon/electron noise (Poisson
statistics) limited, and read noise is insignificant.

Look at the plots on this page:
http://www.astrosurf.org/buil/20d/20dvs10d.htm
The last 4 plots on the page show the linearity of the
response proving the noise source is Poisson statistics
limited from counting electrons. Not other noise
sources besides read noise and Poisson noise is evident in
the data for the 4 systems shown (2 Canon and one Nikon DSLRs).
I hope the above example is clear enough for you.
As shown above, photon noise becomes large enough that ADC
errors are insignificant.
The signal from a CCD or CMOS sensor is definitely a voltage.
Photons excite an electron which become trapped in a potential
well. The collected electrons are amplified and the analog
voltage is digitized by the ADC.
Good question. It is well documented by multiple people, and well
understood in the digital astrophotography community. Check out
digital_astro in Yahoo groups. The 12-bit limited DSLR is easily
demonstrated by taking pictures at ISO 100 then 400 at the same
exposure. If the iso 400 image shows more shadow detail, than
the iso 100 image, the camera is ADC limited, not read noise limited.
Try this on the Canon and Nikon 12-bit systems and you will see
they are bit limited. Then can you get more shadow detail at
ISO 800 versus 400? If so, what conclusion would you draw
concerning digitizing all the signal requirements? Exercise left
Exactly. That is why log functions (like gamma), and dodging
and burning are used to compress dynamic range for printing
and display.
The eye can see over a 10,000 range in contrast detection
thresholds in any one fixed view, see:
Blackwell, J. Optical Society America, v 36, p624-643, 1946, and
Clark, R.N., Visual Astronomy of the Deep Sky, Cambridge University Press and
Sky Publishing, 355 pages, Cambridge, 1990.

Note the eye scans and dynamically adapts, so as you look at a
scene, like bright clouds to dark shadows, the eye can see
much larger ranges.

Here is a simple experiment you can do. Go out with a star
chart on a clear night with a full moon. Wait a few minutes
for your eyes to adjust. Now find the faintest stars you can
detect when the you can see the full moon in your field of view.
Try and limit the moon and stars to within about 45 degrees
of straight up (the zenith). If you have clear skies away
from city lights, you will probably be able to see magnitude
3 stars. The full moon has a stellar magnitude of -12.5.
If you can see magnitude 2.5 stars, the magnitude range you
are seeing is 15. Every 5 magnitudes is a factor of 100,
so 15 is 100 * 100 * 100 = 1,000,000. Thus, the dynamic
range in this relatively low light condition is about 1
million to one, perhaps higher!
Well, I guess I'll just have to continue working on spacecraft
instrument design and calibration, then flying them to other planets.

Your criticism is remarkably lacking in references and any proof
of any values you cite and you use statements like
"given what it does *not* say" to try and raise doubt without
adding any concrete data to your arguments.
Now you are starting on personal attacks.

Roger

Roger N. Clark (change username to rnclark), Oct 21, 2005
8. ### Floyd DavidsonGuest

There might well be 4096 levels, with 12 bit digitization (and
16384 with 14 bit), however that is not to say that the
quantizer can *distinguish* that many with any certainty. With
a read noise of 7.5 and a well capacity of 50,000, the dynamic
range of the sensor is 76.5 dB, and the maximum range of a 12
bit digital signal is 72+ dB. But the smallest accurately
distinguishable quantum is about 2.7 times the read noise, or
roughly 20.25 electrons, which divided into the maximum well
capacity of 50,000 gives us 2469 distinct levels. (I'll leave
it to you to determine the actual dynamic range of the
quantizer..., and what the significance of poking 2469
distinguishable levels into 4096 data levels does in practice.)

And switching from a 12 bit ADC to a 14 bit ADC is not going
to increase the number of *valid* levels that are actually
discrete.

Your 1% is fairly accurate (though I doubt that you actually
realize why it is). At a signal level of 500 the read noise is
10 dB below the Poisson noise, and from that point on up can be
considered relatively negligible. However, percentage doesn't
give a good indication of how significant that is. It is, out
of a 13 f/stop range, affecting the lower 6 f/stops. That is
not insignificant!
Which does *not* demonstrate your claim that read noise is
insignificant at higher levels. It may not affect the SNR
of the input analog signal, but it very definitely has an
effect on the design, and hence output, of the entire system
including at higher brightness levels (where adjacent levels
are going to be incorrectly detected because the range of each
level is smaller than the uncertainty caused by noise).
It demonstrates why understanding basics, such as logarithms and
dB, are important.
For a part of the total dynamic range.
But that voltage is *not* what is being used in the equation!
Then you need to provide *voltage* values, if you want to use
the formula that works with voltages...
Wonderful... but you still have nothing but pointless speculation.
So you are left with the fact that it is *not* significant,
because you cannot display the detail.
So your original statement is not correct. The difference is
not "much greater". It is greater for commonly used sensors in
DSLRs, and is about 3 dB better than the one you cited figures
for above. But of course there are sensors available for
scientific cameras and for large format cameras that have at
least that much more again in sensitivity over the human eye.
Can you see those faint stars when they are within a few
degrees of the moon? Or only when the are "in your field of
view"? Or better yet, if you look directly at the full moon,
and then look at the star, how long will it take before you
can see the star?
I do assume you are a talented person, with a great deal of
expertise in some field.
No need to fabricate. I have cited a number of sources and do
not find it necessary to repeat them at every step. You may
have noticed that others were complaining because I cited too
many references!

What you present above is just one more variation on the "I
personally" theme that is both common and invalid.
That fact that you cite something which says nothing did seem
to be significant.
That may be your bag, but not mine.

Floyd Davidson, Oct 21, 2005
9. ### David J TaylorGuest

Roger N. Clark (change username to rnclark) wrote:
[]
Yes, he did that with me as well.

David

David J Taylor, Oct 21, 2005
10. ### Floyd DavidsonGuest

Which he did *not* demonstrate to have ever happened!
And neither can you.

When you post personal opinion or information as "evidence" and
someone attacks that opinion, it is *not* a personal attack on
you. There is nothing gratuitous about it. Discussion of what
you post is quite proper and does not constitute a "personal
attack".

I do *not* discuss anything about you that *you* do not first
make reference to.

Hence people who say the definition of "distortion" relates to
linear vs. non-linear *in their mind*, opens up discussion of
just how clear their mind is, or not. Likewise anyone who can't
deal with dB calculations for dynamic range probably should not
cite their "engineering" background as a valid reason to believe
that camera design teams built system design errors into the top
DSLRs.

Floyd Davidson, Oct 21, 2005
11. ### Roger N. Clark (change username to rnclark)Guest

Here are a few gleaned from this thread, responding to
multiple people. A consistent trend.
---------------------------------------------------------

Floyd Davidson wrote:

Your 1% is fairly accurate (though I doubt that you actually
realize why it is).

Instead, you are babbling again

You've been, and continue, to spout techno babble. Big words
that you don't understand.

You don't know enough about signals and signal processing to
discuss this.

You probably can't explain even simple things....

I'll be damned. You actually passed the test.
It is complete nonsense. Just like most of what you've been
posting.

That is simply bullshit.

So the only argument you have left is an attempt a gratuitous
insults. You aren't even accurate at insulting people!

Your lack of exposure to the details is not significant until
you start mouthing off, particularly with gratuitous insults for
those who are familiar with the details.

Your narrow experience doesn't mean the rest of the world is so
constrained....

There are an unknown large number of people who don't understand
it.

What you "know", and the way all of this works seems to be
distinctly different.

That's some of the best babble anyone has come up with in a thread
that has been filled with it.

The problem you have is that I *do* have hands on experience
with how these things work at the base level.

*You*, on the other hand, cannot corroborate anything you've
claimed.

And just
as clearly if *you* don't understand the basics of digital
communications you will continue to misunderstand how it works
with imaging.

More babble. Is that supposed to relate to some useful topic?

Incredible ignorance on your part.

Ahhh... you don't have a clue as to what the term means.

That's Babble.

Yes, under the circumstances that I noted (which confused you
greatly).

I'm not sure why you would be so upset that someone tried to
trap you with a little psuedo babble, given that is exactly what
you've been posting for quite some time.

It is *all* designed to make fun of you and provide
others with a a laugh or 10.

Well well... you can actually translate binary numbers to
another base. The question though, is if you understand the
significance.

Roger N. Clark (change username to rnclark), Oct 22, 2005
12. ### Floyd DavidsonGuest

Not one is a gratuitious insult or personal attack. All of them
relate *directly* to the context that someone else posted (and
that you have removed).

Which is not to say that my comments might not be insulting!
Just that none are gratuitious... the comments derive from and
are directed at the text I was responding to. If you are
insulted by accurate critiques of your statements, perhaps you
should consider ways to be more accurate yourself!

Because when *you* are left with nothing to debate but my
writing style, there is little doubt that your technical
discussion wasn't up to snuff.
....

Floyd Davidson, Oct 22, 2005
13. ### Roger N. Clark (change username to rnclark)Guest

Floyd,
Just my opinion: I've seen some excellent helpful
posts from you, and then others where you might have a
small technical point, but so do others. You then,
along with them, beat it to death, but you quickly
go into an insulting tone that puts everyone off.
You dismiss a technical web page if it doesn't
suit your idea or pre-conceived notions
(e.g."However, *something* is suspect about those
numbers..." no the numbers are well reproduced).
If you would just cool down, I think you could help
a lot more.

For example, what I would like to know is about ADCs
regarding some of the things you've said.
I don't understand this from manufacturers specifications, nor
from an observational specification. Manufacturers spec sheets
do not indicate this level of noise. In fact they say
an RMS accuracy of an ADC is 1/sqrt(12) * LSB = 0.289 LSB so a
12 bit ADC has 0.007% RMS accuracy, which for a digital camera
input of 50,000 electrons would be 3.5 electrons. Where does your
20.25 electrons come from?

Some references:
http://en.wikipedia.org/wiki/Analog_to_digital_converter

http://en.wikipedia.org/wiki/Analog_to_digital_converter

http://www.absoluteastronomy.com/encyclopedia/a/an/analog-to-digital_converter.htm

A typical application in sciences is to use an ADC with a couple
of more bits than you really need, and just throw away the lowest
few bits. In high end DSLRs, the throughput is around 100 megapixels
per second, to that is the required speed of ADCs. Here is a
16-bit ADC that runs at 100 MSPS (mega samples /sec), just introduced
in May of this year, but production didn't start until September:

http://www.analogzone.com/acqp0509.htm

This one chip uses 2.3 watts, which in a digital camera is
significant but not outrageous. Can you cite 14-bit converters
that do 100 MSPS that have been in production for at least 3 years?
That might be a factor in Canon not putting 14-bit converters
in their high end cameras.

You questioned excellent references on the dynamic range of the eye.
The Blackwell, 1946 reference shows *thresholds* and one can
see thresholds over 10,000 to one. But much of what we observe
is well above the threshold. Your question:
The answer is yes. The whole original statement is that many people
can see stars with the full moon in the sky in the same
field of view at the same time. It is not difficult
to see magnitude 3 stars. Also remember, stars are at the lower
limit of vision. Brighter subjects make it easier. One can also
see stars as well as distant street lights that are brighter than the
full moon--all in the same field of view at the same time.

But as you roam you vision around looking at a scene, from shadows
ti highlights, you eye dynamically adjusts, making the dynamic range
even more impressive.

Experiment: when the are bright clouds visible out a window,
look at the bright clouds, and then in shadows in your
house (maybe construct some by the window). Then meter,
with a camera or spot meter, the brightest things you can see and
the darkest shadows where you can see detail. You will
find how impressive the eye really is!

Roger

Roger N. Clark (change username to rnclark), Oct 22, 2005
14. ### Floyd DavidsonGuest

Three articles in a row that have no point other than to discuss
my person and my writing style based on out of context
distortions, as if that is going make up for all of your
technical errors that I've been correcting.

Get over it. The topic is not me, and it is not you or your
wounded ego.

Floyd Davidson, Oct 22, 2005