Scanner: Dynamic Range, A/D bit-depth - another point of view

Discussion in 'Digital Darkroom' started by og, Jun 22, 2003.

  1. og

    og

    After reading the very informative article by Bob Atkins about Dynamic
    Range (http://www.photo.net/learn/drange/) I made some comments. I got
    no reactions, but probably because few persons are going back to check
    updates on this page. I post it here, in order to get some feedback (a
    summarized version of my initial comments):

    1: The Dynamic-Range of the scanner is not related at all to the
    bit-depth of the A/D converter (the DR is related to the range of
    light the CCD is able to measure accurately [min;MAX], the bit-depth
    A/D converter gives the number of steps to store the data (tonal
    gradations)). A scanner with a DR=4.0 and a 2 bits A/D converter can
    exist.

    2: Most scanners have a real Dynamic-Range between 3.2 and 3.8. If you
    scan color negatives (DR up to 2.0), you don't take advantage of all
    the bit-depth of your A/D converter. I suppose that you lose about 4
    bits. This means that in order to get a good 3x8 bits image, you need
    a 12 bit A/D converter. A 14 or 16 bits A/D converter will give more
    room for further tonal modifications, but you could also consider a 20
    bit A/D converter to store the image accurately in a 3x16 bits file.
    (this is not the case with slides).

    Bob's article (and all the linked documents at the end of the page)
    doesn't support this at all. Still, I think there is something
    interesting there, and I would like your feedback (please: read the
    article first, then my comments).

    Thank you all,
     
  2. I think the argument is that the number of bits determines the distance between notch 0 and notch 1 and therefore the darkest value it can detect before going to black. You're right this is a very brain dead assumption though especially if the A/D converter does not have a linear response.
     
  3. With regard to your point 1, from my fairly-non-scientific understanding, the 4.0 CCD and the 2-bit converter could coexist, but you couldn't get all of the detail the CCD could read into the file. If that's not the case, then I don't get it.
     
  4. og

    og

    If I consider the example (DR=4.0, 2 bits A/D): I have 4 values (2 bits) to store the useful input from the CCD which goes from 1 to 10000 units (unit = very small amount of volt). I decide to define the levels: below 2 (0=pure black), between 2 and 5000 (1=grey-dark), between 5000 and 9999 (2=grey-light) and above 9999 (3=pure white). I use 2 and 9999 because I know that my CCD can correctly handle the information - I even add 1 unit for more confort.

    In my digital data, if I see 0, I know that the signal was below 2. if I see 3, I know that the signal was above 9999. So I am still able to handle the range of signal (and light) going from 2 to 9999, which is a Dynamic Range of 3.7.

    I agree that the precision of the A/D converter and the way it is designed does limit the final Dynamic Range (here, I have decided to go from 4.0 to 3.7 by choosing these particular lower/upper values).

    I also agree that if I use a simple linear A/D converter (0 = from 0 to 2500 units, 1 = from 2500 to 5000, 2 = from 5000 to 7500, 3 = from 7500 to 10000), I loose all the information of the first step, and get a final DR=0.6. This is what is considered by a lot of people, and in this situation the final Dynamic Range is limited by the theoritical Dynamic Range of the A/D converter.

    But my point of view is that an A/D converter of a scanner is not designed like this (simpler linear). However, I know that I can be completely wrong... an A/D specialist, somewhere ? : )
     
  5. Here's an answer.
    Forget about the number of bits.
    Lets start with a question. How many different levels of light can your scanner see?

    Suppose your scanner measures 255 (or 65536 for 16 bit encoding) for the lightest light available. What is the smallest light level measurable?
    Well its equivalent to the all the electronic noise given by the scanner. In other words you cannot measure a light level which gives a signal lower than the noise produced by the scanner. Suppose the noise measured on the same scale as above is less than 1. Then with 8 bit conversion you will loose information: you cannot measure more than 255 linear levels.
    Now most scanners would measure on this scale a noise level roughly equal to 1 or less, therefore needing more than 8 bits to encode the entire scale. Again, most scanners encode with more bits than needed (12-14 is already EXCEPTIONNAL performance for non cooled CCDs), so the numbers of levels is the number of times the smallest level measurable (given by the scanners noise) can be packed on the scale.
    This gives the density range, D , which the log of the maximal signal measured divided by the noise. A very good scanner can give a D of 3 to 3.3 maximum (yes... get measuring yourself please), depending on the color also.
    But is it all? No

    Equally important is the possibilty to "put" the lightest level of your slide onto the "255" level of the scanner! This means that for a good scan not only must the D (density range) of your slide be lower than that of the scanner (in order to fit in all the levels...) but the Dmin (the lightest spot) be put right on 254-255 (without saturation). This feature was "the" big problem of some earlier scanners like the Minolta Dual Scan.

    Now to answer the comment above: a 12-bit converter is almost always sufficient for your scanner.... but not for you since you will loose detail for slide (D roughly up to 3.4-3.6) or most BW negatives. For color negatives everything is OK.
    So I ensure you, a 12 bit converter is sufficient and a 14 or 16 bit converter WILL NOT give you more room for further tonal modifications since you do not have more light levels.
     
  6. og

    og

    Al, thanks for your feedback. I agree with your first part (DR, Noise) and second part ('fitting' of both DR) - (and my DR of 3.8 is probably optimistic, I know... but that's what Minolta announced for its new Elite 5400, and it's the first time I see something more realistic than the usual '16bits=>DR=4.8', so I tend to believe them. But I hope to be able to test this soon : ).

    On the conclusion (last part), however, I think we don't have the same 'interpretation' (sorry if my english doesn't make it :(

    In the 'real world', variation of luminosity is continuous, therefore the analog part of the scanner (CCD) also gives a continuous response = infinite number of light levels. Then the A/D converter 'decides' to classify everything in 256 levels (8bits) or 4096 (12bits) or... But I still have an infinite number of light levels I can take advantage of with a higher A/D converter (16bits, 128bits... whatever).

    Now, if I consider a color negative with a DR=2 (1 to 100), it DOESN'T mean that a 7bits A/D converter (0 to 127) will be able to get everything from the film. Nor does a slide with a DR=3.6 require at least a 12 bits A/D converter .

    It rather works the other way: I want to show my pictures on prints with 3x8 bits. Because I work from a color negative, I need to do a LOT of tonal adjustment (get rid of Orange Mask, increase contrast by 2 or 3 stops) = I need, may be, 3x4 more bits. The scanner is also not using the full CCD dynamic-range, but perhaps just 75% of it. So I would need a (8+4)/0.75 = 16 bits A/D converter.

    With a slide, you do less tonal adjustment and the scanner is using the full CCD dynamic-range: a 10 bits A/D converter would already be a lot for usual needs. No ? But you can find Drum scanners with a 12 bits A/D converter and a real Dynamic Range of 3.8 (higher than the theoritical Dynamic Range of the A/D converter = 3.6).

    Olivier

    PS:

    1 - Thanks for all comments here, I really think that helps.

    2 - If you know where to find the Dmin and Dmax (not just the DR) of various slides, negatives and scanners: I would be very happy to bring more accurate conclusions...
     
  7. I am one of the few people who commented on Bob's article, and can assure you that most people in this forum aren't very interested in such technical issues. My own article on Jpeg Compression http://www.photo.net/learn/jpeg/ hardly received any comments, though Jpeg is used every day by millions of people...

    Sampling the levels of a continuous variable does pose some problems, which I queried (in my own confused way). Talking about Dynamic Range in the absence of clear definitions, and without considering noise, is not very constructive. Bob's point was to treat manufacturers exaggerated claims with some scepticism.

    Most people have no idea that all real-world images are gamma-encoded, and not in a linear-colour space. This has implications which complicate the discussion. The most important point I made was that gamma conversion should be done early in the process, to avoid truncation to 8-bits when exporting from the scanner to the image eidtor.
     
  8. Actually, light is discrete, not continuous -- it comes in indivisible units called photons. Of course, each one has energy E = h nu ... which (for visible wavelengths) is so small we will never measure it in a desktop device, so we need jillions of 'em and it winds up looking like we are measuring something continous. :)
    All that said, CCDs are inherently linear-response devices, and it is easier to make a good wide linear A/D converter and dump its values to the computer and have the computer apply good curve adjustments to get good logarithmic responses, than it is to have the scanner do most of the work -- not for any inherent reason except that the computer has more memory and compute-horsepower than the scanner, and can do more complex transformations. At some point we may have scanners that have Pentium 64000 processors in them (as is happening, minus the hyperbole, with disk drives now), but until then, I would rather have "lots of bits" in a fast dumb scanner, than "8 bits" in a "smart" scanner that turns out not to be as smart as it thinks it is.
     
  9. Olivier, may be you want in French... your reasoning is not right.
    In the "real world" you may have an infinite number of levels (thats even wrong... since light is quantified! but still... lets suppose!) but you scanner CANNOT distinguish between two levels if the light difference is higher than the noise!!! It is no point adding extra bits. Its like measuring with a ruler without your glasses... there is no point adding a micrometer scale if the millimeter scale is blurred!!
    I understand your need for tonal adjustements. But there it is. The Minolta scanner boasts 3.8 DR. I doubt very much. Personnaly I have measured Dmax on a Nikon 4000. It is 3.4 in the red (less in the green but I won't tell people here!). Its easy to measure. An excellent article, pity written in French is on www.galerie-photo.com.
    If you need to make big tonal adjustements try scanning twice for different light intensities. This is why DR is not necessarily the most important feature. Other feature like light adjustement and gain control are equally important. Still bear in mind above.

    Now a CCD is "physically" linear. The problem is the human eye is non-linear in its "apreciation" of light (its a statistical physionomical measurement). This means that the linear light levels need to be converted by a "gamma" (to simplify) curve. This curve translates, for exemple, the fact that human eye is very sensitive to tones in the dark areas of your picture.
    In other words, you need more linear levels in the dark areas as in the middle grey. Since the CCD gives a linear response, you therefore need more levels over the whole scale... and this means a higher DR.
    This is why it appears that a high DR scanner "gives more detail in the shadows" than lower DR scanners: the primary effect of high DR is the tonal resolution, as it appears to the human eye, of dark areas.

    Got it? Tout compris?
     
  10. Voilà, here is an exemple (not perfect, but the best on the net so far) on how to measure your dynamic range.
    Good luck! http://www.galerie-photo.com/canoscan-d2400uf.html
     
  11. og

    og

    Al, I see your point (also by reading the linked article). Correct me if I am wrong:

    Random noise is always added to the ("theoritical good") signal, thus it makes this signal unaccurate (or 'blurred') by an amount = sigma(noise) - from a statistical point of view. So, it is useless to measure this signal with steps narrower than the sigma(noise).

    The number of those 'smallest useful' steps corresponds to a certain amount of useful bit-depth converter. If you get a higher A/D converter, you will not get more useful information. So, with a real 'measured' dyn=3.8, I would be already happy with a 14 bits A/D converter. Ok...

    Agreed on the nature of light and on the greater importance of accuracy in the darker tones, due to the eye response. I still need to read more about gamma and color profiles, though... (PN, NormanKoren and other sites will help).

    I will choose my scanner mostly on price, speed, overall quality (ICE included...) but I wanted to understand more about Dyn and bit-depth. Good. Thanks everyone - interesting thread for me : )

    (I will post this conclusion (+link here), on the article page by Bob Atkins).
     
  12. og

    og

    Another update : what has been discussed here so far is about "Dark Noise"(electronic noise), but there are other types of noise: "Read Noise"(quantification noise) and "Photon Noise". You will find more information here.
    I found (in forums) that 1.5bits should be substracted for quantification noise, and that photon noise will show mainly in the shadows of Scans from Color Negatives (longer exposure without clipping may help). It is also said that 8bits bit-depth might be all you need for color negatives... Hmm. Maybe.
     
  13. og

    og

    About Photon noise (and grain), here is a very interesting post by Kennedy McEwen [on the alt.comp.periphs.scanners NewsGroup]:

    Question (by Hesham): "Hello all, I have a Minolta Dual III… So far, I have scanned three negtives: a Kodak gold 200, Fuji Superia 200 and Fuji Superia 100. I notice a lot of grain aliasing with the Superia 100. This was a shocker since it is one of my favorite negative films and since I knew it is a very fine grain film..." (excerpt)

    Answer : "Some of your comments are quite common from people who have just started to make high performance scans, particularly those not coming from a chemical printing background, and mainly just amount to reaching the limits of the medium and the equipment you are using.

    First off, it is very easy with a scanner to get "sucked into the
    pixels" when you are looking at the image on screen rather than
    the results on paper - which is, after all, the only comparison you
    can really make with what you have previous experience of. So a
    lot of new scanner owners find themselves zooming in to the
    detail and discovering that either the image isn't as sharp as or,
    in your case, the noise is worse than, they might expect when in
    reality it isn't that different from what they have seen from
    chemically produced results. That isn't necessarily a deficiency
    of either the scanner or the film - its just that its a lot more
    convenient to look at a higher magnification image using
    software than under the old hardware route. So before you get
    too depressed at the results from your new toy, make a few
    prints at similar sizes (or even a little larger, but not too different)
    than you are familiar with and judge the results on that - not on
    the zoomed image on the monitor. If you are scanning the full
    negative then you will probably find the results are not
    significantly different from what you are used to and have hitherto
    regarded as exceptional or at the limits of the film and camera.

    Having said all of that, you won't get quite as good results from
    negative stock using the Dual III as you will from a high quality
    chemical process lab or even from different film stock, and there
    are several reasons for this. The first, as you have correctly
    ascertained, is related to the granular structure of the image
    recorded on the film.

    If the grain was faithfully reproduced by the scanner (and printer)
    then printed images would not look any different from chemically
    produced prints of the same size - which is why this is ALWAYS
    the first port of call when comparing your new toy to the
    conventional prints you have had in the past. The problem is that
    grain is NEVER reproduced perfectly and, without going to
    extreme expense, it cannot be done with any resolution of
    scanner.

    One reason for this is that the grain that you see in a chemical
    image is just the random grouping of the small silver crystals (or
    dye clouds, in the case of colour emulsion) and not the crystals
    themselves, which are much smaller. So real grain in the film is
    actually a range of fine details the largest of which are
    represented by the clusters and the finest of which are the
    crystals themselves. That is how the chemical process
    reproduces the grain, as a range of detail, but when you scan it
    the smallest detail that the scanner can produce is one pixel in
    size - and even if this isn't as large as the biggest of the random
    clusters, it is certainly a lot larger than single silver crystals and
    dye clouds. Hence the grain looks coarser on the scan than it
    does in the chemical image - and this is often described as
    grain aliasing. Scanning at higher resolution doesn't mean less
    aliasing occurs, but the limiting size of the grain is smaller and
    thus less objectionable.

    So your scanner results will always be coarser grained than the
    chemical print (unless you scan in an oversampled system), but
    the higher the resolution of your scanner and the smaller the
    final print then the less obvious that difference becomes. This is
    true irrespective of whether the film is negative or slide media - it
    is simply the geometry of the scanner's optical system.

    The second effect is related to the negative film stock and, whilst
    it is also present on chemical prints, its effect is to amplify the
    deficiencies of the scanned image. As you know, the image on a
    slide is similar to the original scene, often with increased
    contrast. Blacks are very dense and whites are almost perfectly
    clear on good film stock. With a negative image, the blacks are
    reproduced on a base orange mask, but even the densest areas
    of the negative are considerably more transparent than the
    densest regions of a slide image. This reduced contrast is one
    reason why negative film stock has a much larger latitude than
    equivalent slide stock, but it also means that the image scanned
    from a negative must be contrast stretched relative to that of the
    slide film - and this exaggerates the visibility of any image
    deficiency, including real grain and aliased grain. In a chemical
    print, real grain is also contrast stretched, but aliased grain only
    exists in scanned images. So the difference in grain between a
    scanned image and a chemical print is usually significantly
    greater with negative stock than with slide stock.

    A third effect is shot noise, or photon noise, which is a problem
    with all CCD based scanners, and is particularly relevant to
    negative film stock and, especially, under-exposed images.

    We like to think of light as a perfect continuum - that it can be
    made brighter or darker on an infinitely fine scale. But, in fact,
    the brightness of the light in any part of your image is just a
    measure of how frequently the individual photons are being
    emitted or reflected from it. A bright image has billions of billions
    of photons per second reaching your eye, while a dark image
    may only have a few hundred per second. Unless the light
    source is a laser or something similar, then the light is
    uncorrelated, which means that the arrival of any photon at your
    eye (or the CCD in the scanner) is unrelated to the arrival of any
    other - in short, the arrival time of the photons from any part of the
    image is random. If the CCD collects photons for a fixed period
    of time then there is a random variation in the number of photons
    it collects even for equally bright parts of the image - in fact, even
    from the same part of the image if multiple samples are taken!
    This has nothing to do with the quality of the CCD or the
    electronics - it is just a consequence of the quantum nature of
    light discovered by Max Planck just over a century ago.

    Being perfectly random, it turns out that the average photon
    noise level is just the square root of the total number of photons
    the CCD accumulates in the given exposure time. This is just
    the same as tossing a coin - sometimes you get heads,
    sometimes tails. Repeat that for 10 times and you get, on
    average, 5 heads and 5 tails - but the average variation (or
    noise) is the square root of the number of time you toss the coin
    - so there is a fairly average chance that you could get 8 heads
    and 2 tails or 2 heads and 8 tails. (It is perfectly possible to get
    10 heads and no tails of course, but the probability of that is
    lower and would not be considered an 'average' result.)

    The problem is that all CCDs can only accumulate a certain
    number of photons before they reach saturation - each photon is
    converted to an electron and stored in the cell as an electric
    charge. Consequently, to avoid saturation, the exposure time is
    limited and the photon noise is effectively the square root of the
    number of photons which the CCD can accumulate. Even the
    CCDs in a good scanner, such as those used in the Minolta,
    Canon or Nikon series might only be capable of storing about
    50,000 electrons before saturating - resulting in a saturated
    photon noise of only 223. Thus, at saturation, the signal to noise
    ratio of the CCD is only 223, equivalent to a little less than 8-bits!

    Of course, you need a much greater precision of ADC than this to
    capture the full contrast range because the dark areas of the
    image might only cause a few photons to reach the cell during
    the exposure time, so don't get the idea that the 16-bit ADC in
    your scanner is of no value! Nevertheless the lighter parts of the
    scanned image have a signal to noise ratio which much poorer
    than the ADC, the rest of the electronics and, indeed, the CCD as
    measured - and this is significant when it comes to scanning
    underexposed negatives. These are very light images which
    cause most of the cells in the CCD to approach saturation levels
    - so
    the signal to noise ratio of the final image may only be just over 8
    bits, even though the scanner has a 16-bit ADC. The low
    contrast of the negative image means that few, if any, of the CCD
    cells have a low photon count, so the signal to noise of the final
    image is dominated by photon noise. By comparison, with slide
    film, the high base density can be compensated by increasing
    the exposure time so that the contrast is enhanced on the CCD
    itself - recovering as much signal to noise as exists on the
    image itself - without saturating the CCD. The low contrast of the
    negative film stock means that the image recorded by the CCD
    scanner has, by definition, a much poorer signal to noise ratio
    than that from a slide - and this limitation is exacerbated by
    underexposing the negative film (and also by overexposing slide
    film, but this is usually less of an issue).

    The only solution to this problem is to artificially increase the
    storage capacity of the CCD by taking multiple exposures in
    each position of the scan - multi-sampling. Of course, this also
    averages out all of the other noise sources in the scanner but,
    as shown previously, in the case of underexposed negatives
    other noise sources are usually insignificant to photon noise.
    Doubling the number of scan passes increases the signal to
    noise by a factor of 1.4, or adds approximately half a bit extra
    resolution to the image. So if your original underexposed
    scanned negative image has a signal to noise (and these
    figures are real measure results from a Nikon CS4000 I own, not
    theoretical estimates) of 200, or a little less than 8-bits, then 4
    times multi-sampling will give a signal to noise of about 400, or
    almost 9-bits, while 16x multi-sampling will result in almost 10
    effective bits in the image. That is a lot better than the single
    pass, but probably still quite a bit less than the real signal to
    noise ratio on the negative that could be reproduced in a
    chemical print.

    After all of that you are probably thinking that scanning from
    negatives is doomed and a complete waste of time. Not so - the
    differences that I am talking about between digital and chemical
    produced images from negatives can be quite small, especially
    if you take care to use fine grain film, high resolution scans and
    print at scales where the pixels are imperceptibly small to
    minimise the visibility of aliased grain. They can, unfortunately,
    also be significant so understanding these issues might help
    you overcome some of the limitations you have already
    encountered in the future.

    Instead of getting "sucked into the pixels" though, stand back
    and admire the full image. After all, that is probably the first thing
    you would do if it was a big chemical print you had just made!

    Kennedy McEwen"

    And of course, I do not fully agree : )
     

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