What is the Characteristic Curve function of a digital camera?

Discussion in 'Digital Photography' started by Roger N. Clark (change username to rnclark), Jan 30, 2005.

  1. Hi,
    I'm looking for the equation for the conversion of
    the linear intensity data from a digital camera sensor
    to the image output. Please see the measured
    curve in Figure 7 at:

    This is not a standard Gamma function. Using a
    Gamma function, the Gamma appears to be around 8
    at the high end, dropping to 1 at the low end.

    Does anyone know the mathematical formula for
    curves like this one? I've tried several, but
    can't quite match the data over the whole range.

    Roger N. Clark (change username to rnclark), Jan 30, 2005
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  2. It sure looks like the curve is linear from 100-2000,
    logarithmic from 2000-10000 with one constant, logarithmic
    from 10000-30000 with another constant and then it

    I would be surprised if you find one simple formula for this.

    BTW - did I get you right that the RAW values are linear?
    I.e. that this non linear behaviour is only an artefact of the

    In that case - why destroy the initial linear response?

    Roland Karlsson, Jan 30, 2005
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  3. Roger N. Clark (change username to rnclark)

    Marvin Guest

    There are several linearizing functions for p[hotographic emulsions that have been known for many years. I published a
    little note on the subject:
    Remarks on Linearization of Characteristic Curves in Photographic Photometry.
    M. Margoshes
    Appl. Optics 8, 818 (1969)

    The one that worked best for me was the Kaiser transform. I incorporated it into a Basic computer program, published in
    Application of Digital Computers in Spectrochemical Analysis;
    Computational Methods in Photographic Photometry.
    M. Margoshes and S. D. Rasberry
    Spectrochim. Acta 23B, 497-513 (1969)

    The functions were developed by trial, not based on emulsion theory. It would be suprising if they worked equally well for
    digital camera sensor. I can't tell from the curves in the reference you cited whether they do or don't come close to the
    curves for photographic emulsions. The conventional Hurter-Driffield characteristic curve for photo emulsions is logarithmic
    in both axes, and the one in the Web page you cited is linear in both axes.

    Get back to me if you can't find the papers. I don't have time now to dig out my copies.
    Marvin, Jan 30, 2005
  4. "Roger N. Clark (change username to rnclark)" <>
    wrote in message SNIP
    It looks like a Rodbard function fits rather well (although I don't
    have the full dataset to fully check). "Rodbard" is a four parameter
    general curve fit function proposed by David Rodbard at NIH:
    y = d + (a-d) / (1 - (x/c)^b) ,

    with the following approximated Parameters:
    a = 0.018343439862814965
    b = 1.0643295289322523
    c = 6262.359656339619
    d = 72032.53123412121

    Bart van der Wolf, Jan 30, 2005
  5. Marvin,
    Thanks! This does help, sort of. I started web searches
    for papers online. I am an Optical Society of America member,
    and have subscribed to Applied Optics since 1975. So I am
    shocked to find when I want a paper online, I must
    pay $15 for it. This is simply not acceptable. Other journals
    I subscribe to give you access, for example, as an AAS member,
    it costs $12/year for online access to all papers,
    so $15 for one paper for a member is just ridiculous.
    Scientific journals are really becoming a hindrance
    to scientific information and research.

    Sorry for the rant--this is simply becoming a larger problem
    community wide. The National Institute of Health is putting
    all health articles online for free access. Other sciences need
    to do the same. I do realize it does cost to create and maintain
    such databases, but times are changing, and things should get
    easier, not harder. More and more libraries are not stocking
    journals because they are becoming too expensive and too many.

    Several other searches for things like
    Hurter-Driffield characteristic curve resulted
    in the same must pay a subscription issue.

    Anyway, if you could find a reprint and post the equation,
    I would be greatfull. Please also include the full reference
    for me to cite. In the meantime, I'll keep searching.

    Roger N. Clark (change username to rnclark), Jan 30, 2005
  6. Thanks, I'll play with this.

    Here is some raw data from my 1D Mark II
    (from Figure 7 at
    http://clarkvision.com/imagedetail/dynamicrange2 )
    This, by the way, shows that the jpeg and raw are
    saturating at about the same point and below the
    linear sensor saturation. Noise in the linear data
    are dominated by read noise at the low end
    (~ 10 DN) and photon statistics at the high end
    (~ sqrt(DN*(52300/65535)). The data were random
    lines extracted from the data in figure 7 above.
    Why the raw saturates at 65499 I do not know.

    Inteni- scaled
    sity Raw JPEG
    DN camera image
    Value Value Value
    16-bit 16-bit 16-bit
    64 710 1028
    103 1162 1285
    180 1731 1542
    245 2273 1542
    297 2761 2056
    348 3266 3341
    478 4437 3855
    736 6732 5911
    1007 9158 8481
    1356 12107 10280
    2105 17493 16448
    3153 23694 23387
    4311 28998 29555
    5499 33421 33667
    7967 40916 41120
    12366 48857 47802
    17866 54049 54227
    22868 57562 57568
    27723 60346 59624
    32204 61927 61166
    36840 63391 62194
    41378 64293 63222
    47811 65135 64507
    52789 65450 65278
    55180 65499 65278
    57366 65499 65535
    58254 65499 65535
    59034 65499 65535
    59559 65499 65535
    65535 65499 65535

    Roger N. Clark (change username to rnclark), Jan 30, 2005
  7. SNIP

    Using the above Rodbard function, those scaled Raw camera values, as a
    function of DN, are best fitted with parameters:
    a = 307.5908474937861
    b = 1.091885854762039
    c = 6101.1148040713415
    d = 71370.43124044045

    The scaled JPEG values, as a function of DN, are best fitted with
    a = 143.93060451108522
    b = 1.1247148320126283
    c = 6038.784118954135
    d = 70505.46445427524

    The reducing slope at the top is obviously the hardest part to
    accurately fit in a simple function. A piecewise set of functions
    would be more accurate.

    Bart van der Wolf, Jan 30, 2005
  8. Roger N. Clark (change username to rnclark)

    JPS Guest

    In message <>,
    Blackpoint subtracted (36)?

    The following chart is messed up here, because you used TAB characters.
    TAB characters are useless on usenet.
    JPS, Jan 31, 2005
  9. Just a question about your experimental protocol.
    I'm not sure how you established the correct exposure for
    the scene. One of the well-known characteristics of color
    negative film is its ability to handle overexposure well.
    So I think it might be instructive to see the results for
    this class of film with a variety of different exposures.
    My personal (non-scientific) experience is that there is a
    lot more detail in the shadows than is commonly expected if
    the film is exposed at a lower nominal ISO. I commonly shoot
    200 ISO film at 100-125 to compensate for lens vignetting and
    have no problems.
    Many pictures taken with ultra wide angle lenses show the sun
    in the frame and it is surprising how close to the sun there
    is still sky detail.
    Robert Feinman, Jan 31, 2005
  10. In my tests on the above page, I did a series of exposures
    from -3 to +3 stops. The one I chose saturates the
    highlights and was over exposed compared to the meter
    reading. Si I am quite sure I captured the entire
    latitude of the film.

    Roger N. Clark (change username to rnclark), Jan 31, 2005
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