dynamic range and A/D conversion

Discussion in 'Digital Photography' started by Johannes, Sep 21, 2006.

  1. Johannes

    Johannes Guest

    Hi,

    Can anybody explain to me if there is a connection between the dynamic
    range of a digital camera/CCD-/CMOS-Sensor and the bit width of the
    analog-to-digital converter - does a 12 bit converter lead to a larger
    dynamic range than a 8 bit converter - and if so why?

    Thanks a lot for your input!
    Johannes
     
    Johannes, Sep 21, 2006
    #1
    1. Advertising

  2. Johannes wrote:
    > Hi,
    >
    > Can anybody explain to me if there is a connection between the dynamic
    > range of a digital camera/CCD-/CMOS-Sensor and the bit width of the
    > analog-to-digital converter - does a 12 bit converter lead to a larger
    > dynamic range than a 8 bit converter - and if so why?
    >
    > Thanks a lot for your input!
    > Johannes


    You might find these articles and their references interesting reading:

    http://www.clarkvision.com/imagedetail/dynamicrange/index.html

    http://www.clarkvision.com/imagedetail/digital.signal.to.noise/

    David
     
    David J Taylor, Sep 21, 2006
    #2
    1. Advertising

  3. Johannes

    Johannes Guest

    David J Taylor schrieb:

    > You might find these articles and their references interesting reading:
    > http://www.clarkvision.com/imagedetail/dynamicrange/index.html
    > http://www.clarkvision.com/imagedetail/digital.signal.to.noise/
    > David


    Thanks for the reference but my english is not good enough for this
    kind of technical matter. My question is, to be more precise, if I
    loose dynamic range when the information containing, say, 12 stops of
    range is digitized by a converter that uses only 8 bit.

    Best regards!
    Johannes
     
    Johannes, Sep 21, 2006
    #3
  4. Johannes wrote:
    > David J Taylor schrieb:
    >
    >> You might find these articles and their references interesting
    >> reading:
    >> http://www.clarkvision.com/imagedetail/dynamicrange/index.html
    >> http://www.clarkvision.com/imagedetail/digital.signal.to.noise/
    >> David

    >
    > Thanks for the reference but my english is not good enough for this
    > kind of technical matter. My question is, to be more precise, if I
    > loose dynamic range when the information containing, say, 12 stops of
    > range is digitized by a converter that uses only 8 bit.
    >
    > Best regards!
    > Johannes


    Johannes,

    If the digitising is only eight bits linear, then a lot of dynamic range
    will be lost. As one f/stop is half the light, full scale on 8 bits is
    256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
    only 8 stops. There is no tonal contrast in the lowest level, it is
    either on or off, so the dynamic range may be considered as well less than
    8 stops *depending how you define dynamic range).

    With the JPEG images, a non-linear encoding is used so that the digital
    value is more nearly the square-root of the light level: 1, sqrt (0.5),
    sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
    range possible, at the expense of a less accurate representation of a
    particular light level (in the highlights).

    Does that help?

    Cheers,
    David
     
    David J Taylor, Sep 21, 2006
    #4
  5. David J Taylor wrote:
    > Johannes wrote:
    >> David J Taylor schrieb:
    >>
    >>> You might find these articles and their references interesting
    >>> reading:
    >>> http://www.clarkvision.com/imagedetail/dynamicrange/index.html
    >>> http://www.clarkvision.com/imagedetail/digital.signal.to.noise/
    >>> David

    >>
    >> Thanks for the reference but my english is not good enough for this
    >> kind of technical matter. My question is, to be more precise, if I
    >> loose dynamic range when the information containing, say, 12 stops of
    >> range is digitized by a converter that uses only 8 bit.
    >>
    >> Best regards!
    >> Johannes

    >
    > Johannes,
    >
    > If the digitising is only eight bits linear, then a lot of dynamic
    > range will be lost. As one f/stop is half the light, full scale on 8
    > bits is 256, 1 stop down is 128 and so forth, 2 stops is 64, so the
    > total range is only 8 stops. There is no tonal contrast in the
    > lowest level, it is either on or off, so the dynamic range may be
    > considered as well less than 8 stops *depending how you define
    > dynamic range).
    > With the JPEG images, a non-linear encoding is used so that the
    > digital value is more nearly the square-root of the light level: 1,
    > sqrt (0.5), sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a
    > greater dynamic range possible, at the expense of a less accurate
    > representation of a particular light level (in the highlights).
    >
    > Does that help?
    >
    > Cheers,
    > David


    I should also have added that the typical ADC is 12-bit accuracy, followed
    by the non-linear encoding to 8-bit JPEG.

    David
     
    David J Taylor, Sep 21, 2006
    #5
  6. Johannes

    Johannes Guest

    David J Taylor schrieb:
    > Johannes,
    >
    > If the digitising is only eight bits linear, then a lot of dynamic range
    > will be lost. As one f/stop is half the light, full scale on 8 bits is
    > 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
    > only 8 stops. There is no tonal contrast in the lowest level, it is
    > either on or off, so the dynamic range may be considered as well less than
    > 8 stops *depending how you define dynamic range).
    >
    > With the JPEG images, a non-linear encoding is used so that the digital
    > value is more nearly the square-root of the light level: 1, sqrt (0.5),
    > sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
    > range possible, at the expense of a less accurate representation of a
    > particular light level (in the highlights).
    > Does that help?



    Dear David,

    As I´m more deeply involved into analog photography than into digital
    I define dynamic range as the range of f-stops between the brightest
    and the darkest area that still shows perceptable details. As far ist
    this definition is concerned I fully understand your explanation: There
    is no visible detail in the lowest level and in the highest level
    (which are white and black) - that´s a true loss of valuable
    information - and so the dynamic range is remarkable decreased after 8
    bit conversion. That helped a lot!

    Johannes
     
    Johannes, Sep 21, 2006
    #6
  7. Johannes wrote:
    []
    > Dear David,
    >
    > As I´m more deeply involved into analog photography than into digital
    > I define dynamic range as the range of f-stops between the brightest
    > and the darkest area that still shows perceptable details. As far ist
    > this definition is concerned I fully understand your explanation:
    > There is no visible detail in the lowest level and in the highest
    > level (which are white and black) - that´s a true loss of valuable
    > information - and so the dynamic range is remarkable decreased after 8
    > bit conversion. That helped a lot!
    >
    > Johannes


    Johannes,

    I'm glad that helped. Certainly 8-bit linear conversion would produce an
    image with notably degraded dynamic range, but 8-bit gamma 2.2 corrected
    encoded data (such as found in JPEGs or analog/digital/analog TV) can
    produce greyscale images which exceed the limits of the human eye (in
    terms of dynamic range, under typical display conditions). Once you get
    into colour, there is some evidence that 24-bit RGB images can be seen to
    limit the displayed colour differences, but these effects are quite
    subtle.

    For processing, of course, you may want 12-bit raw data if you are going
    to alter the levels appreciably. The very best DSLRs are now showing that
    12-bit ADCs are not quite enough, and 14 bits would be required to capture
    the full range available from the sensor.

    Cheers,
    David
     
    David J Taylor, Sep 22, 2006
    #7
  8. Johannes

    bugbear Guest

    David J Taylor wrote:
    > Johannes,
    >
    > If the digitising is only eight bits linear, then a lot of dynamic range
    > will be lost. As one f/stop is half the light, full scale on 8 bits is
    > 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
    > only 8 stops. There is no tonal contrast in the lowest level, it is
    > either on or off, so the dynamic range may be considered as well less than
    > 8 stops *depending how you define dynamic range).
    >
    > With the JPEG images, a non-linear encoding is used so that the digital
    > value is more nearly the square-root of the light level: 1, sqrt (0.5),
    > sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
    > range possible, at the expense of a less accurate representation of a
    > particular light level (in the highlights).


    Even if all the encoders were linear, surely there's a simple
    interplay between dynamic range and accuracy?

    A single bit (potential values 0 and 1!) can encode
    a massive dynamic range (e.g. 10 f-stops worth).
    It just won't have much accuracy within that range :)

    The same logic applies to 8 bit versus 12 bit.

    There's no reason they shouldn't have the same dynamic range,
    with different accuracy.

    Of course, one could use 8bit to have the same accuracy
    as 12 bit, but a smaller range.

    Or any compromise between these 2 extremes.

    (non-linear encoding improves perceievd accuracy,
    since (IIRC) the eye is a log (ish) device,
    but I believe this is a separable issue)

    BugBear
     
    bugbear, Sep 22, 2006
    #8
  9. Johannes wrote:
    > Hi,
    >
    > Can anybody explain to me if there is a connection between the dynamic
    > range of a digital camera/CCD-/CMOS-Sensor and the bit width of the
    > analog-to-digital converter - does a 12 bit converter lead to a larger
    > dynamic range than a 8 bit converter - and if so why?
    >
    > Thanks a lot for your input!
    > Johannes


    There are many factors that in total determine dynamic range. Bit
    width is ONE of them. The quantization (number of bits) sets a hard
    limit- the dynamic range cannot be greater than the range allowed by
    the quantization steps of the number of bits.

    However, signal-to-noise-ratio is another big factor. If the S/N is
    less than appropriate for the quantization steps, then you will merely
    be digitizing the noise in greater detail :)

    The S/N, on the other hand, depends on the chip, the quality of the
    preamp electronics and the quality of the A/D chip. It will also vary
    with exposure.
     
    Don Stauffer in Minnesota, Sep 22, 2006
    #9
  10. Johannes wrote:
    > David J Taylor schrieb:

    snip
    >
    > Dear David,
    >
    > As I´m more deeply involved into analog photography than into digital
    > I define dynamic range as the range of f-stops between the brightest
    > and the darkest area that still shows perceptable details. As far ist
    > this definition is concerned I fully understand your explanation: There
    > is no visible detail in the lowest level and in the highest level
    > (which are white and black) - that´s a true loss of valuable
    > information - and so the dynamic range is remarkable decreased after 8
    > bit conversion. That helped a lot!
    >
    > Johannes


    There is another type of dynamic range that is sometimes called the
    small-signal dynamic range. Sometimes there is a big floor to the
    black level, which would seem to be a limit on DR. However, if the
    noise is low, that floor can be subtracted out. What is more
    bothersome is the noise in medium exposures- variations in level in
    medium intensity. So one definition is the ratio of that noise in the
    middle of the exposure range to the numerical value of the maximum
    possible signal.
     
    Don Stauffer in Minnesota, Sep 22, 2006
    #10
  11. bugbear wrote:
    []
    > Even if all the encoders were linear, surely there's a simple
    > interplay between dynamic range and accuracy?
    >
    > A single bit (potential values 0 and 1!) can encode
    > a massive dynamic range (e.g. 10 f-stops worth).
    > It just won't have much accuracy within that range :)
    >
    > The same logic applies to 8 bit versus 12 bit.
    >
    > There's no reason they shouldn't have the same dynamic range,
    > with different accuracy.
    >
    > Of course, one could use 8bit to have the same accuracy
    > as 12 bit, but a smaller range.
    >
    > Or any compromise between these 2 extremes.
    >
    > (non-linear encoding improves perceievd accuracy,
    > since (IIRC) the eye is a log (ish) device,
    > but I believe this is a separable issue)
    >
    > BugBear


    Yes, if you talk about different encoding choices, there are many
    different possibilities. That's why I suggested that you needed a
    definition of dynamic range to discuss this in a quantitative way.

    I don't think that you should ignore any of the components of the vision
    chain, in particular the characteristics of the display device and the
    human eye, when choosing an optimum encoding for a given number of bits on
    the storage or transmission channel.

    Cheers,
    David
     
    David J Taylor, Sep 22, 2006
    #11
  12. Johannes

    Guest

    David J Taylor wrote:

    > Johannes,
    > If the digitising is only eight bits linear, then a lot of dynamic range
    > will be lost. As one f/stop is half the light, full scale on 8 bits is
    > 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
    > only 8 stops. There is no tonal contrast in the lowest level, it is
    > either on or off, so the dynamic range may be considered as well less than
    > 8 stops *depending how you define dynamic range).


    Well, that´s kind of interesting.
    I always thought the tonal values stay the same after a/d conversion
    and they´re only spread out on a smaller strap. Isn´t it the case
    that, when doing a/d conversion with 8 bit or whatever bit width, the
    lowest level holds the darkest tonal level that the image contains and
    the highest level the brightest and that only the sampling is
    closer/smaller? What you imply is a lowest level as total black and a
    highest as total white. Is that really so?

    > With the JPEG images, a non-linear encoding is used so that the digital
    > value is more nearly the square-root of the light level: 1, sqrt (0.5),
    > sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
    > range possible, at the expense of a less accurate representation of a
    > particular light level (in the highlights).


    Maybe I haven´t recognized any loss due to that gamma conversion.

    Greetings!
    Mr.Adams
     
    , Sep 23, 2006
    #12
  13. wrote:
    > David J Taylor wrote:
    >
    >> Johannes,
    >> If the digitising is only eight bits linear, then a lot of dynamic
    >> range will be lost. As one f/stop is half the light, full scale on
    >> 8 bits is 256, 1 stop down is 128 and so forth, 2 stops is 64, so
    >> the total range is only 8 stops. There is no tonal contrast in the
    >> lowest level, it is either on or off, so the dynamic range may be
    >> considered as well less than 8 stops *depending how you define
    >> dynamic range).

    >
    > Well, that´s kind of interesting.
    > I always thought the tonal values stay the same after a/d conversion
    > and they´re only spread out on a smaller strap. Isn´t it the case
    > that, when doing a/d conversion with 8 bit or whatever bit width, the
    > lowest level holds the darkest tonal level that the image contains and
    > the highest level the brightest and that only the sampling is
    > closer/smaller? What you imply is a lowest level as total black and a
    > highest as total white. Is that really so?
    >
    >> With the JPEG images, a non-linear encoding is used so that the
    >> digital value is more nearly the square-root of the light level: 1,
    >> sqrt (0.5), sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a
    >> greater dynamic range possible, at the expense of a less accurate
    >> representation of a particular light level (in the highlights).

    >
    > Maybe I haven´t recognized any loss due to that gamma conversion.
    >
    > Greetings!
    > Mr.Adams


    The ADC should linearly convert light levels from the sensor to digital
    values, yes, and the accuracy depends on the number of bits. However
    general photographic usage talks in terms of "stops" which are not a
    linear scale, but a logarithmic one (corresponding more closely to how the
    eye responds to light). Alternatively, you could look on it as a
    percantage accuracy. With a linear conversion, the quantisation step
    expressed as a fraction of the signal in a 256-level signal is 1/256 at
    the white end, but "100%" at the black end (1/1). With a gamma-corrected
    JPEG, the percentage steps are more equal across the whole range.

    Does that help?

    David
     
    David J Taylor, Sep 24, 2006
    #13
  14. Johannes

    Guest

    David J Taylor schrieb:
    > The ADC should linearly convert light levels from the sensor to digital
    > values, yes, and the accuracy depends on the number of bits. However
    > general photographic usage talks in terms of "stops" which are not a
    > linear scale, but a logarithmic one (corresponding more closely to how the
    > eye responds to light). Alternatively, you could look on it as a
    > percantage accuracy. With a linear conversion, the quantisation step
    > expressed as a fraction of the signal in a 256-level signal is 1/256 at
    > the white end, but "100%" at the black end (1/1). With a gamma-corrected
    > JPEG, the percentage steps are more equal across the whole range.
    > Does that help?


    Not really because I still do not see why it should compress the
    dynamic range in the picture. In a 256-level signal the spot at 1/256
    represents the brightest part of the image and 1/1 the darkest. The
    logarithmic range in between can well be 10 stops, it´s only that I
    have just 8 different values to represent them. But: The spot at 1/256
    is still 10 stops brighter than at 1/1 or am I on the totally wrong
    track here?

    Greetings!
    Tony Adams
     
    , Sep 24, 2006
    #14
  15. wrote:
    > David J Taylor schrieb:
    >> The ADC should linearly convert light levels from the sensor to
    >> digital values, yes, and the accuracy depends on the number of bits.
    >> However general photographic usage talks in terms of "stops" which
    >> are not a linear scale, but a logarithmic one (corresponding more
    >> closely to how the eye responds to light). Alternatively, you could
    >> look on it as a percantage accuracy. With a linear conversion, the
    >> quantisation step expressed as a fraction of the signal in a
    >> 256-level signal is 1/256 at the white end, but "100%" at the black
    >> end (1/1). With a gamma-corrected JPEG, the percentage steps are
    >> more equal across the whole range.
    >> Does that help?

    >
    > Not really because I still do not see why it should compress the
    > dynamic range in the picture. In a 256-level signal the spot at 1/256
    > represents the brightest part of the image and 1/1 the darkest. The
    > logarithmic range in between can well be 10 stops, it´s only that I
    > have just 8 different values to represent them. But: The spot at 1/256
    > is still 10 stops brighter than at 1/1 or am I on the totally wrong
    > track here?
    >
    > Greetings!
    > Tony Adams


    Tony,

    If you have a linear ADC, the spot with digital value 255 is 255 times
    brighter than the spot with digital value 1. As a stop is a factor of 2
    in light-level, an 8-bit linear representation will have a maximum range
    of 8 stops (255 to 128, to 64, to 32, to 16, to 8, to 4, to 2, to 1), but
    at the lowest recordable brightness level, the digital value is either on
    or off, i.e. there is no "tonal range" at 8 stops down from the maximum
    brightness. In a typical camera, a 12-bit linear convertor is used, which
    provides a 12-stop range (if you accept zero tonal range at the lowest
    brightness level).

    [I am ignoring non-linearity, dark current, readout noise etc.]

    Once you start encoding the data - either for a RAW file or for JPEG, you
    can do various tricks to represent 12-bit linear data with fewer bits, by
    some sort of algorithm chosen so that the eye cannot see the brightness
    errors which result from using the algorithm.

    Any closer?

    David
     
    David J Taylor, Sep 24, 2006
    #15
  16. wrote:
    >David J Taylor schrieb:
    >> The ADC should linearly convert light levels from the sensor to digital
    >> values, yes, and the accuracy depends on the number of bits. However
    >> general photographic usage talks in terms of "stops" which are not a
    >> linear scale, but a logarithmic one (corresponding more closely to how the
    >> eye responds to light). Alternatively, you could look on it as a
    >> percantage accuracy. With a linear conversion, the quantisation step
    >> expressed as a fraction of the signal in a 256-level signal is 1/256 at
    >> the white end, but "100%" at the black end (1/1). With a gamma-corrected
    >> JPEG, the percentage steps are more equal across the whole range.
    >> Does that help?

    >
    >Not really because I still do not see why it should compress the
    >dynamic range in the picture. In a 256-level signal the spot at 1/256
    >represents the brightest part of the image and 1/1 the darkest. The
    >logarithmic range in between can well be 10 stops, it´s only that I
    >have just 8 different values to represent them. But: The spot at 1/256
    >is still 10 stops brighter than at 1/1 or am I on the totally wrong
    >track here?


    With a "linear" conversion, the number of bits determines the
    ratio of the maximum level to the minimum level that can be
    recorded, and that is of course the dynamic range.

    Your statement is true for what is called a "non-linear" A-D
    conversion. The 8-bit representation used by JPEG is an example
    of exactly that. The non-linear transfer function determines
    the "dynamic range".

    A significant point to remember though, is that "dynamic range"
    does not necessarily equate to the same thing as "useful dynamic
    range". Hence while it is true that JPEG technically can record
    a greater dynamic range in 8 bits that a linear 12 bit file can,
    there is a very significant difference between them, and the
    _useful_ dynamic range of the 12 bit linear file is greater.

    Here is a chart showing the "normalized" values for 12 bit linear
    data:

    | Number of | |
    | Levels | |
    Fstop | 12 bit | Normalized Pixel |
    Range | Linear | value |
    ------|------------|------------------|
    1 | 2048 | 1.0 |
    2 | 1024 | 0.5 |
    3 | 512 | 0.25 |
    4 | 256 | 0.125 |
    5 | 128 | 0.0625 |
    6 | 64 | 0.03125 |
    7 | 32 | 0.015625 |
    8 | 16 | 0.007812 |
    9 | 8 | 0.003906 |
    10 | 4 | 0.001953 |
    11 | 2 | 0.0009765 |
    12 | 1 | 0.0004883 |


    The brightest level would be 1.0, represented by a binary value
    of 1111 1111 1111. Half that brightness would be 0.5,
    represented by a binary value of 0111 1111 1111. One quarter of
    the maximum brightness, 0.25, would be represented by 0011 1111
    1111.

    The range above 0.5 all the way to 1.0 is divided into 0111 1111
    1111 equal parts, which in decimal is 2048 levels. Hence the
    accuracy with which a brightness value in the highest f/stop can
    be recorded is extremely good. (Probably more than 10 or 20
    times better than the eye can distinquish, and hence using 2048
    levels to represent only a 1 f/stop range is a real waste.)

    Down at the 9th f/stop, the range is divided into 0000 0000 0111
    values (decimal 8), which is not particularly very good accuracy.

    As it happens, 8 levels in an entire f/stop is just about the
    minimum that is useful. Fewer than that and the image will
    appear to be "posterized".

    Hence even though a 12 bit file technically has a dynamic range
    of 12 f/stops, the actualy useful dynamic range is about 9
    f/stops.

    However, with JPEG the increments are not linear (and there is
    no waste of 2048 values just for 1 f/stop), and instead a "gamma
    correction" curve is applied (basically that is a lookup table
    which quantizes a larger number of values into a smaller number,
    in a non-linear way). Here is the same chart as above, with
    some extra columns showing JPEG data too:

    +------- 12 bit linear -------+ +----------- 8 bit JPEG ------------+
    / \/ \
    | No. Levels | Normalized Pixel | 8 bit | No. Levels |
    Fstop | 12 bit | value | Gam.Cor. | 8 Bit JPEG |
    Range | Linear | Linear * Gam.Cor. | Value | Gamma Corr |
    ------|------------|-------------*--------------|----------|------------|
    1 | 2048 | 1.0 * 1.0 | 255 | 69 |
    2 | 1024 | 0.5 * 0.72974 | 186 | 50 |
    3 | 512 | 0.25 * 0.53252 | 136 | 37 |
    4 | 256 | 0.125 * 0.38860 | 99 | 27 |
    5 | 128 | 0.0625 * 0.28358 | 72 | 20 |
    6 | 64 | 0.03125 * 0.20694 | 53 | 14 |
    7 | 32 | 0.015625 * 0.15101 | 38 | 10 |
    8 | 16 | 0.007812 * 0.11020 | 28 | 8 |
    9 | 8 | 0.003906 * 0.08042 | 21 | 6 |
    10 | 4 | 0.001953 * 0.05868 | 15 | 4 |
    11 | 2 | 0.0009765 * 0.04282 | 11 | 3 |
    12 | 1 | 0.0004883 * 0.03125 | 8 | 2 |
    13 | | 0.0002415 * 0.02269 | 6 | 2 |
    14 | | 0.0001207 * 0.01656 | 4 | 1 |
    15 | | 0.0000604 * 0.01208 | 3 | 1 |
    16 | | 0.0000302 * 0.00882 | 2 | 0 |
    17 | | 0.0000151 * 0.06434 | 2 | 1 |
    18 | | 0.0000075 * 0.00470 | 1 | 1 |
    19 | | 0.0000038 * 0.00343 | 1 | 0 |

    Some want to claim that the above means that 8 bit gamma
    corrected JPEG data can represent a dynamic range of 19 f/stops!
    Technically that appears to be true, but obviously none of the
    f/stops below the 8th one are of any usefulness, since those all
    have fewer than 8 values per fstop.

    For all practical purposes, the lowest 15 values in a 12 bit
    linear file are of no significance and the lowest 21 values in a
    JPEG file are equally useless. A 12-bit linear file can retain
    a 9 f/stop dynamic range. A JPEG gamma correctd file can retain
    8 f/stops of dynamic range.

    --
    Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
    Ukpeagvik (Barrow, Alaska)
     
    Floyd L. Davidson, Sep 24, 2006
    #16
  17. Johannes

    Guest

    David J Taylor wrote:

    > Tony,
    > If you have a linear ADC, the spot with digital value 255 is 255 times
    > brighter than the spot with digital value 1. As a stop is a factor of 2
    > in light-level, an 8-bit linear representation will have a maximum range
    > of 8 stops (255 to 128, to 64, to 32, to 16, to 8, to 4, to 2, to 1), but
    > at the lowest recordable brightness level, the digital value is either on
    > or off, i.e. there is no "tonal range" at 8 stops down from the maximum
    > brightness. In a typical camera, a 12-bit linear convertor is used, which
    > provides a 12-stop range (if you accept zero tonal range at the lowest
    > brightness level).
    > [I am ignoring non-linearity, dark current, readout noise etc.]
    > Once you start encoding the data - either for a RAW file or for JPEG, you
    > can do various tricks to represent 12-bit linear data with fewer bits, by
    > some sort of algorithm chosen so that the eye cannot see the brightness
    > errors which result from using the algorithm.


    Ah, so the maximum contrast ratio that a 8 bit encoded image can hold
    is 256:1 or 8 stops. If the sources ratio is higher tonal values are
    cut and I loose shadow details?!

    Tony Adams
     
    , Sep 24, 2006
    #17
  18. wrote:
    []
    > Ah, so the maximum contrast ratio that a 8 bit encoded image can hold
    > is 256:1 or 8 stops. If the sources ratio is higher tonal values are
    > cut and I loose shadow details?!
    >
    > Tony Adams


    Yes, if the encoding is linear.

    If (as in all camera JPEGs) the encoding is not linear (typically it is
    gamma corrected) the dynamic range (or number of f/stops) would be
    greater.

    David
     
    David J Taylor, Sep 24, 2006
    #18
  19. Johannes

    Guest

    "Johannes" <> writes:

    > David J Taylor schrieb:


    >> You might find these articles and their references interesting reading:
    >> http://www.clarkvision.com/imagedetail/dynamicrange/index.html
    >> http://www.clarkvision.com/imagedetail/digital.signal.to.noise/


    > Thanks for the reference but my english is not good enough for this
    > kind of technical matter. My question is, to be more precise, if I
    > loose dynamic range when the information containing, say, 12 stops
    > of range is digitized by a converter that uses only 8 bit.


    You have two factors, the max and min of the range, and the number
    of levels between them.

    If you had a 1 bit converter, it is on or off. It does not tell you
    the range from off to on, it could be a dim night light, or a 500KW
    arc lamp. But for both only one step.

    A 2 bit unit has 4 levels, but again it does not tell you the total
    range from 0 to 2^N.

    Your 12 stops with an 8 bit AD need not lose range at all but the step
    per data number will be bigger than say a 12 bit converter if the 12
    bit unit has the same max and min range.

    --
    Paul Repacholi 1 Crescent Rd.,
    +61 (08) 9257-1001 Kalamunda.
    West Australia 6076
    comp.os.vms,- The Older, Grumpier Slashdot
    Raw, Cooked or Well-done, it's all half baked.
    EPIC, The Architecture of the future, always has been, always will be.
     
    , Sep 24, 2006
    #19
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. Hoffa
    Replies:
    0
    Views:
    720
    Hoffa
    Oct 25, 2006
  2. Hoffa
    Replies:
    1
    Views:
    1,625
    Walter Roberson
    Oct 25, 2006
  3. Roger N. Clark (change username to rnclark)

    Dynamic range of digital and film: new data

    Roger N. Clark (change username to rnclark), Nov 7, 2004, in forum: Digital Photography
    Replies:
    32
    Views:
    884
    Roger N. Clark (change username to rnclark)
    Nov 14, 2004
  4. Robert Feinman

    Scene range vs dynamic range

    Robert Feinman, Jun 30, 2005, in forum: Digital Photography
    Replies:
    2
    Views:
    700
    Marvin
    Jul 4, 2005
  5. Diego Balgera
    Replies:
    5
    Views:
    7,793
    Johann Lo
    Feb 8, 2008
Loading...

Share This Page