Noise, exposure, and the right side of the historgram?

Discussion in 'Digital Photography' started by dperez@juno_nospam.com, Nov 1, 2004.

  1. Of late I've seen a couple discussions of the importance of not underexposing a
    digital image. Specifically this is because if you have 6 stops of dynamic
    range, half the data, or 2048 levels, of a 12-bit capture is in the brightest
    stop, 1024 in the next stop and so on until the lowest stop, which hsa 64
    levels...

    SO, based on this, the argument is to be as bright as possible without blowing
    out the highlights. And that being 1 stop underexposed would cause a loss of
    50% of the possible levels in the image...

    My question is which is BETTER - to underexpose an image at a given ISO or to
    increase the ISO and expose the image more brightly? For example - if I'm at
    ISO 100 and my histogram is pushed toward the dark end - NOT clipped, just
    "dark" am I better to increase the ISO to say 400 so my histogram is pushed
    toward the bright end - AGAIN NOT CLIPPED...

    Now to make things more interesting - in the RAW converter, I'd MOST LIKELY
    increase the exposure of the "darker" image by 1 stop, and MOST LIKELY reduce
    the exposure of the "brighter" image by 1 stop... So, in the end, IN THEORY,
    both images would be identical once they got to Photoshop after conversion...

    From what I understand, increasing the exposure on the "darker" image causes
    more noise in the shadows... And the higher ISO has more noise everywhere...
    But, is there a difference in the noise? Is one more obvious than the other?
    Is one luminance noise versus color noise?

    How about an even more extreme example - am I better leaving the ISO at 200 and
    having the camera underexpose a shot by 1 stop (actual underexposure so I'm
    perhaps clipping just a bit on the dark end) or bumping the ISO to 800, which I
    KNOW causes visible noise and getting the exposure much brighter?

    I"ll most likely run some tests just to see how things look in both cases, but
    I'm curious what others have found...
     
    dperez@juno_nospam.com, Nov 1, 2004
    #1
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  2. dperez@juno_nospam.com

    Robertwgross Guest

    dperez wrote:
    >Of late I've seen a couple discussions of the importance of not underexposing

    a
    >digital image. Specifically this is because if you have 6 stops of dynamic
    >range, half the data, or 2048 levels, of a 12-bit capture is in the brightest
    >stop, 1024 in the next stop and so on until the lowest stop, which hsa 64
    >levels...
    >SO, based on this, the argument is to be as bright as possible without
    >blowing
    >out the highlights. And that being 1 stop underexposed would cause a loss of
    >50% of the possible levels in the image...
    >My question is which is BETTER - to underexpose an image at a given ISO or to
    >increase the ISO and expose the image more brightly? For example - if I'm at
    >ISO 100 and my histogram is pushed toward the dark end - NOT clipped, just
    >"dark" am I better to increase the ISO to say 400 so my histogram is pushed
    >toward the bright end - AGAIN NOT CLIPPED...
    >Now to make things more interesting - in the RAW converter, I'd MOST LIKELY
    >increase the exposure of the "darker" image by 1 stop, and MOST LIKELY reduce
    >the exposure of the "brighter" image by 1 stop... So, in the end, IN THEORY,
    >both images would be identical once they got to Photoshop after conversion...
    >From what I understand, increasing the exposure on the "darker" image causes
    >more noise in the shadows... And the higher ISO has more noise everywhere...
    >But, is there a difference in the noise? Is one more obvious than the other?
    >Is one luminance noise versus color noise?
    >How about an even more extreme example - am I better leaving the ISO at 200
    >and
    >having the camera underexpose a shot by 1 stop (actual underexposure so I'm
    >perhaps clipping just a bit on the dark end) or bumping the ISO to 800, which
    >I
    >KNOW causes visible noise and getting the exposure much brighter?
    >I"ll most likely run some tests just to see how things look in both cases,
    >but
    >I'm curious what others have found...


    Results are highly variable. You'll get better opinions if you identify what
    camera you use. Different cameras have different abilities to pull image detail
    out of shadows.

    ---Bob Gross---
     
    Robertwgross, Nov 1, 2004
    #2
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  3. dperez@juno_nospam.com

    andre Guest

    dperez@juno_nospam.com wrote:
    > Of late I've seen a couple discussions of the importance of not underexposing a
    > digital image. Specifically this is because if you have 6 stops of dynamic
    > range, half the data, or 2048 levels, of a 12-bit capture is in the brightest
    > stop, 1024 in the next stop and so on until the lowest stop, which hsa 64
    > levels...
    >
    > SO, based on this, the argument is to be as bright as possible without blowing
    > out the highlights. And that being 1 stop underexposed would cause a loss of
    > 50% of the possible levels in the image...
    >
    > My question is which is BETTER - to underexpose an image at a given ISO or to
    > increase the ISO and expose the image more brightly? For example - if I'm at
    > ISO 100 and my histogram is pushed toward the dark end - NOT clipped, just
    > "dark" am I better to increase the ISO to say 400 so my histogram is pushed
    > toward the bright end - AGAIN NOT CLIPPED...
    >
    > Now to make things more interesting - in the RAW converter, I'd MOST LIKELY
    > increase the exposure of the "darker" image by 1 stop, and MOST LIKELY reduce
    > the exposure of the "brighter" image by 1 stop... So, in the end, IN THEORY,
    > both images would be identical once they got to Photoshop after conversion...
    >
    > From what I understand, increasing the exposure on the "darker" image causes
    > more noise in the shadows... And the higher ISO has more noise everywhere...
    > But, is there a difference in the noise? Is one more obvious than the other?
    > Is one luminance noise versus color noise?
    >
    > How about an even more extreme example - am I better leaving the ISO at 200 and
    > having the camera underexpose a shot by 1 stop (actual underexposure so I'm
    > perhaps clipping just a bit on the dark end) or bumping the ISO to 800, which I
    > KNOW causes visible noise and getting the exposure much brighter?
    >
    > I"ll most likely run some tests just to see how things look in both cases, but
    > I'm curious what others have found...


    If i can afford it I always leave mine at ISO 100 to get the lowest
    possible noise. If you can keep it at ISO 100 and just change the
    exposure to expose correctly do it. If not you should increase ISO
    (after all thats what it is there for).
    Even though your picture gets noisy you will still get the full dynamic
    range of the sensor.
    The noise that you are seing on a high iso setting is nothing more than
    low levels amplified stronger. Since the noise always stays constant,
    but the levels of your ccd output are lower you need more amplification
    and that brings out more noise.
    What you are proposing only shifts the problem to the computer for
    amplification, but you do not eliminate the noise. So again bu
    increasing exposure on the raw file you will essentialy end up with the
    same result, but you will also have to live with your reduced dynamic
    range (vs. the shot with higher iso)

    Hope that was somewhat clear.

    Andre

    --
    ----------------------------------
    http://www.aguntherphotography.com
     
    andre, Nov 1, 2004
    #3
  4. dperez@juno_nospam.com

    Guest

    dperez@juno_nospam.com wrote:
    > Of late I've seen a couple discussions of the importance of not underexposing a
    > digital image. Specifically this is because if you have 6 stops of dynamic
    > range, half the data, or 2048 levels, of a 12-bit capture is in the brightest
    > stop, 1024 in the next stop and so on until the lowest stop, which hsa 64
    > levels...


    > SO, based on this, the argument is to be as bright as possible
    > without blowing out the highlights. And that being 1 stop
    > underexposed would cause a loss of 50% of the possible levels in the
    > image...


    > My question is which is BETTER - to underexpose an image at a given
    > ISO or to increase the ISO and expose the image more brightly?


    It depends on the camera, but my answer is for high-end DSLRs.
    Usually, such cameras have a "normal" range that goes from 100ish to
    800ish ISO, and a "boost" range. Within the "normal" range, you
    should set the ISO to what you need rather than deliberately
    underexpose. Within the "boost" range I think it's just scaling up
    the recorded image data, so it doesn't help much.

    > For example - if I'm at ISO 100 and my histogram is pushed toward
    > the dark end - NOT clipped, just "dark" am I better to increase the
    > ISO to say 400 so my histogram is pushed toward the bright end -
    > AGAIN NOT CLIPPED...


    > Now to make things more interesting - in the RAW converter, I'd MOST
    > LIKELY increase the exposure of the "darker" image by 1 stop, and
    > MOST LIKELY reduce the exposure of the "brighter" image by 1
    > stop... So, in the end, IN THEORY, both images would be identical
    > once they got to Photoshop after conversion...


    No, because the ISO setting in a DSLR can affect the gain before the
    A-D convereter.

    > How about an even more extreme example - am I better leaving the ISO
    > at 200 and having the camera underexpose a shot by 1 stop (actual
    > underexposure so I'm perhaps clipping just a bit on the dark end) or
    > bumping the ISO to 800, which I KNOW causes visible noise and
    > getting the exposure much brighter?


    I think you'll get better results by setting the ISO appropriately.

    > I"ll most likely run some tests just to see how things look in both
    > cases, but I'm curious what others have found...


    Interesting. Let us know.

    Andrew.
     
    , Nov 1, 2004
    #4
  5. dperez@juno_nospam.com

    Guest

    In message <xolhd.3207$>,
    andre <> wrote:

    >What you are proposing only shifts the problem to the computer for
    >amplification, but you do not eliminate the noise. So again bu
    >increasing exposure on the raw file you will essentialy end up with the
    >same result, but you will also have to live with your reduced dynamic
    >range (vs. the shot with higher iso)


    Yes, but if you try to bring up the shadows in software, they get ugly
    within 2 or 3 stops, and it's not just because of the noise. Pushing
    ISO 100 3 stops to ISO 800 on my 10D is *much* uglier than using ISO 800
    from the start. The reason is that the noise and shadow detail are
    involved in intermodulation distortion from the quantization. At the
    lower ISO range, I think very little image noise is readout/amplifier
    noise, and going to ISO 200 or 400 (on a DSLR) to get the shutter speed
    you need, and DOF you need, and saturate the RAW dynamic range at the
    same time.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #5
  6. wrote:
    > The reason is that the noise and shadow detail are
    > involved in intermodulation distortion from the quantization.

    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    What the heck does this mean?

    Roger
     
    Roger N. Clark (change username to rnclark), Nov 2, 2004
    #6
  7. dperez@juno_nospam.com wrote:

    > Of late I've seen a couple discussions of the importance of not underexposing a
    > digital image. Specifically this is because if you have 6 stops of dynamic
    > range, half the data, or 2048 levels, of a 12-bit capture is in the brightest
    > stop, 1024 in the next stop and so on until the lowest stop, which hsa 64
    > levels...
    >
    > SO, based on this, the argument is to be as bright as possible without blowing
    > out the highlights. And that being 1 stop underexposed would cause a loss of
    > 50% of the possible levels in the image...
    >
    > My question is which is BETTER - to underexpose an image at a given ISO or to
    > increase the ISO and expose the image more brightly? For example - if I'm at
    > ISO 100 and my histogram is pushed toward the dark end - NOT clipped, just
    > "dark" am I better to increase the ISO to say 400 so my histogram is pushed
    > toward the bright end - AGAIN NOT CLIPPED...
    >
    > Now to make things more interesting - in the RAW converter, I'd MOST LIKELY
    > increase the exposure of the "darker" image by 1 stop, and MOST LIKELY reduce
    > the exposure of the "brighter" image by 1 stop... So, in the end, IN THEORY,
    > both images would be identical once they got to Photoshop after conversion...
    >
    > From what I understand, increasing the exposure on the "darker" image causes
    > more noise in the shadows... And the higher ISO has more noise everywhere...
    > But, is there a difference in the noise? Is one more obvious than the other?
    > Is one luminance noise versus color noise?
    >
    > How about an even more extreme example - am I better leaving the ISO at 200 and
    > having the camera underexpose a shot by 1 stop (actual underexposure so I'm
    > perhaps clipping just a bit on the dark end) or bumping the ISO to 800, which I
    > KNOW causes visible noise and getting the exposure much brighter?
    >
    > I"ll most likely run some tests just to see how things look in both cases, but
    > I'm curious what others have found...


    A lot depends on your camera.
    Here are two pages that might help give some relevant information:
    http://clarkvision.com/imagedetail/digital.signal.to.noise

    http://clarkvision.com/imagedetail/dynamicrange

    I rarely change my metering pattern, but I do adjust the over/under
    exposure based on the subject. I'll often set the iso low and meter
    at -1 or even -2 stops when there are white subjects on dark backgrounds.
    For example, see my bird gallery:

    http://clarkvision.com/galleries/gallery.bird

    and check out the images with white birds. The technical info is there
    so you can see the results. Unless you have a low noise DSLR, you'll probably
    not be able to do this, instead, you will need to get the highest
    signal you can or the image signal to noise will be too low.
    It is always best to maximize signal without blowing the highlights,
    regardless of camera.

    Roger
     
    Roger N. Clark (change username to rnclark), Nov 2, 2004
    #7
  8. dperez@juno_nospam.com

    Confused Guest

    On Mon, 01 Nov 2004 19:05:08 -0700
    In message <>
    "Roger N. Clark (change username to rnclark)" <>
    wrote:

    > wrote:
    > > The reason is that the noise and shadow detail are
    > > involved in intermodulation distortion from the quantization.

    > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    > What the heck does this mean?


    This means that the noise and coarse detail gets mixed together when
    the numbers are added, multiplied, and divided. (He is mixing music
    synthesis and radio terms to explain image manipultion...probably
    because there is no clear way to explain it in technical terms.)

    If the same photo had been shot at ISO 800 then more detail would have
    visible despite the noise. Kinda like the difference between punching
    up an old 78 record vs listening to a 11kHz CD. :)

    Jeff
     
    Confused, Nov 2, 2004
    #8
  9. Confused wrote:

    > On Mon, 01 Nov 2004 19:05:08 -0700
    > In message <>
    > "Roger N. Clark (change username to rnclark)" <>
    > wrote:
    >
    >
    >> wrote:
    >>
    >>>The reason is that the noise and shadow detail are
    >>>involved in intermodulation distortion from the quantization.

    >>
    >> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    >>What the heck does this mean?

    >
    >
    > This means that the noise and coarse detail gets mixed together when
    > the numbers are added, multiplied, and divided. (He is mixing music
    > synthesis and radio terms to explain image manipultion...probably
    > because there is no clear way to explain it in technical terms.)
    >
    > If the same photo had been shot at ISO 800 then more detail would have
    > visible despite the noise. Kinda like the difference between punching
    > up an old 78 record vs listening to a 11kHz CD. :)
    >
    > Jeff


    from: http://www.its.bldrdoc.gov/fs-1037/dir-019/_2813.htm

    intermodulation distortion: Nonlinear distortion characterized
    by the appearance, in the output of a device, of frequencies that
    are linear combinations of the fundamental frequencies and all
    harmonics present in the input signals. (188) Note: Harmonic
    components themselves are not usually considered to characterize
    intermodulation distortion. When the harmonics are included as
    part of the distortion, a statement to that effect should be made.

    This HTML version of FS-1037C was last generated on Fri Aug 23 00:22:38 MDT 1996

    "intermodulation distortion from the quantization" is jargon that
    seems meaningless. There is no intermodulation in this case. There
    are no linear combination of frequencies. There are no harmonic
    components. It is simple quantization due to the use of integer
    numbers. The general digital photography term is "posterization."
    If more bits were used in processing, like scaled 16-bit values
    (scaled up from 12-bit camera numbers) posterization can be reduced
    usually to the point of being negligible. This is discussed on
    my dynamic range page:

    http://clarkvision.com/imagedetail/dynamicrange

    Roger
     
    Roger N. Clark (change username to rnclark), Nov 2, 2004
    #9
  10. dperez@juno_nospam.com

    Confused Guest

    On Mon, 01 Nov 2004 20:52:26 -0700
    In message <>
    Roger N. Clark wrote:

    > Confused wrote:
    >
    > > On Mon, 01 Nov 2004 19:05:08 -0700
    > > In message <>
    > > "Roger N. Clark (change username to rnclark)" <>
    > >
    > >> wrote:
    > >>
    > >>>The reason is that the noise and shadow detail are
    > >>>involved in intermodulation distortion from the quantization.
    > >>
    > >> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^


    Roger trolled:

    > >>What the heck does this mean?


    I took the bait. Ironically, I hit the nail exactly on the head,
    only based on my background with additive and FM digital
    synthesis in the 70's, a SW radio monitoring hobby, a career
    in software engineering, and my take on trying to describe something
    that is very confusing.

    But I try and see the simple answers. In this case, there is both an
    objective and subjective similarity in the development of both audio
    and visual reproduction in a quickly changing digital environment.

    > > This means that the noise and coarse detail gets mixed together when
    > > the numbers are added, multiplied, and divided. (He is mixing music
    > > synthesis and radio terms to explain image manipultion...probably
    > > because there is no clear way to explain it in technical terms.)


    See (1)

    > > If the same photo had been shot at ISO 800 then more detail would have
    > > visible despite the noise. Kinda like the difference between punching
    > > up an old 78 record vs listening to a 11kHz CD. :)


    See (2)

    > > Jeff


    (1.a)

    Thanks for making my point. However, I wouldn't consider a government
    agency involved in radio and audio transmission an authority on image
    processing as related to state of the art photographic manipulation.
    The technical details are tightly held by a small number of very
    competitive companies, and any related up to date information held by
    a government would be classified. We the people are left to our best
    interpretation of limited and filtered information.

    http://www.its.bldrdoc.gov/

    "ITS is the research and engineering branch of the
    National Telecommunications and Information Administration (NTIA,
    a part of the U.S. Department of Commerce (DOC)."

    But then, the DOC/NTIA/ITS is involved with audio and radio
    engineering, so, again, thanks for validating my comment!


    > from: http://www.its.bldrdoc.gov/fs-1037/dir-019/_2813.htm
    >
    > intermodulation distortion: Nonlinear distortion characterized
    > by the appearance, in the output of a device, of frequencies that
    > are linear combinations of the fundamental frequencies and all
    > harmonics present in the input signals. (188) Note: Harmonic
    > components themselves are not usually considered to characterize
    > intermodulation distortion. When the harmonics are included as
    > part of the distortion, a statement to that effect should be made.
    >
    > This HTML version of FS-1037C was last generated on Fri Aug 23 00:22:38 MDT 1996
    >
    > "intermodulation distortion from the quantization" is jargon that
    > seems meaningless. There is no intermodulation in this case. There
    > are no linear combination of frequencies.


    (1.b)

    Linearity is not a requirement of "intermodulation"... an
    understandable mistake in a small focused sub-agency of a government
    commerce department. If one were to dig deep enough my statement
    could be made to look false. However, frequency modulation, and
    intermodulation, is not dependent on any concept of linearity. The
    intermodulation of frequencies in music is usually non-linear, and
    when computers enter the picture quantization is a constant problem
    when realizing sound, just as it is in realizing light reflection.
    The sheer number of variables eliminates linearity of distortion
    (modulation, intermodulation, anyothertypeofmodulation of the data at
    hand).

    > There are no harmonic
    > components. It is simple quantization due to the use of integer
    > numbers. The general digital photography term is "posterization."
    > If more bits were used in processing, like scaled 16-bit values
    > (scaled up from 12-bit camera numbers) posterization can be reduced
    > usually to the point of being negligible. This is discussed on
    > my dynamic range page:
    >
    > http://clarkvision.com/imagedetail/dynamicrange


    (2)

    Great examples of ancient 78's and undersampled audio vs photographic
    imaging. Neither sounds good, neither looks good; all garbage. The
    graphs are kinda spiffy... what do they mean? It looks like they are
    saying, "We have to wait for technology to advance another year or
    three before we will be happy with the dynamic range of digital
    scanners and cameras."

    Jeff
     
    Confused, Nov 2, 2004
    #10
  11. dperez@juno_nospam.com

    Guest

    In message <>,
    "Roger N. Clark (change username to rnclark)" <>
    wrote:

    > wrote:
    >> The reason is that the noise and shadow detail are
    >> involved in intermodulation distortion from the quantization.

    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    >What the heck does this mean?


    The noise and signal are summed at the analog sensor level before they
    are (over-) digitized. That makes it harder to separate them afterward.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #11
  12. dperez@juno_nospam.com

    Guest

    In message <>,
    "Roger N. Clark (change username to rnclark)" <>
    wrote:

    >"intermodulation distortion from the quantization" is jargon that
    >seems meaningless. There is no intermodulation in this case. There
    >are no linear combination of frequencies. There are no harmonic
    >components. It is simple quantization due to the use of integer
    >numbers.


    Yes, but quantization leaves one with the impression that there just
    aren't enough levels, whereas in reality, irreversible damage has been
    done which is much worse than if the component elements had been
    quantized separately and combined. How do you think that this should be
    conveyed?
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #12
  13. dperez@juno_nospam.com

    Guest

    In message <>,
    "Roger N. Clark (change username to rnclark)" <>
    wrote:

    >It is always best to maximize signal without blowing the highlights,
    >regardless of camera.


    True, but you have to watch the RAW converter you use, as they tend to
    use a single gamma over most of the mid-range, and change it near the
    ends. For example, I took the same shot last night on a tripod, at
    three different ISOs with the same exact absolute exposure (same shutter
    speed and aperture), and used ACR2.3 to pull one image down by 1 stop,
    and another by 2, to make them look the same. The ISO 400 image had
    slighlty brighter highlights, and two of them (I forgot which) had
    brighter shadows, which you could only see after you boosted the
    shadows. I'm going to make a uncompressed .DNGs of these images, scale
    the RAW data in the DNGs, re-write them, and see how much of the effect
    is from the RAW data, and how much is from the converter, when I get a
    chance.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #13
  14. In article <>,
    says...
    > Yes, but quantization leaves one with the impression that there just
    > aren't enough levels, whereas in reality, irreversible damage has been
    > done which is much worse than if the component elements had been
    > quantized separately and combined. How do you think that this should be
    > conveyed?


    NASCAR analogy.
    --
    http://www.pbase.com/bcbaird/
     
    Brian C. Baird, Nov 2, 2004
    #14
  15. dperez@juno_nospam.com

    Bill Hilton Guest

    >From: dperez@juno_nospam.com

    >I've seen a couple discussions of the importance of not underexposing a
    >digital image.


    Like this one, probably ...
    http://www.luminous-landscape.com/tutorials/expose-right.shtml

    >My question is which is BETTER - to underexpose an image at a given ISO or to
    >increase the ISO and expose the image more brightly?


    Why not run a test and see for yourself?

    After reading Reichmann's article I thought I'd give it a real-world workout to
    see if you can actually see the differences so I photographed a stormy grey sky
    at ISO 400 dead on and with -2 and +2 exposure compensation. I don't really
    start to see noise with my Canon dSLRs until around ISO 320 or so (I think I
    used the 1Ds for this test), which is why I picked 400. In looking at the
    files at 100% it's clear that +2 (overexpose by two stops) has much less noise.

    Then I tested the same scene at ISO 100, same three exposures. The difference
    wasn't as great since there is less noise to begin with.

    Then I checked the ISO 400 shot at + 2 against the ISO 100 shot at 0 and they
    were pretty much the same. Which I think answers your question, lowering ISO
    or adding exposure compensation seemed to have about the same effect.

    Bill
     
    Bill Hilton, Nov 2, 2004
    #15
  16. wrote:
    > In message <>,
    > "Roger N. Clark (change username to rnclark)" <>
    > wrote:
    >
    >
    >>"intermodulation distortion from the quantization" is jargon that
    >>seems meaningless. There is no intermodulation in this case. There
    >>are no linear combination of frequencies. There are no harmonic
    >>components. It is simple quantization due to the use of integer
    >>numbers.

    >
    >
    > Yes, but quantization leaves one with the impression that there just
    > aren't enough levels, whereas in reality, irreversible damage has been
    > done which is much worse than if the component elements had been
    > quantized separately and combined. How do you think that this should be
    > conveyed?


    It's called posterization. See my previous post.

    Roger
     
    Roger N. Clark (change username to rnclark), Nov 2, 2004
    #16
  17. dperez@juno_nospam.com

    Guest

    wrote:
    > The reason is that the noise and shadow detail are
    > involved in intermodulation distortion from the quantization.


    IMD from the quantization? I would certainly hope not. The readout
    noise should be quite enough to linearize the conversion.

    Andrew.
     
    , Nov 2, 2004
    #17
  18. wrote:

    > In message <>,
    > "Roger N. Clark (change username to rnclark)" <>
    > wrote:
    >
    >
    >> wrote:
    >>
    >>>The reason is that the noise and shadow detail are
    >>>involved in intermodulation distortion from the quantization.

    >
    > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    >
    >>What the heck does this mean?

    >
    >
    > The noise and signal are summed at the analog sensor level before they
    > are (over-) digitized. That makes it harder to separate them afterward.


    You can't separate noise from signal. You can reduce noise by smoothing,
    but it also reduces image information. In some cases reducing
    image information is not too harmful, like smoothing the noise in
    the sky: the sky has little spatial information.

    The key to reducing posterization (the integer quantization of a digital
    image) is to record raw data and work in 16-bit. Even if you
    do not record raw, the first processing step should be to convert
    to 16-bit.

    Roger
     
    Roger N. Clark (change username to rnclark), Nov 2, 2004
    #18
  19. dperez@juno_nospam.com

    Guest

    In message <>,
    lid wrote:

    > wrote:
    >> The reason is that the noise and shadow detail are
    >> involved in intermodulation distortion from the quantization.

    >
    >IMD from the quantization? I would certainly hope not. The readout
    >noise should be quite enough to linearize the conversion.


    At low ISOs, the noise is actually quite small, and the low quantization
    of the shadows is mainly what makes it visible. ISO 100 pushed to 800
    looks like garbage compared to ISO 800 on both my 10D and 20D. Not just
    because there are too few levels for the image, but because the noise
    can be exaggerated several times by quantization, and there is an effect
    where it looks brighter than it is, much like a fine black-and-white
    checkerboard looks brighter than a grey surface emitting or reflecting
    the same amount of total light.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #19
  20. dperez@juno_nospam.com

    Guest

    In message <>,
    dy (Bill Hilton) wrote:

    >Then I checked the ISO 400 shot at + 2 against the ISO 100 shot at 0 and they
    >were pretty much the same.


    But the ISO 400 image is represented by 4x as many RAW levels, so you
    can do more with it without posterizing.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
     
    , Nov 2, 2004
    #20
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