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Author Topic: 16-bit DSLR  (Read 62817 times)
Panopeeper
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« Reply #60 on: December 03, 2007, 04:56:14 PM »
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If you only count one photon in 500, then the minumum signal you can count is 500 photons and the dynamic range is 60000/500 = 120, capturable as you state in 7 bits.  But if every photon is counted, then the dynamic range increases to 60000/1 and requires 16 bit to capture.  This is a real increase in dynamic range, by any reasonable definition.

These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.

But I can start counting with photon 1, then 501, 1001, etc.
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Gabor
EricV
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« Reply #61 on: December 03, 2007, 05:46:35 PM »
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These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.

But I can start counting with photon 1, then 501, 1001, etc.
[a href=\"index.php?act=findpost&pid=158018\"][{POST_SNAPBACK}][/a]
Do you really think you can increase the dynamic range of a sensor by a factor of ten simply by counting {1, 11, 21, ...} instead of {0, 10, 20, ...}?  Doesn't that sound ridiculous?  

We all agree that it is possible to concoct non-linear counting schemes which provide extended dynamic range with fewer bits.  But such a scheme is mathematically impossible for a linear sensor (with an offset of zero).

Here is a thought exercise which might prove interesting.  Imagine an array of 120 light bulbs which I want to photograph.  The light bulbs have different brightness, with a linear progression.  The first light bulb is turned off.  The second light bulb emits 500 photons/second into the aperture of my camera.  The third light bulb emits 1000 photons/second.... The last light bulb emits 60000 photons/second.

Now I photograph this array of light bulbs with two different cameras in three different scenarios:
1) Camera full well 60000 photons, linear scale counting one photon out of 500, exposure 1 second.  This camera images the light bulbs with readings {0, 1, 2, ... 120).
2) Camera full well 60000 photons, linear scale counting single photons, exposure 1 second.  This camera images the light bulbs with readings {0, 500, 1000, ... 60000}.
3) Camera full well 60000 photons, linear scale counting single photons, exposure 1/500 second.  This camera images the light bulbs with readings {0, 1, 2, ... 120}.

Would you agree with the following statements?
1) The first camera uses its entire dynamic range to capture the scene.  If the scene had brighter light bulbs, this camera could not capture them.  If the scene had light bulbs with intermediate brightness, this camera could not resolve them.
2) The second camera, with the same exposure, captures the scene with resolution to spare.  The captured dynamic range appears to be the same (because the dimmest bulb still has brightness 500 photons/second), but there is plenty of unused space to capture light bulbs with intermediate brightness.
3) The second camera, with reduced exposure, captures the scene with the same resolution (in fact the same numerical output values), but with dynamic range to spare.  This camera could capture a larger range of light bulb brightness.

Clearly the two cameras do not have the same dynamic range!
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John Sheehy
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« Reply #62 on: December 03, 2007, 06:32:17 PM »
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You could say something about the captured dynamic range if the encoding is the same as it currently is in most cameras, ie. 256, 128, 64 , 32, 16, 8, 4, 2, 1, which represents 8 intervals or just 8 stops of DR. The posterization would be horrendous in the lower mid-tones, producing a practically useful DR of considerably less than the theoretical maximum of 8 stops.
[a href=\"index.php?act=findpost&pid=157823\"][{POST_SNAPBACK}][/a]

8 bits linear is actually sufficient for most cameras (except for the D3 and Canon DSLRS) at ISO 1600, even ISO 800 for some.  Again, you never (visibly) see this RAW quantization in practice, because of noise.
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John Sheehy
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« Reply #63 on: December 03, 2007, 06:36:13 PM »
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The subject was not the practicability but the principle. I could have said "20 stops in ten bits", which is still impractical, but not as extreme as "50 stops with one bit".

The concept of "stop" is relevant mainly for photographers, it plays no role whatsoever in the principle.
[a href=\"index.php?act=findpost&pid=157978\"][{POST_SNAPBACK}][/a]

It's an arbitrary factor, but one that is familiar to photographers, so why not use it?  Now decibels, that is something foreign to the photographer.
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John Sheehy
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« Reply #64 on: December 03, 2007, 06:46:52 PM »
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In another case only every 200th photon counts, but we can go even further: only every 500th photon counts (i.e. the numerical intensity is the number of photons divided by 500). The 60000 photons of a well would be converted in the numerical range of 0-120, which requires just 7 bits.

That would have less than 7 stops of DR in the absence of analog noise at the pixel level, and more at the image level with a little noise, but greater limitations with ample noise.

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The numerical representation is linear in all these cases, but the number of levels covering the entire range is vastly different, and so is the number of required bits to store the data.

In all these cases the covered dynamic range is the same.

[a href=\"index.php?act=findpost&pid=157991\"][{POST_SNAPBACK}][/a]

That's only true if the analog noise is the bottleneck to DR.  Otherwise, your 200 levels becoming 1 level in the deepest shadows throws away DR.
« Last Edit: December 03, 2007, 06:47:24 PM by John Sheehy » Logged
John Sheehy
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« Reply #65 on: December 03, 2007, 06:50:42 PM »
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These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.

But I can start counting with photon 1, then 501, 1001, etc.
[a href=\"index.php?act=findpost&pid=158018\"][{POST_SNAPBACK}][/a]

Your "counting" has no effect on DR.  Bottom line is that without noise, your bottom 500 electrons are squished into 1 level, at the image level, and noise won't help at all at the pixel level.
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Panopeeper
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« Reply #66 on: December 03, 2007, 09:50:13 PM »
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Do you really think you can increase the dynamic range of a sensor by a factor of ten simply by counting {1, 11, 21, ...} instead of {0, 10, 20, ...}?  Doesn't that sound ridiculous?

That would be  truly a ridiculous claim. However, the subject is not the dynamic range of the sensor but the recording of the sensed data. As I posted above:

a given scenery, a given lens and a given sensor

so that only the numeric representation is variable.

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But such a scheme is mathematically impossible for a linear sensor (with an offset of zero).

That's your belief.

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an array of 120 light bulbs which I want to photograph.  The light bulbs have different brightness, with a linear progression.  The first light bulb is turned off.  The second light bulb emits 500 photons/second into the aperture of my camera.  The third light bulb emits 1000 photons/second.... The last light bulb emits 60000 photons/second

The demonstration-idea is excellent. Let's see, what it demosntrates:

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3) Camera full well 60000 photons, linear scale counting single photons, exposure 1/500 second.  This camera images the light bulbs with readings {0, 1, 2, ... 120}

...

3) The second camera, with reduced exposure, captures the scene with the same resolution (in fact the same numerical output values), but with dynamic range to spare.  This camera could capture a larger range of light bulb brightness

Right. So, you have demonstrated, that the second camera can capture the entire original dynamic range in seven bits (just like the first camera), and it can capture even more, in more bits.

In effect you have shown, that if a camera counts only every xth photon, it requires x times higher exposure in order to gain the same number of levels, than another camera, which counts every photon.

No objection here, but this has nothing to do with the numerical representation (bit depth vs. DR).

However, if you are saying, that the first camera is "throwing away" large part of its capability, then consider following:

1. nobody disputed, that greater bit depth results in more levels,

2. in 1/500 sec the first 499 measurable levels are eliminated. They won't be captured with the first camera either, because you conveniently decided, that it counts 0, 500, 1000, etc. (your proof requires this). However, if the first camera counts 5, 505, 1005, etc. then everything looks differently.

The demostration with discrete values leaves much leeway, which can be used so or so.
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Gabor
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« Reply #67 on: December 04, 2007, 06:54:16 AM »
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However, if the first camera counts 5, 505, 1005, etc. then everything looks differently.

Panopeeper a sensor that assigns the first non-zero level (level 1) after counting 5 photons, and next level (level 2) when the count reaches 505 photons is simply NOT LINEAR; and present cameras are linear so they have the bit depth limitation for DR.

I think we all agree that the limitation of DR with bit depth is only a matter of the encoding system used, and as long as this is linear, N bits can encode a maximum of N f-stops of DR.

In fact we already discussed that typical 12-bit linear sensors loose around the first 256 levels because of the black offset level due to sensor electronics. That means an even more reduced limit to the DR of:

log(4096-256)/log(2)=11,9 f-stops
« Last Edit: December 04, 2007, 09:03:24 AM by GLuijk » Logged

Panopeeper
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« Reply #68 on: December 05, 2007, 12:01:12 AM »
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Panopeeper a sensor that assigns the first non-zero level (level 1) after counting 5 photons, and next level (level 2) when the count reaches 505 photons is simply NOT LINEAR

So, you think if you measure something with a measure tape and start on the 50cm mark, then the measurement is not linear?

Beside, you can start at 1, then 501, etc., it is really irrelevant.

I made a simple drawing showing the relationship (or the lack thereof) between DR and bit depth:

« Last Edit: December 05, 2007, 12:09:58 AM by Panopeeper » Logged

Gabor
Jonathan Wienke
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« Reply #69 on: December 05, 2007, 12:28:44 AM »
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So, you think if you measure something with a measure tape and start on the 50cm mark, then the measurement is not linear?

Yes, unless you have a completely new definition of "linear". If 0 = 0 and 1 = 5 and 2 = 505, then the encoding function is by definition not linear. Or did you fail algebra?
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B-Ark
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« Reply #70 on: December 05, 2007, 07:29:36 AM »
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Just to shift the topic a bit, from counting photons, defining levels, etc ...

The dynamic range is ultimately limited by the noise floor - has anyone made any effort to lower the noise floor by cooling the sensor and other analog electronics? Pelletier devices are quite small (albeit rather power hungry), and would probably fit into a DSLR.
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Jonathan Wienke
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« Reply #71 on: December 05, 2007, 08:28:21 AM »
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That is commonly done with cameras designed for astronomy, but the cost/benefit ratio isn't high enough for mainstream DSLR manufacturers to add it as a design feature.
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Ray
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« Reply #72 on: December 05, 2007, 11:24:13 AM »
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.

I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
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John Sheehy
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« Reply #73 on: December 05, 2007, 02:02:53 PM »
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.

I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
[a href=\"index.php?act=findpost&pid=158438\"][{POST_SNAPBACK}][/a]

Here's what you get with a population of 100, averaging 25 electrons:

28 17 22 25 26 35 21 31 36 19
25 36 19 26 27 23 25 13 22 30
21 18 22 17 26 24 23 27 30 23
24 20 18 38 39 22 22 25 22 26
32 18 20 26 17 19 21 20 25 23
21 28 25 29 19 38 20 29 24 30
25 19 21 28 19 25 36 22 31 21
26 25 30 34 33 27 26 29 28 25
25 29 34 38 22 22 30 26 23 37
13 26 31 25 35 23 30 26 26 24

Minimum of 13 and maximum of 39 in this sample, a spread of 26 photons.

With a population of 20 million, I get 4 to 56.
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EricV
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« Reply #74 on: December 05, 2007, 09:18:30 PM »
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.

I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
Yes, this is all correct.  To be precise, the statistics apply to electrons produced by interacting photons, since that is what the sensor actually "counts".

Shot noise has a Poisson distribution, with fluctuations (standard deviation) equal to the square root of the mean.  So in your example, a light level which produces 25 electrons on average will have statistical noise of 5 electrons.  You can generate the full distribution if you have Excel handy (look up the Poisson function), but it is essentially a bell shaped curve with a peak at 25 and a width of 5.

Most sensors are dominated by electronics readout noise at very low light levels so shot noise is rarely a concern.  For example, if a sensor has readout noise of 20 electrons, shot noise will be smaller until the signal reaches 400 electrons, and by then this noise is only 5% of the signal.
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John Sheehy
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« Reply #75 on: December 05, 2007, 09:53:09 PM »
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Here's what you get with a population of 100, averaging 25 electrons:
[a href=\"index.php?act=findpost&pid=158474\"][{POST_SNAPBACK}][/a]

That should be, "what you might get"!

You probably won't get exactly this.
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Ray
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« Reply #76 on: December 05, 2007, 10:35:38 PM »
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That should be, "what you might get"!

You probably won't get exactly this.
[a href=\"index.php?act=findpost&pid=158567\"][{POST_SNAPBACK}][/a]

John,
Just to be sure I've understood the significance of these figures, are you saying that the above variation amongst a group of 100 pixels might apply if we were photographing a dark grey patch where, ideally, each pixel would receive 25 photons if it were free of shot noise? (and hypothetically free of all noise from other sources).
« Last Edit: December 05, 2007, 10:39:21 PM by Ray » Logged
Ray
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« Reply #77 on: December 05, 2007, 10:51:07 PM »
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Most sensors are dominated by electronics readout noise at very low light levels so shot noise is rarely a concern.  For example, if a sensor has readout noise of 20 electrons, shot noise will be smaller until the signal reaches 400 electrons, and by then this noise is only 5% of the signal.
[a href=\"index.php?act=findpost&pid=158562\"][{POST_SNAPBACK}][/a]

So the implication here is that both shot noise and read noise will not be reduced by cooling, just thermal noise, or is read noise partly thermal?
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Panopeeper
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« Reply #78 on: December 05, 2007, 11:03:40 PM »
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Yes, unless you have a completely new definition of "linear". If 0 = 0 and 1 = 5 and 2 = 505, then the encoding function is by definition not linear. Or did you fail algebra?

Do you mind pointing out, where I posted 0 = 0 and 1 = 5 and 2 = 505? Or did you fail reading comprehension?
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Gabor
John Sheehy
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« Reply #79 on: December 06, 2007, 07:57:41 AM »
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John,
Just to be sure I've understood the significance of these figures, are you saying that the above variation amongst a group of 100 pixels might apply if we were photographing a dark grey patch where, ideally, each pixel would receive 25 photons if it were free of shot noise? (and hypothetically free of all noise from other sources).
[a href=\"index.php?act=findpost&pid=158576\"][{POST_SNAPBACK}][/a]

Yes.  I just wanted to demonstrate how far off values are.  The "standard deviation" can seem a bit more conservative than what it really means.  The majority of values are often in a spread about 3x as great as the standard deviation.
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