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Thread started 19 Jul 2014 (Saturday) 11:30
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Chances of 5D MkIV have better dynamic range?

 
Shadowblade
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Jul 24, 2014 20:39 |  #46

CRCchemist wrote in post #17054216 (external link)
Honestly, I have a background in semiconductor fabrication process, and my undergraduate was in electrical engineering. Experience in semiconductor and analog circuit design leads me to believe that this is the case.

Such a low amount of noise is extremely rare, and it's even LESS likely from a circuit design that is densely packed devices. It's an annoying design consideration but it's the reality of the game. It's obviously a lot to explain here and requires about 30 minutes to explain to somebody why transistors work the way they do, but that sample of no noise innately from the Nikon is a little unbelievable to me. There is always a noise floor.

I don't know what, but something is going on under the hood that is a part of the process of the output data of all the millions of devices on the sensor substrate.

The noise floor from the sensor is no less than that of Canon sensors. If anything, Canon's sensor has a slightly better SNR than the D800 sensor (although not the D810) which gives it a slightly better image at extremely high ISOs. But Canon uses an off-sensor A/D conversion chip, which seems to introduce a lot of read noise, both random and pattern.

Even if Nikon's black point were clipped, so what? You need to clip the blacks at the noise floor anyway, because anything below it is random noise anyway and not part of the camera's dynamic range - it's just like applying a low-pass filter to the intensity reading of individual photosites. May as well use the 16384 levels of a 14-bit file to record actual detail, not just noise.




  
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Footbag
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Jul 25, 2014 11:04 |  #47

Shadowblade wrote in post #17054281 (external link)
The noise floor from the sensor is no less than that of Canon sensors. If anything, Canon's sensor has a slightly better SNR than the D800 sensor (although not the D810) which gives it a slightly better image at extremely high ISOs. But Canon uses an off-sensor A/D conversion chip, which seems to introduce a lot of read noise, both random and pattern.

Even if Nikon's black point were clipped, so what? You need to clip the blacks at the noise floor anyway, because anything below it is random noise anyway and not part of the camera's dynamic range - it's just like applying a low-pass filter to the intensity reading of individual photosites. May as well use the 16384 levels of a 14-bit file to record actual detail, not just noise.

Why must you clip the blacks?

I don't think it's a huge deal for most types of photography, but the astro-imaging guys are pulling their hair out trying to calibrate with bias and flat frames. They are the ones who seem to have discovered what is going on and they have or are the process of creating a patch.

Astro-imagers need a linear sensor for calibration to work. I suspect the 7D does a very similar thing as users have similar troubles with calibration.


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pwm2
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Jul 25, 2014 12:37 |  #48

Blacks needs not to be clipped.

Noise + weak exposure results in noise with a higher average value, which represents a lighter tone of dark gray than the dark gray from noise without exposure.

For spatial capture - like our two-dimensional photos, the integration over the surface area allows for signals that are reaching, or even below, the average noise level. Low-pass filtering of the image is working on averages, so low-pass filtering to reduce noise will still allow the capture of shadow shades down to - and into - the noise floor. It's just that we can't expect small features to be retained. And the noise must obviously not contain strong patterns that survives and looks like structures.

When having temporal capture - like a single parameter captured over time - it's also sometimes possible to perform signal analysis to work with data well below the noise level by looking at the frequency distribution of the signal+noise.


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Shadowblade
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Jul 25, 2014 12:47 |  #49

Footbag wrote in post #17055335 (external link)
Why must you clip the blacks?

I don't think it's a huge deal for most types of photography, but the astro-imaging guys are pulling their hair out trying to calibrate with bias and flat frames. They are the ones who seem to have discovered what is going on and they have or are the process of creating a patch.

Astro-imagers need a linear sensor for calibration to work. I suspect the 7D does a very similar thing as users have similar troubles with calibration.

A sensor only produces raw data, not an image.

To get an image out of the data, you need to firstly define what value represents 'black' and what value represents 'white'. Between your defined 'black' and 'white' points, you then have 16384 levels of luminance (for 14-bit output).

'White' is obvious - you define it as whatever value represents a saturated photosite. Set it any lower and you lose dynamic range as whites are clipped, set it any higher and even a maxed-out photosite won't give you white.

'Black' is slightly less obvious. Every sensor has some random noise from the electronics (some more than others) and there is a certain level of exposure below which 'real' signal from incoming photons becomes indistinguishable from electronic noise. It makes sense to set this as the black level. Set it any higher and you start to lose information, since you'd be clipping shadows that are, nevertheless, above the noise threshold. Set it lower, however, and not only would the sensor would never record a 'true' black, since the background noise level would represent a dark grey, but the darkest few shades of grey would also contain absolutely no detail, since they would be below the noise threshold.

Setting 'white' as the saturation point of photosites and 'black' as the noise threshold makes the most of the sensor, by putting the most levels of luminance within the sensor's dynamic range (where they are actually recording real detail) and giving the most contrast to the final image.

It's just like adjusting the points on the Levels tool in Photoshop.




  
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pwm2
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Jul 25, 2014 12:55 |  #50

But the only reason for clip any noise in the raw file is to keep down the size of the raw file (assuming the camera uses compression and not like some old cameras wrote an uncompressed file). And he noise level depends on the temperature and length of exposure.

If you have a 14-bit conversion and considers the last bit to be noise, then you end up with just a 13-bit file and a bit of an issue if claiming 14-bit to the customers.

It would normally be the photographer who should decide on what curves adjustments to do - which in this case relates to what level of the edited data to map to jpeg intensity level 0 and what level of the edited data to map to jpeg intensity level 255. And before that step is performed, different operations can have made use of the noisy bottom-level stop.


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Footbag
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Jul 26, 2014 08:58 |  #51

Clipping the black point in itself will not improve the image, though.

I think what may be happening is that any pixels below a certain value are replaced with black pixels or possibly the adjacent pixels are averaged or bias removed.

That kind of bothers me. It would explain why the D800 appears so much better, but it's pretty much in camera processing. It would be nice if Nikon offered the option to turn it off.


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pwm2
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Jul 26, 2014 10:44 |  #52

Footbag wrote in post #17057170 (external link)
Clipping the black point in itself will not improve the image, though.

I think what may be happening is that any pixels below a certain value are replaced with black pixels or possibly the adjacent pixels are averaged or bias removed.

That kind of bothers me. It would explain why the D800 appears so much better, but it's pretty much in camera processing. It would be nice if Nikon offered the option to turn it off.

Even if there is software processing in camera, the truth is that the D800 apperas so much better because there really is a very distinct difference in what sensor+readout electronics can manage.


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Footbag
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Jul 26, 2014 11:04 |  #53

pwm2 wrote in post #17057358 (external link)
Even if there is software processing in camera, the truth is that the D800 apperas so much better because there really is a very distinct difference in what sensor+readout electronics can manage.

Well, that's the million dollar question... Is it just the sensor and adu converter? Or is something else going on?

Sony had been putting out very high MP, low noise CCD sensors that implies they are improving readout noise. I just wonder if the differing philosophy for photography allows them some room to do some funky things with the data.


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rick_reno
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Jul 26, 2014 11:20 |  #54

I switched to Nikon a couple of years back, but it had nothing to do with all that techno babble about DR, ISO or sensors. That stuff is just incomprehensible to me. I made the change for the cookie Julian gave me, it tasted soooo good. :D




  
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Shadowblade
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Jul 26, 2014 11:25 |  #55

Footbag wrote in post #17057170 (external link)
Clipping the black point in itself will not improve the image, though.

I think what may be happening is that any pixels below a certain value are replaced with black pixels or possibly the adjacent pixels are averaged or bias removed.

That's exactly what setting the black point means. You set a certain level as 'black', and anything below it comes out as black. Set it at the noise threshold and you get the greatest possible contrast with the least noise. Set it lower than that and all you're doing is getting less contrast and more noise, with no added detail.

That kind of bothers me. It would explain why the D800 appears so much better, but it's pretty much in camera processing. It would be nice if Nikon offered the option to turn it off.

You can easily reverse it. Just set the darkest possible value as 2/5/10/whatever in Photoshop. There's no detail distinguishable from noise below the noise threshold anyway - all you'll be doing is losing contrast, without getting any extra detail.




  
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Footbag
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Jul 27, 2014 11:24 |  #56

Shadowblade wrote in post #17057433 (external link)
That's exactly what setting the black point means. You set a certain level as 'black', and anything below it comes out as black. Set it at the noise threshold and you get the greatest possible contrast with the least noise. Set it lower than that and all you're doing is getting less contrast and more noise, with no added detail.

You can easily reverse it. Just set the darkest possible value as 2/5/10/whatever in Photoshop. There's no detail distinguishable from noise below the noise threshold anyway - all you'll be doing is losing contrast, without getting any extra detail.

I understand what clipping the blacks does, I just don't understand why it's "necessary".
As well, clipping the blacks will not allow you to boost the shadows itself. They would have to clip the blacks and add back a truncated low end signal to bring the black point back down. Otherwise, you wouldn't have any contrast in the blacks.


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Jul 27, 2014 12:07 |  #57

Footbag wrote in post #17059211 (external link)
I understand what clipping the blacks does, I just don't understand why it's "necessary".
As well, clipping the blacks will not allow you to boost the shadows itself. They would have to clip the blacks and add back a truncated low end signal to bring the black point back down. Otherwise, you wouldn't have any contrast in the blacks.

When you have a camera with 12 or 14-bit sensor data, and want to create a JPEG image with 8-bit data, you have to define a transfer function - how to convert from your 12 or 14 bits into 8 bits. That's the "curves" adjustments that all photo editing software supports.

So you define a dark value in the raw data and map it to a dark value in the 8-bit JPEG color space.
And you define a light value in the raw data and map it to a light value in the 8-bit JPEG color space.
And you then create a curve which might be straight or might have s-shape or whatever you feel gives the best result. And this curve is used to transform all raw values between the dark and light raw points you specified into values between the dark and light JPEG points you specified.

This process means that a clipping point is defined for black, and another clipping point for white. So like it or not - the JPEG conversion has to perform some form of clipping. You could define the black raw value to be a value below the noise level of the sensor readout. But that isn't to your advantage, since it means that a few of your precious 256 intensity levels in the JPEG file gets wasted to try to represent noise.


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Footbag
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Jul 27, 2014 14:29 |  #58

pwm2 wrote in post #17059305 (external link)
When you have a camera with 12 or 14-bit sensor data, and want to create a JPEG image with 8-bit data, you have to define a transfer function - how to convert from your 12 or 14 bits into 8 bits. That's the "curves" adjustments that all photo editing software supports.

So you define a dark value in the raw data and map it to a dark value in the 8-bit JPEG color space.
And you define a light value in the raw data and map it to a light value in the 8-bit JPEG color space.
And you then create a curve which might be straight or might have s-shape or whatever you feel gives the best result. And this curve is used to transform all raw values between the dark and light raw points you specified into values between the dark and light JPEG points you specified.

This process means that a clipping point is defined for black, and another clipping point for white. So like it or not - the JPEG conversion has to perform some form of clipping. You could define the black raw value to be a value below the noise level of the sensor readout. But that isn't to your advantage, since it means that a few of your precious 256 intensity levels in the JPEG file gets wasted to try to represent noise.

I understand that, but in-camera truncation appears to be what's going on in RAW's. An S-shaped curve would explain a lot of what's going on, but a curve in camera RAW's? It seems like Sony/Nikon is cheating a bit to make it look like the sensor has less noise. I'm fine with it in post, though I don't convert to 8-bit until thew last step.

My camera has 16-bit ADU conversion. :p


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Jul 27, 2014 14:45 |  #59

CRCchemist wrote in post #17052800 (external link)
It's too bad this conversation is in the rumor section of the website. A lot of people ignore the rumors section because 90% of the time it's all wrong. This is good stuff to know.

All that would take would be changing the thread title to something on topic with where the discussion has gone, and to remove the rumored camera body, and maybe a few post deletions.

It may be a while before I get back to this thread, so I'd recommended a PM to remind me if you all decide this should be moved and renamed.


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Jul 27, 2014 16:05 |  #60

Heh! Interesting, I didn't even realize that this had ended up in the "rumors" section, but sure, why not?

But seriously, the discussion seems to be about black point "clipping" and "truncating", but I don't see how that relates to the difference between the Canon technology and the Sony/Nikon technology. I was under the impression that the difference was in the signal-to-noise ratio in lower ISOs -- maybe as "amplified" by the DTA converters, but not to do with "clipping"...am I missing something?


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Chances of 5D MkIV have better dynamic range?
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