[falconeye]

First of all, it's good to see we agree that there is trade-off between DR and resolution

We do not agree.

But obviously you have a problem to understand the sentence I wrote above with the invitation to ask questions in case of trouble to understand. So why not ask questions?

I said that DR is a function of spatial frequency. This is totally diferent from saying that there is a trade of between DR and resolution. A trade of suggests a choice to be made. There isn't. A digital sensor records an image at all spatial frequencies up to the limit given by the pixel pitch. Its DR is a function, not a value.

What you believe is like saying to an audio engineer: Please, cut the recording frequency at 1kHz because I don't want it to become noisy.

Furthermore, if you want to build a capture device based on 1-bit pixels you will find that it won't come out this way. Instead of the nice dithering that conspires to a greytone at a downsampled resolution, you end up with whole areas clipped to black or white.

Not true.
If you read my paragraph about the ideal camera above and add a little margin to avoid clipping the noise, you'll see that a sensor with about 10 trillion 1-bit pixels at ISO 100 would just work perfectly.
There are no whole black or whole white or even whole grey areas. Just nicely dithered due to the shot noise. This thought experiment doubles as a great explaination why shot noise is inevitable too: How could the number of photons ever match exactly the number of pixels in an area of the sensor?

Why is it this hard for you to step back and try to think over your proposition?

I think I've now done my best and give up at this point. In case you ask a sincere question, I may well still try to answer it though.

---

Last edited by falconeye; 07-29-2015 at 02:18 PM.

[jsherman999]

1. Correct
2. False, DR is no function of QE

Falk, question on this ^

If sensor efficiency affects read noise and the noise floor and that affects DR, how does sensor efficiency not affect DR?

Or am I exchanging the concepts of QE and 'efficiency' too liberally?

QE seems to play a role in this overview: DR and factors
.

[Ikarus]

We do not agree.

But obviously you have a problem to understand the sentence I wrote above with the invitation to ask questions in case of trouble to understand. So why not ask questions?

I said that DR is a function of spatial frequency. This is totally diferent from saying that there is a trade of between DR and resolution. A trade of suggests a choice to be made. There isn't. A digital sensor records an image at all spatial frequencies up to the limit given by the pixel pitch. Its DR is a function, not a value.

The 'choice' in your laser printer example is the stepping back from the image so that you don't see the individual low DR pixels anymore. That's equivalent to resampling at a lower resolution, and yes, the DR increases by doing this and no, this does not come for free, since the price you pay is a drop in resolution. If you stick to your choice of looking at the picture at the original resolution, there will be no way of missing the fact that the individual pixel has only a 1-bit DR. I really don't understand how one can deny that this is a directly observable property of the individual pixel and has nothing to do with the size of the sensor as a whole, but I guess we'll have to just agree to disagree here.

What you believe is like saying to an audio engineer: Please, cut the recording frequency at 1kHz because I don't want it to become noisy.

Funny you would bring this up, because I know a thing or two about audio. The equivalent in the audio domain is a widely used method for improving the SNR of an A/D converter by oversampling and subsequent downsampling, in other words, trading time resolution for dynamic range. If you're interested, check out the little article here. Feel free to ask questions, if things are unclear.

[Rondec]

The 'choice' in your laser printer example is the stepping back from the image so that you don't see the individual low DR pixels anymore. That's equivalent to resampling at a lower resolution, and yes, the DR increases by doing this and no, this does not come for free, since the price you pay is a drop in resolution. If you stick to your choice of looking at the picture at the original resolution, there will be no way of missing the fact that the individual pixel has only a 1-bit DR. I really don't understand how one can deny that this is a directly observable property of the individual pixel and has nothing to do with the size of the sensor as a whole, but I guess we'll have to just agree to disagree here.

Funny you would bring this up, because I know a thing or two about audio. The equivalent in the audio domain is a widely used method for improving the SNR of an A/D converter by oversampling and subsequent downsampling, in other words, trading time resolution for dynamic range. If you're interested, check out the little article here. Feel free to ask questions, if things are unclear.

It really is a big picture question, though. Which sensor can absorb/capture the most pixels before its photon wells become full? It turns out that the answer is not a sensor with few but large pixels, but one with many more smaller pixels.

For some reason or other, people insist on looking at these things on a per pixel basis, but that isn't particularly useful, because in the end, you will have some goal image size -- however it is viewed and whatever size it is, the D810 will demonstrate better dynamic range than your A7s/D4s type sensor.

[falconeye]

Falk, question on this ^

If sensor efficiency affects read noise and the noise floor and that affects DR, how does sensor efficiency not affect DR?

Or am I exchanging the concepts of QE and 'efficiency' too liberally?

Quantum efficiency (QE) and read noise are independent properties of a sensor. You may combine them and call it sensor efficiency although that term isn't defined.

Read noise does affect DR, QE does not.

Note that read noise can't become smaller than shot noise which is 1 e- for a 1 e- signal. And QE can't become larger than 1 after color discrimination.

QE doesn't affect DR because it cancels out in the max signal to noise ratio. With lower QE, you would simply expose longer to reach the exact same pixel well fill level or signal, i.e., you reach the same DR but the base ISO will be smaller. Just think of an ND filter and how it wouldn't affect DR too. The article you cite does not say otherwise. It just fails to mention how QE cancels out of formulas when computing DR at the lowest possible ISO. Of course, at any fixed, predefined value for ISO, QE does affect DR indeed. But not its absolute maximum value as scored by DxO.


---------- Post added 30-07-15 at 02:04 ----------

The equivalent in the audio domain is a widely used method for improving the SNR of an A/D converter by oversampling and subsequent downsampling, in other words, trading time resolution for dynamic range.

I hope you are aware that you just confirmed what I said so far, i.e., that a higher than required resolution sensor is nothing but the analogon of an oversampling technique. Where nothing is given up or traded.

[SpecialK]

Well, I hoped that wrapped it up for the OP.

[Parallax]

Well, I hoped that wrapped it up for the OP.

The OP hasn't participated in the thread for 2 weeks. I think it was wrapped up for him with his last post.

[normhead]

Is this thread still going? I go away for 10 days... nothing ever changes.

[Parallax]

You were gone?

[normhead]

You were gone?

Now you hurt my feelings...



I was visiting the Flintstones.

[Ratcheteer]

700hp 12-cylinder monster

never use engines that depend on reciprocating mass to make a point about power when a wankel rotary engine can make 174hp per litre with no pumping loss

This entire confusion is probably due to my inability to explain properly.

even if you could explain properly without resorting to math, the self proclaimed engineer Ikarus most likely will never take you seriously on the subject and still find something to disagree on

Originally posted by Ikarus
Furthermore, if you want to build a capture device based on 1-bit pixels you will find that it won't come out this way. Instead of the nice dithering that conspires to a greytone at a downsampled resolution, you end up with whole areas clipped to black or white.

depends on whether or not it has a bayer style filter to give you gradients, say one pixel has a filter that only blocks 11% of the light the other 33% and the next 66%

[Ian Stuart Forsyth]

This entire confusion is probably due to my inability to explain properly.

First, if you read my post #215 again, I explained that even the ideal camera has a finite dynamic range which is a function DR(ISO,f) of both (equivalent) base ISO and spatial frequency f. DR(ISO,f) is both strictly monotonically decreasing with arguments ISO and f. And because it is physics (optics), there is no CMOS or sensor tech involved. No need to even read those papers.

Therefore and of course, a larger pixel (which corresponds to a smaller spatial frequency) has a larger DR. I need not to read the paper because it says the same. If you look at the dynamic range of a pixel, you look at DR(ISO,f=sensorheight/pixelpitch). However, we photographers only care about DR of the image (i.e., at some predetermined f), not DR of a single pixel.

If anybody wonders how sensor tech could influence these findings: not much, provided read noise isn't much larger than shot noise and quantum efficiency isn't much smaller than 1. Both being true now for the best sensors out there.

Therefore, any understanding of dynmic range starts with understanding its property for the ideal camera (*).

This is what I advocated and it is where equivalence (the attempt to really understand it) does actually help.

Therefore, I won't point to further literature of SNR properties of actual pixels on sensors. It would be misleading. Understanding is always more important than finding a reference for a claim.

__
(*) An ideal camera is a camera registering angle and energy of each photon passing its aperture and shutter, with the finite precision allowed by the quantum nature of photons including Heisenberg's uncertainty principle for the complementary variables of angle/aperture-size and energy/shutter-duration. Aperture and shutter may be virtual, i.e., they are just boundary constraints for photon wavefunctions in space and time, resp. The angle/aperture-size uncertainty is known as lens diffraction to photographers. The energy/shutter-duration uncertainty means that at ultra-short exposure times, color becomes uncertain. This is a non-issue for photographers (so far). The quantum nature of photons means images must be noisy with a finite exposure time.

An ideal camera does never clip, it has a native base ISO of 0. In order to define an ideal camera with a finite base ISO, I define an ideal camera with equivalent base ISO 100 to clip after 1.5E12 (1.5 trillion) registered photons. That's roughly the ISO standard as applied by DxO (for a 24x36mm sensor).

Therefore, DR(ISO=100,f=1/0.5) = 0.75 trillion:1 = 237 dB = 39.5 EV. And DR(ISO=100,f=2309) = 17.5 EV. For the ideal camera, using DxO's 8MP value for f when assuming a 3:2 aspect ratio. An ideal camera has no limit for f except for its diffraction limits. f=1/0.5=2 corresponds to an image half white and half black.

Thank you

[Ikarus]

I hope you are aware that you just confirmed what I said so far, i.e., that a higher than required resolution sensor is nothing but the analogon of an oversampling technique.

Had you put it this way I may have agreed, but you also said that the DR of the individual pixel (sample) is irrelevant. In order for the whole oversampling business to work out, though, one relies on the assumption that the DR of the individual pixels covers the DR of the captured scene, but while the larger, higher-DR pixels are still fine, the smaller, lower-DR pixels may already be clipping, in which case highlight information is irretrievably lost.

[falconeye]

while the larger, higher-DR pixels are still fine, the smaller, lower-DR pixels may already be clipping, in which case highlight information is irretrievably lost.

I leave it as an exercise to you, my friend, to find the error in this sentence.

[Ikarus]

I leave it as an exercise to you, my friend, to find the error in this sentence.

I'm sensing confusion, so here's a simple example: large pixel photo site with capacity 4 vs. small pixel sensor area covered by 4 pixels with capacity 1 each (unrealistically assuming a fill factor of 100%). Photons hit small pixel sensor area with distribution 2/1/1/0, pixel #1 clips, result after downsampling to low resolution is 3, large pixel result is 4.