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CFA Strength and Color Discrimination
- Thread starter Jack Hogan
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- nikon d50
Jack Hogan
Veteran Member
Ah, yes, poor wording. How do you see it Erik?
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Iliah Borg
Forum Pro
Real life overall CFA variation (averaged over 60% sensor area, central part) is 6%, slightly less than 0.1 stop; checked on multiple Canon bodies of the same model. It is harder to check on Nikons, but I do not expect much difference, as I measured SMI on different Nikon bodies of the same model (without any lenses), and it is easily reaches ±3 units.So before my crash we had the Gaussian approximation to the Quantal CMF-derived CFA yielding SMI of 97.9 and eQE of 29.7, 32.1 and 16.4% for r,g,b respectively, pretty close to the ideal (peaks are at 596, 557, 446nm throughout this post).
I played a bit with the widths before deciding to just concentrate on the green channel only. Here is what the sequence looks like if the standard deviation of the green channel is varied between 35 and 65nm with the others unchanged:
Only the green filter stDev was varied, the others were left at the original ideal value
Naturally as the standard deviation is increased more photons will make it through and the eQE of the green filter improves. The improvement in eQE from peak SMI (eQE = 32.1%) when SMI drops below 88 * (eQE = 37.1%) is about 0.2 Stops. That's from the green filter only. So if we assume that after some tweaking we would be able to get a similar improvement from the other two filters, we may indeed get an increase in the overall signal of 0.2 stops in the raw data. Here is one attempt at increasing the standard deviation of all three filters:
The improvement here is about 0.3 stops compared to the benchmark, mainly due to the blue channel. One would need to play a lot more with the variables to see what happens but if these are the improvements that we are talking about my feeling is that in these days of clean sensors I would gladly give up 0.3 stops sensitivity for optimum color. In fact, as a landscaper, I really do not care for additional sensitivity
Jack
* I used an SMI of 88 as the threshold because I went back and calculated the SMI of my own Nikon D610 with a matrix fine tuned to yield the best deltaE, not deltaE2000. This is the method relevant to this thread. You guessed it, SMI was 88.
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http://www.libraw.org/
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Jack Hogan
Veteran Member
Joofa you are definitely talking past me. I am talking about Power/Energy Density Functions vs Photon Density Functions. They don't look the same when plotted against wavelength, do they? Photons are quanta, adjective 'quantal', as used by folks that deal with this sort of stuff. You can download quantal data from the learned site I referenced. So what are you going on about?
Both are fine as long as one mentions which they are using. I use quanta because, as mentioned earlier, I look at the processes before and after the photodiode as separate and independent - and I find it easier that way. You may not and that's fine also.
Jack out.
Both are fine as long as one mentions which they are using. I use quanta because, as mentioned earlier, I look at the processes before and after the photodiode as separate and independent - and I find it easier that way. You may not and that's fine also.
Jack out.
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I know what you are talking about. You are presenting some real facts mixed with psuedoscience here. And, I'm trying to make you see through it.
The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Keep in mind the EM spectrum falling on the sensor has not changed at all whichever view point regarding 'energy-based' or 'quanta-based' view you take.
I have been playing with such data for years. How do you think I drew the graph in the message before?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
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Dj Joofa
http://www.djjoofa.com
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Iliah Borg
Forum Pro
Forget CFA for a moment, take a b/w sensor, or a photodiode.
Why log (why quantal is another question)? And log of what? That would not look nice near zero.
Raw data is not probability or proportional to it, as I saw it in another post. The curves you have are the contribution of energy or photon count (depends how you define them) on the RAW values. In that sense, the RAW values are not really proportional to the photons unless you fix their frequency but then there is no much need to discuss this in the first place. If you stick with the same units, you should be fine. I would download the Energy linear data and not divide by the wavelength after that. I believe that this is what Joofa is saying as well although I did not follow the whole discussion carefully.
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Erik Kaffehr
Veteran Member
Hi Jack,
Do we assume that each photon causes a single electron charge or may it cause more than one electron charge depending on energy?
Best regards
Erik
Jack Hogan
Veteran Member
Thank you Erik. In the visual range relevant to unmodified consumer digital cameras we assume 1 e- per interacting photon independent of the energy of the photon. The number of interacting photons divided by the number of incident photons is QE.
Jack
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Jack Hogan
Veteran Member
I don't get your point about the raw data but, as always, I am ready to accept it gladly if and when I get it. For now I don't. If the raw values represent energy then why do 100 incident photons at 400nm produce the same raw count as 100 incident photons at 600nm, absent a CFA and with the same QE?The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
In my procedure I need stuff that relates to quanta, not energy - and since it makes a difference we need to specify that clearly so that everybody knows what they are looking at, hence quantal.
Jack
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Of course you won't see. You are fixated at using the term 'quantal cmf', even when you are doing nothing regarding a photon-based calculation. Not sure what the use of that term will buy you in any case!I don't get your point about the raw data but, as always, I am ready to accept it gladly if and when I get it. For now I don't.The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
I'm sorry you are not doing a photon calculation in your script. That is just your imagination. Kindly look at lines 16-18 of your own code. It is energy calculation pure and simple! Not photon calculation (despite your round-tripping to mult and div by wavelength, which does nothing, except canceling out.)
--
Dj Joofa
http://www.djjoofa.com
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bobn2
Forum Pro
The way I see it, the raw data is a quantum count, regardless of how it's coded. The number representation doesn't make any difference to the reality (though I realise that 'reality' is a problematic term when it comes to QM).I don't get your point about the raw data but, as always, I am ready to accept it gladly if and when I get it. For now I don't. If the raw values represent energy then why do 100 incident photons at 400nm produce the same raw count as 100 incident photons at 600nm, absent a CFA and with the same QE?The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
In my procedure I need stuff that relates to quanta, not energy - and since it makes a difference we need to specify that clearly so that everybody knows what they are looking at, hence quantal.
Jack
Jack Hogan
Veteran Member
Log of the linear Energy data divided by the wavelength and normalized. I don't use the log version, I use the linear version divided by wavelength in my procedure.Why log (why quantal is another question)? And log of what? That would not look nice near zero.
See my reply to Joofa just above.
I agree.
If you don't divide by wavelength then how do you answer this question using the following CFA Spectral Sensitivity Function - the answer to which is a key step in the process I am following in this thread: What percentage of photons at 445nm make it through the following CFA SPD vs 600nm?
Do you have to divide by the wavelength to get the answer? If you do you may find the following CFA SSF representation more helpful and intuitive:
Answer: 41%
Expressing the curve in relation to quanta also makes all effective QE calculations (the main driver of this thread) much easier and intuitive.
Jack
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bobn2
Forum Pro
The photon energies we're dealing with are far from producing a significant 2 e- interaction. The 2 e- interaction is a double step process, the initially released electron has sufficient energy to free a second (sometimes).
--
Tinkety tonk old fruit, & down with the Nazis!
Bob
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Jack Hogan
Veteran Member
Suggest proper terminology then, as you know I am not a scientist. But other than getting bogged down by form, why don't you answer the question of substance that I asked you?
Jack
It is getting tiring for me. I'm not going to respond further if you keep on ignoring stuff. How many times I have to inform you that the calculation you are doing is energy-based. Got your terminology now. OK.
It is not photon-based. Your own script says something else and you say something else. lines 16-18, how many times I have to repeat are just energy-based calculation, notwithstanding your div / mult by wavelength. Which does absolutely nothing except cancel out. But, in your mind it gets you the rights to claim so called 'quantal cmf'.
The answer was right there in the graph that I presented in a response to Jack Hogan earlier. But, he is too occupied to claim (falsely though) a 'quantal cmf' to have noticed the significance of it.The way I see it, the raw data is a quantum count, regardless of how it's coded. The number representation doesn't make any difference to the reality (though I realise that 'reality' is a problematic term when it comes to QM).I don't get your point about the raw data but, as always, I am ready to accept it gladly if and when I get it. For now I don't. If the raw values represent energy then why do 100 incident photons at 400nm produce the same raw count as 100 incident photons at 600nm, absent a CFA and with the same QE?The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
In my procedure I need stuff that relates to quanta, not energy - and since it makes a difference we need to specify that clearly so that everybody knows what they are looking at, hence quantal.
Jack
bobn2
Forum Pro
I can't see an answer in that graph. The reality of the physics of a camera is that the sensor is counting quanta, not measuring energy directly, and the difference can be quite large in extreme cases. You can see this quite easily by thinking of single wavelength light sources. A 400nm photon has an energy of about half an exajoule, a 600nm photon about one third of an exajoule (if I got all the sums right). So, if you illuminate your sensor with the same energy of a narrow band 400 and 600nm light, and look at the blue channel, the DN of the 600nm will be 50% greater than the 40nm.The answer was right there in the graph that I presented in a response to Jack Hogan earlier. But, he is too occupied to claim (falsely though) a 'quantal cmf' to have noticed the significance of it.The way I see it, the raw data is a quantum count, regardless of how it's coded. The number representation doesn't make any difference to the reality (though I realise that 'reality' is a problematic term when it comes to QM).I don't get your point about the raw data but, as always, I am ready to accept it gladly if and when I get it. For now I don't. If the raw values represent energy then why do 100 incident photons at 400nm produce the same raw count as 100 incident photons at 600nm, absent a CFA and with the same QE?The point that is not getting across to you is that photon and energy-based are two valid but different representations derived from the same EM spectrum. Then how come a real digital camera gives us a single raw number? Shouldn't it give two if it can sense a difference between energy-based or photon-based representation? Think about it.
If it selects one of the two, for the sake of the argument say photon-based, then does that mean that the energy-based representation is wrong?
If the other is wrong, then how come we claim that both view points are valid in the first place?
Exposing incorrect assumptions in your thinking. But, we know you have made up your mind on using the term 'quantal CMF'. Nothing can convince you now if history is a guide.
Again, how come a real digital camera gives us the same raw number, when the two representations of EM spectral intensity falling on a sensor are different?
No, for the Nth time, how hard you might try, the 'raw -number' you think you have is not the photon representation. It is the energy representation.
In my procedure I need stuff that relates to quanta, not energy - and since it makes a difference we need to specify that clearly so that everybody knows what they are looking at, hence quantal.
Jack
All of the curves you're arguing about are somewhat hypothetical, since in no case do you actually know the mix of wavelengths that is actually incident of the sensor, and the relationship between energy and photon count depends on that. So far as the physics of the sensor goes, it is measuring charge, which is generated by a quantum effect, so what it is actually measuring is photons, not energy. Concerning the actual graphs, unless you know how they were derived in great detail (that is, what was the illuminant and how was the response measured) the it's really rather hard to say whether they reflect energy or photon count. Since it's very likely that the measuring device also used the photoelectric effect, then the original data on which they are based will be 'quantal', very likely weighted to produce an 'energy' response using much the same kind of fudge factor that Jack is using.
It does not matter. You talking about one step of the whole process only. View the whole as a black box, assuming linearity. Photons at different wavelengths (or continuous spectral distributions) mapped to numbers. In other words, a linear functional of the spectral density. Now, you need to decide whether the spectral density represents energy or photon count distribution and be consistent.
There is no way to answer that question because the RAW data depends on the silicone conversion sensitivity/rate, as well, etc. The relevant question is what is the effect of X photons at a certain wavelength on the RAW value. Or Y energy concentrated there.If you don't divide by wavelength then how do you answer this question using the following CFA Spectral Sensitivity Function - the answer to which is a key step in the process I am following in this thread: What percentage of photons at 445nm make it through the following CFA SPD vs 600nm?
Do you have to divide by the wavelength to get the answer? If you do you may find the following CFA SSF representation more helpful and intuitive:
Answer: 41%
Expressing the curve in relation to quanta also makes all effective QE calculations (the main driver of this thread) much easier and intuitive.
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