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#321
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
In article , Eric Stevens
wrote: I will give it one more try. stop digging yourself a deeper hole. All that the ADC is doing is digitizing the level of charge which has accumulated in the sensor cells. It does not need to know, nor would it care, how much exposure to light that represents. Hook it up to a sensor and a 14-bit ADC will give you a 14-bit datastream which will tell you all about the charge and nothing about the anount of light which has fallen on the individual cells. As far as the ADC is concerned, the important thing is whether or not the maximum voltage of the charges to which it is exposed is within the ADC's input range. If it is the ADC will digitize it. How much light it required does not matter. The ADC will as happily digitize 2 levels of charge as it will 20,000 levels of charge. No matter what it digitizes it will output a 14-bit data stream. THat's why it won't care whether it is digitizing a collection of charges with a dynamic range of 5 or 50. All will be OK as long as the peak is within the range of the ADC's input. still not getting it. |
#322
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
On Sun, 13 Jan 2019 10:05:17 -0500, Alan Browne
wrote: On 2019-01-12 20:18, Eric Stevens wrote: On Sat, 12 Jan 2019 09:26:28 -0500, Alan Browne wrote: On 2019-01-11 18:20, Eric Stevens wrote: On Fri, 11 Jan 2019 12:51:28 -0500, Alan Browne wrote: On 2019-01-11 10:28, nospam wrote: In article , Eric Stevens wrote: The problem is clearly DXO's testing methods. No matter how you look at this, you have to be able to imagine all kinds of sources of inaccurate measurements, especially if they are slight. I have to agree with nospam and Alan. You can't get DR outside of the limits of the ADC because that is the output you see, but you can certainly get test results outside of that limit. But the digital DR of the output of the ADC is not the same as the analog DR of the sensor. Nor is there any reason why it should be. nobody said it was, however, it's always going to be limited by the adc. Got that Eric? What is 'it'? The DR of the sensor or the DR of the output of the ADC? Obvious. The ADC is the limiting factor. Always. There is NO WAY for DxO to probe the sensor directly (and it would be meaningless to everyone even if they could...) The ADC won't be the limiting factor if it has a better dynamic range than the sensor. The limiting factor is most likely the sensor and the engineers select the ADC appropriately. They could have sampled at 16 bits - but that would just be sampling more noise in the bottom 2+ bits. I suppose if you needed an entropy source that's as good as any if you use a couple dozen sensor sites ... But the ADC is digitizing the analog output of one sensor site at a time. If there is noise it will not become apparent until after the digital value of the ADC outputs are collated. I think at this point I should stop and ask you whereabouts in the pipeline the sensor DR range should be measured. As perhaps an extreme example, do you think it should be measured by what is output to a memory card? ... or should it be closer to the sensor? If so, where (and how)? Where delivered to the end user: the stored image file (raw). That is the only thing that counts and is useful to the user of the product. Nothing else matters. But the raw file is not the end product. It has to be decoded by the raw decoder and at that point whatever DR is inherent in the initial image will be restored. And if, as you propose, some compression has occurred between the ADC and the stored value, then the hocus pocus has simply exchanged 1 form of noise for another. No, I'm saying some form of compression occurs in the act of digitizing. To the engineer deciding on what bit depth to use, intimate knowledge about the sensors would have him (in the design/systems engineering phase) do tests on the sensors to determine its statistics and appropriately decide on the bit depth. Agreed. There is no point in using 14 bits on a sensor with a dynamic range of (say) 5. -- Regards, Eric Stevens |
#323
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
In article , Eric Stevens
wrote: But the digital DR of the output of the ADC is not the same as the analog DR of the sensor. Nor is there any reason why it should be. nobody said it was, however, it's always going to be limited by the adc. Got that Eric? What is 'it'? The DR of the sensor or the DR of the output of the ADC? Obvious. The ADC is the limiting factor. Always. There is NO WAY for DxO to probe the sensor directly (and it would be meaningless to everyone even if they could...) The ADC won't be the limiting factor if it has a better dynamic range than the sensor. The limiting factor is most likely the sensor and the engineers select the ADC appropriately. They could have sampled at 16 bits - but that would just be sampling more noise in the bottom 2+ bits. I suppose if you needed an entropy source that's as good as any if you use a couple dozen sensor sites ... But the ADC is digitizing the analog output of one sensor site at a time. If there is noise it will not become apparent until after the digital value of the ADC outputs are collated. false. I think at this point I should stop and ask you whereabouts in the pipeline the sensor DR range should be measured. As perhaps an extreme example, do you think it should be measured by what is output to a memory card? ... or should it be closer to the sensor? If so, where (and how)? Where delivered to the end user: the stored image file (raw). That is the only thing that counts and is useful to the user of the product. Nothing else matters. But the raw file is not the end product. it is as far as the camera is concerned, and more importantly, many photographers. measuring dynamic range from a jpeg is not a good idea (although some review sites have done that). It has to be decoded by the raw decoder and at that point whatever DR is inherent in the initial image will be restored. no. And if, as you propose, some compression has occurred between the ADC and the stored value, then the hocus pocus has simply exchanged 1 form of noise for another. No, I'm saying some form of compression occurs in the act of digitizing. that would be incorrect. To the engineer deciding on what bit depth to use, intimate knowledge about the sensors would have him (in the design/systems engineering phase) do tests on the sensors to determine its statistics and appropriately decide on the bit depth. Agreed. There is no point in using 14 bits on a sensor with a dynamic range of (say) 5. that part is true. |
#324
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
On Sun, 13 Jan 2019 10:06:13 -0500, Alan Browne
wrote: On 2019-01-12 21:15, Eric Stevens wrote: On Sat, 12 Jan 2019 09:25:23 -0500, Alan Browne wrote: On 2019-01-11 18:18, Eric Stevens wrote: The recorded output of the ADC is limited by the capabilities of the ADC. But these have no effect on the capabilities of the sensor. If If there is no way to encode the information, then that is the mootest of moot points. That may well be but, as I have several times said, it is possible to scale the dynamic range of the sensor to fit the narrower dynamic range of the ADC. To which I've replied numberous times. In a nutshell, you're trading one form of noise for another. The 'scaling' is done during the conversion of analog to digital in the ADC and involves no more noise than is inherent in any analog to digital conversion. -- Regards, Eric Stevens |
#325
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
On Sun, 13 Jan 2019 10:13:45 -0500, Alan Browne
wrote: Yep. An interesting point: according to DxO, when the test uses a paper target, some of the noise may actually be the texture of the paper of the target. That's not noise. To claim it's noise absurd. It's part of the image being recorded. Agreed, but DxO appear to suspect that some testers are making that error. -- Regards, Eric Stevens |
#326
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
In article , Eric Stevens
wrote: The recorded output of the ADC is limited by the capabilities of the ADC. But these have no effect on the capabilities of the sensor. If If there is no way to encode the information, then that is the mootest of moot points. That may well be but, as I have several times said, it is possible to scale the dynamic range of the sensor to fit the narrower dynamic range of the ADC. To which I've replied numberous times. In a nutshell, you're trading one form of noise for another. The 'scaling' is done during the conversion of analog to digital in the ADC and involves no more noise than is inherent in any analog to digital conversion. no. it's linear. no scaling. |
#327
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
On Sun, 13 Jan 2019 10:15:15 -0500, Alan Browne
wrote: On 2019-01-12 21:51, Eric Stevens wrote: On Sat, 12 Jan 2019 12:11:32 -0500, Alan Browne wrote: On 2019-01-12 11:22, nospam wrote: In article , Eric Stevens wrote: Now if I said, for example f/11 you would understand it. But that is not an EV. as i said, you don't understand it. here's a hint: what's the difference between f/8 and f/11 ? a) f/3 b) 1 stop c) 1 EV d) a and b. e) b and c. C'mon _E_ric! I'm rooting for you! Try (8*f - 11*f)/88 or (-3*f)/88 The correct reply was e). But thanks for playing. Seriously this time: b) is always correct. c) is also correct if the lens is mounted in a camera with (presumably) a shutter speed and ISO set. But, without a shutter speed and ISO there can be no EV. -- Regards, Eric Stevens |
#328
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
On Sun, 13 Jan 2019 10:25:18 -0500, Alan Browne
wrote: On 2019-01-12 22:39, Eric Stevens wrote: AS you say, they (probably) have to take it from an image file. Although they could tether the camera. Tethering the camera is just getting the image file w/o storing it on the camera's memory card. I'm done. Really. This is pointless. You have no knowledge of electrical engineering from what I can see. I'm not an engineer (but worked for many and many worked for me) and what I and nospam have been pointing out to you is pretty basic systems engineering reality. I may need to produce those diagrams yet. :-( TANSTAAFL. Look that up. It really embodies the entire thing. snipped the rest because it's just blather that doesn't address the issue at all. Icing to disguise a bad cake, really -- Regards, Eric Stevens |
#329
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
In article , Eric Stevens
wrote: AS you say, they (probably) have to take it from an image file. Although they could tether the camera. Tethering the camera is just getting the image file w/o storing it on the camera's memory card. I'm done. Really. This is pointless. You have no knowledge of electrical engineering from what I can see. I'm not an engineer (but worked for many and many worked for me) and what I and nospam have been pointing out to you is pretty basic systems engineering reality. I may need to produce those diagrams yet. :-( for entertainment purposes, by all means do so. for educational purposes, do not. |
#330
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Finally got to the point where no new camera holds my interest (waiting for specific offering)
In article , Eric Stevens
wrote: Now if I said, for example f/11 you would understand it. But that is not an EV. as i said, you don't understand it. here's a hint: what's the difference between f/8 and f/11 ? a) f/3 b) 1 stop c) 1 EV d) a and b. e) b and c. C'mon _E_ric! I'm rooting for you! Try (8*f - 11*f)/88 or (-3*f)/88 The correct reply was e). But thanks for playing. Seriously this time: b) is always correct. c) is also correct if the lens is mounted in a camera with (presumably) a shutter speed and ISO set. But, without a shutter speed and ISO there can be no EV. false. |
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