Luminous Landscape Forum
Equipment & Techniques => Cameras, Lenses and Shooting gear => Topic started by: wmchauncey on September 02, 2016, 12:32:26 pm
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I do a fair amount of macro work...if image size is of less importance than is acute image sharpness,
would I see a more acutely sharp IQ @ 300% by using a lower MP sensor (older) as opposed to one of the newer 30-50 MP beasts?
Am heavy into PS CC.
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I do a fair amount of macro work...if image size is of less importance than is acute image sharpness,
would I see a more acutely sharp IQ @ 300% by using a lower MP sensor (older) as opposed to one of the newer 30-50 MP beasts?
Am heavy into PS CC.
No.
You just won't be magnifying it as much.
Downsample a 50MP image to the same resolution as an old 12MP camera and the downsized image will be sharper, cleaner and have less aberrations and aliasing artifacts than the low-resolution sensor.
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Hi,
I don't understand your question. In principle a smaller pixel sensor will always give better rendition than a large pixel sensor when seen at the same scale. But, if you stop down to say f/11 or beyond and use extension diffraction will take it's toll. This because the real aperture will be smaller.
On the other hand, Tim Parkin has found that even at f/16 a correctly sharpened image from a Nikon D800 is sharper than an f/8 image from a Sony A900.
https://www.onlandscape.co.uk/2012/08/the-diffraction-limit-how-small-is-too-small/
A good way to measure sharpness is MTF. By and large, the MTF of the system is a product of the MTF of the lens and the MTF of the sensor. The MTF of the lens is affected by diffraction while the MTF of the sensor is not dependent of diffraction and will increase with smaller pixel size.
So, the finer pixels will always have some benefits.
On the other hand, if you take an image with say a 12 MP camera and with a 48 MP camera and look at both images at actual pixels the 12 MP image will look sharper because it is less magnified.
That said using small working apertures there may be little advantage of a high pixel camera.
Best regards
Erik
I do a fair amount of macro work...if image size is of less importance than is acute image sharpness,
would I see a more acutely sharp IQ @ 300% by using a lower MP sensor (older) as opposed to one of the newer 30-50 MP beasts?
Am heavy into PS CC.
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After 100% you are zooming into the sensors dicencted ability, and smaller sensors would be sharper at 300%, but with all other things done optimally to match up the maximum ability of lens, distance, subject, etc.
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I don't understand your question. In principle a smaller pixel sensor will always give better rendition than a large pixel sensor when seen at the same scale. But, if you stop down to say f/11 or beyond and use extension diffraction will take it's toll. This because the real aperture will be smaller.
On the other hand, Tim Parkin has found that even at f/16 a correctly sharpened image from a Nikon D800 is sharper than an f/8 image from a Sony A900.
https://www.onlandscape.co.uk/2012/08/the-diffraction-limit-how-small-is-too-small/
A good way to measure sharpness is MTF. By and large, the MTF of the system is a product of the MTF of the lens and the MTF of the sensor. The MTF of the lens is affected by diffraction while the MTF of the sensor is not dependent of diffraction and will increase with smaller pixel size.
So, the finer pixels will always have some benefits.
On the other hand, if you take an image with say a 12 MP camera and with a 48 MP camera and look at both images at actual pixels the 12 MP image will look sharper because it is less magnified.
That said using small working apertures there may be little advantage of a high pixel camera.
To paraphrase an ex-president, it all depends on what the definition of sharpness is.
The answer to the OP's question depends on the sharpness/acutance metric and the test protocol. For slanted edge MTF testing, the MTF curves are determined by the size and shape of the light-sensitive area on the sensor, and, counter-intuitively, not their density. Acutance derived from the MTF curves will also work that way. So sensors with 4 um square apertures will have the same sharpness in any raw plane if the pitch is 4 um, 8 um, or (gulp) 16 um.
For a constant percentage fill factor, what you say is right even for slanted edge testing, Erik.
Jim
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Hi Jim,
Thanks for making that point. It took me some pondering but now I think I got it into my brain. Need to read up a bit on Jack's fine articles on MTF.
Best regards
Erik
To paraphrase an ex-president, it all depends on what the definition of sharpness is.
The answer to the OP's question depends on the sharpness/acutance metric and the test protocol. For slanted edge MTF testing, the MTF curves are determined by the size and shape of the light-sensitive area on the sensor, and, counter-intuitively, not their density. Acutance derived from the MTF curves will also work that way. So sensors with 4 um square apertures will have the same sharpness in any raw plane if the pitch is 4 um, 8 um, or (gulp) 16 um.
For a constant percentage fill factor, what you say is right even for slanted edge testing, Erik.
Jim
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To paraphrase an ex-president, it all depends on what the definition of sharpness is.
The answer to the OP's question depends on the sharpness/acutance metric and the test protocol. For slanted edge MTF testing, the MTF curves are determined by the size and shape of the light-sensitive area on the sensor, and, counter-intuitively, not their density. Acutance derived from the MTF curves will also work that way. So sensors with 4 um square apertures will have the same sharpness in any raw plane if the pitch is 4 um, 8 um, or (gulp) 16 um.
Hi Jim,
And to add, not only the shape of the light sensitive area on the sensors, but also microlenses (gapless or not) and their radial offset amount to better accommodate oblique ray incidence, especially for short lens flange distances (i.e. more likely with mirrorless camera bodies).
And as always, denser sampling will positively affect absolute limiting resolution (e.g. in cycles/mm), and it will boost the MTF response at lower spatial frequencies, for a given lens.
Cheers,
Bart
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Hi,
I think the pixel aperture sort of includes the effect of the micro lenses.
Increasing the effective sampling area decreases sharpness but reduces aliasing, to my understanding.
Best regards
Erik
Hi Jim,
And to add, not only the shape of the light sensitive area on the sensors, but also microlenses (gapless or not) and their radial offset amount to better accommodate oblique ray incidence, especially for short lens flange distances (i.e. more likely with mirrorless camera bodies).
And as always, denser sampling will positively affect absolute limiting resolution (e.g. in cycles/mm), and it will boost the MTF response at lower spatial frequencies, for a given lens.
Cheers,
Bart
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Hi,
I think the pixel aperture sort of includes the effect of the micro lenses.
Hi Erik,
In that case I'd rather say microlens + photosite aperture, because both are distinct features in image sensor design.
Increasing the effective sampling area decreases sharpness but reduces aliasing, to my understanding.
Yes, that's the result (and with sharpness defined as MTF response). The relatively (optically and physically) larger sampling area also averages more instead of being more of a point sample. So microdetail gets lower contrast, which reduces the limiting resolution of the system, and it reduces the MTF response at higher spatial frequencies than Nyquist, which reduces aliasing. That is also sometimes referred to as a 'better' behaved roll-off of the MTF curve towards the Nyquist frequency.
Cheers,
Bart
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And as always, denser sampling will positively affect absolute limiting resolution (e.g. in cycles/mm), and it will boost the MTF response at lower spatial frequencies, for a given lens.
I don't think that's true for slanted edge testing. Dropping every other column sample and every row sample (and making the region of interest twice as long to pick up more samples if that's necessary) should produce the same MTF curve, whether the horizontal axis is cy/px or cy/mm. If you don't tell the software that you've dropped those samples, cy/ph will be unaffected as well.
The slanted edge method forms a averaged, interpolated edge and examines the characteristics of that edge, but knows nothing about the sampling density (or assumes that it's 100%, if you want to look at it that way).
I know this is completely unintuitive, but think about dropping samples out of, say, a 5um grid and you'll see what I'm getting at. Of course, you need enough samples for the slanted edge method to achieve the desired accuracy. You should be able to drop out samples whose location is random with the same effect.
Jim
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I don't think that's true for slanted edge testing.
Hi Jim,
I was thinking along the lines of this blog post (http://blog.kasson.com/?p=5920) of yours. Which shows the positive effects of denser sampling at different apertures (levels of diffraction) in the plane of best focus.
When I plot a similar lens scenario for 2 different sampling densities I get the result in the attached graph. The shape is similar (and same amplitude at Nyquist), but the Nyquist frequency and limiting resolution, and the amplitudes (at same cy/mm frequencies), are different.
Maybe we're talking about different things though.
Dropping every other column sample and every row sample (and making the region of interest twice as long to pick up more samples if that's necessary) should produce the same MTF curve, whether the horizontal axis is cy/px or cy/mm. If you don't tell the software that you've dropped those samples, cy/ph will be unaffected as well.
I don't think so, because you've swapped 'area like' samples for sparsely sampled 'point like' (quarter area) samples of the edge. Precision is affected, like in a Nearest Neighbor downsample.
There is one other potentially relevant issue (although I think this is something different from what you mention) that is addressed in the relevant ISO Standards, but that I almost never see mentioned in e.g. the posts at DPReview by Jack or Frans. Maybe because MTF mapper (using 10 bins per pixel) as a tool already takes care of it, dunno. The ISO states that one should only bin (they use 4 bins, 1/4th of a pixel resolution) "full phase rotations" if I recall their phrasing correctly. Meaning that if e.g. the slope of the slant is ArcTan(1/10) degrees, one should use a multiple of 10 edge pixels for the SFR evaluation. So a cropped Region of Interest (ROI) should have multiples of 10 pixels in dimension in the length. The data is also filtered with (if I recall correctly) a Hamming filter to avoid ripple effects.
I'll try to find a copy that I used to have of the Standards (one for digital camera resolution, and two for scanner resolution) that I used to have, and see if relevant parts can be shared without copyright issues.
So IMHO, dropping rows and columns could disrupt that slightly more robust averaging process at particular slant angles. It may amount to a small difference when many more sparse samples are used (like a Monte Carlo simulation), but still.
The slanted edge method forms a averaged, interpolated edge and examines the characteristics of that edge, but knows nothing about the sampling density (or assumes that it's 100%, if you want to look at it that way).
I agree, but it will influence the sampling accuracy of the interpolation, especially if there is some lens distortion involved, and we need to include the sampling density in cy/mm to see the real differences. The Cy/px metric will show simila shapes, but the spatial frequencies of the projected image are at different detail sizes.
I know this is completely unintuitive, but think about dropping samples out of, say, a 5um grid and you'll see what I'm getting at. Of course, you need enough samples for the slanted edge method to achieve the desired accuracy. You should be able to drop out samples whose location is random with the same effect.
No, not unituitive, but with issues for precision, and no accurate representation of physical resolution (unless recalibrated for the different sampling density).
I do understand the similar curve shape despite the sampling density, but it represents different levels of detail (cy/mm), and thus different amplitudes (as the attached graph shows).
Cheers,
Bart
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I was thinking along the lines of this blog post (http://blog.kasson.com/?p=5920) of yours. Which shows the positive effects of denser sampling at different apertures (levels of diffraction) in the plane of best focus.
Well, that's one of my favorite bloggers, so I can't dis him too badly. But those simulations all held the fill factor constant, so the finer pitches got smaller active areas.
When I plot a similar lens scenario for 2 different sampling densities I get the result in the attached graph. The shape is similar (and same amplitude at Nyquist), but the Nyquist frequency and limiting resolution, and the amplitudes (at same cy/mm frequencies), are different.
Did you hold the aperture area constant? That would mean that, if a pitch of 4um had a fill factor of 100%, a pitch of 8um would have a fill factor of 25%.
Maybe we're talking about different things though.
I don't think so, because you've swapped 'area like' samples for sparsely sampled 'point like' (quarter area) samples of the edge. Precision is affected, like in a Nearest Neighbor downsample.
There is one other potentially relevant issue (although I think this is something different from what you mention) that is addressed in the relevant ISO Standards, but that I almost never see mentioned in e.g. the posts at DPReview by Jack or Frans. Maybe because MTF mapper (using 10 bins per pixel) as a tool already takes care of it, dunno. The ISO states that one should only bin (they use 4 bins, 1/4th of a pixel resolution) "full phase rotations" if I recall their phrasing correctly. Meaning that if e.g. the slope of the slant is ArcTan(1/10) degrees, one should use a multiple of 10 edge pixels for the SFR evaluation. So a cropped Region of Interest (ROI) should have multiples of 10 pixels in dimension in the length. The data is also filtered with (if I recall correctly) a Hamming filter to avoid ripple effects.
I'll try to find a copy that I used to have of the Standards (one for digital camera resolution, and two for scanner resolution) that I used to have, and see if relevant parts can be shared without copyright issues.
So IMHO, dropping rows and columns could disrupt that slightly more robust averaging process at particular slant angles. It may amount to a small difference when many more sparse samples are used (like a Monte Carlo simulation), but still.
I agree, but it will influence the sampling accuracy of the interpolation, especially if there is some lens distortion involved, and we need to include the sampling density in cy/mm to see the real differences. The Cy/px metric will show simila shapes, but the spatial frequencies of the projected image are at different detail sizes.
No, not unituitive, but with issues for precision, and no accurate representation of physical resolution (unless recalibrated for the different sampling density).
I do understand the similar curve shape despite the sampling density, but it represents different levels of detail (cy/mm), and thus different amplitudes (as the attached graph shows).
I do agree that fewer samples adversely affects the accuracy, but in my original post, I suggested making the edge longer to compensate. In practice, I've found that edges a couple of hundred pixels long are pretty accurate, in that adding more pixels to the edge doesn't reduce the noise much.
We can continue this discussion after I see your answers to the questions up above. I have the feeling that we're not talking about quite the same thing. Every time I have a conversation with you, I learn something, so it's possible that my understanding is flawed in some way.
Jim
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Well, that's one of my favorite bloggers, so I can't dis him too badly. But those simulations all held the fill factor constant, so the finer pitches got smaller active areas.
Right, as in actual practice.
Did you hold the aperture area constant? That would mean that, if a pitch of 4um had a fill factor of 100%, a pitch of 8um would have a fill factor of 25%.
No, the fill factor was assumed to be 100% for the respective photosite areas, which have different pitch and area.
I do agree that fewer samples adversely affects the accuracy, but in my original post, I suggested making the edge longer to compensate.
Yes, that should work, if it were not for lens distortion. The longer the edge segment (= more samples), the more distortion from a straight edge we'll have. It depends on the used algorithms if and how much that weighs in on the total score. Imatest apparently approximates the curvature of the edge, and uses that to compensate (so Norman Koren seems to have found it significant enough to address). I'm not sure what MTF Mapper does.
In practice, I've found that edges a couple of hundred pixels long are pretty accurate, in that adding more pixels to the edge doesn't reduce the noise much.
Yes, same for me. It's a balance between short edge to reduce distortion effects (if not compensated for in algorithms), and long edge for more robust statistics.
We can continue this discussion after I see your answers to the questions up above. I have the feeling that we're not talking about quite the same thing. Every time I have a conversation with you, I learn something, so it's possible that my understanding is flawed in some way.
At least there seems to be a possible difference in the use of the fill factor. I use an assumed 100% fill factor (theoretical perfect micro-lenses) for specific camera sensel pitches (as far as we know the exact pitch, but close enough is close enough). Your sensor simulation model is more accurate than what I used, so that should be reliable enough if real life parameters are used.
Cheers,
Bart
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Hi,
Learning a lot from this discussion!
:-) Erik :-)
Right, as in actual practice.
No, the fill factor was assumed to be 100% for the respective photosite areas, which have different pitch and area.
Yes, that should work, if it were not for lens distortion. The longer the edge segment (= more samples), the more distortion from a straight edge we'll have. It depends on the used algorithms if and how much that weighs in on the total score. Imatest apparently approximates the curvature of the edge, and uses that to compensate (so Norman Koren seems to have found it significant enough to address). I'm not sure what MTF Mapper does.
Yes, same for me. It's a balance between short edge to reduce distortion effects (if not compensated for in algorithms), and long edge for more robust statistics.
At least there seems to be a possible difference in the use of the fill factor. I use an assumed 100% fill factor (theoretical perfect micro-lenses) for specific camera sensel pitches (as far as we know the exact pitch, but close enough is close enough). Your sensor simulation model is more accurate than what I used, so that should be reliable enough if real life parameters are used.
Cheers,
Bart
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No, the fill factor was assumed to be 100% for the respective photosite areas, which have different pitch and area.
Then we are talking at cross-purposes. The point I was making was that the MTF curves calculated from slanted edge testing are, with sufficient sampling of the edge, primarily a function of the light-sensitive apertures or the sensor, not of the pixel pitch per se.
Jim
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Then we are talking at cross-purposes. The point I was making was that the MTF curves calculated from slanted edge testing are, with sufficient sampling of the edge, primarily a function of the light-sensitive apertures or the sensor, not of the pixel pitch per se.
Hi Jim,
Maybe it would be helpful to pick a few specific scenarios?
1. A regular sensor with a certain pitch and aperture (100% fill factor).
2. A sensor with half pitch shift capability? Sample area per capture is the same, offset is half pitch. This results in overlapping samples with a relatively 'large' aperture for the pitch, although the same aperture as in case 1.
3. A sensor with half the pitch of the earlier single exposures. Sample area is 1/4th of the prior examples.
As long as there is a lens involved (OTF), the combined (OTF * sensor MTF =) system MTF will be different in all three cases. The better the lens is, the more the sensor (as lowest common factor) will determine the system resolution.
I'm open to other scenarios if you prefer.
Cheers,
Bart
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Hi Jim,
Maybe it would be helpful to pick a few specific scenarios?
1. A regular sensor with a certain pitch and aperture (100% fill factor).
2. A sensor with half pitch shift capability? Sample area per capture is the same, offset is half pitch. This results in overlapping samples with a relatively 'large' aperture for the pitch, although the same aperture as in case 1.
3. A sensor with half the pitch of the earlier single exposures. Sample area is 1/4th of the prior examples.
As long as there is a lens involved (OTF), the combined (OTF * sensor MTF =) system MTF will be different in all three cases. The better the lens is, the more the sensor (as lowest common factor) will determine the system resolution.
I'm open to other scenarios if you prefer.
Two that I suggested before are a 4um pitch with 100% fill factor and 8um pitch with 25% rectangular fill factor.
When I do MTF calculations, I usually look at one raw color plane at a time.
Jim
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Hi,
My take is that the the pixel aperture is the PSF of the sensor. If fill factor is low, the PSF of the sensor will be small but will sample irregular or fake info if information in the sampled image contains excess detail.
Let's assume a telephone wire hanging into the image filling one third of the pixel vertically. A large aperture pixel will report 0.33 as the wire blocks around 1/3 of the light.
A small aperture pixel will report 1.0, 0.0 or something in between. Not even kitchen physics, I know!
Best regards
Erik
Hi Jim,
Maybe it would be helpful to pick a few specific scenarios?
1. A regular sensor with a certain pitch and aperture (100% fill factor).
2. A sensor with half pitch shift capability? Sample area per capture is the same, offset is half pitch. This results in overlapping samples with a relatively 'large' aperture for the pitch, although the same aperture as in case 1.
3. A sensor with half the pitch of the earlier single exposures. Sample area is 1/4th of the prior examples.
As long as there is a lens involved (OTF), the combined (OTF * sensor MTF =) system MTF will be different in all three cases. The better the lens is, the more the sensor (as lowest common factor) will determine the system resolution.
I'm open to other scenarios if you prefer.
Cheers,
Bart
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1. A regular sensor with a certain pitch and aperture (100% fill factor).
2. A sensor with half pitch shift capability? Sample area per capture is the same, offset is half pitch. This results in overlapping samples with a relatively 'large' aperture for the pitch, although the same aperture as in case 1.
3. A sensor with half the pitch of the earlier single exposures. Sample area is 1/4th of the prior examples.
I'm open to other scenarios if you prefer.
You left out the one that Jim has been talking about Bart: same pixel pitch and aperture in both scenarios but in the second case only every other pixel is used for MTF estimation. Other than noise, the slanted edge method produces the same result :)
Jack
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You left out the one that Jim has been talking about Bart: same pixel pitch and aperture in both scenarios but in the second case only every other pixel is used for MTF estimation. Other than noise, the slanted edge method produces the same result :)
Hi Jack,
Besides me having a hard time seeing that as a common camera option, wouldn't sampling at every other pitch position reduce the resolution by half? That may not change the shape of the MTF but it is at a completely different scale, so the limiting resolution (or Nyquist) and MTF response for given feature sizes in cy/mm would change.
I'm not sure what's being overlooked.
Cheers,
Bart
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Hi Jack,
Besides me having a hard time seeing that as a common camera option, wouldn't sampling at every other pitch position reduce the resolution by half? That may not change the shape of the MTF but it is at a completely different scale, so the limiting resolution (or Nyquist) and MTF response for given feature sizes in cy/mm would change.
I'm not sure what's being overlooked.
Hey Bart,
What is being overlooked is that the slanted edge method produces a radial slice of the 2D MTF of the imaging system as a whole by first effectively projecting the 2D edge onto the 1D edge normal (ESF, its derivative being the LSF). The LSF is therefore a 1D super-resolution projection of the 2D PSF.
The fact that it is super-sampled means that, within limits dictated by acceptable noise in the resulting curves, pitch more or less drops out of the equation, as Jim mentioned. Incidentally, sampling aperture doesn't. So when we show MTF in c/p or calculate Nyquist, we do so based on a priori knowledge of the layout of the sensor (e.g. a grid of known pitch).
Jack
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Maybe because MTF mapper (using 10 bins per pixel) as a tool already takes care of it, dunno. The ISO states that one should only bin (they use 4 bins, 1/4th of a pixel resolution) "full phase rotations" if I recall their phrasing correctly. Meaning that if e.g. the slope of the slant is ArcTan(1/10) degrees, one should use a multiple of 10 edge pixels for the SFR evaluation. So a cropped Region of Interest (ROI) should have multiples of 10 pixels in dimension in the length. The data is also filtered with (if I recall correctly) a Hamming filter to avoid ripple effects.
Hi Bart,
Frans is the better person to ask, but MTF Mapper actually bins supersampled ESF data into bins 1/8th of a pixel wide. The key is making sure that, for the given edge length and angle, there are enough samples to fill the bins. MTF Mapper used to use a hamming window but has since evolved to better methods, including importance sampling and other neat improvements (see his blog for details). With its recent releases I see cleaner curves than the ISO standard, sfrmat3 and probably (http://mtfmapper.blogspot.it/2016/04/mtf-mapper-vs-imatest-vs-quick-mtf.html) (because I do not use it) Imatest.
Jack
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Besides me having a hard time seeing that as a common camera option, wouldn't sampling at every other pitch position reduce the resolution by half? That may not change the shape of the MTF but it is at a completely different scale, so the limiting resolution (or Nyquist) and MTF response for given feature sizes in cy/mm would change.
I'm not sure what's being overlooked.
I think Jack is doing a better job of making my point than I was, but I do want to chime in with one explanation.
I initially introduced dropping samples as a thought experiment; a way to get to the comparison of the 4 um 100% FF and the 8 um 25% FF comparison with a single 4 um 100% FF sensor and a little post processing.
I now fear that that thought experiment has confused the issue. So let's go back to the 4um 100%/8um 25% comparison.
If the resolution of the binning of the slanted edge pixels is fixed at, say, 1 um (quarter pixel in the 4 um case and eighth pixel in the 8 um case), and the population of the 8 um sensor image is enough to average out the noise sufficiently, shouldn't the statistics of the binned edge be the same? I realize that the 4 um case will have less noise, since there are 4 times the number of samples.
Jim
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let's go back to the 4um 100%/8um 25% comparison.
If the resolution of the binning of the slanted edge pixels is fixed at, say, 1 um (quarter pixel in the 4 um case and eighth pixel in the 8 um case), and the population of the 8 um sensor image is enough to average out the noise sufficiently, shouldn't the statistics of the binned edge be the same? I realize that the 4 um case will have less noise, since there are 4 times the number of samples.
Hi Jim,
I think the easiest way to think of this kind of question is to rely on the model (http://www.strollswithmydog.com/resolution-model-digital-cameras-i/) for guidance. Ignoring phase (reasonable in your ideal example) the system MTF can be modeled as the product of the MTFs due to diffraction (a known function of N and lambda) and due to pixel aperture (a function of pixel width and shape) convolved with the sampling grid (typically a comb, function of sampling interval):
MTFsys = MTFdiff x MTFpxAp ** COMBsamp
MTFdiff we know and it is what it is for the given setup;
MTFpxAp varies with the size and shape of the pixel, in your example size would be 4um in one case and 8um in the other; and
COMBsamp is not pixel pitch in the slanted edge method but the pitch of sampling-bin spacing (1/4th and 1/8th of a pixel in your example). Either way it is immaterial in the interval of interest to us (o-1 c/p) because it pushes images of the system MTF out to 4 and 8 c/p respectively.
Pixel pitch is worked back into MTF when we specify the frequency axis. Here (http://www.strollswithmydog.com/units-of-discrete-fourier-transform/) is a clumsy attempt at explaining how this is done.
Jack
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Hi Jim,
I think the easiest way to think of this kind of question is to rely on the model (http://www.strollswithmydog.com/resolution-model-digital-cameras-i/) for guidance. Ignoring phase (reasonable in your ideal example) the system MTF can be modeled as the product of the MTFs due to diffraction (a known function of N and lambda) and due to pixel aperture (a function of pixel width and shape) convolved with the sampling grid (typically a comb, function of sampling interval):
MTFsys = MTFdiff x MTFpxAp ** COMBsamp
MTFdiff we know and it is what it is for the given setup;
MTFpxAp varies with the size and shape of the pixel, in your example size would be 4um in one case and 8um in the other; and
COMBsamp is not pixel pitch in the slanted edge method but the pitch of sampling-bin spacing (1/4th and 1/8th of a pixel in your example). Either way it is immaterial in the interval of interest to us (o-1 c/p) because it pushes images of the system MTF out to 4 and 8 c/p respectively.
Pixel pitch is worked back into MTF when we specify the frequency axis. Here (http://www.strollswithmydog.com/units-of-discrete-fourier-transform/) is a clumsy attempt at explaining how this is done.
Makes sense to me. Let's see if it does to Bart.
Jim
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Hi Bart,
Frans is the better person to ask, but MTF Mapper actually bins supersampled ESF data into bins 1/8th of a pixel wide.
Thanks for the correction, I knew it was more than Imatest (and the ISO standard) uses, so apparently 8 bins are used.
The key is making sure that, for the given edge length and angle, there are enough samples to fill the bins.
Yes, although 8 bins will need twice as many samples to populate to a similar statistical stability as 4 bins do. And longer edges means increased sensitivity for lens distortion. So I assume that Frans has considered that.
MTF Mapper used to use a hamming window but has since evolved to better methods, including importance sampling and other neat improvements (see his blog for details). With its recent releases I see cleaner curves than the ISO standard, sfrmat3 and probably (http://mtfmapper.blogspot.it/2016/04/mtf-mapper-vs-imatest-vs-quick-mtf.html) (because I do not use it) Imatest.
From what I've read, MTF Mapper's algorithms produce quite good results. I just don't know all the implementation details, so I was wondering if it could be even a hair better.
Cheers,
Bart
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P.S. This is what I could find after a short perusal in my files about what the ISO describes (short fragments from a draft version of ISO 16067-2) about the SFR function:
The selected region is converted from digital code values to an edge spread image of normalised photopic intensities via the OECF and colour weighting coefficients a, b, & c.
<<formula>>
Each row of the edge spread image is an estimate of the camera edge spread function (ESF). Each of these ESFs is differentiated to form its discrete line spread function (LSF). The position of the centroid (C) of each of R LSFs is determined along the continuous variable x.
<<formula>>
The slope of the best fit relating the x positions of the centroids to the r index of each row is computed.
<<formula>>
This slope m, is used to compute a shift S(r) to be applied to each row to bring each ESF to coincidence around a common origin at x=0. It effectively takes out the tilt out of the edge.
<<formula>>
The slope is also used to truncate the number of rows of data to the largest number R' that will have an integer number of full phase rotations. For example, if the fit to the centroid moves 0.1 pixels per row, then the largestmultiple of ten rows that is less than R will be used. A check is made that there is at least one full phase rotation.
The next step is the super-sampling and averaging. This step performs a composite requantized edge spread function (ESF) over the discrete variable j, where j is four times more finely sampled than p but is not a continuous variable like x. The super sampling factor is 4, so N=4X bins are created, each with width 0.25 pixels.
... followed by more formulas and conversions between cycles/pixel and cycles/mm.
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Yes.