Help me understand resolving limits

[mtomalty]

Looking for a brief summary of what to expect here.

Say lens A is at its resolving limit on a 22 Mp sensor.
Lens A is then used on a 40 Mp sensor of same dimension but with smaller capture pixels.

How,then, is the lack of resolving ability presented visually?
Does it only resolve up to the look of its 22 Mp 'experience' or is it actually visually worse
because of its inability to resolve with the smaller capture pixels of the 40 Mp sensor ?

Thanks
Mark

[ondebanks]

Mark, it would be the former case. Net image quality would be about the same in both cases. There will be a slight advantage to the 40MP though, not in terms of pure resolution but in terms of less de-Bayering artefacts & moire, and the potential for deconvolution processing with a better sampled PSF.

Ray

[mtomalty]

Thanks Ray


Mark

[ErikKaffehr]

Hi!

I agree with Ray, except that I would say that it is not so simple that a lens resolves 22 MP, it's more like diminishing returns. The finer the details the more the image deteriorates.
Ideally the lens would resolve like the sensor and past that point the lens would not resolve any detail. The detail that the lens resolves past sensor resolution is causing the aliasing artifacts, with Moiré being the best known such artifact.

Best regards
Erik

Mark, it would be the former case. Net image quality would be about the same in both cases. There will be a slight advantage to the 40MP though, not in terms of pure resolution but in terms of less de-Bayering artefacts & moire, and the potential for deconvolution processing with a better sampled PSF.

Ray

[theguywitha645d]

+1

Resolving power is additive. Each component of the system adds to the final image. There are different ways to look at it, but one of the simplest is adding the reciprocal of the resolving power. So a lens that resolves 100 l/mm and a sensor that with a resolving power of 100 l/mm has a systemic resolution of 50 l/mm--1/100 + 1/100 = 2/100 or 1/50. So if you then use a sensor with a resolving power of 200 l/mm with the same lens, you have a systemic resolving power of 66 l/mm-- 1/100 + 1/200 = 3/200 or 1/66.6. So you can see that a higher rez sensor will have a positive impact on the final image.

And if the 20MP image produces a sharp image under specific viewing distances, the 40MP one will too.

And I am sure folks like Erik and Ray can give you some more sophisticated models of systemic resolution or MTF.

[mtomalty]

[quote ]
And I am sure folks like Erik and Ray can give you some more sophisticated models of systemic resolution or MTF.
[/quote]

I can't handle the truth !!

I'm better suited to explanations like,  "Man, that's sharp"  or  "Seems sorta fuzzy"   

[ondebanks]

+1

Resolving power is additive. Each component of the system adds to the final image. There are different ways to look at it, but one of the simplest is adding the reciprocal of the resolving power. So a lens that resolves 100 l/mm and a sensor that with a resolving power of 100 l/mm has a systemic resolution of 50 l/mm--1/100 + 1/100 = 2/100 or 1/50. So if you then use a sensor with a resolving power of 200 l/mm with the same lens, you have a systemic resolving power of 66 l/mm-- 1/100 + 1/200 = 3/200 or 1/66.6. So you can see that a higher rez sensor will have a positive impact on the final image.

And if the 20MP image produces a sharp image under specific viewing distances, the 40MP one will too.

And I am sure folks like Erik and Ray can give you some more sophisticated models of systemic resolution or MTF.

"TheGuy" (I keep forgetting your first name...Mark, isn't it?) -

The formula you gave works well with lens + film, because film itself, like a lens, has a "bell-shaped" impulse response function/point spread function - if you could focus an infinitely narrow point of light onto film, it would record as a small rounded blob, bright in the centre and tapering off radially. The combination of lens PSF and film PSF is given by their mathematical convolution - you basically get a broader bell-curve than either the film alone or the lens alone. The 1/film + 1/lens = 1/system is a reasonable approximation to the width/resolution of this curve. Contrast at all middle and high spatial frequencies is poorer, because all parts of the system bell-curve are broader than either the film alone or the lens alone.

It's a bit different with digital. The response function of a sensor is not a sloping bell-curve, but a single rectangular peak, or "top-hat", whose width is the width of the pixel. When this is convolved with the lens PSF bell-curve, what you get is the sampled lens PSF. Unlike with film, this system response is not broader than the (analog) lens PSF [unless the pixels are too large in comparison to the lens PSF - undersampling]. The system resolution and contrast is therefore determined by the lens alone - with one exception: the highest spatial frequencies, greater than the sampling frequency. The top-hat sensor response takes over at these frequencies, and as it cannot resolve them (there is zero contrast for any detail falling _within_ a pixel), there is a sudden hard limit to the resolution.

If you want to get the most from your lenses on a digital sensor, oversampling the PSF delays the onset of this hard limit. In other words, small pixels are good in the sense that they can render finer detail more like film does, gradually tapering off in contrast out to the limit of the lens, rather than the limit of the sensor.

I sketched this out and scanned it - I hope that the attachment makes the concepts and distinction between film and digital clearer. (My digital system curve sketch is pretty crappy - it should be a much closer match to the lens curve on the left - but I hope you get the idea regardless that the shape/width is the same, Z ~ X)

Ray

[hjulenissen]

I sketched this out and scanned it - I hope that the attachment makes the concepts and distinction between film and digital clearer. (My digital system curve sketch is pretty crappy - it should be a much closer match to the lens curve on the left - but I hope you get the idea regardless that the shape/width is the same, Z ~ X)

The lower figure is only correct when pixels are small relative to the lense PSF, right? If you slap a state-of-the-art Zeiss lense onto a 6MP EOS 300D, I would assume that z was dominated by pixel pitch (y), not lense PSF(x)?

I think this can be understood easily in the frequency domain as well. If you connect two audio equalizers in series, and set one to attenuate 6dB from 12kHz upwards, and the other to attenuate 4 dB from 10kHz upwards, what is the total response? It is hard to say because the two parameters are not enough to describe the filters, but it seems safe to assume that we will have more high-frequency attenuation than either equaliser in isolation.

-h

[ondebanks]

The lower figure is only correct when pixels are small relative to the lense PSF, right? If you slap a state-of-the-art Zeiss lense onto a 6MP EOS 300D, I would assume that z was dominated by pixel pitch (y), not lense PSF(x)?

Yes, that is the undersampling case. Even more so with my 13-micron 2MP DCS-720x! 

I think this can be understood easily in the frequency domain as well. If you connect two audio equalizers in series, and set one to attenuate 6dB from 12kHz upwards, and the other to attenuate 4 dB from 10kHz upwards, what is the total response? It is hard to say because the two parameters are not enough to describe the filters, but it seems safe to assume that we will have more high-frequency attenuation than either equaliser in isolation.

Yes, that's right too. Decibels are a log scale, so your audio example would mean that you add the dB attenuation to get the total attenuation at 12 KHz. It's not just 6+4=10dB, because the 4dB at 10KHz will become a larger attenuation at 12 KHz, and how much larger depends on the filter order (6 dB per decade, 12 dB or whatever). But given the full filter specs, one could work out and sum the dB attenuations at all frequencies.

Optical colour filters work in exactly the same way - they attenuate certain frequencies, of light. The net response of two optical filters in series (one stacked in front of the other) is the product of their individual spectral transmission curves. Multiply linear units, add log units.

Ray

[PierreVandevenne]

in series (one stacked in front of the other) is the product of their individual spectral transmission curves.

I've always found that depressing. Think about the number of photons wasted by 15 lenses zoom designs... 

[BJL]

I've always found that depressing. Think about the number of photons wasted by 15 lenses zoom designs... 

This multiplicative idea does not apply to compound lenses, where the aberration effects of successive elements are "correlated", or better, "anti-correlated": the extra lens elements are there in part to correct for aberrations and improve MTF at high spatial frequencies.

[PierreVandevenne]

It definitely does.

If the first piece of glass has 98% transmission and the next one 97% transmission, you'll get 95% behind them. A zoom with 14 pieces of glass, each with 98% transmission (typical figures for fluorite glass is 99%, but it is becoming rarer and rarer today given the toxicity issues) will lose about 25% of the photons on the way. This can be alleviated to some extent but the best three components apos rarely pass more than 97% of the incoming light.

[ondebanks]

This multiplicative idea does not apply to compound lenses, where the aberration effects of successive elements are "correlated", or better, "anti-correlated": the extra lens elements are there in part to correct for aberrations and improve MTF at high spatial frequencies.

Yeah, you seem to be confusing throughput (electromagnetic frequencies) with correction (spatial frequencies). A multicoated singlet lens has brilliant 98% throughput, but appalling correction!

Ray

[BJL]

If the first piece of glass has 98% transmission ...

Sorry; the topic of this thread is resolution, and the way the resolution limits (MTF curves) combine: see the subject line. So I was replying thinking that you were talking about that, not transmissivity (as indicated by the difference between F-stops and T-stops with cinema lenses).

[PierreVandevenne]

My apologies as well: I shouldn't have jumped in with a related but not identical issue. Let me take the blame for the confusion.

[theguywitha645d]

"TheGuy" (I keep forgetting your first name...Mark, isn't it?) -
Ray

Actually, my first name is Will, but Mark is close enough.

Thanks very much, that is really helpful. I really appreciate you taking the time.

[ondebanks]

Will! Sorry, Will.

I should have guessed Eric/Erik - I'd have a 50% chance of being right on this forum 

Ray

[mtomalty]

Actually, my first name is Will, but Mark is close enough.

Maybe, but you don't want to be a Mark.  He is a bit dense and has a low transmission rate compounded with a number of aberrations 

Mark

[Fine_Art]

"TheGuy" (I keep forgetting your first name...Mark, isn't it?) -

The formula you gave works well with lens + film, because film itself, like a lens, has a "bell-shaped" impulse response function/point spread function - if you could focus an infinitely narrow point of light onto film, it would record as a small rounded blob, bright in the centre and tapering off radially. The combination of lens PSF and film PSF is given by their mathematical convolution - you basically get a broader bell-curve than either the film alone or the lens alone. The 1/film + 1/lens = 1/system is a reasonable approximation to the width/resolution of this curve. Contrast at all middle and high spatial frequencies is poorer, because all parts of the system bell-curve are broader than either the film alone or the lens alone.

It's a bit different with digital. The response function of a sensor is not a sloping bell-curve, but a single rectangular peak, or "top-hat", whose width is the width of the pixel. When this is convolved with the lens PSF bell-curve, what you get is the sampled lens PSF. Unlike with film, this system response is not broader than the (analog) lens PSF [unless the pixels are too large in comparison to the lens PSF - undersampling]. The system resolution and contrast is therefore determined by the lens alone - with one exception: the highest spatial frequencies, greater than the sampling frequency. The top-hat sensor response takes over at these frequencies, and as it cannot resolve them (there is zero contrast for any detail falling _within_ a pixel), there is a sudden hard limit to the resolution.

If you want to get the most from your lenses on a digital sensor, oversampling the PSF delays the onset of this hard limit. In other words, small pixels are good in the sense that they can render finer detail more like film does, gradually tapering off in contrast out to the limit of the lens, rather than the limit of the sensor.

I sketched this out and scanned it - I hope that the attachment makes the concepts and distinction between film and digital clearer. (My digital system curve sketch is pretty crappy - it should be a much closer match to the lens curve on the left - but I hope you get the idea regardless that the shape/width is the same, Z ~ X)

Ray

Very interesting.

Since this is about resolving limits is there a table that shows diffraction limits for Aperture including 2 mirror systems? Assuming a glass lens is an aperture with secondary = 0

X axis 2,3,4,5,6,7,8,9,10 inch primary diameter
Y axis 0, 15%,20%,25%30%,35%,40%,45%,50% of primary diameter

I calculated my 10" f4 with 30% secondary at 5.5 microns ("diffraction limited optics")
My camera has 4.77 micron pixels so it super samples the lens output.

I went on the assumption that the Rayleigh limit formula theta radians = 1.22 lambda (550nm) / Aperture is for diffraction of an outer lens edge only. I used circumference, then mutiplied by the ratio of both circumferences based on that, given the limit is a function of the length of both diffracting edges in a mirror lens system.

If there is a table of diffraction hit as a function of Apertures in microns I'd really like to see it (and know if i messed it up).

[ondebanks]

Very interesting.

Since this is about resolving limits is there a table that shows diffraction limits for Aperture including 2 mirror systems? Assuming a glass lens is an aperture with secondary = 0

X axis 2,3,4,5,6,7,8,9,10 inch primary diameter
Y axis 0, 15%,20%,25%30%,35%,40%,45%,50% of primary diameter

I calculated my 10" f4 with 30% secondary at 5.5 microns ("diffraction limited optics")
My camera has 4.77 micron pixels so it super samples the lens output.

I went on the assumption that the Rayleigh limit formula theta radians = 1.22 lambda (550nm) / Aperture is for diffraction of an outer lens edge only. I used circumference, then mutiplied by the ratio of both circumferences based on that, given the limit is a function of the length of both diffracting edges in a mirror lens system.

If there is a table of diffraction hit as a function of Apertures in microns I'd really like to see it (and know if i messed it up).

It's much simpler than that. The secondary mirror obstruction has two effects:
1) It obviously blocks some light throughput.
2) It reduces MTF contrast at low spatial frequencies. Since you're talking about an astronomical telescope, these frequencies occur on subtle extended detail (the maria of the moon, the cloud belts of the planets, mottling in the disks of galaxies). This is why planetary observers rave about refractors, and reflectors with very small central obstructions.

The presence of the secondary mirror does not reduce the Rayleigh or diffraction-limited resolution. (Another way of saying this is that contrast is unaffected at high spatial frequencies in the MTF). It throws some light from the centre of the PSF (Airy disk) into the surrounding Airy ring, but it doesn't broaden the angular diameter of the Airy disk.

So all you need to do is calculate the resolution limit of your 10" primary mirror, and leave it at that.

There is a very approximate way to describe the effect of the MTF reduction at low spatial frequencies. The contrast of a large obstructed telescope is similar to the contrast of a smaller unobstructed telescope:  approximately,
D_unobstructed = D_primarymirror - D_secondaryobstruction (each D is a diameter)
So your 10" with the 30% central obstruction has the planetary contrast of a 7" unobstructed telescope...and the binary-star splitting resolution of any other 10" telescope.

BTW if you are getting a PSF width of 5.5 microns, then 4.7 micron pixels are undersampling the PSF. But you'll probably find that atmospheric turbulence and tracking errors broaden the PSF a lot anyway, before it reaches the camera.

Ray