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The Kell factor

Doug Kerr

Well-known member
We note that, even with a "very good lens", and an aperture such that the effects of diffraction can essentially be ignored, for a digital camera sensor with, say 4000 pixels per picture width, the measured "resolution" is notably less than the 2000 cycles per picture width we might expect.

I often mention the "Kell factor" in this connection.

This situation comes from several closely-related mechanisms (which are actually different ways of looking at the same thing). Although the basic matter was noted during early work on facsimile, it was first well characterized in 1934 by Raymond D. Kell of RCA during work on early television systems.

Outlook A

Imagine that, with the sensor described above, we had a vertical-line test pattern consisting of 2000 line pairs per picture width (imaged on the senor by a "perfect" lens). Could it be "captured" by the sensor?

First imagine that in one test (probably by accident) the alignment of the pattern was such that each black line and each white line fell along one row of sensels. Then the pattern would be captured in good style by the sensor.

Now imagine that in a second test the alignment of the pattern was such that the boundaries between black and white stripes fell along the centerlines of a row of sensels. Now every sensel would pick up a "mid-gray" result (it being illuminated half by white and half by black, if you will excuse the conceit of being "illuminated by black"). The delivered image would be pure gray.

Thus we certainly could not say that this sensor could resolve a pattern of 2000 line pairs per picture width. Sometimes it could, and sometimes it couldn't at all, and sometimes it "sort-of-could" (that is, with a degraded contrast).

A simplistic explanation of the concept of the Kell factor is that "well, overall, for random orientation of the pattern, the 'average' attainable resolution would be on the order of 75% of that suggested by the sensel pitch." That value, 75%, is the "Kell factor" associated with this situation.

Outlook B

The problem with that outlook is that the notion of the "average resolution" attained over the range of pattern orientation doesn't really have much meaning insofar as the user experience.

But let's do a test (using that same sensor) with a "signal" (it now helps to think in terms of raster scan in a video context) whose frequency is slightly less than 2000 cycles per picture width. As we go along, the average "phase" of the pattern (compared to the phase of the sensels) continues to shift. The result might be like this:

Kell_factor1.png


Courtesy of Wikimedia Commons​

We see a progressive shift from the "picked up OK" to the "all we get is gray" situations. And this is often described as a "beat frequency effect" between the frequency of the sensels and the frequency of the lines. The gray swaths are regions in which the phase between the pattern and the sensel layout is nearly the "adverse one".

And yes, this is in fact a manifestation of aliasing, a matter that is inextricably linked with the matter of the Kell factor.​
Now suppose that we do the same thing but with a pattern whose horizontal line frequency is only 66% of 2000 cycles per picture width (we can of course say that its frequency is 66% of the Nyquist frequency). The result might be like this:

Kell_factor2.png


Courtesy of Wikimedia Commons​

Note that here there is a small appearance of the "beat frequency" phenomenon, but it is not prominent.

So perhaps here we can safely operate with a pattern at a frequency of 66% the Nyquist rate; that is, the usable resolution of this system is 66% the Nyquist rate. The Kell factor here seems to be 66%.

But what about Shannon-Nyquist?

The Shannon-Nyquist sampling theorem tells us that it we sample some phenomenon (the voltage of an audio waveform, the luminance of a path across an image) at a frequency of fs, then all frequency components of the phenomenon with frequency less than fs are "captured" by the succession of samples, and from the succession of samples the original phenomenon can be precisely reconstructed.

Why do we not seem to enjoy this promise in our little example?

Because in digital photography, we do not have a system which end-to-end completely plays the "Shannon-Nyquist game".

In a digital audio system, we sample the waveform, and then, at the "receiving" end, we expose a train of pulses matching the samples to a low-pass filter cutting off at the Nyquist rate. This is the "reconstruction filter", and is a vital part of the whole game. Out of it comes the original waveform.

But in digital photography (as in television), there is really not a "reconstruction filter". We do not, for example, create little spots of light on a display, one for each sample value, and then place in front of this an optical low-pass filter, out of which will come the original image. Rather, we create little blobs of light, one for each sample, and then just look at that array.

The consequence of this "shortcut" is that, for a test pattern a a little below the Nyquist frequency, we see these spurious "beat frequency" components in the delivered image.

In effect, the Kell factor tells us how much the practical resultion of the system has been compromised by this "shortcut".

************

Time for breakfast. Today, five kinds of fruit, scrambled eggs, bacon, and greaseless hashed brown potatoes.

Best regards,

Doug
 

Jerome Marot

Well-known member
But what about Shannon-Nyquist?

The Shannon-Nyquist sampling theorem tells us that it we sample some phenomenon (the voltage of an audio waveform, the luminance of a path across an image) at a frequency of fs, then all frequency components of the phenomenon with frequency less than fs are "captured" by the succession of samples, and from the succession of samples the original phenomenon can be precisely reconstructed.

Why do we not seem to enjoy this promise in our little example?

Because in digital photography, we do not have a system which end-to-end completely plays the "Shannon-Nyquist game".

In a digital audio system, we sample the waveform, and then, at the "receiving" end, we expose a train of pulses matching the samples to a low-pass filter cutting off at the Nyquist rate. This is the "reconstruction filter", and is a vital part of the whole game. Out of it comes the original waveform.


So the grey bands actually come from lack of low-pass filtering before display? I wonder how the result would be influenced by various display technology, e.g. inject prints which use complicated raster systems to emulate continuous colours.
 
Thus we certainly could not say that this sensor could resolve a pattern of 2000 line pairs per picture width. Sometimes it could, and sometimes it couldn't at all, and sometimes it "sort-of-could" (that is, with a degraded contrast).

Hi Doug,

That's why I try to use the description of being able to reliably resolve the detail.

A simplistic explanation of the concept of the Kell factor is that "well, overall, for random orientation of the pattern, the 'average' attainable resolution would be on the order of 75% of that suggested by the sensel pitch." That value, 75%, is the "Kell factor" associated with this situation.

Indeed, although we must also recognize that our input signal to the sensor will be blurred (by residual optical aberrations and diffraction), and usually not exactly aligned with the sampling lines/grid.

But what about Shannon-Nyquist?

The Shannon-Nyquist sampling theorem tells us that it we sample some phenomenon (the voltage of an audio waveform, the luminance of a path across an image) at a frequency of fs, then all frequency components of the phenomenon with frequency less than fs are "captured" by the succession of samples, and from the succession of samples the original phenomenon can be precisely reconstructed.

Why do we not seem to enjoy this promise in our little example?

Because in digital photography, we do not have a system which end-to-end completely plays the "Shannon-Nyquist game".

Exactly. We have a somewhat low-pass filtered input signal, and we can tailor the reconstruction filter in a variety of ways.

But in digital photography (as in television), there is really not a "reconstruction filter". We do not, for example, create little spots of light on a display, one for each sample value, and then place in front of this an optical low-pass filter, out of which will come the original image. Rather, we create little blobs of light, one for each sample, and then just look at that array.

The consequence of this "shortcut" is that, for a test pattern a a little below the Nyquist frequency, we see these spurious "beat frequency" components in the delivered image.

In effect, the Kell factor tells us how much the practical resolution of the system has been compromised by this "shortcut".

Yet, the actual situation does not have to be all that grim. Here are two samples of a more practical situation, spatial detail in all sorts of angles versus the display grid:

Without pre-filtering:
Star_500px.png


and with 0.7 sigma Gaussian blur (very sharp lens and a large fill-factor sensel grid):
Star_500px_GB070.png


Cheers,
Bart
 

Doug Kerr

Well-known member
Hi, Jerome,

So the grey bands actually come from lack of low-pass filtering before display?
Well, I'm not really certain how to best characterize it. But I believe that is part of the story.

But properly recovering components very near the Nyquist frequency (as in the case of the first graphic example I showed) puts very challenging requirements on the reconstruction filter. In the digitization of audio or video waveforms, we ordinarily avert that issue by removing (or at least severely attenuating) such frequencies with the bandpass (antialising) filter.

For example, in digital audio (at least as implemented "per the theory"; there are often compromises in filtering there, too), we do not experience (significantly) anything such as "beat frequency patterns" between a source frequency at almost the Nyquist frequency and the the frequency of the "sampling pattern".

But then neither do we normally have any frequencies near the Nyquist frequency, those having been removed (or at least greatly attenuated) by the bandpass (antialising) filter.

I wonder how the result would be influenced by various display technology, e.g. inject prints which use complicated raster systems to emulate continuous colours.
Indeed I'm sure that various spatial properties of the display system (even a "basic" one) can serve somewhat the role of the "reconstruction filter".

I don't know nearly as much about the details of this as I would like to!

Thanks for your observations.

Best regards,

Doug
 

Doug Kerr

Well-known member
Hi, Jerome,

We can get some intuitive insight into the issue by imagining a system for the transmission of audio waveforms using a sampling frequency of 8 kHz (as is commonly done in telephone transmission). The Shannon-Nyquist theorem tells us that we should be able to, for example, "capture" and later reconstruct a source signal with a frequency of 3999 Hz (assuming of course we take the proper steps to bring that prospect to fruition).

Suppose we in fact allow a signal with a frequency of 3999 Hz into the sampler, and examine the train of sample pulses. The pulses will have alternating polarities, but the overall amplitude of the train will slowly rise and fall, essentially following a sine function. The whole pattern will repeat once a second! (That is, 4000 - 3999, where 4000 is half the sampling frequency; that is, the difference between our signal frequency and the Nyquist frequency.)

It doesn't at all "look like" a picture of a 3999 Hz signal.

How will the reconstruction filter deliver for us (at the "distant" end), from that train of pulses, a 3999 Hz sine wave?

If we look at the frequency spectrum of that train of pulses, it has frequency components at 3999 Hz and 4001 Hz, 7999 Hz and 8001 Hz, 15,999 Hz and 16001 Hz, and so forth (theoretically to infinity).

If our reconstruction filter has little attenuation at 3999 Hz, but substantial attenuation at 4001 Hz (not at all easy to do, of course), and of course substantial attenuation at the higher frequencies mentioned above (easy to do), then what will come out is just the 3999 Hz component - the original signal we aspire to have reconstructed. Success!

But if the reconstruction filter does not have substantial attenuation at 4001 Hz (even though it does at the higher frequencies), then what will come out is a signal with two components, at 3999 Hz and 4001 Hz. That will look like a 4000 Hz signal amplitude-modulated at a rate of 1 Hz. That is, it will sound like a 4000 Hz signal that waxes and wanes, reaching a maximum amplitude twice per second, and going to zero at instants halfway in between.

And those instants where it goes to zero are like the gray bands in the image-sampling example I showed.

Best regards,

Doug
 

Asher Kelman

OPF Owner/Editor-in-Chief
Now I have another disease to worry about, LOL!

In truth, thanks Doug for an explanation that's so clear, I do not have to readmit over and over again.

Asher
 
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