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The slant edge SFR method has become the standard for measuring resolution of lenses and camera systems. It works by scanning a five degree slanted edge to calculate a line spread function. This is differentiated to produce an edge spread function which is in turn passed through a fast Fourier transform to produce an MTF curve (rough description).
EDIT - for the purpose of this question assume there is no anti-aliasing filter since that is a limit independent of the Nyquist Limit.

This article by Peter Burns (the originator) better describes the method.

See the graphs below for an example of a measurement conducted on a Nikon D7000

The measurements would seem to be limited by the Nyquist Limit of the sensor in the camera. See this discussion. But, because the edge is slanted by five degrees it is, in effect, being by super-sampled during the scan.

So my question: does this super-sampling of a five degree edge allow us to measure lens resolution beyond the Nyquist Limit of the camera sensor?

enter image description here
enter image description here
Measurements were made on this test image for the Nikon D7000 from DPReview.com.

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    \$\begingroup\$ I guess this begs the question...how exactly do we measure the resolution of a lens? I guess I always assumed that the medium used to measure lens MTF always had a higher limit than the lens itself. \$\endgroup\$
    – jrista
    Apr 10, 2011 at 20:46
  • \$\begingroup\$ According to imatest.com/docs/sharpness.html#calc "The four bins are combined to calculate an averaged 4x oversampled edge. This allows analysis of spatial frequencies beyond the normal Nyquist frequency." So it seems the answer to your question may be yes, but I don't yet understand the method well enough to know why. \$\endgroup\$
    – Sean
    Apr 10, 2011 at 22:08
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    \$\begingroup\$ @Sean It appears to be an averaging phenomenon. If the line is slanted a bit from the vertical, it might be fair to think of each successive row as sampling the same horizontal signal but shifted a little. This effectively oversamples a single signal. At 5 degrees the slope is about 12, giving about a 12:1 sampling ratio. This should increase the horizontal resolving ability by Sqrt(12) = about 3.5. I suspect that's why the algorithm uses four bins per pixel ("4x oversampled edge"). Thus the answer is definitely "yes." \$\endgroup\$
    – whuber
    Apr 11, 2011 at 18:49
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    \$\begingroup\$ @jrista Try this gedankenexperiment: imagine your sensor is a single enormous pixel, but it has highly precise and repeatable output (around 36 bits should do). Focus a single sharp point of light in the middle. Now plot the sensor's response as you slowly shift it sideways until the focused dot is completely off the edge of the sensor. If the lens is perfect, the sensor's response is constant until the dot falls of the edge, then drops to zero. In reality, lens aberrations will spread the dot, causing a spread in the response curve: the amount of spread is the lens resolution. \$\endgroup\$
    – whuber
    Apr 11, 2011 at 18:55
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    \$\begingroup\$ @All: It might be useful if someone packaged up summaries of the most appropriate references and provided an answer to this question. This was a great question, but it never really received any answers. \$\endgroup\$
    – jrista
    May 12, 2011 at 17:01

1 Answer 1

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This reply expands on the discussion in the comments.

The averaging idea turns out to be the right one, as ably explained by Douglas Kerr in a nice little online paper. The basic ideas are two:

  1. The lens "resolution" is most fully described by considering the mathematical relationship between the light leaving the subject and what reaches the sensor. This relationship, the "modulation transfer function," can be deduced from the simplest of all possible targets: a perfectly dark half-plane on a perfectly bright homogeneous background. Obviously the image on the sensor should be a region of light abruptly terminating along a perfect line. It never is perfect, though, and the imperfections affect the resolution. Ultimately the MTF is determined by looking at how the light intensity varies as we move straight out from the boundary (in both directions, into the dark and into the light) across the sensor.

  2. It is a statistical fact that averages can be more precise than the measurements of which they are constituted. For typical measurement error, the precision follows an inverse square root law: to double the precision, you need four times as many measurements. In principle you can get as precise as you want by averaging enough independently repeated measurements of the same thing.

    This idea can be exploited (and is) in two ways. One is actual repetition, achieved by taking multiple images of the same scene. This is time consuming. The slanted-edge MTF analysis creates repetition within a single image. It does this by slanting the line slightly. This does not change the MTF in any material way and it guarantees that the patterns of the lens's response do not align perfectly with the sensor's pixels.

    Imagine the line being nearly vertical. Each row of pixels serves (almost) as an independent set of measurements of the MTF. The rows march outward from the line, almost perpendicularly. The pixels are registered with respect to the (ideal) line location in varying ways, producing slightly different patterns of response. Averaging these patterns over many rows has almost the same effect as taking multiple images of the line. The result can be adjusted for the fact that the pixels aren't quite perpendicular to the line.

In this way, the slanted-edge method can detect frequencies in the MTF that exceed the limiting frequency of a single image. It works due to the simplicity and regularity of the test pattern.

I have left out many details, such as checking that the line really is straight (and adjusting for slight deviations from linearity). Kerr's article is accessible--there's almost no math there--and well illustrated, so check it out if you would like to know more.

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