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 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 is a clumsy attempt at explaining how this is done.

Jack