Confusion about line pairs per millimeter

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I don't understand why a spatial frequency like 1/pixel_width gives line pairs per millimeter rather than lines per millimeter.
Hello,

I'm having some difficulties in getting my head around the relationship between pixel size and lines/mm or line pairs/mm.

My thinking is the following: I'm assuming we have a sensor with square pixels of side length p (given in mm) and we have a line pattern, so a pattern with white and black lines of equal width (eg a Ronchi ruling). In the ideal case, where our pixels are exactly aligned with the lines, we can capture one line per pixel. So the thinnest line that we can capture has width p. That corresponds to a frequency of 1/p lines per millimeter or 1/2p line pairs per millimeter. If the lines are offset by eg half a pixel then there's aliasing and we can't accurately resolve the pattern. So the best resolution we can reliably get is at the Nyquist frequency of 1/2p lines per millimeter or 1/4p line pairs per millimeter. I'm assuming a perfect lens, a perfect sensor and no diffraction.

To me that sounds like it would make sense. But it doesn't seem to square with some of the things that I read online.
  • For example, on page 3 of https://isl.stanford.edu/~abbas/ee392b/lect09.pdf it says: "Assuming a square pixel with width (pitch) p, the spatial Nyquist rate in each dimension is 𝑓_𝑁𝑦𝑞𝑢𝑖𝑠𝑡 = 1/2𝑝 and is typically reported in line pairs per millimeter (lp/mm)". To me that sounds like 1/2𝑝 gives line pairs per millimeter when in my logic it gives lines per millimeter.
  • Another example is the MTF of an ideal sensor. According to eg https://www.opto-e.com/media/docs/Resources/metrology-white-paper.pdf the MTF of an ideal sensor is given by |sinc(w)|, which is the Fourier transform of a rectangular 'pixel' signal. On page 5 it says that the cutoff frequency is 1/p. They give an example where a 5um pixel results in a cutoff frequency of 200 line pairs per millimeter. In my logic 1 / 0.005 results 200 lines per millimeter (so 100 line pairs per mm).
Can somebody please explain to me where I'm going wrong? My logic seems to be off by a factor 2 and I can't figure out why.

Thanks for your help!
 
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klasdfjllgr said:
[...]
Can somebody please explain to me where I'm going wrong? My logic seems to be off by a factor 2 and I can't figure out why.
Since the modulation transfer function is defined for sinusoids but targets are difficult to manufacture, the alternative 'contrast transfer function', defined for square waves, is more commonly used. In this context, a line pair corresponds to a period of the square wave similar to the use of cycles/mm for the MTF.

Does that help?
 
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Hey, thanks for your reply!

I'm sorry but I don't really understand the point you're making. Would you be able to elaborate?

I used the line pattern in my question but we can use a sinusoidal pattern instead. The smallest period of the sinusoidal pattern that our sensor can resolve is 2 pixels wide (when the peak is roughly on one pixel and the trough is roughly on the adjacent pixel). So the period, which corresponds to a line pair, is 2p. This corresponds to a wave frequency of 1/2p. So we get 1/2p line pairs per mm. The Nyquist frequency is then 1/4p lp/mm. Those are the same numbers as I got for the line pattern (which is different to the two examples). Here's a little drawing ...

Screenshot_20250417_165949_Samsung Notes.jpg
 
Something you need to keep in mind is the difference between a continuous detector and a sampled detector- you are thinking about the extreme case, where the difference between a linear shift-invariant system and your system is a maximum. I would characterize your example as a highly undersampled image.

How about this- a line pair is the 0th-order approximation to a sinusoid (shift your '1 period' bracket half a bar).
 
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So I made a mistake in how I interpreted the Nyquist frequency because I wrongly assumed that I can accurately resolve a pattern at the Nyquist frequency of 1/2p when that frequency is really the last frequency where I can't resolve the pattern.

Thanks for your help!