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A medical Image has a size of 8x8 inches. The sampling resolution is 5 cycles/mm. How many pixels are required? Will an image of size 256x256 be enough?

I know sampling in 1D signal but cannot get the intuition for a 2D image.

I first converted 5 cycles/mm to cycles/inch which came to be about 127 cycles/inch.

So as I understand it, this means we take 127 values for each inch of the input image and each of this is a pixel. So I figured for a 8x8 inch image, we would need

127*8x127*8 = 1016x1016 pixels

Am I right in my thinking or will Nyquist theorem have to be used?

Any kind of help is appreciated.

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  • $\begingroup$ // Am I right in my thinking or will Nyquist theorem have to be used? // ------ you are right in thinking that. But what you need is another factor that is greater than 2 for both vertical and horizontal. Nyquist says you need at least 2033 x 2033 pixels. and practicality says you need a bit more. $\endgroup$ Commented Apr 25, 2019 at 20:34
  • $\begingroup$ maybe 2048 x 2048 would be just enough and it would make any FFT or DCT a bit more happy. $\endgroup$ Commented Apr 25, 2019 at 20:38
  • $\begingroup$ i would, for practicality reasons, toss in another factor of two and make it 4096 x 4096 pixels. are you doing any FFT or DCT transformation to the image? $\endgroup$ Commented Apr 25, 2019 at 20:43
  • $\begingroup$ Thanks, I got it. Can you point me to a resource which explains samping theorem for images? $\endgroup$ Commented Apr 25, 2019 at 20:43
  • $\begingroup$ i just did a quick googley thing and found this. $\endgroup$ Commented Apr 25, 2019 at 20:44

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It is given that the sampling resolution is 5 cycles/mm. Therefore, for better quality, 2 pixels per cycle are required. This is because sampling theorem states that the sampling frequency should be greater than twice the maximum signal frequency. This means that 10 pixels per mm are required. So, the pixel size is 0.1 mm. You may note here that this double rates in both directions are called Nyquist rates.

The given image size (since 1 inch = 2.54 cm) is 8 * 2.54 * 8 * 2.54 cm = 20.32 * 20.32 cm

Therefore, the minimum number of required pixels = 2032 * 2032 pixels. So, an image of size 256 * 256 is not enough.

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If the spatial process that results in that 5 cycles/mm were to act like a "brick wall" filter, where no signals with spatial frequencies higher than 5 cycles/mm get through, then @Void's answer is correct.

However, with more real systems, the answer is more complex. It is likely that the system blurs the image in a way that does not drop off sharply in frequency. In that case, then just as with sampling behind an anti-aliasing filter, you would need to make a tradeoff between the sampling rate (i.e. the pixel density) and the amount of aliasing that you allow. You would need to recognize that in theory there would never be a pixel density that would completely eliminate aliasing -- you would have to choose and acceptable amount of aliasing, then find the pixel density that gives it to you.

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