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I got this problem wrong, but I'm not sure why. I know that the FT of a signal is fully determined within 1 unit cell of a reciprocal lattice and outside this range are replicas. Here's the problem:

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Edit: I didn't add it originally because I thought it wouldn't display well, but here's a screenshot of the exact solution I was given:

enter image description here

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I dont know, if my attempt is correct, but here could be one argumentation (which actually conflicts with both solutions):

Intuitively, I would have agreed to your solution, because the spectrum is equally wide in height and width. But, then what about this argumentation:

  • The circles have radius $a/2$, and their center is at $\pm (\frac{a}{2},\frac{a}{2})$.
  • If I rotate the spectrum by 45 degrees clockwise, I get the center of the circles at $(\pm\frac{\sqrt{2}}{2}a,0)$. Then, the spectrum lies between $\pm\frac{1+\sqrt{2}}{2}a$ in x-dimension and $\pm \frac{a}{2}$ in y-direction.
  • In this case, I can use two different sampling frequencies in the new x- and y-direction, namely $a$ in the y-direction and $a(1+\sqrt{2})$ in the x-direction.

  • This idea could explain the non-symmetric sampling that your prof suggests. However, I do not get the same values as he/she does.

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  • $\begingroup$ Thanks. I added a screenshot of the exact solution in case it helps. I took each circle to be a different replica of the same FT, but is it possible that both circles make up one FT? $\endgroup$
    – Austin
    Mar 19, 2017 at 0:36

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