Power spectrum of 2D image: result interpretation
it does not make any sense to still use circles for calculating the histogram. I'd have to use ellipses, wouldn't I?


That's right. For radial averaging on a rectangular grid, the locus of equidistant points from the origin would look like an ellipse instead of a circle.


  there are four areas (area of the square minus the area of the circle) in each corner, from which I don't get any contribution


That's right, and there's nothing we can do about that. If you care about those high frequencies, you should re-acquire your image at a higher sampling rate.


  they used a crooked square, i.e. a rhombus, with each corner on one of the axis. ... Does this approach make any sense? What about the four triangle-shaped areas in this case?


In this case the locus represents points where the sum of the two spatial frequencies is a constant, i.e. instead of $\sqrt{f_1^2 + f_2^2} = \mbox{constant}$ we follow the points where $|f_1| + |f_2|=\mbox{constant}$. There's no reason why one is better or worse than the other. They are both valid methods of collapsing a 2D function of two spatial frequencies into a 1D function of one "combined" spatial frequency.

As with the circle/ellipse, there's nothing we can do about the bins outside the triangle area that never get analyzed.

When using a DFT, distance can be measured as the number of steps taken on a discrete 2D frequency grid. (See the discussion at the bottom of the page here: http://iubemcenter.indiana.edu/beyondNyquist.html). Constant frequency "ellipses" in this case are really "rectangles" of grid cells centered around the origin. This way all grid points get accounted for and there are no "dead" zones when computing the 1D collapsed PSD from the 2D DFT.