Different results from radial averaging and averaging along x axis of fft2

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I have an image(which is a height profile), and I apply 2D fft (fft2) on the data. Then I compute 2D power spectrum:
win=tukeywin(N,0.25)*tukeywin(M,0.25)';
z_win=z.*win;
Hm=fft2(z_win);
Cq=(a^2/(2*pi*N)^2).*((abs(fftshift(Hm))).^2);
Now, I radially average my power spectrum in order to get a 1D power spectrum from the 2D data.
Q1 : Why the value obtained here, differs with the value I gain if I do average data in x or y direction.
Q2 : What is the unit for radially averaged PSD? I should mention that the unit of my PSD is m^4. Is it m^4 or m^3 ? why? [frequency unit : 1/m]

Answers (1)

Matt J
Matt J on 10 Jun 2013
Edited: Matt J on 10 Jun 2013
Q1: Because the PSD is not circularly symmetric? Is there a reason to expect it to be?
Q2: m^4. Averaging over quantities should not change their units, because it is done by adding things up with dimensionless weighting coefficients.
  5 Comments
Matt J
Matt J on 11 Jun 2013
I don't really understand what you're saying these results reveal/explain. If your PSD is perfectly circularly symmetric, as you claim, I would expect to see identical results along x and y, possibly differing by floating point noise.
Mona Mahboob Kanafi
Mona Mahboob Kanafi on 12 Jun 2013
First, floating point error is not sth that could be considered a huge error. Second, I may forgot to mention the differences in results, slope of two curves I obtain from each methods are 0.8 for Rad Ave and 0.3 for the other. Anyway, I now understand the papers I previously read on this topic, so if you are interested you can also refer to the article below, or many other similar papers in this field. Thanks for everything.
'Using the Power Spectral Density method to characterise the surface topography of optical surfaces'

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