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

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Mona Mahboob Kanafi 2013-6-10

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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]

回答（1 个）

Matt J 2013-6-10

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编辑：Matt J 2013-6-10

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.

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 2013-6-12

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.

http://proceedings.spiedigitallibrary.org/proceeding.aspx?articleid=1347552

'Using the Power Spectral Density method to characterise the surface topography of optical surfaces'

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