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

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]

Answer 1

[Matt J]

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.

[Mona Mahboob Kanafi]

Hi Matt, Thanks for your answer. About Q1, my surface(or image) is isotropic and the power spectrum image is totally circular. [even with Gaussian distribution].

[Matt J]

Okay. But when you're not averaging along x or y, you presumably have to interpolate the PSD at points along your radial line that do not precisely intersect pixels. The interpolation error would introduce small differences. You haven't said how big the differences are...

[Mona Mahboob Kanafi]

Dear Matt,

Today I worked with my data, and I think I got why it is not giving me the same result.

If I average 2D PSD along each line in x direction, what I get has the same slope in loglog scale to the case when I calculate a 1D PSD from each line of height data and average these 1D PSDs.

After doing this, I tested integrating 2D PSD along y direction and what I got was exactly overlapping the result of 1D PSD.

This is what I mean by integrating:

cx=zeros(1,m);

for i = 1 : m

      cx(1,i) = sum(((2*pi)/(n*a))*(Cq(:,i)));

end

Do you agree with the result?

[Matt J]

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]

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'