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

3 次查看（过去 30 天）
Mona Mahboob Kanafi
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]

请先登录，再回答此问题。

回答（1 个）

Matt J
Matt J 2013-6-10
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.
  5 个评论
显示 3更早的评论
Matt J
Matt J 2013-6-11
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 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'

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by