Imagine I have a simple 1D height profile which is NOT symmetric. Now, what is truly important for me is to know what are the frequency content

of the top profile (i.e. a cut profile above the average line).

Target: I need the power spectral density of the top profile.

It shouldn't be this simple to just cut the profile from the mean line to have the top profile values and zero otherwise. Then when I apply FFT

in MATLAB, I believe those zero sections will be a problem, like the case for a step function that a wide range of sine waves will be added to the result while actually there are no such frequency content in my real top profile.

I have uploaded two images for you, to see what I mean; and both explanations in the images are from the same author. In the first image, the writer says that, simply cut the profile and calculate power spectrum for each top and bottom profile.


But the writer some years later, says something else that, for calculating top power spectrum, you need to cut the surface from the mean plane and then replace the bottom profile with some profile which has the same statistical properties as the profile above the average plane:


I hope I could have explained my point. I would really appreciate your help..


I am not sure if I get your question right, but I'll try to explain what I think is going on. What the author seems to try to grasp is the 'roughness' of the curve and this can be done using Fourier methods. Generally energy in higher Fourier coefficients corresponds to high frequency content ('rough' curves), while lower coefficients carry low frequency ('smooth' curves) information.

Now here is what you should try to do: take the profile curve and subtract its mean to have some values above and some below the mean line (i.e. positive and negative values, do not set anything to 0!). Then you perform the Fourier transform and take squares of absolute values to get the power spectrum.

This is by the way completely equivalent to not removing the mean and just ignoring the 0-th coefficient of your Fourier transform.

Hope this helps...


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