Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Having a 2D map filled uniformly by random values (Figure:top-left) to demonstrate a disordered phenomena, the next maps are kernel averaged with a kernel of sizes 3x3, 5x5, ..., 11x11.
The questions are:
What are the patterns appeared in the kernel averaged maps?
What is the physical explanation of existence of such patterns?

For image please see this link.

To generate the maps a kernel based averaging system type C was applied on the original data.

share|cite|improve this question
up vote 3 down vote accepted

What you are seeing are artifacts caused by the hard-wall type kernel you are using. These patterns consist of horizontal and vertical streaks with a definite periodicity, which gets more pronounced as you make the averaging box bigger. The streaks occur because the Fourier transform of a hard-wall box has zeros at certain wavenumbers. To get rid of them, you can average with an approximately Gaussian kernel, described below.

When you apply an averaging, you are multiplying the Fourier transform of the noise by the Fourier transform of the averaging kernel. This is called the "convolution theorem", and "convolution" is just a term for averaging with a kernel. In Fourier space, convolution is just multiplication.

In your case, you are convolving with a box function, which is the function $f(x,y) = 1 $ for |x|<5 |y|<5, and f(x,y)=0 otherwise.

This box function is the product of two step boxes:

$$ f(x,y) = h(x)h(y) $$

where h(x) is the function which is 1 for |x|<5 and 0 otherwise. The Fourier transform of this box function (for continuous x) is

$$ \tilde{h}(k) = {\sin(5k)\over k} $$

Which has zeros when $k = \pm {\pi\over 5}$, $k= \pm {2\pi\over 5}$, $k=\pm{3\pi\over 5}$, $k=\pm{4\pi\over 5}$. This formula is not right, it is making the continuum approximation that 5 is approximately infinite, but the zeros are in the same place for the correct Fourier transform, and this is all I am using. The Fourier space for a lattice repeats outside the region $-\pi$ to $\pi$, so these are all eight zeros.

The Fourier transform of f is the product of the Fourier transform in x and in y:

$$\tilde{f} = \tilde{h}(k_x) \tilde{h}(k_y) $$

This has zeros on eight vertical lines, and eight horizonal lines.

The Fourier transform of the random noise is another random noise in Fourier space (this is only strictly true for Gaussian noise, you are probably using a uniform random number between -1 and 1. The difference doesn't matter for the purposes of this discussion). When you multiply by the Fourier transform of the kernel, you get a random product, but you get zero on the special lines. These zeros produce the horizontal/vertical streaks.

You should be able to generate a similar pattern just by generating random noise, zeroing out the eight lines, and Fourier transforming. The proper pattern also modulates the Fourier stuff away from the zeros by a smooth amplitude, but this is probably less noticible.

To get rid of this, you can use a Kernel which falls off smoothly, so that it doesn't have zeros in the Fourier transform, but falls off smoothly. The easiest is a Gaussian Kernel:

$$f(x,y) = e^{-{(x^2+y^2)\over 2a^2}}$$

Where a is the analog of 1,2,3,4,5. If you don't want to use this, you can use anything smooth that falls off without sharp corners.

share|cite|improve this answer
Thank you so much. You opened new aspects of data analysis and interpreting to me. That was as a lecture to me. – Developer Oct 12 '11 at 11:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.