On 8/20/06, Mathieu Bouchard matju@artengine.ca wrote:
Why do you say "fractal" ?
1/f^a noise is a fractal because it is self similar under different size windows.
1/(kf)^a = 1/k^a * 1/f^a, so the magnitude of spectrum viewed over a different window is just scaled by a certain amplitude. The same applies to the time domain as well (but it has to be interpreted as a probability function).
Note: the fractal-ness of this noise is compromised by having a sampled signal It's a true fractal only when considered as a continuous function. But for large enough windows (where the sampling effects can be basically neglected), it acts like a fractal.
Chuck
P.S. sorry Mathieu for the redundant messages
On Sun, 20 Aug 2006, Charles Henry wrote:
1/f^a noise is a fractal because it is self similar under different size windows. 1/(kf)^a = 1/k^a * 1/f^a, so the magnitude of spectrum viewed over a different window is just scaled by a certain amplitude. The same applies to the time domain as well (but it has to be interpreted as a probability function).
Then by this standard, the 1/x function is self-similar, and so are all hyperbolas. That is, as long as similarity is defined as modulo the group of diagonal matrices conjugated by rotation matrices.
Isn't the definition of fractal requiring some kind of noninteger Hausdorff dimension?
_ _ __ ___ _____ ________ _____________ _____________________ ... | Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju | Freelance Digital Arts Engineer, Montréal QC Canada
Mathieu Bouchard wrote:
On Sun, 20 Aug 2006, Charles Henry wrote:
1/f^a noise is a fractal because it is self similar under different size windows. 1/(kf)^a = 1/k^a * 1/f^a, so the magnitude of spectrum viewed over a different window is just scaled by a certain amplitude. The same applies to the time domain as well (but it has to be interpreted as a probability function).
Then by this standard, the 1/x function is self-similar, and so are all hyperbolas. That is, as long as similarity is defined as modulo the group of diagonal matrices conjugated by rotation matrices.
I don't see that. Zooming in and out of 1/x or a hyperbola just makes the curve look bigger or smaller, whereas noise looks the same at all scales.
Isn't the definition of fractal requiring some kind of noninteger Hausdorff dimension?
From B. Mandelbrot, The Fractal Geometry of Nature, Freeman, 1983, p.15: "A fractal is by definition a set for which the Hausdorff Besicovitch dimension strictly exceeds the topological dimension" "Every set with a noninteger D is fractal" "However, a fractal may have a noninteger D ... the trail of Brownian motion is fractal because D=2, while Dt = 1" (i.e. a randomly meandering line will eventually completely fill a plane)
Martin
_ _ __ ___ _____ ________ _____________ _____________________ ... | Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju | Freelance Digital Arts Engineer, Montréal QC Canada
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On Sun, 20 Aug 2006, Martin Peach wrote:
Then by this standard, the 1/x function is self-similar, and so are all hyperbolas. That is, as long as similarity is defined as modulo the group of diagonal matrices conjugated by rotation matrices.
I don't see that. Zooming in and out of 1/x or a hyperbola just makes the curve look bigger or smaller, whereas noise looks the same at all scales.
What? No, you have to zoom out the x while you zoom in the y by the same amount, or the other way around. The product of the zoom factors of x and y should be 1.
"A fractal is by definition a set for which the Hausdorff Besicovitch dimension strictly exceeds the topological dimension"
I stand corrected.
"However, a fractal may have an integer D ... the trail of Brownian motion is fractal because D=2, while Dt=1" (i.e. a randomly meandering line will eventually completely fill a plane)
Yeah, and another famous example with D=2 and Dt=1 is the Hilbert curve: http://en.wikipedia.org/wiki/Image:Hilbert_curve.png
_ _ __ ___ _____ ________ _____________ _____________________ ... | Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju | Freelance Digital Arts Engineer, Montréal QC Canada
Oh crap, There goes my point...it should have gone like this:
"However a fractal may have an integer D... the trail of Brownian motion is a fractal because D=2 while Dt=1"
Martin
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Charles Henry -
Martin Peach -
Mathieu Bouchard