Entries from Stephan Paukner

I always wondered why it didn’t work to compute the dual Gabor atom by using the image-to-vector methods I explained previously [1,2,3]. Dr. Kaiblinger showed me that the correct way was to use that special isomorphism that walks along the diagonal of the image. Because width and height have to be relatively prime, that path spans the whole image space. And because there are no jumps over pixels, 2D-modulations stay 1D-modulations. This is not yet proved formally, but I can already show first experiments:

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A 2D-frequency is given as a tensor product of two 1D-frequencies with signal lengths p and q, respectively. If their modulation parameters are given as k_p and k_q, then the corresponding 2D-modulation is given by a 1D-modulation of length N and parameter k_N = k_q p − k_p q (mod N) ; as already said, a proof is not given yet.

So those two 2D-frequencies are really identical. The plot of the second one is identical to the first one, so we skip it here.

Now we want to see if the 2D-dual of a separable 2D atom obtained by that isomorphism is identical to the tensor product of the two 1D-duals.

Ibanez, my favorite guitar manufacturer, has created a model with an unusual guitar body shape which is commonly favored by heavy metal guitarists. Having Gibson’s innovative “Flying V” and “Explorer” body shapes from 1958(!) in mind, guitar companies started to release models with similar designs in the late seventies because heavy metal guitarists found that those radical body shapes match their music style and appearance on stage. Ibanez, being favored by virtuosic guitarists such as Paul Gilbert, Joe Satriani or Steve Vai, has lacked guitars of such an uncommon style so far and has finally come up with a V-shaped and an X-shaped guitar model. (I once heard that such untraditional shapes tend to bad resonance behavior, but I think that’s nonsense.) The white V-shaped one would even appeal me, as it’s not so typically “heavy metal” as the black one.

The video promoting the X-model on their website features statements of a guitar body designer and a virtuosic heavy metal guitarist from Germany. The video is underlayed with musical sequences from the guitarist’s band; extremely fast, technical and virtuosic metal music which is considered to be “neo-classical” because it tries to imitate the complex structures of traditional classical music. Unfortunately, when I saw their website and their entry on Wikipedia, they turned out to be a “technical brutal death metal” band, having “death growls” as singing voice and songs about “mutilating stillborn children” and similar. I already asked myself why they chose a name associating them with necrophilia.

Now, does this have to be? Does Ibanez really want to identify themselves with disgustingly gory necrophilic musicians who make guttural sounds while they imagine to slash rotten corpses? This is ridiculous! What do these abject topics have to do with virtuosity on the guitar? What does that have to do with classical music? I’m really agitated. Why can’t virtuosity keep being linked to virtuosic music, and be it of heavy metal style?

However, on the one hand there are awesome metal guitarists using traditional guitar body shapes, and on the other there are “ordinary” rockers such as Lenny Kravitz who use a Gibson Flying V. And companies like Dean Guitars devote themselves to radical body shapes that are appreciated by comparatively “harmless” metal giants such as Dave Mustaine or the late Dimebag Darrel. I therefore consider Ibanez’ choice of that one guitarist for their promo as an accident.

And by the way, if that guy is so keen on appearing evil on stage, why has he got that boring short hair cut?

Apple juice!

Der Basketballspieler befindet sich in Riesenhaft.
Der Däumling befindet sich in Zwergenhaft.
Der Diktator befindet sich in Boshaft.
Der Untalentierte befindet sich in Tölpelhaft.
Der Maler befindet sich in Meisterhaft.
Der Neurotiker befindet sich in Zwanghaft.

So, jetzt ist’s aber gut. Wenn mir irgendwann nochwas einfällt, kommt’s einfach hier drunter dazu.

Irgendwie kommt mir das wie ein sich selbst erhaltendes System vor: Jemand (er-)findet ein Fachgebiet und unterrichtet Schüler darin. Diejenigen, die das interessiert und gut darin sind, beschäftigen sich intensiver mit der Materie und studieren es vielleicht sogar, und werden dann Lehrer, um das Fachgebiet der nachfolgenden Schülergeneration beizubringen, aus der dann die nächsten Lehrer hervorgehen, die das Fach ihrerseits wieder weitergeben. Wozu gibt es dann das Fachgebiet? Aus Spaß an der Freud’? Weil’s grad interessant ist? Wozu Mathematik? Wozu Physik? Wozu Chemie, Biologie? Da muss doch mehr dahinter stecken?

Ja, die Leute, die in dem Fachgebiet aktiv arbeiten, sei es in Forschung oder Anwendung, die rechtfertigen das Fach. Wer also Interesse an einem Gebiet zeigt, es intensiv studiert und dann lediglich Lehrer wird, hat eigentlich versagt. Sicher, “irgendwer muss’s ja machen”, nämlich diejenigen, die nur mittelmäßiges Interesse daran haben, das sie dann allerdings im Laufe der Unterrichtsjahre aus Frust sowieso wieder verlieren.

Das ist nämlich der Grund warum die Schulmathematik garnichts mit der eigentlichen Mathematik zu tun hat. Kein Mathematiker hat eine Formelsammlung. Kein Mathematiker macht Rechenbeispiele. Mathematiker rechnen nicht. Lehrer rechnen. Ja, liebe Schüler, dass das Volumen eines Kreiskegels als V=r²πh/3 zu berechnen ist, braucht tatsächlich keine Sau!

I still have some difficulties understanding sampling lattices in the 4D position-frequency space of images. Here’s a flow of thoughts on this:

A 1D-signal of length N can be interpreted as an element of C^N. (This was always confusing me: an N-dimensional vector is only a 1D-signal!) But in TFA it is actually considered as an infinite periodic vector with period N. Therefore the index set is not just {1,...,N} but rather Z_N:=Z/NZ. This set is a finite group under addition modulo N. The TF-domain of that 1D-signal space is Z_N×Z_N and therefore 2D; it still has the group structure (by components). A sampling subset Λ of this TF-plane takes out certain time-locations and frequencies. If one has a fixed window function g with same signal length, the Gabor family with regard to that sampling subset is the set of those shifts and modulations of the Gabor atom g where the time-locations and frequencies are given by Λ. If Λ has enough elements, namely |Λ|>N, then...??? No, the redundancy doesn’t tell anything about whether the Gabor family is a frame! It is automatically a frame (in the finite setting) as soon as it spans the whole signal space. So what does that redundancy tell us? And why is it automatically large enough as soon as the Gabor family is a frame? Or what?

Another open question is why or when one should have a subgroup as sampling subset. What advances does one get when the sampling is done on a subgroup? If Λ is a subgroup, then there is a dual? Is there no dual when Λ is no subgroup? Or does it just depend on the redundancy then? It might have something to do with the fact that the Gabor frame operator commutes with TF-shifts along that subgroup.

The things get even trickier for 2D-signals. An image of size p×q is also considered as an infinite periodic signal with periods p in one direction and q in the other. The signal space has N=pq dimensions. The index set is Z_p×Z_q. The TF-domain (actually position-frequency domain) becomes (Z_p×Z_q)×(Z_p×Z_q) and is therefore 4D. What’s unclear here is the term of separability with regard to the sampling subset. In the 1D-case (TF-plane is 2D) one names the sampling subset a separable lattice when the shifts and modulations are defined by the multiples of a divisor of N. I.e., one gets a rectangular grid (lattice) in the TF-plane in this case: For every time-shift there is the same set of modulations. A non-separable case could be given by a set of random sampling points. But these don’t form a subgroup in general. A non-separable subgroup could look like a rotated version of a separable lattice. For a rotation by 45° (π/4) one gets the special case of a quincunx lattice (if the correct distances are taken). And now to separabililty of a 4D-lattice: What does a 4D quincunx look like? Or another 4D non-separable lattice? Does this mean that whenever I can split into two 2D-lattices, one talks about separable lattices, independently from the question whether these two are separable again? I think separable means here that one either has a position-lattice ((x₁,x₂)-lattice) and a frequency lattice ((ω₁,ω₂)-lattice) or an (x₁,ω₁)- and an (x₂,ω₂)-lattice, independently from whether these are separable themselves or not. But how does one construct a 4D-set out of two 2D-sets with MATLAB/Octave?

I currently wonder what string gauge (diameter) I should use for my electric guitar. The standard is .009 (i.e. 0.009″ for the high e-string). Thinner strings allow easier string bending, but one has to play with less finger pressure to avoid detuning. As I like to tune all strings down by one halftone, the strings get even “softer”. So one should take a higher string gauge when tuning down. The question is now: Do downtuned .010’ers correspond to normally tuned .009’ers? What gauge should one use when one wants to tune down e.g. by a whole tone and have the “softness” of .009’ers? Here’s my try of a calculation:

Does a downtune by one octave correspond to a loss of half the tension? Whatever amount the tension will get, it doesn’t decrease linearly with the halftones—remember the different distances between the frets! How does one calculate this scale? You can’t just divide the half of the string length by 12 to get the steps between the frets.

Calculating the fret stepping of a guitar
For the 12th halftone, you arrive at ½ of the string length, for the 24th halftone you arrive at ¼ of the length, etc.; the 12th divides the length by 2, the 24th divides it by 4, the 36th divides it by 8, etc. Abbreviating d(n) for the divisior at halftone n, we have d(0)=1, d(12)=2, d(24)=4, d(36)=8, etc., i.e. d(12n)=2ⁿ and therefore d(n)=2^n/12. For the n-th halftone, the string length becomes

.

If you don’t believe my derivation, maybe you believe a (modified) function plot:

The blue crosses indicate the first two octaves, occurring when ½ (=50%) and ¾ (=75%) of the length are removed. The pink cross indicates the fifth fret at about ¼ (=25.085%) of the length. Every guitarist should recognize the fret stepping here!

Gauge stepping when tuning down
I found a link which explains that the relation between string diameter δ, tension t and frequency f is

,

where C is a constant depending on the material. The aforementioned scaling is still valid for the frequencies, i.e., when the frequency f is tuned down by n halftones, the resulting frequency f_n is given as

,

what can be verified for f=440Hz: The next lower tunes are 415.3Hz (n=1) and 392Hz (n=2), and the next higher tunes are 466.2Hz (n=−1) and 493.9Hz (n=−2). Replacing f by f_n in the previous equation yields that when the tension is to be kept, one has to take strings with diameter . As example, .009’er strings that are tuned down by one whole tone should be replaced by .010’ers to keep the tension of the .009’ers.

The other way round, fixing δ_n=0.010 (the taken string gauge) and δ=0.009 (the desired string gauge “feel”), one derives

what results in n=1.82 in this example, a downtune of slightly less than a whole tone. Tuning down .010’ers a “complete” whole tone corresponds to a string gauge of .0089’ers, so really almost .009’ers.

As a final rule of thumb, the steps between the string gauges correspond to tune changes of a whole tone. The following table shows how certain string gauges “feel” when they are tuned down: