User:Arseniiv/Timbres

Here are some approaches to picking harmonics for timbres for this and that purpose, aside of just taking out entire sequences of multiples of, say, 5 from a harmonic timbre.

Golden-harmonic timbres

When you want the golden ratio interval (≈833.1 ¢) to sound nice, you can take a timbre with harmonics 1 : φ : φ² : φ³ : ..., but this set of harmonics looks pretty scarce. What can you populate it with to still handle φ interval nicely but also to be more interesting and to make the timbre more adjustable?

Note that to construct a harmonic timbre from a “bare octave-allowing timbre” 1 : 2 : 4 : 8 : ..., one can just take sums of various subsets of {1, 2, 4, 8, ...} and take all of them as the new timbre. One then recovers all the natural numbers: 3 = 2 + 1, 5 = 4 + 1, 6 = 4 + 2, 7 = 4 + 2 + 1 and so on (of course you know your binary). We can apply the same sums-of-subsets construction here, but with a caveat: as φⁿ = φ^{n − 1} + φ^{n − 2}, we probably should disallow subsets like {φ², φ³, φ⁴}: in this one, φ⁴ effectively contained twice, and its sum is “incorrect”. (That’s easy to do: just disallow subsets which contain {φⁿ, φ^{n + 1}} for some n.) Proceeding this way from powers of φ, we get intervals

1, φ, φ + 1 ≡ φ², φ + 2, 2φ + 1 ≡ φ³, 2φ + 2, 3φ + 1, 3φ + 2 ≡ φ⁴, 3φ + 3, 4φ + 2, 4φ + 3, 4φ + 4, 5φ + 3 ≡ φ⁵, 5φ + 4, 6φ + 3, 6φ + 4, 6φ + 5, 7φ + 4, 7φ + 5, 8φ + 4, 8φ + 5 ≡ φ⁶, ... (G1)

We can note that neighboring intervals in this list differ either by 1 or φ − 1 ≈ 0.68, so they are spaced quite nicely to not be immediately a dissonant mess. (As in harmonic timbres they are all spaced by 1 and that sounds nice, given the greater harmonics are very quiet in regard to the small ones. And 0.68 is pretty close to 1 and is rarer encountered.)

Now multiply an interval r from this list by φ. As it’s a sum of powers of φ with no exponents differing by just 1, so is r φ. We can place other rules on powers in these sums, given these rules behave well under multiplication by φ.

We can slightly depart from a sums-of-subsets approach, filtering all possible m φ + n intervals in another way: as earlier, include each power of φ, and also as earlier make differences between adjacent intervals 1 or φ − 1, but no other constraints. Though I feel the intervals picked, considered as points (m, n) in the plane, should be close to the polygonal chain with vertices φ^k.

The following ASCII art illustrates such a planar representation for the interval list (G1) constructed above. It’s easily seen we can change an angle here and there, e. g. add 2φ + 3 while leaving out 3φ + 1.

  | 0 1 2 3 4 5  n
--+--------------→
0 | o-@   .   .
  |  /    .   .
1 | @-@-o .   .
  |    /  .   .      @ — powers of φ
2 |   @-o .   .      o — other intervals
  |    /  .   .      - — adding 1
3 |   o-@-o   .      / — adding (φ − 1)
  |      /    .
4 |     o-o-o .
  |        /  .
5 |       @-o .
  |        /  .
6 |       o-o-o
  |          / 
7 |         o-o
  |          / 
8 |         o-@-...
  |
m ↓

Initially I came to this scheme by taking base-Fibonacci numeral system but treating each Fibonacci number as a power of φ. I tried to compact the description but it might have gone hard to understand, so feel free to comment.

And I think something in this vein may be possible for any other interval which is a root x of a low-degree polynomial equation xⁿ = ... with integer coefficients (or even rational ones?). And I hope very much such a timbre sounds well — hadn’t tested that yet.

Another timbre

Now I think (G1) has its harmonics too close. We can fix this without remorse if we treat 1 as somewhat distinct from all others and start really adding two chosen differences only from φ. In that case we can choose 1 and φ (we may just scale all of (G1) by φ, effectively skipping some harmonics that are too close to their neighbors):

1, φ, φ + 1 ≡ φ², 2φ + 1 ≡ φ³, 3φ + 1, 3φ + 2 ≡ φ⁴, 4φ + 2, 4φ + 3, 5φ + 3 ≡ φ⁵, 6φ + 3, 6φ + 4, 7φ + 4, 8φ + 4, 8φ + 5 ≡ φ⁶, ... (G2)

Other findings without structuring

We can use a similar approach to build a simple “√2-enduring” timbre:

1, √2, 2, (√2 + 1), 2√2, √2 + 2, 4, (√2 + 3), 2√2 + 2, (3√2 + 1), 4√2, 3√2 + 2, 2√2 + 4, √2 + 6, 8, ... (S1 and S2)

Here we also can either use differences √2 − 1 ≈ 0.4 and 2 − √2 ≈ 0.6 right from the start, or we can start adding 2 − √2 and 2√2 − 2 ≈ 0.8 just after reaching 2, effectively skipping half of the harmonics of the first timbre each time we go from an even power of √2 to the next odd power.

News 2021-10

Regarding (G1)

Finally I’ve written some code to generate all that stuff. Let’s start with (G1), as it’s pretty easy to implement, if I’m not mistaken that it’s specifically (G1) I made. You just take a fibonacci-base system and represent consecutive natural numbers in it: 1, 10, 100, 101, 1000, 1001, 1010, 10000, 10001, 10010, 10100, 10101, 100000, … — and then treat these as if they are base-φ. That does the trick, and rarefied timbres like (G2) might be filtered out by simply banning some digit combinations (that’s not new at this stage: you get these digit strings for (G1) by forbidding consecutive 11 in all possible binary strings).

Well, theory aside I have two recordings for you:

A showcase of this timbre at intervals of several φ down and up. Harmonics are smeared a bit like in Paul Naşca’s PADSynth, and it makes the timbre sound less harsh IMO.

Something like a chord (φ + 2) : 3φ : (φ + 5) : (3φ + 3). I picked these more or less at random, just for the intervals to be not too small nor too wide.

Hopefully I’ll make it somehow more streamlined in the following days.

Regarding (G2)

Okaey it was shockingly easy to render (G2) this night too. 🎻 It turned out I myself wrote above that all you need is multiply all ratios of (G1) by φ (and add 1 at the start of it all), so I just did that. Here is the comparison:

Showcase of this one like the one before.

The same chord, though I tried on a lower note (2000¢ lower, in fact). Still a pretty nice chord eh?

Regarding a silver timbre

This shouldn’t be neither (S1) nor (S2): first, the harmonics seem to sit too far apart, and second, I based the timbre on the silver ratio s = √2 + 1, not just plain √2. Listen here:

Sampled in silver ratios.

Sampled in octaves which shares some partials. Come to that, I should’ve tried octaves for golden timbres too: they should handle this about right too.

Chord, this time (s + 2) : (2s + 1) : (s + 5) : (3s + 3). I think I picked these intervals in a bad way, the chord for golden timbres sounded nicer for me.

First ten partials:

— a bit over-rarified.

A late thought: oh of course that would be too rarified: s > 2! (Not a factorial.) I probably was wise to pick √2 when I mused about these things the first time, but that was too long ago, I forgot half the metadata. So I think a proper (S2) timbre is coming soon, but first I sleep.

Simple intervals played with (G1)

Here are about a dozen samples added to the same pack at Freesound as the others above. Intervals played are 2:1, 3:2, 4:3, 5:3, 5:4, 6:5, 7:4, 7:5, 7:6, 8:7 and the “timbral equivalents” of those for (G1). The first eight partials of (G1) are

1:  1       1       1.00000
2:  φ       φ       1.61803
3:  φ+1     φ²      2.61803
4:  φ+2     φ²+1    3.61803
5:  2φ+1    φ³      4.23607
6:  2φ+2    φ³+1    5.23607
7:  3φ+1    φ³+φ    5.85410
8:  3φ+2    φ⁴      6.85410

and here I call 4x:3x an interval between 4th and 3rd partials, (φ+2):(φ+1) ≈ 1.38197, and like that for all others. Note this convention conflates many JI intervals into the same one, like 2x:1x = 3x:2x = 5x:3x = 7x:4x = φ.