[PD] tabread4~~
Mathieu Bouchard
matju at artengine.ca
Fri Nov 23 22:25:36 CET 2007
On Fri, 23 Nov 2007, Charles Henry wrote:
> First off, we need a loose definition of timbre--timbre is the quality
> by which two sounds may be distinguished, where pitch, loudness, and
> onset time are the same. (in terms of signals, we have just described
> a non-linear space in the first place { s(t) such that |s(t)|^2 = E
> }... but we're not just talking about signals, we mean the sound
> experience itself)
I think that you are right that it can't be all done within a linear
framework: there needs some slight mangling of a linear space to do the
work.
Oh, well, you could do most of the work using a linear space, and then at
the last possible moment, divide the space by products by a positive real,
so that there is one element per possible direction (or by any real, so
that there is one element per possible axis). This could be called R[x]/R+
or R[x]/R respectively. Also, this could be called "spherical space" and
"projective spherical space", respectively.
You could also suppose that the fundamental's amplitude is always 1, which
is another way to give you exactly one element for all possible loudnesses
of a sound that is otherwise the same. This is also better because then it
ensures that it's a unique timbre, as you can't set all odd harmonics to 0
in such a situation (this would have allowed you to pretend a 440 Hz sound
is also a 220 Hz sound and such). Also, sometimes affine spaces are easier
to work with than quotient spaces even if you use those quotient spaces as
little as possible.
A neato aspect of R[x]/R+ is that even though vector addition doesn't work
on it, vector multiplication by matrix works quite well, and for example
R^42/R+ can be acted upon by SO(42,R) and most any other matrix group...
although SO(42,R) is the most tightly fitting matrix group in this
case: SO matrices preserve the L2-norm of vectors, so what nicer thing can
there be for a set of pseudo-vectors in which L2-norm has been made
irrelevant?
> As such, it cannot be silence. And if silence is not a timbre in our
> space, what is the additive identity? The additive identity of x(t)
> most closely resembles x(t) itself, since loudness is irrelevant.
If you are using the affine space, you can't simply add and you can't
simply multiply by a scalar: instead, the fundamental operation is the
convex sum of "vectors": as a single operation, you add together any
number of vectors, weighted, where the total weight has to be 1, so that
the amplitude of the fundamental sticks to 1.
> OK, so how about linearity? If we take two timbres x(t) and y(t),
> then we can construct a timbre z(t,a)=ax(t)+(1-a)y(t) (0<=a<=1)
> which interpolates between x and y.
This is a special case of the "convexity" requirement.
> And let's take a particularly bad example. We'll take x(t) to be a
> harmonic series. Then, we'll let y(t) be the same harmonic series,
> with a single mis-tuned partial, while keeping pitch constant. Then
> z(t) becomes dissonant moving between x(t) and y(t), even though
> dissonance was not significant in x(t) or y(t).
I don't quite understand how this works. Can you make a version of this
example with actual figures?
> Can we move the central moment of spectral density all the way to
> infinity while keeping pitch constant?
What do you mean "to infinity" ?
Anyway, it depends on how "perceptual" you are trying to be, supposing
that we don't argue on the meaning of "to infinity".
> If the space of timbres is bounded, then it cannot be a vector space
> (because it fails to be closed under scalar multiplication).
You mean bounded how? bounded in amplitude or in frequency? if it's
bounded in frequencies, it's still linear. but you sound like you mean
it'd be bounded in amplitude, which wouldn't be as much linear, but the
spherical space above would make this issue moot anyway.
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| Mathieu Bouchard - tél:+1.514.383.3801, Montréal QC Canada
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