Charles Henry czhenry at gmail.com
Fri Nov 23 17:16:28 CET 2007

```> > I feel absolutely certain that I can convince you that timbre is *not* a
> > vector space, using only the defining properties of a vector space.
>
> Ok, let's do that. How do you prove it?

With another little thought experiment.  If I can't convince you, I'll
eat my words (yum)

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)

Key properties of a vector space are:
a) associative
b) commutative
c) There exists an additve identity
d) There exists an additive inverse
e) linearity (the space is closed under addition and scalar multiplication)

No prob: we will define x(t) and y(t) as timbres.  O(t) is the origin timbre.
Already, we run into a little problem as a vector space.  It's clear
we can define an additive inverse of a timbre x(t) as O(t)-x(t).  But
what is O(t)?  Since O(t) is a point in our space, it must have
loudness and pitch.  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.

Now if we include an inner product on our vector space, we can create
orthogonal dimensions of timbre, which depends on the location of
O(t).  We can have a space which is the orthogonal complement of a
vector x(t).  How can we be certain that O(t) is in the center of our
space?  If we move O(t), we would also be changing the orthogonal
complement of x(t).  (this is not such a bad problem)

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.
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).

Lastly, is our space bounded?  Can we find a certain maximum
dissonance/consonance?  Can we move the central moment of spectral
density all the way to infinity while keeping pitch constant?  If the
space of timbres is bounded, then it cannot be a vector space (because
it fails to be closed under scalar multiplication).

These are just some of the things I have been thinking about when it
comes to timbre.  My general view of the auditory system is that it is
a huge mish-mash of non-linearities.  Now, the approach of
psychoacoustics has often been to treat those non-linearities one at a
time, as if they don't intersect with each other.  That's fine for
proving an effect occurs (in fact it's preferrable to having umpteen
million variables to consider), but when you start putting them all
together from experiments that weren't designed as such, you're bound
to miss the ways the non-linearities interact with each other.

>
> > However, getting from A to B, and showing this is true would take an
> > exquisitely designed experiment, a real work of art :P
>
> That's a detail :-P
>
> Especially as I believe that timbre is a vector space. This is as long as
> we agree that timbre is a reduced form of the spectrum of a periodic
> sound, and not the more complicated things that happen with attacks, nor
> the whole range of an instrument.
>
>
>   _ _ __ ___ _____ ________ _____________ _____________________ ...
> | Mathieu Bouchard - tél:+1.514.383.3801, Montréal QC Canada

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