On Jan 23, 1:04�am, HardySpicer <gyansor...@gmail.com> wrote:
> On Jan 23, 5:41�pm, David Bernier <david53...@gmail.com> wrote:
>
> > On Jan 21, 11:27�pm, c...@kcwc.com (Curt Welch) wrote:
>
> > > HardySpicer <gyansor...@gmail.com> wrote:
> > > > On Jan 22, 12:39=A0pm, c...@kcwc.com (Curt Welch) wrote:
> > > > > "Jeffery Tomas" <Jeffery_To...@Gmail.com> wrote:
> > > > > > The idea is very basic
>
> > > > > > The outputs(the mic's) are a linear combination of the input sources.
> > > > > > T=
> > > > he
> > > > > > coefficients give the relative magnitude of each source contributing
> > > > > > to the mic. If a source is further away it will contribute less to
> > > > > > the output. In fact, the coefficient should be 1/distance^2. If the
> > > > > > distanc=
> > > > e
> > > > > > is large 1/distance^2 is small and vice versa.
>
> > > > > > O_i =3D sum(1/a_k^2*S_k)
>
> > > > > > or
>
> > > > > > O =3D A*S
>
> > > > > > where O is the output's(mics), A is the "conversion matrix" and S is
> > > > > > th=
> > > > e
> > > > > > sources(which is what we want to recover). From basic linear algebra
> > > > > > we know S =3D A^-1*O.
>
> > > > > > There are statistical and deterministic methods to determine A since
> > > > > > it is essentially the "transfer function". The important thing is to
> > > > > > recognize is: If A is known and the sources and sinks(mics) are time
> > > > > > independent(or at least slowly changing) then we can recover the
> > > > > > source=
> > > > s.
> > > > > > The more accurate we know A the more accurate our decomposition will
> > > > > > be=
> > > > .
>
> > > > > > I'll demonstrate for two sources and sinks but I'll use inverted
> > > > > > linear distance wlog
>
> > > > > > O1 =3D a*S1 + b*S2
> > > > > > O2 =3D c*S1 + d*S2
>
> > > > > > Inverting gives
>
> > > > > > S1 =3D (dO1 - bO2)/(ad - cb)
> > > > > > S2 =3D -(cO1 - aO2)/(ad - cb)
>
> > > > > > Now the problem is to determine the coefficients.
>
> > > > > > This is simple a problem of plugging in values:
>
> > > > > > Analogy: y =3D a + bx + cx^2
>
> > > > > > How do we find a b and c? Well if we know 3 pairs of points we just
> > > > > > plu=
> > > > g
> > > > > > in and solve the linear system of equations.
>
> > > > > > Unfortunately we do not know the Source vector. BUT if we "listen"
> > > > > > for common situations we can easily narrow down the search and
> > > > > > possibly arrive at correct coefficients.
>
> > > > > > Now since we know the outputs we know there ratio. Assume O1/O2 =3D x
> > > > > > a=
> > > > nd
> > > > > > S1/S2 =3D y then
>
> > > > > > y =3D -(dx - b)/(cx - a)
>
> > > > > > now this is like the polynomial sampling problem I gave above. We
> > > > > > just don't have our points (x,y) to determine the coefficients.
>
> > > > > > There are several ways to potentially get y such as negative
> > > > > > feedback, guestimating, kalman filters, monte carlo, etc...
>
> > > > > > e.g., if we guess at the coefficients we arrive at a certain
> > > > > > geometrica=
> > > > l
> > > > > > scenario that will either coincide with the real one or be
> > > > > > "off"(producing estimated outputs that do not jive with what we are
> > > > > > actually getting... we can perturb our coefficients until they
> > > > > > "wiggle" there way to the correct values using appropriate adaptive
> > > > > > methods).
>
> > > > > > If we know certain properties of the sources then we can include
> > > > > > those =
> > > > in
> > > > > > the algorithms to reduce the complexity/increase the effectiveness of
> > > > > > t=
> > > > he
> > > > > > algorithms.
>
> > > > > > For example, by knowing that S1 and S2 are uncorrelated we can
> > > > > > determin=
> > > > e
> > > > > > that when S2 =3D 0 then we end up with a 1 source system making it
> > > > > > much easier to solve. e.g., our equation reduces to
>
> > > > > > S1 =3D aO1 + bO2
>
> > > > > > and instead of 4 coefficients we only have to find 2. This results in
> > > > > > a triangular geometry instead of a quadrilateral. Moreover, by
> > > > > > symmetric =
> > > > we
> > > > > > can do the same for S2 and we end up with another easy system to
> > > > > > solve. This applies to any number of sources. =A0With uncorrelated
> > > > > > sources we should be able to treat the system as just one source a
> > > > > > time.
>
> > > > > > For such indepence the coefficients are directly related to the
> > > > > > distanc=
> > > > es
> > > > > > and can be easily calculated from the mic intensities. (a =3D 1/d1^2
> > > > > > an=
> > > > d b
> > > > > > =3D 1/d2^2 but a is the level from O1 and b is the level from O2)
>
> > > > > > In general though it is not the case that the sources are
> > > > > > uncorrelated =
> > > > or
> > > > > > time independent and more advanced tricks or general approximations
> > > > > > may have to be used.
>
> > > > > > The more I think about it the more it seems like it might be quite
> > > > > > accurate for most signals and setups. One problem though is frequency
> > > > > > response as we are assuming a lossless transmission which is usually
> > > > > > no=
> > > > t
> > > > > > the case and could cause problems with some algorithmic approaches.
>
> > > > > Yes, but none of that has much to do with the how the Blind Source
> > > > > Separation solutions actually work. =A0It's fare more interesting than
> > > > > wh=
> > > > at
> > > > > you are suggesting above. =A0It makes none of the assumptions you talk
> > > > > ab=
> > > > out
> > > > > above and instead directly calculates all 4 coefficients without
> > > > > knowing anything about the sources.
>
> > > > > I believe what it is doing, is using the fact that when you sum
> > > > > separate non-correlated signals together, the sum tends to approach a
> > > > > Gaussian pow=
> > > > er
> > > > > distribution. =A0The more uncorrelated signals you sum together, the
> > > > > clos=
> > > > er
> > > > > it becomes to a pure Gaussian distribution.
>
> > > > > As such, by measuring how far away from a Gaussian distribution a given
> > > > > signal is, it can pick the four coefficients which maximize the
> > > > > non-Gaussian characteristics of the two extracted sources signals.
> > > > > =A0The=
> > > > �two
> > > > > sources signals will always be less Gaussian than the combined signal.
> > > > > =
> > > > =A0The
> > > > > only limitation to the approach is that the signals must be generally
> > > > > non-correlated, and they must not be Gaussian to start with (which is
> > > > > generally true for all signals a human would care about to start with).
>
> > > > > Hardy points out that in real life acoustic situations the summation is
> > > > > never just linear. =A0However, I don't know how relevant that actually
> > > > > is=
> > > > .
> > > > > If the source signals are arbitrary distorted BEFORE they are mixed,
> > > > > this sort of approach can still separate the signals (back to their
> > > > > pre-mixed distorted version). =A0And likewise, filtering =A0or
> > > > > distorting =A0the su=
> > > > mmed
> > > > > signal likewise, will not prevent their separation. =A0So the fact that
> > > > > a microphone and the recording system is non-linear is not by itself, a
> > > > > problem, as long as the same transform is effectively applied to each
> > > > > source _interdependently_. =A0But, I might be stupid here, because by
> > > > > definition, a transform which is not linear, might, by definition,
> > > > > imply the transform is creating dependency. =A0I'm not sure about that.
>
> > > > > If the recording system (and the type of transform Hardy is talking
> > > > > about=
> > > > ),
> > > > > causes cross-modulation of the signals, then the resulting distribution
> > > > > means the signals become at least partially correlated, which reduces
> > > > > how well they can be separated.
>
> > > > > But, as long as the linear combination dominates over any non-linear
> > > > > effects, I would think this sort of a approach could still do a good
> > > > > job =
> > > > of
> > > > > separation, meaning that the separated signals would be far better than
> > > > > t=
> > > > he
> > > > > mixed signals you started with. =A0And I would think that any audio
> > > > > recor=
> > > > ding
> > > > > application that was wroth using, the linear effects would strongly
> > > > > dominate over the non-linear effects.
>
> > > > > --
> > > > > Curt Welch =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0
> > > > > =
> > > > =A0 =A0 =A0 =A0 =A0 =A0 =A0http://CurtWelch.Com/
> > > > > c...@kcwc.com =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0
> > > > > =A0=
> > > > �=A0 =A0 =A0 =A0 =A0http://NewsReader.Com/
>
> > > > It makes none of the assumptions you talk about
> > > > above and instead directly calculates all 4 coefficients without
> > > > knowing
> > > > anything about the sources.
>
> > > > Doesn't actually, that is impossible. You see the amplitudes of what
> > > > you estimate are not known because the gains are as well.
> > > > This is more akin to decorrelation than estimating the unknown matrix.
> > > > Remember ICA doesn't give absolute amplitudes, these are scaled later.
> > > > Also the order of the estimates are unknown.
>
> > > Ah, Ok, after thinking about it I see why that's true.
>
> > > With two signals, you have a mixing matrix with 4 constants, which will
> > > both re-scale the input signals, and mix them in different ratios across
> > > the two output signals. �To extract the original signals, you only need to
> > > know the ratio, and it's impossible to extract the starting amplitude,
> > > because different amplitude source signals could produce the same mixed
> > > output signals, just by adjusting the 4 constants. �There's just not enough
> > > information there to know what starting amplitude was.
>
> > > So ICA gives you in effect, 2 out of the 4 values - which is all you need
> > > to separate the signals, but not enough to know what the original
> > > amplitudes were.
>
> > [...]
>
> > In multi-antenna radio-communications ,
> > such as the new 802.11n IEEE Wifi standard,
> > there's an important concept known as:
> > "quasi-static fading channel".
>
> > All it means, essentially, is that the
> > "mixing maxtrix" say A, e.g. 2x2 complex-valued
> > matrix if 2 receive, 2 transmit anttennas,
> > changes slowly over time.
>
> > With sound waves, if the mixing matrix or
> > mixing coefficients change by little
> > in 100 milliseconds or 0.1 second, one
> > might make an analogy to
> > a quasi-static fading channel, as understood
> > in radio wave propagation. �Being
> > quasistatic is better than being a fast-fading
> > channel for radio waves when the entropy
> > of the two source ...
>
> > read more �
>
> But they don't! Of course it depends on the environment but you can
> never have just a matrix. It has to be a transfer function matrix
> which is time-varying and more than often non-minimum phase. This is
> why it isn't easy.
[...]
Perhaps it might be useful to have a mental
picture of multiple echos in very mountainous
terrain, or a deserted city with lots of
sky-scrapers ...
For sound propagation indoors, I've heard of
but not studied how some concert halls
(say in Sydney, Australia) have good design for
music, others not so good. If I'm not
mistaken, the acoustics in a concert hall
differs according to whether there's lots
of people inside (during a concert), or just
a handful of people ...
David