Sun Sep 2 17:42:47 CEST 2012

Stochastic vs. Deterministic


Find a good mathematical model for representing real-valued, discrete
time "low pass" signal s[n] encoded in fixed sample rate binary signal

This consists mainly of finding a morphism between operations on b[n]
and real-valued continuous time signals, i.e. map the signals + a
collection of meaningful operations.


The main heuristic is that "proper decorrelation" is a good idea to
build an interesting morphism.

  b[n] = s[n] + e[n]

b[n] and s[n] are related through some kind of linear smoothing /
low-pass filtering operation.

An important property is that s[n] is recoverable from b[n] such that

  R (b[n]) = s[n] + e'[n]  with |e'| << |e|

meaning that a recovery function function R takes the binary sequence
b and pruces an approximation to s that is much better than b itself.
Ideally |e'| = 0.

Currently it is assumed that R is a linear low pass filter, which
means that e is a high pass signal: all the noise is in the higher
frequency bands.

Representing b[n] \in {1,+1} allows the use of the real multiplication
operator to represent the boolean function AND.  To make
multiplication of two binary signals b1[n] and b2[n] meaningful, the
correlation of their noise components should be meaningful:

     b1 = s1 + e1
     b2 = s2 + e2
     b1 b2 = s1 s2  +  s1 e2 + s2 e1 + e1 e2
     b12   = s12    +  e12

In some sense, the resulting signal should have the same properties,
such that R(b1 b2) ~= s1 s2

Sticking to the linear filter reconstruction property, we need e12 to
be high pass so it can be filtered out by R.  This gives 3 conditions:

     R(s1 e2) ~= 0
     R(s2 e1) ~= 0
     R(e1 e2) ~= 0

of course still satisfying the reconstruction properties for the
signals themselves, next to the reconstruction of the product above:

     R(e1)    ~= 0
     R(e2)    ~= 0

If R is linear, these equations are weighted infinite sums.

The question is now: given arbitrary s1 and s2, is it possible to
construct b1 = s1 + e1 and b2 = s2 + e2 such that the constraints are

My guess is that it is not possible to do this exactly and thus this
needs an error measure to quantify the approximation "~=".