To allow Bob to distinguish Alice's signal from the noise, Alice must choose a signal that is linearly independent
from the two noise signals. While Alice and Bob could theoretically agree on any linearly independent vector (as long
as it is linearly independent, a correct solution is possible), the computions are easiest if the vector is orthogonal
to the span of the two noise vectors.
The two noise polynomials can be represented as vectors in
R^{3}, and the noise plane
P can be defined as their span:
We want to find a unit vector orthogonal to the plane
P. It is sufficient to find a solution to the following system of
equations:
Solving the above yields the vector:
Next, we must determine the coefficients of the curve Bob observed.
This curve is a linear combination of the
noise polynomials and Alice's polynomial (Alice's polynoial is represented by
u).
| | (1)^{2} | (1) | 1 | (2)^{2} | (2) | 1 | (3)^{2} | (3) | 1 |
| |
| ⋅ | | | |
| = | |
| = | |
Finally, we must remove the contribution of the noise vectors from the observed vector in order
to isolate Alice's signal. There are two ways to approach this (they are equivalent).
One approach is to find the orthogonal projection of the observed vector onto the span of the noise vectors,
and then to subtract this projection from the observed vector. Another approach is to project the observed
vector directly onto span {
u}; we use the latter.
( | | ⋅ (1/√(11)) ⋅ | | ) ⋅ (1/√(11)) ⋅ | |
| |
| = | |
Since
u is a unit vector, the above vector's length represents the scalar by which
u was multiplied,
so its length is the scalar
s in
s ⋅
u.