Digital Signal Processing Reference
In-Depth Information
x
T
†
2
μ
x
(
n
)
μ
δ
+
w
o
(
n
)
=
(
n
)
2
e
(
n
)
=
2
x
(
n
)
e
(
n
),
(4.26)
δ
+
x
(
n
)
x
(
n
)
which is the regularized NLMS update.
Now, the NLMS can be seen as an algorithm that at each time step computes the
new estimate by doing the orthogonal projection of the old estimate onto the plane
generated by
e
p
(
1
2
e
2
δ
+
x
(
n
)
μ
x
(
n
)
n
)
−
−
(
n
)
=
0. Notice that when
μ
=
1 and
δ
=
0,
the projection is done onto the space
e
p
(
0, which agrees with the interpretation
found in Sect.
4.2.1
. Looking at the NLMS as a projection algorithm will be actually
extended later to the family of affine projection algorithms. In Chap.
5
we will analyze
more deeply the concept of orthogonal projections and their properties.
n
)
=
4.2.4 One More Interpretation of the NLMS
Another interesting connection can be made between NLMS and LMS algorithms.
Consider the data available at time
n
, i.e.,
d
(
n
)
and
x
(
n
)
, and an initial estimate
w
)
an iterating repeatedly
with the same input-output pairs
, the final estimate will be
the same as the one obtained by performing a single NLMS update with step size
equal to one. Although a proper proof is provided in [
3
], an intuitive explanation is
provided here.
Since
d
(
n
−
1
)
.In[
3
], it is shown that using an LMS with step size
μ
, starting at
w
(
n
−
1
are fixed, an error surface can be associated, which depends
only on the filter coefficients. If the time index is dropped (to emphasize that the
input and output are fixed) this surface can be expressed as:
(
n
)
and
x
(
n
)
e
2
d
2
w
T
xx
T
w
2
d
w
T
x
J
(
w
)
=|
|=
+
−
.
(4.27)
The LMS will perform several iterations at this surface. Using the subscript
i
to
denote the iteration number, then
w
i
−
1
x
w
i
=
w
i
−
1
+
μ
x
(
d
−
),
w
0
=
w
(
n
−
1
).
(4.28)
If
μ
is small enough to guarantee the stability of the algorithm, i.e., if
μ
is chosen
)
−
2
,(
4.28
) can be interpreted as an SD search on the surface
(
4.27
). In the limit, its minimum will be found. This minimum will satisfy
so that
μ<
2
x
(
n
xx
T
w
min
=
d
x
.
(Footnote 4 continued)
x
T
†
x
(
n
)
(
n
)
=
.
2
x
(
n
)
