Digital Signal Processing Reference
In-Depth Information
From this we see that the optimum noise-cancelling transform is
−
R
12
R
−
1
A
opt
=
(20
.
61)
22
Substituting into Eq. (20.57) we see that the minimized error is
E
opt
=Tr
.
22
R
12
R
11
−
R
12
R
−
1
(20
.
62)
This completes the solution to the problem. Let us consider two extreme special
cases. First, if the noise components are identical (
e
1
=
e
2
)then
R
12
=
R
11
=
R
22
, whence
E
opt
=0
,
as expected. This is complete noise cancellation. At the other extreme, if the
noise components are orthogonal, that is,
R
12
=
0
,
then
A
opt
=
−
I
and
A
opt
=
0
and
E
opt
=Tr
R
11
.
That is, knowledge of
e
2
does not help to reduce the effect of the noise
e
1
.
The reconstruction error is
e
1
itself, so the mean square error is the trace of its
autocorrelation
R
11
.
This example is included here only to demonstrate another application of
matrix calculus. In this particular example the use of the orthogonality principle
leads to the same result rather quickly. See Problem 12.6.
20.6 Being careful with interpretations ...
The main purpose of this section is to emphasize the need for careful attention in
the application of notations such as
∂/∂
Z
and
∂/∂
Z
∗
.
For example, even though
some of the real examples summarized in Table 20.1 appear to be special cases
of the complex examples in Table 20.3, the formulas in the latter table must
be correctly interpreted to get those in the former. Consider for example the
formulas
∂
(
Z
†
AZ
)
∂z
rs
∂
(
Z
†
AZ
)
∂z
rs
=
Z
†
A
I
rs
,
T
=
I
rs
AZ
(20
.
63)
in Table 20.3. Suppose we want to recover the special case when everything
is real, that is, find
∂
(
X
T
AX
)
/∂x
rs
. fwelet
Z
=
X
(i.e., set
Y
=
0
in
Z
=
X
+
j
Y
), and similarly let
z
rs
=
x
rs
,
then Eqs. (20.63) reduce to
T
T
∂
(
X
AX
)
∂x
rs
∂
(
X
AX
)
∂x
rs
T
rs
AX
=
X
A
I
rs
,
=
I
(20
.
64)
That is, we get two different answers for the same real derivative! Something
is clearly wrong. Similar contradictions can be found even in the first entry
of Table 20.3, which has
∂z/∂z
=1
,∂z/∂z
∗
= 0. If we set
z
=
x
=
real
“everywhere” we get
∂x/∂x
=1
,∂x/∂x
=0
,
which again is a contradiction.
So we have to be careful in order to derive the real-valued special cases from
the complex formulas of Tables 20.3 and 20.4. Here are the points to be noted
carefully, in order to make correct use of the complex gradient operators:
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