Biomedical Engineering Reference
In-Depth Information
8.4.3 Deriving Causal Relationships in the Frequency Domain
In order to derive causal relationships between two channels, we should distinguish
between the amount of the signal power coming from its own past and the amount
of the power coming from the past of the other channel. On the basis of Eq. (
8.46
),
we are able to do this separation. That is, using Eq. (
8.46
), we have
x
(
f
)
=
H
xx
e
x
(
f
)
+
H
xy
e
y
(
f
).
(8.52)
On the right-hand side of the equation above, the first term is the intrinsic term
representing the influence of the past of
x
(
t
)
, and the second term represents the
influence of the past of
y
(
t
)
. We then decompose the power of the signal
x
(
f
)
, such
that
H
xx
H
xy
2
H
xx
Σ
x
H
xx
+
H
xy
Σ
y
H
xy
+
|
x
(
f
)
|
=
2
ʳ
.
(8.53)
On the right-hand side above, the first term represents the intrinsic term and the second
term expresses the influence from the past of
y
, i.e., the causal term. However,
it is unclear how much amount of the information represented by the third term is
attributed to the intrinsic or the causal terms. Therefore, to attain the intrinsic/causal
factorization, we first transform Eq. (
8.46
) into a domain where the innovation terms
are uncorrelated, and, in this domain, we factorize the total signal power into the
intrinsic and causal terms.
Let us factorize
(
f
)
2
|
x
(
f
)
|
in this manner. The transformation is performed in this
case using
ʠ
defined such that
10
−
ʳ
∗
/Σ
x
1
ʠ
=
.
(8.54)
Using Eq. (
8.46
) and
10
ʳ
∗
/Σ
x
1
ʠ
−
1
=
,
we have
x
H
xx
H
xy
H
yx
H
yy
10
ʳ
∗
/Σ
x
1
10
−
ʳ
∗
/Σ
x
1
e
x
(
(
f
)
f
)
=
.
(8.55)
y
(
f
)
e
y
(
f
)
We then have
⊡
⊣
⊤
⊦
x
ʳ
∗
H
xx
+
Σ
x
H
xy
H
xy
e
x
(
f
)
(
f
)
=
.
(8.56)
−
ʳ
∗
y
(
f
)
ʳ
∗
Σ
x
e
x
(
f
)
+
e
y
(
f
)
H
yx
+
Σ
x
H
yy
H
yy
