Geology Reference
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Thus,
2
M
−
2
j
+
2,2
j
,2
1
2 (
M
2
F
B
P
3
=
+
−
2)
j
=
1
2
M
−
2
j
,1
+
2,1
1
2 (
M
2
j
j
,1
+
γ
∗
2,2
j
=
+
γ
2,2
.
(2.496)
+
2,1
+
−
2)
j
=
1
As before, the free parameter is chosen to minimise the prediction error power. For
a minimum of
P
3
the partial derivatives with respect to the real and imaginary parts
of γ
2,2
vanish. Then,
M
−
2
j
,1
j
,1
j
,1
j
,1
∂
P
3
∂Reγ
2,2
=
1
2
(
M
B
F
∗
j
+
2,1
+
γ
∗
2,2
B
∗
B
∗
F
j
+
2,1
B
+
+
γ
2,2
−
2
)
j
=
1
2,1
2,1
2,1
2,1
j
B
∗
j
,1
F
∗
j
F
∗
j
j
,1
+
γ
∗
2,2
j
+
+
γ
2,2
+
=
0,
(2.497)
+
+
+
+
and
M
−
2
j
,1
j
,1
j
,1
j
,1
∂
P
3
∂Imγ
2,2
=
i
2 (
M
F
∗
j
+
γ
∗
2,2
B
∗
B
∗
j
−
+
γ
2,2
+
2,1
+
2,1
−
2)
j
=
1
j
+
2,1
j
+
2,1
j
+
2,1
j
+
2,1
F
B
∗
j
,1
F
∗
F
∗
B
j
,1
+
γ
∗
2,2
F
−
+
γ
2,2
+
=
0.
(2.498)
Adding
i
×
(2.498) to (2.497) gives
j
+
2,1
m
−
2
j
,1
j
,1
j
+
2,1
B
∗
F
j
+
2,1
B
F
B
∗
j
,1
F
∗
+
γ
2,2
+
+
γ
2,2
=
0.
(2.499)
j
=
1
Solving for the value of γ
2,2
that minimises
P
3
,weget
2
M
−
2
j
,1
M
−
2
j
+
2,1
j
,1
2
F
B
∗
F
B
γ
2,2
=−
2
j
+
2,1
+
.
(2.500)
j
=
1
j
=
1
Successive coe
cients of the prediction error filter can be found in this manner
so that after
N
+
1 recursions it is found that
2
M
−
N
−
1
j
,
N
M
−
N
−
1
j
+
N
+
1,
N
j
,
N
2
j
+
N
+
1,
N
B
∗
γ
N
+
1,
N
+
1
=−
2
+
.
(2.501)
j
=
1
j
=
1
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