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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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