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adaptive filter theory(Simon.Haykin)(answer)
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更新于2023-03-03
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For courses in Adaptive Filters. Haykin examines both the mathematical theory behind various linear adaptive filters and the elements of supervised multilayer perceptrons. In its fourth edition, this highly successful book has been updated and refined to stay current with the field and develop concepts in as unified and accessible a manner as possible.
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1
CHAPTER 1
1.1 Let
(1)
(2)
We are given that
(3)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
1.2 We know that the correlation matrix R is Hermitian; that is
Given that the inverse matrix R
-1
exists, we may write
where I is the identity matrix. Taking the Hermitian transpose of both sides:
Hence,
That is, the inverse matrix R
-1
is Hermitian.
1.3 For the case of a two-by-two matrix, we may
r
u
k() Eun()u
*
nk–()[]=
r
y
k() Eyn()y
*
nk–()[]=
yn() un a+()un a–()–=
r
y
k() Eun a+()un a–()–()u
*
nak–+()u
*
na– k–()–()[]=
2r
u
k() r
u
2ak+()– r
u
2a– k+()–=
R
H
R=
R
1–
R
H
I=
RR
H–
I=
R
H–
R
1–
=
R
u
R
s
R
ν
+=
2
For R
u
to be nonsingular, we require
With r
12
= r
21
for real data, this condition reduces to
Since this is quadratic in , we may impose the following condition on for nonsingu-
larity of R
u
:
where
1.4 We are given
This matrix is positive definite because
r
11
r
12
r
21
r
22
σ
2
0
0 σ
2
+=
r
11
σ
2
+ r
12
r
21
r
22
σ
2
+
=
det R
u
() r
11
σ
2
+()r
22
σ
2
+()r
12
r
21
0>–=
r
11
σ
2
+()r
22
σ
2
+()r
12
r
21
0>–
σ
2
σ
2
σ
2
1
2
---
r
11
r
22
+()1
4∆
r
r
11
r
22
+()
2
1–
--------------------------------------–
>
∆
r
r
11
r
22
r
12
2
–=
R
11
11
=
a
T
Ra a
1
,a
2
[]
11
11
a
1
a
2
=
a
1
2
2a
1
a
2
a
2
2
++=
3
for all nonzero values of a
1
and a
2
(Positive definiteness is stronger than nonnegative definiteness.)
But the matrix R is singular because
Hence, it is possible for a matrix to be positive definite and yet it can be singular.
1.5 (a)
(1)
Let
(2)
where a, b and C are to be determined. Multiplying (1) by (2):
where I
M+1
is the identity matrix. Therefore,
(3)
(4)
(5)
(6)
From Eq. (4):
a
1
a
2
+()
2
0>=
det R() 1()
2
1()
2
–0==
R
M+1
r 0()
r
r
H
R
M
=
R
M+1
1–
a
b
b
H
C
=
I
M+1
r 0()
r
r
H
R
M
a
b
b
H
C
=
r 0()a r
H
b+1=
ra R
M
b+ 0=
rb
H
R
M
C+ I
M
=
r 0()b
H
r
H
C+ 0
T
=
4
(7)
Hence, from (3) and (7):
(8)
Correspondingly,
(9)
From (5):
(10)
As a check, the results of Eqs. (9) and (10) should satisfy Eq. (6).
We have thus shown that
bR
M
1–
ra–=
a
1
r 0() r
H
R
M
1–
r–
------------------------------------=
b
R
M
1–
r
r 0() r
H
R
M
1–
r–
------------------------------------
–=
CR
M
1–
R
M
1–
rb
H
–=
R
M
1–
R
M
1–
rr
H
R
M
1–
r 0() r
H
R
M
1–
r–
------------------------------------+=
r 0()b
H
r
H
C+
r 0()r
H
R
M
1–
r 0() r
H
R
M
1–
r–
------------------------------------– r
H
R
M
1–
r
H
R
M
1–
rr
H
R
M
1–
r 0() r
H
R
M
1–
r–
-------------------------------------++=
0
T
=
R
M+1
1–
0
0
0
T
R
M
1–
a
1
R
M
1–
r
r
H
R
M
1–
–
R
M
1–
rr
H
R
M
1–
+=
0
0
0
T
R
M
1–
a
1
R
M
1–
r–
1 r
H
R
M
1–
–[]+=
5
where the scalar a is defined by Eq. (8):
(b) (11)
Let
(12)
where D, e and f are to be determined. Multiplying (11) by (12):
Therefore
(13)
(14)
(15)
(16)
From (14):
(17)
Hence, from (15) and (17):
(18)
Correspondingly,
R
M+1
R
M
r
BT
r
B*
r 0()
=
R
M+1
1–
D
e
H
e
f
=
I
M+1
R
M
r
BT
r
B*
r 0()
=
D
e
H
e
f
R
M
Dr
B*
e
H
+ I=
R
M
er
B*
f+ 0=
r
BT
e r 0()f+1=
r
BT
D r 0()e
H
+ 0
T
=
e – R
M
1–
r
B*
f=
f
1
r 0() r
BT
R
M
1–
r
B*
–
---------------------------------------------=
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