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《自适应滤波器原理》中文第四版课后题答案.pdf
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CHAPTER 1
1.1
Let
ru k( )
ry k( )
[
E u n( )u* n
k–(
)
]
=
[
E y n( )y* n k–(
)
]
=
We are given that
y n( )
=
a+(
u n
) u n
–
a–(
)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
(1)
(2)
(3)
ry k( )
=
E u n
a+(
[
(
) u n a–(
–
)
) u* n
(
k–+(
a
)
(
u* n a–
–
k–
)
)
]
=
2ru k( )
–
(
ru 2a
)
k+
(
–
ru
2a–
k+
)
1.2 We know that the correlation matrix R is Hermitian; that is
RH
R=
Given that the inverse matrix R-1 exists, we may write
R 1– RH
I=
where I is the identity matrix. Taking the Hermitian transpose of both sides:
RR H–
I=
Hence,
R H–
R 1–
=
That is, the inverse matrix R-1 is Hermitian.
1.3
For the case of a two-by-two matrix, we may
Ru
=
Rs Rν+
1
=
=
r11 r12
r21 r22
+
σ2 0
0 σ2
σ2+
r11
r21
r12
σ2+
r22
For Ru to be nonsingular, we require
(
det Ru
)
=
(
r11
σ2+
) r22
(
σ2+
)
–
r12r21
0>
With r12 = r21 for real data, this condition reduces to
(
r11
σ2+
) r22
(
σ2+
)
–
r12r21
0>
Since this is quadratic in
larity of Ru:
σ2
, we may impose the following condition on
σ2
for nonsingu-
σ2
>
1
(
--- r11
2
r22+
)
1
–
4∆
--------------------------------------
)2 1–
(
r11
r
r22+
where
∆
r
=
r11r22
2–
r12
1.4 We are given
R
=
1 1
1 1
This matrix is positive definite because
aT Ra
=
[
a1,a2
] 1 1
1 1
=
2
a1
+
2a1a2
+
a1
a2
2
a2
2
(
=
a1
a2+
)2
0>
for all nonzero values of a1 and a2
(Positive definiteness is stronger than nonnegative definiteness.)
But the matrix R is singular because
det R(
)
=
1( )2
–
1( )2
=
0
Hence, it is possible for a matrix to be positive definite and yet it can be singular.
1.5
(a)
RM+1
=
r 0( )
r
rH
RM
Let
1–
RM+1
=
a
b
bH
C
where a, b and C are to be determined. Multiplying (1) by (2):
IM+1
=
r 0( )
r
rH
RM
a
b
bH
C
where IM+1 is the identity matrix. Therefore,
r 0( )a
+
rHb
1=
ra RMb
+
0=
rbH RMC
+
IM=
r 0( )bH rHC+
0T=
From Eq. (4):
3
(1)
(2)
(3)
(4)
(5)
(6)
b
–=
1– ra
RM
Hence, from (3) and (7):
a
=
1
------------------------------------
1– r
rHRM
r 0( )
–
Correspondingly,
b
–=
1– r
RM
------------------------------------
1– r
rHRM
r 0( )
–
From (5):
C
=
1–
RM
–
1– rbH
RM
=
1–
RM
+
1– rrHRM
1–
RM
------------------------------------
1– r
rHRM
r 0( )
–
As a check, the results of Eqs. (9) and (10) should satisfy Eq. (6).
(7)
(8)
(9)
(10)
r 0( )bH rHC+
=
–
1–
r 0( )rHRM
------------------------------------
1– r
rHRM
r 0( )
–
+
1–
rHRM
+
1–
1– rrHRM
rHRM
-------------------------------------
1– r
rHRM
r 0( )
–
0T=
We have thus shown that
1–
RM+1
=
=
0
0
0
0
0T
1–
RM
0T
1–
RM
+
a
1
1– r
RM
1–
rHRM
–
1– rrHRM
RM
1–
+
a
1
1– r
RM
–
[
1
–
1–
rHRM
]
4
where the scalar a is defined by Eq. (8):
(b)
RM+1
=
RM
rBT
rB*
r 0( )
Let
1–
RM+1
=
D
eH
e
f
where D, e and f are to be determined. Multiplying (11) by (12):
IM+1
=
RM
rBT
rB*
r 0( )
D
eH
e
f
Therefore
RMD rB*eH
+
I=
RMe
+
rB* f
0=
rBT e
r 0( ) f
+
1=
rBT D r 0( )eH
+
0T=
From (14):
e
=
1– rB* f
– RM
Hence, from (15) and (17):
f
=
---------------------------------------------
1– rB*
r 0( )
1
rBT RM
–
Correspondingly,
5
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
e
–=
1– rB*
RM
rBT RM
–
---------------------------------------------
1– rB*
r 0( )
From (13):
D
=
1–
RM
–
1– rB*eH
RM
=
1–
RM
+
1– rB*rBT RM
1–
RM
---------------------------------------------
1– rB*
r 0( )
–
rBT RM
As a check, the results of Eqs. (19) and (20) must satisfy Eq. (16). Thus
(19)
(20)
rBT D r 0( )eH
+
=
1–
rBT RM
+
1– rB*rBT RM
1–
rBT RM
------------------------------------------------
1– rB*
rBT RM
r 0( )
–
–
0T=
We have thus shown that
1–
r 0( )rBT RM
rBT RM
–
---------------------------------------------
1– rB*
r 0( )
1–
RM+1
=
1–
RM
0T
0
0
+
f
1– rB*rBT RM
RM
–
rBT RM
1–
1–
–
1– rB*
RM
1
=
1–
RM
0T
0
0
+
f
1– rB*
R– M
1
[
–
rBT RM
1–
]
1
where the scalar f is defined by Eq. (18).
1.6
(a) We express the difference equation describing the first-order AR process u(n) as
u n( )
=
v n( ) w1u n 1–(
+
)
where w1 = -a1. Solving this equation by repeated substitution, we get
u n( )
=
v n( ) w1v n 1–(
+
) w1u n
+
2–(
)
6
…=
=
v n( ) w1v n 1–(
+
) w1
+
2v n 2–(
) … w1
+
+
n-1v 1( )
(1)
Here we have used the initial condition
u 0( )
0=
or equivalently
u 1( )
v 1( )
=
Taking the expected value of both sides of Eq. (1) and using
[
E v n( )
]
µ=
for all n,
we get the geometric series
[
E u n( )
]
=
µ w1
+
µ w1
+
2µ … w1
+
n-1µ
+
=
n–
µ 1 w1
---------------
1 w1–
µn,
,
1≠
w1
w1
1=
This result shows that if
AR process u(n) is not stationary. If, however,
condition:
, then E[u(n)] is a function of time n. Accordingly, the
the AR parameter satisfies the
µ 0≠
1<
or
a1
1<
w1
then
[
E n( )
]
→
µ
--------------- as n
1 w1–
∞→
Under this condition, we say that the AR process is asymptotically stationary to order
one.
(b) When the white noise process v(n) has zero mean, the AR process u(n) will likewise
have zero mean. Then
7
var v n( )
[
]
=
σ
2
v
var u n( )
[
]
=
E u2 n( )
[
].
Substituting Eq. (1) into (2), and recognizing that for the white noise process
[
E v n( )v k( )
]
=
σ
2
v
0,
n
n
k=
k≠
we get the geometric series
(2)
(3)
var u n( )
[
]
=
σ
2 1 w1
(
2 w1
+
+
v
4 … w1
+
+
2n-2
)
=
σ
2n
2 1 w1
–
------------------
v
2–
1 w1
,
σ
2n,
v
1≠
w1
w1
1=
When |a1| < 1 or |w1| < 1, then
var u n( )
[
]
≈
2
σ
v
---------------
2–
1 w1
=
2
σ
v
--------------
2–
1 a1
for large n
(c) The autocorrelation function of the AR process u(n) equals E[u(n)u(n-k)]. Substituting
Eq. (1) into this formula, and using Eq. (3), we get
[
E u n( )u n k–(
)
]
=
σ
2 w1
(
k w1
+
v
k+2 … w1
+
+
k+2n-2
)
=
σ
2w1
v
2n
k 1 w1
–
------------------
2–
1 w1
,
1≠
w1
σ
2n,
v
w1
1=
8
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