A COMPLETE SOLUTION MANUAL FOR ADAPTIVE
FILTER THEORY ONLY 4TH EDITION
CHAPTER 1
1.1 Let
r u (k ) = Eu(n)u * (n – k ) (1)
r y (k ) = E y(n)y * (n – k ) (2)
We are given that
y(n) = u(n + a) – u(n – a) (3)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
r y (k ) = E(u(n + a) – u(n – a))(u * (n + a – k ) – u * (n – a – k ))
= 2r u (k ) – r u (2a + k ) – r u (– 2a + k )
1.2 We know that the correlation matrix R is Hermitian; that is
H
R = R
Given that the inverse matrix R-1 exists, we may write
–1 H
R R = I
where I is the identity matrix. Taking the Hermitian transpose of both sides:
–H
RR = I
Hence,
–H –1
R = R
That is, the inverse matrix R-1 is Hermitian.
1.3 For the case of a two-by-two matrix, we may
1
,Ru = Rs + R
2
, r11 r12 2
= + 0
r21 r22 2
0
2
r11 + r 12
=
2
r21 r22 +
For Ru to be nonsingular, we require
2 2
det(R u ) = (r 11 + )(r 22 + ) – r 12 r 21 0
With r12 = r21 for real data, this condition reduces to
2 2
(r 11 + )(r 22 + ) – r 12 r 21 0
2 2
Since this is quadratic in , we may impose the following condition on for nonsingu-
larity of Ru:
2 1 4r
--(r11 + r22) 1 ---------------------
2 (r + r ) – 1
2
11 22
2
where = r r –r
r 11 22 12
1.4 We are given
R = 11
1 1
This matrix is positive definite because
a
a Ra = a1 ,a2 11 11 a1
T
2
2 2
= a1 + 2a 1 a 2 + a 2
3
, 2
= (a 1 + a 2 ) 0 for all nonzero values of a1 and a2
(Positive definiteness is stronger than nonnegative definiteness.)
But the matrix R is singular because
2 2
det(R) = (1) – (1) = 0
Hence, it is possible for a matrix to be positive definite and yet it can be singular.
1.5 (a)
H
r(0) r
RM +1 = (1)
r RM
Let
a H
–1 b
R M +1 = (2)
b C
where a, b and C are to be determined. Multiplying (1) by (2):
H H
r(0) r a b
IM +1 =
r RM b C
where IM+1 is the identity matrix. Therefore,
H
r(0)a + r b = 1 (3)
ra + RM b = 0 (4)
H
rb + R M C = IM (5)
H H T
r(0)b +r C = 0 (6)
From Eq. (4):
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FILTER THEORY ONLY 4TH EDITION
CHAPTER 1
1.1 Let
r u (k ) = Eu(n)u * (n – k ) (1)
r y (k ) = E y(n)y * (n – k ) (2)
We are given that
y(n) = u(n + a) – u(n – a) (3)
Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get
r y (k ) = E(u(n + a) – u(n – a))(u * (n + a – k ) – u * (n – a – k ))
= 2r u (k ) – r u (2a + k ) – r u (– 2a + k )
1.2 We know that the correlation matrix R is Hermitian; that is
H
R = R
Given that the inverse matrix R-1 exists, we may write
–1 H
R R = I
where I is the identity matrix. Taking the Hermitian transpose of both sides:
–H
RR = I
Hence,
–H –1
R = R
That is, the inverse matrix R-1 is Hermitian.
1.3 For the case of a two-by-two matrix, we may
1
,Ru = Rs + R
2
, r11 r12 2
= + 0
r21 r22 2
0
2
r11 + r 12
=
2
r21 r22 +
For Ru to be nonsingular, we require
2 2
det(R u ) = (r 11 + )(r 22 + ) – r 12 r 21 0
With r12 = r21 for real data, this condition reduces to
2 2
(r 11 + )(r 22 + ) – r 12 r 21 0
2 2
Since this is quadratic in , we may impose the following condition on for nonsingu-
larity of Ru:
2 1 4r
--(r11 + r22) 1 ---------------------
2 (r + r ) – 1
2
11 22
2
where = r r –r
r 11 22 12
1.4 We are given
R = 11
1 1
This matrix is positive definite because
a
a Ra = a1 ,a2 11 11 a1
T
2
2 2
= a1 + 2a 1 a 2 + a 2
3
, 2
= (a 1 + a 2 ) 0 for all nonzero values of a1 and a2
(Positive definiteness is stronger than nonnegative definiteness.)
But the matrix R is singular because
2 2
det(R) = (1) – (1) = 0
Hence, it is possible for a matrix to be positive definite and yet it can be singular.
1.5 (a)
H
r(0) r
RM +1 = (1)
r RM
Let
a H
–1 b
R M +1 = (2)
b C
where a, b and C are to be determined. Multiplying (1) by (2):
H H
r(0) r a b
IM +1 =
r RM b C
where IM+1 is the identity matrix. Therefore,
H
r(0)a + r b = 1 (3)
ra + RM b = 0 (4)
H
rb + R M C = IM (5)
H H T
r(0)b +r C = 0 (6)
From Eq. (4):
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