ALL CHAPTERS COVERED
, CHAPTER 1
1.1 Let
r u ( k ) = E [ u ( 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
Ru = Rs + Rν
1
, r 11 r 12 2
= + σ 0
r 21 r 22 2
0 σ
2
r 11 + σ r 12
=
2
r 21 r 22 + σ
For Ru to be nonsingular, we require
2 2
det ( R u ) = ( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0
With r12 = r21 for real data, this condition reduces to
2 2
( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0
2 2
Since this is quadratic in σ , we may impose the following condition on σ for nonsingu-
larity of Ru:
2 1 4∆ r
σ > --- ( r 11 + r 22 ) 1 – --------------------------------------
2 ( r + r ) – 1
2
11 22
2
where ∆ r = r 11 r 22 – r 12
1.4 We are given
R = 1 1
1 1
This matrix is positive definite because
a
a Ra = [ a 1 ,a 2 ] 1 1 1
T
1 1 a2
2 2
= a 1 + 2a 1 a 2 + a 2
2
, 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
R M+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
r(0) r a b
H
I M+1 =
r RM b C
where IM+1 is the identity matrix. Therefore,
H
r ( 0 )a + r b = 1 (3)
ra + R M b = 0 (4)
H
rb + R M C = I M (5)
H H T
r ( 0 )b + r C = 0 (6)
From Eq. (4):
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, CHAPTER 1
1.1 Let
r u ( k ) = E [ u ( 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
Ru = Rs + Rν
1
, r 11 r 12 2
= + σ 0
r 21 r 22 2
0 σ
2
r 11 + σ r 12
=
2
r 21 r 22 + σ
For Ru to be nonsingular, we require
2 2
det ( R u ) = ( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0
With r12 = r21 for real data, this condition reduces to
2 2
( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0
2 2
Since this is quadratic in σ , we may impose the following condition on σ for nonsingu-
larity of Ru:
2 1 4∆ r
σ > --- ( r 11 + r 22 ) 1 – --------------------------------------
2 ( r + r ) – 1
2
11 22
2
where ∆ r = r 11 r 22 – r 12
1.4 We are given
R = 1 1
1 1
This matrix is positive definite because
a
a Ra = [ a 1 ,a 2 ] 1 1 1
T
1 1 a2
2 2
= a 1 + 2a 1 a 2 + a 2
2
, 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
R M+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
r(0) r a b
H
I M+1 =
r RM b C
where IM+1 is the identity matrix. Therefore,
H
r ( 0 )a + r b = 1 (3)
ra + R M b = 0 (4)
H
rb + R M C = I M (5)
H H T
r ( 0 )b + r C = 0 (6)
From Eq. (4):
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