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CHAPTER 1
1.1 Let r u ( k ) = E [ u ( n )u * ( n – k ) ] r y ( k ) = E [ y ( n )y * ( n – k ) ] We are given that y(n) = u(n + a) – u(n – a) 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 R
H(1) (2)

(3)

= R

Given that the inverse matrix R-1 exists, we may write R R
–1 H

= 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

2 r 11 r 12 σ 0 = + 2 r 21 r 22 0 σ

=

r 11 + σ r 212

r 12 r 22 + σ
2

For Ru to be nonsingular, we require det ( R u ) = ( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0 With r12 = r21 for real data, this condition reduces to ( r 11 + σ ) ( r 22 + σ ) – r 12 r 21 > 0 Since this is quadratic in σ , we may impose the following condition on σ for nonsingularity of Ru: 4∆ r   2 1 σ > -- ( r 11 + r 22 )  1 – ------------------------------------- 2 2  ( r + r ) – 1
11 22 2 2 2 2 2 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 T a Ra = [ a 1 ,a 2 ] 1 1 1 1 1 a2 = a 1 + 2a 1 a 2 + a 2
2 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 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) r(0) r R M+1 = r RM Let
–1 H 2 2

2

(1)

R M+1 =

a b

b C

H

(2)

where a, b and C are to be determined. Multiplying (1) by (2): r(0) r I M+1 = r RM
H

a b

b

H

C

where IM+1 is the identity matrix. Therefore, r ( 0 )a + r b = 1 ra + R M b = 0 rb + R M C = I M r (0 )b + r C = 0 From Eq. (4):
H H T H H

(3) (4) (5) (6)

3

b = – R M ra Hence, from (3) and (7): 1 a = ----------------------------------H –1 r ( 0 ) – r RM r Correspondingly, RM r b = – ----------------------------------H –1 r ( 0 ) – r RM r From (5): C = R M – R M rb
–1 –1 H –1

–1

(7)

(8)

(9)

–1 H –1 R M rr R M –1 = R M + ----------------------------------H –1 r ( 0 ) –r RM r

(10)

As a check, the results of Eqs. (9) and (10) should satisfy Eq. (6). r ( 0 )r R M H – 1 r R M rr R M r ( 0 )b + r C = – ----------------------------------- + r R M + -----------------------------------H –1 H –1 r ( 0 ) – r RM r r ( 0 ) – r RM r
H H H –1 H –1 H –1

= 0

T

We have thus shown that
T 1 –r R M 0 0 + a –1 0 R–1 R M r R – 1 rr H R – 1 M M M H –1

–1 R M+1 =T 1 0 0 = +a –1 0 R–1 –R M r M

[ 1 –r R M ]

H

–1

4

where the scalar a is defined by Eq. (8): r R M+1 = BT r(0) r Let
–1

(b)

RM

B*

(11)

D e
H

R M+1 =

e f

(12)

where D, e and f are to be determined. Multiplying (11) by (12): r I M+1 = BT r(0) r Therefore RM D + r RM e + r r r
BT B* H

RM

B*

D e
H

e f

e

= I

(13) (14) (15)
T

B*f = 0

e + r(0) f = 1 D + r ( 0 )e
H

BT

= 0

(16)

From (14): e = – RM r
– 1 B*

f

(17)

Hence, from (15) and (17): 1 f = -------------------------------------------BT – 1 B* r ( 0 ) – r RM r Correspondingly, (18)

5

RM r e = – -------------------------------------------BT – 1 B* r ( 0 ) – r RM r From (13): D = RM – RM r
–1 – 1 B* H

– 1 B*

(19)

e

– 1 B* BT– 1 RM r r RM –1 = R M + -------------------------------------------BT – 1 B* r ( 0 ) – r RM r

(20)

As a check, the results of Eqs. (19) and (20) must satisfy Eq. (16). Thus
BT H BT BT – 1 B* BT – 1 BT – 1 r ( 0 )r R M –1 r R M r r R M R M + ----------------------------------------------- – -------------------------------------------BT – 1 B* BT – 1 B* r ( 0 ) – r RM r r ( 0 ) – r RM r...
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