Filtros
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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