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Chapter_01

# Chapter_01 - CHAPTER 1 1.1 Let ru(k = E[u(n)u(n k r y(k = E...

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1 CHAPTER 1 1.1 Let (1) (2) We are given that (3) Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get 1.2 We know that the correlation matrix R is Hermitian; that is Given that the inverse matrix R -1 exists, we may write where I is the identity matrix. Taking the Hermitian transpose of both sides: Hence, That is, the inverse matrix R -1 is Hermitian. 1.3 For the case of a two-by-two matrix, we may r u k ( ) E u n ( ) u * n k ( ) [ ] = r y k ( ) E y n ( ) y * n k ( ) [ ] = y n ( ) u n a + ( ) u n a ( ) = r y k ( ) E u n a + ( ) u n a ( ) ( ) u * n a k + ( ) u * n a k ( ) ( ) [ ] = 2 r u k ( ) r u 2 a k + ( ) r u 2 a k + ( ) = R H R = R 1 R H I = RR H I = R H R 1 = R u R s R ν + =

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2 For R u to be nonsingular, we require With r 12 = r 21 for real data, this condition reduces to Since this is quadratic in , we may impose the following condition on for nonsingu- larity of R u : where 1.4 We are given This matrix is positive definite because r 11 r 12 r 21 r 22 σ 2 0 0 σ 2 + = r 11 σ 2 + r 12 r 21 r 22 σ 2 + = det R u ( ) r 11 σ 2 + ( ) r 22 σ 2 + ( ) r 12 r 21 0 > = r 11 σ 2 + ( ) r 22 σ 2 + ( ) r 12 r 21 0 > σ 2 σ 2 σ 2 1 2 -- r 11 r 22 + ( ) 1 4 r r 11 r 22 + ( ) 2 1 ------------------------------------- > r r 11 r 22 r 12 2 = R 1 1 1 1 = a T Ra a 1 , a 2 [ ] 1 1 1 1 a 1 a 2 = a 1 2 2 a 1 a 2 a 2 2 + + =
3 for all nonzero values of a 1 and a 2 (Positive definiteness is stronger than nonnegative definiteness.) But the matrix R is singular because Hence, it is possible for a matrix to be positive definite and yet it can be singular. 1.5 (a) (1) Let (2) where a, b and C are to be determined. Multiplying (1) by (2): where I M +1 is the identity matrix. Therefore, (3) (4) (5) (6) From Eq. (4): a 1 a 2 + ( ) 2 0 > = det R ( ) 1 ( ) 2 1 ( ) 2 0 = = R M +1 r 0 ( ) r r H R M = R M +1 1 a b b H C = I M +1 r 0 ( ) r r H R M a b b H C = r 0 ( ) a r H b + 1 = r a R M b + 0 = rb H R M C + I M = r 0 ( ) b H r H C + 0 T =

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4 (7) Hence, from (3) and (7): (8) Correspondingly, (9) From (5): (10) As a check, the results of Eqs. (9) and (10) should satisfy Eq. (6). We have thus shown that b R M 1 r a = a 1 r 0 ( ) r H R M 1 r ------------------------------------ = b R M 1 r r 0 ( ) r H R M 1 r ------------------------------------ = C R M 1 R M 1 rb H = R M 1 R M 1 rr H R M 1 r 0 ( ) r H R M 1 r ------------------------------------ + = r 0 ( ) b H r H C + r 0 ( ) r H R M 1 r 0 ( ) r H R M 1 r ------------------------------------ r H R M 1 r H R M 1 rr H R M 1 r 0 ( ) r H R M 1 r ------------------------------------ + + = 0 T = R M +1 1 0 0 0 T R M 1 a 1 R M 1 r r H R M 1 R M 1 rr H R M 1 + = 0 0 0 T R M 1 a 1 R M 1 r 1 r H R M 1 [ ] + =
5 where the scalar a is defined by Eq. (8): (b) (11) Let (12) where D , e and f are to be determined. Multiplying (11) by (12): Therefore (13) (14) (15) (16) From (14): (17) Hence, from (15) and (17): (18) Correspondingly, R M +1 R M r BT r B * r 0 ( ) = R M +1 1 D e H e f = I M +1 R M r BT r B * r 0 ( ) = D e H e f R M D r B * e H + I = R M e r B * f + 0 = r BT e r 0 ( ) f + 1 = r BT D r 0 ( ) e H + 0 T = e R M 1 r B * f = f 1 r 0 ( ) r BT R M 1 r B * -------------------------------------------- =

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6 (19) From (13): (20) As a check, the results of Eqs. (19) and (20) must satisfy Eq. (16). Thus We have thus shown that where the scalar f is defined by Eq. (18).
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