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adaptive filter theory(Simon.Haykin)(answer)

adaptive

filter

theory;answer

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For courses in Adaptive Filters. Haykin examines both the mathematical theory behind various linear adaptive filters and the elements of supervised multilayer perceptrons. In its fourth edition, this highly successful book has been updated and refined to stay current with the field and develop concepts in as unified and accessible a manner as possible.

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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

k() Eun()u

nk–()[]=

k() Eyn()y

nk–()[]=

yn() un a+()un a–()–=

k() Eun a+()un a–()–()u

nak–+()u

na– k–()–()[]=

k() r

2ak+()– r

2a– k+()–=

1–

H–

1–

For R

to be nonsingular, we require

With r

= r

for real data, this condition reduces to

Since this is quadratic in , we may impose the following condition on for nonsingu-

larity of R

where

1.4 We are given

This matrix is positive deﬁnite because

0 σ

+ r

det R

() r

+()r

0>–=

+()r

0>–

---

+()1

4∆

+()

1–

--------------------------------------–







∆

–=

Ra a

[]

++=

for all nonzero values of a

and a

(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):

+()

0>=

det R() 1()

1()

–0==

M+1

r 0()

M+1

1–

M+1

r 0()

r 0()a r

b+1=

ra R

b+ 0=

C+ I

r 0()b

C+ 0

(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

1–

ra–=

r 0() r

1–

r–

------------------------------------=

1–

r 0() r

1–

r–

------------------------------------

–=

1–

–=

1–

r 0() r

1–

r–

------------------------------------+=

r 0()b

C+

r 0()r

1–

r 0() r

1–

r–

------------------------------------– r

1–

r 0() r

1–

r–

-------------------------------------++=

M+1

1–

–

1–

r–

1 r

1–

–[]+=

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,

M+1

r 0()

M+1

1–

M+1

r 0()

+ I=

f+ 0=

e r 0()f+1=

D r 0()e

+ 0

e – R

1–

r 0() r

1–

–

---------------------------------------------=

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zhuyi490

2014-01-23

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