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Chapter_04

# Chapter_04 - CHAPTER 4 4.1(a For convergence of the...

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108 CHAPTER 4 4.1 (a) For convergence of the steepest-descent algorithm: where is the largest eigenvalue of the correlation matrix R . We are given The two eigenvalues of R are λ 1 = 0.5 λ 2 = 1.5 Hence . The step-size parameter µ must therefore satisfy the condition We may thus choose µ = 1.0. (b) From Eq. 4.9 of the text, With µ = 1 and we therefore have 0 µ 2 λ max ----------- < < λ max R 1 0.5 0.5 1 = λ max 1.5 = 0 µ 2 1.5 ------- < < 1.334 = w n 1 + ( ) w n ( ) µ p Rw n ( ) [ ] + = p 0.5 0.25 = w n 1 + ( ) w n ( ) 0.5 0.25 1 0.5 0.5 1 w n ( ) + =

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109 That is, Equivalently, we may write (c) To investigate the effect of varying the step-size parameter µ on the trajectory, we find it convenient to work with v ( n ) rather than with w ( n ). The k th natural mode of the steepest descent algorithm is described by Specifically, For the initial condition, we have From the solution to Problem 2.2: 1 0 0 1 1 0.5 0.5 1 w n ( ) 0.5 0.25 = 0 0.5 0.5 0 w n ( ) 0.5 0.25 = w 1 n +1 ( ) w 2 n +1 ( ) 0 0.5 0.5 1 w 1 n ( ) w 2 n ( ) 0.5 0.25 = w 1 n +1 ( ) 0.5 w 2 n ( ) 0.5 = w 2 n +1 ( ) 0.5 w 1 n ( ) 0.25 = ν k n +1 ( ) 1 µλ k ( k n ( ), = k 1 2 , = ν 1 n +1 ( ) 1 0.5 µ ( 1 n ( ) = ν 2 n +1 ( ) 1 1.5 µ ( 2 n ( ) = v 0 ( ) ν 1 0 ( ) ν 2 0 ( ) = Q H w o =
110 Hence, That is, For n > 0, we have , k = 1,2 Hence, Solution - This represents an oscillatory trajectory. w o 0.5 0 = Q 1 2 ------ 1 1 1 1 = v 0 ( ) 1 2 ------ 1 1 1 1 0.5 0 = 1 2 ------ 0.5 0.5 = ν 1 0 ( ) ν 2 0 ( ) 0.5 2 ------- = = ν k n ( ) 1 µλ k ( ) n ν k 0 ( ) = ν 1 n ( ) 1 0.5 µ ( ) n ν 1 0 ( ) = ν 2 n ( ) 1 1.5 µ ( ) n ν 2 0 ( ) = µ 1 = ν 1 n ( ) 0.5 ( ) n ν 1 0 ( ) = ν 2 n ( ) 0.5 ( ) n ν 2 0 ( ) =

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111 This second solution represents a damped trajectory. The transition from a damped to an oscillatory trajector occurs at Specifically, for 0 < µ < 0.667, the trajectory is damped. On the other hand, for 0.667 < µ < 1.334 the trajectory is oscillatory. 4.2 We are given (a) The correlation matrix R = r (0). Hence, λ max = r (0) Correspondingly, (b) Time constant of the filter is (c) µ 0.1 = ν 1 n ( ) 0.59 ( ) n ν 1 0 ( ) = ν 2 n ( ) 0.85 ( ) n ν 2 0 ( ) = µ 1 1.5 ------- 0.667 = = J w ( ) J min r 0 ( ) w w o ( ) 2 + = µ max 2 λ max ----------- 2 r 0 ( ) ---------- = = τ 1 1 µλ 1 --------- 1 µ r 0 ( ) ------------- = Slope = 2 r (0)( w - w o ) J ( w ) 0 w o w w J min
112 4.3 (a) There is a single mode with eigenvalue λ 1 = r (0), and q 1 = 1, Hence, where v 1 ( n ) = q 1 ( w o - w ( n )) = ( w o - w ( n )) (b) 4.4 The estimation error e ( n ) equals where d ( n ) is the desired response, w ( n ) is the tap-weight vector, and u ( n ) is the tap-input vector. Hence, the gradient of the instantaneous squared error equals 4.5 Consider the approximation to the inverse of the correlation matrix: where µ is a positive constant bounded in value as where is the largest eigenvalue of R . Note that according to this approximation, we have R -1 (1) = µ I . Correspondingly, we may approximate the optimum Wiener solution as

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