Chapter_06 - CHAPTER 6 6.1 (a) We note that J ( n +1 ) = E...

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150 CHAPTER 6 6.1 (a) We note that where Hence, Dropping the terms that are not a function of µ ( n ), and differentiating J ( n +1) with respect to µ ( n ), we get Setting the result equal to zero, and solving for µ ( n ): Jn +1 () Een +1 2 [] = en +1 dn +1 w H n +1 u n +1 = +1 Edn +1 w H n +1 u n +1 2 = σ d 2 w n 1 2 -- µ n ()∇ n H p p H w n 1 2 µ n n = w n 1 2 µ n n H Rw n 1 2 µ n n + ∂µ n -------------- +1 n 1 2 µ n H n p 1 2 µ n p H n + = 1 2 µ n H n n 1 2 µ n n 1 2 µ n w n 1 2 µ n n   H R n } 1 2 H n p H n = 1 2 H n n w H n R n + µ n H n R n +
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151 (1) (b) We are given that Hence, Eq. (1) simplifies to Using instantaneous estimates for R and : we find that the corresponding value of is Correspondingly, we have which is recognized as the normalized LMS algorithm. µ o n () 1 2 -- H n Rw n w H n R n () ∇ H n pp H n + H n R n --------------------------------------------------------------------------------------------------------------------------------------- = n 2 n p = µ o n H n ()∇ n H n R n --------------------------------- = n R ˆ u n u H n = n 2 u n u H n w ˆ n u n d * n [] = 2 u n e * n = µ o n µ ˆ o n u H n u n en 2 u H n u n 2 2 --------------------------------------------------- = 1 u H n u n --------------------------- 1 u n 2 ------------------ == w ˆ n +1 w ˆ n µ ˜ u n 2 u n e * n + =
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152 6.2 6.3 The second statement is the correct one. The justification is obvious from the solution to Problem 2; see also Eq. (6.10) of the text. By definition, Since Replace all of these terms into the NLMS formula and look at each element in the vector. We then find that the correct answer is 6.4 δ w ˆ n +1 () w ˆ n +1 - w ˆ n = 1 u n 2 ------------------ u H n u n w ˆ n
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This note was uploaded on 09/11/2010 for the course EE EE254 taught by Professor Ujin during the Spring '10 term at YTI Career Institute.

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Chapter_06 - CHAPTER 6 6.1 (a) We note that J ( n +1 ) = E...

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