From the Publisher: Haykin examines both the mathematical theory behind various linear adaptive filters with finite-duration impulse response (FIR) and the elements of supervised neural networks. The Third Edition of this highly successful book has been updated and refined to keep current with the field and develop concepts in as unified and accessible a manner as possible. Book Description Softcover. Condition: New. Brand New, International Edition, Paperback, ISBN and Cover image may differ but contents similar to U.S. Edition, Printed in Black & White. End Chapter Exercises may differ.

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Adaptive filter theory 5th edition haykin solutions manual • 1. 21 Adaptive Filter Theory 5th Edition Haykin SOLUTIONS MANUAL Full download at: haykin-solutions-manual/ Chapter 2 Problem 2.1 a) Let wk = x + j y p(−k) = a + j b We may then write f =wkp∗ (−k) =(x+ j y)(a − j b) =(ax + by) + j(ay − bx) Letting where f = u + j v u = ax + by v = ay − bx • 22 Hence, ∂u ∂u = a = b ∂x ∂y ∂v ∂v = a ∂y ∂x = −b • 23 PROBLEM 2.1.

K From these results we can immediately see that ∂u ∂v = ∂x ∂y ∂v ∂u ∂x = − ∂y In other words, the product term wkp∗ (−k) satisfies the Cauchy-Riemann equations, and so this term is analytic. B) Let Let f =wkp∗ (−k) =(x− j y)(a + j b) =(ax + by) + j(bx − ay) with f = u + jv u = ax + by v = bx − ay Hence, ∂u ∂u =a ∂x ∂y ∂v ∂v =b ∂x ∂y = b = −a From these results we immediately see that ∂u ∂v = ∂x ∂y ∂v ∂u ∂x = − ∂y In other words, the product term w∗ p(−k) does not satisfy the Cauchy-Riemann equations, and so this term is not analytic.

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• 24 PROBLEM 2.2. D d d Problem 2.2 a) From the Wiener-Hopf equation, we have w0 = R−1 p (1) We are given that 1 0.5 R = 0.5 1 0.5 p = 0.25 Hence the inverse of R is 1 0.5 −1 R−1 = = 0.5 1 1 1 −0.5 −1 0.75 −0.5 1 Using Equation (1), we therefore get 1 1 −0.5 0.5 w0 = 0.75 −0.5 1 0.25 1 0.375 = 0.75 0 0.5 = 0 b) The minimum mean-square error is Jmin =σ2 − pH w0 =σ2 − 0.5 0.25 =σ2 − 0.25 0.5 0 • 25 PROBLEM 2.2. C) The eigenvalues of the matrix R are roots of the characteristic equation: (1 − λ)2 − (0.5)2 = 0 That is, the two roots are λ1 = 0.5 and λ2 = 1.5 The associated eigenvectors are defined by Rq = λq For λ1 = 0.5, we have 1 0.5 q11 = 0.5 q11 0.5 1 q12 q12 Expanded this becomes q11 + 0.5q12 = 0.5q11 0.5q11 + q12 = 0.5q12 Therefore, q11 = −q12 Normalizing the eigenvector q1 to unit length, we therefore have 1 q1 = √ 2 1 −1 Similarly, for the eigenvalue λ2 = 1.5, we may show that 1 q2 = √ 2 1 1 • 26 PROBLEM 2.3.