Adaptive Filter Theory 5th Edition Haykin SOLUTIONS MANUAL Full download at: Chapter 2 Problem 2.1 a) Let wk = x + j y p(−k) = a + j b We may then write f =wk p∗ (−k) =(x + j y)(a − j b) =(ax + by) + j(ay − bx) Letting f = u +jv where u = ax + by v = ay − bx 21 Hence, ∂u =a ∂x ∂v =a ∂y ∂u =b ∂y ∂v = −b ∂x 22 PROBLEM 2.1. Free rock band songs download. From these results we can immediately see that ∂u ∂v = ∂x ∂y ∂v ∂u =− ∂x ∂y In other words, the product term wk p∗ (−k) satisfies the Cauchy-Riemann equations, and so this term is analytic. B) Let f =wk p∗ (−k) =(x − j y)(a + j b) =(ax + by) + j(bx − ay) Let f = u + jv with u = ax + by v = bx − ay Hence, ∂u =a ∂x ∂v =b ∂x ∂u =b ∂y ∂v = −a ∂y From these results we immediately see that ∂u ∂v = ∂x ∂y ∂v ∂u =− ∂x ∂y In other words, the product term w∗kp(−k) does not satisfy the Cauchy-Riemann equations, and so this term is not analytic. 23 PROBLEM 2.2. Problem 2.2 a) From the Wiener-Hopf equation, we have w0 = R−1 p (1) We are given that R = 1 0.5 0.5 1 p = 0.5 0.25 Hence the inverse of R is R−1 = 1 0.5 0.5 1 −1 1 1 = −0.5 0.75 −0.5 1 −1 Using Equation (1), we therefore get 1 1 −0.5 0.5 −0.5 1 0.25 0.75 1 0.375 = 0 0.75 0.5 = 0 w0 = b) The minimum mean-square error is Jmin =σd2 − pH w0 =σd2 − 0.5 0.25 0.5 0 =σd2 − 0.25 24 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 0.5 1 q11 q = 0.5 11 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 1 q1 = √ −1 2 Similarly, for the eigenvalue λ2 = 1.5, we may show that 1 1 q2 = √ 2 1 25 PROBLEM 2.3.

Adaptive Filters > Adaptive Filter Theory. PreK–12 Education. Haykin, McMaster University, Ontario Canada. Instructor resource file download. Adaptive Filter Theory, Simon O. Haykin, May 16, 2013, Technology & Engineering,. This is the eBook of the printed book and may not include any media, website.

Accordingly, we may express the Wiener filter in terms of its eigenvalues and eigenvectors as follows:! 2 X 1 qi qHi p w0 = λ i i=1 = = = = = = 1 1 q1 qH1 + q2 q2H p λ1 λ2 1 1 1 1 −1 + 1 1 −1 3 1 1 1 1 1 −1 0.5 + −1 1 0.25 3 1 1 4 2 − 0.5 3 3 2 4 0.25 − 3 3 4 1 − 6 6 1 1 − + 3 3 0.5 0 0.5 0.25 Problem 2.3 a) From the Wiener-Hopf equation we have w0 = R−1 p We are given R = (1) 1 0.5 0.25 0.5 1 0.5 0.25 0.5 1 and p = 0.5 0.25 0.125 T 26 PROBLEM 2.3. Hence, the use of these values in Equation (1) yields w0 =R−1 p 1 0.5 0.25 = 0.5 1 0.5 0.25 0.5 1 −1 0.5 0.25 0.125 1.33 −0.67 0 = −0.67 1.67 −0.67 0 −0.67 1.33 w0 = 0.5 0 0 0.5 0.25 0.125 T b) The Minimum mean-square error is Jmin =σd2 − pH w0 =σd2 − 0.5 0.25 0.125 0.5 0 0 =σd2 − 0.25 c) The eigenvalues of the matrix R are λ1 λ2 λ3 = 0.4069 0.75 1.8431 The corresponding eigenvectors constitute the orthogonal matrix: Q= −0.4544 −0.7071 0.5418 0.7662 0 0.6426 −0.4544 0.7071 0.5418 Accordingly, we may express the Wiener filter in terms of its eigenvalues and eigenvectors as follows:! 3 X 1 qi qHi p w0 = λ i i=1 27 PROBLEM 2.4. W0 = CHAPTER 2.

−0.4544 1 0.7662 −0.4544 0.7662 −0.4544 0.4069 −0.4544 −0.7071 1 −0.7071 0 −0.7071 0 + 0.75 0.7071 0.5418 1 0.6426 + 1.8431 0.5418 w0 = 0.5 418 0.6426 0.5418 0.2065 −0.3482 0.2065 1 −0.3482 0.5871 −0.3482 0.4069 0.2065 −0.3482 0.2065 1 + 0.75 0.5 0 −0.5 0 0 0 −0.5 0 0.5 0.2935 0.3482 0.2935 1 + 0.3482 0.4129 0.3482 1.8431 0.2935 0.3482 0.2935 = 0.5 0 0 Problem 2.4 By definition, the correlation matrix R = E[u(n)uH (n)] Where u(n) = 0.5 × 0.25 0.125 u(n) u(n − 1). U(0) Invoking the ergodicity theorem, N 1 X u(n)uH (n) R(N ) = N + 1 n=0 28 0.5 × 0.25 0.125 PROBLEM 2.5. Likewise, we may compute the cross-correlation vector p = E[u(n)d∗ (n)] as the time average N 1 X u(n)d∗ (n) p(N ) = N + 1 n=0 The tap-weight vector of the wiener filter is thus defined by the matrix product!−1 N!