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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.
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808 drum samples torrent. 1 2 i Accordingly, we may express the Wiener filter in terms of its eigenvalues and eigenvectors as follows: 2 1! W0 = X qiqH p λ i i=1 1 1 = q1 qH + λ1 q2qH p λ2 = 1 1 −1 + 1 1 1 1 0.5 −1 3 1 0.25 = 1 −1 + 1 1 1 0.5 −1 1 3 1 1 0.25 4 2 = 3 − 3 0.5 2 4 − 3 3 4 1 0.25 = 6 − 6 1 1 − 3 + 3 0.5 = 0 Problem 2.3 a) From the Wiener-Hopf equation we have w0 = R−1 p (1) We are given 1 0.5 0.25 R = 0.5 1 0.5 0.25 0.5 1 and p = 0.5 0.25 0.125 T • 27 PROBLEM 2.3.