Adaptive filter theory simon haykin 3rd edition solution manual click here adaptive. Adaptive filter tue 29 jan 2019 025400 gmt solution manual haykin 3rd edition pdf. Immediately see that u x v y v x u y in other words the product term w kp k.

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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. • 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.

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. D d d T i Hence, the use of these values in Equation (1) yields w0 =R−1 p 1 0.5 0.25 −1 0.5 = 0.5 1 0.5 0.25 0.5 1 0.25 0.125 1.33 −0.67 0 0.5 = −0.67 1.67 −0.67 0.25 0 −0.67 1.33 0.125 w0 = 0.5 0 0 b) The Minimum mean-square error is Jmin =σ2 − pH w0 0.5 =σ2 − 0.5 0.25 0.125 0 0 =σ2 − 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: −0.4544 −0.7071 0.5418 Q = 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 1! W0 = X qiqH p λ i i=1 • 28 PROBLEM 2.4. 0.6426 0.5418 0.6426 0.5418 × 0.25 0.5418 0.125. 1 −0.4544 w0 = 0.4069 0.7662 −0.4544 0.7662 −0.4544 −0.4544 1 + 0.75 −0.7071 0 0.7071 −0.7071 0 −0.7071 1 + 1.8431 0.5418 0.5 1 0.2065 −0.3482 0.2065 w0 = 0.4069 −0.3482 0.5871 −0.3482 0.2065 −0.3482 0.2065 0.5 0 −0.51 + 0.75 0 0 0 −0.5 0 0.5 1 0.2935 0.3482 0.2935 0.5 + 0.3482 0.4129 0.3482 × 0.25 0.5 = 0 0 1.8431 0.2935 0.3482 0.2935 0.125 Problem 2.4 By definition, the correlation matrix R = E[u(n)uH (n)] Where u(n) u(n) = u(n − 1) u(0) Invoking the ergodicity theorem, R(N ) = 1 N + 1 NX u(n)uH (n) n=0 • 29 PROBLEM 2.5. Α Likewise, we may compute the cross-correlation vector p = E[u(n)d∗ (n)] as the time average p(N ) = 1 N + 1 NX u(n)d∗ (n) n=0 The tap-weight vector of the wiener filter is thus defined by the matrix product w0(N ) = NX u(n)uH (n) n=0!−1 N!

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X u(n)d∗ (n) n=0 Problem 2.5 a) R =E[u(n)uH (n)] =E[(α(n)s(n) + v(n))(α∗ (n)sH (n) + vH (n))] With α(n) uncorrelated with v(n), we have R =E[ α(n) 2 ]s(n)sH (n) + E[v(n)vH (n)] =σ2 s(n)sH (n) + Rv (1) where Rv is the correlation matrix of v b) The cross-correlation vector between the input vector u(n) and the desired response d(n) is p = E[u(n)d∗ (n)] (2) If d(n) is uncorrelated with u(n), we have p = 0 Hence, the tap-weight of the wiener filter is w0 =R−1 p =0 • 30 PROBLEM 2.5. Α =E v p c) With σ2 = 0, Equation (1) reduces to R = Rv with the desired response d(n) = v(n − k) Equation (2) yields p =E[(α(n)s(n) + v(n)v∗ (n − k))] =E[(v(n)v∗ (n − k))] v(n) v(n − 1).