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Dec 30, 2017 - Download at: adaptive filter theory haykin pdf free download. Filter theory 4th edition pdf adaptive filter theory simon haykin p. ∂v ∂u ∂x = − ∂y In other words, the product term wkp∗ (−k) satisfies. Jan 24, 2018 - Simon Haykin is an electrical engineer, noted for his pioneering work in with emphasis on applications in radar and communications.
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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 = Eu(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 = Eu(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! X u(n)d∗ (n) n=0 Problem 2.5 a) R =Eu(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) + Ev(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 = Eu(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).
(v∗ (n − k)). V(n − M + 1) =E rv(n) rv(n − 1). Rv(k − M + 1), 0 ≤ k ≤ M − 1 (3) where rv(k) is the autocorrelation of v(n) for lag k. Accordingly, the tap-weight vector of the (optimum) wiener filter is w0 =R−1 p =R−1 where p is defined in Equation (3). D) For a desired response d(n) = α(n) exp(− j ωτ). 31 PROBLEM 2.6.
Α α α 0 v σ2 d. The cross-correlation vector p is p =Eu(n)(d∗ n) =E(α(n)s(n) + v(n)) α∗ (n) exp(− j ωτ) =s(n) exp(j ωτ)E α(n) 2 =σ2 s(n) exp(j ωτ) =σ2 =σ2 1 exp(− j ω).
Exp((− j ω)(M − 1)) exp(j ωτ) exp(j ω(τ − 1)). Exp(j ωτ) exp((j ω)(τ − M + 1)) The corresponding value of the tap-weight vector of the Wiener filter is w =σ2 (σ2 s(n)sH (n) + R )−1 exp(j ωτ) exp(j ω(τ − 1)).α α exp((j ω)(τ − M + 1)) −1 exp(j ωτ) exp(j ω(τ − 1)) = s(n)sH (n) + 1 R α exp((j ω)(τ − M + 1)) Problem 2.6 The optimum filtering solution is defined by the Wiener-Hopf equation Rw0 = p (1) for which the minimum mean-square error is Jmin = σ2 − pH w0 (2). 32 PROBLEM 2.6.