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Solutions Manual for Adaptive Filter Theory 5th Edition by Haykin IBSN 9780132671453

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Solutions Manual for Adaptive Filter Theory 5th Edition by Haykin IBSN 9780132671453 Full download: https://goo.gl/ZNQPPH adaptive filter theory simon haykin 5th edition pdf adaptive filter theory 5th edition pdf adaptive filter theory simon haykin 4th edition pdf adaptive filter theory haykin pdf adaptive filter theory 4th edition pdf adaptive filter theory haykin pdf free download adaptive filter theory pdf adaptive filter theory simon haykin 3rd edition pdf free download
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    Solutions Manual for Adaptive Filter Theory 5th Edition by Haykin IBSN 9780132671453 Full download: http://downloadlink.org/p/solutions-manual-for-adaptive-filter-theory-5th-edition-by-haykin-ibsn-9780132671453/  Chapter 2   Problem 2.1   a)   Let w k = x +  j y  p(− k) = a +  j  b  We may then write f = w k     p ∗   ( − k  )   =(x +  j y)(a −    j  b )   =(ax +  by) +  j(ay −    bx )  Letting where f = u +  j v   u = ax +  by v = ay −    bx Hence,     ∂ u ∂ u   = a =  b   ∂ x ∂ y  ∂ v ∂ v  = a   ∂ y ∂ x   = −  b    From these results we can immediately see that ∂ u ∂ v  =   ∂ x ∂ y  ∂ v ∂ u   ∂ x = −   ∂ y In other words, the product term w k     p ∗   (− k) satisfies the Cauchy-Riemann equations, and so this term is analytic.  b)   Let Let f = w k     p ∗   ( − k  )   =(x −    j y)(a +  j  b )   =(ax +  by) +  j(bx −   ay )   with f = u +  j v   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.    d   Problem 2.2   a)   From the Wiener-Hopf equation, we have w 0 = 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  w 0 =  0.75  −0.5 1 0.25   1   0 . 375   =  0.75 0  0 . 5   = 0   b)   The minimum mean-square error is J min   = σ 2 −  p H   w 0   = σ 2   −   0 . 5   0 . 25   = σ 2 − 0.25 0 . 5   0
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