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Solution Manual - Digital Signal Processing: A Computer-Based Approach Fourth Edition ( Sanjit K. Mitra (Author) latest Editon

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Digital Signal Processing: A Computer-Based Approa

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Digital Signal Processing

Solution Manual Digital Signal Processing: A Computer-Based Approach Fourth Edition Sanjit K. Mitra (Author)

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a computer based approach
solution manual digital signal processing
fourth edition sanjit k mitra author

Institution Digital Signal Processing: A Computer-Based Approa
Course Digital Signal Processing: A Computer-Based Approa

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Kylaexcell work 1 Solution Manual Digital Signal Processing: A Computer -Based Approach Fourth Edition Sanjit K. Mitra (Author) Kylaexcell work 2 0, 0,  0, 0, 0,  2 Chapter 2 2.1 (a) x1 1 = 22.85, x1 2 = 9.1396, x1  = 4.81, (b) x2 1 = 18.68, x = 7.1944, 2 x2  = 3.48. 2.2 [n] = 1,  n  0, n  0. Henceforth , [−n − 1] = 1,  n  0, n  0. n Thus, x[n] = [n] + [−n − 1]. 2.3 (a) Consider the sequence defined by x[n] =  [k]. k =− If n < 0, then k = 0 is not included in the sum and Henceforth , x[n] = 0 for n < 0. On the other hand, for n  0, k = 0 is included in the sum, and as a result, x[n] =1 for n  0. Therefore, x[n] = n [k] = 1, n  0, = [n].  k =− 0, n  0, (b) Since [n] = 1,  n  0, n  0, it follows that [n − 1] = 1,  n  1, n  1. Henc eforth , [n] − [n − 1] = 1,  n = 0, = [n]. n 0, 2.4 Recall [n] − [n − 1] = [n]. Henceforth , x[n] = [n] + 3[n − 1] − 2[n − 2] + 4[n − 3] = ([n] − [n − 1]) + 3([n − 1] − [n − 2]) − 2([n − 2] − [n − 3]) + 4([n − 3] − [n − 4]) = [n] + 2[n − 1] − 5[n − 2] + 6[n − 3] − 4[n − 4]. 2.5 (a) (b) c[n] = x[−n + 2] = {2 0 − 3  d[n] = y[−n − 3] = {− 2 7 8 − 2 1 0 − 1 5 − 4}, − 3 6 0 0},  (c) e[n] = w[−n] = {5 − 2 0 − 1 2 2 3 0 0},  (d) u[n] = x[n] + y[n − 2] = {− 4 5 1 − 2 3  − 3 1 0 8 7 − 2}, (e) v[n] = x[n]  w[n + 4] = {0 15 2 − 4 3 0  − 4 0}, (f) s[n] = y[n] − w[n + 4] = {− 3 4 − 5 0  0 10 2 − 2}, (g) r[n] = 3.5 y[n] = {21 − 10.5  − 3.5 0 2.8 24.5 − 7}. 2.6 (a) x[n] = −4[n + 3] + 5[n + 2] + [n + 1] − 2[n] − 3[n − 1] + 2[n − 3], y[n] = 6[n + 1] − 3[n] − [n − 1] + 8[n − 3] + 7[n − 4] − 2[n − 5], w[n] = 3[n − 2] + 2[n − 3] + 2[n − 4] − [n − 5] − 2[n − 7] + 5[n − 8], (b) Recall [n] = [n] − [n − 1]. Henceforth , x[n] = −4([n + 3] − [n + 2]) + 5([n + 2] − [n + 1]) + ([n + 1] − [n]) − 2([n] − [n − 1]) − 3([n − 1] − [n − 2]) + 2([n − 3] − [n − 4]) Kylaexcell work 3 h[0] + w[n] + z x[n _ 1] _ 1 11 z + w[n _ 1] _ 1 12 + z _ 1 z _ 1 x[n _ 2] 21 w[n 2] _ 22 = −4[n + 3] + 9[n + 2] − 4[n + 1] − 3[n] − [n − 1] + 3[n − 2] + 2[n − 3] − 2[n − 4], 2.7 (a) x[n] FROM the above figure it follows that y[n] y[n] = h[0]x[n] + h[1]x[n − 1] + h[2]x[n − 2]. (b) x[n] y[n] FROM the above figure we get w[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2]) and y[n] = w[n] + 12 w[n − 1] + 22w[n − 2]. we arrive at Making use of the first equation in the second y[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2]) + 12h[0](x[n − 1] + 11x[n − 2] + 21x[n − 3]) + 22h[0](x[n − 2] + 11x[n − 3] + 21x[n − 4]) = h[0](x[n] + (11 + 12 )x[n − 1] + (21 + 1211 + 22 )x[n − 2] + (1221 + 2211 )x[n − 3] + 2221x[n − 4]). (c) Figure P2.1(c) is a cascade of a first -order section and a second -order section. The input -output relation remains unchanged if the ordering of the two sections is interchanged as shown below. x[n] y[n] z _ 1 x[n-1] z _ 1 x[n-2] h[0] h[1] h[2] + + + w[n] 0.6 + u[n] + y[n+1] _ 1 _ 0.8 z z _ 1 0.3 0.4 + + w[n _ 1] z _ 1 _ 0.5 0.2 w[n _ 2] Kylaexcell work 4 The second -order section can be redrawn as shown below without changing its input - output relation. x[n] y[n] The second -order section can be seen to be cascade of two sections. Interchanging their ordering we finally arrive at the structure shown below: x[n] 0.6 _ 1 s[n] + + u[n] + _ 1 y[n+1] _ 1 z x[n _ 1] _ 1 0.3 + _ 0.8 + z u[n _ 1] _ 1 z 0.4 y[n] z x[n _ 2] 0.2 _ 0.5 z u[n _ 2] Analyzing the above structure we arrive at s[n] = 0.6x[n] + 0.3x[n − 1] + 0.2x[n − 2], u[n] = s[n] − 0.8u[n − 1] − 0.5u[n − 2], y[n + 1] = u[n] + 0.4 y[n]. FROM u[n] = y[n + 1] − 0.4 y[n]. Substituting this in the second equation we get after some algebra y[n + 1] = s[n] − 0.4 y[n] − 0.18y[n − 1] + 0.8y[n − 2]. Making use of the first equation in this equation we finally arrive at the desired input -output relation y[n] + 0.4 y[n − 1] + 0.18y[n − 2] − 0.2y[n − 3] = 0.6 x[n − 1] + 0.3x[n − 2] + 0.2 x[n − 3]. (d) Figure P2.19(d) is a parallel connection of a first -order section and a second -order section. The second -order section can be redrawn as a cascade of two sections as indicated below: x[n] y 2[n] + w[n] 0.6 + u[n] + y[n+1] _ 1 z _ 1 z _ 1 _ 0.8 z 0.3 0.4 + w[n _ 1] + z _ 1 z _ 1 _ 0.5 w[n _2] 0.2 + w[n] _ 1 _ 1 _ 0.8 z z 0.3 + w[n _ 1] + z _ 1 z _ 1 _ 0.5 w[n _2] 0.2

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Solution Manual - Digital Signal Processing: A Computer-Based Approach Fourth Edition ( Sanjit K. Mitra (Author) latest Editon

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