By Mitra A.

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2 n= 0 n =0 ∑ − j2πn(k +3) /24 − j12, k = 3, j12, k = 21, ˜ Hence X 2 [k] = 36, k = 4, 20, otherwise. 36 Since ˜p[n] is periodic with period N, then from Eq. 34, 1 ˜p[n] = N N−1 ∑ P˜ [k] e− j2πkn / N where using Eq. 168b) we get P˜ [k] = k =0 Hence ˜p[n] = N −1 ∑ p˜[n]e− j2πkn / N = 1. n= 0 1 N N−1 ∑ e− j2πkn / N . 37 X ω=2πk /N ∞ ∑ = X(e j2πk /N ) = x[n] e− j2πkn / N , – ∞ < k < ∞. n= −∞ ˜ [k + l N] = X(e j2π( k+ l N) / N ) = X(e j2πk / N e j2πl ) = X(e j2πk / N ) = X ˜ [k]. (a) X 1 (b) x˜ [n] = N = N−1 ∑ k =0 N−1 ∞ ˜ [k]e j2πkn / N = 1 ∑ ∑ x[l ]e − j2πkl /N e j2πkn / N X N k =0 l = −∞ N −1 − j2πkr 1 N−1 ∞ 1 ∞ j2πk(n− l ) /N ˜ x[l] e .

73 Let A n = n (λ i ) . Then = λ i . Now lim = 1. Since λ i < 1, there exists An n n→ ∞ n A 1 + λi ∞ a positive integer N o such that for all n > N o , 0 < n +1 < < 1. Hence ∑ n =0 A n An 2 converges. 74 (a) x[n] = {3 −2 0 1 4 5 2 } , −3 ≤ n ≤ 3. r xx[l ] = ∑ n =−3 x[n] x[n − l ] . Note, r xx[−6] = x[3]x[−3] = 2 × 3 = 6, r xx[−5] = x[3] x[−2] + x[2]x[−3] = 2 × (−2) + 5 × 3 =11, r xx[−4] = x[3]x[−1] + x[2] x[−2] + x[1] x[−3] = 2 × 0 + 5 × (−2) + 4 × 3 = 2, r xx[−3] = x[3] x[0] + x[2]x[−1] + x[1]x[−2] + x[0] x[−3] = 2 ×1 + 5 × 0 + 4 × (−2) +1 × 3 = −3, r xx[−2] = x[3]x[1] + x[2] x[0] + x[1] x[−1] + x[0]x[−2] + x[−1]x[−3] =11, r xx[−1] = x[3]x[2] + x[2] x[1] + x[1] x[0] + x[0]x[−1] + x[−1]x[−2] + x[−2]x[−3] = 28, r xx[0] = x[3] x[3] + x[2] x[2] + x[1]x[1] + x[0]x[0] + x[ −1] x[−1] + x[−2] x[−2] + x[−3]x[−3] = 59.

Jω 2 X(e ) dω = 2π ∑ x[n] 2 = 168π. ∫−π (Using Parseval's relation) n= −∞ (e) π 2 dX(e jω ) dω = 2π −π dω ∫ ∞ ∑ n ⋅ x[n] 2 = 1238π. 27 Let G1 (e jω ) denote the DTFT of g1 [n]. (b) g 2 [n] = g1[n] + g1 [n − 4]. Hence, the DTFT of g 2 [n] is given by G 2 (e jω ) = G1 (e jω ) + e− j4ω G1(e jω ) = (1+ e − j4ω )G1(e jω ). (c) g 3 [n] = g 1[−(n −3)]+ g 1[n − 4]. Now, the DTFT of g1 [−n] is given by G1 (e − jω ) . Hence, the DTFT of g 3 [n] is given by G 3 (e jω ) = e − j3ω G1(e − jω ) + e − j4ω G1(e jω ).