By Mitra A.

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**Additional info for Digital Signal Processing (SOLUTIONS MANUAL)**

**Example text**

N α µ[n] = (n + 1)x1 [n] = n x1 [n] + x1 [n]. ( + 1 1 − αe – jω = 1 (1 − αe – jω ) 2 using the differentiation-in-frequency Now asuume, the it holds for m. Consider next x m+1 [n] = (n + m)! (m)! 1 n + m (n + m − 1)! n n + m = α µ[n], = x m[n] = ⋅ n ⋅ x m[n] + x m [n]. (m − 1)! m m 1 d 1 1 α e– jω 1 X m +1 (e jω ) = j + = + – jω m – jω m – jω m +1 m dω (1 − α e ) (1 − α e ) (1 − αe ) (1 − α e– jω ) m 1 = . (1 − α e – jω ) m +1 50 1 ] (1− e− jω(2N+1) ) sin(ω N + 2 ) = (1 − e − jω ) sin(ω / 2) otherwise.

Sin ω αsin ω . Therefore, θ(ω ) == tan −1 – . 1 − α cosω 1 − α cosω (a) y[n] = µ[n] = y ev [n] + y od[n], where y ev [n] = ( y[n] + y[−n]) = ( µ[n] + µ[−n]) = + δ[ n], 2 2 2 2 1 1 and y od [n] = (y[n] − y[−n]) = (µ[n] − µ[−n]) = µ[n] − – δ[n] . 2 2 2 2 ∞ ∞ 1 1 1 Now, Yev (e jω ) = 2π δ(ω + 2πk) + = π δ(ω + 2πk) + . 2 2 2 k= –∞ k= –∞ 1 1 1 ∑ 1 1 ∑ 1 1 1 Since y od [n] = µ[n] − + δ[n], y od [n] = µ[n − 1] − + δ[n −1]. As a result, 2 2 2 2 45 1 1 1 2 − jω 1 2 1 2 1 2 y od [n] − y od [n −1] = µ[n] − µ[n − 1] + δ[n −1] − δ[n] = δ[n] + δ[n −1].

N= 0 N−1 ˜ [k + l N] = X ∑ ˜x[n] e − j2π(k+ l N)n / N n =0 N−1 = ∑ ˜x[n] e n =0 − j2πkn / N − j2πl n e N−1 = ∑ x˜[n]e − j2πkn / N = X˜ [k]. 35 (a) x˜ 1[n] = cos = e jπn / 4 + e − jπn / 4 . The period of x˜ 1[n] is N = 8. 4 2 7 7 j2πn / 8 − j2πkn / 8 − j2πn / 8 − j2πkn / 8 ˜ [k] = 1 X e e + e e 1 2 n= 0 n=0 7 7 1 = e− j2πn(k −1) / 8 + e −j2πn( k+1) / 8 . Now, from Eq. 28) we observe 2 n=0 n= 0 { ∑ ∑ ∑ 7 ∑ n =0 } ∑ { 8, for k = 1, e− j2πn(k −1) /8 = 0, otherwise, and { k = 1, 7, ˜ [k] = 4, X 1 0, otherwise.