By Sanjit Mitra

«Digital sign Processing: A Computer-Based process» is meant for a two-semester path on electronic sign processing for seniors or first-year graduate scholars. in keeping with person suggestions, a couple of new issues were extra to the second one variation, whereas a few extra themes from the 1st variation were got rid of. the writer has taken nice care to arrange the chapters extra logically through reordering the sections inside chapters. extra worked-out examples have additionally been incorporated. The e-book comprises greater than 500 difficulties and a hundred and fifty MATLAB workouts.

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Ex. 2 = 4 (m) y[n] = β n u[n] ∗ ∞ p=0 δ[n − 4p], |β| < 1 for n < 0 6 y[n] = 0 for n ≥ 0 n = 0, 4, 8, ... n 4 β 4k y[n] = k=0 1 − β 4( 4 +1) 1 − β4 n = 1, 5, 9, ... n y[n] = for n ≥ 0 n−1 4 β 4k−1 y[n] = k=0 n−1 1 − β 4( 4 +1) 1 − β4 1 y[n] = β for n ≥ 0 n = 2, 6, 10, ... n−2 4 β 4k−2 y[n] = k=0 y[n] = for n ≥ 0 n−2 1 − β 4( 4 +1) 1 − β4 1 β2 n = 3, 7, 11, ... 34. Consider the discrete-time signals depicted in Fig. 34. Evaluate the convolution sums indicated below. (a) m[n] = x[n] ∗ z[n] for n + 5 < 0 n < −5 m[n] = 0 for n + 5 < 4 −5 ≤ n < −1 n+5 m[n] = 1=n+6 k=0 for n − 1 < 1 −1 ≤ n < 2 3 n+5 m[n] = 1+2 k=0 for n + 5 < 9 1 = 2n + 8 k=4 2≤n<4 3 n+5 m[n] = 1+2 k=n−1 for n − 1 < 4 4≤n<5 3 8 m[n] = 1 = 15 − n 1+2 k=n−1 for n − 1 < 9 1=9+n k=4 k=4 5 ≤ n < 10 8 1 = 20 − 2n m[n] = 2 k=n−1 for n − 1 ≥ 9 n ≥ 10 m[n] = 0 0 n+6 2n + 8 m[n] = 9+n 15 − n 20 − 2n 0 (b) m[n] = x[n] ∗ y[n] for n + 5 < −3 n < −8 8 n < −5 −5 ≤ n < −1 −1 ≤ n < 2 2≤n<4 4≤n<5 5 ≤ n < 10 n ≥ 10 m[n] = 0 −8 ≤ n < −4 for n + 5 < 1 n+5 m[n] = 1=n+9 k=−3 for n − 1 < −2 −4 ≤ n < −1 0 n+5 1− m[n] = k=−3 1 = −n − 1 k=1 −1 ≤ n < 0 for n + 5 < 5 0 n+5 1− m[n] = k=n−1 for n − 1 < 1 1 = −2n − 4 k=1 0≤n<2 0 4 1− m[n] = k=n−1 for n − 1 < 5 1 = −n − 2 k=1 2≤n<6 4 m[n] = − 1=n−6 k=n−1 for n − 1 ≥ 5 n≥6 m[n] = 0 0 n+3 −n − 1 m[n] = −2n − 4 −n − 2 n−6 0 n < −8 −8 ≤ n < −4 −4 ≤ n < −1 −1 ≤ n < 0 0≤n<2 2≤n<6 n≥6 (c) m[n] = x[n] ∗ f [n] for n + 5 < −5 n < −10 m[n] = 0 for n − 1 < −5 −10 ≤ n < −4 m[n] = for n + 5 < 6 k=−5 1 k = −5n − 55 + (n + 10)(n + 11) 2 1 2 n+5 k= 7 21 (n − 1) + 2 2 k= 1 1 (7 − n) (n − 1) + (6 − n) 2 2 k=n−1 1≤n<7 m[n] = for n − 1 ≥ 6 n+5 −4 ≤ n < 1 m[n] = for n − 1 < 6 1 2 1 2 5 k=n−1 n≥7 9 m[n] = 0 0 1 −5n − 55 + 2 (n + 10)(n + 11) 7 21 m[n] = 2 (n − 1) + 2 1 1 2 (7 − n) (n − 1) + 2 (6 − n) 0 n < −10 −10 ≤ n < −4 −4 ≤ n < 1 1≤n<7 n≥7 (d) m[n] = x[n] ∗ g[n] for n + 5 < −8 n < −13 m[n] = 0 for n − 1 < −7 −14 ≤ n < −6 n+5 m[n] = 1 = n + 14 k=−8 for n + 5 < 4 −6 ≤ n < −1 −2 1 = −n m[n] = k=n−1 for n − 1 < −1 −1 ≤ n < 0 −2 n+5 m[n] = k=n−1 for n − 1 < 4 1 = −2 1+ k=4 0≤n<5 n+5 m[n] = 1=n+2 k=4 for n − 1 < 11 5 ≤ n < 12 10 1 = 12 − n m[n] = k=n−1 for n − 1 ≥ 11 n ≥ 12 m[n] = 0 0 n + 14 −n m[n] = −2 n+2 12 − n 0 n < −13 −13 ≤ n < −6 −6 ≤ n < −1 −1 ≤ n < 0 0≤n<5 5 ≤ n < 12 n ≥ 12 (e) m[n] = y[n] ∗ z[n] The remaining problems will not show all of the steps of convolution, instead ﬁgures and intervals will be given for the solution.

91 Integration is preferred over differentiation for two reasons: (i) Integration tends to attenuate high frequencies. Recognizing that noise contains a broad band of frequencies, integration has a smoothing effect on receiver noise. (ii) Differentiation tends to accentuate high frequencies. Correspondingly, differentiation has the opposite effect to integration on receiver noise. 92 From Fig. 92, we have i ( t ) = i1 ( t ) + i2 ( t ) 1 t v ( t ) = Ri 2 ( t ) = ---- ∫ i 1 ( τ ) dτ C –∞ This pair of equations may be rewritten in the equivalent form: 1 i 1 ( t ) = i ( t ) – ---v ( t ) R 1 t v ( t ) = ---- ∫ i 1 ( τ ) dτ C –∞ 37 Correspondingly, we may represent the parallel RC circuit of Fig.

Y(t ) = x (t ) , Narrowband filter centered on the frequency pω. 21, we start with the pair of inputs: 1 ∆ x 1 ( t ) = ---u t + --- ∆ 2 1 ∆ x 2 ( t ) = ---u t – --- ∆ 2 The corresponding outputs are respectively given by 33 1 y 1 ( t ) = --- 1 – e ∆ 1 y 2 ( t ) = --- 1 – e ∆ ∆ – α t + --- 2 ∆ – α t – --- 2 ∆ ∆ cos ω n t + --- u t + --- 2 2 ∆ ∆ cos ω n t – --- u t – --- 2 2 The response to the input x∆ ( t ) = x1 ( t ) – x2 ( t ) is given by 1 ∆ ∆ y ∆ ( t ) = --- u t + --- – u t – --- ∆ 2 2 –α t + ∆--- ω n ∆ ∆ 2 1 – --- e cos ω n t + --------- u t + -- ∆ 2 2 –e ∆ – α t – --- 2 ω n ∆ ∆ cos ω n t – ---------- u t – --- 2 2 As ∆ → 0, x ∆ ( t ) → δ ( t ) .