a4_solution - Solutions to Assignment 4 Problem 10.1...

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Unformatted text preview: Solutions to Assignment 4 Problem 10.1 Expressing the difference equation as y[k ] 2 y[k 1] x[k 1] 2 y[k 1] 2u(k 1) 2 y[k 1] 2 y[k 1] 2 x[k ] 2u[k ] k 1 k 1 Starting from k = 0, we iterate to calculate the output as y[0] 2 y[ 1] 4 y[1] 2 y[0] 2 10 y[2] 2 y[1] 2 22 y[3] 2 y[2] 2 46 y[4] 2 y[3] 2 94 y[5] 2 y[4] 2 190 Problem 10.5 (a) The graphical approach is illustrated in Fig. 10.5(a). Case (k < 6), Fig. S10.5(a) subplot (v): x1[k ] x2 [k ] 0 . Case (6 ≤ k ≤ 2), Fig. S10.5(b) subplot (vi): x1[k ] x2 [k ] k 4 m 2 1 (k 7) . 1 ( 6 k ) . 2 Case ( ≤ k ≤ 6), Fig. S10.5(b) subplot (vii): x1[k ] x 2 [k ] mk 4 Case (k > 6), Fig. S10.5(b) subplot (viii): x1[k ] x2 [k ] 0 The convolution sum is plotted as a function of k in subplot (ix) of Fig. S10.5(a). (b) From Problem 10.4, we know that a k u[k ] b k u[k ] Setting a = 0.8 and b = 0.5, we get 1 ( a b ) (a k 1 b k 1 )u[k ] . 0.8k u[k ] 0.5k u[k ] 10 (0.8k 1 0.5k 1 )u[k ] . 3 Using the time shifting property, 0.8k 5 u[k 5] 0.5k u[k ] 10 (0.8k 51 0.5k 51 )u[k 5] , 3 or, 0.8k 5 u[k 5] 0.5k u[k ] 10 (0.8k 4 0.5k 4 )u[k 5] , 3 2 Assignment 5 x1[m] u[m 2] u[m 3] x2 [m] u[m 4] u[m 5] m 5 4 3 2 1 0 1 2345 5 4 3 2 1 0 1 2345 m (i) Sequence x1[m] x2 [m] u[m 4] u[m 5] (ii) Sequence x2[m] x2 [k m] x2 [(m k )] )] m k k 1 k 1 k 1 k 2 k 3 k 4 k 5 k 5 k k+4 k k k k k k 5 4 3 2 1 0 1 2345 m (iii) Sequence x2[m] x1[m] x2 [k m] (iv) Sequence x2[km] x1[m] x2 [k m] 2 1 0 1 2345 m k k k+4 k k k 2 1 0 1 234 5 m (v) Convolution x1[m] * x2[m] for k < 6. x1[m] x2 [k m] (viii) Convolution x1[m] * x2[m] for ( ≤ k ≤ 6) 2 3 4 5 which implies or, (c) The graphical approach is illustrated in Fig. S10.5(c). Case (k ≤ 6), Fig. S10.5(c) subplot (v): x1[k ] x2 [k ] k 4 k+4 k k k k (vi) Convolution x1[m] * x2[m] for (6 ≤ k ≤ 2) x1[m] x2 [k m] 2 1 0 1 m k2 k3 k3 k4 2 1 0 1 2345 m k k k 1 k 2 k 3 k 4 k 4 2345 (ix) Convolution x1[m] * x2[m] as a function of k. Fig. S10.5(a) Convolution sum for part (a) in Problem 10.5. 0.8k u[k 5] 0.5k u[k ] 0.85 10 (0.8k 4 0.5k 4 )u[k 5] , 3 0.8k u[k 5] 0.5k u[k ] 1.09227 (0.8k 4 0.5k 4 )u[k 5] . k (viii) Convolution x1[m] * x2[m] for k > 6. 4 3 2 x1[k ] x2 [k ] 1 1 k 7 6 5 4 3 2 1 0 1 234567 m 0.4 k m 7 m 0.4 k m ( k 4) (2 / 35)m 33 353 7k . 16 Assignment 4 3 x1[m] 7 m u[m 2] x2 [m] 0.4 m u[m 4] m 4567 m 6 5 4 3 2 1 0 1234567 6 5 4 3 2 1 1 2 3 (i) Sequence x1[m] x2 [m] 0.4 m u[m 4] (ii) Sequence x2[m] x2 [k m] 0.4 k m u[k m 4] m 6 5 4 3 2 1 0 1234567 k5 k6 k6 k k k k k k k k k k1 k2 k3 k4 k4 k k k k k k m (iii) Sequence x2[m] x1[m] x2 [m] (iv) Sequence x2[km] x1[m] x2 [m] k 5 k 6 k 6 k 7 k 8 k 4 k 4 m 4567 m 4567 k 4 k 4 6 5 4 3 2 1 1 2 3 6 5 4 3 2 1 1 2 3 k 5 k 6 k 6 k 7 k 7 k 8 k 9 k10 k11 k11 k11 k12 k12 k12 k13 k14 (v) Convolution x1[m] * x2[m] for k ≤ 6 (vi) Convolution x1[m] * x2[m] for k ≥ 6 y[k ] 1.33 0.53 0.19 0.004 0.02 k 6 5 4 3 2 1 0 1234 567 (vii) Convolution x1[m] * x2[m] as a function of k. Fig. S10.5(c) Convolution sum for part (c) in Problem 10.5. Case ( ≤ k ≤ 6), Fig. S10.5(b) subplot (vii): x1[k ] x2 [k ] m 2 0.4k m 7 m 0.4k m 2 (2 / 35) m 353 0.4k . 132 The convolution sum is plotted as a function of k in subplot (viii) of Fig. S10.5(c). Problem 10.18 For each of the following impulse responses, determine if the DT system is (i) memoryless; (ii) causal; and (iii) stable. (a) (b) (c) (d) h[k] = u[k + 7] – u[k – 8]; h[k] = sin(k/8) u[k]; h[k] = 6ku[k]; h[k] = 0.9|k|; 4 Assignment 5 (e) h[k ] m (1) m (k 2m). h[k ] 1 0 7k 7 otherwise. Solution: (a) Expressing we observe that: (i) (ii) h[k] ≠ 0 for k ≠ 0. Therefore, the system is NOT memoryless. h[k] ≠ 0 for k < 0, e.g., h[1] = 1. Therefore, the system is NOT causal. sin( k / 8) k 0 h[k ] 0 k 0. (iii) |h[k]| = 15 < ∞. Therefore, the system is bounded-input-bounded-output (BIBO) stable. (b) Expressing we observe that: (i) (ii) h[k] ≠ 0 for k ≠ 0. Therefore, the system is NOT memoryless. h[k] = 0 for k < 0. Therefore, the system is causal. (iii) |h[k]| = ∞. Therefore, the system is NOT stable. (c) Expressing we observe that: (i) (ii) h[k] ≠ 0 for k ≠ 0. Therefore, the system is NOT memoryless. h[k] ≠ 0 for k < 0, e.g., h[2] = 1/36. Therefore, the system is NOT causal. k h[k ] 6 0 k 0 k 0. (iii) |h[k]| = 1/(1 – 1/6) = 1.2 < ∞. Therefore, the system is stable. (d) Expressing we observe that: (i) (ii) h[k] ≠ 0 for k ≠ 0. Therefore, the system is NOT memoryless. h[k] ≠ 0 for k < 0, e.g., h[2] = 0.81. Therefore, the system is NOT causal. k h[k ] 0.9 k 0.9 k 0 k 0. (iii) |h[k]| = 21 < ∞. Therefore, the system is BIBO stable. (e) Expressing h[k ] m (1) m (k 2m) (k 4) (k 2) (k ) (k 2) (k 4) we observe that: (i) (ii) h[k] ≠ 0 for k ≠ 0. Therefore, the system is NOT memoryless. h[k] ≠ 0 for k < 0, e.g., h[2] = 1. Therefore, the system is NOT causal. Assignment 4 5 (iii) Since h[k] includes an infinite number of impulse functions, therefore, |h[k]| = ∞.Therefore, the system is NOT stable. 6 Assignment 5 Assignment 4 7 8 Assignment 5 Assignment 4 9 10 Assignment 5 Assignment 4 11 12 Assignment 5 Assignment 4 13 14 Assignment 5 Assignment 4 15 ▌ ...
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This note was uploaded on 12/11/2010 for the course CSE CSE 3451 taught by Professor A during the Fall '10 term at York University.

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