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# Discrete-Time Signal Processing (2nd Edition) (Prentice-Hall Signal Processing Series)

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EE141 Digital Signal Processing (Fall 2003) Solution Manual Prepared by Xiaojun Tang and Zhenzhen Ye Text: A.V. Oppenheim, R.W. Schafer, and J. R. Buck, Discrete-Time Signal Processing , 2 nd edition, Prentice-Hall, 1999. HW#1: P2.1 (a), (c), (e), (g); P2.4; P2.24 HW#2: P2.5; P2.18; P2.29 (a), (c), (e) HW#3: P2.40; P2.41; P3.27 (a), (c) HW#4: P3.6 (d), (e); P3.20; P4.1; P4.3 HW#5: P4.5; P4.7; P5.2; P5.3 HW#6: P5.10; P5.12; P5.15 HW#7: P6.7; P6.8; P6.11; P6.25; P7.15 HW#1: P2.1 (a), (c), (e), (g); P2.4; P2.24 P2.1 (a) T(x[n]) = g[n]x[n]; Stable if g[n] is bounded; Causal – output is not decided by future input; Linear – T(ax[n] + by[n]) = ag[n]x[n] + bg[n]y[n] = aT(x[n]) + bT(y[n]); Time variant – T(x[n-m]) = g[n]x[n-m]; Memoryless – output only depends on x[n] with same n; (c) + = = 0 0 ] [ ]) [ ( n n n n k k x n x T ; Stable; Causal only if n 0 = 0, else non-causal; Linear – T(ax[n] + by[n]) = aT(x[n]) + bT(y[n]); Time Invariant – + = + = = = 0 0 0 0 ' ] ' [ ] [ ]) [ ( n n n n k n m n n m n k m k x k x m n x T ; Memoryless only if n 0 = 0; (e) ] [ ]) [ ( n x e n x T = ; Stable; Causal; Nonlinear – T(ax[n] + by[n]) aT(x[n]) + bT(y[n]); Time Invariant – ] [ ]) [ ( m n x e m n x T = ; Memoryless – output only depends on x[n] with same n; (g) T(x[n]) = x[-n];

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Stable; Non-Causal – output depends on future input; Linear – T(ax[n] + by[n]) = aT(x[n]) + bT(y[n]); Time Variant; Not Memoryless; P2.4 As ] 1 [ 2 ] 2 [ 8 1 ] 1 [ 4 3 ] [ = + n x n y n y n y , the Fourier transform is jw jw jw jw jw jw jw e e X e e Y e e Y e Y = + ) ( 2 ) ( 8 1 ) ( 4 3 ) ( 2 When 1 ) ( ] [ ] [ = = jw e X n n x δ , thus = + = jw jw jw jw jw jw e e e e e e Y ) 4 / 1 ( 1 1 ) 2 / 1 ( 1 1 8 ) 8 / 1 ( ) 4 / 3 ( 1 2 ) ( 2 ] [ 4 1 2 1 8 ] [ n u n y n n = P2.24 As h[n] = [1 1 1 1 –2 –2] for n from 0 to 5 and x[n] = u[n-4], the system response is: y[n] = x[n] * h[n] = [0 0 0 0 1 2 3 4 2 0 …. .]; The sketch is shown as follows: HW#2: P2.5; P2.18; P2.29 (a), (c), (e) P2.5 (a) The roots for polynomial 0 6 5 1 2 1 = + z z are 2 and 3, so the homogeneous response for the system is: n n A A n y 3 2 ] [ 2 1 + =
(b) As ] 1 [ 2 ] 2 [ 6 ] 1 [ 5 ] [ = + n x n y n y n y and ] [ ] [ n n x δ = , the impulse response of the system is: ] [ ) 2 3 ( 2 ] [ 2 1 1 3 1 1 2 6 5 1 2 ) ( 2 n u n h e e e e e e H n n jw jw jw jw jw jw = = + = (c) As ] 1 [ 2 ] 2 [ 6 ] 1 [ 5 ] [ = + n x n y n y n y and ] [ ] [ n u n x = , the step response of the system is: () ( ) ] [ ) 1 2 3 ( ] [ 3 , 3 1 3 2 1 4 1 1 1 6 5 1 2 ) ( ) ( ) ( 2 1 1 1 1 1 2 1 1 n u n y z z z z z z z z z X z H z Y n n + = > + = + = = + + P2.18 (a) ] [ ) 2 / 1 ( ] [ n u n h n = Causal, the output of the system does not depend on future input; (b) ] 1 [ ) 2 / 1 ( ] [ = n u n h n Causal, the output of the system does not depend on future input; (c) n n h ) 2 / 1 ( ] [ = Non-Causal, the output of the system depends on future input; (d) ] 2 [ ] 2 [ ] [ + = n u n u n h Non-Causal, the output of the system depends on future input; (e) ] 1 [ 3 ] [ ) 3 / 1 ( ] [ + = n u n u n h n n Non-Causal, the output of the system depends on future input; P2.29 As x[n] = [1 1 1 1 1 1/2] for n from –1 to 4, (a) x[n-2] = [1 1 1 1 1 1/2] for n from 1 to 6; The sketch is:

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(c) x[2n] = [1 1 1/2] for n from 0 to 2; The sketch is: (e) x[n-1] δ [n-3] = x[2]; The sketch is: HW#3: P2.40, P2.41, P3.27 (a), (c) P2.40
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01 - EE141 Digital Signal Processing (Fall 2003) Solution...

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