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20102ee113_1_hw7_sol_2010

# 20102ee113_1_hw7_sol_2010 - EE113 Digital Signal Processing...

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EE113: Digital Signal Processing Spring 2010 Prof. Mihaela van der Schaar Homework #7 Solutions Problems from Chapter 15 15.1. i. x ( n ) = δ ( n ) X ( e jw ) = 1 hence, X ( k ) = X ( e jw ) | w = 2 πk N = 1 The plot of X ( k ) for k = 8 is shown in Figure 1. Figure 1: A plot of X ( k ) for N = 8. ii. x ( n ) = u ( n ) - u ( n - N ) = ( 1 0 n N - 1 0 otherwise From the example in Page 150, we have X ( k ) = ( N k = 0 0 k = 1 , 2 , ··· ,N - 1 The plot of X ( k ) for k = 8 is shown in Figure 2. iii. x ( n ) = δ ( n - n o ) + δ ( n - N + n o ) , n o < N X ( k ) = N - 1 X k =0 x p ( n ) e - j 2 πkn N = e - j 2 πkno N + e - j 2 πk ( N - no ) N = e - j 2 πkno N + e j 2 πkno N = 2cos ± 2 πk N n o , 0 k N - 1 The plot of X ( k ) for N = 8 and n o = 2is shown in Figure 3.

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Figure 2: A plot of X ( k ) for N = 8. Figure 3: A plot of X ( k ) for N = 8 and n o = 2. iv. x ( n ) = - δ ( n - n o ) + δ ( n - N + n o ) , n o < N from (iii), we see that X ( k ) = - e - j 2 πno N k + e j 2 πno N k = 2 j sin ± 2 πn o N k The plot of X ( k ) for k = 8 and n o = 2 is shown in Figure 4. v. x ( n ) = cos ± 2 πk o N n , k o < N = 1 2 e j 2 πko N n + 1 2 e - j 2 πko N n X ( k ) = 1 2 N - 1 X n =0 h e j 2 πko N n e - j 2 πk N n + e - j 2 πko N n e - j 2 πk N n i = N 2 [ δ ( k - k o ) + δ ( k - N + k o ))]
X ( k ) /j for N = 8 and n o = 2. The plot of X ( k ) for N = 8 and k o = 3 is shown in Figure 5. Figure 5: A plot of X ( k ) for N = 8 and k o = 3 vi. x ( n ) = sin ± 2 πnk o N , k o < N Similar to (v), we can show that X ( k ) = - j 2 N - 1 X n =0 h e - j 2 πn ( k - ko ) N - e - j 2 πn ( k + ko ) N i = - j N 2 [ δ ( k - k o ) - δ ( k - N + k o )] The plot of X ( k ) for N = 8 and k o = 3 is shown in Figure 6. 15.8. Let

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20102ee113_1_hw7_sol_2010 - EE113 Digital Signal Processing...

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