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20082ee113_1_hw7_sol

# 20082ee113_1_hw7_sol - EE113 Digital Signal Processing Prof...

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EE113: Digital Signal Processing Spring 2008 Prof. Mihaela van der Schaar Homework #7 Solutions Prepared by Martin Andersen and Hyunggon Park 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 N = 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 N = 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 πkn o N + e - j 2 πk ( N - n o ) N = e - j 2 πkno N + e j 2 πkno N = 2 cos 2 πk N n o , 0 k N - 1

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Figure 2: A plot of X ( k ) for N = 8. The plot of X ( k ) for N = 8 and n o = 2 is shown in Figure 3. 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 ) /j for N = 8 and n o = 2 is shown in Figure 4. vi) x ( n ) = sin 2 πnk o N , k o < N
Figure 4: A plot of X ( k ) /j for N = 8 and n o = 2. Similar to (v), which was covered in the lecture, we can show that X ( k ) = - j 2 N - 1 X n =0 h e - j 2 πn ( k - k o ) N - e - j 2 πn ( k + k o ) N i = - j N 2 [ δ ( k - k o ) - δ ( k - N + k o )] The plot of X ( k ) /j for N = 8 and k o = 3 is shown in Figure 5. Figure 5: A plot of X ( k ) /j for N = 8 and k o = 3 15.8. Let y ( n ) = x 1 ( n ) * x 2 ( n ) = k = -∞ x 1 ( k ) x 2 ( n - k ).

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