20082ee113_1_hw7_sol

20082ee113_1_hw7_sol - EE113: Digital Signal Processing...

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Unformatted text preview: 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 ) = ( n ) X ( e jw ) = 1 Hence, X ( k ) = X ( e jw ) | w = 2 k 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 k = 1 , 2 , ,N- 1 The plot of X ( k ) for N = 8 is shown in Figure 2. iii) x ( n ) = ( n- n o ) + ( n- N + n o ) , n o < N X ( k ) = N- 1 X k =0 x p ( n ) e- j 2 kn N = e- j 2 kn o N + e- j 2 k ( N- n o ) N = e- j 2 kno N + e j 2 kno N = 2cos 2 k N n o , k N- 1 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 ) =- ( n- n o ) + ( n- N + n o ) , n o < N From (iii), we see that X ( k ) =- e- j 2 no N k + e j 2 no N k = 2 j sin 2 n 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 nk 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 n ( k- k o ) N- e- j 2 n ( k + k o ) N i =- j N 2 [ ( k- k o )- ( k- N + k o )] The plot of X ( k ) /j for N = 8 and k o = 3 is shown in Figure 5....
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20082ee113_1_hw7_sol - EE113: Digital Signal Processing...

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