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20102ee113_1_HW 3 Solution Spring 2010

# 20102ee113_1_HW 3 Solution Spring 2010 - EE113 Digital...

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EE113: Digital Signal Processing Spring 2010 Prof. Mihaela van der Schaar Homework #3 Solutions Prepared by Nick Mastronarde, Yu Zhang, and Shaolei Ren Problems from Chapter # 6 6.3 i. The sequence y ( n ) = u ( n + 1) - u ( n - 2) has three nonzero samples at n = - 1 , 0 , 1. i.e., y ( n ) = { 1 , 1 , 1 } , where the boxed sample is the sample at n = 0. Let z ( n ) = x ( n ) * y ( n ), since x ( n ) is nonzero for - 2 n 6 and y ( n ) is nonzero for - 1 n 1, then z ( n ) is nonzero for - 3 n 7. We can write y ( n ) alternatively as y ( n ) = δ ( n + 1) + δ ( n ) + δ ( n - 1) Then z ( n ) = x ( n + 1) + x ( n ) + x ( n - 1) which can be evaluated as follows: x ( n + 1) 3 2 1 1 3 3 2 1 1 x ( n ) 3 2 1 1 3 3 2 1 1 x ( n - 1) 3 2 1 1 3 3 2 1 1 z ( n ) 3 5 6 4 5 7 8 6 4 2 1 Since z ( n ) = 0 for n < - 3 And n > 7, we get z ( n ) = { 3 , 5 , 6 , 4 , 5 , 7 , 8 , 6 , 4 , 2 , 1 } ii. The samples of h ( n ) = 1 2 x ( n + 2) - 3 2 δ ( n ) + u ( n - 3) are calculated as follows: 1 2 x ( n + 2) 3 / 2 1 1 / 2 1 / 2 3/2 3 / 2 1 1 / 2 1 / 2 - 3 2 δ ( n ) -3/2 u ( n - 3) 1 1 1 1 1 · · · h ( n ) 3 / 2 1 1 / 2 1 / 2 0 3 / 2 1 3 / 2 3 / 2 1 1 1 · · · and the sequence h 1 ( n ) = ( 1 2 ) n h ( n ) u ( n ) is given by h 1 ( n ) = 0 n 0 3 / 4 n = 1 1 / 4 n = 2 3 / 16 n = 3 3 / 32 n = 4 ( 1 2 ) n n 5 which could be written as h 1 ( n ) = 3 4 δ ( n - 1) + 1 4 δ ( n - 2) + 3 16 δ ( n - 3) + 3 32 δ ( n - 4) + 1 2 n u ( n - 5) 1

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Then, y ( n ) = x ( n ) * h 1 ( n ) = 1 3 n u ( n ) * 3 4 δ ( n - 1) + 1 4 δ ( n - 2) + 3 16 δ ( n - 3) + 3 32 δ ( n - 4) + 1 2 n u ( n - 5) = 3 4 1 3 n - 1 u ( n - 1) + 1 4 1 3 n - 2 u ( n - 2) + 3 16 1 3 n - 3 u ( n - 3) + 3 32 1 3 n - 4 u ( n - 4) + X k = -∞ 1 3 k u ( k ) 1 2 n - k u ( n - k - 5) | {z } v ( n ) The last term can be evaluated as follows v
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20102ee113_1_HW 3 Solution Spring 2010 - EE113 Digital...

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