20102ee113_1_HW 2 Solution Spring 2010

20102ee113_1_HW 2 Solution Spring 2010 - EE113: Digital...

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EE113: Digital Signal Processing Spring 2010 Prof. Mihaela van der Schaar Homework #2 Solutions Prepared by Nick Mastronarde, Yu Zhang, and Shaolei Ren 4.6. x ( n ) = δ ( n ) y ( n ) = S [ δ ( n )] = 0 . 5 n δ ( n ) + δ ( n - 1) = δ ( n ) + δ ( n - 1) The system is linear Reason: Since S [ a 1 x 1 ( n ) + a 2 x 2 ( n )] = 0 . 5 n ( a 1 x 1 ( n ) + a 2 x 2 ( n )) + ( a 1 x 1 ( n - 1) + a 2 x 2 ( n - 1)) = a 1 (0 . 5 n x 1 ( n ) + x 1 ( n - 1)) + a 2 (0 . 5 n x 2 ( n ) + x 2 ( n - 1)) and a 1 y 1 ( n ) + a 2 y 2 ( n ) = a 1 (0 . 5 n x 1 ( n ) + x 1 ( n - 1)) + a 2 (0 . 5 n x 2 ( n ) + x 2 ( n - 1)) The response of system to δ ( n - 3) is given by x ( n ) = δ ( n - 3) y ( n ) = S [ δ ( n - 3)] = 0 . 5 n δ ( n - 3) + δ ( n - 4) = 0 . 5 3 δ ( n - 3) + δ ( n - 4) 4.10. Starting from y ( n ) = x (2 n ), this means that the samples that contribute to y ( n ) are {··· ,x ( - 4) ,x ( - 2) ,x (0) ,x (2) ,x (4) , ···} Now shifting x ( n ) to x ( n - k ) these samples will be shifted to {··· ,x
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20102ee113_1_HW 2 Solution Spring 2010 - EE113: Digital...

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