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20102ee113_1_HW 4 Solution Spring 2010

# 20102ee113_1_HW 4 Solution Spring 2010 - EE113 Digital...

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EE113: Digital Signal Processing Spring 2010 Prof. Mihaela van der Schaar Homework #4 Solutions Prepared by Nick Mastronarde, Yu Zhang, and Shaolei Ren 8.7 Consider the difference equation y ( n ) = 3 4 y ( n - 1) - 1 8 y ( n - 2) + x ( n ) . 1. The modes of the system are the roots of the charateristic polynomial: λ 2 - 3 4 λ + 1 8 = 0 λ 1 = 1 2 λ 2 = 1 4 . 2. The solutions to the homogeneous equation are: y h ( n ) = α 1 λ n 1 + α 2 λ n 2 = α 1 1 2 n + α 2 1 4 n , for all n. 3. Particular solutions: i. For x ( n ) = u ( n ): Letting y ( n ) = Ku ( n ) yields Ku ( n ) = 3 4 Ku ( n - 1) - 1 8 Ku ( n - 2) + u ( n ) and for n 2 we get K = 8 3 and y p ( n ) = 8 3 u ( n ) , n 2 . ii. For x ( n ) = nu ( n ): Letting y ( n ) = ( K 1 + K 2 n ) u ( n ) yields ( K 1 + K 2 n ) u ( n ) = 3 4 ( K 1 + K 2 ( n - 1)) u ( n - 1) - 1 8 ( K 1 + K 2 ( n - 2)) u ( n - 2) + nu ( n ) and for n 2 we get K 1 + nK 2 = 3 4 ( K 1 + nK 2 - K 2 ) - 1 8 ( K 1 + nK 2 - 2 K 2 ) + n or equivalently, n ( 3 4 K 2 - 1 8 K 2 - K 2 + 1) + 3 4 ( K 1 - K 2 ) - 1 8 ( K 1 - 2 K 2 ) - K 1 = 0 . Solving the equations 3 4 K 2 - 1 8 K 2 - K 2 + 1 = 0 , 3 4 ( K 1 - K 2 ) - 1 8 ( K 1 - 2 K 2 ) - K 1 = 0 yields K 2 = 8 3 and K 1 = - 32 9 . Thus, y p ( n ) = 8 3 n - 32 9 u ( n ) , n 2 . 1

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iii. For x ( n ) = ( 1 3 ) n u ( n ): Letting y ( n ) = K ( 1 3 ) n u ( n ) yields K 1 3 n u ( n ) = 3 4 K 1 3 n - 1 u ( n - 1) - 1 8 K 1 3 n - 2 u ( n - 2) + 1 3 n u ( n ) and for n 2 we get K 1 3 n - K 3 4 1 3 n - 1 + K 1 8 1 3 n - 2 = 1 3 n or equivalently, K 1 3 n " 1 - 3 4 1 3 - 1 + 1 8 1 3 - 2 # = 1 3 n Solving for K yields K = " 1 - 3 4 1 3 - 1 + 1 8 1 3 - 2 # - 1 = - 8 and hence y p ( n ) = - 8 1 3 n u ( n ) , n 2 . 4. For the complete solutions we will assume that the system is relaxed since no initial conditions are given. i. For x ( n ) = u ( n ): y ( n ) = y h ( n ) + y p ( n ) = α 1 1 2 n + α 2 1 4 n + 8 3 u ( n ) , n 0 . Since the system is relaxed, we get y (0) = 1 and y (1) = 7 4 , and consequently, 1 = α 1 + α 2 + 8 3 , 7 4 = α 1 2 + α 2 4 + 8 3 . Solving for α 1 and α 2 yields α 1 = - 2 and α 2 = 1 3 , and y ( n ) = - 2 1 2 n + 1 3 1 4 n + 8 3 u ( n ) , n 0 . ii. For x ( n ) = nu ( n ): y ( n ) = y h ( n ) + y p ( n ) = α 1 1 2 n + α 2 1 4 n + 8 3 n - 32 9 u ( n ) , n 0 .
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