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Unformatted text preview: 1 EE113 Homework 6 Problem 8.37 The input sequence x ( n ) = u ( n 1) is first processed by the causal system y ( n ) = 2 y ( n 1) + x ( n ) , y ( 1) = 1 The result, y ( n ) , is then processed by an LTI system with impulse response sequence h ( n ) = { 1 , 1 } What is the output of this second system? Solution: First let us find out the output sequence y ( n ) for the first system. The particular solution for x ( n ) = u ( n 1) is assumed to be y p ( n ) = Ku ( n 1) . Then, for any n 2 , we can get Ku ( n 1) = 2 Ku ( n 2) + u ( n 1) = K = 1 which leads to the particular solution for any n 2 as: y p ( n ) = u ( n 1) The characteristic root of the homogeneous equation is = 2 Thus, the general solution should be y h ( n ) = C 2 n The complete solution then can be written as y ( n ) = C 2 n u ( n 1) , n 2 Propagating the initial condition y ( 1) = 1 to n = 1 yields y ( n = 1) = 2 y ( n = 0) + x ( n = 1) = 4 y ( n = 1) + u (0) = 5 Then, plugging this condition into the complete solution, we can solve for C as y ( n = 1) = 2 C 1 = 5 = C = 3 Thus the complete solution is y ( n ) = 3 2 n u ( n 1) , n 1 November 8, 2011 DRAFT 2 Because the system is initialized at n = 1 , we have to modify the result to accommodate y (0) and y ( 1) by introducing delta functions at time n = 0 and n = 1 such as y ( n ) = 3 2 n u ( n 1) + c 1 ( n ) + c 2 ( n + 1) , n  1 Substituting y ( 1) = 1 and y (0) = 2 yields y ( n = 1) = 3 2 + c 2 = 1 = c 2 = 1 2 y ( n = 0) = 3 + c 1 = 2 = c 1 = 1 Therefore, the complete response should be y ( n ) = 3 2 n u ( n + 1) u ( n 1) ( n ) 1 2 ( n + 1) = 3 2 n u ( n + 1) u ( n ) 1 2 ( n + 1) Now, the output of the second system, denoted as z ( n ) , can be calculated by z ( n ) = y ( n ) h ( n ) = y ( n ) [ ( n ) ( n 1)] = y ( n ) ( n ) y ( n ) ( n 1) = y ( n ) y ( n 1) = 3 2 n u ( n + 1) u ( n ) 1 2 ( n + 1) 3 2 n 1 u ( n ) + u ( n 1) + 1 2 ( n ) = 3 2 n u ( n + 1) 3 2 n 1 u ( n ) 1 2 ( n + 1) 1 2 ( n ) = 3 2 n u ( n ) 3 2 n 1 u ( n ) ( n + 1) 1 2 ( n ) = 3 2 n 1 u ( n ) ( n + 1) 1 2 ( n ) = 3 2 n 1 u ( n 1) ( n + 1) ( n ) November 8, 2011 DRAFT 3 Problem 8.41 Find the complete response of the causal system y ( n ) 2 3 y ( n 1) 1 3 y ( n 2) = x ( n ) , y ( 2) = 1 , y ( 1) = 0 to the input sequence x ( n ) = 1 2 u ( n 2) + ( 1 4 ) n 1 u ( n ) Check your answer. Solution: First of all, we can separate the input sequence x ( n ) into two parts: x 1 ( n ) = 1 2 u ( n 2) , x 2 ( n ) = ( 1 4 ) n 1 u ( n ) Then, the output sequence y ( n ) can be separated into three parts: y ( n ) = y zi ( n ) + y zs , 1 ( n ) + y zs , 2 ( n ) Let us find the general solution now. The characteristic roots are 1 = 1 , 2 = 1 3 The general solution can thus be written as y h ( n ) = C 1 + C 2 ( 1 3 ) n , n To find the zeroinput response, we substitute in the initial conditions:...
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This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.
 Fall '08
 Walker

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