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Homework_6_solution

Homework_6_solution - 1 EE113 Homework 6 Problem 8.37 The...

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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 λ = 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
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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 δ ( n ) + c 2 δ ( 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) - δ ( n ) - 1 2 δ ( n + 1) = 3 · 2 n u ( n + 1) - u ( n ) - 1 2 δ ( 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 ) [ δ ( n ) - δ ( n - 1)] = y ( n ) ⋆ δ ( n ) - y ( n ) ⋆ δ ( n - 1) = y ( n ) - y ( n - 1) = 3 · 2 n u ( n + 1) - u ( n ) - 1 2 δ ( n + 1) - 3 · 2 n 1 u ( n ) + u ( n - 1) + 1 2 δ ( n ) = 3 · 2 n u ( n + 1) - 3 · 2 n 1 u ( n ) - 1 2 δ ( n + 1) - 1 2 δ ( n ) = 3 · 2 n u ( n ) - 3 · 2 n 1 u ( n ) - δ ( n + 1) - 1 2 δ ( n ) = 3 · 2 n 1 u ( n ) - δ ( n + 1) - 1 2 δ ( n ) = 3 · 2 n 1 u ( n - 1) - δ ( n + 1) - δ ( n ) November 8, 2011 DRAFT
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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 λ 1 = 1 , λ 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 zero-input response, we substitute in the initial conditions: y zi ( n = - 2) = C 1 + 9 C 2 = 1 , y zi ( n = - 1) = C 1 - 3 C 2 = 0 Solving for C 1 and C 2 yields C 1 = 1 4 , C 2 = 1 12 Hence, the zero-input response is y zi ( n ) = [ 1 4 + 1 12 ( - 1 3 ) n ] u ( n + 2) The impulse response sequence h ( n ) can be determined by the general solution by substituting the initial conditions h ( - 1) = 0 and h (0) = 1 : h ( n = - 1) = C 1 - 3 C 2 = 0 , h ( n = 0) = C 1 + C 2 = 1 Solving for C 1 and C 2 yields C 1 = 3 4 , C 2 = 1 4 Thus, the impulse response sequence h ( n ) is h ( n ) = [ 3 4 + 1 4 ( - 1 3 ) n ] u ( n ) November 8, 2011 DRAFT
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4 The zero-state response for x 1 ( n ) , denoted as y zs , 1 ( n ) , can be found by y zs , 1 ( n ) = x 1 ( n ) ⋆ h ( n ) = 1 2 u ( n - 2) [ 3 4 + 1 4 ( - 1 3 ) n ] u ( n ) = k = −∞ [ 3 4 + 1 4 ( - 1 3 ) k ] u ( k ) · 1 2 u ( n -
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