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Homework_5_solution(1) - 1 EE113 Homework 5 Problem 6.7...

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Unformatted text preview: 1 EE113 Homework 5 Problem 6.7 Evaluate the convolution sum nβ1 1 3 u(βn + 2) β Β· u(n β 2) using both the analytical and graphical methods. Compare your results. Solution: Here we only give the analytical solution. = = β nβ1 1 3 x(n) = u(βn + 2) β Β· u(n β 2) 1 3 u(βn + m + 2) m=ββ β Β· u(m β 2) mβ1 1 3 m=max{2,nβ2} mβ1 If n < 4, then it is β x(n) = = m=2 1 3 1β 1 = 2 If n β₯ 4, then it is 1 3 β x(n) = m=nβ2 = = mβ1 1 3 1 3 mβ1 1 nβ3 3 1 1β 3 nβ4 1 2 1 3 Therefore, x(n) = October 27, 2011 1 u(βn + 3) + 2 1 3 nβ4 u(n β 4) DRAFT 2 Problem 6.9 Evaluate nβ1 1 2 1 3 Β· u(n) β n Β· u(n β 2) β u(n + 1) using the distributivity property of the convolution sum. Solution: By the distributivity, we can get nβ1 1 2 x(n) = β = nβ2 m=0 1 2 1 3 n n =2 1 3 1 3 n =4 =4 2 3 1 2 =2 October 27, 2011 mβ1 1 2 m=ββ = Β· u(n) β mβ1 nβ2 1β 3 2 1β 3 2 n β nβm 1 3 Β· 1 3 1 3 nβm nβ1 Β· u(n) β u(n + 1) Β· u(n β m β 2) β n+1 Β· u(n β 2) β Β· u(n β 2) β 2 Β· u(n β 2) β 2 1 Β· u(n β 2) β 4 1 β 1 2 m=ββ mβ1 1 2 mβ1 Β· u(m) Β· u(n β m + 1) Β· u(n + 1) Β· u(n + 1) Β· u(n + 1) 1 2 β 1 Β· u(n β 2) β 4 1 β β m 1 2 m=0 1 n+2 β2 1β 1 2 n 1 2 m=0 n+1 m nβ1 3 2 nβ1 1 2 Β· u(n β 2) β Β· u(m) Β· 3 2 m=0 n 1 3 n+2 Β· u(n + 1) n+2 Β· u(n + 1) DRAFT 3 Problem 6.23 Consider two possibly complex-valued sequences x(n) and h(n). Their cross-correlation is the sequence rxh (n) whose samples are deο¬ned as follows: β rxh (n) = k=ββ x(k )hβ (k β n) β (a) Verify that rxh (n) = x(n) β h (βn). (b) Use the graphical method to evaluate the correlation of the two sequences: x(n) = {β2, 1, β1, 2} and h(n) = {0, 1, 2} Solution: (a) By deο¬nition, we can write rxh (n) = x(n) β hβ (βn) = = β k=ββ β k=ββ x(k )hβ (β(n β k )) x(k )hβ (k β n) which completes the proof. (b) The result is rxh (n) = {β4, 0, β1, 3, 2, 0} October 27, 2011 DRAFT 4 Problem 6.31 A causal LTI system is described by the difference equation y (n) = 1 y (n β 1) + x(n β 1) 4 Find its impulse response sequence. Find also the response to the input sequence shown in Fig. 6.18 using the convolution sum method. Solution: Plugging the impulse input x(n) = Ξ΄ (n) into the difference equation yields y (n) = 1 y (n β 1) + Ξ΄ (n β 1) 4 That is y (0) = 0 y (1) = 1 y (2) = y (3) = 1 4 1 4 2 1 4 nβ1 . . . y (n) = From Fig. 6.18, we can write the input sequence as x(n) = β2Ξ΄ (n β 1) + 2Ξ΄ (n β 2) β Ξ΄ (n β 3) + 2Ξ΄ (n β 5). The response is thus y (n) = x(n) β h(n) = 1 4 = β2 October 27, 2011 nβ1 1 4 Β· u(n β 1) β [β2Ξ΄ (n β 1) + 2Ξ΄ (n β 2) β Ξ΄ (n β 3) + 2Ξ΄ (n β 5)] nβ2 u(n β 2) + 2 1 4 nβ3 u(n β 3) β 1 4 nβ4 u(n β 4) + 2 1 4 nβ6 u(n β 6) DRAFT 5 Problem 7.9 Determine the solution of each of the following homogeneous equations with initial conditions: (a) y (n) + 9y (n β 2) = 0, y (0) = 0, y (β1) = 1. (b) y (n) + 9y (n β 2) = 0, y (0) = 0, y (β1) = j . Solution: The corresponding characteristic function is Ξ»2 + 9 = 0 with distinct modes at Ξ»1 = 3j, Ξ»2 = β3j Therefore, the general solution to the homogeneous equation is y (n) = C1 (3j )n + C2 (β3j )n for arbitrary constants {C1 , C2 }. Substituting the initial conditions leads to speciο¬ed values for {C1 , C2 }. (a) If y (0) = 0, y (β1) = 1, then 0 = C + C 1 2 Thus the solution for this case is 1 = C1 3j β C2 3j =β C = 1 3j 2 C2 = β 3j 2 (3j )n+1 + (β3j )n+1 2 Ο j Ο (n+1) 2 e + eβj 2 (n+1) = 3n+1 Β· 2 Ο n+1 =3 Β· cos (n + 1) 2 y (n) = (b) If y (0) = 0, y (β1) = j , then 0 = C + C 1 2 Thus the solution for this case is j = C1 3j β C2 3j =β C = β 3 1 2 C2 = 3 2 3 3 y (n) = β (3j )n + (β3j )n 2 2 Ο Ο e j 2 n β e βj 2 n = βj Β· 3n+1 Β· 2j Ο n = βj Β· 3n+1 Β· sin 2 October 27, 2011 DRAFT 6 Problem 7.11 Find the impulse-response sequences of the following causal LTI systems: (a) y (n) + y (n β 1) β 5y (n β 2) = x(n). (b) y (n) = β4y (n β 2) + x(n β 1). (c) y (n) β 4y (n β 2) = 2x(n). Solution: (a) The characteristic function for the homogeneous equation is Ξ»2 + Ξ» β 5 = 0 with distinct modes at β β1 + 21 Ξ»1 = , 2 β β1 β 21 Ξ»2 = 2 Thus the general solution is β β1 + 21 2 y (n) = C1 n + C2 β β1 β 21 2 n for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (β1) = 0 and y (0) = 1. Therefore, we can write Cβ + 1 β1+ 21 2 C2 β β1β 21 2 =0 =β C1 + C2 = 1 The impulse response is thus 1 y (n) = β 21 β β1 + 21 2 n+1 C = 1 C2 = 1 ββ 21 β β1+ 21 β 2 21 β 1+ 21 β 2 21 β β1 β 21 2 n+1 (b) The characteristic function for the homogeneous equation is Ξ»2 + 4 = 0 with distinct modes at Ξ»1 = 2j, Ξ»2 = β2j Thus the general solution is n n y (n) = C1 (2j ) + C2 (β2j ) for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (0) = 0 and y (1) = 1. Therefore, we can write 2j Β· C β 2j Β· C = 1 1 2 C1 + C2 = 0 October 27, 2011 =β C = 1 1 4j C2 = β 1 4j DRAFT 7 The impulse response is thus 1 1 nβ1 nβ1 Β· u(n β 1) (2j ) + (β2j ) 2 2 Ο = 2nβ1 cos (n β 1) Β· u(n β 1) 2 y (n) = (c) The characteristic function for the homogeneous equation is Ξ»2 β 4 = 0 with distinct modes at Ξ»1 = 2, Ξ»2 = β2 Thus the general solution is n y (n) = C1 2n + C2 (β2) for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (β1) = 0 and y (0) = 2. Therefore, we can write C1 β 2 The impulse response is thus C2 2 =0 C1 + C2 = 2 =β C = 1 1 C2 = 1 n y (n) = [2n + (β2) ] Β· u(n) October 27, 2011 DRAFT 8 Problem 7.19 Consider the relaxed and causal system 1 5 y (n) + y (n β 1) β y (n β 2) = x(n β 2) 6 6 (a) Is the system BIBO stable? (b) Find its impulse-response sequence. 1 (c) Find its response to x(n) = ( 2 )n u(n β 2). Solution: Let us ο¬rst ο¬nd out the impulse response sequence h(n) for this system. The characteristic function for the homogeneous equation is 5 1 Ξ»2 + Ξ» β = 0 6 6 with distinct modes at Ξ»1 = β1, 1 6 Ξ»2 = Thus the general solution is n 1 6 h(n) = C1 (β1)n + C2 for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (1) = 0 and y (2) = 1. Therefore, we can write β C + 1 C1 + The impulse response is thus C2 6 C2 36 =β =1 6 36 n (β1) + 7 7 h(n) = C = 1 =0 C2 = 6 7 36 7 n 1 6 Β· u(n β 2) Then, we can show that the impulse response h(n) is not absolute summable: β n=2 |h(n)| = β n=2 β n 1 6 36 6 n (β1) β 7 7 β₯ β n=2 36 6 n (β1) β 7 7 n=2 1 6 n ββ Thus we conclude that this system is not BIBO stable. The response to x(n) = ( 1 )n u(n β 2) is 2 1 2 y (n) = = n β 6 7 m=ββ 1 2 + = = October 27, 2011 6 7 1 2 n nβm m=2 n Β· u(n β 2) m Β· u(n β m β 2) Β· (β1) Β· u(m β 2) β 36 7 m=ββ nβ2 1 6 36 6 n (β1) + 7 7 Β· u(n β 2) β m (β2) 1 2 nβm Β· u(n β 4) + 2 24 4 β (β2)nβ1 + 7 7 1 β 9 1 3 36 7 nβ1 1 2 n nβ2 m=2 1 2 m 1 6 Β· u(n β m β 2) Β· Β· u(m β 2) 1 3 m Β· u(n β 4) n Β· u(n β 4) DRAFT 9 Problem 7.31 Find the solution y (n) of the so-called Fibonacci difference equation y (n) β y (n β 1) β y (n β 2) = 0 with y (β1) = 0 and y (0) = 1. Solution: The characteristic function for the homogeneous equation is Ξ»2 β Ξ» β 1 = 0 with distinct modes at β 1β 5 Ξ»2 = 2 β 1+ 5 , Ξ»1 = 2 Thus the general solution is y (n) = C1 β 1+ 5 2 n + C2 β 1β 5 2 n for some constants C1 and C2 . The initial conditions y (β1) = 0 and y (0) = 1 are β Cβ + Cβ = 0 C = 1+ 5 2 β 1+ 1 5 1 1β 5 25 2 2 =β β C1 + C2 = 1 C2 = β 1β 5 β 25 Therefore, y (n) is 1 y (n) = β 5 October 27, 2011 β 1+ 5 2 n+1 1 ββ 5 β 1β 5 2 n+1 DRAFT ...
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