Homework_5_solution(1)

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 n−1 1 3 u(−n + 2) ⋆ · u(n − 2) using both the analytical and graphical methods. Compare your results. Solution: Here we only give the analytical solution. = = ∞ n−1 1 3 x(n) = u(−n + 2) ⋆ · u(n − 2) 1 3 u(−n + m + 2) m=−∞ ∞ · u(m − 2) m−1 1 3 m=max{2,n−2} m−1 If n < 4, then it is ∞ x(n) = = m=2 1 3 1− 1 = 2 If n ≥ 4, then it is 1 3 ∞ x(n) = m=n−2 = = m−1 1 3 1 3 m−1 1 n−3 3 1 1− 3 n−4 1 2 1 3 Therefore, x(n) = October 27, 2011 1 u(−n + 3) + 2 1 3 n−4 u(n − 4) DRAFT 2 Problem 6.9 Evaluate n−1 1 2 1 3 · u(n) ⋆ n · u(n − 2) − u(n + 1) using the distributivity property of the convolution sum. Solution: By the distributivity, we can get n−1 1 2 x(n) = ∞ = n−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 m−1 1 2 m=−∞ = · u(n) ⋆ m−1 n−2 1− 3 2 1− 3 2 n − n−m 1 3 · 1 3 1 3 n−m n−1 · u(n) ⋆ u(n + 1) · u(n − m − 2) − n+1 · u(n − 2) − · u(n − 2) − 2 · u(n − 2) − 2 1 · u(n − 2) − 4 1 − 1 2 m=−∞ m−1 1 2 m−1 · u(m) · u(n − m + 1) · u(n + 1) · u(n + 1) · u(n + 1) 1 2 − 1 · u(n − 2) − 4 1 − ∞ m 1 2 m=0 1 n+2 −2 1− 1 2 n 1 2 m=0 n+1 m n−1 3 2 n−1 1 2 · u(n − 2) − · u(m) · 3 2 m=0 n 1 3 n+2 · u(n + 1) n+2 · 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 defined as follows: ∞ rxh (n) = k=−∞ x(k )h∗ (k − n) ∗ (a) Verify that rxh (n) = x(n) ⋆ h (−n). (b) Use the graphical method to evaluate the correlation of the two sequences: x(n) = {−2, 1, −1, 2} and h(n) = {0, 1, 2} Solution: (a) By definition, we can write rxh (n) = x(n) ⋆ h∗ (−n) = = ∞ k=−∞ ∞ k=−∞ x(k )h∗ (−(n − k )) x(k )h∗ (k − n) which completes the proof. (b) The result is rxh (n) = {−4, 0, −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 − 1) + x(n − 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) = δ (n) into the difference equation yields y (n) = 1 y (n − 1) + δ (n − 1) 4 That is y (0) = 0 y (1) = 1 y (2) = y (3) = 1 4 1 4 2 1 4 n−1 . . . y (n) = From Fig. 6.18, we can write the input sequence as x(n) = −2δ (n − 1) + 2δ (n − 2) − δ (n − 3) + 2δ (n − 5). The response is thus y (n) = x(n) ⋆ h(n) = 1 4 = −2 October 27, 2011 n−1 1 4 · u(n − 1) ⋆ [−2δ (n − 1) + 2δ (n − 2) − δ (n − 3) + 2δ (n − 5)] n−2 u(n − 2) + 2 1 4 n−3 u(n − 3) − 1 4 n−4 u(n − 4) + 2 1 4 n−6 u(n − 6) DRAFT 5 Problem 7.9 Determine the solution of each of the following homogeneous equations with initial conditions: (a) y (n) + 9y (n − 2) = 0, y (0) = 0, y (−1) = 1. (b) y (n) + 9y (n − 2) = 0, y (0) = 0, y (−1) = j . Solution: The corresponding characteristic function is λ2 + 9 = 0 with distinct modes at λ1 = 3j, λ2 = −3j Therefore, the general solution to the homogeneous equation is y (n) = C1 (3j )n + C2 (−3j )n for arbitrary constants {C1 , C2 }. Substituting the initial conditions leads to specified values for {C1 , C2 }. (a) If y (0) = 0, y (−1) = 1, then 0 = C + C 1 2 Thus the solution for this case is 1 = C1 3j − C2 3j =⇒ C = 1 3j 2 C2 = − 3j 2 (3j )n+1 + (−3j )n+1 2 π j π (n+1) 2 e + e−j 2 (n+1) = 3n+1 · 2 π n+1 =3 · cos (n + 1) 2 y (n) = (b) If y (0) = 0, y (−1) = j , then 0 = C + C 1 2 Thus the solution for this case is j = C1 3j − C2 3j =⇒ C = − 3 1 2 C2 = 3 2 3 3 y (n) = − (3j )n + (−3j )n 2 2 π π e j 2 n − e −j 2 n = −j · 3n+1 · 2j π n = −j · 3n+1 · 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 − 1) − 5y (n − 2) = x(n). (b) y (n) = −4y (n − 2) + x(n − 1). (c) y (n) − 4y (n − 2) = 2x(n). Solution: (a) The characteristic function for the homogeneous equation is λ2 + λ − 5 = 0 with distinct modes at √ −1 + 21 λ1 = , 2 √ −1 − 21 λ2 = 2 Thus the general solution is √ −1 + 21 2 y (n) = C1 n + C2 √ −1 − 21 2 n for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (−1) = 0 and y (0) = 1. Therefore, we can write C√ + 1 −1+ 21 2 C2 √ −1− 21 2 =0 =⇒ C1 + C2 = 1 The impulse response is thus 1 y (n) = √ 21 √ −1 + 21 2 n+1 C = 1 C2 = 1 −√ 21 √ −1+ 21 √ 2 21 √ 1+ 21 √ 2 21 √ −1 − 21 2 n+1 (b) The characteristic function for the homogeneous equation is λ2 + 4 = 0 with distinct modes at λ1 = 2j, λ2 = −2j Thus the general solution is n n y (n) = C1 (2j ) + C2 (−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 · C − 2j · C = 1 1 2 C1 + C2 = 0 October 27, 2011 =⇒ C = 1 1 4j C2 = − 1 4j DRAFT 7 The impulse response is thus 1 1 n−1 n−1 · u(n − 1) (2j ) + (−2j ) 2 2 π = 2n−1 cos (n − 1) · u(n − 1) 2 y (n) = (c) The characteristic function for the homogeneous equation is λ2 − 4 = 0 with distinct modes at λ1 = 2, λ2 = −2 Thus the general solution is n y (n) = C1 2n + C2 (−2) for some constants C1 and C2 . For the impulse response, it is equivalent to the initial condition y (−1) = 0 and y (0) = 2. Therefore, we can write C1 − 2 The impulse response is thus C2 2 =0 C1 + C2 = 2 =⇒ C = 1 1 C2 = 1 n y (n) = [2n + (−2) ] · u(n) October 27, 2011 DRAFT 8 Problem 7.19 Consider the relaxed and causal system 1 5 y (n) + y (n − 1) − y (n − 2) = x(n − 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 − 2). Solution: Let us first find out the impulse response sequence h(n) for this system. The characteristic function for the homogeneous equation is 5 1 λ2 + λ − = 0 6 6 with distinct modes at λ1 = −1, 1 6 λ2 = Thus the general solution is n 1 6 h(n) = C1 (−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 − C + 1 C1 + The impulse response is thus C2 6 C2 36 =⇒ =1 6 36 n (−1) + 7 7 h(n) = C = 1 =0 C2 = 6 7 36 7 n 1 6 · u(n − 2) Then, we can show that the impulse response h(n) is not absolute summable: ∞ n=2 |h(n)| = ∞ n=2 ∞ n 1 6 36 6 n (−1) − 7 7 ≥ ∞ n=2 36 6 n (−1) − 7 7 n=2 1 6 n →∞ Thus we conclude that this system is not BIBO stable. The response to x(n) = ( 1 )n u(n − 2) is 2 1 2 y (n) = = n ∞ 6 7 m=−∞ 1 2 + = = October 27, 2011 6 7 1 2 n n−m m=2 n · u(n − 2) m · u(n − m − 2) · (−1) · u(m − 2) ∞ 36 7 m=−∞ n−2 1 6 36 6 n (−1) + 7 7 · u(n − 2) ⋆ m (−2) 1 2 n−m · u(n − 4) + 2 24 4 − (−2)n−1 + 7 7 1 − 9 1 3 36 7 n−1 1 2 n n−2 m=2 1 2 m 1 6 · u(n − m − 2) · · u(m − 2) 1 3 m · u(n − 4) n · u(n − 4) DRAFT 9 Problem 7.31 Find the solution y (n) of the so-called Fibonacci difference equation y (n) − y (n − 1) − y (n − 2) = 0 with y (−1) = 0 and y (0) = 1. Solution: The characteristic function for the homogeneous equation is λ2 − λ − 1 = 0 with distinct modes at √ 1− 5 λ2 = 2 √ 1+ 5 , λ1 = 2 Thus the general solution is y (n) = C1 √ 1+ 5 2 n + C2 √ 1− 5 2 n for some constants C1 and C2 . The initial conditions y (−1) = 0 and y (0) = 1 are √ C√ + C√ = 0 C = 1+ 5 2 √ 1+ 1 5 1 1− 5 25 2 2 =⇒ √ C1 + C2 = 1 C2 = − 1− 5 √ 25 Therefore, y (n) is 1 y (n) = √ 5 October 27, 2011 √ 1+ 5 2 n+1 1 −√ 5 √ 1− 5 2 n+1 DRAFT ...
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