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Electrical Engineering 20N - Spring 2000 - Final Exam

# Electrical Engineering 20N - Spring 2000 - Final Exam -...

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Unformatted text preview: EECS 20. Final Exam Solution May 15, 2000. 1. (a) Linear: all except S3 . S , S , and S . Causal: S , S and S . Since the system is causal, hn = 0 for n 0. In addition, h satisﬁes hn = n + n , 1 , hn , 1 (b) Time invariant: (c) 2. (a) 1 1 2 3 3 6 (just let the input be an impulse). Thus, h0 h1 h2 h3 h4 hn = 1 = 1  = ,  1 ,  , 1 ,  =  = =  , , 1 2 3 , n,1 1  ,  so hn = , n, un , 1 + , where un is the unit step function. 1 n un;  (b) Although we could calculate the DTFT of the impulse response, it is easier to just let the input be a complex exponential, xn = ei!n : The output then will be yn = H !ei!n : Hence, the following equation must be satisﬁed, H !ei!n + H !ei!n,1 = ei!n + ei!n,1 : We can factor out ei!n and divide through by it, getting H !1 + e,i!  = 1 + e,i! : Hence, ,i! + H ! = 11+ ee,i! : (c) The output will be zero if the frequency ! of the sinusoid is such that H !  = 0. This occurs if e,i! = ,1, which occurs if ! =  . Thus, the following input will yield zero output: xn = cosn: 1 (d) omega = [0: pi/400: pi]; H = (1 + exp(-i*omega))./(1 + alpha*exp(-i * omega)); plot(omega, abs(H)); (e) A reasonable choice for the state s is sn = xn , 1; yn , 1 T : With this choice, A= (f) 3. "  0 , 1 ; b= "  "  ; c = , ; d = 1: If = 1, the frequency response becomes H !  = 1 and the impulse response becomes hn = n. 0 1 1 1 (a) False. The output frequency may not be the same as the input frequency. (b) False. You only know the response to one frequency. (c) True. The frequency response is the DTFT of the impulse response. (d) True. The impulse response is frequency response. yn , yn , 1, from which you can determine the (e) True. If the system were LTI, the response to the delayed impulse would be the delayed impulse response. (f) False. The system might be LTI with impulse response given by hn = y n , y n , 2 + y n , 4 , y n , 6 +   . 4. (a) The fundamental frequency is ! 0  radians/second : = 10 (b) The Fourier series coefﬁcients are A ;A 0 =0 k =0 1 ;A =1 2 ;A =1 3 ; Ak = 0 for k =1 3 ; and for all k: (c) The sampled signal is yn n=10 + cos20n=10 + cos30n=10 = 1 + 2 cosn: The fundamental frequency is therefore ! =  radians/sample. = cos10 0 (d) The DFS coefﬁcients are A 0 ;A =1 1 ; =2 1 =0 : There are no more coefﬁcients, since the period is p = 2. (e) The “smoothest” (lowest frequency content) interpolating signal is wt = 1 + 2 cos10t: 2 (f) Yes, there is aliasing distortion. The 10 Hz cosine has been aliased down to DC, and the 15 Hz cosine has been aliased down to 5 Hz, overlapping the 5 Hz cosine. (g) Sampling at twice the highest frequency will work. The highest frequency is 15 Hz, so sampling at 30 Hz will avoid aliasing distortion. 5. The sawtooth signal has period p = 1 second, so its fundamental frequency is 2 radians/second, considerably above the passband of the ﬁlter. Thus, only the DC term gets through the ﬁlter. The DC term is the average over one period, which is 1/2, so the output is yn = 1=2: 6. (a) False. (b) True. (c) True. (d) True. (e) False. 7. The machine is shown below: {0}/0 {1}/0 {1}/0 e f {0}/0 {0}/0 The simulation relation is fa; e; b; f ; c; g; d; gg: 3 {1}/1 g ...
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