06FMT3SOL

06FMT3SOL - ECE 301 Midterm#3 7:00—8:00pm Tuesday Nov 28...

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Unformatted text preview: ECE 301, Midterm #3 7:00—8:00pm Tuesday. Nov. 28, PHYS 112, . Enter your name, student ID number, e—mail address, and signature in the space provided on this page, NOW! . This is a closed book exam. . This exam contains 5 questions. For multiple-choice and short—answer questions, there is no need to justify your answers. You have one hour to complete it. I will suggest not spending too much time on a single question, and work on those you know how to solve. . The sub—questions of a given question are listed from the easiest to the hardest. The best strategy may be to finish only the sub—questions you know exactly how to solve. . There are a total of 12 pages in the exam booklet. Use the back of each page for rough work. . Neither calculator nor crib sheet is allowed. . Important! There is no question identical to the exercises, and therefore trying to duplicate those solutions from your memory will be given zero credit. . Read through all of the problems first, and consult with the TA during the first 15 minutes. After that, no questions should be asked unless under special circum— stances, which is at TA’s discretion. You can also get a feel for how long each question might take after browsing through the entire question set. Good luck! gotutftb 1/\ Name: Student ID: E—mail: Signature: The continuous Fourier transform pair: 1 0° ‘ X (w)ej“’tdw 27" wz—oo / x(t)e*j“’tdt t =-oo The discrete Fourier transform pair: ll 1 , é; 27f X(w)e”’”dw Z x[n]e_j“’” n=—oo Question 1: [30%] Multiple—choice / short-answer questions 1. [Outcomes 4, 5, and 6: 5%] Briefly explain the difference between AM—DSB and AM—SSB. What is the advantage of AM-SSB. 2. [Outcomes 4, 5, and 6: 5%] Briefly explain the difference between synchronous and asynchronous demodulations for AM signals. 3. [Outcomes 4, 5, and 6: 5%] The following signal is an AM—SSB system. Is the system retaining the upper or the lower sideband. (Hint: This is the hardest sub— question and you may want to skip this sub—question and come back to it later.) (If you do not know the answer, explain how to derive the spectrum Y(w) from X (w) given y(t) = 33(15) cos(wt), which will make you satisfy the ABET requirement for this sub—question.) 4. [Outcomes 1, 4, 5, and 6: 5%] Suppose “radiowav” is a wav file containing two radio stations. We would like to “tune” to the radio station centered at 5.5kHz by the following MATLAB codes. duration=8; f_sample=44100; t=(((O-4)*f_sample+0.5):((duration-4)*f_sample-O.5))/f_sample; [radio, f_sample, N]=wavread(’radio’); radio= radio’; w=1500*2*pi; wa=????; wb=????; wc=????; hd=1/(pi*t).*(sin(wa*t)-sin(wb*t)); he=1/(pi*t).*(sin(w*t)); y=ece301conv(radio, hd); z=ece301conv(y.*cos(wc*t), he); wavplay(z,f_sample); What is the meaning of wa? What value should wa be? 5. [Outcomes 4, 5, and 6: 5%} What are the values of wb and wc? 6. [Outcomes 1, 2, 4, 5, and 6: 5%] We found out that the sound volume of this particular radio station is unexpectedly small. Is there anything wrong with the code? (If your answer is yes, explain why / where is the error? If your answer is no, any possible explanation in terms of the amplitude modulation?) 1 ‘Dgg : USQ Cam/dle side—bani ‘par flashes/0N ty Acacia] miepiyaa. Omani) AM“SSB ; U392 gamer The side/WMM Oflép [Hider 01/ an: DIP/Der. Admwfigq : Com Supper—t Mane [email protected] S'éotffuus“ (Next/Mk the , EMde‘EA. 2. @ndmmous Demooh The Kecém/e/y has ‘60 Rica/er OK Question 2: [32%] Short~answer / multiple—choice questions: 1, [Outcomes 4, 5, and 6: 5%] What are the two conditions of the sampling theo— rem, and what are their physical meanings in terms of the corresponding frequency spectrum? 2. [Outcomes 4, 5, and 6: 6%] m(t) = cos(77rt) is sampled at time instants t = g for n = 0, i1, i2, . . .. We use 33,,(t) to denote the corresponding impulse train sampling. What are the values of xix-71;) and 33,,(02)? (a) 0, (b) 008(1), (0) cos(1)oo, (d) cos(77T/5), (e) cos(77r/5)oo. 3. [Outcomes 4, 5, and 6: 5%] For perfect reconstruction, xp(t) will be passed through a low—pass filter. What is the cut—off frequency of the low—pass filter? (a) 3.57r, (b) 47r, (c) 57r. Work-out questions 1. [Outcomes 4, 5, and 6: 5%] Let the original signal :r(t) = Ll(t+1/3) —L{(t—2/3) and the sampling period be T5 = 0.5. What are the sampled values xs[n] constructed from 53(15)? (Hint: We may have many zero values for different 2. [Outcomes 4, 5, and 6: 7%] Suppose we use the band—limited perfect reconstruction to reconstruct 33(25) from (ms[n],Ts), which corresponds to low—pass filters in the frequency domain or superposition of sinc functions in the time domain. Let xr(t) denote the reconstructed signal. Find out 3341/3). (Hint: Use the superposition of sinc functions.) 3. [Outcomes 4, 5, and 6: 4%] Is xr(1/3) = 513(1/3)? If so, explain why? If not, explain why? >1” 2 y; :(OTL vac) ' 4* %[OJ C W 1: X0321, m 965%] = ’XCOS :4. ( I O 6&0szch {1% a“ Offier [’1‘ 5 mt): may WQMD'W mt +xgtij k S"?A(_1KC"{Z~OL§)) flittwls) MEL): fine-ifs“ gzntém) ‘1?" + --———.-—-— 1A ‘TC/s _ 3 ,E XE “ 3} “3? ‘f l 2 ) \ CUE ‘ 4m Question 5’: [Work—out Question 18%] Suppose 33(t) is sampled with sampling period Ts, namely, :r(t) is converted into (xs[n],Ts). One student tries to construct a “digital differentiator” by setting ys = 2;} (xs[n + 1] — xs[n — And then use (ys[n], T3) to reconstruct y(t). 1. [Outcomes 1, 2: 5%] What is the impulse response hs[n] of the discrete system mslnl H yslnl? 2. [Outcomes 4, and 5: 5%] What is the corresponding frequency response H3011)? 3. [Outcomes 4, 5, and 6: 8%] What is the corresponding frequency response H (w) of the end—to—end system :c(t) I—a y(t)? (Hint: You may want to start with plotting Hs(w) and carefully mark the axis. Then plot Hp(w) of the system: asp(t) 1—) yp(t). And in the end, find H (w) by considering the LPF used in the reconstruction. The intermediate steps will give you partial credits.) Question 4: [Work-out Question 13%] 1. [Outcome 3: 6%] Let x1[n] = 6[n —- 2] — 26]n +1] and x2[n] = Ll[n +1] — Ll]n — 1]. Find out x1[n] >I< $2 2. [Outcome 3: 7%] Let = Ll(t + 1) — Ll(t — 1) and 324(t) = 6(t —— 3) + 6(t + 3). Find out 303(t) * 3:4(t). 7c? [A] 96 [n3 $0" M u At] 0(1th‘96 X11313 = Question 5: [Multiple-choice Question 15%] Consider the following two systems: 231(t) H y1(t) and 332(t) r—> y2(t) satisfying $1(t)cos(wct) cos (wot + 152(0) . 1/105) 92“) (5) (6) . [Outcomes 1, 4, and 5: 4%] Are these two systems linear? . [Outcomes 1, 4, and 5: 4%] Are these two systems time-invariant? . [Outcomes 1, 4, and 5: 4%] Are these two systems memoryless? . Outcomes 1, 4, and 5: 3%] Is the first system invertible? Justify your answer by one sentence. Sfem 1— gs‘éam ;_ Linear NON] P’W‘F We “Vaglg‘afi Tine _ Vary)? \/]6ny {65.5 xii, ~SXWL) My 961d) (MW Gnu/e giétj=€3~ Of you QM answer £6;$%TIM Follows. 015 /\ Tux/«fifth 3 T3406 System i F3 Q {J/S‘fiim. :41 as or? M «23%! «\s a“ rt TS Newbie, (We Can rwmstmei 1%: flew PM) bomb/*[fmffng fie’fl‘ ...
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This note was uploaded on 02/07/2011 for the course ECE 301 taught by Professor V."ragu"balakrishnan during the Fall '06 term at Purdue.

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06FMT3SOL - ECE 301 Midterm#3 7:00—8:00pm Tuesday Nov 28...

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