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examII_s04 - EE 350 EXAM II 1 March 2004 Last Name...

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Unformatted text preview: EE 350 EXAM II 1 March 2004 Last Name: gsglu‘ngni First Name: ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score I Instructions 1. You have two hours to complete this exam. asap This is a closed-book exam. You are allowed to use both sides of a 8.5” by 11” note sheet. . Calculators are not allowed. . Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number, for example, “Problem 2.1) Continued.” No credit will be given to a solution that does not meet this requirement. . Do not remove any pages from this exam. Loose papers will not be accepted and a grade of zero will be assigned. . The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. Problem 1: (25 Points) 1. (12 points) A LTI system with input f(t) and response y(t) is represented by the ODE 1') + 4W) = f0)- Find the impulse response representation of the system and, if your solution contains a term with a delta functional, simplify that term. Funk the «Zero-{tactic Vfll‘t-C'bp mpg“ 66*?) 'Hfien huh :J&/£+:. — Char. E60. : )z'l' “1:0. Rodhs 7_ Char. £60", 2 = i: 2). Form % nectar-fl. (aroma: ;h(t) 2 ”cg: 2t 4- 6J1!) 2-f- Form # par-legv- “399059;. apt-e) :— cc. (_ a». ronJ'éun‘L') 50'! cy- . " ”(PC °“ 37* ”W =1 ‘9 arm“? t3° SolV\na_'Fur~ 0 qua 6 (a C°3 =¥C°7 =0) 0 ;C°3=-0: fi-q-{T fis-yu/ all 82-0 3073 =0 :2 2.6 3 Ezra—shack; un‘fiasfif rospoffigfi act) 2 #(l‘ COS 1t) WLt) Mb) : if = J; $102.1: Mt) +- ‘Lw'Cl’WS “3 He) W a when #50 2. (8 points) A LTI system, different from the one considered in part 1, has the impulse response h(t) : [1 — sin(2t)] u(t). Is this system BIBO stable ? Show work to justify your answer. b(t) 3. (5 points) The block diagram in Figure 1 contains three LTI systems whose impulse response functions are h1(t), MO), and h3(t). The input to the overall system is f(t), and the output is y(t). Represent the composite system using a. single impulse response function h(t) so that W) = f (t) * W), and express h(t) in terms of h1(t), h2(t), and h3(t). f(t) Figure 1: System comprised of several subsystems specified by their impulse response function. (0) got) : «~01:an - a.C£=7>rh3(t3 (O) 3.64:3 : -F(-b) at. in Hz) Cb) Wm = germ-Ht) s was: ‘FHn-usz-B-F“) subwuey can WV. (A) ("to (:2) 3M; 3C4.) : [:Ftt) 9: hot.) -F(£)]-)the) -- [-Fz-tw-t‘J—gfl» In It) at .— 'Fft) ark, (flaring-t) - “HQ-H419 - {it-53+ k‘H-J sham ofl—t} :1 {(45).}‘E h.(t5*‘12(‘a -h.b[-Q 45063 5-h3 (t}] W hlt) Problem 2: (25 points) 1. (14 points) Use the graphical convolution method to determine an expression for the zero-state response y(t) : f(t) * h.(t) for N) h(t) —2 u(t) + 4 u(t — 1) 3 e—(t ’ 2) u(t — 2). In order to receive credit: 0 Specify regions of integration by writing the convolution integral with appropriate limits of integration and specifying the range of t for which the integral is valid. 0 For each convolution integral, substitute for f and h using the expressions given above and simplify the integrand by eliminating unnecessary unit-step functions. 0 For each region of integration, provide a sketch showing the position of f and h. 0 Do not evaluate the convolution integral for each region of integration. Firs‘l: sketch 46-h) amp. 2115):" {-‘c-e 5“" .. . 3- e -Z 2 yet): I -[.‘(t)h(t~t3oQ—'C ‘09 {-ct) hue—c) ""3 1. 1'. ’C -2 '11-'2- Realon 1' 13-2. <0 or t 4 2. when t'Z. <0) -F(‘c')h({:"c) :0 Cor W02 C) omfl. 't > 3 -(4:-c— z) p—J ~7— I 3e = SHOE“: 4:)ch 4' Sfctwflt- ficflc ° I 2. (5 points) Another system has the impulse response h(t) and input f (t) h(t) = "(t+2)—u(t—3) m) = e-3(t-1)u(t)_e—3(t—1),,(t_4). What is the width (duration) of the zero-state response y(t) = f(t) * h(t) ? To receive credit, justify your result or show how you applied an appropriate property. b (t) Lmflih (guru/é MA % h Lt) .4": 5. may; Cot/mam % FHA 75 ‘4 (1.5m; the, W\£+; (gufw‘bmv PPOPQV—EJJ :4: a: m = fit) a: hut-A, flan .— uwfltkj a“) — (4].;ch % -FC¢) 4, owjfi % 5/16) (#1th awry‘hm) ’ a (-8) IS 7 I 3. (6 points) Another system has the impulse response h(t) and input f(t) Mt) : 3 cos(t) [u (i + g) “ “- (t _ 1)] 2 f(t) = 6(t+g)+6(t—g). Neatly sketch f(t), h(t), and y(t) = f(t) II: h(t). Use the. resulb ‘Flbvk 8'“? '7): {It-Tl 8 (t +15) arm) + 80c— 1;)...Hfl gob): .Fct)a<\\lt)= : th+1§zD + th’g) Problem 3: (25 points) 1. (9 points) The circuit shown in Figure 2 has an input voltage f(t) and an output voltage y(t). Deter- mine the frequency response function, H (1w) 2 Y / F, and express your answer in standard form |"<' bm(Jw)"‘ + l>m_1(1w)"“1 ‘ ' - + In (w) + be HOw) = ~ : (Jw)" + (In-1(Jw)"_1 + “1(Jw) + “0 "a Figure 2: Passive RC filter circuit. Skeflk 'tko. Circuit in fl'Q— [Adar final}: ml 053 ‘9 S'vfce- J ‘EflmsCoIMU-vizmq {=0 fin/L} thfl- Ctrcmf' R'flL/(R|+p-z,> RIHKL = 1»: R1 ‘7. 4" (n+2; fl 3 9’ 'gqu ’5 05:3 voHaaac, ‘jl’ét‘Sl-On V = “2* W £2— E --—--f———;;"‘ 8.71:. K31. ywo + 3.13:1; 7 (R5 + I/wc.) Ra. . file; v z _______¢_________————— w /F R; CR.+R,_) + CR.+¢D/du/¢. +R.R.. a d—WCRJRL‘I" (2’; H R-HQ éthflsck.+¢p)+fl.flz3¢ + a 2— 10 2. (9 points) Consider a LTI system with input f(t) and output y(t) that has the frequency response function 0 001 J . w H : —~——-. (3”) 1+ 10.001w Using phasor analysis techniques, determine the sinusoidal steady-state response of the system for the input f(t) = 7+ —% cos(1000t+ 15°). From your result, determine whether the system is a lowpass filter or a highpass filter. gCt) : 7 ”90) -+ T—Li: [H9 1000” C05 (loco-l: 4- 15° dell-“(710003) wkcrq, H993 z:- 0. I [H(alm)l:F = [i = L 000 = ”L -| H 0‘! 3 ”7 LVN»: 210,1— Annf-H QAQ Jo #- (+3 :: '12-: (as (loao‘t +60°) J'lzem 75 ~ IOWA/cu“; WCHIr- Lemon. tllq— DC- ferm % tkelbpflé 35 m4: flaifél' {1,7}; #0. tau-lam. 11 3. (7 points) Write code for a MATLAB m-file that generates the Bode magnitude and phase plots for the frequency response function 30.001 w H = ———-——. (7”) 1+ 90.001“; over three decades starting at 10 rad /sec using a frequency vector containing five hundred logarithmi- cally equally spaced points. omega. = floaspcec i, LI, 500)) P = E 0.005 0‘]; Q : 0.00], I]; £0909. C an r] G?) omea a); 12 \ Problem 4: (25 points) 1. (12 points) Consider two real-valued orthonormal signals ¢1(t) and ¢2(t) that are defined over the interval [—T,T]. Form two new signals (1130.) and (3540,) defined as ‘75305) a ‘75105) — 5 ‘75205) ‘75403) a ¢1(t) + b (7520), where a, and b are non-zero and real-valued. Determine all values of a and b for which ¢3(t) and ¢4(t) are orthonormal. We need. ho choose. a uni b Judo tha" <¢3J¢a> 9b <¢VJ¢~1>=lJ «ma <¢39 4’7) :0. 75‘56 Confirm-4'5 (51/295; the ‘Followly relw‘Etoaslth babe“ a “fig 5" I = «33.4533 —- <«¢.—L¢z, “5,. 591.): q lat/p,» z.b<¢/¢Q+b’-<¢2/¢z> I = q 7-+b’- (0) I = <4“, a.» = <14 +b¢z a¢.+b¢z> = q "‘(abyiy qu <¢fg+k<¢z¢g 1‘ “1+5; (Jame as (0)) 2. (13 points) Consider the set of three mutually orthonormal signals $105), $203), $305) in Figure 3 that are defined over the interval [—1, 1]. These three orthnormal signals are combined to produce a signal 905) = C1 1P105) + 62 1P205) + Ca 1P3“), where c1, c2, and C3 are real-valued constants. Determine the value of c1, C2, C3 that minimizes the energy of the signal g(t) — sin(7rt) over the interval [—1, 1]. w1( t) we w3(t) Figure 3: Three mutually orthonormal signals defined over the interval [—1,1]. “€160 the Sid/sag 5.1% 7S ortk°v90“£J ‘Lo MIflIMIu thfl- anew”? 4 the. affroumwfimn erNr- #C-b)~5m(rr'¢) choose. c; = < (1,135”. (IT-6)) 3 < 4?, 5m ("10> <. “’1'; 95‘), 2| hr...“ the 5:7,...9 5&1: vrtl‘onunmua- JL 9. C — l \[o n , I I ~ ‘Hm mun) i-b- 1:. 2 fimhfilt .-~_2._, “(n-1;)! -I 2 r21" 0 all an o L———————-—-—-—-l even mt? NHL 14 ' WOJL Cm C’- = j V119 Sm (”a «84; ‘I to“). Fe» W Quen fin 73% 00.645 ”1‘11?er 7.5 (flan'LLxcfiQ ’£° the“ ”ne '4”— CI eyéép'é’ ‘Fur' a, man shun?) arise 5° ‘ even fin C 1: ”fl 3 J V; (+3 5m (Wt)o€1‘: "‘I \l I 2- g (-45.3 Sm (M) 47-9 0 \l O \.——-—————J will?” 15 ...
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