examI_s02 - EE — 350 Exam 1 ‘ 4 February 2002.a Last...

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Unformatted text preview: EE — 350 ,, ' Exam 1 ‘ 4 February 2002 .a ' Last Name SOL/800$ . First Name Student # / Section D N P 0L T Problem Wei _ht 25 25 25 25 Score 31.. wNv—l E, Test Form A 19m You have 2 hours to complete the exam. Calculators are [not allowed. This exam is closed book. You may use one 8.5 " x 11" note sheet. 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 1.2 Continued". NO credit will be given to solutions that do not meet \ this requirement. ‘ - ~ 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. ‘ 6. If you introduce a voltage or current in the analysis of a circuit, you must clearly label the new variable in the circuit diagram and indicate the voltage polarity or current direction. If you fail to clearly define the voltages and currents used in your analysis, you will receive ZERO credit. , 75‘ “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. " PWP?‘ Problem 1 (25 points) 1.1 (15 points) Determine if the following system, with input f(t) and output y(t), is zero—state linear, time invariant and causal. Y(t)=cos(t)f(t+1) W We» 6951,99 = (osc-t) ace“) 4 (a as.» Ohm = (om-h 13 (t-H) 2. (+3 $1424!) ; «3,149 at. «C (a 0: cos , +'/5£Z(Q'E_3 + 3 coscofi(t+t) +£7wa the. aero-S'EW'LQ {05122. “06¢”. 66> + Mire dug/0; the £51207 is zero —-5 11$ Hm. sastem time, Invonnarflli ? 43 (t3 = .F (i???) @ gal-m: (osLQ-flCt'l-I) '2. I In iewns 62 411-9) 521-3 -_-, (05 {-b‘) ‘9'lc‘t‘k—r + D we. :L-‘i- the 5ast2m 75 4;qu "'7"°""°""LJ ihe” :— H: T), ‘H‘UWQVQ—V‘) 2 (4:3 :2 0% (4: +T) a; {$11.9 “Ct + % 7+1) 7L avg—t} = cos (+3 filfi‘r +1) \C-lz-t—fl = Costly-T) 41‘ (11+ I 9 ‘9.— So the. s stem 3; not 't'lm-e/“lflvaflan L. n V mgs’k hCLV-Q. 1.2 (10 points) Express the following function in terms of the unit step function, u(t), and polynomials in t. m-berce pt 4; J; (a = (- 2t -2) ( M—h‘l’?) - «(H03 ‘ W if it —2.(t4'-l \ e. :1 5°? 0 otVQrwlSG- M 42 rce? i: l m : (.ch m (mm mm» W/ 5:: ,. l l < 1: <2. Pa 3, O o’H‘QrwnW 1C (19 = 59. H57 + file) Problem 2 (25 points) 2.1 (9 points) Determine if the signal f (t) = 3e'2’u (—t) is a power signal, energy signal or neither. If the signal is a power or energy signal, find the appropriate measure. Rt) 2.2 (7 points) A system is modeled by the linear ordinary differential equation dfdf)+20f<t>=f’%,,ii’+25y<t> i) Is the system asymptotically stable? Justify your answer. ii) Determine the rise time of the system's step response. iii) Determine the settling time of the system's step response. 1) .7 ga- Foot # 't’Q, aka. faLLQ VI 54739 e Dyan-Elvn Jwfls‘plej QC?“ r— 7s+ 23‘ so ’2\| <0) tho. amQ. ‘ 7h: -25‘, (Se/CW 2.1.1) Tfle mSvltS '1!» parts (Li) uni can) 082- ‘ 4; s 0):!) gas; w—ttofls jQuQ’Qa/ci 'er ~PVS£ 'ofooQV‘ $65 em whom, no -_—. o , $2.ch m :1 #0,, this Sarnéem/ the ttfva. vet/[v.99 % fr qmfl, *5 WI“ achéV‘ ‘Pram those ytven aboVQa 2.3 (9 points) Write the Matlab code necessary to generate a plot of y (t) = e‘a' cos (2t)u (t) on the interval 0 S t S 3. Include appropriate labels on the x and y axis. 13 = anSFaLQ. (o, 33', 5L :- QXFbBaz-fldk CoSCZek-D') #lab‘8‘ ( \ anplft/OQQ' I) Problem 3 (25 points) 3.1 (8 points) For the linear circuit shown, find the ordinary differential equation relating the input signal f(t) to the output signal y(t). Place your answer in standard form. 3.2 (5 points) Find the characteristic equation of the system and its root(s). 79E, c\\aro.cLerxs—Hg eat/03:10!) 75 R‘a—R'I— = O L’\="\+ 62A / L 7% rod-l: % the, c aruc'L'e hS’éjC 25V 0.1le \5 3.3 (12 points) Consider a system, with input f(t) = 3t and output y(t), that is modeled by the ordinary differential equation dfii’)+6y<r>=4f<r>. Given the initial condition y(0+) = 3, find the total response using the classical solution method. Ch OeV‘cLGLQ Y‘ 543%.. Q agate {an O! n be (*5 foo‘t ‘. 6? (fi\ :- 7‘ + 6 = O :2) fl’ = '- 6 TRQ, xvrm % ‘UQ 20.07074090th s'gle'lllofl i5 -6'l: 044mg: 0&3.th = C16 1: 20. Tlie, £9,” % (the. par’éicJar Su’v'élofl 35 og—t—é’t. be, 430 ’Fmo. vaS‘Lt‘Ev't-Q ap :3 0C, + at Info thQ— 0 0C. empo— é. g!) + 65p :' 12+: c—v) t9.+6oa+6/3t 2122s Far- the. Qusi: efi‘JQIt-éd '50 th .~ l3 4" 6°C.. ‘2 O a) g 1: 2 Q 0; = ”é 6/9 :: 1'2. " V * — J- t :2 0. gr) (t3 -— 3 4* 2t F: a C ‘ {:3 = h ”:5 + (1’45). m , 7R2, bring So'v‘lZIo/l :3 g t g 5/ Using 1'th Ifll‘L‘ld-Q— canopt-émm -L a lo '60 J_+ 2.0 7:) c|:3.3_ 7 Problem4 (25 points) 4.1 (9 points) For the circuit shown, determine the ordinary differential equation model of the A Circuit's input—output relation. TRQ. Volta? VA 75 'tkafi d—C r055 K ‘- Vfl- 1 Ra- GE) - U$|na_ KCL oil 00on— 9: *‘Fufl 4' .LLl-D + La (-0 +5; [-D=O S U‘OS'Ll‘LJh‘ “a- .. CJVA- : p‘gCJV-v “ i (Ltt) = (L(o+3+ JESYQITS f 0 ”I: = LL(o+3+ :5. 3 War dank/es (Larry + —%—S:;g(r)aac + RC3? + £1.” :3 Hi) 10 4.2 (8 points) A system's input-output relationship is given by 39f(t)= (D2 +6D+13)y(t). Let f (t) = 311 (t), y (0+ ) = 3 , and dig) = —9. Determine the zero-state response. t=0+ (9(3): zz+67i+13 = 2%- wanZ-I-wg' ,. 3 Con = n? 1’: 19:; — 77.3— (mvj- w9.=wn»/z—r2- MAW; 43..-}. «.175 AS the 35432") 7.5 vnoOQr-gampeeQ (because LP4") .. -t > #hca : éfwflkCAroswlf+ GJMuago'éB: Q 3 (Bras fig++ (ismfizt) t—Q 82¢qu 'the. qury ~Cvnc‘5un IS a- COOSt‘Cl/i'i) ,Oe'é if 1°C. 2?]:- 6%4- ’33P: 3.37 :9 5f: 7 #20 (:5— (0+ (+5 = 9:31: (fires “74: + 65,.) FOUL—3 + 7 £20 3;: — a” a? a9 J’ém/IS (a): . 032°. 0410052 9 6! m9. 6 4:0 5 we; 3% ti? a ”Vet Co A ‘ 0"“ 2-5 I .2 3 9+7 :7) ’9:_7 0 5‘03 0 [173.0 «- Bh‘o‘ C05 ‘00) aC®:o : -3("7 +03 ~1— (*7/735 (3: “27/17:, 11 4.3 (8 points) A system's input-output relationship is given by 39f(t)=(D2+6D+13)y(t). Let f(t)=3u(t), J’(0+)=3,and dig) = —9. Detennine the zero-input response. t=0+ To {met the urO-lhpvb rasfonSe. «Ll-(=3 7: ref '50 ECWOJ ans. So It ‘FoNoM 5” = O: 5/: 6%04— (Bay :. 57.0 a) a? :O The «Ramayana,» .93,me ,Qqs 15?» same, [urn Q5 :0 pavt 1-2) amoa So geif‘bfi = a}, H.) +07%) _-.- “to Saa‘bcf? ’the, am'éwQ anva’élonS Q—3t (Ace: [174: + [55m I73 4:) t 20 CM 032. A aan 6 12 ...
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