midterm_solns

# midterm_solns - Department ef E-lchl'ieal 3L5 Cnmputer...

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Unformatted text preview: Department ef E-lchl'ieal 3L5 Cnmputer Engineering Eli—3F. EB Unchreitry ef (Jalii'n1'11ia1 Santa, Barbara Wrintcr iﬂll Shynk HAUL #15 I'VIIDTE RM EXAM SOLUTIONS __.——_________ INSTRUCTIONS i. This exam is open hunk anci epcn notes. ‘r’eu may use :1. calculator. ‘2. It ceneiete [11' 3 problems and is “'1:th e meccirnum of Eli} uninte. The problems are net, of equal dilhculty! so use c'liecrcLiun in ellecuting 1mm lime. Answer all questions in an}; order. 3. Show your anewcrs in the epacce pa'nvided: and mm the back side cf the exam pagen' if you nccrl auiclitmrlel work spam. Show ycrur reasoning and the essential steps clearlyr and concisely. Last Name, Final, Name: Scores: 1- _ 2. 3. _ rlllml: ECE 213 Winter ﬂﬂll 1. TIME—DOMAIN CIRCUIT ANALYSIS {30 points} AHHutm-‘i that tlm Switch in tho circuit helm? Inn-i iii-1H1] Murat-iii ﬂir H. Icing> time IJHl—{Jflj I'. = [3. {HJ Write a, second—order differential equatinn {DE} ﬁrm the mpm‘itm‘ voltage LIEU.) for t 3 U. [13} Find tho L'r‘LITIlFIETC transform of the DH: write the Ciirmapi'nuling uheu-mnﬂriath: H]11a1.i[1l!l._ and determine [-er l ypr: of dzunpwi Lain'nii. after the switch opens. [c] Using a, partial fraction expansion: ﬁnd the Hiihitmil viii] fur I, :3 ﬂ: iliclii-i'Iing :11] parameters and [fﬂﬂﬁitiiﬂTll'-5. 1001'} 100'!) “JV Solution: . . J nPﬁ'E’ {1.) 1:?0 2“; 56%an {ELL- E} (1% gag; Egg; {HIE}: {EELS-q? '1 ‘- = ‘ ' mint; KVL- n3 : 3421+ L‘izi + ”as; Hum M / {if Eu ism-51'“ n :1 Qéﬂc. . a x13: 3RL\$E¥LC§EQ+VL d‘ \$3. a»: in? ~ ..1- : cg}; ﬁzﬁvr‘ﬁh gubﬁf L”:— {ll-‘2‘ 361A VHWES cLVci. #1 gm; V: =V5 r-'~ 3‘ "' r-f ﬁjr 3:: :5 +3: :7: “5éiz+won_c+w*w '0‘”: a“ 61:1 ch: 9‘ e) IV‘CUSLE It: -. n l - it]? (1,) Layman WWSGW MoTﬁ‘. étarn rmTuﬁ‘l wit) g mnaiﬁI‘m-aﬁ - E? ’— EQUC‘J,‘ iﬂﬂSVci‘Iﬂ 154'; Pig-— ‘5. un<lerrclmﬁel Ch’H'i“~ I"! 5: DE” g-9+1aoa+1020 6201315.: 'lﬂDi—IN |ﬁquFJGSF ._—- h-S‘ﬂ'le-Gh" i-‘Qﬂ I} :5 I “55’“ _. ngnj: 4: f.‘ Effie, vi..— 2 IE5 : 1-0 ..___..-—- ‘* '—,—'—'—“—'—' _ o 1 - a (gulf 1 am: + mi: ) S-(ﬁ+§ﬂ+a So.ﬁ§3>(g+§a_& sci/{3a) :- 6 JH" I“? + 3* S mg‘oﬁsoﬂﬂ Ema? Sam I. 4P" 7:“ “r5 :10 F_____. I; g :- m ‘*‘ _ 10‘; ‘i’WJﬁ’D/HQ HM _ 3 .2512: ‘3‘- a. . (CS-+50 O) I'ni ‘1) gL-HSO'FéSOF‘i—iq (533+i‘G-‘3ﬁj (—Qdfé/Cé-‘QIB \._.d—-—-—_.‘,s--—.__ﬂ x H36: 1'; — S "' 05550) ~ 40.3 _ at , ~ *‘ “a ____ ‘3‘“?3 Q 2 ‘c§+ﬂ 3 6 ”5’ 5' 5: IL d m EL £543.51“; E’. .-'.~ ﬂ L ' 1 a w m w; I; + , n r ”I“ _ 1:5 F 3 g+§o+d§o 301 52.15ch 0 30! Fir—Um \ L WHJEL Hui-5R —-§(}E (3&4: Kiwi 15‘ ) m uttl {1(5050 E. c 05. (Sea/(TIE?! TFQ-q‘EEBUIE) 426C ﬂint}: [110+ was a CJSC313-ExSTf-aﬂ'233]Lift) ﬂ-H EUR 2]] Winter 2011 2. S—DUMATN CIRCUIT ANALYSIS [3|] points} [:1} Find the transfer funrtion H [5} = Elie-‘1!” 1r"; {3] for the circuit below. (Hint: “Trim an oxln-He-asirm for Vat-5‘) by hunting thn Thévenin (quixralcnt r-it'rnit tn the left. :31' Cg.) [1)] Sketch the 10010.5 and ”zit-11»; nl' Hta} on the s—plmm fur H; = R2 = Eﬂﬂﬂ £3 and C] —' (T2 = 2 ”F. {It} Find the impulm rtah'pmnse far this circuit! Lu, determine unlit] when unﬁt} --. 5(1). ”ll: [3' R2 C2 'u'DI: t) F *3 I at f’ '" Solution: L01 ‘1 C13 F _.._ 1123,} Iii-3‘8 gnﬁﬁtVGul'T LT? }ﬂ~1 : Rti’ﬂa-fﬂtga‘clg cf qo‘lWEﬁ“ é _ Rgcagjt 1 MT“ 3 52?“ Wits} 62 at; T“ ﬂags >V1m unﬁt? % US 36‘3“ n tilﬁlﬁm “43L“ — UCﬁLS Han/15) RaCtS 15;) H13]: EH21; m 10:35; va'+'3,5. 4_ 1 , 33+3¢=a§+ my -—;;;1.'3 agﬁa mPﬁ 34:1" ((1 rmﬁulm Reaftmae: Him -—-——‘3-: inf-c) Ha lac»: my“: 1 r5} Jr :3 HM —= erI ' SFTBéIE 3+3‘R-a~ 34:35:38 €3'33’»3~ Q): ‘IEJOS _-—-———-——_- 3 ‘Ll‘T-I S+3€a~ java-E #3331: -3613 MICHEIFHE 4% ”7.1-5: jum] Ff 1F.- 2E 1I.-"u"inter Eﬂll 3. DIODE MODELS [2U puitlte} Let I3 = 113' 12 [HA and VT = ‘35 111%". [it] Dltutt'llliﬂt’. the current in lhreugh the diede in the circuit helew using, the expuirnetltiel model. J'teetlnle initially that if}; = U ttereee the diode. {Three itertttiene of the nenlineer equal-i011 are euﬂieient.) (h) T he piecewise linear triedel ef the diede is given by _ _ ”1 up <1: 1].? “3' _ { eve + b. m :3 [1.7 “3 Fur e. = [Ll E'Lt'll'l l) = --ll.[l?t ﬁnrl the current in through the t'lltitle. {Hittte Sinee the JIIIIHll-‘il i5 lirlHEtl'. yum will eel-we explitdllj; fur in.) [e] Returning Lti the Expuilltlll-lﬂl tiltitlﬂ JIltitlL‘l 111 Ill-ELIE {all determine here your answer changes if a. secend identical diede i5 plaemi in series.- with the ether eireuit elements. [de Explain What happens when three identieal diedee are phat-ed in eeriee in the eireuit. h 2‘“ e seen <3 1WS L0“ {:31 EgunTiiﬂi: JG 1 a n Mt, “0 Solution: ) W f%-—--_ _ “a x . {U} Irina”; _ 1- '-\ 1TE-t-HTiM'} n e JD HE) 2 VT in [4D}IS) HR (3 C) Ll ".l l I, one“ @959”! l a, Qi'i'ltlgl‘ a-S‘m ll 3 c} f? I “Hi 62*ng ,e'r 9“" ﬁ . _ (b1 RtgﬂﬁrCtL 'm egn bur-T17 VD — 3640-5-07? "qt gout =2 Reiniﬂron :5) JD: Still": ewe .1 LC) Ewmiww TuLﬂJ aﬂcxfrr RfFHHX;L*} T] WITH-1} gnaw-DI _ 4‘33 W) mm {TPFﬂTiﬂFp V9 )0 _ o o q I OFTJSH FLOWS & ﬁ-éqgi “338 3 Dréqé‘ﬁ rue-HG Pa‘r‘ 3 Eircrglﬁ‘éj L043. {HIM EXfrCTﬁvg.H;-l HM.) CUP‘r-C’MT bELM‘ZﬂL {.ﬁnc‘lq VD 5.; 9,7 ) J a ﬁ 1’13ch anzubvo :7) Ugégvzvb M ...
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