{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam 2 soln opt - EECS 314 Winter 2005 Exam 2 Questions...

Info iconThis preview shows pages 1–26. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 20
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 314 // Winter 2005 // Exam 2 // Questions Thursday March 17, 2005, 9:30 — 11:00 AM Students will take exam in the following rooms, according to the first letter of their last names: 220 Chrysler Auditorium (our lecture room) — A through L 1504 G.G.Brown (Lee Iacocca Lecture Hall) — M through S 1303 EECS — T through z Students whose doctors recommend “1 1/2 time” on the exam will take it in 2246 EECS. ' Instructor: Alexander Ganago W 1"“ $0M P i¢ W Exam rules 5 “wt 1. O“ 9 Under the Honor Code, do NOT discuss this exam with anybody until 5 am. Friday March 18, 2005 1. The exam costs 50 points and includes two groups of multiple-choice questions. Questions 1 through 10 are 2 points each; questions 11 through 20 are 3 points each. 2. You will be given 80 minutes to work on the exam (besides the time you need to fill in your name and UM ID number on the answer sheet). If you have a class right after the exam, plan to arrive early for an earlier start. You must sign the attendance form before you take the exam. The answers to the exam should be written on the standard sheet for multiple-choice questions provided by the University of Michigan. 5. Bring your calculators, good 2B pencils and erasers! 6 . The exam will be graded automatically; the scanning machine will evaluate your answers based on the darkest spots on your answer sheet. Ifyou believe your answer sheet gets so dirty that the machine might misread your answers please ask your instructor for a blank sheet. Note that the machine cannot accept “double answers” such as “A and E.” 7. This exam booklet lists all questions and provides workspace. You may keep it for reference. If you need more workspace, ask your instructor for extra pages. Calculators are allowed. PDA, cell phones, and laptop/notebook computers are not allowed on the exams. 0. Books and notebooks are NOT allowed. Two “cheat sheets” of standard size 8.5x11” are allowed (with notes made on both sides). 11. Hand in your exam to your instructor. Do NOT leave until your instructor makes a written note certifying the receipt of your answer sheet. PE” “2°.” Abide by the Engineering Honor Code: “I have neither given nor received any aid on this exam, nor have I concealed any violation of the Honor Code.” Alexander Ganago Page 1 0% February 28 — March 9, 2005 26 Version 1.1 EECS 314 // Winter 2005 // Exam 2 // Questions Section One: Z-point questions These waveforms belong to both questions 1 and 2. l The circuit shown on the diagram below C + VIN R Vour is fed with an input signal, which is a square wave at a low frequency, Which of the output waveforms is observed? A. Waveform A B. Waveform B C. Waveform C E. None of the above. 2 The circuit shown on the diagram below L + VIN 9 VOUT is fed with an input signal, which is a square wave at a low frequency. Which of the output waveforms is observed? A. Waveform A B. Waveform B C. Waveform C D. Waveform D E. None of the above. Alexander Ganago Page 2 0% Zé February 28 — March 7, 2005 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 3 ‘ In the series RLC circuit shown on V . . . V — C the diagram, the Input Signal from + L + the function generator, which is a low~frequency square wave, and the output waveform are shown on L C + the sketch below. VIN R VR Which of the voltages is measured as the output? A. VOUT: VL C. VOUT : VR D. VOUT can be either VC or VR (they are the same) E. VOUT can be none of the above voltages VL , VC , VR 7 Alexander Ganago Page/4 [1/ February 28 - March 7, 2005 3 a4 26 Version3.1 SECS 3’14 Q/Kam 2 W'wdcex 2005‘ La.) Wit/Lu; 0'3 VM cut +9-2- time WM Vt“ W war/{'83 C6) 4421 46243.0— Uo‘Qua VM Lt 90c) +641 15. agree/rad +3142. M of; earl, W— PZ/riekat Vin. (a) KUL ad: LL1— vaz Lula“ Vin, m M W Li: 2 O1") -0L KVLaJE-L 0° W3C Rafi-g ’T' W’Pe/r: Win-“14:3 U‘a (‘5 waging/“oats gut" (fan/(- Clad. & 2245;905ng @9 Written M“ f C L‘ 0 9 200$ by Alexander Ganago we 1:3 Zé BECquq flaw 2 WLWZOOS "'3 rqutr abacuajt out i=22+ (5rd“ “44% m 41““3‘“) .prm+0.SUtp —-0.IU Uh: +O.S\/ dew-4:0 WWW $04M KUL) 6M: 0'1“ —U‘4 50M = —0.5v—(+o.sv) U‘opvt : ~¢V m 14 Wax/e/LQVIM 3) Written W VCL 9 , 20057337 Alexander Ganago 9%fe%% y ’W WK 4 513 : LIA “i: ,Pr KVL) U 0 SU 0 WA +£9.45 2 l t = .— 4263 2 ' ‘:S)C 7M k'fl—Jt’e {TL : LR:L¢)- LLch AML'toKQL ‘HAAAS Lg: O’ M ('pVB'I—n Ohm! eat/£49011 - v=L%= o Mubarak-4T; wen-aw KVL U"c_=U“ _ ) um. tr-O‘l' W’r avgt‘e/r U1k (“MM £~rm — 0 S l/ to +0 5V W‘MOMS Written MM CL 9 iZOOS’by Alexander Ganago €466? «(fizé Written W 04% ZOOfiy Alexander Ganago fag/1 8 6% 26 EECS 314 // Winter 2005 // Exam 2 // Questions 4 According to the principle of 10 Q j15 Q superposition, the voltage Vx can be calculated as the sum of voltages, each of which is calculated in a circuit with one of the sources “killed.” Which of the voltages will you add 3445" A to calculate Vx ? 109 i159 A- VX=V1+V2 B. Vx =HV1 + V3 C. Vx = V2 + V3 D. Vx = V3 + V4 10 Q j15 Q -j5 Q + _ Alexander Ganago Page/1 of/ j I g :24 Version 3.1 “.3 (MCI/’20. 0%. {u/erfajlébh, ”Kill” M Sourre, (Lac—p +64 0&2”) +0144 Written W54 :2 ,ZOQfgy Alexander Ganago £461 10 .4 EECS 314 // Winter 2005 // Exam 2 // Questions 5 (Submitted by James Kim; modified) In the circuit shown on the diagram, what is M C the nominal value of capacitance C that cTo'rAL 2 F _.::::_. 4 F ensures the total capacitance equal 2 F ? W A.lF C. 6F D. 9F E. No solution is possible, because any capacitance C produces C TOTA L > 2 F. C(1Fqu) :6? (in parauel Jada!) C-GF = 2:- C+6F ’ CTOTAL= C'éFr-C'ZF4H12FZ 6 (Submitted by Philip Choi; modified) 2009 ? What is the cutoff frequency . A. 636.62 Hz Vin Vout B. 4000 Hz 0. 25,133 Hz SOmH D. 0.25 Hz E. 1.571 H2 - Eff. Q:Jw}‘ LTLVJl/cwe. wc=B—— R +J ‘5’ L 4 + ‘13- L’ w¢ cod 2 “05% = 4,000 £9.41; - 41¢: “J‘- = 634,42 Hz: Alexander Ganago Page/6' of K February 28 - March 7, 2005 EECS 314 // Winter 2005 // Exam 2 // Questions 7 In the circuit shown on the diagram, with the inductance L = 2 mH and resistance R = 15 Q, the resonant frequency equals 12.345 kHz = = 12,345 Hz. Thus the capacitance C equals A. 6.446 mF B. 3.281 uF C. 522.17 nF D. 166.2 nF < E. 83.105n? > m #25014wa {reticent/raj ”Res: J——-—- ,C ~ 60 _ '1 RES- 2“ - 2Tl'\’L.C 9. _ 4 __> d: “9 mm; <1 Airch‘ urges 4 C - 411:2“ 2¥iO-S*@.23‘l5“ 10“); C = 8.3105*¢o‘3 F = 83.105 M: A “-5 wu® Alexander Ganago Page / of/wf February 28 — March 7, 2005 42. 0% 2 6 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 8 V At the resonani frequency + VL _ + C _ a) RES = the current V L ' C L C + through the capacrtor VIN R VR A. Is not related to the current through the inductor B. Has the same magnitude as the-current through the inductor but a different phase C. Is exactly equal to the current through the inductor D. Has the same phase as the current through the inductor but a different magnitude B. Is related to the current through the inductor but the relationship depends on the bandwidth. Dug. ’to KQL.) +(u. (burr-emf [S M We ‘Hw'ocf 4,2,2 dire/99H- awe/“.26: Comma/+2.4 Lu series ([+- W myt- Mam/F, out remuaucz O’Y‘ Wot)- 9 At the resonant frequency + VL _ + VC _ l a) RES = rm the voltage across th 'to L.C L C + e capa01 r VIN R VR A. Has a different magnitude and a different phase compared to the voltage across the inductor; the exact relationship depends on the quality factor B. Has the same phase as the voltage across the inductor but a different magnitude C. Is exactly equal to the voltage across the inductor D. Has the same magnitude as the voltage across the inductor but a different phase B. Is not related to the voltage across the inductor. Sag WM “flaw aura-3,214 Alexander Ganago Page/ of )1 February 28 — March 7, 2005 130% Zé Version 3.1 Z2ij+jLJC 4 ajt Y‘eSonquce Z {S PUM’Q’I‘Q red MB (A) L. : .1...— LJC > Time. (Lawrence ‘Har‘oa’xa/é‘. acne/94. are/‘14}- W {S '&1 Ma :E: VHU : {Vial '836 Z R a; Y‘ZSOVLamce ) A Vc:I‘J'¢A)CL: Wml . R 4 Quit wame.’ (AL:— JC9+ Pa”? 4“ 6% 26 Written M ;, 14905— by Alexander Ganago EECS 314 // Winter 2005 // Exam 2 // Questions 10 At the resonan’i frequency + VL _ + VC _ a) RES = the complex '\/ L ' C L C + power absorbed by the capacitor VIN R VR A. Is equal to the complex conjugate to the complex power absorbed by the inductor B. Has the same real part as the complex power absorbed by the inductor but a different imaginary part, which depends on the bandwidth C. Is exactly equal to the complex power absorbed by the inductor D. Has the same phase as the complex power absorbed by the inductor but a different magnitude E. Has a different real part but the same imaginary part as the complex power absorbed by the inductor. S12. (493,2 44 (3w WA mam 0/2. +84 aurraui‘ M (1%. A f E“ E S UL} N‘L < H > % :1 lel .1... 5(9‘E) A z -J'E Sci: '1' lVNl . L Q. a 2. R?— wC’. A ' W V l J(9+_ s : 1’ IN 'WL'e 2. L. 2 K g : .1— IVIMlZ.J___-.e‘)l: ‘— 2 R” wc’ " Alexander Ganago Page/z of )4 February 28 — March 7, 2005 45-4 2 6 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions Section Two: 3—point questions 11 (Submitted by James Kim; modified) In the circuit shown on the diagram, what is the energy stored in 10-mH inductor? Assume DC steady state. A. 2 u] B. 5 m] C. 20 m] D.45mJ E. 80 m] At :DC M 917212. —-> :SHORT Thus no wrr'e-ki’ #804415th lZJLa—«othz resis'tws. The circaqi- Ezochs: Alexander Ganago Page ,(ofys’ February 28 — March 7, 2005 4-6 6L 2 6 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 12 (Submitted by James Kim; 5 nF modified) In the circuit shown on the diagram, what is the energy stored in lO-nF 20 "F capacitor? Assume DC steady state. A. 1620 n] B. 720 nJ C. 259.2nJ 10 nF 3 £2 D. 180nJ E. 64.8 n] 48“9 Tvd VC jg: 3"“ _ I: g1+3JL+3JL:2A 3J7. EWfifl ‘2 —4— Q 'ch : é» iOnF-x-CGV)2 E W31 = 1‘30 MJ A’MSWO/r CD Alexander Ganago Page 46% February 28 — March 7, 2005 i70£ 2 6 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 1 3 (Contributed by Philip Choi) In the circuit shown on the diagram, at what time will the resistor voltage “1 _vC(t=O) sv VRU)‘ 2 A. t = 060211115 B. t =1.833ms C. t = 0.7959ms D. t = 3.219ms Pram KVL.) ——U‘C_+U‘K: nutt70 4444-45 U-R=U‘d TMS we. Eco/(L ~93? (fact): JZ—U‘dct=o) It is NOT mes-50.13 'ta anuIMe ”DC 96% .s'l-aJt‘e @1444.wa 4,6 t20‘“ umse m. 010 Mt rm 02.60% Vs =5u_ -"-/c: 010:): 0‘th=0)-€ R = 17:.U-cée=o) Alexander Ganago Page W February 28 — March 9, 2005 18 9;; 2.1;; Version 1.1 EECS 314 // Winter 2005 // Exam 2 // Questions 14 The time constant ’17 of the circuit shown on this diagram equals 50 Q 20 Q . A. 4.726 ms B. 847.3 us Vs 9 8.2 mH C. 164 .0 us D. 82.00 us ' "Ti/U. Giraud-i- Caa... 82. 3'2.ch E cits WM“ 2%)de WW +1“ WW eq/W (Ca: 3.2MH RT quk va resistance “U To 42(va Rm ”(L{u# ‘HA-L Scour-Ce RT = 2.0.11. + Cause)». = + SO'SO-Q-z3g. 5 RT ZOJL SD+3O 7 J1. ’C = 9'2 MH 2 O. 2446 MSec = 244.6/MS 39.7572. 1414st @ Alexander Ganago Page February 28 — March 7, 2005 19 ¢ Zé Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 15 (Submitted by Rob Bai, modified) The resonant frequency CURES of the circuit shown C L on the diagram equals R é Total. +R) +J u.) L 4' R J “J C ad: resonance 3% {5 real i -— «- JUC. " ~ ij+R (1 w 2-—LC_) «jwac’ (1-w1LC)+ijC (1 w LC)— —jo~>RC ij*(1-c~> LC) —J'eokc. R +R€ALNUMBERS W5 wL (1—0.; LC): (aw-c! Ele‘w: 0 0r L(1-w7-LC_)= Alexander Ganago Page W‘- February 28- March 7,2005 g2_o 2Q Version3.l GIOIUHM em. ,0 O4 1 EECSBAQ Mama 2, [email protected]: L.‘(i—¢OZLC>= RZC L _. wZLzC' = 22C floPUU; W we @045 944 Mae 0&0 we)? PARA/2 tom EECS 314 // Winter 2005 // Exam 2 // Questions 16 R1 The circuit shown on the diagram is a l A. Low-Pass filter Input C C. Band-Pass filter D. Band—Reject filter E. None of the above \NPUT LH AL . . 3‘6” l2; om 1- ms to A H? I o QM [In MM , .1" l— MW % Alexander Gango Page W February 28 —— March 7, 2005 22. 0£ 26 Version 3.1 EECS 314 // Winter 2005 // Exam 2 // Questions 17 In the circuit shown on the diagram, the output voltage Vow equals (in volts)... A. 180\/2 +j18W2 B. 180w/2“j180\/2 12/.45" * D. —18W2 —j180\/2 E. None of the above. —-A 11: (GE +JéU—2)A :fCQE—J'éfz‘fi 11'I2:JI‘ZEA ‘ VOJ=C1S+54S)1CMZE>V th {3490M - 430E) V Vau/C='(-490\FZ +5490£)V fl[email protected] Alexander Ganago W February 28 —— March 9, 2005 2 g 9% 2.6 ‘ Version 1.1 EECS 314 // Winter 2005 // Exam 2 // Questions 18 In the circuit shown on the diagram, the component values are: L = 1 mH C = 1 nF The input signal is a low—frequency square wave (see the sketch). The output signal is measured across the capacitor; that is VOUT : VC Voltage The output waveform shown on the sketch can be observed if . .. A. R is between 2 kg and 3 k9 B. R is 3 kg or more , C. Risleorless ’ D. R can be any of the above E. Additional information must be provided. w : O< :- “"""‘ ° \ICE: ’ Z-LJ (,Jo : 4 2 ioéfii‘é. W Sec —3 L<io R<2L.106= 2*10 «log 2L R < 2 kfl. IOU/{SWM @ Alexander Ganago Page Lé-ef-H—a February 28— March 9,2005 24 9% 2Q: Version1.1 EECS 314 // Winter 2005 // Exam 2 // Questions 19 1244st -j5 £2 The output voltage VOUT in this circuit equals (in volts). .. B _ 48 78- -J11 69 C, 112.9 +198-24 D. 159.7 +j39.93 E. None of the above. 4 4 42.0 ' J’” V0¢,[ . + . 0] 40—J40+5E+‘)é Z Vewt = Z72.é+Jé8.17 units A“; WM Alexander Ganago Page m February 28- March 9,2005 Yéi/J > 6 Versionl. EECS 314 // Winter 2005 // Exam 2 // Questions 20 The average power absorbed by both resistors in this circuit equals A. 300 W 40w+2ow = 60w Answer-CG) Alexander Ganago Page W February 28 — March 7, 2005 Zé 4 2 6 Version 3.1 ...
View Full Document

{[ snackBarMessage ]}