2009+Fall+Final+Solutions (1)

2009+Fall+Final+Solutions (1) - EECS 215 Fall Term 2009...

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Unformatted text preview: EECS 215 Fall Term 2009 Final Exam Name: 50 (M31: 04,5 Rules: Friday, December 18, 2009 Closed Book, etc. 2 Pages (8.5”xl 1”, double sided) of notes are allowed Calculators are needed and allowed. No devices with full alphanumeric keyboards are permitted. Exam is given under the College of Engineering Honor Code principles and practices. . No communications of any kind are allowed. Use of cell phones, cameras, personal data assistants, computers, or any other electronic devices, besides approved calculators, will be treated as an Honor Code violation. 7. Work is to be done in the Exam booklet. Turn in all pages of the exam. Do not unstaple the pages. 8. DO NOT WRITE ON THE BACKS OF PAGES. Work on backs of pages will NOT be graded. 9. Show your work and briefly explain major steps to maximize partial credit. (For example: i3=il+i2, node A, KCL.) NO CREDIT WILL BE GIVEN IF NO WORK IS SHOWN. 10. WRITE YOUR FINAL ANSWERS IN THE AREAS PROVIDED 9539’.— as» Sign the College of Engineering Honor Code Below. (N0 credit will be given for the exam without a signed pledge): I have neither given nor received aid on this examination. Do not write below this line *************************************************************************** Problem 1 [ ] Problem 2 [ ] Problem 3 [ ] Problem 4 [ 1 Problem 5 [ ] Total Problem 1: Short Answers (5 points each) (a) The two circuits below have one key difference. Distinguish this difference and indicate the circuit functionality for each circuit. Circuit 1 Po $;’rcve Circuit 2 (b) For the circuit below, the capacitor is charged at t = 0, while the inductor is fully discharged at t: 0. How will the circuit behave for t > 0? What can the circuit be used for? 030([41‘66/ sflmsma/g/ 1:" V ost/ar‘for —9 Fray/army Riga/rang (c) Electricity is typically transmitted through power lines at high-voltage (kV) rather than the 120 V rms value that we use. Why is this done (circle all that apply)? 0 fl 3 rovides a better power factor k '7] her V —> ,9 SS I.) {a u/él’ ’2 I“: o A downed hi h voltage wire would be safer t Quc (d) What is a zener diode used for? Be; +: Pier" (Cd/Mark fit 60C) 1/4”an VQ firm/Ice {Vet/arse lbw/S) (e) The equivalent circuit for a MOSFET resembles (selectene): __ ccvs cccs G H D 4.. vcvs - 5’” VV‘ Problem 2: Second Order Circuit (20 points) For the circuit below, determine vc(t) for t > 0. R=4Q L: 12.5 mH C=O.5 mF - K=0.2 S vs(t) = [5 + 10u(t)] V 7+k7/Z :0 (5)441: {52 Emuflé+k>~64§€t 7%“ Mekflifij +/¢>L J£*L( 421?; (Ll—2) W 47—978 +654) $- 3’; wry/5'14 +>0 20L “/62 6/3320 A,= W5): 317% 443.3 ”(1563) A}: - WNW z'M/a ~2$3,8- (—53,» Problem 3: Maximum Power (20 points) For the circuit shown, choose the load impedance ZL so that the power dissipated in it is a maximum. How much power will that be? 1.4 m 20001,, <9. Problem 4: ac Power (20 points) 452 29 n Vms=IOOAZV I 139 Load Z : 3 kW @ pf = 0.7 leading Use available information to determine: E23: PF; 0'7” (‘75 (QQ/ 92,, aos”(a7>?4“7 kit/Vila / (df’tgi‘fii’e 0 . = ngaél-eia/Iéééz = 392, I7+g4zzi 5'25: 4aa.444é;,-’0i1,. \/ v.2 3 VW fig [41,491.93 V2 == \/5 (Z//(2+;3>> 4+(Z//{PL+}3)) 3§§éé 2M (32%) s a. (44,: 4% 0' Problem 5: Freguencx Response (20 points) For the op-amp circuit shown below: (a) Obtain an expression for H(o)) = V0 /V5 in standard form. (b) Generate spectral plots for the magnitude of H(o)), given that R. = R2 = 100 Q, C. = 10 11F, and C2 = 0.4 ”F. (c) What type of filter is it? What is its maximum gain? 93*}357; l +$ZJR1 Vizd‘l’5Z7RR) VG jab, m 1%: m. aw, (a) (b) (C) 10 2 PM) fés‘om‘nce / “Madam ll 12 §u38 wu =—-__.__—_—__--————__-————___-_ B I-_______lg§____ in . 1 51‘— W __ Z l ‘ l §.R& -________-—______ Anu______-________-_ Le r i- 1' Factor Constant K Zero (E, Origin Um)“ Pole (d; Origin (jun-N Simple Zero 0 + jmlwc)” Simple Pole 1 .v (I 4- ju/(uc) Quadratic Zero Bode Magnitude slope = ZON (IE/decade 0 dB 1 a) 0) slope = ~20Nddeecade : slope = 201V dBa‘decade 0 dB (0 me we as : slope = —20NdB/decade slope = 401V dB/decade [1 +12§mlmc + (jmfwc)2}N 0 dB Quadratic Pole slope = ~40N dBa’decade [l +125ww¢+<jomcflv \ Bode Phase :180' ifK< C 0’ ifK > 0 upon-ac... l3 ...
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