examI_f08

examI_f08 - EE 350 EXAM I 25 September 2008 Last Name...

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Unformatted text preview: EE 350 EXAM I 25 September 2008 Last Name (Print): 3 olv E! OHS First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO EL 4 25 _ Total 100 __l_ Test Form B INSTRUCTIONS 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You may use one 8.5” X 11” note sheet. 3. Calculators are not allowed. 4. 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. 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. (7 points) Determine if the signal f(t) = 2e"ll [u(t + 1) — u(t — 1)] is an energy signal, power signal, or neither. If the signal is an energy or power signal, calculate E f or Pf, respectively. "F09 "\ +1 BQ-Cowse Em l-pCt)l’—90J 24 um” be— pin/£2. lit-)0 00 °° t” —t _" _. 2?: {lzceiesl Zak =q 6 0% --l: ’0 Onotzfieigflae When-£20 1C (1:) 2. (11 points) A system with input f(t) has the zero—state response W) = f(2t— 1). o (4 points) Is the system zero-state linear or nonlinear? To receive partial credit you must justify your ' fim 613—» (W = neat-o has) a? #266 = fiatfl) {3306) = l '- -F (Z‘E’I) ,9] (—9 t) - OC-fittb-r {afim #3 3 No‘ka— #3619: 19" o—az ‘ché‘ranJ ‘finle ‘FL It) Q +oll0uq§ 1955 the. 552me IS CQrO'S‘éu—ée Menu“. 0 (4 points) Is the system time—invariant 0r time—varying? To receive partial credit you must justify your -F,(+.) E—a g. It) got) “Goa—1°) agape/e) No'ée 'tl'uu'é (9,112) = 44313) = "F. (Zt'l-T) -7‘ 941(1) =£(1—1) any» $.69:- ‘flfZ-E -D _—_9 $5155-73 g. 41' (1“: ,T) ,1) _._. G (24: .4 -27) 82mm. 714;) .- -F|(Lt'l-T) 755M477) = «She—psz the £34217) ‘15 ’lee.~vou*imiP o (3 points) Is the system causal or noncausal? To receive partial credit you must justify your answer. Suepon we Calculaiaq aa at) oak; amt 4; =10 sec“ the” (m: 430°!) :évr‘ & J Ira N) a“ thug}; bl “(U/{aim % of; {um 15:10 sec- 3“ -f—( ) i 3. (7 points) Figure 1 shows the input f (t) and corresponding zero—state response y(t) of a linear time-invariant system. f0) 31(1) -2 0 2 0 2 Figure 1: The input f (t) and resulting zero—state response y(t) of a LTI system. A new input f1(t) is applied to same system, and figure 2 shows the resulting zero—state response y1(t). MU) Figure 2: The input f1 (t) to the LTI system results in the zero-state response yl Find an expression for f1(t) in terms of f(t), and then carefully sketch f1(t) in figure 3. 8&1 “pea-L,” - 2. 0:3- 2 (72-) Because, {the SdSiLEo-n 75 LTI, 05 J TS cussva [fl let/EM J £uy‘ a" L_ T]: 5d s'bem 4m —{:}~9 alts &035 not guaran‘léfle, ‘Szct/ ;) (*5/92) ~ Anti (:9) -—-;,:: ~ a ‘n (Gucci mi My: Jae/z) 0 mg? k v m ' 6‘ Figure 3: Sketch of the input f1 Problem 2: (25 points) 1. (10 points) When solving for the zero—state response of a particular LTI system, it is determined that the homogeneous solution has the form yh(t) : 6—“ {A cos(4t) + B sin(4t)} + Cest for t 2 0, where the coefficients A, B, and C are nonzero real-valued constants. o (5 points) Using figure 4, indicate the location of the roots of the characteristic equation in the A-plane. .. —\T"l: - 744:; ‘IJZrm egtiAcosma-ibamm-YS: e {Awsw£)+ Kwanzaa] @02on Complem cirydfiwlzo. routs: 7sh-L: '5 '3':ng 4: ' ten" Caz; Wueavk a, rec/t “E ()3 = +3 Figure 4: Location of the characteristic roots in the A—plane. o (5 points) State the characteristic equation in the standard form An+an_1)\”_l +~~~+a1A+ao :0 by specifying the coefficients ai. Qm= (a-mm-m cw w = (7\+6‘7~/)(7~+€+/~1)(7“3) = (21+ lo7\ -r HOW-3) 7‘3 + (Io-37 79' + (—30 + V137 ' '23 QC» = 7‘3 + 77?- +[l7l "[23 :0 2. (15 points) A LTI system with input f(t) and output y(t) has the ODE representation :9 + ay+ (2a —4>y = (2a — 4m where a is a real—valued constant. c (3 points) Specify the roots of the characteristic equation in terms of the parameter a. = ’32. \lmz __ —or-ivacz_¢+.5 “1,2 3 + _(2¢-‘I) _ 7’ 8 - - - —cc-°C-41 : we: (cc-~07— _ 2—1311 z Jeri—i; —————- 2 " z. 2. 7.. 2|='2J 322—m‘l‘1- o (2 points) For What values of a is the system unstable? 0 (2 points) For what values of a is the system asymptotically stable? fie sastem is dual mp'ba't1cg9 stable. when both 7“ CWQ 77— are’ Yuaw'tlw. For a; = -d‘ + 2' < " o (2 points) For what value of a is the system marginally stable? \ :. 'tMS we siéte'n 5 may] 0Q”? «Ul‘m n A 17-" 2' qnfi 7"- o) (dfl‘lL Lon hows when 10L=+Zl o (2 points) For what value of a is the zero—state unit-step response critically damped? 7%9. rasponse Ts crdmcdul? gain/kale taken 7“ uni 37, are. real) flejvb'elV‘Q—J am9— [flent( This (Mafia-4.1m kdfls “Fur” , when fl‘ =. '/\7__ = “Z. 0 (2 points) For what values of a is the zero-state unit-step response overdamped? anQ Juana") this (means W‘ a— oC. > Z) excep'é at. = 4“! _ (0C = 4“! is exolchnD— because. flier mat-f “VG-— 19522.ch— ) o (2 points) For what values of a, if any, is the zero—state unit-step response underdamped? WM 9» “luck 1. ME fume, Comflleit Cavfiafig With an. M “flue, reui pa/‘é. Problem 3: (25 points) 1. (12 points) The circuit in Figure 5 is driven by two independent voltage sources7 each having a constant strength Vs. Prior to the switch closing at time t : 0, the current and voltages within the circuit have reached steady—state values. The current y(t) through the inductance L represents the circuit output. Figure 5: The switch in the passive RL circuit is closed at time t 2 0. o (3 points) Find an expression for the current y(0+) in terms of the circuit parameters. Justify your answer. t=—o— mgr/OE.» apnea-,5 «.5 a. Shur‘fi 'a(0*)=a(o') bum Curran-l: Cuom'l: change ansbunbuneb-dla. in L. o (3 points) What is the steady-state output current, g/(oo)7 in terms of the circuit parameters? Justify your answer. In .s-Leawo "shit; ;_ bec°me$ ax. Confewn'é- C(n‘Q. So the Iflca/c‘éufi “'FPQMJ 4L3- ov shark cwcvs‘t‘: Cl L7... 3-0”) = c. + CZ %_ F—CL— c- ,S _ I/ \ Lot V5 "42: ’3' 5? ,Obms V Vs igfifi :1 0i 0 (6 points) Find an ODE that describes the behavior of for t 2 0 and express your answer in the standard form dy E + (10 31W ~ be by specifying the coefficients a0 and be in terms of R, L , and Vs. 2. (13 points) A LTI system with input f(t) and output y(t) is represented by the ODE The system is driven by the input and the initial value of the output is y(0) = 3. o (6 points) Determine the zero-input response yzi(t) for t Z O. _ at 1+ 2.=—o =‘> ’7‘.='2- ‘1'“9‘ yhc'gr'c’e 7» grape tzo 1172$, Char. 96 n. : pavhmuw “khan: arc-1;): @. 4) éé+ zzr/g =0 32:33: ahit3+gffifl = Ca'Zt + o t 20. To swing-I 53C (0) :3.) [games 32:” t 20‘ o (6 points) Determine the zero—state response yzs(t) for t 2 0. ._.. "Z‘b 4.5 a—LDO‘Q'J #5053” C— e 133°- graz): cc. +GL't + b’tz J £20 7.. 2. a t a 2m F.,_——Jz’ a? +Zyp 2 2t.“ *9 Z745? + +22 72+ 2+(t) if. yr: rd a :=> iz“*’”‘° ’M‘l’ -[+3_¢, =0 =>d—=’7' V=l \b = Y(2.t‘) + 2.(7+/5- t + CONN-*3 at :0 =0 arm: J5 --I=+t7- tZo #aS(-e)= ghfls)+y/(£>: E e _c.-'-“‘ 798: gisav): O 2. .. 2. #3566): “is 21— + — +1: 153" Problem 4: (25 points) 1. (8 points) The RLC network in figure 6 is driven by a constant current source with strength [0. The switch has been opened long enough so that all the voltage and currents within the circuit reach steady-state values before the switch is closed at time zero. Figure 6: The switch in the passive RLC circuit is closed at time t = O. o (2 points) Determine the voltage, y(0+), across the capacitor immediately after the switch is closed. Clrcglt out: i=0“: (Micah ail/tears K3 Shari: QK‘EJ C‘s/)cLU-L" as 0/2,, 0.5140. 05ml) Lou-u ‘lk E —9 one 02 V09) RH'QI %~RL o (2 points) Determine the derivative of the voltage, y(0+), across the capacitor immediately after the switch is closed. 36- (Wr'n47 J‘VI‘5I9n) van; the ohun when) (Lfo’ = ——K—‘ 19 ‘—‘- Ezra ‘ ‘L(o+)= (Lro‘) = Jf‘lo J as fur/211$ cannot de’ye’ 'nsbun'ecmm‘d' r" L UM? KCL ‘17 Name, 14 in 1:ng 6 afi firm tzo+ vials“ —Io + (is? 4- {009*} + (Lid’) =0 -Ia + 351:: + (C(O'r) + Sir—Ia $0 => Cb(o+)=o 0 (4 points) Find an ODE that describes the response of the capacitor voltage y(t) for t Z O, and express your answer in the standard form y+aiy+aoy=bo by specifying the coefficients a0, a1, and 170 in terms of the circuit parameters R, L, C, and IO. 10 A (LCE I0 6 L VCL 01E hog-Q, :4 d‘dfls (Mb) 4 E _ Io + 2) + c.3112, + LLfo+)+ {Saccxgt =67 D‘¥X’o’€nfiw‘[fie Ext“ Suges % b“. gal-Vi: a «drawn mg rQ§pafE '57 {Sher 11 2. (9 points) A circuit with input voltage f (t) and output voltage y(t) is represented by the ODE dgy dy — 6— 9 t=1 t. dt + dt+ m 8m The circuit is driven by the input voltage f(t) = (1 + 2e‘t) u(t), and the initial state of the system is specified by y(0) = 1 and = —1. Determine the response y(t) for t Z 0. ' E m: final-67+? = (7+332'=o kHz-“=01; + Czte‘ £20- 5 % I ' Becqum O an -| an rah ray-Es 04: the. charm-Lerufio eava'élofl, awe): «+1959 1:20, 54. sable. was) [2.5 cm ODE £0 99 09,4: — — - 4.- et *ééet + 7¢+7@2t, =- l3+ 362, a"? Jr W a. :2 12 3. (8 points) Figure 7 shows an m-file for generating the zero-state response, y(t), of a circuit driven by some input. A = 1000; B = conv( [1,16,100], [1,5] ); D = [0: 0.0001 : 31’; C = 2*ones(size(D)); y = lsim(A, B, C, D); Figure 7: MATLAB m—file for calculating the zero-state response y(t). 0 (3 points) What numeric values would the MATLAB command roots(B) return? L112, 2. acne-dies ihe pobnomwQ, Q(;\) (a, GM ace Q(7~\: ('7‘z+l€>’7‘+loo3C'/W+S) :- BC?) ‘- 7‘3 4- 2| 7cz‘ 44802 4- :00 \r/ the 005 , fin. (ammungv rth (8) return: the, oWw‘éfiflS'bt rifts! 7[ + w are. m “A: W rans pl the, Schn-Q—aruoqf fierm) a + 67 l J r——‘ .. r. - i: 6 21,1:- ‘8': 67‘109 -'-' '8: 56 8 fl‘ 1 «Jule. the. '90-‘35» (3.9;) ha) a. roy‘t ’33 = —$‘. find, 50 WI” roéwA root (8) o (2 points) Over what range of time values is the zeros-state response calculated? T/me, 'U'Wr‘q, ,OMQ. genera. J the 15th 9654* VSQ-Q/g~ QS'M' I b .1: E0: o.ool -‘ 3’] Rafi. So the. 434,02. Pct/1&9, % 116$,mgloc'étvfl :3 oft £3 fl. «Q.» to. 14/50 nafe that tho, (.011: ,01371. fun, the (7 (49/7/‘Om‘ 0'55) [9(b)—..hoc>J thQ. the. 'I'Wrth ,Qmo, specnqe, fig IflflJé ~51 Hil ; LvLI-t), o (3 points) Assuming that the response y(t) reaches a steady-state value over the simulation time interval, what is the expected steady-state value of the zero-state response y(t)? 7%., ODE- represen-Luém % 1k. Lop/c.4294 (as-’3'me Q‘Dlgzpmypj or C O. a 4» ll? +|soa:/ ffoafl : 200.9 In J'Lea-gi-5£a—£ea 524:0) ghee-505006155 :25“) “W55 13 ...
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examI_f08 - EE 350 EXAM I 25 September 2008 Last Name...

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