Fall2010-F - ESE 271 Final Exam Name: Fall, 2010 ID Number:...

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Unformatted text preview: ESE 271 Final Exam Name: Fall, 2010 ID Number: Each problem is worth 25 points. Do not write your answers on this cover page. Prob. 1: Prob. 2: Prob. 3: Prob. 4: Prob. 1; This circuit is in the AC steady state. Do a modal analysis to find the node voltage v1(t) as a. cosinusoid. 2 (I (t) waif: fleck Pharar ?an'tlig'e-$ are ant: P f Augean around ZI, dependent Vaitu,e source. at a. 3-V . ’ V‘V =2]; 37' _.'=' 6J—2Jll/ IWJI('QIE ballogn; ' a J N. baffle». kc; m, 2 V4,? V‘v V2 4- .Lgp- + ’1 i z lg”? -3v, + ,4 J J 'J Thin) S‘t'mrjg'fte: fa - 3 (3 3": + V2 J Ad4 Oquut’m‘ti‘ Q) and @1 V(;+J'3) =J‘7‘ ' 0 50, JQ 7Z?a° ’ Q'y‘f‘é /’¢ Prob. 2: Determine Z L so that the average power dissipated in Z L is a maximum. Then, assuming that Z L is a resistor in series with either an inductor or a capacitor, determine the value of those elements. Finally, determine the maximum average power dissipated in Z L. ._L_a UL 5- rfiu’j l A I 2L” J— H J’ FINAL”) 2 if 9 [Voci 4-57 I M 2 .0525— W F ~" / ' fl /6 Prob. 3: Using the Laplace transform and using the mesh currents shown, find i1(t) for 0 < t < 00 All initial conditions are 0 at t = 0+. (Kt) "alts TRANSFonnen (mun-r,- THEREFOR}, A» I I2: @--—-*> A»! I<VL Areowvo OVH'D‘ ‘D’P’ 1: _ v, '9' I, + ’51.? + T - 0 I, + (“+7013 =/ '— 2 \ 3)“ ®/ 1., + ffljjifimz, :f A A”! :3, I ‘ AvI — A +- “Q. ’ ‘ 4(A-H) ‘ A ’H A:- 3:“ 2 / 76m; Prob. 4: Determine the convolution f(t) of the two functions g(t) = tu(t) and h(t) : e"‘(cost)u(t): N) = gmxha) 2? MEN.— (( ) l H ) 4"." 4+! " = -— {A : M = .l _____.. 7 42 “HOW-I (A + I—J)(A+I+J) A+I 1—7») : W A," {A +1 —j)(A +1 +j) *- B A A B M FrA): A: 4‘ Ai +A+l—J + A+I+J A+I ' A2 - (A+/)‘+/ " 2 4:0 A °’ W“ - = a 2 I‘ JA- [A+/)2+I AID — ((A+')L+/) [OI—:0 B * A?" I :: A‘IAH-r-J) ’33.”; 44 7;- ) M? 8:70 t- 50’ 2‘ 218,53-“ ca—t(pf+m¢3) (H) ~ ‘2" + 01:! ,p= r ' “t (aw-70") {(2‘) = ‘3" +39 CW Prob. 4: Determine the convolution f(t) of the two functions g(t) = tu(t) and h(t) : €“(Cos t)u(t): f0) = g(t)xh(t) :7 , EN“- ‘ 4+! (7‘(4): .1... : 4+ = I. . 5‘0) 42+] .. W FM) ~ 4(1)“ ’j)(/-‘f’+J) as B Fax): A: —+ ’6' +——~o———-A+"J + A+I+J A+I I A2 - (A+/)‘+/ “ 2 4:0 ‘4 A“ (AMY-H -(,A+/)2/A+,) : 0 A) ‘ (IA- [A+I)1+I AID ((A+I),_+I)z flip b = A?" I :: A2{A+’ +J.) A124,” 4 ,8): '47,.— ) an? ’3: 70 So _o(fw(pt+wa) )fn) - ~11 + 2/818 2 «=1 , [3 = NH = 3- rée’tmhmw) 2 ...
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This note was uploaded on 09/26/2011 for the course ESE 271 taught by Professor Zemanian during the Spring '08 term at SUNY Stony Brook.

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Fall2010-F - ESE 271 Final Exam Name: Fall, 2010 ID Number:...

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