FinalEaxm_f03_new

# FinalEaxm_f03_new - ESE 271 Final Exam Name: Fall, 2003 ID...

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Unformatted text preview: ESE 271 Final Exam Name: Fall, 2003 ID Number: Do not. place your answers on this front page. Prob. 1: (30 points) Prob. ‘2: (20 points) Prob. 3: (30 points) Prob. 4: (‘20 points) Prob. 1. (30 points): This Circuit is in the DC steady state at t : 0%. The switch is open fort < 0 and as closed for t > 0. (a) (10 points) Find z'(0+) and v(0+). (b) (‘20 points) Find the Laplace transform of the transient voltage v(t) when It > 0. Write your answer as a. polynomial over a polynomial. in 2/! (LT-l {47‘ ‘5 ’5", we MH-JE o_._ __ .'.“-'_ _ ‘ 4-—'l'- _l_ L..- 3 IKE} . k "T II r .a in» ; H1 “w. #911 r 1* "if-J r '; _- I}. .- rFr'lQ-rl ‘ '1" LE3 F0“ ‘:3 n ,V ‘J . H 1’” 1.: _ __,_‘____'_‘5-|I"—_ ' _J—TII- II I | _!—I_ ' ._ J1 we}; I _ ,I,_ ______-- REL “J I) w J' t | 3—” 4— : 5' PI -'r|'.'}-v'JI- o—l + E;- H 2 I. K I: -'I. I—'-. ' 1 Jr” |;_q\+. _n 2'? X If .- __I/__:I {ﬂiLt-r a.) 1L 'L/J. L .. n 2 V(I+: +4) * -‘-+ N 3 “1+2- V: —.——a—---————'_ Prob. 2: (‘20 points) Find the initial value f(0+), the initial slope f(1)(0+), and the ﬁnal value ﬂee) of the function f(t) whose Laplace transform is 252 — 33 — 4 F(.s) = 2 3(3 + 2) fat/53+} = 25, F r' ,1 -: 2 .a—rm .r’i p '1: {fan} : LM .5 [,5 r M“: ~ (n+1; _______ _ _ 41.40" i. J. {A+1_} ' ﬂ:- : Ilia-54"»? «i- FA .3: i n _' _ L ~ . . - Q“; 4 u' ﬂag} at: ﬂ _: :f‘Ji-is-i “‘~~- -- 114m: 41' +Q41+¥A AI a ,r} A? —)’1 21 Z ILA— --— —-—- : _ '1‘“ Axes» éi+¥nl+¢rﬁ¢ fling] “ELM; [‘5 :- ——/ Prob. 3. (30 points) This circuit is in a. transient state. The transformer’s parameters are L1 = 2 H, L2 = 1 H, and M = 1 H. The initial conditions are z';(0+) = 0 and {2(0+) : 2. Using Cramer’s rule, ﬁnd an expression for the Laplace transform 11(3) of the transient i103). Your answer should he a. determinant over a determinant. The entries of those determinants may be functiOns of s. :- -.'r':l."' L1" r _ _ f gr [I _5 \ __1_ “ll _;_r___ _._-. p a (I 3' a; -' 4 r - r, 'r_ «:1‘ -' ' z I“? £fﬁL}J.}f r- v 1" .— 1r 3 .- _. I , n ‘ _,. 1-7.5"; /—. _-- + 3 .r‘L I. + J!» —‘ n? — —'_ (I l?) - .-"-{l f' ' J J. -|- ’ E :: - I —— --ﬁ- -|'.-’ - " i ‘ _-b--t| i": ” 4." ' ' — 5—_ “165' Ion-'3 4.1—; -;{,-:'-2_ :5 3 .. _".__:; 41/ _ J'l? I . I .' I :3 .1. ' i? .; a i | :I‘Nb HE.” L I, .' 1 I'll- naﬁfpl'lzrxd'" I 'H .57 _-_ \II, _‘_ _'_ (-4 1”“ r.’) Jl 1L gird ‘INI II.“ : _'.u 44;! A}? +r’y..._~;r 2 4: +3 .r_ I -. |-| 1 :1..- ‘g | I, ____. -— r». .- -' A 4H ' Eden. “yaw”; I: .- 73,-FI._._- .. : 7 h 3* 4c , l _._ | ' - i. I I Prob. 4. (20 points) This circuit. is in the AC steady state. For what value of the turns ratio n = Ng/Nl and for what. value of the inductance L will the average power absorbed in the 100 Q resistor be a, maximum? N iuV‘nJ‘ I M turns 2 4 "J "‘ -_—_._____Nﬁ_,_____—.r‘l|r_\___——— #- Jr———— —' ' l . - ‘ . .ﬂ '. ’ (i) ' 5 .' J __— - ‘l I“ .I Ii ____ I i j] J! .L |____--__ _|'_-J L____"—l '. :' —; a r If?! 21 H -- ' 5-27 . . E’. _ , - . we "H 'i " 3'3 33""? '4’" , :- if +f ‘I‘r , ‘1» HM: -. j: : z‘ujrj 7'1; 2 3- 1.1 _ — I /-:-" .1 - '_ do ‘9' "'4' H” : “7—— 4, J. P11 ,‘wl" fo/ 2 KP” — J— y’] 1' ___—— n; _:' HI - 5!" —'_:_ .I'f' r a —" '-' '1 (xiii? fol : ___.___, 39’ hnf/Qj‘ z. _ . _ K AWN- s _‘.-r_-"'fI ‘JK ﬁlial-.2! _ Hi/‘F_J m.— T_ H .Irg'l ,.-_ z] :_ "Li; .m-‘IJ 2 (_ 5"“: Hf {Ix ' x39; h _— J {it 3'; 5’qu _ 2:5} 2’. :_F._I;_—-o# ...
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## This note was uploaded on 06/05/2011 for the course ESE 271 taught by Professor Zemanian during the Summer '08 term at SUNY Stony Brook.

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FinalEaxm_f03_new - ESE 271 Final Exam Name: Fall, 2003 ID...

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