Act Missmatched Line

3.8 recital Soluti ons {rom portion .3 3.8.4 proceeding: !Vlismatched Une Given : I = 250 m L = 02· 10- 6 .!:!. 50 . m U(t,O) = Ul(t ) = Ül sin(wt) f = 300· 103 Hz [~ l B = 0 let out a R + jw L G + jwC Q.O = loss-less gunstock means that R and G = 0 =[& 0.2·1O- 6 H l~ . 1O- 9 F -V _ 10 .2 . 18· 10- 6 . kg . m 2 . kg . m 2 10- 9 . A 2 . S2 . A 2 . s4 0.2 . 18 . 10- 6 . kg 2 . m 4 10- 9 . A4 . s6 = 60 kg· m 2 A2 = 60n . s3 I /1 ~ r l~(~~\ 50 1 H 51 1F = = 1 Wb = 1 kg:i m~ A ~ 1Q V = 2 1 A 0 4 S 52 1 n = 1;::3 kC> ln 2 5 132 3.8 execution Solu tions from Part :1 I D_~~ QI L U h 11.~ = 11.~5.( cos [ß(l - z) ] + j .2:0 . i~ . sin [ß (1 - z)] (3 .8. 1) Question: 11.(2) = ? (O utpu t voltage) act 3.8.5 : Exercise: l\IIismatched Line, Part b 133 from Part:3 + j. wont z 1 R = 0 since J term factor } =J 1 2 2 1 j 134 3.8 Exercise Solutions from Part ;) Part c 1/\ U2{i)/ = ? !. i ! .... , . . U 2~ ,I=. U(t,z =IJ li.2({) i= li.2 . time phase factoT yi elds t ime dep en dence =Im{ real scaling factoT I ·e -j~ . e jwt } V~-t,· ,...- .J sin !.l ) Ui(t )/= - 2 . Ul .

cos(wt) I 135 3.8 Exercise -cas (xl x A -0.5 (1 3.8.6: Exercise: Misrnatched Line, Part c - 1 c- 2 output / z t 0 3.8.8 kink over the 136 -20, I t 0 Figure 3,8.9: Forward Part d T -2 = Zo =v 1 3 {jj CI = 60 n ! 120 fl 1.:2 = 120 n 60 n 60 fl 137 3,8 Exercise Solutions hom Part :J Part e - up(~\= A e- jßz . = ~ , (i Zo + Q2) , eJ.t e-Jßz \zit ­ 1 2 =-, (f!:.2 R Z0 +,) :!L2 ,e jßl e - jßz 2 1 ( u2--+u,) ) e j ß l e , 60n , 2 120n -­ -jßz jßz =4:!hJ,e 3 , 3...If you lack to get a full essay, guild it on our website: OrderCustomPaper.com

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