Fall_2006 Final - IEineIEnfiiA I EEL elite I...

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Unformatted text preview: IEineIEnfiiA I - EEL elite. ' - I FellZIJ-flfi" '- . '3 Drt'Srit't'tstetitt - fifljflfitffifflfi“ ' ..-_'._____'_____..'_.__'..;_'H- Last Name First Name Seetten # - ”(1) wet wet efthe feltuwtng' queetiette clearly in the speeePIeIt'idei ._ 'SI I .{e} Wt' 15 the etiterien for two eletnettte te be' m eetiee? “23% W. my“? .- 5' 42b} State the 'I'ltevenin sTheetettt for DC eireuits III I I III. I III SEE-H flake «4mm. flcrtrwumge} m gem“, iI-IIEI waj—md’: E’PE at 'IDC VBW.. WM I in :_WI MI aet- 212%ng - 'A.‘ 5 I _{e) IShew that art iIttduetet. ae'te IeIs a short eiflmt in a DC etetttlf,»r etete. \i I' 2 L d ' L = 4:? . L- 'd 4: 'ngfieetE mew .5 'I I [dJI Wt 1e meant Ihy e phaeer dtagrem‘? 12]. C1111~$i1ieri the nil-£1111 5111111111 "5 . (a)1111111111111111111511111111]:111111111'11111111151111111bilabiaflngmem.‘ -' '1}— fifl'figfi_'r = Q a G- I 10' =: {1:} Fimi 1111: ThEvenin 11 eq11i1111111111 1:ir1_:1111 51:11.11 by 1113 25-9 I'ESiSIflr. - . . 5 (1:) [1511:1111- 11111111: 111111115 11: f" 1111 puwér in the: 25-11 1135151111. 19”: 1:111 3_g.__3<::CiW 211-91; 11.1 (3) Theffoniamg circuit was in a DC steadyjstaté'befnm magma ginsed at ;_= n. . Firidfile. followingquafitifies': ""5 "'IiaJIirfil-ifl}; . PS. " I I 1(4) Fimi tha fullawmg quantities in the circuit shuwfi The scurces opt-11111:: at- 151'] Hz. . . 1-6. ._VI.-- . -r-,I'_}2_1:Ia-f'_i-j:r'5€1 _:1._1;:.1-1.l=__j '5' I (1:)1'111: 111111111111111111 1311111111 in th: dependent snurc'e 11.1 t = 5 ms.- '-P'-.._._:(1gfifi.~} in — .171 1391 araC'hwt +1g3,11)—11—' " -- ' ' - 111123-1-[19191-1?“ #6:":st ~__— H.361 1: 1:: Ming) [- 1: r1355 'Pfi1fi3wfi .-- . 5 (511;; the'lfallcwing circuit; [ccwer'is transferrec tc tIt'J-zuttrei}r rcaiatit'e 1am Rb . -5 I -(aI)I ILIFirIidI the matching T.r'alue .chRL fcIr maximum average pcwertranafer. ' E :n 'JII’IQE? ii, C 52’3—‘135'3-33 ' “Clpcadi—J'IDGUJL . 2;. 2:3,». 53' LEE. 97?? J'LI . T, n' ' 912. JL‘I. QE_ t ETE.Ii: _IaIE. E - . [SI - '(b); Ccnnect the ahctre matching lead tc A anti B and find the equivalent ' - impedance seen by the source. Give yeti: answer- in the rectangtflar farm: Fig- I 73%?" ' I"‘:_:Iifi{c? u [562. +I (319;. n 11% £93 ”_'I.._-;I 413mg 113.1:3? 2?+__)__ “332 393]. FQHca “Jag tear/~51.” lbwtfiifggyi 3;-“ ' - SI I .. . (c) What is the newer factor. of the impedance calculated' in part {H3 I - . ' ."bf: (leaf—ES 5-1”): ca, ‘3th Ci£30iM ' . 5 I (cl) With the matching lead ccnnected find the Iccmpieit pcwer previdetl II-IIIII. theI filtt'i‘finttiflurcer . _ . . 3 - IIII'Iir’i-M' %e_E-I ~ _ m . ' - -- ' - 7. far 5 a Cl) 1-59 re 4e33t’5‘t ' ' e: III-9b 5—343 2-. I135; 51-- HHVFI (44.2. i-—_] BIEEUVPY ...
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