110_1_hw03_sol - 8W:P‘CO‘OU/MS Qb AS in any true design...

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Unformatted text preview: 8W :P‘CO‘OU/MS Qb AS in any true design probicm, there i5 more than one possible soluiion. MOdCl answers foliow: (a) Y z I 2,111 8 a! a) m I radfis. Comtruct this using a l S conductance in pm‘aliel with an inductance L web that Ime 4, 01' L = 250131;}. (b) Y ’3' 200 13:15,} (purely real at (o = l rad/s). This can be constructed using a 200 1an cmu‘luctmme (R = 5 Q), in parallel with an inducier L and capacitor C such that mC ----- lme m 0. Arbifi‘az‘ily seiécting L = 1 H, we find {hat C = 1 F. Om: mimic“ therefore is: a 5 £2 resistor in parallei with a 1 F capacitor in p‘amllei with a E II inductor. (c) Y = 7.180" ILLS u G +jB at mm 100 radis. G = Rcfk’} = 7603800 = 1.21% (an 822,? (:5). use rcsis‘tor). B = Enz{‘Y} = 7’sriné300 == 6.894'15. We may realize this sus-cepizmce by placing a capacitor C7 in {Jamil-e} with the resistor such thatij = 146.89th1' C‘ = 68.94 M? git-F. , I“ pe‘EMfi 9w Maid whm‘ga VH9?) 01% “PM jugdfow bewfi'mem Jrke «9W2 Jewmdewf‘ gamma, C bQfiM}: ’0‘) “Jm : gig garb/AK #7 ”J 9—32 :3 "(2.. 5); mgcArg ML 0% +36": 370.} (Vlz‘) V» ”M“ “a” M? $ng 48%; 4: W + ’ /D{?’ «”5 /o,é “jg, B)! Uin‘Ag KC!) 024 %6 37253;?» CV4) 3WM 9 we «PM W— \LL :2” M! Aw3.séw (énvegwh‘vg £70ka 3172 Mme "We $wmm) was) :— ?%9 cm; 00% 4335‘UMV CD Pmbkam D) £11736?de Anmhjsig EXAM?\€. . (he beat WM % omMyR a cart GWVNCMW ‘5 h, 0 evmwx-e om elemeM‘ M“ a’mme 3’) 58‘ 56‘" 0s Tefmence, 0-,? ‘HM RAVMJV W’é‘f and +1: W\0v€ ’huwovxd 444.1 30w c2, . ’ M'Sx‘w Su raga—van (Mad ‘0 mUC loo (A79 (lxlosé ’9') V J ‘ E 3 “ 5 w: 1X (0“ rod/3 ' B‘j evaoLh‘vtj emch wmyonm’r FesPed +0 W . we. com fine 'HmQ recbuxred awed-Ton gawk”? h, am‘xwe GM UMDUL. (Le) ‘7; : (QR) ‘ ‘7? ° Then quj SNAPN‘CWQ amalu’eis, m WRW News Law @ node 2. I] :3 VI‘ "fill : V3! ‘1 ° ‘3‘ ’ 47ml, ‘3' New to] Scum“. 2 , to: Z‘Mug V‘chs 'F—ev) Oof‘é‘iff‘l. 91p Q5> Considering the a) I 2x104 radf 5; source first, we make the following replaccments: 100 cos (2x10? + 3") v a 100 43“ V 33 up A.) 71.515 [2 :12 “H «+1124 0 92 “F —> ~1‘05435 (2 Then (Vl' -» 100 z 3“)z47x103+ V17 (11:.515)+ (VJ-4’5)" <56><103 W443) 2 0 W (V3. — V! ')/ (56x103 +1448) + V37" {-j0.5435)R0 {2] Solving, we find that + V; /(4.7 M03) Vg' : 3.223 A ~87°1nV and \"z' 1 31.28 .1 4770 nV Thus, VIII) : 3.223 cos (2x103? .. 87”) mV and v;'(!) == 31.28 003(12x104‘1 - 177°) 11V Considering the effects ofthe a) = 2><105 radf 3 source next, 100 cos (2x10? — 3°) V w) 100 A3" V 33 HP «4, 70,2515 :2 1;: pH +1224 :2 92 pF‘—> 10.05435 Q Then __ V~'/(41xw")+w/-j0.1515 +£V’,"—V3")/ (56x103+ 1448) = 0 [3] m" ~ Va")! (56le +_;44.8) + (w — 100 z 3")!" 47x103 ~+~ vz'v (70.05435) : o [41 Soix'ing, we find that 71" = 312.8 K 1770 53V and Vg" 3 115.7 4 -93° HV Thus. v1”(r) : 312.8 (:05 (2)40“? + 177°)13V and 113"“): 115.7 cos(2><105r- 93°) LN Adding. we find W) : 3.223x10’3009(2x104t~87°)+312.8%10’12 808(2XI057'1' 177°)V and Mr) =3L2w8f10‘9008(2x104(~l77°‘)+ 115.7x10"”cos<2x105r~93°) V ...
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