ELEG205HW8 - (a) (ds=-1000ij1000s'l; (e) 12(1): 8 + 2 CBS...

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Unformatted text preview: (a) (ds=-1000ij1000s'l; (e) 12(1): 8 + 2 CBS t‘mV camiot be attributed a single complex frequency. In a circuit analySIS problem, superposition will need to be invoked, where the original fimction L is expressed as v(z‘) = v16) + 1220‘) with 1210?) = 8 mV and 122(1) = 2 cos th The com I i , . p] ‘ frequency of Mr) 1 and the complex frequency of 1:20!) i E Q = 9443“ ,uc, s = j207r. Thus, g L: 9cos(20m: + 43°) ,uC. (3)Att=1,q(1)“—* q(1.)=._9cos(207r+43°) pc 6582;10- - (‘0) Maximum (0 -NO. The 'i'ndication would be enegafive real part in the cefilpllex frequency. 10. (a) “(0.1) f—* (20 — 1'30) e('2+j50)(0'1) = (36.06 2' -5631“) cm W : 36.06e‘0-Zz [56.310 +j5(180)/ 7:] = 29.52 42302" (or 29.52 21—1298” V). (b) Re{ vs } ={ 36.06 62’ cos (50c— 56.31") V. (C)R6{Vs(0.1) } = 29.52 cos (230.2”) = £18.89 V (d) The complex frequency of this waveform is 3 =52 +j50 s"1 (c) 5* z (—2 +j50)* = -2 — 1'50 6'1 ' is: I "Le; vSM = (102:? Let ifarcgd : Ime . 1 ' ‘ _ _ L __ R: difivrced (a) V303 +I ' Ext-9 .52};st '- WV I , a supetpoSit'iOn of our actual voliages and currents with corresponding imaginaiy ccmpcnefiié. Substituting, 1047356“ = Me“ +Lse"1 - [1] 1043” d 104:5" R_+sL 100+(—2+j10)2x10‘3. = 0.11299" orI: Thus, i(t) = Rcfies‘} : 0.1521005 (102: + 2.990) A. (b) By Ohm’s law, 10‘) 2 1005(1): 10 e‘zf‘cos (10: + 299°) V. We obtain 1220‘) by recognising from Eq. [1] that V; e“ = LseS'I , 01‘ V2 = (2x10’3)(92 +j10)(0.142.99°) = 2.0441043" mV - Thus W) = 2.046” cos (10-: + 104.30) mV . ' V37=10€-2: co-S(10t+30°)V 5-5 = ‘2+j10: Vs =104309 V (b) _ _ _ w. _1_.5 .128"? 25 j125_ ] _ N 600+fl°°0 10 5 —1—j5m_—5—j25 _ 5_ (—25—;125)/26 C_—-2+j10 —-1+j5 26 Z - '— (—5'—j25+I30)/26 , Z 26 _12"5—125 _ 5—11 — _ __ 1 104300 (—5—j25)/26- 10430° —5—j25 5430o msm 125 150° —1—j5 ‘- x-——— —___. W — 4+j4 5+(—5—j25)/26”94+j4 130—5—j25_2+j2 125—j25"2+j2 S—jl —j1 Z.” =5+0.5(—2+j10)—j1:4+j4g | H I\ U.) C 0 ice) 03536;” cos(10t—105°) A 1‘ , = 203'” cos 4: A, 1' r: 3063‘ sin4tA s. I 52 .T-Sl=2040°,132 _ —j30,§=-—3+j4 .226: 10. —3—j4 —3+;4 —3wj4 v =20_5(7.2+j6.4) X ~6+j8 (—6+j8)(3.8-j1.6) " —2.2+ 1‘64 4.2+ j6.4 4-2+ j6.4 _ —600+ jSOO —- j30(~—22.8+12.8+ 1304+ j9.6) _ —600+ j800— j30(—10+ 1'40) 7 . ‘7 - —2.2+j6.4 '_ ~2_2+ j6.4 _ ='185.15‘z—47.58°V “2-2+j6»4 ' g =0.4(—3—j4) 24.2—11.6, 2L = —6+j8 —j30. “1722+ 16.4 vx(t) = 185-135“ cos(4t —47.53°)V 26?? . - = \ _— '- 40+—~ = 20+500s ) 1|. (40+5005) Zm [20+2x10f3s] M 2x10‘3s] ( ‘1 _. . 805’ +' 3000§+25566$ _ W 50 16(O-25) 50 16s r15s2+50s+4000 ' (a) ZER=—+—«—_=~+ .s 16+O.25 s s+80 s2+803 (b) Z,.n(j8) = "1024+ 4090+J4°0 = 0.15842— j4.666 Q _ ~64+j640 _ (c) Zita (“24—16) ___ 5-85014w114399 r —32-—_}24—160+}480 ~ , d 50 0.2 sR 0.2125"- +10s + 50R ( ) z +—~——= 2: , s R + 0.28 0.25 + Rs . zines) = aggro}? 55R = 50, R , (e) R ...
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This note was uploaded on 04/03/2009 for the course ELEG 205 taught by Professor Weile during the Fall '08 term at University of Delaware.

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ELEG205HW8 - (a) (ds=-1000ij1000s'l; (e) 12(1): 8 + 2 CBS...

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