hw12_solutions - '313 .11. _ J’Tfflac H...

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Unformatted text preview: '313 .11. _ J’Tfflac H \lzrx500'x106x4zrx10‘7 x'ixlO7r 6 = 2.6902><10"‘5 R: 2 =ww—_2—4T—7=0.0354Q/m -' wéac 0.3x2.6902-x10,x7x10 pod _ 41: x 10‘7x 1.2x 10-2 L: .I— 03 _ =510.26lnH/m _2 ~ 221 pF/m 1 xf/uco; Jxxsoxlofixmtxm-Vx5.28x107 27.744161 0'6 A R _ . . ._ 27: .7441110"6 x5.28x10’ 2569.09 = 0.6359 Q/m , 1 1 + 1 [1 1] 0.8x10" 2.611104] _.103(1.25+Q.3836) = -—-+-—-— 27r§ac g b —7 Lz—y—IniElJ’U‘IO 1113322. 7x10”7H/m 271' a 271: 0.8 -5 _ G = 2—”3 = 27”") = 5.33x10'5 S/m . 2.6 M—e— In W 0.8 -9 27326351210 = ——.—35~’£ =1.65x10 1° F/m 2.6 Wfl 111W 0.8 31-6 . R G . R G 2 wJLC 1+—-——-+——§ (tn/LC 1+ + J ij ij J [ 2 M 2 ij] Zuzl. m...» é~i50x2.8x103=1.4x10m _ —9 106 = 71.43 pF/m z 50 4:15 .—.—~+ L== 8.=1.786x10‘7H/m' ur - r 2.8x10 ——-—-—"‘““—'—"‘ , 7/ = ‘KR + jcoL)(G+ij) _' R + ij G +ij r20 = R + ij = (0.01 ‘4)(50 + 10) = 0.5 + j200 R = 0.5 film 200 - 200 L=—=——a——-———6=3§979x a) 27rx800x10 321 .4 JRG=6><10'2 —_+ 6:36:12 =I8x10" S/m From (1) and (3), 1 10‘8 uZo=--=75><2.8x10“ ——~> ' =4.761x10'“F/m C _8 ._.____.__ L=»Z—°= 75 3:2.678x10‘7H/m u 2.8x10 ———~—-——~— Prob. 11.16 ‘ _IR+ij.-_ 6.5+j2n x 2x 106 x 3.4x 1046 = 6.5+ j42.73 G+ij = 8.4x10"3 +j27rx 2x106 x 21.5x10~12 = (8.4+ j0.27)x10"3 {12+ ij 6.5+ j42.73 Za :: —‘—“-‘I : —.-———_3'- G+ JGJC (8.4+ j0.27)x 10 .7 2,, 7: 71.714 39750 = 55.12 + j45.85.0 'y = 1[(R + jco L)(G1 jco C) = "(43.1948I.14°)(8.4 x 10441.84") 7 = 0.45 + jO.4/m haw 0.39x5.6 a) '— 275x2><106 :0~—"—'——'178.3fls ., 7' rob. 11.17 ‘ w‘w"‘v"m"wvv‘rw'vwv-r‘rwvvm:vvrvv-ww’vm-wwww- Zsc '0 t4n([31),zm = — 1'2; cot(Bl) -0.56 "' 8 = ~0.24s+ j1.0548_ _ cos2+ ' .2 J _ . 323 cosh(x + jy) = cosh(x)cos(y) + j sinh(x)sin (y) = —0.4831 + j0.5362 z. = "' — 0.4831 + j0.5362 :113 + 12.7269 Prob. 11.22 - 120-50 (3) 1“: 24$ 20 g = Z +2 170 L o .411 Z“ = 2.4 For resistive load, 5 = E; __ , v a m 2L + jZo tan(fil) = z _____ (b) 2“" " 20 + jZLtanQBI) fll=~—. =60” 23 116 (65 + j38)(— 0.2454 + 1'1 .0548) zm = 50 120 + 15W“ 60 = 34.634 - 40.65°Q 50+ j120tan 60" m 342 ' he s = 4 circle, moe 90° from Vmin towards the load and obta 0.46 -— j0.88 at. 7_ —j0.88) = 27.6—1523 Q (Exact value = 28.2353 452.9412 at. = 270° or -90“ r = 054-90" Prob. 11.51 t=21—12=9cm'=fl—+ix7200-=1620' 40 40 At P, zL = 2.6— jl.2 72L =szo =50(2.6—j1.2)=130—-j60§2 (Exact valuea= 130.49 — j58.219 Q) 5 ob. 11.52 B 23"“) =1.'764 GHz 17 ——-~———— I: 3.2 cm: 3—3—3. —> 1355" 3.43 ‘ zL=1.4—j0.8 = 2L = 50(1.4-jo.8) = 70—;409 944.5o Prob. 11.53 _Rg-Ra _ 0~50 _'_ ’ 123+}?a 0+50 1" FL, RL —R0 W 80—50 a = _ v “0.3 . RL+R0 :80+50 Prob. 11.54 _ZL—Za _ 0520—20 _ 17 L _._ V = ZL - Z 3 + Z L _ The voltage and current bounce diagrams are shown below 0.5 V =— 27 =5.4 V, I v 345 216.05mA 344 (Voltage bounce diagram) —(Guuent--bounce-diagrm) _ __ .... __ _ From the bounce diagrams, we obtain V(0,t) and I(0,t) as shown below: ' 347" S cc 1(1,;)' we obtain I(l,t) by scaling V(l,t) down by 150. The resfilt is sh ' a below. _' Lazo “150-50 - zL+za 150+150 2ng 2 50(12) 2 SV 75 FL ml .2: 348 2.5115 . 2.5M 3 .5 us 3 .Sns (Voltage) (Current) ' _‘ The bounce diagrams are for the leading pulse. The bounce diagrams for the second pulse is delayed by Ins and negated because of -12V. ‘ ‘ (b) For each time interval, we add the contributions of the two pulses together. For 0 <t< Ins, V(0,t) -—- 8V For 1 < t < 2m, V(0,t) = -3 + 4 — 4/3 = -5.331v For 2 < t < 3 p3, V(0,t) ~—— —(4-4/3)-2/3 + 2/9 = -2.667 -o.444 = -3.11v For 3 < t < 4143, V(0,t) = 0.444 + 1/9 —l/27 = 0.444+0.074l = 0.518V For 4 < t < Bus, V(0,t)== —0.0741 — 0.0124 = -0.0864V We do the same thing at the load end. For 0 <t < 0.5;”, . mar) = 0 For 0.5 < t < 1.5m, V(e,t)=8+4 = 12 For 1.5 < t < 2.5m, V(e,t)=-12 + (-4/3 -2/3')=-12-2=-14 For 2.5 < t < 3.5,us, V(t’,_t)=2+2/9 +1/9=2.333 For 3.5 <t<4.5ys, V(t’,t)=-0.333-1l27-l/54=—0.3886V The results are shown below. “349 -0.3886V. mgr) = mgr) Since I (E, t) = ZL 150 , we scale V(£,t) by a factor of 1/150 as shown below. 350 mm) 80 __ 2L — Zn 0 — 50 __ Z L _+ 20 ' 0 + 50 The voltage bounce diagram is shown below FL -1 ...
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This note was uploaded on 02/04/2012 for the course ECE 311 taught by Professor Peroulis during the Fall '08 term at Purdue University-West Lafayette.

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hw12_solutions - '313 .11. _ J’Tfflac H...

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