HW3solution

HW3solution - ’ EE335 HW3 solution 7.19. (a) __60+j180...

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Unformatted text preview: ’ EE335 HW3 solution 7.19. (a) __60+j180 “L. so Z 0.211 + 0.31 = 0.521 ' Lgpoint B 10.0 7 0‘ I; -- - , a'” A," '\ 7 . z(—o.3uj)=;3£€ Egg-3; r=o.3634165 fl Rotate on Smith chart r ' 290.3112) =4 + 1'3 . Moving across the connector, Z-=.Z!/ZP + RI =15.1+j7.8 z (01)=?%=0202+ 10.104 ' Lapoint C . Rotate on Smith chart L’ 0315 1' 0-22 = 0236 point D 24—02211): 4.1 + 11.7 21.5 = 307 +j123 (actual 302 + j154) (b) q" '= L = 82 +139 = 81-312.9'. (actual 31.543.4'} Zn. +Z.a . =1.2+j3.6 => bointA, I"=0.863432' mm... ..__._. ‘__...,.._._:- 1.... ...t. ..‘_.....mm..»......._............................._............_........m w».....,......,.__.._._._.... ‘fi’v'Eh’iZ'Tfl‘EFfiWEE .I ..r “L wufi‘hx‘wny‘hug I “- ....:£IE 'l'lil mull-l- =r ....,....-.,—._ ._. 1- 2 I I; = 93 watts (actual 8.95 watts) V in = V"; aim-“11 + 11—22;,” L point D 013 Smith chart 0.659 £10u :> V' =44/1-80' "‘1 moi): 17.64— 553° K(0)=V;[1+11(q)] I l—P point C on Smith chart 0.659 41165" Going across the reSistive network, Z(—0.31/’L) = 4 + j8 23/360 v 0 —————~— =8.5— '3.5=9.1z-.2.1.0' "[ ‘)[Z//300+11] J . 1/; z = V": emm-mfi + 11—03110] L.- pom’t B on Smith chart => V+ =32.7z§174° "‘2 VL = v;[1+ 132(03] L. point A on Smith chart 1/; = 58.581 "1712“ P. = 1M2 Re[i] = 28.6 Watts m 2 z L ‘1} HW 4 solution 7.20. Use a Smith chart. ‘ .153 - .055 = .0981 "(if 'y=*j.7 w Take the negative of imaginaxy side Point C; move to y = on Point D 3: = .403» .25 21531 . V“'WR==3 , A =2(20cm)=40cm d .- E— = .125 -—> Rotate this towards load from minimum (a) zt=0.6—j0_.8 —-> Zg=BO—-j40 (b) Reflect Point A to Point B assuming stub in paraliel Rotate Point B to Point C; Point C = 1 + j 1.17 d 1125-1711665 = 0.29151 = 11.6 cm i . _ / Position negative imaginary at-j 1.17 and rotate to = on 3’ er=.25-.135=.114A=4.56mn 7.23. 7.22. “3311'” CHAR-1'" ZL='60+j80, 20:509, d: 2:I =1.2+j1.6 Point A, Rotate on constant conductance to Point C: y “C = 0.3 + j0.28 Rotate on constant Fcircle to Point D: yflD = 1+ j1.38 ., y = "anywzjflfi8, ym2=1+jO-yw=—j1.38 "l Rotate from y = on to stub susccptanccs of yM and y“ toward generator I 2 , . £,1=.25+_O95=.345;L, 282 =.25—.15=.10). .3Fa6 "SMITH CHART" —1-=.1 PointA; yflL=10 PointB R0 toward generator to g = I circle, Point C, _1.-j.28 ==> d3 =25 - .201 = 0.0493, Take negative of imaginary portion = 1128 and plot Point D; rotate toward load to y = w -. e, :.2s+.195=.445.1 3'7 ‘ *s'n? cw}: 4-2;? at“;— 7.25. "SM [TEE CHART" ...
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This note was uploaded on 08/13/2010 for the course EE EE 335 taught by Professor Xaio during the Spring '10 term at Cal Poly.

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HW3solution - ’ EE335 HW3 solution 7.19. (a) __60+j180...

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