ESE319 SAMPLE PROBLEM SET 5

ESE319 SAMPLE PROBLEM SET 5 - SAMPLE. PeoBLEM $51”; A...

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Unformatted text preview: SAMPLE. PeoBLEM $51”; A 45011 ah“ fiangm‘35‘b" 1““8 1's {germinated 2.” a load 1m9d'““ 2!. <4)de cvrresfanals. 1:5; Pat-n1. ZLN Show" a.” “as: $2ch Smi-Hs Ckwt. " c-nVen first the ’GYanSMu’ssp’an 1s'ne 1's 7 xo-zésa long) find w Muwrng using the Sand“ Char-f. 1‘) z,_ «2) 13 V) Z5“ vi) # a? vol‘tage maxn‘ma and a; we moves 532 1.054 fianei {I iffie inpuf flme Volfage w'm'ma encaunfem! \ vii) fine pracb'm of er Ema—wenge fumbled power £ecfeal bad: and fl H43 Hmeqvergge Power which 4:5 12-? whack {swbsorkeat by H»: 1034. @ The Complete Smith Chart ‘ Black Magic Design L . n ‘ . 0N ‘NT (G/Yn "iST'AN - RADIALLY SCALED PARAMETERS 4’» J), TUWARDLOAU -> . -1uw.\kuuu.\zr-.x/\Ton ‘. ‘° - . . . . . . 1' 4’ wow 20 10 5 4 1 25 2 IS IS H ll ll 1 IS l0 7 5 4 3 2 1 £500. .._.,,.-_‘_{L*H__...,£third4 .g...,_L.‘,..J...,v.K...H...“., g . , .. .. . . . V .. ... ‘8? {0‘5} no 30 20 15 10 8 6 5' 4 3 2 I H! H 1.2 L1 H l6 LB 2 LI 1 5 [0 20 m 00 04):?) O l 3 4 5 6 7 8 9 ll) 12 H 20 311,40 ' (N 0.6 (Hi | l 5 2 3 l 5 IU 15: Qkflo _.;._-fi_‘_+.‘.,1.,z., ..J.}.l.‘v1-‘Lr4——L1.=.I,lf5-‘K51..A‘._§._‘| 1....,:4..,n.g 1.‘....,...:.1 §,,..4.. . e . . . . ‘ ‘ "o l 0,9 018 0.7 0.6 0.5 0.4 0.3 0,2 0‘] 005 am 030 I.) L6 I.5 {.6 I 7 l B I 9 Z 2 S 3 I 5 IOU 0’! l 0.9 0.8 0.7 0.6 0.5 [L4 0.3 0.2 (LI 0‘1 0.99 0.95 U“) 0! 0.7 [Ni 0,5 04 (I3 02 (LI 0 .._..A_..,.l,:.,.L4_L___Leh..;.A,J.L..J_.VL.V:J‘J,‘L_...,.A.,.A..A....l .. . H, . .. . . . , . CENTER ".0 (ll 02 03 04 U5 0.6 0.7 0.8 0.9 | Lt LZ I] H I75 Hi |.7 IX 1.!) 2 ‘_L-,L__Le....-_...,L_J__4#___.4,.. . . L.LJ.L_J_;_L. 1 . . , . l , . . e . . . . . e l . . . . V . . , . , ORIGIN @ The Complete Smith Chart ' Black Magic Design Load plane 15 3‘6 0.0735 cm Wig 0.12 0.13 16 0.11 0.14 .8 o. 0) “39 0 3 90 37 0.36 Q15 0 8 0 “$9 “A 10 % V, 0 .35 (216 a» \1“ (gm 3 N v» 70 034 gt “2' a o .4 a]; '1. o ‘-‘ a” o 9‘ <19 “5 N 6'0 ‘9‘? b 10 to ’- \*3/ N 0] L— e 6’ .90. we?) $3 a (9 0 0° def 01 N. .3? 23.5 c: w 5 60 $0 w 9\ “v a] Q’ 5. 0“ 1. 0 ‘9 v. y» 19 § V3 / (s Q5" 0 0 0A 8 v“? 0&3 0 w \x's ’7. ~ 0 w 3‘ 9 .5 e 0 8‘ $0 .9?3 w g? ’3’? 9000 .5 g r 4 <6 / g (‘10 W {‘0‘ re e S 5' 0 § 9 I\ \ 9 e g v V' 9 P7 2 § § 0:5] a} E3 c9 5” 3 e 3' g “9‘ Q 5~ 6,. 10 xx 0 o g a. S a Q . N 8 a? 8 5 .9 3% 5‘1" °° V1 0 O m o 5 e N I . E N ’0 vi a a a 3 7° :0 5: N S" 6 03’ '3 °‘ 3 < V, r‘ 3" 3'0 E 5% A —L— 1 505 d .— '—— H N "a v, we m. h. w, 62 N v ‘9 as 6. =2 . c 5 E *— QW‘Q c o o c o c: a H .—‘ v—< H N 6') <5 In 1—1 N In 2 O far-c o 53 ~ 8 o a g n _ I l 1 7 1 m ‘— 7—1 RESISTAN ONENT (R/Zo,ORCONDUCTANCEcowéNENnG/Yo 095 g __ -.é q, a g __ w 2 3 .o“ 2.; a; A 2 a 2 93° G <5? 1 Q 0 g g . I 39 a. a: W Go? r Q‘ 9’ ‘5 § 2 E "3 V ‘% a. s ‘ é) % 30° 31% OI. (\3 4 - 5 ° 4; \ 9;: 0] m g) 4 9p 2. 9w 0’ 12; 00 ‘9. 95 O 6‘" 4'" a & 0% cg: m f 90 r. (50' ~ L a: \o o a o _ ‘9: 0 a 0.67;. 2 a 5/: § .2! g a 1‘0 ’ ‘1 C) .9 - a 0. 15:. 0 Q,‘l/OO 5‘3. Q Q 0 Q ‘90 0. 4‘" an" ‘3’ 4/530 ‘3 “<1 1 b. . . ‘00 9e, 20 1 x“ é" $9 5441 0 q’ 0 ‘o ’02,}, .o. 9‘ '60 0 g l\ D r 6’00 3L \ a 4: 0°: ‘3“ ,3, °=. " 521 .g 0 \“9. ° °5 9 ‘3‘: 99‘ 600 011. 0‘": "‘ 0; 91: ‘3‘“ [15‘0 . \ ‘8" - 93% I0 001» 06_ 08' 91“ V « [1‘ 1'0 ’0 0 21-0 510 v 9” . .1: 62" 2‘0 0 890 1.90 9 a 0.3435 on wig 55’ RADIALLY SCALED PARAMETERS TOWARD LOAD —> <— TOWARD GENERATOR w10040 20 10 5 4 3 2.5 2 1.8 1.6 1.4 121.1 1 15 10 7 5 4 3 2 1 111.m1.y.1.. . y..y.m..y...1..1.y..yI... .1..1......1H...H........H..1.... 1 .1......:11H|1.11.1.11H..11.1.., 111..1H...1.1..H...H..rx....m1.x 9° 0 30 20 15 10 8 6 4 3 2 1 11 1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 w 0 l 2 3 5 6 7 8 9 10 12 14 20 30°90 0.1 0.2 0.4 0.6 0.8 I 1.5 2 3 4 5 6 1015m {'x‘TJ-r'u‘ '1'1' '1'1'1'11'1‘1‘1' HHH ‘1’111'1‘1‘1'111T-‘I-+I‘1‘{“i'1"” .11‘1-i-H1':u‘xhuhlkn'n‘ul k'lLr‘v'i'u'u'u'i‘u'i'1'1’1'i'n‘x'n‘uLu‘1'1'11'1'1'1‘101'1 1 0.9 0,8 0.7 0.6 0.5 0. 0.3 0.2 0.1 0.05 0.01 00 1.1 1.2 1.3 1.4 1.5 1.6 171.519 2 2.5 3 4 5 10m 1 . 0.8 0.7 0,6 0.5 0.4 0.3 0.2 0.1 01 0.99 0.95 0.9 0,8 0.7 0.6 0.5 0.4 0.3 0.2 0,1 0 “1.1.1.1...HHH .11....H...”.11.,11111HH1. 11......1L. 1 . 1.....1..._._....1 CENTER QX§ . 0.1 0.2 0.3 0.4 0.5 0,6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 <§ AHMHUHHHHH1.1...1H,“...[.144y.1.11.1.1.,.L_,_._1H.1.1....1”...H.1......1.1.y.......1.11.. ORIGIN ‘1": 0' 62 593+!TJ52N @ ¢) ZLN= 0'3 +J' 0-5 —~ ZL: 202W: 55.52511 (gnsflgr "‘9 IL=‘I’J<TL= 0'62 ‘23'50 €45.2va wt) fins??? 156i) 5:: 4—3 (6 4'5 Hse value 03C V” af \{nax Pat 7 4- Lacs-km of 1hpui’ Plane Pm“ L931 Plane .M wraééyr ' _,\o .0002; / '26‘57‘“ ‘— eZus‘V- loca'b‘m OJ: inpu'f plane ¥YM {owl {Jane flmms‘m *ucafim 0; W! M a“ “*3 “‘16 0-2650 {6,3137% E wig elach‘w‘ 03C 1‘0?ch Pk” ‘9“ “1‘3 scale . 41.4w” EM: aezé—W‘ v) zmr 0-7—un a zazmm " / :60 (0‘7’J“25) Alma/er fl .....-.. W.) VOHGge maxl'ma = 3L! # voltage mMi‘ma :20 .. 2 V“) PW“ : “mg: 0.62 :0334 414.5qu Rina Pabs —- \’ Pfgfie. ; ‘r, 0_3<0’/4 2 Pm‘nc ...
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This note was uploaded on 03/16/2011 for the course ESE 319 taught by Professor Dhadwal during the Fall '08 term at SUNY Stony Brook.

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ESE319 SAMPLE PROBLEM SET 5 - SAMPLE. PeoBLEM $51”; A...

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