EXAM 2 - Quiz 4 A NAME: Fall 20 1 O 1009 lossless air...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Quiz 4 A NAME: Fall 20 1 O 1009 lossless air transmission line of length 100.63m is driven by a steady-state sinusoidal voltage source of frequency 150MHz. The transmission line is terminated in an unknown load. What is known is that the impedance looking to the right at the point which is 20.16m from the load plane, and thus 80.47m from the input plane, is 25-1509. Using a Smith Chart or otherwise, find: 1) 2) 3) 4) 5) Zi (input impedance of the line) 2;, (load impedance) 1‘], (voltage reflection coefficient at the load plane) VSWR S # of voltage maxima as well as voltage minima between the input and load planes of the line ‘3 5L: logyglogHz imjf \fA-t'SXiOg/M/s E)": 3%“) m/s Issue? Vs W 52047”; = 40,2351 and 100.68mE—60-‘3i5fi ZC~-Io-IOSA):25—Jsga ‘ l . -1 . Q l ZU=IOOA, ‘42 I 'm . i 5 ZL. \ l l ( 1 :‘2m 20~t6m 5 “"6” = (0.0%), 2m ) \ \ N i ‘ ' ‘ I ’c‘aFui‘ observal-{m Leaf! Plese Plane Plane 2- o —— 50.315}, .- ((9.033 E (40.08%) : 0. 35.4035 . The” '9 '1) Ewkr “‘6 smkfi“ Charf SA. ZNG—W'MQ) z Zo~ . 1,565 m “Hus circle 4mm the, constant lfll cé‘rc’le Pagst‘nfl ’Hmmgh Hu‘s Point, 40M EU or “Nabth Hm Jr (AWL .43 503554003 = 40.2351 figfirflggenqawlj zit 36 thmarcis ganemtw Alwsi- Me wig {tale {1’ 90 35mm ZN paint szw P ' (9.2 irrifl/flinvi . y ' scale 0.424 <— obsermtv‘m (>0th Z=-‘0'ogz f.” “f9 +0.13 5 (— oL'stahL'E 131 7\ F move 4ch ?M “Lt/((39 M O ‘ g r: 0‘ l 59 (m mtg 5561?; O - (" 4.“ F1Qhe on a) 7 (53:1: (11“ g I _ ' . see 5m."H1dTar‘H ZN- o.é(+i\37 C -. . . L ZCN as Hm "(Family seen E9 e A “Ms, Z; = ZOZL-N : waxibiiéilgl-Bfl 2) 7L“ «(res m il’izcimst Circle 3": distance of W‘og)‘ _ i . , . . . . ~- use oat/Z Scalp. (0-083 45 e ain'cauy qw‘u’alent ’W 04737}. #6. 757*61 <— olasewai-fm (lane on wile rim—(e I + 0.0%0 <— Jt‘st «‘n a {v move 3cm all?” 1e ‘7 (if? r OJSG 4&- gqafifif’lahg on wtfl Sta . r F . Br .— 177677.73 (533 Sim-Ha Cilat‘ib “‘9 ZL: ZOZL~~ 6Q_J‘30_(1 4’95F_ 4 - ZLN~ t 3] I f; af 3% Pollni' m TL. FOV ?(LY‘P05€ 03C (“sfntt‘n'm o“ {4“ gym“? Cbth L .{alte (Ill at va,L Mnt, W5) {rm lI‘l scale) (11;; 0,5?! Pram time angle offeifigcb‘m COL’tc-F pscale/ @L:“34'$o IL: 0.634243“ finalize? , . . . n‘ ..y.» The Complete Smith Chart (@9112 4 (613£11"jve)/ P1) 3 e 2 501151111ch Black Magic Design I ’5 111911112 af rig-:59 on scale l .9 ‘ II I 1.... ' - . h. 1111111111-§=E:§Ea=3’€§:a:éaaaafié “'fiiaié’iité; . I’ll-Ia - ' a \ I a - "‘ 9.2-4 _ . cacow N III-- I I I Vmax Po‘n [0-424 on . e wtg scale Load plane at 0-66 on cote scale and 3‘7 0° on {+35 scale mmvscAuDPAmm $20k? 4% J1,» mwm LOAD—> <_ TOWARDCENERATOR $9 4‘ 44 v -1oo1o 211 1o 5 1 a 2.5 2 1.11 1.11 1.1 1.211 1 15 1o 1 s 1 3 2 1 ‘ 0:5; e??:%fi' ~40 311 211 15 1o 11 a s 1 3 z 1 11 1.1 1.2 1.1 1.4 1.11 1.1 2 3 1 s 111 211 - ‘I'é’ '00 O o 1 2 3 4 s s 1 a 9 1o 12 11 211 311 o 0.1 112 11.1 0.5 112 1 1.5 2 3 4 s a 1111s- ' ‘ d} ‘99, ,4 57“" ‘v 'v 4. 1 0.11 11.1 0.7 0.5 11.5 11.1 11.3 0.2 11.1 0.115 11.111 1111 1.1 1.2 13 1.1 1.5 1.11 1.11.1192 2.5 3 1 s 111- 'v' ‘1 “‘2 1 0.9 11.2 0.1 0.6 11.5 11.1 113 02 11.1 01 11.99 11.95 0.9 11.2 11.1 11.11 11.5 11.1 11.3 0.2 0.1 o ’x (P cm J 110 0.1 02 0.3 0.1 as 11.1 11.7 o: as 1 1.1 1 2 1.3 1.1 1.5 1.5 1.1 1.11 1.9 2 ’~ 011mm trim-6‘3 0294?. ff WW. _ 5012:- Raw. 4) 8:49 (an? r cozw‘vgm’ée % V51”? 09 “"3 me‘ 90'” N m %e “ES:me cc‘rflz) \m :: \ D 5) # Vouaje may a 0 finsmer IUI H 4f: VDI-fafle w‘m‘ma 1% “1‘55 % iv make acmPkie cc'kfle armmaf fie ‘IJ‘ICWKL circle, ‘30?» 5':— (00 Ema; on: 306$ Mum»? H12 {Thanh/'5. drag, The 34%”{w’ma'l OBIS) mum! Hug {JUL-z: ' arch goes Hm?th “‘6 V'nm‘n Pfli'fi‘jc (and: mjc fig Vmw. Pomh‘ flaws fi‘e #- 0; vci'bgg 'ml’m’ma 45 \m warfi a; meme, max-cm W Quiz 5 NAJVE: The voltage at any point z on a 509 air transmission line of length L=100.35m has the instantaneous expression v(z,t) = -10 sin(12n.103t - 41:2 — 30°) + 5 cos(12n.103t + 41rz + 300), V. Find the following without using a Smith Chart: i) wavelength A and frequency f of the waves on the line ii) the voltage reflection coefficient I‘L load plane iii) the input impedance of the line iv) the phasor expression for the voltage at any point 2 on the line v) the instantaneous expression for the current at any point 2 on the line 4314:“: wt 1:373} imam 75 4/ hi w~33v§ 5:: 6X‘03HZ 0*” 60 MHZ 4’15“? w=i3fi$W 'fib 5 u, ~ V, I ' 4:) find amt wart the 4’M/twiwevcu 97(9Y‘855lm 51:! VCZ/ To TL One +1” Joy W Quasar form: ‘ at B ‘ ech (4 the “Presg‘m 39‘” Viz/fl» cm 3' flflswr 4n . O h” v 9 9? «gamma ACMHSO ) V r V54: 3m 9 + ‘3 6 J Now the general GOCPresu‘m :9! VC:: (ts—5&2 _ jkz) « "i ~32 ’" “‘2 c V 8 +3931 6 VCZ) : V e + v e w ' L T: 9‘ , : EBA-300 : 54,310 : 0.5é300 flngef‘ 46) L 0“" ‘ €300 “<90.SUA,300 i0 <6oo ’ * -1300 g 0 M?" 0’: l3 €430"E géov and O+=Jw 6 "-‘JW '30 ‘ ._ f.- _ J2le . _ J... (if) Frm ‘99»: ~Sizww‘l‘a PCz): I‘CO) 6 om 3655 24f TL 6 2 AIL . ~J ,‘10036 46) 1.:(3'54’500X e i W ‘ :43: $2007 .3 23C (40‘ 4) : e-J = e , fixo 4 ‘ (40$ +0 4) "J 27‘ 40‘ “J2 ’ e—Jflfi ' : e [W - Liz/V £50,“, gjoswsw — - ° -144“ -3x44” ' E1o—O‘5130x1 :8 : 0.; 474" ‘ 0 ~‘ — ' I74" .._ .5 Sta Z _ l+ 1140 \+ 0.5 «1740 _ fl+ 0:8 COSS74') J8 CN‘ : WI..— - ' ‘ 7“” + 0356:1(740 l—I’o (k- o‘ggtm" <\_o,bcasl4) J W2 E W)- WM .csw Cos V74": ~ 0‘99; gym i740: OJ“; 0 0;}- 0‘995) 'J' O'SJ‘O'WE‘ % ((—0-5)~10r09 ZW (\+0v5$a.99‘3) “*J'O‘ng’”); (H'O-‘UJrJUOS Ougjjoms 06 3 j, I _/ ~— [.5+J'o-0r; ‘3 3 Macaw Z; - 202m = 532;; A+ A- v) icz): y, 554’ g 6sz 2:, 2‘7 : ‘0450‘a 6"th . E1300 ejkz ‘30 80 L, = +30) ~JCVZ-6oo) Ckz Angie.” I ._——€4 ._- 6 g Quiz 6 NAME: A 509 lossless air transmission line length 25.25}: is terminated in a load impedance ZL=100+jO Q. The line is driven at its input plane by a sinusoidal steady-state voltage source vs(t)=2 cos(41t.108t), V which has an internal impedance Zs=25+j0 9. Find the following without using a Smith Chart: i) the input impedance Zi of the line ii) the voltage reflection coefficient n at the load plane iii) the phasor expression for voltage at any point z on the line iv) the phasor expression for current at any point z on the line, and v) the time-average power absorbed by the load ZO The 5251252 (or? TL ’65 e c/Eru'collg gnaw—[9'15 R, 1’70 3 2.»ng TL 4'56)?” 419.5949” @CPYBSSCW "Icz a + 5g 8 ) Volbfle I J VC-%)=\/(efl Ektgeyj) .3: 'TL A+‘° ,L" 4'3 A+ Af'J’ "J’ : / '— —J' V @ :\/(52+-J3_9 2) V (J 3) 3 A5250 4's grit/en by Hm sawY‘CQ cmoliygmj (-6., HM; ampecfance seen lay the voumje SWY‘CG; _, _ —’ 2‘; ° : 00 w WM — A vs = f~ m K _., , 4‘ Z~+Z 15*29 My? Zg V925 ” i . 2 Mr 3 ) ‘ FPM (D 3% @/ J Z i V ~— A+ . O)" V :: "I g r, the Fifi-saw agaregsrfij: thage‘atang 991% z cm TL 43 ‘ ’ .VCZJ: “J; 6 J .t as“)! \/ ‘ ' ' ' (675%)? w) By MOSPQCi‘Ol/S/ 1(2) : 13%] C€~J‘kz_ L eJ'kz) ’4 _ 405W ,: L W2 9 3 I ~ - -§‘:" “J 2% 2 L27 : "g? z 3750*“, Penn; = lle PGKG':4‘EUW '° Beats: a?" gm?” 0 ...
View Full Document

Page1 / 6

EXAM 2 - Quiz 4 A NAME: Fall 20 1 O 1009 lossless air...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online