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ESE319 SAMPLE PROBLEM SET 6

# ESE319 SAMPLE PROBLEM SET 6 - sAiNxPL'E‘W'PRoiB'LEW/h...

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Unformatted text preview: sAiNxPL'E‘W'PRoiB'LEW/h SE"? 6 ....I.IOIQO|OO|C\$§0.0.0.0000... [1] Find the shortest length of a lossless, open-circuited transmission line of characteristic impedance Zo =300.Q which at its input end looks like an inductor of 1H at 100MH:. Assume the line has a phase velocity up =0. 80 where c is the speed of light in vacuum. [2] A 509 lossless air transmission line of characteristic impedance Zo=50.(2 is terminated in a load impedance ZL=25-j0.0. The line is M4 long and is connected at its input terminals to a generator vé= 3 sin (2007:. 105t) V with internal impedance ZS =50-j O .(2 z,_=15+:' 0-0- i) Find the voltage reﬂection coefﬁcient 17-1/ 4) at the input terminals of the line. ii) Find the impedance 21-174) seen by the generator. iii) Find the expressions for the voltage and current phasors V(z) and 1(2)), respectively, for any point z on the transmission line. iv) Find the expressions for the instantaneous voltage v(z,t) and current i(:, i(z, t) at any point z on the line. v) . Find the time-average powers carried by the i- z propagating waves in the transmission line? What are the time-average values of the power incident at, reﬂected from and absorbed by the load? [3] The circuit connecting a 300.0 antenna to two 300.0 FM receivers is shown. The voltage source represents the antenna and the loads ZL2 and 21.3 represent the input impedances of the two receivers. It is given that each of the three transmission lines has a length of 1501 and that the characteristic impedance of each of the lines is 300.0. i) Reduce the given circuit to one in which the circuit to the right of TL] is replaced by a single load 2;, connected at the end of TL1. Hence ﬁnd the expressions for the voltage and current phasors V1 (2) and 11(2) at any point on TL]. ii) Find the time-average power incident at, reflected by, and absorbed by ZL. What is the time—average power absorbed by each of the two receivers? .séw 990.89% .3“ 6.. wwev “L 11; 11(5): 4’70 =1wen£i 1 ' . [i] ﬁk/Zzw‘h’w ZL+ZD ZLg'ao—i-JM L A+ 1H4: jzkz [t + EL ] 2| 'WA. °?'€*"—°“"“.“"* *"M’mt’m “9(2): \/ e e in ! W2 A 3‘. r ‘iliz u - -"L J *5 1(2):%e‘z[&—1‘Le J Info/d” 0 2&2 ﬂane , 2C1) : V9) --Z H—I‘L eJ I #5116 _ — _ _ 1+ 5 ZLn- ZC L) — ZC . ”JZkL l' e . 'L -' 4L 7H1 N _ iii—3;}:- ' Eff-1E / “(L -jLL 32 \$6.ka . Favvmh {1 input ‘ 7 - ' (V \Ztnld :: ”J COfLL AMPeJMfe 5‘}.— an. oPen—cu’rcm'ied TL [4 Ema-H» L am {min-{W We wm'f Hue oPen—Ciwcw'iéol "TL ’h 1;“: Qt‘ne 17.217 alt Hz; «60‘»; 2M“) 5a 2”: J‘caL =52me e: —,sz @ka W k )0 Wd’ “(w/t- L~L {‘7 Jae gamer» 3.} and I 531 Hﬂ-‘ag LL<0 S)“ 6 mix 1‘4 231'. am : -w - - g 3m , ‘ I- —s‘ —7 W’mkh ._ ‘32qu _—_4,775on Wu '» SrL ’ ' '7 s ﬁnes 231—47751do 2: “IE- . "o’ m k: 213‘ M L-zn _. 7\: 0.2”” 7*: 9'4 "L Orgc 1 - a" f 8 K¥_~(O Hz ' 23C ‘ _ " '7'L ’ 3T, 0" L322— :(ygm AWN?" 1% skate-9f ’Qmafk 4” Open'c‘lnw’fd TL 6M infwl' Aknpecgmue 4-5 .M ‘4' W (Hind?! 1H E\$?ressaan far W’bbge 9‘1“” my £5 03“ H94 35 VG) = 0+ :k [1 +363) 92142] : 0+ [ gjsiz+ I’CO) éjkz] 0-6., \7CZ) : f:— [e-Jnkzﬂ é" 9‘47] Ajzmar Q) The 9% I’m-“4’0“ :sz ovwnwf 901x354 ECZ} 4‘5 am' H!” ‘97 ‘- - 3-— ‘JLY JLZ 1%-}on +356 ) w .J» \$42 , J, x1 7 A ”'1’;th @j 3 3%mCW'f—Lz) _. _ £60: (Ou*f-§<2)~ :w(mf+kz) ﬂyicw/f @ w P. I l— \v*\$ ‘ 3/5 _ .3, W 405W 6) m 2 ‘ __— 7.) - Aw _D. 20 2 ‘30 E? : \T‘ZP :- 1. P _ .L J) £ 4“ 3 M" _ 400W @ngw,” 63, A " - 2 g, i. _ g — - - 'L W Mary @3 ”Pure; : 0,447.49046 .— L a)“ ; 043;va Many ‘W , WW Few dosage}? 53 6609‘ t} W PM ""E’W‘m : 2 E359 : 5:3“ ...
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ESE319 SAMPLE PROBLEM SET 6 - sAiNxPL'E‘W'PRoiB'LEW/h...

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