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quiz5_solution[1] - Name[TA Student Number ELEC3909 Quiz 5...

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Unformatted text preview: Name: [TA Student Number: ELEC3909 Quiz 5 June 18, 2008 1. For your fourth year project you have built an FM radio transmitter with a carrier frequency of 100 MHz. The transmitter has been designed to operate in a 50 9 system. All you have available to test your transmitter is a conventional FM radio receiver with 300 :2 input impedance. You need to build a matching network to connect the 300 Q receiver to your 50 (2 system. You decide to do this using sections of 50 Q coax cable. The cable has a dielectric with 3r = 2.25. Show your work on the admittance Smith chart attached. a) What is the normalized conductance of the radio receiver input in the 50 Q system? Plot this on the Smith chart. ' l [25" f Imark ”ii-L? - — Tia/H L Ya I / s D b) Using the Smith chart, what is the voltage reflection coefficient F of the receiver in the 50 9 system? What fraction of the incident signal power is reflected? I mark FrlPJejg “Z 0.?2 . power {attuned ll?!l : (74722 : 05h? c) What is the shortest length D of cable (in meters) required to transform the receiver input to 50 Q in parallel with a susceptance? Is this susceptance capacitive or inductive? 4 marks : ..—--~ _-; .__§___ — 3(19? .a 45 304%} ‘ reconfirm ’ 2th D: 048% —0 :arrm :_ an“ :Dflém Bataan trawl? (n tartar” Home), its sweftm is (figural/t. d) What matching element- an inductor or a capacitor— must be connec ed in parallel wi the line to complete the match? What value of inductor or capacitor is required? 2 marine A“ 1hdMC-{of muff he CofllLflCfQVl Til P‘Wkllfl WWI 'Utl Elite. /\ - Yr : kiwi. ‘_ I Eli—2.0 . Y b 7 Ta *1)” ‘_ ht; ‘2r0 In: §$;:aqz 9:0 ~ I “ “‘wa l 3 98 (0—2 L t w r.- G—bu ZliquXU/{OM 2 X H e) Suppose the matching element found in (d) is not available as a discrete component, but instead must be synthesized using a section of 50 ohm line of length I. To minimize I, should the stub have a short circuit or open circuit termination? Explain. What length l is required? 2 marks Short aren't lawman slam be and in 7th Std As Wt 212a! a CflmPl’tle an ind-actor, 13h short CTle'ilrt can acted as an rid-mar With a starter (alga tlan, tL Of’m C'IYCull. (f 0324A —- 0~2§A :: 0.0M "I M4 Um r mam Name: Smith Chart ELEC 3909 - EM Waves I 5.62629. gnu" I—u-z a ‘1’. a . “7. a t: I a ‘l . r - Is Ancficomgwmr'wzamra éonau . H I Ncncomww‘r {inn Student No: _..._.—.._. \ RADIALL‘I‘ SEALED PARAMETERS ‘9, - \- TOWARDLOAD—> <—TOWARDGENERATUR 4‘ £01» % -1qu III II] 5 I 3 2.5 2 13 Is III 'I' 5 I 3 2 1 III/£516? \ QQD‘QF—Hw-H-Il'l'.‘ : -. 1-4-\-"r'."'."'."I-M-‘r- -.-..I.-+,-.-I-. 4.x— 4, Q. Q 5% “-10 an 20 :5 ID a 6 5 I I.I L2 L3 LI l.6 III 2 3 I s In an .. ‘2' $33 {bf-asa/ 0 I 2 :1 a 5 s ? s s In I2 0 0.1 0.2 In 0.5 0.3 l l5 2 3 l 5 a 10 IS»- efiz , é “fig“; :I-H—I—I++I-+-. .'i.:.'. . ' H-hr—IHr'.'.I'..'.“.'.':I.-I-H......'..I.'..W+.H- A" ,--' (9/ “a 1 n.» as 0.2 as us a»: 0.: 02 0.1 am am on LI L2 1.3 LI Is LE 1.7 La La 2 25 :I I 5 In. ‘r ,- 6s}; ‘3? I .. Ina [LS DJ 0.6 I]: 04 0.: 02 DJ . 0| 0.59 0.95 0.5 M . 11'? 5.5 as 0.1 ."E’ 0.2 0.1 I) _’ (9/ CENTER ¢§/ 0.0 III D3 EL] ”.4 [I5 0.5 “.7 I13 0.9 1 1.: L2 L3 L4 '5 I.“ LT LE 1.9 2 , ORFBTN ...
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