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Homework_3_Solutions - Wow/cage a w estaeism Ensign...

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Unformatted text preview: Wow/cage a w estaeism Ensign thetimiaett waitigsie titles. Using the line parametets L a 354.55 iiH-m‘i and C s 325 eBai“? the chaeactei‘isttc impedance sad the phase vetocit}? of each has car: he found as g a 2'0: g __ 364.5 X 10*9 ~54ti 0 m x 1042 ' l 1 = = ~—-—-——-—————-~—-——— 2 14.8 - ‘ x/LO «(364.5 x t0‘9)(125 x 30“”) cm “8 The one—way time delay of each iine can be found as td = 5/11}, :(40 cm)/(l4.8 cm—tts”) 22.7 its: The reflection coefficients at the source— and the load—ends of the lines and at the junction point where the tines meet are ‘ l ”:0 RS+Z0 * s+54 " r5 = —0.8 _&W%W%w%_ ft” RL+Z0 ‘ es+54 ”M _(Zo“Ze)-Z0_27--54_ i I13‘”(:vzouzo)+.zo " 27+54 " 3 tespectiveiy. The initiai voitage launched at the source end of the line is Zo 54 +- : 0V1~RS+ZOV° 6+54 (5 V) =45 v Using these values, the bounce diagram and the voltages “V5 and °VL are sketched as shown. As seen from the bounce diagram, at t = 3t; : 8.1+ us, the voltage W112 = Tj°V+A2=(—ll3)(l.2 V) =—0.4 V reflected from the junction back onto tine A propagating toward the source end caacei completely with the two voltage disturbances transmitted to tine A from lines B and C, each having an amplitude given by Vi]; = (l + Fj)°V§1=(2/3)(0.3 V)=0.2 V (where ”V131 is a voltage disturbance reflected back to line B from the load RL). As a result, no reflections from the solace end occurs for the time interval 5.4m s t S 16.2ns. This cancellation plays a significaot sole iii minimizing the effects ot reflections due to impedance mismatches in the circuit, and is the basis in which the configuration in hand is coitsiciei'eo to be ‘optimized’. E‘tg. 22.2% Etgtrre tor Erehlerrt 22-32%. Bounce ttiegrem and “”515 aha “EFL verses ts Esta. A. digitat E6 thtereohrteett (a) At t z t), the driver gate chahges from LOW to HIGH state launching a voltage disturbance of attryhtnde 0V; = “+505 V) 2 3.91 v at the driver end of the interconnect traveling toward the load. When this voltage arrives the load end of the interconnect at t = id = 1.5 ns, a reflected voltage of amplitude °Vf = IE!” 2 (+1)(3.91 V) = 3.91 V is launched back on the interconnect traveling toward the driver gate. Soy at t = 1.5 nsa the load end voltage of the interconnectchanges from ”l/‘L = 0to°VL 2 7.8l V > 3.75 V and as a result of this change, the load gate turns ON. (h) When the reflected voltage arrives the driver end of the interconnect at t = id = 3 Its, a new voltage of amplitude l4~50 V2 “W 2144-50 (3.9l V) 2 (—0.563)(3.9l V) 2 «2.20 V is launched on the interconnect traveling toward the load gate. This voltage arrives at the loarl gate at t = 4.5 ms when a new voltage ot’ amplitude 1/2“ = “W 2 —2.20 V is launched on the interconnect toward the driver gate. As a result, the load—end voltage drops from ‘t/L 2 7.8t ‘V to ”Q 2 3.412 V < 3.75 V and so the load gate turns @FF. This process continues on. As seen from the hotteee dtagrara atrrt the sheteh ef the load vottage ”th versus 73, the loa& gate tame tat eertttanehtty {tea ti >» 3.75 V); at t z 535:; x 7’5 he. sen, Bigitet Ea cisenitm At t 2 ti? en incinent soitsge of steniitntte St) ”W ._ V __ V E.- sessofi )~ 2'5 is launched at the driver end of the tine, This voltage strives tile innction of ttte tln‘ee lines at t =(2 tt)t2 ns-ft‘1)=4 ns when it is completely ensoiroetl since the jnnction reflection coefficient in this case is given 33y (100 I] 100) —- 50 _ (we || 100)~t-50 ‘ The two voltages transmitted on tines Z and 3 strive at the load gates at different times (one at i=2: ns and the other at i=3 ns) where tetiections occur. The load reflection coefficient at each load is given by l1 2 +1 since Zin >> 20. When the reflected voltages strive back at the junction joining the tines lines, both reflections and transmissions occtn‘ since 13—923 = (100 u 50) — too _ 1“,, =1“q :——-——-—-—-.——._._...__ 2 ‘3 3 ‘2 noonsmuoo ~05 This process continues on. Using the above values, the bounce diegt’atn and the variation of the voltages “i”; and °tfg vetsns 2? see es shown. Accottiing to these figures, gate :3 turns {til @fimafiéfiifiy st t :2 22 ns sect gets: :22 tents KEN netnienentiy at t rs: is}? new @433 j'i‘iisee inenticni notennest Tits intent inteedsnce oi the ttanstnission tines connected to the antennas At? A2, anti is} see nil the some given by Z _ Z _ (too)2 2‘ ‘ 2 ‘ 3 ” {Silt—3'59) mid ttie corresponding ndtnittences are {50 + jfitl) : = Y 2 W YE E72 3 {ifiifiz At the junction of transmission lines ’2: sect 3 (connected to antennas A2 and A3}, the eqniysient admittance seen is the parallel coineinntion of Y2 and Y3 as (i00+j1o0) = Y :: Y” Y“ 3 (too)2 At the middle junction, the equivalent impedance seen can be mitten as (loo)2 2123 = {Zt H Zzsm) = [Z1 M 223 ot the comesponding admittance follows as s ... Y Z23 3/123 =Yi + 3’2334] —- 1+ (100)2 _ (564—3758) 1 th+j)+(t~~g): t (toot2 + {lee ~2~ jinn) ”” too too i t ’i‘hereioiet7 om :2 'iiltiiii With 44233 the innnt irngsedance seen at the sottree end of the lira/’4 iong itttiiii tine is za3o Zm :- 3123 e l one. iising ng the sonreeeend aoitage ll; can he toned as Zia ice V :———~—-—-——————— 2 : Rs-t-Zin 0 loo—rimfi ) 6V t4: and from V5,, the time—average power input to the transmission line at the source end (which is the total power delivered to antennas A1, A2, and A3 since the tines are lossless) can be calculated as _i 6: Next, we want to find out how this power splits between the three antennas. At the middle junction, the time—average power input to each one of the two lines (one connected to Al and the other connected to A2 and A3 via two other lines) can be expressed as E 2 Zin l Rn=§ (ice): 0. isw 2 l V 231:2 2-2; %€{Z1} 1 V- 2 = _ __ 1 P23 2 Z23“) 934323”; where V- is the voltage phasor across the terminals of the middle junction Noting that Zr 2 100— j 1th t2 and Zzgfiq = 100 + 3100 it, one can conclude that the input power splits into two equal parts at the middle junction and so P1 = 0. 89 W and P23 = O. 09 W (i e., antenna Al gets 0 ()9 W and antennas A2 and A3 get the other 0 09 W). Similarly? the power at the junction of transmission lines connected to antennas A2 and A3 splits into two eqnal parts? ie power delivered to antenna A? is P2 =83. 8345 W and to antenna A3 13 P3“ _ (l. 845 W respectiveiy. a, Matching with a single lininnerl elernento (a) This is a single thinned shnnt element matching problem discussed in Section 3.5.i. The idea is to find the minimum distance away front the load position where the real part (i.e., the conductance) of the line admittance Y1 at that position (i.e., B-B’) looking toward the load is given by Y1 = n —jB =0.02-—jB s-m—1 such that the total admittance Y2 seen from the line on the left hand side at position A-A’ is Y2=Y1+Yi=Yii~jB+jB=Yb ire, the lumped shunt eiernent is used to eliminate the imaginary (susceptance) part of Y}. We first start hy calcnlating the load reflection coefficient as ZL-Zo_ ltl—-50 ___._._.__ N —j7r ZL i- a, to ~i~ 50 “‘ Q6676 FL: Using {3.443 with p 2 0.667 and 1/1 = —7r rads the distance l for the position of the shunt elernent can he found as H )r -1 )i _1 .— 47? [mp cos (—p)] _ EE—n —- cos (—0.667)] __ 6.0669A,0.433}\ zilfifl rad irorn which the minimum distance follows as 5mm 2: titir‘Bo‘Etfit. A} {ieéeg e}; :2»: 39 e3e~£mfé gee f s E GEE, we have A m cwjj" :2 361‘? cm me? SE? flee ecem? eeéee eff gee geemem eéeeenee fefifiewe ee 5mm a {fiflefiéfififi} e: 2.53%. cm. Since flee minimum éiiemeee 5mm fie eeiainecfi free: 91 : ces'R-flfié’?) 2 E320 9: 2.3% mi file unknown 33mm eiemem needed is e capacfiier (fie, B > G), Usieg [3.45], me eermefiizefi eescepfimee of Ehe ehunt capacitei to be connectee in paeafiel mm the 5G9 fine a9: the minimum dieéaece pesitien 5mm 2 9.066% can be calculated es .« { B=+——12?p4¥—» 24mm ~+i.79 1- p2 x/a —(O.66‘7)2 ‘ The value cf me shunt capacitor can then be evaluateé a3 ._ (3.79) 5:33: CZ ... Cgmgaee e 0 {W ) 0 > {23: x WWW} E} {2"} Ueiflg {3.5%}, we have A A ~—1 l = Em — cos p) 2 EI—W — 2 0.18M, 0.31M 2i0.841 rad Therefore, the minimum distance fofilows as 1mm ~ 0 183A ~ (0 183)(30 5 49 ‘ . _ . _ . ._ . _ . 2 . cm. S mmxmum dssiance 1mm IS obtemed from 61 : cos“1(0.667) 2 482° 2: (3)843: red, the ugfiigg series element needed is again a capacia‘or (fie, X < 9). The value of Ehe series capacitor can be calcuiaiecfi as w 2 X=~——-—-—-p :~—i.792~& Ewpz 020’ 50 -> CN 24.45 a}? ” {1.7%272- x 109) 3-5339 ggeeetieeeeeee meiehéimg. (e) For fine fame circeii wéfia the singée gamer-wave ieenefeweee we find 222 = VZORL = c/(soxeee) 9: 14mg: 3501‘ icee Seeeiécé c‘éfcmi: imefivmg Ewe gummy-«wave eecfiene cascaded mgefihea fiche input imgedeiiee ef me Ewe Eraesfecmem can be wmi‘en as ZQ22 and RL Zinz = 2 Z 2 3 1 S =‘g—2RL=ZO é Z‘ 1: m Zinz ZQ2 eeeufitfing in 3Q; /EQ; : g Zfl/RL. Bu: 23,5: is glee given that ZQEZQE : ZQRL. 'E‘Elerefece, selving fiaeee We equafiieee eémefieeeeuséy, we maze E F“ ‘ ‘ ” w i @8888) 8 8 £388 8 «ii/eegeeei c I L V - yiehiii’ig ZQ; 2 84.182 and ZORL ~ (smegma) ZQI “ 84.1 2 238g} respectiveiy. (b) At 15% above the design frequency we have for the first circuit: zin=zQW _ N6 21 929.330 QZQ + jRL tan[(27r)(1/4)(115)1 0 e and Zia — 50 1 + In“ I 17- = 2 0.278 S'— « _ 1 ml Zin + 50} —+ 418: ~1 768 and for the second circuit we have Z __ 2Q z; +jZQ1tan[(27r)(l/4)(115)] ‘“ ZQ1+7’Zntan[(27r)(1/4)(1.15)] where Z, = z BL+jZtaai1[{27r)(1/4)(1.15)] ‘“ Q2 ZQ2 + jRL tan[(27r)(1/4)(1.15)] subsiituting values we find Zin : 44155813930 and —5€) Z::+ 50 1+1" in i —— IE“! I Hal: 2 1.144; ‘2 6.0673 —> 3: Similar analysis for 15% beiow the design frequency gives S ~ 1.22 for the fuse circuit and S_ N l ()2 fair the second ciieuit. ...
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