Question

1) You have a transmission line with Zo = 500 that you wish to match to a (100-j150) load. The phase velocity on the line is 2.0x108 m/s and the frequency of operation is 91.9MHz (KSDB-FM). Use the Smith chart to find: a. The shortest distance from the load that you can insert a quarter wave transformer to match the line to the load. b. The characteristic impedance and length of the quarter wave transformer from part a. C. The shortest possible distance from the load that a capacitor or inductor could be inserted to match the line to the load d. The value of the capacitor or inductor needed at the point found in c. e. The length of the shortest possible shorted stub line needed to match the line instead of using the lump element you found in d. f. The length of the shortest possible open stub line needed to match the line instead of using the lump element you found in d.

Answer 1

0.13 0.14 0.1 0.17 0.13 LHHHtth 038 037 0.35 0.15 0.15 etttttt 0.36 0.35 15 0.1 0.16 0.39 0.09 OA 0.34 100 ... 2 0 .18 01 tattu 0.33 0.32 0.2 01 0.41 033 0.0% JUI 0.02 0.42 U> 0.43 0.06 0.44 0.05 0.3 tartott 1902 0.3 WANAN 0.45 0.21 0.04 0.04 06 0.29 0.79 0.46 0.22 E 0.28 047 OF WWARD GENERATOR --> ttttttttt 0.27 0.23 AVELENGTHS TOWARD GENE .48 RERECTION COEF 0.490 170 0.26 0.24 0.24 Hutt WAVELENG ULILLLU TON COEFFCIENTO .. dia 0.75 0.25 DR 6409 6 WAVAER Ezo 970 .490 -170 0.23 0.27 .48 TOVARD LOAD - 091 0.22 10 0.47 : 1 0.05 act. 0.29 0.46 40 0.3 0.05 0.45 140 0.19 0.06 50 031 0.44 130 0.18 0.07 0.43 -120 -60 0.34 0.08 0.17 -110 70 0.42 0.9 16 0.09 TTTTTTTTTTT 0.33 100 08. 90 0.41 0.15 0.15 0.34 34 0.11 0.14 0.12 0.13 0.4 -- - 0.13 --- 0.35 0.39 HHHH --- 0.36 190 .38 0.37

Z Z 20 L Song z = 100 -jlson, 2=50r, Op = 2x 108 m/s f = 91.9 mitz de Vp - 24.108 = 2.18m † 91.98106 a) Normalige ze = Ex = ? = 100-jiso = 2-3; plot it as point n' on caart. Taking im' as center of imp' as radices, draw & circle. More from 'p' by a distance 'l, on s-circle in such clockwise direction such that į is real. (point Q) l;= Are PQ=0.2148= 0.214x2.18 =10.467m 5) p = Z1 = 0.15 » z = 50(0.15) = 7.51 20T = JZ z = | sox75 = (19.365) lq = /2 = 2.18 = (0.545m] () Take diameetric opposite of 'p on s-circle as 'Pi. move from 'p' from cepto intersection of s-circle with 8=1 circle cat 'R'. id= Are P, R = 0.1554 = 0.155x2.18 = 10.338m) . Also R = 1+2 2 - Y = -2.2; į such that Zin=1 - Zinazo. Y = -221 - > La senyerapiot 39.360 14) Zin to

e move from o (open circuit impedance or short circuit admittane) cepto I (-j2:2) on s-circle in clockwise direction lse = Are OT = 0.0594 Fint & - 0.059x2:18 Zinazo = 10:129m) f) move from Q (Short Circuit impedance or open circuit cadmittance, cepto T (-;2-2) on s-circle in Clockwise direction... 2 loc = Arc QT = 0.3096 - 0:309x2.18 =10.674m) Zin Zo b