Question

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A 75-12 lossless transmission line has l = 0.752. An inductive load, Zl, is terminated where ZL = 75 + 75 12. (a) Find the standing wave ratio S on the line. (b) Find the load reflection coefficient, r(O), and the reflection coefficient seen at input, 7(1). (c) Find the input impedance of the line, Zin, for four different line lengths: 11 = 1/8, 12 = 9/4, 13 = 34/8, and 14 = 1/2, respectively. V(z), 7(z) ZO.B

Answer 1

All the values of reflection coefficient and input impedance depends on length of the transmission line and by varying the length, the values are changed
Given transmission line rce V (2), I (2) TO 7 Zo,ß Zin - (a) Zo = 752 length l = 0.151, ZL = 15 + 375.2 Standing wave ratio s & WH sa Vmax V* Vmin 1+ 1 I-Irl ZL- Zo ZL + Zo 75 + 75-75 75+;75+75 j75 - 150+; 15 24J L90° J5 | 2656 = 0.45|63.44 irl = 0.45 of = 63.44

s !+ 0.45 1-0.45 15 - 2.636 20 coefficient r(o) = 2 2 - 2 10 ZLt Zo (b) Load reflection 110) = 0.45 L63.444 input, r(e) = 1r &(or-apes Reflection coefficient seen at input, rel) - laulē j (63-44-21 ILI = 0.45 Or = 63.44 - o.u5c ß = 27 l = 0.15A :. ßl = 21 x 0.151 - 31.- 210 reso. (63.44-270) [r(e) = 0.45 08206.56 (© Input impedance of the line is given by ZIN = 2 ZL +j Zotanße Zorj ZL tanßl (i) For length li= 1/8 tanßle = tan ( 24 x / ) = tan = 1. ZIN = 20. Zl+jZo Zo+j ZL = 75, 75 +; 75 + j 75 = 75. 75+(75 +j75) . 75+; 150 = 75. 75 +; 75-75 - 75 + 7150 ZN, 150-3152)

(i) For length ls - - - - tanplz = tan ( 2A + 2) = tan 2. 7 ZL + j Zo tan a ZIN - Zoo Tot jz tan - 2. tan. I things tj 70) tanah (Tan + j ZL) X an N NNE - 75 +;75 75 i tj 31 (iii) For [Z5N₂ = 37.5 -8 37.5 2. length lze se tanßl3 = tan 24 x 3/2 = tan 37 = -1. Zin - 20 ZL - jzo Zo-j ZL 75.. 75+j15 - 375 = 75-j (75 +; 75) 15 = 75. 150-jT5 75 [ Zeng = 30+; 15 on

(iv) For length ly = 1 / 2 tan (Blu) = tan (2x + 2) = tană = 0 ZIN - 20. 2 + = 2.