Question

Q1. (20pts) Consider a lossless transmission line with characteristic impedance Zo=50 Q is used to connect a sinusoidal source with internal resistance Rs=50 2, and 10 Volts peak value to feed a load of ZL=300 Q. At an operating signal wavelength of do, the length of the transmission line is L=2.25o. Determine the followings: a) Standing wave ratio on the line b) Average power delivered to the load in dBm. c) The voltage and current phasor expressions on the transmission line RS Vs ZO ZL 2 — Z=-Z Z=0

Answer 1

Solution:

6) Average Power delivered to the load in dom. Given dala! characteristica impedance Zo's:"50$ Sowrie Resistonca Rss 5or Input voltage Vin Load line 9th of length 3oor - Jo sin(?*ft) Impedance zu 2.2.500 L Rs ZL=300 2.2500 2:0 Zin ZEL Wave 92 Stording It is given by ratio on the line 1+1 & whese Tires the 'Reflection. Coefficient of line. ZL-Zo 250 ZL + Zo 300-50 300 +.59 35 ㅈ + 517 SE 517 s 18 = 6 Zin = zo 2 otj ZL L Where B B Let us Jinsi calculate input power average. input Impedance of line will be given by- zi fjz, tang? tanpl 2x a B.: 2* *.2.252 = 4.57 300+; 50x tonlusnij 50 50+; 3 boxton (4.51) 50 X 50 6 7 j tanl4.500) tj 6 tanly.51) + j 139.53 1 + j 837.17 (64;13.9.53771-8.37.172 (1) ? +(837,17) 2 = 0.0035.(116816.33 -74883.99) So Zin . G - 2500 2503 . + 502 Vin Zin Vp -10V - (408.85-j 17.0992 There fone equivalent circuit Zin Vin Up x. Zint 50 Vin =10x 408.85-317.09 458.85-j17.09 8.911 volt Hence : Pincang) Vin 8.911 50 Vin Peak) = I Vin Iin 2 Iin : zo 2 -0.79 4 W 50 = Pin laug) = 1 / 2 x (8.911) for lossless line Pelang, Pincaug) we can write pv 0.794 W s? Since Iw = 30 dBm 0.7-94 X 30 dBm PL PL 23. 82 dBm (c) Voltage Phasos, expression - 14 ý འ༡༤ 317,3%70 i 3X10 2.25 ne previously got. Vin 8.911 peak Vinit) = 8.911 sin [TB +) and Jiniti 3. Diniti so Iinit) .= 0.1782 sin ( 20 ft)