A 50 ohm lossless air transmission line is terminated in a load impedance Z L =

Question

A 50 ohm lossless air transmission line is terminated in a load impedance ZL = 12.5 + j20 ohm. Find the following using a Smith Chart:

i)              the load admittance YL.

ii)            the closest distance to the load plane where the normalized admittance on the line is 1 + jbN.

iii)           the length of a short-circuited stub of characteristic impedance Zos=100 ohm which, upon placement in shunt with the line at the 1 + jbN admittance point, produces impedance matching on the line so that there exists no reflected wave to the left of the short-circuit placement plane.

just print it out and take a picture of your work and upload that as an answer and make sure you explain how you got your answer

Explanation / Answer

ZL = 12.5 + j20 ohm.

Rl = 50 ohm

i)              the load admittance YL.= Zl/Rl

putting the value we ve

Yl = 12.5 + j20 ohm/ 50 ohm

= 0.25 + j0.4

now using the smith chart locate the 0.25 on the horizontal axis and 0.4 on the imaginary axis to get the answer....

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