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A lossless transmission line of length 0.57 m is connected to a vector network analyser. When terminated in a short circuit the input impedance was measured to be: Zin (s/c) = j45 Ω. When terminated in an open circuit the input impedance was measured to be: Zin (o/c) = - j125 Ω. From other measurements it is known that the length of the line is 3λ.

Find the phase constant (β) of the transmission line in the other measurement.

The answer should be 33.1 rad/m but I am not sure how to get there. I would normally work out the propogation constant first using the R, G, L and C values and then deduce the phase constant from there, but the values aren't given in the question. Any explainations are appreciated.

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Because this looks like a homework problem I will refrain from solving the question in its entirety, but I'll give a framework for the solution that you can apply to get a final answer.

You know first that multiplying the magnitudes of Zin(s/c) and Zin(o/c) gives you \$Z_0^2\$ as justified in the footnote1; this value is 5625 so you know that this is a 75-ohm line.

Next, given that the line length is approximately 3λ, we need to account for the fact that the actual s/c and o/c impedances aren't perfect shorts/opens. With the help of a smith chart (e.g. by trial and error), or by using the mathematical statement relating the reflection coefficient seen at the input with the input impedance, you can find that the actual line length is slightly higher than 3λ (the Zin(s/c) and Zin(o/c) are satisfied for lengths just above 0, λ/2, λ, 3λ/2, ...).

After that, it's simple geometry and accounting for length - you have an electrical length in wavelengths, multiply by 2*pi to get length in radians, and divide by your physical length to get your phase constant.

1They are reciprocals of each other so their product is 1 when measured as normalized impedances, and the equation is simply scaled by multiplying both sides by Z_0.

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  • \$\begingroup\$ Thanks, but it isn't a homework problem, I am just doing a past paper which the lecturer has released to prepare for my exam on friday but I appreciate the advice. \$\endgroup\$ Commented Jan 11, 2022 at 15:49
  • \$\begingroup\$ @asifkhan123 no problem, feel free to add a comment if you need any more assistance or clarity at any point. \$\endgroup\$
    – nanofarad
    Commented Jan 11, 2022 at 15:51

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