# Wave Propagation Velocity?

Discussion in 'General Electronics Chat' started by lam58, Jan 10, 2015.

1. ### lam58 Thread Starter Member

Jan 3, 2014
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0
Hello, I'm stuck on how to find the propagation velocity of the signal in the Line as stated in Question C on the attached image below.

So far I've only attempted the solution for the overhead line, I've found $Z_0$ and $\gamma$, however, I'm not entirely sure how to find the wave propagation velocity. I've attempted it, but to no avail.

$Z_0 = \sqrt{\frac{Z}{Y}} = \sqrt{\frac{0.03 + j0.38}{0 + j5{x10^{-6}}}} = \sqrt{76{x10^3} \angle-4.5^o}$

$= 275.7\angle-2.25^o \Omega = 275.5 - j10.82 \Omega$

$\gamma = \sqrt{ZY} = \sqrt{(0.03 + j0.38)(0 + j5x10^{-6})} = \sqrt{(0.4 \angle85.5^o)(5x10^{-6} \angle90^o)}$

$= 1.41x10^{-3} \angle{87.75^o} = 5.5x10^{-5} + j1.41x10^{-3}$

At this point I'm not sure how to get the wave propagation velocity. I've tried making the assumption that propagation velocity $U = \frac{1}{\sqrt{LC}}$ the only problem here is that this is for a lossless line and I'm not entirely sure if this assumption is allowed considering this isn't really a lossless line. In any case I get:

$U = \frac{1}{\sqrt{\frac{(0.38)(5x10^{-6})}{1000}}} = 725.5x10^3 ms^{-1}$ Note that I divided by 1000 because the stated values in the table are given per km.

Am I on the right track here or am I way off?

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