# Wave propagation velocity?

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.

1. Homework Statement

Table one (in the attached image) shows typical values of Z and Y for an overhead line and underground cable. Please note that this is an engineering problem, thus all complex numbers are expressed in a+jb rather than a+ib.

## Homework Equations

Using simplifying assumptions calculate $$Z_0, \gamma$$ and the wave propagation velocity for each case.

## The Attempt at a Solution

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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## Answers and Replies

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TSny
Homework Helper
Gold Member
Aren't the imaginary parts of Z and Y equal to ωL and ωC, respectively? So, for example, 0.38 Ω/km is the value of ωL rather than the value of L. Or am I mistaken?

If you include the units given for Z and Y in the table when you calculate U, do you get m/s?