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Homework Help: Wave propagation velocity?

  1. Jan 10, 2015 #1
    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. The problem statement, all variables and given/known data

    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.

    2. Relevant equations
    Using simplifying assumptions calculate [tex]Z_0, \gamma[/tex] and the wave propagation velocity for each case.

    3. The attempt at a solution

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

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

    [tex]= 275.7\angle-2.25^o \Omega = 275.5 - j10.82 \Omega[/tex]

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

    [tex]= 1.41x10^{-3} \angle{87.75^o} = 5.5x10^{-5} + j1.41x10^{-3}[/tex]

    At this point I'm not sure how to get the wave propagation velocity. I've tried making the assumption that propagation velocity [tex] U = \frac{1}{\sqrt{LC}}[/tex] 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:

    [tex] U = \frac{1}{\sqrt{\frac{(0.38)(5x10^{-6})}{1000}}} = 725.5x10^3 ms^{-1}[/tex] 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?

    Attached Files:

    • wave.png
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    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2


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    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?
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