Wave Propagation Velocity?

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

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    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?