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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.
Using simplifying assumptions calculate [tex]Z_0, \gamma[/tex] and the wave propagation velocity for each case.
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} \angle4.5^o}[/tex]
[tex]= 275.7\angle2.25^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?
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 [tex]Z_0, \gamma[/tex] 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 [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} \angle4.5^o}[/tex]
[tex]= 275.7\angle2.25^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?
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