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Transmission line Secondary Coefficients

  1. Jan 31, 2017 #1

    David J

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    Gold Member

    1. The problem statement, all variables and given/known data
    A transmission line has the primary coefficients as given below.

    ##R=2\Omega/m##
    ##L=8 nH/m##
    ##G=0.5 mS/m##
    ##C=0.23 pF/m##

    Determine the lines secondary coefficients ##Z0##. ##\alpha## and ##\beta## at a frequency of ##1 GHz##

    2. Relevant equations

    In my notes I am given

    ##\alpha=\frac{R}{2}\sqrt\frac{C}{L} +\frac{G}{2}\sqrt\frac{L}{C}## and ##\beta=\omega\sqrt{LC}##

    3. The attempt at a solution

    ##\alpha=\frac{2}{2}\sqrt\frac{0.23 X 10^-12}{8 X 10^-9} +\frac{0.5 X 10^-3}{2}\sqrt\frac{8 X 10^-9}{0.23 X10^-12}##

    ##\alpha=1 X \sqrt{2.875 X 10^-5} + (2.4 X 10^-4)\sqrt{34,782.6}##

    ##\alpha=\sqrt{2.875 X 10^-5} + (2.4 X 10^-4)\sqrt{34,782.6}##

    ##\alpha=(5.362 X 10^-3) +0.044760 = 0.050121902## nepers per meter

    I think this is correct. I am unsure how to input the single multiplication sign `X` in LaTeX form. I think my " to the power of`s" are correct for ##R, L, G## and ##C## but I am unsure about the final result in nepers per meter

    For the second part I got the following:-

    ##\beta=\omega\sqrt{LC}##

    ##(2\pi)(1 X 10^9) \sqrt{(8 X 10^-9)(0.23 X 10^-21)}##

    So I have ##6,283,185,307\sqrt{1.84 X 10^-21}##

    So ##6,283,185,307(4.289522 X 10^-11)=0.269518623## radians

    So ##\beta= 0.269518623## radians

    This second answer I am not so sure as I have some very large numbers but I have followed the examples in my notes.

    Any comments on the two attempts above would be appreciated.

    Thanks




     
  2. jcsd
  3. Jan 31, 2017 #2

    gneill

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    Staff: Mentor

    The values look okay. ##\beta## should be radians per meter. You'll want to round to the appropriate number of significant figures to match your "givens".

    For multiplication in LaTeX you can use \times or \cdot : ##a \cdot b = a \times b##.
     
  4. Jan 31, 2017 #3

    David J

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    Gold Member

    Thanks a lot for your help with this
     
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