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Transmission Line (simple Prob)

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Incase formatting of numerical answers below is weird, please see attached picture of question.

    "Consider a distortionless transmission line with impedance 50 Ω, capacitance 0.1
    nF/m and attenuation 0.01 dB/m. Find:
    (a) the resistance, inductance and conductance of the line
    (b) the wave propagation velocity in the line
    (c) the relative amplitude of a voltage after it has traveled a distance of 2 km in
    the line.
    [Ans. (a) R = 0.057 Ω/m, L = 2.5 x 10
    -7
    H/m, G = 2.3 x 10
    -5
    S/m; (b) v = 2 x 10
    8
    m/s;
    (c) 10%]."

    2. Relevant equations

    Zo = SQRT(L/C)

    For distortionless line

    R/L = G/C

    3. The attempt at a solution

    Part b) is easy (from u = 1/Sqrt(LC))

    However, I assume Zo = 50 ohms, as it lets me find L = 502*0.1n = 2.5E-7

    But then what the devil do they mean in the numerical answer given for R = 0.057ohm/m?
    and what is with the conductance G? that isn't 1/any R value. Is that conductance and R wrong?

    More pressingly though, my log is really rusty! this is where I'm most confused, I've been messing around with e^-attenuationconstant*distance and a loss = 0.01dB = 20 log(V1/V0) formula but I'm not sure if that's the right approach as I'm not getting anywhere?

    Could anyone offer an opinion and if possible show me how the loss is 10% over 2km?

    Thanks heaps!
     

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    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2

    rude man

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    If you're given attenuation you can immediately get R since you know Z_zero.
    Then, what relates R,L,C and G in a distortionless line?

    Knowing just L and C enables you to find the phase (wave propagation) velocity.
    Part (c) should be evident from the attenuation number. You know the attenuation constant and the line length ...
     
  4. Apr 27, 2014 #3
    So as I said, R = Zo right?
    Well why is that R = 0.057 ohm/m and not 50ohm/m? In the numerical solution (in picture)
    And G, what R is that the inverse of?


    Thanks
     
    Last edited: Apr 27, 2014
  5. Apr 27, 2014 #4

    rude man

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    Because R = α Ro, not Ro, where Ro = Re{Zo}.

    G is not the inverse of anything significant. G is the conductive leakage per unit length of one conductor to the other. Units are Siemens/m.

    BTW what are the units of α in the above expression?
     
  6. Apr 27, 2014 #5
    Oh, R = α Ro; Alpha is Neper/m, which is 0.115129dB*0.01*50

    that makes sense, I'm really confused as to why I'm only learning that is what R is now and why I've never seen that written anywhere before.

    So conductance is RC/L.

    Thanks, you really cleared that up!
     
  7. Apr 27, 2014 #6

    rude man

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    You're on top of it now!
     
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