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Lossless Transmission Line

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data

    A lossless transmission line with

    characteristic impedance Z0 = 50Ω , β = 5∏x10-3,
    and length l =80 meters, is terminated into a load ZL = 80Ω. The transmission line is powered by a source with 120 V and ZG = 12 Ω. Calculate the load voltage.



    2. Relevant equations

    V(z) = V+(e-jβz + [itex]\Gamma[/itex]ejβz

    [itex]\Gamma[/itex] = (ZL - Z0)/(ZL + Z0)

    Zinput = Z0(ZL+Z0tanh(jβL))/(Z0+ZLtanh(jβL))

    V+ = VGZinput/((eL+[itex]\Gamma[/itex]e-jβL)(Zinput+ZG))

    3. The attempt at a solution

    [itex]\Gamma[/itex] = (80-50)/(80+50) = 3/13

    Zinput = 50(80+50tanh(j*.4*pi)/(50+80tanh(j*.4*pi)

    I can't find the input independence because my calculator can't evaluate the imaginary argument in the tanh function. I also have to same problem with solving for V+ and the imaginary exponents in the denominator. Does anyone know a way around this, or am I completely doing the wrong thing here?
     
  2. jcsd
  3. May 30, 2014 #2
    I'd like to add that I'm very, very confused by the concepts, and I'm basically just juggling the equations derived in class.
     
  4. May 30, 2014 #3

    rude man

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    Homework Helper
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    There is a simple equation for E2 for a lossless transmission line, involving only β, l, Z0 and load impedance ZLand E1.

    What is tanh(jx) ) in terms of a non-hyperbolic trig function?

    Unfortunartely for you, E1 is the voltage at the driven end of the line so you have to take care of the source impedance ZG. Hint: ZG forms a voltage divider with Zinput.

    I can't post the equation for E2. Would violate the terms of use of this forum. It can be found in many places.

    BTW the nice way to handle this problem is with ABCD matrices but I take it you haven't had those yet.
     
    Last edited: May 30, 2014
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