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Transmission Line derivation

  1. Sep 24, 2011 #1
    Eqn. (2.1)
    [tex]\frac{\partial v}{\partial z} = -L \frac{\partial i}{\partial t}[/tex]
    Eqn. (2.2)
    [tex]\frac{\partial i}{\partial z} = -C \frac{\partial v}{\partial t}[/tex]
    Eqn. (2.5)
    [tex]v(z,t)=V^+f(t-\frac{z}{v_p})+V^-f(t+\frac{z}{v_p})[/tex]

    From the book I'm reading:
    "By substituting Equation (2.5) into Equations (2.1) and (2.2), we
    determine that the current has the form:"
    [tex]i(z,t)=\frac{V^+}{Z_0}f(t-\frac{z}{v_p})+\frac{V^-}{Z_0}f(t+\frac{z}{v_p})[/tex]

    So, I've seen this derivation about four different ways, and I know this is the result, but I can't see what moves I need to substitute 2.5 into 2.1 and 2.2 to get 2.5. Can someone please enlighten me? Thanks all!
     
  2. jcsd
  3. Sep 25, 2011 #2

    uart

    User Avatar
    Science Advisor

    Just some preliminaries first. If you differentiate 2.1 wrt "t" and 2.2 wrt "z" (and then repeat the other way around) you get the following second order PDE's.

    [tex]\frac{\partial^2 v}{\partial z^2} = L C \frac{\partial^2 v}{\partial t^2}[/tex]

    [tex]\frac{\partial^2 i}{\partial z^2} = L C \frac{\partial^2 i}{\partial t^2}[/tex]

    These separable PDE's give the motivation for the form given in 2.5 and tell us that the phase velocity is:

    [tex]v_p = \frac{1}{\sqrt{L C}}[/tex]


    Since the above second order PDE's tell us that i(z,t) will be in the same form as v(z,t) we can therefore write:

    [tex]i(z,t)=I^+ \, f(t-\frac{z}{v_p})+I^- \, f(t+\frac{z}{v_p})[/tex]

    By differentiating the above wrt "t" and differentiating 2.5 wrt "z" and substituting into 2.1 we can equate coefficients to show the following:

    [tex]I+ = \frac{V^+}{L v_p} = \frac{V^+}{Z_0}[/tex]

    [tex]I- = - \frac{V^-}{L v_p} = - \frac{V^-}{Z_0}[/tex]

    Hope that helps.
     
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