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How do you relay EM wave parameters to Transmission line parameters?

  1. Mar 4, 2009 #1
    I am studying EM wave and transmission lines. I see both derive equations for propagation constant [tex]\gamma[/tex]:

    Plane wave velocity is 1/[tex]\sqrt{\mu\epsilon}[/tex] and [tex]\eta[/tex] = [tex]\sqrt{\mu/\epsilon}[/tex]

    Transmission line velocity is 1/[tex]\sqrt{LC}[/tex] and Z0=[tex]\sqrt{L/C}[/tex].

    From that the book just to say the velocity of both are the same and [tex]\mu[/tex] [tex]\epsilon[/tex] = LC

    I see they both are propagation constant, but I don't see they are the same!! Can anyone explain to me how they relate together?

  2. jcsd
  3. Mar 5, 2009 #2
    The first pair of equations is the physics representation, using the permeability of the medium per unit length and the permitivity per unit length. The latter two equations are the engineering equivalents, for example in a coaxial transmission line, where L and C are the inductance per unit length and capacitance per unit length. As an exercise, calculate L and C for RG-8 cable, with the velocity of propagation equal to about 2/3 the velocity of light, and Z = 50 ohms (impedance of free space = 377 ohms).
  4. Mar 5, 2009 #3
    Thanks for the reply.

    I know the result is true, I am not question the validity of the equation, just I cannot find the physics to link the two. I can derive the formulas of the transmission line and plane wave and arrive the equations in the book. Just that the book simply claim the two propagation constants are the same........WHY? What is the equation linking L,C of unit length to [tex]\epsilon[/tex] and [tex]\mu[/tex]?

    Can you point me to material that link the two? Not just the Helmholtz's equation or the general wave equations that give the propagation constant.
  5. Mar 5, 2009 #4
    Take your derivations for L and C per unit length, form the ratios and products as per the expressions for the propagation velocity and impedance, and cancel out all the 2 pi's etc. You should get the physics expressions.
  6. Mar 19, 2009 #5
    Thanks, I think I got it.
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