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Transmission Line - Solve for frequency subject to phase angle

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A transmission line has the following properties:

    [tex]L_{0} = 1 mHm^{-1}[/tex]
    [tex]C_{0} = 10 \mu F m^{-1}[/tex]
    [tex]R_{0} = 10 \Omega m^{-1}[/tex]
    [tex]G_{0} = 0 \Omega^{-1} m^{-1} [/tex]

    That is, inductance per meter, capacitance per metre etc. The line is 10m long.

    The problem is to find the angular frequency ([tex]\omega[/tex]) such that the propagation constant ([tex]\gamma[/tex]) and characteristic impedance ([tex]Z_{0}[/tex]) have phase angle [tex]\frac{-\pi}{6}[/tex]

    2. Relevant equations
    [tex]\gamma^{2} = (R_{0} + i\omega L_{0})(G_{0} + i\omega C_{0}) \cdots (1)[/tex]
    [tex]Z_{0} = \sqrt{\frac{R_{0} + i\omega L_{0}}{G_{0} + i\omega C_{0}}} \cdots (2)[/tex]

    and geometric representation of complex numbers will come into it.

    3. The attempt at a solution

    I worked out that the ratio of the imaginary part of the propagation constant and the real part is -sqrt(3). Ie;

    [tex] \gamma = \alpha + \beta i[/tex]
    [tex] \frac{\beta}{\alpha} = -\sqrt{3}[/tex]

    Then went to find an equation for omega from eqn (1) above by substituting in known values and taking the square root of both sides. However, you end up with a pretty hideous equation with omega distributed through it everywhere and no way to solve for it, because to take the square root you must put the complex number in polar form. I ran into the same problem with solving for characteristic impedance.

    It's obvious I'm approaching this from the wrong way, any suggestions?
    Last edited: Jun 18, 2009
  2. jcsd
  3. Jun 19, 2009 #2
    Okay, here's how much further I've got with it.

    [tex] \gamma^{2} = (R + i\omega L)(i \omega C)[/tex]

    [tex] \gamma^{2} = R C \omega i - \omega^{2} L C [/tex]

    [tex] \gamma^{2} = \sqrt{R^{2} C^{2} \omega^{2} + L^{2} C^{2} \omega^{4}} e^{(\pi - arctan(\frac{R}{L \omega}))i}[/tex]

    [tex] \gamma = (R^{2} C^{2} \omega^{2} + L^{2} C^{2} \omega^{4})^{\frac{1}{4}} e^{0.5(\pi - arctan(\frac{R}{L \omega})) i}[/tex]

    So the phase angle is all the stuff before the i in the exponent of e. If it's equal to -pi/6 the following must be true:

    [tex] 0.5(\pi - arctan(\frac{R}{L \omega})) = \frac{- \pi}{6}[/tex]

    re arranging it all:

    [tex]\omega = \frac{R}{L tan(\frac{4 \pi}{3})} [/tex]

    which is all well and good, but the calculated value of omega (5774 rad per second) does not give the right phase angle when substituted back into the original formula.

    I think the problem is that the units in the argument of the arctan aren't right.

    Any help?
  4. Jun 19, 2009 #3


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    Hi Andrusko! :smile:

    Why the π ?

    Why not just [tex] \gamma^{2} = \sqrt{R^{2} C^{2} \omega^{2} + L^{2} C^{2} \omega^{4}} e^{(arctan(\frac{-R}{L \omega}))i}[/tex] ?
  5. Jun 19, 2009 #4
    Oh good point. I figured the vector representing gamma would be sitting in the 90 - 180 degrees of the unit circle... but it's not necessarily. That is sneaky...
  6. Jun 21, 2009 #5
    It's still not working. Changing the arctan argument yields the same answer.

    Buggered if I know.
  7. Jun 21, 2009 #6


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    Yes, 60º instead of 240º … no real difference. :redface:

    hmm … looking back, I'm not clear as to what the original question was asking …
    (btw, you can get capital gamma by typing \Gamma instead of \gamma :wink:)

    I don't see how Gamma and Z0 can have the same phase angle … so what does the question actually mean? :confused:
  8. Jun 21, 2009 #7


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    1. Isn't G the conductance? If so, how can G be 0? Doesn't that mean that the transmission line doesn't conduct at all? Also, shouldn't G be the reciprocal of R? (G = 1/R). Was that just a typo? That will change your equations a lot.

    2. I think tiny-tim is right. If you look at your equations as given, they can be written like this:


    A = R + iωL = aexp(),

    B = G + iωC = bexp()

    Then we see that

    γ = (AB)½ = (ab)½exp(i(α + β)/2)


    Z = (A/B)½ = (a/b)½exp(i(α - β)/2)

    The only way the arguments of these two complex numbers can be the same is if

    β = -ββ = 0

    However, since β = arctan(ωC/G), this can only be true if ω = 0.
    Last edited: Jun 21, 2009
  9. Jun 21, 2009 #8
    Hey thanks for the replies.

    The conductance (G) was deemed negligible in the question. I think its a 2 part question. Ie; solve for frequency for the phase angle of gamma and then separately solve for frequency for the phase angle of impedance, not that they are both pi/6 at the same time. Or otherwise it's quite trivial (omega = 0).

    The only thing I can think of is that the argument of the arctan isn't unitless. Wolfram alpha reports ohms/henries*angular frequency as a spatial angular frequency, that is units of angle^(-1). Surely that must be where the mistake is. But I'm at a loss to know what to do about it.

    I went and asked my lecturer about this today and he couldn't get it out, and he wrote the question. Another piece of evidence to suggest he is crazy.
  10. Jun 21, 2009 #9


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    Let me repeat myself:

  11. Jun 22, 2009 #10


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    (I still don't understand the question. :redface:)

    There's some detailed calculations at http://en.wikipedia.org/wiki/Transmission_line#Telegrapher.27s_equations … do they help? :smile:
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