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EM Waves in Transmission lines

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data

    (a) Find the capacitance per unit length and find the inductance per unit length of a transmission wire above a infinite conducting plane. I used the image method

    (b) Show that EM waves can propagate, and find their
    speed and the characteristic impedance of the line.

    (c) Find the characteristic impedance for no reflected power?

    (d) The frequency of incoming wave multiplied by 4, find fraction reflected power and the phase difference between the incident and reflected waves.


    2. Relevant equations

    3. The attempt at a solution

    Part (a)

    Part (b) and (c)
    I got an imaginary value for the impedance, not sure if it's right. It implies there is a ∏/2 phase difference between voltage and current?


    Part (d)

    For part (d), I'm not sure why there would be a phase difference? If it is a short circuit, then there would be a phase change of ∏ (Vr/Vi = -1) if it is an open circuit then it would be completely reflected. What has ramping up the input frequency got to do with the phase difference between incoming and reflected wave?
  2. jcsd
  3. Jan 4, 2014 #2


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    V should be the potential difference between the two actual conductors, not between the wire and its image. Likewise, the flux ##\Phi## should be for a region between the plane and the wire.

    L’ is the ratio of the absolute value of the flux and the current. So, it should be positive. The characteristic impedance should be a real, positive number for a lossless transmission line.
    Think about the condition to have no reflection at the point of connection of the infinite transmission line with the ##\lambda /4## transmission line so that all the power can be delivered to the load. This link might help
  4. Jan 5, 2014 #3
    By the way, the answer to part (c) is [tex]Z_{in} = \frac {Z^2}{Z_L} [/tex], right?

    for [tex] l = λ/4, [/tex] the additional matching network need to be added is:

    [tex] Z_{in} = \frac {V_{(-l)}}{I_{(-l)}} = \frac {V_i e^{jkl} + V_r e^{-jkl} } {V_i e^{jkl} - V_r e^{-jkl} - V_r e^{-jkl} }Z = \frac { (Z_L + Z)e^{jkl} + (Z_L - Z)e^{-jkl} } {(Z_L + Z) e^{jkl} - (Z_L - Z)e^{-jkl} } Z [/tex]

    For [tex] l = λ/4 [/tex] , [tex] Z_{in} = \frac {Z^2}{Z_L} [/tex]

    For part (d):
    As the input frequency is increased by 4 the input wavelength divides by 4, now it would imply that the matching line length becomes [tex] \frac {λ}{4} → λ [/tex]

    This results in a mismatch, causing a reflected wave. I have found that the new input impedance [tex] Z'_{in} = 0[/tex]

    This is equivalent to a short circuit, and I have found the new [tex] \frac {V_r}{V_i} = -1 [/tex] so the phase difference is ∏.. and all power is reflected!
    Last edited: Jan 5, 2014
  5. Jan 5, 2014 #4


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    Yes. You can simplify this using the relationship given between Z and ZL

    OK, but some typing errors appear in the denominator of the middle expression.

    The length of the matching line becomes λ.

    I get a different answer. Can you show your calculation?
  6. Jan 5, 2014 #5
    When [tex]l = λ, e^{jkl} = e^{j \frac{2∏}{λ} λ} = 0[/tex] so [tex] Z_{in} = \frac {0}{0} Z [/tex]
  7. Jan 5, 2014 #6


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    What is the value of ej2##\pi## ?

    Note e is never zero. It's a point on the unit circle in the complex plane.
  8. Jan 5, 2014 #7
    latest calculation below.
    Last edited: Jan 5, 2014
  9. Jan 5, 2014 #8


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    I don't get that result for Zin.
  10. Jan 5, 2014 #9
    Sorry, I got [tex] Z_{in} = Z_L [/tex]

    So on the left side, we have impedance of [tex] \frac {1}{4} Z_L [/tex] and on the right side we have impedance of [tex] Z_L [/tex].

    So [tex] \frac {V_r}{V_i} = + \frac {3}{5} [/tex] (No phase difference)

    So proportion of power reflected = [tex] (\frac {V_r}{V_i})^2 = \frac {9}{25} [/tex]

    As for phase difference, I don't think there's a phase difference? it should be zero. (Anyway all cases include 0 or ∏)
    Last edited: Jan 5, 2014
  11. Jan 5, 2014 #10


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    I think that's right.
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