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Standing Wave pattern on an open circuited transmission line with length λ/4

  1. May 19, 2010 #1
    Here's a scenario about which I have doubt. Consider a transmission line which has a length λ/4 and which has open circuit at load end. The source resistance is zero. The source voltage be Vs and the source frequency is sufficiently high to be considered as high frequency application . The line is perfectly loss-less (sorry for considering an ideal situation, I just am asking this to clear my basic concepts). My question is what will be the voltage standing wave patterns on the line ?

    Now as we know the open circuit at load end will give us ΓL (which is reflection coefficient at load end) = 1. So the voltage standing wave has voltage maxima , Vmax = 2|V+| and voltage minima , Vmin= 0 . where |V+| is the magnitude of the forward travelling wave. But interestingly in this scenario the voltage maxima occurs at load end and thus the voltage minima occurs at source end (as the length is λ/4) . So because of this Vmin cant be zero but equals a finite quantity i.e the source voltage Vs . Also this leads to Vmax and |V+|being equal to infinite . So my doubt is what would be the voltage at other points on the standing wave. I personally think it is infinite at all pts except at the source end where it is Vs.

    Please click on image to see it in original size

    I tried to simulate the situation on this http://www.amanogawa.com/archive/StandingWavePattern1/StandingWavePattern1-2.html" [Broken] but since it can't deal with infinities my doubt remained unsolved.On the applet ,|V+| I got was a very large number instead of getting infinity.

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    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 19, 2010 #2


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    If you assume everything is perfect, then you will get crazy results.
    In reality, the input impedance will not be zero (if it was, how would you feed it?) and there will not be infinite voltage at the other end.
    Computers can't deal with infinity.

    The shape of the actual waveform will be like the first quarter wave on a sinewave. Starting at zero (or very small) at the input end of the transmission line, and rising to a maximum (but not infinite) at the open circuited end.

    Have a look on Google for the program RJELINE2 by Reg Edwards. This calculates realistic values for transmission lines.
    One good site is this one:
    It carries most or all of the wonderful programs by the late Reg Edwards.

    The general formula for this is :
    line impedance = Square root of ( load impedance * input impedance)
    (this is the formula for a geometric mean)

    Input impedance = Line impedance * line impedance / load impedance

    So for a 100 ohm line and a load of 600 ohms, the input impedance will be:
    100 * 100 / 600 or 16.7 ohms.

    So, if there was 100 watts entering the line,
    the voltage at the input would be SQRT( 100 * 16.667) or 40.8 volts
    At the load end, the voltage would be SQRT( 100 * 600) or 244 volts.

    This is quite a step up in voltage but nowhere near infinite.
    Last edited: May 19, 2010
  4. May 20, 2010 #3
    I agree with the previous poster; you need to excite the quarter wave line with a source and series resistor that matches the impedance of the line. Then you'll get zero volts at the feedpoint.
  5. May 20, 2010 #4
    Thanks vk6kro for solving my doubt. I wanted to confirm whether the scenario will lead to crazy results as you say.

    I guess , as you say there will be either some source resistance or some input impedance or some loss in the line so we won't get infinities in practice.

    In fact , I can't fathom a situation in physics where actual infinities are encountered.But that isn't pertinent to this thread:biggrin:. My doubt was in relation to the particular scenario which you guys have addressed well
    Last edited: May 20, 2010
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