reflection coefficient +1 and -1 for open and short

Discussion in 'Wireless & RF Design' started by donut, Jun 19, 2012.

  1. donut

    Thread Starter Member

    May 23, 2012
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    could someone please explain to me why reflection coefficient of +1 is the result of a open circuit and why reflection coefficient of -1 is the result of a short circuit? I am reading that it has to do with the voltage magnitude at different locations along the transmission line but it is still not clear to me. The attached files come from the book that explains short termination and open circuit termination in respects to magnitudes of voltages.
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    There are a number of ways to answer. The rigorous way is to apply boundary conditions formally, but I'm guessing you want a simple circuit theory explanation that makes intuitive sense.

    I like the idea of thinking about turning on a DC voltage source and driving a transmission line, with the end terminated a long distance away. Assume the voltage source gives a voltage of V when the load is open. Let's say the supply has a 50 ohm source impedance, and you connect a 50 ohm transmission line. Consider 3 cases as follows.

    If the line termination is a 50 ohm resistor, you expect no reflection and the voltage at the power supply will be V/2, where V is the applied voltage. Why? Because it's like a voltage divider with 50 ohms source impedance and 50 ohm load impedance.

    If the line is shorted at the far end, then initially the voltage at the supply will be V/2, because there is no time for the reflection to go and come back. Hence, the power supply thinks you have a nicely matched termination. However, once the reflection does go and come back, you expect the voltage to go to zero because you have a short. This means that the reflected voltage would need to be a negative value with equal magnitude (R=-1) compared to the incident voltage so that they will sum to zero in steady state.

    If the line is open at the far end, then initially the voltage at the supply will be V/2, because there is no time for the reflection to go and come back. Hence, the power supply thinks you have a nicely matched termination. However, once the reflection does go and come back, you expect the voltage to go to V because you have an open load and no loading of the supply. This means that the reflected voltage would need to be a positive value with equal magnitude (R=+1) compared to the incident voltage so that they will sum to double the value in steady state.
     
    Last edited: Jun 19, 2012
    t_n_k likes this.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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