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Quarter wavelength confusion

  1. Jan 5, 2015 #1
    Reading up on quarter wave transmission lines and I'm confused about the actual length of the line.

    Is a quarter wavelength line equal to the length of the incident and reflected wave?

    Or is it only equal to the length of the incident wave?
  2. jcsd
  3. Jan 5, 2015 #2


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    The incident wave and the reflected wave will have the same frequency.

    In the same medium, they will travel at the same speed, so they will have the same wavelength.
  4. Jan 5, 2015 #3


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    so --- λ/4 ( in metres) = 300/freq(MHz) /4 x VF ( Velocity Factor) of the transmission line

    300/1296 = 0.23m (free space wavelength)
    0.23/4 = 0.057 (free space λ/4 wavelength)
    0.057 x 0.66 = 0.038m electrical λ/4

    The 0.66 VF I used is just a common velocity factor of a number of coaxial cables ( there are many others)

  5. Jan 7, 2015 #4
    Ok, now I understand that part (thanks btw)..But what about when people say the quarter wavelength transforms an open circuit to a short circuit and vise versa?

    Can someone explain that?
    What I don't understand, say you have an open circuit, a quarter wavelength transforms it into a short circuit....so does this mean the source only sees a short and maximum current flows through the entire circuit?
  6. Jan 7, 2015 #5


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    If you take an electrical 1/4 wave transmission line and short circuit one end, the other end appears as an open circuit when looking into the line. Conversely, if you leave one end open, the other end will appear to be a short circuit.
  7. Jan 8, 2015 #6
    Imagine looking at the start and end of a 1/4 wave line. This corresponds to a phase shift of 90 degrees.
    If there is an impedance mismatch at the 1/4 wave point, it will send back a reflection. By the time reflected wave returns, it will have shifted an additional 90 degrees.
    Now, if the mismatch is an open, then there will be no place for the energy in the line's inductive component (current) to go, and so it will charge the capacitive part forming a voltage that reflects in-phase.
    If the mismatch is a short, the capacitive component of the line will see a short, and the energy will be transferred to the inductive component. This will form a 180 degree out of phase reflection.
    By the time the wave from the mismatch travels back up the 1/4 wavelength, the signal will have traveled a total of 180 degrees (down and back). The open mismatch starts in phase, but appears as 180 degrees out of phase. The short will be 180 degrees out of phase but it will have an additional 180 degrees from the delay of the line for a total of 360 degrees. Thus the returned signal appears as the incoming signal in this case.
    A short appears an open and an open appears as a short.
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