1/2 Wave center fed resonant dipole --- why is the impedance 72 Ohms

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by K9AXN, May 25, 2019.

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  1. K9AXN

    K9AXN Premium Subscriber QRZ Page


    This thread will speak to the antenna internals used to explain the reasoning that a 1/2 wavelength, electrically resonant, center fed dipole will present a 72 Ohm feed point impedance. It will be followed by the same process for the end fed wire and why it presents a 5000 Ohm impedance.

    Fact 1: The surge impedance of the wire is the foundation that wire antennas are anchored.

    Fact 2: A Sin shaped wave in a wire will loose approximately 12% of its Voltage and Current per wave length.

    Fact 3: Surge/Displacement current will be non-reactive even though it is composed of inductance and capacity. It is always accelerating to the speed of light.

    Fact 4: The Q due to radiation of a 1/2 wave dipole is approximately 8.3333

    Fact 5: The end of a wire that has no other connection will have a surge impedance of approximately 600 Ohms until the incident wave reaches the end and returns to the source end.

    Fact 6: Electron flow and displacement/Surge current can be intuitively understood using the Newtons cradle.

    Fact 7: The combined unsigned voltage and current in a resonant wire will the same anywhere on the wire.

    Important Concepts 1: If you are 5000 feet from a person that raises his hand, it will be 5000 ns before you know that he has done so. If you apply a voltage to the end of a wire, the source cannot know what's at the other end until a reflection does or does not return. The 600 Ohm surge impedance is the limiting factor until the reflection occurs or no response is seen due to matched 600 Ohm load.

    For now, please do not challenge these assertions until the proofs are posted.

    Next post will begin with the cycle by cycle spin up of the antenna using a 100 watt source, 72 Ohm twin lead and 1/2 wave, electrically resonant, wire in space.

    Regards Jim K9AXN
  2. AA5CT

    AA5CT Ham Member QRZ Page

    Axiomatic fact.

    And unless you're prepared to utilize FDTD (Finite-difference time-domain method) methods to (help) analyze the physics, I'm out.

    (Too many other productive things to do, and its TOO EASY to run off into a ditch without some guardrails.)

    Besides, you won't address other oddities seen in antennas that allow 'funny' designs to radiate efficiently, so your "rules" won't be universally applicable anyway. I want to see Maxwell adhered to, without any new physics involved as well.

  3. AG6QR

    AG6QR Subscriber QRZ Page

    Either these two don't mean what it sounds like they mean, or the axioms are self-contradictory.
    AH7I likes this.
  4. K9AXN

    K9AXN Premium Subscriber QRZ Page

    Thanks both WPA and QR for the responses but I will continue with the proofs.
    For WPA, I could use your expertise to move this along. Please explain the logic and reflections that result when a 72 Ohm twin lead is connected to a 1/2 wave length center fed resonant diploe. Do this cycle by cycle using a string of equal amplitude sin waves and do not use any of the common equations not even Algebra --- words than anyone can understand.
    Feed the wire with 100 Watts. If you have trouble let me know and we will work through it. If you insist on leaving I will understand.
    Regards Jim
    Last edited: May 25, 2019
  5. W9WQA

    W9WQA Ham Member QRZ Page

    new peeing match...stay "tuned"...i will.
  6. WR2E

    WR2E Ham Member QRZ Page

    I'm not going to let a little urine stand in the way of learning something! ;)
    K3GM, K9AXN and WQ4G like this.
  7. AH7I

    AH7I Ham Member QRZ Page

  8. AH7I

    AH7I Ham Member QRZ Page

    Can you post the proofs prior to "spin up of the antenna" ?
  9. KM3F

    KM3F Ham Member QRZ Page

    The height above earth is being ignored.
    The reflection from earth does a Phase Addition Subtraction along the element lengths to arrive at the feed impedance of a half wave dipole..
    That is why there is a height vs impedance graph.
    It affect pattern as well.
    Why go through all the other mechanics?
    Otherwise, yes the impedance is about 72 ohms at resonance near a half wave above earth.
    To match as 50 ohms, the frequency has to be changed or the element lengths changed to arrive at 50 ohms per the frequency, assuming the height has not changed.
    K3XR likes this.
  10. KX0Z

    KX0Z Ham Member QRZ Page

    On to the next thread

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