Question about a dual end fed antenna - possible?

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by W4LLZ, Aug 5, 2020.

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  1. KM1H

    KM1H Ham Member QRZ Page

    Mike and Rich, you will never win a word battle with AXN and he is impervious to ever admitting he is wrong. Think a new version of AG6K
    If he is ignored long enough he fades away to maybe create another web page laugher.
     
    AK5B likes this.
  2. WA7ARK

    WA7ARK Ham Member QRZ Page

    I'm a slow learner...

    Best I can hope for is that other folks learn something from my postings..., then look at his, and draw their own conclusions...
     
  3. K9AXN

    K9AXN Premium Subscriber QRZ Page

    Mike, I forgot about this thread.
    First, my compliments for being one of the few that are truly qualified to use the these tools on QRZ or EHAM. I believe that we can carry on a reasonable and friendly dialog if it is limited to two people so I will not respond to any other people on this thread. I hope you will be willing to help me through this project. We may disagree on some logic and principals but believe we can approach it with open minds.

    I'm sure you know that I do not use any of the Electronic CAD tools but do not discourage their use in any way. I would like to pursue this issue, through to the explanation of the logical reasoning surrounding the feed point impedance of a center fed electrically resonant 1/2λ wire, end fed electrically Anti-resonant 1/2λ wire, then center fed 1λ electrically Anti-resonant wire. Then we can compare results.

    First, to answer the question above please download the schematic at http://k9axn.com/attachments/Finished_3_jpg_final.jpg

    Short Photo of center of voltage divider http://k9axn.com/attachments/Single_wire_good_1.JPG

    Long photo of voltage divider http://k9axn.com/attachments/Single_wire_good_2.JPG

    Video clip of adjusting the voltage divider to measure the Z0 of the wire http://k9axn.com/attachments/Single_wire_good.avi

    Download them and let me know, then we can refer to them and explain the reasoning and proofs for this part.

    The measurements and proofs will be the Z0 of a 54 ft wire, Capacitance, Inductance, Velocity factor, and approximate loss due to radiation using a DC pulse with a 20ns rise time followed by 95ns steady state.

    Thanks and hope we can make this work.

    Regards Jim K9AXN
     
  4. WA7ARK

    WA7ARK Ham Member QRZ Page

    Congratulations. You have re-invented a TDR...

    Velocity of propagation of a pulse along a wire is ~1.07ns per foot. It would take the pulse 54ft*1.07ns/ft = 58 ns to reach the far end, reflect and another 58 ns to arrive back at the driven end, at which point the reflection causes the step in voltage at 2x58ns = 116ns.

    What does this have to do with antennas?
     
  5. W9XMT

    W9XMT Ham Member QRZ Page

    The spectral components of a baseband pulse train having rise- and fall-times of 20 nanoseconds will simultaneously occupy a bandwidth of nearly O Hz to about 175 MHz. Defining such a pulse train to be "DC" is a totally inept description for it.

    In any case, it is a very poor choice as a source to determine/measure the operating characteristics of the radiating conductor(s) of a practical antenna system.
     
  6. K9AXN

    K9AXN Premium Subscriber QRZ Page

    My compliments to you for having copied a canned measurement regarding Velocity factor.
    IT was measured using this experiment --- C= 109.8039632ns for a travel of 108 ft and 115.75ns measured; for a velocity factor of 94.863%

    The Velocity factor is one of the properties used to calculate the Capacitance and Inductance (Actually acceptance and Permeability) of a wire or transmission line!

    --------------------------------------------------------------------------------------------------------------------------------------------------------------------
    The Capacitance of the Wire:
    The voltage divider resistors are varied until the steady state line is centered at 4 Volts. The resistance is then measured to be approximately 600 Ohms, (4 ÷ 600 = .006666A) flowing into the wire until it returns to the source, where it increases to 8V because current ceases. Ignoring the loss due to radiation and resistance, the wire is now fully charged to 8V @ .006666A for .000000115.75ns for 6.1732716-o9 Joules.

    Calculate the Capacitance:
    First: Calculate the speed that the wave travels through the wire 299792458 • .94863 = 284392119 Meters per second.
    Second: Calculate the current/sec that enters the wire with 1 Volt applied. .0066666A ÷ 4 = .00166665A
    Third: Divide the current --- .00166665 by 284392119 = 5.860394465pF is the capacity/Meter of the wire

    Calculate the Inductance:
    First, the measured speed. 284392119 Meters/sec
    Second: The Z0 of the wire --- 600 Ohms
    Third: Divide the Z0 of the wire by 284392119 = 2109.763105 nH

    Now prove the calculations:

    √(2109.763105-6 ÷ 5.86039465-12) = 600 Ohms
    1/√(2109.763105-6 x 5.86039465 -12) = 284393536 Meters/sec

    Important: The Capacitance and Inductance are actually Acceptance and Permeability both of which are real properties NOT imaginary --- the same as the Impedance of space..

    I trust you will not continue with sarcastic comments.

    FYI, two things are needed to calculate antenna properties ignoring the resistance and environmental properties are the Z0 of the wire and Q due to radiation.

    Another comment: EZNEC did not state in their limitations statement that they do not support END FED because of the notion that the impedance at the end was Infinite. That is gossip.

    There's a lot more to these gossip notions for later.

    Regards Jim
     
  7. WA7ARK

    WA7ARK Ham Member QRZ Page

    It is circuit theory (Kirchhoff) that makes it impossible to drive a current into the end of a wire. That has nothing to do with NEC or even antenna theory. The forcing function (source) has two terminals; if one terminal is connected to the wire, what do you connect the other terminal to??? NEC understands Kirchhoff, so if you try to build a model like that it just says that the feedpoint impedance is infinite, the current is zero, the radiation is zero, no antenna!

    If you take a free-space wire, cut it into two pieces, you can insert the source (voltage or current) between the two wire segments because now both terminals of the source have something to connect to. If you cut a half-wavelength long wire into two equal pieces and insert a source, we (antenna designers) call that a "center-fed-dipole". If you cut a wire at ~10%/90%, it radiates just like the CFdipole, but it has a different feed-point impedance than the CFdipole, and I call it a "near-end-fed-half-wave". I have yet to see anyone build a true end-fed antenna..., it is a physical impossibility.

    Free-space antenna wires do not behave like your TDR wire because your generator is effectively connected to an infinite ground plane. Now, if we take your 54ft wire, and model it in NEC using a swept RF voltage generator where one terminal of the generator is connected to a "perfect ground plane", the other generator terminal is connected to the 54ft wire, and the wire is oriented at right angles to the ground plane, we find that the wire takes power from the generator (radiation) and that power varies with frequency (feedpoint impedance).

    If we measure the phasor current and voltage into the wire, we see that the ratio of V to I (Complex Impedance) also varies with frequency. That is a Marconi antenna... If your TDR experiment says anything relevant about antennas, which I doubt, then it is mimicing a Marconi (monopole driven against a ground), not a wire in free space...
     
    Last edited: Sep 8, 2020
  8. K9AXN

    K9AXN Premium Subscriber QRZ Page

    If you look at the schematic, it uses the charge pool (Capacity and neutral ground charge pool to add or take charge from the signal generator. That is a moot point. You and others have said that DC steady state current cannot flow into a single unterminated wire ---- you are mistaken!

    Well you just saw 4V @.006666A flow into that wire for 115.75ns. You say, it needs a return path because it has but one connection. What is that return path? How can it become charged to 6.1732716-o9 Joules?

    Once it is charged, it can be dis-connected and it will retain the charge. If you should touch that wire to another wire, current will flow and the charge will equalize.

    Your speaking to closed circuit conditions and totally ignoring relativity. If a charged object is attached to a wire that is not terminated, what impedance does it present to the charged wire? 600 Ohms real, and the 600 Ohms does not represent a loss, they represent Admittance and Permeability, only the resistance of the wire is a loss.

    I disagree with your understanding regarding EZNEC. It uses Maxwell's time varying current equations. Those equations require the end opposite the Perfect ground to be near infinite impedance for any accuracy. The equations would be (P= I2 x R) . If the Maxwell equation used for the (time varying acceleration of charge) was used, the voltage at the max voltage end would be used for calculations (P=E2 /R).

    Something to think about! In your post #34 there is a rendition speaking to a 1wl center fed resonant wire. What you did is take a 1/2wl resonant center fed wire and stretch it out to 1wl creating a pair of end fed 1/2wl wires. Interestingly the numbers that EZNEC rendered are nearly exactly those for a 1/2wl end fed wire, not 1wl. What does that mean? Now add the beam shapes for a horizontal version. Looks to be approximately 30 degree narrower and has 2dB of gain over a dipole. That contradicts common thinking that the ends of a dipole radiate very little.

    WERE THERE ANY ERRORS IN THE EARLIER CALCULATIONS? We have now covered the measurement of the Velocity factor, Capacitance, and Inductance of a wire using a DC pulse.

    Next, what part the 20ns rise time represents and how it is used to calculate the radiation resistance.

    Regards Jim K9AXN
     
  9. W9XMT

    W9XMT Ham Member QRZ Page

    If the above posit of K9AXN is true, he/anyone would be able to get a flashlight bulb to illuminate with only one of its terminals connected to a DC power source.
    How can a single value for the radiation resistance of a fixed-length conductor be determined by applying a waveform having spectral components ranging from nearly O Hz to about 175 MHz?

    The radiation resistance of such a conductor depends on its physical dimensions in terms of wavelengths at the operating frequency — which value is NOT constant for a "DC" amplitude transition time of 20 nanoseconds.
     
    Last edited: Sep 9, 2020
  10. K9AXN

    K9AXN Premium Subscriber QRZ Page

    Did you read this in that post???? "Next, what part the 20ns rise time represents and how it is used to calculate the radiation resistance". I will increase the size of the font to make it more readable --- thanks for letting me know the difficulty you are having reading the material.

    Regards Jim K9AXN

     

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