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. K9AXN

    K9AXN Premium Subscriber QRZ Page

    KB7WG: Seems your getting the point far better than XMT or ARK.

    For XMT and ARK,

    Please go back and read what I posted.

    K9AXN: Well of course it does ---- on one .5λ wire. We are talking about a pair of .5λ wires combined to be a center fed 1λ dipole. All of the calculations found on post #19 are for a pair of .5λ wires combined to be a 1λ dipole. Check to see what NEC says about 500 watts to each of the .5λ halves. You may be surprised to find your same error result because NEC configures it as a .5λ radiating wire and a .5λ non radiating counterpoise which would render the same error.

    For XMT and ARK: There was a limitation statement in the EZNEC doc at some point that suggested they did not support a truly end fed 1/2wl wire. They however, support an OCFD wire using two connections. The favored point on the OCFD wire is at the 2500 Ohm point which is used almost universally for OCFD antennas. This incorrectly suggests that the impedance of a truly end fed 1/2wl wire is 2500 Ohms. The impedance at the very end of a 1/2wl wire (Anti-resonant point) is 5000 Ohms and was proven in post #19.

    Please point to whatever error that you find and test the translations for reciprocity in post #19. Then get back.

    There two possibilities for your conclusions that the impedance of a 1wl center fed wire is 5000 Ohms. EZNEC has a problem with feeding the anti-resonant point on a wire and assumes it to be 2500 Ohms or you have not discovered how to use the tool properly.

    For XMT or ARK, Would either of you like to a crack at explaining how the impedance of an anti-resonant 1/2wl wire becomes 5000 ohms or 2500 using your theory --- cycle by cycle? If not stand by and let me finish the logic macros for those who might wish to develop an intuitive understanding of transmission and antenna internals.

    Regards Jim
  2. KB7WG

    KB7WG Ham Member QRZ Page

    I must apologize. I forget where I am. Of course most here are concerned with performance.

    My only interest is in the true charge mechanics. After that, the rest is easy.

    New emission and detection methods are my present study of interest. The shortened piezo radiators and direct antenna modulation has interesting possibilities. And I believe some interesting revelations.

    What would you think of just a small area "pole" for emission, without any concern for physical wavelength........and maybe even direct antenna demodulation? Just saying.

    Review and add those studies to your present antenna understanding. Very interesting.

    Amazing future in piezo. Back to crystal rocks.
  3. W9XMT

    W9XMT Ham Member QRZ Page

    There is only one reason for a conclusion that the feedpoint impedance of a center-fed, 1WL horizontal dipole over average Earth soil is in the region of 5,000 Ω — because that is what Physics tells us. The graphic below shows this reality.

    K9AXN: This time please NOTE that the vertical line in the center panel below is the Z axis of the 3D space occupied by the wire model. The top of the Z axis marker happened to extend nearly all the way to the horizontal conductor in this particular 3D layout sketch, but it is graphical only, and no part of the antenna system.

    If that vertical line was a physical conductor, its color would be black, like the horizontal conductor there.

  4. WA7ARK

    WA7ARK Ham Member QRZ Page

    Negative Jim. A dipole is a dipole is a dipole, regardless of how long it is....

    Start with a 1/2wl wire in free space, insert a RF source right at the wire center. This is a classic textbook dipole such as is shown in every antenna theory book I have ever studied. From an antenna theory standpoint, I see nothing wrong with the Wikipedia entry for "Dipole Antenna", which summarizes dipole theory quite well. It agrees with every antenna textbook (Kraus, Terman, Balanis, and Jascik) I have ever studied (before antenna simulation existed).

    Theory says that the feed impedance of a resonant 0.5 wl dipole in free space should be 73.1 Ohms. What does NEC say?
    This is for a center-fed, 10m long wire in free space. The calculus shown in Wiki says ~73.1 Ohms, NEC says 73.02 Ohms at 14.546MHz, so I would say that NEC did quite well. 1wl at 14.546MHz is 20.61m, so NEC actually showed us what every ham that has ever trimmed a dipole knows; that is that at resonance the dipole is slightly shorter than 0.5wl, in this case 10m/20.61m = 0.485wl (remember 468/f???).

    I ask EzNec to show us the current distribution along the wire:
    The green line is the dipole. The vertical distance from the green dots to the magenta line is proportional to the magnitude of the standing current wave along the dipole. Compare that to what is shown in the Wiki!

    So let us double the length of the wire from 10m to 20m (0.5wl to 1wl), and ask NEC about resonance, impedance, and current distribution:
    First, notice that the resonant frequency shifted lower. Notice that the voltage/current required to drive 100W into the 1wl dipole is radically different than when it was only 0.5wl long... Finally, here is the current distribution on the longer wire:

    With the 1wl wire, the current at the feedpoint is only 140mA (compared to 1.17A for the 0.5wl wire), but it is not zero! Since the antenna takes power and radiates, the current cannot be zero; it has to be finite, but small... Notice that the peak current at 1/4 and 3/4 of the way along the wire has a peak amplitude of about 1/2 of what it was with the 0.5wl dipole. Hm, theory says that radiation is proportional to current times distance. Since the dipole length doubled, wouldn't you expect the peak current to be halved???

    I see nothing to suggest that NEC is not showing us reality.

    I will do one more NEC simulation run to show what happens if you start with a 0.5wl dipole and gradually lengthen it while asking NEC to solve for the feed impedance, and the current distribution for several steps between 0.5 and 1wl:
    L is the dipole length in meters. I show that to make the 1wl dipole resonate at 14.55MHz, you would have to make it ~19.5m long..., and we see that the feedpoint Z is close to 5000 Ohms...

    Here is are the current distributions for five steps of dipole length from 10m to 20m (0.5wl to 1.0wl):

    Parting shot: If I make the dipole 10m long, NEC calculates that the feedpoint Z will be 107.35 + j0 Ohms (i.e. resonant) at 44.41MHz, meaning that we are driving it at its third harmonic. This trick is known to any ham that has used a 7MHz dipole at 21MHz. Here is what NEC shows us when a 10m dipole is driven first at 14.547MHz and then at 44.41MHz:

    So Jim, why would NEC do so well in predicting the behavior of a 1/2wl dipole in full agreement with accepted theory and ham experience, and not be able to predict what happens as the dipole is gradually lengthened through 1wl????
    Last edited: Aug 11, 2020
  5. K9AXN

    K9AXN Premium Subscriber QRZ Page

    Do either of you understand the following logic map used to calculate the feed point impedance values for a center and end fed 1/2wl wire?
    If not please explain where you disagree and explain your reasoning. Point to the specific sentence or thought and I'll explain exactly the logic. EZNEC may have a problem in that I believe they do not support end fed antennas only OCFD. If you cannot find an error or can't explain your conflict I'm moving on to explaining why and how a 1/2wl wire impedance at the center is approximately 72 Ohms and end 5000 Ohms. Different thread --- Transmission line internals.

    BTW, if anyone else does not understand the calculations below, I'll work with you to develop an intuitive understanding. There are no mysteries here.

    Regards Jim K9AXN

    Last edited: Aug 11, 2020
  6. WA7ARK

    WA7ARK Ham Member QRZ Page

    I showed you the scientific consensus about the FP Z of a 1/2wl dipole in the Wiki reference. NEC is consistent with that.

    The impedance at the end of a half-wave wire in free space is infinite (also supported by scientific consensus). It is impossible to build a radiating antenna without displacing the feedpoint some finite distance along the wire. It is impossible in the real world, and it is impossible to make a NEC model. That should add to the confidence in NEC!

    A source is a two terminal device that has the same current entering one terminal as leaving the other terminal. You cannot just leave a source terminal float. It has to connect to something capable of storing charge. This is a fundamental law in circuit analysis (Kirchhoff); you cannot wish it away!

    The reason that NEC does not support end-fed antennas is because they cannot exist in nature!

    I believe I just did!

    Before you add anything else, you need to respond to my question: "why would NEC do so well in predicting the behavior of a 1/2wl dipole in full agreement with accepted theory and ham experience, and not be able to predict what happens as the dipole is gradually lengthened through 1wl?"
  7. K9AXN

    K9AXN Premium Subscriber QRZ Page

    Again, if the two of you folks XMT or ARK do not understand or disagree with the math and logic below, let me know. Point to the sentence, paragraph, and logic point and We can have a friendly discussion to reconcile the differing statements. You don't get to say it's wrong but You don't know why or its because EZNEC says so. EZNEC is a super tool when used by people who are qualified to develop the correct input and correctly interpret the output. If you can't explain your logic point and supporting math to state your position, your wasting my time.


    Here is a 20,000 ft overview of the design process from the paleontological era. It will be drilled into later.

    Only two things need to be known in order to design a dipole driven from either the center or end: The Z0 of the wire and Q.

    The Z0 of the wire is 600 Ohms and Q of a 1/2wl dipole driven from either end, which is 8.3333.

    We will provide a teaching experiment developed in 1964 to measure the capacitance, inductance, radiation ratio, and velocity fraction of the wire. In the next iteration we cover a comprehensive cycle by cycle explanation of both the resonant and anti-resonant wires.

    This is for a .5λ wire with 500 Watts applied.

    First we use the Z0 of the wire, 600 Ohms to calculate a base. (500W÷ 600 = .83333) (√ .83333 = .91287A) (.91287 x 600 = 547V) (.91287 x 547 = 500W)

    Now we take the Q of a .5λ wire = 8.3333 (This will be proved when we drill down to the next level). Now render the √ of 8.3333) (√8.3333 = 2.8867)

    We now have everything needed for final calculations.


    First calculate the voltage, current and feed point impedance of the resonant (Center fed wire). Voltage (547 ÷ 2.8867 = 189V) Current (.91287 x 2.8867 = 2.635A)

    (189 ÷ 2.635 = 72 Ohms) (189V x 2.635 = 500 watts).

    END FED:

    (547 x 2.8867 = 1579V) (.91287 ÷ 2.8867 = .31623A) (1579 x .31623 = 500 Watts) (1579V ÷ .31623 = 5000 Ohms)

    THERE YOU HAVE IT. If you don't understand the translations or logic I'll be happy to respond. Otherwise I need to move on the teaching experiment and transmission line logic.

    AGDay to ya
    Regards Jim
  8. WA7ARK

    WA7ARK Ham Member QRZ Page

    Tell us what you mean by "The Z0 of the wire is 600 Ohms".

    What kind of instrument?

    Measured how?

    Measured where?

    Measured between which two points?
  9. W9XMT

    W9XMT Ham Member QRZ Page

    K9AXN: To agree with your "math and logic below" would negate Kirchhoff's 1st Law, which requires that current flowing into a node (or a junction) must be equal to current flowing out of it.

    That cannot occur with only one connection from a 2-terminal source (e.g., a transmitter) to any point along a single conductor — including either absolute, physical endpoint.

    Kirchhoff's 1st Law was true in 1964, and it is still true today.
  10. K8JD

    K8JD Ham Member QRZ Page

    Keep in mind that two crossed dipoles , fed together, will divide the power between them.
    So to try to get omnidirectional operation, you will only have half the radiated power of a single dipole to get out.
    AK5B likes this.

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