On Wed, 8 Oct 1997 20:01:07 +0000 w8jitom@postoffice.worldnet.att.net
writes:
>> From: km1h@juno.com (km1h @ juno.com)
>> Subject: Re: TopBand: Folded unipoles
To: <topband@contesting.com>
>> Date: Tue, 7 Oct 97 15:44:28 +0000
>
>Hi Carl,
>
>Please excuse me for the long post, but the good point is this is the
>last one from me on this topic since it is much too time consuming.
I agree with just about everything Tom except for the ground resistance.
If the radials are a constant...no changes...then I do not understand how
their RF resistance becomes transformed when you vary the radiating
element.
Using a 4:1 toroid xfmr for example. The coax shield, the radials, the
other end of the folded element and the cold end of both the 50 and 200
Ohm windings are all at one common point.
The RF resistance of the radials/earth has not changed.
Look at it this way, the only variable in the ground system is the folded
end of the wire. Since it was agreed that the ground resistance was 12
Ohms with the monopole would it not then be correct to say that the
folded monopole far end is now placed in PARALLEL with the already
existing ground?
Yes, the far end of the fold is transformed to 48 Ohms, no argument
there.
All laws are satisfied....if the ground resistance was a perfect zero
placing 48 Ohms in parallel still equals zero.
Placing the 48 in parallel with 12 gives us 9.6 Ohms effective which
raises the efficiency of the radiated versus the heat wasted signal.
That is in line with all the too many to ignore reports of the perceived
effectiveness of the folded monopole versus a standard monopole with a
poor ground.
Does this make some sense?
I'm not saying you are wrong, but I'm just not convinced....yet.
If I placed a noise bridge at the 50 Ohm xfmr input what would I measure?
Assume again pure resistance, no reactive part.
If you dont wish to continue I understand...this takes time; I've blown
the morning on this. Maybe Peter, Ian, Earl, Frank, John (the lurker),
etc have some insights to add.
73 Carl KM1H
>> I'm confused Tom....
>> Your first assumption was that the 1/4 wave monopole was over
>perfect
>> ground....correct?
>
>Correct. It was an example intended to give a "feel" for how the
>current behaves. If you follow the description of currents, you will
>understand how the system behaves. The point was the ground current
>doesn't change a bit.
>
>> Take a REAL system with 35 Ohms Z and 12 Ohms of ground loss; not
>> uncommon in many residential installations. The VSWR looks real good
>but
>> 1/3 of the power is warming the worms( ~ 2dB). No matching section
>> involved or required here.
>
>OK, in that case let's assume the vertical's total common mode base
>current (the current flowing UP into the vertical) is one ampere. In
>that case, if the feedline does not radiate, the current flowing into
>the ground connection is also exactly one ampere. Remember
>Kirchoff says the current entering a junction HAS to equal the
>current leaving that junction. It's important to remember that is
>Kirchoff's law, not Kirchoff's "guess".
>
>Anytime that current is not exactly equal, the feedline is part of
>the antenna and is radiating.
>
>> Next fold that wire and we have a 4X impedence transformation so the
>> radiation resistance is now 140 Ohms. The ground loss is still 12
>Ohms
>> but that is only 8.6% of the total.
>
>That's was another point I tried to make. We all too often mix
>different resistances measured at different points together, and try
>to use it to reach a conclusion.
>
>The FEEDPOINT terminal resistance is simply the radiation resistance
>and loss resistance transformed by the skirt wires. If we fail to
>equally transform the ground resistance, we can no longer compare
>the resistances at the new and very different feedpoint terminals!
>
>Rather than get confused, let's look at the currents since we all
>know power is I*I *R.
>
>Of course we have NO idea how much of that radiator power is
>radiated, and how much is lost in the surrounding environment. We
>also have NO idea how much of the 12 watts of ground power is
>converted to heat, and how much contributes to a radiated field.
>
>So let's just ASSUME the example is perfect, and all ground
>resistance represents conversion to heat (loss), and all antenna
>resistance represents conversion to radiated power.
>
>In the simple monopole example, ground loss would be 1*1*12=12
>watts. Radiator power would be 1*1*35=35 watts.
>
>We have 12 watts heat and 35 watts signal. The feedpoint impedance is
>47 ohms, the voltage is 47 volts, and the current is one ampere. We
>obviously have 47 watts of applied power. Faraday, Ohm, Maxwell, and
>Volta are all satisfied, and you knew that.
>
>Let's say we split the radiator into two equal diameter conductors.
>Now the FEEDPOINT impedance changes. The **reason** for the
>impedance change is now the feedline connects in series with half of
>the radiator, so it "sees" half of the SYSTEM's current. We simply
>sliced the radiator down the middle, and fed one half of it! We have
>half the current in each wire, or 1/2 ampere in EACH wire. But the
>total current is still one ampere, or the signal would be much
>weaker!
>
>Keeping the applied power fixed, since the current is each wire
>is half, we need twice the voltage for the same power. That means the
>feedline terminal voltage is now 94 volts. Working that out, 1/2
>ampere at 94 volts is 47 watts. The feedpoint impedance is now 188
>ohms, exactly four times what it was before.
>
>(If we broke that feedpoint impedance down into the resistances, as
>viewed ***from that point in the system***, it would amount to 140
>ohms of radiation caused resistance and 48 ohms of ground loss caused
>resistance. Looking at it from the feedline terminals: The power
>loss would be .5*.5*48= 12 watts. The radiated power .5*.5*140= 35
>watts. The applied power would be .5*.5*188= 47 watts. )
>
>The radiator current is still the total of two one half ampere wires
>in parallel, or one ampere. Since ONLY charge acceleration causes
>radiation, and the NET current and length of the radiator is the
>same, the total charge acceleration and radiation intensity is
>exactly the same.
>
>The current flowing into our "ground loss resistance" is one half
>ampere from the directly fed conductor, and one half ampere
>from the actual feedpoint. The total ground current is one ampere.
>The sum of currents flowing up into the radiator equals the sum of
>currents flowing into the ground, as it does in any series circuit.
>
>The ground loss is still 1*1*12 (I squared R), or twelve watts. The
>radiated power is still 1*1*35, because splitting the conductor does
>not change the REAL IRE defined radiation resistance or the pattern
>of the antenna. The applied power is still 47 watts, the
>radiated power still 35 watts, and lunch is not free.
>
>> Assuming a quality matching network
>> the loss in the ground now is only a fraction of a dB.
>
>Not according to Kirchoff, Faraday, Maxwell, and Ohm. ALL four would
>have to be incorrect for that to be true, since they all agree with
>each other.
>
>> Perhaps someone here will model this and report one way or the
>other. I'm
>> just going with what I have thought was the correct assumption.
>
>If you model it, be sure to model it correctly. Add the ground
>resistance as a load on a single wire. Build the folded monopole
>above that single wire that represents the lossy ground. Connect the
>source between that single wire and the antenna (of course with the
>programs earth connection at the OTHER end of the load). Then tie the
>second unbroken and unfed wire back to the same common point as the
>source. (After all, you do connect the shield to the base of the
>grounded element, and the ground "resistance" is some distance away
>from that point.)
>
>At least two popular magazine icons clearly don't understand what
>radiation resistance is, and how it should be applied. One W6 has
>the incorrect theory in his Handbooks, and it even appears in the
>ARRL Antenna Compendium (but thankfully not the ARRL
>Antenna Handbook).
>
>73, Tom W8JI
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