TopBand: Folded unipoles
w8jitom@postoffice.worldnet.att.net
w8jitom@postoffice.worldnet.att.net
Wed, 8 Oct 1997 20:01:07 +0000
> From: km1h@juno.com (km1h @ juno.com)
> Subject: Re: TopBand: Folded unipoles
> 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'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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