Peter,
I am somewhat surprised to receive such questions from your good self. They
are not so far from the realms of Ohm's Law as to cause YOU any
difficulties. Perhaps after the festivities you are feeling too lazy to
satisfy your own curiosity by exercising your brain cells. ;o)
You must be aware, even without thinking about it, a lumped radiation
resistance must always be associated with a definite location on an antenna
at which the current is known. This by no means need be at the feedpoint.
But I guess this is the first occasion on which you have been confronted
with the *distributed* variety and have been brought to a sudden dead stop.
Let's stay with the well-known resonant 1/2-wave dipole. The objective is
to directly compare radiation resistance with wire loss resistance. To do
this means the same current must flow through both just as if they were in
series with each other.
{ Many people are familiar with the simple equation, efficiency = Rrad /
( Rrad + Rloss ) and state it whenever an appropriate occasion arises. It
sounds very learned of course. But in the whole of North America I venture
to say hardly a single radio amateur knows from where Rloss and Rrad can be
obtained (except perhaps ground loss with verticals) and what its value is.
It follows that few have ever used the equation presented in Handbook
articles, etc. }
We have a choice. 1. Lump both the radiation resistance and conductor
resistance together at one point after transforming from the distributed to
lumped value of wire loss.
Or 2, leave the wire resistance where it is and distribute the radiation
resistance along the wire. We have no choice about the type of
istribution - it must be the same as the wire resistance is distributed -
i.e., uniformly.
Whatever we do we cannot avoid transforming from a lumped to distributed
resistance value, or vice-versa. Electrical engineers do it all the time.
In the case of a dipole there are several ways. But its a simple process
and the result is amazingly even more simple.
I prefer to begin with the accurate assumption of a sinewave distribution of
current along the dipole wire with the maximum of 1 amp at the dipole
centre. Then integrate P = I squared R from one end of the wire to the
other to find the total power dissipated in the wire.
The equivalent lump of resistance located at the centre (where 1 amp flows)
turns out to be exactly half of uniformly distributed end-to-end resistance
of the wire. In fact, that's exactly how the radiation resistance of the
usual 70-ohm lump got itself into a dipole's feedpoint. It is exactly half
of 140 ohms. If radiation resistance itself had any say in the matter I am
sure it would prefer to be nicely spread along the length of the wire
instead of being stuck in a lump next to the feedpoint.
If the end-to-end wire loss resistance is R ohms then the ficticious
equivalent lump at the centre feedpoint is exactly R/2 ohms. So easy to
remember, eh?
Another way of obtaining exactly the same result is to calculate the input
impedance of a 1/4-wave, open-circuit, transmission line, which of course is
the same as half of a half-wave dipole. It even has a 1/4-sinewave current
distribution along its length. The input resistance at resonance is always
half of the conductor loss resistance. With a good impedance bridge this can
be measured to keep Roy happy.
In fact, it is the pair of 1/4-wave, open-circuit, single-wire lines
constituting the dipole which transform the uniformly distributed wire loss
resistance to the equivalent lumped 1/2-value input resistances as measured
at the dipole centre. And, of course, the antenna performs exactly the same
transformation on an antenna's uniformly distributed radiation resistance.
I sometimes feel sorry for things which find themselves securely locked in,
constrained for ever to obey the irresistible laws of nature, helpless to do
othewise, for ever.
See how the interlocking bits of the jig-saw puzzle now fit very nicely
together.
Your general question - yes it would be possible to 'assume' any arbitrary
mathematical distribution of radiation or loss resistance and then find an
equivalent lumped value which would radiate/dissipate the same power when
located at a particular current point. But it would not be of any practical
use - it would never correspond to an actual antenna. When calculating
efficiency of wire antennas it seems only a uniform distribution of
resistance is of any use. An investigator has no choice in the matter.
Calculating the efficiency of coil loaded antennas gets complicated. The
current distributions of the upper and lower sections are different and so
are their efficiencies. But efficiencies are so high in the conductors
themselves ball-park guesses are good enough. However it is still necessary
to transform various effects, including those due to the coil, to the common
base feedpoint in order to calculate input impedance.
---
Best Wishes, Reg, G4FGQ
===================================
Reg:
[snip]
For calculating convenience, we assume the radiation resistance, Rrad,
is
uniformly distributed along the length of the wire and is 140 ohms which
has
been calculated from its dimensions. It only has two - Length and
Diameter. But for a half-wave dipole it is always about 140 ohms. Wire
diameter has a relatively small effect on Rrad.
[snip]
Reg, in your model, is your *assumption* "for calculating convenience"
that
radiation
resistance is uniformly distributed along the antenna structure, i.e. the
transmission line
that represents the antenna in your model, supported by any theory or is
it
just a
mathematical *fit* to the data?
For example, one could *assume* literally any analytic distribution of
radiation resistance
along an antenna's length, for instance sinusoidal, catenary, exponential,
triangular, etc...
and come up with a value/function for that particular distribution that
has
the equivalent
effect of a lumped value placed at the antenna feedpoint. What is so
unique
about uniform?
Why do you think *uniform* is any better than any other distribution of
Rrad?
I have no axe to grind here, just curiosity...
Best Regards for the New Year.
--
Peter K1PO
Indialantic By-the-Sea, FL.
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