Roy Lewallen wrote in message ...


I disagree on this point. You can think of an antenna as a type
of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned
antenna is "matching" 50 to 377 ohms, and will therefore have no
reflections, ideally.

Thinking of it doesn't make it so. All the energy applied to the antenna
is radiated, less loss, regardless of the antenna's feedpoint impedance.
How does that fit into your model.


What if the antenna was a wire shorting the transmission line?
Then there would be very little radiated power, and a lot of reflected
power.

You can make a transformer to match 50 Ohms to, say, 200 Ohms or
so. Why not for 50 to 377? It's just another number.

Except in the case of free space, with a given permeability and
permittivity, you have the impedance of free space, which doesn't need
a transmission line.




ok, so perhaps the way to think of it is: when an antenna is
matched, the I and V curves

what curves?


Well, a carrier with a single frequency will be a sinewave,
correct? Feeding a real resistive impedance, the V and the I
sinewaves will be in phase (no reactance).


will be in phase (no reactance), and the
product of I*V (integrated) will be the power transmitted.

The average power radiated is always the real part of V*I(conjugate). If
V and I are in phase, this is simply equal to V*I. But what does this
have to do with the confusion between a resistor and resistance?


It doesn't really, but it does relate to this discussion. In the
sense that V and I will be in phase for either a matched antenna or an
ideal dummy load.





I suppose what this all means is that if you have a matched
antenna, it's V and I curves
will be IN PHASE and will have the exact
same RMS values as if you had a truly resistive dummy load instead.

Yes and no. If you're feeding the antenna with a 450 ohm line, it's
matched to the line only if its impedance is 450 + j0 ohms, so it looks
like a 450 ohm dummy load, not a 50 ohm one. On the other hand, you can
feed a 50 + j0 ohm antenna with a half wavelength of 450 ohm line, and
get a perfect 50 ohm match to a transmitter at the input end of the
line, while running a 9:1 SWR on the the transmission line. Then either
the antenna or the input of the line looks like a 50 ohm dummy load.


You could make that half wavelength transmission line almost any
other characteristic impedance, like 25 or 200 Ohms, and you would
still wind up back at 50 Ohms at the input of the line (but you
couldn't change frequencies unless they were multiples of 2 of the
fundamental).

But the point is, you will still be at 50 Ohms at the input, so
the V and I sinewaves should be in phase.


Therefore, you can consider the center of the Smith Chart (or the
entire real (non-reactive) impedance line) as a real resistance like
an ideal dummy load.

Do you agree with this statement Roy?

Yes.


Cool. But a network analyzer looking into a black box will not
be able to tell you whether the 50 Ohms it is reading is radiated
resistance or dissipated resistance. This seems to be the crux of my
question.



Now tell me, why can't you just make some 377 ohm transmission line
(easy to make) open circuited, and dispense with the antenna altogether?

When you figure out the answer to that one, you might begin to see the
error with your mental model.

Ok by me. I'll be waiting for your patented no-antenna 377 ohm feedline.

Roy Lewallen, W7EL


Why would you need a 377 Ohm feedline for free space, when free
space itself is the transmission line??

What's wrong with thinking of an antenna as a type of series
Inductor, with a distributed shunt capacitance, that can be thought of
as a type of distributed "L" matching network that transforms from 50
Ohms to 377?

This is related to how you need to bend the ground radials of a
1/4 WL vertical whip at 45 deg angles down, to get the input impedance
closer to 50 Ohms (as opposed to 36 Ohms or something like that if you
leave them horizontal).


Slick