Roy Lewallen wrote in message ...


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.

Yes, there would. But every watt delivered to that wire would be either
radiated or dissipated.


Actually mostly reflected back to the source, so not radiated or
dissipated assuming ideal lossless transmission lines.



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.

That's right. You can make a transformer that converts the circuit V/I
ratio to 377 ohms. But this doesn't make a plane wave whose E to H field
ratio is 377 ohms. But you can have an antenna with a feedpoint
impedance of, say, 3 - j200 ohms, and a couple of wavelengths from the
antenna, the ratio of E to H field produced by that antenna will be 377
ohms. Another antenna, with feedpoint impedance of 950 + j700 ohms, will
also produce a fields whose E to H ratio is 377 ohms. In fact, you can
have any antenna impedance you'd like, and a few wavelengths away, the
ratio of E to H field will be 377 ohms, provided the antenna is immersed
in something resembling free space. But the fields in, on, or around a
377 ohm transmission line are unlikely to have an E to H ratio of 377 ohms.

You're trying to say that the 377 ohm E/H ratio of free space is the
same thing as a V/I ratio of 377 ohms. It isn't. Any more than 10 ft-lbs
of torque is the same as 10 ft-lbs of work or energy.



Interesting. I'll have to look this up more.



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.

Sorry, I can't make any sense out of that.



Well, my point was that you don't need a transmission line for
free space because otherwise it wouldn't be wireless. But your above
point is well taken, that there is no current flowing in free space,
in an expanding EM wave, while there definitely is current in a
transmission line.




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.

You're right. The network analyzer can't tell you what's happening to
those watts going into the antenna. Too bad. If it could, we'd have a
really cool, easy way to measure antenna efficiency, wouldn't we?


That would be excellent.



If we insist on separating all resistance into "dissipative" and
"dissipationless" categories, we have to consider time. Within a small
fraction of a second, most of the energy going into an antenna is
dissipated -- mostly in the ground or (at HF) the ionosphere. So we'll
have to consider that portion of the radiation resistance as
"dissipative". The stuff that goes into space will take longer to turn
into heat, but it will eventually.


Certainly the EM wave will heat up ever so slightly any bits of
metal it comes across on the way to outer space.



So the remainder of the radiation
resistance is one that's initially dissipationless but becomes
dissipative with time. Just think, we can have a whole new branch of
circuit theory to calculate the time constants and mathematical
functions involved in the transition between dissipationless and
dissipative states! And of course it would have to be cross-disiplinary,
involving cosmology, meteorology, and geology at the very least. There
are textbooks to be written! PhD's to earn! Just think of the potential
papers on the resistance of storage batteries alone! High self-discharge
rate means a faster transition from dissipationless to dissipative. . .


And the EM wave will theoretically continue forever, even if it is
in steradians (power dropping off by the cube of the distance?), so
perhaps eventually most of it will be dissipated as heat.

But, as you know, a capacitor also never fully charges...



Sorry, I digress. I just get so *excited* when I think of all the
possibilities this opens up for all those folks living drab and boring
lives and with so much time and so little productive to do. . .

But has this ******* creation really simplified things or enhanced
understanding?


Very interesting stuff. And it's certainly enhanced MY
understanding.




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?

Because that's not what it does, and thinking of it that way leads you
to impossible conclusions. The antenna is converting power to E and H
fields. The ratio of E to H, or the terminal V to I are immaterial to
the conversion process. You're continuing to be suckered into thinking
that because the ratio of E to H in free space has the dimensions of
ohms that it's the same thing as the ratio of V to I in a circuit. It
isn't. A strand of spaghetti one foot long isn't the same thing as a one
foot stick of licorice, just because the unit of each is a foot.



But an antenna must be performing some sort of transformer action.
If you were designing an antenna to radiate underwater, or though
Jell-o, or any other medium of a different dielectric constant than
free space, you would have to change it's geometry. Even if it is E
to H, and not V to I.

If an antenna is not a transformer of some type, then why is it
affected by it's surroundings so much? They obviously are, just like
the primary's impedance is affected by what the secondary sees in a
transformer. Certainly having lots of metal in close proximity will
affect the impedance of your antenna.



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).

There are a number of ways in which an antenna and transmission line are
similar. But don't take the analogy too far. Start with a quarter
wavelength transmission line, start splitting the conductors apart until
they're opposed like a dipole, and tell me how you've ended up with an
input impedance of 73 ohms.


Well, I'm not sure, but you would start off at an open, which
would be transformed to a virtual short. But from there, it sounds
like a complex mathematical derivation to get the 73 Ohms.

My point is that the 73 Ohms is dependant on the dipole's
surroundings, depending on how far from the ground and such, so it is
a transformer of some sort.



There are plenty of texts you can read, on all different levels, if
you're really interested in learning about antennas, fields, and waves.

Roy Lewallen, W7EL


Which one's can you recommend?


Slick