Dr. Slick wrote:
Roy Lewallen wrote in message ...

So, in theory, if you have achieved a perfect match with your
antenna, you will have matched the impedance to the 377 Ohms of free
space, you will not have reflections at the matching point, and the
energy will radiate in whatever pattern you have designed for.

No, an antenna doesn't "match" the impedance of free space. The input
impedance of an antenna is the ratio of V to I. The impedance of free
space is the ratio of the E field to the H field of a plane wave. They
both happen to have units of ohms, but they're different things and
there's no "matching" going on. If you apply 100 watts to an antenna,
resonant or not, 100 watts will be radiated, less loss, regardless of
the antenna's input impedance.



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.


Have you heard of something called mis-match loss? If the
antenna's input impedance is not matched to the tranmission line (or
the final PA), then the radiated power will be significantly less than
100 watts.

I have indeed, and have posted several times about this
often-misunderstood and misused term. You can find the postings by going
to http://www.groups.google.com and searching this group for postings by
me containing "mismatch loss".




The funny thing about this, is that you cannot say that the 50

Ohms in the center of the chart is a "resistive" 50 Ohms, as there is
very little real resistance in the average antenna.

Sure you can. You're confusing resistance with resistors. Resistance is
a dimension, like length. There are lots of things which have dimensions
of resistance but aren't resistors, like transresistance, characteristic
resistance of a transmission line, or radiation resistance to name just
a few. People who have a shaky understanding of basic electric circuit
theory seem to have trouble dealing with this, but it becomes easier to
deal with as you learn more about basic electricity. A mechanical
example is torque and work, which have the same dimensions (force times
distance) but are definitely different things.




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

what curves?

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?




This "resistive"

50 Ohms is really what people call the "radiation" resistance, which
is something of a misnomer again, because this is trying to equate the
successful impedance matching and subsequent non-reflected EM
radiation with a truly real resistance like an ideal dummy load.

Sorry, that doesn't make a whole lot of sense. Yes, it's called the
radiation resistance, but it's not a misnomer at all. (I suppose it
would be if you called it a "radiation resistor", but nobody I know of
has ever called it that.) If you calculate the power "consumed" by this
resistance (that is, the power flow into it), it's the power being
radiated. If the radiation and loss resistance of an antenna were zero,
no energy would flow into it and consequently none would be radiated.
(If the radiation resistance was zero and the loss resistance wasn't,
then energy would flow into the antenna but none would be radiated -- it
would all be dissipated as heat.)



If both the radiation and loss resistance were zero, i would
expect an ideal short, and therefore full -180 degree reflections.

You're assuming that the antenna is fed with a transmission line, but
that's ok.

I suppose what this all means is that if you have a matched
antenna, it's V and I curves

what 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.

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.


Your use of "non-reflected EM radiation" seems to imply that the
radiation from an antenna is somehow bounced back from space if the
antenna feedpoint impedance is reactive. That's one of the rather
bizarre and very wrong conclusions you could draw from the mistaken idea
that the antenna "matched" the characteristic impedance of free space.



I think i'm correct to think of antennas as impedance matching
transformers.

50 Ohms to 377 Ohms.

I like to think of 'em as sort of potato guns, launching RF potato
photons into the aether. But that doesn't make them potato guns.

Feel free to think of them any way you like, as long as you consistently
get the right answer.

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.


. . .


I believe i understand the Chart better than you think, Roy,
enough to know that you do know what you are talking about.

I still think it's ok to consider antennas as impedance
tranformers, but you have brought up some very good points.


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

Roy Lewallen, W7EL

Roy Lewallen wrote:
I'll be waiting for your patented no-antenna 377 ohm feedline.

I wonder if that patent has ever been applied for? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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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

Dr. Slick wrote:
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.

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

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.

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.

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).

Yes.

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.

V and I will be in phase at the feedpoint of a purely resistive antenna.
This is true whether or not it's matched to the transmission line
feeding it, or whether the transmission line is matched to the antenna.
The relative phase of V and I at the antenna terminals has nothing to do
with whether the antenna or transmission line are matched. It also
doesn't matter what fraction of that resistance represents energy
dissipated locally and how much represents energy radiated.




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.

At the input and output of the line, yes.

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.

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?

But it's worse than that. If you connect the network analyzer to two
series resistors, it can't even tell how much power is going to each
one! And it can't even tell the difference between a 50 ohm resistor, a
100 ohm resistor on the other side of a 2:1 impedance transformer, a 100
ohm resistor on the other end of a quarter wavelength of 70.7 ohm
transmission line, or a really long piece of lossy 50 ohm transmission
line. Boy, they sure are stupid.

How come nobody complains that the impedance of a light bulb, an LED, or
a loaded electric motor is resistive? It's too bad it is, too.
Otherwise, the power company wouldn't charge us for the power we're
using to run those things. Neither the power company nor the network
analyzer knows or cares how much of the power going to a light bulb or
LED is converted to light and how much to heat, or how much of the power
going to an electric motor is doing work.

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. 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. . .

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?

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??

By golly, you're right. Design your transmitter for 377 ohm output and
do away with the transmission line, too. Just let them joules slip right
out, perfectly matched, right straight to free space. Voila!

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.

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.

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

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

Dr. Slick wrote:
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.


No. "Reflected" power isn't delivered to the wire. It's an analytical
function that exists only on the feedline. If the feedline has no loss,
the same amount of power entering the line exits the line. You can have
any amount of "reflected power" you want by simply changing the
characteristic impedance of the line -- with no effect on the power
either exiting or leaving the line.


. . .


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.


Yes, that's right. But if you want to dig deeper, you'll find that a
"displacement current" can be mathematically described which
conveniently accounts for some electromagnetic phenomena. It is, though,
a different critter from the conducted current on a transmission line.

An antenna can reasonably be viewed as a transducer. It converts the
electrical energy entering it into electromagnetic energy -- fields. As
is the case for any transducer, the stuff coming out is different than
the stuff going in. Think in terms of an audio speaker, which converts
electrical energy into sound waves, and you'll be on the right track.

. . .


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.

Not transformer, transducer. More like a speaker than a megaphone.

If an antenna is not a transformer of some type, then why is it
affected by it's surroundings so much?

Egad, how can I answer that? If a fish isn't a transformer, then why is
it affected by its surroundings so much? If an air variable capacitor
isn't a transformer, then why is it affected by its surroundings so
much? What does sensitivity to surroundings have to do with being a
transformer?

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.

It is true that the equations describing coupling between two antenna
elements are the same as the for the coupling between windings of a
transformer. But a single antenna element isn't a transformer any more
than a single inductor is a transformer. When you apply V and I to an
antenna, it creates E and H fields. When you apply V and I to an
inductor, it creates E and H fields. Both the antenna and inductor are
acting as transducers, converting the form of the applied energy. If you
put a secondary winding in the field of the primary inductor, the field
induces a voltage in the secondary. If (and only if) the secondary is
connected to a load, causing current to flow in it, that current
produces a field which couples back to the primary, altering its
current. Coupled antennas, or an antenna coupled to any other conductor,
work the same way -- although localized currents can flow in the absence
of an intentional load if the antenna is a reasonable fraction of a
wavelength long.

But the single antenna isn't a transformer, any more than the single
inductor is. Each is a transducer, and the secondary winding, or coupled
conductor, is another transducer.


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.

And it's really, really tough and requires some *really* creative (read:
bogus) math to derive it from simply transmission line phenomena.

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.


Not by itself it isn't. But if you can make a transformer by putting two
antenna elements close together -- put V and I into one and extract it
in a different ratio from the other. It's going to be a pretty lossy
transformer, though, due to energy lost to radiation. (You'll find extra
resistance at the "primary" feedpoint that'll nicely account for this.)



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?

One of my favorites is King, Mimno, and Wing, _Transmission Lines,
Antennas, and Waveguides_, and that's probably the one I'd choose if I
had to select just one. It was reprinted as a paperback by Dover in
1965, and the paperback be found as a used book pretty readily and
inexpensively. Kraus' _Antennas_ is my favorite text on antennas, and is
certainly one of the most, if not the most, highly regarded. It's now in
its third edition, and you can often find used copies of earlier
editions at reasonable prices. For transmission lines, there's an
excellent treatment in Johnson's _Transmission Lines and Networks_. I
refer to Kraus' _Electromagnetics_ particularly when dealing with waves
in space. And Holt's _Introduction to Magnetic Fields and Waves_ is
pretty good for both.

There are a lot of others, each with its strong and weak points. But you
can't go wrong with these.

Roy Lewallen, W7EL

On Wed, 16 Jul 2003 18:20:22 -0700, Roy Lewallen
wrote:

An antenna can reasonably be viewed as a transducer. It converts the
electrical energy entering it into electromagnetic energy -- fields. As
is the case for any transducer, the stuff coming out is different than
the stuff going in. Think in terms of an audio speaker, which converts
electrical energy into sound waves, and you'll be on the right track.

Roy:


Great analogy!

The characteristic acoustic impedance of air (standard temp &
pressure) is about 413 Rayleighs (or Pascal-Seconds/cubic meter).

Do we worry about matching 8 ohms of electrical speaker impedance to
413 Rayleighs? C.f. Paul Klipsch and the Horn speaker.

I wonder if much of the antenna radiation resitance/Tline
impedance/reflection/intrisnic impedance of free space confusion stems
from use of the same words to describe things that may be modeled
mathmetically identically, but have different physical modalities?

In heat sink calculations, for example, we use "thermal resistance"
and an Ohm's law model but few would confuse ohms of resistance with
degrees C/watt.


Jack K8ZOA

Dr. Slick wrote:

But an antenna must be performing some sort of transformer action.

Not quite - but there is a word for what it does: it's a transducer.

A transducer is any gadget that converts energy from one form into a
*different* form. Examples include a loudspeaker (electrical energy to
sound/mechanical energy), a microphone (the reverse), a light bulb and a
photocell.

From that point of view, a resistor is a transducer that converts
electrical energy into heat energy... but it also has some useful
electrical properties :-)

An antenna is a transducer that converts electrical energy into E and H
fields, and the reverse.

You'll also notice that all practical transducers convert some of their
input energy into heat energy.

It's a useful word for a useful idea.

(Cecil - can your IEEE Dictionary help us with a formal definition?)


On the other hand, if you insist on using the word "transformer", you'll
keep on believing you can work out new facts about antennas from what
you already know about transformers:
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.

That's a perfect example of the trap, because in reality it's not "just
like". An antenna also has E-field interactions with its environment
that a transformer doesn't have, so any resemblance will literally be
only half-true.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian, G3SEK wrote:
"Examples include a loudspeaker---."

Good transducer example. Its problem is abysmal efficiency, even if
better than the usual incandescent lamp.

The loudspeaker`s efficiency can be improved by a better match to its
medium. The usual loudspeaker is small in terms of wavelength. A result
is that it is capable of exerting much force on a small area of a very
compliant medium, air. Air could better accept power exerted over a much
larger area, especially at low frequencies, with less force required to
make the air move..

We have a high-Z source and a low-Z sink in the loudspeaker and air.
Conversion from electric power to mechanical power can be more efficient
through better impedance matching. Two solutions are often used for a
better match, a larger loudspeaker or a horn between the loudspeaker and
its air load. The larger speaker is directly a better match. The horn is
an acoustic transformer. They both improve energy conversion efficiency.

Best regards, Richard Harrison, KB5WZI

"Ian White, G3SEK" wrote in message ...

A transducer is any gadget that converts energy from one form into a
*different* form. Examples include a loudspeaker (electrical energy to
sound/mechanical energy), a microphone (the reverse), a light bulb and a
photocell.


It's a useful word for a useful idea.

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.

That's a perfect example of the trap, because in reality it's not "just
like". An antenna also has E-field interactions with its environment
that a transformer doesn't have, so any resemblance will literally be
only half-true.


Roy has clarified this adequately already.

Ok, I was half correct then. Two transducers make up one
transformer.

Certainly two dipoles very close to one another will affect each
other's impedance.

And a regular transformer with a core can definitely be affected
by a close EM-field.


Slick