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Old July 16th 03, 08:12 AM
Roy Lewallen
 
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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

  #23   Report Post  
Old July 16th 03, 12:27 PM
Dave Shrader
 
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I would like one of those antennas!!

BTW, it should be patented!! It looks better than some I've read about!!

;-)

DD, W1MCE

W5DXP wrote:

Richard Harrison wrote:

What happens? Our loading, adjusted for 50 watts output, produces 50
volts at an in-phase current of 50 amps.



Hmmmm, 50 watts in, 2500 watts out. How much will you take for
that antenna, Richard? :-)


  #24   Report Post  
Old July 16th 03, 02:58 PM
W5DXP
 
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Dr. Slick wrote:
"You cannot tell if the 50 Ohms reading on a Network analyzer into
a Black Box is a dissipative resistance like a dummy load, or if it is
a radiated resistance of a perfectly matched antenna. You don't have
that information."


Conversion of RF energy to heat can be measured. Conversion of RF energy
to EM radiation can be measured.
--
73, Cecil http://www.qsl.net/w5dxp



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  #25   Report Post  
Old July 16th 03, 07:17 PM
Dilon Earl
 
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On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen
wrote:


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.
Roy Lewallen, W7EL


Roy;
Hope you don't mind if I ask you a couple of questions about your
last sentence.
Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no
losses, now if the antenna's input impedance is changed to say 50 ohms
(non-reactive). You would have some loss with the mismatch between
the coax and antenna?
The transmitter would see 50 ohms?



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Old July 16th 03, 07:38 PM
Roy Lewallen
 
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Dilon Earl wrote:
On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen
wrote:


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.
Roy Lewallen, W7EL



Roy;
Hope you don't mind if I ask you a couple of questions about your
last sentence.
Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no
losses, now if the antenna's input impedance is changed to say 50 ohms
(non-reactive). You would have some loss with the mismatch between
the coax and antenna?


Mismatch does not cause loss (that is, conversion of electrical energy
to heat), except that the loss of a lossy transmission line will
increase (very slightly unless initial loss is great) when SWR is
elevated. In this case, the line SWR would be 1.5:1, which would not
cause a significant amount of extra loss even if the cable were very
lossy when matched. So the answer is no.

The transmitter would see 50 ohms?


The transmitter could see any of a variety of impedances, depending on
the length of the 75 ohm transmission line. Only if the line were an
exact multiple of an electrical half wavelength would the transmitter
see 50 ohms, resistive. If the line were an odd number of quarter
wavelengths, the transmitter would see 112.5 ohms, resistive. At all
other lengths, the transmitter would see a complex (partly resistive and
partly reactive) load.

There is a term called "mismatch loss", which is widely misunderstood in
the amateur community because of its name. It doesn't really represent
loss at all, but a signal reduction for other reasons. I've explained
this before in this newsgroup, so if you're interested, you should be
able to find my earlier postings via http://www.groups.google.com.

Roy Lewallen, W7EL

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Old July 16th 03, 08:54 PM
W5DXP
 
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William E. Sabin wrote:
If a 50 ohm generator is connected to a 50 ohm resistor (or resistance),
the maximum possible power is delivered to the load.


Maybe it should be "maximum *available* steady-state power" which is
not always the same thing as "maximum "possible" power"? :-)

And the 50 ohm resistance can be the V/I ratio looking into a Z0-match
point with a high mismatch loss between the Z0-match and a mismatched
antenna, i.e. maximum available power (minus feedline losses) can still
be delivered to a mismatched load by doing the matching back down the
transmission line, e.g. an antenna tuner.
--
73, Cecil http://www.qsl.net/w5dxp



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Old July 16th 03, 09:34 PM
William E. Sabin
 
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W5DXP wrote:
William E. Sabin wrote:

If a 50 ohm generator is connected to a 50 ohm resistor (or
resistance), the maximum possible power is delivered to the load.



Maybe it should be "maximum *available* steady-state power" which is
not always the same thing as "maximum "possible" power"? :-)


Maximum possible is correct. The maximum
"available" from the generator is a constant
value. The maximum "delivered" is the thing can be
lower than the maximum.

Bill W0IYH

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Old July 16th 03, 09:41 PM
Reg Edwards
 
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"William E. Sabin" wrote
If a 50 ohm generator is connected to a 50 ohm
resistor (or resistance), the maximum possible
power is delivered to the load.

If the load is greater than or less than than 50
ohms resistor (or resistance) the power delivered
to the load is less than the maximum possible.


==============================

Unfortunately, the use of the term "Reflection Loss" to descibe performance
of amateur feedline + antenna systems at HF merely adds to the confusion.

"Maximum Available (or possible) Power" delivered to the load occurs only
when there's a "Conjugate Match" between generator and load.

But a Conjugate Match does not exist. It cannot even be assumed. For the
simple reason the internal impedance of the generator is never known. And
it wouldn't make any difference to how the system is set up and operated if
it was.

Even the generator (transmitter) designer doesn't know what the internal
impedance is. He couldn't care less. It can vary all over the shop depending
on what the load impedance is.
---
Reg, G4FGQ


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Old July 16th 03, 10:22 PM
William E. Sabin
 
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Reg Edwards wrote:
"William E. Sabin" wrote

If a 50 ohm generator is connected to a 50 ohm
resistor (or resistance), the maximum possible
power is delivered to the load.

If the load is greater than or less than than 50
ohms resistor (or resistance) the power delivered
to the load is less than the maximum possible.



==============================

Unfortunately, the use of the term "Reflection Loss" to descibe performance
of amateur feedline + antenna systems at HF merely adds to the confusion.

"Maximum Available (or possible) Power" delivered to the load occurs only
when there's a "Conjugate Match" between generator and load.

But a Conjugate Match does not exist. It cannot even be assumed. For the
simple reason the internal impedance of the generator is never known. And
it wouldn't make any difference to how the system is set up and operated if
it was.

Even the generator (transmitter) designer doesn't know what the internal
impedance is. He couldn't care less. It can vary all over the shop depending
on what the load impedance is.
---


All of this is common knowledge, or should be.

However, this gets beyond the basic idea of
mismatch loss. It was not my intention to get into
an expanded discussion of the various
contingencies, which have been discussed ad
nauseum in this forum. One should start with the
simplest case.

If an equipment is designed to work into a 50 ohm
load, and the load is not 50 ohms, a "return loss"
can be measured with a 50 ohm directional coupler.
No knowledge of the generator impedance is needed
for this. The return loss is also a measure of the
mismatch loss, if the generator resistance (50
ohms) is known. An impedance "transforming"
network can improve the load impedance closer to
50+j0 ohms. The equipment will then perform as
advertised.

In many kinds of circuits mismatch loss and
conjugate match are important.

Bill W0IYH



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