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Old July 15th 03, 08:20 PM
W5DXP
 
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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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Old July 15th 03, 10:26 PM
Dr. Slick
 
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W5DXP wrote in message ...
Dr. Slick wrote:
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.


From the IEEE Dictionary: "resistance (1)(B) The real part of impedance."

Apparently, all the resistance in the average antenna is real. :-)
--



You misunderstand my point, Cecil.

If the antenna is tuned correctly, the "radiation" resistance is a
real 50 Ohms, with the V and I waveforms IN PHASE. But, my point is
that you can take a DC measurement anywhere on the ideal lossless
antenna and you will never see 50 Ohms anywhere, only shorts.

Ideally, the antenna never heats up, and has no resistive losses.
This is not to say that it won't have a "radiation" resistance of 50
Ohms at the resonant, tuned frequency. And to a transmitter at the
resonant frequency, this will electrically appear to be the same thing
as an ideal dummy load of 50 Ohms.

Roy's message has clarified a few things.


Slick
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Old July 15th 03, 11:09 PM
W5DXP
 
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Dr. Slick wrote:

W5DXP wrote:
Apparently, all the resistance in the average antenna is real. :-)


You misunderstand my point, Cecil.


Actually, it was a play on words using your definition of "real" Vs
the IEEE Dictionary and mathematical definition of "real". The Devil
made me do it but I did provide a smiley face.
--
73, Cecil http://www.qsl.net/w5dxp



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Old July 15th 03, 11:15 PM
W5DXP
 
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Dr. Slick wrote:
Why would you need the antenna tuner in this case? Assuming your
XMTR is 50 Ohms on the output:

100W XMTR--50 ohm feedline--+---50 ohm load


Voila! A perfect match. No mis-matchlosses, and no need for a
tuner.


Yes, but a real-world antenna rarely has a 50 ohm feedpoint impedance
even though it is resonant. The feedpoint resistance of a resonant
1/2WL dipole can vary from about 40 ohms to about 100 ohms depending
upon its height above ground. And a resonant dipole won't usually
cover the entire 75m band without a tuner of some kind.
--
73, Cecil http://www.qsl.net/w5dxp



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Old July 16th 03, 02:37 AM
Richard Harrison
 
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Dr. Slick wrote:
"But, my point is that you can take a DC measurement anywhere on the
ideal lossless antenna and you will never see 50 Ohms anywhere, only
shorts."

True. D-C resistance of a lossless wire is zero. But, apply 50 watts r-f
power to an antenna adjusted to present 50-ohms resistance at a
particular frequency.

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

The 50 watts is being radiated. As far as the transmitter is concerned,
the power might just as well be feeding a dummy load.

There are several reasons a d-c ohmmeter doesn`t read radiation
resistance. Once the ohmmeter`s d-c has charged the antenna, current
flow stops. D-C doesn`t radiate. A d-c ohmmeter doesn`t even measure the
right loss resistance value. Some of the loss resistance in a true
antenna with actual losses, comes from skin effect at radio frequencies,
where the conduction is forced to the outside of the conductors,
decreasing their effective cross-section and increasing their loss. R-F
also has the ability to induce eddy currents in surrounding conductors
and to agitate molecular movement in surrounding insulators. R-F thus
increases loss over that measured by passing d-c through the antenna.

Radiation resistance can be readily measured by using an r-f bridge
instrument. The bridge and antenna are excited by a generator operating
at the same frequency the transmitter will use. The bridge will indicate
reactance in the antenna, if it is not resonant. The null detector for
the bridge is typically a good radio receiver.

Radiation resistance is real though it does not heat the antenna. Loss
resistance is real though it does heat the antenna and its surroundings.

A resistance is a volt to amp ratio in which amps are in-phase with the
volts.

A resistor is a special type of resistance in which the electrical
energy applied to the resistance is converted into heat.

Best regards, Richard Harrison, KB5WZI



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Old July 16th 03, 04:48 AM
W5DXP
 
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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? :-)
--
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Old July 16th 03, 06:42 AM
Dr. Slick
 
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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
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Old July 16th 03, 07:09 AM
Dr. Slick
 
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(Richard Harrison) wrote in message ...
Dr. Slick wrote:
"But, my point is that you can take a DC measurement anywhere on the
ideal lossless antenna and you will never see 50 Ohms anywhere, only
shorts."

True. D-C resistance of a lossless wire is zero. But, apply 50 watts r-f
power to an antenna adjusted to present 50-ohms resistance at a
particular frequency.

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



That would be 2500 watts? I assume you mean 50 volts and 1 amp to
get 50 watts.



Radiation resistance can be readily measured by using an r-f bridge
instrument. The bridge and antenna are excited by a generator operating
at the same frequency the transmitter will use. The bridge will indicate
reactance in the antenna, if it is not resonant. The null detector for
the bridge is typically a good radio receiver.

Radiation resistance is real though it does not heat the antenna. Loss
resistance is real though it does heat the antenna and its surroundings.

A resistance is a volt to amp ratio in which amps are in-phase with the
volts.


The key here is IN-PHASE. If you read my other post, you and Roy
have clarified this for me: Both radiation resistance and dissipative
resistance are the real portion of the impedance.


A resistor is a special type of resistance in which the electrical
energy applied to the resistance is converted into heat.


With very little radiated energy.


Thanks for the input,

Slick


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