Dr. Slick wrote:
Do you know of anyone who has mathematically derived the 73 Ohms
of a dipole in free space?

_Fields_and_Waves_in_Communication_Electronics_, Ramo, Whinnery,
& Van Duzer. Pages 647, 648, sections 12.05, 12.06.

The feedpoint current of a dipole is caused by the in-phase superposition
of the forward current and reflected current at the balanced feedpoint
of a standing wave antenna. A 1/2WL dipole and an open 1/4WL stub have
similarities. The deviation of the feedpoint impedance from zero ohms
gives an indication of the losses due to dissipation and radiation.
--
73, Cecil http://www.qsl.net/w5dxp



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Kraus comes up with Z=73 + j42.5. He then goes on to say that an actual
dipole is made a few % shorter, which yields 65 + j0. When I did an EZNEC
calculation on a 1/2 wave dipole at 3MHz, I did not quite get that. For a
#30 wire in free space I got 76.81 + j43.89 at 3 MHz, and 72.88 + j0.3465
at 2.94 MHz.

I let EZNEC tell me what the wavelength was, and used 1/2 of that for the
length of the dipole.

Tam/WB2TT
"W5DXP" wrote in message
...
Dr. Slick wrote:
Do you know of anyone who has mathematically derived the 73 Ohms
of a dipole in free space?

_Fields_and_Waves_in_Communication_Electronics_, Ramo, Whinnery,
& Van Duzer. Pages 647, 648, sections 12.05, 12.06.

The feedpoint current of a dipole is caused by the in-phase superposition
of the forward current and reflected current at the balanced feedpoint
of a standing wave antenna. A 1/2WL dipole and an open 1/4WL stub have
similarities. The deviation of the feedpoint impedance from zero ohms
gives an indication of the losses due to dissipation and radiation.
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 17 Jul 2003 07:49:40 -0400, Jack Smith
wrote:
Do we worry about matching 8 ohms of electrical speaker impedance to
413 Rayleighs? C.f. Paul Klipsch and the Horn speaker.


Hi Jack,

Someone must, or we would see more 600 Ohm speakers.

It is facile by half to simply accept the end product of design and
anoint it as an example of a general solution. There are many
antennas that are NOT 50 Ohms. What they ARE is actually of no
consequence except in the sense of efficiency and mission (the antenna
cares not a whit about either).

I have many examples of 2 Ohm antennas (and lower) and of 600 Ohm
antennas (and easily higher) and ALL can be induced to radiate all of
the power applied to them. The distinguishing factor across the board
is that each hi-Z antenna presents similar, physical characteristics
to all other hi-Z antennas; and each low-Z antenna presents similar,
physical characteristics to all other low-Z antennas. All couple
power to the same load of the æther. Clearly impedance and size are
correlated and it is up to the designer to accommodate losses to
achieve similar performance. The same statement is equally applicable
to speakers of any impedance.

Is the antenna transforming its Z to that of the æther? Of course it
is just as the speaker is. Are they both transducers? Of course they
are when transducer is applied loosely (but strictly speaking - no).
Injecting this notion that transducers are a class distinct from
transformers is simply myopic to force an argument. No sooner is the
notion introduced than we find the correlative transducer of the
receive antenna introduced to recover the power - now transformed (and
very inefficiently one might add). The remainder of that power
becomes part of the background noise of the cosmos (far more of it
than is ever recovered for actual use).

Transducers, as a class, are far more prone to the loss through
resistance than transformers - by definition. The speaker is feeding
a lossy medium of air, and the sonar is feeding the less lossy medium
of water. The difference is in the compression characteristics that
turns power into heat. Core loss of the transformer is not due to
compression, but is a direct analog (and electrons bumping into each
other and atoms does constitute a form of compressive loss). There is
no loss in space/æther but neither are there any phonons, the classic
transport of transducer emission and coupling. If an antenna is to
qualify as transducer, it must be with the proviso that it is
distinctly different from every other transducer in lacking the common
transport mechanism of phonons. This is like say walking is a form of
mass transportation if you simply ignore the word mass. To support
these specious forms requires enormous exaggerations.

Another transducer available as a common example (or perhaps not for
less well-heeled equipment) is found in the Collins mechanical filter
for interstage coupling. It has both input and output transducers
that couple the mechanical (and thus heat-prone) energy into
nickel-steel resonant disks. Nickel-steel is obviously less
compressive than either air or water, and exhibits far higher Q (which
is a factor of both antennas and transducers - in their medium) to the
advantage of the circuit. To any bench tech working on receivers,
they would unhesitatingly call these IF Transformers.

Does an antenna "transform" any Z to another Z? The process is
obviously performed with concomitant and equivalent issues of
efficiency regardless of the term inserted between quotes. Does the
term substitution bring any change, or does it correct any error? No.
It is a tautology to suggest that "transducer" is appropriate when
every presumption finds a corresponding "transducer" necessitated by
the force of discussing fields (how does one know these fields exist
without the absolute necessity of completing the transformer action?).
One may "know" in the purely abstract sense, but such knowledge
through the centuries has rarely preceded the actuality of observation
in the real transformed world.

The distinction between transduction and transformation does not
preclude the sense of an antenna serving as a bridge between two
system impedances. Neither hi-Z nor low-Z structures have a
stranglehold on design, except through economy. We commonly employ
very low-Z sources (transistors) to feed modest-Z loads (a common
quarterwave antenna). The economic factor of that load (a quarterwave
at 160M) is sometimes unsupportable and yet we find very few short,
low-Z antennas designed with direct feed from the same low-Z
transistor. Economy again forces some form of transformation (I would
hesitate to call a Tuner a transducer) in that the commercial market
sees very little sense in building low-Z sources for an incredibly
small niche who would refuse to pay the price. Instead, commercial
design accommodates to one Z and expects the user to transform it
along the way. The same logic extends to, and through the antenna.

73's
Richard Clark, KB7QHC

On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin"
sabinw@mwci-news wrote:

Dilon Earl wrote:


Where does the loss occur? If you have 3 db of mismatch loss, is it
in the coax, tank circuit?


The loss in "mismatch loss" refers only to the
fact that the power delivered by the generator to
the load is less than it would be if the load
resistance were the same value as the generator
resistance, in other words if the load and
generator were "matched".

The best way to get a handle on this subject is to
draw a diagram of a generator with voltage V=10,
an internal resistance of 50 ohms, and a load
resistor of R ohms. Let R vary from 1 ohm to 100
ohms and calculate the power dissipated in the
generator resistance (50 ohms), the power in the
load resistance (R), and the total power. Plot a
graph of the three quantities. The load power goes
through a maximum when R=50 ohms.

The maximum power dissipated in the generator
resistance is 10^2/50=2 W, which occurs when R=0
ohms. The minimum power dissipated in the
generator resistance is 3.33^2/50=0.22 W which
occurs when R=100 ohms. When R=50 ohms, the load
power is 5^2/50=0.5 W (the maximum value), the
dissipation in the generator resistance is
5^2/50=0.5 W and the total power is 10^2/100=1 W.

Bill W0IYH

Bill;
Thanks, that all makes sense. Can you consider a Transmitter to
have an internal resistance like the generator that changes with the
plate and tune controls?
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?
Sorry to ask such simple questions. I did search through Google
on posts on this subject, just never could find the answer I was
looking for.

On Thu, 17 Jul 2003 17:00:55 GMT, Dilon Earl
wrote:
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?

Hi Dilon,

Does it become 3 watts hotter under the same drive conditions without
the reflected power? You would be surprised how few pundits actually
discuss this in these terms. Of course everyone would be surprised if
anyone attempted to perform this chore.

I like to include this jab at those who rave on about the
impossibility of knowing the internal resistance of a transmitter and
are satisfied to squeak out 100W RF for 250W DC in.

73's
Richard Clark, KB7QHC

"Reg Edwards" wrote in message ...

I didn't say they were unimportant. I said they served only to add to the
confusion when considering operation of the usual amateur installation when
the generator internal resistance is unknown.

Indeed, and not only that, the generator (ham transmitter) is commonly
neither a linear system nor time invariant. Also, maximum power
(conjugate-matched load) from a linear generator is generally not the
most efficient case. A great many generators and amplifiers are
distincly NOT designed to deliver power to a matched load, but rather
to deliver power efficiently to a specific load which is mismatched
with respect to the output impedance of the generator/amplifier.

There are times when knowing that a generator is a linear 50 ohm
source (within some small tolerance) is important--I deal with them
all the time in the work I do--but in a typical ham transmitter
application, that's very seldom if ever the case.

Cheers,
Tom

"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

Dr. Slick wrote:



Do you know of anyone who has mathematically derived the 73 Ohms
of a dipole in free space?


Antennas, John Kraus, McGraw Hill 1950, Chapter 5-6, pages 143 to 146
gives a complete derivation for a 1/2 wavelength Antenna.

You need some Calculus and infinite series to understand the derivation.

Deacon Dave, W1MCE

(Richard Harrison) wrote in message ...
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.


But a speaker is designed for broadbanded operation, 20-22kHz, not
just one frequency like many antennas.

I'm sure a speaker optimized for one frequency alone will be much
different than a broadband one.


Slick

Dilon Earl wrote:

Bill;
Thanks, that all makes sense. Can you consider a Transmitter to
have an internal resistance like the generator that changes with the
plate and tune controls?
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?


No.

If the transmitter output is 100 W and the
reflected power is 3 W, then the 100 W is the
difference between 100+3=103 W (forward power) and
3 W (reflected power).

The question "where does the reflected power go?"
never seems to have an acceptable answer. Very
strange.

A good way to look at is as follows: The junction
of the transmitter output jack and the coax to the
antenna is a "node", which is just a "point" or
"location" where the jack and the coax meet. At
this node the voltage is exactly equal to the
voltage output of the amplifier (VPA) and also the
voltage across the input of the coax (VCOAX). The
voltage VCOAX) across the coax is equal to the
phasor sum of a forward voltage wave that travels
toward the antenna and a reverse voltage wave that
is traveling from the antenna backward toward the
transmitter.

Also, at the node, IPA is the current from the PA
and ICOAX is the phasor sum of a current wave that
travels to the antenna and a return current wave
that travels toward the transmitter.

At the node, the IPA current and the ICOAX current
are exactly equal and in the same direction
(toward the antenna). At the node the IPA current
is equal to the ICOAX coax forward current minus
the ICOAX reflected current. In other words there
is an *EQUILIBRIUM* at the node between VPA
voltage and VCOAX voltage, and an *EQUILIBRIUM*
between IPA current and ICOAX (forward and
reflected) current.

This explanation accounts for everything that is
going on at the node. The answer to the question
"where does the reflected power go?" is the
following: "It is a nonsense question that has
caused nothing but misery". The reflected power
does not actually *GO* anywhere. The correct
answer is that forward and reflected coax waves
always combine precisely and exactly with the
voltage and current that is delivered by the PA.
The voltage and current at the junction are
correctly accounted for. The basic principles here
are Kirchhoff's voltage law and Kirchhoff's
current law, as applied to the node. You can study
Kirchhoff's laws in the textbooks.

If we apply these laws and calculate the 100 W
power out of the PA and the 100 W power that is
dumped into the coax, they are exactly equal. They
cannot possibly be unequal. The power delivered is
the real part of the product of VPA and IPA (100
W), which is identical to the real part of the
product of VCOAX and ICOAX (100 W).

Observe carefully the following: We do not need to
know anything about the PA and its circuitry. The
PA is nothing more than an anonymous "black box".
In other words, any 100 W (output) PA will perform
exactly as I have described.

Bill W0IYH