Dr. Slick wrote:
"I don`t have that book."

I don`t have Kraus either and I miss it.

Space has a magnetic permeability and a dielectric constant. The square
root of their ratio is the characteristic resistance of space. It is
376.7 ohms = 120 pi ohms.

The reciprocal of the square root of the product of the permeability and
dielectric consrtant of space is the velocity of EM radiation
propagation. It is 300 million m/sec.

The above is courtesy of King, Mimno, and Wing, in "Transmission Lines,
Antennas, and Wave Guides", on page 73.

The authors must have considered the information important as they
repeated it on page 117. They followed the repetition with a discussion
of the radiation resistance and input resistance of an antenna. They
note that radiation resistance can`t be measured between two terminals
in a circuit. The I squared R of the antenna power does not conveniently
compute as might be expected with circuit terminals, as current is a
variable along an antenna in most cases.

Best regards, Richard Harrison, KB5WZI

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

Dr. Slick wrote:
"What does it say?"

I don`t have Kraus, unfortunately.

I do have Arnold B. Bailey`s "TV and Other Receiving Antennas". Bailey
covers more antenna territory than most, and does an excellent job of
it. Bailey also includes a catalog of antenna types, all sized for 200
MHz for easy comparison.

Bailey says the surge impedance of an antenna is inversely proportional
to the capacitance per unit length. Reminds one of a transmission line.
This is non-uniform, so Bailey has an empirical equation which says the
larger the periphery of the rod, ther smaller the average surge
impedance.

The ratio of the electric field to the magnetic field surrounding an
antenna must be related to the ratio of volts to amps in the antenna
wire (the surge impedance).

The surge impedance of a thin-wire 1/2-wave dipole from page 500 is 610
ohms (average).

The surge impedance of a fat-cylinder 1/2-wave dipole from page 502 is
240 ohms (average).

Pattern and gain are identical for both antennas. But, Dr. Slick may be
on to something after all. The bandwidth of the fat antenna is about 3X
that that of the thin. In antennas, bandwidth is often an indicator of
match.

Best regards, Richard Harrison, KB5WZI

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

W5DXP wrote in message ...
Dr. Slick wrote:
But a Black Box to me implies you have limited
information from it.

Black boxes radiate heat very well. :-)

Only if they're dummy loads! :)


Slick

Roy Lewallen wrote in message ...
I'd be one of the people arguing. Radiation resistance fits every
definition of resistance. There's no rule that a resistance has to
dissipate power. The late Mr. Carr was quite apparently confusing
resistance with a resistor, a common mistake.


Your point has been well taken, Roy. But you have to admit that
radiation resistance is not a easily understood concept (which is why
it may be a common mistake), so for someone to call it a "fictitious"
resistance can make sense, in the sense that it is not a dissipated
resistance. After all, "Imaginary" numbers are well accepted. And
from an arguing sematics point of view (which is unfortunately
necessary sometimes), even you call it "radiation resistance", which
means that it is obviously not the same thing as a dissipative
resistance like a 50 Ohm resistor.

That being said, rest assured, Roy, that you have convinced me!



Slick

Roy Lewallen wrote in message ...
I'd be one of the people arguing. Radiation resistance fits every
definition of resistance. There's no rule that a resistance has to
dissipate power. The late Mr. Carr was quite apparently confusing
resistance with a resistor, a common mistake.


BTW, did Joseph Carr really pass away? Sad, his book is very practical.


Slick

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

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

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

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

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

Yes, he died not long ago, within the last year I believe.

Roy Lewallen, W7EL

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

I'd be one of the people arguing. Radiation resistance fits every
definition of resistance. There's no rule that a resistance has to
dissipate power. The late Mr. Carr was quite apparently confusing
resistance with a resistor, a common mistake.



BTW, did Joseph Carr really pass away? Sad, his book is very practical.


Slick

Most simple derivations for the input impedance of a dipole assume it's
infinitely thin. The general problem of a dipole made from wire of
finite diameter is a lot tougher, and is the topic of the papers by the
authors I listed in another recent posting. With EZNEC, you'll find that
the dipole impedance will continue to change as you make the wire
diameter smaller and smaller, until it gets too small for the program to
handle at all.

Roy Lewallen, W7EL

Tarmo Tammaru wrote:
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

"Tom Bruhns" wrote
"Reg Edwards" wrote

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.

====================================
Tom, To add a bit more -

50-ohm generators as used in laboratories (so that measured reflexion loss,
mismatch loss etc, mean something) are effectively constant voltage
generators in series with a 50-ohms resistor, or constant current generators
in shunt with a 50-ohm resistor. They may be followed by an ampifier whose
output impedance is held constant at 50-ohms by some automatic means. None
of these circuits bear much resemblance to a pair of 807's and a tuned tank.

The best that can be said about Rg of the usual HF radio transmitter is that
Rg is indeterminate. IT EVEN VARIES AS THE LOAD IMPEDANCE IS CHANGED which
most of the Guru's contributing to this newsgroup appear to be unaware of or
at least choose to disregard. So what does "adjusting RL to equal Rg" mean?
To use it in a description of feeder + antenna behaviour further propagates
myths, including those surrounding SWR, forward power, reflected power, SWR
meters, etc.

Does Terman ever bother to mention Rg of a Tx PA? If he doesn't it can't
matter very much to him. The ARRL handbook, when numerically designing a
transistor linear HF PA, makes no mention of Rg.
----
Reg, G4FGQ

Dr. Slick wrote:
"What`s the definition of "surge impedance" versus regular old
"impedance"?"

Arnold B. Bailey treats this better than anybody I`ve seen. But, there
are many treatments. The regular old impedance of an antenna depends
upon its termination.

Surge impedance of an antenna depends on its conductor`s inductance per
unit length and capacitance per unit length. In the antenna these are
not uniform as they are in a transmission line, and average values have
been found useful.

Best regards, Richard Harrison, KB5WZI

On Thu, 17 Jul 2003 12:05:19 -0700, Roy Lewallen
wrote:

Most simple derivations for the input impedance of a dipole assume it's
infinitely thin. The general problem of a dipole made from wire of
finite diameter is a lot tougher, and is the topic of the papers by the
authors I listed in another recent posting. With EZNEC, you'll find that
the dipole impedance will continue to change as you make the wire
diameter smaller and smaller, until it gets too small for the program to
handle at all.

Roy Lewallen, W7EL

Hi All,

The derivation of dipole electrical characteristics comes by neither
thin nor thick (cylindrical) elements but through a simpler
(conceptually, not mathematically) work described by S.A. Schelkunoff
in "Advanced Antenna Theory," John Wiley and Sons, 1952.

Schelkunoff approaches the design as merely the extension of the
transmission line and he answers the issue of the antenna (the thin
wire form) being non-linear (the presumed incremental
inductance/capacitance is not constant along the length of the split
transmission line) by simply employing conical structures.

The Biconical Dipole "develops a transverse spherical (TEM) wave
analogous to that on a conventional transmission line" (reference
"Antennas and Radiowave Propagation," Robert E. Collin, McGraw Hill,
1985). "Thus the biconical antenna theory provides a theoretical
basis for assuming a sinusoidal current distribution on thin-wire
antennas."

Like any transmission line terminated in its own character impedance,
the Biconical Dipole (within limits imposed by size and apex angle)
also presents a wide frequency range exhibiting a constant radiation
resistance (about 160 Ohms across three octaves, by my margin notes).

The easiest validation of this is found in the Discone.
http://www.qsl.net/kb7qhc/antenna/Discone/discone.htm

73's
Richard Clark, KB7QHC

Schelkunoff's method is elegant, and one that lends itself to relatively
simple calculation -- in closed form -- with a computer. However, it
doesn't give results which are in as good agreement with measurements
than some other methods, so some assumptions made in his derivation
aren't completely correct. A good summary of various methods and their
validity appears in David Middleton and Ronold King, "The Thin
Cylindrical Antenna: A Comparison of Theories", _Journal of Applied
Physics_, Vol. 17, April, 1946.

The program for calculation of the "Field Day Special" antenna
(ftp://eznec.com/pub/fdsp~.exe) uses Schelkunoff's method, and it's
perfectly adequate for the purpose.

Of course, these days we can easily do very accurate calculations from
very fundamental equations with a computer using the method of moments
or other methods. There's a very good description of the method of
moments in the second edition of Kraus' _Antennas_. I assume it's also
in the third edition, which I don't yet have.

Roy Lewallen, W7EL

Richard Clark wrote:
On Thu, 17 Jul 2003 12:05:19 -0700, Roy Lewallen
wrote:


Most simple derivations for the input impedance of a dipole assume it's
infinitely thin. The general problem of a dipole made from wire of
finite diameter is a lot tougher, and is the topic of the papers by the
authors I listed in another recent posting. With EZNEC, you'll find that
the dipole impedance will continue to change as you make the wire
diameter smaller and smaller, until it gets too small for the program to
handle at all.

Roy Lewallen, W7EL


Hi All,

The derivation of dipole electrical characteristics comes by neither
thin nor thick (cylindrical) elements but through a simpler
(conceptually, not mathematically) work described by S.A. Schelkunoff
in "Advanced Antenna Theory," John Wiley and Sons, 1952.

Schelkunoff approaches the design as merely the extension of the
transmission line and he answers the issue of the antenna (the thin
wire form) being non-linear (the presumed incremental
inductance/capacitance is not constant along the length of the split
transmission line) by simply employing conical structures.

The Biconical Dipole "develops a transverse spherical (TEM) wave
analogous to that on a conventional transmission line" (reference
"Antennas and Radiowave Propagation," Robert E. Collin, McGraw Hill,
1985). "Thus the biconical antenna theory provides a theoretical
basis for assuming a sinusoidal current distribution on thin-wire
antennas."

Like any transmission line terminated in its own character impedance,
the Biconical Dipole (within limits imposed by size and apex angle)
also presents a wide frequency range exhibiting a constant radiation
resistance (about 160 Ohms across three octaves, by my margin notes).

The easiest validation of this is found in the Discone.
http://www.qsl.net/kb7qhc/antenna/Discone/discone.htm

73's
Richard Clark, KB7QHC

I referred to Terman as "him".

It should, of course, have been "HIM". ;o)
---
Reg.

William E. Sabin wrote:
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).

If the source is a signal generator equipped with a circulator and
load, the generator is putting out 103 watts, and the circulator
load is dissipating 3 watts, is the generator still only putting
out 100 watts by definition?
--
73, Cecil, W5DXP

On Thu, 17 Jul 2003 17:12:33 GMT, Richard Clark
wrote:

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

Richard;
I'm not sure if it does get 3 watts hotter. I was always under
the impression that operating a transmitter with a high reflected
power was unhealthy for my PA.

On Thu, 17 Jul 2003 14:42:37 -0700, W5DXP
wrote:

William E. Sabin wrote:
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).

If the source is a signal generator equipped with a circulator and
load, the generator is putting out 103 watts, and the circulator
load is dissipating 3 watts, is the generator still only putting
out 100 watts by definition?

No, you just said it was putting 103 watts.. :-)

Was that the right answer?

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

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

Anybody, such as an 18-year old Japanese student, can do it who can
integrate the power radiated by each elemental length of wire over a
surrounding sphere.

The radiation pattern comes out in the wash. It's an elementary matter.
Hopefully you are able to do it for yourself.

It is likely Heaviside was the first, mostly in his head, but because it was
so obvious he never bothered to write it down.

The Ph.D's of his age would have ridiculed the idea anyway on the grounds
that some of his other more important work lacked rigor. MFJ-259B's and
Viagra were yet to come.

On Thu, 17 Jul 2003 22:56:41 GMT, Dilon Earl
wrote:

Richard;
I'm not sure if it does get 3 watts hotter. I was always under
the impression that operating a transmitter with a high reflected
power was unhealthy for my PA.


Hi Dilon,

Some would suggest not, but then they wouldn't warrant their own
advice.

73's
Richard Clark, KB7QHC

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

---------------------------------------------------

Rich, I'm reminded of Laurel and Hardy. Here's another fine mess you've
got yourself into.
---
From your favourite Italian Clown.

William E. Sabin wrote:
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


This discussion assumes that we want the 100 W PA
to actually deliver 100 W to the coax. Assume a
Bird model 43 wattmeter in the line. If the PA is
actually delivering 100 W to the coax, then the
forward power must be 103 W and the reflected
power must be 3 W. The PA output power is
103-3=100 W. This is how we use the Bird
wattmeter. The Bird instruction manual tells us this.

Keeping everything simple and not getting into
peripheral issues is desirable at this point. For
example, a circulator will dissipate the 3 W, but
the above discussion does not assume a circulator.
The circulator's job is to force a 50 ohm load on
the PA, despite the fact that the coax input
impedance is not 50 ohms.

Bill W0IYH

I agree equivalent circuits are invalid when it comes to calculating either
efficiency or internal power loss. The circuit in my example was the ACTUAL
circuit, i.e., a 10 volt source with a series 50 ohm resistor, not a Thevenin
nor a Norton equivalent. With this restriction, I don't see anything wrong with
my statement. Depending on what the ACTUAL circuit is in a power amplifier, I
believe the current drawn can either increase of decrease when the load is removed.

Ron

W5DXP wrote:

Ron wrote:

It helps me understand reflected power to think of a 50 ohm source of
10 volts connected to a half wave lossless line. In this situation the
line can be removed from the equation and the load can be considered
connected directly to the 50 ohm 10 volt source. If the load R is
either a short or open circuit, there will be zero power transferred
to the load, but there will be a big difference in the power
dissipated in the source, two watts with the short and zero watts with
the open.


Not with a Norton source. :-) Quoting _Fields_and_Waves_in_Communication_
Electronics_, by Ramo, Whinnery, & Van Duzer, page 721: "It must be
emphasized, as in any Thevenin equivalent circuit, that the equivalent
circuit was derived to tell what happens in the *LOAD* under different
load conditions, and significance cannot be automatically attached to
a calculation of power loss in the internal impedance of the equivalent
circuit." Seems your above assertion violates that sage admonition.

Roy Lewallen wrote in message ...

Understood. But you have to admit that transformers and
transducers have some similarities. You will still have to optimize
the windings and magnets and size/shape of the speaker cone for
optimum power tranfer of a single tone. And you will have to have to
change the design if you decide to transmit sound underwater.


And transducers and automobiles have some similarities. You have to
optimize the engine of a car for optimum acceleration, and change the
tire tread design if you decide you want to drive on wet roads. So
really, an antenna is like an automobile.


Your sarcasm doesn't make your points as well as your logic.

How about this: An antenna is a transducer, and a transformer is
made up of two transducers.

So you need two antennas to make a transformer. And stick them
close together so they couple well, just for arguments sake.

Fair enough.





I see your point, that the primary could be considered one
antenna, and the core material like free space, and the secondary
would be the receive antenna. But i suspect even a single
transducer/antenna can be optimized for maximum lines of flux through
a core at a particular frequency, or max ERP in the case of the
antenna. Otherwise we wouldn't have to tune these things.


Yep, and an automobile can be optimized for maximum acceleration. Good
argument for considering an antenna a type of automobile, no?

Y'see, if you really, really want an antenna to be a kind of automobile,
you can cook up a bunch of reasons to convince yourself that it is. The
same method works for astrology and fortune telling, too.


Shall i call this a Straw man argument? Or putting words in
someone's mouth?

Ok, an antenna is a transducer. But you can still optimize it for
ERP, and that will depend on the impedance of free space or water or
whatever. Why not throw out the whole concept of free space impedance
if it doesn't matter?



Slick

Roy Lewallen wrote in message ...
I agree that it makes some sort of sense to call radiation resistance a
"ficticious resistor". It makes no sense to call it a "ficticious
resistance" -- any more than you'd call the capacitance of a short
antenna a "ficticious capacitance". It only advertises a lack of
knowlege of the principles of basic electricity.

Roy Lewallen, W7EL


Actually, you are correct. I mis-quoted him somewhat. He calls
the radiation resistance of an antenna as the value of a ficticious
resistor that would dissipate the same amount of power that is
radiated by said antenna.

Wouldn't want to denigrate the reputation of a cool dude...


Slick

"Dr. Slick" wrote:

(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.


You can be just as sure that the speaker presents its rated impedance only
at one frequency, too.

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

.. . .


I see your point, that the primary could be considered one
antenna, and the core material like free space, and the secondary
would be the receive antenna. But i suspect even a single
transducer/antenna can be optimized for maximum lines of flux through
a core at a particular frequency, or max ERP in the case of the
antenna. Otherwise we wouldn't have to tune these things.


Yep, and an automobile can be optimized for maximum acceleration. Good
argument for considering an antenna a type of automobile, no?

Y'see, if you really, really want an antenna to be a kind of automobile,
you can cook up a bunch of reasons to convince yourself that it is. The
same method works for astrology and fortune telling, too.



Shall i call this a Straw man argument? Or putting words in
someone's mouth?

Feel free to call it what you want. I believe I've made as valid an
argument for an antenna being an automobile as you did for it being a
transformer, and based on the same criteria.

Ok, an antenna is a transducer. But you can still optimize it for
ERP, and that will depend on the impedance of free space or water or
whatever. Why not throw out the whole concept of free space impedance
if it doesn't matter?

The optimization of an antenna depends on many factors, only one of
which is the nature of the medium in which it's immersed. And among the
medium's important properties are its permeability, permittivity, and
the velocity of a wave propagating in it. The phase velocity and
characteristic impedance can both be calculated from the permeability
and permittivity, so you can't really say any one of these is more
important than the other.

It doesn't make any sense to throw out the concept of free space
impedance just because it confuses people who don't know what it means.
It's an extremely useful and well-understood concept. For example,
reflection of a wave from a plane conductor or the ground can easily be
found by calculating a reflection coefficient based on the impedance of
the reflecting surface and the impedance of the impinging wave. (The
impedance of a wave can be quite different close to an antenna than it
is after it's traveled some distance.) If you look in some of those
texts I recommended, you'll find the impedance of free space cropping up
all over the place.

What needs to be thrown away is the belief that all impedances are the
ratio of a voltage to a current, along with the notion that only
resistors can have resistance.

Roy Lewallen, W7EL

Reg Edwards wrote:
I referred to Terman as "him".

It should, of course, have been "HIM". ;o)

Not really - "Him" will do nicely. Just spell his surname in capitals
:-)

Seriously, people like Terman, Kraus and Jasik do deserve our respect,
for developing textbooks that have become 'standards'. Over several
editions they have been subject to searching examination from thousands
of teachers and students, so there aren't many errors left in there.

That's the valid reason for using those names as touchstones. To
contradict one of those standard texts, you'd better have some good
arguments prepared.

- BUT -

Never quote a textbook as a substitute for doing your own thinking. That
is the ultimate disrespect to the original authors.


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