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