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  #31   Report Post  
Old July 16th 03, 11:56 PM
Reg Edwards
 
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In many kinds of circuits mismatch loss and
conjugate match are important.

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

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.



  #33   Report Post  
Old July 17th 03, 12:05 AM
Dr. Slick
 
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W5DXP wrote in message ...
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.



Agreed. But a Black Box to me implies you have limited
information from it. My point is that if someone gives you an
impedance plot of a resistive 50 Ohms, you will not be able to tell if
it is dissipative (lossy) or radiated resistance.

I was just reading that Joseph Carr calls radiated resistance as
a sort of "ficticious" resistance. I'm sure many here would argue
this description, but it kinda makes sense to me.


Slick
  #34   Report Post  
Old July 17th 03, 12:38 AM
William E. Sabin
 
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Reg Edwards wrote:
In many kinds of circuits mismatch loss and
conjugate match are important.


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

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.



That too is very well understood by just about
everyone.

Bill W0IYH

  #35   Report Post  
Old July 17th 03, 12:40 AM
Roy Lewallen
 
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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.

Why not call radiation resistance "real" resistance and loss resistance
"ficticious"? Makes just as much sense as the other way around -- that
is to say, none.

Roy Lewallen, W7EL

Dr. Slick wrote:
W5DXP wrote in message ...

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.




Agreed. But a Black Box to me implies you have limited
information from it. My point is that if someone gives you an
impedance plot of a resistive 50 Ohms, you will not be able to tell if
it is dissipative (lossy) or radiated resistance.

I was just reading that Joseph Carr calls radiated resistance as
a sort of "ficticious" resistance. I'm sure many here would argue
this description, but it kinda makes sense to me.


Slick




  #36   Report Post  
Old July 17th 03, 01:19 AM
Dr. Slick
 
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Roy Lewallen wrote in message ...


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.


Actually mostly reflected back to the source, so not radiated or
dissipated assuming ideal lossless transmission lines.



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.



Interesting. I'll have to look this up more.



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.




Well, my point was that you don't need a transmission line for
free space because otherwise it wouldn't be wireless. But your above
point is well taken, that there is no current flowing in free space,
in an expanding EM wave, while there definitely is current in a
transmission line.




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?


That would be excellent.



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.



Certainly the EM wave will heat up ever so slightly any bits of
metal it comes across on the way to outer space.



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. . .


And the EM wave will theoretically continue forever, even if it is
in steradians (power dropping off by the cube of the distance?), so
perhaps eventually most of it will be dissipated as heat.

But, as you know, a capacitor also never fully charges...



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?


Very interesting stuff. And it's certainly enhanced MY
understanding.




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.



But an antenna must be performing some sort of transformer action.
If you were designing an antenna to radiate underwater, or though
Jell-o, or any other medium of a different dielectric constant than
free space, you would have to change it's geometry. Even if it is E
to H, and not V to I.

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. Certainly having lots of metal in close proximity will
affect the impedance of your antenna.



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.


Well, I'm not sure, but you would start off at an open, which
would be transformed to a virtual short. But from there, it sounds
like a complex mathematical derivation to get the 73 Ohms.

My point is that the 73 Ohms is dependant on the dipole's
surroundings, depending on how far from the ground and such, so it is
a transformer of some sort.



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



Which one's can you recommend?


Slick
  #37   Report Post  
Old July 17th 03, 01:24 AM
Dr. Slick
 
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"Tarmo Tammaru" wrote in message ...
"Dr. Slick" wrote in message
om...
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.


Look up "Impedance of Free Space"in the Kraus book.

Tam/WB2TT



I don't have that book. What does it say?


Slick
  #38   Report Post  
Old July 17th 03, 02:20 AM
Roy Lewallen
 
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Dr. Slick wrote:
Roy Lewallen wrote in message ...


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.



Actually mostly reflected back to the source, so not radiated or
dissipated assuming ideal lossless transmission lines.


No. "Reflected" power isn't delivered to the wire. It's an analytical
function that exists only on the feedline. If the feedline has no loss,
the same amount of power entering the line exits the line. You can have
any amount of "reflected power" you want by simply changing the
characteristic impedance of the line -- with no effect on the power
either exiting or leaving the line.


. . .



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.





Well, my point was that you don't need a transmission line for
free space because otherwise it wouldn't be wireless. But your above
point is well taken, that there is no current flowing in free space,
in an expanding EM wave, while there definitely is current in a
transmission line.


Yes, that's right. But if you want to dig deeper, you'll find that a
"displacement current" can be mathematically described which
conveniently accounts for some electromagnetic phenomena. It is, though,
a different critter from the conducted current on a transmission line.

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.

. . .


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.




But an antenna must be performing some sort of transformer action.
If you were designing an antenna to radiate underwater, or though
Jell-o, or any other medium of a different dielectric constant than
free space, you would have to change it's geometry. Even if it is E
to H, and not V to I.


Not transformer, transducer. More like a speaker than a megaphone.

If an antenna is not a transformer of some type, then why is it
affected by it's surroundings so much?


Egad, how can I answer that? If a fish isn't a transformer, then why is
it affected by its surroundings so much? If an air variable capacitor
isn't a transformer, then why is it affected by its surroundings so
much? What does sensitivity to surroundings have to do with being a
transformer?

They obviously are, just like
the primary's impedance is affected by what the secondary sees in a
transformer. Certainly having lots of metal in close proximity will
affect the impedance of your antenna.


It is true that the equations describing coupling between two antenna
elements are the same as the for the coupling between windings of a
transformer. But a single antenna element isn't a transformer any more
than a single inductor is a transformer. When you apply V and I to an
antenna, it creates E and H fields. When you apply V and I to an
inductor, it creates E and H fields. Both the antenna and inductor are
acting as transducers, converting the form of the applied energy. If you
put a secondary winding in the field of the primary inductor, the field
induces a voltage in the secondary. If (and only if) the secondary is
connected to a load, causing current to flow in it, that current
produces a field which couples back to the primary, altering its
current. Coupled antennas, or an antenna coupled to any other conductor,
work the same way -- although localized currents can flow in the absence
of an intentional load if the antenna is a reasonable fraction of a
wavelength long.

But the single antenna isn't a transformer, any more than the single
inductor is. Each is a transducer, and the secondary winding, or coupled
conductor, is another transducer.


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.



Well, I'm not sure, but you would start off at an open, which
would be transformed to a virtual short. But from there, it sounds
like a complex mathematical derivation to get the 73 Ohms.


And it's really, really tough and requires some *really* creative (read:
bogus) math to derive it from simply transmission line phenomena.

My point is that the 73 Ohms is dependant on the dipole's
surroundings, depending on how far from the ground and such, so it is
a transformer of some sort.


Not by itself it isn't. But if you can make a transformer by putting two
antenna elements close together -- put V and I into one and extract it
in a different ratio from the other. It's going to be a pretty lossy
transformer, though, due to energy lost to radiation. (You'll find extra
resistance at the "primary" feedpoint that'll nicely account for this.)



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




Which one's can you recommend?


One of my favorites is King, Mimno, and Wing, _Transmission Lines,
Antennas, and Waveguides_, and that's probably the one I'd choose if I
had to select just one. It was reprinted as a paperback by Dover in
1965, and the paperback be found as a used book pretty readily and
inexpensively. Kraus' _Antennas_ is my favorite text on antennas, and is
certainly one of the most, if not the most, highly regarded. It's now in
its third edition, and you can often find used copies of earlier
editions at reasonable prices. For transmission lines, there's an
excellent treatment in Johnson's _Transmission Lines and Networks_. I
refer to Kraus' _Electromagnetics_ particularly when dealing with waves
in space. And Holt's _Introduction to Magnetic Fields and Waves_ is
pretty good for both.

There are a lot of others, each with its strong and weak points. But you
can't go wrong with these.

Roy Lewallen, W7EL

  #39   Report Post  
Old July 17th 03, 03:23 AM
W5DXP
 
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William E. Sabin wrote:

W5DXP wrote:
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.


Point is that "maximum possible power" will cause a lot of transmitters
to exceed their maximum power rating and overheat.
--
73, Cecil http://www.qsl.net/w5dxp



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  #40   Report Post  
Old July 17th 03, 05:02 AM
Tarmo Tammaru
 
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What might be of interest in this discussion is that after he derives the
impedance of free space, he uses that to find the Radiation Resistance of a
short dipole for dW. Where d is the length of the dipole and W is
wavelength. (I did not want to do Greek letters). He ends up with an
equation of the form

R=377k(d/W)**2 or R=790 (d/W)**2.

As a sanity check let d/W=1/2, which violates the , but still gives a
fairly close answer of 197 Ohms, compared to the actual 168 Ohms. Note that
this is not the same as Feedpoint Resistance because it is not referred to
the current maximum. Kraus does not actually say this, but seems that the
near field would be the mechanism for "matching" this to the far field 377
Ohms. The transmitter only sees the feedpoint, the rest of the universe sees
the whole antenna.

If I interpret it correctly, this 197 (168) Ohms in independent of where you
feed the dipole. Kind of hard to boil several pages into one paragraph,
especially since most of this stuff I haven't seen in decades.

Tam/WB2TT
"Dr. Slick" wrote in message
om...

I don't have that book. What does it say?


Slick





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