Owen Duffy wrote in
:

....

If I take a half wave folded dipole immersed in some environment where
the ambient noise temperature is T, and attach a load directly to the
feedpoint, the load power will be maximum when it is about 300 ohms
rather than about 75 ohms, and the noise power density due to ambient
noise would be K*T*300 W/Hz rather than K*T*75 W/Hz. If Rr is the
(virtual) resistance due to coupling of the antenna with distant
space, then surely this example suggests that Rr is 300 rather than 75
ohms.

(If I performed the same experiment with a plain half wave dipole, the
load power will be maximum when it is about 75 ohms, and the noise
power density due to ambient noise would be K*T*75 W/Hz.)

Sorry, that is plainly wrong. Clarity struck whilst having breakfast, the
received power of a matched system should be independent of R.

Noise power density is simply K*T W/Hz. There is no R term.

The text should read...

If I take a half wave folded dipole immersed in some environment where
the ambient noise temperature is T, and attach a load directly to the
feedpoint, the load power (due to ambient noise) will be maximum when it
is about 300 ohms rather than about 75 ohms. If Rr is the (virtual)
resistance due to coupling of the antenna with distant space, then surely
this example suggests that Rr is 300 rather than 75 ohms.

My apologies.

Owen

On Jul 15, 3:01*pm, Owen Duffy wrote:
"J.B. Wood" wrote in news:i1monr$2od$1
@ra.nrl.navy.mil:

On 07/15/2010 04:14 AM, Owen Duffy wrote:
I note some variation in the use of the term 'Radiation Resistance' (Rr)
that suggests that it has different meanings to different folk.

snip
Hello, and I don't find any ambiguities in any of my various EM and
antenna theory textbooks. *FWIW, from the IEEE Standard Dictionary of
Electrical and Electronics Terms:

"Radiation resistance (antenna). The radio of the power radiated by an
antenna to the square of the rms antenna current referred to a specified
point. *Note: *This term is of limited utility in lossy media."

Hmmm. The last statement suggests that, as defined, it is not clear and
unambiguous in the real world because the real world involves "lossy
media".

The "reference to a specified point" suggests that if one gives a value for
Rr, it is necessary to also state the reference point. Is that what it
means?

This is exactly the lack of clarity that is troubling me.

So if we're looking at free (in vacuo) space the radiation resistance is
simply a "load" resistance component that accounts for where the
radiated power goes. *The radiation resistance doesn't include any other
resistive losses in the antenna structure/proximity operating
environment that may also be dissipating source power introduced at the
feedpoint of the antenna.

This does not address the issue of ground reflection that I mentioned.

*An aerodynamic analogy would be the
distinction between "induced" drag (the price paid for "lift") and
"parasite" drag, which are both components of the total drag.
Sincerely, and 73s from N4GGO,

I am not an aerodynamics type, so drawing that analolgy only helps to
confuse. You might as well use optics!

I know you are trying to be helpful John, but the IREE definition doesn't
seem to clarify the issue.

To put some numbers on my first example, if I have an NEC model of a centre
fed half wave dipole with zero conductor losses, mounted over real (ie
lossy) ground, and feedpoint R at resonance is say, 60 ohms, and total
power in the *far field* divided by I^2 is say, 50 ohms, is Rr 50 ohms? Is
the power "radiated" from such a dipole ONLY the power that makes it to
'distant space', or is radiated power input power less dipole conductor
losses?

The IREE definition suggests that I need also to state that Rr is 50 ohms
at the centre, and the term is is of "limited utility" (not unambiguously
clear?) because of the lossy ground reflections.

If indeed the term Radiation Resistance is only applicable in lossless
scenarios as suggested by the IREE dictionary, what it a clear and
unambiguous language for the real world?

Cheers
Owen

In real world terms radiation resistance is measured by the vector
that overcomes radiation resistance or the conveyance of
communication. This compels the measurement of that which is
accelerated as it is an action and reaction type vector. If one
doesn't have a measurement of the mass that is being accelerated then
radiation resistance itself cannot be supplied. What happens
to the accellerated mass has no connection what so ever to the
accelleration vector.To find the accelerating vector one must first
determine the efficiency of the apparatus used and this will vary
dependent on the method used to produce the accelerating vector so
that one can determine the losses. So if we cannot identify that
vector which creates acceleration of charge where the charge is the
measurement of radiation one must first determine what creates
radiation so that the radiation unit can be measured. The bottom line
is
that one must use a superconductor where only the accelerating vector
comprises of the impedance
seen by the time varying current and where the
resistance of the radiating member is divorced from the equation as is
coupling losses in the absence of a magnetic field.
Art

Owen Duffy wrote:
Thanks Roy.

I note you observe similar variation in usage as I note.

Yes, consistency in an application is more important than a common meaning
of the term, but a common meaning of the term assists simpler
communication.

True. But we can't force consistency of a term that's already ubiquitous
in the literature with a variety of meanings. Saying it's so doesn't
make it so.

Regarding say, a base fed folded monopole and efficiency calculations, if
the connection to ground is though of as having some actual value Rg, since
the current flowing in Rg is twice the feedpoint current, consistent
development of the circuit model will reveal the correct efficiency as:

Rr/(Rr+2Rg)

where Rr is the sum of power in the far field divided by feed point current
squared. You don't need to fudge Rr to get the result, proper allowance of
the power due to the actual current in Rg provides the correct result.

Ok, here we go. Remember that efficiency is really a power ratio, not a
resistance ratio. It reduces to the familiar resistance formula only
when the currents in both radiation and loss resistances are the same.

Let's talk about Rg. An unfolded monopole has a single connection to
ground, and we can call this resistance Rg. If Rr is the base radiation
resistance, then the same current flows through Rr and Rg, so efficiency
= Pr/(Pr + Pg) = Rr/(Rr + Rg) and everything's fine. But when we fold
it, there are two connections to ground -- the "cold" side of the
feedline and the non-feed monopole conductor. Each has half the original
current. The "hot" side of the feedline carries the same current as the
"cold" side so its current is half the original value also. You have
your choice for Rg -- you can consider it to be the original ground
system resistance but with twice the current flowing through it as
through the feedpoint resistance; or you can split the original into two
equal parallel resistances of twice the value, each with the same
current as at the feedpoint. In the first case, you get the equation you
posted. In the second, you get Rr/(Rr + Rg). We've basically referred
the ground resistance to the transformed feedpoint.

The surest way to stay out of trouble is to always calculate efficiency
as a ratio of powers. If you use I^2 * R for radiation power and loss
power, you can't go wrong, regardless of where you choose either R to
be, as long as the I is at the same point.

Kraus (Annennas for All Applications) effectively defines Rr as part of his
development of the concept of a pair of conductors transitioning from a
non-radiating transmission line to an antenna to free space radiation.

He does say "... the radiation resistance Rr, may be thought of as a
"virtual" resistance that does not exist physically but is a quantity
coupling the antenna to distant regions of space via a "virtual"
transmission line."

It is his use of "distant regions of space" that suggests in the case of
ground reflection, it is the remaining total power in distant free space
after lossy reflection that is used to calculate Rr. The power lost in
reflection would be a component of feed point R, but not Rr.

Well, we can get carried away with this, too. Nearby ground sucks power
from the near field and that power is never radiated. The longer
distance ground reflection primarily responsible for elevation pattern
development uses power which has been radiated from the antenna
conductor(s). Is that reflection "distant"? When you calculate an
antenna's efficiency, do you include the power radiated from the
conductor before or after the ground reflection? What about power that's
lost by radiation to space? It's just as surely lost for terrestrial
communication as power warming the ground. Answer: It's entirely up to
you. You could even consider all energy which doesn't strike your
receiving antenna as "loss". All you have to do is clearly state what
you're including and what you're not.

He also states a little earlier "... the antenna appears to the
transmission line as a resistance, Rr, called the *radiation resistance*.
It is not related to any in the antenna itself, but a resistance coupled to
the from space to the antenna terminals." This seems fairly clear to me
that he defines radiation resistance to be at the transmission line /
antenna interface.

Kraus is consistent with this, but other respected authors use the term
radiation resistance differently. The few who use the term radiation
resistance when lossy ground is present, though, seem to regard
near-field coupling loss to ground as loss, and not consider far field
reflection in efficiency calculations at all.

Both of these statements by Kraus are simple, but would seem to be capable
of application to real antenna systems. I can't immediately think of
exceptions (game on???).

As I said above, how distant?

In Kraus's language, ground reflections might reasonable be considered part
of the 'antenna' since they influence its pattern and loss, and loss in the
ground reflections is due to resistance "in the 'antenna' itself" and so
excluded from Rr.

Is there anything in Kraus's statements that is wrong, or my
interpretatiohn of them.

Owen

Kraus isn't wrong. Neither are the other respected authors who use the
term differently. I'm sorry, but you're looking for something that
doesn't exist, and I don't see the point in trying to invent a strict
definition just for your own use.

Roy Lewallen, W7EL

Owen Duffy wrote in
:

Rr/(Rr+2Rg)

That should have an exponent in the

Rr/(Rr+2^2Rg)

On 07/15/2010 04:01 PM, Owen Duffy wrote:

"Radiation resistance (antenna). The radio of the power radiated by an
antenna to the square of the rms antenna current referred to a specified
point. Note: This term is of limited utility in lossy media."


Hmmm. The last statement suggests that, as defined, it is not clear and
unambiguous in the real world because the real world involves "lossy
media".


Lossy media is that which absorbs radiation passing through it. IOW it
heats up. This is different than say the outside air being warmed
through conduction from the earth's surface being in turn heated up by
radiation from the sun.

The "reference to a specified point" suggests that if one gives a value for
Rr, it is necessary to also state the reference point. Is that what it
means?

Hello, and yes, you would have to specify where the quantity applies. Rr
is being calculated as I^2 * Rr = Power radiated. The usual reference
point is the feedpoint of the antenna. Note that the antenna feedpoint
could also be defined to include matching networks and even transmission
line. Of course if these other components also radiate they contribute
to the antenna's radiated power.



This is exactly the lack of clarity that is troubling me.

So if we're looking at free (in vacuo) space the radiation resistance is
simply a "load" resistance component that accounts for where the
radiated power goes. The radiation resistance doesn't include any other
resistive losses in the antenna structure/proximity operating
environment that may also be dissipating source power introduced at the
feedpoint of the antenna.

This does not address the issue of ground reflection that I mentioned.


It doesn't matter to the definition of Rr what other agencies may modify
an antenna's characteristics. For example, we measure (at a particular
frequency) the real (resistive) part of its feedpoint impedance. A
portion of that resistance is due to ohmic losses in the earth, antenna
structure, and any other items forward of the feedpoint. The remainder
of the resistance is Rr. In this example the "antenna" consists of the
monopole and its near-field operating environment.

An aerodynamic analogy would be the
distinction between "induced" drag (the price paid for "lift") and
"parasite" drag, which are both components of the total drag.
Sincerely, and 73s from N4GGO,

I am not an aerodynamics type, so drawing that analolgy only helps to
confuse. You might as well use optics!

I know you are trying to be helpful John, but the IREE definition doesn't
seem to clarify the issue.

Well, I've spent a great deal my professional career as an EE dealing
with USN shipboard antennas and just happen to have ham radio as an
"office" related hobby. As I said in my previous post I don't have a
problem with what Rr means. It seems like a rather straightforward and
simple concept. I think you're trying to read more into it then is there.


To put some numbers on my first example, if I have an NEC model of a centre
fed half wave dipole with zero conductor losses, mounted over real (ie
lossy) ground, and feedpoint R at resonance is say, 60 ohms, and total
power in the *far field* divided by I^2 is say, 50 ohms, is Rr 50 ohms? Is
the power "radiated" from such a dipole ONLY the power that makes it to
'distant space', or is radiated power input power less dipole conductor
losses?


The radiated (far field) power is what is relevant to Rr. The radiated
power is the power accepted by the antenna designated feedpoint less the
other ohmic (items that are dissipating heat) losses forward of the
antenna feed and in its (near field) vicinity. Also, by "accepted"
power I mean the actual power into the antenna terminals (incident power
less reflected power).

The IREE definition suggests that I need also to state that Rr is 50 ohms
at the centre, and the term is is of "limited utility" (not unambiguously
clear?) because of the lossy ground reflections.


No it doesn't.


If indeed the term Radiation Resistance is only applicable in lossless
scenarios as suggested by the IREE dictionary, what it a clear and
unambiguous language for the real world?

Cheers
Owen

The definition doesn't say that (cf the word "limited"). Again I think
you're trying to read items, that while possibility contributing to the
measured/calculated Rr value are irrelevant to the basic definition.
IOW those other items such as earth grounds if present really ARE part
of the antenna. The power radiated by the antenna could propagate as
ground wave, sky wave or in combination - it doesn't matter. Sincerely,
and 73s from N4GGO,

--
John Wood (Code 5520) e-mail:

Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337