N0GW wrote:
. . .
So... Yes the antenna is a transducer. No, it does not transform 50
ohms into 377 ohms. 377 ohms refers to the eletrostatic and magnentic
fields as they exist in the near field of an antenna or conductor. It
does not refer to what is going on electrically in the antenna
conductors.
. . .

377 ohms does not describe the E and H fields in the near field. 377
ohms is the ratio of E to H in the *far field* when the medium is free
space or, for practical purposes, air. In the near field, the ratio of E
to H can be not only far from 377 ohms, but it's commonly also complex
(that is, E and H not in time phase). For an illustration, model a short
dipole or small loop with EZNEC or NEC-2, and use the near field
analysis to find E and H at some point close to the antenna (within a
fraction of a wavelength). When you divide E by H, you'll get a wide
variety of results(*) depending on the type of antenna and the
observation point. But as you get farther and farther from *any*
antenna, you'll find that the ratio always converges to 377 ohms, purely
real (that is, the E and H fields in time phase).

(*) The ratio of E to H is called the "wave impedance". In the far
field, and only in the far field, it equals the intrinsic impedance of
the medium. And, as Gary and others have said, this shouldn't be
confused with the ratio of voltage to current at an antenna feedpoint.
They are not at all the same thing, in spite of having the same units of
ohms.

Roy Lewallen, W7EL

Roy Lewallen wrote:

377 ohms does not describe the E and H fields in the near field. 377
ohms is the ratio of E to H in the *far field* when the medium is free
space or, for practical purposes, air. In the near field, the ratio of E
to H can be not only far from 377 ohms, but it's commonly also complex
(that is, E and H not in time phase). For an illustration, model a short
dipole or small loop with EZNEC or NEC-2, and use the near field
analysis to find E and H at some point close to the antenna (within a
fraction of a wavelength). When you divide E by H, you'll get a wide
variety of results(*) depending on the type of antenna and the
observation point. But as you get farther and farther from *any*
antenna, you'll find that the ratio always converges to 377 ohms, purely
real (that is, the E and H fields in time phase).

Yes, I agree with that completely Roy. I apologize for simplifying my
response so much as to not mention this. I was trying to answer the
question at the same level as was asked. I did not mean to offend the
more mathematically astute members of this group.

I will stand by my comment that radiation from antennas, no matter how
well predicted mathematically, is not well understood at a subatomic
level. I personally prefer a model that assumes photons result from
electron acceleration (or deceleration or energy level decrease).
There are obviously competing models.

Gary
N0GW

N0GW wrote:
Roy Lewallen wrote:
377 ohms does not describe the E and H fields in the near field. 377
ohms is the ratio of E to H in the *far field* when the medium is free
space or, for practical purposes, air. In the near field, the ratio of E
to H can be not only far from 377 ohms, but it's commonly also complex
(that is, E and H not in time phase). For an illustration, model a short
dipole or small loop with EZNEC or NEC-2, and use the near field
analysis to find E and H at some point close to the antenna (within a
fraction of a wavelength). When you divide E by H, you'll get a wide
variety of results(*) depending on the type of antenna and the
observation point. But as you get farther and farther from *any*
antenna, you'll find that the ratio always converges to 377 ohms, purely
real (that is, the E and H fields in time phase).

Yes, I agree with that completely Roy. I apologize for simplifying my
response so much as to not mention this. I was trying to answer the
question at the same level as was asked. I did not mean to offend the
more mathematically astute members of this group.

I will stand by my comment that radiation from antennas, no matter how
well predicted mathematically, is not well understood at a subatomic
level. I personally prefer a model that assumes photons result from
electron acceleration (or deceleration or energy level decrease).
There are obviously competing models.

I'm not the least bit offended; I just corrected a statement which
wasn't true.

Intelligent discussion of the subatomic and quantum physical aspects of
electromagnetic radiation are for people mathematically much more astute
than I, so I'll leave that for you.

Roy Lewallen, W7EL

Cecil:

[snip]
"Cecil Moore" wrote in message
om...
Peter O. Brackett wrote:
What? ... What exactly is "f (Zo)"?
Thoughts, comments.

Peter, I for one, have missed your style.

Consider the following:

I(s)
+--------------------------------------------open
|
V(s) 1/4 wavelength, Z0=600 ohms
|
+--------------------------------------------open

Given: The ratio of V(s)/I(s) is 50+j0 ohms. Can you
solve for f(Z0)?
--
73, Cecil http://www.w5dxp.com
[snip]

Heh, heh...

No of course not, you would need more than just this one
experiment/measurement to determine Zo.

The case you depict is at a "singularity" so to speak and is a "pathalogical
case", because with the 1/4 wave line open at the far end, one sees only a
short circuit with Z = V/I = 0.0 [zero] as the driving point impedance and
one needs more equations than just this one singular situation to solve for
Zo of the line.

In fact though if you "vary" the Zo of the line, you would then see a change
in the driving point impendance Z of the line from zero to another value.

From such a thought experiment one should be able to formulate an expression
for Z(Zo).

Thoughts comments...

--
Pete k1po
Indialantic By-the-Sea, FL.

Peter O. Brackett wrote:

"Cecil Moore" wrote:

I(s)
+--------------------------------------------open
|
V(s) 1/4 wavelength, Z0=600 ohms
|
+--------------------------------------------open

Given: The ratio of V(s)/I(s) is 50+j0 ohms. Can you
solve for f(Z0)?

The case you depict is at a "singularity" so to speak and is a "pathalogical
case", because with the 1/4 wave line open at the far end, one sees only a
short circuit with Z = V/I = 0.0 [zero] as the driving point impedance and
one needs more equations than just this one singular situation to solve for
Zo of the line.

Sorry I wasn't clear. One sees 50 ohms, not a short circuit,
and the Z0 of the line is given at 600 ohms. The line is not
lossless and not even low loss. There is enough resistance
in the stub wire to cause a 50+j0 ohm impedance looking into
the stub. I was just wondering what is the nature of your
f(Z0) function.
--
73, Cecil http://www.w5dxp.com

Cecil et al:

[snip]
the stub. I was just wondering what is the nature of your
f(Z0) function.
--
73, Cecil http://www.w5dxp.com
[snip]

I like working with Cecil!

Like a Zen Master's rehtorical approach to facilitating understanding,
Cecil's approach
to this seemingly paradoxical "circuit-to-wave transducer" question is quite
illuminating.

Cecil has neatly sidestepped the fact that, even for the simplest practical
antennas, elementary analytic
formulae for antenna driving point impedances have never been discovered.
Let alone formulae that explicitly
show Zo as an independent variable.

This [obscure?] fact [of delinquint formulae] often comes as a surprise to
most electromagnetic novitiates, I know it did to me.

[Aside: As far as I know, the fact that no one has ever worked out an exact
analytic formula for the driving point
impedance of a simple practical half wave dipole, is not a problem in
practices since other
approximate and/or "sledge hammer" style numerical methods provide
appropriately accurate answers to all
practical Engineering questions about such matters.]

However, as a "seeker of truth", Cecil has noted an easier path as an
approach to the apparently paradoxical
question of the relationship of driving point impedance to the wave or
characteristic impedance of free space
or any other propagating media.

Cecil has zeroed in on an alternative that might give us some insight!

Namely the relatively simple "exact" formula, first revealed by Heaviside
and Kelvin approximately two hundred years ago, the celebrated formula for
the driving point impedance Z = V/I of a lossless transmission line of
characteristic [surge] impedance Zo
terminated in a load impedance ZL. This driving point impedance is given by
the surprising simple relation...

Z(Zo) = Zo[(ZL*cos(theta) + jZo*sin(theta))/(Zo*cos(theta) +
jZL*sin(theta))] (1)

Where theta = 2*pi*(d/lambda) is the relative fractional length of the
transmission line, where d is
the line length and lambda is the wavelength of a sinusoidal signal
supported on the line at the particualr
frequency of interest. Zo of course is the characteristic [surge or wave]
impedance of the line.

It is also well known [Again Kelvin and Heaviside] that Zo can be simply
expressed in terms of the fundamental
transmission line parametric constants [R, L, C, G] by the [equally]
celebrated formula for the characteristic
[surge or wave] impedance of the transmission line as

Zo = sqrt[(R + jwL)/(G + jwC)]; Where, in the lossless case R=G=0.0, Zo -
sqrt(L/C).

[Aside: The terms Impedance, and Reactance were first defined by (Reg
Edward's hero) Oliver Heaviside. I wonder
if an equally simple formula for the driving point impedance in terms of the
Zo of free space for some simple
antenna is lying out there somewhere waiting to be discovered (grin). ]

As can readily be seen, the driving point impedance Z is a function of the
dependent variable Zo and...

although the effect of this relationship is often referred to as an
trasmission line impedance "transformer", the analogy
between the so-called "transmission line transformer" (or should we say
"transducer") described by (1) falls short of
the simple turns ratio relationship where Z = ZL*N^2.

To gain insight here, Cecil has obliquely suggested that, instead of
searching for an antenna formula, that we invert
the celebrated formula (1) and use it to determine unknown characteristic
impedance Zo by assuming ZL known, and
measuring Z.

Inverting formula (1) we obtain the following relationship.

Zo(Z) = ZL[(cos(theta) - jZ*sin(theta))/(Zcos(theta) - jsin(theta))] (2)

[Aside: Apart from the fact that line parameters L,C are also implicit in
the wavelength, Cecil is this right?]

Thus we see that the relationship between Zo and Z is not a simple linear
relationship as for the common transformer,
but instead is, what mathematicians often refer to as, a so-called "bilinear
relationship.

I wonder, is it possible that such a simple relationship exists for some
antennas as well as transmission lines?

An interesting invention... now it will be public domain (smile).

One could clearly construct a sensor to measure unknown Zo's by constructing
a small piece of rigid air dielectric
terminated transmission line and then "immersing" the sensor in substances
of unknown Zo and then determine those unknown
Zo's by measuring the driving point impedance Z. The calibration curve for
this Zo sensor would be the inverse
relationship (2).

Sigh, it's too bad there is not a simple analytical relationship like (1)
for antennas, for perhaps this would
address the OP's question of the relationship between 377 Ohms free space
wave impedance and 73 Ohms driving point
impedance more directly.

On the other hand we now can see that, contrary to Roy's recent assertion up
the thread (grin), that certainly an exact
analytic solution to this problem is likely a challenging Ph.D. thesis
topic.

For... after two hundred or more years [As far as I know...] no one has yet
worked out an exact simple
analytical expression [similar to (1)] for the driving point impedance of
any simple practical antenna.

A Ph.D. thesis indeed!

Thanks Cecil!

Thoughts comments...

--
Pete K1PO
Indialantic By-the-Sea, FL

Peter O. Brackett wrote:
Zo(Z) = ZL[(cos(theta) - jZ*sin(theta))/(Zcos(theta) - jsin(theta))] (2)

[Aside: Apart from the fact that line parameters L,C are also implicit in
the wavelength, Cecil is this right?]

What got me going on this subject is months ago, someone
asked what would be the feedpoint impedance of an infinitely
long dipole in free space. Reg said it would be about 1200
ohms. Since that figure is obviously related directly to Z0,
it got me to thinking about the similarity of dipoles to
transmission lines. In fact, Balanis, in his 2nd edition
"Antenna Theory" illustrates how a dipole is created by
gradually opening up 1/4WL of a transmission line. That's
on page 18. The current distribution on the dipole after
unfolding is the same as the current distribution on the
transmission line stub before unfolding.

For transmission line analysis, we begin with simple lossless
line formulas and then add complexity such as losses per unit
length. For what we call near lossless feedlines, we often
ignore the losses or at least consider them to be secondary
effects. Going where angels fear to tread, I thought, why can't
these same principles be applied to dipole antennas with
admittedly reduced accuracy? Or as one of the r.r.a.a gurus
said: "A wrong answer is better than no answer at all." :-)

My thoughts didn't go to solving for Z0 as you did above.
Using the well known Z0 formula for a single wire transmission
line above ground, we get Z0=600 ohms for #14 wire 30 feet above
ground and it certainly bears a resemblance to an infinite dipole
made of #14 wire 30 feet above ground. Putting a differential
balanced source in the middle of the single-wire transmission
line would result in a balanced feedpoint Z0 impedance of 1200
ohms. 1/2 of this dipole resembles a 1/4WL stub. An infinite
dipole is, of course, a traveling wave antenna.

This is getting long but I think you can see where it is going.
Make each 1/2 of the dipole equal to 1/4WL and we have the
standard standing wave antenna. Analyze the 1/2 dipole as a
lossy 1/4WL stub with a Z0 of 600 ohms not differentiating between
radiation loss and other losses. (For this purpose, we are not
interested in analyzing the radiation.) Hence, the earlier
lossy stub where the impedance looking into the stub was 50
ohms and the Z0 was 600 ohms.

Now quoting Balanis again, page 488 and 489:
"The current and voltage distributions on open-ended wire
antennas are *similar* to the standing wave patterns on open-ended
transmission lines." "Standing wave antennas, such as the dipole,
can be analyzed as traveling wave antennas with waves propagating
in opposite directions (forward and backward) and represented by
traveling wave currents If and Ib in Figure 10.1(a)."

Figure 10.1(a) is very similar to the graphic depicting
a single-wire transmission line over ground whe

Z0 = 138*log(4D/d) D=height, d=wire diameter
--
73, Cecil http://www.w5dxp.com

Cecil:

[snip]
What got me going on this subject is months ago, someone
asked what would be the feedpoint impedance of an infinitely
long dipole in free space. Reg said it would be about 1200
ohms. Since that figure is obviously related directly to Z0,
it got me to thinking about the similarity of dipoles to
transmission lines. In fact, Balanis, in his 2nd edition
"Antenna Theory" illustrates how a dipole is created by
gradually opening up 1/4WL of a transmission line. That's
on page 18. The current distribution on the dipole after
unfolding is the same as the current distribution on the
transmission line stub before unfolding.
[snip]

Well, as we all "know" the current wave on a dipole antenna is exceedingly
close to sinusoidal, but not [exactly] sinusoidal, because if it were
exactly sinusoidal it wouldn't be radiating. That [small] difference
between the actual current distribution on an antenna and the actual current
distribution on a transmission line is the [tell-tale] residual that
separates us from an exact analytic expression for the driving point
impedance of a dipole.

Interesting stuff...

Our man Reg [Edwards, RIP] always said that... "an antenna is just a lossy
transmission line." And of course that is what it looks like approximately.

Heh, heh... everyone always wanted to know the "mathematical" formulae and
theory behind Reg's compact programs. He tantalized us all with a peek or
two at some selections of his Turbo Pascal source code, but essentially left
us all wondering... "How does he do that?"

Apparently, at least one could infer so from his comments, Reg often modeled
antennas as "lossy" transmission lines in some of his
"nutcracker/lightweight" programs. Did he use the Heaviside/Kelvin
formulae? I wonder...

It would be interesting to compare how closely the input impedance of a 1/2
wave lossless feed line of appropriate Zo (Say 600 Ohms?) terminated in a 73
Ohm resistor would approximate that of a "real" dipole. At resonance it
would be 73 Ohms at least.

Such a comparison should be simple to check using EZNEC numerical readouts
for the dipole and comparing to the numerical results obtained from the
formula (1) for the input impedance of the terminated line.

[snip]
For transmission line analysis, we begin with simple lossless
line formulas and then add complexity such as losses per unit
length. For what we call near lossless feedlines, we often
ignore the losses or at least consider them to be secondary
effects. Going where angels fear to tread, I thought, why can't
these same principles be applied to dipole antennas with
admittedly reduced accuracy? Or as one of the r.r.a.a gurus
said: "A wrong answer is better than no answer at all." :-)
[snip]

In many transmission line modeling programs [such as the ones used by the
designers of xDSL modems who, unlike radio amateurs, need models that range
from DC and on up over many decades of frequency range.] the fundamental
transmission line parameters R, L, C, G are often replaced by [empirically
derived] functions of frequency that represent "perturbations" from the
constants to mimic skin and proximity effects. Both R and L are
simultaneously affected by skin and proximity effects. Of course both
Heaviside and Kelvin knew of these effects but could not include them in
their simple derivations.

Speaking of the "L" parameter and proximity... I thought the article by
Gerrit Barrere KJ7KV in the most recent QEX was interesting because he
points out that a large fraction of the "L" parameter in transmission lines
results from the mutual inductance between and because of the proximity of
the two conductors not the individual self inductance of the conductors.

This is not obvious when looking at the "standard textbook"
presentation/derivation of the Heaviside/Kelvin formulation for a
differential section of transmission line. Such standard textbook
derivations almost universally begin with a lumped differential model
consisting of series R, series L, shunt C, shunt G per unit length with no
mention of mutual inductance. In fact of course the "standard" model is
"equivalent" to a model that explicitly exhibits the mutual inductance,
[Because Leq = L1 + L2 + 2M] but it is much more physically satisfying to
see the transmission line inductance presented the way Barrere did.

Thoughts, comments?

--
Pete K1PO
Indialantic By-the-Sea, FL

"Cecil Moore" wrote in message
. ..
Peter O. Brackett wrote:
Zo(Z) = ZL[(cos(theta) - jZ*sin(theta))/(Zcos(theta) - jsin(theta))] (2)

[Aside: Apart from the fact that line parameters L,C are also implicit
in the wavelength, Cecil is this right?]

What got me going on this subject is months ago, someone
asked what would be the feedpoint impedance of an infinitely
long dipole in free space. Reg said it would be about 1200
ohms. Since that figure is obviously related directly to Z0,
it got me to thinking about the similarity of dipoles to
transmission lines. In fact, Balanis, in his 2nd edition
"Antenna Theory" illustrates how a dipole is created by
gradually opening up 1/4WL of a transmission line. That's
on page 18. The current distribution on the dipole after
unfolding is the same as the current distribution on the
transmission line stub before unfolding.

For transmission line analysis, we begin with simple lossless
line formulas and then add complexity such as losses per unit
length. For what we call near lossless feedlines, we often
ignore the losses or at least consider them to be secondary
effects. Going where angels fear to tread, I thought, why can't
these same principles be applied to dipole antennas with
admittedly reduced accuracy? Or as one of the r.r.a.a gurus
said: "A wrong answer is better than no answer at all." :-)

My thoughts didn't go to solving for Z0 as you did above.
Using the well known Z0 formula for a single wire transmission
line above ground, we get Z0=600 ohms for #14 wire 30 feet above
ground and it certainly bears a resemblance to an infinite dipole
made of #14 wire 30 feet above ground. Putting a differential
balanced source in the middle of the single-wire transmission
line would result in a balanced feedpoint Z0 impedance of 1200
ohms. 1/2 of this dipole resembles a 1/4WL stub. An infinite
dipole is, of course, a traveling wave antenna.

This is getting long but I think you can see where it is going.
Make each 1/2 of the dipole equal to 1/4WL and we have the
standard standing wave antenna. Analyze the 1/2 dipole as a
lossy 1/4WL stub with a Z0 of 600 ohms not differentiating between
radiation loss and other losses. (For this purpose, we are not
interested in analyzing the radiation.) Hence, the earlier
lossy stub where the impedance looking into the stub was 50
ohms and the Z0 was 600 ohms.

Now quoting Balanis again, page 488 and 489:
"The current and voltage distributions on open-ended wire
antennas are *similar* to the standing wave patterns on open-ended
transmission lines." "Standing wave antennas, such as the dipole,
can be analyzed as traveling wave antennas with waves propagating
in opposite directions (forward and backward) and represented by
traveling wave currents If and Ib in Figure 10.1(a)."

Figure 10.1(a) is very similar to the graphic depicting
a single-wire transmission line over ground whe

Z0 = 138*log(4D/d) D=height, d=wire diameter
--
73, Cecil http://www.w5dxp.com

Richard:

[snip]
"Richard Clark" wrote in message
...
On Sun, 10 Sep 2006 16:01:34 -0700, Richard Clark
wrote:

In fact, in the near field of an antenna, there is nothing that
resembles 377 Ohms of Z.
[snip]

Correct, but don't we all believe that the wave impedance of "free space" is
approximately 377 Ohms...

Everywhere...

Even in the near field of an antenna.

That is an antenna itself has no effect on the fundamental u and e of the
media in which it is immersed. u and e are defined only in terms of and as
affecting "plane wave" [TEM mode?] propagation, and...

After all the antenna is very small, and free space is very large (grin),
and so a tiny antenna cannot change u and e everywhere!

The fields E and H in the "near region" of an antenna where the waves are
not "plane" on the other hand may not be related by 377 Ohms, simply because
the waves emanating from the "near" antenna are not plane, but...

There might just also be plane waves passing through identically the same
region of space, say emanating from a more distant antenna. The ratio for
those plane E and H fields will indeed be 377 Ohms over the exact same
region of space where Zo is different because of simultaneous but non-planar
waves.

So in fact... the wave impedance of free space can have many values
simultaneously, one [universal?] constant value of ~377 Ohms for plane
waves, while it may have many other [arbitrary] values for waves passing
through the same region of space that are not plane.

Thoughts, comments?

--
Pete K1PO
Indialantic By-the-Sea, FL

The page at:
http://home.comcast.net/~kb7qhc/ante...pole/index.htm
dramatically reveals that the near fields fluctuate wildly from 377
Ohms, and I have restricted my analysis to those values falling at
roughly 100 Ohms or 1000 Ohms (the hot spots marking the feed point
region and the tips of the dipole).

Other antenna design's modification of the 377 near field around them
can be observed at:
http://home.comcast.net/~kb7qhc/ante...elds/index.htm

73's
Richard Clark, KB7QHC

On Tue, 12 Sep 2006 16:00:24 GMT, "Peter O. Brackett"
wrote:

In fact, in the near field of an antenna, there is nothing that
resembles 377 Ohms of Z.
[snip]

Correct, but don't we all believe that the wave impedance of "free space" is
approximately 377 Ohms...

Hi Peter,

Beliefs. -sigh- Is this one of those transcendental statements about
navel gazing?

Everywhere...

Even in the near field of an antenna.

No. Not even in the near field of an antenna.

That is an antenna itself has no effect on the fundamental u and e of the
media in which it is immersed.

Wrong.

After all the antenna is very small, and free space is very large (grin),
and so a tiny antenna cannot change u and e everywhere!

Abstracting from near space to everywhere is the source of your error.

The fields E and H in the "near region" of an antenna where the waves are
not "plane" on the other hand may not be related by 377 Ohms, simply because
the waves emanating from the "near" antenna are not plane, but...

The waves are not plane where the waves are not plane, but... Is this
a Zen "but?"

There might just also be plane waves passing through identically the same
region of space, say emanating from a more distant antenna.

Wrong.

The ratio for
those plane E and H fields will indeed be 377 Ohms over the exact same
region of space where Zo is different because of simultaneous but non-planar
waves.

Wrong.

So in fact... the wave impedance of free space can have many values
simultaneously, one [universal?] constant value of ~377 Ohms for plane
waves, while it may have many other [arbitrary] values for waves passing
through the same region of space that are not plane.

Thoughts, comments?

Wrong.

Peter, are you trying to bust loose a seized bearing? Most of this
reads like the Molly Bloom citation from a technical translation of
"Ulysses."

73's
Richard Clark, KB7QHC