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Old August 10th 03, 11:07 PM
Roy Lewallen
 
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No, it's really more a matter of how the antenna is oriented relative to
the flow of the Earth's Chi.

Roy Lewallen, W7EL

Yuri Blanarovich wrote:
W7EL writes:

The driving point impedance of the antenna depends on where you drive
it, and it bears no relationship I know of to the wave impedance (which
is, I assume, what you mean by "resistive space impedance") close to the
antenna.



We can look at the lowest impedance in particular antenna, which will have
higher impedance points elsewhere along its length. Looking at different
antennas or arrays we can have antennas with higher lowest impedance. Like
folded dipoles and loops. Would that not indicate and provide closer "match" to
free space impedance? Again, K8CFU measured that folded monopole "surprisingly"
gave higher field strengths than simple monopole radiator. Wouldn't that
indicate that there is something "wrong" (good) about higher impedance
antennas? Capture area reflected in here?

Yuri, K3BU



  #22   Report Post  
Old August 11th 03, 02:21 AM
Dr. Slick
 
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"David Robbins" wrote in message ...
actually i would expect that a change in E/H would change the driving point
impedance and also the performance of the antenna. some possible examples
that show this effect are the changes in element sizes when modeling an
antenna printed on a dielectric circuit board material or sandwiched in a
dielectric media. the change in wire length due to insulation is another
example, the dielectric properties of the insulation change the E/H ratio
near the wire. some examples may be found in many electromagnetics texts,
look at things like dielectric waveguides, or dielectrics in waveguides,
wires in dielectric media. even the detailed calculation of fields within a
dielectric filled coaxial cable should show this effect, change the
dielectric and you change the characteristic impedance... a measurable
effect from changing the 'space impedence' between the wires.



Agreed, and emmersing a waterproof antenna into water will also
affect the input impedance.



Slick
  #23   Report Post  
Old August 15th 03, 03:40 PM
William E. Sabin
 
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Roy Lewallen wrote:

In the fourth paragraph, you say that "real power is in the real part of
the impedance", and in the last, that it's "found by integrating the
Poynting vector slightly outside the surface of the antenna". The
impedance is E/H, the Poynting vector E X H. Clearly these aren't
equivalent.

The radiated power is, as you say, the integral of the Poynting vector
over a surface. (And the average, or "real", radiated power is the
average of this.)


Correction "real part of Poynting vector" noted.

The problem remains:

How is the *real* part of the antenna input
impedance, regardless of how it is fed and
regardless of what kind of antenna it is, get
"transformed" to the *real* 377 ohms of free space?

I believe (intuitively) that the reactive E and H
near-fields collaborate to create an impedance
transformation function, in much the same way as a
lumped-element reactive L and C network. In other
words, energy shuffling between inductive and
capacitive fields do the job and the E and H
fields modify to the real values of free space.
The details of this are murky, But I believe the
basic idea is correct.

Bill W0IYH


William E. Sabin wrote:


There seems to more explanation needed.

If a lossless dipole is loaded with 100 W of *real* power, that is the
real power in the far field, and it is also the real power very close
to the antenna, regardless of the type of antenna.

The value of real power is the same everywhere.

Since real power is in the real part of the impedance, then how does
the value of real impedance (not the magnitude of impedance) vary with
distance from the antenna?

It seems that very close to (but slightly removed from) the antenna
the real part of the resistive space impedance is nearly the same as
the real part of the driving point impedance of the antenna. This real
part is then transformed to 377 ohms (real) within the near field,
suggesting that the open space adjacent to the antenna performs an
impedance transformation. The near-field reactive fields perform this
function in some manner.

The real power radiated is found by integrating the Poynting vector
slightly outside the surface of the antenna, and is equal to the real
power into the (lossless) antenna. This value is constant everywhere
beyond the antenna.

Bill W0IYH



  #24   Report Post  
Old August 15th 03, 04:05 PM
W5DXP
 
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William E. Sabin wrote:
I believe (intuitively) that the reactive E and H near-fields
collaborate to create an impedance transformation function, in much the
same way as a lumped-element reactive L and C network. In other words,
energy shuffling between inductive and capacitive fields do the job and
the E and H fields modify to the real values of free space. The details
of this are murky, But I believe the basic idea is correct.


_Optics_, by Hecht, has a section 2.10 - Cylindrical Waves.
There is an interesting statement in that section: "No solutions
in terms of arbitrary functions can now be found as there were
for both spherical and plane waves."

The net reactive impedance component on a standing-wave antenna
is the result of the superposition of forward and reflected waves
on the standing-wave antenna. Presumably, a traveling-wave antenna,
like a terminated Rhombic, doesn't have reactive impedance components.

So my question is: Since the voltage and current are always in phase
in a traveling-wave antenna, is the near field of a traveling-wave
antenna ever reactive?
--
73, Cecil, W5DXP

  #25   Report Post  
Old August 15th 03, 06:21 PM
Reg Edwards
 
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So my question is: Since the voltage and current are always in phase
in a traveling-wave antenna, is the near field of a traveling-wave
antenna ever reactive?

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

Cec, you're leading yourself astray again. What's reactance to do with
anything other feedpoint impedance?

Stand at a distance from a very long Beverage antenna. Focus your attention
on a particular half-wave length of it.

The voltage at one end of the half-wave will whizz up and down at a
frequency of x megahertz.

At the the other end of the half-wave length the voltage will whizz down and
up at x megahertz, ie., in time-antiphase with it.

Therefore, from where you are standing, the half-wavelength of wire will
behave and radiate exactly like a half-wave dipole. You have no means of
knowing whether there are standing waves along the wire or not. And clearly
it doesn't matter. To segregate antennas between standing-wave and
non-standing-wave types can be misleading.

To continue with the Beverage. Adjacent 1/2-wavelengths of wire form a
co-linear array are in antiphase with each other. Therefore there is no
broadside radiation from a long Beverage which contains an even number
number of halfwavelengths. There is a sharp null at an angle of 90 degrees
from the wire and as overall length increases so does the number of lobes in
the general direction of the wire.

This is just the opposite of a co-linear array, a standing-wave antenna,
along which the successive half-wave dipoles are all in time-phase with each
other.

But both types of antenna incorporate radiating 1/2-wave dipoles. And if
the near-field of one type has a reactive near-field (whatever THAT means)
then so must the other.

If there are no standing waves it does NOT mean the voltage along the whole
length of line or antenna is whizzing up and down in simultaneous time-phase
in which case there would indeed be a non-reactive near field. But neither
could there be any length or time delay involved.

Don't confuse instantaneous RF volts over a cycle with the envelope which
may remain constant or vary with time or distance.
----
Reg, G4FGQ.




  #26   Report Post  
Old August 15th 03, 06:39 PM
Richard Clark
 
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On Fri, 15 Aug 2003 09:40:12 -0500, "William E. Sabin"
[email protected] wrote:
The problem remains:

How is the *real* part of the antenna input
impedance, regardless of how it is fed and
regardless of what kind of antenna it is, get
"transformed" to the *real* 377 ohms of free space?


Hi Bill,

Transformation, as a term, seems to be problematic without any more
care for the preferred term of transduction (ignoring the historical
usage it clashes with). How words could have any bearing on the
process itself is more a calmative to the user than a need for the
group.

So, if we were to simply ignore ALL the terms, how many show up at the
table to discuss the PROCESS (I hope that's the right word...)?

If we simply cast off the electrical aspect of it (seeing how
difficult it is to conduct discussion for this topic in that
vernacular), the correlative of the organ pipe would be useful. It
too creates a standing wave at the drive point; and it employs a
resonant structure wherein the wave stands. It conforms to the
transmission line principles of termination in that a close or open at
the end is meaningful, and harmonically related to wavelength in a
media. If this seems an outrage (because the former kidnapping of
terms is ignored) consider the following quote from Reference Data for
Radio Engineers:
"...Maxwell's initial work on electrical networks
was based on the previous work of Lagrange
in dynamic systems."
This reference then tumbles into the discussion of "Acoustic and
Mechanical Networks and their Electrical Analogs"

It can be seen that the structure imposes critical significance in the
harmonic component, but is wholly inert without excitation. In other
words, it is not the causative agent, nor is it the agent of
transmission. The pressure excess would cause air flow with or
without it. Of course, there is an efficiency problem in that lax
attitude and that necessarily brings us back to structure and fields
(pressure in this case).

What has this to do with near field and far field? For the organ
pipe, what is the near field, what is the far field? Here, we get
into issues as we formerly did by looking at dimension and wavelength.
There are two classes of Acoustic Impedance that bear to this
intimately.

Those two classes compute for a spherical wave front, and for a planar
wave front:
"...the acoustic impedance for a spherical
wave has an equivalent electrical circuit
comprising a resistance shunted by an inductance.
In this form, it is obvious that a small spherical
source (r is small) cannot radiate efficiently since
the radiation resistance [formula] is shunted by a small
inductance [formula]."

The plane wave Acoustic Impedance formula does not exhibit this
inductive shunt. The difference between the two cases is simply a
matter of scale, and is as arbitrarily chosen as with the abandoned
antenna. That is to say, the definition of antenna far field being
expressed as residing 10 wavelengths away also finds the correlative
in this difference of Acoustic Impedance.

What is this shunt? The compressibility of the medium which is the
mechanical analog of storage.

What is the difference between the case of the organ pipe and the
antenna? For the pipe, the medium is lossy (and employing a vacuum
brings its own obvious issues for the organ) and we find the loss
expressed in phonons (the heat of jostling material). For the antenna
(especially in the void of a vacuum, a useful medium) we find no such
issue and consequently no related phonons (loss to heat within the
medium). Some would note this also encompasses the traditional
demarcation between transducer and transformer.

Irrespective of the difference, both exhibit a region wherein the
MEDIA supports the transition (and perhaps we should call these
structures transitioners --- only kidding :-).

As I stated in the past, it is absurd to crop the picture such that
the description demands that an antenna ends at the literal tips of
its structure as if virtual clips connect it to the æther.

73's
Richard Clark, KB7QHC
  #27   Report Post  
Old August 15th 03, 07:06 PM
Roy Lewallen
 
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It's a simple matter to model a Beverage with EZNEC and observe the near
field at any point in space you'd like. EZNEC reports phase angles of
the E and H fields, so it won't take long for you to find out.

Roy Lewallen, W7EL

W5DXP wrote:

So my question is: Since the voltage and current are always in phase
in a traveling-wave antenna, is the near field of a traveling-wave
antenna ever reactive?
--
73, Cecil, W5DXP


  #28   Report Post  
Old August 15th 03, 08:57 PM
William E. Sabin
 
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William E. Sabin wrote:
Roy Lewallen wrote:

In the fourth paragraph, you say that "real power is in the real part
of the impedance", and in the last, that it's "found by integrating
the Poynting vector slightly outside the surface of the antenna". The
impedance is E/H, the Poynting vector E X H. Clearly these aren't
equivalent.

The radiated power is, as you say, the integral of the Poynting vector
over a surface. (And the average, or "real", radiated power is the
average of this.)



Correction "real part of Poynting vector" noted.

The problem remains:

How is the *real* part of the antenna input impedance, regardless of how
it is fed and regardless of what kind of antenna it is, get
"transformed" to the *real* 377 ohms of free space?

I believe (intuitively) that the reactive E and H near-fields
collaborate to create an impedance transformation function, in much the
same way as a lumped-element reactive L and C network. In other words,
energy shuffling between inductive and capacitive fields do the job and
the E and H fields modify to the real values of free space. The details
of this are murky, But I believe the basic idea is correct.

Bill W0IYH


For example, consider an EZNEC solution to an
antenna, say a 50 ohm dipole. The far-field 377
ohm solution provided by the program is precisely
the field that I am thinking about. How does
EZNEC, with its finite-element, method-of-moments
algorithm, transform a 50 ohm dipole input
resistance to 377 ohms in free space?

I don't want the equations, I want a word
description (preferably simple) of how EZNEC
performs this magic.

The far-field E and H fields are different from
the near-field E and H fields. What is going on?

Bill W0IYH

  #29   Report Post  
Old August 15th 03, 09:38 PM
Richard Clark
 
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On Fri, 15 Aug 2003 14:57:46 -0500, "William E. Sabin"
[email protected] wrote:


The far-field E and H fields are different from
the near-field E and H fields. What is going on?


Hi Bill,

The continuum of the structure presents a delay (by "moments" to use
the vernacular of MOM) that combines with all "moments" of the
previously existing and "near" separated field(s) to cause local
free-space media fluctuations in Z. At a greater distance, such
differences become trivial.

The local fields present a non-homogenous free-space media, some of
which is transparent, some of which is reflective, much of it
somewhere in between. The antenna distorts the medium it resides in
presenting much the same effect as gravity distorting the space-time
continuum. This is a leap of faith, certainly, but offers a
visualization that may be familiar. In optics it would be something
like dispersion where the structure is smaller than the wavelength
exciting it.

73's
Richard Clark, KB7QHC
  #30   Report Post  
Old August 15th 03, 10:13 PM
Roy Lewallen
 
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EZNEC doesn't do the transformation you describe.

The following description is a very simplified version of how NEC works.
I believe the whole NEC-2 manual is available on the web, for anyone who
wants a deeper and surely more accurate explanation.

First, an impedance is calculated for each segment of each wire, and a
mutual impedance for every segment relative to every other segment. This
is done in a rather complex way by assuming that each segment has sine,
cosine, and constant currents, calculating the field from each segment
arriving at each other segment, and evaluating the current induced on
the other segment by it. These impedances are put into a matrix, then
the currents on each segment are found by solving Ohm's law in matrix
form, where the E is provided by the specified sources. Once the
currents are found, the impedance at each of the sources is known. The
field from each segment is computed from the known current and assumed
current distribution along the segment with an approximate integral
equation that's solved numerically. The impedance of the medium (fixed
at free space in NEC-2 but user selectable in NEC-4) is of course
involved in this calculation, as it is for the mutual impedance calculation.

The fields are summed to obtain the overall field (both E and H) at any
point the user specifies. Both are reported in a near field analysis
output. In a far field calculation, the distance of the observation
point to all segments is assumed to be the same, and only the E field is
calculated.

An excellent and easy to follow description of the method of moments can
be found in Kraus' _Antennas_, Second Ed. I assume it's in the third
edition also, but it's not in the first. The NEC-2 manual recommends
R.F. Harrington, _Field Computation by Moment Methods_ (McMillan, 1968)
but I haven't seen this book.

I've tried to point out on this thread that although the feedpoint
impedance is an impedance with the units of ohms, and the impedance of a
plane wave in free space also has the units of ohms, they're not the
same thing. Feedpoint impedance is the ratio of a current to a voltage.
Wave impedance, or the intrinsic impedance of a medium, is the ratio of
an E field to an H field -- it's also the square root of the ratio of
the medium's permeability to its permittivity. An antenna converts
currents and voltages to E and H fields, it doesn't just transform one
impedance to another. Hence my insistence on calling an antenna a
transducer rather than a transformer.

Any explanation of an antenna as a transformer will have to include
parasitic array elements, which have zero feedpoint impedance, and array
elements that have negative feepoint resistances.

The answer to your last question is beyond my ability to answer. It's
discussed in great detail in most electromagnetics and antenna texts.

Roy Lewallen, W7EL

William E. Sabin wrote:
William E. Sabin wrote:

For example, consider an EZNEC solution to an antenna, say a 50 ohm
dipole. The far-field 377 ohm solution provided by the program is
precisely the field that I am thinking about. How does EZNEC, with its
finite-element, method-of-moments algorithm, transform a 50 ohm dipole
input resistance to 377 ohms in free space?

I don't want the equations, I want a word description (preferably
simple) of how EZNEC performs this magic.

The far-field E and H fields are different from the near-field E and H
fields. What is going on?

Bill W0IYH




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