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#21




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




"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




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 nearfields collaborate to create an impedance transformation function, in much the same way as a lumpedelement 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 nearfield 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




William E. Sabin wrote:
I believe (intuitively) that the reactive E and H nearfields collaborate to create an impedance transformation function, in much the same way as a lumpedelement 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 standingwave antenna is the result of the superposition of forward and reflected waves on the standingwave antenna. Presumably, a travelingwave 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 travelingwave antenna, is the near field of a travelingwave antenna ever reactive?  73, Cecil, W5DXP 
#25




So my question is: Since the voltage and current are always in phase
in a travelingwave antenna, is the near field of a travelingwave 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 halfwave length of it. The voltage at one end of the halfwave will whizz up and down at a frequency of x megahertz. At the the other end of the halfwave length the voltage will whizz down and up at x megahertz, ie., in timeantiphase with it. Therefore, from where you are standing, the halfwavelength of wire will behave and radiate exactly like a halfwave 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 standingwave and nonstandingwave types can be misleading. To continue with the Beverage. Adjacent 1/2wavelengths of wire form a colinear 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 colinear array, a standingwave antenna, along which the successive halfwave dipoles are all in timephase with each other. But both types of antenna incorporate radiating 1/2wave dipoles. And if the nearfield of one type has a reactive nearfield (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 timephase in which case there would indeed be a nonreactive 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




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 Ã¦ther. 73's Richard Clark, KB7QHC 
#27




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 travelingwave antenna, is the near field of a travelingwave antenna ever reactive?  73, Cecil, W5DXP 
#28




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 nearfields collaborate to create an impedance transformation function, in much the same way as a lumpedelement 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 farfield 377 ohm solution provided by the program is precisely the field that I am thinking about. How does EZNEC, with its finiteelement, methodofmoments 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 farfield E and H fields are different from the nearfield E and H fields. What is going on? Bill W0IYH 
#29




On Fri, 15 Aug 2003 14:57:46 0500, "William E. Sabin"
[email protected] wrote: The farfield E and H fields are different from the nearfield 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 freespace media fluctuations in Z. At a greater distance, such differences become trivial. The local fields present a nonhomogenous freespace 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 spacetime 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




EZNEC doesn't do the transformation you describe.
The following description is a very simplified version of how NEC works. I believe the whole NEC2 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 NEC2 but user selectable in NEC4) 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 NEC2 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 farfield 377 ohm solution provided by the program is precisely the field that I am thinking about. How does EZNEC, with its finiteelement, methodofmoments 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 farfield E and H fields are different from the nearfield E and H fields. What is going on? Bill W0IYH 
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