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Supporting theory that Antennas "Match" to 377 Ohms (Free space)
Hi,
I looked through Albert Shadowitz's "The Electromagnetic Field", and found on page 554 support for my original statement that antennas "match" to the impedance of free space (377 Ohms). I didn't buy the book, but professor Shadowitz did write on this page about how creating antennas to most efficiently transfer power to free space is a similar problem to matching a circuits source to its load. He goes on to make a short comparison between source/load impedances to an antenna matching to the impedance of free-space. Food for thought and no doubt, debate. Slick |
And another good reason the folded dipole FM antenna works so great. There
is actually science to support use of a 300 ohm antenna. who'd a thunk it. -- 73 es cul wb3fup a Salty Bear "Dr. Slick" wrote in message m... Hi, I looked through Albert Shadowitz's "The Electromagnetic Field", and found on page 554 support for my original statement that antennas "match" to the impedance of free space (377 Ohms). I didn't buy the book, but professor Shadowitz did write on this page about how creating antennas to most efficiently transfer power to free space is a similar problem to matching a circuits source to its load. He goes on to make a short comparison between source/load impedances to an antenna matching to the impedance of free-space. Food for thought and no doubt, debate. Slick |
Hi, I looked through Albert Shadowitz's "The Electromagnetic Field", and found on page 554 support for my original statement that antennas "match" to the impedance of free space (377 Ohms). I didn't buy the book, but professor Shadowitz did write on this page about how creating antennas to most efficiently transfer power to free space is a similar problem to matching a circuits source to its load. He goes on to make a short comparison between source/load impedances to an antenna matching to the impedance of free-space. Food for thought and no doubt, debate. Slick Yep, I tried to question that few moons back, but was "convinced" here that it is not important. Generally antenna exhibits all kinds of impedances along its length. I was reasoning that antenna having its lowest impedance higher or closer to 377 ohm should have better efficiency in coupling to the space (air). So loops and folded dipoles should be better in that respect. K8CFU et al, when doing experiments with verticals and radials, found that folded monopole measured higher signal levels (over simple monopole) than expected. Any progress since then? Yuri, K3BU |
Dr. Slick wrote:
"---creating antennas to most efficiently transfer power to free space is a similar problem to matching a circuit`s source to its load. (Quote from Shadowitz)" Reciprocity rules in antennas. Kraus has an Apendix D (Absorbing Materials) to his "Antennas For All Applications". Kraus says: "The use of space cloth (Z=377 ohms per square) placed lambda/4 from a reflecting plane was invented by Winfield Salisbury (1) at Harvard Radio Research Laboratory during WW-2 ---." (Shades of stealth) However, a century of antenna experimentation has not revealed a practical need to especially design an antenna to match its radiation to a resistive 377 ohms. Most effective radiation occurs when the antenna circuit is matched and has no loss. 100% of the energy accepted is radiated. Best regards, Richard Harrison, KB5WZI |
An antenna's radiating efficiency has nothing whatever to do with the
impedance of its feedline, or whether it's matched to it or not. Now I suppose somebody will drag in the irrelevant matter of SWR on the feedline. --- Reg, G4FGQ |
Richard Harrison wrote:
Most effective radiation occurs when the antenna circuit is matched and has no loss. 100% of the energy accepted is radiated. In the real world for the same size wire on HF, a folded dipole should be slightly more efficient than a dipole because of lower I^2*R losses. -- 73, Cecil, W5DXP |
Reg Edwards wrote:
An antenna's radiating efficiency has nothing whatever to do with the impedance of its feedline, or whether it's matched to it or not. Now I suppose somebody will drag in the irrelevant matter of SWR on the feedline. Actually, you brought up the subject. :-) The feedline's power transfer efficiency is just as important as the antenna's radiating efficiency. -- 73, Cecil, W5DXP |
Now I suppose somebody will drag in the irrelevant matter of SWR on the
feedline. Actually, you brought up the subject. :-) The feedline's power transfer efficiency is just as important as the antenna's radiating efficiency. -- ============================ Cecil, you forgot the efficiency of the PA DC power supply. There's much more energy wasted there than what's lost in the feeder. Sort out your power budget. ;o) --- Reg. |
On Fri, 8 Aug 2003 10:09:01 -0400, "WB3FUP \(Mike Hall\)"
wrote: And another good reason the folded dipole FM antenna works so great. There is actually science to support use of a 300 ohm antenna. who'd a thunk it. Hi Mike, What you describe is the feedpoint Z not the antenna Z which to all intents and purposes is not far from the original, single-wire dipole. 73's Richard Clark, KB7QHC |
Reg Edwards wrote:
Actually, you brought up the subject. :-) The feedline's power transfer efficiency is just as important as the antenna's radiating efficiency. Cecil, you forgot the efficiency of the PA DC power supply. There's much more energy wasted there than what's lost in the feeder. My DC power supply is a 12 VDC marine battery charged by a solar panel. What's the efficiency of free energy? :-) Sort out your power budget. ;o) Actually, I don't much care about the efficiency of the electronics. 60 Hz energy is cheap. I am much more interested in getting the generated RF into the Æther. -- 73, Cecil, W5DXP |
I am much more interested in getting the
generated RF into the Æther. --- Cec, Good, now we're back on track. But what's the 377 ohms of nothingness to do with a random length of wire which has any impedance you fancy just by connecting to it. Even without making a connection and just using your imagination. Could the person who confidently raised this subject from the dead please give us some clues about calculating the turns ratio. I don't have ready access to the works of Maxwell and there's no mention of it in my 1992 edition of the ARRL Handbook. And I think my smiley is better than yours! |
Tricky Dick Sez -
What you describe is the feedpoint Z not the antenna Z which to all intents and purposes is not far from the original, single-wire dipole. Tricky, After all these years you're catching on. What you really meant to say was that the feedpoint impedance is not the same thing as the radiation resistance. 'Antenna' impedance' in the present context is not a phrase known to radio engineering. Please define. --- Reg |
In article ,
Reg Edwards wrote: Cec, Good, now we're back on track. But what's the 377 ohms of nothingness to do with a random length of wire which has any impedance you fancy just by connecting to it. Even without making a connection and just using your imagination. Could the person who confidently raised this subject from the dead please give us some clues about calculating the turns ratio. I think you have to measure the diameter of free space first, before you can calculate the turns ratio. Do write, when you get to the far side, and let us know how the weather is, OK? ;-) -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Only at some distance from the antenna. You can create local E/H ratios
of nearly any value (magnitude and phase). Roy Lewallen, W7EL W5DXP wrote: The ratio of the radiated E-field to H-field has no other choice. If you stuff EM radiation into free space, the ratio of E-field to H-field is 376.7 ohms. Zero energy is lost from the EM spectrum when an electron throws off a photon (until that photon is annihilated). |
Roy Lewallen wrote:
Only at some distance from the antenna. You can create local E/H ratios of nearly any value (magnitude and phase). Roy Lewallen, W7EL W5DXP wrote: The ratio of the radiated E-field to H-field has no other choice. If you stuff EM radiation into free space, the ratio of E-field to H-field is 376.7 ohms. Zero energy is lost from the EM spectrum when an electron throws off a photon (until that photon is annihilated). The near field has a reactive component to the impedance. But is it true that the real part of that complex impedance must be 376.7 ohms resistive? Bill W0IYH |
William E. Sabin wrote:
Roy Lewallen wrote: Only at some distance from the antenna. You can create local E/H ratios of nearly any value (magnitude and phase). Roy Lewallen, W7EL W5DXP wrote: The ratio of the radiated E-field to H-field has no other choice. If you stuff EM radiation into free space, the ratio of E-field to H-field is 376.7 ohms. Zero energy is lost from the EM spectrum when an electron throws off a photon (until that photon is annihilated). The near field has a reactive component to the impedance. But is it true that the real part of that complex impedance must be 376.7 ohms resistive? Bill W0IYH Not at all. For example, the magnitude of of the wave impedance E/H is much lower than 377 ohms very close to a small loop, and much higher than 377 ohms very close to a short dipole. Interestingly, as you move away from a small loop, the magnitude of E/H actually increases to a value greater than 377 ohms, then slowly approaches 377 ohms from the high side as you move even farther away. The opposite happens for a short dipole -- the E/H ratio drops below 377 ohms some distance away (a fraction of a wavelength), then increases to 377 ohms as you go farther yet. Roy Lewallen, W7EL |
On Sun, 10 Aug 2003 08:54:56 -0500, "William E. Sabin"
sabinw@mwci-news wrote: 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 figures at: http://home.comcast.net/~kb7qhc/ante...pole/index.htm illustrate just how the dipole's near-field reactance maps out (without respect for phase, and expressed in SWR relative to free space Z). Note that employing the term transform and antenna within the same context is not de rigueur. ;-) 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Note that employing the term transform and antenna within the same context is not de rigueur. It is 100 percent correct and appropriate. Bill W0IYH |
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.) The integral doesn't need to be taken slightly outside the surface of the antenna, but can be any closed surface enclosing the antenna. There's no necessity for E/H, or the real part of E/H, to be constant in order to have the integral of E X H be constant. 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. If you find any published, modeled, measured, or calculated support for that contention, I'd be very interested in it. Roy Lewallen, W7EL 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 |
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. "Roy Lewallen" wrote in message ... 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.) The integral doesn't need to be taken slightly outside the surface of the antenna, but can be any closed surface enclosing the antenna. There's no necessity for E/H, or the real part of E/H, to be constant in order to have the integral of E X H be constant. 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. If you find any published, modeled, measured, or calculated support for that contention, I'd be very interested in it. Roy Lewallen, W7EL 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 |
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 |
"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 |
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 |
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 |
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. |
On Fri, 15 Aug 2003 09:40:12 -0500, "William E. Sabin"
sabinw@mwci-news 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 |
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 |
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 |
On Fri, 15 Aug 2003 14:57:46 -0500, "William E. Sabin"
sabinw@mwci-news 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 |
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 |
Reg Edwards wrote:
I answered your original question by saying that the near field due to a halfwavelength of wire along an antenna with standing waves (like a co-linear array of in-phase dipoles) was just the same as a halfwavelength along a wire in an antenna without any standing waves (like a Beverage). Unfortunately, that cannot be true. The maximum H-field amplitude at a current null cannot possibly be the same as the maximum H-field where a current null never exists. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Unfortunately, that cannot be true. The maximum H-field amplitude at a
current null cannot possibly be the same as the maximum H-field where a current null never exists. ============================ Dear Cec, your nitpicking is inexhaustible and at this time in the morning I havn't the energy to sort out what you are talking about. Best to agree to differ, eh? ;o) --- Reg. |
Reg Edwards wrote:
Dear Cec, your nitpicking is inexhaustible and at this time in the morning I havn't the energy to sort out what you are talking about. Best to agree to differ, eh? ;o) I don't see how you can possibly assert that the fields are the same for a standing-wave antenna and a traveling-wave antenna because they obviously are not the same. A magnetic pickup at a current null on a standing-wave antenna will read a low maximum value. A magnetic pickup at the same point on a traveling-wave antenna will read a high maximum value. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote:
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'm looking for a text to help me increase my understanding of antennas beyond what is contained in the ARRL Antenna Handbook. It looks like "Antennas" by Kraus is it. Can anyone recommend any others? Thanks and 73, -- * Do NOT use Reply * Reply only through ARRL forwarding service to K3TD Tad, K3TD |
tad danley wrote:
I'm looking for a text to help me increase my understanding of antennas beyond what is contained in the ARRL Antenna Handbook. It looks like "Antennas" by Kraus is it. Can anyone recommend any others? _Antenna_Engineering_Handbook_, edited by Jasik, contributions by many. _Antenna_Theory_Analysis_and_Design_, by Balanis -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 17 Aug 2003 13:56:11 GMT, tad danley
wrote: Roy Lewallen wrote: 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. This may not be a good analogy, but Specific Impulse of rocket motors helps me to remember that the 'units' of something have to be considered in the context of what is being measured. Specific impulse is a measure of the performance of a rocket motor. It measures the thrust obtained from a single kilogram of propellant burned in one second. The 'units' of Specific Impulse are seconds, but we're not measuring 'time'. 73, Hi Tad, Your point is well taken. ALL physical phenomenon can be expressed through a chain of conversions in the MKS system of units. When someone tells you that their terminology is inconsistent between disciplines (as such offered in this and other threads); it must then be amenable to reduction to MKS terms or one of the two conflicting expressions is invalid. That is to say to the specific matter about the usage of "ohms:" Here, the unit of ohm must be reduced to Meters, Kilograms, and Seconds for both usages (electrical and radiative). At that point, both will have a common basis for comparison and if in fact their reduced terms are identical, then their common usage is also identical. One simple example is with the measurement of body weight on the bathroom scale (a torsion or compression device) as opposed to the weight measured on the doctor's scale (a beam balance). Let's say before you go to the doctor's, you weigh yourself in around 165 pounds. When you arrive at the doctor's, his scale says you weigh around 75 kilograms. Let's remove this same scenario to the moon (you live in one of those futuristic 1990's colonies forecast by the space race back in the 60's). Before you went to the doctor's you weighed in around 33 pounds. When you arrive at the doctor's, his scale says you weigh around 75 kilograms. Here we find the expression "pounds" suffers what appears to be the same plight of "ohms" in that the determination of a value is inconsistent. You may also note constants of proportionality on earth and the moon. These constants when expressed as a ratio also describe the significant differences between the earth and the moon. The problem is that the term "weight" has a hidden association to the constant of Gravity. The expression Gram is one of Mass, not weight. The expression pound is not an expression of Mass unless you expand it to include the term for the particular constant of Gravity. Mass is constant in the Newtonian Universe, and weight is not. If you were to have reduced the pounds to the MKS system both times, you would have found it consistent both times (here on earth, and on the moon). If you reduce the "ohms" to the MKS system both times.... Well I will leave that for further deliberation as some are sure to be surprised. :-) 73's Richard Clark, KB7QHC |
Roy Lewallen wrote in message ...
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. I've agreed with you on the semantics of antennas as transducers, but two transducers DO make a transformer. Ohms are still always Ohms, regardless of what you are measuring. And it's very interesting that the E and H fields have units of Volts/meter and Ampere(turn)/meter, which when you divide one by the other, you get basically Volts/ampere, just like you would in a transmission line. But I don't claim that a wave traveling in a transmission line is the same as a wave traveling through free space. Slick |
tad danley wrote:
Roy Lewallen wrote: 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'm looking for a text to help me increase my understanding of antennas beyond what is contained in the ARRL Antenna Handbook. It looks like "Antennas" by Kraus is it. Can anyone recommend any others? Thanks and 73, Kraus is not only an antenna expert, he is a world-class authority on the entire field of Electromagnetics, based on Maxwell's equations. His mathematics is elegant. Bill W0IYH |
....and his writing is lucid. I read his first edition, a gift from my
Father, and knew where I wanted to go to grad school. He is also a very fine person. Buy and read his books. 73 Mac N8TT -- J. Mc Laughlin - Michigan USA Home: "William E. Sabin" sabinw@mwci-news wrote in message ... snip Kraus is not only an antenna expert, he is a world-class authority on the entire field of Electromagnetics, based on Maxwell's equations. His mathematics is elegant. Bill W0IYH |
On Mon, 18 Aug 2003 11:28:40 -0700, Roy Lewallen
wrote: . . . An antenna is a structure that transforms Radiation Resistance into the Impedance of free space, as shown, and by definition. Both use identical MKS units, both are identical characteristics. Sorry, that's a demonstrably absurd assertion. Hi Roy, 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. So, how do the "ohms" of free space differ from the "ohms" of a quarter wave monopole's Radiation Resistance? The demonstrables you offer do not enlighten us in what physical constants these unique terms of your usage diverge from those in the MKS system. As I pointed out in my posting, whatever derivation for the characteristic Z of free space is, it must ultimately devolve to the identical expression for the common Ohm. I would offer by way of caution that the expression Zc = (µ0 / e0)^0.5 where µ0 is expressed in Henrys per meter and where e0 is expressed in Farads per meter; that that, too, arrives at the same Ohms employed by carbon resistors and Radiation Resistance. This is much like trying to compare miles per gallon and kilometers per liter. When push comes to shove, the reduction to MKS will reveal that the same container of gas will get you down the same stretch of asphalt the same distance no matter what mix of terms you substitute for liquid volume and length. This goes to include offbeat descriptive terms like miles per liter; kilometers per gallon; or furlongs per hogshead for that identical container of gas. Solutions of proportionality are not unique physical constants. 73's Richard Clark, KB7QHC |
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