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Antenna reception theory
Gene Fuller wrote:
If you know anything about physics, you must know that energy is at the core of almost all physical analysis. I couldn't agree more, but that's exactly the topic from which you tried to divert attention in your posting. Here's what you said: "I guess I still need to seek out some "real-world physicists" to figure out that energy is indeed something to "worry about". I never would have imagined such a thing!" If you don't worry about energy, you have nothing to add to the discussion. If you do worry about energy, please read my "WorldRadio" article which tells you more than you (and others) ever wanted to know about energy in an RF transmission line. -- 73, Cecil http://www.qsl.net/w5dxp |
Antenna reception theory
Cecil,
Sorry, I should have written, "I guess I still need to seek out some "real-world physicists" to figure out that energy is indeed something to "worry about". I never would have imagined such a thing!" 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) I thought my intention was obvious, but it seems I failed to communicate. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: If you know anything about physics, you must know that energy is at the core of almost all physical analysis. I couldn't agree more, but that's exactly the topic from which you tried to divert attention in your posting. Here's what you said: "I guess I still need to seek out some "real-world physicists" to figure out that energy is indeed something to "worry about". I never would have imagined such a thing!" If you don't worry about energy, you have nothing to add to the discussion. If you do worry about energy, please read my "WorldRadio" article which tells you more than you (and others) ever wanted to know about energy in an RF transmission line. |
Antenna reception theory
Roy Lewallen, W7EL wrote:
"A short while ago, I explained why your Faraday cage doesn`t separate the E and H fields as you claim." n I am misunderstood. I never used the tem Faraday cage. I understand the Faraday cage to be a completely shielded enclosure which could be a metal automobiole body, a steel rebar reinforced concrete structure or a screened room. These all tend to completely block both the E-field and the H-field components of an electromagnetic wave. I gave an example of the metal rakes (Faraday screens) used at the medium wave broadcast plants where I worked. The function of the Faraday screen was primarily to eliminate capacitive cooupling, and thus the E-field between the transmission line and the antenna. This reduces harmonic radiation. It adds expense and complexity but is worth the price for lightning suppression if for nothing else. Most Faraday screens are used at powerline frequencies to eliminate capaacitive coupling between primary and secondary of a power transformer. I checked the internet for a recent posting and found an article by who wrote: "Beware, this cannot be an auto-transformer. It must be a transformer with two completely isolated windings, and with a Faraday Screen wound between the two." Why would one pay a lot of extra money for a transformer which eliminated capacitive coupling if it didn`t work, especially in Havana, Cuba? Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Gene, W4SZ wrote:
"How do you know that they also negate the fundamental properties of time-varing electric and magnetic fields as expressed by Maxwell`s equations?" I did not make that claim. Pits from the lightning strikes are evidence that the voltage gradient is high between the coil and the Faraday screen. The second coil is clean, not pitted. I explained the operation of a Faraday screen earlier but was misunderstood so I`ll quote a professional writer and consulting engineer, B. Whitfield Griffith, Jr. who writes on page 246 of "Radio-Electronic Transmission Fundamentals": "It is often desirable and practical to allow one type of coupling between circuits and eliminate the other. Figure 28-1, for instance, shows two coils which are coupled but have a special type of shielding between them that causes the coupling to be purely inductive and not capacitive, This shield consists of an array of parallel wires which are grounded at one end only. The lines of electric force from the coils terminate on these grounded wires, so that there are no electric lines of force passing from one coil to the other and hence no capacitive coupling. On the other hand, since the wires do not form closed loops, there is no circulating current in the shield and therefore nothing to stop the penetration of the shield by the magnetic field. This type of shield has long been known as a Faraday screen. It is important that the capacitive coupling be eliminated in many cases, since it has the characteristics of a high-pass filter, tending to accentuate the harmonic content of the transmitted signal." I never tried it, but I`d bet that someone has measured the radiated harmonics both with and without the Faraday screens inplace. That would prove effectiveness. Short-circuiting the open ends of the Faraday screen wires would kill transfer of the signal between primary and secondary of the transformer, and the transmitter would likely complain. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
*Sigh*
Richard Harrison wrote: Roy Lewallen, W7EL wrote: "A short while ago, I explained why your Faraday cage doesn`t separate the E and H fields as you claim." n I am misunderstood. I never used the tem Faraday cage. I understand the Faraday cage to be a completely shielded enclosure which could be a metal automobiole body, a steel rebar reinforced concrete structure or a screened room. These all tend to completely block both the E-field and the H-field components of an electromagnetic wave. Sorry, I meant "Faraday screen", which is the term you used, and I used in my posting explaining its operation. If you block either the E or H field, you also block the other. You can't independently block one or the other. . . . Why would one pay a lot of extra money for a transformer which eliminated capacitive coupling if it didn`t work, especially in Havana, Cuba? Because in order to "work" it doesn't need to "eliminate capacitive coupling". All it needs to do is locally reduce the E/H field ratio, which is what it does. Roy Lewallen, W7EL |
Antenna reception theory
Gene Fuller wrote:
I thought my intention was obvious, but it seems I failed to communicate. Others on this newsgroup have admonished me for worrying about energy and refused to discuss the subject. I thought you were doing the same. Sorry. But do you actually have any references that contradict "Optics", by Hecht? In Dr. Best's article, he superposes V1 with V2 such that constructive interference energy is needed to complete the superposition. On this newsgroup, I asked Dr. Best where that necessary constructive interference energy comes from and he didn't know. That's when I went searching for references and found them in the field of optics. Constructive interference energy can be supplied by local sources as occurs in W7EL's "Food for Thought #1" with its DC example. Or constructive interference energy can be supplied at a point away from the source(s) by destructive interference, e.g. wave cancellation at the non-reflection surface of a layer of thin-film on glass or at a match point in a transmission line. -- 73, Cecil http://www.qsl.net/w5dxp |
Antenna reception theory
After reading this thread it seems there is significant confusion between E and
H fields created locally from electric current and the E and H aspects of a passing RF photon stream. It is easy to see how this can occur. That said the tools for working with them are entirely different. E and H fields from capacitors, inductors, and transformers are predominantly local effects. Photons are not involved, they are created by electric current and are explained by Maxwell equations. They can be created separately, that is a capacitor is pretty much a pure E field and an inductor an H field, albeit none of these are perfect. Photons are not involved in the operation of a On the other hand EM waves and photons which are RF waves and visible light and beyond are entirely different. Although they are referred to as electro-magnetic, they are in fact massless particles with a well defined energy that is a function of frequency. These particles, traveling at the speed of light, create what is referred to as RF including RF fields. The E and M components cannot be separated. Antennas are devices which convert RF currents into photons for transmission and to convert photons into RF currents on reception, all this at the frequencies we are mostly interested in. These are two very different physical constructs. Now at RF frequencies - Faraday shields are local electrical devices, they have nothing to do with photons and antennas. (Although they may be effective at blocking their passage.) Electric currents in equipment create both E and H fields. These fields are not associated with photons and can be separated. Faraday shields are a way to block the E portions of those local fields while allowing the passage of the H. Dan Roy Lewallen wrote: *Sigh* Richard Harrison wrote: Roy Lewallen, W7EL wrote: "A short while ago, I explained why your Faraday cage doesn`t separate the E and H fields as you claim." n I am misunderstood. I never used the tem Faraday cage. I understand the Faraday cage to be a completely shielded enclosure which could be a metal automobiole body, a steel rebar reinforced concrete structure or a screened room. These all tend to completely block both the E-field and the H-field components of an electromagnetic wave. Sorry, I meant "Faraday screen", which is the term you used, and I used in my posting explaining its operation. If you block either the E or H field, you also block the other. You can't independently block one or the other. . . . Why would one pay a lot of extra money for a transformer which eliminated capacitive coupling if it didn`t work, especially in Havana, Cuba? Because in order to "work" it doesn't need to "eliminate capacitive coupling". All it needs to do is locally reduce the E/H field ratio, which is what it does. Roy Lewallen, W7EL |
Antenna reception theory
Richard Clark, KB7QHC wrote:
"From your copy of Bailey, review the text, and reconcile his remarks." Richard Clarks advice is good. Arnold B. Bailey in "TV and Other Receiving Antennas" does a more extensive job of explaining how antennas work than most other authors. Wish everybody interested in antennas could read Bailey.His catalog of antenna types is convenient too. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Dan Sawyeror wrote:
"They can be created separately, that is a capacitor is pretty much a pure E field and an inductor an H field." Agreed. Whenever a current grows or shrinks (with acceleration) it produces both an E-field and an H-field. I didn`t clam a permanent divorce, only an instantaneous separation. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
"dansawyeror" wrote - After reading this thread it seems there is significant confusion between E and H fields created locally from electric current and the E and H aspects of a passing RF photon stream. ======================================== When the geometric dimensions of the structures are short compared with a wavelength, the amps and volts are associated with the NEAR FIELD and Ohm's Law applies. The radiation field is so weak in comparison it can be forgotten about. KISS. Why does everybody HAVE to unnecessarily complicate matters in order to understand what really goes on? What have photons to do with winning a contest? ;o) ---- Reg. |
Antenna reception theory
"Richard Harrison" wrote in message
... Richard Clarks advice is good. Arnold B. Bailey in "TV and Other Receiving Antennas" does a more extensive job of explaining how antennas work than most other authors. Wish everybody interested in antennas could read Bailey.His catalog of antenna types is convenient too. It's out of print, but I've requested a copy through interlibrary loan. I'm curious to see what he has to say on the standard V-shaped rabbit ears that have been in use for decades... although one can readily simulate them and see exactly how they perform, I've yet to find anyone who had a good idea as to why rabbit ears haven't traditionally been oriented purely horizontally! |
Antenna reception theory
Roy Lewallen wrote: *Sigh* Richard Harrison wrote: Roy Lewallen, W7EL wrote: "A short while ago, I explained why your Faraday cage doesn`t separate the E and H fields as you claim." n I am misunderstood. I never used the tem Faraday cage. I understand the Faraday cage to be a completely shielded enclosure which could be a metal automobiole body, a steel rebar reinforced concrete structure or a screened room. These all tend to completely block both the E-field and the H-field components of an electromagnetic wave. Sorry, I meant "Faraday screen", which is the term you used, and I used in my posting explaining its operation. If you block either the E or H field, you also block the other. You can't independently block one or the other. . . . Why would one pay a lot of extra money for a transformer which eliminated capacitive coupling if it didn`t work, especially in Havana, Cuba? Because in order to "work" it doesn't need to "eliminate capacitive coupling". All it needs to do is locally reduce the E/H field ratio, which is what it does. Roy Lewallen, W7EL It might be useful to also point out the means by which these electric and/or magnetic shields do their job. They do it not so much by blocking as much as by diverting the fields. They serve to conduct the field around the object that is being shielded. 73 de ac6xg |
Antenna reception theory
Cecil,
I do not have a copy of Hecht, but I doubt that he has made any serious mistakes. Certainly he should have no mistakes in an area that is as well understand and widely discussed as plane wave interactions with discontinuities in the medium. The classic treatment of this problem, found in virtually every college-level textbook on E&M or optics, is to set up the appropriate wave equations, add the boundary conditions, and crank out the answer. Then there is typically some sort of analysis and discussion that says, "The reflected intensity plus the transmitted intensity is equal to the incident intensity. Energy is conserved." I suspect Hecht provides exactly that sort of description. I know that all of the relevant textbooks I have do so. I believe you are reading too much into something Hecht is saying, perhaps in an effort to somehow reconcile conservation of energy. The beauty of the laws of E&M, as expressed by Maxwell's equations and other fundamental properties, is that conservation of energy is automatic, at least in ordinary circumstances. If one correctly solves for the field equations, the energy conservation will come along for free. Conversely, it is customary to use energy considerations as the primary vehicle for addressing many physical problems in advanced mechanics, quantum mechanics, solid state physics, and other branches of science. The bottom line is that there are a number of tools available to develop correct solutions to physical problems. Steve Best chose one path, and you choose another. You both come up with the same answer in terms of what can be measured. The mathematical constructs underlying the solution may be different, but those constructs are not directly measurable. Don't limit your toolbox. Sometimes a screwdriver is easier to use than a monkey wrench. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: I thought my intention was obvious, but it seems I failed to communicate. Others on this newsgroup have admonished me for worrying about energy and refused to discuss the subject. I thought you were doing the same. Sorry. But do you actually have any references that contradict "Optics", by Hecht? In Dr. Best's article, he superposes V1 with V2 such that constructive interference energy is needed to complete the superposition. On this newsgroup, I asked Dr. Best where that necessary constructive interference energy comes from and he didn't know. That's when I went searching for references and found them in the field of optics. Constructive interference energy can be supplied by local sources as occurs in W7EL's "Food for Thought #1" with its DC example. Or constructive interference energy can be supplied at a point away from the source(s) by destructive interference, e.g. wave cancellation at the non-reflection surface of a layer of thin-film on glass or at a match point in a transmission line. |
Antenna reception theory
Reg Edwards wrote:
What have photons to do with winning a contest? ;o) Just try winning a contest without them. :-) -- 73, Cecil, http://www.qsl.net/w5dxp |
Antenna reception theory
Gene Fuller wrote:
I do not have a copy of Hecht, but I doubt that he has made any serious mistakes. It would be worth your while to visit a local university and check out Hecht's chapters on superposition and interference. "The reflected intensity plus the transmitted intensity is equal to the incident intensity. Energy is conserved." I suspect Hecht provides exactly that sort of description. Much more than that. As you know, irradiance is power/unit-area and Hecht spends many pages on irradiance and energy. I believe you are reading too much into something Hecht is saying, perhaps in an effort to somehow reconcile conservation of energy. Spoken by someone who hasn't even read Hecht? I suspect if you read Hecht, you would perceive the same information as I. Hecht is big on conservation of energy and spends many pages discussing such things involving EM waves. The beauty of the laws of E&M, as expressed by Maxwell's equations and other fundamental properties, is that conservation of energy is automatic, at least in ordinary circumstances. There is still an underlying Q&A about what happens to the energy in those waves. The energy concept is in addition to what's already there, not any kind of replacement for it. The bottom line is that there are a number of tools available to develop correct solutions to physical problems. Steve Best chose one path, and you choose another. Nope, we chose the same path. Steve just fell off the path and down the cliff about 2/3 of the way through his articles. Steve gave us a very good picture of what happens to the energy toward the load but he gave us a distorted view of what happens to the energy toward the source. Instead of Steve's one-sided approach, I presented both sides thus merely expanding what Steve had already done. Don't limit your toolbox. Funny, just above you seemed to recommend limiting the toolbox to Maxwell's equations and tried to discourage me from thinking about energy. -- 73, Cecil, http://www.qsl.net/w5dxp |
Antenna reception theory
Cecil,
The waves you are so worried about are merely convenient, but fictitious, adjuncts to your mathematical model. You need worry only about the energy of real, measurable waves, not those adjuncts that simplify the math. The use of such adjuncts is done frequently in solving real problems. Just don't confuse the internals of the model with physical reality. 73, Gene, W4SZ Cecil Moore wrote: There is still an underlying Q&A about what happens to the energy in those waves. The energy concept is in addition to what's already there, not any kind of replacement for it. |
Antenna reception theory
Gene Fuller wrote:
Cecil, The waves you are so worried about are merely convenient, but fictitious, adjuncts to your mathematical model. You need worry only about the energy of real, measurable waves, not those adjuncts that simplify the math. The use of such adjuncts is done frequently in solving real problems. Just don't confuse the internals of the model with physical reality. 73, Gene, W4SZ Might this be an example, Gene? From "An Energy Analysis at an Impedance Discontinuity in an RF Transmission Line, Part I" "100% wave cancellation means 100% energy reflection." 73 de ac6xg |
Antenna reception theory
Gene Fuller wrote:
The waves you are so worried about are merely convenient, but fictitious, adjuncts to your mathematical model. Let's see: 1. Canceled waves are "fictitious adjuncts to my math model". 2. Therefore, they don't need to be canceled, because they are only "fictitious adjuncts to my math model". 3. Therefore, we can remove whatever is doing the canceling of those "fictitious adjuncts" without changing anything. 4. Darn, how is it that I can see those "fictitious adjuncts to my math model" so well that I can't see anything else? What's wrong with this picture? Could there be a not-so-hidden contradiction accompanied by confusion of cause and effect? You need worry only about the energy of real, measurable waves, not those adjuncts that simplify the math. You don't even know me, Gene. Where do you get the balls to decide what I need to worry about and what I don't need to worry about? I have been worried and needing an energy analysis model for 40 years. I now have one and can now sleep like a baby at night, but no thanks to you. :-) -- 73, Cecil, http://www.qsl.net/w5dxp |
Antenna reception theory
"Joel Kolstad" bravely wrote to "All" (06 Dec 05 09:28:55)
--- on the heady topic of " Antenna reception theory" JK From: "Joel Kolstad" JK Xref: core-easynews rec.radio.amateur.antenna:220898 JK "Richard Harrison" wrote in message JK ... Richard Clarks advice is good. Arnold B. Bailey in "TV and Other Receiving Antennas" [,,,] JK It's out of print, but I've requested a copy through interlibrary JK loan. I'm curious to see what he has to say on the standard V-shaped JK rabbit ears that have been in use for decades... although one can JK readily simulate them and see exactly how they perform, I've yet to JK find anyone who had a good idea as to why rabbit ears haven't JK traditionally been oriented purely horizontally! I think the reason why you don't see rabbit ears oriented horizontally is that they don't seem to work well as dipoles. When standing up they aren't even a V antenna and at first one would think they are vertically polarized but they are somewhat directive with a bipolar pattern. Rabbit ears are a *******ized form of a couple antenna types. Sometimes mainly one element is responsible for most of the signal while the other behaves as a reference or ground. For example on low frequency channels like Ch-3, reception is best if one of the elements is oriented straight up and the other horizontally pointing in the direction of the transmitter. This won't work well with Ch-12 and the standing V is best then. A*s*i*m*o*v .... Men are men and needs must err. - Euripides |
Antenna reception theory
Jim Kelley wrote:
"100% wave cancellation means 100% energy reflection." That's a no-brainer when quoted in the context of an RF transmission line. Shame on you for once again quoting something out of context. If it happened in space, it would say, "100% wave cancellation means 100% energy redistribution." as explained on the web pages below. There are only two directions in a transmission line. If energy rejected by a mismatched load doesn't make it to the source, in a lossless system, it must necessarily be 100% re-reflected. Rho^2 is reflected by the impedance discontinuity at the match point. That's step one. Step two is that 100% wave cancellation. Given those two steps, no reflected energy flows toward the source. Therefore, it must necessarily have been re-reflected back toward the load. Since it joins the forward energy wave, we can measure that is exactly what happens. It's a no-brainer. Here's the in-context quote. The destructive interference energy resulting from wave cancellation at an impedance discontinuity becomes an equal magnitude of constructive interference in the opposite direction. Since there are only two directions in a transmission line, wave cancellation is the equivalent of an energy reflection. 100% wave cancellation means 100% energy reflection. [9] [9] Quotes from two web pages from the field of optical engineering: www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be zero." "In the absence of absorption or scatter, the principle of conservation of energy indicates all ‘lost’ reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." http://micro.magnet.fsu.edu/primer/j...ons/index.html "... when two waves of equal amplitude and wavelength that are 180-degrees ... out of phase with each other meet, they are not actually annihilated, ... All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." Note from W5DXP: In an RF transmission line, since there are only two possible directions, the only "regions that permit constructive interference" at an impedance discontinuity is the opposite direction from the direction of destructive interference. -- 73, Cecil, http://www.qsl.net/w5dxp |
Antenna reception theory
Cecil,
Sorry, I mistakenly thought the discussion was about technology. Silly me! Since it is actually about psychology, not physics, I am outta here. 73, Gene W4SZ Cecil Moore wrote: You don't even know me, Gene. Where do you get the balls to decide what I need to worry about and what I don't need to worry about? I have been worried and needing an energy analysis model for 40 years. I now have one and can now sleep like a baby at night, but no thanks to you. :-) |
Antenna reception theory
Gene Fuller wrote:
Sorry, I mistakenly thought the discussion was about technology. Silly me! We wouldn't have technology without the human need and drive to understand nature. Many great scientists were textbook examples of O-C personalities. -- 73, Cecil http://www.qsl.net/w5dxp |
Antenna reception theory
Roy Lewallen, W7EL wrote:
"A short while ago, I explained why your Faraday cage doesn`t separate E and H as you claim." There are many examples of Faraday screens at work removing the E field while allowing H field coupling. I chose the Faraday sdreens in medium wave broadcasting used to reduce radiation of harmonics of the channel frequency. More numerous examples abound. These are the Faraday screens used in isolation power transformers between primary and secindary coils to prevent transient high-frequency energy coupling. My 19th edition of the ARRL Antenna Book contains yet another example on page 14-2 which says: "Fig 2 - Shielded loop for direction finding. The ends of the shielding turn are not connected, to prevent shielding the loop from magnetic fields. The shield is effective against electric fields. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Roy, W7EL wrote:
"If the wavelength is 1m, the voltage induced in the center of an open-circuited 1m dipole by a 1V/m field is 0.5 volt, not 1 volt." An open-circuited dipole (has a small gap in its middle) is mot resonnant at a wavelength of 2 meters, but its individual pieces are resonant at a wavelength of 1 meter. At 2 meters, not a wavelength 1m, each 0.5 m piece of the half-wavelength, 1 meter long dipole has a high reactance because 0.5 m is too short at a wavelength of 2 meters to be resonant. At longer wavelengths, the reactance rises.. High reactance does not oppose non-existing current in an open-circuit. I agree the voltage induced in 1/2-meter of wire properly placed within a 1V/m uniform field is 0.5 volt, not 1 volt. The induced voltage in a wire within a uniform field sweeping the wire rises uniformly along the wire. It can be assumed to be the summation of tiny increments of voltage all along the wire. The voltages of the too-short dipole halves add just as two cells in some flashlights add. Their vectors are head to tail. But, current will be limited by radiation and loss resistances of the wires. It will also be limited by reactance in the wires. Open-circuit, 0.5 V + 0.5 V = 1V. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Richard (Harrison),
What is the voltage measured between the bottom end and ground, of a 1 metre high vertical antenna, above a perfect ground, when the vertically-polarised field strength is 1 volt per metre, and antenna height is shorter than a 1/4-wavelength. Just a number please. ---- Reg. |
Antenna reception theory
Reg, G4FGQ wrote:
"What is the voltage measured between the bottom end and ground of a 1 metre high vertical antenna above a perfect ground when the verticallly-polarized field strength is 1 volt per metre, and antenna height is shorter than a 1/4-wavelength?" I see no tricks in the question. The field strength is given as "1 volt per metre". From page 23-3 of the 19th edition of the ARRL Antenna Book: "The standard of measure for field intensities is the voltage developed in a wire that is 1 meter long, expressed as volts per meter." While field strength is not the same as the volts delivered to a receiver, because of the voltage division between antenna impedance and receiver input impedance, there is no voltage division when the antenna is loaded with an open circuit. We assume the r-f voltmeter used to measure voltage at the base of the antenna has an infinite input impedance. The antenna used for field strength measurements is often a loop, but we are not concerned with the measurement itself. Reg had a very simple question, "What`s the voltage at the base of a 1-meter high wire? The voltage at the base of an open-circuit 1-metre wire iis one volt because it goes straight to the definition of field intensity in volts/m. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Richard Harrison wrote:
. . . I agree the voltage induced in 1/2-meter of wire properly placed within a 1V/m uniform field is 0.5 volt, not 1 volt. The induced voltage in a wire within a uniform field sweeping the wire rises uniformly along the wire. It can be assumed to be the summation of tiny increments of voltage all along the wire. The voltages of the too-short dipole halves add just as two cells in some flashlights add. Their vectors are head to tail. But, current will be limited by radiation and loss resistances of the wires. It will also be limited by reactance in the wires. Open-circuit, 0.5 V + 0.5 V = 1V. There are two incorrect statements here. First, the voltage induced in the wire doesn't rise uniformly along the wire. It's sinusoidal, even for a very short wire. This is different from the transmitting case but interestingly doesn't interfere with reciprocity. Secondly, the voltage at the center of an open-circuited 1 meter (electrically short) dipole in and parallel a 1 volt/meter field is 0.5 volt as I said earlier, not 1 volt. I'll be glad to provide a number of references. Both these statements can also be verified by modeling. The definition of field strength, incidentally, has nothing to do with the voltage of a dipole immersed in that field. Roy Lewallen, W7EL |
Antenna reception theory
Richard Harrison wrote:
. . . The voltage at the base of an open-circuit 1-metre wire iis one volt because it goes straight to the definition of field intensity in volts/m. Do you have a reference which defines field strength in terms of voltage induced in a wire? Roy Lewallen, W7EL |
Antenna reception theory
Roy Lewallen wrote"
"Do you have a reference which defines field strength in terms of voltage induced in a wire?" Here is more of one I already posted. It is from the 19th edition of the ARRL Antenna Book on page 23-3: "The strength of a wave is measured as the voltage between two points lying on an electric line of force in the plane of the wave front. The standard of measure for field intensity is the voltage developed in a wire that is 1 meter long, expressed as volts per meter. (If the wire were 2 meters long, the voltage developed would be divided by two to determine the field strength in volts per meter.)" Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Richard Harrison wrote:
Roy Lewallen wrote" "Do you have a reference which defines field strength in terms of voltage induced in a wire?" Here is more of one I already posted. It is from the 19th edition of the ARRL Antenna Book on page 23-3: "The strength of a wave is measured as the voltage between two points lying on an electric line of force in the plane of the wave front. The standard of measure for field intensity is the voltage developed in a wire that is 1 meter long, expressed as volts per meter. (If the wire were 2 meters long, the voltage developed would be divided by two to determine the field strength in volts per meter.)" Best regards, Richard Harrison, KB5WZI Thanks. I'll contact Dean Straw to get that corrected. Roy Lewallen, W7EL |
Antenna reception theory
Roy Lewallen , W7EL wrote:
"Do you have a reference which defines field strength in terms of voltage induced in a wire?" Here is a reference from a professional source. B. Whitfield Griffith, Jr. was Director of Advanced Development at General Dynamics Corporation at Garland, Texas when his book, "Radio-Electronicc Transmission Fundamentals" was published by Mc Graw-Hill in 1972. On page 322, Griffith writes: "The strength of an electromagnetic wave is generally measured in terms of the intensity of the electric field; this is expressed in volts per meter, or millivolts or microvolts per meter, as the conditions may indicate. This value may be understood as being the numbeer of volts which would be induced in a piece of wire one meter long placed in the field parallel to the electric lines of force; the induction of voltage would result from the movement of the magnetic flux across the wire. Griffith agrees with Terman. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
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Antenna reception theory
Asimov wrote:
"He clearly contradicts himself in my opinion." Griffith echoes Terman. Terman writes on page 2 of his 1955 edition: "The strength of a radio wave is measured in terms of the voltage stress produced by the electric field of the wave and it is usually expressed in microvolts per meter." Terman does not literally mean that it is the electric field which is measured. In his handbook Terman says that a magnetic loop antenna is usually the field strenngth meter`s antenna. Terman meant that field strength is quoted in volts per meter. It is irrelevant which field is actually measured as the two fields are locked in magnitude by the impedance of space, 377 ohms. If you know one, you know the other. Terman also wrote on page 2: "The strength of the wave measured in terms of microvolts per meter of stress in space is also exactly the same voltage that the magnetic flux of the wave induces in a conductor 1 m long when sweeping across this conductor with the velocity of light." The electric and magnetic fields serve to recreate each other in their flight from their source. Either could be measured to get the strength of the wave. Neither field directly produces volts in a wire. As Terman says, it is the wave sweeping the wire at the velocity of light, the d(phi)/dt that generates the volts. Best regards, Richard Harrison, KB5WZI |
Antenna reception theory
Asimov wrote: This reminds me of a contradiction I find in Einstein's equivalence principle. He states accelerating at 1G is the same as being pulled by the Earth's gravity at 1G. In my opinion, they are similar but they are not quite the same thing simply because in the former there is motion and for the latter there is not. A contradiction which might only appear if you were to confuse little g and big G. Eistein was referring to the former, which is the acceleration a falling body experiences as a result of the force of gravity on Earth - a change of velocity of 9.8 meters per second in one second. Clearly motion is involved. Big G is the universal gravitational constant, which is in units of Newton meters squared per kilogram squared. Clearly, this relates to a force in Newtons which varies as the square of mass and the inverse square of distance, according to Newton's law of gravitation. Motion is only involved if the force is not met with an equal and opposing force - as is obviously the case when standing on the ground. ac6xg |
Antenna reception theory
Roy Lewallen wrote: Richard Harrison wrote: . . . I agree the voltage induced in 1/2-meter of wire properly placed within a 1V/m uniform field is 0.5 volt, not 1 volt. The induced voltage in a wire within a uniform field sweeping the wire rises uniformly along the wire. It can be assumed to be the summation of tiny increments of voltage all along the wire. The voltages of the too-short dipole halves add just as two cells in some flashlights add. Their vectors are head to tail. But, current will be limited by radiation and loss resistances of the wires. It will also be limited by reactance in the wires. Open-circuit, 0.5 V + 0.5 V = 1V. There are two incorrect statements here. First, the voltage induced in the wire doesn't rise uniformly along the wire. It's sinusoidal, even for a very short wire. Interesting. Assuming a plane wave sweeping broadside, with the field being the same at every point along the wire, one might be inclined to argue that the voltage induced on a wire should be the same at every point along a finite length. The "rise" in voltage as the field sweeps past would be with respect to time, rather than with respect to position. Sort of explains why a radio receiver works just fine with only one wire attached when you think about it. ;-) 73, ac6xg |
Antenna reception theory
On Fri, 16 Dec 2005 13:56:12 -0800, Jim Kelley
wrote: one might be inclined to argue that the voltage induced on a wire should be the same at every point along a finite length One might, if it were an exceptionally short piece of wire. Otherwise it must exhibit some greater reactance than nearly 0 and support a difference of potential in proportion to the current through it. There is also the radiation (non-0) resistance to consider (same observation of a potential difference there too). Presumably, this all hinges on what is meant, in practical terms, for "finite length." The "rise" in voltage as the field sweeps past would be with respect to time, With respect to what time? The time for the wave to sweep past? What frequency? The full wave, or its peak? or its average (0)? rather than with respect to position. Position of what? If we could generate a step-wave, and that wave impinges upon the wire broadside, and we were to arbitrarily assign the frequency of 3 MHz (so that the wire length of 1 Meter is one percent long - a short antenna by any definition); then the first degree would immediately demand a response in 1nS. 1nS is a very appreciable dT for 1 Meter's worth of inductance. Still and all, 1nS is hardly the slice of the wave an antenna wire sees, unless it is 1nS thick - or 33cm. This comes close to a tower's size, but receivers work with far thinner stuff and the dV/dT certainly pushes the voltage drop across 1 meter of that wire up higher. 73's Richard Clark, KB7QHC |
Antenna reception theory
Jim Kelley wrote:
Sort of explains why a radio receiver works just fine with only one wire attached when you think about it. ;-) It doesn't work fine when you don't think about it? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Antenna reception theory
Richard Clark wrote:
On Fri, 16 Dec 2005 13:56:12 -0800, Jim Kelley wrote: one might be inclined to argue that the voltage induced on a wire should be the same at every point along a finite length One might, if it were an exceptionally short piece of wire. Otherwise it must exhibit some greater reactance than nearly 0 and support a difference of potential in proportion to the current through it. There is also the radiation (non-0) resistance to consider (same observation of a potential difference there too). You seem to be arguing that the induced potential in the wire is due to IR or IZ. I don't think that's true. According to JC Maxwell, the magnitude of the induced voltage and current should depend on the EM field and on the in situ permitivity and permeability of the wire. As long as those things are constant along the length of that wire, there is no reason to expect anything but uniform voltage and current along the length of the wire. As Roy said, an antenna does not behave the same on receive as it does on transmit apparently. A receive antenna does not appear to behave like a transmission line. Presumably, this all hinges on what is meant, in practical terms, for "finite length." Finite here just means small compared to any curvature of the wave front. Were the wire long enough compared to the distance to the source, one couldn't expect the fields to be uniform along the length of the wire. The "rise" in voltage as the field sweeps past would be with respect to time, With respect to what time? The time for the wave to sweep past? What frequency? The full wave, or its peak? or its average (0)? It just means voltage as a function of time - like the waveform induced by the fields of a passing electromagnetic wave for example. rather than with respect to position. Position of what? It means position along the wire - as in the topic of the conversation you've joined. I apologize for any lack of clarity on my part. 73, ac6xg |
Antenna reception theory
This thread has presented a clear illustration of the danger of quoting
from a book, even an authoritative one, without fully understanding the context. Before I continue, let me stipulate that the conductors being talked about here are all electrically short (much shorter than a wavelength), and are placed parallel to the E field. The following statements won't be generally true if both those conditions aren't met. I've been saying that the voltage at the center of an open circuited 1 m dipole immersed in a 1 V/m field is 0.5 volts. Richard has quoted some references which say that the voltage induced in a 1 m wire immersed in a 1 V/m field is 1 volt, and implying that it follows that the voltage at the center of an open circuited dipole of that length is 1 volt. It isn't, and I'll try to explain why. The voltage drop, or emf, across a very short conductor in a field of E volts/meter is E*L volts, where L is the length of the very short conductor in meters. This is undoubtedly the reason for the statements like the ones Richard has quoted. If we look at the emf generated across each tiny part of a one meter long dipole by a 1 V/m field and add them up, we'll find that they total 1 volt. But what's the voltage across the dipole, from end to end? The answer to that is it's just about anything we want it to be. Any time we try to measure the voltage between two points in space when there's a time-varying H field present, the answer we get depends on the path we take between the two points. Crudely but not entirely accurately put, it depends on how we arrange the voltmeter leads. We can, however, measure the voltage at the dipole center, across a gap that's arbitrarily small. So what's that voltage? I can't think of any line of reasoning which would deduce that it's equal to the total emf along the wire (1 volt for our example). You can casually open various texts and find the answer to that -- it's either 0.5 volts, as I've said, or 1 volt, as Richard has said. To understand the reason for the apparent contradiction requires digging more deeply into the texts. Most of the texts I have analyze field-conductor interactions with a special kind of dipole which has a uniform current distribution -- that is, the current is the same amplitude all along the dipole's length. There are a couple of ways you can physically make a dipole like this. One is to begin with a conventional (but short) dipole and add large end hats, like a two-ended top loaded vertical. A number of authors (e.g., Kraus) show a diagram of such a dipole. Another way to get a distribution like this is to stipulate that the dipole really be only a tiny segment of a longer wire. The short conductor with uniform current is often called a "Hertzian dipole", sometimes an "elemental dipole", sometimes just a short or infinitesimal dipole, or by Terman, a "doublet". As Balanis (_Antenna Theory - Analysis and Design_, p. 109) says, "Although a constant current distribution is not realizable, it is a mathematical quantity that is used to represent actual current distributions of antennas that have been incremented into many small lengths." The voltage at the center of a one meter antenna of this type in a 1 V/m field is 1 volt. But this fictitious dipole, used for conceptual and computational convenience, isn't a real dipole. A real dipole -- that is, just a single, straight wire with a source or load at its center and no end hats -- doesn't have a uniform current distribution. Instead, the current is greatest in the center, dropping to zero at the ends. When transmitting, the current on a short dipole drops nearly linearly from the center to the ends. When receiving, the current distribution is nearly sinusoidal. The net result of this non-uniform current is that the voltage at the center is less than it is for a uniform-current dipole -- exactly half as much, actually. Conceptually, it's because the current near the ends contributes less to the voltage at the center. I'm going to wave my hands over the significance of the different receiving and transmitting current distributions, except to say that reciprocity is still satisfied in all ways, including the transmitting and receiving impedances being the same. To add confusion, the gain, directivity, and effective apertures of both types of dipole are the same -- 1.5, 1.5, and 3 * lambda^2 / (8 * pi) respectively. This means that you can extract the same amount of power from an impinging wave with either type of antenna, provided that you terminate each in the complex conjugate of its transmitting feedpoint impedance. The radiation resistance of the uniform-current dipole is 4 times that of the conventional dipole. (Remember, these are all electrically short.) There's a common term for the relationship between the field strength and the length of a conductor, called the "effective height" or "effective length". The voltage at the center of a dipole in a field of E volts/m is simply E * the effective length. The concept is valid for any length conductor, not just short ones. The effective length of a uniform-current dipole is equal to the wire length. The effective length of a short conventional dipole is 0.5 times the wire length. The effective length for receiving is the same as the effective length for transmitting -- in transmitting, it relates the strength of the field produced to the *voltage* -- not power -- applied across the feedpoint. If you apply 0.5 volts to a standard dipole and 1.0 volts to a uniform-current dipole, the power applied to each will be the same because of the 1:4 ratio of radiation resistance, and the generated fields will be the same. This is consistent with the antenna gains being the same. As I mentioned, most text authors use a uniform-current dipole for analysis. One which directly derives the voltage of a standard short dipole is King, Mimno, and Wing, _Transmission Lines, Antennas, and Wave Guides_. Many others, including Kraus, _Antennas_ (2nd Ed. p. 41), derive the effective length for a short conventional dipole as 0.5 * the physical length, from which the open circuit voltage due to an impinging field can easily be determined. As a last note on a point of contention, electric field strength is usually defined not by the voltage induced in a conductor but from the force between charges using the Lorenz force law. The unit of electric field strength is found to be newtons/coulomb, which is the same as volts/meter. Among the texts using this definition are Kraus (_Electromagnetics_), Terman (_Radio Engineering_), Ida, Majid, and Ramo et al. Roy Lewallen, W7EL |
Antenna reception theory
Jim Kelley wrote:
Interesting. Assuming a plane wave sweeping broadside, with the field being the same at every point along the wire, one might be inclined to argue that the voltage induced on a wire should be the same at every point along a finite length. The "rise" in voltage as the field sweeps past would be with respect to time, rather than with respect to position. Sort of explains why a radio receiver works just fine with only one wire attached when you think about it. ;-) The Earth's (static) electric field is about 120 V/m at ground level on a stormless day. Better not walk too fast. Jogging might be fatal! Roy Lewallen, W7EL |
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