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"Roy Lewallen" bravely wrote to "All" (07 Apr 05 15:08:40)
--- on the heady topic of " VF, low-loss line, high-impedence line - relationship" RL From: Roy Lewallen RL Xref: aeinews rec.radio.amateur.antenna:28088 RL Asimov wrote: Since a portion of the EM field in open wire line is free to travel outside the conductor into the environment then we may safely assume there is an exchange between the environment and the conductor. RL If the conductors are perfectly conducting, no part of the field at RL all exists within the conductor. With good conductors like copper and RL at HF and above, there's very little penetration of the conductor by RL the fields, either electric or magnetic. Is there any electron current in the conductor or not? RL First of all, a mismatch doesn't cause loss. An impedance mismatch in any medium causes a scattering of the energy. Of course, it isn't a net loss as far as the universe is concerned but some of the energy doesn't arrive where it was intended. RL Secondly, as I explained in my last posting, the characteristic RL impedance of a transmission line isn't the same thing as the RL characteristic impedance of free space. May I suggest you make up your mind whether the electric energy is travelling in a conductive medium or not? RL It has to do with the reflective coefficient where the energy is returned. RL Well, no. There isn't a bundle of energy trying to escape the line and RL bouncing off the air, or bouncing off the air as it travels along the RL line, or bouncing off the conductors into the air. So reflection RL coefficient isn't applicable here. What makes you so sure? RL I'm afraid that the conclusions you've reached about loss and RL characteristic impedance are based on a poor understanding of RL fundamental transmission line operation. The result is some RL conclusions that are, and are well known to be, untrue. I think you are only concerned with modeling of transmission lines as lumped constants but models can only go so far in explaining how something works. Models are like analogies and we all know no analogy is perfect even this one. A*s*i*m*o*v .... Anyone not wearing 2,000,000 sunblock is gonna have a REAL_ BAD_ DAY_7 |
On Fri, 08 Apr 2005 06:42:05 -0700, Wes Stewart
wrote: On Fri, 08 Apr 2005 06:05:49 -0700, Wes Stewart wrote: I'll try to scan it to pdf and post is somewhere. http://www.qsl.net/n7ws/Sterba_Openwire.pdf My first scan didn't have the appendix. It's there now. |
On Fri, 08 Apr 2005 10:19:31 -0500, Cecil Moore
wrote: Wes Stewart wrote: http://www.qsl.net/n7ws/Sterba_Openwire.pdf There seems to be a dotted line for feedline radiation going to zero as feedline length goes to zero. :-) So it would seem. QRX, Reg will explain. :-) |
Reg Edwards wrote:
In summary, the system as a whole BEHAVES as if there is NO radiation from the line itself ... How about BPL? (The Devil made me do it.) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Ian, G3SEK wrote:
"Now tell us about radiation from the line." Reg`s problem has a line as long as it is wide. That would make a very small square loop if I read right. The same current flows in all 4 sides. According to Arnold King: Beta is a factor = 2 pi/Lambda Say Lambda=1, then Beta=2pi Radiation resistance= 20 times Beta to the 4th times A squared Say the sides of the tiny square loop are 0.01Lambda, then A, the loop`s enclosed area is 0.0001 Radiation resistance = 20 (1555) (10 tp -8th) =31100 (10 to -8th) =0.00031 ohm Radiated power = Isquared (radiation resistance) Radiated power = negligible Best regards, Richard Harrison, KB5WZI |
On Fri, 08 Apr 2005 08:54:11 -0700, Wes Stewart
wrote: On Fri, 08 Apr 2005 06:42:05 -0700, Wes Stewart wrote: On Fri, 08 Apr 2005 06:05:49 -0700, Wes Stewart wrote: I'll try to scan it to pdf and post is somewhere. http://www.qsl.net/n7ws/Sterba_Openwire.pdf My first scan didn't have the appendix. It's there now. Hi Wes, Could I trouble you for a copy of section IV that continues on beyond page 1180? I am interested in the discussion relating to Figure 12 so you can restrict your efforts to that if this section runs on too long. Thanks, Richard, KB7QHC |
Reg Edwards wrote:
You've told us about radiation from the connections to the generator and the termination. Now tell us about radiation from the line. ================================= Ian, you are falling into the same sort of trap as old wives who imagine most radiation comes from the middle 1/3rd of a dipole because that's where most of the current is. It is self-misleading to consider the various parts of a radiating system to be separate components which are capable of radiating independently of each other. They can't. Actually they can, because that isn't the same as saying... A system's behaviour must be treated as a whole. That is true, of course. Every component of an antenna (or in this case, a parallel-wire transmission line) interacts with every other component. The totality of those interactions is what determines how the RF voltage and current will distribute themselves along the wires. But once you know the magnitude and phase of the current in each small segment of the antenna (which need not depend on theory or modeling - in principle you could go around and measure it) then you have taken complete account of the interactions. The radiated field from the whole antenna is then the sum of the fields from the individual components radiating independently. However, we weren't originally talking about that... We have already discussed that the power radiated from a generator + twin-line + load is a constant and is independent of line length. No, you have only asserted that. Total power radiated is equal to that radiated from a wire having a length equal to line spacing with a radiation resistance appropriate to that length. The location of the radiator, insofar as the far-field is concerned, can be considered to be at the load. The current which flows in the radiator is the same as that flowing in a matched load. And the load current is independent of line length. Only if there are no radiative losses from the line itself - and you have only asserted that, not proved it. Mathematically, the only way for the total power radiated to remain constant and independent of line length is for zero radiation from the line. Well obviously - but that is a circular argument, based entirely on your assertion that the power delivered to the load is independent of the line length. In summary, the system as a whole BEHAVES as if there is NO radiation from the line itself - only from fictitious very short monopoles (or dipoles?) at its ends. Sorry, but the "behaves as if" argument doesn't wash, because those short monopoles are real. Since the line spacing is non-zero, those short transverse sections must always exist, both in practice and in your circuit model. Each section carries RF current, so it radiates - no question about that, but it is entirely an end effect. It has nothing whatever to do with radiation from the main line. Looking edge-on at the line, we have two conductors carrying equal and opposite currents, but one is slightly farther away than the other so their transverse radiated fields do not quite cancel out. The only question is mathematical: how does the small loss of energy through radiation translate into a dB/m or dB/wavelength loss along the transmission line? -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Ian, G3SEK wrote:
"---how does the small loss of energy through radiation translate into dBm or dB/wavelength loss along a transmission line?" It is the reverse of a beverage antenna, which is a sort of single-wire transmission line above the earth in its simple configuration. The Beverage is a horizontal wire sensitive to vertically polarized waves. It is working with the wave throughout its travel along its length. The Beverage is vertically polarized because that is the direction of the electric field between its conductors, the wire and the earth beneath the wire. The direction of the electric field in a parallel-wire transmission line is from wire to wire. The effective radiator length of this polarization is the line spacing. This is short compared to the length of the transmission line in nearly all cases. The radiation is not emerging from the end of the transmission line. It radiates slightly all along the line as the wave navigates the line, much as the Beverage gathers energy slightly as the wave sweeps along its length. Best regards, Richard Harrison, KB5WZI |
"Ian White G3SEK" wrote in message ... Reg Edwards wrote: You've told us about radiation from the connections to the generator and the termination. Now tell us about radiation from the line. ================================= Ian, you are falling into the same sort of trap as old wives who imagine most radiation comes from the middle 1/3rd of a dipole because that's where most of the current is. It is self-misleading to consider the various parts of a radiating system to be separate components which are capable of radiating independently of each other. They can't. Actually they can, because that isn't the same as saying... A system's behaviour must be treated as a whole. That is true, of course. Every component of an antenna (or in this case, a parallel-wire transmission line) interacts with every other component. The totality of those interactions is what determines how the RF voltage and current will distribute themselves along the wires. But once you know the magnitude and phase of the current in each small segment of the antenna (which need not depend on theory or modeling - in principle you could go around and measure it) then you have taken complete account of the interactions. The radiated field from the whole antenna is then the sum of the fields from the individual components radiating independently. However, we weren't originally talking about that... We have already discussed that the power radiated from a generator + twin-line + load is a constant and is independent of line length. No, you have only asserted that. Total power radiated is equal to that radiated from a wire having a length equal to line spacing with a radiation resistance appropriate to that length. The location of the radiator, insofar as the far-field is concerned, can be considered to be at the load. The current which flows in the radiator is the same as that flowing in a matched load. And the load current is independent of line length. Only if there are no radiative losses from the line itself - and you have only asserted that, not proved it. Mathematically, the only way for the total power radiated to remain constant and independent of line length is for zero radiation from the line. Well obviously - but that is a circular argument, based entirely on your assertion that the power delivered to the load is independent of the line length. In summary, the system as a whole BEHAVES as if there is NO radiation from the line itself - only from fictitious very short monopoles (or dipoles?) at its ends. Sorry, but the "behaves as if" argument doesn't wash, because those short monopoles are real. Since the line spacing is non-zero, those short transverse sections must always exist, both in practice and in your circuit model. Each section carries RF current, so it radiates - no question about that, but it is entirely an end effect. It has nothing whatever to do with radiation from the main line. Looking edge-on at the line, we have two conductors carrying equal and opposite currents, but one is slightly farther away than the other so their transverse radiated fields do not quite cancel out. The only question is mathematical: how does the small loss of energy through radiation translate into a dB/m or dB/wavelength loss along the transmission line? -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Asimov wrote:
"Roy Lewallen" bravely wrote to "All" (07 Apr 05 15:08:40) --- on the heady topic of " VF, low-loss line, high-impedence line - relationship" RL From: Roy Lewallen RL Xref: aeinews rec.radio.amateur.antenna:28088 RL Asimov wrote: Since a portion of the EM field in open wire line is free to travel outside the conductor into the environment then we may safely assume there is an exchange between the environment and the conductor. RL If the conductors are perfectly conducting, no part of the field at RL all exists within the conductor. With good conductors like copper and RL at HF and above, there's very little penetration of the conductor by RL the fields, either electric or magnetic. Is there any electron current in the conductor or not? In a perfect conductor, no. In a real but good conductor like copper, it's confined to a very thin layer at the surface. Look up "skin effect" in any electromagnetics or basic text on RF, or google it. RL First of all, a mismatch doesn't cause loss. An impedance mismatch in any medium causes a scattering of the energy. Of course, it isn't a net loss as far as the universe is concerned but some of the energy doesn't arrive where it was intended. If I connect a 50 ohm source to a one wavelength, 300 ohm transmission line and connect the other end of that line to a 50 ohm resistor, there's a 6:1 mismatch at both ends of the line. The power supplied by the source and the power delivered to the load are exactly the same as if I had used a 50 ohm line instead. This is, of course, overlooking resitive loss in the line. If you consider the resistive loss, it can be greater in one line than the other (the 300 ohm line might be less lossy), depending on the physical construction of the line. No loss is caused by the mismatch. No "scattering of energy occurs". All of the energy from the source arrives at the load, where it was intended. RL Secondly, as I explained in my last posting, the characteristic RL impedance of a transmission line isn't the same thing as the RL characteristic impedance of free space. May I suggest you make up your mind whether the electric energy is travelling in a conductive medium or not? I'm sorry, I don't understand what you're asking. No RF energy exists in or travels in a perfect conductor. RL It has to do with the reflective coefficient where the energy is returned. RL Well, no. There isn't a bundle of energy trying to escape the line and RL bouncing off the air, or bouncing off the air as it travels along the RL line, or bouncing off the conductors into the air. So reflection RL coefficient isn't applicable here. What makes you so sure? A basic understanding of electromagnetics derived from an electrical engineering education, extensive additional reading and study, and about 30 years of engineering design experience including design of microwave and very high speed time domain circuitry. RL I'm afraid that the conclusions you've reached about loss and RL characteristic impedance are based on a poor understanding of RL fundamental transmission line operation. The result is some RL conclusions that are, and are well known to be, untrue. I think you are only concerned with modeling of transmission lines as lumped constants but models can only go so far in explaining how something works. Models are like analogies and we all know no analogy is perfect even this one. I have no idea what makes you think that my modeling or understanding of transmission lines is limited to lumped constant models -- it's certainly not true. Indeed, no analogy is perfect, but some are certainly better than others, and some are demonstrably false. I'm afraid that some of the ones you've put forth are in the latter category. Roy Lewallen, W7EL |
"Ian White said - Looking edge-on at the line, we have two conductors carrying equal and opposite currents, but one is slightly farther away than the other so their transverse radiated fields do not quite cancel out. =========================== Ian, Oh yes they do. Next to each half wavelength of line there is another half wavelength of line in which the current is in antiphase with it. And so, in the far field, the fields from adjacent half-wavelengths of line cancel each other out. Now you'll say my logic falls down when the line length is an odd number of half wavelengths. But you must not consider half wavelengths of line to be behaving independently of each other. ---- Reg, G4FGQ |
Reg Edwards wrote:
Ian, Oh yes they do. Next to each half wavelength of line there is another half wavelength of line in which the current is in antiphase with it. And so, in the far field, the fields from adjacent half-wavelengths of line cancel each other out. . . . No, they don't. They cancel only in two directions, directly normal to the plane containing the wires. Radiation occurs in all other directions, because the fields don't add in antiphase. An example of an antenna which uses two closely spaced elements carrying equal out-of-phase currents is the W8JK. Roy Lewallen, W7EL |
"Roy Lewallen" bravely wrote to "All" (08 Apr 05 15:05:11)
--- on the heady topic of " VF, low-loss line, high-impedence line - relationship" RL From: Roy Lewallen RL Xref: aeinews rec.radio.amateur.antenna:28146 RL Asimov wrote: Is there any electron current in the conductor or not? RL In a perfect conductor, no. In a real but good conductor like copper, RL it's confined to a very thin layer at the surface. Look up "skin RL effect" in any electromagnetics or basic text on RF, or google it. If I put an RF voltmeter/ampmeter at any point along the perfect conductor I will get a zero reading is that right? So then if the conductor was perfect then I wouldn't need a conductor at all in the first place? Seems a shame to waste perfect unobtainium just to have the last word. ;-) RL First of all, a mismatch doesn't cause loss. An impedance mismatch in any medium causes a scattering of the energy. RL If I connect a 50 ohm source to a one wavelength, 300 ohm transmission RL line and connect the other end of that line to a 50 ohm resistor, RL there's a 6:1 mismatch at both ends of the line. [,,,] RL No loss is caused by the mismatch. No "scattering of energy occurs". RL All of the energy from the source arrives at the load, where it was RL intended. So why bother with high priced cable when a piece of clothes hanger will do just as well, encantations? I think your are still in April 1 mode, Mr.Roy. We are way past diatribe here, thank you. A*s*i*m*o*v .... Pandora's Law: Never open a box you didn't close yourself |
I hope that most readers will see from the examples I've given that the
assertions you've made are false. And perhaps some will be spurred to do a little reading and studying in order to better understand the topic. I don't think anything else I'm likely to say will change the minds of those who aren't interested in either looking at the evidence or learning more about the subject, so my time is better spent at other pursuits. I'll leave you with your notions undisturbed and intact. Roy Lewallen, W7EL Asimov wrote: . . . So why bother with high priced cable when a piece of clothes hanger will do just as well, encantations? I think your are still in April 1 mode, Mr.Roy. We are way past diatribe here, thank you. A*s*i*m*o*v ... Pandora's Law: Never open a box you didn't close yourself |
On Friday, 08 Apr 2005 22:59:00 -500, In a bit of not so clever
editing "Asimov" wrote in a response to a perfectly clear post by Roy: RL If I connect a 50 ohm source to a one wavelength, 300 ohm transmission RL line and connect the other end of that line to a 50 ohm resistor, RL there's a 6:1 mismatch at both ends of the line. [,,,] RL No loss is caused by the mismatch. No "scattering of energy occurs". RL All of the energy from the source arrives at the load, where it was RL intended. So why bother with high priced cable when a piece of clothes hanger will do just as well, encantations? I think your are still in April 1 mode, Mr.Roy. We are way past diatribe here, thank you. What Roy actually said was: "If I connect a 50 ohm source to a one wavelength, 300 ohm transmission line and connect the other end of that line to a 50 ohm resistor, there's a 6:1 mismatch at both ends of the line. The power supplied by the source and the power delivered to the load are exactly the same as if I had used a 50 ohm line instead. This is, of course, overlooking resitive loss in the line. If you consider the resistive loss, it can be greater in one line than the other (the 300 ohm line might be less lossy), depending on the physical construction of the line." So you see, our shyster poster left out the part where Roy considers that resistive loss plays a part, and by inference makes copper wire a better choice than clothes hangers. |
On Sat, 09 Apr 2005 01:56:47 -0700, Roy Lewallen
wrote: I hope that most readers will see from the examples I've given that the assertions you've made are false. And perhaps some will be spurred to do a little reading and studying in order to better understand the topic. I don't think anything else I'm likely to say will change the minds of those who aren't interested in either looking at the evidence or learning more about the subject, so my time is better spent at other pursuits. I'll leave you with your notions undisturbed and intact. And it is to be hoped, expressed in another forum. |
"Roy Lewallen" wrote Reg Edwards wrote: Ian, Oh yes they do. Next to each half wavelength of line there is another half wavelength of line in which the current is in antiphase with it. And so, in the far field, the fields from adjacent half-wavelengths of line cancel each other out. . . . No, they don't. They cancel only in two directions, directly normal to the plane containing the wires. Radiation occurs in all other directions, because the fields don't add in antiphase. An example of an antenna which uses two closely spaced elements carrying equal out-of-phase currents is the W8JK. Roy Lewallen, W7EL ================================= Roy, I've never head of a W8JK. You are confusing the issue. The problem is concerned with a LONG balanced transmission line and its terminations which form part of the whole radiating system. And as we can agree it is incorrect to consider parts of the system in isolation. To simplify the questions, wthout loss of rigor, it is best to consider the line itself as being lossless with matched terminations. I have stated that power radiated from the system is independent of line length and nobody has disagreed. Indeed, a radiating power calculating formula from reputable authors (of which I was unaware) has confirmed this. The power radiated from the system is identical to that radiated from a monopole or short dipole, of length equal to wire spacing, with a current equal to the current which flows in the terminations (ie., the load). The terminations actually exist. Radiated power = Load current-squared times calculated radiation resistance. That is obviously true down even to zero line length. The implication is that radiation occurs only from the termination(s) and that no radiation occus from the line. But, I repeat, we must NOT consider the parts in isolation as do old wives. You have stated that radiation from the line itself (in isolation) must exist in the plane of the wires because of the finite spacing between the line wires. But we must consider only the far field. Not that in the immediate vicinity of the line and its termination. I suspect that the radiation pattern of a LONG-line SYSTEM converges towards that from a monopole located in the position of the load. Many of us are curious to acquire an idea of what the radiation pattern looks like. You are familiar with programs which produce far-field radiation patterns. Do you know of a program which accurately produces the radiation pattern of a very long close-spaced, zero resistance, pair of wires terminated with a wire of length equal to wire spacing and including a load resistance equal to Zo. Patterns, of course, will change with frequency. It will be necessary to statistically analyse results. Or just look at them from a common sense point of view. From a practical engineering viewpoint it is quite sufficient to know what I innocently stated in the first place - the minute amount of power lost is the load's radiation resistance times load current squared and is independent of line length. ---- Reg, G4FGQ |
Reg Edwards wrote:
"Roy Lewallen" wrote Reg Edwards wrote: Ian, Oh yes they do. Next to each half wavelength of line there is another half wavelength of line in which the current is in antiphase with it. And so, in the far field, the fields from adjacent half-wavelengths of line cancel each other out. . . . No, they don't. They cancel only in two directions, directly normal to the plane containing the wires. Radiation occurs in all other directions, because the fields don't add in antiphase. An example of an antenna which uses two closely spaced elements carrying equal out-of-phase currents is the W8JK. Roy Lewallen, W7EL ================================= Roy, I've never head of a W8JK. You are confusing the issue. The problem is concerned with a LONG balanced transmission line and its terminations which form part of the whole radiating system. And as we can agree it is incorrect to consider parts of the system in isolation. To simplify the questions, wthout loss of rigor, it is best to consider the line itself as being lossless with matched terminations. I have stated that power radiated from the system is independent of line length and nobody has disagreed. Here on the back row, there's always been a hand raised in disagreement on that point. No argument that it's very, very small. But exactly zero - definitely not. Indeed, a radiating power calculating formula from reputable authors (of which I was unaware) has confirmed this. If you mean the Sterba reference, then please re-read it. All the endorsements of the formula that you quote are peppered with caveats such as "an approximation" and "providing that operations are confined to wavelengths other than those within the ultra-short-wave region." This is for the very good reason that some small amount of transverse radiation does exist. Transverse radiation in the plane of the line is small because the vector components of radiation from the two parallel lines are equal in magnitude and almost exactly opposite in phase - but never exactly opposite. I am probably the only person in this discussion who has actually USED parallel-wire lines "within the ultra-short-wave region". If you can maintain good balance, the losses due to transverse radiation are negligibly small for engineering purposes. But to claim they are exactly zero is a physical absurdity... and I'll always disagree with those. (Sorry, I'll have to be out of this discussion again for a while.) -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
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