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50 Ohms "Real Resistive" impedance a Misnomer?
Hello,
I'd like to start a discussion (or light a brush-fire, as the case may be for this NG!), about what a 50 Ohm impedance match really means. On our trusty Smith Chart, assuming it is normalized to 50 Ohms, the center is considered and labeled as the "real resistive" 50 match point. In fact, the entire middle horizontal line is the "real" part of the impedance. I'm sure many of you have read the popular description/model of a transmission line as an infinite chain of alternating series inductors with shunt capacitors, with the resulting characteristic impedance as Z=(L/C)**1/2, where the L and C are distributed inductances and capacitances. So, in theory, if you have achieved a perfect match with your antenna, you will have matched the impedance to the 377 Ohms of free space, you will not have reflections at the matching point, and the energy will radiate in whatever pattern you have designed for. The funny thing about this, is that you cannot say that the 50 Ohms in the center of the chart is a "resistive" 50 Ohms, as there is very little real resistance in the average antenna. This "resistive" 50 Ohms is really what people call the "radiation" resistance, which is something of a misnomer again, because this is trying to equate the successful impedance matching and subsequent non-reflected EM radiation with a truly real resistance like an ideal dummy load. Of course, it's well known that a truly real resistive 50 Ohm dummy load should appear exactly like a properly matched antenna to the transmitter. Why do i ask all this? Well, if you believe that complex impedance measurements (series equivalent) by MFJ antenna analyzers are not completely inaccurate, then it appears that two 1/4 watt 100 Ohm resistors in parallel (lead lengths short) are a much more consistent 50 Ohms over the VHF band than almost all the higher power dummy loads we have tested. Problem is, the high power dummy loads will vary from 52 to 45 "real" ohms depending on the frequency, with the "real" part of the impedance getting lower with increasing frequency, so it doesn't seem to be a "skin effect". The spread gets much worse when you put a 3' jumper coax in between, and even more worse when you add a power/swr meter. Then the "real" Ohms will be from 65 to 35 ohms, with the max and mins not correlating with frequency at all, and the stray reactances will be much more too, but just as varied with frequency. So much for "50 ohm" jumper cables! I suppose they are as close as they can get them for a particular price. My theory is that the "real" part of the impedance is mainly the truly resistive 50 ohms of the dummy load at low frequencies around 10 MHz or so...but as you go up in frequency, the parasitics of the dummy load and the coax jumper cable will cause "radiation" resistance to be mixed in with this truly real 50 ohms, giving us readings all over the map. What do you folks think? Dr. Slick |
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Dr. Slick wrote:
Hello, I'd like to start a discussion (or light a brush-fire, as the case may be for this NG!), about what a 50 Ohm impedance match really means. On our trusty Smith Chart, assuming it is normalized to 50 Ohms, the center is considered and labeled as the "real resistive" 50 match point. In fact, the entire middle horizontal line is the "real" part of the impedance. I'm sure many of you have read the popular description/model of a transmission line as an infinite chain of alternating series inductors with shunt capacitors, with the resulting characteristic impedance as Z=(L/C)**1/2, where the L and C are distributed inductances and capacitances. So, in theory, if you have achieved a perfect match with your antenna, you will have matched the impedance to the 377 Ohms of free space, you will not have reflections at the matching point, and the energy will radiate in whatever pattern you have designed for. No, an antenna doesn't "match" the impedance of free space. The input impedance of an antenna is the ratio of V to I. The impedance of free space is the ratio of the E field to the H field of a plane wave. They both happen to have units of ohms, but they're different things and there's no "matching" going on. If you apply 100 watts to an antenna, resonant or not, 100 watts will be radiated, less loss, regardless of the antenna's input impedance. The funny thing about this, is that you cannot say that the 50 Ohms in the center of the chart is a "resistive" 50 Ohms, as there is very little real resistance in the average antenna. Sure you can. You're confusing resistance with resistors. Resistance is a dimension, like length. There are lots of things which have dimensions of resistance but aren't resistors, like transresistance, characteristic resistance of a transmission line, or radiation resistance to name just a few. People who have a shaky understanding of basic electric circuit theory seem to have trouble dealing with this, but it becomes easier to deal with as you learn more about basic electricity. A mechanical example is torque and work, which have the same dimensions (force times distance) but are definitely different things. This "resistive" 50 Ohms is really what people call the "radiation" resistance, which is something of a misnomer again, because this is trying to equate the successful impedance matching and subsequent non-reflected EM radiation with a truly real resistance like an ideal dummy load. Sorry, that doesn't make a whole lot of sense. Yes, it's called the radiation resistance, but it's not a misnomer at all. (I suppose it would be if you called it a "radiation resistor", but nobody I know of has ever called it that.) If you calculate the power "consumed" by this resistance (that is, the power flow into it), it's the power being radiated. If the radiation and loss resistance of an antenna were zero, no energy would flow into it and consequently none would be radiated. (If the radiation resistance was zero and the loss resistance wasn't, then energy would flow into the antenna but none would be radiated -- it would all be dissipated as heat.) Your use of "non-reflected EM radiation" seems to imply that the radiation from an antenna is somehow bounced back from space if the antenna feedpoint impedance is reactive. That's one of the rather bizarre and very wrong conclusions you could draw from the mistaken idea that the antenna "matched" the characteristic impedance of free space. Of course, it's well known that a truly real resistive 50 Ohm dummy load should appear exactly like a properly matched antenna to the transmitter. At one frequency. Why do i ask all this? Well, if you believe that complex impedance measurements (series equivalent) by MFJ antenna analyzers are not completely inaccurate, then it appears that two 1/4 watt 100 Ohm resistors in parallel (lead lengths short) are a much more consistent 50 Ohms over the VHF band than almost all the higher power dummy loads we have tested. No surprise. It's much harder to make something that's physically big have a consistent impedance at high frequencies than for something physically small, simply because stray inductances and capacitances are both larger for large objects. Let that be a lesson. Problem is, the high power dummy loads will vary from 52 to 45 "real" ohms depending on the frequency, with the "real" part of the impedance getting lower with increasing frequency, so it doesn't seem to be a "skin effect". The spread gets much worse when you put a 3' jumper coax in between, and even more worse when you add a power/swr meter. Then the "real" Ohms will be from 65 to 35 ohms, with the max and mins not correlating with frequency at all, and the stray reactances will be much more too, but just as varied with frequency. So much for "50 ohm" jumper cables! I suppose they are as close as they can get them for a particular price. It's no trick at all to tranform a real resistance to a different value of real resistance using only purely reactive L and C components. It's done all the time. An L network, with two components, is the simplest circuit which can pull off this magical trick. Just pick a point on the real axis of that Smith chart of yours and follow reactance lines around -- first XL, then XC, or vice-versa, until you end up back on the real axis again -- at a different resistance value. The amount of XL and XC you transit along the way are in fact the values you'd need to make an L network to do the transformation. As for the transmission line, start at, say, 45 ohms, then go in a circle around the center, reading off R and X values as you go. Those are the values of R and X you can get with a 50 ohm transmission terminated with a 45 + j0 ohm load. Of course, "50 ohm" lines are often quite a ways off -- I've measured them at up to 62 ohms or so. And you're right, you can get better ones if it's worth a lot to you. Oh, also notice that if you start on the real axis anywhere but the center and go around a half circle, representing 90 degrees of lossless transmission line, you end up at a different place on the real axis. Presto! You've pulled off a transformation of a purely real impedance with a lossless transmission line. Cool, huh? My theory is that the "real" part of the impedance is mainly the truly resistive 50 ohms of the dummy load at low frequencies around 10 MHz or so...but as you go up in frequency, the parasitics of the dummy load and the coax jumper cable will cause "radiation" resistance to be mixed in with this truly real 50 ohms, giving us readings all over the map. Although radiation will cause an increase in terminal resistance (remember, it accounts for the radiated power), it's not at all necessary in order to cause the dummy load resistance (real part of the impedance) to vary. The stray L and C can do that all by themselves, without any radiation at all. What do you folks think? I think you'd benefit a lot from learning how to do some basic operations with a Smith chart. It would broaden your horizons a lot. Roy Lewallen, W7EL |
Dr. Slick wrote:
The funny thing about this, is that you cannot say that the 50 Ohms in the center of the chart is a "resistive" 50 Ohms, as there is very little real resistance in the average antenna. From the IEEE Dictionary: "resistance (1)(B) The real part of impedance." Apparently, all the resistance in the average antenna is real. :-) -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
Dr. Slick wrote:
Have you heard of something called mis-match loss? If the antenna's input impedance is not matched to the tranmission line (or the final PA), then the radiated power will be significantly less than 100 watts. Not if there's an antenna tuner (Z0-match) in the circuit. The following will radiate most of the mismatch loss from the load. 100W XMTR--50 ohm feedline--+--1/2WL 150 ohm feedline--50 ohm load All of the reflected energy is re-routed back toward the load at the '+' Z0-match point through re-reflection and wave cancellation. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Dr. Slick wrote:
"The funny thing about this is that you cannot say that the 50 ohms in the center of the chart is a "resistive" 50 ohms as there is very little real resistance in the average antenna." Resistance is defined as real. That is, current is instantaneously proportional to the voltage. Any efficient antenna has a high ratio of radiation resistance to loss resistance. Resistance is the ratio of in-phase voltage to current accepted by an antenna. Part is made by loss in the antenna. part is made by radiation from the antenna. They are often represented by an equivalent circuit of two resistors in series. Dr, Frederick Emmons Terman says of radiation resistance: "This is the resistance that, when inserted in series with the antenna, will consume the same amount of power as is actually radiated. ---it is customary to refer the radiation resistance to a current maximum in the case of an ungrounded antenna, and to the base of the antenna when the antenna is grounded." Best regards, Richard Harrison, KB5WZI |
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Dr. Slick wrote:
Roy Lewallen wrote in message ... So, in theory, if you have achieved a perfect match with your antenna, you will have matched the impedance to the 377 Ohms of free space, you will not have reflections at the matching point, and the energy will radiate in whatever pattern you have designed for. No, an antenna doesn't "match" the impedance of free space. The input impedance of an antenna is the ratio of V to I. The impedance of free space is the ratio of the E field to the H field of a plane wave. They both happen to have units of ohms, but they're different things and there's no "matching" going on. If you apply 100 watts to an antenna, resonant or not, 100 watts will be radiated, less loss, regardless of the antenna's input impedance. I disagree on this point. You can think of an antenna as a type of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned antenna is "matching" 50 to 377 ohms, and will therefore have no reflections, ideally. Thinking of it doesn't make it so. All the energy applied to the antenna is radiated, less loss, regardless of the antenna's feedpoint impedance. How does that fit into your model. Have you heard of something called mis-match loss? If the antenna's input impedance is not matched to the tranmission line (or the final PA), then the radiated power will be significantly less than 100 watts. I have indeed, and have posted several times about this often-misunderstood and misused term. You can find the postings by going to http://www.groups.google.com and searching this group for postings by me containing "mismatch loss". The funny thing about this, is that you cannot say that the 50 Ohms in the center of the chart is a "resistive" 50 Ohms, as there is very little real resistance in the average antenna. Sure you can. You're confusing resistance with resistors. Resistance is a dimension, like length. There are lots of things which have dimensions of resistance but aren't resistors, like transresistance, characteristic resistance of a transmission line, or radiation resistance to name just a few. People who have a shaky understanding of basic electric circuit theory seem to have trouble dealing with this, but it becomes easier to deal with as you learn more about basic electricity. A mechanical example is torque and work, which have the same dimensions (force times distance) but are definitely different things. ok, so perhaps the way to think of it is: when an antenna is matched, the I and V curves what curves? will be in phase (no reactance), and the product of I*V (integrated) will be the power transmitted. The average power radiated is always the real part of V*I(conjugate). If V and I are in phase, this is simply equal to V*I. But what does this have to do with the confusion between a resistor and resistance? This "resistive" 50 Ohms is really what people call the "radiation" resistance, which is something of a misnomer again, because this is trying to equate the successful impedance matching and subsequent non-reflected EM radiation with a truly real resistance like an ideal dummy load. Sorry, that doesn't make a whole lot of sense. Yes, it's called the radiation resistance, but it's not a misnomer at all. (I suppose it would be if you called it a "radiation resistor", but nobody I know of has ever called it that.) If you calculate the power "consumed" by this resistance (that is, the power flow into it), it's the power being radiated. If the radiation and loss resistance of an antenna were zero, no energy would flow into it and consequently none would be radiated. (If the radiation resistance was zero and the loss resistance wasn't, then energy would flow into the antenna but none would be radiated -- it would all be dissipated as heat.) If both the radiation and loss resistance were zero, i would expect an ideal short, and therefore full -180 degree reflections. You're assuming that the antenna is fed with a transmission line, but that's ok. I suppose what this all means is that if you have a matched antenna, it's V and I curves what curves? will be IN PHASE and will have the exact same RMS values as if you had a truly resistive dummy load instead. Yes and no. If you're feeding the antenna with a 450 ohm line, it's matched to the line only if its impedance is 450 + j0 ohms, so it looks like a 450 ohm dummy load, not a 50 ohm one. On the other hand, you can feed a 50 + j0 ohm antenna with a half wavelength of 450 ohm line, and get a perfect 50 ohm match to a transmitter at the input end of the line, while running a 9:1 SWR on the the transmission line. Then either the antenna or the input of the line looks like a 50 ohm dummy load. Therefore, you can consider the center of the Smith Chart (or the entire real (non-reactive) impedance line) as a real resistance like an ideal dummy load. Do you agree with this statement Roy? Yes. Your use of "non-reflected EM radiation" seems to imply that the radiation from an antenna is somehow bounced back from space if the antenna feedpoint impedance is reactive. That's one of the rather bizarre and very wrong conclusions you could draw from the mistaken idea that the antenna "matched" the characteristic impedance of free space. I think i'm correct to think of antennas as impedance matching transformers. 50 Ohms to 377 Ohms. I like to think of 'em as sort of potato guns, launching RF potato photons into the aether. But that doesn't make them potato guns. Feel free to think of them any way you like, as long as you consistently get the right answer. Now tell me, why can't you just make some 377 ohm transmission line (easy to make) open circuited, and dispense with the antenna altogether? When you figure out the answer to that one, you might begin to see the error with your mental model. . . . I believe i understand the Chart better than you think, Roy, enough to know that you do know what you are talking about. I still think it's ok to consider antennas as impedance tranformers, but you have brought up some very good points. Ok by me. I'll be waiting for your patented no-antenna 377 ohm feedline. Roy Lewallen, W7EL |
Roy Lewallen wrote:
I'll be waiting for your patented no-antenna 377 ohm feedline. I wonder if that patent has ever been applied for? :-) -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote in message ...
Dr. Slick wrote: The funny thing about this, is that you cannot say that the 50 Ohms in the center of the chart is a "resistive" 50 Ohms, as there is very little real resistance in the average antenna. From the IEEE Dictionary: "resistance (1)(B) The real part of impedance." Apparently, all the resistance in the average antenna is real. :-) -- You misunderstand my point, Cecil. If the antenna is tuned correctly, the "radiation" resistance is a real 50 Ohms, with the V and I waveforms IN PHASE. But, my point is that you can take a DC measurement anywhere on the ideal lossless antenna and you will never see 50 Ohms anywhere, only shorts. Ideally, the antenna never heats up, and has no resistive losses. This is not to say that it won't have a "radiation" resistance of 50 Ohms at the resonant, tuned frequency. And to a transmitter at the resonant frequency, this will electrically appear to be the same thing as an ideal dummy load of 50 Ohms. Roy's message has clarified a few things. Slick |
Dr. Slick wrote:
W5DXP wrote: Apparently, all the resistance in the average antenna is real. :-) You misunderstand my point, Cecil. Actually, it was a play on words using your definition of "real" Vs the IEEE Dictionary and mathematical definition of "real". The Devil made me do it but I did provide a smiley face. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Dr. Slick wrote:
Why would you need the antenna tuner in this case? Assuming your XMTR is 50 Ohms on the output: 100W XMTR--50 ohm feedline--+---50 ohm load Voila! A perfect match. No mis-matchlosses, and no need for a tuner. Yes, but a real-world antenna rarely has a 50 ohm feedpoint impedance even though it is resonant. The feedpoint resistance of a resonant 1/2WL dipole can vary from about 40 ohms to about 100 ohms depending upon its height above ground. And a resonant dipole won't usually cover the entire 75m band without a tuner of some kind. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Dr. Slick wrote:
"But, my point is that you can take a DC measurement anywhere on the ideal lossless antenna and you will never see 50 Ohms anywhere, only shorts." True. D-C resistance of a lossless wire is zero. But, apply 50 watts r-f power to an antenna adjusted to present 50-ohms resistance at a particular frequency. What happens? Our loading, adjusted for 50 watts output, produces 50 volts at an in-phase current of 50 amps. The 50 watts is being radiated. As far as the transmitter is concerned, the power might just as well be feeding a dummy load. There are several reasons a d-c ohmmeter doesn`t read radiation resistance. Once the ohmmeter`s d-c has charged the antenna, current flow stops. D-C doesn`t radiate. A d-c ohmmeter doesn`t even measure the right loss resistance value. Some of the loss resistance in a true antenna with actual losses, comes from skin effect at radio frequencies, where the conduction is forced to the outside of the conductors, decreasing their effective cross-section and increasing their loss. R-F also has the ability to induce eddy currents in surrounding conductors and to agitate molecular movement in surrounding insulators. R-F thus increases loss over that measured by passing d-c through the antenna. Radiation resistance can be readily measured by using an r-f bridge instrument. The bridge and antenna are excited by a generator operating at the same frequency the transmitter will use. The bridge will indicate reactance in the antenna, if it is not resonant. The null detector for the bridge is typically a good radio receiver. Radiation resistance is real though it does not heat the antenna. Loss resistance is real though it does heat the antenna and its surroundings. A resistance is a volt to amp ratio in which amps are in-phase with the volts. A resistor is a special type of resistance in which the electrical energy applied to the resistance is converted into heat. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
What happens? Our loading, adjusted for 50 watts output, produces 50 volts at an in-phase current of 50 amps. Hmmmm, 50 watts in, 2500 watts out. How much will you take for that antenna, Richard? :-) -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
Roy Lewallen wrote in message ...
I disagree on this point. You can think of an antenna as a type of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned antenna is "matching" 50 to 377 ohms, and will therefore have no reflections, ideally. Thinking of it doesn't make it so. All the energy applied to the antenna is radiated, less loss, regardless of the antenna's feedpoint impedance. How does that fit into your model. What if the antenna was a wire shorting the transmission line? Then there would be very little radiated power, and a lot of reflected power. You can make a transformer to match 50 Ohms to, say, 200 Ohms or so. Why not for 50 to 377? It's just another number. Except in the case of free space, with a given permeability and permittivity, you have the impedance of free space, which doesn't need a transmission line. ok, so perhaps the way to think of it is: when an antenna is matched, the I and V curves what curves? Well, a carrier with a single frequency will be a sinewave, correct? Feeding a real resistive impedance, the V and the I sinewaves will be in phase (no reactance). will be in phase (no reactance), and the product of I*V (integrated) will be the power transmitted. The average power radiated is always the real part of V*I(conjugate). If V and I are in phase, this is simply equal to V*I. But what does this have to do with the confusion between a resistor and resistance? It doesn't really, but it does relate to this discussion. In the sense that V and I will be in phase for either a matched antenna or an ideal dummy load. I suppose what this all means is that if you have a matched antenna, it's V and I curves will be IN PHASE and will have the exact same RMS values as if you had a truly resistive dummy load instead. Yes and no. If you're feeding the antenna with a 450 ohm line, it's matched to the line only if its impedance is 450 + j0 ohms, so it looks like a 450 ohm dummy load, not a 50 ohm one. On the other hand, you can feed a 50 + j0 ohm antenna with a half wavelength of 450 ohm line, and get a perfect 50 ohm match to a transmitter at the input end of the line, while running a 9:1 SWR on the the transmission line. Then either the antenna or the input of the line looks like a 50 ohm dummy load. You could make that half wavelength transmission line almost any other characteristic impedance, like 25 or 200 Ohms, and you would still wind up back at 50 Ohms at the input of the line (but you couldn't change frequencies unless they were multiples of 2 of the fundamental). But the point is, you will still be at 50 Ohms at the input, so the V and I sinewaves should be in phase. Therefore, you can consider the center of the Smith Chart (or the entire real (non-reactive) impedance line) as a real resistance like an ideal dummy load. Do you agree with this statement Roy? Yes. Cool. But a network analyzer looking into a black box will not be able to tell you whether the 50 Ohms it is reading is radiated resistance or dissipated resistance. This seems to be the crux of my question. Now tell me, why can't you just make some 377 ohm transmission line (easy to make) open circuited, and dispense with the antenna altogether? When you figure out the answer to that one, you might begin to see the error with your mental model. Ok by me. I'll be waiting for your patented no-antenna 377 ohm feedline. Roy Lewallen, W7EL Why would you need a 377 Ohm feedline for free space, when free space itself is the transmission line?? What's wrong with thinking of an antenna as a type of series Inductor, with a distributed shunt capacitance, that can be thought of as a type of distributed "L" matching network that transforms from 50 Ohms to 377? This is related to how you need to bend the ground radials of a 1/4 WL vertical whip at 45 deg angles down, to get the input impedance closer to 50 Ohms (as opposed to 36 Ohms or something like that if you leave them horizontal). Slick |
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Dr. Slick wrote:
Roy Lewallen wrote in message ... I disagree on this point. You can think of an antenna as a type of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned antenna is "matching" 50 to 377 ohms, and will therefore have no reflections, ideally. Thinking of it doesn't make it so. All the energy applied to the antenna is radiated, less loss, regardless of the antenna's feedpoint impedance. How does that fit into your model. What if the antenna was a wire shorting the transmission line? Then there would be very little radiated power, and a lot of reflected power. Yes, there would. But every watt delivered to that wire would be either radiated or dissipated. You can make a transformer to match 50 Ohms to, say, 200 Ohms or so. Why not for 50 to 377? It's just another number. That's right. You can make a transformer that converts the circuit V/I ratio to 377 ohms. But this doesn't make a plane wave whose E to H field ratio is 377 ohms. But you can have an antenna with a feedpoint impedance of, say, 3 - j200 ohms, and a couple of wavelengths from the antenna, the ratio of E to H field produced by that antenna will be 377 ohms. Another antenna, with feedpoint impedance of 950 + j700 ohms, will also produce a fields whose E to H ratio is 377 ohms. In fact, you can have any antenna impedance you'd like, and a few wavelengths away, the ratio of E to H field will be 377 ohms, provided the antenna is immersed in something resembling free space. But the fields in, on, or around a 377 ohm transmission line are unlikely to have an E to H ratio of 377 ohms. You're trying to say that the 377 ohm E/H ratio of free space is the same thing as a V/I ratio of 377 ohms. It isn't. Any more than 10 ft-lbs of torque is the same as 10 ft-lbs of work or energy. Except in the case of free space, with a given permeability and permittivity, you have the impedance of free space, which doesn't need a transmission line. Sorry, I can't make any sense out of that. ok, so perhaps the way to think of it is: when an antenna is matched, the I and V curves what curves? Well, a carrier with a single frequency will be a sinewave, correct? Feeding a real resistive impedance, the V and the I sinewaves will be in phase (no reactance). Yes. will be in phase (no reactance), and the product of I*V (integrated) will be the power transmitted. The average power radiated is always the real part of V*I(conjugate). If V and I are in phase, this is simply equal to V*I. But what does this have to do with the confusion between a resistor and resistance? It doesn't really, but it does relate to this discussion. In the sense that V and I will be in phase for either a matched antenna or an ideal dummy load. V and I will be in phase at the feedpoint of a purely resistive antenna. This is true whether or not it's matched to the transmission line feeding it, or whether the transmission line is matched to the antenna. The relative phase of V and I at the antenna terminals has nothing to do with whether the antenna or transmission line are matched. It also doesn't matter what fraction of that resistance represents energy dissipated locally and how much represents energy radiated. I suppose what this all means is that if you have a matched antenna, it's V and I curves will be IN PHASE and will have the exact same RMS values as if you had a truly resistive dummy load instead. Yes and no. If you're feeding the antenna with a 450 ohm line, it's matched to the line only if its impedance is 450 + j0 ohms, so it looks like a 450 ohm dummy load, not a 50 ohm one. On the other hand, you can feed a 50 + j0 ohm antenna with a half wavelength of 450 ohm line, and get a perfect 50 ohm match to a transmitter at the input end of the line, while running a 9:1 SWR on the the transmission line. Then either the antenna or the input of the line looks like a 50 ohm dummy load. You could make that half wavelength transmission line almost any other characteristic impedance, like 25 or 200 Ohms, and you would still wind up back at 50 Ohms at the input of the line (but you couldn't change frequencies unless they were multiples of 2 of the fundamental). But the point is, you will still be at 50 Ohms at the input, so the V and I sinewaves should be in phase. At the input and output of the line, yes. Therefore, you can consider the center of the Smith Chart (or the entire real (non-reactive) impedance line) as a real resistance like an ideal dummy load. Do you agree with this statement Roy? Yes. Cool. But a network analyzer looking into a black box will not be able to tell you whether the 50 Ohms it is reading is radiated resistance or dissipated resistance. This seems to be the crux of my question. You're right. The network analyzer can't tell you what's happening to those watts going into the antenna. Too bad. If it could, we'd have a really cool, easy way to measure antenna efficiency, wouldn't we? But it's worse than that. If you connect the network analyzer to two series resistors, it can't even tell how much power is going to each one! And it can't even tell the difference between a 50 ohm resistor, a 100 ohm resistor on the other side of a 2:1 impedance transformer, a 100 ohm resistor on the other end of a quarter wavelength of 70.7 ohm transmission line, or a really long piece of lossy 50 ohm transmission line. Boy, they sure are stupid. How come nobody complains that the impedance of a light bulb, an LED, or a loaded electric motor is resistive? It's too bad it is, too. Otherwise, the power company wouldn't charge us for the power we're using to run those things. Neither the power company nor the network analyzer knows or cares how much of the power going to a light bulb or LED is converted to light and how much to heat, or how much of the power going to an electric motor is doing work. If we insist on separating all resistance into "dissipative" and "dissipationless" categories, we have to consider time. Within a small fraction of a second, most of the energy going into an antenna is dissipated -- mostly in the ground or (at HF) the ionosphere. So we'll have to consider that portion of the radiation resistance as "dissipative". The stuff that goes into space will take longer to turn into heat, but it will eventually. So the remainder of the radiation resistance is one that's initially dissipationless but becomes dissipative with time. Just think, we can have a whole new branch of circuit theory to calculate the time constants and mathematical functions involved in the transition between dissipationless and dissipative states! And of course it would have to be cross-disiplinary, involving cosmology, meteorology, and geology at the very least. There are textbooks to be written! PhD's to earn! Just think of the potential papers on the resistance of storage batteries alone! High self-discharge rate means a faster transition from dissipationless to dissipative. . . Sorry, I digress. I just get so *excited* when I think of all the possibilities this opens up for all those folks living drab and boring lives and with so much time and so little productive to do. . . But has this ******* creation really simplified things or enhanced understanding? Now tell me, why can't you just make some 377 ohm transmission line (easy to make) open circuited, and dispense with the antenna altogether? When you figure out the answer to that one, you might begin to see the error with your mental model. Ok by me. I'll be waiting for your patented no-antenna 377 ohm feedline. Roy Lewallen, W7EL Why would you need a 377 Ohm feedline for free space, when free space itself is the transmission line?? By golly, you're right. Design your transmitter for 377 ohm output and do away with the transmission line, too. Just let them joules slip right out, perfectly matched, right straight to free space. Voila! What's wrong with thinking of an antenna as a type of series Inductor, with a distributed shunt capacitance, that can be thought of as a type of distributed "L" matching network that transforms from 50 Ohms to 377? Because that's not what it does, and thinking of it that way leads you to impossible conclusions. The antenna is converting power to E and H fields. The ratio of E to H, or the terminal V to I are immaterial to the conversion process. You're continuing to be suckered into thinking that because the ratio of E to H in free space has the dimensions of ohms that it's the same thing as the ratio of V to I in a circuit. It isn't. A strand of spaghetti one foot long isn't the same thing as a one foot stick of licorice, just because the unit of each is a foot. This is related to how you need to bend the ground radials of a 1/4 WL vertical whip at 45 deg angles down, to get the input impedance closer to 50 Ohms (as opposed to 36 Ohms or something like that if you leave them horizontal). There are a number of ways in which an antenna and transmission line are similar. But don't take the analogy too far. Start with a quarter wavelength transmission line, start splitting the conductors apart until they're opposed like a dipole, and tell me how you've ended up with an input impedance of 73 ohms. There are plenty of texts you can read, on all different levels, if you're really interested in learning about antennas, fields, and waves. Roy Lewallen, W7EL |
I would like one of those antennas!!
BTW, it should be patented!! It looks better than some I've read about!! ;-) DD, W1MCE W5DXP wrote: Richard Harrison wrote: What happens? Our loading, adjusted for 50 watts output, produces 50 volts at an in-phase current of 50 amps. Hmmmm, 50 watts in, 2500 watts out. How much will you take for that antenna, Richard? :-) |
Dr. Slick wrote:
"You cannot tell if the 50 Ohms reading on a Network analyzer into a Black Box is a dissipative resistance like a dummy load, or if it is a radiated resistance of a perfectly matched antenna. You don't have that information." Conversion of RF energy to heat can be measured. Conversion of RF energy to EM radiation can be measured. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen
wrote: No, an antenna doesn't "match" the impedance of free space. The input impedance of an antenna is the ratio of V to I. The impedance of free space is the ratio of the E field to the H field of a plane wave. They both happen to have units of ohms, but they're different things and there's no "matching" going on. If you apply 100 watts to an antenna, resonant or not, 100 watts will be radiated, less loss, regardless of the antenna's input impedance. Roy Lewallen, W7EL Roy; Hope you don't mind if I ask you a couple of questions about your last sentence. Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no losses, now if the antenna's input impedance is changed to say 50 ohms (non-reactive). You would have some loss with the mismatch between the coax and antenna? The transmitter would see 50 ohms? |
Dilon Earl wrote:
On Tue, 15 Jul 2003 01:20:02 -0700, Roy Lewallen wrote: No, an antenna doesn't "match" the impedance of free space. The input impedance of an antenna is the ratio of V to I. The impedance of free space is the ratio of the E field to the H field of a plane wave. They both happen to have units of ohms, but they're different things and there's no "matching" going on. If you apply 100 watts to an antenna, resonant or not, 100 watts will be radiated, less loss, regardless of the antenna's input impedance. Roy Lewallen, W7EL Roy; Hope you don't mind if I ask you a couple of questions about your last sentence. Given a 75 ohm dipole fed 100 watts with 75 0hm coax. Assume no losses, now if the antenna's input impedance is changed to say 50 ohms (non-reactive). You would have some loss with the mismatch between the coax and antenna? Mismatch does not cause loss (that is, conversion of electrical energy to heat), except that the loss of a lossy transmission line will increase (very slightly unless initial loss is great) when SWR is elevated. In this case, the line SWR would be 1.5:1, which would not cause a significant amount of extra loss even if the cable were very lossy when matched. So the answer is no. The transmitter would see 50 ohms? The transmitter could see any of a variety of impedances, depending on the length of the 75 ohm transmission line. Only if the line were an exact multiple of an electrical half wavelength would the transmitter see 50 ohms, resistive. If the line were an odd number of quarter wavelengths, the transmitter would see 112.5 ohms, resistive. At all other lengths, the transmitter would see a complex (partly resistive and partly reactive) load. There is a term called "mismatch loss", which is widely misunderstood in the amateur community because of its name. It doesn't really represent loss at all, but a signal reduction for other reasons. I've explained this before in this newsgroup, so if you're interested, you should be able to find my earlier postings via http://www.groups.google.com. Roy Lewallen, W7EL |
William E. Sabin wrote:
If a 50 ohm generator is connected to a 50 ohm resistor (or resistance), the maximum possible power is delivered to the load. Maybe it should be "maximum *available* steady-state power" which is not always the same thing as "maximum "possible" power"? :-) And the 50 ohm resistance can be the V/I ratio looking into a Z0-match point with a high mismatch loss between the Z0-match and a mismatched antenna, i.e. maximum available power (minus feedline losses) can still be delivered to a mismatched load by doing the matching back down the transmission line, e.g. an antenna tuner. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote:
William E. Sabin wrote: If a 50 ohm generator is connected to a 50 ohm resistor (or resistance), the maximum possible power is delivered to the load. Maybe it should be "maximum *available* steady-state power" which is not always the same thing as "maximum "possible" power"? :-) Maximum possible is correct. The maximum "available" from the generator is a constant value. The maximum "delivered" is the thing can be lower than the maximum. Bill W0IYH |
"William E. Sabin" wrote If a 50 ohm generator is connected to a 50 ohm resistor (or resistance), the maximum possible power is delivered to the load. If the load is greater than or less than than 50 ohms resistor (or resistance) the power delivered to the load is less than the maximum possible. ============================== Unfortunately, the use of the term "Reflection Loss" to descibe performance of amateur feedline + antenna systems at HF merely adds to the confusion. "Maximum Available (or possible) Power" delivered to the load occurs only when there's a "Conjugate Match" between generator and load. But a Conjugate Match does not exist. It cannot even be assumed. For the simple reason the internal impedance of the generator is never known. And it wouldn't make any difference to how the system is set up and operated if it was. Even the generator (transmitter) designer doesn't know what the internal impedance is. He couldn't care less. It can vary all over the shop depending on what the load impedance is. --- Reg, G4FGQ |
Reg Edwards wrote:
"William E. Sabin" wrote If a 50 ohm generator is connected to a 50 ohm resistor (or resistance), the maximum possible power is delivered to the load. If the load is greater than or less than than 50 ohms resistor (or resistance) the power delivered to the load is less than the maximum possible. ============================== Unfortunately, the use of the term "Reflection Loss" to descibe performance of amateur feedline + antenna systems at HF merely adds to the confusion. "Maximum Available (or possible) Power" delivered to the load occurs only when there's a "Conjugate Match" between generator and load. But a Conjugate Match does not exist. It cannot even be assumed. For the simple reason the internal impedance of the generator is never known. And it wouldn't make any difference to how the system is set up and operated if it was. Even the generator (transmitter) designer doesn't know what the internal impedance is. He couldn't care less. It can vary all over the shop depending on what the load impedance is. --- All of this is common knowledge, or should be. However, this gets beyond the basic idea of mismatch loss. It was not my intention to get into an expanded discussion of the various contingencies, which have been discussed ad nauseum in this forum. One should start with the simplest case. If an equipment is designed to work into a 50 ohm load, and the load is not 50 ohms, a "return loss" can be measured with a 50 ohm directional coupler. No knowledge of the generator impedance is needed for this. The return loss is also a measure of the mismatch loss, if the generator resistance (50 ohms) is known. An impedance "transforming" network can improve the load impedance closer to 50+j0 ohms. The equipment will then perform as advertised. In many kinds of circuits mismatch loss and conjugate match are important. Bill W0IYH |
In many kinds of circuits mismatch loss and
conjugate match are important. ============================== I didn't say they were unimportant. I said they served only to add to the confusion when considering operation of the usual amateur installation when the generator internal resistance is unknown. |
W5DXP wrote in message ...
Dr. Slick wrote: "You cannot tell if the 50 Ohms reading on a Network analyzer into a Black Box is a dissipative resistance like a dummy load, or if it is a radiated resistance of a perfectly matched antenna. You don't have that information." Conversion of RF energy to heat can be measured. Conversion of RF energy to EM radiation can be measured. Agreed. But a Black Box to me implies you have limited information from it. My point is that if someone gives you an impedance plot of a resistive 50 Ohms, you will not be able to tell if it is dissipative (lossy) or radiated resistance. I was just reading that Joseph Carr calls radiated resistance as a sort of "ficticious" resistance. I'm sure many here would argue this description, but it kinda makes sense to me. Slick |
Reg Edwards wrote:
In many kinds of circuits mismatch loss and conjugate match are important. ============================== I didn't say they were unimportant. I said they served only to add to the confusion when considering operation of the usual amateur installation when the generator internal resistance is unknown. That too is very well understood by just about everyone. Bill W0IYH |
I'd be one of the people arguing. Radiation resistance fits every
definition of resistance. There's no rule that a resistance has to dissipate power. The late Mr. Carr was quite apparently confusing resistance with a resistor, a common mistake. Why not call radiation resistance "real" resistance and loss resistance "ficticious"? Makes just as much sense as the other way around -- that is to say, none. Roy Lewallen, W7EL Dr. Slick wrote: W5DXP wrote in message ... Dr. Slick wrote: "You cannot tell if the 50 Ohms reading on a Network analyzer into a Black Box is a dissipative resistance like a dummy load, or if it is a radiated resistance of a perfectly matched antenna. You don't have that information." Conversion of RF energy to heat can be measured. Conversion of RF energy to EM radiation can be measured. Agreed. But a Black Box to me implies you have limited information from it. My point is that if someone gives you an impedance plot of a resistive 50 Ohms, you will not be able to tell if it is dissipative (lossy) or radiated resistance. I was just reading that Joseph Carr calls radiated resistance as a sort of "ficticious" resistance. I'm sure many here would argue this description, but it kinda makes sense to me. Slick |
Roy Lewallen wrote in message ...
What if the antenna was a wire shorting the transmission line? Then there would be very little radiated power, and a lot of reflected power. Yes, there would. But every watt delivered to that wire would be either radiated or dissipated. Actually mostly reflected back to the source, so not radiated or dissipated assuming ideal lossless transmission lines. You can make a transformer to match 50 Ohms to, say, 200 Ohms or so. Why not for 50 to 377? It's just another number. That's right. You can make a transformer that converts the circuit V/I ratio to 377 ohms. But this doesn't make a plane wave whose E to H field ratio is 377 ohms. But you can have an antenna with a feedpoint impedance of, say, 3 - j200 ohms, and a couple of wavelengths from the antenna, the ratio of E to H field produced by that antenna will be 377 ohms. Another antenna, with feedpoint impedance of 950 + j700 ohms, will also produce a fields whose E to H ratio is 377 ohms. In fact, you can have any antenna impedance you'd like, and a few wavelengths away, the ratio of E to H field will be 377 ohms, provided the antenna is immersed in something resembling free space. But the fields in, on, or around a 377 ohm transmission line are unlikely to have an E to H ratio of 377 ohms. You're trying to say that the 377 ohm E/H ratio of free space is the same thing as a V/I ratio of 377 ohms. It isn't. Any more than 10 ft-lbs of torque is the same as 10 ft-lbs of work or energy. Interesting. I'll have to look this up more. Except in the case of free space, with a given permeability and permittivity, you have the impedance of free space, which doesn't need a transmission line. Sorry, I can't make any sense out of that. Well, my point was that you don't need a transmission line for free space because otherwise it wouldn't be wireless. But your above point is well taken, that there is no current flowing in free space, in an expanding EM wave, while there definitely is current in a transmission line. Cool. But a network analyzer looking into a black box will not be able to tell you whether the 50 Ohms it is reading is radiated resistance or dissipated resistance. This seems to be the crux of my question. You're right. The network analyzer can't tell you what's happening to those watts going into the antenna. Too bad. If it could, we'd have a really cool, easy way to measure antenna efficiency, wouldn't we? That would be excellent. If we insist on separating all resistance into "dissipative" and "dissipationless" categories, we have to consider time. Within a small fraction of a second, most of the energy going into an antenna is dissipated -- mostly in the ground or (at HF) the ionosphere. So we'll have to consider that portion of the radiation resistance as "dissipative". The stuff that goes into space will take longer to turn into heat, but it will eventually. Certainly the EM wave will heat up ever so slightly any bits of metal it comes across on the way to outer space. So the remainder of the radiation resistance is one that's initially dissipationless but becomes dissipative with time. Just think, we can have a whole new branch of circuit theory to calculate the time constants and mathematical functions involved in the transition between dissipationless and dissipative states! And of course it would have to be cross-disiplinary, involving cosmology, meteorology, and geology at the very least. There are textbooks to be written! PhD's to earn! Just think of the potential papers on the resistance of storage batteries alone! High self-discharge rate means a faster transition from dissipationless to dissipative. . . And the EM wave will theoretically continue forever, even if it is in steradians (power dropping off by the cube of the distance?), so perhaps eventually most of it will be dissipated as heat. But, as you know, a capacitor also never fully charges... Sorry, I digress. I just get so *excited* when I think of all the possibilities this opens up for all those folks living drab and boring lives and with so much time and so little productive to do. . . But has this ******* creation really simplified things or enhanced understanding? Very interesting stuff. And it's certainly enhanced MY understanding. What's wrong with thinking of an antenna as a type of series Inductor, with a distributed shunt capacitance, that can be thought of as a type of distributed "L" matching network that transforms from 50 Ohms to 377? Because that's not what it does, and thinking of it that way leads you to impossible conclusions. The antenna is converting power to E and H fields. The ratio of E to H, or the terminal V to I are immaterial to the conversion process. You're continuing to be suckered into thinking that because the ratio of E to H in free space has the dimensions of ohms that it's the same thing as the ratio of V to I in a circuit. It isn't. A strand of spaghetti one foot long isn't the same thing as a one foot stick of licorice, just because the unit of each is a foot. But an antenna must be performing some sort of transformer action. If you were designing an antenna to radiate underwater, or though Jell-o, or any other medium of a different dielectric constant than free space, you would have to change it's geometry. Even if it is E to H, and not V to I. If an antenna is not a transformer of some type, then why is it affected by it's surroundings so much? They obviously are, just like the primary's impedance is affected by what the secondary sees in a transformer. Certainly having lots of metal in close proximity will affect the impedance of your antenna. This is related to how you need to bend the ground radials of a 1/4 WL vertical whip at 45 deg angles down, to get the input impedance closer to 50 Ohms (as opposed to 36 Ohms or something like that if you leave them horizontal). There are a number of ways in which an antenna and transmission line are similar. But don't take the analogy too far. Start with a quarter wavelength transmission line, start splitting the conductors apart until they're opposed like a dipole, and tell me how you've ended up with an input impedance of 73 ohms. Well, I'm not sure, but you would start off at an open, which would be transformed to a virtual short. But from there, it sounds like a complex mathematical derivation to get the 73 Ohms. My point is that the 73 Ohms is dependant on the dipole's surroundings, depending on how far from the ground and such, so it is a transformer of some sort. There are plenty of texts you can read, on all different levels, if you're really interested in learning about antennas, fields, and waves. Roy Lewallen, W7EL Which one's can you recommend? Slick |
"Tarmo Tammaru" wrote in message ...
"Dr. Slick" wrote in message om... I disagree on this point. You can think of an antenna as a type of tranformer, from 50 Ohms to 377 Ohms of free space. A tuned antenna is "matching" 50 to 377 ohms, and will therefore have no reflections, ideally. Look up "Impedance of Free Space"in the Kraus book. Tam/WB2TT I don't have that book. What does it say? Slick |
Dr. Slick wrote:
Roy Lewallen wrote in message ... What if the antenna was a wire shorting the transmission line? Then there would be very little radiated power, and a lot of reflected power. Yes, there would. But every watt delivered to that wire would be either radiated or dissipated. Actually mostly reflected back to the source, so not radiated or dissipated assuming ideal lossless transmission lines. No. "Reflected" power isn't delivered to the wire. It's an analytical function that exists only on the feedline. If the feedline has no loss, the same amount of power entering the line exits the line. You can have any amount of "reflected power" you want by simply changing the characteristic impedance of the line -- with no effect on the power either exiting or leaving the line. . . . Except in the case of free space, with a given permeability and permittivity, you have the impedance of free space, which doesn't need a transmission line. Sorry, I can't make any sense out of that. Well, my point was that you don't need a transmission line for free space because otherwise it wouldn't be wireless. But your above point is well taken, that there is no current flowing in free space, in an expanding EM wave, while there definitely is current in a transmission line. Yes, that's right. But if you want to dig deeper, you'll find that a "displacement current" can be mathematically described which conveniently accounts for some electromagnetic phenomena. It is, though, a different critter from the conducted current on a transmission line. An antenna can reasonably be viewed as a transducer. It converts the electrical energy entering it into electromagnetic energy -- fields. As is the case for any transducer, the stuff coming out is different than the stuff going in. Think in terms of an audio speaker, which converts electrical energy into sound waves, and you'll be on the right track. . . . What's wrong with thinking of an antenna as a type of series Inductor, with a distributed shunt capacitance, that can be thought of as a type of distributed "L" matching network that transforms from 50 Ohms to 377? Because that's not what it does, and thinking of it that way leads you to impossible conclusions. The antenna is converting power to E and H fields. The ratio of E to H, or the terminal V to I are immaterial to the conversion process. You're continuing to be suckered into thinking that because the ratio of E to H in free space has the dimensions of ohms that it's the same thing as the ratio of V to I in a circuit. It isn't. A strand of spaghetti one foot long isn't the same thing as a one foot stick of licorice, just because the unit of each is a foot. But an antenna must be performing some sort of transformer action. If you were designing an antenna to radiate underwater, or though Jell-o, or any other medium of a different dielectric constant than free space, you would have to change it's geometry. Even if it is E to H, and not V to I. Not transformer, transducer. More like a speaker than a megaphone. If an antenna is not a transformer of some type, then why is it affected by it's surroundings so much? Egad, how can I answer that? If a fish isn't a transformer, then why is it affected by its surroundings so much? If an air variable capacitor isn't a transformer, then why is it affected by its surroundings so much? What does sensitivity to surroundings have to do with being a transformer? They obviously are, just like the primary's impedance is affected by what the secondary sees in a transformer. Certainly having lots of metal in close proximity will affect the impedance of your antenna. It is true that the equations describing coupling between two antenna elements are the same as the for the coupling between windings of a transformer. But a single antenna element isn't a transformer any more than a single inductor is a transformer. When you apply V and I to an antenna, it creates E and H fields. When you apply V and I to an inductor, it creates E and H fields. Both the antenna and inductor are acting as transducers, converting the form of the applied energy. If you put a secondary winding in the field of the primary inductor, the field induces a voltage in the secondary. If (and only if) the secondary is connected to a load, causing current to flow in it, that current produces a field which couples back to the primary, altering its current. Coupled antennas, or an antenna coupled to any other conductor, work the same way -- although localized currents can flow in the absence of an intentional load if the antenna is a reasonable fraction of a wavelength long. But the single antenna isn't a transformer, any more than the single inductor is. Each is a transducer, and the secondary winding, or coupled conductor, is another transducer. This is related to how you need to bend the ground radials of a 1/4 WL vertical whip at 45 deg angles down, to get the input impedance closer to 50 Ohms (as opposed to 36 Ohms or something like that if you leave them horizontal). There are a number of ways in which an antenna and transmission line are similar. But don't take the analogy too far. Start with a quarter wavelength transmission line, start splitting the conductors apart until they're opposed like a dipole, and tell me how you've ended up with an input impedance of 73 ohms. Well, I'm not sure, but you would start off at an open, which would be transformed to a virtual short. But from there, it sounds like a complex mathematical derivation to get the 73 Ohms. And it's really, really tough and requires some *really* creative (read: bogus) math to derive it from simply transmission line phenomena. My point is that the 73 Ohms is dependant on the dipole's surroundings, depending on how far from the ground and such, so it is a transformer of some sort. Not by itself it isn't. But if you can make a transformer by putting two antenna elements close together -- put V and I into one and extract it in a different ratio from the other. It's going to be a pretty lossy transformer, though, due to energy lost to radiation. (You'll find extra resistance at the "primary" feedpoint that'll nicely account for this.) There are plenty of texts you can read, on all different levels, if you're really interested in learning about antennas, fields, and waves. Roy Lewallen, W7EL Which one's can you recommend? One of my favorites is King, Mimno, and Wing, _Transmission Lines, Antennas, and Waveguides_, and that's probably the one I'd choose if I had to select just one. It was reprinted as a paperback by Dover in 1965, and the paperback be found as a used book pretty readily and inexpensively. Kraus' _Antennas_ is my favorite text on antennas, and is certainly one of the most, if not the most, highly regarded. It's now in its third edition, and you can often find used copies of earlier editions at reasonable prices. For transmission lines, there's an excellent treatment in Johnson's _Transmission Lines and Networks_. I refer to Kraus' _Electromagnetics_ particularly when dealing with waves in space. And Holt's _Introduction to Magnetic Fields and Waves_ is pretty good for both. There are a lot of others, each with its strong and weak points. But you can't go wrong with these. Roy Lewallen, W7EL |
William E. Sabin wrote:
W5DXP wrote: Maybe it should be "maximum *available* steady-state power" which is not always the same thing as "maximum "possible" power"? :-) Maximum possible is correct. The maximum "available" from the generator is a constant value. The maximum "delivered" is the thing can be lower than the maximum. Point is that "maximum possible power" will cause a lot of transmitters to exceed their maximum power rating and overheat. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
What might be of interest in this discussion is that after he derives the
impedance of free space, he uses that to find the Radiation Resistance of a short dipole for dW. Where d is the length of the dipole and W is wavelength. (I did not want to do Greek letters). He ends up with an equation of the form R=377k(d/W)**2 or R=790 (d/W)**2. As a sanity check let d/W=1/2, which violates the , but still gives a fairly close answer of 197 Ohms, compared to the actual 168 Ohms. Note that this is not the same as Feedpoint Resistance because it is not referred to the current maximum. Kraus does not actually say this, but seems that the near field would be the mechanism for "matching" this to the far field 377 Ohms. The transmitter only sees the feedpoint, the rest of the universe sees the whole antenna. If I interpret it correctly, this 197 (168) Ohms in independent of where you feed the dipole. Kind of hard to boil several pages into one paragraph, especially since most of this stuff I haven't seen in decades. Tam/WB2TT "Dr. Slick" wrote in message om... I don't have that book. What does it say? Slick |
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