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"Reg Edwards" wrote in message ...
.... ==================================== Tom, To add a bit more - 50-ohm generators as used in laboratories (so that measured reflexion loss, mismatch loss etc, mean something) are effectively constant voltage generators in series with a 50-ohms resistor, or constant current generators in shunt with a 50-ohm resistor. They may be followed by an ampifier whose output impedance is held constant at 50-ohms by some automatic means. None of these circuits bear much resemblance to a pair of 807's and a tuned tank. The best that can be said about Rg of the usual HF radio transmitter is that Rg is indeterminate. IT EVEN VARIES AS THE LOAD IMPEDANCE IS CHANGED which most of the Guru's contributing to this newsgroup appear to be unaware of or at least choose to disregard. So what does "adjusting RL to equal Rg" mean? To use it in a description of feeder + antenna behaviour further propagates myths, including those surrounding SWR, forward power, reflected power, SWR meters, etc. Does Terman ever bother to mention Rg of a Tx PA? If he doesn't it can't matter very much to him. The ARRL handbook, when numerically designing a transistor linear HF PA, makes no mention of Rg. Amen, brother. I was thinking after making my last posting to this thread that the one thing I DON'T bother thinking about when designing a PA is what source impedance it will present. I worry about currents, voltages, efficiency, distortion, a network to present the proper load to the active device(s)... but not Rg. In precision instrumentation systems, the output is commonly levelled or monitored through a levelling splitter (not to be confused with a power divider), so that a virtual zero-impedance point can be established, with a 50-ohm (or other Zo) resistor from that point to each output. And network analyzers are commonly calibrated with precision loads so that the imperfections in their outputs and reflectometers and cabling can be backed out by the calibration software. Cheers, Tom |
"Roy Lewallen" wrote in message ... Most simple derivations for the input impedance of a dipole assume it's infinitely thin. That is why I used #30 wire The general problem of a dipole made from wire of finite diameter is a lot tougher, and is the topic of the papers by the authors I listed in another recent posting. With EZNEC, you'll find that the dipole impedance will continue to change as you make the wire diameter smaller and smaller, until it gets too small for the program to handle at all. When I switched to #14, the impedance did not change more than a few tenths of an Ohm. However, resonance went down about 10 KHz. BTW, when I took my first Junior level EE Fields & Waves course I asked my prof about the same question that Slick brought up. Was convinced that 300 Ohm TV antennas were really 377 Ohms! Tam/WB2TT |
Any transducer I can think of offhand converts between mechanical energy and
electrical energy, as for instance a loudspeaker, microphone, mechanical filter, etc. As for the impedance of free space, one way to build a stealth aircraft is to cover it with material that has a resistivity of 377 Ohms/square. Then there is no reflection Tam/WB2TT |
"Dilon Earl" wrote in message ... On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin" sabinw@mwci-news wrote: If I have a 100 watt transmitter and my wattmeter shows 3 watts reflected. Is 3 watts actually being dissipated in the tank and final PA? That would depend on the output impedance of the transmitter. If it is 50 Ohms, all the reflected power would be absorbed by the transmitter. If it is 0 Ohms or its Norton equivalent, it is all reflected, and none is dissipated. Tam/WB2TT |
This whole discussion points out the fallacy of trying to understand
reflections by sending a CW signal. Consider a 100W pulse radar transmitter. Also, assume the transmitter rurns off between pulses ( a good assumption), and that the antenna mismatch causes 3% of the power to be reflected. When the pulse reaches the antenna, 97W is transmitted, and 3W is returned. All of this 3 W is reflected by the transmitter, and a second pulse is produced where 2.91W is radiated and .09 W is reflected back. This is re-reflected, and a third pulse is produced where .0873 W is radiated and ..0027 W is reflected, etc, etc. Tam/WB2TT "W5DXP" wrote in message ... William E. Sabin wrote: Keeping everything simple and not getting into peripheral issues is desirable at this point. The assumption that the source is outputting (forward power minus reflected power) is essentially saying that all amplifiers are perfectly matched and re-reflect all reflected power. That seems like a stretch to me. -- 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! =----- |
Tarmo Tammaru wrote:
This whole discussion points out the fallacy of trying to understand reflections by sending a CW signal. There is nothing wrong with considering the steady-state solution. We do that all the time. The steady-state is achieved mathematically by considering an infinite sum of forward and reflected waves (for a CW signal) until it converges to a steady state. In the final solution the source impedance is found to have no effect on the standing wave pattern (the SWR). Reference: W.C. Johnson "Transmission Lines and Networks", McGraw-Hill 1950, chapter 4, Equation 4.23. In the steady state the final result is according to my discussion. Bill W0IYH |
On Fri, 18 Jul 2003 09:03:38 -0400, "Tarmo Tammaru"
wrote: Any transducer I can think of offhand converts between mechanical energy and electrical energy, as for instance a loudspeaker, microphone, mechanical filter, etc. That seems to be ignored to complement its use as analogy. As for the impedance of free space, one way to build a stealth aircraft is to cover it with material that has a resistivity of 377 Ohms/square. Then there is no reflection Tam/WB2TT Hi Tam, That material has been around for a couple of decades, but is quite heavy. You might find yourself with a fleet of stealth bumper cars as a result. 73's Richard Clark, KB7QHC |
On Fri, 18 Jul 2003 09:51:35 -0700, W5DXP
wrote: Dick Carroll wrote: You can be just as sure that the speaker presents its rated impedance only at one frequency, too. Is it still 1000 Hz? Hi All, Such easy things to confirm..... I have a pair of Pioneer speakers, looked at their spec.s pasted to the back, rated flat from 10Hz to 25KHz (flat being order of two from nominal 10 Ohms) with a small (BW) peak at 80Hz. Years ago in my metrology lab, we abandoned laboratory grade amplifiers and speakers for commercial ones as the market offered far better product at much cheaper rates. Sound, like light, makes virtual experts of everyone through their misperception of its common features. Eyes and ears are fine instruments clouded by the brain. 73's Richard Clark, KB7QHC |
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On Fri, 18 Jul 2003 09:47:32 -0400, "Tarmo Tammaru"
wrote: Consider a 100W pulse radar transmitter. Hi Tarmo, Was this a speculation or borne of actual experience? (I am not talking about the obvious, exceptionally low power.) I understand the significance of what you wrote following it, .0027 W is reflected, etc, etc. but in my experience with radars (megawatt models that I serviced, calibrated and offered formal training in), this does not happen. 73's Richard Clark, KB7QHC |
On Fri, 18 Jul 2003 08:20:54 +0100, "Ian White, G3SEK"
wrote: The same people produce the same arguments... Hi Ian, Your statement is a self-fulfilling example. What purpose does it serve to merely say t'ain't so in place of commenting on data taken in the field? There are equipment operators. There are bench techs. There are some of both. And then there are cut-and-paste theoreticians. 73's Richard Clark, KB7QHC |
On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru"
wrote: "Dilon Earl" wrote in message ... On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin" sabinw@mwci-news wrote: If I have a 100 watt transmitter and my wattmeter shows 3 watts reflected. Is 3 watts actually being dissipated in the tank and final PA? That would depend on the output impedance of the transmitter. If it is 50 Ohms, all the reflected power would be absorbed by the transmitter. If it is 0 Ohms or its Norton equivalent, it is all reflected, and none is dissipated. Tam/WB2TT Hi Tam, This is painfully obvious. It is also painfully demonstrative. It also appears to be painfully avoided in lieu of providing an actual value (Punchinello seems to wholly ignore his own cries that lacking numbers renders such whining as ignorance). I've read for years that the common RF rig is NOT a 50Ohm source, and absolutely none dare commit themselves to just what value it is (much less offer their own measure). Being a physical reality, the rig must present some real value, but vacuous theory seems to bar that discussion. 73's Richard Clark, KB7QHC |
Dr. Slick wrote:
How do they measure a "surge impedance"? For a transmission line, it is the same as the characteristic impedance. For a traveling wave antenna, it is the same as the feedpoint impedance. So for a center-fed standing wave antenna, like a dipole, you could terminate the ends of the antenna to eliminate reflections and then measure the feedpoint impedance. Standing wave antennas are like transmission lines with standing waves. The impedance at any point is the (forward voltage wave plus the reflected voltage wave) divided by the (forward current wave plus the reflected current wave). The feedpoint impedance of traveling wave antennas is usually about 600-800 ohms according to The ARRL Antenna Book. -- 73, Cecil, W5DXP |
Dick Carroll wrote:
You can be just as sure that the speaker presents its rated impedance only at one frequency, too. Is it still 1000 Hz? -- 73, Cecil, W5DXP |
Roy Lewallen wrote:
What needs to be thrown away is the belief that all impedances are the ratio of a voltage to a current, along with the notion that only resistors can have resistance. I agree, Roy, but what can we do about it? I had been using "virtual impedance" to differentiate a voltage to current ratio from an intrinsic physical impedance. How would you differentiate an intrinsic physical impedance from a voltage to current ratio? -- 73, Cecil, W5DXP |
You can do about it what you like. What I've chosen to do about it is to
try and educate the people who will listen, and ignore those who won't. I find the concepts perfectly understandable without the need for additional adjectives. Roy Lewallen, W7EL W5DXP wrote: Roy Lewallen wrote: What needs to be thrown away is the belief that all impedances are the ratio of a voltage to a current, along with the notion that only resistors can have resistance. I agree, Roy, but what can we do about it? I had been using "virtual impedance" to differentiate a voltage to current ratio from an intrinsic physical impedance. How would you differentiate an intrinsic physical impedance from a voltage to current ratio? -- 73, Cecil, W5DXP |
And the mice are off.
Is it just my perception, or is it the fate of all threads in this newsgroup to end up being Cecil arguing about power waves? Cecil, have you considered starting your own newsgroup, say alt.cecils.power.waves? Roy Lewallen, W7EL W5DXP wrote: Ian White, G3SEK wrote: "Where does the reflected power go?" is even worse because it contains at least two built-in misconceptions: that reflected power waves exist, and that they have to "go" somewhere. Ian, it can be proven that reflected power waves exist. Ramo & Whinnery say they exist. Standing waves in a single source, single feedline, single load system cannot exist without reflected waves. The HP Application Note AN 95-1 defines the power in reflected waves. "|a2|^2 = Power reflected from the load." Saying something is a misconception is only an opinion that differs from a lot of expert opinions. . . |
How about light bulbs, solar panels, thermocouples, batteries, fuel
cells, and fireflies? Roy Lewallen, W7EL Tarmo Tammaru wrote: Any transducer I can think of offhand converts between mechanical energy and electrical energy, as for instance a loudspeaker, microphone, mechanical filter, etc. As for the impedance of free space, one way to build a stealth aircraft is to cover it with material that has a resistivity of 377 Ohms/square. Then there is no reflection Tam/WB2TT |
On Fri, 18 Jul 2003 11:45:13 -0700, Roy Lewallen
wrote: Sigh. I guess one more time. A mouse in the maze. No cheese at the end of that one. I'm firmly in agreement with Bill and Ian on this one. Roy Lewallen, W7EL your's and other opinions merely supports my contention: I've read for years that the common RF rig is NOT a 50Ohm source, and absolutely none dare commit themselves to just what value it is (much less offer their own measure). Being a physical reality, the rig must present some real value, but vacuous theory seems to bar that discussion. 73's Richard Clark, KB7QHC I still find it strange that the savants and pundits will say what it is NOT, but NOT what it IS. Is one Ohm too much? Is ten thousand Ohms too little? Do anyone of you have a simple number that must exist as a physically verifiable entity of a physical example commonly available to any Ham? What is it that heats up in the presence of mismatch that all manufacturers go to great length to protect against? Does it have to be a carbon resistor to qualify? Which one? Why is it so hard to quantify in the face of such firm agreement among you gentlemen? My guess is that you would refuse to warrant your answer in the face of catastrophic failure - evidence contrary to your opinions. Sorry, but puzzles and enigmas do not answer the reality of heat and heat does not arrive through the offices of some virtualized component born of substituted theories of matching. 73's Richard Clark, KB7QHC |
W5DXP wrote:
Point is that "maximum possible power" will cause a lot of transmitters to exceed their maximum power rating and overheat. How much is the maximum possible power? H-Bomb? Gamma ray burst? Big Bang? ;-) 73, ac6xg |
On Fri, 18 Jul 2003 11:54:52 -0700, Roy Lewallen
wrote: How about light bulbs, solar panels, thermocouples, batteries, fuel cells, and fireflies? A light bulb is a transducer? Amusing example of non mechanical translation until you touch it, then the phonons exhibit that mechanical transfer in the form of heat. Common experience as your hand approached that opportunity would have revealed that fate through radiation and that is neglecting another mechanical form of energy transmission offered. That other form is convection. You have not offered counter-examples, you simply ignore positive examples. I've spent nearly 30 years in the field of Electro-Optics with contracts as recent as this year. I've yet to see the trade press or research commonly describe these items as "transducers." That is not to say I would reject any such reference, but I would say few are treading that path to attend a philosophical distinction that is largely semantic. Stick with antenna as transducer, it makes a fine metaphor. 73's Richard Clark, KB7QHC Roy Lewallen, W7EL Tarmo Tammaru wrote: Any transducer I can think of offhand converts between mechanical energy and electrical energy, as for instance a loudspeaker, microphone, mechanical filter, etc. As for the impedance of free space, one way to build a stealth aircraft is to cover it with material that has a resistivity of 377 Ohms/square. Then there is no reflection Tam/WB2TT |
Call it a simple example. I would assume in a real system multiple pulses
are would be highly undesirable and eliminated by discharging the line at the end of the first pulse. Maybe I should have said 100W pulse generator with on output impedance of 0 or infinity, and a lousy dummy load. At a lower power you could hook up a 'scope to the middle of the line and see the multiple reflections. Tam/WB2TT "Richard Clark" wrote in message ... On Fri, 18 Jul 2003 09:47:32 -0400, "Tarmo Tammaru" wrote: Consider a 100W pulse radar transmitter. Hi Tarmo, Was this a speculation or borne of actual experience? (I am not talking about the obvious, exceptionally low power.) I understand the significance of what you wrote following it, .0027 W is reflected, etc, etc. but in my experience with radars (megawatt models that I serviced, calibrated and offered formal training in), this does not happen. 73's Richard Clark, KB7QHC |
On Fri, 18 Jul 2003 15:41:21 -0400, "Tarmo Tammaru"
wrote: Call it a simple example. I would assume in a real system multiple pulses are would be highly undesirable and eliminated by discharging the line at the end of the first pulse. Maybe I should have said 100W pulse generator with on output impedance of 0 or infinity, and a lousy dummy load. At a lower power you could hook up a 'scope to the middle of the line and see the multiple reflections. Tam/WB2TT Hi Tam, I would call your Radar a faulty example. The point of the matter is that real equipment exhibits real dissipation of reflected power in the conventional expectation. Your new example above anticipates this by forcing a failure of that expectation through not matching the load. Such problems that arise as a consequence were written up by NBS, Hewlett-Packard and Steven Adams in the discussion of Mismatch Uncertainty. The fact that various pundits and savants have no actual value to offer in substitution for the oft-repeated refrain "it a'in't 50Ohms" is that any value offered would be immediately demonstrated as being wrong through standard bench top verification. As Ian has characterized this as an exercise in futility, I am content to observe no one stepping up to the bench for validation. 73's Richard Clark, KB7QHC |
W5DXP wrote:
William E. Sabin wrote: If the transmitter output is 100 W and the reflected power is 3 W, then the 100 W is the difference between 100+3=103 W (forward power) and 3 W (reflected power). If the source is a signal generator equipped with a circulator and load, the generator is putting out 103 watts, and the circulator load is dissipating 3 watts, is the generator still only putting out 100 watts by definition? If the sig gen is putting out 100 watts, with 3 watts reflected and 97 watts going to the load, then 3 watts must be going to the circulator. If 100 watts is going to the load and 3 watts is reflected back to the circulator, then the sig gen is putting out 103 watts. But if 97 watts is getting to the load, 3 watts is reflected and there is no circulator or other load, then how much do you think the sig gen is actually putting out? 73, ac6xg |
"Richard Clark" wrote in message ... On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru" wrote: I've read for years that the common RF rig is NOT a 50Ohm source, and absolutely none dare commit themselves to just what value it is (much less offer their own measure). Being a physical reality, the rig must present some real value, but vacuous theory seems to bar that discussion. 73's Richard Clark, KB7QHC Yeah, seems to be a deep dark secret. If you look at the specs of RF power transistors, they will give the output impedance vs frequency - BUT you have to look at the footnote. In virtually all cases what they mean is the conjugate of the load impedance. It is the jX of the transistor (1/jY), in parallel with ((VCC-Vsat)**2) /2P. I have never gotten around to doing this, but I believe the data sheets for tubes like the 811A and 813 do give the plate resistance, which should make it possible to calculate the output impedance at the lower frequencies like 160m. Tam/WB2TT |
On Fri, 18 Jul 2003 16:01:43 -0400, "Tarmo Tammaru"
wrote: Yeah, seems to be a deep dark secret. If you look at the specs of RF power transistors, they will give the output impedance vs frequency - BUT you have to look at the footnote. In virtually all cases what they mean is the conjugate of the load impedance. It is the jX of the transistor (1/jY), in parallel with ((VCC-Vsat)**2) /2P. Hi Tam, Motorola offers quite specific characteristics across frequency. Reference MRF421, MRF433, MRF454 for examples of dirt ordinary power transistors found in more than 20 years of transistorized Ham transmitters. Take their own data, Z transform them through transformers (not transducers) and you find 50Ohms without any more sophisticated math than that required of the standard Technology Certificate of training. Where does it go through after that? A low pass filter designed for 50Ohms to an antenna jack specified to deliver full power to a 50Ohm load. What technical rebuttal do I hear in response to simple engineering data? "It is impossible to determine the output Z of this source." For some I can well imagine they do find it difficult.... 73's Richard Clark, KB7QHC |
Roy Lewallen wrote in message ...
Y'see, if you really, really want an antenna to be a kind of automobile, you can cook up a bunch of reasons to convince yourself that it is. The same method works for astrology and fortune telling, too. Shall i call this a Straw man argument? Or putting words in someone's mouth? Feel free to call it what you want. I believe I've made as valid an argument for an antenna being an automobile as you did for it being a transformer, and based on the same criteria. Well, if you agree that two antennas/transducers in close proximity will make a transformer (albeit a somewhat inefficient one!), then i don't think i was that far off base. The optimization of an antenna depends on many factors, only one of which is the nature of the medium in which it's immersed. And among the medium's important properties are its permeability, permittivity, and the velocity of a wave propagating in it. The phase velocity and characteristic impedance can both be calculated from the permeability and permittivity, so you can't really say any one of these is more important than the other. It doesn't make any sense to throw out the concept of free space impedance just because it confuses people who don't know what it means. It's an extremely useful and well-understood concept. For example, reflection of a wave from a plane conductor or the ground can easily be found by calculating a reflection coefficient based on the impedance of the reflecting surface and the impedance of the impinging wave. (The impedance of a wave can be quite different close to an antenna than it is after it's traveled some distance.) If you look in some of those texts I recommended, you'll find the impedance of free space cropping up all over the place. What needs to be thrown away is the belief that all impedances are the ratio of a voltage to a current, along with the notion that only resistors can have resistance. Roy Lewallen, W7EL You have convinced me that you are correct about both of these points. But i don't think that an antennas impedance will not be affected by the permeability of the medium that surrounds it. An antennas input impedance will be different in free space as opposed to being immersed in water, for example. This indicates to me that the antenna is indeed "matching" 50 Ohms to the impedance of free space, even if it is a different type of impedance. Do you think that the characteristics of a transformer of a specific turns ratio, gauge wire, and core geometry, will NOT depend on the core material? I would say definitely it WILL depend on the material. Slick |
Actually, several people (W8JI among them) have measured the output
impedance of common amateur linear amplifiers by at least a couple of methods. The most credible measurements show, interestingly, a value very close to 50 ohms when the amplifier is adjusted for normal operation. Of course, it doesn't really matter, but people continue to make a big deal out of it. Roy Lewallen, W7EL Tarmo Tammaru wrote: "Richard Clark" wrote in message ... On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru" wrote: I've read for years that the common RF rig is NOT a 50Ohm source, and absolutely none dare commit themselves to just what value it is (much less offer their own measure). Being a physical reality, the rig must present some real value, but vacuous theory seems to bar that discussion. 73's Richard Clark, KB7QHC Yeah, seems to be a deep dark secret. If you look at the specs of RF power transistors, they will give the output impedance vs frequency - BUT you have to look at the footnote. In virtually all cases what they mean is the conjugate of the load impedance. It is the jX of the transistor (1/jY), in parallel with ((VCC-Vsat)**2) /2P. I have never gotten around to doing this, but I believe the data sheets for tubes like the 811A and 813 do give the plate resistance, which should make it possible to calculate the output impedance at the lower frequencies like 160m. Tam/WB2TT |
Tarmo Tammaru wrote:
Roy, You are cheating. In the steady state there is no load on your source. Regardless of what the Bird meter reads. Do one of the following: Why is this cheating? There is reverse power on the line. The source is not absorbing the reverse power. You and others have said, without qualification, that it does. I've shown a case where it doesn't. 1.Short the end of the 1/2 wave line. 2.Use a 1/4 wave open ended line. 3.Get a pulse generator, 0 Ohm output impedance +50 Ohm series resistor. Set the pulse with to 100ns and 1V, and use an arbitrary length of coax, either open or shorted, but longer than 100ns. Grab a 'scope and look at the junction of the coax and the 50 Ohm resistor. You will be able to see the .5V reflected pulse appear across the 50 ohm resistor. ALL of the reflected energy was absorbed, and half of the forward power. When talking of amateurs and transmitters, we're dealing with sinusoidal, steady state conditions. You've just described a transient pulse situation. It's different in several ways, one of the most important being that the source is off when the returning pulse arrives. I'm fully able to discuss TDR phenomena, but it's not relevant, and only adds confusion to a discussion of amateur transmitters and transmission lines. In sinusoidal, steady state conditions, it's absolutely incorrect to say that the reflected power is absorbed by the source, whether the source is matched or not. And it's easy to show it's incorrect, as I've done. Roy Lewallen, W7EL Tam/WB2TT "Roy Lewallen" wrote in message ... Sigh. I guess one more time. A mouse in the maze. 70.7 volt RMS voltage source, 50 ohm series resistor. Connect to an open circuited, half wavelength transmission line. Put your magic lossless Bird wattmeter in the line and measure the forward and reverse power: Pf = 100 watts Pr = 100 watts The "transmitter" is perfectly matched to the line. The "reflected power" is 100 watts. The dissipation of the "transmitter" source impedance is zero. Not 100 watts. Not even one watt. Zero. No, how can anyone possibly say that when the transmitter is matched, the reflected power is absorbed by the transmitter? Any number of other examples can easily be found. You'll find a few others in the "Food For Thought" series available from ftp://eznec.com/pub/food_for_thought/. I'm firmly in agreement with Bill and Ian on this one. Roy Lewallen, W7EL |
Dr. Slick wrote:
Roy Lewallen wrote in message ... Y'see, if you really, really want an antenna to be a kind of automobile, you can cook up a bunch of reasons to convince yourself that it is. The same method works for astrology and fortune telling, too. Shall i call this a Straw man argument? Or putting words in someone's mouth? Feel free to call it what you want. I believe I've made as valid an argument for an antenna being an automobile as you did for it being a transformer, and based on the same criteria. Well, if you agree that two antennas/transducers in close proximity will make a transformer (albeit a somewhat inefficient one!), then i don't think i was that far off base. I agree. The optimization of an antenna depends on many factors, only one of which is the nature of the medium in which it's immersed. And among the medium's important properties are its permeability, permittivity, and the velocity of a wave propagating in it. The phase velocity and characteristic impedance can both be calculated from the permeability and permittivity, so you can't really say any one of these is more important than the other. It doesn't make any sense to throw out the concept of free space impedance just because it confuses people who don't know what it means. It's an extremely useful and well-understood concept. For example, reflection of a wave from a plane conductor or the ground can easily be found by calculating a reflection coefficient based on the impedance of the reflecting surface and the impedance of the impinging wave. (The impedance of a wave can be quite different close to an antenna than it is after it's traveled some distance.) If you look in some of those texts I recommended, you'll find the impedance of free space cropping up all over the place. What needs to be thrown away is the belief that all impedances are the ratio of a voltage to a current, along with the notion that only resistors can have resistance. Roy Lewallen, W7EL You have convinced me that you are correct about both of these points. Good. Then the effort was worthwhile. But i don't think that an antennas impedance will not be affected by the permeability of the medium that surrounds it. An antennas input impedance will be different in free space as opposed to being immersed in water, for example. Indeed it will. This indicates to me that the antenna is indeed "matching" 50 Ohms to the impedance of free space, even if it is a different type of impedance. That's a leap I'm unable to make or to follow. Do you think that the characteristics of a transformer of a specific turns ratio, gauge wire, and core geometry, will NOT depend on the core material? I would say definitely it WILL depend on the material. Actually, an adequate core shouldn't appear as a significant factor in transformer performance. Naturally, an inadequate core will adversely affect it. But I just don't accept that as evidence, let alone "proof" that an antenna is fundamentally an impedance matching device. I see that you won't be swayed from your visualization. But hopefully some of the other readers can see the fallacy of the concept. I think I've done all I can, so I'll leave this topic now. *Chuckle* I was just reminded of something that happened years ago, when my son was a small boy. He learned that I was an engineer, so he couldn't wait to see the train I drove. After a great deal of repeated, patient, explanation, I finally got across (I thought) a description of what I did, and that it had nothing to do with trains. Well, he had occasion to visit me at work quite a long time later. He kept wandering off. When I asked why, he explained that he was trying to find where the train was kept. Yeah, I might not drive trains, but I must have *something* to do with trains. Slick, you've got the right concepts now, but you're still looking for that train. Roy Lewallen, W7EL |
W5DXP wrote:
The answer has been found but most people think they already know everything there is to know. The answer is contained in the following example. The sources are signal generators with circulator loads (SGCL). The (50) or (150) subscript is the ohmic value of the circulator load resistor. The signal generators are phase-locked and can be turned on and off independently. 100W SGCL(50)--1WL 50 ohm feedline--+--1/2WL 150 ohm feedline--33.33W SGCL(150) Pfwd1-- Pfwd2-- --Pref1 --Pref2 Step 1: With the 100W SGCL(50) on and the 33.33W SGCL(150) off, the following conditions exist: Pfwd1 = 100W, Pref1 = 25W, Pfwd2 = 75W, Pref2 = 0W Step 2: Now turn on the 33.33W SGCL(150). At the instant the rearward- traveling signal reaches the impedance discontinuity, the following conditions exist at the discontinuity: Pfwd1 = 100W, Pref1 = 0W, Pfwd2 = 133.33W, Pref2 = 33.33W All anyone has to do is figure out what happened to Pref1 = 25W in Step 1 the instant the 33.33W arrived at the impedance discontinuity in Step 2. This is not a transient buildup condition. If we ignore any distortion in the 33.33W Pref2 wavefront, this is an immediate event and Pref1 is immediately canceled by an equal magnitude and opposite phase Pref2(1-|rho|^2) wavefront at the moment it first arrives. You're sourcing and sinking an additional 33.33 watts, and yet the wattmeter can't discern the difference between this scenario and the 100 watt, single source scenario. The example illustrates perfectly the shortcomings of the idea of power flow, as well as some of the faulty conclusions that can be drawn from measurements made by a directional power meter. 73, AC6XG |
Jim Kelley wrote:
W5DXP wrote: Point is that "maximum possible power" will cause a lot of transmitters to exceed their maximum power rating and overheat. How much is the maximum possible power? H-Bomb? Gamma ray burst? Big Bang? ;-) Maximum power is when you turn your metal 6L6 upside down in a glass of water during a contest. :-) -- 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! =----- |
Jim Kelley wrote:
If the sig gen is putting out 100 watts, with 3 watts reflected and 97 watts going to the load, ... The forward power is 103 watts and the reflected power is 3 watts in the given example. The signal generator equipped with a circulator load is putting out 103 watts. -- 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! =----- |
Jim Kelley wrote:
W5DXP wrote: The answer has been found but most people think they already know everything there is to know. The answer is contained in the following example. The sources are signal generators with circulator loads (SGCL). The (50) or (150) subscript is the ohmic value of the circulator load resistor. The signal generators are phase-locked and can be turned on and off independently. 100W SGCL(50)--1WL 50 ohm feedline--+--1/2WL 150 ohm feedline--33.33W SGCL(150) Pfwd1-- Pfwd2-- --Pref1 --Pref2 Step 1: With the 100W SGCL(50) on and the 33.33W SGCL(150) off, the following conditions exist: Pfwd1 = 100W, Pref1 = 25W, Pfwd2 = 75W, Pref2 = 0W Step 2: Now turn on the 33.33W SGCL(150). At the instant the rearward- traveling signal reaches the impedance discontinuity, the following conditions exist at the discontinuity: Pfwd1 = 100W, Pref1 = 0W, Pfwd2 = 133.33W, Pref2 = 33.33W All anyone has to do is figure out what happened to Pref1 = 25W in Step 1 the instant the 33.33W arrived at the impedance discontinuity in Step 2. This is not a transient buildup condition. If we ignore any distortion in the 33.33W Pref2 wavefront, this is an immediate event and Pref1 is immediately canceled by an equal magnitude and opposite phase Pref2(1-|rho|^2) wavefront at the moment it first arrives. You're sourcing and sinking an additional 33.33 watts, and yet the wattmeter can't discern the difference between this scenario and the 100 watt, single source scenario. But that sourcing and sinking is occurring *INSIDE* the SGCL(150). The net power inside SGCL(150) is 100W dissipated, *exactly* like the other scenario. The example illustrates perfectly the shortcomings of the idea of power flow, as well as some of the faulty conclusions that can be drawn from measurements made by a directional power meter. The shortcoming I notice is your sidestepping of the question: What happened to Pref1=25W? It just seems to have disappeared when we turned on SGCL(150) and the 33.33W wavefront arrived at the impedance discontinuity. What could have possibly made Pref1 disappear? -- 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! =----- |
It's not clear why you'd make that assumption.
A transmission line doesn't radiate; a traveling wave antenna does. The feedpoint impedance of a traveling wave antenna is dictated by both the termination resistance and by radiation. The characteristic impedance of a transmission line isn't a function of either. You can make a transmission line with an arbitrarily high impedance, although because of the nearly logarithmic relationship between wire diameter/spacing and Z0, values of more than a few hundred ohms (for open wire line, and considerably less for coax) become impractical. Of course, you'd have to go to very low frequencies to make use of very high Z0 transmission line, to avoid line radiation or, in the case of coax, propagation of other than TEM waves. And, just out of curiosity, how would you construct a traveling wave dipole? Roy Lewallen, W7EL W5DXP wrote: Reg Edwards wrote: The feedpoint impedance of traveling wave antennas is usually about 600-800 ohms according to The ARRL Antenna Book. Anyway, if you have correctly quoted it, the ARRL Antenna Book is wrong, What is the maximum Z0 possible with transmission line? Seems that would be the feedpoint impedance of a traveling wave dipole. |
"Roy Lewallen" wrote in message ... Tarmo Tammaru wrote: Roy, You are cheating. In the steady state there is no load on your source. Regardless of what the Bird meter reads. Do one of the following: Why is this cheating? There is reverse power on the line. The source is not absorbing the reverse power. You and others have said, without qualification, that it does. I've shown a case where it doesn't. What you have done is to put a resonant parallel tuned circuit on the output end of the 50 Ohm resistor. After 1/2 cycle of RF, the voltage on the two ends of the resistor is equal. Hence no power is delivered from the source, provided it is lossless coax. Tell me if I am wrong, but you could put an HP oscillator and a transformer that delivers 70.7V into an open circuit in place of the 100W amp, and in the steady state you would reach the same conditions. It would just take longer. If the Bird read 100W before, it would still read 100W. Assume it is a perfect WM that does not absorb any power. If you buy the tuned circuit analogy, the WM is in effect measuring circulating current in the tuned circuit. Here is what I think the problem with trying to understand reflected power in transmitters is. On the one hand, you have forward and reflected power in and out of a black box. On the other hand, open up the black box and measure the IMPEDANCE back towards the load. Without knowledge of reflections, we know from the Smith chart how a misterminated line is going to change the load on the collector of the transistor, and hence cause it to change its power output. So, in the impedance analysis there is no concept of absorbing reflected power. You don't like talking about pulses, but that is the way I was taught reflections. We used "Electromagnetic Energy Transmission and Radiation" by Adler, Chu, and Fano. 1960 (gulp!) Tam/WB2TT |
Roy Lewallen wrote:
A transmission line doesn't radiate; a traveling wave antenna does. The energy flowing in a flat transmission line goes somewhere. The energy flowing in a traveling wave antenna goes somewhere. Both systems are flat so they seem similar to me. A terminated Rhombic radiates and has approximately the same feedpoint impedance as wide spaced transmission line. And, just out of curiosity, how would you construct a traveling wave dipole? In my head or in real life? A terminated Sloping-V comes to mind. -- 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:
Why is this cheating? There is reverse power on the line. The source is not absorbing the reverse power. If "the source is not absorbing the reverse power", then the reverse power is being re-reflected to become the forward power. During steady- state, the forward power is not coming from the source so it has to be coming from a re-reflection process. It can only come from one of those two places. In a lossless line, you could disconnect the line at the source and there would still exist forward energy and reflected energy. The reflected energy would be reflected from one open end and the forward energy would be reflected from the other open end. It's essentially a super- conducting ring that has been cut and straightened out. Make the stub one second long and you will have 200 joules stored in the line. The source will have supplied all of that 200 joules with the voltage and current in phase, i.e. it was supplied as joules/sec. -- 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! =----- |
Ian White, G3SEK wrote:
* The output impedance of the transistor doesn't come into the story at all - not when characterizing RF power devices that are not operating in class A. Even the device manufacturer doesn't know or care what it is. Neither need we. Tubes and transistor power amplifiers quite oftem use negative feedback to improve SSB linearity. Improvements of 5 to 10 dB are common. The negative feedback reduces the internal impedance of the tube and transistor amplifiers. The tube/transistor data sheets do not consider this factor. Again, we usually don't really know or care much about the values of the internal impedances. But there is a special case. Voice/music/data tube transmitters operating at low frequencies have a problem called "sideband clipping" where the plate tank selectivity may be too sharp and reduces the modulation bandwidth. The internal impedance tends to broaden the response at resonance. When designing the tank circuit this effect may have to be included. Bill W0IYH |
On Fri, 18 Jul 2003 22:32:13 -0500, W5DXP
wrote: Maximum power is when you turn your metal 6L6 upside down in a glass of water during a contest. :-) What I have learned so far: If you have a 100 watt transmitter, the watt meter shows 3 watts reflected. I deliver 103 watts to the antenna. I now know where the reflected power go's. But where did it come from? If I could find a way to have 100 watts reflected I could put 200 watts to the antenna from a 100 watt transmitter. If my transmitter has an output impedance of 50 ohms, all the reflected power will be absorbed by my PA. I need to find a way to change my output impedance to something other than 50 ohms? If I could make my SB-401 act like a radar transmitter, and time the pulses correctly, I could cancel out the reflected power pulses. For some reason I need a circulator on my SB-401. To get max power out my 6146's I need to turn them upside down in a glass of water? :-) |
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