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#31




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. 
#33




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 
#34




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 
#35




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 
#36




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 ftlbs of torque is the same as 10 ftlbs 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 crossdisiplinary, 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 selfdischarge 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 Jello, 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 
#37




"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 
#38




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 Jello, 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 
#39




William E. Sabin wrote:
W5DXP wrote: Maybe it should be "maximum *available* steadystate 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! = 
#40




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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