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#22




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 ftlbs of torque is the same as 10 ftlbs 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 (nonreactive) 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 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. . . 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 noantenna 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 
#23




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 inphase current of 50 amps. Hmmmm, 50 watts in, 2500 watts out. How much will you take for that antenna, Richard? :) 
#24




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! = 
#25




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 (nonreactive). You would have some loss with the mismatch between the coax and antenna? The transmitter would see 50 ohms? 
#26




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 (nonreactive). 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 
#27




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* steadystate 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 Z0match point with a high mismatch loss between the Z0match 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! = 
#28




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* 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. Bill W0IYH 
#29




"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 
#30




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 
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