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#131
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Cecil Moore wrote: Many examples exist for current drops in distributed networks. That's one thing that makes circuit analysis invalid for distributed network problems. The series current is NOT the same value everywhere in a distributed network. Asserting that there is no such thing as "current drop" in distributed networks simply indicates an invalid choice of models. How much current drop is there at 440 MHz in 100 feet of RG-58 between the source and a 50 ohm load? Answer: A 20 dB power drop equates to a 40 dB current drop. Cecil, You seem to like the "Roach Motel" theory of current flow. The electrons check in, but they don't check out. Here's a clue. Conservation of charge is every bit as fundamental as conservation of energy. Current does not just disappear. So what happens in your abused RG-58 case? Answer: this is not a simple series circuit. At every point along the line the current splits between continuing down the line and shunting to the other half of the transmission line. When the line is lossless, the shunting is purely reactive, and no net current flows. However, when there is loss in the line, there is a small phase shift along with the attenuation, and net current is shunted. The "circuit" model, as you like to call it, is every bit as valid as the "distributed network" model. However, due to the distributed time and space considerations in a transmission line, the "circuit" model is mathematically intractable for many applications. The physical reality remains the same even if we cannot easily do the math. Oh, by the way, in a constant impedance environment the current change corresponding to a power reduction of 20 dB is also 20 dB, not 40 dB. 73, Gene W4SZ |
#132
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On Mon, 25 Oct 2004 15:10:22 GMT, Gene Fuller
wrote: However, due to the distributed time and space considerations in a transmission line, the "circuit" model is mathematically intractable for many applications. AKA Violation of Kirchhoff |
#133
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Richard,
I am not sure why you think there is a violation of Kirchhoff. The current law is unchanged. The analysis of the voltage law is much more complicated, but not incorrect. The situation is not really any different than the use of retarded potentials for radiation. One must carefully keep track of the loop voltages with consideration for time and space differences, but there is no fundamental difference in the physics. For distributed networks, the Kirchhoff voltage calculation is difficult (intractable). It is not impossible, but it is unnecessary due to the existence of the much more friendly transmission line formulations. 73, Gene W4SZ Richard Clark wrote: On Mon, 25 Oct 2004 15:10:22 GMT, Gene Fuller wrote: However, due to the distributed time and space considerations in a transmission line, the "circuit" model is mathematically intractable for many applications. AKA Violation of Kirchhoff |
#134
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Tom Donaly, KA6RUH wrote:
"I know there are people in the past who have attempted to characterize antennas as transmission lines." True. You can include Terman among them. Terman discusses antenna current distribution on page 866 of his 1955 edition: "Under most circumstances the losses are sufficiently low and the ratio of wire length to diameter sufficiently great so that to a first approximation the current distribution can be taken as that for a line with zero losses; it then has the characteristics discussed in Sec, 4-5." Sec.4-5 is found on page 95 and is titled: "The Effect of Attenuation on Voltage and Current Distribution - Lossless Lines". Sec. 4-5 is in Terman`s chapter on Transmission lines. Obviously an open-circuit antenna has the same current distribution as an open-circuit transmission line, and for the same reason. Best regards, Richard Harrison, KB5WZI |
#135
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"Tom Donaly" wrote:
I expect someone on this newsgroup will start talking about "voltage flow" next. The E-field and the H-field are inseparable in an EM RF wave. If the H-field (current component) is flowing, then so is the E-field (voltage component). If the E-field (voltage component) is dropping due to radiation, then so is the H-field (current component). That's one of the differences between a distributed network RF wave and a DC bench circuit. And how could we possibly have a Power Flow Vector without the voltage flowing along with the power and the current? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
#136
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Tom Donaly wrote: If you make clear what you mean and why you think it makes sense there's nothing wrong with using non-standard terminology if it aids communication. If you just pull it out of the air as if it were a normal technical term that all technical people use, then you're just gassing nonsense. I expect someone on this newsgroup will start talking about "voltage flow" next. 73, Tom Donaly, KA6RUH Probably best we distinguish between "flow", and propagation, before we digress into discussions of "standing wave propagation". ;-) 73, Jim AC6XG |
#137
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Richard Harrison wrote:
Tom Donaly, KA6RUH wrote: "I know there are people in the past who have attempted to characterize antennas as transmission lines." True. You can include Terman among them. Terman discusses antenna current distribution on page 866 of his 1955 edition: "Under most circumstances the losses are sufficiently low and the ratio of wire length to diameter sufficiently great so that to a first approximation the current distribution can be taken as that for a line with zero losses; it then has the characteristics discussed in Sec, 4-5." Sec.4-5 is found on page 95 and is titled: "The Effect of Attenuation on Voltage and Current Distribution - Lossless Lines". Sec. 4-5 is in Terman`s chapter on Transmission lines. Obviously an open-circuit antenna has the same current distribution as an open-circuit transmission line, and for the same reason. Best regards, Richard Harrison, KB5WZI You don't even have to go as far back as Terman. Reg Edwards bases his antenna programs on the transmission line behavior of antennas, and his programs do as well as you'd want. That's one way to look at antenna behavior. It's not the only way, and it doesn't mean that Cecil knows what he's talking about when he espouses the theories he made up in his head out there in the hot Texas sun. 73, Tom Donaly, KA6RUH |
#138
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In message , Cecil Moore
writes How much current drop is there at 440 MHz in 100 feet of RG-58 between the source and a 50 ohm load? Answer: A 20 dB power drop equates to a 40 dB current drop. Uh? Are you really REALLY sure? Do you want to change your mind? Ian. -- |
#139
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It's not the only way, and it doesn't mean that
Cecil knows what he's talking about when he espouses the theories he made up in his head out there in the hot Texas sun. 73, Tom Donaly, KA6RUH We get the picture of you Tom, you can stop with that crap, quite enough! Chipster will get jelous for you to take away his crown as troller extraordinaire. Yuri |
#140
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Gene Fuller wrote:
Conservation of charge is every bit as fundamental as conservation of energy. Current does not just disappear. So what happens in your abused RG-58 case? Answer: this is not a simple series circuit. Thanks Gene, that is exactly my point. 17th century simple series DC concepts don't work on RF distributed networks. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
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