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#71
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Jim Kelley wrote:
One point: Isn't the input impedance of a dipole normally specified at a wavelength equal to twice the electrical length of the antenna? As far as I know, dipoles have infinite DC resistance at zero Hertz. ;-) That would be true for an electrical dipole but we are obviously talking about physical poles here, i.e. two infinite conductive fishing poles. :-) -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#72
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Reg Edwards wrote:
Cec, what is your best estimate of the input impedance of an infinitely long dipole. Just a number please. Guesstimate: 900-j175 ohms. -- 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 =--- |
#73
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Reg Edwards wrote:
On a long terminated non-resonant line I guess the current falls off crudely exponentially at a rate of about 1 dB per wavelength. Seems to me if we assume 1 dB per wavelength for both current and voltage and knowing the feedpoint impedance of a one WL dipole, we could calculate the characteristic impedance of the antenna at the feedpoint. Seems to me it would be essentially the same calculation as for a 1/2WL open-circuit stub where the transmission line has losses of 1 dB per wavelength and we measure a stub impedance of 5000 ohms. We could calculate the Z0 of the stub. Couldn't we use a TDR to measure the loss in an open-ended 1/2WL piece of wire? -- 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 =--- |
#74
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| ----------------------------------------------------------
| "Jim Kelley" | wrote in message ... | | [...] | | "X-rays will prove to be a hoax." | | http://zapatopi.net/kelvin/quotes/ | | ac6xg | ---------------------------------------------------------- A I hope you will excuse me the next example. Let f = Sin[x] I choose: x = 2*k*pi and k goes to infinity one by one: 0, 1, 2, ... Definitely then I found correctly f(oo) = 0. Cecil chooses: x = 2*k*pi + pi/2 and k goes to infinity one by one, as before. Definitely he founds correctly f(oo) = +1 Lord Kelvin chose: x = 2*k*pi - pi/2 and k went to infinity one by one, as above. Definitely he founded correctly f(oo) = -1 All of us we are correct in all steps, but the value f(oo) does not exist as a single one. In fact f(oo) takes every value between -1 and +1. f(oo) definitely depends on the way in which each one of us went to infinity. IMHO: this is the kind of behavior of Zinp. B But in addition to that there is one more to say: Zinp is a result of the order in which we consider the limits for the wire radius and the length to wavelength ratio. If a is the wire radius and L/wl is the ratio of length to wavelength then I can imagine five cases: 1 First the a is going to zero, a formula is produced for Zinp, then the L/wl is going to infinity and a number may or may not be the result for Zinp. 2 First the L/wl is going to infinity a formula is produced for Zinp, then the a is going to zero and a number may or may not be the result for Zinp. 3 Simultaneously, both the L/wl is going to infinity and the a is going to zero, and a number may or may not be the result for Zinp. 4 We keep a constant value for L/wl, then a is going to zero and a number may or may not be the result for Zinp. 5 We keep a constant value for a, then L/wl is going to infinity and a number may or may not be the result for Zinp. [ On the occasion I have to confess that the movie at [ http://antennas.ee.duth.gr/ftp/visua...s/fu010100.zip [ 850 KB [ belongs to the last case. For a possible conclusion let me mention a remarkable note from a Mathematical book: "The biggest source of erroneous conclusions have to do with the order we consider the limits" (and which order we tend then to forget ... ) Sincerely, pezSV7BAXdag |
#75
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On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote: Reg Edwards wrote: And remember Lord Kelvin. ---- Reg. "To measure is to know." also "X-rays will prove to be a hoax." http://zapatopi.net/kelvin/quotes/ ac6xg Terrific reference, Jim, I've added it to my 'favorites'. Walt |
#76
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On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote: Reg Edwards wrote: And remember Lord Kelvin. "X-rays will prove to be a hoax." http://zapatopi.net/kelvin/quotes/ Hi Jim, Certainly a trove of complexity. Keeping to the tenor of your choice: "I have not had a moment's peace or happiness in respect to electromagnetic theory since November 28, 1846. All this time I have been liable to fits of ether dipsomania, kept away at intervals only by rigorous abstention from thought on the subject." I sippose that California varietals were not available as an alternative to ether. Oh how we can celebrate the modern march of progress by viewing the problems of electromagnetic theory through the bottom of a wine glass. Such libation allows one to simultaneously to offer solutions and to observe that "rigorous abstention from thought." Lord Kelvinator would be proud. 73's Richard Clark, KB7QHC |
#77
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On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote: "X-rays will prove to be a hoax." http://zapatopi.net/kelvin/quotes/ "Radio has no future." Hi Jim, Almost any lesson can be drawn from the hazard of quote choices: "The wireless telegraphy is one of the most wonderful inventions the world has ever seen." One has to allow that he was either a world class loon, or had the ability to change his mind. No doubt the latter characteristic prevails, but this is hardly the forum for its celebration. Thanx for the link to such a trove. 73's Richard Clark, KB7QHC |
#78
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Richard Clark wrote: On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley wrote: Reg Edwards wrote: And remember Lord Kelvin. "X-rays will prove to be a hoax." http://zapatopi.net/kelvin/quotes/ Hi Jim, Certainly a trove of complexity. Keeping to the tenor of your choice: "I have not had a moment's peace or happiness in respect to electromagnetic theory since November 28, 1846. All this time I have been liable to fits of ether dipsomania, kept away at intervals only by rigorous abstention from thought on the subject." Or as is my case, intervals of rigorous abstention from participating in discussions on rec.radio.amateur.antenna. ac6xg |
#79
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Reg Edwards wrote:
"Ian Jackson" wrote - Are you sure it's as high as that, Reg? I once did a Smith Chart plot of the impedance at the centre of a dipole, the valued being taken from a table 'compiled by Wu' (LK Wu?). These only catered for a lengths up to a few wavelengths. As the plot progressed round and round the Smith Chart, it seemed to be heading for something around 350 to 400 ohms. I've just done a search on 'Wu+dipole+impedance', and one of the results is http://www.fars.k6ya.org/docs/antenn...nce-models.pdf I'll have a read of it today. =================================== The characteristic impedance of an infinitely long wire is Zo. If we cut the line and measure between the two ends we obtain an input impedance of twice Zo. Which is the answer to our problem. Zo is a function of wavelength, conductor diameter and conductor resistance R where R includes the uniformly distributed radiation resistance. On a high Zo line the radiation resistance is small compared with Zo and the only effect of the radiation resistance is to give Zo a small negative angle. Which when estimating Zo can be ignored. (It is conductor resistance which at HF gives Zo of ALL lines a very small negative angle). In the problem posed, the current is also uniformly distributed along the low-loss line and radiation resistance is not the value we are familiar with and what we might do with it. And so we get approximately - Rin = 120 * ( Ln( Wavelength / 2 / d ) - 1 ) At a wavelength of 2 metres and a conductor diameter of 10mm the input resistance = 433 ohms. I cannot guarantee the above formula to be correct. But is it low enough for you? ;o) Mr Wu calculates radiation resistance which is not the same as input impedance unless correctly referenced. It is usual in technical papers to calculate Radres at one end of the antenna. Or it may be the distributed value. I havn't the time to find and study the full text. From past experience, with me, it usually ends up as a wild goose chase. ---- Reg. Sounds reasonable, Reg. To put it for simple people like me, it would mean it's a transmission line of diameter x with an infinite diameter shield. Then we feed 2 of them, balanced, colinear, and that's our R sub r. Did I misunderstand? tom K0TAR |
#80
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Roy Lewallen wrote:
Sure. 42. It might not be better, but it's just as good. Formulas can be made very simple if you simply ignore any inconvenient facts. Like radiation from an antenna. Roy Lewallen, W7EL 42. The answer to the question, of life, the universe, and everything. The movie wasn't bad, but the radio play was still better. tom K0TAR |
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