|
Ham op wrote:
NOTE: an infinitely long wire in free space cannot be terminated. Doesn't need to be terminated. Reflections are eliminated without terminations. A terminated wire in an anechoic chamber is not the same as a free space model. Not exactly the same but close enough for government work. Reflections from the ends and from the walls are reduced to a negligible value. Been there, done that. -- 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 =--- |
| "Reg Edwards" wrote in message
| ... | | You had better go back to your old college days | and start again with new teachers. | | The input impedance of an infinite dipole | is as I have already simply mathematically described. | It changes a little with frequency and wire diameter. | | Anything less than infinite length will be altogether different. | | Now you can stop trying to pull our legs. | ---- | Reg. Dear Mr. Reg Edwards, This is an interesting question indeed! Well, my wife yinSV7DMCdag has developed sometime an AVI movie [ in ZIP form: [ http://antennas.ee.duth.gr/ftp/visua...s/fu010100.zip [ 850 KB which deals among other properties of a filamentary dipole, with its Radiation Resistance relative to the Input Terminals (Driving Point) [ Please take a look if you like [ in the part D of the movie, [ at the bottom right of the screen This is actually the Input Resistance Rinp of a filamentary dipole of perfect conductivity in free space. The ratio of dipole length L to wavelength lambda takes values up to 10, but someone may maintain arguably that either the limit of Rinp exists in general as infinity or does not exist at all. Yours Sincerely, pezSV7BAXdag |
In message , Reg
Edwards writes What is the impedance at the centre of an infinitely long dipole (in free space)? =============================== Its not very different from - Zin = 120 * Ln( Wavelength / d ) ohms. where d = conductor diameter, both measured in metres. Thus, at wavelength = 80 metres with 14 gauge copper wire, input impedance = 1300 ohms approx. If you don't believe me, just measure it. ---- Reg. 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. Cheers, Ian. -- |
| "Ian Jackson"
| wrote in message ... | [...] | it seemed to be heading for something around 350 to 400 ohms. | [...] | Ian. 120*pi maybe ... pezSV7BAXdag |
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. Maybe 377 ohms? Remember that any finite length dipole is a standing wave antenna and the feedpoint impedance is (Vfor+Vref)/(Ifor+Iref) where Vfor is the forward voltage phasor, Vref is the reflected voltage phasor, Ifor is the forward current phasor, and Iref is the reflected current phasor. For a 1/2WL resonant dipole the feedpoint impedance is low: R = (|Vfor|-|Vref|)/(|Ifor|+|Iref|) ~ 73 ohms For a 1WL (anti)resonant dipole the feedpoint impedance is high: R = (|Vfor|+|Vref|)/(|Ifor|-|Iref|) ~ 5200 ohms (EZNEC) An infinite dipole would not be a standing wave antenna. It would be a traveling wave antenna (as in a terminated rhombic). So the feedpoint impedance of an infinite dipole would be Vfor/Ifor=Z0. Since the reflections modify the feedpoint impedance, we might suspect that Vfor/Ifor falls between the feedpoint impedance for a 1/2WL dipole and a one WL dipole. Seems to me, the Z0 of the dipole, i.e. Vfor/Ifor, must be in the ballpark of the square root of the product of those two feedpoint impedances. -- 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 =---- |
"Cecil Moore"
For a 1/2WL resonant dipole the feedpoint impedance is low: R = (|Vfor|-|Vref|)/(|Ifor|+|Iref|) ~ 73 ohms _________________ The 73 ohm radiation resistance value applies to a physical 1/2-wave, thin-wire, linear dipole in free space, however a reactance term of + j42.5 ohms also applies to such a configuration (Kraus 3rd Edition, p. 182). The dipole length needs to be shorted by several percent in order to zero out the reactance term, at which time (according to Kraus), the resistance term will be about 65 ohms. RF |
In message , Cecil Moore
writes 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. Maybe 377 ohms? Remember that any finite length dipole is a standing wave antenna and the feedpoint impedance is (Vfor+Vref)/(Ifor+Iref) where Vfor is the forward voltage phasor, Vref is the reflected voltage phasor, Ifor is the forward current phasor, and Iref is the reflected current phasor. For a 1/2WL resonant dipole the feedpoint impedance is low: R = (|Vfor|-|Vref|)/(|Ifor|+|Iref|) ~ 73 ohms For a 1WL (anti)resonant dipole the feedpoint impedance is high: R = (|Vfor|+|Vref|)/(|Ifor|-|Iref|) ~ 5200 ohms (EZNEC) An infinite dipole would not be a standing wave antenna. It would be a traveling wave antenna (as in a terminated rhombic). So the feedpoint impedance of an infinite dipole would be Vfor/Ifor=Z0. Since the reflections modify the feedpoint impedance, we might suspect that Vfor/Ifor falls between the feedpoint impedance for a 1/2WL dipole and a one WL dipole. Seems to me, the Z0 of the dipole, i.e. Vfor/Ifor, must be in the ballpark of the square root of the product of those two feedpoint impedances. Yes, I did think of 377 ohms (which I understand is 'the impedance of free space'), but I'm no expert in these matters. As you indicate, the impedance must lie somewhere between 73 and 5200 ohms. You suggest that this might be something like the square root of the product of those two feedpoint impedances (the geometric mean), which gives 616 ohms. However, you would see 600 ohms simply by looking into an infinite length of 600 ohm feeder, which has parallel, non-radiating conductors. If the length of the feeder was relatively short (compared with infinity!!), pulling the conductors apart would increase the impedance (probably to a lot more than 616 ohms). The question is, 'when does radiation start to influence the impedance?' If you look at K6OIK's paper at http://www.fars.k6ya.org/docs/antenn...nce-models.pdf and look at, for example, page 22, you can see how the feed impedance at odd halfwaves increases, and at even halfwaves, decreases. I only found this paper this morning, and haven't had time to look to see which (if any) of the many formulas was used to obtain the plot. It must be possible to get close to the infinity condition by entering values for a very, very long dipole. Cheers, Ian. -- |
In King and Harrrison's _Antennas and Waves_, they show a plot of
calculated antenna feedpoint impedance as X vs R up to about 5 wavelengths. Antenna wire radius is 0.008496 wavelength. The Z of an infinite length antenna is indicated by locating the centers of the circles and noting that the center converges. The point of convergence for this particular wire radius is about 250 - j170 ohms. In the chapter on experimental measurements, there's a plot of the calculated admittance of an antenna of radius 0.000635 wavelength up to about 10 wavelengths. Superimposed are measured values from another source which show very good agreement. The theoretical values converge at 214 - j189 ohms, and the measured values at 218 - j174 ohms. Dervivation takes about a chapter of very heavy math, and numerical results were obtained with a computer. Roy Lewallen, W7EL |
Ian Jackson wrote:
"Maybe 377 ohms?" Arnold B. Bailey in "TV and Other Receiving Antennas" shows his calculations of radiation resistance at the current maximum point for a center-fed thin dipole at its various resonances: 1st--------------------------------72 ohms 2nd------------------------------200 ohms 3rd-------------------------------102 ohms 4th-------------------------------260 ohms 5th-------------------------------117 ohms 6th-------------------------------295 ohms 7th-------------------------------127 ohms 8th-------------------------------321 ohms 9th-------------------------------135 ohms 10th-----------------------------340 ohms He also shows the current distributions and feed point resistances which I can`t and won`t. The input resistance of the center-fed antenna at resonance never equals the surge resistance in value, but is related to it. Best regards, Richard Harrison, KB5WZI |
Roy Lewallen wrote:
The theoretical values converge at 214 - j189 ohms, and the measured values at 218 - j174 ohms. Free space? As a data point, I pushed EZNEC to the limit on 40m with a 9000 ft. dipole. Resonant feedpoint resistance at 7.152 is 390 ohms. Anti-resonant feedpoint resistance at 7.092 is 1980 ohms. It appears that EZNEC would converge to something in between those two values for an infinite dipole in free space. I ran into the segment limit at 66 wavelengths. -- 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 =--- |
Richard Harrison wrote:
Ian Jackson wrote: "Maybe 377 ohms?" Arnold B. Bailey in "TV and Other Receiving Antennas" shows his calculations of radiation resistance at the current maximum point for a center-fed thin dipole at its various resonances: 1st--------------------------------72 ohms 2nd------------------------------200 ohms 3rd-------------------------------102 ohms 4th-------------------------------260 ohms 5th-------------------------------117 ohms 6th-------------------------------295 ohms 7th-------------------------------127 ohms 8th-------------------------------321 ohms 9th-------------------------------135 ohms 10th-----------------------------340 ohms Note that those are not the feedpoint impedances. They are based on the current maximum points which often occur somewhere besides the feedpoint. -- 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 =--- |
| "Roy Lewallen"
| wrote in message ... | [...] | The Z of an | infinite length antenna is indicated by locating the centers of the | circles and noting that the center converges. | [...] | Roy Lewallen, W7EL If we discuss here the impedance referenced to the input (base) current - and not to the maximum one - then IMHO: The quoted text above does not prove convergence. The convergence must be independent of the way the length goes to infinity. The centers of whatever circles may converge to a finite complex number but their radii have to simultaneously converge to zero, to have convergence. But the limit for Z exists if and only if both the limits for R and X exist. Therefore if the limit for R is dependent on the way the length goes to infinity then its limit does not exist. A guess for either a non-existent limit for R or an infinite one comes out from: http://antennas.ee.duth.gr/ftp/visua...s/fu010100.zip [850 KB] If either of the above is true for R then the corresponding is true for Z: The limit for Z does not exist or is (in general) the complex infinity. But always and only for the the impedance referenced to the input (base) current. Sincerely, pezSV7BAXdag |
pezSV7BAXdag wrote:
The limit for Z does not exist or is (in general) the complex infinity. As the length of a dipole is increased, for the same power input, more energy is radiated during the first transcient cycle and less is available for reflection from the ends of the dipole. Reflected energy is what is causing the feedpoint impedance to change. As the length of the dipole is incrementally increased, the magnitude of the reflected energy is incrementally decreased. I believe Balanis alludes to this characteristic of standing-wave antennas. The feedpoint impedance is Zfp = (Vfor+Vref)/(Ifor+Iref) using phasor addition. The limit of that equation as Vref and Iref go to zero is Vfor/Ifor. That's what happens for an infinitely long dipole. That's also what happens during the transient phase of a finite dipole. Thus, Vfor/Ifor can be thought of as the characteristic impedance of the dipole. Seems to me, Vfor/Ifor could actually be measured during the transient phase of a long finite dipole. Will a TDR report the ratio of V/I for an RF pulse? -- 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 =---- |
Cecil Moore wrote:
Roy Lewallen wrote: The theoretical values converge at 214 - j189 ohms, and the measured values at 218 - j174 ohms. Free space? As a data point, I pushed EZNEC to the limit on 40m with a 9000 ft. dipole. Resonant feedpoint resistance at 7.152 is 390 ohms. Anti-resonant feedpoint resistance at 7.092 is 1980 ohms. It appears that EZNEC would converge to something in between those two values for an infinite dipole in free space. Forgot to add, EZNEC would also converge to approximately the same reactance value as above. -- 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 =---- |
Cecil, W5DXP wrote:
"Note that those are not the feedpoint impedances.". Well, the imprdance I qupted was the radiation resistance, which is the voltage to current ratio of an antenna at the maximum current point in the antenna. I was just too lazy to post the complete Table given by Bailey. Only when the order of resonance in a center-fed dipole is odd, 1st, 3rd, 5th, etc, is the feedpoint resistance the same as the radiation resistance. In odd-ordered resonances, the antenna feedpoint resistance is te same as the radiation resistance. In odd-orfered resonances, the antenna feedpoint is at s current loop. In even-ordered resonances, the value of the feedpoint resistance is equal to the feedpoint impedance squared, divided by the radiation resistance value. Bailey has worked it all out for us. Best regards, Richard Harrison, KB5WZI |
| -----------------------------------------------------------------
| Subject...: 73 Ohms, How do you get it? | Sent......: Monday, September 19, 2005 5:27 PM | Newsgroups: rec.radio.amateur.antenna | From......: "Cecil Moore" | ----------------------------------------------------------------- | [...] | As the length of a dipole is increased, for the same | power input, more energy is radiated during the first | transcient cycle and less is available for reflection | from the ends of the dipole. Reflected energy is what | is causing the feedpoint impedance to change. As the | length of the dipole is incrementally increased, the | magnitude of the reflected energy is incrementally | decreased. I believe Balanis alludes to this characteristic | of standing-wave antennas. | | The feedpoint impedance is Zfp = (Vfor+Vref)/(Ifor+Iref) | using phasor addition. | | The limit of that equation as Vref and Iref go to zero | is Vfor/Ifor. That's what happens for an infinitely | long dipole. That's also what happens during the transient | phase of a finite dipole. Thus, Vfor/Ifor can be thought | of as the characteristic impedance of the dipole. Seems | to me, Vfor/Ifor could actually be measured during the | transient phase of a long finite dipole. Will a TDR | report the ratio of V/I for an RF pulse? | -- | 73, Cecil http://www.qsl.net/w5dxp | ----------------------------------------------------------------- Do you mean that since the length is infinite there is no reflected wave? But then here it is, once again, one of the most controversial issues... Well, I think we are in the front of a case in which the limit depends on the way we approach it. Every logical way to approach a limit is permissible. And these ways are infinite in number of course. But the convergence is too demanding: "She" wants all these limits to be equal. If just two of them are unequal the convergence simply does not exist. Mathematically this is not a rare case, they say. Sincerely, pezSV7BAXdag |
"Ian Jackson" wrote in message ... In message , Reg Edwards writes What is the impedance at the centre of an infinitely long dipole (in free space)? =============================== Its not very different from - Zin = 120 * Ln( Wavelength / d ) ohms. where d = conductor diameter, both measured in metres. Thus, at wavelength = 80 metres with 14 gauge copper wire, input impedance = 1300 ohms approx. If you don't believe me, just measure it. ---- Reg. 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. Cheers, Ian. -- |
"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. |
pezSV7BAXdag wrote:
Do you mean that since the length is infinite there is no reflected wave? But then here it is, once again, one of the most controversial issues... Yep, it's my digital logic in action. :-) If there's no end, then there cannot be reflections from the ends. And please note that I am not saying that the characteristic impedance of a dipole is constant all up and down the line. I'm only concerned about the apparent characteristic impedance at the feedpoint, i.e. Vfor/Ifor. Consider an open circuit transmission line. At 1/4WL, it exhibits a very low impedance. At 1/2WL it exhibits a very high impedance. As the transmission line length is increased to infinity, because of losses, the "stub" impedance will spiral into the Z0 characteristic impedance point. A similar concept should apply to a dipole. -- 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 =---- |
Reg Edwards wrote:
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. Reg, in the real world, an antenna has radiation losses so the current decays along its length. Is there any formula that includes an attenuation factor for a traveling wave antenna? It would be akin to the attenuation factor for a transmission line but presumably higher. I have estimated that, for a 1/2WL dipole, the reflected voltage and current have dropped by approximately 10% below the forward voltage and current during the round trip to the ends of the antenna and back. -- 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 =---- |
Cecil Moore wrote: Roy Lewallen wrote: The theoretical values converge at 214 - j189 ohms, and the measured values at 218 - j174 ohms. Free space? As a data point, I pushed EZNEC to the limit on 40m with a 9000 ft. dipole. Resonant feedpoint resistance at 7.152 is 390 ohms. Anti-resonant feedpoint resistance at 7.092 is 1980 ohms. It appears that EZNEC would converge to something in between those two values for an infinite dipole in free space. I ran into the segment limit at 66 wavelengths. 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. ;-) ac6xg |
Cec, what is your best estimate of the input impedance of an
infinitely long dipole. Just a number please. Remember Lord Kelvin! ---- Reg. |
Reg Edwards wrote:
=================================== 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). This assumption is correct only when the transmission line conductors are closely spaced. That isn't at all true for the halves of a dipole. 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) I can't see how it can possibly be correct. Unless I'm mistaken, you've completely ignored the effect of radiation in calculating the radiation resistance. It sure makes the calculation a lot simpler, though! 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. It depends on the author. Kraus uses feedpoint resistance and radiation resistance interchangeably when loss is assumed to be zero. It's traditional in AM broadcasting to give radiation resistance referred to a current maximum. The conclusion is that radiation resistance can be referred to any point along an antenna you wish, which means that it's essential to state what point you're using as a reference. Roy Lewallen, W7EL |
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. ;-) No, you can calculate or specify the input impedance of a dipole at any frequency. As frequency approaches zero, a dipole's input resistance approaches zero and its reactance approaches minus inifnity. That is, it looks like a capacitor, and the capacitive reactance gets larger as the frequency gets lower. Which is just what you'd expect from a couple of electrically very short wires having no DC connection. Roy Lewallen, W7EL |
Reg Edwards wrote: 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. Reg, in the real world, an antenna has radiation losses so the current decays along its length. ============================== Yes, I know. Just insert "approximately" before "uniformly" On a long terminated non-resonant line I guess the current falls off crudely exponentially at a rate of about 1 dB per wavelength. The situation is similar to that very long, low, terminated antenna wire whose name I can't remember. And whose input resistance at LF is about 550 ohms. Putting two of them back-to-back gives an input resistance of twice that = 1100 ohms. Which indicates the answer to our infinitely long dipole problem. ---- Reg. |
Roy Lewallen wrote: 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. ;-) As frequency approaches zero, a dipole's input resistance approaches zero and its reactance approaches minus inifnity. That is, it looks like a capacitor, and the capacitive reactance gets larger as the frequency gets lower. Which is just what you'd expect from a couple of electrically very short wires having no DC connection. Roy Lewallen, W7EL I'll give you a Mulligan on that one if you like, Roy. ;-) 73, ac6xg |
Roy, if you don't like my simple approximate formula, can YOU produce
a better one without plagiarising? ---- Reg. |
Reg Edwards wrote:
Roy, if you don't like my simple approximate formula, can YOU produce a better one without plagiarising? ---- Reg. 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 |
"Roy Lewallen" wrote Reg Edwards wrote: Roy, if you don't like my simple approximate formula, can YOU produce a better one without plagiarising? ---- Reg. ====================================== 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 ===================================== Don't be silly. I didn't ignore radiation resistance. I said it was small enough in comparison with Zo, as an approximation, to forget about. And remember Lord Kelvin. ---- Reg. |
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 |
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 =---- |
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 =--- |
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 =--- |
| ----------------------------------------------------------
| "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 |
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 |
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 |
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 |
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 |
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 |
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 |
| All times are GMT +1. The time now is 12:47 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com