|
Tom Donaly wrote:
Balanis uses a highly mathematical approach in most of his book, supplemented by many graphs and charts. Cecil's quote, like his quote of Tom Rauch on loading coils is only a very small part of the total. You want me to quote the total Balanis book?????? Why don't you, instead, just pick one subject upon which you think you and I disagree, and discuss it. The only thing I know for sure that you and I disagree on is the current at each end of a bugcatcher coil. -- 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:
There you (in the plural) go again - using handbooks as bibles. Nope, not a bible, just a quote with which I didn't even say I agreed (or not). But ~4% efficiency sounds about right for an 8 foot center-loaded mobile antenna on 75m. Cec, I'm on deep red, South African Western Cape, Pinotage-Shiraz tonight. You should try some. Makes a change from Californian, Texan and John Wayne, six-shooter politics. I'm chugging Franzia Merlot while cleaning my Colt Python .357 Magnum and Winchester 30-30 Lever-Action Carbine. :-) -- 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 =--- |
Forgotten? How can we forget a "fact" we learned wasn't true in the
first place? According to the many references I have, the equation you quote is a simplified equation that's valid for a single wire over a perfect conducting ground plane, where the height is a very small fraction of a wavelength (i.e., radiation is negligible). Even when you ignore the relatively poor conductivity and the permittivity of real ground, the equation is certainly not valid if the wire is high enough for significant radiation to take place. There are several reasons for this: 1. The field shapes become different from the shapes assumed in deriving the equation. 2. Radiation would make the impedance complex rather than purely real. 3. The voltage between the conductor and ground depends on the path taken to measure it, so "characteristic impedance" takes on a whole different meaning, if it has any at all in this context. There is, of course, also the problem of ignoring the finite conductivity of real ground, which will likewise impact the angle of the impedance. It's surely tempting to take a nice, simplistic equation like this and build from it a whole theory of how things work. The seductive thing about it is that it seems to work, sort of, for some special applications. But it's a house of cards, and is at its root based on invalid assumptions. So all the wonderful theories that follow from it are fatally flawed and not to be trusted. As apparently the only person on this newsgroup to have "learned" this "fact", it would serve you well to un-learn it. That is, if you're really interested in discovering how things really work rather than clinging to possibly mistaken notions about how they do. Roy Lewallen, W7EL Cecil Moore wrote: Apparently, a lot of the otherwise knowledgeable people on this newsgroup have forgotten that the formula for the characteristic impedance of a single-wire transmission line is 138*log(4h/d) where h is the height of the wire above ground and d is the diameter of the wire. There's no difference between that single-wire transmission line and a lot of ham antennas. That single-wire transmission line radiates just like an antenna. 1/2WL of #16 wire 24 feet in the air has a Z0 of 600 ohms. If that center-fed dipole were terminated at each end with a 600 ohm load, it would be a traveling-wave antenna with a feedpoint impedance of 600 ohms. Take away the loads and there's a match to 50 ohm coax at the feedpoint. The only difference in those two antennas is that removing the loads turned the antenna into a standing-wave antenna and reflections are arriving back at the feedpoint, lowering the feedpoint impedance. Any coil installed in a standing wave antenna is going to be subjected to both forward and reflected currents. There is no hope of understanding the current in a loading coil without understanding the component currents flowing both directions through the loading coil. -- 73, Cecil, W5DXP |
Radiation Resistance. Roy, what an excellent, well needed exposition, in plain English, addressed mainly to professionals who should know better, and are guilty (dare I say it) of obtaining money under false pretences. You omitted only "DISTRIBUTED radiation resistance", a term essential to but absent from this newsgroup. It's probably also absent from Terman and Kraus, the latter I have never read. It's an aid to clear logical thought. It should be used whenever radiation resistance is compared with conductor resistance, inductance, capacitance, etc., of elongated wires and loading coils. As an example, it so happens that the distributed, end-to-end, radiation resistance, Rd, of a half-wave diopole is exactly twice (easily proved) the feedpoint resistance of around 72 ohms. The radiating efficiency of a half-wave dipole is then, very simply and accurately - Efficiency = 144 / ( 144 + overall HF conductor resistance ) although some old-wives may wish to argue about it on the grounds that it's far too simple. ---- Reg, G4FGQ |
Sorry, I take issue with this. The radiation resistance, as universally
used in the professional literature, *is* a distributed radiation resistance -- it's the resistance that "consumes" the power radiated from the entire antenna, not just one point on the antenna. But that entire power-consuming property is commonly lumped into a single component -- the "radiation resistance" which can be defined (or "referred to") anywhere on the antenna you'd like, including but not limited to the feedpoint. And when placed at that point, it consumes the amount of power radiated from the entire antenna. It's not one single, absolute value, but a component whose value depends on where you define it on the antenna. A very simple and correct way of looking at it is to realize that if P watts is being radiated from the antenna, the radiation resistance value has to equal P/I^2, where I is the magnitude of the current at the point where you're measuring or defining the radiation resistance. So the radiation resistance always "consumes" P watts. If you want to calculate efficiency, you have to do the same thing with the loss resistance, and make a single R that consumes the same amount of power as the total antenna loss. Again, you can define it anywhere on the antenna including the feedpoint, and it'll have a different value wherever you put it. To calculate efficiency from radiation and loss resistances, both have to be -- correctly -- defined at (or "referred to") the same point. There's no need for additional "essential" fundamental terms -- the simple concept of radiation resistance as I've described it is perfectly adequate to explain and calculate antenna radiation and efficiency. But like other concepts, it does take a little effort to understand it. Roy Lewallen, W7EL Reg Edwards wrote: Radiation Resistance. Roy, what an excellent, well needed exposition, in plain English, addressed mainly to professionals who should know better, and are guilty (dare I say it) of obtaining money under false pretences. You omitted only "DISTRIBUTED radiation resistance", a term essential to but absent from this newsgroup. It's probably also absent from Terman and Kraus, the latter I have never read. It's an aid to clear logical thought. It should be used whenever radiation resistance is compared with conductor resistance, inductance, capacitance, etc., of elongated wires and loading coils. As an example, it so happens that the distributed, end-to-end, radiation resistance, Rd, of a half-wave diopole is exactly twice (easily proved) the feedpoint resistance of around 72 ohms. The radiating efficiency of a half-wave dipole is then, very simply and accurately - Efficiency = 144 / ( 144 + overall HF conductor resistance ) although some old-wives may wish to argue about it on the grounds that it's far too simple. ---- Reg, G4FGQ |
On Thu, 04 Nov 2004 20:38:35 -0700, Wes Stewart
wrote: Or, model a short lossless monopole over perfect ground and determine the feedpoint R. In this case, R is totally due to radiation loss, i.e. "radiation resistance." Add a lossless loading inductance somewhere in the middle and see what happens to R. Hi Wes, The difference between the two (perfect/real) insofar as Z is hardly remarkable. First I will start with a conventionally sized quarterwave and by iteration approach the short antenna and observe effects. I am using the model VERT1.EZ that is in the EZNEC distribution and modifying it by turns. For instance, I immediately turn on the wire loss. 40mm thick radiator 10.3 meters tall: Impedance = 36.68 + J 2.999 ohms which lends every appearance to expectation of Rr that could be expected from a lossless perfect grounded world. Best gain is -0.03dBi next iteration: cut that sucker in half: Impedance = 6.867 - J 301 ohms which, again, conforms to most authorities on the basis of Rr. best gain 0.16dBi How about that! More gain than for the quarterwave (but hardly remarkable). This makes me wonder why any futzing is required except for the tender requirements of the SWR fearing transmitter (which, by the way, could be as easily taken care of with a tuner). next iteration: load that sucker for grins and giggles: load = 605 Ohms Xl up 55% Impedance = 13.43 + J 0.1587 ohms Did I double Rr? (Only my hairdresser knows.) best gain 0.13dBi Hmm, losing ground for our effort, it makes a pretty picture of current distribution that conforms to all the descriptions here (sans the balderdash of curve fitting to a sine wave). I am sure someone will rescue this situation from my ineptitude by a better load placement, so I will leave that unfinished work to the adept practitioners. next iteration: cut that sucker down half again (and remove the load): Impedance = 1.59 - J 624.6 ohms Something tells me that this isn't off the scale of the perfect comparison. best gain: 0.25dBi Hmm, the trend seems to go counter to intuition. next iteration: -sigh- what charms could loading bring us? load = 1220 Ohms Xl up 55% Impedance = 3.791 + J 1.232 ohms more than doubled the Rr? best gain: 0.23dBi Now, all of this is for a source that is a constant current generator; we've monkeyed with the current distribution and put more resistance (Rr?) into the equation with loading; and each time loading craps in the punch bowl. So much for theories of Rr being modified by loading. I would appreciate other effort in kind to correct any oversights I've made (not just the usual palaver of tedious "explanations" - especially those sophmoric studies of current-in/current-out). 73's Richard Clark, KB7QHC |
Roy Lewallen wrote:
According to the many references I have, the equation you quote is a simplified equation that's valid for a single wire over a perfect conducting ground plane, where the height is a very small fraction of a wavelength (i.e., radiation is negligible). Even if everything you say is true, it doesn't nullify the concepts of physics. Even if the Z0 is changing point to point along the length of the wire, as it surely does for a vertical antenna, the idea that standing wave antennas don't possess standing waves is ridiculous. The idea that the net antenna current on a standing wave antenna is not itself a standing wave is ridiculous. The idea that the net antenna current on a standing wave is not the result of the superposition of the forward current and reflected current is ridiculous. The idea that RF waves can stand still is ridiculous. The idea that current flows in only one direction in a standing wave antenna is ridiculous. A Rhombic, for instance, is a traveling wave antenna. Its feedpoint impedance is equivalent to its characteristic impedance which is hundreds of ohms. In a standing wave antenna, it is the reflected wave superposing with the forward wave at the feedpoint that determines the feedpoint impedance, low for 1/2WL center-fed dipoles and high for 1WL center-fed dipoles. The feedpoint impedance depends upon interference between the forward wave and the reflected wave. All you have proven is that this is a difficult subject to quantitize, but we already knew that. It is not a difficult subject to conceptualize. So, Roy, please answer the following true/false questions. Standing wave antennas actually exhibit standing waves as described by Kraus, Balanis, and others. ______ Standing waves are created by the superposition of forward waves and reflected waves. ________ RF waves cannot stand still. _______ If true, it follows that "standing" waves are an artifact of superposition and cannot exist without the two underlying component waves. What is moving at the speed of light is the forward wave and the reflected wave. ________ Hint: RF waves must move at the speed of light. Therefore, RF standing waves have two components, each moving at the speed of light in opposite directions. For what it's worth, here's a quote from The ARRL Antenna Book, 15th edition, page 24-22 under "Single Wire Line": "The characteristic impedance of the single wire line depends on the conductor size and the height of the wire above ground, ranging from 500 to 600 ohms for #12 or #14 conductors at heights of 10 to 30 ft." Nothing said about "perfect ground" or "small fractions of wavelength". Again, the concepts that I discuss fall out perfectly from the laws of physics. That they are difficult to quantify is not a good reason to adopt a closed mind. (This reminds me of my Southern Baptist upbringing where some subjects were forbidden). -- 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 - Sorry, I take issue with this. The radiation resistance, as universally used in the professional literature, *is* a distributed radiation resistance -- ==================================== Yes, I know. Didn't I just say so? It's just that nobody ever refers to it as such. It would avoid a lot of misunderstandings and arguments if they did! ---- Reg. |
Roy Lewallen, W7EL wrote:
"Sorry. I take issue with this. The radiation resistance as universally used in the professional literature "is" a distributed radiation resistance---." One definition, but not universal by any means. Terman is as professional as needed for most purposes. Terman defines radiation resistance and the custom for stating it on page 891 of his 1955 edition: "Unless specifically stated to the contrary, it is customary to refer the radiation resistance to a current maximum in the case of an ungrounded antenna, and to the current at the base of the antenna when the antenna is grounded." Terman`s definition is unequivocal and useful. It is echoed by other authors. Kraus says on page 182 of his final 3rd edition: "Ro = 60 times the intergral, zero to pi, of the square of Cos [(beta L/2) cos theta] - cos(beta L/2) / sin theta, d theta Where the radiation resistance Ro is referred to the current maximum. In this case of a 1/2-wave antenna, this is at the center of the antenna or at the terminals of the transmission line (see Fig. 6-7)." The solution for a thin dipole yields 73 ohms. Of course, the impedance of a standing-wave antenna varies continuously along its length due to interaction of waves traveling in opposite directions. For practical purposes, we can define Ro as the resistive part and transform its value to the antenna input terminals if these don`t correspond to a current maximum. Best regards, Richard Harrison, KB5WZI |
On Fri, 05 Nov 2004 06:29:58 -0600, Cecil Moore
wrote: the idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. That all such ideas were expressed by only you ... is typical. |
Richard Harrison wrote:
"Ro = 60 times the intergral, zero to pi, ... Heh, heh, since you don't know the last digit of pi, Roy probably won't allow you to use it. :-) Where the radiation resistance Ro is referred to the current maximum. The point seems to be that the impedance at the current maximum point includes terms besides radiation resistance. In simplified form, for a resonant antenna, Rfeed = Rrad + Rloss Rloss includes I^2*R losses and ground losses and is sometimes negligible and sometimes not. For some antenna configurations, Rloss is negligible, so the feedpoint resistance can be very close to the radiation resistance, e.g. a dipole in free space. For other antenna configurations, Rloss is much greater than the radiation resistance, e.g. an 8 foot center-loaded 75m mobile antenna. My screwdriver has approximately a 12.5 ohm feedpoint resistance on 75m. I consider approximately 10 ohms of that to be ground loss. -- 73, Cecil, W5DXP |
On Fri, 05 Nov 2004 06:29:58 -0600, Cecil Moore
wrote: the idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. The idea that ... is ridiculous. On Fri, 05 Nov 2004 11:46:32 -0600, Cecil Moore wrote: Kraus and Balanis also express the same ideas :-) |
Richard Clark wrote:
That all such ideas were expressed by only you ... is typical. Kraus and Balanis also express the same ideas which are obvious from the underlying EM wave physics. Although I don't have Terman's book, apparently so does he. -- 73, Cecil, W5DXP |
As another example, the resonant Q of a half-wave dipole is -
Q = Omega * Ld / Rd or The reactance of its DISTRIBUTED inductance divided by its DISTRIBUTED radiation resistance. Perfectionists may DIRECTLY add conductor resistance to the radiation resistance because that, too, is a DISTRIBUTED quantity. --- Reg. |
On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark
wrote: So much for theories of Rr being modified by loading. I would appreciate other effort in kind to correct any oversights I've made Hi All, Nothing to offer? I didn't think so. :-) Well, to enlarge the dialogue (but still reject those who prefer to change the topic to what they CAN prove), any -ahem- "explanations?" 73's Richard Clark, KB7QHC |
I'm chugging Franzia Merlot while cleaning my Colt Python .357 Magnum
and Winchester 30-30 Lever-Action Carbine. :-) -- 73, Cecil ============================ After shooting a rattler do you, Hollywood-fashion, put the weapon to your lips and blow the smoke out of the barrel? ;o) ---- Reg. |
Richard Clark wrote: On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark wrote: So much for theories of Rr being modified by loading. I would appreciate other effort in kind to correct any oversights I've made Hi All, Nothing to offer? I didn't think so. :-) I have a question. Can you express the mathematical and/or physical relationship between Rr and antenna gain? It would sure help to clarify the point you were trying to make. Thanks and 73, AC6XG |
Reg Edwards wrote:
W5DXP wrote: I'm chugging Franzia Merlot while cleaning my Colt Python .357 Magnum and Winchester 30-30 Lever-Action Carbine. :-) After shooting a rattler do you, Hollywood-fashion, put the weapon to your lips and blow the smoke out of the barrel? ;o) I'm not that primitive, Reg. I use commercial ammo with smokeless powder. But a friend of mine indeed does use black powder with his cap and ball revolver. -- 73, Cecil, W5DXP |
Richard,
It is not clear just what you were trying to demonstrate, but there was no obvious connection to Rr. Maximum gain is unrelated to Rr. Were you looking for validation that you correctly loaded the numbers into EZNEC? Looked OK to me. (OK, so my trolling is not as professional as yours, but I tried.) If you really are concerned about Rr, then consider the following. Rr is defined by the equation: Total power radiated = a*I*I*Rr, where I and Rr are referred to the same point on the antenna. For short antennas this reference point is generally taken as the feedpoint, which is also the current maximum point. There is no loss, so all power into the antenna is radiated. Therefore the Rr for each of your perfect world, zero loss examples is proportional to the feedpoint resistance. Apply an appropriate scaling factor (a) if you want the correct numbers. Did you have a question about something? The bottom line, of course, is that none of this matters in a perfect world with zero loss. All the power input is shot into space. Please let us know where we can find that perfect world :-) 73, Gene W4SZ Richard Clark wrote: On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark wrote: So much for theories of Rr being modified by loading. I would appreciate other effort in kind to correct any oversights I've made Hi All, Nothing to offer? I didn't think so. :-) Well, to enlarge the dialogue (but still reject those who prefer to change the topic to what they CAN prove), any -ahem- "explanations?" 73's Richard Clark, KB7QHC |
On Fri, 05 Nov 2004 19:15:01 GMT, Gene Fuller
wrote: Richard, It is not clear just what you were trying to demonstrate, but there was no obvious connection to Rr. Maximum gain is unrelated to Rr. Hi Gene, The tenor of correspondence here would suggest otherwise. The evidence of testing would suggest there is. The disconnect is that the evidence counters the suggestions in correspondence. Were you looking for validation that you correctly loaded the numbers into EZNEC? Looked OK to me. Thanx, but now we may BOTH be wrong. ;-) For short antennas this reference point is generally taken as the feedpoint, which is also the current maximum point. There is no loss I did explicitly state that wire loss had been turned on, and real ground was used. , so all power into the antenna is radiated. Therefore the Rr for each of your perfect world Again, there was no "perfect world." , zero loss examples is proportional to the feedpoint resistance. Apply an appropriate scaling factor (a) if you want the correct numbers. Wholly unnecessary, EZNEC can cope with loss quite well and demonstrates a real world solution (barring my errors of commission or omission) within tolerable limits. Did you have a question about something? I would appreciate other effort in kind to correct any oversights I've made The bottom line, of course, is that none of this matters in a perfect world with zero loss. All the power input is shot into space. Please let us know where we can find that perfect world :-) You are right about your quality of trolling. A "perfect world?" You couldn't supply one? ;-) 73's Richard Clark, KB7QHC |
On Fri, 05 Nov 2004 10:45:57 -0800, Jim Kelley
wrote: I have a question. Can you express the mathematical and/or physical relationship between Rr and antenna gain? It would sure help to clarify the point you were trying to make. Hi Jim, I would have thought someone else could, given the bandwidth of discussion in making the current taper shorter and the constant current section longer. Testing does not bear their facile relationship out however, and for the topic of a short antenna (otherwise, why are we talking about loading coils?) it would seem that antenna gain is immutable over several octaves below a quarterwave length. Of course, I coulda done something wrong. I did use a commonly available design. I did use a commonly available modeler. I even may have done the wrong thing in choosing a design that could be evaluated for free. Perhaps I erred in providing the cogent details of construction. It took all of 20 minutes to accomplish (far less time than that expended in theories of current-in/current-out). These technical hurdles appear to have set the bar too high for my work's refutation in kind. I appreciate that "it's hard work!" ;-) To answer your question, if you just abandon the perfect load, then you stand to achieve a higher gain. If you shorten the antenna, then you stand to achieve a higher gain. There is no change in Rr with the addition of Xl. Hence the mathematical relationship for an antenna shorter than quarterwave would be suggested as: gain ~ 1/Rr gain ~ 1/Xl Rr Z Don't take this gain to the bank however. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Cecil Moore wrote: Kraus and Balanis also express the same ideas :-) Apparently, you are not reading the quotes: From "Antennas For All Applications", by Kraus and Marhefka, 3rd edition: From page 187 regarding standing wave ANTENNAS: "A sinusoidal current distribution may be regarded as the standing wave produced by two uniform (unattenuated) traveling waves of equal amplitude moving in opposite directions along the antenna." From page 465 regarding a dipole ANTENNA longer than 1/2WL: "When the antenna (dipole) is infinitesimally thin, the phase varies as a step function, being constant over 1/2WL and changing by 180 deg. at the end of the 1/2WL interval. This type of phase variation is observed in a PURE STANDING WAVE." emphasis mine. **************** From _Antenna_Theory_, Balanis, Second Edition, Chapter 10, page 488 & 489 "The current and voltage distributions on open-ended wire antennas are similar to the standing wave patterns on open-ended transmission lines ... Standing wave antennas, such as the dipole, can be analyzed as traveling wave antennas with waves propagating in opposite directions (forward and backward) and represented by traveling wave currents If and Ib ..." I got my ideas from Kraus and Balanis, and in particular from Balanis' antenna courses at ASU where he found no fault with my ideas. Do you agree or disagree with Kraus and Balanis that standing wave antennas exhibit standing waves? If they do, they follow the well known laws of physics governing standing waves. The combined effort to suppers the ideas of Kraus and Balanis is getting more ridiculous with each passing day. It can no longer be explained by ignorance. -- 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 =--- |
Cecil Moore wrote:
The combined effort to suppers the ideas of Kraus and Balanis is getting Darn spellchecker. "suppers" should be "suppress". more ridiculous with each passing day. It can no longer be explained by ignorance. -- 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 =--- |
Gene Fuller wrote:
The bottom line, of course, is that none of this matters in a perfect world with zero loss. All the power input is shot into space. Please let us know where we can find that perfect world :-) EZNEC? - lossless antennas and ground planes. -- 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 =--- |
On Fri, 05 Nov 2004 13:57:16 -0600, Cecil Moore
wrote: The combined effort to suppers the ideas of Kraus and Balanis is getting Darn spellchecker. "suppers" should be "suppress". more ridiculous with each passing day. It can no longer be explained by ignorance. Blame the spellchecker? |
Richard Clark wrote:
First I will start with a conventionally sized quarterwave and by iteration approach the short antenna and observe effects. I am using the model VERT1.EZ that is in the EZNEC distribution and modifying it by turns. For instance, I immediately turn on the wire loss. 40mm thick radiator 10.3 meters tall: Impedance = 36.68 + J 2.999 ohms which lends every appearance to expectation of Rr that could be expected from a lossless perfect grounded world. Best gain is -0.03dBi next iteration: cut that sucker in half: Impedance = 6.867 - J 301 ohms which, again, conforms to most authorities on the basis of Rr. best gain 0.16dBi How about that! More gain than for the quarterwave (but hardly remarkable). This is indeed an interesting result, even though it's small. As a dipole or monopole gets shorter than a resonant length, the current distribution changes from sinusoidal to triangular. In the ideal case (and in the absence of loss), this results in the shorter antenna having a slightly fatter pattern and therefore slightly less gain. An infinitesimally short antenna has a gain of less than 0.5 below that of a resonant one, due to this (and again, in the absence of loss). So the trend you observed was backwards. The first two things I checked -- segmentation and wire loss -- proved to not be the reason. It appears to be the effect of the current distribution on the reflected wave from the ground. The radiation from each part of the antenna reflects from the ground, and the nature of the reflection depends on the angle at which it strikes the ground. Radiation from different parts of the antenna strike at different angles, and the shorter antenna has its current, therefore its radiation, apportioned differently. It's this difference that causes the slight backward trend in gain. I wouldn't put too much weight on it, however, both because of its being small, and that EZNEC and similar programs use a pretty simple ray-tracing method of calculating ground reflection rather than a more complex method which includes diffraction and other effects. You'll see the theoretical trend if you change the ground type to Perfect, and likely a slightly different trend if you change the ground constants. This makes me wonder why any futzing is required except for the tender requirements of the SWR fearing transmitter (which, by the way, could be as easily taken care of with a tuner). next iteration: load that sucker for grins and giggles: load = 605 Ohms Xl up 55% Impedance = 13.43 + J 0.1587 ohms Did I double Rr? (Only my hairdresser knows.) best gain 0.13dBi Hmm, losing ground for our effort, it makes a pretty picture of current distribution that conforms to all the descriptions here (sans the balderdash of curve fitting to a sine wave). I am sure someone will rescue this situation from my ineptitude by a better load placement, so I will leave that unfinished work to the adept practitioners. As I pointed out above, going from a sinusoidal to a triangular distribution changes the gain less than 0.5 dB, so this change in distribution changes the gain even less. And again, the interaction with real ground reverses the trend. With Perfect ground, you'll see that there's actually a 0.02 dB gain improvement with the new distribution. This is, of course, insignificant, but it does show that things are working as they should. next iteration: cut that sucker down half again (and remove the load): Impedance = 1.59 - J 624.6 ohms Something tells me that this isn't off the scale of the perfect comparison. best gain: 0.25dBi Hmm, the trend seems to go counter to intuition. That's because you didn't understand the reason for the trend in the first place. What's happening now is that the gain reduction caused by the changed distribution is no longer overwhelmed by the gain increase caused by the interaction with real ground. Again, go to Perfect ground and you'll see that the trend of less gain as the antenna shortens is continuing as exepected. next iteration: -sigh- what charms could loading bring us? load = 1220 Ohms Xl up 55% Impedance = 3.791 + J 1.232 ohms more than doubled the Rr? best gain: 0.23dBi And with Perfect ground, the gain is the same within 0.01 dB with and without the load -- the modified current distribution wasn't enough to make any significant difference in the pattern and hence the gain. Now, all of this is for a source that is a constant current generator; we've monkeyed with the current distribution and put more resistance (Rr?) into the equation with loading; and each time loading craps in the punch bowl. So much for theories of Rr being modified by loading. Everything you've done shows that Rr is modified by loading. Rr at the feedpoint is simply the resistance at the feedpoint. And it clearly changes when you insert the load. What makes you think that the Rr isn't changing? I would appreciate other effort in kind to correct any oversights I've made (not just the usual palaver of tedious "explanations" - especially those sophmoric studies of current-in/current-out). Perhaps you're expecting the gain to vary in the same way as Rr. But you see, gain will change with Rr only if the pattern shape stays the same. And each time you insert the load, the current distribution changes, which changes the shape of the pattern, which changes the gain. This is the gain change you're seeing as you change the antenna length and add loading. The main relationship between Rr and gain is the efficiency. The antennas you've modeled are so close to being 100% efficient that increasing Rr has no appreciable effect on gain. Check it yourself -- make the antenna 100% efficient by removing the wire loss and notice that there's no change in gain (maybe 0.01 dB with the full size antenna). But now repeat your experiments with, say, a 10 ohm resistive load at the base to simulate ground system loss. And you'll see that the gain improves dramatically as your loading increases Rr (the feedpoint resistance). This is due to the improved efficiency you gain by increasing Rr. Roy Lewallen, W7EL |
Richard Clark wrote: On Fri, 05 Nov 2004 10:45:57 -0800, Jim Kelley wrote: I have a question. Can you express the mathematical and/or physical relationship between Rr and antenna gain? It would sure help to clarify the point you were trying to make. Hi Jim, I would have thought someone else could, given the bandwidth of discussion in making the current taper shorter and the constant current section longer. Testing does not bear their facile relationship out however, and for the topic of a short antenna (otherwise, why are we talking about loading coils?) it would seem that antenna gain is immutable over several octaves below a quarterwave length. Of course, I coulda done something wrong. I did use a commonly available design. I did use a commonly available modeler. I even may have done the wrong thing in choosing a design that could be evaluated for free. Perhaps I erred in providing the cogent details of construction. It took all of 20 minutes to accomplish (far less time than that expended in theories of current-in/current-out). These technical hurdles appear to have set the bar too high for my work's refutation in kind. I appreciate that "it's hard work!" ;-) To answer your question, if you just abandon the perfect load, then you stand to achieve a higher gain. If you shorten the antenna, then you stand to achieve a higher gain. There is no change in Rr with the addition of Xl. Hence the mathematical relationship for an antenna shorter than quarterwave would be suggested as: gain ~ 1/Rr gain ~ 1/Xl Rr Z Don't take this gain to the bank however. Point being that antenna gain has spatial implications which Rr by itself could not provide in the solutions. One should conclude from your results (very nice work, by the way) that the modeler apparently doesn't figure Xl as contributing to the radiation resistance. But isn't that basically the crux of the argument here - whether or not that is in fact the case? If not, then we would have to conclude that loading coil has zero physical and electrical length, and that the impedance is constant from one end to the other. Obviously I could be wrong, but I think those are the assumptions made in your modeling software. On the other hand, if the impedance of the coil is not constant from one end to the other, and it in fact does have some real physical and/or electrical length, then I think the radiation resistance of the antenna would have to be effected by its presence in the circuit. That is, if indeed Ro is the integral of disributed r along the entire physical and/or electrical length of the antenna (credit for that formula going to an esteemed contributor to this newsgroup earlier today). Or perhaps more concisely put, if the loading coil itself contributes to the field radiating from the antenna, then it should likewise have a Rr associated with it. The converse would of course still be true. 73, Jim AC6XG |
On Fri, 05 Nov 2004 13:06:40 -0800, Roy Lewallen
wrote: What makes you think that the Rr isn't changing? Hi Roy, With what? I have offered a very ascetic report of very simple actions from very simple terms. You have chosen to change those terms to fit your own answer (I do not choose to put this into the context of a perfect world). Please offer effort in kind, or point out what error I've made instead of what error I might have made if something were changed from my model. 73's Richard Clark, KB7QHC |
On Fri, 05 Nov 2004 13:37:27 -0800, Jim Kelley
wrote: Point being that antenna gain has spatial implications which Rr by itself could not provide in the solutions. Hi Jim, Not sure where this is going, so I will stand with my own statements. One should conclude from your results (very nice work, by the way) that the modeler apparently doesn't figure Xl as contributing to the radiation resistance. Again, words. I would offer that Xl does not adjust Rr (insofar as it does adjust drivepoint Z). Let me add - much (there are miniscule differences). But isn't that basically the crux of the argument here - whether or not that is in fact the case? If not, then we would have to conclude that loading coil has zero physical and electrical length, and that the impedance is constant from one end to the other. As I say above, it is not a difference of ZERO it is a miniscule difference (with a negative correlation). I will leave it to others to recapture (or diminish) gain through helix building. I am satisfied those returns will be equally diminutive. Obviously I could be wrong, but I think those are the assumptions made in your modeling software. We both could be wrong. No one has yet to step up to the bar and offer work in kind. On the other hand, if the impedance of the coil is not constant from one end to the other, and it in fact does have some real physical and/or electrical length, then I think the radiation resistance of the antenna would have to be effected by its presence in the circuit. That is, if indeed Ro is the integral of disributed r along the entire physical and/or electrical length of the antenna (credit for that formula going to an esteemed contributor to this newsgroup earlier today). Or perhaps more concisely put, if the loading coil itself contributes to the field radiating from the antenna, then it should likewise have a Rr associated with it. The converse would of course still be true. This returns us to matters of degree. By simple observation comparing the standard full sized radiator to ANY of the iterations, it is obvious that nothing significant can be said of the physicality of the load, much less its contribution. In other words, is there some magical coil or magical placement that would create a short super-radiator that exceeds the performance of the standard quarterwave? [no here can recognize a eh/cfa claim?] Let's put a handicap on this. Is there some magic combination of coil/position that would even EQUAL the performance of the standard quarterwave? 73's Richard Clark, KB7QHC |
It might be helpful to elaborate a bit more about radiation resistance.
Consider an antenna that has no loss. If we apply 100 watts, say, to this antenna, the radiation resistance "consumes" that 100 watts. That is to say, all 100 watts is radiated. In the case of a resonant lossless quarter wave vertical, for example, the current at the base will be about 1.67 amps if 100 watts is being radiated. We can solve for Rr at the base from P = I^2 * Rr, where I is the base current, to get Rr = P / I^2, with the result that Rr is about 36 ohms. This is the radiation resistance referred to the base -- it "consumes" the 100 watts. (We could have calculated Rr at some other point along the antenna, where I is different, and gotten a different value. But P still has to equal I^2 * Rr, where I is the current at the point to which Rr is being referred.) Now, what determines the current we get at the base, for a given applied power and radiator length? The answer is the current distribution -- that is, the way the current varies along the length of the conductor. (I'm only considering a simple single wire antenna here. When other conductors are involved, mutual coupling between conductors also plays a role.) Putting a loading coil at the bottom of the antenna doesn't change the current distribution, it only changes the feedpoint reactance. So it doesn't change the feedpoint current for a given power input, so the radiation resistance doesn't change. But if you put a loading coil part way up the antenna, the current distribution does change. This alters the base current for the same power input and therefore the radiation resistance changes. Remember that Rr = P / I^2, where Rr and I are measured at the same point (in this case the base feedpoint), so if I changes, Rr changes. Likewise, top loading alters the current distribution and consequently the radiation resistance. For people who would like to see this graphically, the demo version of EZNEC is adequate. Just look at the View Antenna display after running a pattern, source data, or current calculation, and you'll see how the current varies along the antenna. If you set a fixed power level in the Options menu (Power Level selection), you can also see, by clicking Src Dat, exactly how the current at the source changes as the current distribution does. If you have a fixed amount of loss, say at the base of the antenna due to ground system loss, the amount of power dissipated in that loss as heat is Ploss = I^2 * Rloss, where I is the current flowing through that loss, in this case the current at the antenna base. So for a given amount of applied power, you minimize the power lost when you minimize the base current. This is exactly equivalent to saying you're raising the radiation resistance referred to the base. That's why mobile antenna users consider higher radiation resistance a virtue -- it means lower feedpoint current for a given power input, and therefore less power lost in the necessarily imperfect ground system. While the principles are all the same, wire loss has to be treated a bit differently because altering the current distribution changes the amount of wire loss (which is usually combined into a single loss resistance referred to the feedpoint, or the same point where the radiation resistance is referred). Also, changing the wire length alters the wire loss, as does the total current in the wire which increases as the wire is shortened for a given power input. All these can be dealt with analytically or with a modeling program, but it's easy to lose track of exactly what's happening when all these factors are present at the same time. Fortunately, wire loss is insignificant for the vast majority of typical amateur applications. With modeling, it's easy to determine when it is and isn't significant, simply by turning wire loss on and off and observing how much the results change. Roy Lewallen, W7EL |
I'm sorry, but I can't make any sense out of your response, so I'm
unable to answer your question. Roy Lewallen, W7EL Richard Clark wrote: On Fri, 05 Nov 2004 13:06:40 -0800, Roy Lewallen wrote: What makes you think that the Rr isn't changing? Hi Roy, With what? I have offered a very ascetic report of very simple actions from very simple terms. You have chosen to change those terms to fit your own answer (I do not choose to put this into the context of a perfect world). Please offer effort in kind, or point out what error I've made instead of what error I might have made if something were changed from my model. 73's Richard Clark, KB7QHC |
On Fri, 05 Nov 2004 15:20:26 -0800, Roy Lewallen
wrote: I'm sorry, but I can't make any sense out of your response, so I'm unable to answer your question. What makes you think that the Rr isn't changing? With what? |
Loading.
You wrote: So much for theories of Rr being modified by loading. Being a simple-minded, non-liberally educated person, I took this to mean that you'd shown that Rr isn't modified by loading, disproving theories that Rr is modified by loading. (Otherwise, I figured, you would have said "So much for theories of Rr not being modified by loading.") Did you mean instead that there are theories that Rr isn't modified by loading? If so, who on earth is proposing those theories? As a basic understanding of Rr shows, and as your modeling results also show, there's no basis for such a "theory". What's next, a "theory" that V = I/R? Roy Lewallen, W7EL Richard Clark wrote: On Fri, 05 Nov 2004 15:20:26 -0800, Roy Lewallen wrote: I'm sorry, but I can't make any sense out of your response, so I'm unable to answer your question. What makes you think that the Rr isn't changing? With what? |
Richard Clark wrote:
On Fri, 05 Nov 2004 13:37:27 -0800, Jim Kelley wrote: Point being that antenna gain has spatial implications which Rr by itself could not provide in the solutions. Hi Jim, Not sure where this is going, so I will stand with my own statements. I wouldn't have it any other way, Richard. The examination of the underlying theory taking place in this discussion resolves in different ways. Any other 'work in kind' would depend largely on which way one resolves the theory. Horse first, then cart. Reasonable questions have been posed here. Some of the points being raised remain to an extent, unreconciled with certain aspects of conventional treatments. Obviously, thanks in no small part to empirical methods, this poses no greater hinderance to the design of efficient antennas than does the lack of a precise understanding of the nature of turbulent flow to the design of efficient fluid pumps. I find any contention that no further investigation is necessary to be less than convincing. On the contrary, the work done by the oft cited E&M engineers clearly indicates there is more to the story - a point that said engineers would undoubtedly concede without hesitation. On the other hand, if the impedance of the coil is not constant from one end to the other, and it in fact does have some real physical and/or electrical length, then I think the radiation resistance of the antenna would have to be effected by its presence in the circuit. That is, if indeed Ro is the integral of disributed r along the entire physical and/or electrical length of the antenna (credit for that formula going to an esteemed contributor to this newsgroup earlier today). Or perhaps more concisely put, if the loading coil itself contributes to the field radiating from the antenna, then it should likewise have a Rr associated with it. The converse would of course still be true. This returns us to matters of degree. By simple observation comparing the standard full sized radiator to ANY of the iterations, it is obvious that nothing significant can be said of the physicality of the load, much less its contribution. In other words, is there some magical coil or magical placement that would create a short super-radiator that exceeds the performance of the standard quarterwave? [no here can recognize a eh/cfa claim?] Let's put a handicap on this. Is there some magic combination of coil/position that would even EQUAL the performance of the standard quarterwave? Absent a thorough understanding of the phenomena, one might be inclined to conjecture about magical things. But I'm inclined toward taking satisfaction from persuing a thorough understanding of the phenomena rather than from strutting the pretense of already having one. 73, Jim AC6XG |
Concerning current distribution -- at least on a monopole above a perfectly
conducting ground. I have noticed that in any antenna, of any length, inductively loaded, or not, the: Integral of I(z)dz (where the units of dz are in fractions of a wavelength, and I(z) the current distribution) is virtually a constant. Assuming the same input power in all cases. Suppose it seems pretty obvious since the total radiated power from any structure is also essentially constant. I must admit I have not seen gain increasing, with decreasing size, but then I have not seriously tried. Regards, Frank "Roy Lewallen" wrote in message ... It might be helpful to elaborate a bit more about radiation resistance. Consider an antenna that has no loss. If we apply 100 watts, say, to this antenna, the radiation resistance "consumes" that 100 watts. That is to say, all 100 watts is radiated. In the case of a resonant lossless quarter wave vertical, for example, the current at the base will be about 1.67 amps if 100 watts is being radiated. We can solve for Rr at the base from P = I^2 * Rr, where I is the base current, to get Rr = P / I^2, with the result that Rr is about 36 ohms. This is the radiation resistance referred to the base -- it "consumes" the 100 watts. (We could have calculated Rr at some other point along the antenna, where I is different, and gotten a different value. But P still has to equal I^2 * Rr, where I is the current at the point to which Rr is being referred.) Now, what determines the current we get at the base, for a given applied power and radiator length? The answer is the current distribution -- that is, the way the current varies along the length of the conductor. (I'm only considering a simple single wire antenna here. When other conductors are involved, mutual coupling between conductors also plays a role.) Putting a loading coil at the bottom of the antenna doesn't change the current distribution, it only changes the feedpoint reactance. So it doesn't change the feedpoint current for a given power input, so the radiation resistance doesn't change. But if you put a loading coil part way up the antenna, the current distribution does change. This alters the base current for the same power input and therefore the radiation resistance changes. Remember that Rr = P / I^2, where Rr and I are measured at the same point (in this case the base feedpoint), so if I changes, Rr changes. Likewise, top loading alters the current distribution and consequently the radiation resistance. For people who would like to see this graphically, the demo version of EZNEC is adequate. Just look at the View Antenna display after running a pattern, source data, or current calculation, and you'll see how the current varies along the antenna. If you set a fixed power level in the Options menu (Power Level selection), you can also see, by clicking Src Dat, exactly how the current at the source changes as the current distribution does. If you have a fixed amount of loss, say at the base of the antenna due to ground system loss, the amount of power dissipated in that loss as heat is Ploss = I^2 * Rloss, where I is the current flowing through that loss, in this case the current at the antenna base. So for a given amount of applied power, you minimize the power lost when you minimize the base current. This is exactly equivalent to saying you're raising the radiation resistance referred to the base. That's why mobile antenna users consider higher radiation resistance a virtue -- it means lower feedpoint current for a given power input, and therefore less power lost in the necessarily imperfect ground system. While the principles are all the same, wire loss has to be treated a bit differently because altering the current distribution changes the amount of wire loss (which is usually combined into a single loss resistance referred to the feedpoint, or the same point where the radiation resistance is referred). Also, changing the wire length alters the wire loss, as does the total current in the wire which increases as the wire is shortened for a given power input. All these can be dealt with analytically or with a modeling program, but it's easy to lose track of exactly what's happening when all these factors are present at the same time. Fortunately, wire loss is insignificant for the vast majority of typical amateur applications. With modeling, it's easy to determine when it is and isn't significant, simply by turning wire loss on and off and observing how much the results change. Roy Lewallen, W7EL |
On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark
wrote: Good day Richard, Sorry this isn't too timely, but I've been off visiting a dear friend and soon to be silent key. |On Thu, 04 Nov 2004 20:38:35 -0700, Wes Stewart |wrote: | |Or, model a short lossless monopole over perfect ground and determine |the feedpoint R. In this case, R is totally due to radiation loss, |i.e. "radiation resistance." Add a lossless loading inductance |somewhere in the middle and see what happens to R. | |Hi Wes, | |The difference between the two (perfect/real) insofar as Z is hardly |remarkable. True. | |First I will start with a conventionally sized quarterwave and by |iteration approach the short antenna and observe effects. I am using |the model VERT1.EZ that is in the EZNEC distribution and modifying it |by turns. For instance, I immediately turn on the wire loss. | |40mm thick radiator 10.3 meters tall: | Impedance = 36.68 + J 2.999 ohms |which lends every appearance to expectation of Rr that could be |expected from a lossless perfect grounded world. |Best gain is | -0.03dBi | |next iteration: | |cut that sucker in half: | Impedance = 6.867 - J 301 ohms |which, again, conforms to most authorities on the basis of Rr. |best gain | 0.16dBi |How about that! More gain than for the quarterwave (but hardly |remarkable). This makes me wonder why any futzing is required except |for the tender requirements of the SWR fearing transmitter (which, by |the way, could be as easily taken care of with a tuner). | |next iteration: | |load that sucker for grins and giggles: | load = 605 Ohms Xl up 55% | Impedance = 13.43 + J 0.1587 ohms |Did I double Rr? (Only my hairdresser knows.) |best gain | 0.13dBi |Hmm, losing ground for our effort, it makes a pretty picture of |current distribution that conforms to all the descriptions here (sans |the balderdash of curve fitting to a sine wave). I am sure someone |will rescue this situation from my ineptitude by a better load |placement, so I will leave that unfinished work to the adept |practitioners. | |next iteration: | |cut that sucker down half again (and remove the load): | Impedance = 1.59 - J 624.6 ohms |Something tells me that this isn't off the scale of the perfect |comparison. |best gain: | 0.25dBi |Hmm, the trend seems to go counter to intuition. | |next iteration: | |-sigh- what charms could loading bring us? | load = 1220 Ohms Xl up 55% | Impedance = 3.791 + J 1.232 ohms |more than doubled the Rr? |best gain: | 0.23dBi | |Now, all of this is for a source that is a constant current generator; |we've monkeyed with the current distribution and put more resistance |(Rr?) into the equation with loading; and each time loading craps in |the punch bowl. | |So much for theories of Rr being modified by loading. I would |appreciate other effort in kind to correct any oversights I've made |(not just the usual palaver of tedious "explanations" - especially |those sophmoric studies of current-in/current-out). Sure. Your calculations are impeccable so far (although it would be better to use more than 10 segments IMHO). I believe it is your contention that loading to resonance with an arbitrarily positioned inductor, or not loading at all, does not affect the gain, and the radiation resistance is not the same as the changing feedpoint resistance. I am in the other camp, along with Hansen, Devoldere, et. al. who say that the current distribution does affect the radiation resistance (and in the real world, the gain/efficiency). I hope you would agree that the normalized gain would be a good proxy for efficiency. For example if we use the lossless 1/4 wavelength monopole over perfect ground as a reference, then gain with respect to that (5.15 dBi) would be an indicator of efficiency. I believe that you will agree that the efficiency can be determined by: Rr eta = ------------- Eq.1 Rr + Rg + Rl where Rr = radiation resistance Rg = ground resistance Rl = all other resistances (conductor, etc) I think you would also agree that for the full-sized monopole over perfect ground the feedpoint resistance of ~36 Ohm = radiation resistance. As an old (sorry [g]) metrologist, you're very familiar with substitution, so let's set Rl = 0 (lossless case) and eta to 0.5 (-3 dB). Per Eq. 1, Rg = Rr. So in our model, if I add a simulated ground resistance, Rg, that reduces the gain by 3 dB, I have by substitution, determined the radiation resistance. Sure enough, if I add a 36 Ohm load at the bottom of the perfect 1/4 wave monopole, the gain drops to 2.14 dBi, and the feedpoint resistance doubles. I will let you try this with the other cases. I trust you will find that the radiation resistance does decrease with shorter radiators and/or lower loading points. I too I would appreciate other effort in kind to correct any oversights I've made. Wes |
On Sat, 06 Nov 2004 01:31:40 GMT, "Frank"
wrote: Assuming the same input power in all cases. Hi Frank, What would you say to "assuming the same current in all cases" and either/both the Rr and drivepoint Z have changed to no particular evidence in the far field? I have been posited with a "theory" that V = I/R? to which I would wryly note that nothing has been said of V (and certainly no interest has been shown), I is known and observations of P (as evident in the far field, which has occurred to me may be the issue - had to say it for myself, didn't I?) for a changing R (somewhere through the iterations). 73's Richard Clark, KB7QHC |
Ah, yes, that's one of those simple observations that lures people into
making false generalizations. I think you'll find that your observation isn't true for any length, but only for lengths of a quarter wavelength (for a monopole) and shorter. And only for an antenna consisting of a single straight wire. Under those conditions, the phase of the current is nearly constant along the wire, so the fields from the various parts of the antenna add together in phase at a distant point broadside to the wire. The maximum field strength is, therefore, proportional to the sum of the fields from the individual segments which, in turn, is proportional to the integral of the currents on the segments. Since the maximum field strength (or gain) doesn't change much from a very short wire to a quarter wavelength one, the integral of the current stays pretty constant. But don't, for heaven's sake, think you've discovered a rule that applies for all antennas. Not even all straight, single wire monopoles. Roy Lewallen, W7EL Frank wrote: Concerning current distribution -- at least on a monopole above a perfectly conducting ground. I have noticed that in any antenna, of any length, inductively loaded, or not, the: Integral of I(z)dz (where the units of dz are in fractions of a wavelength, and I(z) the current distribution) is virtually a constant. Assuming the same input power in all cases. Suppose it seems pretty obvious since the total radiated power from any structure is also essentially constant. I must admit I have not seen gain increasing, with decreasing size, but then I have not seriously tried. Regards, Frank |
Oh, I would never make such an assumption that it applies to all antennas --
well maybe it did cross my mind!. My guess was that probably everybody knew it anyway, but thought it was interesting. I have also noticed that the phase of the current is constant over thin wire monopoles of less than a 1/4 wave Regarding Richard's comments "Assuming the same currents". Not sure I understand, since the feed-point current varies with constant power -- as does the voltage. Also confused by the expression V=I/R. 73, Frank "Roy Lewallen" wrote in message ... Ah, yes, that's one of those simple observations that lures people into making false generalizations. I think you'll find that your observation isn't true for any length, but only for lengths of a quarter wavelength (for a monopole) and shorter. And only for an antenna consisting of a single straight wire. Under those conditions, the phase of the current is nearly constant along the wire, so the fields from the various parts of the antenna add together in phase at a distant point broadside to the wire. The maximum field strength is, therefore, proportional to the sum of the fields from the individual segments which, in turn, is proportional to the integral of the currents on the segments. Since the maximum field strength (or gain) doesn't change much from a very short wire to a quarter wavelength one, the integral of the current stays pretty constant. But don't, for heaven's sake, think you've discovered a rule that applies for all antennas. Not even all straight, single wire monopoles. Roy Lewallen, W7EL Frank wrote: Concerning current distribution -- at least on a monopole above a perfectly conducting ground. I have noticed that in any antenna, of any length, inductively loaded, or not, the: Integral of I(z)dz (where the units of dz are in fractions of a wavelength, and I(z) the current distribution) is virtually a constant. Assuming the same input power in all cases. Suppose it seems pretty obvious since the total radiated power from any structure is also essentially constant. I must admit I have not seen gain increasing, with decreasing size, but then I have not seriously tried. Regards, Frank |
| All times are GMT +1. The time now is 01:26 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com