| Page 1 of 2 | 1 | 2 |
|
|
Measuring Antenna Efficiency
When the subject of antenna efficiency comes up, it often involves a
discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? |
Measuring Antenna Efficiency
On 5 Mar, 08:04, "Wayne" wrote:
When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? In my opinion NO! You have two resistances, one for Radiation and one for wire resistance ie skin resistance However, you can measure the resistance as a loss and then utelise your equation for a dipole that is resonant. If more than one element is being measured then each element must be resonant in situ before you can repeat that particular method. My thought behind this is and I can be in error, that some "radiation" in its formative period may well cancel each other in the near field. Art |
Measuring Antenna Efficiency
There's no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? |
Measuring Antenna Efficiency
On 5 Mar, 09:04, Roy Lewallen wrote:
There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts?- Hide quoted text - - Show quoted text - If you adopt my expansion of gauss it can be calculated. If you look at the example in Chapter 21 of the Rutgers book on fields on the net maybe, but maybe, you will think a bit different but I expect you to fight off the idea of change Art Art |
Measuring Antenna Efficiency
"Wayne" wrote in
news:5cXGh.391$iD4.256@trnddc06: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a The reason for focus on ground loss on (HF) verticals is that, unless you have taken extreme measures with a ground system, ground loss dwarfs other losses and in that case dominates considerations of efficiency. Next would come loading coils. dipole? Could one calculate "power out/power in" by measuring the Half wave dipoles made from practical materials are usually very high efficiency, losses commonly range in the area of 1% to ~3%. Loading coils are a significant loss element in loaded dipoles. Some folk (eg ARRL) have a mind that linear loading (folding the conductors back on themselves) is lossless, but my analysis of the Cobra shows that is not the case, see http://www.vk1od.net/cobra/index.htm . Even other lengths of unloaded dipoles may be very efficient, but the feedpoint impedance may drive huge losses on the feedline and so whilst the radiator is efficient, the antenna system may be inefficient. Components of an antenna system interact with each other in a complex way, and it is important to analyse the entire antenna system (radiator, earth, transmission line, balun, ATU etc) to obtain a correct understanding of how the system works overall. VSWR and declaring that everything not reflected was transmitted? It Roy has already explained to you that you have some misconceptions about VSWR, "forward power", and "reflected power". There has been another raging discussion here about what happens to the "reflected power", it isn't necessarily, and isn't usually lost (ie dissipated as heat), but as I have stated above the feedpoint impedance may drive huge losses on the feedline, it may also reduce the power available from the transmitter and may reduce the transmitter efficiency. would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? In terms of efficiency on the larger scale, a significant of power is lost in the process of reflecting some rays from real ground. Owen |
Measuring Antenna Efficiency
Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? "Roy Lewallen" wrote in message ... There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL Thanks for the reply. My dipole example is intended to avoid transmission line issues by not having one, and the elements are assumed to be reasonably low loss. If I do some quick back-of-the-envelope calculations, for a VSWR of 1.3:1, I get an efficiency of about 98.3% (using the equation 1-gamma^2). Assuming a resistance of 1 ohm in the dipole conductors the efficiency I calculate is about 98.6% (72/73). Are there any other loss issues missing in this example. |
Measuring Antenna Efficiency
"art" wrote in message oups.com... On 5 Mar, 09:04, Roy Lewallen wrote: There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts?- Hide quoted text - - Show quoted text - If you adopt my expansion of gauss it can be calculated. If you look at the example in Chapter 21 of the Rutgers book on fields on the net maybe, but maybe, you will think a bit different but I expect you to fight off the idea of change Art Art-- Thanks for your previous reponse. I don't know if the comment above is directed to me or Roy. But I can tell you that I am not the person who is going to understand an expansion of gauss :) Wayne |
Measuring Antenna Efficiency
"Wayne" wrote in
news:7L%Gh.506$Ih.268@trnddc02: Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? "Roy Lewallen" wrote in message ... There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL Thanks for the reply. My dipole example is intended to avoid transmission line issues by not having one, and the elements are assumed to be reasonably low loss. If I do some quick back-of-the-envelope calculations, for a VSWR of 1.3:1, I get an efficiency of about 98.3% (using the equation 1-gamma^2). Assuming a You probably mean 1 less the magnitude of the reflection coefficient squared, which I would write as 1-|Gamma|^2 or 1-rho^2. Your implication is that "reflected power" is (necessarily) lost, that is wrong. If you do not have a transmission line, why are you trying to use transmission line concepts to solve the problem? The power delivered to the antenna is the real part of V^2/Z where V is the feedpoint voltage and Z is the feedpoint impedance. resistance of 1 ohm in the dipole conductors the efficiency I calculate is about 98.6% (72/73). The number you guess could be reasonable, and if it is truly the equivalent R at the feedpoint (rather than R/unit length * length), it does impact the antenna efficiency in the way you calculate. Owen |
Measuring Antenna Efficiency
Wayne wrote:
Thanks for the reply. My dipole example is intended to avoid transmission line issues by not having one, and the elements are assumed to be reasonably low loss. If I do some quick back-of-the-envelope calculations, for a VSWR of 1.3:1, I get an efficiency of about 98.3% (using the equation 1-gamma^2). Assuming a resistance of 1 ohm in the dipole conductors the efficiency I calculate is about 98.6% (72/73). Are there any other loss issues missing in this example. There is no relationship between SWR and efficiency. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
Owen Duffy wrote in news:Xns98EB57182E4B6nonenowhere@
61.9.191.5: the antenna is the real part of V^2/Z where V is the feedpoint voltage the antenna is the real part of V^2/Z where V is the feedpoint RMS voltage |
Measuring Antenna Efficiency
On 5 Mar, 12:58, Owen Duffy wrote:
"Wayne" wrote innews:5cXGh.391$iD4.256@trnddc06: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a The reason for focus on ground loss on (HF) verticals is that, unless you have taken extreme measures with a ground system, ground loss dwarfs other losses and in that case dominates considerations of efficiency. Next would come loading coils. dipole? Could one calculate "power out/power in" by measuring the Half wave dipoles made from practical materials are usually very high efficiency, losses commonly range in the area of 1% to ~3%. Loading coils are a significant loss element in loaded dipoles. Some folk (eg ARRL) have a mind that linear loading (folding the conductors back on themselves) is lossless, but my analysis of the Cobra shows that is not the case, seehttp://www.vk1od.net/cobra/index.htm. Even other lengths of unloaded dipoles may be very efficient, but the feedpoint impedance may drive huge losses on the feedline and so whilst the radiator is efficient, the antenna system may be inefficient. Components of an antenna system interact with each other in a complex way, and it is important to analyse the entire antenna system (radiator, earth, transmission line, balun, ATU etc) to obtain a correct understanding of how the system works overall. VSWR and declaring that everything not reflected was transmitted? It Roy has already explained to you that you have some misconceptions about VSWR, "forward power", and "reflected power". There has been another raging discussion here about what happens to the "reflected power", it isn't necessarily, and isn't usually lost (ie dissipated as heat), but as I have stated above the feedpoint impedance may drive huge losses on the feedline, it may also reduce the power available from the transmitter and may reduce the transmitter efficiency. would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? In terms of efficiency on the larger scale, a significant of power is lost in the process of reflecting some rays from real ground. Owen Oh my !!!!!! My dipole has a 1:1 swr. Neigbour next door puts a fence around his lot. My SWR goes to 3:1. No problem, swr has no connection to efficiency somebody said so everything is O.K. Except what is the definition of efficiency? Radiation resistance, effective use of the radiation, cancellation of radiation????? You better get that bit straightened out before you enlarge and expouse other statements which may cancel the legitamacy of your response. Another point to ponder on the other side. At what point can we separate the formation of radiation, at the radiation surface, the beginning of the near field the exit from the near field? If you are going to talk efficiency then you must have a closed border around which equilibrium is determined which brings us back to Gauss. Pretty neat!! Art Art |
Measuring Antenna Efficiency
art wrote:
. . . Except what is the definition of efficiency? Radiation resistance, effective use of the radiation, cancellation of radiation????? The definition of efficiency with respect to antennas is very well defined, understood, and agreed upon in all the amateur and professional literature. Virtually every text and professional paper uses an identical definition, and that is: the fraction of the power applied to an antenna which is radiated. I know that Art often uses the term to mean several other different things which I don't believe he's ever clarified (at least not that I could understand), so some caution should be used in making assumptions about its meaning when used in this newsgroup. . . . Roy Lewallen, W7EL |
Measuring Antenna Efficiency
"Wayne" wrote in message news:5cXGh.391$iD4.256@trnddc06... When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? As an example consider a horizontal 75 m dipole, constructed of #14 AWG copper, at various heights above an average ground. The radiation efficiency, according to NEC, is shown below: height 10 ft, efficiency 14%; height 30 ft, efficiency 54%, height 90 ft efficiency 82% The above results do not include the ground wave, which can be considered lost power. Ground absorption also increases with proximity to the ground. Regards, Frank |
Measuring Antenna Efficiency
"Frank" wrote in news:f71Hh.16901$lY6.7593
@edtnps90: "Wayne" wrote in message news:5cXGh.391$iD4.256@trnddc06... When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts? As an example consider a horizontal 75 m dipole, constructed of #14 AWG copper, at various heights above an average ground. The radiation efficiency, according to NEC, is shown below: height 10 ft, efficiency 14%; height 30 ft, efficiency 54%, height 90 ft efficiency 82% The above results do not include the ground wave, which can be considered lost power. Ground absorption also increases with proximity to the ground. Frank, What is the figure for free space? I suspect closer to 99%. If that is the case, don't your figures include loss (or absorbption) in rays reflected by the ground? Owen |
Measuring Antenna Efficiency
On 5 Mar, 14:47, Roy Lewallen wrote:
art wrote: . . . Except what is the definition of efficiency? Radiation resistance, effective use of the radiation, cancellation of radiation????? The definition of efficiency with respect to antennas is very well defined, understood, and agreed upon in all the amateur and professional literature. Virtually every text and professional paper uses an identical definition, and that is: the fraction of the power applied to an antenna which is radiated. I know that Art often uses the term to mean several other different things which I don't believe he's ever clarified (at least not that I could understand), so some caution should be used in making assumptions about its meaning when used in this newsgroup. . . . Roy Lewallen, W7EL Exactly, whose definition, whose limitations, by what authority? The term efficiency is universal in its intent and is measered by the terms imposed. You did not pose any limitations of any kind. The word efficiency is not identified with potato's in one part of science and carrots in another part of science. It has true meaning in all sciences relative to the context that is used. The number of books you have is not a deciding factor tho it probably had a hand in you ASSUMPTIONS Art |
Measuring Antenna Efficiency
"Owen Duffy" wrote in message ... "Frank" wrote in news:f71Hh.16901$lY6.7593 ........ As an example consider a horizontal 75 m dipole, constructed of #14 AWG copper, at various heights above an average ground. The radiation efficiency, according to NEC, is shown below: height 10 ft, efficiency 14%; height 30 ft, efficiency 54%, height 90 ft efficiency 82% The above results do not include the ground wave, which can be considered lost power. Ground absorption also increases with proximity to the ground. Frank, What is the figure for free space? I suspect closer to 99%. If that is the case, don't your figures include loss (or absorbption) in rays reflected by the ground? Owen Correct Owen. NEC shows 97.3% for free space, and 100 %, as expected, with perfect conductors. Certainly the loss does include absorption of the reflected rays. As mentioned before, in previous threads, it is very tedious to determine what percentage of the "Loss" is due to ground wave radiation. One of these days I will write the code necessary to compute the actual TRP including ground wave. Frank |
Measuring Antenna Efficiency
Frank wrote:
Correct Owen. NEC shows 97.3% for free space, and 100 %, as expected, with perfect conductors. Certainly the loss does include absorption of the reflected rays. As mentioned before, in previous threads, it is very tedious to determine what percentage of the "Loss" is due to ground wave radiation. One of these days I will write the code necessary to compute the actual TRP including ground wave. That capability is already built into NEC, as the average gain calculation. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
"Roy Lewallen" wrote in message ... Wayne wrote: Thanks for the reply. My dipole example is intended to avoid transmission line issues by not having one, and the elements are assumed to be reasonably low loss. If I do some quick back-of-the-envelope calculations, for a VSWR of 1.3:1, I get an efficiency of about 98.3% (using the equation 1-gamma^2). Assuming a resistance of 1 ohm in the dipole conductors the efficiency I calculate is about 98.6% (72/73). Are there any other loss issues missing in this example. There is no relationship between SWR and efficiency. Roy Lewallen, W7EL Agreed. For example, SWR into a dummy load. The equation given was just to show the approximation I was trying to make for the dipole case. |
Measuring Antenna Efficiency
"Frank" wrote in message news:_c3Hh.16918$lY6.10683@edtnps90... "Owen Duffy" wrote in message ... "Frank" wrote in news:f71Hh.16901$lY6.7593 ....... As an example consider a horizontal 75 m dipole, constructed of #14 AWG copper, at various heights above an average ground. The radiation efficiency, according to NEC, is shown below: height 10 ft, efficiency 14%; height 30 ft, efficiency 54%, height 90 ft efficiency 82% The above results do not include the ground wave, which can be considered lost power. Ground absorption also increases with proximity to the ground. Frank, What is the figure for free space? I suspect closer to 99%. If that is the case, don't your figures include loss (or absorbption) in rays reflected by the ground? Owen Correct Owen. NEC shows 97.3% for free space, and 100 %, as expected, with perfect conductors. Certainly the loss does include absorption of the reflected rays. As mentioned before, in previous threads, it is very tedious to determine what percentage of the "Loss" is due to ground wave radiation. One of these days I will write the code necessary to compute the actual TRP including ground wave. Frank Interesting. Thanks Frank. |
Measuring Antenna Efficiency
Wayne wrote:
"Roy Lewallen" wrote in message There is no relationship between SWR and efficiency. Roy Lewallen, W7EL Agreed. For example, SWR into a dummy load. The equation given was just to show the approximation I was trying to make for the dipole case. I don't seem to be communicating well, but that's not unusual. It's a shortcoming I have. So I'll try again. There is not even an approximate relationship between SWR and efficiency, so what you calculated was not an approximation of the efficiency. It was simply a number which has no relationship whatsoever to the efficiency. You could have used any equation you might dream up, and the result would be as meaningless as the result you got. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
I should elaborate a little.
The average gain is the ratio of the total power in all directions at a great distance (beyond the point where the surface wave has decayed to a negligible value) to the power into the antenna from all the sources. (There's a factor of two also involved when using a ground plane with NEC but not with EZNEC.) So the average gain is the efficiency if you consider ground reflection and the decay of the surface wave to be part of the loss. Roy Lewallen, W7EL Roy Lewallen wrote: Frank wrote: Correct Owen. NEC shows 97.3% for free space, and 100 %, as expected, with perfect conductors. Certainly the loss does include absorption of the reflected rays. As mentioned before, in previous threads, it is very tedious to determine what percentage of the "Loss" is due to ground wave radiation. One of these days I will write the code necessary to compute the actual TRP including ground wave. That capability is already built into NEC, as the average gain calculation. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
On 5 Mar, 20:41, Roy Lewallen wrote:
I should elaborate a little. The average gain is the ratio of the total power in all directions at a great distance (beyond the point where the surface wave has decayed to a negligible value) to the power into the antenna from all the sources. (There's a factor of two also involved when using a ground plane with NEC but not with EZNEC.) So the average gain is the efficiency if you consider ground reflection and the decay of the surface wave to be part of the loss. Roy Lewallen, W7EL Lets have another look at this. Roy inferred that the radiation field volume is the total useful output He then goes on to say that average gain what ever that means relative to the final radiation field is the efficiency. He also adds a condition relative to the definition of efficiency that this is only true IF you count ground reflection and the decay of the surface wave to be part of the loss Hmmm I don't think anybody would deny that surface wave represents a loss relative to usefull work though some might say it contributes to current flow, but why single out ground reflection as a loss since that can be useful? So Roy is classifying efficiency as something he considers usefull and ground reflection is not usefull. He also throws average gain into the equation without providing a definition of average gain ( like gain is an advance over something he doesn't want to state) Jimminy cricket I agree that you need to provide more elaboration Why is it you can't say the useful result of what you provided is the radiation volume where efficiency is useful output over input times 100? Why does one have to place conditions on : efficiency = useful output/ actual input x 100 ? Seems like efficiency in radiation is not the same as efficiencies in other sciences. Possibly a definition supplied by a related commitee solely for their own interpretation even though it is not in accordance with other diciplines. Also possibly based on the number of books on a particular shelf.And then the following week they placed conditions to clarify what efficiency includes and does not include such as certain portions of radiation, possibly with a different color to the norm phew Roy Lewallen wrote: Frank wrote: Correct Owen. NEC shows 97.3% for free space, and 100 %, as expected, with perfect conductors. Certainly the loss does include absorption of the reflected rays. As mentioned before, in previous threads, it is very tedious to determine what percentage of the "Loss" is due to ground wave radiation. One of these days I will write the code necessary to compute the actual TRP including ground wave. That capability is already built into NEC, as the average gain calculation. Roy Lewallen, W7EL- Hide quoted text - - Show quoted text - |
Measuring Antenna Efficiency
"Roy Lewallen" wrote in message ... I should elaborate a little. The average gain is the ratio of the total power in all directions at a great distance (beyond the point where the surface wave has decayed to a negligible value) to the power into the antenna from all the sources. (There's a factor of two also involved when using a ground plane with NEC but not with EZNEC.) So the average gain is the efficiency if you consider ground reflection and the decay of the surface wave to be part of the loss. Roy Lewallen, W7EL I was using average gain for my calculation of efficiency; i.e. XNDA = 1001, or 1002. I have also been considering the factor of "2" in the results. To be accurate, and to determine the radiation resistance of a structure, you do need to include the surface wave. The only way I can think of doing this is to sum "E X H" close enough to the radiating structure so as to include all its elements. At the moment I am using Excel to compute the Poynting vector. Even then, there is some question as to how much ground absorption effects the results between antenna and the hemispherical radius of computation. I have noticed some weird results if you get too close to the ends of a buried radial system. Regards, Frank (VE6CB) |
Measuring Antenna Efficiency
"art" wrote in message oups.com... On 5 Mar, 20:41, Roy Lewallen wrote: I should elaborate a little. The average gain is the ratio of the total power in all directions at a great distance (beyond the point where the surface wave has decayed to a negligible value) to the power into the antenna from all the sources. (There's a factor of two also involved when using a ground plane with NEC but not with EZNEC.) So the average gain is the efficiency if you consider ground reflection and the decay of the surface wave to be part of the loss. Roy Lewallen, W7EL Lets have another look at this. Roy inferred that the radiation field volume is the total useful output He then goes on to say that average gain what ever that means relative to the final radiation field is the efficiency. He also adds a condition relative to the definition of efficiency that this is only true IF you count ground reflection and the decay of the surface wave to be part of the loss Hmmm I don't think anybody would deny that surface wave represents a loss relative to usefull work though some might say it contributes to current flow, but why single out ground reflection as a loss since that can be useful? So Roy is classifying efficiency as something he considers usefull and ground reflection is not usefull. He also throws average gain into the equation without providing a definition of average gain ( like gain is an advance over something he doesn't want to state) Jimminy cricket I agree that you need to provide more elaboration Why is it you can't say the useful result of what you provided is the radiation volume where efficiency is useful output over input times 100? Why does one have to place conditions on : efficiency = useful output/ actual input x 100 ? Seems like efficiency in radiation is not the same as efficiencies in other sciences. Possibly a definition supplied by a related commitee solely for their own interpretation even though it is not in accordance with other diciplines. Also possibly based on the number of books on a particular shelf.And then the following week they placed conditions to clarify what efficiency includes and does not include such as certain portions of radiation, possibly with a different color to the norm phew "Ground reflection loss" is probably a more precise term. Frank |
Measuring Antenna Efficiency
Frank wrote:
On 5 Mar, 20:41, Roy Lewallen wrote: I should elaborate a little. The average gain is the ratio of the total power in all directions at a great distance (beyond the point where the surface wave has decayed to a negligible value) to the power into the antenna from all the sources. (There's a factor of two also involved when using a ground plane with NEC but not with EZNEC.) So the average gain is the efficiency if you consider ground reflection and the decay of the surface wave to be part of the loss. "Ground reflection loss" is probably a more precise term. Yes, that is what I meant and what I should have said, instead of just "ground reflection". Thanks for the correction. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
On 5 Mar, 10:04, Roy Lewallen wrote:
There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL I believe this to be untrue. If an array is in equilibrium you have skin depth that give you a resistance figure as well as d.c. resistance . You also know power input thus all input and output power is therefore known. Pray tell what energy cannot be accounted for? Remember radiation does not begin to occur until the arbitary border is punctured thus at that time it can be considered as output. Movement of flux cannot begin until the clock starts or time begins So now you have a beginning. An enclosed arbitary border that containes energy and a end section that represents radiation. You could also use the potential momentum theorem to determine the exit proportions of electric and magnetic particles to generate a electromagnetic field as well determining what particles return to the radiator to serve in the formation of skin depth or radiation resistance which serves to augument time changing current as well as accounting for decay. All this requires a smattering of understanding with respect to Einstein law of relativity which most will probably tend to dismiss as hog wash especially those who view the subject of static particles as being useless. The answer Roy gave is only applicable when an array has parasitics that either deflect or attract energy with respect to polarity( Walter note the use of the word "polarity" with respect to antennas) Ofcourse some will drop back to point out Pointings Vector so this could be interesting. If you want to disagree start off with a resonant dipole to ensure applicability of ones auguments. Art Unwin Wayne wrote: When the subject of antenna efficiency comes up, it often involves a discussion of ground losses on verticals. What about, for example, a dipole? Could one calculate "power out/power in" by measuring the VSWR and declaring that everything not reflected was transmitted? It would seem more accurate to actually measure power out and power in, but that introduces inaccuracies by having to calibrate the setup. Thoughts?- Hide quoted text - - Show quoted text - |
Measuring Antenna Efficiency
"art" wrote in message oups.com... On 5 Mar, 10:04, Roy Lewallen wrote: There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL I believe this to be untrue. If an array is in equilibrium you have skin depth that give you a resistance figure as well as d.c. resistance . You also know power input thus all input and output power is therefore known. Pray tell what energy cannot be accounted for? Remember radiation does not begin to occur until the arbitary border is punctured thus at that time it can be considered as output. Movement of flux cannot begin until the clock starts or time begins So now you have a beginning. An enclosed arbitary border that containes energy and a end section that represents radiation. You could also use the potential momentum theorem to determine the exit proportions of electric and magnetic particles to generate a electromagnetic field as well determining what particles return to the radiator to serve in the formation of skin depth or radiation resistance which serves to augument time changing current as well as accounting for decay. All this requires a smattering of understanding with respect to Einstein law of relativity which most will probably tend to dismiss as hog wash especially those who view the subject of static particles as being useless. The answer Roy gave is only applicable when an array has parasitics that either deflect or attract energy with respect to polarity( Walter note the use of the word "polarity" with respect to antennas) Ofcourse some will drop back to point out Pointings Vector so this could be interesting. If you want to disagree start off with a resonant dipole to ensure applicability of ones auguments. Art Unwin Art Roy said "There's no direct way to measure" , what you are describing isnt a direct way. Jimmie |
Measuring Antenna Efficiency
On 22 Apr, 18:18, "Jimmie D" wrote:
"art" wrote in message oups.com... On 5 Mar, 10:04, Roy Lewallen wrote: There's no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating. "Reflected" power is not power that isn't transmitted. You can find the power being applied to the antenna by subtracting the "reverse" or "reflected" power from the "forward" power, but that tells you nothing about what fraction is radiated and what fraction lost as heat. Roy Lewallen, W7EL I believe this to be untrue. If an array is in equilibrium you have skin depth that give you a resistance figure as well as d.c. resistance . You also know power input thus all input and output power is therefore known. Pray tell what energy cannot be accounted for? Remember radiation does not begin to occur until the arbitary border is punctured thus at that time it can be considered as output. Movement of flux cannot begin until the clock starts or time begins So now you have a beginning. An enclosed arbitary border that containes energy and a end section that represents radiation. You could also use the potential momentum theorem to determine the exit proportions of electric and magnetic particles to generate a electromagnetic field as well determining what particles return to the radiator to serve in the formation of skin depth or radiation resistance which serves to augument time changing current as well as accounting for decay. All this requires a smattering of understanding with respect to Einstein law of relativity which most will probably tend to dismiss as hog wash especially those who view the subject of static particles as being useless. The answer Roy gave is only applicable when an array has parasitics that either deflect or attract energy with respect to polarity( Walter note the use of the word "polarity" with respect to antennas) Ofcourse some will drop back to point out Pointings Vector so this could be interesting. If you want to disagree start off with a resonant dipole to ensure applicability of ones auguments. Art Unwin Art Roy said "There's no direct way to measure" , what you are describing isnt a direct way. Jimmie- Hide quoted text - - Show quoted text - I stand corrected. You cannot wet your finger and stick it up into the air expecting that one could obtain a direct measure of radiation. Most observant of you Jimmie Art |
Measuring Antenna Efficiency
Jimmie D reported Roy Lewallen to write:
"There is no direct way to measure the total power being radiated other than sampling the field at many points in all directions and integrating." That sounds right to me. An approximation is sometimes made by taking 36 samples of field strength in volts per meter at 10-fegrees of azimuth intervals at the same distance from the central antenna system. Each of these sample values is squared and the sum of these squared samples is divided by 36, the number of samples, to get their average. The square poot of this quotient is then the average field strength at that distance from the antenna. A true average signal strength should be the same as the value an isotropic antenna would radiate at a given distance. Knowing the field strength, one could calculate the watts per square meter of the envelope of radiation at a given distance and total the watts per square meter of all the squares to get the total power being radiated. Since 1960, I`ve used the Bird wattmeter satisfactorily to get the total power being delivered by the transmitter and radiated by the antenna. It should be the same if the transmission line and antenna have low losses. It is simply the difference between the forward power indication and the reverse power indication. Many lines and antennas have very high efficiencies. Best regards, Richard Harrison, KB5WZI |
Measuring Antenna Efficiency
Owen Duffy wrote:
Firstly you seem to assume that your 36 samples around the azimut circle adequately fulfill Roy's "sampling the field at many points in all directions", surely he mean't all elevation angles as well as all azimuth angles. What is the hourly rental charge on a helicopter? :-) -- 73, Cecil http://www.w5dxp.com |
Measuring Antenna Efficiency
Cecil Moore wrote:
Owen Duffy wrote: Firstly you seem to assume that your 36 samples around the azimut circle adequately fulfill Roy's "sampling the field at many points in all directions", surely he mean't all elevation angles as well as all azimuth angles. What is the hourly rental charge on a helicopter? :-) that's what balloons and blimps are for, or model airplanes.. |
Measuring Antenna Efficiency
Owen wrote:
"Is that true?" Only for limited conditions. Owen is correct that enough samples must be taken to catch all variations in signal strength, in elevation and azimuth if total radiated power is to be determined. A simple vertical antenna rests on the earth and has a null at its tip. It is far from an isotropic radiator. The sampling I described is done on AM broadcast antennas. Only the wave which travels along the earth is usually of any interest. The minimum distance from the broadcast antenna for any field strength measurements is usually one mile to be sure the far field is being measured. The FCC`s ground wave intensity charts assume that if you have ground conductivity such as sea water, a certain power and antenna efficiency deliver 100 millivolts per meter at a distance of one mile from the antenna. Field intensity in millivolts declines linearly with distance so that at 10 miles you might have 10 mv/m if you have 100 mv/m at one mile.Power is proportional to the square of the voltage. Best regards, Richard Harrison, KB5WZI |
Measuring Antenna Efficiency
Richard Harrison wrote:
The FCC`s ground wave intensity charts assume that if you have ground conductivity such as sea water, a certain power and antenna efficiency deliver 100 millivolts per meter at a distance of one mile from the antenna. Field intensity in millivolts declines linearly with distance so that at 10 miles you might have 10 mv/m if you have 100 mv/m at one mile. . . . The attenuation of the ground wave is the same as the free space attenuation? Roy Lewallen, W7EL |
Measuring Antenna Efficiency
Roy Lewallen, W7EL wrote:
"The attenuation of the ground wave is the same as the free space attenuation?" No, it is greater. The attenuation of the ground wave is more rapid because the soil has resistance and imposes a loss on the passing wave. In free space, there is no ground loss and the signal loss is from the spreading of the signal which reduces its density. It declines 6 dB (its power is quartered) every time the distance from the transmitter doubles, that is its volts per meter or signal strength is cut in half. Ground waves decrease more rapidly with distance from the transmitter due to imperfect conductivity in the earth`s surface. Loss goes up with the operating frequency and with higher resistivity in the soil. The FCC`s ground-wave field intensity charts cover limited frequency ranges and each curve is for a specific soil conductivity. Best regards, Richard Harrison, KB5WZI |
Measuring Antenna Efficiency
Richard Harrison wrote:
Roy Lewallen, W7EL wrote: "The attenuation of the ground wave is the same as the free space attenuation?" No, it is greater. The attenuation of the ground wave is more rapid because the soil has resistance and imposes a loss on the passing wave. . . . I should hope so. That's why your statement that the field strength decreases as the inverse of the distance in the context of AM broadcast measurements at ground level was puzzling. Roy Lewallen, W7EL |
Measuring Antenna Efficiency
On Mon, 23 Apr 2007 17:30:20 -0700, Roy Lewallen
wrote: Richard Harrison wrote: Roy Lewallen, W7EL wrote: "The attenuation of the ground wave is the same as the free space attenuation?" No, it is greater. The attenuation of the ground wave is more rapid because the soil has resistance and imposes a loss on the passing wave. . . . I should hope so. That's why your statement that the field strength decreases as the inverse of the distance in the context of AM broadcast measurements at ground level was puzzling. Richard had been reciting decade-for-decade declines of FS faithfully from the FCC charts -over seawater- which he had specifically called out. "Faithfully" except for an inadvertent substitution of decade miles for decade kilometers distance. 73's Richard Clark, KB7QHC |
Measuring Antenna Efficiency
Richard Clark, KB7QHC wrote:
"Faithfully" except for an inadvertent substitution of decade miles for decade kilometers distance." Thank you Richard for such a kind word as "inadvertent". I make mistakes and I apologize for them. Hopefully, none of consequence go uncorrected and mislead anyone. I learn more from my mistakes than from correct postings but it isn`t my favorite way to learn. Thank you to everyone who prompts me to search for the answer to an interesting question. I`m still puzzled about "cluster antennas". Best regards, Richard Harrison, KB5WZI |
Measuring Antenna Efficiency
|
Measuring Antenna Efficiency
I`m still puzzled about "cluster antennas". Best regards, Richard Harrison, KB5WZI I really don't know why it puzzles you. I have determined that coupling of elements which means parasitic coupling is wastfull. To avoid this coupling one must have elements that are all resonant. This means ofcourse that element dimensions are necessarily different from neighboring elements so that coupling per se ( there is no attraction or repeling actions between elements) removed which thus creates an arrangement that is in equilibrium. When elements are associated with each other one can call them a 'collection' of elements or a 'cluster' of elements. Now with respect to efficiency this array does not have coupling associated with reflectors and directors thus if the array was energised then a far field of radiation is formed without any wastefull interaction of particles between elements. So when the far field is formed the only "waste" of energy is that created by skin resistance which is easily calculated, since without interaction between elements non radiating energy collision losses cannot occur. If one is looking for efficiency one usually work with input energy equals output energy plus losses. When one is examining the amount of radiation formed in the far field because one does not want to get into a situation of interactive bombardment of partcles that do not contribut to radiation in the far field prior to the generation of radiation. This analysis I suppose can be considered a simple relationship with Poyntings vector and in a way provides support for Poynting where none existed before but I may be over reaching there in some eyes. Hope that satisfies your curiousity Art |
| All times are GMT +1. The time now is 04:19 PM. |
|
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com