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#1
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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? |
#2
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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 |
#3
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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? |
#4
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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 |
#5
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"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 |
#6
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![]() 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. |
#7
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![]() "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 |
#8
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"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 |
#9
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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 |
#10
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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 |
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