| Home |
| Search |
| Today's Posts |
|
#15
April 29th 07, 06:30 PM
posted to rec.radio.amateur.antenna
|
|||
|
|||
Measuring Antenna Efficiency
On Apr 28, 8:07 pm, art wrote:
On 5 Mar, 14:14, "Wayne" wrote: 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. I would say you are close enough to say you are correct. Because you chose a dipole which is in a state of equilibrium and thus particles projected from the dipole cannot collide with other particles from other parasitic radiators. Aren't you not basically refering to the foundations of Poyntings vector which like Gauss is refering to an item in equilibrium when subjected to a time variable of zero value ? If the item is not in a state of equilibrium collision of particles may well occur without a radiation field reaction thus one cannot calculate the resultant field since energy transfer due to particle collision prevents the return of particles to the mother radiator.- Hide quoted text - - Show quoted text - what is a particle? |
| Thread Tools | Search this Thread |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| Yagi efficiency | Antenna | |||
| Yagi efficiency | Antenna | |||
| measuring antenna resonance with an 8405a | Antenna | |||
| High Efficiency Mobile HF Antenna? | Antenna | |||
Threaded Mode
