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On Tue, 16 Dec 2003 22:38:50 -0800, said...
Sorry for the slow response. s'ok. but i don't think my stinkin' reader alerted me to the response. or I missed it. The voltage at the terminals of a short, lossless dipole of length l which is perfectly terminated with the complex conjugate of its feedpoint impedance is E * l / 2 where E is the impinging field in volts per meter. Any mismatch lowers this value -- it's one of those valid and useful applications of the concept of "mismatch loss", which can be used to quantify the reduction when mismatched. It's one of the contributions to the extensive variation with frequency of the "antenna factor"(*) of test dipoles used in EMI work. (And careful measurement of this factor is a large fraction of what you're paying for when you purchase one.) Doesn't matter how wonderful your field strength meter is unless you have a very good handle on the "antenna factor" of the test antenna. I'll dig into the Antenna Engrs Handbook. I guess we've been lucky. So far no FCC gigs on leakage. Haven't heard about anyone failing a flyover either. Maybe the systems consider the FCC spec and set their spec to account for cheap-assed dipoles. If so, that's smart and I'd be surprised. Some day I'll check that out. In simple terms, the factor of 1/2 is the ratio of the average current to the feedpoint current. The average current along a short dipole is about 1/2 the current at the feedpoint. You could get a factor of 1 if you used a short dipole with large end hats to make the current distribution uniform. Kraus addresses this case in detail in _Antennas_. And no, I don't mean "antenna to free space" match. I mean the match between the antenna feedpoint as a source and the detector as a load. That's a given. I figure Paul already knows that. BRs, Mike (*) Ratio of output voltage when terminated in 50 ohms to the impinging field in volts/meter. Roy Lewallen, W7EL Active8 wrote: On Sat, 13 Dec 2003 17:39:27 -0800, said... It's too bad it isn't that simple. A 1V/m field doesn't result in one volt at the feedpoint of a perfectly matched one meter dipole or monopole, and the value it does induce depends on the quality of the impedance match you mean antenna to free space, right? as well as the fraction of a wavelength the one meter antenna length represents. amplify, very please. por favor. Refresh my ram. | E(uV/m) | V (dBmV) = 20log | --------- / 1000 | | 0.021f(MHz) | plus correction for distance (regulations for limits are for specific measuring distances), etc. I'll mull the above eq over. Gotta figure out where the .021 came from, but not now. My eyes are getting fatigued from this 'puter. And, if one volt does appear at the feedpoint, it's very unlikely that a simple circuit will measure it as one volt. It would have to be calibrated to compensate for the circuit. Maybe that's why it's called a "relative" field strength meter. Relative to another signal or no signal ![]() Probably best to stick with your $1.5 kilobuck meter if you really want to measure field strength. I don't. He does ![]() field strenth using a cheap ass dipole cut to the frequency of interest with or without an external preamp and do it to the satisfaction of the FCC, assuming it's calibrated. I even have a near-field probe, not so cheap. It beats guess work. I wouldn't expect his sniffer to be real accurate but he did ask for guesstimates. Started off as "around 4 feet" for a half watter now we're at 100mW - prob his reference Tx. BRs, Mike Roy Lewallen, W7EL Active8 wrote: On 13 Dec 2003 03:50:41 -0800, said... What you are descrbing is a "signal sniffer", not a signal strength meter. Who are you replying to? Paul did not say Signal Stength Meter, but the guy you replied to kinda hints at it when he mentions power level. I would have to say that my 1GHz Signal Level Meter, which cost $1500 would be the better than a relative field strength meter, but if he measures say, 1V with his Tx off and 2V with it on, then that's 1V and if his Rx antenna is a 1m dipole, that's 1V/m. Mike |
#3
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On Tue, 16 Dec 2003 22:38:50 -0800, said...
Sorry for the slow response. s'ok. but i don't think my stinkin' reader alerted me to the response. or I missed it. The voltage at the terminals of a short, lossless dipole of length l which is perfectly terminated with the complex conjugate of its feedpoint impedance is E * l / 2 where E is the impinging field in volts per meter. Any mismatch lowers this value -- it's one of those valid and useful applications of the concept of "mismatch loss", which can be used to quantify the reduction when mismatched. It's one of the contributions to the extensive variation with frequency of the "antenna factor"(*) of test dipoles used in EMI work. (And careful measurement of this factor is a large fraction of what you're paying for when you purchase one.) Doesn't matter how wonderful your field strength meter is unless you have a very good handle on the "antenna factor" of the test antenna. I'll dig into the Antenna Engrs Handbook. I guess we've been lucky. So far no FCC gigs on leakage. Haven't heard about anyone failing a flyover either. Maybe the systems consider the FCC spec and set their spec to account for cheap-assed dipoles. If so, that's smart and I'd be surprised. Some day I'll check that out. In simple terms, the factor of 1/2 is the ratio of the average current to the feedpoint current. The average current along a short dipole is about 1/2 the current at the feedpoint. You could get a factor of 1 if you used a short dipole with large end hats to make the current distribution uniform. Kraus addresses this case in detail in _Antennas_. And no, I don't mean "antenna to free space" match. I mean the match between the antenna feedpoint as a source and the detector as a load. That's a given. I figure Paul already knows that. BRs, Mike (*) Ratio of output voltage when terminated in 50 ohms to the impinging field in volts/meter. Roy Lewallen, W7EL Active8 wrote: On Sat, 13 Dec 2003 17:39:27 -0800, said... It's too bad it isn't that simple. A 1V/m field doesn't result in one volt at the feedpoint of a perfectly matched one meter dipole or monopole, and the value it does induce depends on the quality of the impedance match you mean antenna to free space, right? as well as the fraction of a wavelength the one meter antenna length represents. amplify, very please. por favor. Refresh my ram. | E(uV/m) | V (dBmV) = 20log | --------- / 1000 | | 0.021f(MHz) | plus correction for distance (regulations for limits are for specific measuring distances), etc. I'll mull the above eq over. Gotta figure out where the .021 came from, but not now. My eyes are getting fatigued from this 'puter. And, if one volt does appear at the feedpoint, it's very unlikely that a simple circuit will measure it as one volt. It would have to be calibrated to compensate for the circuit. Maybe that's why it's called a "relative" field strength meter. Relative to another signal or no signal ![]() Probably best to stick with your $1.5 kilobuck meter if you really want to measure field strength. I don't. He does ![]() field strenth using a cheap ass dipole cut to the frequency of interest with or without an external preamp and do it to the satisfaction of the FCC, assuming it's calibrated. I even have a near-field probe, not so cheap. It beats guess work. I wouldn't expect his sniffer to be real accurate but he did ask for guesstimates. Started off as "around 4 feet" for a half watter now we're at 100mW - prob his reference Tx. BRs, Mike Roy Lewallen, W7EL Active8 wrote: On 13 Dec 2003 03:50:41 -0800, said... What you are descrbing is a "signal sniffer", not a signal strength meter. Who are you replying to? Paul did not say Signal Stength Meter, but the guy you replied to kinda hints at it when he mentions power level. I would have to say that my 1GHz Signal Level Meter, which cost $1500 would be the better than a relative field strength meter, but if he measures say, 1V with his Tx off and 2V with it on, then that's 1V and if his Rx antenna is a 1m dipole, that's 1V/m. Mike |
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