Rich
Full Member
RF Systems Engr (retired)
Posts: 112
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Post by Rich on Jun 30, 2018 16:12:31 GMT
Even expensive, accurately-calibrated instruments still need to be used with due understanding of the operating and measuring environment for the system under test. These days when using MoM software such as NEC, it is possible to accurately calculate the performance of a properly-modeled antenna system. Doing so is a lot less expensive than buying, setting up and measuring "real" hardware — even if the investigator has the budget, knowledge and experience to do so, accurately. Below is an example of using NEC4.2 for this purpose, along with some comments that apply to these scenarios.
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Post by timinbovey on Jul 1, 2018 12:24:18 GMT
I have to agree with Rich that this is indeed possible and a lot cheaper.
The problem is rarely does one know the data needed to use the software. Most critically you have to know the output of the transmitter. As shown 11.43 NANOWATTS is what's required into a half wave dipole to hit the legal limit.
I've yet to test a transmitter that puts out anything close to 11 NANOWATTS. That's virtually so low as to be immeasurable. I had a slug for my Coaxial Dynamics watt meter custom made for me for the broadcast FM band, giving a full scale reading of 100 milliwatts. It's nearly impossible to measure 1 milliwatt (you're talking reading 100 increments on a 4" meter) I can't imagine trying to measure 11 nanowatts! I suppose one could with a more expensive meter.
And I've never seen a transmitter that comes equipped with a dipole. We must remember, to be legally sold in the USA an FM Part 15 transmitter MUST be certified. And it must be certified with the supplied antenna. All of these transmitters rely on a poor, inefficient antenna to get their signal down into legal range. The second you switch it for a different antenna, the certification is gone and you'll also be over the limit.
So you'd have to know the output of the transmitter and the data for the supplied antenna, which is usually a hunk of wire or a telescopic whip attached to the transmitter, or sometimes a "rubber duck" type antenna.
Lack of this data makes using the software difficult.
And as noted in the comments above this does not take into account increases in field strength caused by connecting other necessary cables to the transmitter. Length and angle of both power cables and audio cables made a HUGE jump in field strength in every transmitter I tested.
So even if you had the output data and antenna data, you still wouldn't model output of the transmitter in actual operation, unless you were running it on an internal battery and fed it no audio.
Tim in Bovey
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Rich
Full Member
RF Systems Engr (retired)
Posts: 112
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Post by Rich on Jul 2, 2018 12:18:05 GMT
... The problem is rarely does one know the data needed to use the software. Most critically you have to know the output of the transmitter. ... We must remember, to be legally sold in the USA an FM Part 15 transmitter MUST be certified. And it must be certified with the supplied antenna. All of these transmitters rely on a poor, inefficient antenna to get their signal down into legal range. Thanks for your good comments, Tim.
One "takeaway" that I was hoping for in my original post is the realization that the radiated fields of a VHF (and above) transmit system with an antenna of any physical form located near the Earth do not necessarily have the same, inverse-distance relationship that they have in free space.
For example, using my NEC model set up exactly as in my original post, the field at twice the horizontal distance (6 meters) and the same elevation above the Earth (2.13 meters) is 72.3% of that for the same elevation at 3 meters from the transmit antenna. See the graphic next below.
A free space path (no reflections/obstructions) having double the path length would have a field intensity of 50% of its original value at 3 meters, or 136.2 µV/m in this case.
If the FCC or anybody else accurately measured the field at 6 meters as 196.4 µV/m, then by the inverse distance rule the field at 3 meters horizontal distance and that height would have to be 392.8 µV/m. However that is an inaccurate value for that field in this scenario — by 392.8/272.3 = 44% error.
Of course that can lead to incorrect assumptions needed for such systems to comply with FCC §15.239.
This all reminds me of a saying that used to circulate in the engineering department of my last employer:
"For every simple question there is a simple answer, and usually it is wrong."
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