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I just viewed the story on the home page about the FCC finding a pirate radio station in Olympia WA. The FCC's notice claimed that the individual's station produced a signal of 4,428 microvolts per meter (µV/m) at 436 meters. Is there any way to estimate how many watts (ERP) they were operating at?
At FM broadcast band frequencies, an effective radiated power (ERP) of about 76 milliwatts produces a field intensity of 4,428 µV/m at a distance of 436 meters over a free space path (no reflections, no obstructions).
A 1/2-wave, center-fed, dipole transmit antenna has a peak power gain of about 1.64 X, so the Z-matched power needed at its feedpoint to produce the field measured by the FCC could be on the order of 76/1.64 = 46 milliwatts.
Free Radio Olympia has been around for over 10 years. They used to crowd out one of my stations. I asked them to move down the dial, and they did. That was several years ago.
I believe the pirate's power was more like a little over 100W. 100W into a dipole is +165dB (uV) at the antenna. The FCC measured a field of 4400uV at 430meters which is equivalent to +160dBu. Certainly, that would be enough to get one in hot water.
I believe the pirate's power was more like a little over 100W. 100W into a dipole is +165dB (uV) at the antenna. The FCC measured a field of 4400uV at 430meters which is equivalent to +160dBu
Actually an E field intensity of 4400 µV/m = 20*log(4400) = 73 dB(uV/m), approximately.
However the voltage across the feedpoint of an antenna does not directly translate to the field intensity produced at a given distance, by a given radiated power.
The equation needed for that is:
E =SQRT(49.2 * P) / D
where E = Field Intensity in V/m
P = Radiated Power in watts
D = Distance in meters
So in this case E = SQRT(49.2 * 0.076) / 436 = 0.004435 V/m, or 4,435 µV/m, which essentially is the value reported by the FCC.
The radiated power needed to generate that free-space field at that distance is about 76 milliwatts.
I saw on the link posted by Bill Wolfenbarger that this operator reports using 100 watts, but if he was radiating 100 watts toward the location where the FCC did this measurement, the free-space field would have been around 160 millivolts/meter, or about 104 dB(µV/m).
Note that the difference in decibels between 73 dB(µV/m) and 104 dB(µV/m) is the same as the difference in decibels between 76 milliwatts and 100 watts, ie, ~31 dB.
I'm too lazy to work this out by hand. I just used the power to voltage converter in rfInvestigator v3.5 to determine the dbu (it has a tool to convert Watts to dBu). According to rfSoftware 100W = 165dBu. I also ran the freespace loss model and the computer says that 4400uV = 160dBu. These numbers agree with the stated power on the pirate's own website. I don't really think that the FCC would bother traveling to Olympia for a 73mW pirate. The legal limit is about 9mW so 73mW would be barely detectable. It wouldn't go more than a block or two, even using a good antenna. It's hard to see how they could provide coverage to an entire city with that. I noticed you were using 20log to convert to microvolts, but a voltage conversion would use 10log wouldn't it?
... I just used the power to voltage converter in rfInvestigator v3.5 to determine the dbu (it has a tool to convert Watts to dBu). According to rfSoftware 100W = 165dBu. I also ran the freespace loss model and the computer says that 4400uV = 160dBu. ...etc
I don't have your software, so I can't really advise on why it is not giving the correct result. I suspect it is not designed to do what is needed here.
If "dBu" here really means decibels with respect to a field intensity of 1 µV/m, then 4400 µV/m is 20*log(4400) = 73 dBu, approx. This should make sense, as a field of 70 dBu is 3160 µV/m -- the minimum for COL coverage of an FM station.
I noticed you were using 20log to convert to microvolts, but a voltage conversion would use 10log wouldn't it?
There were quite a few FCC citations in just the past several months given to unlicensed operators radiating powers of less than 100 mW (which radiated power can be calculated using the equation I posted).
There is no legal power limit defined for unlicensed FM operators -- instead they are restricted to a maximum field of 250 µV/m at a distance of 3 meters in any direction from the transmit antenna. See FCC 15.239.
The Z-matched power needed to produce that peak field from a center-fed, 1/2-wave dipole is only 11.43 nW (0.000 000 011 43... watts).
Operating with 100 watts is just asking for trouble. A couple of watts ... and nobody would have noticed.
If they are streaming on the internet, why even bother with over-the-air?
A potentially costly error in judgement.
If "dBu" here really means decibels with respect to a field intensity of 1 µV/m, then 4400 µV/m is 20*log(4400) = 73 dBu, approx. This should make sense, as a field of 70 dBu is 3160 µV/m -- the minimum for COL coverage of an FM station.
That's a weak pirate. There are many relator home sell transmitters that might make it that far. Several gyms use these for people on aerobics machines to allow them to watch television. The ifb use is common as well.
I was told one time if it's available a block or so away that the Part 15 usage is settled. If you can hear it 5 blocks or more, or, there is a repetitive issue, it might be another thing altogether.
Just to note that for those seeking "simple," the equation I posted earlier in this thread leads to the same result as this consulting engineer reportedly found, and with many fewer intermediate calculations.
I'm not so sure that the commission correctly identified the transmitting site, therefore the calculated power is possibly way off. This station has been on the air for over ten years and covers Olympia fairly well. What surprises me is that it has been there for so long and they just now are figuring it out. In the state capital to boot...
Several years ago they were on 101.9, first adjacent to my Aberdeen station, and I asked them to move. And they did move to 98.5. I figured if they didn't bother me, I didn't much care.
This christmas here in the east side of the Netherlands (near the German border) there where several pirate stations using 10.000 to 30.000W transmitters going far over 150KW ERP... (on 8 stacked dipoles)
some of them do a illegal broadcast show every weak and sometimes they do 24/7 for a whole weak ore two weaks.... ;D
This christmas here in the east side of the Netherlands (near the German border) there where several pirate stations using 10.000 to 30.000W transmitters going far over 150KW ERP... (on 8 stacked dipoles)
some of them do a illegal broadcast show every weak and sometimes they do 24/7 for a whole weak ore two weaks.... ;D
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