Hamilton wrote "Since legal FM only gets shouting distance basically requiring ridiculous limitations on an industry..."
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According to FCC Public Notice 14089, the expected range for legal, unlicensed AM and FM under Part 15 are both about 200 feet.
Click *here* for a link to that FCC document.
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Wow, I could stay here all day. It is true that an AM transmitter cetified under Part 15.209 would get just a few hundred feet of range, just like FM. 15.209 is a radiation limitation. But a lot of AM transmitters are certified under Part 15.219. These transmitters generally must be installed by a technician, and have adjustable power and removable antenna. (And also must be installed by someone who knows what a ground is) (couldn't help putting that in) A 15.209 AM transmitter is something like a MR. microphone.
Hamilton wrote: "It is true that an AM transmitter cetified under Part 15.209 would get just a few hundred feet of range, just like FM. 15.209 is a radiation limitation. But a lot of AM transmitters are certified under Part 15.219."
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If you read the Public Notice linked in my last post, you will see that the column for unlicensed AM refers to the use of a tx with 100 mW input power. That is part of the definitions given in Part 15.219, not 15.209.
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That FCC Public Notice document should be removed as the Part15 regs are the definative rule regarding the part 15 rules and documentation. The notice is just that... but it doesn't have any legal binding... I believe any FCC attorney and even the FCC will point that out!
How many times have you seen that 'public notice' mentioned in the NOUOs and NAL's in the FCC Enforcement website? Give me one evidence to that fact it was ever used or referenced?
I think Keith is right and the ultimate 'RULE' will be the FCC agent and what he believes is the 'nature' of the complaint... in the end he will succumb to the Part 15.219 which is the 3 meter rule for the antenna and the ground.
Radiopilot
That's right, the 200 foot statement is notorious and is on the long list of standard questions we get like "how can you advertise you can transmit on 1710 when the FCC band only goes to 1705?" and other standard questions we get almost daily. My explanation comes almost verbatim from the FCC.
Rich, I'd like to point out a few things about that "200 ft"estimate from FCC Public Notice 14089.
Firstly, the FCC's estimates of range are conservative for other services as well. Consider part 95 citizen's band on the 11 meter band. The official FCC estimate of expected range for the 27 mhz citizen's band service on their website is 1 to 5 miles. However realistically that would maybe be for two mobile units not in particularly good condition in hilly terrain or something. Using strictly legal gear in that service, ranges of 30 miles or more are not at all unusual and can be quite reliable except perhaps in the peak of sunspot activity. In fact, part 95 includes a rule specifically prohibiting intentional contacts between stations more than 150 miles distant. No need for such a rule if the FCC actually considered their expected range estimate to be an absolute upper limit. So although their official estimate of the range may be one to five miles, they recognize that in legal operation it is possible to exceed that by a considerable amount.
Secondly, they do not specify what sort or quality of receiver is being used in that estimate, nor what quality of reception they're talking about.
Thirdly.. By your own calculation/estimate when replying to a question about what people with "non grey area" xmitter/antenna systems are getting ranged from:
"5.000 mV/m > 0.0340 miles (good signal to cheap, indoor radio)"
(which is indeed close to 200 ft)
to:
"0.100 mV/m > 1.5865 miles (acceptable, but noisy signal to a good car radio in an area with no overhead wires, no local electrical interference, and no co-channel interference)."
.. which is certainly considerably more than 200 ft.
It is understood that you were giving comparative estimates based on your calculations of a strictly legal by 15.219(b) station for a variety of receiving equipment and qualities of reception. But in light of your own calculations it seems perhaps a trifle polemic to then use the estimates of Public Notice 14089 as a point of argument. Or am I perhaps missing your point here?
Daniel
Daniel wrote: "But in light of your own calculations it seems perhaps a trifle polemic to then use the estimates of Public Notice 14089 as a point of argument. Or am I perhaps missing your point here?"
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The FCC knows that radio waves do not go immediately to zero field strength after a given distance. The 200 feet "maximum" radius in that note is given as approximate, and of course the Notice does not define what value of field strength constitutes acceptable coverage. The notice is given as guidance, not as a legal requirement.
But the unlicensed, maximum AM coverage radius of about 200 feet shown in FCC Public Notice 14089 is not so different from what a strictly legal, ground-mounted Part 15.219 AM antenna and tx should do, for an average indoor receiver in a suburban metro area.
Here is the development of that conclusion (in round numbers). This approach is a bit different than the one I used in my post you quoted. But it is more easily understood and verified by anyone wishing to do so. It gives somewhat shorter ranges than I calculated earlier, but I may have used different assumptions then.
* The standard groundwave field produced by a 1 kW AM broadcast tx using a 1/4-wave vertical monopole with 120 radials each at least 1/4-wave long is 300 mV/m at a 1 km radius. This value is taken from the FCC website, and is supported by thousands of field measurements over many decades.
* The difference in power between 1 kW and the ~80 mW output power of a good Part 15 AM tx is 41 dB.
*The gain of a very good, ground-mounted 3-m Part 15 AM antenna system on 1700 kHz (~25 ohms total in the coil and r-f ground connection) is about 25 dB below that of the 1/4-wave monopole described above.
So the field from the Part 15 system at a 1 km radius will be about 66 dB below that of the broadcast system reference. A voltage reduction of 66 dB is a multiplier of 0.0005, which means the Part 15 field at 1 km will be 300 mV/m x 0.0005 = 0.150 mV/m.
At these short distances the field is inversely related to the distance from the antenna, so we can construct this table:
Distance in km > Field in mV/m
1 > 0.15
1/2 > 0.30
1/4 > 0.60
1/8 > 1.2
1/16 > 2.4
The distance of 1/16 km is 205 feet. Not too coincidentally, it takes a field of about 2 mV/m to provide a fairly noise-free AM signal to a cheap indoor radio in an urban setting.
"Part 15" systems that are capable of producing good signals under these conditions to cheap indoor radios over much greater ranges than this are able to do so only by using a more efficient (elevated) antenna system including a long radiating ground conductor, more tx power output, or possibly both of those.
"At these short distances the field is inversely related to the distance from the antenna, so we can construct this table:"
Shouldn't the inverse relation you're referring to be the famous inverse square proportion used for everything from magnetcs to light? I'd assume so, unless there is something very peculiar in the near field at AM BCB frequencies I've never heard of..
In which case the signal strength at the ranges closer then 1km would be considerably higher than the table you provided and the inverse square proportion will result in a curve function, not a linear one. That it would be the inverse square proportion is in accord with your mentions of how signal strength tapers off as distance increases, which I would agree with due to the inverse square proportion invoplved. But it's not linear, it is a curve. That is why it doesn't just drop off to zero field strength, as you pointed out.
Taking your calculation of .15 mv at 1 km for example, the signal strength in mv at .5km would be .6 mv, not .3
And at 1/4 km it would be 16 times greater than at 1 km, or 2.4mv
By the estimated 2mv needed for usable reception, 1/4 km should definitely be more than sufficient for even a cheap indoor radio in an urban setting.
1/4 km = 250 m. 1 m = approx 3.28 ft. So 250 * 3.28 = 820 ft, not 200 ft.
A city block is generally estimated at around 12 blocks to the mile. 5,280 ft in a mile, and one 12th of that would be 440 ft.
Since we were saying 2 mv for a cheap indoor radio, and the field strength at 820 would be 2.4 mv, it seems reasonable that a 2 block radius for clear reception to the typical cheap indoor radio would be realistic. Guesstimating that the strength would be at least near 2 mv for 880 ft if it is 2.4 mv at 820 will be reasonably close, probably at least as close as the estimate of 2 mv for the cheap indoor radio to begin with.
That's pretty consistent with what we hear people talking about here. A range of a couple blocks in every direction where the signal is strong enough to pick it up reasonably with an inexpensive radio, assuming they took some pains to make a pretty good antenna and ground radials for their station.
Now that drops off fast as you move away, since every time you double your distance from the antenna the power is *quartered*, not halved. n = 1 / d^2 where n is the field strength and d is the distance. No magic in that, it's simple sense, since if the horizontal radiation pattern from the antenna is approximately circular, the area of a circle is pi * r^2. So every time you double the radius, the same amount of power is having to cover considerably more area.
For example, if a transmitter has a "range"of mile from the antenna to some sort of useful receiver..1 squared is still 1, therefore 3.14 * 1 = 3.14 square miles of area that it covers. But if it had a range of 2 miles, 2 squared is 4. and 3.14 * 4 would be an area of 12.56 square miles. If someone says their transmitter is getting out 3 miles, that would be 28.26 (3.14 * 9) square miles of effective coverage off a tenth of a watt? I'm not going to say it's impossible, but I'll say I'd sure have to see it for myself before I'd believe it.
That would be quite a lot to expect to do effectively with a transmitter with a tenth of a watt to the final rf stage and much of it going noplace because the very short antenna isn't efficient at all.
Anyway, back to the matter.. It's an inverse square proportion and as such it drops off fast each time you double your distance from the source. However, it is a curve and the taper becomes more gradual with distance. I haven't done the math in the other direction, but considering how a squared function tapers off to be almost parallel to the axis fairly quickly, if it's .15 mv at 1 km, as you calculated, then your original estimate of 1.5865 miles by the other method may agree closer than it seems at first with the calculations you posted in this thread.
The curve tapers off fast at first but quickly becomes quite gradual.
Assuming your .15 at 1km..
Distance in km > field in mv
1 > .15
1/2 > .6
1/4 > 2.4
1/8 > 9.6
1/16 > 38.4
See how it tapers off very fast at first and then much more gradually as the distance increases?
We're not pulling the inverse square proportion out of thin air, either. The ARRL states that "The transmitted field intensity decreases as the inverse square of the distance". It's basic physics, and unless I've made a miscalculation, that would put the 2 mv level at somewhere just past 1/4 km which is 250 m, not 200 ft.
This is consistent with what we hear most people people saying here. Usable reception on almost anything for a couple blocks, after that you pretty much need a good car radio or some other decent quality reciever. I haven't heard anyone here claim that they can pick up their station at 2 miles on a Radio Shack "Flavoradio", for example. But a couple blocks to those Flavoradios and cheap indoor radios would cover an urban neighborhood nicely, and something well over a km to a good car radio or other quality reciever is nothing to sneeze at for 100 milliwatts into the final rf and an antenna way too short to be efficient.
Daniel
Daniel,
Rich can probably answer this better than I but he is correct that the field strength in volts/meter (an intensity) decreases directly with distance.
The inverse square law applies to power which decreases according to the square of distance. In calculating the field strength in volts/meter it is OK to use the linear decrease with distance.
Your reference to the ARRL stating that the intensity follows the inverse square law caught my attention. I opened my ARRL Antenna Handbook and read "In free space the field intensity of the wave decreases directly with the distance from the source.". Maybe your book is wrong?
Neil
Hiya Neil.
I'll admit it's been a lot of years since I was in physics class and as such I looked for a recent source to confirm that I was correct in recalling that the inverse square does apply to radio. The ARRL is usually considered a good source for such things so far as I know, but my copy of the ARRL's Amatuer handbook is from WWII. So I hit google and the reference comes from their pdf of the 1999 second edition of the article "Antenna Height and Communication Effectiveness".
pg 15 (document pg#, not pdf pg#), paragraph 2
In any case, if it is incorrect when it states in the article that "The transmitted field intensity decreases as the inverse square of the distance", then I stand corrected. Maybe the article is wrong, I found it when looking for a reference to refresh my memory on the topic via the search string: ARRL signal distance "inverse square"
Perhaps someone should inform them?
Daniel
Daniel,
I checked the link and searched the text for "inverse square" and found the statement "The transmitted field intensity decreases as the inverse square of the distance.".
Indeed, you did not misquote nor misinterpret what was written. Unfortunately this statement is not correct.
Boy did we pull this thread off the original post topic. Oh well, we are having fun.
Neil
You guys are right about range, when I talk about range I have in mind a car radio or decent portable. If a listener has a $4 clock radio, inside a brick or block building, in NYC where there will be a high noise floor then the Part 15 range for that listener is going to be sad. People have to keep in mind the noise floor for a particular area. I once had a customer that bought a transmitter on trial saying that he had to get one mile range. Come to find out later that there was a 1KW station that could hardly get 2 miles range because the noise floor was so high in his area! People that get long Part 15 ranges are almost always in the country (low noise floor) on a hill.
Daniel wrote: "In any case, if it is incorrect when it states in the article that "The transmitted field intensity decreases as the inverse square of the distance", then I stand corrected. Maybe the article is wrong, I found it when looking for a reference to refresh my memory on the topic via the search string: ARRL signal distance "inverse square." Perhaps someone should inform them?"
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Daniel,
Neil has already posted an answer to your ARRL quote, but I thought I'd do a Google search for something authoritative to support the fact that field strength is related to inverse distance, only (not the inverse square). I was amazed when I got to the Wikipedia entry for RADIO PROPAGATION that stated exactly what that ARRL document did.
So I edited the Wikipedia entry, and taking the cue from you I also sent an email to ARRL about it.
Here's part of what I wrote on Wikipedia:
Doubling the distance from a transmitter means that the power density of the radiated wave at that new location is reduced to one-quarter of its previous value.
The far-field magnitudes of the electric and magnetic field components of electromagnetic radiation are equal, and their field strengths are inversely proportional to distance. Doubling the propagation path distance from the transmitter reduces their received field strengths by one-half. The reduction of each of these fields by one-half is the result of the power density reduction to one-quarter, over that doubled path length.
As far as something authoritative, below is the equation to calculate field strength (not power) radiated by a 1/2-wave dipole, taken from REFERENCE DATA FOR RADIO ENGINEERS, 6th edition, page 27-7:
E = SQRT(49.2*P)/ R
where P is radiated power in watts, and R is distance in meters
The "R" term is not squared.
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I'd gotten the basic idea from what Neil had replied, but the formula from Rich clears up the confusion perfectly.
Thanks and a tip of the hat to both of you, gentlemen.
Daniel
Hamilton wrote: "The talking house ATU connects to the inside unit with up to 100 feet of coax, outside braid GROUNDED! ...I would guess that the longer the cables, the more your radiation from that unit."
Note that r-f power can be conducted from one place to another via a coaxial cable with its shield grounded, and still radiate essentially none of that power directly via that coax. This is the common configuration for AM, FM and TV broadcast antenna systems, where the tx is in a building near the tower(s), yet the r-f power entering the coax also appears at its termination (less I^2R loss in the coax), where it is radiated by the antenna.
But this is not the case for the exposed conductors used with the AM1000 and similar Part 15 AM txs to connect the r-f output to the antenna or loading coil. Such exposed conductors do radiate, and that is why they must be included in the total radiating length of the antenna, along with the complete conducting path to a true r-f ground reference.
People just need to use common sense, things can be ground, not just the dirt. This is standard engineering.
Standard r-f engineering recognizes that an r-f ground reference does not radiate. In fact that is the chief characteristic of an r-f ground when used as a reference for vertical monopole radiators. This "common sense" needs to defer to the reality of the physics involved here.
Elevated (whip/mast) Part 15 AM installations connecting their "ground lead" to a "massive ground wire," flagpole, mast, billboard etc are not connecting that ground lead to a non-radiating reference point. Those added conducting paths attached to the ground lead also radiate, and at the top tip of that conductor radiates as much as the short ground lead itself at the attachment point. Certainly that does not qualify those attachment points as r-f grounds. If they were, they would not radiate.
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