Someone is asking on another part of radio-info. what is the "ERP" of certain directional AM stations? I have been looking for a formula/s to calculate this. Does someone know how the equivalent power in the main lobe of a directional AM station compares to an omnidirectional AM station, given the same tower height, ground system, etc. for both? I remember reading several years ago that someone said that WMVP AM1000 in Chicago had an ERP of about 200kW in the main lobe at some set distance such as 1kM, but no calculations were shown to back up that claim. Does anyone know how to calculate the "ERP" of an directional AM station? Thanks!
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"ERP" of Directional AM stations
- Thread starter stormy01
- Start date
In international agreements, the reference is somewhat ambiguously stated as 300 mV/m for 1 kW @ 1 km. This corresponds to a little less than 1/4 wave efficiency.
Since many older licensed AM stations operate at less than this efficiency, many at reduced input power, even nondirectional stations, a more consistent reference might be Class B minimum efficiency of 282 mV/m for 1 kW @ 1 km for all classes. This is consistent with the late great Ed Buterbaugh's assessment of CKLW's maxima of "about 300 kW".
Take the maximum field of the DA array, or at any azimuth, divide it by the reference efficiency, and square it. Thus if the inverse field is 2820 mV/m @ 1 km, and the reference is 282, (2820/282)^2=100 kW.
Isotropic (173.2 mV/m for 1 kW @ 1 km) ERPs (iERP) are more impressive, but don't mean much. Half wave dipole efficiency is 222 mV/m for 1 kW @ 1 km.
Since many older licensed AM stations operate at less than this efficiency, many at reduced input power, even nondirectional stations, a more consistent reference might be Class B minimum efficiency of 282 mV/m for 1 kW @ 1 km for all classes. This is consistent with the late great Ed Buterbaugh's assessment of CKLW's maxima of "about 300 kW".
Take the maximum field of the DA array, or at any azimuth, divide it by the reference efficiency, and square it. Thus if the inverse field is 2820 mV/m @ 1 km, and the reference is 282, (2820/282)^2=100 kW.
Isotropic (173.2 mV/m for 1 kW @ 1 km) ERPs (iERP) are more impressive, but don't mean much. Half wave dipole efficiency is 222 mV/m for 1 kW @ 1 km.
Before you quote "AM ERPs" though, realize that it is a nebulous concept not officially defined by the FCC. It is in some international agreements I have read, and plenty of AM engineers talk about the concept. But one thing I would caution against that I have seen repeatedly online is multiplying the number of towers by the input power. It would depend on much more. There is something called pattern gain and loss. If you have an efficient two tower pattern, with slightly above 1/4 wavelength height, by the time you take standard patterns (another debatable concept) into effect, you might end up with a standard pattern maximum that is three times the input power. And many three tower in line arrays are designed so that the center tower acts like two towers. In many cases, if you used the minimum Class B efficiency, it will come out about five times the input for a three tower in line. This is consistent with the maximum theoretical gain of a three element ham band antenna like for 15 meters, often quoted as 7.5 dB, but those usually use parasitics, not three driven elements.
Realize also that with antifade nondirectional antennas between 0.5 and 0.625 wavelength, they may have twice the "ERP" of a less than 1/4 wave minimum height for class antenna. Thus a Class A nondirectional 50 kW with a 195 degree tower already has an "ERP" of close to 100 kW.
Realize also that with antifade nondirectional antennas between 0.5 and 0.625 wavelength, they may have twice the "ERP" of a less than 1/4 wave minimum height for class antenna. Thus a Class A nondirectional 50 kW with a 195 degree tower already has an "ERP" of close to 100 kW.
Schroedingers Cat said:Isotropic (173.2 mV/m for 1 kW @ 1 km) ERPs (iERP) are more impressive, but don't mean much. Half wave dipole efficiency is 222 mV/m for 1 kW @ 1 km.
I never heard those numbers. Do you maybe mean @ 1 MILE rather than @ 1 kM? Your numbers sound awfully close to the minimum inverse-distance fields at 1 mile for Class B and Class A AMs. I am, of course, familiar with the Class B-minimum inverse-distance field of of 281.7 mV/m/kW @ 1 km. Requires a tower of ~54 degrees, assuming a "standard" ground of 120 1/4-wave radials. If you extend the discussion to minimum-efficiency Class C radiators, the number drops to 241 mV/m/kW @ 1 km. This requires a tower shorter than 50 degrees, I believe, but how much shorter, I don't know. If you compare a 225-degree non-sectionalized radiator with a full ground system (440 mV/m/kW @ 1 km) with a minimally efficient Class C radiator (241 mV/m/kW @ 1 km), you discover that the "ERP" of a ND 225-degree radiator is 3.33 times that of the minimally efficient Class C radiator!
Now add the effect of a high-gain DA (six towers or more) with a single lobe. Such patterns can rather easily have a radiation maximum 3.16 times that of an ND radiator of the same height. The effect, then, could theoretically be a 33.3-fold difference in ERP with the same antenna-input power(!)
222 number is right out of Reitz and Milford E & M text. 222/1.60937= 138 mV/m at 1 mile for a dipole.
173.2/1.60937=107.6 mV/m at 1 mile.
If you radiate 1 kW into a sphere of 1 km radius, and use the conversion of 120 PI=377 ohms to convert from power density to electromagnetic field strength, it comes out. This conversion factor is known as the characteristic impedance of free space.
Believe me, I reviewed it before posting.
173.2/1.60937=107.6 mV/m at 1 mile.
If you radiate 1 kW into a sphere of 1 km radius, and use the conversion of 120 PI=377 ohms to convert from power density to electromagnetic field strength, it comes out. This conversion factor is known as the characteristic impedance of free space.
Believe me, I reviewed it before posting.
Schroedingers Cat said:Believe me, I reviewed it before posting.
But if you review your original post AGAIN, you will see that you wrote 1 km and not 1 mile. I was confused by your numbers because they seemed to be correct for 1 mile and not for 1 km. Now you seem to be saying that that is indeed the case. No big deal. But when you make a mistake and catch it, or, as in this case, when someone else catches it, a post acknowledging your error would certainly seem appropriate--just to clear up the confusion. A post suggesting that your original posting was right seems inappropriate. At least one of us is confused about whether your formulas apply to the inverse-distance field at 1 km or 1 mile. I think your numbers are for 1 mile. If so, it seems appropriate for you to acknowlege your error, rather than insisting that you were right.
Dan, I'm sorry if you interpreted it that way. The original post did say 1 km. If I made a mistake, I corrected that immediately, within 5 minutes if I did. I know that you are extremely knowlegeable on these matters. Please don't interpret defending my position as arrogance. Like you, I was long ago exposed to this data expressed in miles, not km, and sometimes I do get confused and post English units. When most of the country's roads are laid out in miles, it is more logical for me to remember data in miles and convert, even though my scientific background is virtually all MKS units.
One mistake I do acknowledge is that the reference should be from "Electromagnetic Fields And Waves", Second Edition by Lorrain and Corson, not Reitz and Milford. On page 636, Problem 14-8 reads:
"Calculate the electric field intensity in millivolts per meter at a distance of one kilometer in the equatorial plane of a half-wave" [dipole] "antenna radiating one kilowatt of power." Problem 14-7 makes it clear that it is a dipole, not a half wave ground plane.
The answer is in the back of the book, on page 697, "222 millvolts /meter".
I can't find the problem in either book (L&C or R&M) for the isotropic source, but I do remember doing the problem, either in the course, or on my own to convince myself of the result for an isotropic source.
One mistake I do acknowledge is that the reference should be from "Electromagnetic Fields And Waves", Second Edition by Lorrain and Corson, not Reitz and Milford. On page 636, Problem 14-8 reads:
"Calculate the electric field intensity in millivolts per meter at a distance of one kilometer in the equatorial plane of a half-wave" [dipole] "antenna radiating one kilowatt of power." Problem 14-7 makes it clear that it is a dipole, not a half wave ground plane.
The answer is in the back of the book, on page 697, "222 millvolts /meter".
I can't find the problem in either book (L&C or R&M) for the isotropic source, but I do remember doing the problem, either in the course, or on my own to convince myself of the result for an isotropic source.
Read the comment by Ermi Roos for another reference to the 173.2 figure.
http://part15.us/node/1639
He refers to these as being from the Antenna reference book by Kraus.
http://part15.us/node/1639
He refers to these as being from the Antenna reference book by Kraus.
A simple and useful equation for calculating the far-field field intensity in the directions of maximum radiation from a center-fed, 1/2-wave dipole in free space is:
F = SQRT(49.2 * P) / D
where F is the field intensity in volts/meter, P is the radiated power in watts, and D is the distance from the dipole in meters.
So the field intensity 1 km away from this dipole when radiating 1 kW is 0.2218 V/m, or about 222 mV/m.
The maximum gain of the above dipole is 1.46 X that of an isotropic radiator (or 2.15 dB).
So the field intensity at 1 km for 1 kW radiated isotropically will be 222 mV/m / 1.46 = 152 mV/m.
The apparent, far-field gains of these antennas when mounted at/near the surface of a perfect ground plane is about 3 dB higher, as all radiation then is confined to one hemisphere.
A 1/4-wave vertical monopole, base-driven against a perfect ground plane has the same intrinsic gain in the horizontal plane as a vertical 1/2-wave dipole in free space. This is due to the fact that the feedpoint Z of such a monopole is 1/2 that of the free-space dipole -- so for equal applied power, twice the currrent will flow on the monopole as on each arm of the dipole.
The reflection from the ground plane of the radiation "launched" by the monopole adds 3 dB to the original radiated value.
So the net, apparent gain of a 1/4-wave monopole + perfect ground system is 3 dB higher than the free-space dipole.
A gain of 3 dB is a multiplier of 1.414, so the theoretical maximum field from a 1/4-wave monopole driven against a perfect ground plane is 222 mV/m x 1.414 = 314 mV/m (approx).
The fact that the radial ground systems used with MW broadcast antennas typically have an ohm or two of r-f resistance accounts for the difference in radiated fields (or the FCC "efficiency") between the theoretical and the practical cases.
The above analysis can be confirmed by reference to standard antenna engineering textbooks and papers written by Brown, Kraus, Balanis, Johnson & Jasik, etc.
//
F = SQRT(49.2 * P) / D
where F is the field intensity in volts/meter, P is the radiated power in watts, and D is the distance from the dipole in meters.
So the field intensity 1 km away from this dipole when radiating 1 kW is 0.2218 V/m, or about 222 mV/m.
The maximum gain of the above dipole is 1.46 X that of an isotropic radiator (or 2.15 dB).
So the field intensity at 1 km for 1 kW radiated isotropically will be 222 mV/m / 1.46 = 152 mV/m.
The apparent, far-field gains of these antennas when mounted at/near the surface of a perfect ground plane is about 3 dB higher, as all radiation then is confined to one hemisphere.
A 1/4-wave vertical monopole, base-driven against a perfect ground plane has the same intrinsic gain in the horizontal plane as a vertical 1/2-wave dipole in free space. This is due to the fact that the feedpoint Z of such a monopole is 1/2 that of the free-space dipole -- so for equal applied power, twice the currrent will flow on the monopole as on each arm of the dipole.
The reflection from the ground plane of the radiation "launched" by the monopole adds 3 dB to the original radiated value.
So the net, apparent gain of a 1/4-wave monopole + perfect ground system is 3 dB higher than the free-space dipole.
A gain of 3 dB is a multiplier of 1.414, so the theoretical maximum field from a 1/4-wave monopole driven against a perfect ground plane is 222 mV/m x 1.414 = 314 mV/m (approx).
The fact that the radial ground systems used with MW broadcast antennas typically have an ohm or two of r-f resistance accounts for the difference in radiated fields (or the FCC "efficiency") between the theoretical and the practical cases.
The above analysis can be confirmed by reference to standard antenna engineering textbooks and papers written by Brown, Kraus, Balanis, Johnson & Jasik, etc.
//
Schroedingers Cat said:Dan, I'm sorry if you interpreted it that way. The original post did say 1 km. If I made a mistake, I corrected that immediately, within 5 minutes if I did.
Oh, my goodness! I followed your link to the thread by Roos and others. All I can say is that it gets into theoretical constructs to which I have never given much--if any--thought. Apparently, the similarity between 173.2 mV/m/kW @ 1 km of your original post and the thread to which you linked and the FCC's old Class II/Class III minimum efficiency of 175 mV/m/kW @ 1 MILE is pure coincidence. So too, is the similarity between your 222 (or 221.5) mV/m/kW @ 1 km and the FCC's old Class I minimum of 225 mV/m/kW @ 1 MILE.
The old Class II/III number was for a short (~54 degree) monopole over a "nearly" ideal 1/4-wave-radius ground plane (of unspecified resistance). As I understand it, the value of the ground resistance is quite important to the accuracy of the FCC curves of antenna efficiency vs tower height. The accuracy of those curves for antennas shorter than ~80 degrees has been challenged several times, but Media Bureau Engineering has simply ignored the challenges.
I don't think there is much argument about the FCC's 225 mV/m/kW @ 1 MILE Class I minimum efficiency because the ground resistance becomes less important as the monopole height increases. (If I'm not mistaken, it becomes completely UNimportant at a monopole hight of 180 degrees--and ONLY at EXACTLY 180 degrees.) The 225 mV/m/kW @ 1 MILE number applies at ~165 degrees.
It's all a little much for my poor addled brain, especially at this early hour.
DanStrassberg said:As I understand it, the value of the ground resistance is quite important to the accuracy of the FCC curves of antenna efficiency vs tower height. The accuracy of those curves for antennas shorter than ~80 degrees has been challenged several times, but Media Bureau Engineering has simply ignored the challenges.
To explore this a bit I did a quick NEC workup for 1 MHz monopoles of various free-space electrical heights, all with an OD of 42 inches, a ground loss of two ohms, and neglecting any matching network losses. A perfect ground plane was assumed.
Radiation Groundwave Field Intensity at 1 km
Height, degrees Resistance, ohms for 1 kW radiated power (mV/m)
90 44 308
80 31 302
70 21 295
60 14 286
50 9 275
The 90-degree antenna system in this scenario has about 25% more ERP than the 50-degree system (308/275)^2.
//
To add another layer to the confusion, electromagnetic texts show the radiation resistance of a perfect resonant dipole as 73.2 ohms, and that of a perfect resonant 1/4 wave ground plane, as half that, or 36.6 ohms, as it radiates into a hemisphere and not a sphere. The radiation resistance is where the inductive and capacitive reactance exactly cancel, another way of saying resonant.
I have seen texts that say that the actual height of a resonant monopole or dipole is less than the calculated height based on the speed of light, because the velocity is somewhat less in a real material. With dipoles, you usually see quoted the velocity of about 0.95 times C.
A friend of mine was making antenna bays from scratch and asked me about calculated dimensions. I told him to make the first bay at the full speed of light and then cut it until he got to resonance. It's a good thing because he was going to start out by cutting it to 95% of the speed of light. For whatever reason, it ended up resonant at about 98% of the speed of light.
I have seen texts that say that the actual height of a resonant monopole or dipole is less than the calculated height based on the speed of light, because the velocity is somewhat less in a real material. With dipoles, you usually see quoted the velocity of about 0.95 times C.
A friend of mine was making antenna bays from scratch and asked me about calculated dimensions. I told him to make the first bay at the full speed of light and then cut it until he got to resonance. It's a good thing because he was going to start out by cutting it to 95% of the speed of light. For whatever reason, it ended up resonant at about 98% of the speed of light.
Mr. Fry, for a real monopole with ground loss, what would the radiation resistance at resonance be? Would the ground loss add to the theoretical 36.6 ohms, or would there be other factors? I see you would hit 36.6 around 84 degrees by linear interpolation.
Schroedingers Cat said:To add another layer to the confusion, electromagnetic texts show the radiation resistance of a perfect resonant dipole as 73.2 ohms, and that of a perfect resonant 1/4 wave ground plane, as half that, or 36.6 ohms, as it radiates into a hemisphere and not a sphere. The radiation resistance is where the inductive and capacitive reactance exactly cancel, another way of saying resonant.
I guess I'd refine that last sentence a bit, because radiation resistance exists whether or not a radiator is naturally resonant (Xl = Xc).
The radiation resistance of a monopole depends not only on its physical height in free-space wavelengths, but its on its width and outline profile as well. Together these factors determine its electrical height/length in degrees, which value results in a certain input Z for a given frequency.
So two towers of equal physical height but different widths will have different radiation resistances on a given frequency.
Kraus shows in Antennas, Vol 3 (p. 182) that the impedance of a thin-wire, center-fed, 1/2-wave dipole in free space is 73 +j42.5 ohms. The dipole length must be shortened by several percent to reduce the 42.5 ohm reactive term to zero. As radiation resistance is a function of the electrical length of such a dipole, at the naturally resonant length near 1/2-wavelength the radiation resistance falls to about 65 ohms. The feedpoint Z of that first, naturally resonant condition then is ~65 +/-j0 ohms.
Radiators having larger ratios of height to width than in this thin-wire case (such as a tower) need to be shortened by a greater percentage of their height in order to become naturally resonant. This also may reduce the radiation resistance at natural resonance by a greater percentage than in the thin-wire case.
In my table of 5 monopole tower heights I ignored the reactive term calculated by NEC. None of those heights was naturally resonant for the conditions I used, yet they all had some value of radiation resistance. In practice the reactive term is canceled by a network at the feedpoint (whose loss I ignored in my calculations).
Schroedingers Cat said:Mr. Fry, for a real monopole with ground loss, what would the radiation resistance at resonance be? Would the ground loss add to the theoretical 36.6 ohms, or would there be other factors?
That will depend on the electrical length of the monopole, which depends on the factors I outlined above in this post.
The radiation resistance is only the real part of the feedpoint impedance related to the coupling of the antenna with space. A measurement made at that point will include any other real series resistances present in the antenna system, such as incurred in the matching network and r-f ground.
The radiation resistances I showed in my table are calculated for the monopole only, and excluded the r-f ground and other loss(es).
//
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