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I have been informed my GRNDWAV3 program is in error - it calculates
the power input to a matched receiver to be 6dB greater than it ought to be, or exactly 4 times the correct power input. Most of my programs calculate results based on what I consider to be fundamental reasoning. But GRNDWAV3 is one of the few where I have adapted formulae from the text books or 'bibles'. My informant is an Icelandic amateur who appears to know what he is talking about and is mathematically very convincing. For various resons, for the time being I propose to leave him out of this discussion. That is, of course, if a discussion should evolve. The problem fundamentally revolves around the gain of short vertical antennas, both transmitting and receiving, above a perfect ground, relative to isotropic. But for present purposes what an isotrope actually is can be forgotten about. It exists only in one's imagination. Numbers cannot be avoided. So let's keep them as simple as possible by starting with the MF standard of 1 Kilowatt, radiated from a short vertical antenna above a perfect ground. Actual antenna height and frequency don't matter. According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts, which (according to the text books) is correctly calculated by my program. To calculate matched reciever input power from field strength it is necessary to state vertical antenna height, frequency and radiation resistance. Again choosing simple values - Antenna height = 1 metre. Frequency = 20 MHz. Calculated radiation resistance = 1.758 ohms. Matched receiver input resistance is also 1.758 ohms. According to requirements antenna height is short compared with a wavelength. I am confident that radiation resistance is correct at 20 MHz for a 1 metre vertical. Antenna reactance is tuned out and disappears from the argument. So we have a simple circuit consisting of a generator with a resistive load of the same value, both equal to 1.758 ohms. According to the text books (as confirmed by Roy) the generator voltage is 300 millivolts. (A 1 metre high antenna with a field strength of 300 mV per metre.) The power available to the receiver is therefore - Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts. Which is the value calculated by my program although it does it in a different way by not involving field strength. It calculates it more directly from the 1 kW transmitter power and the antenna gains of a pair of vertical Tx and Rx antennas relative to isotropic. Nevertheless, I think my informant may be correct. That indeed my program states receiver power input to be 4 times greater than what it actually is. IMPORTANTLY, he says an NEC numerical program confirms his own calculations. NEC programs are not dependent on what a program user's ideas may be about antenna gains relative to isotropic. They calculate directly from fundamental metre-amps and volts. I am presently out of touch with my informant. I do not know which NEC program confirms his calculations. I have recently asked Roy what is the voltage measured between the bottom end of a 1 metre long vertical antenna and ground when the field strength is 1 volt per metre. He says it is 1 volt and no doubt the Bibles agree. It is intriging, if the value should be only 0.5 volts then my program would give the (suspected) correct answer to the simple question - "What is the power input to a matched receiver using a 1 metre vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw transmitter also using a short vertical antenna?" Short is less than 1/4-wavelength. Is it 12.8 milliwatts, or is it 3.2 milliwatts? Is there an NEC numerical program which will do the job? If there is perhaps somebody could use it. Most important, do I have to correct the program bug for the sake of 6 dB when the calculating uncertainty at long distances is plus or minus 10 or 15 dB ? ---- Reg, G4FGQ |
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Reg, G4FGQ wrote:
"He says it is 1 volt and no doubt the bibles agree." OK. You have an antenna with some radiation resistance and a lossless conjugate match to a load. 1/2 the antenna voltage which equals the volts per meter field strength in your 1-meter wire, is dropped across the radiation resistance and the other 0.5 volt appears across the receiver load. The radiation resistance of the antenna becomes the Thevenin equivalent source resistance of the generator feeding the receiver load.. The power lost to reradiation is 0.5 volt times the current in the radiation resistance. The power delivered to the matched receiver load is exactly the same. Best regards, Richard Harrison, KB5WZI |
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Reg Edwards wrote:
I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought to be, or exactly 4 times the correct power input. Most of my programs calculate results based on what I consider to be fundamental reasoning. But GRNDWAV3 is one of the few where I have adapted formulae from the text books or 'bibles'. My informant is an Icelandic amateur who appears to know what he is talking about and is mathematically very convincing. For various resons, for the time being I propose to leave him out of this discussion. That is, of course, if a discussion should evolve. The problem fundamentally revolves around the gain of short vertical antennas, both transmitting and receiving, above a perfect ground, relative to isotropic. But for present purposes what an isotrope actually is can be forgotten about. It exists only in one's imagination. Numbers cannot be avoided. So let's keep them as simple as possible by starting with the MF standard of 1 Kilowatt, radiated from a short vertical antenna above a perfect ground. Actual antenna height and frequency don't matter. According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts, which (according to the text books) is correctly calculated by my program. To calculate matched reciever input power from field strength it is necessary to state vertical antenna height, frequency and radiation resistance. Again choosing simple values - Antenna height = 1 metre. Frequency = 20 MHz. Calculated radiation resistance = 1.758 ohms. Matched receiver input resistance is also 1.758 ohms. According to requirements antenna height is short compared with a wavelength. I am confident that radiation resistance is correct at 20 MHz for a 1 metre vertical. Antenna reactance is tuned out and disappears from the argument. So we have a simple circuit consisting of a generator with a resistive load of the same value, both equal to 1.758 ohms. According to the text books (as confirmed by Roy) the generator voltage is 300 millivolts. (A 1 metre high antenna with a field strength of 300 mV per metre.) The power available to the receiver is therefore - Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts. Which is the value calculated by my program although it does it in a different way by not involving field strength. It calculates it more directly from the 1 kW transmitter power and the antenna gains of a pair of vertical Tx and Rx antennas relative to isotropic. Nevertheless, I think my informant may be correct. That indeed my program states receiver power input to be 4 times greater than what it actually is. IMPORTANTLY, he says an NEC numerical program confirms his own calculations. NEC programs are not dependent on what a program user's ideas may be about antenna gains relative to isotropic. They calculate directly from fundamental metre-amps and volts. I am presently out of touch with my informant. I do not know which NEC program confirms his calculations. I have recently asked Roy what is the voltage measured between the bottom end of a 1 metre long vertical antenna and ground when the field strength is 1 volt per metre. He says it is 1 volt and no doubt the Bibles agree. It is intriging, if the value should be only 0.5 volts then my program would give the (suspected) correct answer to the simple question - "What is the power input to a matched receiver using a 1 metre vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw transmitter also using a short vertical antenna?" Short is less than 1/4-wavelength. Is it 12.8 milliwatts, or is it 3.2 milliwatts? Is there an NEC numerical program which will do the job? If there is perhaps somebody could use it. Most important, do I have to correct the program bug for the sake of 6 dB when the calculating uncertainty at long distances is plus or minus 10 or 15 dB ? ---- Reg, G4FGQ How was your Icelandic amateur doing his NEC calculations? If he calculated the voltage that would generate 1kW into a load, then excited his transmit antenna with a matched generator of that source impedance -- which would drop the power by a factor of four. If he did that and you did your calculations with 1kW going _to_ the transmit antenna that would be your source of error. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com |
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"Tim Wescott" wrote in message ... Reg Edwards wrote: I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought to be, or exactly 4 times the correct power input. Most of my programs calculate results based on what I consider to be fundamental reasoning. But GRNDWAV3 is one of the few where I have adapted formulae from the text books or 'bibles'. My informant is an Icelandic amateur who appears to know what he is talking about and is mathematically very convincing. For various resons, for the time being I propose to leave him out of this discussion. That is, of course, if a discussion should evolve. The problem fundamentally revolves around the gain of short vertical antennas, both transmitting and receiving, above a perfect ground, relative to isotropic. But for present purposes what an isotrope actually is can be forgotten about. It exists only in one's imagination. Numbers cannot be avoided. So let's keep them as simple as possible by starting with the MF standard of 1 Kilowatt, radiated from a short vertical antenna above a perfect ground. Actual antenna height and frequency don't matter. According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts, which (according to the text books) is correctly calculated by my program. To calculate matched reciever input power from field strength it is necessary to state vertical antenna height, frequency and radiation resistance. Again choosing simple values - Antenna height = 1 metre. Frequency = 20 MHz. Calculated radiation resistance = 1.758 ohms. Matched receiver input resistance is also 1.758 ohms. According to requirements antenna height is short compared with a wavelength. I am confident that radiation resistance is correct at 20 MHz for a 1 metre vertical. Antenna reactance is tuned out and disappears from the argument. So we have a simple circuit consisting of a generator with a resistive load of the same value, both equal to 1.758 ohms. According to the text books (as confirmed by Roy) the generator voltage is 300 millivolts. (A 1 metre high antenna with a field strength of 300 mV per metre.) The power available to the receiver is therefore - Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts. Which is the value calculated by my program although it does it in a different way by not involving field strength. It calculates it more directly from the 1 kW transmitter power and the antenna gains of a pair of vertical Tx and Rx antennas relative to isotropic. Nevertheless, I think my informant may be correct. That indeed my program states receiver power input to be 4 times greater than what it actually is. IMPORTANTLY, he says an NEC numerical program confirms his own calculations. NEC programs are not dependent on what a program user's ideas may be about antenna gains relative to isotropic. They calculate directly from fundamental metre-amps and volts. I am presently out of touch with my informant. I do not know which NEC program confirms his calculations. I have recently asked Roy what is the voltage measured between the bottom end of a 1 metre long vertical antenna and ground when the field strength is 1 volt per metre. He says it is 1 volt and no doubt the Bibles agree. It is intriging, if the value should be only 0.5 volts then my program would give the (suspected) correct answer to the simple uestion - "What is the power input to a matched receiver using a 1 metre vertical antenna, at 20 MHz, at a distance of 1 Km from a 1Kw transmitter also using a short vertical antenna?" Short is less than 1/4-wavelength. Is it 12.8 milliwatts, or is it 3.2 milliwatts? Is there an NEC numerical program which will do the job? If there is perhaps somebody could use it. Most important, do I have to correct the program bug for the sake of 6 dB when the calculating uncertainty at long distances is plus or minus 10 or 15 dB ? ---- Reg, G4FGQ How was your Icelandic amateur doing his NEC calculations? If he calculated the voltage that would generate 1kW into a load, then excited his transmit antenna with a matched generator of that source impedance -- which would drop the power by a factor of four. If he did that and you did your calculations with 1kW going _to_ the transmit antenna that would be your source of error. ===================================== Tim, I think the error, if there is one, is most likely at the receiving end. The RADIATED power is 1000 watts. It doesn't matter how it got into the ether from the transmitter. But you have made me think again about what should be done with transmitting antenna gain. ---- Reg. |
Back to fundamentals
On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards"
wrote: The problem fundamentally revolves around the gain of short vertical antennas, both transmitting and receiving, Hi Reggie, This is not what you proposed the first time through this exercise. As your original was stated in terms of a field of 1V/M there is no need to elaborate on about the radiator. Numbers cannot be avoided. So let's keep them as simple as possible by starting with the MF standard of 1 Kilowatt, radiated from a short vertical antenna above a perfect ground. Actual antenna height and frequency don't matter. Then its mention at being both MF (MW?) standard, and being a short vertical is in distinct conflict. Again, there is nothing inherent to the problem short of demonstrable evidence in the field, which returns us to the classic FCC ground wave charts for MF (MW?). According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts, which (according to the text books) is correctly calculated by my program. Are you claiming your program computes field V/M or power delivered by a field of V/M? To calculate matched reciever input power from field strength Here again is a new specification to your original query. Matched? I would have suspected so, but being unmatched or being mismatched brings a spectrum of answers that span to more than 1V/M down to microV/M. it is necessary to state vertical antenna height, frequency and radiation resistance. Again choosing simple values - Antenna height = 1 metre. Frequency = 20 MHz. Calculated radiation resistance = 1.758 ohms. Matched receiver input resistance is also 1.758 ohms. Even more elaboration that goes beyond the original. I suppose for the purposes of discussion it works, but it goes well beyond the premise of classic MF (MW?) work. According to requirements antenna height is short compared with a wavelength. I am confident that radiation resistance is correct at 20 MHz for a 1 metre vertical. When we have theoretical work supported by real field data in the MW band, why take up a frequency outside that? Antenna reactance is tuned out and disappears from the argument. This is to be expected, but was never actually expressed anywhere. So we have a simple circuit consisting of a generator with a resistive load of the same value, both equal to 1.758 ohms. The generator's Z is immaterial to the discussion if you have a defined field. According to the text books (as confirmed by Roy) the generator voltage is 300 millivolts. (A 1 metre high antenna with a field strength of 300 mV per metre.) I have obtained comparable (though not exact) results for real (not perfect) grounds to compare against actual (not theoretical) data. Those not comparable have come from mismatched 1M high antennas against various loads (already discussed above). The power available to the receiver is therefore - Pr = Square( 0.3/2 ) divided by 1.758 = 12.8 milliwatts. Where'd the 2 of the ( 0.3/2 ) come from? Which is the value calculated by my program Nevertheless, I think my informant may be correct. It is quite simple to fool experts. They are by far easier targets than the naive who ask for the missing work (like this fool here who wonders just what your correspondent offered to convince you). Is it 12.8 milliwatts, or is it 3.2 milliwatts? EZNEC computes it at EČ/R 73's Richard Clark, KB7QHC |
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Hi Richard,
You must have got out of bed the wrong side this morning. ;o) ---- Yours, Punchinello |
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Sorry, I forgot "per metre".
I should have said - "According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts per metre." ---- Reg. |
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The answer is 3 mW.
Any version of EZNEC can be used to do this calculation. The demo program will yield slightly less accurate results because of the limited number of segments(*). I modeled two vertical wires, 1 meter high and 1 mm diameter, spaced 1 km apart, at 20 MHz, over perfect ground. The reported feedpoint impedance varies with segmentation, from 1.988 - j952.3 ohms at 10 segments/wire to 1.72 - j882 ohms at 100 segments/wire. Accuracy is likely to degrade with a larger number of segments, since even 100 results in segment length/diameter ratio less than NEC recommendations. I used 100 segments/wire for the test. One of the choices in EZNEC of far field strength reporting is in V/m at 1 kW input and 1 km distance. For this antenna, EZNEC reports 300.8 V/m (RMS) at ground level. EZNEC also permits setting a fixed power input, so this was set to 1 kW. The resulting source voltage and current are 21270 V. and 24.12 A. respectively. A load of 1.72 + j882 ohms was placed at the base of the second vertical. EZNEC reports a power of 3.234 mW being dissipated in this load. Care has to be used when analyzing the current induced in one antenna by another which is distant using numerical calculations. Errors can occur due to truncation and other causes when the ratio of distances between the two antennas is great relative to the segment lengths or to segment distances within one of the antennas. However, EZNEC gets virtually identical results when using mixed and double precision NEC-2 calculating engines, which indicates that the limit hasn't been reached and that numerical problems aren't occurring. (Another check which can be done is to reduce the distance between antennas by a factor of two. The power in the load resistance should increase by a factor of four.) Another critical matter is the setting of the load reactance. The reactance is many times larger than the resistance, so a slight error in setting its value will result in a large difference in load current and therefore load dissipation. For example, if the segmentation is changed from 100 to 50 segments/wire and no other change is made to the model, the reported load power becomes 0.3917 watts. The reason is that the reported source impedance is now 1.756 - j891.4 ohms, while the load is still 1.72 + j882 ohms. Changing the load to the proper conjugately matched value of 1.756 + 891.4 ohms returns the load power to the correct value of 3.24 mW. All given, I'd trust the reported load power to be easily within 10% of the theoretically correct value. (*) Results for 10 segments/wire are 1.988 - j952.3 ohms for the source impedance, 300.71 V/m field strength at 1 km for 1 kW, and 3.24 mW in a conjugately matched load impedance in the distant vertical. Roy Lewallen, W7EL |
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On Fri, 9 Dec 2005 00:04:38 +0000 (UTC), "Reg Edwards"
wrote: You must have got out of bed the wrong side this morning. ;o) Ah hah, Another troll using the Keflavik proxy, eh wot? Yes, we should all beware a straight question from a crooked source. 73's Richard Clark, KB7QHC |
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Roy Lewallen wrote:
. . . Another critical matter is the setting of the load reactance. The reactance is many times larger than the resistance, so a slight error in setting its value will result in a large difference in load current and therefore load dissipation. For example, if the segmentation is changed from 100 to 50 segments/wire and no other change is made to the model, the reported load power becomes 0.3917 watts. . . . Correction: That should be 0.3917 mW. The error doesn't alter the conclusion. Roy Lewallen, W7EL |
The factor of 2
Roy,
My program requires antenna gains relative to isotropic to be entered. To help discover where I was going wrong a numerical type program was needed. Numerical programs do not need the intervention of fallible human ideas about isotropes, mirror images and ground reflections. I did not realise that EZNEC has the ability to calculate voltages and currents induced in elements miles away from the radiating element. But having the correct answer, I still have a problem. Praps you can help me to solve it. Staying with the same example of - Frequency = 20 MHz. Tx power = 1000 watts. Distance = 1 kilometre. Rx antenna height = 1 metre. Rx antenna Rin = 1.944 ohms, including wire resistance. Rx antenna -jXin is not needed. Field strength = 300 millivolts per metre. According to You, Terman and other Bibles, Volts induced in the 1 metre high antenna = 300 millivolts. So we have a generator with open-circuit volts of 300 mV, with an internal resistance of 1.944 ohms, with an Rx load resistance also of 1.944 ohms (which is in excellent agreement with EZNEC). From which, power generated in the receiver = 11.6 milliwatts BUT THIS IS SIX DB GREATER THAN THAT CALCULATED BY EZNEC. From other considerations, and taking EZNEC's small errors into account, it is EXACTLY 6.02 dB too large. THE CALCULATION WOULD BE CORRECT IF THE VOLTAGE INDUCED IN THE RECEIVING ANTENNA WAS EXACTLY HALF OF THE FIELD STRENGTH. OR THE FIELD STRENGTH FROM THE 1KW TRANSMITTER WAS EXACTLY HALF OF THE BIBLICAL VALUE OF 300 mV. Where or how is the above calculation going wrong? A factor of 2 is involved somewhere. Thanks for your time and patience. ---- Reg, G4FGQ. |
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"Roy Lewallen" wrote in message ... The answer is 3 mW. Any version of EZNEC can be used to do this calculation. The demo program will yield slightly less accurate results because of the limited number of segments(*). I modeled two vertical wires, 1 meter high and 1 mm diameter, spaced 1 km apart, at 20 MHz, over perfect ground. The reported feedpoint impedance varies with segmentation, from 1.988 - j952.3 ohms at 10 segments/wire to 1.72 - j882 ohms at 100 segments/wire. Accuracy is likely to degrade with a larger number of segments, since even 100 results in segment length/diameter ratio less than NEC recommendations. I used 100 segments/wire for the test. One of the choices in EZNEC of far field strength reporting is in V/m at 1 kW input and 1 km distance. For this antenna, EZNEC reports 300.8 V/m (RMS) at ground level. EZNEC also permits setting a fixed power input, so this was set to 1 kW. The resulting source voltage and current are 21270 V. and 24.12 A. respectively. A load of 1.72 + j882 ohms was placed at the base of the second vertical. EZNEC reports a power of 3.234 mW being dissipated in this load. Care has to be used when analyzing the current induced in one antenna by another which is distant using numerical calculations. Errors can occur due to truncation and other causes when the ratio of distances between the two antennas is great relative to the segment lengths or to segment distances within one of the antennas. However, EZNEC gets virtually identical results when using mixed and double precision NEC-2 calculating engines, which indicates that the limit hasn't been reached and that numerical problems aren't occurring. (Another check which can be done is to reduce the distance between antennas by a factor of two. The power in the load resistance should increase by a factor of four.) Another critical matter is the setting of the load reactance. The reactance is many times larger than the resistance, so a slight error in setting its value will result in a large difference in load current and therefore load dissipation. For example, if the segmentation is changed from 100 to 50 segments/wire and no other change is made to the model, the reported load power becomes 0.3917 watts. The reason is that the reported source impedance is now 1.756 - j891.4 ohms, while the load is still 1.72 + j882 ohms. Changing the load to the proper conjugately matched value of 1.756 + 891.4 ohms returns the load power to the correct value of 3.24 mW. All given, I'd trust the reported load power to be easily within 10% of the theoretically correct value. (*) Results for 10 segments/wire are 1.988 - j952.3 ohms for the source impedance, 300.71 V/m field strength at 1 km for 1 kW, and 3.24 mW in a conjugately matched load impedance in the distant vertical. Roy Lewallen, W7EL I agree with the E field computation at 1km. NEC2 calculates the normalized peak E field as 425.452 V/m, which gives 300 mV/m at 1 km. The input impedance, with 50 segments (#14 AWG, perfect conductor) is 1.747 - j823.798 ohms. The NEC output files shows the TRP at 1 kW. For some reason I seem to get a different received power. If I model a 1 meter monopole, above a perfectly conducting ground, loaded at the base segment with the complex conjugate of 1.747 + j823.798, and an incident peak E field of of 1V/m. NEC computes the peak base current as 0.28636 A. Dividing by 3.3333, for the equivalent RMS current from 300 mV/m RMS gives: 0.08591 A RMS. power in the load then equals 12.9 mW, which seems to agree with Reg's figure. Frank |
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I agree with the E field computation at 1km. NEC2 calculates the
normalized peak E field as 425.452 V/m, which gives 300 mV/m at 1 km. The input impedance, with 50 segments (#14 AWG, perfect conductor) is 1.747 - j823.798 ohms. The NEC output files shows the TRP at 1 kW. For some reason I seem to get a different received power. If I model a 1 meter monopole, above a perfectly conducting ground, loaded at the base segment with the complex conjugate of 1.747 + j823.798, and an incident peak E field of of 1V/m. NEC computes the peak base current as 0.28636 A. Dividing by 3.3333, for the equivalent RMS current from 300 mV/m RMS gives: 0.08591 A RMS. power in the load then equals 12.9 mW, which seems to agree with Reg's figure. Frank PS -- NEC computes the gain of a 1 meter, ideal conductor, monopole above a perfectly conducting ground, as 4.8 dB (Ground wave). Also note the current in the base, from an incident peak E-field of 1V/m is 0.28636A, which through 1.747 ohms, is 0.5 V peak. This appears to agree exactly with Reg's rational. Frank |
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Reg, G4FGQ wrote:
"According to textbooks, the field strength from 1 KW at 1 kilometre = 300 millivolts per metre." Terman agrees with Reg`s signal strength produced by 1 KW radiated at a distance of 1 kilometer of: 300 millivolts per meter. There are some conditions. The transmitting antenna is vertical and short compared with a 1/4-wavelength. It is nondirectional in a horizontal plane. The height of both antennas is low enough so that the space wave does not dominate propagation between them. Terman`s field strength derives from the equation first given by Sommerfeld and includes a factor aounting for ground losses. Best regards, Richard Harrison, KB5WZI |
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"Richard Harrison" wrote in message ... Reg, G4FGQ wrote: "According to textbooks, the field strength from 1 KW at 1 kilometre = 300 millivolts per metre." Terman agrees with Reg`s signal strength produced by 1 KW radiated at a distance of 1 kilometer of: 300 millivolts per meter. There are some conditions. The transmitting antenna is vertical and short compared with a 1/4-wavelength. It is nondirectional in a horizontal plane. The height of both antennas is low enough so that the space wave does not dominate propagation between them. Terman`s field strength derives from the equation first given by Sommerfeld and includes a factor aounting for ground losses. Best regards, Richard Harrison, KB5WZI ======================================== Richard, thanks for the confirmation. All you have to do now is calculate from the field strength the power available to a matched receiver at 20 MHz with a vertical receiving antenna 1 metre high. If I tell you the antenna's radiation resistance is 1.758 ohms then you can forget about the frequency. ---- Reg. |
The factor of 2
On Fri, 9 Dec 2005 03:43:27 +0000 (UTC), "Reg Edwards"
wrote: Where or how is the above calculation going wrong? A factor of 2 is involved somewhere. Thanks for your time and patience. Asking for another chin lifted for the sucker punch? Reggie, this is precious. |
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On Fri, 9 Dec 2005 00:16:31 +0000 (UTC), "Reg Edwards"
wrote: Sorry, I forgot "per metre". I should have said - "According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts per metre." For an isotropic radiator, is it correct to calculate the power flux density at 1Km at 1000/(4*pi*1000**2), and to find the field strength from FS in V.m = (power flux density * 120*pi)**0.5? That gives 173mV/m. It would be 245mV/m if the power were radiated uniformly in hemisphere. 300mV/m is conditional on the power radiated in a hemisphere and from an antenna with directivity (field proportion to the cosine of the angle of elevation). Does that make sense? Owen -- |
The factor of 2
Reg Edwards wrote:
Roy, My program requires antenna gains relative to isotropic to be entered. To help discover where I was going wrong a numerical type program was needed. Numerical programs do not need the intervention of fallible human ideas about isotropes, mirror images and ground reflections. I did not realise that EZNEC has the ability to calculate voltages and currents induced in elements miles away from the radiating element. There's a limit because of numerical precision, which I cautioned about in my last posting. But this problem is within its capabilities. But having the correct answer, I still have a problem. Praps you can help me to solve it. . . . Yes, I'm very interested by the apparent contradiction. But I'm also seemingly getting some contradictory answers -- I'll have something to offer after I get it sorted out. Roy Lewallen, W7EL |
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On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards"
wrote: I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought to be, or exactly 4 times the correct power input. Most of my programs calculate results based on what I consider to be fundamental reasoning. But GRNDWAV3 is one of the few where I have adapted formulae from the text books or 'bibles'. My informant is an Icelandic amateur who appears to know what he is talking about and is mathematically very convincing. For various resons, for the time being I propose to leave him out of this discussion. That is, of course, if a discussion should evolve. The problem fundamentally revolves around the gain of short vertical antennas, both transmitting and receiving, above a perfect ground, relative to isotropic. But for present purposes what an isotrope actually is can be forgotten about. It exists only in one's imagination. Numbers cannot be avoided. So let's keep them as simple as possible by starting with the MF standard of 1 Kilowatt, radiated from a short vertical antenna above a perfect ground. Actual antenna height and frequency don't matter. According to the text books, the field strength from 1 Kw at 1 kilometre = 300 millivolts, which (according to the text books) is correctly calculated by my program. To calculate matched reciever input power from field strength it is necessary to state vertical antenna height, frequency and radiation resistance. Again choosing simple values - An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? Owen -- |
The factor of 2
The reason for the contradiction is that I got the first result of 1
volt for the base of a 1 meter vertical wire above perfect ground in a 1 V/m field by using NEC with a plane wave excitation source which produced a 1 V/m plane wave field. The 3 mW value I got later for Reg's model was obtained by generating a 1 V/m field by putting a conventional source at the base of a second short vertical. And what I've now determined after a considerable amount of experimentation is: The current reported by NEC-2 or NEC-4 to be induced in a wire (or the voltage in its center or between base and ground) by an impinging field created by another antenna is exactly half the value it is when the same field is created instead by an NEC plane wave excitation source, when a ground plane is present. This doesn't occur in free space models, which seem to produce correct results. Unless there's some problem with interpreting the meaning of the plane wave source's field value, it looks like this is a bug in NEC-2 and NEC-4. I've posted a query on a mailing list frequented by the real experts at using these programs, and I'll report back what I find out from them. We really need a sound theoretical basis for deciding what the value of induced current or voltage should be, for a final determination of which answer is right and which is wrong. I'll try to take a good look at that tomorrow. But in the meantime, we do know that the field strength generated by a short vertical with a source at its base is being reported correctly by NEC. NEC programs have been used very widely for determining induced currents and field strengths, and my guess is that the plane wave excitation feature is relatively rarely used, and less so over ground. Consequently, I'll put my money on the result obtained by exciting a second antenna to generate the field rather than on the plane wave excitation source result. If this conjecture is correct, then I was wrong when I said in an earlier posting that the voltage at the base of a one meter wire over ground was one volt when exposed to a one V/m field -- it should be 0.5 volt. I got the 1 volt result by using an NEC plane wave excitation source -- ironically, after first verifying that I got the known theoretical result in free space. And my more recent posting giving the power in Reg's example antenna load as 3 mW rather than 12 is correct. I'll post more as I find out more. Roy Lewallen, W7EL |
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"Owen Duffy" wrote in message ... On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards" wrote: I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought........ ..........resistance. Again choosing simple values - An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? Owen Then S = 2.387*10**(-4) W/m**2; and A = 0.119*(Lambda)**2, where Lambda = 20m. (p.44) Therefore A = 47.6 m**2; and Pr = 11.36 mW. Since we have done this modeling of incident E-fields before, should there not be some correlation with NEC2? Frank PS, your FORTRAN notation threw me for a second. |
The factor of 2
"A factor of 2 is involved somewhere?"
First place I would look is at the coltage actually applied to the receiver. If the antenna intercepts 1 volt per meter in a 1-meter antenna, only 0.5 volt is applied to the receiver. The other 0.5 volt is lost to reradiation in a matched antenna system. Best regards, Richard Harrison, KB5WZI |
The factor of 2
Reg wrote:
"So we have a generator with open-circuit volts of 300 mV, with an internal resistance of 1.944 ohms, into an Rx load resistance of 1.944 ohms---." You lose 1/2 the open-circuit voltage in a matched antenna`s radiation resistance. 1/2 the voltage in the receiver`s input resistance causes 1/2 the current. Received carrier power is only 1/4, or in other words, 6 db less than twice the voltage would produce were it available across the receiver`s input resistance. 0.15 volts squared over 1.944 ohms = 0.01157 watts on my Chinese calculator Best regards, Richard Harrison, KB5WZI |
The factor of 2
"Richard Harrison" wrote in message ... Reg wrote: "So we have a generator with open-circuit volts of 300 mV, with an internal resistance of 1.944 ohms, into an Rx load resistance of 1.944 ohms---." You lose 1/2 the open-circuit voltage in a matched antenna`s radiation resistance. 1/2 the voltage in the receiver`s input resistance causes 1/2 the current. Received carrier power is only 1/4, or in other words, 6 db less than twice the voltage would produce were it available across the receiver`s input resistance. 0.15 volts squared over 1.944 ohms = 0.01157 watts on my Chinese calculator ===================================== Yes Richard, I fully agree. The trouble is that there are other ways of calculating receiver input power, seemingly equally valid. But they give an input power exactly 1/4 as big or 6 dB less. Which way is correct? Terman, Kraus, Balani, their Bibles and numerous computer programs are all at loggerheads with each other. The very foundations of radio engineering are being undermined. ;o) ---- Reg. |
Back to fundamentals
On Fri, 09 Dec 2005 13:06:17 GMT, "Frank"
wrote: "Owen Duffy" wrote in message .. . On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards" wrote: I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought........ ..........resistance. Again choosing simple values - An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? Owen Then S = 2.387*10**(-4) W/m**2; and A = 0.119*(Lambda)**2, where Lambda = 20m. (p.44) I thought Reg was talking about f=20E6, so Lambda~=15m isn't it? Therefore A = 47.6 m**2; and Pr = 11.36 mW. Since we have done this modeling of incident E-fields before, should there not be some correlation with NEC2? Frank PS, your FORTRAN notation threw me for a second. Several languages use ** as the exponentiation operator, FORTRAN was probably the first. C, the C-like languages and IIRC most of the Algol languages use **. The ^ operator used in VBA is a logical operator in most languages. Owen -- |
Back to fundamentals
Owen Duffy wrote:
Several languages use ** as the exponentiation operator, FORTRAN was probably the first. C, the C-like languages and IIRC most of the Algol languages use **. The ^ operator used in VBA is a logical operator in most languages. How about 10SUP-4/SUP? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
The factor of 2
I've received an authoritative answer about NEC plane wave excitation.
When a 1 V/m incident plane wave is specified via a plane wave source and a ground plane is present, the field strength at all points is 2 V/m, not 1 V/m as I had assumed. The rationale is that the "incident wave" is reflected by the ground plane, doubling its strength. The conclusions from this are that: 1. NEC reports that the voltage from the base of a 1 meter electrically short vertical wire to perfect ground in the presence of a 1 V/m field is 0.5 volt, not 1 volt as I said in my earlier posting in response to a question by Reg. I apologize for the error. 2. The power intercepted by the matched dipole in the problem recently posed by Reg is approximately 3 mW, not 12. The EZNEC calculation I described, which does not use a plane wave source, is correct. The same result can be obtained with NEC by using two antennas as in EZNEC, or with a 212 V/m (peak, equal to 150 V/m RMS) plane wave source which produces a 300 V/m RMS field at the loaded antenna. This is a good place to give an additional caution to people using NEC for calculations. NEC uses peak, not RMS values for all voltages and currents. Power results will be off by a factor of two or four if a user mistakenly assumes RMS values. EZNEC uses RMS values throughout. When Reg posed the dilemma about the factor of four disparity in reported powers, my first thought was that this was the cause. As it turned out, it wasn't, but caution is needed. Results should always be given a reality check, as Reg has done. Any model -- and this doesn't exclude the mathematical models we often consider "theory" -- can be subject to many errors, including but not limited to misapplication, misinterpretation, and limitations of an approximation or numerical calculation. Roy Lewallen, W7EL |
The factor of 2
On Fri, 09 Dec 2005 11:45:32 -0800, Roy Lewallen
wrote: 2. The power intercepted by the matched dipole in the problem recently posed by Reg is approximately 3 mW, not 12. The EZNEC calculation I described, which does not use a plane wave source, is correct. The same result can be obtained with NEC by using two antennas as in EZNEC, or with a 212 V/m (peak, equal to 150 V/m RMS) plane wave source which produces a 300 V/m RMS field at the loaded antenna. To a certain extent, this comes back to a decision about whether ground reflection contributes to the received power, and you are saying that NEC assumes it does under plane wave excitation in presence of a ground plane. In "running the numbers", I note that the radiation resistance indicated by NEC for a short dipole in free space is quite different to that predicted by Kraus for a dipole with uniform current, (Rr=80*pi()**2(L/Lambda)**2)! Owen -- |
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Owen Duffy wrote:
PS, your FORTRAN notation threw me for a second. Several languages use ** as the exponentiation operator, FORTRAN was probably the first. C, the C-like languages and IIRC most of the Algol languages use **. The ^ operator used in VBA is a logical operator in most languages. They say a FORTRAN-trained programmer can write FORTRAN in any language, .AND. of course that's .TRUE. -- 73 from Ian GM3SEK |
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Owen Duffy wrote:
On Fri, 09 Dec 2005 13:06:17 GMT, "Frank" wrote: "Owen Duffy" wrote in message . .. On Thu, 8 Dec 2005 21:44:07 +0000 (UTC), "Reg Edwards" wrote: I have been informed my GRNDWAV3 program is in error - it calculates the power input to a matched receiver to be 6dB greater than it ought........ ..........resistance. Again choosing simple values - An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? Owen Then S = 2.387*10**(-4) W/m**2; and A = 0.119*(Lambda)**2, where Lambda = 20m. (p.44) I thought Reg was talking about f=20E6, so Lambda~=15m isn't it? Therefore A = 47.6 m**2; and Pr = 11.36 mW. Since we have done this modeling of incident E-fields before, should there not be some correlation with NEC2? Frank PS, your FORTRAN notation threw me for a second. Several languages use ** as the exponentiation operator, FORTRAN was probably the first. C, the C-like languages and IIRC most of the Algol languages use **. The ^ operator used in VBA is a logical operator in most languages. Owen -- It's been a while since you've done mathematical stuff in C, hasn't it? C does _not_ use a '**' operator. If you want to raise y to the x power you use "pow(y, x)". -- Tim Wescott Wescott Design Services http://www.wescottdesign.com |
The factor of 2
Owen Duffy wrote:
. . . In "running the numbers", I note that the radiation resistance indicated by NEC for a short dipole in free space is quite different to that predicted by Kraus for a dipole with uniform current, (Rr=80*pi()**2(L/Lambda)**2)! The only way to achieve uniform current on a short dipole is with large capacity hats at the ends of the dipole. Otherwise, the current tapers nearly linearly from a maximum at the center to zero at the ends. If you'll look closely at Kraus' figure of the short dipole he analyzes, you'll see that it has capacity hats. Nearly all other authors analyze just a straight wire which doesn't have those hats, and consequently linear rather than uniform current distribution. And of course get quite a different result. I'll bet you didn't include large capacity hats in your model. I haven't tried it, but you should get results much closer to Kraus' if you do. NEC analysis gives radiation resistance very close to theoretical when analyzing a plain straight wire dipole, but this isn't what Kraus does in his book. It is interesting, though, to see how much effect the wire diameter has on the impedance, and that the wire has to be very thin indeed to approach the theoretical impedance for an infinitesimally thin dipole. Roy Lewallen, W7EL |
Back to fundamentals
Owen Duffy wrote:
An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? I believe that's correct. Note that Kraus uses a plain wire dipole for his aperture correction, but a dipole with end hats (and therefore uniform current) for input impedance calculations. See my other recent posting for further comments. Roy Lewallen, W7EL |
Back to fundamentals
On Fri, 09 Dec 2005 13:49:31 -0800, Tim Wescott
wrote: It's been a while since you've done mathematical stuff in C, hasn't it? C does _not_ use a '**' operator. If you want to raise y to the x power you use "pow(y, x)". It has, I tend to do most my ad-hoc stuff in Perl these days, and it uses the ** operator. Perl is c-like, but as you say Tim, it did not inherit the ** op from C. Languages that lack an exponentiation operator are a right pain in the butt, but there are lots of them. Owen -- |
The factor of 2
On Fri, 9 Dec 2005 15:41:47 +0000 (UTC), "Reg Edwards"
wrote: Which way is correct? Terman, Kraus, Balani, their Bibles and numerous computer programs are all at loggerheads with each other. This is simply through massive missinterpretation by you, who disdain the work you "recite" (a suspect activity) and the complete absence of your using the first principles of Lord Kelvinator, whom you claim to revere for his clarity in approaching problems through demonstrables. Result? YAT (yet another troll) |
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Owen Duffy wrote:
An alternative is to calculate the power collected by a lossless, matched receiver as Pr=S*A. In this case, S=0.3**2/(120*pi) Kraus derives A (the effective apperture) for a short dipole to be 3/8/pi*wavelength**2. This gives the power collected by the receiver as 6.4mW. If the antenna and receiver were disected by the ground plane, wouldn't there be 3.2mW developed in each half of the receiver load? I found another source (Ramo et al) which directly gives the ratio of power in the load of a matched receiving antenna to the power applied to a transmitting antenna, in terms of the effective apertures of the antennas. This doesn't require the intermediate step of calculating field strength. The equation is: Wr/Wt = (Aer * Aet) / (lambda^2 * r^2) whe Wr, Wt are received and transmitted power respectively Aer, Aet are the receiving and transmitting antenna effective apertures lambda = wavelength r = distance between the antennas Note that effective aperture, like other measures of an antenna pattern, is a function of the direction from the antenna. So this equation is correct regardless of antenna orientation as long as Aer and Aet are correctly calculated. Letting K = 3/(8 * pi) ~ 0.1194 we can write the equation cited by Owen for effective aperture of a short dipole in its most favored direction (broadside) in free space as Ae = K * lambda^2 (This is the effective aperture of an infinitesimally short dipole. However, it changes very little with length when the dipole is electrically short. The effective aperture broadside to a half wave dipole is only 10% greater.) Then we get that Wr = Wt * (K^2 * lambda^4) / (lambda^2 * r^2) = Wt * K^2 * lambda^2 / r^2 For Reg's example, lambda = 15 meters and r = 1 km. Since this is a free space analysis involving dipoles so far, I'll apply 2 kW (Wt) to the transmitting dipole, resulting in Wr = 6.412 mW. Now we can split the model exactly in half with a ground plane. On the top of the ground plane, the transmitting antenna has exactly half the applied power, or 1 kW, which is what we had in Reg's example. Half the load power is in the upper plane also, so we have 3.206 mW for the load power in Reg's example setup. This is very close to the 3.234 mW result from the EZNEC model. The antenna's effective height (that is, the ratio of induced voltage to field strength) has been at issue. As Reg pointed out, ~ 3 mW at the load requires an effective height of 0.5 meter for the 1 meter high antenna. (I incorrectly gave it as 1 m in an earlier posting.) I did find an explicit equation for effective height for a vertical over perfect ground, in King, Mimno, and Wing (Dover edition, p. 165). This also confirms that the correct effective height is 0.5 m for the 1 m electrically short vertical antenna over ground. I'm satisfied that we have the answer to Reg's question. It's been an educational process for me -- thanks for posing it. One final note, regarding the NEC applied plane wave. My earlier statement that the resulting field is twice the plane wave source magnitude when a ground plane is present is true only when the plane wave is applied over perfect ground at exactly grazing incidence (zenith angle = 90 deg.). If applied from other angles the resulting field strength will be different. If you apply a vertically polarized wave over a ground plane, I believe the resulting field strength will look like the pattern from a vertical radiator over a perfect ground plane -- strongest when applied at the horizon, decreasing when applied at higher angles, and dropping to zero if applied from directly overhead. I haven't confirmed this, but believe it's necessary in order to get a receiving pattern that's the same as the transmitting pattern. So use it with caution when a ground plane is present, and don't casually make assumptions about the resulting field. Roy Lewallen, W7EL |
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Hi Roy,
A couple of comments on your excellent input into this discussion. Roy Lewallen wrote: I found another source (Ramo et al) which directly gives the ratio of power in the load of a matched receiving antenna to the power applied to a transmitting antenna, in terms of the effective apertures of the antennas. This doesn't require the intermediate step of calculating field strength. The equation is: Wr/Wt = (Aer * Aet) / (lambda^2 * r^2) whe Wr, Wt are received and transmitted power respectively Aer, Aet are the receiving and transmitting antenna effective apertures lambda = wavelength r = distance between the antennas Kraus "Antennas" also describes this equation. He refers to it as the "Friis transmission formula" on pages 48 and 49 of the second edition. One final note, regarding the NEC applied plane wave. My earlier statement that the resulting field is twice the plane wave source magnitude when a ground plane is present is true only when the plane wave is applied over perfect ground at exactly grazing incidence (zenith angle = 90 deg.). If applied from other angles the resulting field strength will be different. If you apply a vertically polarized wave over a ground plane, I believe the resulting field strength will look like the pattern from a vertical radiator over a perfect ground plane -- strongest when applied at the horizon, decreasing when applied at higher angles, and dropping to zero if applied from directly overhead. I haven't confirmed this, but believe it's necessary in order to get a receiving pattern that's the same as the transmitting pattern. So use it with caution when a ground plane is present, and don't casually make assumptions about the resulting field. I believe a better way to describe this situation is that the plane wave field strength does not go to zero, but rather the effective aperture of the antenna goes to zero as the plane wave is applied from overhead. This does not change your conclusion with respect to antenna patterns. The oblique-incidence plane wave equations are slightly messy, but they are well described in treatments of waveguides. 73, Gene W4SZ |
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As expected, the discussion appears to have fizzled out.
Having for the first time allowed myself to be led astray by the academic professors and authors and their Bibles on the subject, I still have the problem of correcting a bug in my computer program. The program itself, GRNDWAV3, is too interesting, useful and educational to simply withdraw it. Lets not be confused by fallible human ideas and notions on - Reflections from the ground. Mirror images in the ground. Antenna gains relative to isotropic. Antenna gains relative to isotropic with a ground plane. Antenna gains relative to isotropic of dipoles. Half hemispheres. What on Earth is an isotropic antenna anyway? And now we are being introduced to waveguides. Thanks to Roy's investigations and clarification, the solution to my problem is perfectly simple - "The effective height of a short vertical antenna is half of its actual height and the voltage induced in it is half of the field strength in volts per metre." Which gives the correct answers from my program. And which was proven and well known 100 years ago by the early radio engineers who were really the first amateurs in the game. Having got that off my chest, I can now finish the second half of a bottle of Australian, Bantock Station, Special Reserve, Cabernet Sauvignon Shiraz which I can thoroughly recommend. ---- Reg, G4FGQ |
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Gene Fuller wrote:
. . . [I wrote:] One final note, regarding the NEC applied plane wave. My earlier statement that the resulting field is twice the plane wave source magnitude when a ground plane is present is true only when the plane wave is applied over perfect ground at exactly grazing incidence (zenith angle = 90 deg.). If applied from other angles the resulting field strength will be different. If you apply a vertically polarized wave over a ground plane, I believe the resulting field strength will look like the pattern from a vertical radiator over a perfect ground plane -- strongest when applied at the horizon, decreasing when applied at higher angles, and dropping to zero if applied from directly overhead. I haven't confirmed this, but believe it's necessary in order to get a receiving pattern that's the same as the transmitting pattern. So use it with caution when a ground plane is present, and don't casually make assumptions about the resulting field. I believe a better way to describe this situation is that the plane wave field strength does not go to zero, but rather the effective aperture of the antenna goes to zero as the plane wave is applied from overhead. This does not change your conclusion with respect to antenna patterns. I'm not sure either one of us quite has it right. In a model experiment, I set up a short open circuited vertical dipole just above a perfect ground plane, and applied a vertically polarized plane wave from the horizon. Let's call the resulting voltage at the dipole center V1. Then I changed the direction of the plane wave so it was coming from an elevation angle of 45 degrees above the horizon, but with the same amplitude. The dipole voltage was about 0.7 * V1, about what we'd expect from the change in effective aperture of the vertical dipole due to the different arrival elevation angle. But if I tilt the dipole back 45 degrees so it's parallel to the incident E field, the voltage drops to about 0.5 * V1, another 3 dB. I believe this indicates that the field in the vicinity of the dipole is oriented normal to the ground plane, and it has a magnitude that's about 0.7 as great as it is when the same amplitude wave is fired from a horizontal direction. A second check was to tilt it 45 degrees the other way, so its end is pointing directly toward the direction of the impinging wave. The result was again about 0.5 * V1, adding proof that the field in its vicinity is normal to the ground plane and not tilted in the direction of the source. So the antenna aperture is indeed changing as we change the orientation of the antenna relative to the field in its vicinity. But that's not the same as the orientation of the antenna relative to the direction from which the plane wave originates. (They are of course the same if the ground plane is absent.) The change in antenna output (in this case) when the source direction is changed is due to the fact that the magnitude of the field strength has changed, not because its orientation relative to the antenna has changed. When the direction of the plane wave is elevated 45 degrees, it has equal horizontally and vertically polarized components. The horizontal components cancel on reflection, while the vertical components reinforce as before. This leaves only the vertical component in the vicinity of the sense dipole, and it's 0.707 * the value when the same amplitude wave is coming in horizontally. The dipole voltage is maximum when it's oriented to be parallel with this field, that is, vertical. At least I think this is a correct interpretation of what I'm seeing. These effects are all tied in together, and I've spent so long looking at the problem from the transmitting direction that I'm having some trouble getting my thinking turned around. But I'm slowly getting there. Roy Lewallen, W7EL |
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