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Back to fundamentals
"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. |
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