"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.