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.

"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

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

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 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.

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.

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
--

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

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 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