"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.

"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

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

"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.

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
--

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

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

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
--

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

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