What I am still not understanding, is . . . .


To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

----
Reg

Cecil,

If you prefer your voltages to flow and your currents to drop, have at
it. :-)

Here's a hint.

In business, politics, crime (sorry for the redundancy), and a bunch of
other stuff there is an old saw, "Follow the money."

In electronics the appropriate dictum is, "Follow the electrons."

There is no law regarding conservation of voltage. There is a
fundamental law about conservation of change, and an equally strong law
dealing with continuity of current.

Go ahead and label a change in current at two points on a wire a "drop"
if you like, but don't confuse this change with a drop in voltage. They
ain't the same thing.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

I am quite familiar with standing waves, thank you. I have no
disagreements with Terman, Kraus, Balanis, or any other legitimate
experts.


What I am still not understanding, is since the exponential equations
for voltage and current in a transmission line are identical except
for the Z0 term, how can something happen to the current without
the same thing happening to the voltage at the same time? How can
something happen to the voltage without also happening to the current
at the same time? In a matched system, the voltage and current arrives
at the load at exactly the same time attenuated by exactly the same
amount. But that voltage didn't flow and that current didn't drop???
--
73, Cecil, W5DXP

Gene Fuller wrote:
Go ahead and label a change in current at two points on a wire a "drop"
if you like, but don't confuse this change with a drop in voltage. They
ain't the same thing.

In the matched exponential transmission line equations, the
attenuation factors for the voltage and current are identical,
i.e. they decrease by exactly the same percentage.
--
73, Cecil, W5DXP

"Frank" wrote in message
news:t4okd.90482$VA5.33610@clgrps13...

"Frank" wrote in message
news:H4hkd.141267$9b.112169@edtnps84...
Modeled #14 AWG, copper conductor, 32ft monopole, 29 radials of 25ft, and
base 6" above (nominal lambda/1000) Sommerfeld/Norton ground of Er = 13,
sigma = 0.013 S/m at 1.8 MHz. All segments 6".

NEC2 computes:
Zin = 2.87 - j1358 Efficiency 92%

RADIALS2 computes (with radials 1mm below ground):
Zin = 1.55 - j1310 Efficiency 23.5%

Not a large amount of difference, but thought I had gotten closer results
with a different monopole, but seem to have deleted the code (Not sure
why such a large difference in efficiency). NEC2 is supposed to provide
a reasonable approximation of a buried radial monopole when at about
lambda/1000 above ground. Be interested in any comments, and what NEC4
provides if anybody has it.

73,

Of course the higher efficiency is due to NEC calculating only the I^2R
losses, and not the TRP. TRP should be fairly easy to calculate since the
pattern is "phi" independent. Have not checked to see if there is a TRP
card.
Note that a 32 ft monopole mounted on a perfect ground has an input
impedance of 1.58 - j1311 Ohms. The efficiency is reduced to 86% due to
increased I^2R losses.

Frank

From the calculated field strength (as a function of Theta) the TRP for 100
W input, which includes copper and ground losses, shows 27.4 W, or 27.4%
efficient. In very close agreement with the RADIALS2 program. The only
noticeable discrepancy appears to be in the real part of Zin.

Frank

Frank, as you say, the height of the radials in the NEC2 model is only 6
inches above ground.

The radials are shallow-buried in the RADIALS2 model. It can't do elevated
radials.

The ground loss resistance as height decreases, as seen by the antenna,
increases very fast percentage-wise as the radials get within a few inches
of the ground. It is due to very close magnetic and electric coupling to
ground. Radials are transmission lines, insulated from but running very
close to a resistive slab of soil.

This would account for the computed higher input resistance of the radials
( 3.5 - j*3.3 ohms ) ( for 29, 25-feet long radials. Rg=77, K=13 ) in
program RADIALS2.

The calculated antenna input impedance in RADIALS2 is that of the antenna
alone. For feedpoint impedance add the input impedance of the radial system.
Presumably, NEC2 does not compute the input reactance of radials.

Efficiency is calculated in the usual way from the sum of antenna input
resistance and radials' input resistance.

If you contrive to change the radials input reactance without changing
frequency or the antenna, you will notice the loading coil tunes it out
along with antenna reactance.

Incidentally, when elevated radials are near the ground their velocity
factor decreases fast which makes a mess of the usual recommendation to
prune them to 1/4-wave free-space length.

When radials are actually lying on the ground surface the velocity factor
decreases to roughly 0.5 of the velocity of light. When buried the
underground VF can fall to as small as 0.15 depending on soil permittivity.
(or moisture content.)
----
Reg, G4FGQ

Reg, thanks for the info. I see I was making an error with RADIALS2. Did
not realize that you had to add the radial impedance to the antenna
impedance; I thought it was computed in the final result. I figured
something was weird since the input impedance was similar to the antenna
modeled over a perfect ground. NEC2 does compute the input impedance of the
complete structure, but as mentioned before it is limited, in that all wires
must be = lambda/1000 above the ground (at 1.8 MHz about 6"). I am just
starting to delve into computational electromagnetics, so do not know that
much about NEC. It uses the "Method of moments" in its computations. The
theory of operation manual is available for download at several web sites.

Anyway, with RADIALS2, I now get an input impedance of 5.1 - j1303, and with
NEC2 2.9 - j1358. I did try entering a negative number for the depth of
the radials, but RADIALS2 did not like it.

Your comments about the effect on the radials of being buried are also very
interesting, and obviously indicate the reason for our slightly different
results. The very low VF of buried radials indicates that the length is
less important.

As for efficiency, NEC2 computes a normalized far E-field at 1 meter. For
phi independent structures it becomes trivial to integrate the power density
over a hemispherical region to arrive at the true total radiated power. 100
Watts into the antenna radiates 27 Watts, again, very close to RADIALS2'
computed efficiency of 23.5%. Come to think of it, I guess I could have
estimated the losses -- as you do -- by comparing the input impedance of an
antenna over a perfect ground with the same antenna over a lossy ground.
Still I think it was more fun playing with Excel spread sheets and coming up
with a similar answer.

Regard,

Frank

"Reg Edwards" wrote in message
...
Frank, as you say, the height of the radials in the NEC2 model is only 6
inches above ground.

The radials are shallow-buried in the RADIALS2 model. It can't do elevated
radials.

The ground loss resistance as height decreases, as seen by the antenna,
increases very fast percentage-wise as the radials get within a few inches
of the ground. It is due to very close magnetic and electric coupling to
ground. Radials are transmission lines, insulated from but running very
close to a resistive slab of soil.

This would account for the computed higher input resistance of the radials
( 3.5 - j*3.3 ohms ) ( for 29, 25-feet long radials. Rg=77, K=13 ) in
program RADIALS2.

The calculated antenna input impedance in RADIALS2 is that of the antenna
alone. For feedpoint impedance add the input impedance of the radial
system.
Presumably, NEC2 does not compute the input reactance of radials.

Efficiency is calculated in the usual way from the sum of antenna input
resistance and radials' input resistance.

If you contrive to change the radials input reactance without changing
frequency or the antenna, you will notice the loading coil tunes it out
along with antenna reactance.

Incidentally, when elevated radials are near the ground their velocity
factor decreases fast which makes a mess of the usual recommendation to
prune them to 1/4-wave free-space length.

When radials are actually lying on the ground surface the velocity factor
decreases to roughly 0.5 of the velocity of light. When buried the
underground VF can fall to as small as 0.15 depending on soil
permittivity.
(or moisture content.)
----
Reg, G4FGQ

Reg Edwards wrote:
To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?
--
73, Cecil, W5DXP

Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V


Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

73, ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

It would be too much of a coincidence for 'a', the attenuation
factor, to be the same whether in series or in shunt. Most of
the attenuation at HF is due to I^2*R losses, a series event.

Transmission lines are distributed networks involved with EM wave
energy transmission. I^2*R losses can cause a decrease in current
just as it can cause a decrease in voltage. The sequence of events
is obvious.

1. The RF voltage drops because of I^2*R losses.
2. The proportional E-field decreases because of the voltage drop.
3. Since the E-field to H-field ratio is fixed by Z0, the H-field
decreases as does the ExH power in the wave.
4. Since the RF current is proportional to the H-field, the current
decreases by the same percentage as the voltage.

The chain of cause and effect is obvious. The current decreases
because of I^2*R losses in the transmission line which is a
distributed network, not a circuit.
--
73, Cecil, W5DXP

Jim Kelley wrote:



Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V



Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?


Depends on whether 'a' is in series or in shunt.

73, ac6xg

Apparently I should have added that distributed resistive losses in
series create distributed voltage drops with a common current through
all, but distributed resistive losses in shunt create a distribution of
Norton current sources, where the shunt current on the transmission line
or radiator at a given point is the sum of all equivalent current
sources between that point and the end of the transmission line/radiator.

73, Jim AC6XG