NEC-List has had a lot on CFA models lately, but I thought I would try
my hand at it. It sounds like Jack Belrose is going in the direction
of a real-world model, with a ground screen, feed wires and coupled
tower. I tried a clean, rotationally symmetric, lossless version to
try to understand the basic operation. This results in the E-plate
cylinder having a closed bottom, while the actual antenna apparently
has an open bottom with a single feed wire, as Jack is modeling it.
However, this should not make a difference in whether or not it works.
I will put the conclusions first, since many may not make it through
all these numbers. They are:
1) NEC is capable of modeling this antenna if you are careful, but due
to the small size and quadrature feed, the radiation may be
considerably less than NEC is able to show from feed point voltage
and current alone. Average gain evaluation is essential to get the
correct radiated power.
2) The "D-plate" is a poor producer of magnetic field and a very poor
radiator (a capacitor).
3) The "E-plate" does most of the radiating, but is definitely a small
antenna.
4) E field dominates in the near field, E/H >> 377.
5) Fields do not interact or combine in free space, they just
superpose. You can excite the separate sources with any voltages
or currents you want, and the resulting fields (near E, near H,
radiated E,H) will just be the linear combination of the fields
that you would get by exciting the sources separately.
6) The CFA has the properties of an electrically small antenna. If
the installation is giving good radiation, the explanation involve
the matching, feed lines or coupling to other structures. Maybe it
just has a very good matching network that manages to get the power
in no matter how small the antenna looks. The quadrature phasing
of the feeds results in very large power flow from one antenna to
the other, as the large reactive field from one antenna interacts
in phase with the other source. What happens to this power that
enters through one source and exits from the other will apparently
depend on the matching network.
NEC results are included below in some detail. Maybe someone familiar
with antenna feeds and matching networks can spot something in the
numbers that would make this antenna work or appear to work.
D-PLATE MODEL
First I tried looking at the D-plate disk alone. The disk was modeled
with 120 radials with 2 m radius, 0.6 m above a PEC ground at 1161
KHz. It had a single feed wire on the axis with 5 segments and a 1
volt source on the bottom segment. Actually this was run with NECGS,
which is optimized for rotational symmetry, but NEC2D would give the
same results. The source current from NEC was
I_in = 4.19611E-08 + j2.23048E-03 A and input impedance
Z_in = 0.00843 - j448.33 ohms.
Obviously not a good radiator by itself. With radiation resistance
smaller than reactance by 1.8e-5, one might wonder if it is dropping
into the solution noise level, but the average gain was 2.001 (should
be 2.0). The error is 0.05 percent, so it seems to be an accurate
model.
I computed the near E and H fields along a line from the axis at a
height of h=0.3 m, halfway between the disk and ground. Since I
cannot show plots here I will try to describe them. The vertical E
field had a broad peak of around V/h =1.66 V/m at a radius of around
r=0.5 m. It sloped downward to around 1.5 V/m at the edge of the disk
and then dropped rapidly outside the plates. Close to the axis it
dropped rapidly to satisfy the boundary condition on the feed wire.
Looks reasonable for a capacitor with a wire down the center.
Magnetic field is really the quantity of interest here. At the
smallest radius for which I computed near fields, r=0.1 m, NEC gave
H_phi = 0.0035286 A/m. The magnetic field H_w due to the wire with
current I_w can be approximated for small r as
H_w = I_w/(2 Pi r). (1)
Using the source current I_in, the magnitude of H_w at r=0.1 m is
0.003549 A/m, in close agreement with the NEC result. This simple
formula (1) works here due to the uniform fields between the plates.
Without the plates you would need to use the near field formula for a
current element and integrate over the current distribution. The
displacement current between the plates is in the opposite direction
to the the wire current, and as the radius r increases there will be
more displacement current within the radius encircling the axis. If
we assume that the electric field between the plates is E_z =
V_in/Height = 1.66 V/m the displacement current can be estimated as
I_d = j*omega*D*Area = j(2*Pi*f)(epsilon_0*Ez)(Pi*r^2)
The net current within the circle of radius r is the wire current
minus displacement current, so the total H_phi can be estimated as
H_tot = [I_in - j(2*Pi*f)(epsilon_0*Ez)(Pi*r^2)] / (2*Pi*r) (2)
For visualizing these fields it helps to multiply by r to suppress the
singularity at r=0. Then r*H_w (equation 1) becomes a constant and
r*H_tot starts equal to r*H_w and goes downward in a parabola with
increasing r. The plot of the NEC result (r*H_phi) followed this
downward parabola, and was on top of the H_tot line to about r=1.8 m.
Then it started leveling off while the approximation of equation (2)
goes to zero. At a distance of r=10 m the H from NEC was about 24 dB
below the value of equation (1). At r=39.9 m it was 34 dB below
equation (1) and was still falling faster than 1/r. From r=36 m to
r=40 m the H from NEC fell about as 1/r^1.54. Far enough into the
radiation region H will fall off as 1/r, but at a much lower value
than predicted by equation (1), which would hold for an infinite wire
at DC.
This shows that NEC is evaluating the magnetic field due to changing
D, and it cancels the magnetic field due to the feed current. The
D-plate seems to be a poor producer of H and a very poor radiator. If
the vertical feed wire is removed from between the plates and
connected to the disks with a horizontal transmission line, as some
people have done, the canceling conduction and displacement currents
will be separated, but there is still cancellation. If separation is
large enough they could produce an interference pattern.
CFA MODEL
To model the complete CFA I entered an outline as shown below and
rotated it to make 45 radial sections.
.
| . CFA D and E plates and feed wires
| . rotated about the z axis.
____ .
_______ |.
||.
The D-plate disk had a diameter of 4 m and was 0.6 m above the PEC
ground. The E-plate cylinder had a height of 2.5 m and diameter of 2
m with bottom 0.6 m above the disk. Segment lengths were about 0.12
m, and circumferential segments were included at all radial segment
ends for a total of 4230 segments. This runs fast in NECGS, but would
take a while in regular NEC. The E and D plate feeds were coaxial
cylinders of radius 0.1 m and 0.2 m, respectively, and each of the 45
wires in each cylinder was fed with a voltage source. I started by
exciting the E-plate and D-plate sources separately to get impedances
and mutual coupling. The results are listed below, including source
voltage and current, input impedance, power, average gain, radiated
field at theta=90 degrees (Erad) and near Ez and Hy fields at several
distances. The letters E and D refer to quantities for the E-plate
and D-plate feeds.
***** Case 1, E plate excited:
VE = 1., VD = 0.
IE = 2.20649E-7 + j1.9749E-3
ID = 3.54973E-8 - j1.0359E-3
ZinE = 0.05657 - j506.354 ohms
PinE = 1.103E-7 Watts
Average Gain = 1.98894
EradE = (4.44372E-3, Angle 179.99 Degrees)
At r = 3.1 m: EzEa = (0.10138, Angle 179.97 Degrees) V/m
HyEa = (3.7954E-5, Angle 89.98 Degrees) A/m
Abs[EzEa/HyEa] = 2671.13 ohms
At r = 10.1 m: EzEb = (6.3557E-3, Angle -179.37 Degrees) V/m
HyEb = (4.7308E-6, Angle 89.70 Degrees) A/m
Abs[EzEb/HyEb] = 1343.47 ohms
At r = 39.7 m: EzEc = (1.1541E-4, Angle -149.10 Degrees) V/m
HyEc = (4.2682E-7, Angle 78.63 Degrees) A/m
Abs[EzEc/HyEc] = 270.39 ohms
***** Case 2, D plate excited:
VE = 0., VD = 1.
IE = 3.60295e-8 - j1.02043e-3
ID = 5.79632E-9 + j2.99238e-3
ZinD = 6.4732E-4 - j334.18 ohms
PinD = 2.89815E-9 Watts
Average Gain = 2.01876
EradD = (7.25312E-4, Angle 179.999 Degrees)
At r = 3.1 m: EzDa = (0.13142, Angle 180. Degrees) V/m
HyDa = (1.4149E-5, Angle 89.99 Degrees) A/m
Abs[EzDa/HyDa] = 9288.29 ohms
At r = 10.1 m: EzDb = (1.3910E-3, Angle -179.53 Degrees) V/m
HyDb = (8.5048E-7, Angle 89.73 Degrees) A/m
Abs[EzDb/HyDb] = 1635.55 ohms
At r = 39.7 m: EzDc = (1.9186E-5, Angle -149.71 Degrees) V/m
HyDc = (7.0052E-8, Angle 78.69 Degrees) A/m
Abs[EzDc/HyDc] = 273.88 ohms
>From these results we can get the short-circuit admittance matrix for
the two excitation ports:
Ymat = (Y11, Y12) = (2.20649E-7 + j1.9749E-3, 3.54973E-8 - j1.0359E-3)
(Y21, Y22) (3.60295e-8 - j1.02043e-3, 5.79632E-9 + j2.99238e-3)
The matrix should be symmetric, y12=y21. It is almost, but there is a
difference of about 1.5 percent. Doesn't look too bad.
Now we will try exciting both sources with equal currents and the E
plate leading the D plate by 90 degrees. That apparently is the
excitation that should result in E and H fields that combine to form
an outward Poynting's vector. I used the currents that Jack Belrose
has used, although our impedances and input powers are somewhat
different.
IE = (10.23, Angle +90 degrees)
ID = (10.23, Angle -90 degrees)
Since I was running a basic NEC code that does not let you enter
current sources, I solved for the excitation voltages from the matrix
equation:
(VE) = Inverse[ymat].(IE)
(VD) (ID)
which gave VE = 2917.47 - j6004.52
VD = -1422.4 - j4464.95
When these voltages were fed into NEC the results were
***** Case 3, E plate leading D plate by 90 degrees:
VE = 2917.47 - j6004.52, VD = -1422.40 - j4464.95
IE = 7.30274 + j7.21170 = (10.26, Angle 44.6)
ID = 7.14087 - j7.27879 = (10.20, Angle -45.55)
PinE = -10998.6 Watts
PinD = 11171.1 Watts
Pin_total = 172.5 Watts
Eradp = (32.2126, Angle 111.73 Degrees)
At r = 3.1 m: Ezpa = (1200.3, Angle 95.22 Degrees) V/m
Hypa = (.30485, Angle 17.27 Degrees) A/m
Abs[Ezpa/Hypa] = 3937.35 ohms
At r = 10.1 m: Ezpb = (47.353, Angle 111.08 Degrees) V/m
Hypb = (0.034579, Angle 21.06 Degrees) A/m
Abs[Ezpb/Hypb] = 1369.41 ohms
At r = 39.7 m: Ezpc = (0.83723, Angle 42.52 Degrees) V/m
Hypc = (3.0956E-3, Angle 10.35 Degrees) A/m
Abs[Ezpc/Hypc] = 270.46 ohms
It looks like 172.5 watts of power is being radiated, but about 11000
watts is leaving the D plate and going into the E plate source (and
matching network). This is typical of an electrically small antenna
with sources fed in quadrature. The self and mutual admittances have
small real parts, so a voltage applied to a source will produce a
current nearly 90 degrees out of phase, and little power input. But
when the two sources are fed 90 degrees out of phase, mutual
admittance coupling from each source induces a current in phase with
the other voltage. This results in lots of power flow between drive
points, but not any more radiation. Reactive power goes into the
antenna and back to the source, and this quadrature power goes from
one source to the other. In either case the power is assumed to be
recycled by the source when no loss is specified in the model, but
that probably will not be the case in a real feed network.
Also, note that the ratio of Ez/Hy at r=3.1 m is 10 times the 377
value that we wanted. Since H is small anyway, why not try reversing
the phase, with the E-plate current lagging the D-plate current. That
gives
***** Case 4, D plate leading E plate by 90 degrees:
VE = -2917.47 - j6004.52, VD = 1422.40 - j4464.95
IE = 7.30157 - j7.21467 = (10.27, Angle -44.66)
ID = 7.14065 + j7.27830 = (10.20, Angle +45.55)
Pin1 = 11009.3 Watts
Pin2 = -11170.2 Watts
Pin_total = -160.9 Watts
Erad = (32.2126, Angle 111.73 Degrees)
At r = 3.1 m: Ez2a = (1200.3, Angle 95.22 Degrees) V/m
Hy2a = (.30485, Angle 17.27 Degrees) A/m
Abs[Ez2a/Hy2a] = 3937.35 ohms
At r = 10.1 m: Ez2b = (47.353, Angle 111.08 Degrees) V/m
Hy2b = (0.034579, Angle 21.06 Degrees) A/m
Abs[Ez2b/Hy2b] = 1369.41 ohms
At r = 39.7 m: Ez2c = (0.83723, Angle 42.52 Degrees) V/m
Hy2c = (3.0956E-3, Angle 10.35 Degrees) A/m
Abs[Ez2c/Hy2c] = 270.46 ohms
Now we really have something; more power coming out than going in! We
can run the station and sell power to the electric company. What is
going on? The clue is found in the Average Gain:
Case 3: Average Gain = 0.0668
Case 4: Average Gain = -0.0716
Since there is no loss and the antenna is over PEC ground the average
gain should be 2.0. Average gain returned by NEC is the result of
integrating the far field power over a region set on the RP command,
dividing by input power computed from V and I at the source and also
multiplying by 4Pi/(solid angle integrated), since it is assumed that
the antenna radiates the same into regions not covered by the
integration as in the region integrated. If we integrate over the
upper hemisphere over PEC ground, the average gain should be 2.0 when
no losses are present.
The radiation of the CFA is so small that it has fallen below the
error level of the solution. In cases 3 and 4 the sum of powers into
the two feed points is about 1.4 percent of the power leaving or
entering either port. But 1.4 percent is about the error level of the
solution, as shown by the asymmetry of the admittance matrix. The
actual power radiated is about Prad = (Average Gain/2) X (Power
computed from V and I at the sources) Making this correction for cases
3 and 4:
Case 3: Prad = (172.5)(.0668/2) = 5.76 Watts
Case 4: Prad = (-160.9)(-.0716/2) = 5.76 Watts
So it does not matter whether the E-plate leads or lags.
The integral of far field power is a variational form for radiated
power, integrating over the current to get radiated field and then
integrating .5*Abs[radiated E]^2/eta over the hemisphere to get total
radiated power. It works well with assumed sinusoidal currents, and
should give a very accurate result for radiated power with the much
more accurate currents from NEC.
So 5.76 watts should be an accurate result for the total radiated
power for the currents in cases 3 and 4. I don't know about the
interpretation of radiation resistance for a small antenna with two
feed ports. However, we can divide the radiated power by Abs[I_in]^2
at each port to get a resistance to compare with that of a circuit
needed to match this antenna. Using Rrad = Prad/(0.5*Abs[I_in]^2)
(0.5 in the denominator since NEC gives peak current)
Case 3: RradE = 0.109 ohms
RradD = 0.107 ohms
That looks like a small antenna.
Another thing we can see from these results is that the field produced
when the antenna is excited with VE volts on the E-plate source and VD
volts on the D-plate source is just the linear combination of the
fields when the sources are excited separately:
Erad = VE*EradE + VD*EradD.
where EradE and EradD come from cases 1 and 2. The same holds for
near E and H, as can be checked from the above numbers. Fields do not
interact or combine in free space. They just superpose and continue
doing what they were doing in the absence of the other field.
In all of these cases the ratio of Ez to Hy is much greater than the
desired value of 377 ohms in the near field. However, the ratio of Ez
to Hy is different when we excite the E-plate or the D-plate, so we
can solve for excitation voltages that will produce Ez/Hy = -376.73
ohms for an outward Poynting's vector at some distance r using the
linear combination equations. We will try this for the distance r =
3.1 m, which seems like it should be in the "interaction zone". The
result is:
***** Case 5, Excitation to produce Ez= 1 V/m, Hy= -1/376.7 A/m at r=3.1 m
VE = 3.91979 + j98.1712
VD = -10.5933 - j75.7328
IE = -0.271158 + j0.0185699 = (0.27179, angle 176.082 degrees)
ID = 0.328316 - j0.0357566 = (0.330358, angle -6.21553 degrees)
Pin1 = 0.380072 Watts
Pin2 = -0.385002 Watts
Pin_total = -4.9306E-3 Watts
Average Gain = -0.327868
Corrected radiated power = 0.001616 Watts
RradE = 0.0437 ohms
RradD = 0.0296 ohms
Erad = (0.38144, Angle 91.48 Degrees)
At r = 1.5 m: Ez = (105.93, Angle 81.79 Degrees) V/m
Hy = (9.9376E-4, Angle -74.30 Degrees) A/m
Ez/Hy = (1.066E5, angle 156.09 Degrees) ohms
At r = 2.9 m: Ez = (3.0025, Angle 61.05 Degrees) V/m
Hy = (2.8216E-3, Angle -179.68 Degrees) A/m
Ez/Hy = (1.6411E3, angle 240.73 Degrees) ohms
At r = 3.1 m: Ez = (0.99893, Angle -0.01 Degrees) V/m
Hy = (2.6544E-3, Angle 180.0 Degrees) A/m
Ez/Hy = (376.33, angle 179.99 Degrees) ohms
At r = 3.3 m: Ez = (1.6809, Angle -65.50 Degrees) V/m
Hy = (2.4830E-3, Angle 179.75 Degrees) A/m
Ez/Hy = (667.96, angle 114.75 Degrees) ohms
At r = 4.9 m: Ez = (2.6926, Angle -89.08 Degrees) V/m
Hy = (1.4269E-3, Angle 178.95 Degrees) A/m
Ez/Hy = (1887.03, angle 91.97 Degrees) ohms
At r = 10.1 m: Ez = (5.1874E-1, Angle -90.46 Degrees) V/m
Hy = (4.0013E-4, Angle 178.33 Degrees) A/m
Ez/Hy = (1295.51, angle 91.21 Degrees) ohms
At r = 39.7 m: Ez = (9.8815E-3, Angle -60.46 Degrees) V/m
Hy = (3.6608E-5, Angle 167.16 Degrees) A/m
Ez/Hy = (269.73, angle 132.38 Degrees) ohms
Well, it worked at r=3.1 m, but the ratio of Ez/Hy does not stay -377
for long. It is small near the feed, has a large peak at r = 1.5 m,
under the D-plate, then drops sharply to -377 at r = 3.1 m and then
climbs to a peak of around 1900 ohms around r = 4.9, before gradually
converging to -377 ohms in the far field. For radiation it still
looks like a small antenna.
Jerry Burke
LLNL
Received on Tue Apr 27 1999 - 14:36:14 EDT
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