Owen Duffy wrote:
. . .
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)!

The only way to achieve uniform current on a short dipole is with large
capacity hats at the ends of the dipole. Otherwise, the current tapers
nearly linearly from a maximum at the center to zero at the ends. If
you'll look closely at Kraus' figure of the short dipole he analyzes,
you'll see that it has capacity hats. Nearly all other authors analyze
just a straight wire which doesn't have those hats, and consequently
linear rather than uniform current distribution. And of course get quite
a different result.

I'll bet you didn't include large capacity hats in your model. I haven't
tried it, but you should get results much closer to Kraus' if you do.
NEC analysis gives radiation resistance very close to theoretical when
analyzing a plain straight wire dipole, but this isn't what Kraus does
in his book. It is interesting, though, to see how much effect the wire
diameter has on the impedance, and that the wire has to be very thin
indeed to approach the theoretical impedance for an infinitesimally thin
dipole.

Roy Lewallen, W7EL

Owen Duffy wrote:


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?

I believe that's correct. Note that Kraus uses a plain wire dipole for
his aperture correction, but a dipole with end hats (and therefore
uniform current) for input impedance calculations. See my other recent
posting for further comments.

Roy Lewallen, W7EL

On Fri, 09 Dec 2005 13:49:31 -0800, Tim Wescott
wrote:


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)".

It has, I tend to do most my ad-hoc stuff in Perl these days, and it
uses the ** operator. Perl is c-like, but as you say Tim, it did not
inherit the ** op from C.

Languages that lack an exponentiation operator are a right pain in the
butt, but there are lots of them.

Owen
--

On Fri, 9 Dec 2005 15:41:47 +0000 (UTC), "Reg Edwards"
wrote:

Which way is correct? Terman, Kraus, Balani, their Bibles and
numerous computer programs are all at loggerheads with each other.

This is simply through massive missinterpretation by you, who disdain
the work you "recite" (a suspect activity) and the complete absence of
your using the first principles of Lord Kelvinator, whom you claim to
revere for his clarity in approaching problems through demonstrables.

Result? YAT (yet another troll)

Owen Duffy wrote:

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?

I found another source (Ramo et al) which directly gives the ratio of
power in the load of a matched receiving antenna to the power applied to
a transmitting antenna, in terms of the effective apertures of the
antennas. This doesn't require the intermediate step of calculating
field strength. The equation is:

Wr/Wt = (Aer * Aet) / (lambda^2 * r^2)

whe
Wr, Wt are received and transmitted power respectively
Aer, Aet are the receiving and transmitting antenna effective apertures
lambda = wavelength
r = distance between the antennas

Note that effective aperture, like other measures of an antenna pattern,
is a function of the direction from the antenna. So this equation is
correct regardless of antenna orientation as long as Aer and Aet are
correctly calculated.

Letting

K = 3/(8 * pi) ~ 0.1194

we can write the equation cited by Owen for effective aperture of a
short dipole in its most favored direction (broadside) in free space as

Ae = K * lambda^2

(This is the effective aperture of an infinitesimally short dipole.
However, it changes very little with length when the dipole is
electrically short. The effective aperture broadside to a half wave
dipole is only 10% greater.)

Then we get that

Wr = Wt * (K^2 * lambda^4) / (lambda^2 * r^2)
= Wt * K^2 * lambda^2 / r^2

For Reg's example, lambda = 15 meters and r = 1 km. Since this is a free
space analysis involving dipoles so far, I'll apply 2 kW (Wt) to the
transmitting dipole, resulting in Wr = 6.412 mW.

Now we can split the model exactly in half with a ground plane. On the
top of the ground plane, the transmitting antenna has exactly half the
applied power, or 1 kW, which is what we had in Reg's example. Half the
load power is in the upper plane also, so we have 3.206 mW for the load
power in Reg's example setup. This is very close to the 3.234 mW result
from the EZNEC model.

The antenna's effective height (that is, the ratio of induced voltage to
field strength) has been at issue. As Reg pointed out, ~ 3 mW at the
load requires an effective height of 0.5 meter for the 1 meter high
antenna. (I incorrectly gave it as 1 m in an earlier posting.) I did
find an explicit equation for effective height for a vertical over
perfect ground, in King, Mimno, and Wing (Dover edition, p. 165). This
also confirms that the correct effective height is 0.5 m for the 1 m
electrically short vertical antenna over ground.

I'm satisfied that we have the answer to Reg's question. It's been an
educational process for me -- thanks for posing it.

One final note, regarding the NEC applied plane wave. My earlier
statement that the resulting field is twice the plane wave source
magnitude when a ground plane is present is true only when the plane
wave is applied over perfect ground at exactly grazing incidence (zenith
angle = 90 deg.). If applied from other angles the resulting field
strength will be different. If you apply a vertically polarized wave
over a ground plane, I believe the resulting field strength will look
like the pattern from a vertical radiator over a perfect ground plane --
strongest when applied at the horizon, decreasing when applied at higher
angles, and dropping to zero if applied from directly overhead. I
haven't confirmed this, but believe it's necessary in order to get a
receiving pattern that's the same as the transmitting pattern. So use it
with caution when a ground plane is present, and don't casually make
assumptions about the resulting field.

Roy Lewallen, W7EL

Hi Roy,

A couple of comments on your excellent input into this discussion.

Roy Lewallen wrote:

I found another source (Ramo et al) which directly gives the ratio of
power in the load of a matched receiving antenna to the power applied to
a transmitting antenna, in terms of the effective apertures of the
antennas. This doesn't require the intermediate step of calculating
field strength. The equation is:

Wr/Wt = (Aer * Aet) / (lambda^2 * r^2)

whe
Wr, Wt are received and transmitted power respectively
Aer, Aet are the receiving and transmitting antenna effective apertures
lambda = wavelength
r = distance between the antennas


Kraus "Antennas" also describes this equation. He refers to it as the
"Friis transmission formula" on pages 48 and 49 of the second edition.



One final note, regarding the NEC applied plane wave. My earlier
statement that the resulting field is twice the plane wave source
magnitude when a ground plane is present is true only when the plane
wave is applied over perfect ground at exactly grazing incidence (zenith
angle = 90 deg.). If applied from other angles the resulting field
strength will be different. If you apply a vertically polarized wave
over a ground plane, I believe the resulting field strength will look
like the pattern from a vertical radiator over a perfect ground plane --
strongest when applied at the horizon, decreasing when applied at higher
angles, and dropping to zero if applied from directly overhead. I
haven't confirmed this, but believe it's necessary in order to get a
receiving pattern that's the same as the transmitting pattern. So use it
with caution when a ground plane is present, and don't casually make
assumptions about the resulting field.


I believe a better way to describe this situation is that the plane wave
field strength does not go to zero, but rather the effective aperture of
the antenna goes to zero as the plane wave is applied from overhead.
This does not change your conclusion with respect to antenna patterns.

The oblique-incidence plane wave equations are slightly messy, but they
are well described in treatments of waveguides.

73,
Gene
W4SZ

As expected, the discussion appears to have fizzled out.

Having for the first time allowed myself to be led astray by the
academic professors and authors and their Bibles on the subject, I
still have the problem of correcting a bug in my computer program. The
program itself, GRNDWAV3, is too interesting, useful and educational
to simply withdraw it.

Lets not be confused by fallible human ideas and notions on -

Reflections from the ground.
Mirror images in the ground.
Antenna gains relative to isotropic.
Antenna gains relative to isotropic with a ground plane.
Antenna gains relative to isotropic of dipoles.
Half hemispheres.
What on Earth is an isotropic antenna anyway?
And now we are being introduced to waveguides.

Thanks to Roy's investigations and clarification, the solution to my
problem is perfectly simple -

"The effective height of a short vertical antenna is half of its
actual height and the voltage induced in it is half of the field
strength in volts per metre."

Which gives the correct answers from my program.

And which was proven and well known 100 years ago by the early radio
engineers who were really the first amateurs in the game.

Having got that off my chest, I can now finish the second half of a
bottle of Australian, Bantock Station, Special Reserve, Cabernet
Sauvignon Shiraz which I can thoroughly recommend.
----
Reg, G4FGQ

Gene Fuller wrote:
. . .

[I wrote:]
One final note, regarding the NEC applied plane wave. My earlier
statement that the resulting field is twice the plane wave source
magnitude when a ground plane is present is true only when the plane
wave is applied over perfect ground at exactly grazing incidence
(zenith angle = 90 deg.). If applied from other angles the resulting
field strength will be different. If you apply a vertically polarized
wave over a ground plane, I believe the resulting field strength will
look like the pattern from a vertical radiator over a perfect ground
plane -- strongest when applied at the horizon, decreasing when
applied at higher angles, and dropping to zero if applied from
directly overhead. I haven't confirmed this, but believe it's
necessary in order to get a receiving pattern that's the same as the
transmitting pattern. So use it with caution when a ground plane is
present, and don't casually make assumptions about the resulting field.



I believe a better way to describe this situation is that the plane wave
field strength does not go to zero, but rather the effective aperture of
the antenna goes to zero as the plane wave is applied from overhead.
This does not change your conclusion with respect to antenna patterns.

I'm not sure either one of us quite has it right. In a model experiment,
I set up a short open circuited vertical dipole just above a perfect
ground plane, and applied a vertically polarized plane wave from the
horizon. Let's call the resulting voltage at the dipole center V1. Then
I changed the direction of the plane wave so it was coming from an
elevation angle of 45 degrees above the horizon, but with the same
amplitude. The dipole voltage was about 0.7 * V1, about what we'd expect
from the change in effective aperture of the vertical dipole due to the
different arrival elevation angle. But if I tilt the dipole back 45
degrees so it's parallel to the incident E field, the voltage drops to
about 0.5 * V1, another 3 dB. I believe this indicates that the field in
the vicinity of the dipole is oriented normal to the ground plane, and
it has a magnitude that's about 0.7 as great as it is when the same
amplitude wave is fired from a horizontal direction. A second check was
to tilt it 45 degrees the other way, so its end is pointing directly
toward the direction of the impinging wave. The result was again about
0.5 * V1, adding proof that the field in its vicinity is normal to the
ground plane and not tilted in the direction of the source. So the
antenna aperture is indeed changing as we change the orientation of the
antenna relative to the field in its vicinity. But that's not the same
as the orientation of the antenna relative to the direction from which
the plane wave originates. (They are of course the same if the ground
plane is absent.) The change in antenna output (in this case) when the
source direction is changed is due to the fact that the magnitude of the
field strength has changed, not because its orientation relative to the
antenna has changed.

When the direction of the plane wave is elevated 45 degrees, it has
equal horizontally and vertically polarized components. The horizontal
components cancel on reflection, while the vertical components reinforce
as before. This leaves only the vertical component in the vicinity of
the sense dipole, and it's 0.707 * the value when the same amplitude
wave is coming in horizontally. The dipole voltage is maximum when it's
oriented to be parallel with this field, that is, vertical.

At least I think this is a correct interpretation of what I'm seeing.
These effects are all tied in together, and I've spent so long looking
at the problem from the transmitting direction that I'm having some
trouble getting my thinking turned around. But I'm slowly getting there.

Roy Lewallen, W7EL