Hi Paul,

"Paul Keinanen" wrote in message
news
On Fri, 29 Sep 2006 17:13:28 -0700, "Joel Kolstad"
wrote:

No. The problem is that wavelength of a, say, 10kHz signal is 3km in free
space.

This must be a typo, since the free space wavelength at 10 kHz is 30
km.

Oops, sorry... yeah, you're correct -- my mistake.

On Sat, 30 Sep 2006 09:56:11 +0300, Paul Keinanen wrote:

On Fri, 29 Sep 2006 16:05:21 +0300, Ceriel Nosforit
wrote:

Hey all,

I read about people using their sound card to catch transmissions from
dedicated senders such as SAQ by just hooking a roll of wire to the
mic-in and apparently finding some success in this. This got my
wondering, what aside from laws and fines is stopping me from hooking my
80W stereo amp up to some sort of antenna for global transmission fun?
Somebody must have thought of this before...

At least previously, the frequency tables started at 9 kHz, so anything
below that would not cause any interference to any other service.

However, the problem with VLF is that any practical antenna is going to
be very short compared to wavelength and since the radiation resistance
is proportional to the square of frequency for antennas well below 1/4
wavelength, most of the power injected into an antenna is going to be
dissipated in resistive losses.

At the LF aeronautical beacon band with 90 m antennas, the antenna
efficiency based on measurements flown around these beacons seems to be
about 1 %. In Europe, the maximum _radiated_ power limit on the 135 kHz
amateur radio band is 1 W, but generating that kind of radiated power
with reasonable sized antennas (30 m) would require at least 1 kW of
transmitter power, indicating that the practical antenna efficiency is
about 0.1 %. At 13 kHz, the efficiency would be about 0.001 %.

The near field distance for a simple antenna extends to about 1/6
wavelength, so at VLF, the practical communication range for amateur
communication systems would be well within the near field.

Since you are apparently from Finland and since the Finnish
telecommunication law only grants the jurisdiction to the
telecommunication authorities for "freely propagating" electromagnetic
radiation, my interpretation of the law is that it does not cover any
near field i.e. magnetic or electrostatic communication systems, in
which the near field communication systems work.

Of course, if you are able to generate huge magnetic or electric fields
that cause interference to other systems, this may cause problems to
you.

But otherwise, go ahead with your experiments, but unfortunately the
laws of physics will hit you sooner or later :-).

Paul OH3LWR

Woha. A lot of info to assimilate. Thank you.

A few quick questions; the efficiency at higher frequencies does only
increse logarithmically, correct? Around where is the 'knee' where
increment of frequency no longer provideas a significant increase in
efficiency? - Ballpark numbers and guesstimates are perfectly OK for me,
since I'm only curious of the general idea.
Finally, what exactly do you mean by 'near field'? Is it an arbitrary line
in the sand that separates near fields from normal fields, or a completely
different physical phenomena?

--
Nos

"Ceriel Nosforit" wrote in message
news
A few quick questions; the efficiency at higher frequencies does only
increse logarithmically, correct?

Sort of, but not really. It's linear for electrically short antennas, but
it's actually oscillatory in nature (albeit this occurs after you've hit a
half wavelength -- not something you're likely to do at VLF).

Finally, what exactly do you mean by 'near field'? Is it an arbitrary line
in the sand that separates near fields from normal fields, or a completely
different physical phenomena?

Any element that you're pumping energy into creates an electric and magnetic
field. In a non-radiating component, that energy just "flows" around the
element, e.g., the magnetic field diagrams you're probably familiar with for
something like a solenoidal inductor. "Near" and "far" field aren't
necessarily precisely defined terms, but the idea is that that energy that's
just flowing around the element is in the "near field" whereas the energy
that's actually radiating out towards Alpha Centauri is in the "far field."
As a ballbark estimate, the near field of an element is within about a
wavelength away (in distance) from it... hence the mention that, at 10kHz with
a 30km wavelength, anywhere in your home is within the near field.

Mathematically, if you look at the equations for a dipole, the electric and
magnetic field look something like E(r) or H(r) = foo/r+bar/r^2+bar/r^3+...
(where r is the radial distance from the dipole), and hence the power in the
field (the Poynting vector) is something like P(r) = baz/r^2+quux/r^3+...
From this equation, you can see that over large distance the only term that
matters is bar/r^2 -- this is the far field radiation. (Another way to define
near field vs. far field, in fact, is to solve P(r) for r when the baz/r^2
terms equals quux/r^3+... -- reasonable enough, as at that point, the far
field energy flux is equal to the near field energy flux.)

The above is a fair amount of hand-waving and probably some of it is just
plain wrong :-) -- you really should pick up a book on the topic or start
Googling. Krauss' books are excellent, by the way, in that the exercises are
often "real world"-based, meant to demonstrate either how something really
does work or, if it isn't practical, why not.

The Germans call near-field and far-field the Fresnel-region and the
Fraunhofer-region, which is really a lot more colorful IMHO.

---Joel

On Sun, 01 Oct 2006 01:23:51 +0300, Ceriel Nosforit
wrote:

A few quick questions; the efficiency at higher frequencies does only
increse logarithmically, correct? Around where is the 'knee' where
increment of frequency no longer provideas a significant increase in
efficiency? -

You can model the antenna system as a series connection of a loss
resistance and the radiation resistance. The loss resistance will
convert the RF power created with your expensive gear into heat :-),
while the power "dissipated" by the radiation resistance will actually
disappear as EM fields.

For full sized antennas, the loss resistance is small compared to the
radiation resistance and most of the RF power generated will actually
radiate. However, even with 1/4 wave vertical antennas, the ground
losses can be significant, if not ground planes are used, but the
antenna is grounded with grounding electrodes.

When the antenna size is below perhaps 1/10 wavelength, the radiation
resistance drops by the square of frequency, so the radiation
resistance can quite easily be well below 1 ohm at LF and below. The
loss resistance (include grounding and loading coil losses) can be
several ohms, thus the majority of the generator power is dissipated
in the losses and only a very small part is actually radiated by the
very small radiation resistance.

While the metallic conductor skin effect losses are proportional to
the square_root_ of frequency, the ground losses are harder to
predict, so the total efficiency at VLF frequencies is almost
proportional to the square of frequency (not logarithmic). The 'knee'
would be at 1/4 wavelength for a vertical antenna. At even higher
frequencies, the radiation resistance would vary cyclically.

Paul OH3LWR

ORIGINAL MESSAGE:

On Sat, 30 Sep 2006 00:12:10 +0300, Ceriel Nosforit
wrote:

Hehe. Well yeah, I figured as much, but why doesn't it work? Too high
impendance? Why is that something the amp can't handle, if I maybe resist
the urge to crank it up to eleven?

I'm approaching the issue from the engineering perspective, but I don't
have much knowledge in this particular field. If you can enlighten me I'd
be much obliged.

------------ REPLY FOLLOWS ------------

Here's what is working against you:

1. Frequency is too low for the power level you are using. At higher
frequencies in the shortwave region, 80 watts is plenty to talk around
the world, but not down in the audio range.

2. If you did have lots of power, you would still need an efficient
antenna. Unless you own an entire county, you won't have enough room
to put one up.

The US military does use very low frequencies in the audio range to
communicate with submarines world wide, but they have a pretty good
budget for it. :-)

It can be done, but not by you or me.

Bill, W6WRT

Paul Keinanen wrote:
. . .
When the antenna size is below perhaps 1/10 wavelength, the radiation
resistance drops by the square of frequency, so the radiation
resistance can quite easily be well below 1 ohm at LF and below. The
loss resistance (include grounding and loading coil losses) can be
several ohms, thus the majority of the generator power is dissipated
in the losses and only a very small part is actually radiated by the
very small radiation resistance.
. . .

Neglecting ground effects, the radiation resistance of a four-sided loop
100 meters on a side at 100 kHz is 45 milliohms. The radiation
resistance of a four-sided loop 10 meters on a side at 10 kHz is about
400 picoohms (4 X 10^-10 ohms). If you used 2 mm diameter wire to
construct the loops, the first would have an efficiency of 0.7%, and the
second of 0.00000016 percent. These are very optimistic, since they
don't account for the considerable loss you'd incur by induction into
the ground and objects for quite some distance around. They also don't
account for losses in the required impedance matching network. And a
receiving antenna would have the same efficiency.

You can easily get these numbers with the free EZNEC demo program from
http://eznec.com. Or use a calculator and the simple equations you'll
find in any antenna text.

The OP wondered if anybody had ever thought of this before. The answer
is yes, the first time probably well over a hundred years ago. Anyone
doing the simple calculations sees immediately why it's not a great idea.

Roy Lewallen, W7EL

On Sun, 01 Oct 2006 13:12:54 -0700, Roy Lewallen wrote:

Paul Keinanen wrote:

snip

The OP wondered if anybody had ever thought of this before. The answer
is yes, the first time probably well over a hundred years ago. Anyone
doing the simple calculations sees immediately why it's not a great idea.

Yes, I'm beginnig to see that now. However, I am learning a great deal
in the process, and the equations I have found to explain the issue in
detail don't seem simple at all. But if it was simple it wouldn't be
interesting, now would it?

Right now I'm looking at;
http://en.wikipedia.org/wiki/Near-field
http://en.wikipedia.org/wiki/Radiation_pattern
http://en.wikipedia.org/wiki/Fresnel_zone

Lucky for me I'm trying to build up an interest in math. Cos here's plenty
of it...

Again, thank you all for your replies. It'll be a while until I really
have an intuitive understanding of the subject like you all do, but I am
patient.

--
Nos

To get the numbers I came up with, all you need is the equation for the
radiation resistance of a small loop and the resistance of copper wire.
It shouldn't be hard to find the equation for the small loop, and it's
very simple. At those frequencies, the resistance of the wire should be
nearly the same as the DC resistance, so all you need is a copper wire
table(*). Both should be readily available on the web.

Of course, if you want to know about other kinds of antennas and more
detailed effects like the field intensity in the presence of ground, the
equations can get difficult indeed, and some are too difficult to solve
directly. That's what the modeling programs are for.

(*) At higher frequencies or with much larger diameter wires, you'll
need to account for skin effect. This requires calculation of a square
root and can be done on a pocket calculator as can the others.

Roy Lewallen, W7EL

Ceriel Nosforit wrote:
On Sun, 01 Oct 2006 13:12:54 -0700, Roy Lewallen wrote:

Paul Keinanen wrote:

snip

The OP wondered if anybody had ever thought of this before. The answer
is yes, the first time probably well over a hundred years ago. Anyone
doing the simple calculations sees immediately why it's not a great idea.

Yes, I'm beginnig to see that now. However, I am learning a great deal
in the process, and the equations I have found to explain the issue in
detail don't seem simple at all. But if it was simple it wouldn't be
interesting, now would it?

Right now I'm looking at;
http://en.wikipedia.org/wiki/Near-field
http://en.wikipedia.org/wiki/Radiation_pattern
http://en.wikipedia.org/wiki/Fresnel_zone

Lucky for me I'm trying to build up an interest in math. Cos here's plenty
of it...

Again, thank you all for your replies. It'll be a while until I really
have an intuitive understanding of the subject like you all do, but I am
patient.

On Sun, 01 Oct 2006 20:36:23 -0700, Roy Lewallen
wrote:

To get the numbers I came up with, all you need is the equation for the
radiation resistance of a small loop and the resistance of copper wire.
It shouldn't be hard to find the equation for the small loop, and it's
very simple. At those frequencies, the resistance of the wire should be
nearly the same as the DC resistance, so all you need is a copper wire
table(*). Both should be readily available on the web.

Small magnetic loops with gains in -30 ..-60 dB range can be usable
for receiving due to the extreme noise levels on LF and VLF, but for
transmitting, they are far to lossy.

To get any significant communication distance, you would need a
vertical polarised signal. The popular antenna among 135 kHz
experimenters as well as in LF aeronautical beacons is a vertical
tower with as much top capacitance as you can put up.

Look at the antenna systems of old ships using the LF band, these have
multiple parallel wires erected between the masts in the bow and
stern. These wires form the top capacitance and a vertical wire going
directly from it isolator on the radio room to the top wires, which is
the actual vertical radiator. The top loading will increase the
current in the vertical conductor and hence vertical radiation.

Paul

You can use ferrit antenna rods. That works for receivers very fine in VLF
band and for transmitters too - as long as the ferrit won't go into
saturation.
That is a magnetic antenna and it is much shorter than the equivalent
mentioned.

- Henry


"Paul Keinanen" schrieb im Newsbeitrag
news
On Fri, 29 Sep 2006 17:13:28 -0700, "Joel Kolstad"
wrote:

No. The problem is that wavelength of a, say, 10kHz signal is 3km in
free
space.

This must be a typo, since the free space wavelength at 10 kHz is 30
km.

Paul OH3LWR