Percy Picacity wrote in
:

However, this does not change the fact that standing waves do not 'use
up' any of the power fed to the aerial

Is that like potential vs kinetic energy? After all, a filter could be said
to 'store' energy in an eternal oscillation if it had no losses, and nothign
drawing output from it. The moment you do, you lose energy, the 'note' fades.

Given that if you produce a standing wave in a tank of liquid such that one
bulge exists above the rim, the standing wave can be considered a form of
storage (potential energy), because that tank will hold more liquid that it
would if brim full without the wave.

rickman wrote:
On 11/1/2014 1:03 PM, wrote:
gareth wrote:
Ignoring, for the moment, travelling wave antenna, and restricting
discussion to standing wave antennae ...

An antenna is an antenna.

Deep thoughts...


A wave is launched, and radiates SOME of the power, and suffers
both I2R losses and dielectric and permeability losses associated
with creating and collapsing the near field.

Nope, voltage is applied to an antenna causing currents to be created
which in turn cause an electromagnetic field to be created.

As antennas are made of real materials they have a resistance and the
current through that resistance leads to losses.

I thought there were *real* materials with no resistance. Isn't that
what a superconductor is?

Well, to be pendatic, there are no real materials with zero resistance
that can be used to build antennas.

As all the current existing superconductors require a bunch of supporting
equipment to keep them cold, they can't be used for antennas.

If room temperature superconductors are ever invented...

However, those are like a cure for the common cold, practical fusion
power, and peace in the Middle East, all just around the corner for
the past half century or so.



--
Jim Pennino

gareth wrote:
"gareth" wrote in message
...
Ignoring, for the moment, travelling wave antenna, and restricting
discussion to standing wave antennae ...

A wave is launched, and radiates SOME of the power, and suffers
both I2R losses and dielectric and permeability losses associated
with creating and collapsing the near field.

Of course, it goes without saying that the wave was already travelling up
the feeder
and it diffracts along the elements of the antenna, rather than being
launched from
the feedpoint!

Nope; there is an electric field in a feed line (other than wave guide)
but no electromagnetic field.

As a problem for the student, how big would a wave guide have to be to
be able to transfer 7 Mhz?

About the only antenaa where a "wave is launched" is a dielectric lens
antenna with a wave guide feed.

Of course, at the other end of the wave guide is an antenna to which
voltage is applied, which causes current flow in the antenna, which
causes an electromagnetic field to be created in the wave guide which
then flows to the antenna.


--
Jim Pennino

On Sat, 01 Nov 2014 16:05:53 -0500, Lostgallifreyan
wrote:

Jeff Liebermann wrote in
:

Without a
reflection, there can be no standing waves.

That's the one bit that comes naturally to my own understanding, such as it
is. How far does this parallel with an optical laser cavity? I'd find it
easier to understand if someone here who knows both can point out a few
essential similarotes and differences.

Sigh... topic drift again.

The parallel with a longitudinal mode laser cavity is fairly close.
http://en.wikipedia.org/wiki/Longitudinal_mode
The transmission line is some multiple of 1/2 wavelength long. The
signal bounces back and forth between the ends, reinforcing itself
with every bounce, until it spews forth from from one end or edge.
Obviously, without reflections, there would not be any laser action.

Also, in the ringing of a resonant audio filter (or any electronic filter),
there seem to be parallels there too. After all you can only have ringing, a
note produced, while energy remains in the system.

Not quite. If you apply energy to a resonant circuit (electrical or
mechanical), that then remove the input, you'll get a damped wave
(i.e. exponential decay) output where the rate of decay is determined
by the losses in the system. You could build a transmission line
oscillator, which would exhibit some rather small damped wave output
when turned off, but in most cases, there's no connection with
reflected or standing waves because there is usually no transmission
line.

Might as well be part of the problem. What I do in my spare time. I
recorded these in about 1998. Please forgive my screwups, plagerism,
lack of coherent style, sloppy fingering, etc:
http://802.11junk.com/jeffl/music/


--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

"Brian Reay" wrote in message
...
He is confusing the current and voltage distribution plots for waves.

No, there is no confusion on my part. Please explain why you think
that, for I fear that there may be confusion on your part.

Plus,
an RF wave has a magnetic component.

Well, i think we all knew that.


That can't exist IN the antenna
element as it is conductor.

Yes, and no, for it is the magnetic componentry in the wire
that causes the skin effect.

"Brian Reay" wrote in message
...
It is nonsense, they can be no wave in the element due to it being a
conductor.

You seem to be unaware that a travelling wave around a wire is what
causes the wave to move along the wire, and not the electrons inside,
which only oscillate a very short distance about their mean.

He is confusing the I and V plots for waves.

There is no confusion on my part. Perhaps you could explain where you
think I am confused, for I had not mentioned the separated I and V
waveforms.

Perhaps you are confused yourself, perhaps, by the current maximum
at the centre of a dipole, for it is not a DC maximum but rises and falls
in magnitude?

On Sat, 1 Nov 2014 21:14:48 +0000, Percy Picacity
wrote:

However, this does not change the fact that standing waves do not 'use
up' any of the power fed to the aerial (in principle, increased current
intensity increases resistive losses, but this loss can be made
arbitrarily low by having a lower wire resistance). Standing waves do
not in principle use 'power' at all and certainly do not dissipate
energy that otherwise would be radiated. They require a signal to be
applied to the transmission line but, whether the power is radiated at
the other end or the signal merely meets a mismatch, say an open
circuit, the standing wave does not affect, or need to use, any of the
power that leaves the other end. Indeed they work just as well if no
power whatever is used, as in the open circuit case.

I'll make it even easier. An RF signal can only do three things:
- Radiate (as in an antenna)
- Conduct (pass through as in a transmission line)
- Dissipate (convert to heat)
Real transmission line and antenna systems involve combinations of
these three mechanisms. If you run into something that doesn't quite
fit into one or more of these mechanisms, it's probably wrong.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

gareth wrote:
"Brian Reay" wrote in message
...
It is nonsense, they can be no wave in the element due to it being a
conductor.

You seem to be unaware that a travelling wave around a wire is what
causes the wave to move along the wire, and not the electrons inside,
which only oscillate a very short distance about their mean.

You seem to be unaware that current is the total, net movement of all
the electrons in a wire, not just a single electron.

He is confusing the I and V plots for waves.

There is no confusion on my part. Perhaps you could explain where you
think I am confused, for I had not mentioned the separated I and V
waveforms.

Yeah, right.

Perhaps you are confused yourself, perhaps, by the current maximum
at the centre of a dipole, for it is not a DC maximum but rises and falls
in magnitude?

Only a very confused individual would babble on about the instantaneous
current or voltage.



--
Jim Pennino

On 2014-11-01 22:02:48 +0000, Jeff Liebermann said:

On Sat, 1 Nov 2014 21:14:48 +0000, Percy Picacity
wrote:

However, this does not change the fact that standing waves do not 'use
up' any of the power fed to the aerial (in principle, increased current
intensity increases resistive losses, but this loss can be made
arbitrarily low by having a lower wire resistance). Standing waves do
not in principle use 'power' at all and certainly do not dissipate
energy that otherwise would be radiated. They require a signal to be
applied to the transmission line but, whether the power is radiated at
the other end or the signal merely meets a mismatch, say an open
circuit, the standing wave does not affect, or need to use, any of the
power that leaves the other end. Indeed they work just as well if no
power whatever is used, as in the open circuit case.

I'll make it even easier. An RF signal can only do three things:
- Radiate (as in an antenna)
- Conduct (pass through as in a transmission line)
- Dissipate (convert to heat)
Real transmission line and antenna systems involve combinations of
these three mechanisms. If you run into something that doesn't quite
fit into one or more of these mechanisms, it's probably wrong.

If 'conduct' includes the case where the signal goes to the other end
of the transmission line but does not go beyond it to any other
component, I'll agree with you. In that case, neglecting losses, no
power is used (apart from a truly tiny amount transiently as the wave
builds up and energy is stored in the first few microseconds)

--

Percy Picacity

gareth wrote:
"Brian Reay" wrote in message
...
He is confusing the current and voltage distribution plots for waves.

No, there is no confusion on my part. Please explain why you think
that, for I fear that there may be confusion on your part.

Plus,
an RF wave has a magnetic component.

Well, i think we all knew that.


That can't exist IN the antenna
element as it is conductor.

Yes, and no, for it is the magnetic componentry in the wire
that causes the skin effect.

Magnetic fields can exist in a conductor.

Electromagnetic fields can not exist in a conductor.



--
Jim Pennino