On Sat, 1 Nov 2014 15:26:52 -0000, "gareth"
wrote:

***** Without the remnants of non-radiated power, there could NOT be
a standing wave!

That's quite true. Standing waves require a transmission line. If
all the RF has been radiated, and there are no "remnants" left in the
transmission line, there can be no standing waves because there is no
RF.

Think about the other boundary conditions.

If you unplug the coax cable and antenna, and then transmit into an
open circuit, there are no standing waves. All the RF power is
converted to heat in the output stage. There's no transmission line
upon which to produce standing waves and there's no antenna to
radiate. Without a transmission line or antenna, there can be no
radiation and therefore, not standing waves.

The other extreme is also true. If you have an infinitely long
lossless coaxial cable, with either an open, short, or black hole at
the far end, there are no reflections because the wave will never
quite reach the open or short to produce a reflection. Without a
reflection, there can be no standing waves.



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

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.

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.

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

Jeff Liebermann wrote in
news
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.


Well, doesn't that still mean there's energy there to power that decay? I get
the exponential bit (had to explore that a lot to code a synth. so the
energy never entirely vanishes, sort of like half-life in fissile materials.

Point taken about usually no transmission line. I remember CB'ers talking of
SWR meters and aiming for as little standing wave as possible. That seems to
go with what Jim (Pennino) said about not having a standing wave without a
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/


Nice. Though you do what I tend to do, caught in a singular sort of chord
sequence, expression, whatever it is. I got it from listening to Tangerine
Dream. It's a nice trap (and yours is more elegant than mine too) but it is a
trap. I found a way out of it though, if I can make the effort and maintain
the habit... When walking, I hum, or whistle, and the tunage is far more
varied than when I'm in front of an instrument. I have a small USB memoery
thinger with audio recording (and annoyingly short battery life) that I
usually fail to carry with me. Shame, because a couple of days ago I
improvised for about a half hour on a Cuban tune, 'Chan Chan', very
effectively, and had never done that before. The more I learn about Beethoven
and Schubert, the more I hear about how good they were at improvisation! Most
people seem to think that 'vclassical' was all ablout written formalism. Of
course it wasn't... it was closer to jazz in the sense that it was about
capturing improvisation. Their skill was in doing exactly that. Mozart heard
Allegri's 'Misereri' at one sitting, and recalled enough to write it down
later. I can't do that, but with a bit of technical help (if I can be
bothered to persist in using it), I might make a few things, at least enough
to test the instrument I'm trying to byuild. I have no web storage, but if
you're interested I can try to put three of four bits into a Usenet test
binary group somewhere.. They're not very elaborate, just a few very varied
ideas.

On 2014-11-01 20:44:55 +0000, Jeff Liebermann said:

On Sat, 1 Nov 2014 15:26:52 -0000, "gareth"
wrote:

***** Without the remnants of non-radiated power, there could NOT be
a standing wave!

That's quite true. Standing waves require a transmission line. If
all the RF has been radiated, and there are no "remnants" left in the
transmission line, there can be no standing waves because there is no
RF.

Think about the other boundary conditions.

If you unplug the coax cable and antenna, and then transmit into an
open circuit, there are no standing waves. All the RF power is
converted to heat in the output stage. There's no transmission line
upon which to produce standing waves and there's no antenna to
radiate. Without a transmission line or antenna, there can be no
radiation and therefore, not standing waves.

The other extreme is also true. If you have an infinitely long
lossless coaxial cable, with either an open, short, or black hole at
the far end, there are no reflections because the wave will never
quite reach the open or short to produce a reflection. Without a
reflection, there can be no standing waves.

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.



--

Percy Picacity

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.

On 11/1/2014 5:23 PM, Lostgallifreyan wrote:
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.

What? For a wave to have a "bulge" above the top of the tank means
there is a trough well below the top of the tank. The amount of liquid
does not change because you make waves in the tank.

--

Rick

rickman wrote in :

What? For a wave to have a "bulge" above the top of the tank means
there is a trough well below the top of the tank. The amount of liquid
does not change because you make waves in the tank.


I didn't say it did. Anyway, I've been looking at images on Google,
apparently the single half-wave form is concave in a tank of liquid, not
convex. Maybe that too is possible, I don't know. If it is, then you can add
liquid to what was a brim-fill tank before it overflows once the wave is set
up, which would indicate that a form of storage has been set up.

Look at it another way... Any standing wave shape in a tank will not be flat,
and when the small amount of energy maintaining it is stopped, it will give
back energy until it is flat, so it looks like a way to store energy. A 'tank
ciruit' has to have a way to store energy too.

(As to why I go into these apparently off-topic variations, it's because I
think a person can know something by being very specific, but probably cannot
understand it unless they look at as many other things as possible in which
similar action occurs.)

On 11/2/2014 5:02 AM, Lostgallifreyan wrote:
rickman wrote in :

What? For a wave to have a "bulge" above the top of the tank means
there is a trough well below the top of the tank. The amount of liquid
does not change because you make waves in the tank.


I didn't say it did. Anyway,

What did you mean by, "that tank will hold more liquid that it
would if brim full without the wave"???


I've been looking at images on Google,
apparently the single half-wave form is concave in a tank of liquid, not
convex. Maybe that too is possible, I don't know. If it is, then you can add
liquid to what was a brim-fill tank before it overflows once the wave is set
up, which would indicate that a form of storage has been set up.

You are aware that a standing wave still moves up and down, no? So a
trough turns into a crest and vice versa.

--

Rick

rickman wrote in :

You are aware that a standing wave still moves up and down, no? So a
trough turns into a crest and vice versa.


I've seen waves on water in a tank that don;t. You can see them in the
laminar flow froma tap set low, as the stream hits the ceramic. Totally
static...

About the other thing, I put it badly, but imaguine a tank brim-full. Now
imagine a static half-wave, convex, supported by the sides with walls
extended up from the tank (I didn't mention that before, stupidly). Now I'm
not even sure that ONE halfwave can be static, convex, maybe it can only be
stable concave due to gravity. I really don't know. I was just thining that
if statiuc convex IS possible, then once set up, the edns of the takk have
water reaching them below the brim, ergo you could chuck a bit more water in
to top it up. Hence storage. Same (likely dubuious) logic as a flyhweel,
in that to store extra anagery you can either rasie speed for fixed mass, or
raise mass for fixed speed. If it's recirpocal, then an ability to add mass
to that tank seemed to indicate storage capability being raised.

There may well be a flaw in all that, but I'm expressing it as best I can.