On 01/11/14 21:38, wrote:
As a problem for the student, how big would a wave guide have to be to
be able to transfer 7MHz?

I'll guess at 34.8488m x 15.7988m by scaling the dimensions for 5.85 to
8.2GHz. (C Band)

--
;-)
..
73 de Frank Turner-Smith G3VKI - mine's a pint.
..
http://turner-smith.co.uk
..
Ubuntu 12.04
Thunderbirds are go.

Frank Turner-Smith G3VKI wrote:
On 01/11/14 21:38, wrote:
As a problem for the student, how big would a wave guide have to be to
be able to transfer 7MHz?

I'll guess at 34.8488m x 15.7988m by scaling the dimensions for 5.85 to
8.2GHz. (C Band)

Sounds in the ball park to me.

For further reading enjoyment and why there is no EM field inside of RG-8:

http://en.wikipedia.org/wiki/Cutoff_frequency


--
Jim Pennino

On 11/1/2014 11:26 AM, gareth wrote:
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.

At first, there is no standing wave, until the wave reaches the point of
reflection
in the antenna and heads back the way it has come (because not all has been
radiated*****)
On the way back, it againn suffers the losses described above, as well as
radiating a
bit more.

It then reaches the other end and suffers further reflections ad infinitum.

An interesting conclusion is, therefore, that the I2R losses are repeated,
each tiome with a smaller loss, as the wave decrements.

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

I think the subject says it all.

--

Rick

On Sat, 01 Nov 2014 15:32:03 -0400, rickman wrote:

On 11/1/2014 11:26 AM, gareth wrote:
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.

At first, there is no standing wave, until the wave reaches the point of
reflection
in the antenna and heads back the way it has come (because not all has been
radiated*****)
On the way back, it againn suffers the losses described above, as well as
radiating a
bit more.

It then reaches the other end and suffers further reflections ad infinitum.

An interesting conclusion is, therefore, that the I2R losses are repeated,
each tiome with a smaller loss, as the wave decrements.

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

I think the subject says it all.

Yes, antennae radiate all the power fed to them.
The larger part as electromagnetic wave,
the smaller part as infrared radiation.

w.

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.