Gene Fuller wrote:

Isn't superposition wonderful!

73,
Gene
W4SZ

Yup, it's why I find religion so amusing.

tom
K0TAR

wrote:

Funny thing is that I'm down with the slow-wave loading coil thing.
Maxwell's equations are your friend... not a very nice friend when
Bessel functions are involved, but...

Faster than light, though? Maybe the phase velocity in some tricky
cryogenic world, but I dunno...


And light has now gone faster than light. Well, sort of. It did get
there before it left in this case.

http://www.scienceblog.com/cms/light...rds-10590.html

tom
K0TAR

Cecil Moore wrote:
Gene Fuller wrote:

Why would anyone try to prove that the basic math of adding sinusoidal
functions is incorrect? To the contrary, you are the one who insists
that a standing wave and its constituent traveling wave components are
somehow different and unique.


Actually, it was you who made that assertion and thanks for the
opportunity to quote you once again:

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

So standing waves are "somehow different" from traveling waves
according to your own assertions. The traveling wave possesses
phase characteristics and the standing wave doesn't.


Cecil,

You keep making the same mistake. Yes, you can analyze traveling waves
instead of standing waves if you so choose. However, there is not one
bit of additional physical information in the traveling waves that is
not in the standing wave. Any "phase characteristic" is simply a
function of the mathematical manipulations you use.

Perhaps someday you will actually understand superposition, but I won't
hold my breath.

73,
Gene
W4SZ

Gene Fuller wrote:
However, there is not one
bit of additional physical information in the traveling waves that is
not in the standing wave.

I agree with you but W8JI and W7EL have rejected the concept that
there is any phase information in the standing wave current magnitude.
They have rejected any use of the arc-cosine function in calculating
that phase. The following graphs show the difference in the standing
wave current and the traveling wave current.

http://www.qsl.net/w5dxp/travstnd.GIF

The standing wave current phase contains zero phase information
as you have stated. As you say, all the standing wave current
phase information is contained in the magnitude but the arc-cosine
function for obtaining that phase information has been rejected by
the experts. For the traveling wave, there is phase information
contained in the phase, none in the magnitude.

Every time you make a technical assertion, you support my argument.
Seems your argument is really with the side that rejects the arc-
cosine function for obtaining phase information.
--
73, Cecil http://www.qsl.net/w5dxp

En/na Roy Lewallen ha escrit:
EA3FYA - Toni wrote:

I know that a coax cable does not radiate (if common mode currents
properly suppressed) because both conductors are apparently "in the
same place" (wouldn't know how to express it in more technical terms).

Here's why it doesn't radiate: In a coaxial cable with a solid shield,
the differential mode current is entirely inside the shield. Current and
fields penetrate only a very small distance from the inner surface of
the shield, and no significant amount ever makes it through to the
outside. This is assuming that the shield is at least several skin
depths thick, which is a good assumption at HF and above.

When you say "Current and fields penetrate only a very small
distance...", I agree for the current part, but I'm not so sure for the
fields part:

As I understand it you can not "stop a field" in no way, though you can
certainly nullify it with an identical but opposite field.

Then the question is whether the two fields (the one from the current
flowing in the shield + the one from the current flowing in the inner
conductor) nullify at all points in the immediate vicinity of the
shield. I certainly believe it but would like to understand why this is so.

I guess the mathematical proof would involve assuming the braid is an
infinite number of conductors equally spaced around the center
conductor, each having it's infinitesimal share of the shield current,
and integrating all of their fields at the point of interest (Would
probably be able to do so back when I was at university but now it is
too strong math for me). Would this be a good approximation of the problem?

--
Toni

Thanks four your answers.

I was forgetting you normally use coax in a unbalanced configuration
where the braid is supposed to be at 0 voltage so only currents matter.

Would all this still hold if you used the coax as a _balanced_
transmission line? (unusual but -I think- possible). In this case
wouldn't voltages develop on the braid that could capacitively couple to
other conductors?

(assuming perfect solid shield, ...)

--
Toni

the voltage on the braid is not zero on the inside, it varies along with the
wave traveling along the inside of the coax. and the currents are exactly
balanced inside the coax also, they have to be or it wouldn't work. this
notion of balanced vs un-balanced transmission lines is totally confusing to
most people, in a proper system, say just with a dummy load on a coax the
currents on the shield exactly balance the current on center conductor. so
why do we go through all this stuff with bal-uns?? and coax chokes?? the
currents are already balanced, so WHY?? come on you gurus out there,
explain this one!


"Toni" wrote in message
...
Thanks four your answers.

I was forgetting you normally use coax in a unbalanced configuration where
the braid is supposed to be at 0 voltage so only currents matter.

Would all this still hold if you used the coax as a _balanced_
transmission line? (unusual but -I think- possible). In this case wouldn't
voltages develop on the braid that could capacitively couple to other
conductors?

(assuming perfect solid shield, ...)

--
Toni

Dave wrote:
so
why do we go through all this stuff with bal-uns?? and coax chokes?? the
currents are already balanced, so WHY?? come on you gurus out there,
explain this one!

Water comes out of a hose whether the hose is leaky or not.
So why ever bother patching or replacing a leaky hose?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Gene Fuller wrote:

However, there is not one bit of additional physical information in
the traveling waves that is not in the standing wave.


I agree with you but W8JI and W7EL have rejected the concept that
there is any phase information in the standing wave current magnitude.
They have rejected any use of the arc-cosine function in calculating
that phase. The following graphs show the difference in the standing
wave current and the traveling wave current.

http://www.qsl.net/w5dxp/travstnd.GIF

The standing wave current phase contains zero phase information
as you have stated. As you say, all the standing wave current
phase information is contained in the magnitude but the arc-cosine
function for obtaining that phase information has been rejected by
the experts. For the traveling wave, there is phase information
contained in the phase, none in the magnitude.

Every time you make a technical assertion, you support my argument.
Seems your argument is really with the side that rejects the arc-
cosine function for obtaining phase information.

Cecil,

You still don't get it.

When I said the phase information was gone, I meant it. Any phase
information you think you find by looking at the constituent traveling
waves is merely an artifact of the math. It has no physical meaning or
reality. If there is anything interesting left in the traveling wave
analysis, then the standing wave is not the complete representation of
the electromagnetic phenomena. This is a different problem.

Yes, you can apply modulation, insert directional couplers, look at
startup transients, or perform other tricks to get "real" phase
information. However, that again becomes a different problem, not the
original simple steady-state combination of traveling waves into a
standing wave.

73,
Gene
W4SZ

Toni wrote:
En/na Roy Lewallen ha escrit:
EA3FYA - Toni wrote:

I know that a coax cable does not radiate (if common mode currents
properly suppressed) because both conductors are apparently "in the
same place" (wouldn't know how to express it in more technical terms).

Here's why it doesn't radiate: In a coaxial cable with a solid shield,
the differential mode current is entirely inside the shield. Current
and fields penetrate only a very small distance from the inner surface
of the shield, and no significant amount ever makes it through to the
outside. This is assuming that the shield is at least several skin
depths thick, which is a good assumption at HF and above.

When you say "Current and fields penetrate only a very small
distance...", I agree for the current part, but I'm not so sure for the
fields part:

As I understand it you can not "stop a field" in no way, though you can
certainly nullify it with an identical but opposite field.

You bet you can stop a field. It can be stopped either by reflection,
absorption, or a combination of the two. Inside an anechoic chamber,
absorbing materials stop internal fields to prevent reflections. A
screen room or metallic shield reflects external fields.

Then the question is whether the two fields (the one from the current
flowing in the shield + the one from the current flowing in the inner
conductor) nullify at all points in the immediate vicinity of the
shield. I certainly believe it but would like to understand why this is so.

Indeed they do. Look up Ampere's Law. If you draw a boundary through the
middle of the shield or outside the shield, you'll find that the sum of
currents within that boundary is zero. According to the law, that means
that no net field penetrates the boundary. Because of the physical
symmetry, no net field means no field at all.

I guess the mathematical proof would involve assuming the braid is an
infinite number of conductors equally spaced around the center
conductor, each having it's infinitesimal share of the shield current,
and integrating all of their fields at the point of interest (Would
probably be able to do so back when I was at university but now it is
too strong math for me). Would this be a good approximation of the problem?

No, it's not that complicated, but a path or surface integration is
required to use Ampere's law.

Roy Lewallen, W7EL