Cecil Moore wrote:

Over the years, I have challenged anyone on this newsgroup to create
a standing wave in a single source system without having the existence
of a forward wave and a reflected wave. Nobody has furnished any proof
that standing waves are possible in a single source system without the
existence of forward and reflected waves.

Cecil,

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. No one denies the simultaneous existence
of standing waves and traveling waves.

Isn't superposition wonderful!

73,
Gene
W4SZ

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.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

So it was you who asserted that standing wave current is "somehow
different" from traveling wave current and I agree with you. It's
obvious they are "somehow different" because they have different
mathematical equations. Have you changed your mind since your
above quoted posting?

No one denies the simultaneous existence
of standing waves and traveling waves.

Of course they do, Gene, that is the whole point. Here is a quote
from K8LV's article:

"I wish to emphasize the fact that the forward and reverse
waves really do not exist separately ..."

That certainly *denies* the separate existence of the underlying
traveling waves so your above assertion is false. I believe that
W7EL also denies the separate existence of forward and reverse
waves and introduced the technical term, "sloshing", to explain
what happens to the energy in a transmission line with reflections.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil wrote:
This is just one more example of the dumbing down of amateur radio
accompanying the dumbing down of the US educational system in
general. Unfortunately, it seems to be a trend that cannot be
reversed because it is the biased view being pushed by the ARRL
and its supporters.

Hmmm...not sure I agree that the folks at ARRL are deliberately being
dumb (or maybe I just misunderstood you). Seems more unintentional to
me. After all, the technical editor of QEX let publish that bizarre
article that claimed to prove by math that phasing SSB receivers were
not possible. Can''t imagine a political motivation for that though I
have to wonder bigtime how that one got by. QEX really really needs
for some good peer review.

Ah, to have Ham Radio magazine back again. Loved that thing. Learned
most of my radio from it.

73,
Glenn AC7ZN

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:

Isn't superposition wonderful!

73,
Gene
W4SZ

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

tom
K0TAR

En/na Dave ha escrit:
... Come on, just a little fight???

Just a question about coax cables:

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).

Now the question is: This is true for big distances from the coax, but
is it also true when you get very close to the coax? Imagine a conductor
taped to the outside of a coax for some meters. The capacitive coupling
to the braid is much higher than the coupling to the inner conductor.
Would it pick some of the current in the coax. If not, why not?

(apart from fun I'm really interested in the answer as I'm not quite
sure if a coax running parallel to unshielded and not twisted computer
cables would pick harmonics from it on RX or create interferences on TX)

--
Toni

no... and yes.... there would be some pickup, but only from leakage through
an imperfect braid. because the current is on the inside of the shield it
would not couple to the cable on the outside. the electric and magnetic
fields are contained completely inside the shield.


"EA3FYA - Toni" wrote in message
...
En/na Dave ha escrit:
... Come on, just a little fight???

Just a question about coax cables:

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).

Now the question is: This is true for big distances from the coax, but is
it also true when you get very close to the coax? Imagine a conductor
taped to the outside of a coax for some meters. The capacitive coupling to
the braid is much higher than the coupling to the inner conductor. Would
it pick some of the current in the coax. If not, why not?

(apart from fun I'm really interested in the answer as I'm not quite sure
if a coax running parallel to unshielded and not twisted computer cables
would pick harmonics from it on RX or create interferences on TX)

--
Toni

EA3FYA - Toni wrote:

Just a question about coax cables:

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.

Common mode current, by contrast, flows on the outside of the shield,
and its field radiates outward from there.

Now the question is: This is true for big distances from the coax, but
is it also true when you get very close to the coax? Imagine a conductor
taped to the outside of a coax for some meters. The capacitive coupling
to the braid is much higher than the coupling to the inner conductor.
Would it pick some of the current in the coax. If not, why not?

Again assuming a solid shield -- the center conductor carries a current
and therefore creates a field. The inner surface of the shield carries
an equal and opposite current and also creates a field. But those fields
are equal and opposite, and cancel at all points beyond a thin layer on
the inner surface of the shield. Since there's no significant field at
any point outside the shield, it doesn't matter where you look, you
won't find any, and there isn't any field to couple to anything else.

In reality, any shield other than a completely solid one (such as the
shield of hard line or semi-rigid coax) will leak some because of gaps
or holes. And the field will couple more strongly to wires which are
close than those which are far away. Whether the amount of leakage is
significant or not depends on the application.

(apart from fun I'm really interested in the answer as I'm not quite
sure if a coax running parallel to unshielded and not twisted computer
cables would pick harmonics from it on RX or create interferences on TX)

You might get enough leakage through the shield of ordinary coax to
cause problems in both cases. It depends on the transmit power level,
the signals in the wires, the length over which they're bundled, the
frequencies involved, the quality of the shield, and so forth.
Separating them even a small distance would reduce the coupling
considerably. But you're likely to have more trouble with common mode
current.

Roy Lewallen, W7EL

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

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