wrote:
So current flowing on the coax braid = Current flowing from X to B +
Current flowing from Y to B.

+-----A------------X-----------------------------
|
+-----B------------Y


As you probably know, standing waves in distributed
networks modify the conditions. The current "flowing"
from X to B may not be the same amplitude or phase at
X and B. The current "flowing" from Y to B may not be
the same amplitude or phase at Y and B. This is NOT
a DC circuit! So how can any sane person claim that
the common mode current at X is identical to the
common mode current at B?
--
73, Cecil http://www.qsl.net/w5dxp

We're talking about the instantaneous common mode current of the real
system. That's the current actually flowing in the same direction in
both legs of the stub.

wrote:
We're talking about the instantaneous common mode current of the real
system. That's the current actually flowing in the same direction in
both legs of the stub.

Hi Dan,

Cecil now has to sign his Mensa membership card over to you.
You get the point and totally understand the problem. He doesn't.

73 Tom

wrote:
We're talking about the instantaneous common mode current of the real
system. That's the current actually flowing in the same direction in
both legs of the stub.

But standing waves of common mode current can
and probably do exist.

The amplitude of the common mode current varies
depending upon its location in the distributed
network. At a common mode standing wave node, the
common mode current may measure zero. 1/4WL away, it
may measure an appreciable magnitude. I have
measured such on my own transmission lines.

I suspect a dead short forces a common mode
standing wave node. What else could possibly
happen?
--
73, Cecil http://www.qsl.net/w5dxp

wrote:
wrote:
We're talking about the instantaneous common mode current of the real
system. That's the current actually flowing in the same direction in
both legs of the stub.

Cecil now has to sign his Mensa membership card over to you.
You get the point and totally understand the problem. He doesn't.

Seems that neither one of you guys realize that common
mode current standing waves usually exist so the standing
wave common mode current amplitude is NOT constant from
one point to another in a distributed network. The lumped
constant model strikes (out) once again. When are you guys
going to realize that RF is NOT DC?
--
73, Cecil http://www.qsl.net/w5dxp

In the case where the J-pole is isolated (crossbanding HT at the
feedpoint?) there's absolutely a node for the common mode current at
the short, just like there's a node for the current on a wire at the
end where the current can't flow.

If you put a feedline on it, though, the standing wave pattern of
common mode current changes.

The feedline + jpole develops some pattern of standing waves on it.
The common mode current is induced on the entire length of coax and
stub...

The sum of the common mode currents at point B is always as stated
before... but the magnitude of ALL THREE of them depends on the details
of the coax, whether or not there is a choke balun on it, the length
and boundary condition on the other end of the coax.

That is to say the common mode impedance at the top of the stub where
the halfwave is connected depends on the details of the feedline.

Stub vs. Stub + Coax: different common mode standing wave pattern,
different common mode impedance at the top of the stub, different
magnitude of current induced on the stub+coax than on the plain stub...


Dan

wrote:
If you put a feedline on it, though, the standing wave pattern of
common mode current changes.

+-----A------------X-----------------------------
|
+-----B------------Y


On my diagram, the coax is always connected
at A and B so *nothing has changed*. I just don't see
any reason for the common mode current at 'X' to be
exactly the same as the common mode current at 'B'.
Do you?
--
73, Cecil http://www.qsl.net/w5dxp

wrote:
Cecil Moore wrote:
On my diagram, the coax is always connected
at A and B so *nothing has changed*. I just don't see
any reason for the common mode current at 'X' to be
exactly the same as the common mode current at 'B'.
Do you?

No, but no one said that except you.
What I said was:

Sorry, that's not what you said. What you said was that there is
as much common mode current flowing down the feedline (at point B)
as is flowing into the half-wave end-fed section of the antenna
(at point X).

+--A--------X-------------------
|
+--B--------Y

The half-wave end-fed antenna starts at point X. The feedline is
at point B. Here's what you said:

wrote:
1.) You have a half-wave end-fed antenna. There has to be as much
common mode current leaving the end of that point and flowing down the
feedline as there is flowing out onto the antenna at that point. There
isn't any exception to this rule.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

The half-wave end-fed antenna starts at point X. The feedline is
at point B. Here's what you said:

wrote:
1.) You have a half-wave end-fed antenna. There has to be as much
common mode current leaving the end of that point and flowing down the
feedline as there is flowing out onto the antenna at that point. There
isn't any exception to this rule.

The STUB is a balanced transmission line. If you quoted the entire post
everyone would see what I meant.

Your constant little games of pulling things from context just waste
time and ruin the purpose of technical forums. You really should quit
that, and try to use the forums for education.

73 Tom

wrote:
Cecil Moore wrote:
The half-wave end-fed antenna starts at point X. The feedline is
at point B. Here's what you said:

wrote:
1.) You have a half-wave end-fed antenna. There has to be as much
common mode current leaving the end of that point and flowing down the
feedline as there is flowing out onto the antenna at that point. There
isn't any exception to this rule.

The STUB is a balanced transmission line. If you quoted the entire post
everyone would see what I meant.

That is a complete quote of your item number 1. What you said is
self explanatory. You said the current flowing onto the antenna
is equal to the current flowing down the feedline. You were wrong.
--
73, Cecil http://www.qsl.net/w5dxp