In message , rickman
writes
On 7/29/2015 4:05 PM, Ian Jackson wrote:




The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?

Maybe 'standing wave' is the wrong description. What I'm referring to is
the approximately sinusoidal current and voltage distribution along the
length of the antenna (high voltage at the ends, high current at the
centre feedpoint).

However, this does result from the outgoing AC wave meeting the wave
bouncing back from the ends of the antenna. If you accept that both legs
of the antenna have this 'waveform' (although where there's no balun,
they are probably unequal), and the shape of the ;waveform; is the
vectorial summation of the go-and-return RF signals, then it's pretty
easy to see why a fair proportion of the returning signal should head
down the outside of the shield. It's probably easier to visualise this
than trying to work out why some of the forward-going RF signal (on the
inside of the shield) should chose to do an about-turn at the antenna
feedpoint, and immediately come back down on the outside of the shield -
instead of flowing into the antenna wire.

There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.


--
Ian