On 8/3/2015 4:40 AM, Jeff wrote:

So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Maybe you can draw a diagram and show yourself where the current flows.
The earth connection on the shield outside is in parallel with the
earth connection on the shield inside. You can't have any current flow
on the outside because any current flowing on the shield inside balances
with the current flowing in the center conductor. The resistor requires
the two currents are equal with none left to travel down the shield
outside.


Indeed, as you state there is no current flow on the outer of the coax;
the same situation exists for a perfectly balanced dipole attached
directly to the coax, as far as the coax is concerned the situation is
no different, all it sees is a perfectly matched load, just like the
resistor, all of the power is dissipated in the antenna and none flows
anywhere else including the coax outer.

But there is a difference. With the resistor load the current flowing
from the inner lead of the coax *must* (by Kirchoff's 1st law) balance
the current flowing on the shield inner surface. So there is no current
remaining to flow on the shield outer surface. No diagram needed.

For the case of the dipole, the load is no longer balanced with the
shield outer surface connected to one side and there is no requirement
for the current in the two arms to be equal. So in this case current
flows on the shield outer surface in an amount inversely proportional to
the impedance seen by the current in the antenna element and the shield.

--

Rick