On Thursday 08 January 2009 12:54:30 pm Lee K7TJR wrote:
> I cannot wrap my mind around what the termination resistance
> is in the forward direction.
The following hand waving exercise might help. Note that there
is no mathematical rigor here. When I started thinking about
this, I started with energy flow from the receiver end towards
the radiation resistance.
The termination resistance is whatever impedance is presented by
the receiver, feedline, and transformer attached to the 'forward'
(as opposed to the 'reverse' or "normal") port, transformed by
the 2 wire Beverage as a regular transmission line, transformed
by the impedance of the reflection transformer or lack thereof.
The missing reflection transformer has nearly "infinite" impedance,
so theoretically, no energy is dissipated, and there is no signal
impedance transformation. All the energy is reflected, and again
transformed by the Beverage as transmission line, etc. The
receiver sees the open circuited transmission line (through the
matching transformer) as a stub with whatever impedance you get
from an open circuit stub (short from quarter wave or odd
multiple, open from half wave or multiple, and something in
between otherwise).
If one of the wires is shorted to ground (which looks like a couple
hundred ohms resistive and a little reactance), then not all of
the energy is available on the shorted wire to perfectly support
a proper reflection. In other words, some of the reflected
energy is common mode (same polarity on both wires) and some is
differential mode (opposite polarity). The common mode energy
would now be available to be dissipated by the radiation
resistance.
If there is a reflection transformer that is perfectly matched
and lossless, then the current on the wires would be the same
as the open wire and shorted wire case. You get some partial
conversion from differential mode to common mode because of the
ground loss.
Now I'm thinking about the energy flow from the received signal
towards the receiver.
In the case of less than perfection (real life), then some of the
energy is reflected as a common mode signal, which can degrade the
"forward" versus "reverse" port isolation (or front to back ratio,
if you want to think that way). However, as long as the impedance
of the shorted wire is a lot different than the impedance of the
open wire, most of the energy will be reflected in differential
mode, which is what you want. Since the open wire has an extremely
high resistance compared to the shorted wire, you have that big
difference. The only way I can imagine defeating that would be
excessive capacitive coupling of the end of the open wire to
something near ground potential.
Hope the hand waving helps to visualize what might be happening.
Maybe this will trigger someones memory of a more rigorous
treatment. Also, I'm tired of typing.
Victor, K1LT
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