On Sun, 23 May 2004 02:07:13 GMT, "Henry Kolesnik"
wrote:

I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR


Hi Hank,

Round Two.

A short or an open certainly reflects. However, as you have observed
through your question, so does a poorly matched antenna; thus you must
agree that it presents neither a short nor an open. As such,
reflection is not confined to these two conditions.

We can display a condition where we have a 2:1 mismatch. This is
fairly commonplace as a consideration, if not as a reality. Here, the
reflected power is less than the total applied (some 12% if dead
reckoning is correct). That is, if the antenna presents a 2:1
mismatch to the power applied to it, nearly 90% of that power will
proceed to be radiated with a trivial portion returned to the source
(or the tuner). If we boost that mismatch to 10:1, that increases the
reflection substantially (let's call it 90% in this spirit of dead
reckoning) and naturally less is radiated. The math is quite as
simple as a balance ledger.

10:1 for a 50 Ohm source would mean either the load presents a Z of
500 Ohms, or 5 Ohms. From this you can see that we are approaching
either an open (hi-Z) or a short (lo-Z) and either perform the same
job of reflection - short of total reflection. As such, there is no
distinction to the power whether it encounters either, the reflection
ensues by virtue of the simple mismatch, not by the literal condition.
The computation of mismatch defines how reflective the interface is.

73's
Richard Clark, KB7QHC