On Mon, 4 Sep 2006 20:44:17 +0100, "David" nospam@nospam wrote:

RF transmitter output has impedance of 50 ohms and is connected to dipole
with a feedpoint impedance of 50 ohms via feeder with characteristic
impedance of 50 ohms. System is perfectly matched. I expect SWR meter to
show perfect match of 1:1.

Accepting the figures as an example and not necessarily a reality...

You have just described a point which is a junction between:
- a load where the ratio of voltage to current is 50+j0;
- a feedline whe
* the ratio of the voltage to current due to the forward travelling
wave must each be in the ratio 50+j0;
* the ratio of the voltage to current due to the reflected
travelling waves must each be in the ratio 50+j0.

Having regard for the sign of the traveling waves, the only solution
to those constraints / conditions is that the reflected travelling
wave must have zero amplitude.


Dipole has a standing wave on it. Ends of dipole are at high voltage. Dipole
is centre-fed with centre being high current point. Standing wave means that
a reflected wave exists. Wave is reflected from open ends of dipole. What
happens to the reflected wave? How does it vanish at centre of dipole? Why
does reflected wave not go along feeder into transmitter output? There
cannot be a reflected wave on feeder because SWR is 1:1.

Lets be clear that we are now talking about a single frequency.

At any point, the forward and reflected waves resolve to a single
voltage at that point, and a single current flowing at that point, and
the ratio of voltage to current is the impedance (and these are all
complex quantities, ie they have real and imaginery parts).

If the point you consider is the feedpoint, and the ratio of voltage
to current is 50+j0, then that is the impedance, it fully describes
the load at that frequency.

You talk of the "reflected wave" as if it has inertia, that it must
keep travelling when it reaches the junction of the feedline and the
antenna? You are not alone in speaking that way, but thinking that way
will get in the way of understanding what is happening. Next thing,
you will be thinking that the reflected wave must travel back to the
PA anode and will be absorbed there causing overheating.

Owen
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