I haven't taught this subject at the University level for a few years now,
so my terminology is a little rusty. VSWR = (1 + abs(rho))/(1 - abs(rho))

When the reflection coefficient rho is +/- 1 the VSWR is infinite. When rho
= 0 the VSWR is 1:1. A VSWR of 1:1 corresponds to the unit circle at the
center of the Smith Chart.

wrote in message
oups.com...

Bob Agnew wrote:

So how does a 1/2WL piece of transmission line driving a
50 ohm load wind up with the voltage and current in phase
no matter what the SWR?
--

If the characteristic impedance of the transmission line is 50 ohms, then
there are no reflections; furthermore the current and voltage are in
phase
at every point along the line, There are no standing waves in this case.

In fact it doesnt matter how long the line is as long as it is terminated
in
its charcteristic impedance. This case corresponds to the circle in the
middle of the Smith Chart on which the impedance is constant.

It doesn't matter what the characteristic impedance of the
transmission line is as long as it is an electrical 1/2WL. As such,
you can have reflections, feeding a 50 ohm resistive load, and the
voltage and current will be in phase 1/2WL back fron the load. Antenna
matching with transmission line transformers use different impedance
transmission lines to wind up with a perfect match.
73 Gary N4AST