Dave wrote:
"Cecil Moore" wrote:
An open 1/4WL stub detours all the associated frequency
energy into the stub and then reflects it back to the source.

a new theory!

Not a new theory. I learned that at Texas A&M back in the 50's.
I had a pretty smart RF prof. It's the same thing that happens
without a load.

XMTR---------feedline---------------open

so does all the energy go down to the end
of the stub and have to make a u-turn to get back?
Energy is normally completely reflected at a physical open
circuit in a piece of transmission line. Stubs are no exception
to transmission line theory.

oh, oh! i know this one!! but if the voltage at the mouth of the stub is
zero, how does any wave flow down into the stub?

If the standing-wave minimum voltage is zero, why is the current
at that exact point at its maximum value. Maybe you should review
"Plumber's Delight" beam construction rules and standing-waves.

there is no voltage there
to drive it?? how does the incoming wave know to not flow past the stub
connection and get reflected instead??

Because reflections only occur at a physical impedance discontinuity.
There is no physical impedance discontinuity until the open is reached.

XMTR---one wavelength feedline---+---1/4WL---open
|
load

Note the load is connected at a "Plumber's Delight" V=0 point.

The voltage 1/4WL from the open end is zero. What do you think the
current is 1/4WL from the open end? What do you think the voltage
is at the open. Sounds like you believe that there is no current
or voltage on the entire length of the above since it is an odd
number of 1/4WL's long.

doesn't that make it a virtual short
and just reflect back to the source from there??

Nope, in the above example, there is a virtual short at the XMTR output.
Does that mean to you that there's no voltage or current on the entire
feedline? Virtual impedances don't cause reflections.

On a 50 ohm feedline with an SWR of 2:1, an impedance of 100 ohms
repeats every 1/2WL but it doesn't cause any reflections at the
100 ohm points because there is not a physical impedance discontinuity
there.
--
73, Cecil http://www.qsl.net/w5dxp