Cecil, W5DXP wrote:
"You switch off one of the signals. How long does it take for me to
sense any energy?"

Yes, if the transit time is one year, that`s how long it takes to sense
any change in received signal.

I was wrong when I wrote waves would be exhibitable on demand were they
annhilated. Cecil was right. It makes no difference whether two equal
and opposite signals are received or no signal is received. Same result.

What happens at an open circuit on a transmission line? The current is
interrupted and must reverse direction as it has nowhere else to go. A
changed direction is a reversed polarity so incident and reflected
currents add to zero.

Energy in the magnetic field is eliminated by the addition to zero of
the incident and reflected currents. This canceled energy goes to the
only place it can go, into the electric field. This results in doubling
the voltage at the open-circuit end of the line.

In a short-circuit on a line, the current doubles and the voltage goes
to zero.

1/4-wave back from a short, a near open circuit exists. !/4-wave back
from an open or a short, conditions are inverted on a low-loss line.
Where there is an open-circuit at the end of a line, a near short
circuit exists 1/4-wave back.

I think that at the open circuit at the end of a line, the excess
voltage launches the reflected wave. At the short on a line, excess
current launches a reverse wave. Voltage produces current and current
produces voltage. Voltage and current are dominnated by the Zo of the
line in all movement through the line.

Cecil has argued that the high impedance at the open end of a shorted
1/4-wave stub does not inhibit current into the stub. The reflection
point is at the short, not at the high impedance point back 1/4-wave
from the short. That sounds reasonable to me, but I wonder if it makes
any difference.

Best regards, Richard Harrison, KB5WZI