It seems reasonable that if I have an open ended transmission line that the
current would reflect back and change phase. After all, the electrons at the
end have nowhere to go but back down the line. However, the voltage is a
different matter. There is no phase reversal (or polarization change) There
doesn't seem to be an intuitive reason for this. However, if one examines
the equations for current and capacitive voltage, then it falls out of the
math. Still, where is the non-math that indicates this is true?

Now suppose instead the line is short circuited. The voltage returns down
the line, and the current does not. In this case, there doesn't seem to be
any non-math or intuitive feel for why the short should cause this--either
for voltage or current. Can one clue me in on what's really happening above
(voltage) and here (voltage and current)? Inquiring minds want to know.

Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"I often quote myself - it adds spice to
my conversation." - George Bernard Shaw

Web Page: home.earthlink.net/~mtnviews

W. Watson wrote:
"It seems reasonable that if I have an open ended transmission line that
the current would reflect back and change phase."

That`s about what happens. In fact, the incident and reflected waves at
an open circuit have voltages of the same phase and magnitudes. These
correspond to a reflection coefficient of 1 on an angle of zero. The
result is a doubling of the incident value of voltage at the open
circuit. I recall this being called "the Ferranti effect".

It stands to reason that energy is not created or destroyed but the
ebergy associated with the H-field must go somewhere when the current
stops. The only place it can go is into the E-field, so this accounts
for the instantaneous voltage doubling at the open circuit. The current
reverses direction and its incident and reflected values add to zero at
the open end of the line.

When either voltage or current has a phase reversal, but not both are
reversed, a reversal of wave travel direction is indicated. The same
thing happens at a short circuit but it is the voltages which add to
zero, indicating its phase reversal at the short.

When both current and voltage are reversed in phase, no change in travel
direction is indicated as this happens regularly in the cycle of the
wave.

Best regards, Richard Harrison, KB5WZI

W. Watson wrote:
It seems reasonable that if I have an open ended transmission line that
the current would reflect back and change phase. After all, the
electrons at the end have nowhere to go but back down the line. However,
the voltage is a different matter. There is no phase reversal (or
polarization change) There doesn't seem to be an intuitive reason for
this. However, if one examines the equations for current and capacitive
voltage, then it falls out of the math. Still, where is the non-math
that indicates this is true?

The forward current hits an open-circuit. The net current is zero.
Therefore, the reflected current must be equal in magnitude and
opposite in phase to the forward current at the open-circuit.
Since the net current goes to zero, i.e. the magnetic field goes
to zero, all the energy existing at that point must
migrate into the electric field thus increasing the voltage. And
indeed, the voltage doubles at an open-circuit indicating that
the reflected voltage is equal in magnitude and phase to the
forward voltage. A simple RF voltage measurement at the open-
circuit will prove that the above is true.

Now suppose instead the line is short circuited. The voltage returns
down the line, and the current does not. In this case, there doesn't
seem to be any non-math or intuitive feel for why the short should cause
this--either for voltage or current. Can one clue me in on what's really
happening above (voltage) and here (voltage and current)? Inquiring
minds want to know.

It's (surprise) the reverse of an open-circuit. The net voltage goes
to zero at the short indicating that the electric field is zero at
that point. Therefore, all the energy existing at that point migrates
into the magnetic field. The reflected voltage is therefore equal in
magnitude and 180 degrees out of phase with the forward voltage. And,
indeed, the current at the short is double the magnitude of the forward
current indicating that the forward current and reflected current are
equal in magnitude and phase at a short circuit. RF voltage and current
measurements prove it to be true.
--
73, Cecil, http://www.qsl.net/w5dxp