Roy Lewallen wrote:

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?

Now suppose instead the line is short circuited. The voltage returns
down the line, and the current does not.


That's not true. Both are totally reflected.
Thanks for the response. Why is any wave reflected at all? In the open case,
I can easily visualize the electrons (drifting cloud) having nowhere to go
when they reach the end of the line, so the current literally has to go in
reverse. Perhaps the shorted line represents a reactive circuit to the
incident wave?

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.


The math is dictated by the boundary conditions, and those can be used
to gain an intuitive feel for what's happening. When both forward and
reflected waves are present, the voltage and current at any point on the
line are the sum of the forward and reverse waves. When the line is open
circuited, the current at the end of the line is of course zero. So the
sum of the forward and reverse current wave is zero at the end, and the
only way that can be is for the forward and reverse waves to be equal in
magnitude and out of phase. Likewise, when the line is shorted, the
voltage is zero, and the only way that can happen is for the reverse
voltage wave to be equal in magnitude and opposite in phase to the
forward voltage wave.
Yes, this makes sense that one can deduce the result from the boundary
conditions. I'll stop here for the moment until I understand the answer to
the question above.

Some confusion can result when discussing the reverse wave. The reverse
current wave can be, and usually is, defined as positive in the same
direction as the forward wave. If you use this definition, the current
changes phase at a shorted end, since the sum of the two current waves
has to equal zero when both are defined as being positive in the same
direction. However, you can also say that the current simply reverses
direction and remains the same in phase. Those two viewpoints are
equivalent. I've been using, and will continue to use, the convention
that the direction of positive current is toward the load for both
waves, for this discussion.

At an open end, the forward and reverse voltages are equal and in phase,
and at a shorted end, the forward and reverse currents are equal and in
phase. This can be seen by realizing that the voltage/current ratio of
the forward wave is the same as for the reverse wave -- both equal the
line Z0, but because of the current convention, the sign of the current
reverses for the reflected wave. So we know that Vf/If = -Vr/Ir, where
If and Ir are defined as positive when flowing toward the load. Now
let's apply what we know about shorted ends to find what happens at an
open end. At an open end, we know that If = -Ir. Since Vf/If = -Vr/Ir,
it follows that Vr = Vf. Likewise, at an open end, we know that Vf =
-Vr, so from the same equation If = Ir.

Roy Lewallen, W7EL



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
--
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