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

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
--
"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:

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?
Well, I looked at some physics books that I have access to, Waves by
Crawford, in particular. He spends an entire chapter, called
Reflections(!), and seems to hit every aspect of how all this works in every
conceivable circumstance. One somewhat length section deals with
transmission between different media using a variety of explanations using
dash pots, strings, and so on. He gets all this out of the way by developing
reflection coefficients and other tools, including some boundary analysis, I
believe. Then in about 2-3 pages dispenses with transmission lines by
looking at boundary conditions using what was developed earlier. This is
non-trivial material, and would likely takes many pages of explanation and
diagrams to reduce it to a simpler form.

I think I'll pass on going any further with my question via this thread. I
can probably drag out enough satisfactory info on this from the book and a
few other sources to keep me happy. We could develop a very long thread
otherwise. Nevertheless, thanks for your interest and insights via shedding
some light on boundary conditions.

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

Wayne Watson wrote:
"The voltage returns down the line, and the current does not."

What happens on reflection is that either the phase of the current is
reversed with an open circuit, or the phase of the voltage is reversed
with a short circuit, in the event of a complete reflection from an open
or a short. In either case, there is a reversal in phase of only one of
the components of the electromagnetic wave, not both. However, the total
wave is reflected.

A transverse electromagnetic wave consists of two components, an
electric field and a magnetic field. An incident wave has both parts
in-phase until it is reflected.

A reflected wave has its two parts 180-degrees out-of-phase upon its
first reflection. This fact is how the directional coupler in the Bird
Wattmeter distinguishes between the waves traveling in opposite
directions through the wattmeter.

Both voltage and current return upon reflection. The difference is zero
phase difference in one direction, and 180-degrees difference in the
opposite direction.

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

Wayne,



*** Imbedded comments:



"W. Watson" wrote in message
nk.net...

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?



Wayne,

Welcome to the wide-world of the inquisitive.

*** Please understand, I'm not trying to be, snooty, a wise-guy or know
it-all. I'll say that I think you are trying to use an intuitive view when
you don't have the proper background/experience to base the intuition on. I
think that you are trying to use "standard" concepts when trying to
understand waves. Intuition comes from some kind of experience that tells
us how things work, but, in regular life, we only gain experience from
mostly "normal" things. Waves are a bit different and it takes some new
experiences to gain that intuition. That's why they teach those wave tanks
in high school -- it give us some understanding of this wave phenomenon.

I can liken it to trying to understand multiplication when all you know is
addition and subtraction. Moving to multiplication or division isn't really
hard, it just takes some new concepts we hadn't seen before,. However, once
we understand it, it's "simple" and the concepts become second nature. We
need to realize that some NEW concepts are needed to understand waves, in
general.



On to waves... Someone may be able to use pure math and "explain" waves,
but there are more intuitive (gut feel) ways which help. Besides, I'd have
to work really hard trying to reconstruct all that math from college. ( I
couldn't solve a triple integral now to save my life)

Waves can be hard to explain in terms of ONLY current and voltage - when we
try to use our purely physical experience. I work hard to form mental
models of how electronics works and found that there are some areas that are
hard to understand with a simple Newtonian understanding (though I do my
best to keep my "mental models", as I call my understanding, as simple as
possible) - and it has served me well when trying to understand this stuff
and explain it to others.

The best analogy for waves is waves on water. Thought there is a major
shortcoming in that electromagnetic waves in space don't need a medium to
carry them, the rest works out pretty well. On a transmission line there
are waves and they behave like water waves.

Some important concepts to possibly help.

Water has mass and it takes a force to move it.

When a force ( a rock or wind) pushes on the water, it moves in a certain
way - very analogous to electrical current.

In short, a body of water has something analogous to a characteristic
impedance. "Push" it with a voltage and it responds with an amount of
current according to the analogous "ohm's law" of water.

When the water in one place moves, it pushes on the next bit of water. That
push causes a motion in a manner in a way only water can respond (with its
"ohm's law"). This is what makes waves happen. There *IS* "something"
moving along with that wave. Now put a wall in the water - the edge of the
tank. The wave reflects--we CAN SEE it with our own eyes. While you can do
some force, mass, acceleration and velocity math and make it all work out,
I'll stop here.



Going more general on you here... Some time go I realized that thinking
about water pushing on water leads to something like a "characteristic
impedance" of water. This concept allows the idea of the motion of the
water being transferred to adjacent bits of water and this leads to a wave
"propagating" / traveling` over the water.

Now, with this in mind, whenever the stuff that is pushed on by the water
does not respond just like water, something different happens. Since the two
extremes are a short and open, it is reasonable to assume (and correct in a
very general cense) that whatever does happen is somewhere between the
"open" and "short" behavior.

Unfortunately, with water, it is hard to come up with both a short and an
open, so this model falls apart a bit. However, the Slinky toy, stretched
out on the floor can be observed and have either a short or open placed at
the end---as well as other various "loads".

For the "short" hook it to a solid, stationary object (a wall). For the
open, put a long string between the far (load) end and the wall. For the
others, you can put dashpots and other springs (of different size and
therefore "characteristic" impedance).



I know this is a digression off your question, but I think you should study
waves in other media which are more directly observable to feel better about
transmission line waves.



Hence your following comment and Lew's explanation is how we look at it.
There is a summing of things going on at the "load".

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.



*** Switching to voltage here, we realize that there is no such constraint
on the voltage to be zero at the open, but what can it be? It's probably
SOMETHING.. If we believe in waves and reflections, one possibility is that
it is in phase, therefore the sum is double, but alas, I can't muster, at
this time, an explanation of why in-phase. Perhaps someone has a better
"mental model" that I.



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.



*** Switching to current at the short, leads to my reasoning just like the
above and the same lack of an "Intuitive" explanation.



Lew uses the characteristic impedance view below and that works for me, but
then, I think I understand all this already! I think that if you 'believe',
so to speak, in the Characteristic Impedance concept, this makes more
"intuitive" sense. Good luck. Remainder un-edited for continuity.



73, Steve, K9DCI




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
--
"I often quote myself - it adds spice to
my conversation." - George Bernard Shaw

Web Page: home.earthlink.net/~mtnviews