Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 5:03 pm, Keith Dysart wrote:
I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Of course, what you say is only true for net energy. If we have a
forward Poynting vector of 100 watts and a reflected Poynting vector
of 100 watts, no net energy is flowing at any point.
--
73, Cecil, w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:

On Fri, 11 Jan 2008 14:03:10 -0800 (PST), Keith Dysart
wrote:


On Jan 11, 1:44 pm, Richard Clark wrote:

On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:


They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.


Hi Keith,

If you cannot identify the source, then you cannot proclaim a
separation.

The best term for this is "the enigma effect." I hate to turn this
loose on the world because I know many here will pick it up and run to
their perceived goal for a touch-down.

Before anyone tackles me for a touchback, I'm going to pass this off
and watch the quarterbacks wrestle with religion in the huddle.

73's
Richard Clark, KB7QHC

To single out a particular spot as the site of a lack of energy
movement along an equipotential system in which there is no tranfer of
energy anyway seems to me to be an arbitrary convenience.

Maxwell explained how electromagentic waves 'bounce' if they encounter
matter, and I haven't seen any evidence that it was an incomplete
description.

ac6xg

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 13:59:47 -0800 (PST)
Keith Dysart wrote:

On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............



I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? P = (v^2)/r = (i^2)r or P = (v^2)/r + (i^2)r

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! Too bad we can't see and feel the wave with our own eyes and hands!


At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)





Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............

I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? *P = (v^2)/r = (i^2)r * or * P = (v^2)/r + (i^2)r

P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! *Too bad we can't see and feel the wave with our own eyes and hands!



At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB- Hide quoted text -

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 5:25*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 14:03:10 -0800 (PST), Keith Dysart





wrote:
On Jan 11, 1:44*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 08:15:09 -0800, Roger Sparks
wrote:

They do stop in the sense that power is not delived through the center point.

Hi Roger,

I'm not sure if this is original to you, or one of those conceits that
is being passed around; but how, short of tagging energy (power?), can
you tell its origin? *In other words, delivery by return, or
completion of the path needs to be separable by some information about
the origin that to this point has been missing from the discussion.

If energy can be tagged with a return address, what is that tag?

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Hi Keith,

If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line, then the energy on the left half
of the line originated with the source on the left side
and ditto for the right.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 18:51:07 -0800 (PST), Keith Dysart
wrote:

If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line,

Hi Keith,

If indeed.

The difficulty is you've already allowed that the energy is not
tagged, so its source is unknown at the load. So "if" fails through
this immutable correlation. The separation is not demonstrated.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 6:56*pm, Cecil Moore wrote:
On Jan 11, 5:03 pm, Keith Dysart wrote:

I suggest that since the current at the middle of the line
is always 0, the power is always 0 and therefore no energy
crosses the middle. While tracking mixed energy is difficult,
when no energy crosses a boundary it seems easy to keep the
energy on each side of the boundary separate.

Of course, what you say is only true for net energy. If we have a
forward Poynting vector of 100 watts and a reflected Poynting vector
of 100 watts, no net energy is flowing at any point.

You don't need Poynting vectors to realize that when
the instantaneous power is always 0, no energy is flowing.
And when the instantaneous power is always 0, it is
unnecessary to integrate and average to compute the
net energy flow, because no energy is flowing at all.

And if by your response you really do mean that energy
can be flowing when the instantaneous power is always 0,
please be direct and say so.

But then you will have to come up with a new definition
of instantaneous power for it can not be that it is
the rate of energy transfer if energy is flowing when
the instantaneous power is zero.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Fri, 11 Jan 2008 18:49:20 -0800 (PST)
Keith Dysart wrote:

On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)





Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:

snip............

I would really appreciate seeing some other possible explanations.

One other one which I have seen and am not confortable with is the
explanation that energy in the waves pass through the point in
each direction and sum to zero. But this is indistinguishable from
superposing power which most agree is inappropriate. As well, this
explanation means that P(t) is not equal to V(t) times I(t),
something that I am quite reluctant to agree with.

The other explanation seen is that the voltage waves or the
current waves travel down the line superpose, yielding a total
voltage and current function at each point on the line which
can be used to compute the power. With this explanation, P(t)
is definitely equal to V(t) time I(t), which I do appreciate.
The weakness of this explanation is that it seems to deny
that the wave moves energy. And yet before the pulses collide
it is easy to observe the energy moving in the line, and if
a pulse was not coming in the other direction, there would
be no dispute that the energy travelled to the end of the
line and was absorbed in the load. Yet when the pulses
collide, no energy crosses the middle of the line. Yet
energy can be observed travelling in the line before
and after the pulses collide.

So...

I can give up on pulses (or waves) moving energy. I am not
happy doing that.
I can give up on P(t) = V(t) * I(t). I am not happy doing
that either.

So the (poorly developped) "charge bouncing" explanation
seems like a way out, but I certainly would appreciate
other explanations for consideration.

Hi Keith,

I think you are putting too much emphasis on the measurement of energy in the pulse, or maybe not enought, depending upon how you are making and using the measurements. *

I hope you agree that the transmission line you are pulsing has no standing waves. *

"Standing waves" seem to more be a consequence of sinusoidal
excitation than pulse.
So I expect no "standing waves".

I don't even like using the words "traveling wave" to describe the
pulse situation
since "traveling wave" also seems to bring a lot of baggage.

If not, then the pulse must be a treveling wave. *If it is a traveling wave, the impedance of the wave will be the impedance of the transmission line, except where the two pulses cross paths. *At the crossing point, the impedance will jump to infinity, Z = v/i = v/zero = infinity.

I do not recommend computing this ratio and calling it impedance. I
find it leads
to difficulties. Better just to think of the V and I independantly.

How about the energy/power levels? *Except at the center, as the pulse passes, current and voltage can be measured, so power is being delivered from one section of the line to a second section more distant from the source. *The energy delivered by the power equation can be measured by U = (v^2)/r = (i^2)*r where r is the Zo if the transmission line.

More accurately P(t) = V(t) * I(t). it happens that for the pulses in
question,
except when colliding, that V and I of the pulses observed on the line
are related
by Z0 so by substitution, the expressions above can be derived. But
sometimes it is
better to stick with the more basic expression P=VI.

Where the two pulses cross, the energy level must be the sum of the two energy levels, which would be 2U = 2*(v^2)/r (assuming two pulses of equal voltage). *As you have pointed out, power would not flow at the point of wave first meeting, spreading from the meeting point back towards the two sources a distance equal to the square pulse width. Energy is present however, measured as U = (v^2)/r.

The transmission line can be considered as a lineal capacitor, so U = (CV^2)/2, where C is the capacitance per unit length. *If the waves "bounce", and occupy physical space for a period of time, they should comingle, stop, and fill the capacitor during that time. If so,

2U = C*V'^2
* *= 2*(V^2)/r where v is the voltage of a single pulse and V' = 1,414*v = voltage observed at point of crossing. *

But the voltage at the crossing is 2V. A puzzle? When the pulse is
"travelling",
there is energy in the capacitance and in the inductance of the line.
When the
current stops, the energy in the inductance is added to the energy in
the
capacitance, thus the voltage reaches 2V.

Which is correct? *P = (v^2)/r = (i^2)r * or * P = (v^2)/r + (i^2)r

P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

Correct.

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

When the capacitance is in phase with the magnetic field, both are measured to peak at the same time. This "resistive" condition is found in a transmission line when there are no standing waves. Under resistive conditions, the electric field and inductive field merge into one electromagnetic field that carries only one energy. The identical charges carry both electric and magnetic fields, with the fields in a ratio defined as impedance. Each field is a separate way of looking at the same energy.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

You may have been reading about traveling waves. They offer an explaination for how magnetic and electric fields may get out of phase. When the two fields are out of phase, the U = (CV^2)/2 + (LI^2)/2 calculation will ALWAYS sum to more energy than can be found in the resistive wave. This occurs because more energy actually is resident on the line with reflections, due to waves traveling in two directions. The reflection effectively doubles the length of the transmission line available for energy storage.


If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

I learned long ago that it DOES have a precondition. The power measured will only be resistive if measured across a pure resistance. If the measurement is across resistance plus reactance, then reactive power is also measured so P(t) = V(t) * I(t) becomes the sum of an unknown mix of two or more waves.


The traveling wave is considered to have a pure resistive impedance.

This going by measurements makes things hard! *Too bad we can't see and feel the wave with our own eyes and hands!



At a disconinuity, we find a doubling of the voltages, not an increase of 1.414. *This leads me to believe that the crossing pulses never stop in a physical sense. *They do stop in the sense that power is not delived through the center point.

...Keith

73, Roger, W7WKB- Hide quoted text -

...Keith

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 11, 10:06*pm, Richard Clark wrote:
On Fri, 11 Jan 2008 18:51:07 -0800 (PST), Keith Dysart

wrote:
If you cannot identify the source, then you cannot proclaim a
separation.

That is hardly difficult. If no energy every crosses
the middle of the line,

Hi Keith,

If indeed. *

The difficulty is *you've already allowed that the energy is not
tagged, so its source is unknown at the load. *So "if" fails through
this immutable correlation. *The separation is not demonstrated.

Take two flashlights.
I claim that the energy for the light produced by each
originates in the battery in each flashlight.
Is your claim otherwise?

Connect the negative terminals of the batteries together.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Connect the positive terminals of the batteries together
through an ammeter. For the purposes of this experiment,
the batteries have exactly the same voltage so the ammeter
indicates that no current ever flows between the batteries.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

GO GUYS GO!

over 700 messages in this thread now, it would probalby be over 1000 already
if it hadn't had some contributions sucked away by the 'standing morphing
power ranger waves and other stupid stuff' thread!