Keith wrote:
"2) At quarter wave points along this line, voltages and currents which
are always 0 can be observed -- the standing wave."

This is proof positive in Keith`s own words that the first 0-volt point
encountered by either forward or reflected wave is completely
ineffective in halting or abating progress of the waves.

Keith also wrote:
"6) From 2) and 5), the power (rate of energy flow) at quarter wave
points will be 0."

The contradiction is obvious. Were energy flow or power reduced by SWR
zeros, you would not have multiple zeros along the line.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Keith wrote:
"2) At quarter wave points along this line, voltages and currents which
are always 0 can be observed -- the standing wave."

This is proof positive in Keith`s own words that the first 0-volt point
encountered by either forward or reflected wave is completely
ineffective in halting or abating progress of the waves.

Keith also wrote:
"6) From 2) and 5), the power (rate of energy flow) at quarter wave
points will be 0."

The contradiction is obvious. Were energy flow or power reduced by SWR
zeros, you would not have multiple zeros along the line.

Your analysis overlooks that we are discussing an ideal line that
has reached steady state. Before the line reaches steady state, energy
does cross the quarter wave points, but, of course, while this is
happening the voltages and currents at the quarter wave points are not
yet zero.

As an aside, were we to discuss a line that was not open or shorted
but rather terminated in other than its characteristic impedance, we
would still find voltage and current minimas. At these minima, rather
than the energy flow being 0 as it is for a shorted or open line, the
energy flow (power) is exactly the energy being delivered down the
line.

Something similar occurs while an opern or shorted line is approaching
steady state.

....Keith

Keith wrote:
"Is it step 7)?"

"7) From 6), the energy crossing quarter wave points is zero"

False, as are several other of Keith`s speculations.

Standing waves throughout a long transmission line with a hard short or
open-circuit at its end are proof enough that SWR nulls other than the
one at the actual discontinuity don`t bring the energy crossing 1/4-wave
points to zero. SWR nulls other than the one at the actual open or short
have no effect on traveling waves, absent an additional actual
discontinuity. SWR can`t exist without energy flow in both directions.
Energy flow is continuous. It doesn`t start and stop at SWR nulls. This
continuity is proved by measurements at the nulls, taken in one
direction at a time.

No speculation or even math is needed to observe this behavior on an
actual line. It is usless to try to conform the observable to some
theory. It is far more productive to conform theory to the observable.

Best regards, Richard Harrison, KB5WZI

Keith wrote:
"At these minima, rather than the energy flow being 0 as it is for an
open or shorted line, the energy flow (power) is exactly the energy
being delivered down the line."

The power is not zero at an SWR zero, provided that the line has power
flow. The power flow at the specified null is equal in forward and
reflected wave directions as shown by the complete zero.

Power delivered by the source and to the load is the forward power minus
the reflected power. In the case of a complete reflection, the
difference between forward and reflected power is zero, so the load is
rejecting all the forward power, which is turned around and becomes the
reflected power.

In a lossless line, forward and reflected powers flow unabated from end
to end of the line and are thus have their same amplitudes wherever they
are measured in the line, including SWR null points.

Best regards, Richard Harrison, KB5WZI

wrote:
Your analysis overlooks that we are discussing an ideal line that
has reached steady state. Before the line reaches steady state, energy
does cross the quarter wave points, but, of course, while this is
happening the voltages and currents at the quarter wave points are not
yet zero.

Please stop discussing NET energy and start discussing component energies.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Keith wrote:
"Why are you convinced the reflection point had to move?"

This is the case when a line is extended to greater length, and the
short or open also is moved to the new end of the line.

Ever had a short or open somewhere in the middle of a line?

Actually, no.

I have, and
can testify that the signal doesn`t get past a real hard short or open
in the line.
As energy can`t be destroyed it had to be reflected by the
hard short or open.

Or just stopped and stored, perhaps. This is certainly what happens
when you excite the line with a step function rather than a sinusoid.
When excited by a sinusoid, no energy moves at the quarter wave points
where the voltage or current is always 0. As you move away from the
quarter wave points, more and more energy moves on each cycle until
a maximum amount of energy is moved at the point half way between the
quarter wave points. And then it decreases back towards the next
quarter wave point.

All of this is easily visualized by observing the amplitude of the
p(t) function at various points along the line.

In the case where the line short is moved to a point nearer the source,
or where the short or open is moved to a place more distant on the line
from the source, we know the energy travels all the way to the actual
short or open where it is reflected because nulls occur at several
points along a long transmission line.

"we know" is rather strong. I would strongly suggest that no energy
crosses those points in the line where the voltage and current are
always zero since p(t) is always zero at these points.

If the energy were turned around
before it reached the end of the line, nulls more distant from the
source than the turnaround point would not exist.

Not so, the line has reached steady-state. Now the nulls exist. They
did not exist before the line reached steady-state.

There would be no
energy at the actual short or open at the end of the line were the
energy turned around before it reached the end of the line.

Not necessarily. Only once the line has been charged, does the energy
move back and forth between the quarter wave points, while not
crossing them.

Try visualizing how a step function charges the line. How the voltage
step propagates down the line. How the voltage step is reflected at the
open end and starts travelling back towards the start. How between
the start of the line and the returning voltage step, energy is
flowing to charge the line, but between the returning voltage step
and the open end of the line, the energy previously delivered is
statically stored in the capacitance of the line. How once the line
is completely charged, no current flows, there is no power, and
the energy delivered during charging is stored in the capacitance.

Sinusoidal excitation is more difficult to visualize, but the
prinicipal is the same.

....Keith

Richard Harrison wrote:

Keith wrote:
"Is it step 7)?"

"7) From 6), the energy crossing quarter wave points is zero"

False, as are several other of Keith`s speculations.

I do not see any flaw in step 7).

Assuming that step 6)
"6) From 2) and 5), the power (rate of energy flowing) at quarter
wave points will be 0"
is correct, then step 7)
"7) From 6), the energy crossing quarter wave points is 0"
must also be true, since,

If the rate of energy flowing is zero, then there is no energy
flowing so there can not be any energy crossing the point.

I do not find that step 7) is in error.

If there is an error in the final conclusion then I do not think
that it is an error in step 7) which causes the final error,
but, rather, the error must be in one of the earlier steps.

....Keith

wrote:
When excited by a sinusoid, no energy moves at the quarter wave points
where the voltage or current is always 0.

You keep saying that and it keeps being a false statement. There is
absolutely nothing magic about sinusoids.

"we know" is rather strong. I would strongly suggest that no energy
crosses those points in the line where the voltage and current are
always zero since p(t) is always zero at these points.

What about Ramo and Whinnery's forward Poynting vector and reflected
Poynting vector? Why do you choose to ignore them?

Not necessarily. Only once the line has been charged, does the energy
move back and forth between the quarter wave points, while not
crossing them.

That has been shown to be a false assertion regarding component waves.
There is no impedance discontinuity to cause any reflections. Therefore,
the waves do not move back and forth. Do you really believe that the
energy in a bright interference ring is trapped inside the ring? Get
serious!
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
I do not see any flaw in step 7).

Try opening your eyes.

If the rate of energy flowing is zero, then there is no energy
flowing so there can not be any energy crossing the point.

There is no NET energy flow. There is plenty of component energy
flow as defined by Ramo & Whinnery's forward Poynting vector and
reflected Poynting vector. Why do you continue to ignore those
vectors? Why do you continue to ignore the wealth of knowledge
contained in light interference patterns?
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
When excited by a sinusoid, no energy moves at the quarter wave points
where the voltage or current is always 0.

You keep saying that and it keeps being a false statement. There is
absolutely nothing magic about sinusoids.

Seems to me that the sinusoidal standing wave with minima and maxima
at the quarter wave points can only arise with single frequency
sinusoidal excitation of the line. Are there other signals which
will produce this result?

"we know" is rather strong. I would strongly suggest that no energy
crosses those points in the line where the voltage and current are
always zero since p(t) is always zero at these points.

What about Ramo and Whinnery's forward Poynting vector and reflected
Poynting vector? Why do you choose to ignore them?

I haven't used them because I don't need them to arrive at an answer.
Basic electricity, a dash of circuit theory, a bit of knowledge of
trigonometry, some basic calculus and the ability to think is all
that is required.

Why make the solution more complex than necessary?
Just to scare off the neophyte?

Not necessarily. Only once the line has been charged, does the energy
move back and forth between the quarter wave points, while not
crossing them.

That has been shown to be a false assertion regarding component waves.

Perhaps. Or maybe component waves are not the answer.

There is no impedance discontinuity to cause any reflections. Therefore,
the waves do not move back and forth.
Do you really believe that the
energy in a bright interference ring is trapped inside the ring? Get
serious!

In this context, we are discussing transmission lines. I make NO
assertions about light, how rings happen, or don't, or whether the
theory and practice of optics is in way analogous to what happens
on a transmission line.

Transmission lines and their understanding can stand on their own
without the help of optics.

....Keith