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:
"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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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

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:
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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What is missing here, IMO, is that the physics of energy flow has not
been adequately explained. Cecil provided a reference to it but did not
elaborate for the general readership.

The energy in forward and reflected waves has been well documented for
many years. From Kraus, Electromagnetics, McGraw Hill, 1953, Chapter 9,
Section 9-13, "Energy Relations in a Standing Wave":

EQ 9-145 We = 2eEo^2[cos^2wt*sin^2Bx]

EQ 9-147 Wm = 2uHo^2[sin^2wt*cos^2Bx]

Note: The energy in the E field [We] is a function of the sin^2(Bx).
The energy in the H field [Wm] is a function of the cos^2(Bx).

You will remember from trigonometry the the maxima [or minima] of a sin
and cos are displaced by 90 degrees. Conclusion: when the E field is
zero the H field is maximum; when the H field is zero the E field is
maximum. Ergo! Energy is conserved and propagates through the zero E
field as an H field; also, when the H field is zero the energy is in the
E field. This is what Cecil is referring to when he refers to the
Poynting vector.

It is analogous to a parallel tuned circuit. When the instantaneous
voltage across the capacitor is zero we don't claim there is no energy
in the circuit. We know that the energy is stored in the inductor.
Conversely, when the instantaneous current in the inductor is zero we
don't claim there is no energy in the circuit. We know the energy is
stored in the capacitor.

In a TEM wave the energy cycles between the E field and the H field and
the energy components are 90 degrees out of phase.

Deacon Dave, W1MCE

Keith wrote:
"I do not find that step 7) is in error."

Step 7) declares power is zero at quarter wave points where volts or
amps are zero in the standing wave pattern.

Power flow does not stop at an actual short or open in a line. It merely
changes direction. At zero volt or amp points, where there is no actual
short or open, no reversal of direction occurs. Power flow is affected
in no way by standing wave nulls or maxima where no impedance
discontinuity exists.

Energy exists in every SWR null. It is in the two waves which produce
the null as these experience no change in energy due to standing waves.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Keith wrote:
"I do not find that step 7) is in error."

Step 7) declares power is zero at quarter wave points where volts or
amps are zero in the standing wave pattern.

Unfortunately, probably due to my poor description, you have missed
the point I was trying to make.

You are, I think, disagreeing with the RESULT of step 7.

Even though the result may be wrong, step 7) can be correct.

Step 7) again:
"7) From 6), the energy crossing quarter wave points is 0"

This is saying that if the result of step 6) is correct, then
the result of step 7) is correct. The transformation from 6)
to 7) merely integrated power to get energy; a common and
correct thing to do.

So if you disagree with the result of step 7), it is in step 6)
which you must search for the error since step 7) is correct.

If step 6) is correct, then search in step 2) and step 5) for
the error.

I suspect that like Cecil, you will end up at step 2) as the
source of what you perceive to be an error.

Then, when we resolve whether 2) is in error, we will know
whether the result of 7) is in error.

....Keith