On 2 Jan, 16:38, Roy Lewallen wrote:
I'm top posting this so readers won't have to scroll down to see it, but
so I can include the original posting completely as a reference.

Keith, you've presented a very good and well thought out argument. But
I'm not willing to embrace it without a lot of further critical thought.
Some of the things I find disturbing a

* *1. There are no mathematics to quantitatively describe the phenomenon.
* *2. I don't understand the mechanism which causes waves to bounce.
* *3. No test has been proposed which gives measurable results that will
be different if this phenomenon exists than if it doesn't. (I
acknowledge your proposed test but don't believe it fits in this category.)
* *4. I'm skeptical that this mechanism wouldn't cause visible
distortion when dissimilar waves collide. But without any describing
mathematics or physical basis for the phenomenon, there's no way to
predict what should or shouldn't occur.
* *5. Although the argument about no energy crossing the zero-current
node is compelling, I don't feel that an adequate argument has been
given to justify the wave "bouncing" theory over all other possible
explanations.

None of these make an argument with your logical development, although I
think I might be able to do that too. But I'm very reluctant to accept a
view of wave interaction that's apparently contrary to established and
completely successful theory and one, if true, might have profound
effects on our understanding of how things work. So frankly I'm looking
hard for a flaw in your argument. And I may have found one.

A large part of the argument seems to revolve around a single point in a
perfect transmission line, where the current is exactly zero. This is an
infinitesimal point on a perfect line, so some anomalous things might be
expected to happen there.

Let's consider a transmission line as a huge number of series inductors
and shunt capacitors, each an ideal lumped device. In the ideal case, of
course, there would be an infinite number of each, and each would have
an infinitesimal value. However, the LC product and ratio must remain
correct even in the limiting case. Each L and C is an ideal device, so
the current into one terminal of an inductor has to equal the current
out of the other.

That is not correct. If both the capacitor and the inductance are
ideal
they cannot return the same amount of energy. This can only happen
when the
summation of all vectors are parallel to the axis of the conductor.
One only has to look at the vectors involved with out assigning length
to the vectors to see that the vectorial summation cannot be parallel
to the radiator axis. This is exactly why for maximum horizontal gain
the radiator is tipped away from the earth's surface. This response
is clearly shown when a time varient is added to Gaussian law.




A consequence of this is that either we have a whole
inductor with zero current, or the zero current point occurs between
inductors, at a node to which a capacitor is connected. I think we'll
get the same result using either scenario, but let's consider the second.

If we analyze this situation carefully, we'll find that the inductor on
each side of the zero-current point does have a finite current, equal in
amplitude and flowing in opposite directions. So for half of the cycle,
both are putting positive charge in the capacitor, and for the other
half of the cycle, both are removing charge.

That can only be true for a full wave radiator which emulates
a tank circuit. A half wave radiator is a series circuit,
a whole different ball game !





The capacitor voltage goes
up and down as a result, as we can also see by looking at the voltage at
this zero-current point. So current from both sides is contributing to
the capacitor charge, and turning off either one would change the line
conditions.

Not at the change over point where voltages are equal since a
capacitor
has a slight delay before discharging because of initial inductance
delay.



Any change in the current from the inductor on one side
would change the capacitor voltage, and hence the current on the other
side. So there is an interchange of information from one side to the
other. Each inductor is conveying energy to the capacitor, which is
storing and returning it.

It takes energy to propagate does it not?
Why would you ignore that?

Ok, so let's break the capacitor into two, each being half the original
value, and constrain each inductor to deliver charge only to "its"
capacitor. The wire between the capacitors carries no current because
the capacitors always have equal voltages, and can be cut with no effect.
Again that cannot be. The filling of all the capacitors fill up in a
series format.
snip
Art



Roy Lewallen, W7EL



Keith Dysart wrote:
On Dec 30 2007, 6:18 pm, Roy Lewallen wrote:
Keith Dysart wrote:

I predict that the pulse arriving at the left end will
have the same voltage, current and energy profile as
the pulse launched at the right end and the pulse
arriving at the right end will be similar to the
one launched at the left.
They will appear exactly AS IF they had passed
through each other.
The difficulty with saying THE pulses passed
through each other arises with the energy. The
energy profile of the pulse arriving at the left
will look exactly like that of the one launched
from the right so it will seem that the energy
travelled all the way down the line for delivery
at the far end. And yet, from the experiment above,
when the pulses arriving from each end have the
same shape, no energy crosses the middle of the
line.
So it would seem that the energy that actually
crosses the middle during the collision is
exacly the amount of energy that is needed to
reconstruct the pulses on each side after the
collision.
If all the energy that is launched at one end
does not travel to the other end, then I am
not comfortable saying that THE pulse travelled
from one end to the other.
But I have no problem saying that the system
behaves AS IF the pulses travelled from one
end to the other.
You said that:

* What will happen? Recall one of the basics about
* charge: like charge repel. So it is no surprise
* that these two pulses of charge bounce off each
* and head back from where they came.

Yet it sounds like you are saying that despite this charge repulsion and
bouncing of waves off each other, each wave appears to be completely
unaltered by the other? It seems to me that surely we would detect some
trace of this profound effect.

. . .
Is there any test you can conceive of which would produce different
measurable results if the pulses were repelling and bouncing off each
other or just passing by without noticing the other?

There are equations describing system behavior on the assumption of no
wave interaction, and these equations accurately predict all measurable
aspects of line behavior without exception. Have you developed equations
* based on this charge interaction which predict line behavior with
equal accuracy and universal applicability?

No equations. I expect that such equations would be more complex
than those describing the behaviour using superposition. Since
the existing equations and techniques for analysis are tractable
and produce accurate results, I am not motivated to develop an
alternate set with lower utility.

And yet the "no interaction" model, while accurately predicting
the behaviour has some weaknesses with explaining what is
happening. It is, I suggest, these weaknesses that help lead
some so far astray.

To illustrate some of these weaknesses, consider an example
where a step function from a Z0 matched generator is applied
to a transmission line open at the far end. Charge begins to
flow into the line. The ratio of the current to voltage on
the line is defined by the distributed inductance and
capacitance. The inductance is resisting the change in current
which causes a voltage to charge the capacitance. A voltage
step (call this V for later use) propagates down the line
at the speed of light. In front of this step, the voltage,
current and charge in the line is zero. After the step, the
capacitance is charged to the voltage and charge is flowing
in the inductance.

The step function eventually reaches the open end where
the current can no longer flow. The inductance insists
that the current continue until the capacitance at the
end of the line is charged to the voltage which will stop
the flow. This voltage is double the voltage of the step
function applied to the line (i.e 2*V). Once the
infinitesimal capacitance at the end of the line is
charged, the current now has to stop just a bit earlier
and this charges the inifinitesimal capacitance a bit
further from the end. So a step in the voltage propagates
back along the line towards the source. In front of this
step, current is still flowing. Behind the step, the
current is zero and voltage is 2*V. The charge- Hide quoted text -

- Show quoted text -...

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