Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
People who are having trouble with the concept of a -1 voltage
reflection coefficient for a perfect voltage source might benefit from
the following exercise:

Look at my first analysis, where the perfect source was connected
directly to the transmission line. Make no assumptions about the
impedance or reflection coefficient it presents. Then, when the
reflection of the initial forward wave returns, calculate the value the
re-reflected wave must have in order to make the sum of the waves
present, which is the line input voltage, equal to the perfect source
voltage. The voltage of the perfect source can't change, by definition.
The ratio of the re-reflected wave to the returning wave is the voltage
reflection coefficient (since we're dealing with voltage waves).

I'll do it for you:

The forward wave was vf(t, x) = sin(wt - x)
The returning wave was vr(t, x) = sin(wt + x)

The returning wave will strike the input end of the line and create a
new forward wave with value vf2(t, 0) = Gs * vr(t, 0) at the input,
where Gs is the source voltage reflection coefficient.

Before the first forward wave returns, we have only vf(t, 0) = sin(wt)
at the input end of the line. This is of course the source voltage.


After the wave arrives and re-reflects, we have at the input end

vtot(t, 0) = vf(t, 0) + vr(t, 0) + vf2(t, 0)
= vf(t, 0) + vr(t, 0) * (1 + Gs)

This must equal the source voltage, which is the line input voltage, and
cannot change. So plugging in values:

sin(wt) = sin(wt) + sin(wt) * (1 + Gs)

Solving for Gs = Gs = -1


I have made no statement about the "impedance" of the perfect source.
The only thing I've required is that the voltage remains constant, which
is the very definition of a perfect source. You can do a similar
exercise to show that the voltage reflection coefficient of a perfect
current source is +1.

Roy Lewallen, W7EL

I see your example as identical to Keith's example of two wave pulses
traveling in opposite directions.

At the point of interaction, Keith's example has a reflection factor of
1 or zero, depending upon whether the waves bounce or pass. Keith's
example is not a short circuit because two pulses of identical polarity
are interacting so a reflection factor of -1 could never exist.

In your example, the presence of voltage from the ideal source creates
conditions identical to Keith's example for the returning reflected
wave. Accepting this premise, then the reflection factor must be either
1 or zero, depending upon whether the waves bounce or pass.

By assuming that the waves reflect at the ideal source, you proved that
the reflection factor is -1, which is the factor for a short circuit.
This can not be the case, so waves must not reflect at the ideal voltage
source, they must pass.

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)

Someone will certainly say the the vtot(t,0) at the source location is
the source voltage, because it is defined that way. The conditions at
point (t,0) itself is actually unknown (because vf mysteriously appears,
and vr disappears by going off the transmission line), but point (t, 0)
is defined by assumptions. Therefore, at

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)


73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Can photons explain the state of a transmission
line driven with a step function after the line
has settled to a constant voltage?

Of course, photons can be used to explain all
EM wave action. A step function accelerates
electrons which then emit photons as EM waves.
Hint: electrons cannot move at the speed of
light. EM waves move at the speed of light.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

If I say I am not going to look for a reference
to "creation" in The Bible, are you going to assert
there are no references to creation in The Bible?
Good luck on your ridiculous assertions.

So that settles it, then.

No, it is not settled. Please reply to my posting
where I proved the reflection coefficient is plus
or minus one, the exact opposite of what you assert.

And the arguments that I have seen between Mr. Maxwell
and Dr. Bruene are on a completely different matter.

If you think that, it's prima facie evidence that you
are really confused.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
My attempt to confuse!? We were discussing the determination
of reflection coefficients for Thevenin equivalent circuits.

No, we were discussing destructive and constructive
interference. Sorry about your confusion.

But in another post, you have agreed that there
is a complete lack of references supporting your
position, ...

No, I said I am not going to look for them. Sorry
about your confusion.

Enjoy the new ability to solve problems that
were previously outside your grasp.

I'm sorry, Keith, delusions of grandeur are a problem
outside of my field of expertise. Perhaps a professional
shrink could help you better than I.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Thu, 3 Jan 2008 06:28:19 -0500, "David J Windisch"
wrote:

You were writing of Tesla over Edison, a-c over d-c power transmission, and
I was reminded that Prez Kennedy started some work years ago somewhere in
the West, involving hvdc transmission. I wasn't intentionally obscuring
things in that earlier post. 73 Dave N3HE

Hi Dave,

I am well practiced mixing it up with the masters of obscurity, and I
certainly don't confuse you with Prior Art, or Cecileo. However, the
HVDC project you allude to is new to me (even if it is/was decades
old).

As an aside, I have worked with the lawyer who was instrumental in
closing down WPPS (a nuclear power consortium we generally called
WHOOPS). That consortium agonized that Washington's credit rating
would go down the toilet if we defaulted - barely a blip in the
interest rate resulted when we mothballed several nuclear reactors.

Even further aside, I lived in Japan in the early 50s when they used
DC residential power (and we had to be careful to buy AC/DC
appliances). Woe to those living at the end of the block where the
street lights were dim.

Edison used to portray AC as being the killer current (eventually
selected for use on death row). Actually there were a mix of
characteristics that lent either the death potentiality. DC will
cause the muscles to clamp, and if you seized a hazardous wire, you
could never let go. AC, on the other hand (no pun), would cause
fibrillation, and you stood some chance of releasing the same
hazardous wire. AC, on the third hand (again, no pun), would also
cause the sweat glands to excrete (due to the same fibrillation) and
lower your path resistance (more lethal current).

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Excellent. So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

That is probably a false statement. I just haven't
wasted my time looking for a reference that uses
those exact words.

There are many references that do.

I seriously doubt that they say what you are
asserting. Please produce those references.
In another thread, I proved your assertion wrong.
A Bird wattmeter placed at the output of your source
will read forward power = reflected power. The
reflection coefficient can be calculated from
that. rho = SQRT(Pref/Pfor) = plus or minus 1.0
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 3 Jan, 09:51, Cecil Moore wrote:
Keith Dysart wrote:
If such analysis is appropriate, then it seems
to me that a pulse can be viewed as a chunk of
charge moving down the line.
Q2. Is this an appropriate view?

No.

Q3. If so, then what happens when two such chunks
of charge collide in the middle of the line?

They don't "collide". Clouds of photons collide
and their behavior is well known.

Q5. If no charge crosses the mid-point, then how
do the pulses, made up of chunks of charge.
pass the mid-point?

How can two water waves pass through each other
while the water molecules are only moving up and
down?
--
73, Cecil *http://www.w5dxp.com

Cecil
The current only changes direction at the behest
of the frequency. I know you think that the
current changes at the behest of the
length of radiator used i.e. at the top but
frankly that is lunacy. It is only when the capacitor charge
starts to move to the inductance and starts to store energy
which generates a diamagnetic field can propagation can occur
You must get a pencil and paper and think things out for
yourself instead of relying on books. With your IQ it should
be a cake walk because unlike some others you are able to get
back to first principles instead of learning by rote.
No personal offence intended. Being stubborn is not all that bad
but only if you are willing to think out alternatives given.
Otherwise it is a not invented in my back yard aproach which
amounts to an over inflation of one's ability
Best regards
Art

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The example was carefully chosen to illustrate the
point, of course. But that is the value of particular
examples.

When the pulses are not identical, the energy that crosses
the point is exactly sufficient to turn one pulse
into the other.

The remainder of the energy must bounce
because it does not cross the mid-point.
...Keith

So it really is almost as though the pulses travel through one
another, rather than bounce off one another.

I have seen the concept that energy doesn't cross nodal points alluded
to in some texts. However there are so many exceptions to it found in
physical systems as to render it a dubious notion at best. Useful
perhaps for illustration purposes.

In the discussion of standing waves on a string, Halliday and Resnick
says "It is clear that energy is not transported along the string to
the right or to the left, for energy cannot flow past the nodal points
in the string, which are permanently at rest. Hence the energy
remains "standing" in the string, although it alternates between
vibrational kinetic energy and elastic potential energy."

So the idea is valid for a simple harmonic oscillator in which there
are no losses. In such a case, once the system begins oscillating, no
further input of energy is required in order to maintain oscillation.
Clearly there is no flow of energy into or out of such a system.
What is clear is that energy doesn't pass through the nodes. It is
less clear that there exists an inherent mechanism which prevents the
movement of energy.

And so it appears in cases where there is no transfer of energy that
one might claim that waves bounce off of one another. There are no
other examples, and no supporting mechanism for it of which I am
aware, and so one might be equally justified in claiming that waves
pass through each other in all cases.

73, ac6xg

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Good answers. Exactly as I expected. Now please
explain the applicability of EM waves to the
state of an open circuited line excited with a
step function, especially after it settles to a
constant voltage (where only an E field will be
present).

Before it settles to a constant voltage, there is
acceleration of electrons that results in an EM
photonic wave.

After it settles to a constant voltage, there is
no acceleration of electrons and the EM photonic
wave disappears.

Please see Maxwell's equations for further
enlightenment.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 1:48*pm, Cecil Moore wrote:
Keith Dysart wrote:
Can photons explain the state of a transmission
line driven with a step function after the line
has settled to a constant voltage?

Of course, photons can be used to explain all
EM wave action. A step function accelerates
electrons which then emit photons as EM waves.
Hint: electrons cannot move at the speed of
light. EM waves move at the speed of light.

Please describe the final state of the step
excited open circuited line using photons.

Thanks,

Keith