wrote:

W5DXP wrote:

wrote:

So what happens to these two pulses? They bounce off of each other
and return whence they came.

This has been disproved many, many times. Do a web search for
"superposition" and learn how those waves flow unaffected right
through each other.

So are you saying that like charge does not repel?

No, I'm not saying that at all. In fact, there may not be any charges
crossing the NET voltage = zero point and there doesn't have to be for
energy to be flowing in both directions. Two waves flowing in opposite
directions in a constant Z0 environment superpose but have no effect
on each other. You are continuing to be confused by the NET values.
Did you take a look at that Java-driven web page that I posted a couple
of days ago. If so, I don't see how you could still be confused.

As an analogy, consider two equal magnitude Tsunami waves flowing in
opposite directions in the ocean. According to you, these waves will
reflect off of each other. But it is known that those two waves will
simply flow through each other and continue their original paths
unabated. A single carrier water molecule may only move vertically
and not horizontally at all during this process.

You continue to confuse NET charge carriers with component energy.
--
73, Cecil, W5DXP

W5DXP wrote:

wrote:
There is certainly no power at the zero crossings.

There is no NET power at the zero crossings.

I think you mean there's no instantaneous power at the zero crossings.

73, ac6xg

Jim Kelley wrote:

W5DXP wrote:
There is no NET power at the zero crossings.

I think you mean there's no instantaneous power at the zero crossings.

Well, since the NET voltage is always zero at a voltage node when
the forward power and reflected power are equal, the instantaneous
voltage is always zero, i.e. the steady-state voltage is always
zero. If we have equal power flow vectors in opposite directions,
the NET power is zero at all points up and down the line.
--
73, Cecil, W5DXP

W5DXP wrote:

Jim Kelley wrote:

W5DXP wrote:
There is no NET power at the zero crossings.

I think you mean there's no instantaneous power at the zero crossings.

Well, since the NET voltage is always zero at a voltage node when
the forward power and reflected power are equal, the instantaneous
voltage is always zero, i.e. the steady-state voltage is always
zero.

I've never actually seen the voltage at a node in a standing wave
pattern referred to as an instantaneous voltage - especially considering
that it doesn't vary with time. Instantaneous usually means the
solution to a function f(t) at time t (not f(x) and position x.) Nodes
and zero crossings aren't necessarily the same thing.

73, Jim AC6XG

Jim Kelley wrote:
Nodes and zero crossings aren't necessarily the same thing.

They are for standing waves on lossless unterminated lines,
by definition.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

Jim Kelley wrote:
Nodes and zero crossings aren't necessarily the same thing.

They are for standing waves on lossless unterminated lines,
by definition.

I think not. Standing waves are spatial. At certain points
on the line the (NET) voltage is always zero: nodes.
At other points on the line, the (NET) voltage is sinusoidal
and has 2 zero crossings per cycle. The amplitude of these
sinusoids varies spatially along the line resulting in the
standing wave.

In an ideal line terminated by Zo, no matter where you attach
your oscillograph to the line, you will observe a sinusoid
of the same amplitude. This sinusoid will have zero crossings
and power at the time of these zero crossings will be zero;
no energy will be flowing at the time of the zero crossing.

And going back to the comment that started this sub-thread,
it is this cyclical variation in energy flow which prompted
the power dudes to invent three phase lines in which the
energy flow does not vary cyclically; power is constant.

....Keith

W5DXP wrote:

wrote:
So are you saying that like charge does not repel?

No, I'm not saying that at all. In fact, there may not be any charges
crossing the NET voltage = zero point and there doesn't have to be for
energy to be flowing in both directions.

This does seem to be the sticking point, doesn't it. Starting with
p = v * i,
i = charge_moved/time,
if no charge moves, there can be no power, and from
p = work/time,
if there is no power, no work is being done,
and if no work is being done, no energy has moved.
Ergo, no energy crosses the boundary.

If you find an error in the above derivation, I will happily allow
you that energy flows in both directions.

Two waves flowing in opposite
directions in a constant Z0 environment superpose but have no effect
on each other. You are continuing to be confused by the NET values.

Actually, I am not confused by NET at all. I contend that NET is the
only thing of importance when determining if energy flows.

Did you take a look at that Java-driven web page that I posted a couple
of days ago. If so, I don't see how you could still be confused.

Are you referring to
http://www.mellesgriot.com/products/optics/oc_2_1.htm?

These dudes do seem to have it right. They sum amplitudes to get
the resultant (NET) amplitude and then compute the intensity (power)
from the resultant (NET) amplitude. They do NOT state that energy
flows across a point with zero resultant amplitude nor do they sum
the power (intensity) to determine interference patterns.

As an analogy, consider two equal magnitude Tsunami waves flowing in
opposite directions in the ocean. According to you, these waves will
reflect off of each other. But it is known that those two waves will
simply flow through each other and continue their original paths
unabated.

How can you tell whether it reflected, or flowed through?
The look and feel will be the same.

Remember the collision of two identical balls. Superficial examination
of the collision might cause one to conclude that the energy was
transferred between the balls, especially if you viewed this collision
after viewing the collision of a moving ball with a stationary one
where energy is indeed transferred. Only by looking at the details of
the collision, in particular at the interface where the collision
occurs,
and realizing that f * d is always 0 do you learn that no energy was
transferred.

Similarly for transmission lines at points where v(t) * i(t) is always
0 and, from the web page, apparently for light as well.

....Keith

wrote:

W5DXP wrote:

Jim Kelley wrote:
Nodes and zero crossings aren't necessarily the same thing.

They are for standing waves on lossless unterminated lines,
by definition.

I think not. Standing waves are spatial. At certain points
on the line the (NET) voltage is always zero: nodes.

For the infinite number of times a snapshot of the voltage is
not zero, the node *IS* a zero-crossing point. That is more than
obvious from the JAVA applets on this web page.

http://www.gmi.edu/~drussell/Demos/s....html#standing

In an ideal line terminated by Zo, ...

That configuration is not covered by my statement above which
applies only to standing waves on lossless unterminated lines.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

W5DXP wrote:

wrote:
So are you saying that like charge does not repel?

No, I'm not saying that at all. In fact, there may not be any charges
crossing the NET voltage = zero point and there doesn't have to be for
energy to be flowing in both directions.

This does seem to be the sticking point, doesn't it.

There is no sticking point. It is all explained in _Fields_and_Waves_...
by Ramo, Whinnery, and Van Duzer. It is also explained in _Optics_,
by Hecht.

If you find an error in the above derivation, I will happily allow
you that energy flows in both directions.

That a forward Poynting vector and a reflected Poynting vector exists
is not enough for you? Here's my argument analogous to yours:

My GMC pickup is white. Until you can prove my GMC pickup is not white,
your argument is invalid. Your above argument is just as irrelevant
as the color of my pickup.

Actually, I am not confused by NET at all. I contend that NET is the
only thing of importance when determining if energy flows.

What about the forward Poynting vector and the reflected Poynting vector?
Net is the thing to consider if all you are worried about is net energy.
Net doesn't hack it when you take a look at the forward and reflected
Poynting vectors.

How can you tell whether it reflected, or flowed through?
The look and feel will be the same.

Not exactly. Each Tsunami has its own modulation signature. Many, many
experiments over hundreds of years have proved that waves flow right
through each other.

Similarly for transmission lines at points where v(t) * i(t) is always
0 and, from the web page, apparently for light as well.

You really believe that the energy in a bright interference ring is
trapped between the dark rings by some magical force that exists in
your mind? You really need to read the superposition and interference
chapters in _Optics_.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
We have a choice of two rho for this situation:

Correction: We have a choice of two reflection coefficients
each with its own unique definition.

black box - 0, computed from the surge impedance of the line and
the steady state impedance of the load

Actually, Sqrt(Pref/Pfwd), the definition of rho.

open box - 0.5, computed from the surge impedance of the line and
the surge impedance of the load

Actually, (150-50)/(150+50), the definition of s11.

In any case, what we have in this experiment is a case where there
IS an impedance discontinuity and yet there is no reflection (if
you use the "black box" rho, as is often done).

This is technically not true. The NET reflections are zero. There
are two non-zero component reflections as seen from the s-parameter
equation:

b1 = s11*a1 + s12*a2

These three terms are all reflections. b1 is the NET reflections
toward the source. Since b1 = zero, s11*a1 = -s12*a2, i.e. the two
component reflections are of equal magnitude and opposite phase and
therefore cancel. This is explained in the last three paragraphs on
the Melles-Griot web page.

So, if we are allowed to say in the first experiment that rho is 0
despite an impedance discontinuity, we are equally allowed to say
for the second that rho is -1 despite the absence of a discontinuity.

There is NOT an absence of a discontinuity. The discontinuity is as
large as it can possibly be, an unterminated transmission line. There
is an infinite SWR on the 1/4WL line. The apparent zero impedance
at the input is simply a V/I ratio where Vnet=Vfwd+Vref=0
The forward wave is carrying Vfwd^2*Z0 watts associated with the
forward Poynting vector and the reflected wave is carrying Vref^2*Z0
watts associated with the reflected Poynting vector. A directional
wattmeter will indicate the same thing.

There is NO physical discontinuity at the black box. The only physical
discontinuity is the open end of the stub. That's where 100% of the
reflection action takes place.
--
73, Cecil http://www.qsl.net/w5dxp



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