Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
It is worth doing to convince yourself. Then examine
P(t) to understand how the instaneous energy transfer
varies with time.

Oh, I know how to integrate P(t). But I don't
comprehend the utility of the following:

The instantaneous value of voltage is 10 volts.
The instantaneous value of current is 1 amp.
The voltage and current are in phase.

The instantaneous power is 10 joules per 0 sec?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
You keep writing, over
and over, that the Sun just sits there in space. We both know that
isn't actually correct either. The Sun rotates and is moving through
space quite rapidly.

Of course, the sun's rotation and movement through
space have nothing to do with the sun rising, traveling
across the sky, and then setting.

I've asked a number of times how you refer to the
phenomena of what most of us call "Sunrise" and "Sunset."

I have responded four or five times now. If you are
too dense to get it this time, I will probably not reply
again. The sun does NOT "rise", does NOT travel across
the sky, and does NOT "set".
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
. . .

This does raise the question of the description of the traveling wave
used in an earlier posting.

The example was the open ended 1/2 wavelength transmission line, Zo = 50
ohms, with 1v p-p applied at the source end. The term wt is the phase
reference. At the center of the line, (using the source end as a
reference), you gave vr(t) =-.05*sin(wt-90 deg.) and ir(t) =
0.01*sin(wt-90 deg.)

By using the time (wt-90), I think you mean that the peak occurs 90
degrees behind the leading edge.

The leading edge and peak of what? The function sin(wt -90) looks
exactly like the function sin(wt) except that it's delayed 90 degrees in
phase. So the forward voltage wave is delayed by 90 degrees relative to
the source voltage. This is due to the propagation time down 90
electrical degrees of transmission line.

I think we are in sync here, but something is missing. When I think of
a traveling sine wave, it must have a beginning as a point of beginning
discussion. I pick a point which is the zero voltage point between wave
halves. It follows that the maximum voltage point will be 90 degrees
later. I think you are doing the same thing, but maybe not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.


This posting was certainly correct if we consider only the first
reflected wave. However, I think we should consider that TWO reflected
waves may exist on the line under final stable conditions. This might
happen because the leading edge of the reflected wave will not reach the
source until the entire second half of the initial exciting wave has
been delivered. Thus we have a full wave delivered to the 1/2
wavelength line before the source ever "knows" that the transmission
line is not infinitely long. We need to consider the entire wave period
from (wt-0) to (wt-360.

If these things occur, then at the center of the line, final stable
vr(t) and ir(t) are composed of two parts, vr(wt-90) and vr(wt-270) and
corresponding ir(wt-90) and ir(wt-270). We should be describing vr(t)
as vrt(t) where vrt is the summed voltage of the two reflected waves.

Sorry, it's much worse than this.

Unless you have a perfect termination at the source or the load, there
will be an *infinite number*, not just one or two, sets of forward and
reflected waves beginning from the time the source is first turned on.
You can try to keep track of them separately if you want, but you'll
have an infinite number to deal with. After the first reflected wave
reaches the source, its reflection becomes a new forward wave and it
adds to the already present forward and reflected waves. The general
approach to dealing with the infinity of following waves is to note that
exactly the same fraction of the new forward wave will be reflected as
of the first forward wave. So the second set of forward and reflected
waves have exactly the same relationship as the first set. This is true
of each set in turn. Superposition holds, so we can sum the forward and
reverse waves into any number of groups we want and solve problems
separately for each group. Commonly, all the forward waves are added
together into a total forward wave, and the reverse waves into a total
reverse wave. These total waves have exactly the same relationship to
each other that the first forward and reflected waves did -- the only
result of all the reflections which followed the first is that the
magnitude and phase of the total forward and total reverse waves are
different from the first pair. But they've been changed by exactly the
same factor.

It's not terribly difficult to do a fundamental analysis of what happens
at each reflection, then sum the infinite series to get the total
forward and total reverse waves. When you do, you'll get the values used
in transmission line equations. I've gone through this exercise a number
of times, and I recommend it to anyone wanting a deeper understanding of
wave phenomena. Again, the results using this analysis method are
identical to a direct steady state solution assuming that all
reflections have already occurred.

The infinite number of waves could, of course, be combined into two or
more sets instead of just one, with analysis done on each. If done
correctly, you should get exactly the same result but with considerably
more work.

I do want to add one caution, however. The analysis of a line from
startup and including all reflections doesn't work well in some
theoretical but physically unrealizable cases. One such case happens to
be the one recently under discussion, where a line has a zero loss
termination at both ends (in that case, a perfect voltage source at one
end and an open circuit at the other. In those situations, infinite
currents or voltages occur during runup, and the re-reflections continue
to occur forever, so convergence is never reached. Other approaches are
more productive to solving that class of theoretical circuits.

Again, we seem to be in complete agreement except for the statement "In
those situations, infinite currents or voltages occur during runup".
For many years I thought that "initial current into a transmission line
at startup" would be very high, limited only by the inductive
characteristics of the line. With this understanding, I thought that
voltage would lead current at runup. It was not until I saw the formula
Zo = 1/cC that I realized that a transmission line presents a true
resistive load at startup. Current and voltage are always in phase at
startup.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.
. . .

which is the total power (rate of energy delivery) contained in the
standing wave at the points 45 degrees each side of center. If we want
to find the total energy contained in the standing wave, we would
integrate over the entire time period of 180 degrees.

So think I.

I haven't gone through your analysis, because it doesn't look like
you're including the infinity of forward and reverse waves into your two.

. . .

Roy Lewallen, W7EL

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assume a constant power source and you will get the results
that Roy is talking about.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assume a constant power source and you will get the results
that Roy is talking about.
No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

Now the question here is "Do the waves bounce off one another?" which
would result in a doubling of observed voltage, or "Do the wave pass
through one another?" which would allow a condition of energy entering
the system equal to energy leaving the system.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Assume a constant power source and you will get the results
that Roy is talking about.

No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

No, because a *constant power source* is pumping joules/second
into the system no matter what voltage or current it requires
to move those joules/second into the system. It's like the
power source in "Forbidden Planet".

Now the question here is "Do the waves bounce off one another?"

Waves do NOT "bounce" off one another. At a physical
impedance discontinuity, the component waves can
superpose in such a way as to redistribute their energy
contents in a different direction. (Redistribution of energy
in a different direction in a transmission line implies
reflections.) In the absence of a physical impedance
discontinuity, waves just pass through each other.

www.mellesgriot.com/products/optics/oc_2_1.htm

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
How can the current in the middle of the line be 0.4 amps
when the current at both points '+' is zero? Does that
0.4 amps survive a cut at point '+'?

This should have been: Does that 0.4 amps survive a cut
at both points '+' where the current is zero?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Assume a constant power source and you will get the results
that Roy is talking about.

No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

No, because a *constant power source* is pumping joules/second
into the system no matter what voltage or current it requires
to move those joules/second into the system. It's like the
power source in "Forbidden Planet".

Now the question here is "Do the waves bounce off one another?"

Waves do NOT "bounce" off one another. At a physical
impedance discontinuity, the component waves can
superpose in such a way as to redistribute their energy
contents in a different direction. (Redistribution of energy
in a different direction in a transmission line implies
reflections.) In the absence of a physical impedance
discontinuity, waves just pass through each other.

www.mellesgriot.com/products/optics/oc_2_1.htm

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html

There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

Does this place us at a logical impasse, with current reversing and
voltage doubling at in one argument (at the open end), but not doubling
at the source end? No, the voltage will double at the source end when
stability is reached after one full cycle (in the 1/2 wave example).

Logically then, we must recognize that our source voltage WILL NOT
remain constant following the arrival of the reflected wave. Certainly
this is what we find when we retune our transmitters after changing
frequency.

What would be the logic of insisting that the input voltage be held
constant to the 1/2 wave example after it is shown that the reflected
wave must interact with the incoming wave give a very high impedance at
the source?

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
You keep writing, over and over, that the Sun just sits there in
space. We both know that isn't actually correct either. The Sun
rotates and is moving through space quite rapidly.

Of course, the sun's rotation and movement through
space have nothing to do with the sun rising, traveling
across the sky, and then setting.

I've asked a number of times how you refer to the phenomena of what
most of us call "Sunrise" and "Sunset."

I have responded four or five times now.

You certainly have. Each of the times I've asked the questions of what
you call Sunrise and Sunset, you've not provided an answer.

If you are
too dense to get it this time, I will probably not reply
again.

I'm not at all dense, Cecil, as you know from previous exchanges through
the years. I've asked questions. What I get from you in response, does
not contain answers. If you choose not to answer, just say so. A
threat not to reply isn't really much of a threat.

The sun does NOT "rise", does NOT travel across
the sky, and does NOT "set".

Let me ask you once again: If you awaken before dawn and observe the
sun when it first becomes visible, what do *you*, Cecil Moore, call the
phenomenon you are seeing? Please note that I now say and have
previously written that I do not believe that the Sun rises, sets or
travels across the sky. I've stated that the terms Sunrise and Sunset
are commonly used to describe the Sun's first appearance and last
appearance of each day.

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

I think we are in sync here, but something is missing. When I think of
a traveling sine wave, it must have a beginning as a point of beginning
discussion. I pick a point which is the zero voltage point between wave
halves. It follows that the maximum voltage point will be 90 degrees
later. I think you are doing the same thing, but maybe not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.

Yes, you're describing some of the properties of a sinusoidal traveling
wave. I generally describe them mathematically.

Again, we seem to be in complete agreement except for the statement "In
those situations, infinite currents or voltages occur during runup". For
many years I thought that "initial current into a transmission line at
startup" would be very high, limited only by the inductive
characteristics of the line. With this understanding, I thought that
voltage would lead current at runup. It was not until I saw the formula
Zo = 1/cC that I realized that a transmission line presents a true
resistive load at startup. Current and voltage are always in phase at
startup.

They are provided that Z0 is purely resistive. That follows from the
simplifying assumption that loss is zero or in the special case of a
distortionless line, and it's often a reasonable approximation. But it's
generally not strictly true.

But that doesn't have anything to do with my statement, which deals with
theoretical cases where neither end of the line has loss. For example,
look at a half wavelength short circuited line driven by a voltage
source. Everything is fine until the initial traveling wave reaches the
end and returns to the source end.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assuming you're talking about the half wavelength open circuited line
driven by a voltage source -- please do the math and show the magnitude
and phase of the initial forward wave, the reflected wave, the wave
re-reflected from the source, and so forth for a few cycles, to show
that what you say is true. My calculations show it is not. I'd do it,
but I find that the effort of showing anything mathematically is pretty
much a waste of effort here, since it's generally ignored. It appears
that the general reader isn't comfortable with high school level
trigonometry and basic complex arithmetic, which is a good explanation
of why this is such fertile ground for pseudo-science. But I promise
I'll read your mathematical analysis of the transmission line run-up.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 27, 10:53*am, Cecil Moore wrote:
Keith Dysart wrote:
It is worth doing to convince yourself. Then examine
P(t) to understand how the instaneous energy transfer
varies with time.

Oh, I know how to integrate P(t).

But did you know that your favourite V * I * cos(theta)
was derived from the function describing instantaneous
power?

But I don't
comprehend the utility of the following:

The instantaneous value of voltage is 10 volts.
The instantaneous value of current is 1 amp.
The voltage and current are in phase.

The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful. With the function describing the
intantaneous values with respect to time, you
can integrate. You can find the total energy
transfered. You can find when it is transferred.
Is it steady? Or does it vary? There is lots
to learn.

You can even learn that when the instantaneous
power at some point is 0 for all instances,
then no energy is transferred. That would be
a useful learning.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

I think we are in sync here, but something is missing. When I think
of a traveling sine wave, it must have a beginning as a point of
beginning discussion. I pick a point which is the zero voltage point
between wave halves. It follows that the maximum voltage point will
be 90 degrees later. I think you are doing the same thing, but maybe
not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.

Yes, you're describing some of the properties of a sinusoidal traveling
wave. I generally describe them mathematically.

Again, we seem to be in complete agreement except for the statement
"In those situations, infinite currents or voltages occur during
runup". For many years I thought that "initial current into a
transmission line at startup" would be very high, limited only by the
inductive characteristics of the line. With this understanding, I
thought that voltage would lead current at runup. It was not until I
saw the formula Zo = 1/cC that I realized that a transmission line
presents a true resistive load at startup. Current and voltage are
always in phase at startup.

They are provided that Z0 is purely resistive. That follows from the
simplifying assumption that loss is zero or in the special case of a
distortionless line, and it's often a reasonable approximation. But it's
generally not strictly true.

But that doesn't have anything to do with my statement, which deals with
theoretical cases where neither end of the line has loss. For example,
look at a half wavelength short circuited line driven by a voltage
source. Everything is fine until the initial traveling wave reaches the
end and returns to the source end.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assuming you're talking about the half wavelength open circuited line
driven by a voltage source -- please do the math and show the magnitude
and phase of the initial forward wave, the reflected wave, the wave
re-reflected from the source, and so forth for a few cycles, to show
that what you say is true. My calculations show it is not. I'd do it,
but I find that the effort of showing anything mathematically is pretty
much a waste of effort here, since it's generally ignored. It appears
that the general reader isn't comfortable with high school level
trigonometry and basic complex arithmetic, which is a good explanation
of why this is such fertile ground for pseudo-science. But I promise
I'll read your mathematical analysis of the transmission line run-up.

Roy Lewallen, W7EL

OK. I think I should tweak the example just a little to clarify that
our source voltage will change when the reflected wave arrives back at
the source end. To do this, I suggest that we increase our transmission
line to one wavelength long. This so we can see what happens to the
source if we pretended that we had not moved it all.

We pick our lead edge at wt-0 and define it to be positive voltage. The
next positive leading edge will occur at wt-360. Of course, a half
cycle of positive voltage will follow for 180 degrees following points
wt-0 and wt-360.

Initiate the wave and let it travel 540 degrees down the transmission
line. At this point, the leading edge wt-0 has reflected and has
reached a point 180 degrees from the full wave source. This is the point
that was originally our source point on the 1/2 wave line.
Mathematically, wt-0 is parallel/matched with wt-360, but because the
wt-0 has been reflected, the current has been reversed but the voltage
has not been changed.

Lets move to wave point wt-1 and wt-361 so that we will have non-zero
voltage and current.

vt = vr(wt-1) + vf(wt-361) = 2*.5*sin(1)

where 1v p-p has been originally applied and vf(t) = vr(t) and vt(t) is
total voltage at any time point. Notice that the total voltage is now
2vr(t) = 2vf(t). This doubling continues as the wave moves forward, with

vt(t) = vr(wt-2) + vf(wt-362) ......... = 2vf(t) = 2vr(t)

The current is similar except, very important, the current was reversed
when reflected from the open end.

it(t) = vf(wt-362) - vr(wt-2) ...... = vf(t) - vr(t) = 0


The effect on the old source point is to make the impedance infinitely
high for all ongoing wave forward motion, which is not stopping the
wave, only indicating that power no longer moves past this point.

At time 720 degrees, the reflected wave (wt-0) reaches our revised
source point where it matches with wt-720 and begins raising the source
impedance, stopping power movement into the system. From this time on,
no further power enters the revised system because of the high impedance.

We should notice here that no power leaves the system after this time as
well. The high impedance works both ways, for forward and reflected
wave. There is no need for additional reflection analysis because both
source and full wave systems are stable and self contained after this
720 degree point. The source is effectively "turned off", and the full
wave system isolated.

If the source was parallel with a 50 ohm resistor (assuming a 50 ohm
transmission line), then the reflected wave would be matched with the
resistor and absorbed. Power would be continually moved through the
transmission line, giving hot spots on the line at 90 and 270 degree
points. We would still see the doubled voltage points at the end (360
degrees) and 180 degree points.

If we move to the shorted transmission line case, the math is identical
except that the voltages are reversed but currents add. The result at
the source is a low impedance where power can no longer be applied
because the voltage is always zero. Could we make current flow through
a resistor in parallel to the transmission line at the source in this
case? I think not.

So I think.

Thanks for taking time to consider these words. It takes real time to
carefully consider the arguments (and to present them).

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

This is a common confusion of cause and effect. Take a
short-circuit 1/4WL stub for instance. A very high
resistance is established at the mouth of the stub but
that high resistance has zero effect on the forward current
which keeps on flowing into the stub. Without the forward
current flowing uninhibited into the stub, the very high
resistance could not be maintained. In fact, the current
is a maximum at the shorted end of a 1/4WL stub. If you
don't believe it, measure it. Where did that current come
from if current cannot flow into the stub?

The stub impedance is the result of the ratio of voltage to
current. It is a virtual impedance and since it is an effect,
it cannot be the cause of anything. The people who say that
a virtual impedance is the same thing as an impedor have not
read the definitions of those things in the IEEE Dictionary.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Please note that I now say and have
previously written that I do not believe that the Sun rises, ...

Good, that is all I was trying to get you to admit -
that the "rising of the sun" is an illusion caused by
the rotation of the earth.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:
The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful.

But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Please note that I now say and have
previously written that I do not believe that the Sun rises, sets or
travels across the sky.

Very good, that is all I was trying to get you to admit -
that the "rising of the sun" is an illusion which was
my original point.

Incidentally, Webster's says the sun does actually rise. :-)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

[...]

OK Cecil, good job well done ... again, you logic is flawless--I BELIEVE!

Time to come home now. (to reality)

Heils' got a "hardon", let him be--time'll fix it ... maybe ... ;-)

He still ain't done nothin' I ain't ever done.

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

...
But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?

TIME?

I thought we already dealt with that; ain't no such thing--there IS
movement ... you will argue these points forever--until you STOP
believing in time (like Santa.)

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

This is a common confusion of cause and effect. Take a
short-circuit 1/4WL stub for instance. A very high
resistance is established at the mouth of the stub but
that high resistance has zero effect on the forward current
which keeps on flowing into the stub. Without the forward
current flowing uninhibited into the stub, the very high
resistance could not be maintained. In fact, the current
is a maximum at the shorted end of a 1/4WL stub. If you
don't believe it, measure it. Where did that current come
from if current cannot flow into the stub?

Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.


The stub impedance is the result of the ratio of voltage to
current. It is a virtual impedance and since it is an effect,
it cannot be the cause of anything. The people who say that
a virtual impedance is the same thing as an impedor have not
read the definitions of those things in the IEEE Dictionary.

Measured in ohms, virtual, and impedance is not the same as impedor. OK
An effect caused by earlier events. Agreed.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 27, 6:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
The instantaneous power is 10 joules per 0 sec?

There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful.

But that instantaneous instant is NOT one second
long. Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?

Time for some calculus review. Look up differentiation.

Perhaps try googling "in the limit as t approaches 0..."

But at least you now see the utility.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 27, 10:42*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
A schematic shows exactly what is happening. There is no
path from SGCL1 to R1. There is no path from SGCL2 to R2.

SGCL1---1---2------2---1---SGCL2
* * * * *\ / * * * *\ /
* * * * * 3 * * * * *3
* * * * * | * * * * *|
* * * * * R1 * * * * R2

There is nothing in the circuit to cause any reflections.
So the power dissipated in R2 comes from SGCL1 and the
power in R1 comes from SGCL2.

Can not happen after cutting the branches.

The inclusion of circulators in the example
ensures that it is a distributed network example.
Cutting the branches is not a valid action in
distributed network examples because technically
it is a zero current "point" and not a zero current
"branch", i.e. the current is not zero throughout
the entire branch. See below.

The beauty of the distributed model is that it has
a lot of very, very (say infinitely) small branches.
(Review your calculus, it is much the same). One
can indeed therefore, cut the branch at a point.

If you wish, feel free to view the branch as having,
the width of a point.

Sorry, the lumped circuit model is known to fail for
distributed network problems. That's probably why
the distributed network model still survives today
but has been discarded and forgotten by many in the
rather strange rush to use a shortcut method at all costs.

Or are you disuputing the validity of cutting branches
with zero current?

Of course, it is obviously invalid in distributed
network problems. We can add 1/2WL of lossless
transmission line to the example to see why it is
invalid.
* * * * * * * * * *1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
* * * * * \ / * * * * * * * * * * * * \ /
* * * * * *3 * * * * * * * * * * * * * 3
* * * * * *| * * * * * * * * * * * * * |
* * * * * *R1 * * * * * * * * * * * * *R2

Your zero current "branch" is now 1/2WL long and in
the center of that zero current "branch", the current
is at a maximum value of 0.4 amps for 50 ohm signal
generator voltages of 10 volts as in your original example.

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

Of course the branch in the middle of your line is not
one of them.

How can the current in the middle of the line be 0.4 amps
when the current at both points '+' is zero? Does that
0.4 amps survive a cut at point '+'?

Absolutely, if the line is lossless. Cut both "+" and the
current in middle of the line still remains.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
[...]

I believe RF knows length, it has to be aware of movement--time? Time,
I think it has little use of that ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Where did that current come
from if current cannot flow into the stub?

Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.

EM RF current is stored in the stub? In what form?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

Cecil Moore wrote:
1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
\ / \ /
3 3
| |
R1 R2

Your zero current "branch" is now 1/2WL long and in
the center of that zero current "branch", the current
is at a maximum value of 0.4 amps for 50 ohm signal
generator voltages of 10 volts as in your original example.

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

The point is that in the above example, there are absolutely
no reflections. When you cut the line you cause reflections
where none existed before. It is obviously invalid to completely
change the operation of the circuit in that manner. We could
be sending data from SG1 to R2 and from SG2 to R1. Those data
streams stop when you cut the line.

How can the current in the middle of the line be 0.4 amps
when the current at both points '+' is zero? Does that
0.4 amps survive a cut at point '+'?

Absolutely, if the line is lossless. Cut both "+" and the
current in middle of the line still remains.

We both know that is a physical impossibility. Sometimes you
are just forced to accept reality and get on with it.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Perhaps try googling "in the limit as t approaches 0..."

So how many joules can pass a point in zero seconds?

But at least you now see the utility.

Nope, I don't. I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
\ / \ /
3 3
| |
R1 R2

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

Assume your life depends upon the information transmitted
from SGCL1 to R2. Would you make the cut?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
Please note that I now say and have previously written that I do not
believe that the Sun rises, sets or travels across the sky.

Very good, that is all I was trying to get you to admit -
that the "rising of the sun" is an illusion which was
my original point.

You've danced and dodged, Cecil.

What do you call the first and last appearances of the Sun's rays each
day?

Incidentally, Webster's says the sun does actually rise. :-)

My local TV meteorologist talks of Sunrise and Sunset. My logging
program uses the terms. NASA uses the terms as well.


Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 1:02*am, Cecil Moore wrote:
Keith Dysart wrote:

* Cecil Moore wrote:

* * * * * * * * * *1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
* * * * *\ / * * * * * * * * * * * * \ /
* * * * * 3 * * * * * * * * * * * * * 3
* * * * * | * * * * * * * * * * * * * |
* * * * * R1 * * * * * * * * * * * * *R2

Your zero current "branch" is now 1/2WL long and in
the center of that zero current "branch", the current
is at a maximum value of 0.4 amps for 50 ohm signal
generator voltages of 10 volts as in your original example.

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

The point is that in the above example, there are absolutely
no reflections. When you cut the line you cause reflections
where none existed before.

This is what makes the example so fascinating.

Before the cut, there is a distribution of voltage, current
and power on the line. The functions representing these can
be written as

V(x,t)
I(x,t)
P(x,t)

After the cut(s), the voltage, current and power distributions
are exactly the same: V(x,t), I(x,t), P(x,t).

The cuts changed nothing about the conditions in the circuit.

And yet the claim is made that before the cuts there are no
reflections and after the cut there are a bunch. And yet the
conditions in the circuit are EXACTLY the same.
But befonone, after:bunch. But conditions are exactly
the same.

It is obviously invalid to completely
change the operation of the circuit in that manner.

Just cutting a branch. Completely legal.

Is the operation of the circuit completely different?
Perhaps it is reasonable to view a virtual open as
producing a reflection. You have to work this out for
yourself. But as you do so, keep front and center the
fact that the voltage, current and power distributions
are identical before and after the cut(s). The before
and after circuits are behaving identically.

If you were merely provided with V(x,t), I(x,t) and
P(x,t) you could not determine whether there were cuts
or not.

We could
be sending data from SG1 to R2 and from SG2 to R1. Those data
streams stop when you cut the line.

The specification of the conditions in the experiment mean
that there can be no data stream. If there were data, the
current would not be always 0.

How can the current in the middle of the line be 0.4 amps
when the current at both points '+' is zero? Does that
0.4 amps survive a cut at point '+'?

Absolutely, if the line is lossless. Cut both "+" and the
current in middle of the line still remains.

We both know that is a physical impossibility. Sometimes you
are just forced to accept reality and get on with it.

It is completely possible for a lossline line.
It is no different than a connected capacitor and inductor
which will ring for ever given the appropriate initial
conditions.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 1:38*am, Cecil Moore wrote:
Keith Dysart wrote:
Perhaps try googling "in the limit as t approaches 0..."

So how many joules can pass a point in zero seconds?

You have descended into silliness. You are out on your
Harley doing 60 miles/hour. How far do you travel in 0
seconds?

So your point was?

But at least you now see the utility.

Nope, I don't. I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.

Ahhh. The difficulty is because you don't "think" that
the concept is mentioned in Hecht. Could that be because
at 5E14 Hz, it is difficult to measure? Try a circuit
analysis or transmission line book. You will have more
luck. There is a good chance that your Ramo and Whinery
mention it. Look near where they derive
Pavg = Vrms * Irms * cos(theta)

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 1:44*am, Cecil Moore wrote:
Keith Dysart wrote:
* * * * * * * * * *1/2WL 50 ohm
SGCL1---1---2--+--lossless line--+--2---1---SGCL2
* * * * *\ / * * * * * * * * * * * * \ /
* * * * * 3 * * * * * * * * * * * * * 3
* * * * * | * * * * * * * * * * * * * |
* * * * * R1 * * * * * * * * * * * * *R2

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

Assume your life depends upon the information transmitted
from SGCL1 to R2. Would you make the cut?

With the current always 0, data is not being successfully
transmitted. So with the experiment at hand, life is over.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 28, 1:38 am, Cecil Moore wrote:
Keith Dysart wrote:
Perhaps try googling "in the limit as t approaches 0..."
So how many joules can pass a point in zero seconds?

You have descended into silliness. You are out on your
Harley doing 60 miles/hour. How far do you travel in 0
seconds?

So your point was?

But at least you now see the utility.
Nope, I don't. I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.

Ahhh. The difficulty is because you don't "think" that
the concept is mentioned in Hecht. Could that be because
at 5E14 Hz, it is difficult to measure? Try a circuit
analysis or transmission line book. You will have more
luck. There is a good chance that your Ramo and Whinery
mention it. Look near where they derive
Pavg = Vrms * Irms * cos(theta)

...Keith

Ramo, Whinnery, and Van Duzer, _Fields and Waves in Communication
Electronics_: p. 16 (in derivation of Eq. 3).

Van Valkenburg, _Network Analysis_: Eq. 14-2, p. 420.

Pearson and Maler, _Introductory Circuit Analysis_: Eq. 5.42, p. 251.

Weidner & Sells, _Elementary Classical Physics, Vol. 2_: Eq. 30-10, p. 912.

_IEEE Standard Dictionary of Electrical and Electronics Terms, Third
Ed._: "Power, instantaneous (two-wire circuits)"

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 8:37*am, Roy Lewallen wrote:
Keith Dysart wrote:
On Dec 28, 1:38 am, Cecil Moore wrote:
Keith Dysart wrote:
Perhaps try googling "in the limit as t approaches 0..."
So how many joules can pass a point in zero seconds?

You have descended into silliness. You are out on your
Harley doing 60 miles/hour. How far do you travel in 0
seconds?

So your point was?

But at least you now see the utility.
Nope, I don't. I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.

Ahhh. The difficulty is because you don't "think" that
the concept is mentioned in Hecht. Could that be because
at 5E14 Hz, it is difficult to measure? Try a circuit
analysis or transmission line book. You will have more
luck. There is a good chance that your Ramo and Whinery
mention it. Look near where they derive
Pavg = Vrms * Irms * cos(theta)

...Keith

Ramo, Whinnery, and Van Duzer, _Fields and Waves in Communication
Electronics_: p. 16 (in derivation of Eq. 3).

Van Valkenburg, _Network Analysis_: Eq. 14-2, p. 420.

Pearson and Maler, _Introductory Circuit Analysis_: Eq. 5.42, p. 251.

Weidner & Sells, _Elementary Classical Physics, Vol. 2_: Eq. 30-10, p. 912..

_IEEE Standard Dictionary of Electrical and Electronics Terms, Third
Ed._: "Power, instantaneous (two-wire circuits)"

Roy Lewallen, W7EL

That last one will be of particular interest to Cecil.
Its in the Seventh Edition, as well, on page 562.

Of possibly even greater interest to Cecil is the
second entry following 'instantaneous power'...

"instantaneous Poynting vector (P
(t,
r)) (of an
electromagnetic wave) The vector product of the
instantaneous electric and magnetic field vectors.
The integral of P
(t,
r) over a surface is the
instantaneous electromagnetic power flow through
the surface."

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Where did that current come
from if current cannot flow into the stub?

Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.

EM RF current is stored in the stub? In what form?

Come on Cecil! Let's not go around in circles! You know very well how
it happens.

On the remote chance that you are serious, I suggest you read CAREFULLY
my other postings. If you want even more information, read your own
postings from the past.

For my part, I have learned from you and your examples. For better or
worse, I feel much more comfortable in my knowledge base and ability to
communicate to others. As to the validity of my postings, each reader
will need to decide for himself, just as the reader must do for your
postings.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The cuts changed nothing about the conditions in the circuit.

This reminds me of the guru who asserted that he could
replace his 50 ohm antenna with a 50 ohm resistor
without changing the conditions.

And yet the claim is made that before the cuts there are no
reflections and after the cut there are a bunch. And yet the
conditions in the circuit are EXACTLY the same.

No, conditions are not exactly the same. Before the
cut, there were no reflections. After the cut, there
are reflections. Conditions have changed. I'll bet
the change in the natural noise pattern, which exists
in every system, could be detected at the time of
the cut.

It is no different than a connected capacitor and inductor
which will ring for ever given the appropriate initial
conditions.

If we are talking about things that can happen only in your
mind, why stop with irrelevant ringing assertions? Why not
assert that you can leap tall buildings at a single bound
(in your mind)?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

You are out on your
Harley doing 60 miles/hour. How far do you travel in 0
seconds?

So your point was?

That was my point. You made it for me.

I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.

Ahhh. The difficulty is because you don't "think" that
the concept is mentioned in Hecht.

How did you manage to reply to a canceled posting?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
With the current always 0, data is not being successfully
transmitted. So with the experiment at hand, life is over.

Come to think of it, in a real-world system, the current
is indeed never zero. So if your life depended on it,
would you cut that connection that is keeping you alive?

If you really want to prove your point, you could demand
that the experiment be performed in total darkness with
the two sources turned off.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Ramo, Whinnery, and Van Duzer, _Fields and Waves in Communication
Electronics_: p. 16 (in derivation of Eq. 3).

Van Valkenburg, _Network Analysis_: Eq. 14-2, p. 420.

Pearson and Maler, _Introductory Circuit Analysis_: Eq. 5.42, p. 251.

Weidner & Sells, _Elementary Classical Physics, Vol. 2_: Eq. 30-10, p. 912.

_IEEE Standard Dictionary of Electrical and Electronics Terms, Third
Ed._: "Power, instantaneous (two-wire circuits)"

You guys are replying to a canceled posting.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Of possibly even greater interest to Cecil is the
second entry following 'instantaneous power'...

What should be of interest to you guys is that you are
replying to and continuing replies to a posting that
was canceled.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Roger wrote:
Stored in the 1/4 WL between the short and mouth. No more current
needed once stability is reached.

EM RF current is stored in the stub? In what form?

Come on Cecil! Let's not go around in circles! You know very well how
it happens.

Here's an example using a circulator and load in
a 50 ohm system. Please think about it.

SGCL---1---2------------------------------+
\ / | 1/4
3 | WL
| everything is 50 ohms | shorted
R | stub

Are there any reflections at point '+'?

If not, how is energy stored in the stub?

If so, what causes those reflections?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 28, 10:00*am, Cecil Moore wrote:
Keith Dysart wrote:
You are out on your
Harley doing 60 miles/hour. How far do you travel in 0
seconds?

So your point was?

That was my point. You made it for me.

Ahhh. So the utility is the same utility one gets from
useful numbers like 60 miles/hour. OK.

I don't think the concept of instantaneous
power is mentioned at all in "Optics", by Hecht.
Ahhh. The difficulty is because you don't "think" that
the concept is mentioned in Hecht.

How did you manage to reply to a canceled posting?

I use http://groups.google.com to view usenet and
your post is still there.

Cancelling is a very iffy thing. Better to decide
before clicking send. Once it has entered the net
it is likely there forever. Readers and servers
may, or may not, honour future requests.

...Keith