Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
It is obvious that the power dissipated in circuilator
1 must be provided by signal generator 1 and the power
dissipated in circulator 2 must be provided by signal
generator 2.

Nope, that's not obvious at all. Make the signal generators
the same phase locked frequency with slightly different
modulation. The modulation for signal generator #1 will
appear across the load resistor in signal generator #2 and
vice versa.

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.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The expression you really mean is

Pavg = Vrms * Irms * cos(A)

Yep, that's what I meant.

No. I mean multiply the instantaneous value by
the instantaneous value, ...

It is not clear to me what physical meaning, if any,
can be attached to such a product.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Tom Donaly wrote:
That's o.k., Cecil doesn't really understand them. If he did,
he wouldn't need to parrot them out of a book.

When I don't parrot them out of a book, Tom, you accuse
me of making them up. You cannot be satisfied.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
The terms are used by meteorologists and scientists as well as laymen.
Yet I know of no one who believes the Sun is moving across the sky.

But lots of people believed the sun was moving across the
sky when the term "sun rising" was coined. The rising of
the sun was and is an illusion. The sun does not rise -
it just sits there in space.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

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.

. . .

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

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 27, 12:36*am, Cecil Moore wrote:
Keith Dysart wrote:
It is obvious that the power dissipated in circuilator
1 must be provided by signal generator 1 and the power
dissipated in circulator 2 must be provided by signal
generator 2.

Nope, that's not obvious at all. Make the signal generators
the same phase locked frequency with slightly different
modulation. The modulation for signal generator #1 will
appear across the load resistor in signal generator #2 and
vice versa.

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.

And since cutting branches with zero current does not
alter the circuit conditions, it is not happening
before cutting the branches.

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 27, 12:39*am, Cecil Moore wrote:
Keith Dysart wrote:
The expression you really mean is

Pavg = Vrms * Irms * cos(A)

Yep, that's what I meant.

No. I mean multiply the instantaneous value by
the instantaneous value, ...

It is not clear to me what physical meaning, if any,
can be attached to such a product.

When V(t) is the function describing the instaneous
voltage and I(t) is the function describing instaneous
current then

P(t) = V(t) * I(t)

is the function describing the instantenous power,
that is, the rate at which energy is being transferred
at any particular instant.

You can then integrate P(t) over the time of interest,
call it the interval from t0 to t1, divide by (t1-t0)
and obtain the average power for that interval. For
periodic functions, one period is an appropriate
interval to integrate over.

If you substitute
V(t) = Vpeak sin(wt)
I(t) = Ipeak sin(wt+alpha)
compute P(t), integrate and divide, you will obtain
Pavg = Vrms * Irms * cos(alpha)
which is how that convenient expression is derived.

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

Even for a line without reflections, it is valuable
to understand that the energy flow is not continuous
but varies with a period of twice the frequency of
the voltage or current sinusoid.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
if(t) + ir(t) = i(t) = 0.02 * cos(wt - 180) * sin(45) = 0.01414 *
cos(wt - 180)

Roy, since you seem to know how to calculate the amplitude
and phase of the total current, please do that for different
points along the line and then explain how that constant
phase current can be used to measure the delay through
a 75m bugcatcher loading coil.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
The terms are used by meteorologists and scientists as well as laymen.
Yet I know of no one who believes the Sun is moving across the sky.
My logging program tells me both Sunset and Sunrise times for distant
locations. It references no illusion. It simply uses those terms.
What do you call the period when the Sun first becomes visible each day?
What do you call the period at the end of each day, when the Sun ceases
to be seen? Do you actually refer to the illusion of Sunrise or the
illusion of Sunset?

But lots of people believed the sun was moving across the
sky when the term "sun rising" was coined. The rising of
the sun was and is an illusion. The sun does not rise -
it just sits there in space.

Though I realize you might have been around at the time the terms were
first used, it is evident that nearly all still use those terms
today--even those in the scientific community. 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. I've asked a number of times how you refer to the
phenomena of what most of us call "Sunrise" and "Sunset." There must
be a reason that you don't respond to that.

You choose to spend much of your time here tap dancing and engaging in
Vaudevillian banter.

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

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.

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.

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 '+'?

There are no reflections anywhere in the system. Since
the voltages are equal for the signal generators, we
can only conclude that 0.2 amps of traveling wave current
is flowing from SGCL1 to R2 and that 0.2 amps of traveling
wave current is flowing from SGCL2 to R1. The two current
nodes at the '+' points do NOT indicate that zero current
is flowing in the 1/2WL line. They only indicate that the
two traveling wave currents are equal in amplitude and
opposite in phase at the '+' points. Any cut that disrupts
the flow of those traveling wave currents is invalid.
--
73, Cecil http://www.w5dxp.com