Standing-Wave Current vs Traveling-Wave Current
 

Tom Donaly wrote:
All of them should
go back to school and study the whole elephant, so they won't
keep making the same mistakes the three blind men made.

The following all discuss lossless systems in their
writings - Slater, Chipman, Ramo & Whinnery, Johnson,
Kraus, and Balanis. Which of those people are the
three blind men?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Tom Donaly wrote:
Cecil Moore wrote:
Good Grief, Gene, I don't have time to teach you
quantum electrodynamics. Go read a book that tells
you about the nature of photons. It is also the
cornerstone of relativity.

Cecil, you couldn't teach anyone quantum electrodynamics if
they put a gun to your head. Quit pretending.

Probably true, but Gene obviously knows less about the
subject than I do and that's pretty sad.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
"Keith Dysart" wrote in message
...
On Dec 23, 10:12 am, "Dave" wrote:

Are you really prepared to throw away P = VI?

yes, when the V and I are the superimposed voltage and current that you
insist are the real current and voltage on the line.
. . .

Amazing! I suppose Ohm's law is next?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
"Keith Dysart" wrote in message
...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

It does not fail. Present any example of a transmission line terminated
with a load. Choose any line and load impedance and any transmission
line length. I'll tell you the V and I at any point on the line,
calculate the instantaneous power from V(t) * I(t) at that point, and
from that the average power. Then I'll show that this average power
equals the power in the load and the power delivered by the source. As
an added bonus, I can tell you the impedance (ratio of V/I) at any point
along the line. I'll provide both equations that always work and
numerical results.

Then you can show where I've made an error and why my calculations are
invalid.

Can I be any more fair than that?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Dave wrote:
"Keith Dysart" wrote:

"Dave" wrote:
Are you really prepared to throw away P = VI?

yes, when the V and I are the superimposed voltage and current that
you insist are the real current and voltage on the line.

Amazing! I suppose Ohm's law is next?

Hopefully, you realize that P = VI only works for DC.
AC and RF, with their associated phase angles, are much
more complicated than P = VI. As a matter of fact, if
you use P = VI for standing waves, you are infinitely
wrong. There is no power in standing waves no matter
what the voltage and current magnitudes.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Dave wrote:
"Keith Dysart" wrote:
Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere
all the time. use standing waves and it fails. period, end of story.

It does not fail.

Good grief, Roy. The net voltage is 100v. The net current
is 1 amp. The load is infinite. P does NOT equal 100 watts.
P equals zero. P does not equal V * I. P = V*I*cos(A) where
A = 90 degrees.

What the hell is wrong with you people?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
This photonic limitation is something that exists only in your head.

Good Grief, Gene, I don't have time to teach you
quantum electrodynamics. Go read a book that tells
you about the nature of photons. It is also the
cornerstone of relativity.

Cecil,

I recall from some previous messages that you may have read Feynmann's
QED. Why don't you add some Feynmann diagrams into this discussion and
edumacate us all? (I can answer that one: you wouldn't know what to do
with a Feynmann diagram if it bit you on the behind.)

More to the point. If you would stop the silly business about "net"
everything you might begin to understand that standing waves have
effects that move at the speed of light just in the same manner that
traveling waves do. Do you really believe that the fields from the
standing waves don't propagate?

Perhaps you should review the difference between static and stationary.
The standing wave is stationary. It is not static.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
With such profound statements as, "a pure standing wave is technically
NOT an EM wave", it you might either offer some sort of reference or
start planning your trip to Stockholm.

Gene, here is a true/false quiz for you. If you have a
reference that disagrees with the obvious answers, please
quote it.

1. Is EM wave energy photonic in nature? ________

2. Do photons move at the speed of light in a medium? ______

3. Do standing waves move at the speed of light? _______

If the answers are yes, yes, and no, then standing waves
have been eliminated from the set of EM waves.

Cecil,

Too easy.

1. Yes, but who cares? If you want to go beyond ordinary classical
models, then most of your standard formalism about transmission lines
would need a bit of work. Classical works pretty well for HF.

2. Yes.

3. Yes.


Do I win the prize?

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
. . .
You give a good example Keith. It would be correct for measurement at
the load and at every point 1/2 wavelength back to the source from the
load, because the standing wave has the same measurements at these
points. At the 1/4 wavelength point back from the load and every
successive 1/2 wave point back to the source, the equation would also be
correct as demonstrated in Roy's example earlier today.

Excepting for these points, we would also be measuring a reactive
component that could be described as the charging and discharging of the
capacity or inductive component of the transmission line. (Imagine that
we are measuring the mismatched load through a 1/8 wave length long
transmission line, using an Autek RX VECTOR ANALYST instrument) The
inclusion of this reactive component would invalidate the power reading
if we were assuming that the measured power was all going to the load.
. . .

Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms. V at the input is fixed by the source at 100
volts RMS. I = V / Z = 1.6 - j1.2 amps = 2.0 amps magnitude with a phase
angle of -36.87 degrees.

Now I'll translate V and I into time domain quantities. (I could have
calculated I directly in the time domain, but this was simpler.)

Using w for omega, the rotational frequency,

If V(t) = 141.4*sin(wt) [100 volts RMS at zero degrees] then
I(t) = 2.828*sin(wt-36.87 deg.) [2 amps RMS at -36.87 degrees]

Multiplying V * I we get:

V(t) * I(t) = 400 * sin(wt) * sin(wt - 36.87 deg.)

By means of a trig identity, this can be converted to:

= 200 * [cos(36.87 deg) - cos(2wt - 36.87 deg.)]

cos(36.87 deg) = 0.80, so

V(t) * I(t) = 160 - 200 * cos(2wt - 36.87 deg.)

This is a waveform I described in my previous posting. The cosine term
is a sinusoidal waveform at twice the frequency of V and I. The 160 is a
constant ("DC") term which offsets this waveform. The fact that the
waveform is offset means that the power is positive for a larger part of
each cycle than it is negative, so during each cycle, more energy is
moved in one direction than the other. In fact, the offset value of 160
is, as I also explained earlier, the average power. It should be
apparent that the average of the first term, 160, is 160 and that the
average of the second term, the cosine term, is zero.

Let's see how this all squares with the impedance I calculated earlier.

Average power is Irms^2 * R. The R at the line input is 40 ohms and the
magnitude of I is 2.0 amps RMS, so the power from the source and
entering the line is 2^2 * 40 = 160 watts. Whaddaya know. Let's try
Vrms^2 / R. In this case, R is the shunt R. The line input impedance of
40 + j30 ohms can be represented by the parallel combination of 62.50
ohms resistive and a reactance of 83.33 ohms. The 62.50 ohms is what we
use for the R in the V^2 / R calculation. Vrms^2 / R = 100^2 / 62.50 =
160 watts.

We can also calculate the power in the load from its voltage and current
and, with the assumption of a lossless line I've been using, it will
also equal 160 watts.

P = V(t) * I(t) always works. You don't need power factor or reactive
power "corrections", or to have a purely resistive impedance.

This is really awfully basic stuff. Some of the posters here would come
away with a lot more useful knowledge by spending their time reading a
basic electric circuit text rather than making uninformed arguments.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
I recall from some previous messages that you may have read Feynmann's
QED. Why don't you add some Feynmann diagrams into this discussion and
edumacate us all?

They would actually be easier to do than some of the ASCII
schematics that I have drawn.

More to the point. If you would stop the silly business about "net"
everything you might begin to understand that standing waves have
effects that move at the speed of light just in the same manner that
traveling waves do. Do you really believe that the fields from the
standing waves don't propagate?

Yes, the standing wave fields do not propagate so that should
be a clue that they are somewhat unrelated to reality. That the
standing wave fields do not propagate can be seen in the phase
of the standing wave current as reported by EZNEC - it doesn't
change by more than a negligible amount over the entire length
of a 1/2WL dipole just as Kraus reported:

http://www.w5dxp.com/krausdip.gif

Perhaps you should review the difference between static and stationary.
The standing wave is stationary. It is not static.

I don't recall uttering the word "static". Perhaps you would like
to quote me and prove me wrong. If you cannot, would you like
to admit that you are just blowing smoke?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
"Tom Donaly" wrote in message
. net...
Yuri Blanarovich wrote:
"Dave" wrote in message
news:cwPbj.1073$ML6.117@trndny04...
you can do it when it makes physical sense. it does not make sense in
standing waves for all the obvious reasons that i have pointed out. it
does make sense in the individual traveling waves. just accept what
your little swr meter tells you, it shows the forward power and
reflected power, that is all you need and the only powers that make
sense.


Little SWR meter shows forward AND reflected power in one direction, and
reflected power only in reverse direction. Why is the Bird wattmeter
calibrated in Watts, measuring power (forward and reverse) and has chart
to calculate SWR, when there are no standing waves and no power in them?
Laying waves or sitting waves???

Seems to me that the PROBLEM is that some consider standing wave to be
some imaginary, stopped, frozen wave, no good, while some of us consider
standing wave to be the result of superposition of forward and reverse
waves, that can be (their components) measured, current heats when
flowing through resistance, voltage "burns" when poor dielectric.
Like there is standing wave current, but no standing wave, huh????
Or are we forgetting that we are dealing with electromagnetic waves?
Can someone sort out the terminology and definitions?

Yuri, K3BU
Hi, Yuri,
Cecil and Dave aren't taking resistance into account when they
talk about waves. In other words, they're not writing about real
transmission lines, coils, and such. The only lines they care about are
the ones in their minds: the simple ones where the attenuation constants
are always zero, and current and voltage are always either in phase, or
90 degrees out of phase, and there is never any dielectric breakdown, no
matter the voltage. You can only feel sorry for guys like that.
73,
Tom Donaly, KA6RUH

using any decent coax of a reasonable length and typical amateur power
levels the assumptions we have stated are very close to the actual results.
if you want to examine lossy lines in detail then go ahead, the formulas get
much messier and without proper formula rendering on a newsgroup they are
almost impossible to discuss... and for the concepts that have been proposed
the ideal lossless line case is perfectly acceptable.


Not when you put the kind of stress on it that Yuri does. Yuri has a
choice: he can either believe you, or what he can see with his own eyes,
in which case your fantasy line is not at all "perfectly acceptable."
73,
Tom Donaly, KA6RUH

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
3. Do standing waves move at the speed of light? _______

3. Yes.

Do I win the prize?

Of course not. Please explain how standing waves can
move at the speed of light yet only change phase by 1/10
of one degree in 90 degrees of transmission line or antenna?
Please be technically specific and include the math.

We cannot wait to hear your answer to that one.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms.

Whoops, that is not possible for your V & I scalar values.
Please don't tell us that you are using ASCII characters,
normally reserved for scalar values, for phasor values,
without telling anyone what you were doing. Do you think
that is ethical?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

compare and comment on the relation between pf, pr, and p.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
. . .
You give a good example Keith. It would be correct for measurement at
the load and at every point 1/2 wavelength back to the source from the
load, because the standing wave has the same measurements at these
points. At the 1/4 wavelength point back from the load and every
successive 1/2 wave point back to the source, the equation would also
be correct as demonstrated in Roy's example earlier today.

Excepting for these points, we would also be measuring a reactive
component that could be described as the charging and discharging of
the capacity or inductive component of the transmission line.
(Imagine that we are measuring the mismatched load through a 1/8 wave
length long transmission line, using an Autek RX VECTOR ANALYST
instrument) The inclusion of this reactive component would invalidate
the power reading if we were assuming that the measured power was all
going to the load.
. . .

Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms. V at the input is fixed by the source at 100
volts RMS. I = V / Z = 1.6 - j1.2 amps = 2.0 amps magnitude with a phase
angle of -36.87 degrees.

Now I'll translate V and I into time domain quantities. (I could have
calculated I directly in the time domain, but this was simpler.)

Using w for omega, the rotational frequency,

If V(t) = 141.4*sin(wt) [100 volts RMS at zero degrees] then
I(t) = 2.828*sin(wt-36.87 deg.) [2 amps RMS at -36.87 degrees]

Multiplying V * I we get:

V(t) * I(t) = 400 * sin(wt) * sin(wt - 36.87 deg.)

By means of a trig identity, this can be converted to:

= 200 * [cos(36.87 deg) - cos(2wt - 36.87 deg.)]

cos(36.87 deg) = 0.80, so

V(t) * I(t) = 160 - 200 * cos(2wt - 36.87 deg.)

This is a waveform I described in my previous posting. The cosine term
is a sinusoidal waveform at twice the frequency of V and I. The 160 is a
constant ("DC") term which offsets this waveform. The fact that the
waveform is offset means that the power is positive for a larger part of
each cycle than it is negative, so during each cycle, more energy is
moved in one direction than the other. In fact, the offset value of 160
is, as I also explained earlier, the average power. It should be
apparent that the average of the first term, 160, is 160 and that the
average of the second term, the cosine term, is zero.

Let's see how this all squares with the impedance I calculated earlier.

Average power is Irms^2 * R. The R at the line input is 40 ohms and the
magnitude of I is 2.0 amps RMS, so the power from the source and
entering the line is 2^2 * 40 = 160 watts. Whaddaya know. Let's try
Vrms^2 / R. In this case, R is the shunt R. The line input impedance of
40 + j30 ohms can be represented by the parallel combination of 62.50
ohms resistive and a reactance of 83.33 ohms. The 62.50 ohms is what we
use for the R in the V^2 / R calculation. Vrms^2 / R = 100^2 / 62.50 =
160 watts.

We can also calculate the power in the load from its voltage and current
and, with the assumption of a lossless line I've been using, it will
also equal 160 watts.

P = V(t) * I(t) always works. You don't need power factor or reactive
power "corrections", or to have a purely resistive impedance.

This is really awfully basic stuff. Some of the posters here would come
away with a lot more useful knowledge by spending their time reading a
basic electric circuit text rather than making uninformed arguments.

Roy Lewallen, W7EL

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P =
V * I that Keith was referencing is not appropriate at all points on the
line. If I understood Keith correctly, he would have calculated 200
watts input for your example (100 volts at 2 amps).

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

compare and comment on the relation between pf, pr, and p.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.





Do it yourself, Dave. Or, better yet, ask Cecil to do it. Anyway, you
made a mistake in the question. Can you discover what it is?
73,
Tom Donaly, KA6RUH

(P. S. It's Christmas Eve. I've got a quantity of nut-brown ale, a buxom
wench, a roast beef feast, and an insatiable appetite for all three, so
the digladiation will have to wait.)

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P =
V * I that Keith was referencing is not appropriate at all points on the
line. If I understood Keith correctly, he would have calculated 200
watts input for your example (100 volts at 2 amps).

Huh? I calculated 160 watts using P = V(t) * I(t). Why do you think
Keith would have calculated 200 from the same equation? I have no doubt
that his math skills exceed mine, and I used nothing more than complex
arithmetic and high school trig.

I calculated power at the input end of the line, but I can calculate it
from the same equation P = V(t) * I(t) at any point on the line, and
will get exactly the same result. At what point(s) do you think it's not
"appropriate" to use?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

No problem.

Let's use the voltage source as the phase reference, so it will be
(using w to represent the rotational frequency omega):

v(t) at the line input = 0.5 * sin(wt)

At the center of the line, i(t) = 0.02 * sin(wt - 90 deg.) and v(t) = 0.

p(t) = v(t) * i(t) = 0

Note that this isn't the average power (although the average power is
also zero), but the instantaneous power -- meaning that the power at
that point is zero at all times.

These aren't "standing wave" voltage, current, and power, but simply the
total voltage and current, and the power, at that point.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

At the center of the line,

vf(t) = 0.5 * sin(wt - 90 deg.)
if(t) = 0.01 * sin(wt - 90 deg.)
vr(t) = -0.5 * sin(wt - 90 deg.)
ir(t) = 0.01 * sin(wt - 90 deg.)

where I've taken the positive direction of ir to be forward, the same as if.

compare and comment on the relation between pf, pr, and p.

Calculation of P(t) is as follows:

p(t) = v(t) * i(t) = [vf(t) + vr(t)] * [if(t) + ir(t)]

= 0 * 0.02 sin(wt - 90 deg.) = 0 as calculated before.

I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid. There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line. There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero, indicating the movement of energy back
and forth but no net energy flow.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

Sure.

At the far end of the line,

v(t) = sin(wt - 180 deg.)
i(t) = 0

p(t) = v(t) * i(t) = 0 at all times.

vf, if, vr, and ir, are the same everywhere on the line, so see the
previously calculated values. Likewise, the calculation of p(t) from the
forward and reverse traveling waves is the same as before, with the same
result.

I've shown that I can calculate the correct power at two points along
the line by using p(t) = v(t) * i(t) as I can for any point on any line
with any termination. Your question implies that the results I got (zero
at all times at both points) are incorrect. What power (instantaneous
and average) do you calculate for those two points?

(*) In fact, calculation of the power at the source will show that the
power there is also zero at all times. The energy moving back and forth
in the line at all points except the quarter wavelength nulls was put
into the line during the turn-on process. After steady state has been
reached, the source can be turned off (converted to a short circuit)
with no change in what's happening on the line.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

Thanks for the example of P = V(t) * I(t).

I think the example illustrates why the instantaneous power equation P
= V * I that Keith was referencing is not appropriate at all points on
the line. If I understood Keith correctly, he would have calculated
200 watts input for your example (100 volts at 2 amps).

Huh? I calculated 160 watts using P = V(t) * I(t). Why do you think
Keith would have calculated 200 from the same equation? I have no doubt
that his math skills exceed mine, and I used nothing more than complex
arithmetic and high school trig.

Why is it the same equation? I understand your P = V(t) * I(t) to be V
and I as functions of time, but Keith to be using what ever he reads
from his voltmeter and ammeter.

I calculated power at the input end of the line, but I can calculate it
from the same equation P = V(t) * I(t) at any point on the line, and
will get exactly the same result. At what point(s) do you think it's not
"appropriate" to use?

Roy Lewallen, W7EL

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

Why is it the same equation? I understand your P = V(t) * I(t) to be V
and I as functions of time, but Keith to be using what ever he reads
from his voltmeter and ammeter.

If you'll look back through Keith's postings you'll see that he was
referring to the functions of time. It looks like he left off the
explicit (t) at some places which would lead to confusion.

But I hope you realize that you can also find the average power just
fine by calculating Pavg = Vrms * Irms * cos(theta) at any point along
the line, where Vrms and Irms are the total voltage and current at the
point, and theta is the angle between the two. This is true regardless
of the SWR. Also, you can calculate power as Irms^2 * Rser or Vrms^2 /
Rpar where Rser and Rpar are the series and parallel equivalent
resistive parts of the impedance at any point. Like v(t) and i(t), Vrms
and Irms can be found if desired by summing the forward and reflected
waves to find the total value at the point of interest; superposition
applies. It doesn't apply to power, so always do the summation of
voltages and currents before calculating power.

One property of the P(t) = V(t) * I(t) equation is that it also applies
to non-sinusoidal and even non-periodic waveforms -- it can *always* be
used. And you can always find the average power by integrating it then
dividing by the integration period. The average value calculation
reduces to Vrms * Irms * cos(theta) for pure sine waves but not other
waveforms.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Correction:

Roy Lewallen wrote:
. . .
vf, if, vr, and ir, are the same everywhere on the line, so see the
previously calculated values. Likewise, the calculation of p(t) from the
forward and reverse traveling waves is the same as before, with the same
result.

This isn't true. The phase angles of these terms are different at
different places along the line, so the p(t) results are different. At
the load end of the line:

vf(t) = 0.5 * sin(wt - 180 deg.)
if(t) = 0.01 * sin(wt - 180 deg.)
vr(t) = 0.5 * sin(wt - 180 deg.)
ir(t) = -0.01 * sin(wt - 180 deg.)

At this point, the result for p(t) is still zero at all times because of
zero ir + if rather than vr + vf as at the center of the line. At other
points along the line, as I mentioned in the original posting, p(t) will
end up being non-zero -- it'll be a sinusoidal function with rotational
frequency 2wt. But it will have an average value of zero, indicating
movement of energy back and forth but no net energy flow over an
integral number of periods.

I apologize for the error.

May everyone have an enjoyable and peaceful holiday.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
These aren't "standing wave" voltage, current, and power, but simply the
total voltage and current, and the power, at that point.

Good Grief, Roy, no wonder you are clueless about standing
waves. In a lossless open-circuit line, the total voltage and
current *ARE* the standing-wave voltage and current because
traveling wave current and voltage doesn't exist. Good Grief!
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Why is it the same equation? I understand your P = V(t) * I(t) to be V
and I as functions of time, but Keith to be using what ever he reads
from his voltmeter and ammeter.

There is a conspiracy to confuse you by using the same
symbol for completely different values - don't fall for
that Tar Baby.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

Why is it the same equation? I understand your P = V(t) * I(t) to be
V and I as functions of time, but Keith to be using what ever he reads
from his voltmeter and ammeter.

If you'll look back through Keith's postings you'll see that he was
referring to the functions of time. It looks like he left off the
explicit (t) at some places which would lead to confusion.

But I hope you realize that you can also find the average power just
fine by calculating Pavg = Vrms * Irms * cos(theta) at any point along
the line, where Vrms and Irms are the total voltage and current at the
point, and theta is the angle between the two. This is true regardless
of the SWR. Also, you can calculate power as Irms^2 * Rser or Vrms^2 /
Rpar where Rser and Rpar are the series and parallel equivalent
resistive parts of the impedance at any point. Like v(t) and i(t), Vrms
and Irms can be found if desired by summing the forward and reflected
waves to find the total value at the point of interest; superposition
applies. It doesn't apply to power, so always do the summation of
voltages and currents before calculating power.

One property of the P(t) = V(t) * I(t) equation is that it also applies
to non-sinusoidal and even non-periodic waveforms -- it can *always* be
used. And you can always find the average power by integrating it then
dividing by the integration period. The average value calculation
reduces to Vrms * Irms * cos(theta) for pure sine waves but not other
waveforms.

Roy Lewallen, W7EL

Thanks again Roy. Your last two postings have been very helpful. While
I am aware of these relationships, I do not use them often and each use
seems like a first time. So I try to be very careful, but still make
bad errors as you saw yesterday.

I try not to be too critical of any postings, but (as Cecil warns) the
standards of knowledge and explanation are very high on this news group,
and misuse (even accidental) of terms or equations are noted and
frequently flamed or worse. Participants need a thick skin and
persistence, but a great deal can be learned here. I find it to be
quite a learning experience and proving ground.

Merry Christmas!

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 1:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Except that V(x,t) and I(x,t) are not, in general, related by Z0.

*From "Fields and Waves ..." by Ramo & Whinnery, 2nd edition:

V(x,t) = V*e^j(wt-kx) + V'*e^j(wt+kx)

I(x,t) = [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

You are quite unfair to Ramo & Whnnery when you quote them
out of context. It makes them look like they do not have a
clue.

Just because you find the same string ["V(x,t)"] in their
text does not mean that they are talking about the same
thing. (It better not in this case, or you should throw
their book away.)

You, yourself, have made the point in other posts that
the current and voltage on an open circuited line are
in quadrature, so they can not be in the ratio of Z0.

No. The two wave view is merely an alternate set of expressions
which, when summed (i.e. using superposition), provide the actual
voltage and current on the line. These alternate expressions are
obtained by algebraic maniupulation of the more fundamental
descriptive equations.

Methinks you are confusing cause and effect. The standing
wave is not the cause of the two traveling waves.

Of course not. "Standing wave" is just a short hand description
of the distribution of voltage and current on the line.

In my analysis, P(x,t) = V(x,t) * I(x,t) is the equation that means
the power at any point and time can by obtained by measuring the
actual voltage and current on the line at the point and time of interest..

Make that the *NET* power and you will have it nailed.

Are you sure you want to throw away this ability? Are you sure you
want to claim that instantaneous power can NOT be obtained by
multiplying the instaneous measured voltage by the instanteous
measured current?

When the instantaneous voltage is the sum of two more elementary
voltages (same for current) then you are reporting the *NET*
results, not the underlying component results. The *NET* results
do not dictate reality.

Sorry Cecil. The observed voltage and current on the line
are the reality.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

but he missed the definition of pr and pf...

dave in problem statement:
now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point,
where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

so he conveniently skipped that step and instead writes this cop out:

roy:
I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid.

pr and pf are, as i stated, the power in the forward and reflected waves.
There is of course power in these two waves and it is indeed 'sloshing' back
and forth in the line. These are the waves that can be measured by any of
the simple devices such as neon bulbs or bird watt meters that clearly show
equal and opposite powers in the waves. so you can indeed have power in the
traveling waves, but no power in the standing waves... which will always be
the case.

i will give him this point as being correct for a lossless open (or shorted)
line:

There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line.

but then he loses it again:

There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero,

the traveling waves will have power EVERYWHERE on the line, the special
cases are are just the ones where the standing wave is most easily shown as
having no power. obviously if there is power in a wave at one point on a
line it is not going to stop and bypass the quarter wave points, the forward
and reflected waves continue end to end and their power goes with them... it
is at those 'special' points where the voltage or currents in the forward
and reflected waves always cancel each other so if you measure with a simple
tool you will see the voltage or current nulls at those points. that does
not mean there is no power passing those points, only that the voltage or
current in the traveling waves has conveniently canceled each other out at
those points.

and then he has to end up with an obvious contradiction:

indicating the movement of energy back
and forth but no net energy flow.

how does energy not flow if it is moving back and forth???

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:
These aren't "standing wave" voltage, current, and power, but simply
the total voltage and current, and the power, at that point.

Good Grief, Roy, no wonder you are clueless about standing
waves. In a lossless open-circuit line, the total voltage and
current *ARE* the standing-wave voltage and current because
traveling wave current and voltage doesn't exist. Good Grief!

Hmmm, Like Roy, I thought there was still a traveling wave in this
situation. The voltage at the far end of the line must reverse polarity
as time passes, so the waves must continue to travel, or so I would think.

Maybe it could be said better, but I thought Roy was trying to say that
although power could not be detected at the center or ends, it was
flowing as a result of the initial impetus charged into the system. I
would understand that this power would be the power needed to charge the
impedance and capacity of the line as it continually reversed polarity.
This would be real power from energy stored (but constantly moving) on
the 1/2 wavelength line so long as the system is active. We know we
have power present because we find energy distributed as V and I on the
time plot (viewed as a "standing wave" on the time plot).

Is this a "tongue-in-cheek" comment?

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:

Keith Dysart wrote:
Except that V(x,t) and I(x,t) are not, in general, related by Z0.

From "Fields and Waves ..." by Ramo & Whinnery, 2nd edition:

V(x,t) = V*e^j(wt-kx) + V'*e^j(wt+kx)

I(x,t) = [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

You are quite unfair to Ramo & Whnnery when you quote them
out of context. It makes them look like they do not have a
clue.

It's not out of context. Those are their equations for
standing wave voltage and standing wave current. It is
net voltage and net current because each equation is
the sum of two component values.

Just because you find the same string ["V(x,t)"]

Actually, it wasn't the same string. R&W used 'z'
instead of 'x' for the length of the wire as was
common a half-century ago when I had their textbook
for both undergraduate and graduate level courses.

in their text does not mean that they are talking about
the same thing. (It better not in this case, or you should
throw their book away.)

But they *are* talking about the same thing. The first equation
above is the total (standing wave) voltage. The second equation
is the total (standing wave) current. The only term difference
between the two equations is the Z0 term. There is a sign difference
in the current equation that shifts the reflected current by 180
degrees putting the net voltage and net current in quadrature.

Methinks suggesting that Ramo & Whinnery should be thrown away
is "delusions of grandeur".

You, yourself, have made the point in other posts that
the current and voltage on an open circuited line are
in quadrature, so they can not be in the ratio of Z0.

Nobody said they are "in the ratio of Z0" and that's not
what you said before either. You said they are "not related
by Z0" and they are related by Z0 just as the above equations
demonstrate. You didn't say "ratio", you said "related".

The ratio of standing wave voltage to standing wave current
above is:

V(x,t) V*e^j(wt-kx) + V'*e^j(wt+kx)
------ = --------------------------------- = Z
I(x,t) [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

This is the old familiar SWR circle on the Smith Chart.
Smith Charts are normalized to Z0 = 1.0. The impedance
on the Smith Chart must be multiplied by Z0 to get the
actual impedance. Those impedances are indeed
"related to Z0". The s-parameter signals are normalized
to the square root of Z0. Vfor/Ifor = Vref/Iref = Z0.
Virtually everything about a transmission line is "related"
to Z0 including the standing wave voltage and current.

Methinks you are confusing cause and effect. The standing
wave is not the cause of the two traveling waves.

Of course not. "Standing wave" is just a short hand description
of the distribution of voltage and current on the line.

Yes, a "short hand description" that exists only in the human
mind and gets some folks into trouble - like trying to use
the illusion of moving standing wave current to "measure"
the delay through a coil. There is no standing wave current
movement through a coil or a wire. Standing wave current
doesn't move - it just stands there, oscillating in place.
EM current that doesn't move is obviously an illusion (and
a violation of the laws of physics).

Here's what Eugene Hecht said about standing waves in "Optics".

"It [the standing wave phasor] doesn't rotate at all, and
the resultant wave it represents doesn't progress through
space - it's a standing wave."

Applied to a wire or a loading coil, we can paraphrase -
standing wave current "doesn't progress through" a wire or
a loading coil - "it's a standing wave". EZNEC supports this
concept. Take a look at the current phase for a 1/2WL wire
dipole. From one end of the dipole to the other, the total
current phase varies by only a couple of degrees. Kraus agrees:

http://www.w5dxp.com/krausdip.jpg

Sorry Cecil. The observed voltage and current on the line
are the reality.

The net voltage and net current are real but their independent
existence apart from the underlying traveling waves is just
an illusion. They are not independent. They are the *result*
of superposition of the forward and reverse waves. They are
merely an *effect* and not the cause of anything.

There are lots of real illusions. That beautiful Texas
"sunrise" this morning was just an illusion, an effect of
the rotation of the earth.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

Power is the measure of the energy flow past a measurement
point. There is no net energy flow in pure standing waves.
Since the standing wave current phasor is always 90 degrees
different from the standing wave voltage phasor,

Power = V*I*cos(90) = 0 watts Note that V and I do not
have to be zero. There is simply no real power in a pure
standing wave. However, there are plenty of joules in
standing waves that can be converted to real power at
the expense of the standing waves after key-up.

All of the standing wave energy is contained in

Reactive Power = V*I*sin(90) in units of VARS

From the IEEE Dictionary:

"reactive power - For sinusoidal quantities in a two-wire
circuit, reactive power is the product of the voltage,
the current, and the sine of the phase angle between them."
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Hmmm, Like Roy, I thought there was still a traveling wave in this
situation. The voltage at the far end of the line must reverse polarity
as time passes, so the waves must continue to travel, or so I would think.

It is common practice to divide the waves into standing waves
and traveling waves so they can be discussed separately. For
instance, if the forward voltage is 200 volts and the reflected
voltage is 100 volts, we have a mixture of standing waves and
traveling waves.

It is common practice to allocate 100 volts of the forward wave
to the standing wave in order to exactly match the reflected
voltage with the resultant *pure standing wave*.

The other 100 volts is allocated to the *pure traveling wave* that
is making its way from the source to the load. That's how my profs
at Texas A&M chose to teach the subject. It's a sort of reverse
superposition technique that makes everything crystal clear in
a system of mixed standing waves and traveling waves.

Is this a "tongue-in-cheek" comment?

You think? :-)

Note that in a pure standing wave, the magnitudes of the forward
wave and reflected wave are identical.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

Hmmm, Like Roy, I thought there was still a traveling wave in this
situation. The voltage at the far end of the line must reverse polarity
as time passes, so the waves must continue to travel, or so I would think.

In the case of an open circuited line, the voltage traveling wave
maintains the same polarity upon reflection from the end of the line.
Mathematically, the voltage reflection coefficient is +1. The voltage
and current traveling waves continue to travel after reflection, which
results in the interference pattern known as a standing wave.

Maybe it could be said better, but I thought Roy was trying to say that
although power could not be detected at the center or ends, it was
flowing as a result of the initial impetus charged into the system.

It's easy to confuse power and energy, and I've been careful to use
those terms correctly. Power is the rate of energy flow, and I said
nothing about power flowing. (That's Cecil's concept, and careless
application of it leads to irreconcilable problems.)

I
would understand that this power would be the power needed to charge the
impedance and capacity of the line as it continually reversed polarity.

Capacitance is charged by charge, which is the integral of current, so
you can if you want track the charging and discharging of the line's
capacitance by integrating the current. Impedance is the ratio of V to I
and isn't something that's charged.

This would be real power from energy stored (but constantly moving) on
the 1/2 wavelength line so long as the system is active. We know we
have power present because we find energy distributed as V and I on the
time plot (viewed as a "standing wave" on the time plot).

The existence of both voltage and current at any point along the line
tells us that there is instantaneous power at that point, which means
that energy is flowing past that point. At any point along the open
circuited line except the ends and middle (where the power is always
zero), we can see that the power is in the form of a sine wave having
twice the period of v(t) or i(t) and no offset. This means that at any
point, energy flows first one direction, then back, equal amounts and
for equal amounts of time, twice during each v(t) or i(t) cycle. The
result is no net energy flow in either direction.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

It's sad to see that the response to my analysis and equations is insult
and derision rather than any coherent counter argument, but I'm
unfortunately not surprised. I don't see in it any evidence that my
posted calculation was in error -- the only objections I see are that it
doesn't support a flawed theory, so it therefore must be wrong. The
calculations I made are based on solid theory which has been
successfully used for more than a century, and you won't be able to make
any measurement which will refute them. They're also entirely self
consistent with all other transmission line phenomena which can be
calculated or measured.

So I wouldn't bother to respond at all except that it does provide the
opportunity to elaborate a bit on what I posted. If further responses
are as devoid of substance as this one, I'll probably end up plonking
Dave as I did Cecil some time ago, for the same reason.

Dave wrote:
so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

I did what I claimed to be able to do -- correctly calculate the power
from v(t) * i(t). Yuri takes exception to many things I say, but frankly
that bothers me not in the least.

but he missed the definition of pr and pf...

Indeed I did. Can you define them for me please?

p(t) = v(t) * i(t)
v(t) = vf(t) + vr(t)
i(t) = if(t) + ir(t)

Therefore,

p(t) = [vf(t) + vr(t)] * [if(t) + ir(t)]
= vf(t) * if(t) + vr(t) * ir(t) + vf(t) * ir(t) + vr(t) * if(t)

Of these four terms, which, if any, are pf(t) and pr(t)? What are the
two remaining terms called?

dave in problem statement:
now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point,
where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

so he conveniently skipped that step and instead writes this cop out:

No, I did calculate all except pf(t) and pr(t), which you didn't define.
As soon as you do (see the question above), I'll be glad to calculate
them also. Or you could do it for us -- it won't involve more than
simple arithmetic.

roy:
I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid.

pr and pf are, as i stated, the power in the forward and reflected waves.

What would that be, then, vr(t) * ir(t) and vf(t) * if(t)? Where does
the power in those remaining two terms come from or go?

There is of course power in these two waves and it is indeed 'sloshing' back
and forth in the line.

Please note that I didn't say that power was "sloshing" back and forth.
I said that energy was. Power is not the same as energy -- they bear the
same relationship as speed and distance.

These are the waves that can be measured by any of
the simple devices such as neon bulbs or bird watt meters that clearly show
equal and opposite powers in the waves. so you can indeed have power in the
traveling waves, but no power in the standing waves... which will always be
the case.

Unfortunately, people assume that the units indicated on a meter are the
quantity actually being measured, which often they're not. But this has
been explained many times before here.

i will give him this point as being correct for a lossless open (or shorted)
line:

There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line.

but then he loses it again:

There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero,

the traveling waves will have power EVERYWHERE on the line, the special
cases are are just the ones where the standing wave is most easily shown as
having no power. obviously if there is power in a wave at one point on a
line it is not going to stop and bypass the quarter wave points, the forward
and reflected waves continue end to end and their power goes with them... it
is at those 'special' points where the voltage or currents in the forward
and reflected waves always cancel each other so if you measure with a simple
tool you will see the voltage or current nulls at those points. that does
not mean there is no power passing those points, only that the voltage or
current in the traveling waves has conveniently canceled each other out at
those points.

Here's just one of the problems with assigning powers to the traveling
waves, attempting to keep track of them separately, and applying
superposition to nonlinear quantities. The conclusion that there is
power at the ends of the line, for example, is demonstrably not true.
There is no current at the far end of the line at any time, and
therefore no power, as I showed. My comment at the end further shows why
the power at the end of the line must be zero.

and then he has to end up with an obvious contradiction:

indicating the movement of energy back
and forth but no net energy flow.

how does energy not flow if it is moving back and forth???

Hopefully most of the readers were more astute than this and noticed the
word "net". If equal amounts of energy flow in each direction during a
cycle (as indicated by a power waveform with no offset), there is no net
energy flow. It means that energy is being stored at some location in
one direction, then returned during the other half of the power cycle.
This is true in any purely reactive circuit, for example a tank circuit,
where exactly the same calculation I made can be done with the same result.

Consider our open ended line for a moment, and imagine it laid out from
left to right with the open end to the right, so I can name directions.
There's no place to the right of the line to store energy, so no energy
can be moved past that point to the right. If we consider the positive
direction of energy flow as being to the right, this means that the
power at the end of the line can never go positive, even for an instant
-- if it did, it would mean energy is moving past the end of the line
during that time. But since there's no storage mechanism beyond the end,
this can't happen. And since we can't have net energy moving past the
end, either, the power therefore can't go negative at any time either.
So the power at the end of the line must be zero at all times. This is
of course the result I got from v(t) * i(t).

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:


The net voltage and net current are real but their independent
existence apart from the underlying traveling waves is just
an illusion.

Cecil,

This sums it up pretty nicely. Reality is an illusion. No wonder many
people have a hard time accepting your nonsense. You have built your own
little world where reality and illusion are randomly intertwined as
suits the needs of the moment.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

but he missed the definition of pr and pf...

dave in problem statement:
now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point,
where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

so he conveniently skipped that step and instead writes this cop out:

roy:
I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid.

pr and pf are, as i stated, the power in the forward and reflected waves.
There is of course power in these two waves and it is indeed 'sloshing' back
and forth in the line. These are the waves that can be measured by any of
the simple devices such as neon bulbs or bird watt meters that clearly show
equal and opposite powers in the waves. so you can indeed have power in the
traveling waves, but no power in the standing waves... which will always be
the case.

i will give him this point as being correct for a lossless open (or shorted)
line:

There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line.

but then he loses it again:

There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero,

the traveling waves will have power EVERYWHERE on the line, the special
cases are are just the ones where the standing wave is most easily shown as
having no power. obviously if there is power in a wave at one point on a
line it is not going to stop and bypass the quarter wave points, the forward
and reflected waves continue end to end and their power goes with them... it
is at those 'special' points where the voltage or currents in the forward
and reflected waves always cancel each other so if you measure with a simple
tool you will see the voltage or current nulls at those points. that does
not mean there is no power passing those points, only that the voltage or
current in the traveling waves has conveniently canceled each other out at
those points.

and then he has to end up with an obvious contradiction:

indicating the movement of energy back
and forth but no net energy flow.

how does energy not flow if it is moving back and forth???




I see we are back to the old business about colliding waves that
apparently carry vector power. At least it seems that way since the
counter-traveling power can cancel at some points and add up at other
points. I thought we trashed this idea a couple of years ago.

Remember, calculate fields first and then worry about power or energy.
Any change in that calculation order will surely lead to "traveling
waves of average power" and other such gems.

I hope no Joules are hurt in the collision.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:

...
Remember, calculate fields first and then worry about power or energy.
Any change in that calculation order will surely lead to "traveling
waves of average power" and other such gems.

I hope no Joules are hurt in the collision.

73,
Gene
W4SZ

If two boards of a length are secured upon the surface of a body of
water, their parallel distances to each other computed to "contain"
"standing waves" of a certain frequency, and a device to "impart energy"
placed against one, to "strike" the surface of the water--much can be
seen without the necessity of "complicating maths."

Or, a substantial mud puddle, two boards and a timing circuit ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
It's easy to confuse power and energy, and I've been careful to use
those terms correctly. Power is the rate of energy flow, and I said
nothing about power flowing. (That's Cecil's concept, and careless
application of it leads to irreconcilable problems.)

Roy, God Himself appeared to me in a vision and said that
if you don't stop bearing false witness against me, you are
going to end up in a very bad place.

For the record, here's what I said in a WorldRadio article
more than three years ago: "The author has endeavored to satisfy
the purists in this series of articles. The term "power flow" has
been avoided in favor of "energy flow". Power is a measure of that
energy flow per unit time through a plane. Likewise, the EM fields
in the waves do the interfering. Powers, treated as scalars, are
incapable of interference. Any sign associated with a power in
this paper is the sign of the cosine of the phase angle between
two voltage phasors."

The existence of both voltage and current at any point along the line
tells us that there is instantaneous power at that point, ...

Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There
is no power, instantaneous or otherwise, in pure standing waves.
The cosine of 90 degrees is *always* zero.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
... I'll probably end up plonking
Dave as I did Cecil some time ago, for the same reason.

Roy, you ploinked me because I proved you wrong about the
delay through a 75m loading coil and you didn't want to
lose face on this newsgroup. You have refused to look at
any of the EZNEC files that prove you wrong. You have
threatened to recall my copy of EZNEC because you cannot
fact the technical facts reported by your own creation.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 25 Dec, 11:11, Gene Fuller wrote:
Dave wrote:
so roy has correctly calculated the standing wave 'power' to be zero at two
points on the line. *i am sure that yuri will take great exception to this
result showing there is no power in the standing wave.

but he missed the definition of pr and pf...

dave in problem statement:
now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point,
where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

so he conveniently skipped that step and instead writes this cop out:

roy:
I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid.

pr and pf are, as i stated, the power in the forward and reflected waves..
There is of course power in these two waves and it is indeed 'sloshing' back
and forth in the line. *These are the waves that can be measured by any of
the simple devices such as neon bulbs or bird watt meters that clearly show
equal and opposite powers in the waves. *so you can indeed have power in the
traveling waves, but no power in the standing waves... which will always be
the case.

i will give him this point as being correct for a lossless open (or shorted)
line:

There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line.

but then he loses it again:

There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero,

the traveling waves will have power EVERYWHERE on the line, the special
cases are are just the ones where the standing wave is most easily shown as
having no power. *obviously if there is power in a wave at one point on a
line it is not going to stop and bypass the quarter wave points, the forward
and reflected waves continue end to end and their power goes with them.... it
is at those 'special' points where the voltage or currents in the forward
and reflected waves always cancel each other so if you measure with a simple
tool you will see the voltage or current nulls at those points. *that does
not mean there is no power passing those points, only that the voltage or
current in the traveling waves has conveniently canceled each other out at
those points.

and then he has to end up with an obvious contradiction:

indicating the movement of energy back
and forth but no net energy flow.

how does energy not flow if it is moving back and forth???

I see we are back to the old business about colliding waves that
apparently carry vector power. At least it seems that way since the
counter-traveling power can cancel at some points and add up at other
points. I thought we trashed this idea a couple of years ago.



Remember, calculate fields first and then worry about power or energy.
snip-

You can't do that unless you break up the rise in voltage from
beginning to end
in segments. Each segment is different to the other when you are
considering
a tank circuit. This is why these arguments take so long, since one is
talking
about the fields at one segment and all the other corresponders are
refering
to other segments. It is quite possible that many correspondents are
correct
but only with respect to the segment they are referring to.You should
all
get on the same stage when entertaing such that the audio does not
get
to the deafining level while at the same time entertaining the same
audience
Art

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
The net voltage and net current are real but their independent
existence apart from the underlying traveling waves is just
an illusion.

This sums it up pretty nicely. Reality is an illusion. No wonder many
people have a hard time accepting your nonsense. You have built your own
little world where reality and illusion are randomly intertwined as
suits the needs of the moment.

Reality can certainly contain illusions. Sunrise and sunset
are a couple of examples. Magician tricks are another.

If you want to prove that standing waves are not an illusion,
take away the component forward and reverse traveling waves
and show us what you have left. As I said above, to which
you objected, standing waves cannot exist independently
of their forward and reverse components. I stand by that
statement. Prove it wrong if you can.
--
73, Cecil http://www.w5dxp.com