Standing-Wave Current vs Traveling-Wave Current
 

Hi Walt,

I'm a little confused here. I hope you can straighten me out.

Walter Maxwell wrote:

It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves. But you're saying there are currents and voltages "in"
the standing wave. Are you referring to the total current and voltage at
any point along the line? If so, why are they always in quadrature?
Certainly, the total V and I are in quadrature if the line is terminated
by an open, short, or purely reactive load. But not in any other case.
Or do you regard a line as having a "standing wave" with its own voltage
and current which are different from the total V and I? If so, how do
you define a "standing wave"? Are there separate equations for "standing
wave" V and I that are different than for total V and I?

. . .

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
you can measure the 'standing' wave voltage,
that has been known for a long time... but the effects are NOT due to power
in standing waves.

Are the effects due to energy in the standing waves?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Haste makes waste, and errors as well. The standing wave power
equation is incorrect. It should read "Power = V^2 / Zo + I^2 * Zo"

I'm afraid you will find that those are the equations for
power associated with a traveling wave. Actually should be
Power = V^2/Z0 = I^2*A0. There is no net power transfer
associated with a standing wave. For a pure standing wave,
V*I*cos(A) = 0 watts.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Walter Maxwell wrote in
:

....
It appears to me that even with all the successive posts on the
subject of power in the standing wave, you all seem to be missing the
ingredient that proves why there is no useable power in the standing
wave. It is because the current and voltage in the standing wave are
90° out of phase. Multiplying E x I under this condition results in
zero power.

Walt,

I am trying to make sense of this and the first issue is what you mean by
the term "standing wave".

The only meaning that seems possible is that it is the magnitude of the
time alternating voltage or current at some displacement along the
transmission line.

If that is the meaning, then the situation you describe of 90° phase
difference between E and I is rather specific, it can only occur with a
distortionless line AND a load that is (s/c OR o/c OR purely reactive).

Is that the case?

If so, should you have stated the assumptions and how does the case you
discuss help in explanation of general principles?


In addition to another comment above that implies that reflected power
is reactive power, this is not true--reflected power is as real as
forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in
phase in the forward direction, they are traveling 180° out of phase
in the rearward direction. Multiplying voltage and current while 180°
different in phase results in the same power as when they are in
phase.

You seem to be inferring that it is legitimate (in a general sense) to
calculate the power of forward and reflected waves as voltage times
current (eg Vf*If).

Isn't the instantaneous power at a point a function of time, and it is p
(t)=v(t)*i(t) and the expansion of that equals Vf*If-Vr*Ir ONLY when the
other two terms of the expansion cancel, and that is the special case of
a distortionless line.

Are you illustrating general principles with a special case without
stating the underlying assumptions.

Why is it that so many attempts to explain transmission line behaviour,
particularly regarding real and imaginary components of power at a point,
aren't consistent with basic AC circuit theory?

Owen

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Where do you get so many goofy ideas? Do you have any references at all
that support your contention that standing wave energy does not meet the
definition of EM energy? I have been in the wave business professionally
for about 40 years, and I have read many technical papers, reference
books, and text books. I have yet to encounter anything that indicated
the inferior nature of standing waves in the energy community.

I guess the authors of the textbooks never thought anyone
would be so ignorant as to believe that EM waves can stand
still. :-)

EM waves are photonic in nature must travel at the speed of
light in the medium. A standing wave stands still and oscillates
in place. Therefore, A standing wave is not an EM wave - It is
something else, by definition.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

"Power waves" is a standing joke around here. The last person to
seriously consider such things is Cecil, and he now denies ever saying
such.

That's a false statement. I supported power waves back in
the 1990s. I changed my mind around 1998, almost ten years
ago. I have not supported power waves in this 21st century.
Some other posters on this news still support power waves.
However, I do support EM energy waves.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
Roy Lewallen wrote:

...
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important
information is lost. (As was the case of the statistician who drowned
crossing a creek whose average depth was only two feet.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Oddly, "John", another source tells us:

Another hazard is the resonance of the magnetron tube itself. If the
microwave is run without an object to absorb the radiation, a standing
wave will form. The energy is reflected back and forth between the tube
and the cooking chamber.

http://en.wikipedia.org/wiki/Microwave_oven

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
Roy Lewallen wrote:

...
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important
information is lost. (As was the case of the statistician who drowned
crossing a creek whose average depth was only two feet.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Another source, "John", says:

Why is food cooked in a microwave oven sometimes not cooked uniformly?

Inside the microwave oven, the microwaves bounce off the metal internal
walls and set up complex 'standing wave' patterns. As with any wave,
microwaves have peaks and troughs and the intensity of the microwaves is
greatest in the peaks and troughs and lowest at points in between.

So if some food is near one of the peaks it will absorb lots of
microwaves and get really hot, while if it is midway between peaks and
troughs it may receive hardly any microwaves and so not get very hot at
all.

http://www.bbc.co.uk/food/tv_and_rad...icrowave.shtml

So you're telling us that what cooks food in a microwave oven is the
standing waves and that it isn't because the food itself is the load for
the output of the magnetron? You'd have us believe that standing waves
which result from operating a microwave oven without such a load are
present and actually cooking food?

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

We can make Keith's assertion true by the addition of one
word.

"Therefore, the forward and reverse waves cannot be transferring
*net* energy across these points. As Ramo and Whinnery say about
the forward and reflected Poynting vectors:

If Pz+ = Pz- then Pz+ - Pz- = 0
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Walter Maxwell wrote:
It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

In addition to another comment above that implies that reflected power is reactive power, this is not
true--reflected power is as real as forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in phase in the forward direction, they are
traveling 180° out of phase in the rearward direction. Multiplying voltage and current while 180° different in
phase results in the same power as when they are in phase.

Seems to me everything would be made clear by the addition of the
word "net". There is no net power in pure standing waves. There is
no net energy transfer in pure standing waves.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.

It's too bad you don't have the foggiest notion as to just
what it is the equation of. Much of the thousands of posts
could have been avoided if that were the case.
73,
Tom Donaly, KA6RUH

Standing-Wave Current vs Traveling-Wave Current
 

Owen Duffy wrote:
Walter Maxwell wrote:
It appears to me that even with all the successive posts on the
subject of power in the standing wave, you all seem to be missing the
ingredient that proves why there is no useable power in the standing
wave. It is because the current and voltage in the standing wave are
90° out of phase. Multiplying E x I under this condition results in
zero power.

I am trying to make sense of this and the first issue is what you mean by
the term "standing wave".

Walt is right.
Let's look at an arbitrary example of forward and reflected
voltage and current instantaneous phasors at one point on a
particular line.

Vfor = 100v at 45 degrees, Ifor = 2 amps at 45 degrees

The forward voltage and forward current are in phase.

Vref = 100v at -45 degrees, Iref = 2 amps at 135 degrees

The reflected voltage and reflected current are 180 degrees
out of phase.

Now calculate the total voltage and total current.

Total voltage = 2*100cos(45) = 141.4v at 0 deg

Total current = 2*2sin(45) = 2.83a at 90 deg

For a pure standing wave, the instantaneous voltage is
*always* 90 degrees out of phase with the instantaneous
current. There are no V*I*cos(A) watts in a pure standing
wave. There are only V*I*sin(A) VARS.

However, the VARS in the standing wave require energy
which can be converted to watts by I^2*R losses.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Tom Donaly wrote:
Cecil Moore wrote:
Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.

It's too bad you don't have the foggiest notion as to just
what it is the equation of. Much of the thousands of posts
could have been avoided if that were the case.

The technical content of your posting is noted. Here is what
it is the equation of:

http://www.chemmybear.com/standing.html
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

I'm surprised that standing waves seem so hard for people to understand.
They're simply a spatial pattern formed by the interference between
forward and reflected waves or, if you prefer, from the solution to a
general transmission line problem with boundary values applied. I'd hope
that even a brief look at a text would make it clear what standing waves
are, and what they are not.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

...
I guess the authors of the textbooks never thought anyone
would be so ignorant as to believe that EM waves can stand
still. :-)

EM waves are photonic in nature must travel at the speed of
light in the medium. A standing wave stands still and oscillates
in place. Therefore, A standing wave is not an EM wave - It is
something else, by definition.

Guess if you told me we were all communicating on "entangled antennas"
(and, yes, a 1:1 relationship to entangled particles)--I'd have to, at
least, give it a thought! ROFLOL!

Merry Xmas OM,
and warm regards,
JS ;-)

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:

...
Inside the microwave oven, the microwaves bounce off the metal internal
walls and set up complex 'standing wave' patterns. As with any wave,
microwaves have peaks and troughs and the intensity of the microwaves is
greatest in the peaks and troughs and lowest at points in between.
...

Actually, a bit more than that, even ...

Water molecules are slightly magnetic, they are spinning like hell on
those "humps of the standing wave"--friction cooking, you will excuse my
"artistic authors' license" ... or not, in ALL of this ... ;-)

However, the standing wave is much more appreciated, by me--at this
point, thanks to Cecil. And, my attention is always drawn towards
"little oddities" which can serve as diversion. And, the standing wave
IS cooking the turkey--in my humble opinion ... you could say, "I
believe in standing waves."

But then, tomorrow night, I'll be up very late with cookies and milk.
Yanno, I've never seen 'em--yet? ;-)

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
Dave Heil wrote:
[...]

forgot ...

Merry Xmas Heil,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen, W7EL wrote:
"As any text can tell you, the value of Z (ratio of V to I) varies along
a line which has a reflected waves (i.e., has a standing wave)."

Not exactly, maybe the apparent Z. Uniform line is assumed and it has a
Zo determined only by line structure. Zo is identical for a signal
traveling in either direction. However, a directional coupler must be
used to measure voltage and current traveling in one direction while
ignoring voltage and current traveling in the opposite direction. A Bird
wattmeter uses a directional coupler.

Voltage to current ratio on a line with reflections and standing waves
is of little practical value except for determinimg whether the capacity
of the line is exceeded or nearly so.

I would argue that a microwave oven uses the real power delivered to its
contents and not the standing wave as an article attached to another
poster`s comments stated.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:

...
I would argue that a microwave oven uses the real power delivered to its
contents and not the standing wave as an article attached to another
poster`s comments stated.

Best regards, Richard Harrison, KB5WZI


Nice thing about microwave ovens? They can be filled with a material
and the standing waves seen visually, no math needed ... rather handy,
really!

Regards,
Merry Xmas,
and to all a goodnight,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Cecil, W5DXP wrote:
However the VARS in the standing wave require energy which can be
converted to watts---."

VARS is an acronym for Volt Amps Reactive.
Apparent power can include real power and VARS. I would think that VARS
all have volts and amps in quadrature (at 90 degrees). If so, power is
VI cos theta. WI cos 90 degrees = VI (0)= 0, thus the power in VARS is
0.

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"As any text can tell you, the value of Z (ratio of V to I) varies along
a line which has a reflected waves (i.e., has a standing wave)."

Not exactly, maybe the apparent Z. Uniform line is assumed and it has a
Zo determined only by line structure. Zo is identical for a signal
traveling in either direction. However, a directional coupler must be
used to measure voltage and current traveling in one direction while
ignoring voltage and current traveling in the opposite direction. A Bird
wattmeter uses a directional coupler.

I stand by my statement that Z (the ratio of V to I) varies along a line
which has reflected waves.

Voltage to current ratio on a line with reflections and standing waves
is of little practical value except for determinimg whether the capacity
of the line is exceeded or nearly so.

It's of essential use in the design of stub matching, impedance
transformation, and a host of other transmission line applications. A
Smith chart is a tool which shows this impedance, so the impedance
(voltage to current ratio) is of practical value in any application for
which the Smith chart is used.

. . .

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection. Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay. Terman shows the vector diagrams of
incident and reflected waves combined to produce a voltage distribution
on an almost lossless transmission line (Zo=R) for an open circuit case
and for a resistive load case where the load is Zo in Fig. 4-3 on
page 91 of his 1955 opus. Indeed, the angle between the incident and
reflected voltages is 90-degrees in either case.

In Fig. 4-4 on page 92, Terman shows voltage and current distributions
produced on low-loss transmission lines by different load impedances and
in every case volts and amps are in quadrature.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

Owen Duffy wrote:
"If that is the meaning, then the situation you describe of 90-degree
phase difference between E and I is rather specific, it can only occur
with a distortionless line AND a load that is (s/c OR o/c OR purely
reactive).

No. Walter is exactly right, and don`t drag any distortionless line into
the discussion. That is a device for audio circuits. For RF, you only
need a low-loss (Zo = R) transmission line.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:UTAbj.1170$OH6.803@trndny03...

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:mozbj.844$si6.691@trndny08...

"Yuri Blanarovich" wrote in message
...
So you are trying to say that there is standing wave voltage but no
standing wave current and therefore no power associated with
current???

i am saying that there is standing wave voltage, and standing wave
current, but there is no physical sense in multiplying them to
calculate a power. hence the concept of standing power waves is
meaningless.


K3BU found out that when he put 800W into the antenna, the bottom of
the coil started to fry the heatshrink tubing, demonstrating more
power to be dissipated at the bottom of the coil, proportional to the
higher current there, creating more heat and "frying power" (RxI2).
This is in perfect

right R*I^2 makes perfect sense. you are talking about ONLY the
current standing wave which makes perfectly good sense. and the R*I^2
losses associated with it make perfect sense. BUT, resistive losses
ARE NOT a result of power in the standing wave, they are resistive
lossed resulting from the current. remember the initial assumptions of
my analysis, a LOSSLESS line, hence there are not resistive losses. If
this requirement is changed then you can start to talk about resistive
heating at current maximums and all the havoc that can cause.

They seem to be real current and voltage, current heats up resistance,
voltage lights up the neons and power is consumed, portion is
radiated. The larger the current containing portion, the better
antenna efficiency.
Where am I wrong?

again on the voltage, it is exactly analogous to the current above.
voltage waves capacitively coupled to a neon bulb are losses and not
covered in my statements. But again note, that type of measurement is
measuring the peak voltage of the voltage standing wave, and any power
dissipated is sampled from that wave and is not a measure of power in
the standing wave.

I have a hard time to swallow statement that there is no power in
standing wave, when I SAW standing wave's current fry my precious coil
and tip burned off with spectacular corona Elmo's fire due to standing
wave voltage at the tip.

YOU HAVE IT RIGHT! the standing wave current causes heating. the
standing wave voltage causes corona. but neither one represents POWER
in the standing waves.


So we have better than perpetual motion case - we can cause heating
without consuming power. I better get patent for this before Artsie gets
it! There must be some equilibrium somewhere.

...like, there is standing wave current, there standing wave voltage,
but no standing wave and no power in it?

Richard, wake up your english majorettes and snort it out :-)


remember the relationship that must apply within the coax. and can be
extended to antennas, though it gets real messy taking into account the
radiated part and the change in capacitance and inductance along the
length of the radiating elements.

P=V^2/Z0=I^2*Z0

now remember that you only have to look at one or the other and at all
times the relationship in each wave must obey V=I*Z0.

Now consider the infamous shorted coax. at the shorted end the voltage
standing wave is always ZERO, by your logic the power would always be
zero at that point, but how can that be?? conversly, at a point where
the current standing wave is always zero there can be no power in the
standing wave, but at that point the voltage is a maximum so would say
the power was a maximum... an obvious contradiction.


I know one thing, when standing wave current is high, I get loses in the
resistance, wires heating up - power being used.
When standing wave voltage is high, I get dielectric loses, insulation
heats up and melts - power being used, ergo there is a power in standing
wave and is demonstrated by certain magnitude of current and voltage at
particular distance and P = U x I. I don't know what you are feeding
your standing wave antennas, but I am pumping power into them and some
IS radiated, some lost in the resistive or dielectric loses.

73 Yuri



you are SO CLOSE... open your eyes, turn off your preconceived notions
and read what i wrote again slowly and carefully.

FIRST remember the assumption was a LOSSLESS line so there are no
dielectric or resistive losses. But this is only useful because it makes
it easier to see that the power given by V*I in the standing waves
doesn't make sense.

YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE
CORRECT when you say the V^2/R loss in dielectric is REAL.

Where you lose it is that the V*I for the standing waves is not
correct... this is because V and I are related to each other and you
can't apply superposition to a non-linear relationship. If you think you
have a way to do it then please take the given conditions of a lossless
transmission line, shorted at the end, in sinusoidal steady state, and
write the equation for power in the standing wave at 1/4 and 1/2 a
wavelength from the shorted end of the line.


So let me get this straight:
I describe real antenna situation, which is a standing wave circuit, with
real heating of the coil, which is consuming power demonstrably
proportional to the amount of standing wave current.
You are not answering rest of my argument.
You bring in lossless transmission line to argue that there can not be
power and no standing waves.
I still don't get it.

73 Yuri


ok, last try.. YOU ARE RIGHT!!! "consuming power demonstrably proportional
to the amount of standing wave current". LISTEN, YOU ARE RIGHT!!! the loss
in your coil is proportional to the square of the standing wave CURRENT. as
long as you keep saying that you are RIGHT. now repeat after me... the loss
in your coil is proportional to the square of the standing wave CURRENT.
emphasize CURRENT every time. DO NOT start talking about POWER in the
standing wave, then you will be wrong. Use the CURRENT young Yuri, Use the
CURRENT... forget the POWER of the standing wave.

Standing-Wave Current vs Traveling-Wave Current
 

Cecil, W5DXP wrote:
"Therefore, A standing wave is not an EM wave - it is something else, by
definition."

Cecil hit the nail on its head. A standing wave is an interference
display of two waves of the same frequency traveling in opposite
directions.

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection.

Yes.

Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay.

?? Which waves? Forward voltage and reverse current? Forward and reverse
voltage?

Terman shows the vector diagrams of
incident and reflected waves combined to produce a voltage distribution
on an almost lossless transmission line (Zo=R) for an open circuit case
and for a resistive load case where the load is Zo in Fig. 4-3 on
page 91 of his 1955 opus. Indeed, the angle between the incident and
reflected voltages is 90-degrees in either case.

Please look carefully at those diagrams. The horizontal axis is the
distance along the line. The diagrams are showing the relationship
between voltage and current envelopes as a function of position. The
graphs aren't showing the time phase of V and I, which is the matter
under discussion.

In Fig. 4-4 on page 92, Terman shows voltage and current distributions
produced on low-loss transmission lines by different load impedances and
in every case volts and amps are in quadrature.

I don't have that diagram in my 1947 Third Edition, but I'm sure that if
you'll study the diagram and accompanying text you'll find that it's
also a graph of peak amplitude vs position, not V or I as a function of
time.

Let me pose a very simple problem. Suppose you have a quarter wavelength
of 50 ohm transmission line terminated in 25 ohms and driven with a 100
volt RMS sine wave source. Consider the phase of the voltage source to
be the reference of zero phase angle.

1. What is the current at the line input? [Answer: 1 ampere, at 0 degree
phase]
2. What is the ratio of V to I at the line input? [Answer: 100 at an
angle of zero divided by 1 at an angle of zero = 100 + j0 ohms]
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

Feel free to repeat this with any other line length. I'll wait very
patiently for the length which produces V and I in quadrature at the
input. A very interesting result of that would be that no power would be
consumed from the source, so if any reaches the load then we've created
power.

I'd be glad to post the equation relating Z (the ratio of V to I) at the
line input or any point along the line to the load and characteristic
impedances. But I'm afraid it would be wasted effort, since there's a
great reluctance here to actually work an equation or understand its
meaning. But good and accurate graphs of what Richard has claimed the
Terman graphs show (but don't) can be found in the _ARRL Antenna Book_.
In the 20th and 21st Editions, the graphs are Fig. 12 on p. 24-9. In
other editions, they're probably also Fig. 12 in the Transmission Lines
chapter. In the graphs, the angle between the I and E vectors is the
relative phase angle between the two, and also the angle of the
impedance of the point where the vectors are shown.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On *The voltage is
indeed real, as i have said. *you can measure the 'standing' wave voltage,
that has been known for a long time... but the effects are NOT due to power
in standing waves.-

OK, that was the least word infested post I can find...

1. A standing wave is not 'standing' in time... It phase rotates at
the same rate as the excitation frequency...
If the phase is rotating then V and I are changing - else Feynman is
rotating in his grave...
To do so requires the surface electrons at that point on the line to
oscillate back and forth or whatever the heck surface electrons do at
rf frequency... To excite these electrons from one energy state to
another requires power/energy/joules/whatever-you-want-to-call-it...

2. It is real because I can measure it with a volt meter and I can
extract power from it with a lamp, simultaneously..

Also: it will perforate the insulating jacket on the line if the power
level is high enough... I used to maintain a herd of 100KW RF
generators, and they would blow a hole through the side of a quarter
inch thick copper bar in an instant when the load failed in my
youthfull ignorance, I thought it was the standing wave RATIO that
blew the line, silly me ...



Try this thought... A Tesla coil (automobile spark coil) with no load
on the output is all standing wave voltage and no current - so
according to some has no power... Touch your finger to it...
Also try: If an open ended line has 100.000 volts of DC on it, the
standing wave DC contains no power because the current is zero?

And, for whoever it was accused me of bashing someone - reread my
post, carefully.... I absolutely do not bash or flame anyone...

cheers ... denny

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I'm surprised that standing waves seem so hard for people to understand.
They're simply a spatial pattern formed by the interference between
forward and reflected waves or, if you prefer, from the solution to a
general transmission line problem with boundary values applied. I'd hope
that even a brief look at a text would make it clear what standing waves
are, and what they are not.

Roy, if you would take a "brief look at a text", you
would know NOT to try to use standing wave current phase
to measure the delay through a 75m loading coil. At any
point in time, the standing wave current phase is essentially
the same value all up and down a 1/2WL dipole including
through any loading coils. Phase does NOT equate to delay
in a standing wave environment.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 6:49*am, "Dave" wrote:
"Yuri Blanarovich" wrote in message

...







"Dave" wrote in message
news:UTAbj.1170$OH6.803@trndny03...

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:mozbj.844$si6.691@trndny08...

"Yuri Blanarovich" wrote in message
...
So you are trying to say that there is standing wave voltage but no
standing wave current and therefore no power associated with
current???

i am saying that there is standing wave voltage, and standing wave
current, but there is no physical sense in multiplying them to
calculate a power. hence the concept of standing power waves is
meaningless.

K3BU found out that when he put 800W into the antenna, the bottom of
the coil started to fry the heatshrink tubing, demonstrating more
power to be dissipated at the bottom of the coil, proportional to the
higher current there, creating more heat and "frying power" (RxI2).
This is in perfect

right R*I^2 makes perfect sense. *you are talking about ONLY the
current standing wave which makes perfectly good sense. *and the R*I^2
losses associated with it make perfect sense. *BUT, resistive losses
ARE NOT a result of power in the standing wave, they are resistive
lossed resulting from the current. *remember the initial assumptions of
my analysis, a LOSSLESS line, hence there are not resistive losses. *If
this requirement is changed then you can start to talk about resistive
heating at current maximums and all the havoc that can cause.

They seem to be real current and voltage, current heats up resistance,
voltage lights up the neons and power is consumed, portion is
radiated. The larger the current containing portion, the better
antenna efficiency.
Where am I wrong?

again on the voltage, it is exactly analogous to the current above.
voltage waves capacitively coupled to a neon bulb are losses and not
covered in my statements. * But again note, that type of measurement is
measuring the peak voltage of the voltage standing wave, and any power
dissipated is sampled from that wave and is not a measure of power in
the standing wave.

I have a hard time to swallow statement that there is no power in
standing wave, when I SAW standing wave's current fry my precious coil
and tip burned off with spectacular corona Elmo's fire due to standing
wave voltage at the tip.

YOU HAVE IT RIGHT! *the standing wave current causes heating. *the
standing wave voltage causes corona. *but neither one represents POWER
in the standing waves.

So we have better than perpetual motion case - we can cause heating
without consuming power. I better get patent for this before Artsie gets
it! There must be some equilibrium somewhere.

...like, there is standing wave current, there standing wave voltage,
but no standing wave and no power in it?

Richard, wake up your english majorettes and snort it out :-)

remember the relationship that must apply within the coax. and can be
extended to antennas, though it gets real messy taking into account the
radiated part and the change in capacitance and inductance along the
length of the radiating elements.

P=V^2/Z0=I^2*Z0

now remember that you only have to look at one or the other and at all
times the relationship in each wave must obey V=I*Z0.

Now consider the infamous shorted coax. *at the shorted end the voltage
standing wave is always ZERO, by your logic the power would always be
zero at that point, but how can that be?? *conversly, at a point where
the current standing wave is always zero there can be no power in the
standing wave, but at that point the voltage is a maximum so would say
the power was a maximum... an obvious contradiction.

I know one thing, when standing wave current is high, I get loses in the
resistance, wires heating up - power being used.
When standing wave voltage is high, I get dielectric loses, insulation
heats up and melts - power being used, ergo there is a power in standing
wave and is demonstrated by certain magnitude of current and voltage at
particular distance and P = U x I. I don't know what you are feeding
your standing wave antennas, but I am pumping power into them and some
IS radiated, some lost in the resistive or dielectric loses.

73 *Yuri

you are SO CLOSE... open your eyes, turn off your preconceived notions
and read what i wrote again slowly and carefully.

FIRST remember the assumption was a LOSSLESS line so there are no
dielectric or resistive losses. *But this is only useful because it makes
it easier to see that the power given by V*I in the standing waves
doesn't make sense.

YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE
CORRECT when you say the V^2/R loss in dielectric is REAL.

Where you lose it is that the V*I for the standing waves is not
correct... this is because V and I are related to each other and you
can't apply superposition to a non-linear relationship. *If you think you
have a way to do it then please take the given conditions of a lossless
transmission line, shorted at the end, in sinusoidal steady state, and
write the equation for power in the standing wave at 1/4 and 1/2 a
wavelength from the shorted end of the line.

So let me get this straight:
I describe real antenna situation, which is a standing wave circuit, with
real heating of the coil, which is consuming power demonstrably
proportional to the amount of standing wave current.
You are not answering rest of my argument.
You bring in lossless transmission line to argue that there can not be
power and no standing waves.
I still don't get it.

73 Yuri

ok, last try.. *YOU ARE RIGHT!!! *"consuming power demonstrably proportional
to the amount of standing wave current". *LISTEN, YOU ARE RIGHT!!! *the loss
in your coil is proportional to the square of the standing wave CURRENT. *as
long as you keep saying that you are RIGHT. *now repeat after me... the loss
in your coil is proportional to the square of the standing wave CURRENT.
emphasize CURRENT every time. *DO NOT start talking about POWER in the
standing wave, then you will be wrong. *Use the CURRENT young Yuri, Use the
CURRENT... forget the POWER of the standing wave.- Hide quoted text -

It seems to me that a bit more precision in the use of
language might help. So, strictly:

It is consuming power proportional to the square of
the RMS current at that point on the line, and the
constant of proportionality is the resistance of
the line over the length of interest.

When an author writes "standing wave current", do
they always mean the RMS current at a point on the
line? Or do they mean the envelope of the RMS
current at each point along the line? Or the
envelope of the peak current at each point? Or the
RMS value of the spatially distributed peak currents
along the line? Or? What is THE standing wave
current? No wonder there is so much dispute.

And could someone who likes to write "standing
wave power" (Yuri perhaps?) please provide an
unambiguous definition? It does not have to be
the "right" definition, or agreed by all, just
any definition which is unambiguous.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

"Denny" wrote in message
...
On The voltage is
indeed real, as i have said. you can measure the 'standing' wave voltage,
that has been known for a long time... but the effects are NOT due to
power
in standing waves.-

1. A standing wave is not 'standing' in time... It phase rotates at
the same rate as the excitation frequency...
If the phase is rotating then V and I are changing - else Feynman is
rotating in his grave...

true, it is 'standing' in space. it does not move along the line. your
requirement that the 'phase rotates' is another example of why a 'standing'
wave is not the same as a real wave. note the point in the standing wave
where the voltage is zero. it is always zero, there is no 'phase rotation'
at that point. now, if this was a real wave then there would be 'phase
rotation' all along the wave in both time and space.

2. It is real because I can measure it with a volt meter and I can
extract power from it with a lamp, simultaneously..

you can measure the superimposed voltage of the two real waves, the Vf+Vr at
each point on the line. and you can extract power from the superimposed
combination of those waves.

Also: it will perforate the insulating jacket on the line if the power
level is high enough... I used to maintain a herd of 100KW RF
generators, and they would blow a hole through the side of a quarter
inch thick copper bar in an instant when the load failed in my
youthfull ignorance, I thought it was the standing wave RATIO that
blew the line, silly me ...

yep, silly you. it is the superposition of the forward and reflected waves
that can create hot spots in the line. where the moving waves happen to
always be in phase you get peak voltage in the standing wave, where they
always are out of phase you get no voltage.

why do i even bother... time to start plonking more of the ones who refuse
to learn and reduce the noise level on here even more.

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
And, the standing wave
IS cooking the turkey--in my humble opinion ... you could say, "I
believe in standing waves."

If one thinks about it, one will realize that an unchanging
steady-state standing wave cannot cook the turkey. Cooking
the turkey would require the standing wave to give up energy
and if it does, it is no longer a steady-state standing wave.
All of the joules/sec delivered to the load during steady-
state is from traveling-wave energy whether it be in a
microwave oven or in a transmission line.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

...

On Dec 22, 8:43 am, Denny wrote:
Nice graphic, Cecil.. But the thread has drifted beyond recognition..
Part of the original dispute across a couple of threads as I
remember it, was the contention that there is no energy contained
within the reflected wave and therefore no energy contained within the
standing wave, i.e. a mere artifact...

I'd suggest this is a mischaracterization of the contention. I have
seen no disagreement with the notion that the line contains energy.

Assertions about a lack of energy in reflected waves is not
inconsistent with the line containing energy.

I simply wanted to point out that the standing wave on a line does
contain energy and it is a childishly simple exercise to prove it,
therefore the reflected wave must contain energy...

Prepare yourself to rethink this connection.

As far as the questioner, where does the energy go between the
standing wave peaks - oy vey....
If it is a real question - as opposed to a rhetorical device which I
hope was the intent -

It was not rhetorical, but an educational question that followed
from the claim. With the claim that a lit flourescent bulb
demonstrates the presence of energy, it is entirely reasonable
to question what a dark lamp means and the original post did
not suggest this understanding.

then the profound ignorance

There is no need to descend to the level of insult commonly
used by some of the more prolific posters.

of basic physics is
vastly beyond the limited space I have to go over it... See ANY
introductory level, physics textbook for details...

--------------------------------------------------------------

Let us consider a transmission line....

There IS a voltage and current distribution on this line. For
the moment attempt to forget standing waves, travelling waves,
forward waves, reflected waves, .... *Just that:

There IS a voltage and current distribution on this line.
These distributions can be expressed as functions of distance
along the line and time:

*V(x,t)
*I(x,t)

These are the instantaneous real voltage and current
at a particular location (x) and time (t). They can be
measured with a voltmeter and ammeter, though this gets
more challenging at higher frequencies.

Now we know from basic electricity that Power is Volts
times Amps, so we have:

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

P(x,t) is the instantaneous power at any point and time
on the line. Power being the rate of energy flow, P(x,t)
is the instantaneous energy flow at that point and time
on the line.

If you disagree with any of the above please read no
further and post any objections now.

Good! Agreement.

So let's consider the specific example of sinusoidal
signal applied to a transmission line that is open
at the end. After settling, there is a voltage and
current distribution on this line, but how can we
describe it? Now some of you are immediately
thinking "standing wave", and you'd be right. Its
an excellent description, but we need to look at
the details.

So V(x,t) = A cos(x) cos(wt)
where w is radians/second and x is measured in
degrees back from the open end.

Consider t=0. The spatial voltage distribution
is a sinusoid with a maximum at the open end. As
time advances, this spatial sinusoid drops in
amplitude until the voltage everywhere on the
line is 0, then the amplitude heads towards
minus max. Noting that the zero crossings are
always in the same place and the shape is
sinusoidal leading to the name "standing wave".

From a time perspective, every point on the line
has a sinusoidal voltage, but the amplitude
changes with position. The peaks and zero crossings
occur at the same time everywhere, thus the
claim that there is no phase shift as one
moves down the line.

The current is also a sinusoid, but shifted 90
degrees from the voltage sinusoid, thus there
is a current zero where-ever there is a voltage
maximum.

Now power is really interesting. Recall that

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

At certain values of t, the voltage everywhere
on the line is 0, so at these times, no energy
is flowing anywhere on the line. Similarly for
current.

And at certain positions on the line (n*180+90)
the voltage is always 0, so the power is always
0 at these points. No energy is ever flowing
at these points. Similarly for current at points
(n*180).

This is where your argument falls apart. *you can not apply superposition to
power as it is a non-linear relationship with the voltage and current. *this
is where lots of the arguments on this group begin and get stuck forever.
the argument you state in the above paragraph is a contradiction on the most
basic level...

I am not sure what lead you to think I am superposing power. The only
powers I compute are from the actual (in other words, total) voltage
and current present at one point on the line.

take a step back and consider this:

V(x,t)=Z0*I(x,t)

In my use, V(x,t) and I(x,t) are the actual voltage and current and
current at a point on the line and are only related by Z0 in very
exceptional circumstances, an example being when the line is
terminated
in Z0.

which also must be true at every point on the line for the forward and
reflected waves, each taken separately. *so you can write equations like:

Vf=Z0*If
and
Vr=Z0*Ir
(i'll leave off the (x,t) for now)

I would suggest not leaving off (x,t). I understand the short-hand but
it leads many to start thinking in terms of average or RMS. Keeping
(x,t) re-inforces the idea that the measurement is at one point and
one time.

and by superposition you can also do:
V=Vf+Vr
I=If+Ir
Which work just fine if you do fancy graphs and animations to create the
illusion of 'standing' waves along the line. *And as long as you keep the
two equations separate everything is simple.

BUT, lets look at the a place where the standing current wave is always zero
and the voltage standing wave is of course a maximum. *if you plug those
into ohm's law to find the impedance at that point you get:
Z=V/I *(where V is large, and I is zero) *so you get an infinite impedance.
does this surprise anyone? *it shouldn't, this is what the smith chart shows
you should happen every half wave along the line. *the result of
superimposing the waves results in changes in the measured impedance as you
move along the line. *note this is NOT a change in Z0, that is forever a
constant and property of the physical line independent of the waves imposed
on it.

So what does this really mean? *on the surface it would tend to support the
assertion that there is no power flow at the points where the current is
zero, and the impedance measures is infinite. *BUT there is a catch.

P=VI * *is a basic representation for instantaneous power at a point given
voltage and current, again (x,t) left off but that doesn't matter.
but also you have to keep the relationship:
V=Z0*I

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

so you can rewrite the P equation 2 different ways:
P=V^2/Z0 = I^2*Z0
now, obviously these have to hold for any wave that exists in the line. *so
for the forward wave they hold up just fine, and for the reflected wave they
hold up just fine.. BUT if you look at the 'standing' wave they fall apart..
i.e. for the spot where the current is zero and the voltage is a maximum you
get:

P=V^2/Z0 which is a large number
AND
P=I^2*Z0 which is a small number
you can't have it both ways at one point at the same time!

now, what is really happening?
Given:
1. you have 2 waves.

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.

Just as in basic ciruit theory, the partial results, which are
eventually superposed to arrive at the final results, have no
particular meaning. For sure, they are not what is "really
happening".

None-the-less, this "two wave" model is extremely useful.

2. these 2 waves are traveling in opposite directions.
3. each of these waves obeys ohms law.
4. the voltage and/or current of these waves obeys the superposition
principle.
5. you only need to look at voltage OR current since they are linearly
related to each other at every point in each wave.

If you dispute any of the above, do not pass go, do not collect 200$, go
back to school and take fields and waves 101 all over again.

Tsk. Tsk. Insults. Unbecoming.

Lets think voltage waves for now, this is an arbitrary decision as noted
above. *and lets consider a lossless line with a short or open end so there
is 100% reflection.

So, as these two waves travel along the line they periodically go in and out
of phase with each other, there are animations that show this very nicely.
lets think about the time where the two waves completely cancel each other
so the voltage along the line is zero... did the 2 waves dissappear? *no,
because obviously an instant later they are back out of phase and the
voltage doesn't cancel on the line. *is the power zero?? no, since energy
can neither be created nor destroyed, it didn't just dissappear, so neither
can the flow of energy that is called power. *and since we know that when
this voltage minimum occurs there is a current maximum in the superimposed
waves the power represented by that superposition would contradict the power
represented by the voltage minimum... they can't both be correct at the same
time!

So what do you take away from this?
1. Standing waves have no physical significance, they do not represent power
or energy, they do not obey ohms law, they are ONLY a result of
superposition of the voltage and/or current waves in the line.
2. can you measure standing waves? *Yes, of course. *that is how they got
their name, you could measure them and they didn't seem to move on the line.
but this is only because simple measurement tools can't distinguish the
forward and reflected components that make them up.
3. if you want to talk about power and energy you MUST use the individual
traveling waves.

now what does that mean for this lossless line that has 100% reflection??

1. the reflected wave magnitude is equal to the forward wave just traveling
in the opposite direction. (voltage and/or current)
2. the power in the forward wave is equal to the power in the reflected wave
but traveling in the opposite direction.
3. there is energy stored in the line equal to the sum of the integrated
power in the two waves each taken separately. *DO NOT sum them first then
integrate, that is NOT a legal operation since as has been shown the
superposition principle does not apply to power as it is a non-linear
relationship!
4. in steady state the net power flowing past any point in the line is
zero... that is, there is as much power flowing one direction as the other..

ok, i'm done with my lecture... you guys can now ignore me if you want and
go back to your regularly scheduled misconceptions and circular arguments.

4.

Are you really prepared to throw away P = VI?

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.

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?

Throwing this away will invalidate much.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:
Cecil, W5DXP wrote:
However the VARS in the standing wave require energy which can be
converted to watts---."

VARS is an acronym for Volt Amps Reactive.
Apparent power can include real power and VARS. I would think that VARS
all have volts and amps in quadrature (at 90 degrees). If so, power is
VI cos theta. WI cos 90 degrees = VI (0)= 0, thus the power in VARS is
0.

It takes joules of energy for VARS to exist. Any time one
wants to give up the VARS, they can be converted to watts,
just like the energy stored in a capacitor can be converted
to watts by connecting a resistor.

The VARS stored in the standing waves in a transmission line
can be converted to watts by connecting a load equal to the
Z0 of the line. Of course, the standing waves cease to exist
in the process. One cannot have one's cake and eat it too.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

"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.

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.

try to look at it this way. when you look at the forward and reflected
waves separately it is intuitively obvious how the power calculation shows
the flow along the line with each wave. however, when you look at standing
waves you get spots every 1/4 wave where either V(x,t) or I(x,t) is ALWAYS
zero... by V*I this means the power at that point is ALWAYS zero. since
power is just the measure of the flow of energy, and energy can neither be
created nor destroyed then in the traveling wave there can be no energy flow
past those points. where it is obvious from the individual Vf(x,t) and
Vr(x,t) or If(x,t) and Ir(x,t) that are ALWAYS related by Z0 at every point
on the line that power does flow both directions. an obvious contradiction
and if you can't see it by this point i give up.

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?

i want to throw away this falicy and replace it with the real physically
correct calculation.

Throwing this away will invalidate much.

only in your mind.

I have said it enough times now, and hate repeating myself... so you can
live with your poor misguided assumptions and formula. i have shown the
obvious errors

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 23, 11:33*am, Roger wrote:
Keith Dysart wrote:

clip ....



In the setup above used for "standing waves"
it can be seen that there is zero power in
the line every 90 degrees back from the open
end. At a zero power point, no energy is
being transferred. Therefore, the forward
and reverse waves can not be transferring
energy across these points. Conclusion:
forward and reverse waves do not always
transport energy.

....Keith

Hi Keith,

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.

Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I stand by my statement that Z (the ratio of V to I) varies along a line
which has reflected waves.

What you say is true but it is important to note which
definition of "impedance", Z, that you are using. You
are using (1)(B) below, commonly referred to as the
"virtual impedance" definition because the R in the
R+jX impedance doesn't dissipate any power.

I don't think that you and Richard H. are using the same
definition of "impedance".

From the IEEE Dictionary:

"impedance -

(1)(A) The corresponding impedance function with p
replaced by jw in which w is real. Note: Definitions
(A) and (B) are equivalent.

(1)(B) The ratio of the phasor equivalent of a steady-
state sine wave voltage ... to the phasor equivalent
of a steady-state sine wave current ...

(1)(C) A physical device or combination of devices
whose impedance as defined in definition (A) or (B)
can be determined. Note: This sentence illustrates
the double use of the word impedance ... Definition
(C) is a second use of "impedance" and is independent
of definitions (A) and (B).
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

"Keith Dysart" wrote in message
...
On Dec 23, 11:33 am, Roger wrote:

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.

Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

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.

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 23, 2:34*pm, (Richard Harrison)
wrote:
Keith Dsart wrote:

"Therefore, the forward and reverse waves can not be transferring energy
across these points."

A wave is defined as a progressive vibrational disturbance propagated
through a medium, such as air, without progress or advance of the parts
or particles themselves, as in the transmission of sound, light, and an
electromagnetic field. Light, for example, is also calld luminous or
radiant energy. Sound and radio waves are also examples of energy in
motion.

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

Best regards, Richard Harrison, KB5WZI

But in the example there was a transmission line
on which the actual instantaneous voltage and
current can be measured.

And P = V * I seems rather fundamental, so V or
I is always 0, then P must always be 0.

Jumping to disussing waves does not alter this.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection. Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay.

Seems you two are arguing about two different things.

If Z0 is purely resistive: Pure standing waves are *ALWAYS*
in quadrature, i.e. the sine of the angle between V and
I is always 1.0. Pure traveling waves are are *ALWAYS* in
phase or 180 degrees out of phase, i.e. the cosine of the
angle between V and I is always 1.0.

In a mixed environment of standing waves and traveling
waves, the angle between V and I can assume any value.
--
73, Cecil http://www.w5dxp.com