"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.

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

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

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

"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

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

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

"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.

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

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