Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
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.

Please cease the obfuscation. It cannot be explained any
better than these two web pages. Please note that "intensity"
is a power density (watts/unit-area) and it is *energy* that
is redistributed, not voltage or current.

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

"Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists
between reflections of p, then reflected wavefronts interfere
destructively, and overall reflected intensity is a minimum.
If the two reflections are of equal amplitude, then this
amplitude (and hence intensity) minimum will be zero."
[Referring to 1/4 wavelength thin films.]

"In the absence of absorption or scatter, the principle of
conservation of energy indicates all 'lost' reflected intensity
will appear as enhanced intensity in the transmitted beam. The
sum of the reflected and transmitted beam intensities is always
equal to the incident intensity. This important fact has been
confirmed experimentally."

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

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... All of the photon energy present in
these waves must somehow be recovered or redistributed in a new
direction, according to the law of energy conservation ... Instead,
upon meeting, the photons are redistributed to regions that permit
constructive interference, so the effect should be considered as
a redistribution of light waves and photon energy rather than
the spontaneous construction or destruction of light."
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
[...]

Frankly, I like things "simple."

In the standing wave experiment, involving creating a standing wave on a
string with a vibrational apparatus--where would you think one would be
tempted/inclined to attempt a mechanical "contact/connection" with the
string to extract [work,power,energy?]

If anyone elses' inclinations are similar to my own, it would be a
"hump" ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

... 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.

The logic in your argument is flawless, in fact, I believe it ...

However, the voltage/current/power/energy/joules given out by my car
alternator cannot exist independently of the petrol powered motor.
Still, I consider that electrical voltage/current/power/energy/joules a
"SEPARATE entity" ...

Yanno, he didn't come last night ... the cookies and milk still sit
before the hearth, no unaccounted gifts clutter the space beneath the
tree ... yet still, I believe! :-P

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
[ramblings]

I remember, back in the early 60's I built a coaxial tank circuit for
the fm band (88-108mc) for a selective receiver ... I wonder ... sound
of books opening--case coming off the old ti-86 ???

JS

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
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.

There aren't really any sunrises or sunsets, Cecil?

Magician tricks are another.

Magic tricks aren't illusion, but reality? Those guys are really
working magic? Wow, my entire belief system is crumbling.

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Reality can certainly contain illusions. Sunrise and sunset
are a couple of examples.

There aren't really any sunrises or sunsets, Cecil?

The sun doesn't rise or set. That is just an illusion
caused by the planet upon which we are standing rotating
on its axis.

And Dave, neither does the Sun God ride his flaming
chariot across the sky each day.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roy Lewallen wrote:

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.

These comments in total are very interesting, by both authors. Thank
you for them.

It is clear that there is not a standard way of describing energy that
is in the process of being stored in either an inductor or capacitor.
Clearly it is stored, not used (converted) as in a resistor.

Storage of energy in a capacitor (for instance) occurs over time and
requires power to complete. So how do we describe that power if it is
always zero because voltage and current are 90 degrees out of phase?
Should we recognize that only the peak voltage and peak current is 90
degrees out of phase, with the entire charging time occurring within
those two time extremes? Between those two time extremes, the voltage
and current are in phase but at changing impedance, with power flowing
into the capacitor. So think I.

I think of the "standing wave" as being equivalent to the graph on my
power bill that shows power used daily over a month. Granted that the
standing wave is recorded in V or I, but we only need to know the R that
the standing wave is acting through to determine the power that it
represents.

I certainly find this interesting.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

... neither does the Sun God ride his flaming
chariot across the sky each day.

Finally, we are in total agreement!

That is Santa, damn idiots believing in some sun god, DUH!

You think he supports Mrs. Santa and the elves off a single days work
per year? ;-)

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
So how do we describe that power if it is
always zero because voltage and current are 90 degrees out of phase?

Here's how power engineers solve the problem.
Voltage and Current are phasors.

V*I = volt-amps

V*I*cos(A) = Watts, real power

V*I*sin(A) = VARS, reactive power
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
Reality can certainly contain illusions. Sunrise and sunset
are a couple of examples.

There aren't really any sunrises or sunsets, Cecil?

The sun doesn't rise or set. That is just an illusion
caused by the planet upon which we are standing rotating
on its axis.

Sorry, Cecil, it isn't an illusion. Sunrise and sunset are simply names
we've given to that which you've described. They really take place and
are not at all an illusion.

And Dave, neither does the Sun God ride his flaming
chariot across the sky each day.

Then again, you're the only fellow writing about such. I've made no
mention of anything like that.

What about the magic tricks?

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Sorry, Cecil, it isn't an illusion. Sunrise and sunset are simply names
we've given to that which you've described. They really take place and
are not at all an illusion.

You can believe that the sun rises and sets if you want to,
Dave. Most of us know that the sun is fixed at the center
of the solar system. The names "sun rise" and "sun set" were
coined back when humans believed that the sun actually made
a trip across the sky every day over a flat earth. Do you also
believe in a flat earth?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Roy Lewallen wrote:

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.

These comments in total are very interesting, by both authors. Thank
you for them.

It is clear that there is not a standard way of describing energy that
is in the process of being stored in either an inductor or capacitor.
Clearly it is stored, not used (converted) as in a resistor.

Sure there is. The energy stored in a capacitor is 1/2 * C * V^2, and in
an inductor, 1/2 * L * I^2. The rate of energy flow into or out of an
inductor or capacitor is called "power". It's exactly the same stuff as
energy moving into or through a resistor, and it's measured and
described in exactly the same way.

Storage of energy in a capacitor (for instance) occurs over time and
requires power to complete.

Yes, power is the rate of energy transfer.

So how do we describe that power if it is
always zero because voltage and current are 90 degrees out of phase?

See the following example which shows just that situation.

Should we recognize that only the peak voltage and peak current is 90
degrees out of phase, with the entire charging time occurring within
those two time extremes? Between those two time extremes, the voltage
and current are in phase but at changing impedance, with power flowing
into the capacitor. So think I.

The whole concept of impedance gets a bit flaky when working in the time
domain. I recommend sticking to voltage, current, power, and energy when
doing so. Go through the following example, and I believe you'll get a
much better idea of what's happening on a basic level.

But you've made some good observations. The answer is that, as I
mentioned, the instantaneous power (p(t)) is not zero except at the ends
and middle of the example line. Some of the posters are confusing
average and instantaneous power, which is leading to incorrect
statements and conclusions. Instantaneous power can easily be non-zero
while maintaining zero average power, as I'll illustrate.

Let's look, for example, at the half wavelength open circuited line
described by "Dave", being driven by v = 0.5 sin(wt), 1/8 wavelength
from the end. At that point,

vf(t) = 0.5 sin(wt - 135 deg.)
vr(t) = 0.5 sin(wt - 225 deg.)
if(t) = 0.01 sin(wt - 135 deg.)
ir(t) = -0.01 sin(wt - 225 deg.)

considering the positive direction of ir(t) as the same as the positive
direction of if(t) so we can add if(t) and ir(t) to get i(t).

Using elementary trig identities for the addition, and dropping the
explicit identification of degrees for simplicity:

vf(t) + vr(t) = v(t) = sin(wt - 180) * cos(45) = 0.7172 * sin(wt - 180)
if(t) + ir(t) = i(t) = 0.02 * cos(wt - 180) * sin(45) = 0.01414 *
cos(wt - 180)

One important thing to notice is that, unlike the ends or center of the
line, neither the voltage nor the current is zero at this point.
Consequently, their product is also non-zero.

The voltage and current are, of course, 90 degrees out of phase, as they
are everywhere along an open circuited or short circuited line, or one
terminated in a pure reactance. And when the average power is computed
from the instantaneous power, the result will be zero. But that doesn't
mean that the instantaneous power is zero as some are claiming. Let's
see what it does mean. Using another trig identity,

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

This is the instantaneous power which, as I've described before, is a
sinusoidal function with rotational frequency 2w. It has no offset, so
the average value is zero. But it shows that energy moves back and forth
past this point every cycle. Equal amounts move each direction each
cycle, so no net power flows either direction.

There are simpler ways to get the same answer, but this one shows very
basically where the energy is going at every instant. If instead you
start out by throwing away the time information and just looking at
averages, you lose a lot of information along with the basic
understanding of energy movement in the system. I'm afraid some people
have taken this simpler approach without realizing what information and
insight they've lost by doing so. The average value can always be
determined from the instantaneous function, but never the other way around.

I think of the "standing wave" as being equivalent to the graph on my
power bill that shows power used daily over a month. Granted that the
standing wave is recorded in V or I, but we only need to know the R that
the standing wave is acting through to determine the power that it
represents.

Not quite. The power bill graph is a graph of the power, much like the
p(t) function I calculated. The energy, which you pay for, is the
integral of that graph. If you used a constant power of 1 kW for half
the month, then sent 1 kW back into the power grid for the other half of
the month, you'd technically owe nothing, since you'd use no net power.
You might even have stored it somewhere and given the very same energy
back rather than using some and generating an equal amount later. That's
exactly what's happening at the point on the line 1/8 wavelength from
the end.

Roy Lewallen, W7EL

-- I've had many opportunities when doing this to make arithmetic or
trigonometric errors, and I haven't taken the time to thoroughly check
my work -- although the conclusion is correct. I'd appreciate anyone
pointing out any mathematical errors he's found.

Standing-Wave Current vs Traveling-Wave Current
 

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.
--
************************************************** *******************************8

Cecil, your technical and mathematical skills vastly exceed mine... My
poor little monkey brain has to work in the concrete, not the
abstract...
I sort of understand the concept of the standing wave being the
instantaneous vector product of two colliding EM wave fronts...
I accept the fact that MATHEMATICALLY; P = V*I* cos (theta).... And
being that theta is DEFINED as 90 degrees the cos = 0, so the
instantaneous power is zero...
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...
To prove that mathematically there is no power contained in a standing
wave collides with my, farm boy, real world, physics experiments...
The voltage peak on that line when that standing wave forms is real,
it has mucho power (esp when driven by the dragon breath of a 100KW
triode with 15KV on the plate), and it blows the hole in the line at
the same distance from the load every time (proof of standing wave
theory)...

This is why i have a problem accepting that a standing wave contains
no power - regardless of the prominence of the men who wrote the
textbooks..

cheers ... denny

Standing-Wave Current vs Traveling-Wave Current
 

"Denny" wrote in message
...
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.
--
************************************************** *******************************8

Cecil, your technical and mathematical skills vastly exceed mine... My
poor little monkey brain has to work in the concrete, not the
abstract...
I sort of understand the concept of the standing wave being the
instantaneous vector product of two colliding EM wave fronts...
I accept the fact that MATHEMATICALLY; P = V*I* cos (theta).... And
being that theta is DEFINED as 90 degrees the cos = 0, so the
instantaneous power is zero...
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...
To prove that mathematically there is no power contained in a standing
wave collides with my, farm boy, real world, physics experiments...
The voltage peak on that line when that standing wave forms is real,
it has mucho power (esp when driven by the dragon breath of a 100KW
triode with 15KV on the plate), and it blows the hole in the line at
the same distance from the load every time (proof of standing wave
theory)...

This is why i have a problem accepting that a standing wave contains
no power - regardless of the prominence of the men who wrote the
textbooks..

cheers ... denny

Same here, I burn the coil with "no power" on standing wave circuit -
quarter wave resonant loaded whip.
Cecil is telling us that the current becomes automatically traveling wave,
while EZNEC model shows drop along the coil and rest of the circuit with
nice standing wave.
I grew up in the city, around cars and I am having problem swallowing that
we have SW current and voltage but no power, while we pump power into the
circuit and see nasty things happening requiring power.

three cheers... Yuri oK3BU

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
Sorry, Cecil, it isn't an illusion. Sunrise and sunset are simply
names we've given to that which you've described. They really take
place and are not at all an illusion.

You can believe that the sun rises and sets if you want to,
Dave. Most of us know that the sun is fixed at the center
of the solar system. The names "sun rise" and "sun set" were
coined back when humans believed that the sun actually made
a trip across the sky every day over a flat earth.

Sunrise and sunset are simply names we've given to that which you've
described. My logging program provides me with sunrise and sunset times
for various parts of the world. The National Weather Service uses those
terms. You may choose to say at the beginning of each day that the
earth has rotated so that the star at the center of our solar system is
visible. I don't know anyone who spouts that kind of mouthful.

Do you also
believe in a flat earth?

Where are you getting the "also"?

Now, what about the magic tricks?

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
The voltage and current are, of course, 90 degrees out of phase, ...

You have just contradicted yourself. There *cannot* be any
real power associated with a voltage and current that are
90 degrees out of phase.

Power = V*I*cos(90) = 0 watts

I will attempt to check your math. I'll bet your dimensions
are volt*amps and NOT watts.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Denny wrote:
But here is where the math breaks down...
On my herd of 100KW RF generators... Every time the load failed for
even a split second, those fire breathing dragons would blow a hole
through the quarter inch thick slabs of copper that made up the line,
instantaneously...

The math didn't break down, Denny, your logic did.

1. The standing wave voltage is *ALWAYS* 90 degrees out of
phase with the standing wave current, by definition.

2. It is *IMPOSSIBLE* to "blow a hole through the quarter
inch thick slabs of copper" with a voltage and current that
are 90 degrees out of phase.

3. Therefore, the voltage and current that blew the holes
in the copper slabs were NOT 90 degrees out of phase and
therefore were not standing waves. In-phase voltage and
current is required to blow holes in copper slabs.

4. Standing waves are a steady-state phenomenon. Blowing
holes in copper slabs is a transient phenomenon. At the
end of steady-state, the standing wave energy is always
converted into real watts and either radiated or dissipated.

The energy that blew holes in your copper slabs did NOT
meet the definition of "standing wave energy". During
the transient state, it was converted from standing wave
energy (with its 90 degree phase difference between voltage
and current) to traveling wave energy (with its in-phase
relationship between voltage and current).
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Yuri Blanarovich wrote:
Cecil is telling us that the current becomes automatically traveling wave,

Yes, it is impossible to get any watts out of a voltage
and current that are 90 degrees apart simply because
there are zero watts available.

Standing wave voltage and current are 90 degrees apart,
by definition.

Therefore, by definition, any wave that yields watts
is NOT a standing wave.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
The voltage and current are, of course, 90 degrees out of phase, ...

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

Those are not watts, those are volt*amps. Therefore that's
not real power. Real power is:

p(t) = v(t) * i(t) * cos(90) = ZERO
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave Heil wrote:
Sunrise and sunset are simply names we've given to that which you've
described.

Those words were coined when the human race believed
that the sun actually traveled across the sky each
day. The sun "rising" means the sun is moving but
that is just an illusion.

You seem to believe that illusions don't exist in
reality. That very belief is an illusion.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:

... The sun "rising" means the sun is moving but
that is just an illusion.

You seem to believe that illusions don't exist in
reality. That very belief is an illusion.

I shudder to think of having ran into you that day, long ago, in Ft.
Bragg CA, when that young girls hand was in mine, and we watched the
beauty in the "sunset" on that sandy beach and then ...

I married her, yanno? :-P

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

"Cecil Moore" wrote in message
...
Yuri Blanarovich wrote:
Cecil is telling us that the current becomes automatically traveling
wave,

Yes, it is impossible to get any watts out of a voltage
and current that are 90 degrees apart simply because
there are zero watts available.

Standing wave voltage and current are 90 degrees apart,
by definition.

Therefore, by definition, any wave that yields watts
is NOT a standing wave.
--
73, Cecil http://www.w5dxp.com

You mean that when having quarter wave resonant antenna, with standing wave
(current node at the tip, voltage node at the base) that is your 90 degrees
apart?

Yuri

Standing-Wave Current vs Traveling-Wave Current
 

Yuri Blanarovich wrote:
You mean that when having quarter wave resonant antenna, with standing wave
(current node at the tip, voltage node at the base) that is your 90 degrees
apart?

No, on a lossless wire or transmission line with pure
standing waves, at any *PLANE* intersecting the wire or
transmission line, the measured net voltage will be 90
degrees out of phase with the measured net current.

Vtot(t) = Vfor(t) + Vref(t) and |Vfor|=|Vref|

Itot(t) = Ifor(t) - Iref(t) and |Ifor|=|Iref|

The negative sign on Iref(t) puts Itot(t) 90
degrees out of phase with Vtot(t). You can easily
prove that for yourself.

By definition, there cannot be any real average or
real instantaneous power (watts) in a standing wave.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
Cecil Moore wrote:
Roy Lewallen wrote:

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.

These comments in total are very interesting, by both authors. Thank
you for them.

It is clear that there is not a standard way of describing energy that
is in the process of being stored in either an inductor or capacitor.
Clearly it is stored, not used (converted) as in a resistor.

Sure there is. The energy stored in a capacitor is 1/2 * C * V^2, and in
an inductor, 1/2 * L * I^2. The rate of energy flow into or out of an
inductor or capacitor is called "power". It's exactly the same stuff as
energy moving into or through a resistor, and it's measured and
described in exactly the same way.

Storage of energy in a capacitor (for instance) occurs over time and
requires power to complete.

Yes, power is the rate of energy transfer.

So how do we describe that power if it is always zero because voltage
and current are 90 degrees out of phase?

See the following example which shows just that situation.

Should we recognize that only the peak voltage and peak current is 90
degrees out of phase, with the entire charging time occurring within
those two time extremes? Between those two time extremes, the voltage
and current are in phase but at changing impedance, with power flowing
into the capacitor. So think I.

The whole concept of impedance gets a bit flaky when working in the time
domain. I recommend sticking to voltage, current, power, and energy when
doing so. Go through the following example, and I believe you'll get a
much better idea of what's happening on a basic level.

But you've made some good observations. The answer is that, as I
mentioned, the instantaneous power (p(t)) is not zero except at the ends
and middle of the example line. Some of the posters are confusing
average and instantaneous power, which is leading to incorrect
statements and conclusions. Instantaneous power can easily be non-zero
while maintaining zero average power, as I'll illustrate.

Let's look, for example, at the half wavelength open circuited line
described by "Dave", being driven by v = 0.5 sin(wt), 1/8 wavelength
from the end. At that point,

vf(t) = 0.5 sin(wt - 135 deg.)
vr(t) = 0.5 sin(wt - 225 deg.)
if(t) = 0.01 sin(wt - 135 deg.)
ir(t) = -0.01 sin(wt - 225 deg.)

considering the positive direction of ir(t) as the same as the positive
direction of if(t) so we can add if(t) and ir(t) to get i(t).

Using elementary trig identities for the addition, and dropping the
explicit identification of degrees for simplicity:

vf(t) + vr(t) = v(t) = sin(wt - 180) * cos(45) = 0.7172 * sin(wt - 180)
if(t) + ir(t) = i(t) = 0.02 * cos(wt - 180) * sin(45) = 0.01414 *
cos(wt - 180)

One important thing to notice is that, unlike the ends or center of the
line, neither the voltage nor the current is zero at this point.
Consequently, their product is also non-zero.

The voltage and current are, of course, 90 degrees out of phase, as they
are everywhere along an open circuited or short circuited line, or one
terminated in a pure reactance. And when the average power is computed
from the instantaneous power, the result will be zero. But that doesn't
mean that the instantaneous power is zero as some are claiming. Let's
see what it does mean. Using another trig identity,

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

This is the instantaneous power which, as I've described before, is a
sinusoidal function with rotational frequency 2w. It has no offset, so
the average value is zero. But it shows that energy moves back and forth
past this point every cycle. Equal amounts move each direction each
cycle, so no net power flows either direction.

There are simpler ways to get the same answer, but this one shows very
basically where the energy is going at every instant. If instead you
start out by throwing away the time information and just looking at
averages, you lose a lot of information along with the basic
understanding of energy movement in the system. I'm afraid some people
have taken this simpler approach without realizing what information and
insight they've lost by doing so. The average value can always be
determined from the instantaneous function, but never the other way around.

I think of the "standing wave" as being equivalent to the graph on my
power bill that shows power used daily over a month. Granted that the
standing wave is recorded in V or I, but we only need to know the R
that the standing wave is acting through to determine the power that
it represents.

Not quite. The power bill graph is a graph of the power, much like the
p(t) function I calculated. The energy, which you pay for, is the
integral of that graph. If you used a constant power of 1 kW for half
the month, then sent 1 kW back into the power grid for the other half of
the month, you'd technically owe nothing, since you'd use no net power.
You might even have stored it somewhere and given the very same energy
back rather than using some and generating an equal amount later. That's
exactly what's happening at the point on the line 1/8 wavelength from
the end.

Roy Lewallen, W7EL

-- I've had many opportunities when doing this to make arithmetic or
trigonometric errors, and I haven't taken the time to thoroughly check
my work -- although the conclusion is correct. I'd appreciate anyone
pointing out any mathematical errors he's found.
Roy, thanks for this posting and it's immediate predecessors. They are
very helpful both to my understanding and toward understanding how you
are thinking of things.

Your comment about the power bill was particularly helpful because you
are exactly right about the sequence of power, first received, and then
given back. The standing wave is doing exactly that, with the only
difference being that power is moving from one side of the system to the
other, with one side containing "negative power" (negative only in the
sense that polarity is reversed) and the other positive power.

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

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

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

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

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

Then vrt(t) = vr(wt-90) + vr(wt-270) = 0
and irt(t) = ir(wt-90) + ir(wt-270) = 0

Below I will use the following terms:

vf = 1v/2 = 0.5v
vr = vf = 0.5v (for open circuit line)
ir = 0.5/50 = 0.01a

At the ends, vrt(t) = 2*vr(wt-90) = -2*vr(wt-270) = 1v

The voltage is doubled because of reflection. vf(wt-270) is negative.

Now consider power at the 45 and 135 degree points under stable
conditions, 1/8 and 3/8 wavelength in from the source. This would be
the power in the standing wave at the 45 and 135 degree locations.

At the 45 and 135 degree transmission line points, we have two
components present because the peak of each wave will be in alignment
with the leading edge of the reflected part of the same wave.
(Reflected waves being reflected again) Peak current and peak voltage
will be matched with near the near zero leading edge voltage and current.

Pr(left side) = vr(wt-90)*ir(wt-90) + vr(wt-180)*ir(wt-180)
= Pr(right side)= vr(wt-270)*ir(wt-270) + vr(wt-360)*ir(wt-360)
= 0.005w

Notice that vr(wt-90) and ir(wt-90) are different polarity from
vr(wt-270) and ir(wt-270). but the products of v and i are positive.

Find total power in the standing wave at the 45 degree points from
(ignoring the zero leading edge component)

Prt = Pr(left side) + Pr(right side)
= vr(wt-90)*ir(wt-90)+ vr(wt-270)*ir(wt-270)
= 0.005w + 0.005w = 0.01w

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

So think I.

As you can see, by considering that a full wave or complete cycle may
exist on the 1/2 wave transmission line, we can explain power moving
from side to side over the transmission line. I could not have offered
this explanation without your help and guidance and Cecil's as well. I
hope you will agree that it is a correct analysis.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 8:15*pm, Roger wrote:
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.

My apologies for being not being completely clear.

My voltmeter is always one that is appropriate for the situation.
A d'Arsonval movement would work at sufficiently low frequencies
but for the ones under discussion an oscilloscope would probably
be more appropriate. It would be a voltmeter that would be able
to measure the actual voltage at a point on the line at a particular
time. I.e. V(x,t)

Again, apologies.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 2:38*pm, Cecil Moore wrote:
Keith Dysart wrote:
And for a challenging use case, please consider two
circuits connected together. The circuits are in black
boxes so you do not know their details, but the voltage
on the connection between the circuits is measured as
10 V RMS at 4 MHz. The current is measured as 0.

How much energy is being transferred between the
circuits?

Inside each black box is a 50 ohm signal generator equipped
with a circulator and a 50 ohm load resistor. The signal
generators are outputting identical 10 V RMS phase-locked
signals.

Signal generator #1 "sees" the 50 ohm resistor in
signal generator #2 as it's load and supplies 200
milliamp, thus heating up the load resistor.

Signal generator #2 "sees" the 50 ohm resistor in
signal generator #1 as it's load and supplies 200
milliamp, thus heating up the load resistor.

The net voltage is measured as 10 V RMS at 4 MHz.
The net current is measured as 0.

How much energy is being transferred between the
circuits?

Install a one wavelength 50 ohm lossless transmission
line between the two signal generators. Nothing changes
but the standing waves become very visible and measurable.

An excellent example.

Allow me to provide an alternate analysis.

The current between the black boxes is 0.

From circuit analysis theory, any branch with a current
that is always zero can be cut without altering the
conditions anywhere in the rest of the circuit.

Cut the branches connecting the two black boxes.

No conditions change in either of the now two cicuits.

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 2:50*pm, Cecil Moore wrote:
Keith Dysart wrote:
When does P(x,t) not equal V(x,t) * I(x,t)?

Any time the angle between V(x,t) and I(x,t) is
not 0 or 180 degrees. The correct equation is:

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


This is non-sensical. V(x,t) and I(x,t) are functions
representing the instanteous values of the voltage
and current with respect to place and time.

The expression you really mean is

Pavg = Vrms * Irms * cos(A)

You have to keep the time domain equations separate
from the phasor equations.

If you write the above equation as:

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

then you have implied the *dot product* of those
terms.

No. I mean multiply the instantaneous value by
the instantaneous value, or more correctly,
the function representing the instantaneous
value of voltate by the function representing
the instantaneous value of current.

No vectors or phasors; just time domain functions.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 25, 11:10*am, Cecil Moore wrote:
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.

Apologies to Ramo, Whinery and Cecil. I mis-read the
quoted equations.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 25, 11:10 am, Cecil Moore wrote:
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.

Apologies to Ramo, Whinery and Cecil. I mis-read the
quoted equations.

...Keith

That's o.k., Cecil doesn't really understand them. If he did,
he wouldn't need to parrot them out of a book.
73,
Tom Donaly, KA6RUH

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Dave Heil wrote:
Sunrise and sunset are simply names we've given to that which you've
described.

Those words were coined when the human race believed
that the sun actually traveled across the sky each
day. The sun "rising" means the sun is moving but
that is just an illusion.

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

You seem to believe that illusions don't exist in
reality.

On the contrary, Cecil. I've often seen own attempts at being an
illusionist. You have told a number of folks about your being a 300,000
foot tall alien.

That very belief is an illusion.

....except I've not expressed such a belief.

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

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

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

A schematic shows exactly what is happening. There is no
path from SGCL1 to R1. There is no path from SGCL2 to R2.

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

There is nothing in the circuit to cause any reflections.
So the power dissipated in R2 comes from SGCL1 and the
power in R1 comes from SGCL2.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The expression you really mean is

Pavg = Vrms * Irms * cos(A)

Yep, that's what I meant.

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

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

Standing-Wave Current vs Traveling-Wave Current
 

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

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

Standing-Wave Current vs Traveling-Wave Current
 

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

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

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
. . .

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

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

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

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

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

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

Sorry, it's much worse than this.

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

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

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

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

. . .

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

So think I.

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

. . .

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

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

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

A schematic shows exactly what is happening. There is no
path from SGCL1 to R1. There is no path from SGCL2 to R2.

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

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

Can not happen after cutting the branches.

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

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

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

Pavg = Vrms * Irms * cos(A)

Yep, that's what I meant.

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

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

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

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

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

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

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

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

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

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

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

Standing-Wave Current vs Traveling-Wave Current
 

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

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

Though I realize you might have been around at the time the terms were
first used, it is evident that nearly all still use those terms
today--even those in the scientific community. You keep writing, over
and over, that the Sun just sits there in space. We both know that
isn't actually correct either. The Sun rotates and is moving through
space quite rapidly. I've asked a number of times how you refer to the
phenomena of what most of us call "Sunrise" and "Sunset." There must
be a reason that you don't respond to that.

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

Dave K8MN

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:
A schematic shows exactly what is happening. There is no
path from SGCL1 to R1. There is no path from SGCL2 to R2.

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

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

Can not happen after cutting the branches.

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

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

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

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

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

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

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