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

"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

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

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

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

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

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

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

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

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