W5DXP wrote:

Roy Lewallen wrote:
Magid has the most rigorous derivation of power and energy flow on
transmission lines I've seen, as well as other extensive transmission
line information. One conclusion that pricked my ears was that on a
line with a pure standing wave (e.g., a lossless line terminated with
an open or short circuit), ". . . power (and therefore, energy) is
completely trapped within each [lambda]/4 section of this lossless
line, never able to cross the zero-power points and thus constrained
forever to rattle to and fro within each quarter-wave section of this
line."

He's obviously talking about net energy. There is no impedance
discontinuity
in a continuous piece of transmission line so there is nothing to cause
reflections at the zero-power points. Net energy doesn't cross the zero-
power points but equal forward energy and reflected energy must cross the
zero-power points. That is easy to prove by observing ghosting on a TV set
being fed by 1000 feet of ladder-line. If energy is completely trapped
within each 1/4WL section, ghosting would be impossible.

I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field. If the voltage is zero, the H-field is at a maximum. If the current
is zero, the E-field is at a maximum. The energy still exists, just in one
field or the other. V*I*cos(theta) is NET energy. The power in the forward
wave is Vfwd*Ifwd and the power in the reflected wave is Vref*Iref. Those
values are constant all up and down the line for a lossless feedline.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.
Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows. (My light bulb is not drawing any current but
there could still be energy flowing in it!?)
Think carefully before going there.

....Keith

wrote:
W5DXP wrote:
I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

It means there is no NET power transfer. There are power flow vectors
in both directions that are equal in magnitude. Reference: _Fields_
and_Waves_in_Communications_Electronics_ by Ramo, Whinnery, & Van Duzer,
section 6.10, page 350, where they describe Pz-, the reflected wave
Poynting vector and Pz+, the forward wave Poynting vector.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

There is a forward power flow vector and a reflected power flow vector.
There is no net power flowing past any point. There are, however, equal
magnitude component constant power flow vectors flowing in both directions.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.

No, it doesn't. It only requires acceptance of Ramo, Whinnery, & Van Duzer.

However, to reject energy flow across a zero voltage or zero current
point requires a confusion of cause and effect. Energy flow in both
directions is the *CAUSE* of the standing waves. You simply cannot
turn around and say that standing waves eliminate their own cause
but continue to exist anyway.

Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go? They cannot disappear or
stand still. If they are moving, power exists.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
W5DXP wrote:
I forgot to add. At the "zero-power points", either voltage or current is
zero. All that means is that all the energy is contained in the opposite
field.

Not quite all. It also means that there is NO power since P = V x I.

It means there is no NET power transfer.

Do not be afraid to admit that you have changed the definition of P = V
x I
and therefore do not accept the standard definition.

When I learned Pinst = Vinst x Iinst there were no caveats about how
Pinst meant Pnet. Instantaneous energy is flowing or it is not. When
Pinst
is 0 for all time, then there is no energy flowing.

To satisfy your theory (and minimize double think), you have had to
change
this to Pnet is zero to allow these cancelling powers to flow. So be it.

There are power flow vectors
in both directions that are equal in magnitude. Reference: _Fields_
and_Waves_in_Communications_Electronics_ by Ramo, Whinnery, & Van Duzer,
section 6.10, page 350, where they describe Pz-, the reflected wave
Poynting vector and Pz+, the forward wave Poynting vector.

There can be lots of energy present but none of it is flowing past
the zero voltage or zero current point; hence no power.

There is a forward power flow vector and a reflected power flow vector.
There is no net power flowing past any point.

True, sort of. At the quarter wave points where voltage or current are
always
zero, there is no energy flowing. Period.

At other points, energy flows in one direction for a quarter cycle and
then in the other direction for the next quarter cycle, producing a net
of zero. A true instantenouse power meter (one which measures V and I
and
displays V x I) will easily demonstrate this. As a thought experiment,
move such a true power meter along a shorted or open line and think if
its indications in the time domain, then do the averages. It will be
quite instructive. Repeat for a line terminated in other than its
characteristic impedance.

By the way, since energy flows forward for a quarter cycle and backwards
for the next, the maximum distance travelled by this energy is one
quarter
wavelength on the line. It is not flowing all the way to the end of the
line and then back. There is not enough time for this to happen (on a
multi-wavelength line) since it changes direction every quarter cycle.

There are, however, equal
magnitude component constant power flow vectors flowing in both directions.

To believe that energy is flowing across a zero voltage or zero current
point requires the rejection of the view that instantaneous power is
equal to instantaneous voltage multiplied by instantaneous current.

No, it doesn't. It only requires acceptance of Ramo, Whinnery, & Van Duzer.

However, to reject energy flow across a zero voltage or zero current
point requires a confusion of cause and effect. Energy flow in both
directions is the *CAUSE* of the standing waves. You simply cannot
turn around and say that standing waves eliminate their own cause
but continue to exist anyway.

Not quite. Standing voltage and current waves (which are not waves in
the normal sense) can be observed on the line. They can be measured with
real voltage and current instruments; as can real energy flows with
a real (V x I) power meter (but not a 'Bird watt' meter which
is doing something quite different). It happens that if you assume the
existence of forward and reverse voltage and current waves, mathematical
functions can be derived that will produce the same distribution of
voltage and current as observed on the line. This is extraordinarily
convenient some analysis but does not mean that these assumed waves are
real.

A mechanical analogue would be to look at a guy wire on a pole. You can
analyze the forces as two vectors at 90 degrees (or any other angle of
convenience!), but never make the mistake of assuming that there are
actually two guy wires present. Just because it is mathematically
convenient to assume the existence of two vectors does not mean they
exist.

Rejection of
P = V x I would have wide impacts on our understanding of electrical
power and energy flows.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go?

This energy is indeed stored in the stub. None of it moves across zero
voltage
or current boundaries. Between these boundaries, it is indeed moving as
it changes from being stored in the capacitance and inductance of the
line.
This time variation of the location of the stored energy produces the
observed
voltages and currents on the line.

They cannot disappear

Absolutely. They do not disappear.

or stand still.

With AC excitation they do not stand still, but when similar analysis is
done
for a line excited with a DC source, the energy does indeed stand still.
An
open line stores the energy in the capacitance and a shorted line stores
it
in the inductance.

If they are moving, power exists.

Yes, but it never moves more than a quarter wavelength.

....Keith

Keith wrote:
"At the quarter wave points where voltage and current are always zero,
there is no energy flowing. Period."

Yes there is energy flowing, and it is flowing in both directions if it
is flowing in one direction. Otherwise there would be no standing wave.

The waves flow right through each other so long as the impedance is
uniform. Energy flow is unabated at those points of illusory zeros.
Those zeros are not zeros at all in the forward and reverse waves. They
only appear as zeros when your sensor can`t separate the forward from
the reverse. Get a proper directional sensor and the forward wave is
sensed as full amplitude as is the reverse wave.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Keith wrote:
"At the quarter wave points where voltage and current are always zero,
there is no energy flowing. Period."

Yes there is energy flowing, and it is flowing in both directions if it
is flowing in one direction. Otherwise there would be no standing wave.

This view that there is energy flowing in both directions at the same
time
leads to some strange conclusions.

The directional wattmeter uses the following expressions to compute its
displayed value:
Vf = (V + IR)/2
Vr = (V - IR)/2
where V and I are the instantaneous voltage and current at the same
point
on the line and then computes power from:
Pf = average(Vf**2/R)
Pr = average(Vr**2/R)
Appropriate scaling (which we'll ignore for simplicity) is needed
depending on the characteristics of the voltage and current sensors.

Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait
for the transient to die:
Vf = (9 + 0)/2 - 4.5 V
Vr = (9 - 0)/2 - 4.5 V
Pf = 0.405 Watt
Pr = 0.405 Watt
Disconnect the battery and the capacitance of the line remains charged
to
9 Volts. Are you really quite comfortable with the notion that this
line,
charged to 9 Volts, with 0 current everywhere has 0.405 Watts flowing
forward canceling at all points the 0.405 Watts flowing in reverse?

Others have stated that lamp cord has an impedance of about 100 Ohms.
Assume this to be correct.
With a table lamp plugged in but the lamp turned off:
Vf = (120 + 0)/2 - 60 V
Vr = (120 - 0)/2 - 60 V
Pf = 36 Watt
Pr = 36 Watt
Again, are you comfortable with this conclusion?

How about phone lines (phone on hook) at 600 Ohms and 48 Volts:
Vf = 24 V
Vr = 24 V
Pf = 0.96 W
Pr = 0.96 W
In a 400 pair cable there is 384 Watts flowing in the forward and
reverse direction. From a hundred thousand line central office we
get 960 Kilowatts forward and reverse, when the current is zero,
everywhere. Comfort level?

With the forward and reverse power view, to completely understand
the behaviour of the circuit, we need to know Z0 or we can not
compute these forward and reverse powers which seem to be
fundamental.

Consider a flashlight with the lamp off: the wire twists and turns
and has a very non-constant Z0. While computationally tedious, it
is possible to determine Pf and Pr at every point along the wire.
Isn't it necessary to do this to obtain a complete understanding
of what is happening in the circuit, if Pf and Pr are flowing?
If not, why not? The necessity for doing this is the logical
conclusion of the Pf and Pr approach.

Presented with a wire charged to high static voltage, what is Pf and
Pr? Presented with a sphere charge to a high static voltage, what
is the Pf and Pr?
Presented with a simple length of wire, how much power is flowing?

Similarly for a capacitor. The plates of a capacitor must have power
reflecting in all directions even when it is just charged to a
constant voltage (according to the Pf and Pr theory).

Given these bizarre results it would seem wise to limit the application
of the watts indicated by a directional 'watt'meter to those things
which are proper and not assume that it necessarily means that there
is real power flowing.

....Keith

wrote:
Do not be afraid to admit that you have changed the definition of P = V
x I and therefore do not accept the standard definition.

Well, I was taking 'x' as a multiplication sign. Did you mean it as
a cross product sign? In any case, your Vinst is *NET* voltage and
your Iinst is *NET* current. We know that the feedline Z0 forces
Vfwd*Ifwd to be a constant value for a lossless line. We know that
the feedline Z0 forces Vref*Iref to be a constant value for a lossless
line. Vfwd+Vref is the *NET* voltage. Ifwd+Iref is the *NET* current.
Of course, their product, in any form, is going to be *NET* power.

When I learned Pinst = Vinst x Iinst there were no caveats about how
Pinst meant Pnet. Instantaneous energy is flowing or it is not. When
Pinst is 0 for all time, then there is no energy flowing.

But RF energy cannot stand still so if it exists, it must necessarily
flow. If there is no energy flowing, then there is no RF. If there is
no RF, then your statements are irrelevant to this newsgroup. :-)

To satisfy your theory (and minimize double think), you have had to
change this to Pnet is zero to allow these cancelling powers to flow.
So be it.

Your Vinst is *NET* voltage equal to Vfwd+Vref. Your Iinst is *NET*
current equal to Ifwd+Iref. Of course, their product will be *NET*
power. It cannot be anything else.

True, sort of.

Ramo, Whinnery, and Van Duzer will be impressed that you "sort of"
agree with them. :-)

It is not flowing all the way to the end of the
line and then back. There is not enough time for this to happen (on a
multi-wavelength line) since it changes direction every quarter cycle.

Just as I suspected, you are confusing the carriers of the energy with
the energy itself in the waves which moves at the speed of light. A
quarter cycle of time is very, very large compared to the speed of light.

Since water molecules cannot travel at the speed of sound in the ocean,
I assume you would argue that the energy in a tsunami wave cannot travel
at 500 mph, right?

Not quite. Standing voltage and current waves (which are not waves in
the normal sense) can be observed on the line. They can be measured with
real voltage and current instruments; as can real energy flows with
a real (V x I) power meter (but not a 'Bird watt' meter which
is doing something quite different). It happens that if you assume the
existence of forward and reverse voltage and current waves, mathematical
functions can be derived that will produce the same distribution of
voltage and current as observed on the line. This is extraordinarily
convenient some analysis but does not mean that these assumed waves are
real.

So please amaze me with a model that produces standing waves without actual
forward and reflected waves (in a single source, single feedline, single load
system).

A mechanical analogue would be to look at a guy wire on a pole. You can
analyze the forces as two vectors at 90 degrees (or any other angle of
convenience!), but never make the mistake of assuming that there are
actually two guy wires present. Just because it is mathematically
convenient to assume the existence of two vectors does not mean they
exist.

Nobody is rejecting it. If the lossless stub is one second long, it takes
two seconds of *POWER* to bring it to steady-state. If the stub contains
no moving energy, where did all those joules go?

This energy is indeed stored in the stub. None of it moves across zero
voltage or current boundaries.

What exactly, keeps energy from crossing the boundary? Here is an example.

source-------------50 ohm coax--------------+----1/4WL stub-----open

What mechanism of physics keeps energy from crossing the '+' point? Note
that there is no physical impedance discontinuity at point '+'.

This is what I (and Ayn Rand) call "primacy of consciousness" type thinking.
If you believe it strongly enough, your math model will dictate reality.
Something about being able to move mountains with the faith of a grain of
mustard seed. Something about being able to change the SWR by changing the
normalization of a Smith Chart on a sheet of paper. OTOH, I believe in
"primacy of existence", where reality dictates my math models. They may
be wrong but are as close to reality as I can get.

All you have to do to convince me that you are right is explain exactly
how standing waves can be sustained without a forward wave and a reflected
wave (in a system with a single source, single feedline, and single load).
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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