On Mon, 14 Apr 2008 09:10:20 -0500
Cecil Moore wrote:

Keith Dysart wrote:
All the elements
of the system are completely specified in Fig 1-1 and we used
circuit theory to compute the energy flows. Not surprisingly, they
completely balanced:
Ps(t) = Prs(t) + Pg(t)

Yes, but that is only *NET* energy flow and says nothing
about component energy flow. Everything is already known
about net energy flow and there are no arguments about it
so you are wasting your time. Your equation above completely
ignores reflections which is the subject of the thread.

You object to me being satisfied with average energy flow
while you satisfy yourself with net energy flow. I don't see
one iota of conceptual difference between our two positions.

After hundreds of postings, all you have proved is that
Eugene Hecht was right when he said instantaneous powers
are "of limited utility", such that you cannot even tell
me how many joules there are in 100 watts of instantaneous
power when it is the quantity of those very joules that
are required to be conserved and not the 100 watts.

The limit in your quest for tracking instantaneous energy
is knowing the position and momentum of each individual
electron. Good luck on that one.

I am going to summarize the results of my Part 1 article
and be done with it.

In the special case presented in Part 1, there are only
two sources of power dissipation in the entire system,
the load resistor and the source resistor. None of the
reflected energy is dissipated in the load resistor
because the chosen special conditions prohibit reflections
from the source resistor. Therefore, all of the energy not
dissipated in the load resistor is dissipated in the source
resistor because there is no other source of dissipation
in the entire system. Only RL and Rs exist. Pr is not
dissipated in RL. Where is Pr dissipated? Even my ten year
old grandson can solve that problem and he's no future
rocket scientist.
--
73, Cecil http://www.w5dxp.com

This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. My view is that any source must both absorb and deliver power at some none zero impedance. As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source.

While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers power into the source. Charging a battery with zero internal resistance is a good example. Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator.

Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. Some would call that a world of science fiction.
--
73, Roger, W7WKB

Roger Sparks wrote:
While it can be argued that the ideal voltage source would
have zero internal resistance, that argument does not address
the fact that power flowing in the reverse direction (into the
source, against the source supplied voltage) delivers power
into the source.

I thought I had already addressed that topic when I added the
one-wavelength of transmission line to the example in between
the source and source resistance.

But here's an example that may allow better tracking
of the energy flow. Let's modify my Part 1, Fig. 1-1
to add a 50 ohm circulator and load to the ground
leg of the source. Everything else remains the same.

Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL
\ /
3
|
50 ohms
|
GND

How much power is dissipated in the circulator
resistor?

How much power does the source have to supply to
maintain 50 watts of forward power on the transmission
line?

Does this example answer your questions?
--
73, Cecil http://www.w5dxp.com

On Mon, 14 Apr 2008 16:54:47 GMT
Cecil Moore wrote:

Roger Sparks wrote:
While it can be argued that the ideal voltage source would
have zero internal resistance, that argument does not address
the fact that power flowing in the reverse direction (into the
source, against the source supplied voltage) delivers power
into the source.

I thought I had already addressed that topic when I added the
one-wavelength of transmission line to the example in between
the source and source resistance.


I thought the addition of a one wavelength transmission line did not address the issue, and only added more reflections. We still need a reason to assume that a voltage source should *not* absorb power.

But here's an example that may allow better tracking
of the energy flow. Let's modify my Part 1, Fig. 1-1
to add a 50 ohm circulator and load to the ground
leg of the source. Everything else remains the same.

Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL
\ /
3
|
50 ohms
|
GND

How much power is dissipated in the circulator
resistor?

How much power does the source have to supply to
maintain 50 watts of forward power on the transmission
line?

Does this example answer your questions?
--
73, Cecil http://www.w5dxp.com

No, I'm sorry but no. I offered the examples of two real sources that will absorb power when the returning voltage exceeds the output voltage (a battery and a generator turned into a motor). I think that we must allow our voltage source to have that same real property.

I do understand that when we allow the source to receive power, then we need to address source impedance. If we assign a single impedance, then we expect reflections from the source. The simple solution that I propose is to add a source property of absorbing all reflections. This can be accomplished in the real world by making the transmission so long that reflections never return from the source over any reasonable time, or by making the tranmission line sufficiently lossy to absorb reflections. Your example uses the first method.

Does the idea of source receiving power run counter to what you were planning to write in Parts 2 and 3? I am trying to understand why you have such great reluctance to accept that the source could receive power for part of a cycle, especially when it could easily bring the instantaneous power and energy calculations into balance.
--
73, Roger, W7WKB

Roger Sparks wrote:
I offered the examples of two real sources that will absorb
power when the returning voltage exceeds the output voltage
(a battery and a generator turned into a motor). I think that
we must allow our voltage source to have that same real property.

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy. A motor converts electrical
energy into physical work, i.e. it transforms the electrical
energy. An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

I do understand that when we allow the source to receive
power, then we need to address source impedance.

The series source impedance is zero. It acts like a short
circuit to reflections, i.e. there are no reflections.
However, there seem to be 100% reflection from the GND on
the other side of the source.

Does the idea of source receiving power run counter to what
you were planning to write in Parts 2 and 3?

The source will be shown to adjust its output until an
energy balance is achieved. It will throttle back when
destructive interference occurs at the source resistor
and will gear up when constructive interference requires
more energy.

I am trying to
understand why you have such great reluctance to accept that
the source could receive power for part of a cycle, especially
when it could easily bring the instantaneous power and energy
calculations into balance.

There is no known mechanism that would allow an ideal
source to dissipate or store energy. Consider that the
energy you see flowing back into the source is reflected
back through the source by the ground on the other side
and becomes part of the forward wave out of the source.
That would satisfy the distributed network model and
explain why interference exists in the source.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy.

Gee, I'm just a silly old engineer, but I thought batteries converted
chemical energy to electrical energy. Unless you are perhaps speaking
of charging it.

tom
K0TAR

Tom Ring wrote:
Cecil Moore wrote:
A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy.

Gee, I'm just a silly old engineer, but I thought batteries converted
chemical energy to electrical energy. Unless you are perhaps speaking
of charging it.

Here was the preceding comment in context which you trimmed:
"I offered the examples of two real sources that will absorb power when
the returning voltage exceeds the output voltage (a battery and a
generator turned into a motor)."

The context was a battery absorbing power when the
charging voltage exceeds the battery's output voltage.
A battery converts electrical energy to chemical energy
during that charging process.
--
73, Cecil http://www.w5dxp.com

On Tue, 15 Apr 2008 01:22:11 GMT
Cecil Moore wrote:

Roger Sparks wrote:
I offered the examples of two real sources that will absorb
power when the returning voltage exceeds the output voltage
(a battery and a generator turned into a motor). I think that
we must allow our voltage source to have that same real property.

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy. A motor converts electrical
energy into physical work, i.e. it transforms the electrical
energy. An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

No, the separation of Vs and Rs was made to better understand why no interference would occur in Figure 1-1. found at http://www.w5dxp.com/nointfr.htm.

Here is a quote from Part 1.
"4. Since the transmission line is 1/8 wavelength (45 degrees) long and the load is purely resistive, the reflected wave incident upon the source resistor will be 2(45) = 90 degrees out of phase with the forward wave at the source resistor. This is the necessary and sufficient condition to produce zero interference at the source resistor."

The problem is that the source and reflected waves behave as two power sources out of time by 90 degrees. As a result, the current flows as the result of two sine waves, and can be described by only one sine wave. The one sine wave description necessarily shows that power *does* flow into the source during part of the cycle. Interference techniques are used to combine the two sine waves into one wave so it would appear that statement 4 is incorrect.

I do understand that when we allow the source to receive
power, then we need to address source impedance.

The series source impedance is zero. It acts like a short
circuit to reflections, i.e. there are no reflections.
However, there seem to be 100% reflection from the GND on
the other side of the source.

It is not the reflections from the source that is the root of the problem. The root is the way two sine waves combine into one wave that runs at a third phase compared to either of the source waves.

Does the idea of source receiving power run counter to what
you were planning to write in Parts 2 and 3?

The source will be shown to adjust its output until an
energy balance is achieved. It will throttle back when
destructive interference occurs at the source resistor
and will gear up when constructive interference requires
more energy.

I am trying to
understand why you have such great reluctance to accept that
the source could receive power for part of a cycle, especially
when it could easily bring the instantaneous power and energy
calculations into balance.

There is no known mechanism that would allow an ideal
source to dissipate or store energy. Consider that the
energy you see flowing back into the source is reflected
back through the source by the ground on the other side
and becomes part of the forward wave out of the source.
That would satisfy the distributed network model and
explain why interference exists in the source.

I can understand a voltage source that throttles up and down but I can't understand why the throttle all has to be on the plus side. What logic prevents the power from returning to the ideal source from whence it just left?

Our real limit is that only one current can flow for only one voltage for each instant at any place in the circuit. This is how we justify a "one sine wave" description. It is why whenever we have a reflection, we also have interference. It is also the reason that we must have power flowing back into the source for part of the cycle.
--
73, Roger, W7WKB

Roger Sparks wrote:
On Tue, 15 Apr 2008 01:22:11 GMT
Cecil Moore wrote:
An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

No, the separation of Vs and Rs was made to better understand
why no interference would occur in Figure 1-1.

I wasn't talking about my article. I was talking about Vs & Rs
models in general. In the real world, Vs is not separated from
Rs. That only occurs in the ideal model. In the ideal model,
all dissipation is confined to Rs and there is none in Vs.

The problem is that the source and reflected waves behave as
two power sources out of time by 90 degrees.

Not quite correct. The problem is that the forward waves
and reflected waves flowing through the source behave as
two power sources out of time by 90 degrees. The source
wave is the net superposition of the forward wave and
reflected wave. An ideal 50 ohm directional wattmeter
in the circuit will not read the source power. It will
read a forward power which is a different magnitude
than the source power. In any case, only Rs and RL
dissipate power in the system.

I can understand a voltage source that throttles up and
down but I can't understand why the throttle all has to be
on the plus side.

It is not all on the plus side. Whatever energy flows, flows.
Sometimes the flow is forward and sometimes it is backwards.
That's the way AC works. If destructive interference is
present, the source reduces its output power. If constructive
interference is present, the source increases its output
power. But the ideal source does not dissipate power, i.e.
doesn't heat up. All of the heat generated in the entire
system comes from Rs and RL.

Our real limit is that only one current can flow for only one
voltage for each instant at any place in the circuit.

You are, of course, talking about the *net* voltage and the
*net* current after superposition of all the components. But
this discussion is not about net voltage and net current.

This is
how we justify a "one sine wave" description. It is why whenever
we have a reflection, we also have interference. It is also the
reason that we must have power flowing back into the source for
part of the cycle.

I don't know where you got the idea that energy doesn't
flow back into the source for part of the cycle. Since
it is AC, it does flow forward and backward but none is
dissipated, i.e. none is turned into heat in an ideal
source. An equal amount of destructive and constructive
interference occurs during each complete cycle.
--
73, Cecil http://www.w5dxp.com

On Apr 14, 12:06*pm, Roger Sparks wrote:
On Mon, 14 Apr 2008 09:10:20 -0500

Cecil Moore wrote:
Keith Dysart wrote:
All the elements
of the system are completely specified in Fig 1-1 and we used
circuit theory to compute the energy flows. Not surprisingly, they
completely balanced:
* *Ps(t) = Prs(t) + Pg(t)

Yes, but that is only *NET* energy flow and says nothing
about component energy flow. Everything is already known
about net energy flow and there are no arguments about it
so you are wasting your time. Your equation above completely
ignores reflections which is the subject of the thread.

You object to me being satisfied with average energy flow
while you satisfy yourself with net energy flow. I don't see
one iota of conceptual difference between our two positions.

After hundreds of postings, all you have proved is that
Eugene Hecht was right when he said instantaneous powers
are "of limited utility", such that you cannot even tell
me how many joules there are in 100 watts of instantaneous
power when it is the quantity of those very joules that
are required to be conserved and not the 100 watts.

The limit in your quest for tracking instantaneous energy
is knowing the position and momentum of each individual
electron. Good luck on that one.

I am going to summarize the results of my Part 1 article
and be done with it.

In the special case presented in Part 1, there are only
two sources of power dissipation in the entire system,
the load resistor and the source resistor. None of the
reflected energy is dissipated in the load resistor
because the chosen special conditions prohibit reflections
from the source resistor. Therefore, all of the energy not
dissipated in the load resistor is dissipated in the source
resistor because there is no other source of dissipation
in the entire system. Only RL and Rs exist. Pr is not
dissipated in RL. Where is Pr dissipated? Even my ten year
old grandson can solve that problem and he's no future
rocket scientist.
--
73, Cecil *http://www.w5dxp.com

This thread has one assumption that I find very frustrating, a voltage source that is a steady source of power but can not absorb power. *My view is that any source must both absorb and deliver power at some none zero impedance. *

I am not sure why you desire a non-zero impedance. The usual
definition of an
ideal voltage source is that it provides or sinks what ever current is
needed to
hold the desired output voltage. When it is sourcing current then it
is providing
energy. No statement is made about where this energy comes from. When
it is
sinking current, it is absorbing energy. No statement is made about
where
this energy is going.

A non-zero impedance is not required to make any of the above
behaviour work.

If you include a non-zero impedance, then you have a more real world
source
which can often be modeled using the Thevenin equivalent circuit; an
ideal
voltage source (zero impedance) in series with a resistor representing
the impedance of the real world source.

As justification for this view, I offer that current always flows from high voltage to lower voltage, so a real voltage source would have to absorb energy if the external voltage exceeded the voltage of the voltage source.

This is true.

While it can be agrued that the ideal voltage source would have zero internal resistance, that argument does not address the fact that power flowing in the reverse direction (into the source, against the source supplied voltage) delivers *power into the source. *Charging a battery with zero internal resistance is a good example. *Another example is the observation that a generator becomes a motor when the externally suppied voltage exceeds the voltage supplied by the generator.

Yes, we can make the assumption that the voltage source can not absorb power at any time, but the assumption takes us into an unreal world and gives answers that are impossible to duplicate with measurements. *Some would call that a world of science fiction.

...Keith

Keith Dysart wrote:
When it is sourcing current then it is providing
energy. No statement is made about where this energy
comes from.

The question is: Is that energy being created or
dissipated as needed according to your omnipotent
whims?
--
73, Cecil http://www.w5dxp.com