The Rest of the Story
 

Roger Sparks wrote:
. . . But how can we have a source with zero resistance, zero
capacitance, and zero inductance because in the real world, any
source has impedance? The short has "zero resistance, zero
capacitance, and zero inductance but it does not emit energy nor have
a reverese voltage, both properties of the voltage source. It is not
reasonable to assign the properties of the short to the voltage
source, ignoring the reverse voltage situation, and expect reflectons
from the source to be identical to reflections from a short.

None of the ideal components we use for linear circuit analysis exist in
the real world. We use ideal resistances which have no inductance or
capacitance, capacitances which have no resistance or inductance, ideal
inductances which have no resistance or capacitance, ideal controlled
sources which operate over an infinite range of control and output
values. And try making anything even vaguely resembling an ideal
transformer. So what's the problem in accepting an ideal voltage source
as another model element? If you want a better approximation of
something you can build in the real world, add an ideal resistance to
the ideal voltage source, and you'll have a much better representation
of most real sources.

As for the way the source reacts to an impinging wave, note that the
voltage across a short circuit doesn't change when a wave hits it.
Neither does the voltage across an ideal voltage source. Consequently,
they do have exactly the same effect on waves.

Roy Lewallen, W7EL

The Rest of the Story
 

On Apr 18, 3:27*am, Roy Lewallen wrote:
Roger Sparks wrote:
. . . But how can we have a source with zero resistance, zero
capacitance, and zero inductance because in the real world, any
source has impedance? *The short has "zero resistance, zero
capacitance, and zero inductance but it does not emit energy nor have
a reverese voltage, both properties of the voltage source. *It is not
reasonable to assign the properties of the short to the voltage
source, ignoring the reverse voltage situation, and expect reflectons
from the source to be identical to reflections from a short.

None of the ideal components we use for linear circuit analysis exist in
the real world. We use ideal resistances which have no inductance or
capacitance, capacitances which have no resistance or inductance, ideal
inductances which have no resistance or capacitance, ideal controlled
sources which operate over an infinite range of control and output
values. And try making anything even vaguely resembling an ideal
transformer. So what's the problem in accepting an ideal voltage source
as another model element? If you want a better approximation of
something you can build in the real world, add an ideal resistance to
the ideal voltage source, and you'll have a much better representation
of most real sources.

As for the way the source reacts to an impinging wave, note that the
voltage across a short circuit doesn't change when a wave hits it.
Neither does the voltage across an ideal voltage source. Consequently,
they do have exactly the same effect on waves.

Roy Lewallen, W7EL

Perhaps it would help clarify the thinking to plot some voltage-
current
curves.

If we plot V versus I for a resistor (with V on the vertical access)
we
get a line with a slope equal to R. This line passes through the
origin.

For a short, the line is horizontal (i.e. slope and R are zero) and
for
an open the line is vertical (i.e. slope and R are infinite).

Now plot the V-I characteristic of a resistor in series with an ideal
voltage source. Again it is a line with a slope equal to R, but it
does
not pass through the origin, it crosses the vertical axis at the
voltage
provided by the source. So the y-axis crossing is controlled by the
voltage source and the slope is controlled by the resistor. If you
reduce the resistor to zero, you get a horizontal line crossing
the y-axis at the voltage of the source. The line being horizontal
means that no amount of current will change the voltage.

We often talk of resistance as V/I, but there are many situations in
which it is better to think of it as deltaV/deltaI (or, in the limit,
dV/dI); that is, the change in voltage that accompanies a change
in current. This is exactly the slope of the V/I curve at that point
and works for computing resistance regardless of whether there is a
voltage offset present.

And this is how the source resistance of a battery or power supply
would be measured. Measure the voltage and current at one load,
change the load, measure the new voltage and current, compute the
source resistance from the changes in the voltage and current.

This dV/dI view of resistance is extremely useful and can be
applied to devices with very complex V-I curves. Consider a
tunnel diode (http://en.wikipedia.org/wiki/Tunnel_diode).
Over part of its V-I curve, the slope (i.e. resistance) is
negative. If we put a tunnel diode in a circuit with appropriate
bias such that the tunnel diode only operates over this range
of its V-I curve, then for the purposes of that circuit it
can be modelled as a resistor with negative resistance.

Think R=dV/dI, not R=V/I.

...Keith

The Rest of the Story
 

On Thu, 17 Apr 2008 18:43:41 GMT
Cecil Moore wrote:

Roger Sparks wrote:
Yes, in nature the returning power will affect the
frequency of the source.

If the reflected wave is a sine wave coherent with
the forward wave, why would the frequency change
when they are superposed?


I grant that the case for true frequency change was overstated. There is a one time phase change, causing a frequency "bump" for one cycle. This phase change is a one time event, occuring for each successive and diminishing reflected wave.

GND--Vs--x--Rs--------45deg 50 ohm----short
100v 50ohm

This circuit has an impedance of 73.5 + J44.1 ohms
at the point x. Energy sinking into the source occurs
during the cycle. It is not an illusion.

I got a different impedance value but for purposes
of discussion, all that matters is that the impedance
is NOT 50 ohms.

Right, in my haste, I picked up the impedance for a 12.5 ohm load. Sorry. The impedance should be 50 + 50j ohms.

Let's assume your impedance is correct. That means
the ratio of reflected power to forward power at
point 'x' is 0.1452 in the 50 ohm environment.
Why would reflected energy flowing back through
the source be considered "energy sinking". Why
cannot the reflected energy flow back through the
source unimpeded by the 0+j0 impedance and be
reflected by the short to ground on the other side?
That same thing happens all the time to standing
wave current on the outside of the coax braid.
Why do the laws of physics change inside a source?

It is the arbitrariness that I object to. As you say, you chose 50 ohms to calculate the reflection factor. So we have a 50 ohm transmission line zero length long, reflecting from a short circuit. This does nothing for us. We still have the source absorbing power, but now we have now added a mechanism for how the absorbed power behaves within the source.

In this example, we are delivering 100 watts to Rs.
With a power reflection coefficient of 0.1452, the
forward power has to be 117 watts making the reflected
power equal to 17 watts at point 'x'. That reflected
power of 17 watts is not absorbed by the source. It
flows back through the source, unimpeded by the 0+j0
impedance, and is reflected by the short to ground.
It joins the 100 watts being generated by the source
to become the forward wave of 117 watts. No absorbing
of energy required - just good old distributed network
physics.

Once this 17 watt wave is tracked back through the
source, reflected, and superposed with the 100 watts
being generated by the source, all will become clear
without any power absorption by the source being
necessary. Here's a diagram of what's happening.

100w 100w
GND--------Vs----------Rs--------45deg----------short
17w-- 117w-- 50w--
--17w --17w --50w

All energy flows balance perfectly and there is no
absorption by the source.
--

This is much too arbitrary for me.
--
73, Roger, W7WKB

The Rest of the Story
 

On Fri, 18 Apr 2008 00:27:48 -0700
Roy Lewallen wrote:

Roger Sparks wrote:
. . . But how can we have a source with zero resistance, zero
capacitance, and zero inductance because in the real world, any
source has impedance? The short has "zero resistance, zero
capacitance, and zero inductance but it does not emit energy nor have
a reverese voltage, both properties of the voltage source. It is not
reasonable to assign the properties of the short to the voltage
source, ignoring the reverse voltage situation, and expect reflectons
from the source to be identical to reflections from a short.

None of the ideal components we use for linear circuit analysis exist in
the real world. We use ideal resistances which have no inductance or
capacitance, capacitances which have no resistance or inductance, ideal
inductances which have no resistance or capacitance, ideal controlled
sources which operate over an infinite range of control and output
values. And try making anything even vaguely resembling an ideal
transformer. So what's the problem in accepting an ideal voltage source
as another model element? If you want a better approximation of
something you can build in the real world, add an ideal resistance to
the ideal voltage source, and you'll have a much better representation
of most real sources.

As for the way the source reacts to an impinging wave, note that the
voltage across a short circuit doesn't change when a wave hits it.
Neither does the voltage across an ideal voltage source. Consequently,
they do have exactly the same effect on waves.

Roy Lewallen, W7EL

I think we have had a discussion about this previously. I can see that within this thread we have at least three expectations of what happens within the source when a reflection arrives; absorbed, controled reflection, and acts like a short. Vigourus arguments are presented for each expectation, but who can measure what happens within an imaginary device? The seed for an endless argument!
--
73, Roger, W7WKB

The Rest of the Story
 

Keith Dysart wrote:
So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

"Work differently" is a loaded expression. Since both
lumped circuit and distributed network models exist,
it is safe to say that the lumped circuit model and
the distributed network model indeed "work differently".
If they didn't "work differently", there would be no
need for both of them to exist.

Your previous error is obvious. You were using the
lumped circuit model on the left side of Rs and
using the distributed network model on the right
side of Rs. If it is necessary to use the distributed
network model for part of the network, then it is
absolutely necessary to be consistent for all of
the network.

When you switched to the lumped circuit model on
the right side of Rs, the energies balanced. When
you switch to the distributed network model on
the left of Rs, the energies will also balance.

The lumped circuit model is a subset of the distributed
network model. See: http://www.ttr.com/corum/ and
http://www.ttr.com/TELSIKS2001-MASTER-1.pdf

Here's some quotes: "Lumped circuit theory fails because
it's a *theory* whose presuppositions are inadequate.
Every EE in the world was warned of this in their first
sophomore circuits course. ... Lumped circuit theory
isn't absolute truth, it's only an analytical *theory*.
.... Distributed theory encompasses lumped circuits and
always applies."

Where does the energy being absorbed by these ideal voltage
sources go?

0+j0 ohms cannot absorb energy. As Eugene Hecht said:
"If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

You seem to have discovered that "limited utility".
You have an instantaneous power calculated over an
infinitesimally small amount of time being dissipated
or stored in an impedance of 0+j0. Compared to that
assertion, the Virgin Birth seems pretty tame. :-)
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
This is much too arbitrary for me.

It's not arbitrary at all - it's the result of very
carefully chosen boundary conditions including specifying
a 100% 50 ohm system with ideal components. It agrees with
Roy's posting about ideal source impedances and short-circuits.

The bottom line is: If the distributed network model
is used to the right of Rs, it must also be used to
the left of Rs. Once the distributed network principles
are applied to the source, the results are the posted values.
Since no other values are possible, it is certainly not
arbitrary.

What *is* arbitrary is willy-nilly using the distributed
network model for part of the network and using the
lumped circuit model for the other part.

I have updated the diagram to the values that I calculated
for the shorted 45 degree line.

100v 50ohm 45deg
GND--------Vs----------Rs--------50ohm----------short
25w-- 125w-- 50w--
--25w --25w --50w

The net voltages and currents are the result of the
superposition of the component voltages and currents.
The forward and reflected power readings are what
an ideal directional wattmeter would indicate.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
I can see that within this thread we have at least
three expectations of what happens within the source
when a reflection arrives; absorbed, controled
reflection, and acts like a short.

Since a short causes a reflection, it seems that
(2) and (3) are essentially the same thing and
since 0+j0 cannot absorb energy and always
reflects energy, seems (1) can be discarded as
an over-simplification, something for which the
lumped circuit model is famous. For instance:

Given a loading coil installed in a traveling
wave antenna - reckon why EZNEC gives completely
different delays through the inductance when it
is modeled as a lumped inductance vs when it is
modeled using EZNEC's helix feature?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:

Perhaps it would help clarify the thinking to plot some voltage-
current
curves.

If we plot V versus I for a resistor (with V on the vertical access)
we
get a line with a slope equal to R. This line passes through the
origin.

For a short, the line is horizontal (i.e. slope and R are zero) and
for
an open the line is vertical (i.e. slope and R are infinite).

Now plot the V-I characteristic of a resistor in series with an ideal
voltage source. Again it is a line with a slope equal to R, but it
does
not pass through the origin, it crosses the vertical axis at the
voltage
provided by the source. So the y-axis crossing is controlled by the
voltage source and the slope is controlled by the resistor. If you
reduce the resistor to zero, you get a horizontal line crossing
the y-axis at the voltage of the source. The line being horizontal
means that no amount of current will change the voltage.

We often talk of resistance as V/I, but there are many situations in
which it is better to think of it as deltaV/deltaI (or, in the limit,
dV/dI); that is, the change in voltage that accompanies a change
in current. This is exactly the slope of the V/I curve at that point
and works for computing resistance regardless of whether there is a
voltage offset present. . .

Anyone interested in learning more about this and its application can
look up "dynamic resistance" on the web or in an appropriate text.

Roy Lewallen, W7EL

The Rest of the Story
 

Roger Sparks wrote:

I think we have had a discussion about this previously. I can see
that within this thread we have at least three expectations of what
happens within the source when a reflection arrives; absorbed,
controled reflection, and acts like a short. Vigourus arguments are
presented for each expectation, but who can measure what happens
within an imaginary device? The seed for an endless argument!

We can calculate the voltage and current of an imaginary resistance,
imaginary capacitance, imaginary inductance, imaginary
current-controlled voltage source, imaginary ideal transformer, and
ideal lossless transmission line. And we can calculate them for a
perfect voltage source, too. From the voltage and current, we know the
power, and from its sign the direction of energy flow. Certainly we
can't know what happens to the energy which goes into the source. But
why should we care? I can tell you, using the fact that an ideal voltage
source reflects like a short circuit, the voltage and current in every
forward and reflected voltage and current wave from turn-on to steady
state of a transmission line/source/load system. At all times, the
voltage across the ideal source will be invariant, as it must be,
regardless of the current into and out of it. And I can show exactly
where every erg of energy is at every instant in any component and at
every point along the transmission line. I submit that any analysis
technique which can't do this without knowing what happens to energy
entering the source is inferior.

Roy Lewallen, W7EL

The Rest of the Story
 

Roy Lewallen wrote:
Anyone interested in learning more about this and its application can
look up "dynamic resistance" on the web or in an appropriate text.

Another name for "dynamic resistance" is "virtual resistance".
It is an *EFFECT* of superposition, not a cause of anything.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roy Lewallen wrote:
And I can show exactly
where every erg of energy is at every instant in any component and at
every point along the transmission line. I submit that any analysis
technique which can't do this without knowing what happens to energy
entering the source is inferior.

I nominate this for "Quote of the year", by acclamation.
Anyone who responds to the question: Where does the energy
go? - with "We do not know and we do not care." does not
deserve to be in the discussion.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 18 Apr 2008 10:57:24 -0500
Cecil Moore wrote:

Roger Sparks wrote:
This is much too arbitrary for me.

It's not arbitrary at all - it's the result of very
carefully chosen boundary conditions including specifying
a 100% 50 ohm system with ideal components. It agrees with
Roy's posting about ideal source impedances and short-circuits.

It is arbitrary because when using the short circuit model, you limit the discussion to only one example of transmission line termination--the example of a low impedance on one end of the line and a higher impedance on the other. As you well know, two additional examples of transmission line termination are possible--the transmission line impedance is higher (or lower) than the termination at either end.

It is also arbitrary because the reflected wave information is useless. Any reflection from the source is folded into the sine wave to form one wave, which becomes the forward wave. The only real accomplishment from the exercise to to shift the frequency for a one cycle "bump", and that accomplishment does more to introduce confusion than contribute to better understanding.

--
73, Roger, W7WKB

The Rest of the Story
 

On Apr 18, 10:48*am, Cecil Moore wrote:
Keith Dysart wrote:
So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

"Work differently" is a loaded expression. Since both
lumped circuit and distributed network models exist,
it is safe to say that the lumped circuit model and
the distributed network model indeed "work differently".
If they didn't "work differently", there would be no
need for both of them to exist.

Your previous error is obvious. You were using the
lumped circuit model on the left side of Rs and
using the distributed network model on the right
side of Rs. If it is necessary to use the distributed
network model for part of the network, then it is
absolutely necessary to be consistent for all of
the network.

When you switched to the lumped circuit model on
the right side of Rs, the energies balanced. When
you switch to the distributed network model on
the left of Rs, the energies will also balance.

The lumped circuit model is a subset of the distributed
network model. See:http://www.ttr.com/corum/andhttp://w...1-MASTER-1.pdf

Here's some quotes: "Lumped circuit theory fails because
it's a *theory* whose presuppositions are inadequate.
Every EE in the world was warned of this in their first
sophomore circuits course. ... Lumped circuit theory
isn't absolute truth, it's only an analytical *theory*.
... Distributed theory encompasses lumped circuits and
always applies."

It is a good thing I checked the original references, otherwise
I would have had to assign Corum and Corum immediately to the
flake bucket where they could join some of the other contendors
on this group. But no, it turns out they have the appropriate
qualifications on all their statements about when it is
appropriate to use a lumped analysis and when it is not. And
when is lumped okay, when the physical dimensions of the
elements can be measured in small fractions of a wavelength.
Nothing new there, most of us know that.

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

Firstly, there are no inductors to cause any sort of difficulty.
Secondly, it is constructed of ideal components, which have the
luxury of being infinitely small.
And thirdly, it is a DC circuit so the two points above do not
matter any way.

So we are back to the question you keep dodging...

Where does the energy being absorbed by these ideal voltage
sources go?

0+j0 ohms cannot absorb energy.

Now that is a non-sequitor. The element absorbing energy is an
ideal voltage source, not a resistor.

As Eugene Hecht said:
"If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

It is a DC circuit. This means the instantaneous value is
the average value.

(But poor Hecht, here he is saying power is more useful than
cumulative energy, and you misinterpret him to be comparing
instantaneous to average. And is it the distribution over
the area that is being averaged, or the distribution over
time?)

You seem to have discovered that "limited utility".
You have an instantaneous power calculated over an
infinitesimally small amount of time being dissipated
or stored in an impedance of 0+j0. Compared to that
assertion, the Virgin Birth seems pretty tame. :-)

Again, it is a DC circuit since we have had to go back
to learning the fundamentals of ideal voltage sources.

But back to the simple question...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

...Keith

The Rest of the Story
 

On Apr 18, 4:23*pm, Cecil Moore wrote:
Roy Lewallen wrote:
Anyone interested in learning more about this and its application can
look up "dynamic resistance" on the web or in an appropriate text.

Another name for "dynamic resistance" is "virtual resistance".
It is an *EFFECT* of superposition, not a cause of anything.

Are you suggesting that the negative "dynamic resistance" of a
tunnel diode is an *EFFECT* of superposition?

Please feel free to expand on this claim.

...Keith

The Rest of the Story
 

Roger Sparks wrote:
It is arbitrary because when using the short circuit
model, you limit the discussion to only one example of
transmission line termination ...

It is not arbitrary - it is what it is. All of the
source impedance, Rs, is separated from Vs. That
leaves only a 0+j0 dead short impedance possible
for the series source. If it was some other value,
the distributed network model would still work.

Any reflection from the source is folded into the sine
wave to form one wave, which becomes the forward wave.

Yes, in exact accordance with the distributed network
model. What it means is that the source is not delivering
the folded-in reflected energy at the time the reflected
energy joins the source signal. Since I don't see that
energy in Keith's equations, chances are that is why
they didn't balance.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
On Apr 18, 4:23 pm, Cecil Moore wrote:
Another name for "dynamic resistance" is "virtual resistance".
It is an *EFFECT* of superposition, not a cause of anything.

Are you suggesting that the negative "dynamic resistance" of a
tunnel diode is an *EFFECT* of superposition?

Sorry, I was mistaken about the definition of "dynamic"
in the context of physical components.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sat, 19 Apr 2008 13:02:08 GMT
Cecil Moore wrote:

Roger Sparks wrote:

Any reflection from the source is folded into the sine
wave to form one wave, which becomes the forward wave.

Yes, in exact accordance with the distributed network
model. What it means is that the source is not delivering
the folded-in reflected energy at the time the reflected
energy joins the source signal. Since I don't see that
energy in Keith's equations, chances are that is why
they didn't balance.
--

You are missing an important observation here. When the energy from the reflected wave folds into the forward wave, any further analysis is a replication of the first analysis. Nothing new is learned.

I think what you want to see is a source matched to the load, so that all the energy flows one way, to the resistor Rs. Your proposal is to allow/force reflectons from the source Vs so that effectively, the resistor Rs becomes the only load. This is then a demonstration/proof for what?

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
You are missing an important observation here. When the
energy from the reflected wave folds into the forward wave,
any further analysis is a replication of the first analysis.
Nothing new is learned.

I'm posting steady-state values. If it is already
steady-state, nothing new is needed. The point is
that some of the steady-state forward energy is
not being delivered by the source.

Your proposal is to allow/force reflectons from the source
Vs so that effectively, the resistor Rs becomes the only load.

No, the resistor Rs is not the only load. The resistor
plus the rest of the network is the load. With a 45 deg
shorted stub the load is 50+j50 as you previously
reported.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 19, 11:23*pm, Cecil Moore wrote:
Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?

Your explanations are predicated on a misunderstanding of
the behaviour of ideal voltage sources. Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

The discussion needs to digress to address this fundamental
misunderstanding since this misunderstanding renders all
else moot. And so...

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

And we are back to the question you keep dodging...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

Once it is agreed that ideal DC voltage sources can indeed
absorb energy, we can discuss the same for ideal AC voltage
sources. Once that is agreed, we can return to the
discussion of your 'interference free' circuit.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Where does the energy that is flowing into the ideal
DC voltage source on the right go?

I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

From the "IEEE Dictionary":

"Absorption: (2) A general term for the process by which
incident flux is converted to another form of energy,
usually and ultimately to heat. (4) The irreversible
conversion of the energy of an EM wave into another form
of energy as a result of wave interaction with matter."

By what mechanism does an ideal source with an impedance
of 0+j0 manage to dissipate heat? Since the energy absorption
by the ideal source is *irreversible*, where is the heat
(or other form of energy) stored and for how long?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 21, 1:23*am, Cecil Moore wrote:
Keith Dysart wrote:
Where does the energy that is flowing into the ideal
DC voltage source on the right go?

I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

...Keith

The Rest of the Story
 

On Apr 21, 3:25*am, Cecil Moore wrote:
Keith Dysart wrote:
Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

*From the "IEEE Dictionary":

"Absorption: (2) A general term for the process by which
incident flux is converted to another form of energy,
usually and ultimately to heat. (4) The irreversible
conversion of the energy of an EM wave into another form
of energy as a result of wave interaction with matter."

By what mechanism does an ideal source with an impedance
of 0+j0 manage to dissipate heat? Since the energy absorption
by the ideal source is *irreversible*, where is the heat
(or other form of energy) stored and for how long?

Just as ideal voltage sources can provide energy to a circuit,
they can also remove energy from a circuit.

Feel free to substitute the word of your choice for 'remove'.

Dissipate is not a good choice since it usually implies
conversion to heat.

Absorb is not a good word for you, since you can find absorption
in the IEEE dictionary and it also suggests conversion to heat.

The thesaurus (http://thesaurus.reference.com/ suggests 'consume',
'assimilate', 'digest', 'imbibe',
'take up', 'sop up', and 'devour'.

Pick the word that you find least confusing.

Recalling that an ideal voltage source can provide (deliver,
furnish, supply, transfer) energy to a circuit, we need a
non confusing word to describe the concept that an ideal
voltage source can also remove energy from a circuit.

A word that gives no hint about where this energy goes would
be best, since, just as we do not know where the energy
that an ideal voltage source delivers to a circuit comes
from, we do not know where the energy that an ideal voltage
source removes from a circuit goes.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

No, I can quit wasting my time thinking about it
because "we do not care". We not caring takes away
any reason or purpose for continuing the discussion.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

The discussion is moot unless you can prove that an
ideal source in a single-source system can absorb
AVERAGE power. So there's your challenge. If you
cannot do that, my article about average power
stands as written.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Feel free to substitute the word of your choice for 'remove'.

That's the first time you have used the word "remove".
Have you changed your mind about energy being "absorbed",
by the source, i.e. turned into heat?

Dissipate is not a good choice since it usually implies
conversion to heat.

Whoa there Keith, "absorb" is equally not a good choice
since it usually implies conversion to heat as in the IEEE
definitions. If the source only removes energy, then that
is a plus for my side of the argument. If the source has
the ability to remove the destructive interference and
supply it back 90 degrees later as constructive interference,
the entire mystery of where the reflected power goes is
solved. When I previously offered that as a solution, you
turned it down flat. Now you seem to be agreeing with it.

Absorb is not a good word for you, since you can find absorption
in the IEEE dictionary and it also suggests conversion to heat.

That's why I have been arguing loud and long against the
absorption of energy by the source. It would imply that
the source is heating up or has an infinite ability to
"irreversibly convert the energy of an EM wave into another
form of energy". That irreversible energy conversion is
what I have been objecting to. There is no way an impedance
of 0+j0 can cause an irreversible energy conversion.

A word that gives no hint about where this energy goes would
be best, ...

So you can sweep it under the rug and "not care where it
went"? As I said, further discussion is pointless. You
have a magic source that obeys your every whim. Why didn't
you just say that in the first place?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 21, 12:11*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

No, I can quit wasting my time thinking about it
because "we do not care". We not caring takes away
any reason or purpose for continuing the discussion.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

The discussion is moot unless you can prove that an
ideal source in a single-source system can absorb
AVERAGE power. So there's your challenge. If you
cannot do that, my article about average power
stands as written.

The best analogy I can think of is someone saying:
"Until you prove the earth is flat, I will not consider
any evidence that it is round".

Another question remains, since it is difficult to
discern from your writings: Have you grasped the
behaviour of an ideal voltage source when current
is flowing into the source?

You now understand that it removes energy from the
circuit?

If you have that for DC ideal voltage sources, we
can move on to discover what happens with AC ideal
voltage sources. After that, we can go back to what
is happening in your circuit.

...Keith

The Rest of the Story
 

On Apr 21, 12:26*pm, Cecil Moore wrote:
Keith Dysart wrote:
Feel free to substitute the word of your choice for 'remove'.

That's the first time you have used the word "remove".

You need to read more carefully. Not the first time at all.

Have you changed your mind about energy being "absorbed",
by the source, i.e. turned into heat?

What I have said is that an ideal voltage source removes energy
from a circuit and that we do not know what it does with the
energy it removes. In practice, devices which are designed to
approximate ideal voltage sources do simply dissipate the energy
they remove from the circuit.

But that is not part of the definition of an ideal voltage source,
for which no statement about where the energy removed goes is
made.

Dissipate is not a good choice since it usually implies
conversion to heat.

Whoa there Keith, "absorb" is equally not a good choice
since it usually implies conversion to heat as in the IEEE
definitions. If the source only removes energy, then that
is a plus for my side of the argument. If the source has
the ability to remove the destructive interference and
supply it back 90 degrees later as constructive interference,
the entire mystery of where the reflected power goes is
solved. When I previously offered that as a solution, you
turned it down flat. Now you seem to be agreeing with it.

No. We do not know what an ideal voltage source does with
energy it removes. We can not say that it stores and then
returns it, though a particular implementation might do
so. Another implementation might not. So this can not be
used as an explanation.

Absorb is not a good word for you, since you can find absorption
in the IEEE dictionary and it also suggests conversion to heat.

That's why I have been arguing loud and long against the
absorption of energy by the source. It would imply that
the source is heating up or has an infinite ability to
"irreversibly convert the energy of an EM wave into another
form of energy". That irreversible energy conversion is
what I have been objecting to. There is no way an impedance
of 0+j0 can cause an irreversible energy conversion.

I am sorry that the occasional use of the word 'absorb' so
mislead you. I avoided 'dissipate' for that reason. It is
not so obvious why you were mislead by the use of 'remove'.

A word that gives no hint about where this energy goes would
be best, ...

So you can sweep it under the rug and "not care where it
went"?

No. Because the definition of an ideal voltage does not specify
where the energy goes. Therefore we had better not care when
we use an ideal voltage source.

As I said, further discussion is pointless. You
have a magic source that obeys your every whim. Why didn't
you just say that in the first place?

No. My ideal voltage source just obeys the definition of an
ideal voltage source. It provides energy to the circuit when
the circuit conditions demand that it do so and similarly it
removes energy when the circuit conditions demand that it do
so. The definition does not tell us where the energy it
provides comes from, nor does it tell us where the energy
it removes goes.

A fairly simple defintion: The voltage at the terminals is
maintained at the desired value, regardless of the current
flow needed to do so.

No magic in the definition whatsoever. And no need to obey
whims.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Another question remains, since it is difficult to
discern from your writings: Have you grasped the
behaviour of an ideal voltage source when current
is flowing into the source?

You now understand that it removes energy from the
circuit?

It follows that it is futile to try to track any
movement of instantaneous energy. You have convinced
me that you are correct - "We don't care where the
(instantaneous) energy goes."

If you have that for DC ideal voltage sources, we
can move on to discover what happens with AC ideal
voltage sources. After that, we can go back to what
is happening in your circuit.

Since there is an instantaneous leak in the closed
system, it is useless to proceed. You say you don't
care what happens to the energy. I said a couple of
months ago that I didn't care what happens to
instantaneous power. And indeed, you have convinced
me that any attempt to track instantaneous power is
doomed to failure.

My part 1 article based on a single source and
*AVERAGE* powers doesn't have those conceptual
problems and stands as written. Here's the second
paragraph from that article:

"Please note that any power referred to in this paper
is an AVERAGE POWER. Instantaneous power is beyond the
scope of this article, irrelevant to the following
discussion, and "of limited utility" according to
Eugene Hecht. [4]

Your challenge is to prove that a single source
removes an average amount of energy from the
network.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
What I have said is that an ideal voltage source removes energy
from a circuit ...

Sorry, you specifically said that an ideal voltage source
"absorbs" energy, i.e. irreversibly converts energy to
another form, the most common form of which is heat.

I am sorry that the occasional use of the word 'absorb' so
mislead you. I avoided 'dissipate' for that reason.

The IEEE Dictionary says that, in this context, "absorb"
and "dissipate" are virtual synonyms.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Keith Dysart wrote:
What I have said is that an ideal voltage source removes energy
from a circuit ...

Sorry, you specifically said that an ideal voltage source
"absorbs" energy, i.e. irreversibly converts energy to
another form, the most common form of which is heat.

In case you have forgotten :-), here is what you posted
over the past few days:

When it is sinking current, it is *absorbing* energy.

You will find that for some of the time
energy is being *absorbed* by the source.

But we do know that when the sign of Ps(t) is
negative, the source is *absorbing* energy from
the system,

When current flows into a voltage source, the
voltage source is *absorbing* energy.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

This certainly implies that you consider the dissipation
of "absorbed" energy to be a distinct probability.

The source provides or *absorbs* energy.

When 1.5 amps is flowing into the positive terminal,
the ideal voltage source is *absorbing* 15 joules per
second from the circuit.

The ideal voltage source on the right is *absorbing*
5 joules/second from the circuit.

An ideal source provides or *absorbs* energy to
satisfy its basic function which is to hold the
voltage across its terminals at the desired value.

When it is providing energy we do not know where
this energy comes from and when it is *absorbing*
energy we do not know where this energy goes.

But was this because you have learned that you were
in error and now better understand the behaviour of
sources when they are *absorbing* energy?

The ideal voltage source on the right, after the
circuits settle, will be absorbing 50 joules/s
in both cases.

Where does the energy being *absorbed* by these ideal
voltage sources go?

The element *absorbing* energy is an ideal voltage
source, not a resistor.

Despite your protests to the contrary, ideal voltage
sources can, and do, *absorb* energy.

You know, Keith, if you were just ethical enough to
answer my questions, I wouldn't have to treat you
this way.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 22, 8:11*am, Cecil Moore wrote:
Keith Dysart wrote:
Another question remains, since it is difficult to
discern from your writings: Have you grasped the
behaviour of an ideal voltage source when current
is flowing into the source?

You now understand that it removes energy from the
circuit?

It follows that it is futile to try to track any
movement of instantaneous energy. You have convinced
me that you are correct - "We don't care where the
(instantaneous) energy goes."

The wordy non-sequitor leaves us with nagging doubts about
whether the fundamentals of DC ideal voltage sources have
been grasped.

If you have that for DC ideal voltage sources, we
can move on to discover what happens with AC ideal
voltage sources. After that, we can go back to what
is happening in your circuit.

Since there is an instantaneous leak in the closed
system, it is useless to proceed.

Doubts now confirmed.

You say you don't
care what happens to the energy. I said a couple of
months ago that I didn't care what happens to
instantaneous power. And indeed, you have convinced
me that any attempt to track instantaneous power is
doomed to failure.

My part 1 article based on a single source and
*AVERAGE* powers doesn't have those conceptual
problems and stands as written. Here's the second
paragraph from that article:

"Please note that any power referred to in this paper
is an AVERAGE POWER. Instantaneous power is beyond the
scope of this article, irrelevant to the following
discussion, and "of limited utility" according to
Eugene Hecht. [4]

Your challenge is to prove that a single source
removes an average amount of energy from the
network.

But since the whole work is predicated on a misunderstanding
of the behaviour of ideal voltage sources, any conclusions
retain no value.

...Keith

The Rest of the Story
 

On Apr 22, 8:18*am, Cecil Moore wrote:
Keith Dysart wrote:
What I have said is that an ideal voltage source removes energy
from a circuit ...

Sorry, you specifically said that an ideal voltage source
"absorbs" energy, i.e. irreversibly converts energy to
another form, the most common form of which is heat.

I did use the word 'absorb' to convey the concept of removing
energy from a circuit without knowing where the energy went.
You misinterpreted my use of 'absorb' to mean something else.

These things happen. Most readers, upon learning the intended
meaning of a word was different than expected, would simply
re-read the affected passages to glean the intended meaning.

This seems difficult for you. May I suggest that substituting
'remove' for 'absorb' might help you gain an understanding
of the intent.

I am sorry that the occasional use of the word 'absorb' so
mislead you. I avoided 'dissipate' for that reason.

The IEEE Dictionary says that, in this context, "absorb"
and "dissipate" are virtual synonyms.

Don't worry. Knowing your difficulty with detecting the
intended meaning of words, in future I shall avoid 'absorb'
when I mean "removing energy from a circuit without knowing
where the energy goes."

...Keith

The Rest of the Story
 

On Apr 22, 1:12*pm, Cecil Moore wrote:
Cecil Moore wrote:
Keith Dysart wrote:
What I have said is that an ideal voltage source removes energy
from a circuit ...

Sorry, you specifically said that an ideal voltage source
"absorbs" energy, i.e. irreversibly converts energy to
another form, the most common form of which is heat.

In case you have forgotten :-), here is what you posted
over the past few days:

With your difficulty using the word 'absorb' to represent the
abstract concept of removing energy without knowing where it
goes, for a better understanding please re-read these passages
substituting 'remove' for 'absorb'.

When it is sinking current, it is *absorbing* energy.

You will find that for some of the time
energy is being *absorbed* by the source.

But we do know that when the sign of Ps(t) is
negative, the source is *absorbing* energy from
the system,

When current flows into a voltage source, the
voltage source is *absorbing* energy.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

This certainly implies that you consider the dissipation
of "absorbed" energy to be a distinct probability.

It certainly is one of the many possible things that might
happen to the energy that is removed. As I wrote previously,
"In practice, devices which are designed to approximate
ideal voltage sources do simply dissipate the energy they
remove from the circuit."

Still, since we do not *know* what happens to the energy removed
by an ideal voltage source, we can make no assumptions.

The source provides or *absorbs* energy.

When 1.5 amps is flowing into the positive terminal,
the ideal voltage source is *absorbing* 15 joules per
second from the circuit.

The ideal voltage source on the right is *absorbing*
5 joules/second from the circuit.

An ideal source provides or *absorbs* energy to
satisfy its basic function which is to hold the
voltage across its terminals at the desired value.

When it is providing energy we do not know where
this energy comes from and when it is *absorbing*
energy we do not know where this energy goes.

But was this because you have learned that you were
in error and now better understand the behaviour of
sources when they are *absorbing* energy?

The ideal voltage source on the right, after the
circuits settle, will be absorbing 50 joules/s
in both cases.

Where does the energy being *absorbed* by these ideal
voltage sources go?

The element *absorbing* energy is an ideal voltage
source, not a resistor.

Despite your protests to the contrary, ideal voltage
sources can, and do, *absorb* energy.

You know, Keith, if you were just ethical enough to
answer my questions, I wouldn't have to treat you
this way.

Perhaps. But I doubt that it would actually alter your
behaviours.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Your challenge is to prove that a single source
removes an average amount of energy from the
network.

But since the whole work is predicated on a misunderstanding
of the behaviour of ideal voltage sources, any conclusions
retain no value.

Keith, if you can prove that a single source removes
an average amount of energy from the network during
steady-state conditions, I will worship at your feet.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
I did use the word 'absorb' to convey the concept of removing
energy from a circuit without knowing where the energy went.
You misinterpreted my use of 'absorb' to mean something else.

No, I assumed the standard IEEE Definition of the word.
You mistakenly neglected to state the non-standard definition
that you were using.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Perhaps. But I doubt that it would actually alter your
behaviours.

Well, let's try one more time and see. Assuming a 50 ohm
environment (by definition) in the following example, what
will an ideal 50 ohm directional wattmeter indicate at
point 'x' for forward power (Pf1) and reverse power (Pr1)?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----short
| --Pr1 Rs --Pr2=50w
| 100w
Vs=100v
|
Gnd

It's a simple question with easily calculated and
quantifiable answers.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Wed, 23 Apr 2008 07:58:49 -0500
Cecil Moore wrote:



Well, let's try one more time and see. Assuming a 50 ohm
environment (by definition) in the following example, what
will an ideal 50 ohm directional wattmeter indicate at
point 'x' for forward power (Pf1) and reverse power (Pr1)?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----short
| --Pr1 Rs --Pr2=50w
| 100w
Vs=100v
|
Gnd

It's a simple question with easily calculated and
quantifiable answers.

Your circuit is incomplete. The circuit parameters are not fully defined.

Is this the same circuit that you wish to analyze?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----+
| --Pr1 Rs --Pr2=50w |
| 100w |
Vs=100v |
| |
Gnd Gnd

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

What reflection characteristics should we assume for the voltage source Vs?
--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----+
| --Pr1 Rs --Pr2=50w |
| 100w |
Vs=100v |
| |
Gnd--------------------------Braid------------+

The coax is shorted at the end, and Gnd is 45
degrees away from the short.

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

Assume that Rs is 100% of the source impedance.

What reflection characteristics should we assume for the voltage source Vs?

Assuming that Rs is 100% of the source impedance
should answer that question.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 23, 12:03*pm, Cecil Moore wrote:
Roger Sparks wrote:
* * * Pf1-- * 50 ohm * * Pf2=50w--
* *+----x-----/\/\/\/\-------45 deg 50 ohm-----+
* *| *--Pr1 * * Rs * * * --Pr2=50w * * * * * |
* *| * * * * * *100w * * * * * * * * * * * * * |
*Vs=100v * * * * * * * * * * * * * * * * * * * |
* *| * * * * * * * * * * * * * * * * * * * * * |
*Gnd--------------------------Braid------------+

The coax is shorted at the end, and Gnd is 45
degrees away from the short.

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

Assume that Rs is 100% of the source impedance.

What reflection characteristics should we assume for the voltage source Vs?

Assuming that Rs is 100% of the source impedance
should answer that question.

Is this a standard ideal voltage source which provides energy
to the circuit when the product of its voltage and current is
positive and removes energy from the circuit when the product
of its voltage and current is negative?

Or have you assigned it some other non-standard definition?

...Keith