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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

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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

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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

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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

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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

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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

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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

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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

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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

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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