On Wed, 16 Apr 2008 13:36:05 -0500
Cecil Moore wrote:

Roger Sparks wrote:
Cecil Moore wrote:
That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

How can a device with a zero impedance, i.e. zero resistance,
zero capacitance, and zero inductance, absorb energy? We can
certainly allow it to magically absorb energy but of what
use is that?

It gives a tool to check our work inside the system. We think we know what happens once the energy arrives into the known components, but we don't know what exactly happens in the voltage source itself. We solve that by equating "energy in equals energy out". Then we refine that equation to "energy in equals energy returned plus energy disipated plus energy stored during some time period". Ein = Er + Ed + Es

This is the "conservation of energy" principle at work.

A third way is to allow the energy to be stored in constructive
and destructive interference some place in the system.

"A third way is to allow the energy to be stored in constructive
and destructive interference SOME PLACE IN THE SYSTEM." In this quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need to understand where interference is located. If interference is located within the voltage source, it is really located *OUTSIDE* the system because we do not completely understand the voltage source itself.

As you know, this is the one I prefer. Another thing that
neither you nor Keith has done is to account for the
reverse-flowing energy through the source. I suspect if
that was done, every thimbleful of energy would be
accounted for. So far, net energy calculations have been
used on one side of Rs and component energy calculations
on the other. That would work only if Rs was not dissipating
power.

Are you expecting additional reflections? Any additonal reflections from the source would only be mechanical copies of the reflection at the short, and would add or subtract to the forward wave following the rules of sine wave addition.

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.

--
73, Roger, W7WKB

Roger Sparks wrote:
"A third way is to allow the energy to be stored in constructive
and destructive interference SOME PLACE IN THE SYSTEM." In this
quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need
to understand where interference is located.

Actually, all we need to realize is that, in a
distributed network, the current through the source
is NOT equal to the current being provided by the
source. It is the source current superposed with
reflected current and re-reflected current. The fact
that the *NET* power is negative is NOT because the
source is sinking power. It is caused by the other
"sources" of energy in the system, i.e. the energy
in the reflected waves.

Are you expecting additional reflections?

Once steady-state is reached, the magnitude of the
reflections are also steady-state. The total energy
flowing toward the load is the forward energy. The
total energy flowing in the other direction is the
reflected energy. Both of those magnitudes are fixed
during steady-state.

I have asked multiple times that this simple mental
experiment be performed and my request has been ignored.

Please install an ideal 50 ohm directional wattmeter
at point 'x' and tell us what are those readings
for forward power and reflected power.

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

Once the forward and reverse energy flows at point
'x' are understood, it will become clear that all
of the net current through the source is NOT being
provided by the source. That the source is sinking
energy is an illusion caused by reflections.
--
73, Cecil http://www.w5dxp.com

On Thu, 17 Apr 2008 09:46:55 -0500
Cecil Moore wrote:

Roger Sparks wrote:
"A third way is to allow the energy to be stored in constructive
and destructive interference SOME PLACE IN THE SYSTEM." In this
quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need
to understand where interference is located.

Actually, all we need to realize is that, in a
distributed network, the current through the source
is NOT equal to the current being provided by the
source. It is the source current superposed with
reflected current and re-reflected current. The fact
that the *NET* power is negative is NOT because the
source is sinking power. It is caused by the other
"sources" of energy in the system, i.e. the energy
in the reflected waves.

Yes, in nature the returning power will affect the frequency of the source. The ideal voltage source absorbs the reflected power rather than allowing a change in frequency.

Are you expecting additional reflections?

Once steady-state is reached, the magnitude of the
reflections are also steady-state. The total energy
flowing toward the load is the forward energy. The
total energy flowing in the other direction is the
reflected energy. Both of those magnitudes are fixed
during steady-state.

I have asked multiple times that this simple mental
experiment be performed and my request has been ignored.

Please install an ideal 50 ohm directional wattmeter
at point 'x' and tell us what are those readings
for forward power and reflected power.

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

Once the forward and reverse energy flows at point
'x' are understood, it will become clear that all
of the net current through the source is NOT being
provided by the source. That the source is sinking
energy is an illusion caused by reflections.

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.

Think of things this way. Any signal must contain energy. A signal without energy is not a signal. Nature must have some method of signaling between source and load when the load does not match the impedance of power flow. This is simple logic based on the known fact that the apparent impedance of a transmission line depends upon the load at the end of the transmission line.

Power returning to the source is the signal nature requires to throttle the output of the source. Power returning to the source will change the frequency of the source in nature, but in our ideal voltage source, nature is suspended and the frequency is held constant. Our problem as engineers is to isolate each effect and to make valid predictions so that we can build more intelligently.
--
73, Roger, W7WKB

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?

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.

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?

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.
--
73, Cecil http://www.w5dxp.com

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

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

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

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

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

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