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:
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

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

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