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