On Fri, 21 Mar 2008 17:08:39 -0700 (PDT)
Keith Dysart wrote:

On Mar 21, 9:12*am, Roger Sparks wrote:
On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)

Keith Dysart wrote:
On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

For the first 90 degrees of time, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * /
* * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor
* * * *100v RMS * * * * * * * * * * * * * * * * */
* * * * * *| * * * * * * * * * * * * * * * * * * \
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

After 90 degrees of time has passed, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * Vs * * * * * * * * * * * * * * * * * *---
* * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd
HEAVILY CLIPPED
We differ on how to utilize the term Pg(t). *I think it describes the standing wave, not the voltage across Rs. *

I describe Vg(t) as the actual voltage that would be measured by an
oscilloscope at the point g; i.e. the beginning of the transmission
line. It is certainly not the voltage across Rs. Perhaps it IS what
you mean by standing wave. In any case the voltage across the resistor
is
Vs(t) = Vrs(t) + Vg(t)
Vrs(t) = Vs(t) - Vg(t)

Pg(t) is the power being delivered to the transmission line at any
instance t.
Pg(t) = Vg(t) * Ig(t)

...Keith

You gave a very thoughtful reply Keith. It looks to me like two perspectives, both right so far as the analysis has proceeded.
--
73, Roger, W7WKB