Keith Dysart wrote:
So forward and reverse powers each converge to 100 W and 200 W is
dissipated in the source resistor, which suggests that your
equation for computing dissipation in the source resistor is
incorrect.

I remembered what I did when I used that equation. It is
the same technique that Dr. Best used in his article. If
we add 1WL of 25 ohm line to the example, we haven't
changed any steady-state conditions but we have made the
example a lot easier to understand.

Rs 50w-- 100w--
+--/\/\/--+------------------------+------------+
| 25 ohm --50w --100w |
| |
Vs 1WL 1/2WL |Short
70.7v 25 ohm 50 ohm |
| |
+---------+------------------------+------------+

Now the forward power at the source terminals is 50w and
the reflected power at the source terminals is 50w. So the
ratio of Rs/Z0 at the source terminals is 1.0. Therefo

P(Rs) = 50w + 50w + 2*SQRT(50w*50w) = 200w

To take the tedium out of these calculation you might like the
Excel spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

My firewall/virus protection will not allow me to download
EXCEL files with macros. Apparently, it is super easy to
embed a virus or worm in EXCEL macros.
--
73, Cecil http://www.w5dxp.com