So what do YOU get? As I recall, it comes out to a voltage equivalent
to 112.5W for Pf2, and 12.5W for Pr1. Maybe I can head you off at the
pass, and give you the values for further rounds of the echo:
round Pf2 Pr1
1 112.5 12.5
2 153.125 3.125
3 175.78125 0.78125
4 187.69531... 0.19531...
5 193.79883... 0.04883...
6 196.88721... 0.01221...
7 198.44055... 0.00305...
8 199.21951... 0.00076...
9 199.60957... 0.00019...
10 199.80474... 0.00005...

Don't see anything worth discussing about it at the moment.


Cheers,
Tom

(Cecil Moore) wrote in message . com...
(Tom Bruhns) wrote in message om...
For instantaneous values,
Vf2 = (2*Z2*Vf1+(Z1-Z2)*Vr2)/(Z1+Z2)
Vr1 = ((Z2-Z1)*Vf1+2*Z1*Vr2)/(Z1+Z2)

My ISP doesn't seem to be making it lately so I am posting from Google.
Your equation represents Vf1 times the transmission coefficient plus
Vr2 times the reflection coefficient. This is perfectly logical but
IMO doesn't represent what is happening in reality. I hope I am wrong.
Re-arranging the terms:

Vf2 = 2*Z2*Vf1/(Z1+Z2) + Vr2(Z1-Z2)/(Z1+Z2)

For the example presented, I don't think there is enough interference
to support the above superposition. Again, I hope I am wrong but please
bear with me while you prove I am wrong. If I am wrong, it will resolve
something I have been wrestling with for three years.

Please calculate the two voltages, superpose them, and calculate the
resulting Pfwd2 forward power and then let's discuss the results.