On Mar 21, 12:53*pm, Cecil Moore wrote:
Keith Dysart wrote:
In my terms, this leads to
* Vrs(t) = Vs(t) - Vg(t)
* * * * *= Vs(t) - Vf.g(t) - Vr.g(t)

How about expanding those equations for us?

Vs(t) = 141.4*cos(wt) ????
Vg(t) = ____*cos(wt+/-____) ????
Vf.g(t) = ____*cos(wt+/-____) ????
Vr.g(t) = ____*cos(wt+/-____) ????

If you ever did this before, I missed it.

I did. And they are also conveniently in the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection

For your convenience, in the example of Fig 1-1, 100 Vrms
sinusoidal source, 50 ohm source resistor, 45 degrees of
50 ohm line, 12.5 ohm load, after the reflection returns...

Vs(t) = 141.42135623731*cos(wt)
Vg(t) = 82.46211251*cos(wt+30.96375653)
Vf.g(t) = 70.71067812*cos(wt)
Vr.g(t) = 42.42640687*cos(wt+90)

And for completeness...

Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Given the correct voltage equations, I can
prove what I am saying about destructive and
constructive interference averaging out to zero
over one cycle is a fact.

Not that anyone has disputed that. But it would
be good to see anyway.

...Keith