On Feb 5, 8:53*am, Roger Sparks wrote:
On Mon, 4 Feb 2008 12:32:39 -0800 (PST)
Keith Dysart wrote:
[snip]
The voltage at the generator is, therefore,

Vg = Vf.g + Vr.g
* *= 50 /_ 0d + 25 /_ (-45d -45d)
* *= 50 /_ 0d + 25 /_ -90d
* *= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

Whoa! *You started with 50 volts applied to the forward line, then froze the wave by finding the time phase of the wave. *Mathematically, you want to find the voltage at the generator which is the sum of the terms Vf.g and Vr.g. *The sum of those terms is 50 + 0 = 50, not *55.902 /_ -26.565d..

Vr.g is the reflection of Vf from the load so it is
a delayed version of Vr.l, the delay being another
45 degrees.

The reflected voltage at the load is
25 /_ -45d

Another 45 degrees makes it
25 /_ -90d
by the time it gets back to the generator terminal.

So the total voltage at the generator is
Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-45d -45d)
= 50 /_ 0d + 25 /_ -90d
= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

[snip]

Similarly for Ig and Ig(t) and Pg.avg and Pg(t).

[snip]

With the circuit and conditions presented, the reflected voltage and current return out of phase in time with the source voltage and current. *A standing wave condition of 3:1 would always exist throughout the system.

Agreed, though it might be slightly more correct to say
that there is a 3:1 VSWR on the transmission line.

...Keith