On Mar 6, 11:12*am, Cecil Moore wrote:
Keith Dysart wrote:
So using the same 12.5 ohm example,
X.inst = 50 + 50 cos(2wt)
Y.inst = 18 + 18 cos(2wt)
X.inst + Y.inst = 68 + 68 cos(2wt)

Where do you take into account that the forward wave
is 90 degrees out of phase with the reflected wave?

Z.inst = 68 + 68 cos(2wt - 61.9degrees)

Where does the 61.9 degrees come from?

The computation of all of these can be seen in the spreadsheet at
http://keith.dysart.googlepages.com/...d%2Creflection

Recall that: cos(a)cos(b) = 0.5( cos(a+b) + cos(a-b) )

The power in the resistor is computed by multiplying the voltage
Vrs(t) = 82.46 cos(wt -30.96 degrees)
by the current
Irs(t) = 1.649 cos(wt -30.96 degrees)
yielding
Prs(t) = 68 + 68 cos(2wt -61.92 degrees)

For reflected power at the generator
Vr.g(t) = 42.42 cos(wt +90 degrees)
Ir.g(t) = 0.8485 cos(wt -90 degrees)
Pr.g(t) = Vr.g(t) * Ir.g(t)
= -18 + 18 cos(2wt)

Note that I previously made a transcription error in the sign
of one of the terms.

So X.inst + Y.inst is actually 68 + 32 cos(2wt).

...Keith