Gene Fuller wrote:
Cecil Moore wrote:
This is steady-state after the wave interaction. What
you guys don't seem to realize is that s11(a1) and
s12(a2) are continually changing, continually
interacting, and a1 & a2 are rotating phasors
changing with time.

Utter nonsense. Any setup that includes t0 or t=0 is not steady state.

Gene, you obviously misunderstood what I said. There is
no t0 or t=0 in my above statement. It is just that the
delta-t doesn't change from the transient state to the
steady-state.

In an s-parameter analysis:
The normalized voltage, a1, equals Vi1/SQRT(Z0) where
'i' stands for incident voltage. a2 equals Vi2/SQRT(Z0).
These voltages are normally represented in phasor
notation but they can just as easily be represented
in exponential notation where a1 = Vi1*e^jwt+X and
a2 = Vi2*e^jwt+Y, where X and Y are constant phase
angles. Thus, the s-parameter equation becomes:

b1 = Vi1[cos(wt+X)]/SQRT(Z0) + Vi2[cos(wt+Y)]/SQRT(Z0) = 0

adding the delta-t "tick" gives:

b1 = Vi1{cos[w(t+delta-t)+X]}/SQRT(Z0) +
Vi2{cos[w(t+delta-t)+Y]}/SQRT(Z0) = 0

Vi1 obviously has to be an equal magnitude to Vi2 and
X and Y obviously have to be 180 degrees apart. Given
that, for every delta-t "tick" of time, the two real
normalized voltages sum to zero. The square of the
normalized RMS value gives average power in each
wave. The equation can be turned into a differential
equation by having delta-t approach zero in the limit.
--
73, Cecil http://www.w5dxp.com