On Jun 12, 2:37*pm, Cecil Moore wrote:
On Jun 12, 9:00*am, K1TTT wrote:

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.