Jim Kelley wrote:

Cecil Moore wrote:
It is impossible for a "partially reflective surface"
to reflect 100% of the intensity.

But that's wrong. If it was right, then a partially reflective surface
couldn't be used to prevent reflections either.

That's faulty logic born out of ignorance.
Assume s11 = 0.707 in the S-Parameter reflected
voltage equation. a1 is the normalized forward
voltage from the source. Let's assume a1 = 10.

b1 = s11(a1) + s12(a2)

the initial transient state reflection is

b1 = s11(a1) = 0.707(10) = 7.07 normalized volts

and *that term remains constant* throughout the transient
state and throughout steady-state. The impedance discontinuity
with s11=0.707 reflects 70.7% of the incident voltage, period,
no more and no less. The magnitude of a1 reflected by that
impedance discontinuity *DOES NOT CHANGE* from the very
first incidence of a1. So your statement above is obviously
false. Physical impedance discontinuities do not change their
reflection coefficients based on your whim.

So how does b1 wind up at zero? Not by changing s11(a1) as
you imply. b1 is eventually canceled by the buildup to steady-
state of s12(a2) from zero to a magnitude equal to s11(a1) and
a phase opposite of s11(a1). That's *wave cancellation* in action.
What happens to the energy in the canceled waves?

So your premise is completely flawed. s11(a1) doesn't change.
s12 doesn't change. s21 doesn't change. s22 doesn't change.
What changes is the s12(a2) term which is the reflections
from the load. b1 decreases increment by increment due to
the wave cancellation between the fixed value of s11(a1)
and the ever increasing value of s12(a2) until steady-state
is reached and b1 has become zero. At steady-state, s11(a1)
is still equal to 7.07 normalized volts. It has not changed.
Contrary to your assertion, it will not change as long as
a1 is applied.

s11(a1) = 7.07 unchanging throughout the initial transient
build-up and through steady-state. Anything else would
require magic.

b1 is initially equal to 7.07 because a2 is zero.

s12(a2) will eventually build up from 0 to 7.07 at which
point the *net reflections are eliminated by wave cancellation*.

As b1 is decreasing to zero at steady-state, b2 is increasing
to its steady-state value in the other direction. b1 is
undergoing increasing destructive interference and b2 is
undergoing increasing constructive interference until the
time when b1 = 0 and therefore |b1|^2 = 0, i.e. net reflections
are eliminated.
--
73, Cecil http://www.w5dxp.com