As the total energy returning from the second interface is 0.098X, not
0.11X, ...

The total energy returning from the second interface is 0.125 watts.
Your solution is incorrect. Why don't you take a poll to see who
agrees with you?

Even if an infinite summation of ALL reflections kept any energy from
transiting the second interface (a generous allowance not likely to be
observed anywhere), then it cannot exceed what energy was initially
introduced into the system.

It cannot exceed the energy. It can certainly exceed the steady-state
source power (joules/sec) and it does. Some energy is sourced during
the transient stage that does not reach the load. It is stored in the
second medium as standing waves. The forward power in the first medium
is one watt with no reflections. The forward power in the second medium
is 1.125 watts. The reflected power in the second medium is 0.125 watts.
The forward power in the third medium is one watt with no reflections.

Have you never seen the forward power in a transmission line exceed
the source power? If the system is Z0-matched and the SWR on a
1/4WL transformer is 5.83:1, the steady-state forward power on the
1/4WL transformer is *double* the steady-state source power.

Either way, "total cancellation" is not total.

In your n=1,2,4 example, reflected wave cancellation is total in
medium 1 (assuming a lossless system). If n2=sqrt(n1*n3), the
reflected wave cancellation is total.
--
73, Cecil http://www.qsl.net/w5dxp

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