I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. This is a straight-
forward question for which I 'm dead serious. The answer will involve
a resulting reflection coefficient.

Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient Vñ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or
57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match. To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

Recall the voltage reflection coefficient Vñ of the line impedance
prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current
reflection coefficient Iñ = 0.5 at +120°

We now look at the voltage reflection coefficient of the stub, which
is Vñ = 0.5 at +60°, and the corresponding current reflection
coefficient of the stub which is Iñ = 0.5 at -120°.

Observe that resultant angle of the voltage coefficients is 0° and
resultant angle of the current coefficients is 180°, all with the same
magnitude--0.5. It is well known that when these respective
coefficients exist at this point, the matching point, this condition
establishes a virtual open circuit to rearward traveling waves at this
point, prohibiting any further rearward travel of the reflected waves,
thus reversing their direction toward the load.

Consequently, the result is total re-reflection of the two sets of
reflected waves at the match point.

Now the question--what is the resultant voltage coefficient of
reflection at the match point? My position is that ñ = 1.0, which
indicates total reflection. However, I've been criticized on this
value, my critics saying that because this virtual open circuit was
established by wave interference, ñ cannot equal 1.0. They assert
that ñ = 1.0 can be obtained only through a physical open circuit.

So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? I'm holding my breath for your
answers.

Walt