walt wrote:
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

OK, now that I understand the series stub configuration, I have to
ask: "the" reflection coefficient looking into WHAT? Looking back
toward the source in its feedline? Looking back toward the source but
including the series stub? Looking toward the load, including the
series stub?

The way I look at this problem is as a measurements person: once I
know where I'm looking, how will I measure the value of the reflection
coefficient? The way I would normally do it is with a network
analyzer. If I want to do it in the presence of significant power
coming from a source, I can set up directional couplers to allow me to
introduce a test signal (that I must be able to distinguish from the
power coming from the source) and observe the reflection.

You can also replace each piece of transmission line (including the
piece that's the stub) with an S-parameter two-port, and the source
and load as S-parameter one-ports, and apply standard S-parameter math
to see what the reflection coefficient is looking into any particular
port.

I suppose this is going to give you an answer different from the one
you're hoping for...but I believe it's consistent with the definitions
for linear system analysis that I've always used.

Consider the case of two identical sources generating the same signal
but 180 degrees out of phase. Each source is connected to a length of
transmission line, and the two lines are identical. Just for fun, the
line impedance equals the source impedance. The free ends of the
lines are connected in parallel and to a resistive load. Of course,
we find that the signals cancel and there is no power dissipated in
the load. Would we say that the reflection coefficient magnitude
looking back into either line toward the source is unity? I would
not! I would say it is zero, since any signal I send in that
direction is absorbed in the source. Note in this example that you
can disconnect the load and not change anything. Then you simply have
the signal from each source traveling to the other source and being
absorbed; if you put the sources on two different frequencies, you'll
see no standing wave for either frequency (though you'll have to
average for a while to be sure about that, if the frequencies are
close together). And note, please, that "absorbed" is not in general
the same thing as "dissipated."

Cheers,
Tom