Hello Walt,

On Monday, 13 June 2011 05:05:50 UTC+10, 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 Vρ 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 Vρ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or

This is wrong...

57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

as your know!


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.

This is vintage "Reflections" stuff Walt, where you try to explain what is happening with one foot in the time domain and the other in the frequency domain.

To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

I can see you heading to re-re-reflections etc.

Thing is that in the steady state (and your Smith chart works ONLY in the steady state), the reflection on the first mentioned (lossless) line (with the 150 ohm load) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

Sounds boring, but the reflection on the second mentioned (lossless) line (with s/c termination) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

The ratio of V/I at the node where both lines are joined is given for a shunt stub by the inverse of the sum of the inverses of V/I for each.

You can think about reflections from on line entering the other, and energy sloshing around, but the Smith chart is not going to help you because is is simply a graphic computer for finding Gamma, calculating Gamma along the line, adding B, G etc by following lines of constant G, B etc.

If we tried to explain the common microwave three screw tuner using re-reflections, virtual open circuits to rearward traveling waves, conjugate mirrors etc, total re-reflection etc, we would complicate something that is easily explained using a Smith chart (and the underlying maths - The Telegrapher's Equation) and quite conventional AC circuit theory.

In fact, a three stub tuner with practical (ie lossy) lines can be explained fairly easily using a Smith chart, even if a little fiddly iteration is needed to deal with lossy elements.

If the total re-reflection etc concepts are hard to translate from mental models to mathematical expressions for simple scenarios, they become near impossible for more complicated topologies and 'real' components.


Owen