Cecil Moore wrote:
AI4QJ wrote:
Going further, I am still trying to consider how the extra angle can
also be absorbed "into" an impedance discontinuity.

I have started a phasor diagram of it but it is not
finished yet. Maybe a Smith Chart explanation will
work. All lines are lossless.

On a Smith Chart normalized to 100 ohms, lay out the
10 degrees of 100 ohm line from the infinity point,
i.e. the open-circuit point. The reactance value
is tan(90-10) = 5.67. That means the reactance value
is 5.67*100 = -j567 ohms which has to be the value
at the impedance discontinuity.

Now on a Smith Chart normalized to 600 ohms, lay out
the x degrees of 600 ohm line from the zero point
to the point where -j567/600 is located. Read the
number of degrees required. It is Arctan(567/600)
which is equal to ~43 degrees.

The phase shift at the impedance discontinuity is
therefore 90-10-43 = 37 degrees.

Wrong. In the first place, you obviously don't know the criterion for
resonance. In the second place you just assume the number
90 without any reason. In the third place, the number 37
has only your assumption for the necessity of a 90 degree
phase shift to justify its existence. As I wrote before, this
is pretty poor shooting for a professional symbol slinger.
73,
Tom Donaly, KA6RUH