AI4QJ wrote:
I think I see why it no longer surprizes me after going through the smith
chart.
The 100 ohm line (10 degrees) is open. The 600 ohm line has a load impedance
of -j567 ohms, it is not open. The fact that it is terminated with an
impedance (the 100 ohm line) adds degrees on the chart. We should expect the
reactance of the 100 ohm line to add phase angle at the termination similar
to a "discreet component". Hope this makes sense; the smith chart makes it
very clear.

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open

The Smith Chart does make it clear what is happening.
Here is the math to go with it. The impedance at the
junction of the two lines is:

-j100*tan(90-10) = -j100*tan(80) = -j567 ohms
-j600*tan(43.4) = -j600*tan(43.4) = -j567 ohms

The phase shift at the junction of the two lines is:
80-43.4 = 36.6 degrees

Time permitting, I will work up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143
--
73, Cecil http://www.w5dxp.com