Tom Donaly 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.

Do you want to work that out mathematically?

For the stub to be electrically 1/4WL, the following
must hold where L1 and L2 are in degrees.

-jZ01*tan(L1) = -jZ02*cot(L2) = -jZ02*tan(90-L2)

-j600*tan(43.4) = -j100*cot(10) = -j567 ohms at the junction

When L1 = L2, the stub is half Z01 and half Z02.
Such a stub is very close to 1/2 the physical length
of a single-Z0 stub when Z01/Z02 = 6.
--
73, Cecil http://www.w5dxp.com