Cecil Moore wrote:


---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

So how many nanoseconds does that 36.6 degree phase shift represent?

8-)

73,
Gene
W4SZ