AI4QJ wrote:
"Cecil Moore" wrote in message
. net...
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.

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?
73,
Tom Donaly, KA6RUH