On Dec 8, 3:15 pm, "AI4QJ" wrote:
"Gene Fuller" wrote in message

...





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-)

In this example, we have transmission lines, not an antenna or antenna coil.
The total phase shift is 90 degrees or 62.5 nsec.

Only with great stretching.

The 10 degree 100 ohm line contributes 6.94nsec,

Correct.

the 43 degree 600 ohm line contributes 29.86 nsec.

Correct.

But now think in the time domain for a bit.
29.86 nsec after the signal is first applied it reaches the
discontinuity. 29.86 nsec later the first reflection arrives
back at the start. 13.8 nsec later the first reflection from
the end of the 100 ohm section arrives back at the start.
It takes many more reflections of reflections before the
impedance at the input starts to look like a short.

Nowhere in here will you be able to find anything that
happens in 62.5 nsec.

This is quite unlike an actual physical 1/4WL stub
where the first reflection does arrive back in
2 * 62.5 nsec. And the impedance at the input
behaves like a short after exactly 125 nsec.

Of course the ultimate is an actual short, where
Cecil's 90 degrees happens immediately.

These 90 degrees that Cecil insists are "always"
present are quite difficult to locate.

....Keith