Walter Maxwell wrote in
:

....
Re your worked solution (above), I agree that the normalised
admittance looking into 30deg of line with load 16.667+j0 is about
1-j1.1547 (not the different sign).

Yes Owen, you're right. I added the y values at the last moment, and
didn't catch the errors. Both the line and stub signs are reversed.
Sorry 'bout that.

Ok.

I make the normalised admittance looking into the stub about 0+j1.15
(and the reflection coefficient about 0.5-98, how do you get 1+j1.15?

Normalized y looking into the stub directly is y = 0 + 1.1547, but
looking at the stub while on the line at the 30° point is y = 1 +
1.1547. To view the stub separately on the line the line is terminated
in 50 ohms, because the real component of the line impedance at the
match point is 50 ohms.

You have a junction where three current paths appear in parallel, we can
add the admittances of each of those paths.

We are agreed that admittance of the load+30deg line is 1-j1.15, and that
of the stub is 0+j1.15, so the only place the additional 1+j0 can come
from is the source+line branch.

If that is the case, then your explanation of the stub (which I assume to
be a steady state explanation because you are talking about frequency
domain admittances), depends on the source admittance (or impedance). If
the equivalent source impedance at the junction figures in the calcs, you
are saying that the VSWR on the line from source to junction depends on
the source impedance... I thought we got over that error.

My view is that the stub in shunt with the 30deg line+load results in an
equivalent impedance of approximately 50+j0 at the junction, irrespective
of what is on the source side of the junction.

Owen