Ok, now take the capacitor off, and measure the voltage at the end
of the inductor. What do you get?

Afraid to measure it, eh? Go ahead, we can ignore the 15 pF for now,
as
the load was 100pF. What do you get?

You asked for an open. 15pf is about 1000 ohms at 10 Mhz. This is a long
way
from an open. In any case, I suggest it might be your turn to produce
some
experimental results.

Go ahead and measure it anyways, i'd like to know what you get. If you
think about it, this would be a
clue as to the forward Vi voltage.




Classic Rho gives -1, which is a short, and conjugate Rho gives
+1j, which is ALSO a short.

As I recall, the purpose of rho was to compute the reflected voltage so
that net voltage could be computed using:
Vnet = Vfwd + Vref
Using rho = -1 produces Vref = -Vfwd yielding Vnet = 0 as expected for
a short.
Using rho = +1j produces Vref = +1j * Vfwd which does not produce
Vnet=0.
This is not the expected result for a short.



You don't understand what i wrote. Rho = +1j is a short if Zo=50+j50 and
Zl= - j50

For our example, rho= -1 , but this is
NOT a short!! As it shouldn't be!

Hint: What is the center of the Smith Chart when it is normalized
to Zo=50+j50?

If you can't answer this question, that's
ok, most people don't understand the Smith very well.



Riddles just do not cut it.


My questions are indeed riddles for those whose don't understand the
concepts.




Feel free to point out the flaws. If you can find none, question why
you hold so tenaciously to revised rho when it does not work.


i just did point out the flaws.

And i will let go of this argument as
soon as someone can give me evidence
that i should. You certainly have not.


I will try the experiment when i get the chance.

Excellent. There is nothing better than seeing it with your own
eyes.


I don't need to see it with my eyes,
when i know you can't get a larger RMS voltage reflected from a capacitor, than
the RMS that charges it.


Slick