Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50
ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the
current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?
73,
Tom Donaly, KA6RUH