Keeping in mind simplicity, and , using equal length of 50 0hm
to this power splitter- all antennas should be in phase (all
Left elements , to properly phase), and that it gives 4-50 ohm loads
for one 50 ohm source. Why reinvent the wheel?? Jim NN7K


Owen Duffy wrote:
Jim-NN7K . wrote in news:9ch1l.9958$yr3.334
@nlpi068.nbdc.sbc.com:

Owen Duffy wrote:

Firstly, I didn't write the following, Jim did.

More likely, 2 - 1/4 wave (with velocity factor)50 ohm coax's to a
"Tee" fitting-- Each end also to a "Tee" fitting . ( all 50 ohm coax)
(power devider)

2x50 -----------------

2X Quarter wave | "T" fitting source 50 Ohm

2x50 -----------------

IF this is clear enough-- Jim NN7K


Jim, are you introducing another scheme, or were you trying to explain
Jerry's scheme. We sorted Jerry's scheme, he just overlooked some vital
details in his first description. (I haven't said it to date, but I
dislike Jerry's scheme, principally over its use of the balun.)

Yours is another scheme.

There are a lot of ways to do it.

The original question was over an article's diagram that stated that
unequal lines are "WRONG!".

Yours and Jerry's responses have not dealt with the original posting, but
if anything offered alternatives that might be seen to suggest the
original configuration is flawed.

Owen