On Dec 7, 3:02 pm, wrote:
On Dec 7, 1:00 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?
-jcot(5) = -j11.43 normalized to Z0=100 ohms
-j100(11.43) = -j1143 ohms at the junction
-j1143/600 = -j1.905 normalized to Z0=600 ohms
arctan(1.905) = 62.3 degrees of Z0=600 ohm line
Would the phase shift at the junction still be
36.6 degrees?
The new phase shift would be 90-5-62.3 = 22.7 deg.
62.3 + 5 + 22.7 = 90 degrees
This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com
The smith chart shows this well. If I go up only 5 degrees on the
outer circle of open-end (infinite impedance on the 100 ohm line),
lower values of electrical angle corresponds to higher reactance, in
this case -j1143 for the 5 degree line instead of -j567 for the 10
degree line at the junction. The new phase shift at the junction
should be, and is, now lower since the 100 ohm line has a higher
capacitive reactance at the junction. As the 100 ohm line is shortened
to 0 degrees, we have a 600 ohm transmission line that is open and now
the 600 ohm line must be lengthened to the full 90 degrees for 1/4W.
This would correspond to a coil with no stinger.- Hide quoted text -
- Show quoted text -
Ooops, I am posting from the web. This is AI4QJ.