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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. |
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