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1. [AMPS] Feedpoint impedance vs SWR (score: 1): Author: aa6eg@hh.tmx.com (Pat Barthelow); Date: Sat, 2 Oct 1999 14:23:52 -0700 (PDT); I think I am understanding well the measured SWR vs impedance along the line question. My next question is: Given, say, a 100 ohm resistive antenna impedance (at the antenna terminals) connected to (; /archives//html/Amps/1999-10/msg00053.html (7,701 bytes)
2. [AMPS] Feedpoint impedance vs SWR (score: 1): Author: measures@vcnet.com (measures); Date: Sun, 3 Oct 1999 03:04:40 -0700; ? My guess is that a quarter wave 50-ohm line would transform the 100-ohms to [100^2 + 50^2]^0.5 =[10k + 2.5k]^0.5 = 111.8-ohms. However, with a half wavelength of any Z coax, the 100-ohms would not; /archives//html/Amps/1999-10/msg00054.html (7,940 bytes)
3. [AMPS] Feedpoint impedance vs SWR (score: 1): Author: n7rt@doitnow.com (Hardy Landskov); Date: Sat, 02 Oct 1999 17:37:17 -0700; My guess is 25 ohms -- FAQ on WWW: http://www.contesting.com/ampsfaq.html Submissions: amps@contesting.com Administrative requests: amps-REQUEST@contesting.com Problems: owner-amps@contesting.com Sea; /archives//html/Amps/1999-10/msg00058.html (8,234 bytes)
4. [AMPS] Feedpoint impedance vs SWR (score: 1): Author: jono@enteract.com (Jon Ogden); Date: Sun, 3 Oct 1999 21:43:54 -0500; Your GUESS? YOUR GUESS? Via the Smith Chart, one can easily see that traveling on a constanct VSWR circle from the load clockwise to the source that 1/4 wavelength trip will put you at the 25 Ohm poi; /archives//html/Amps/1999-10/msg00083.html (8,310 bytes)
5. [AMPS] Feedpoint impedance vs SWR (score: 1): Author: jono@enteract.com (Jon Ogden); Date: Sun, 3 Oct 1999 21:38:58 -0500; How can one calculate it mathematically? Gee, that's tough! Use a computer! :-) What do you mean by most "extreme" impedance? OK, with a normal Smitch Chart NORMALIZED to 50 Ohms: 100 Ohms is on the; /archives//html/Amps/1999-10/msg00085.html (8,791 bytes)

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