VNA measurements of an antenna return Z, RL, Xs, Rs... and also phase. I usually check Z=50 and Xs=0 to reach the max Return Loss and the lowest SWR 1.0:1, without having care of phase values.

However I noticed that in some cases even if Z and Xs are 50 snd 0 the SWR is high (even 6:1... 10:1) becouse of a particular high value of phase.

What's the best value of phase I have to reach for a perfect matching?

Do I need to have always care of the phase like Z and Xs?

  • \$\begingroup\$ 'phase' indicates TIME DELAY, in cables or across the air. \$\endgroup\$ – analogsystemsrf Oct 28 '18 at 11:09
  • \$\begingroup\$ The VNA measurement is made directly at the antenna connector, so what does "time delay" mean? What's the best value I need to reach? If the antenna is perfectly "matched" Zo=50 and Xs=0 why the SWR is high? \$\endgroup\$ – Kevin White Oct 28 '18 at 12:30

VNA measurement results are just complex numbers. Complex numbers can be expressed in either "magnitude (Z) + phase (P)" or "resistance (Rs) + reactance (Xs)".

You don't want to mix these complementary notations of complex numbers. So Z=50 and Xs=0 is not what you are looking for.

For SWR=1 with a load of Rl=50 Ohms you need Z=50,P=0 or Rs=50,Xs=0, which is the same.

  • \$\begingroup\$ Thanks Stefan Wysss, so when I want to adjust an antenna for the perfect match and for the max power transfer it's sufficient I reach i.e. Z=50 and P=0 that automatically means Xs=0? The complex number I consider(ed) necessary and sufficient is (Z,Xs) without giving importance to phase (that automatically reach the optimum value). \$\endgroup\$ – Kevin White Oct 28 '18 at 13:35
  • \$\begingroup\$ No, Z=50 and Xs=0 is not sufficient. Let's make an example: You try to match an antenna impedance of ZL=50 Ohms with some matching elements and your VNA displays an impedance of Zm=-50 Ohms which is Z=50,P=pi or Rs=-50,Xs=0. So you have Z=50,Xs=0 but the matching is not correct because it has P=pi or Rs=-50. \$\endgroup\$ – Stefan Wyss Oct 28 '18 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.