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The 50 ohm characteristic impedance of my coax is only "characteristic" if it's terminated with $ 50 + 0j$ load. So when the coax is connected to an antenna that is fairly reactive like $ 23-j300 $ then the coax is no longer matching my radio.

If I place a trans-match between the rig and the feed, then I'm tuning the feed line. Where as, if I place the match at the antenna, then I'm tuning the antenna.

It then occurred to me, that I could perhaps analysis this with Two Port Network Theory.

By Brews ohare - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=3296726

If the impedance of the antenna is mostly reactive, then it would seem like there's not much lost if I keep the trans-match at the rig.

Is this a valid understanding ? If not, how would I determine the impedance of the coax and antenna ?

If $I_1 = I_2$,

$$ Z_{tot} = Z_{coax} + Z_{antenna} = (50 + 0j) + (23 + 300j) $$ $$ Z_{mag} = \sqrt{73^2 + 300^2} \sim 309 \Omega $$

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    $\begingroup$ A key synonym for "characteristic impedance" is "surge impedance". Definitionally that impedance doesn't change whether the first reflection has returned or not, because it's defined as the impedance seen before that time. So the characteristic impedance specifically is still considered to be 50Ω even after standing waves are set up. It's whatever the impedance was during the initial "surge". But you are also right that your radio will quickly see a different impedance than that, and that impedance becomes very relevant in practice too! $\endgroup$ Jun 30, 2022 at 20:10

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50 ohm characteristic impedance of my coax is only "characteristic" if it's terminated with 50+0j load

That wording is incorrect. The characteristic impedance of your coax is a constant; it doesn't depend on what you attach. It depends on the shape and the material of the coax.

What does change is the impedance of the load attached to your radio.

So when the coax is connected to an antenna that is fairly reactive like (23−j300)Ω then the coax is no longer matching my radio.

The cable is still matched to your radio. Your antenna isn't matched to the cable!

Now, you're still right, we can "count" the cable towards either the radio or the antenna. If we count it towards the radio, then your antenna is "mismatched to your radio system", if we count it towards the antenna, then "the impedance of the antenna system gets transformed by the cable".

If not, how would I determine the impedance of the coax and antenna ?

In case you remember Smith charts. We use the radio impedance=cable impedance as reference impedance $Z_0$.

If we

  • "include" it into the radio, then the length of the cable transforms the impedance of the radio (normalized, so that's $Z_r=1.0$) on a circle of radius $|Z_r - Z_0|=|1-1|=0$. No matter how far we walk on a circle of radius 0, we stay in the same point! So, from the antenna, we see the cable (and whatever is attached to it) as perfect 50Ω load.
  • "include" it into the antenna. Antenna has impedance $\tilde Z_A= (23−j300 )Ω$, which, normalized to 50Ω reference impedance, is ca $Z_A=0.5 - j6$. We now walk on a circle of radius $|Z_A-Z_0|=\sqrt{0.5^2 + 6^2}\approx 6$, for as many half-wavelengths as your cable is long.

Result:

  1. In a system where we can consider the load matched to the cable, the cable length doesn't matter: from the perspective of the antenna, what is attached is always 50Ω, purely resistive.
  2. In a system where the load is not matched to the cable, cable length matters, and you can transform your mostly reactive impedance to any degree of purely-resistive, capacitive or reactive impedance by walking on a circle in the Smith chart. That's the whole point of that chart! So, you can even transform your heavily reactive load to a purely resistive one through the right length of 50Ω coax, and then use e.g. a resistive voltage divider or a transformer to do the matching. More often, you achieve a match by having some discrete components on either end of the cable, transforming the impedance such that it can matched through a length of different (e.g. 75Ω) coax.
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  • $\begingroup$ I was missing the normalize step and now it makes more sense. I suppose if I calibrate my VNA with the transmission line attached, then that is the same thing. $\endgroup$
    – wbg
    Jun 30, 2022 at 20:11
  • $\begingroup$ @wbg Yep! Part of calibrating a VNA is "zeroing out" its calculations relative to where you want to measure, so you could put open/short/load at the far end of any coax you want to "ignore" (with some corresponding loss of dynamic range or other measurement precision I imagine). You might find youtu.be/ImNRca5ecF0?t=315 and youtu.be/9thbTC8-JtA?t=277 to be helpful too. $\endgroup$ Jun 30, 2022 at 20:26
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A couple of things are missing here.

  • If the antenna is not 50+0i then there will be SWR. If there is SWR, then loss in the coax is magnified.
  • If the impedance is not 50+0i, then the impedance along the coax will rotate as it travels, so the length of the coax affects the phase of the impedance you will see at the other end of the coax
  • You can put an antenna tuner / match anywhere along the coax so that the radio sees 50 ohms. It's convenient to put it at the radio, but you get less loss in the coax if you put it at the antenna. There's other questions here that address that directly.
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  • $\begingroup$ "you get less loss in the coax if you put it at the antenna." That is really a big part of my curiosity. Why that is so and how much ? I'll look for those here. $\endgroup$
    – wbg
    Jun 27, 2022 at 20:07
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    $\begingroup$ The transmatch at the radio doesn't fix SWR in the coax, it just fixes it for the radio. See point 1 why that matters. Better details at ham.stackexchange.com/questions/19939/… $\endgroup$
    – user10489
    Jun 27, 2022 at 23:29

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