Here is my measurement to a piece of RG6 cable. It is around 30cm cable. On the shield is written COAXIAL CABLE 5C-2V 75 ohm. From my measurement I got data as below: Electronic Measurement

From that table, we can see that there is significant impedance of the cable. If we calculate the inner and the outer impedance using formula Z=R+j(XL-XC), where XL=j2pifL, and XC=j/(2pifC), L in Henry, C in Farad, f in Hertz, and pi=3.14159265358979. Calculation with the data below and using 2,100 MHz frquency, I got as below:

  • Z inner=5+j2638.9 ohm, and
  • Z outer=6+j1319.5 ohm.

Then my question is:

  • What is the 75 ohm meaning (written at the the cable shield)?
  • What is those two impedance meaning to the antenna?
  • If we got data in real cable, what to do to match the impedance?

Edit: The cable was just used for C-Band parabolic antenna.


  • 1
    $\begingroup$ I meant to say this when you first posted your antenna question... Antennas made no sense to me until after I understood transmission lines. Then they became obvious. Find a good book on transmission lines and read it carefully; do the exercises. This will all make sense, I promise. $\endgroup$ – Chris K8NVH Aug 6 at 11:27
  • $\begingroup$ Recommend you follow the procedures here, re: "Measuring Self-Inductance and Self-Capacitance of a Coaxial Cable." You should find that $\sqrt{\frac{L}{C}}=75\Omega$. $\endgroup$ – Brian K1LI Aug 6 at 11:27
  • $\begingroup$ @BrianK1LI, if you just simply put the above data into calculation, the inner is around 0.12 ohm while the outer is 0.06 ohm. Very far, right? Is that mean that my cable absolutely unusable? $\endgroup$ – Sitorus Aug 6 at 11:33
  • $\begingroup$ Btw @BrianK1LI, let me read your link. That seem interesting. $\endgroup$ – Sitorus Aug 6 at 11:34
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    $\begingroup$ @ChrisK8NVH Good idea! The Belden datasheet for RG6 specifies $0.318\mu H/m$ and $53.2pF/m$. The square root of the ratio of these values is $77\Omega$. $\endgroup$ – Brian K1LI Aug 6 at 11:50

see a description of 50 vs 75 Ω here:


and the math behind it is here:


  • 2
    $\begingroup$ Welcome to Stack Exchange. While these links may answer the question, the goal of Stack Exchange is to contain answers, not links to answers (especially as they may break). It's good to have links for “further reading”, but please make sure that the text of your question answers the question asked. $\endgroup$ – Kevin Reid AG6YO Aug 6 at 13:46

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