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For a lossless line, you have \$Z_0=\sqrt{\frac{L}{C}}\$ and \$VF=\frac{1}{c\sqrt{LC}}\$.

For a line with R, L, G and C parameters, we have \$Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}\$. But how the formula changes for the VF parameter?

Considering the coaxial cable example, one could say the VF not only depends on \$Z\$, but also in the geometry of the cable, hence a more real expression for VF do not exist?.

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For a lossy transmission line

\$ Z_0 = \frac{\sqrt{R+j\omega L}}{\sqrt{G + j\omega C}} \$

Note that for \$ R << \omega L\$ and \$ G << \omega C\$ this reduces to the expression in your question.

Finding the phase velocity of a lossy line is difficult as the propagation constant \$\beta \$ will, in general, be a non linear function of frequency. (Except for a distortionless line that satisfies the Heaviside Condtion). This means that the velocity factor is not a very useful metric, except over a narrow band or small distance.

In fact it means that different frequency components of a signal will travel at different speeds, resulting in distortion of the signal and inter-symbol interference in the case of digital communications.

The geometry of the cable determines the R, L, G and C parameters as well as Z and the phase velocity.

Edit: I'm not sure if a better general expression than this can be found but the complex propagation constant will be:

\$ \gamma = j\omega \sqrt{LC} \sqrt{1 - j(\frac{R}{\omega L} + \frac{G}{\omega C}) - \frac{RG}{\omega^2LC}}\$

Then the phase velocity is

\$ v_p = \frac{\omega}{Imaginary(\gamma)}\$

Thus the velocity factor is

\$ \frac{\omega}{c Imaginary(\gamma)} \$

And I don't think the Gamma expression can be decomposed into real and imaginary parts without making approximations.

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  • \$\begingroup\$ My mistake. I am refering to the VF parameter. Check the edit. \$\endgroup\$
    – Brethlosze
    May 25, 2019 at 23:52
  • \$\begingroup\$ One would expect G close to zero. Also, one would expect to have a VF for a given specific frequency. GIven that, which would be the approximation for the VF? \$\endgroup\$
    – Brethlosze
    May 25, 2019 at 23:57
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    \$\begingroup\$ The only approximation I've seen used is the one in your original expression \$\endgroup\$
    – jramsay42
    May 26, 2019 at 0:02
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Wikipedia has the fully-general formula for the characteristic impedance at a given angular frequency \$\omega\$:

https://en.wikipedia.org/wiki/Characteristic_impedance

\$Z_0=\sqrt{\frac{R+j\omega{}L}{G+j\omega{}C}}\$

Note that for a lossless line (R=0, G=0), the frequency factor cancels out, but for a lossy line it does not.

I'm not sure about the velocity factor.

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  • \$\begingroup\$ My mistake. I am refering to the VF parameter. Check the edit. \$\endgroup\$
    – Brethlosze
    May 25, 2019 at 23:52
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Not sure, but I think the velocity factor of coax cable is related to relative dielectric constant only.

$$V_F=\frac{1}{\sqrt{\varepsilon_R}} $$

EDIT: In general case, you would need to calculate C and L for the specific geometry of the transmission line. See example

Also the distance from ground has also some effects on velocity factor. For example an insulated wire used as antenna, has a velocity factor:

$$V_F= \sqrt{\dfrac{\varepsilon\cdot ln\dfrac{4h}{D}+ln\dfrac{D}{d}}{\varepsilon\cdot ln\dfrac{4h}{d}}}$$

Reference: G4FGQ and KB6NU , my rearrangement

Where \$h\$ is the height above the ground, \$d\$ is conductor diameter, \$D\$ is overall diameter (conductor + insulator).

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  • \$\begingroup\$ Yes, i found that relationship as described in the link. I dont know if that is truth or how this is related with \$Z_0\$ for the general case. \$\endgroup\$
    – Brethlosze
    May 26, 2019 at 8:28

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