Transmission lines are mathematically described by a set of parameters.
Transmission lines are assumed to be quasi-TEM, (transverse electromagnetic) where the electric and magnetic fields are at all points perpendicular to each other.
In true TEM, there are no field components in the direction of propagation, however with the conductors having finite resistance there must be an E-field component along the length of the wire.
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Per-Unit-Length Parameters
There are four basic per-unit-length parameters: R, L, G and C.
R is derived entirely from the conductor cross-section and its conductivity.
L, G and C are dependent on the geometric arrangement of the conductors and the material properties of the space between them.
For a homogeneous material, L, G and C can be derived from one another.
$$L = \mu\epsilon C^{-1} \\
G = \frac{\sigma}{\epsilon}C = \mu\sigma L^{-1} \\
C = \mu\epsilon L^{-1}$$
Two Wire Line
Two circular conductors in a homogeneous medium (i.e. no insulation).
Wire radius r1 and r2, with center-to-center spacing s.
$$C = \frac{2\pi\epsilon}{\cosh^{-1}{\frac{s^2 – r_1^2 – r_2^2}{2r_1r_2}}} \hspace{5 mm} F/m\\
L = \frac{\mu}{2\pi}\cosh^{-1}{\frac{s^2 – r_1^2 – r_2^2}{2r_1r_2}} \hspace{5 mm} H/m\\
G = \frac{2\pi\sigma}{\cosh^{-1}{\frac{s^2 – r_1^2 – r_2^2}{2r_1r_2}}} \hspace{5 mm} S/m$$
For equal wire size radius r:
$$C = \frac{\pi\epsilon}{\cosh^{-1}{\frac{s}{2r}}} \hspace{5 mm} F/m\\
L = \frac{\mu}{\pi}cosh^{-1}{\frac{s}{2r}} \hspace{5 mm} H/m\\
G = \frac{\pi\sigma}{\cosh^{-1}{\frac{s}{2r}}} \hspace{5 mm} S/m$$
Wire Over a Ground Plane
Circular conductor in a homogeneous medium (i.e. no insulation).
Wire radius r with height from plane to center of wire h.
Note that this is identical to 2x a two-wire line with s = 2h.
This is a result of the method of images, where the ground plane acts as a mirror. The additional 2x comes from the fact that we are still looking at a different capacitance; not from wire to wire but from wire to the mid-point plane.
If the total wire-to-wire capacitance is C, then the midpoint capacitance is 2C, since two series capacitors add to one half.
$$C = \frac{2\pi\epsilon}{\cosh^{-1}{\frac{h}{r}}} \hspace{5 mm} F/m\\
L = \frac{\mu}{2\pi}cosh^{-1}{\frac{h}{r}} \hspace{5 mm} H/m\\
G = \frac{2\pi\sigma}{\cosh^{-1}{\frac{h}{r}}} \hspace{5 mm} S/m$$
Coaxial Cable
Two concentric conductors, with a center, solid cylindrical wire and an external cylindrical shell (shield).
Homogeneous material between conductors.
Center wire radius rc with shield wire radius rs.
$$C = \frac{2\pi\epsilon}{\ln{\frac{r_s}{r_c}}} \hspace{5 mm} F/m\\
L = \frac{\mu}{2\pi}\ln{\frac{r_s}{r_c}} \hspace{5 mm} H/m\\
G = \frac{2\pi\sigma}{\ln{\frac{r_s}{r_c}}} \hspace{5 mm} S/m$$