Coaxial cables as multiconductor transmission lines

When using coaxial cables, it is often assumed that all of the current passed through the inner conductor returns through the shield. This seems like a good assumption, since where else would the current go? It turns out, however, that if the conductors have a finite impedance to some third conductor (for example the ground), this will cause currents to pass to that third conductor and not necessarily return symmetrically through the coax. Such a situation may occur for instance when the coaxial cable is used to feed an antenna. The unbalanced currents (common mode currents) in the coax in turn cause the feedline itself to act as an antenna, and to radiate power unpredictably. Antennas are thus typically connected to the feedline through some form of balun, which presents itself as a large series impedance for any unbalanced currents.

There is a lot of discussion regarding unbalanced currents in coaxial cables on amateur radio forums, and Modern Antenna Design by Thomas A. Milligan quickly mentions three-wire transmission lines in their discussion of baluns. However, I haven't found any computational examples of what is going on in these situations. The following is my attempt at that.

Assume a coaxial cable running on top of a perfectly conducting ground plane. Assume also, that the cable isn't too far away from the plane, so that the fields between the coax and the ground plane can be considered to propagate in the TEM mode. This situation can be thought of as a multiconductor transmission line, with two signal conductors and a reference conductor. We can draw the equivalent circuit of a short line segment of length \(\Delta x\) as shown in the figure below.

Inductances \(L_1\) and \(L_2\) are the self-inductances of conductor 1, i.e. the inner conductor with return through the ground plane, and of conductor 2, i.e. the shield with return through the ground plane. The inductance \(L_{12}\) is the mutual inductance between the two signal-bearing conductors. The capacitances bear a similar nomenclature, and their meaning is more clear from the figure. However, note that the capacitance \(C_1\) from the inner conductor to the ground plane is zero, because the electric field would need to pass through the shield first.

In the limit of \(\Delta x \rightarrow 0\) we get the telegrapher's equations for the situation as

\( \begin{align} \frac{\partial V_1}{\partial x} &= -L_1 \frac{\partial I_1}{\partial t} - L_{12} \frac{\partial I_2}{\partial t} \\ \frac{\partial I_1}{\partial x} &= -C_{12} \frac{\partial V_1}{\partial t} + C_{12} \frac{\partial V_2}{\partial t} \\ \\ \frac{\partial V_2}{\partial x} &= -L_2 \frac{\partial I_2}{\partial t} - L_{12} \frac{\partial I_1}{\partial t} \\ \frac{\partial I_2}{\partial x} &= -(C_2 + C_{12}) \frac{\partial V_2}{\partial t} + C_{12} \frac{\partial V_1}{\partial t} \\ \end{align} \)

Equivalently in matrix form as

\( \begin{align} \frac{\partial \boldsymbol{I}}{\partial x} &= -\boldsymbol{C} \frac{\partial \boldsymbol{V}}{\partial t} \\ \frac{\partial \boldsymbol{V}}{\partial x} &= -\boldsymbol{L} \frac{\partial \boldsymbol{I}}{\partial t} \\ \end{align} \)

where \(\boldsymbol{V} = ( V_1, V_2 )\), \(\boldsymbol{I} = (I_1, I_2)\), while \(\boldsymbol{L}\) is the inductance matrix of the system and \(\boldsymbol{C}\) is the capacitance matrix of the system

\begin{align} \boldsymbol{L} &= \begin{pmatrix} L_1 & L_{12} \\ L_{12} & L_2 \end{pmatrix} \\ \boldsymbol{C} &= \begin{pmatrix} C_{12} & -C_{12} \\ -C_{12} & C_2 + C_{12} \end{pmatrix} \end{align}

With a bit of manipulation, we can derive a vector valued wave equation for the voltage as

\( \begin{align} \boldsymbol{L} \boldsymbol{C} \frac{\partial^2 \boldsymbol{V}}{\partial t^2} &= \frac{\partial^2 \boldsymbol{V}}{\partial x^2} \end{align} \)

This describes two waves propagating at velocities \(1/\sqrt{\lambda_1}\) and \(1/\sqrt{\lambda_2}\), where \(\lambda_1\) and \(\lambda_2\) are the eigenvalues of matrix \(\boldsymbol{L} \boldsymbol{C}\).

Determining the capacitance and inductance parameters requires solving the electrostatic and magnetostatic problems in the appropriate geometries. The approach I took, was to use FEM to compute the electric potential for the electrostatic problem, and the magnetic vector potential for the magnetostatic problem. Computing the capacitance is straightforward from relating the electric field energy of the FEM computation to the energy stored in a capacitor. That is, we have the electric field energy from FEM as

\( \begin{align} W = \int_\Omega \boldsymbol{E} \cdot \boldsymbol{D}\ \mathrm{d}x. \end{align} \)

On the other hand, we have the energy stored in a capacitor as

\( \begin{align} W = \frac{1}{2} CU^2. \end{align} \)

This is then simply computed for the two cases:

1. inner conductor at potential \(V\), shield and ground at potential \(0\)
2. inner conductor and shield at potential \(V\), ground at potential \(0\)

Computing the inductance also follows a similar idea, but there is some additional complication to compute the mutual inductance. The energy in the magnetic field is given by

\( \begin{align} W = \int_\Omega \boldsymbol{H} \cdot \boldsymbol{B}\ \mathrm{d}x. \end{align} \)

On the other hand, we have the energy stored in an inductor as

\( \begin{align} W = \frac{1}{2} LI^2. \end{align} \)

Here we have three different cases that need to be considered.

1. Current \(I\) through inner conductor, current \(0\) through shield and current \(-I\) through ground plane.
2. Current \(0\) through inner conductor, current \(I\) through shield and current \(-I\) through ground plane.
3. Current \(I\) through inner conductor, current \(-I\) through shield and current \(0\) through ground plane.

The inductances \(L_1\) and \(L_2\) in the multiconductor transmission line are the inductance of the inner conductor with return through the ground plane and the inductance of the shield with return through the ground plane, respectively. They can thus be directly computed through the magnetic field energy of cases 1 and 2. The inductance \(L_{12}\), however, is the mutual inductance between the inner and outer conductors. To compute that, we need to consider case 3, where we connect the inner conductor to the shield at the far end. This is equivalent to connecting inductances \(L_1\) and \(L_2\) in series together complete with the mutual inductance. Since the currents are opposing, the mutual inductance acts to decrease the total inductance, and we can write the inductance of the series systems as

\( \begin{align} L_\mathrm{series} = L_1 + L_2 - 2 L_{12} \end{align} \)

Thus we can compute \(L_\mathrm{series}\), \(L_1\) and \(L_2\) from the magnetic field energy, and then solve the mutual inductance \(L_{12}\) from the above equation.

Computation domain for the parameters in the example

Placing a coax of roughly RG316 dimensions and materials over a groundplane at a height of 20mm gives the following values (note: the skin effect on the current distribution is not perfectly accounted for, that's a whole new can of worms).

\( \begin{align} L_1 & = 1023.02 \mathrm{nH}/\mathrm{m} \\ L_2 & = 800.58 \mathrm{nH}/\mathrm{m} \\ L_{12} & = 800.58 \mathrm{nH}/\mathrm{m} \\ C_2 & = 14.05 \mathrm{pF}/\mathrm{m} \\ C_{12} & = 105.05 \mathrm{pF}/\mathrm{m} \end{align} \)

Interestingly, the mutual inductance and the inductance of the shield evaluate as very nearly the same value. This means that the voltage induced in the shield due to changing differential currents is basically zero! This can't simply be a lucky coincidence, but I haven't yet worked out the details.

For a visualization of the results, below is a solution of the telegrapher's equations (again using FEM) for a cable length of 2.1 meters. On the left side, the inner conductor is driven with a time harmonic input of amplitude 1V oscillating at 210 MHz, while the shield is grounded. On the right side, the inner conductor is connected to the shield through a 47 ohm resistance. No connection to ground is made on the right side.

Contrary to intuition, the shield really does remain at ground potential! This is in fact exactly what I've observed every time I've tried to measure the potential on the coax shield.

So, what happens if we drive the shield at the left side and ground the inner conductor? On the right side, inner remains terminated to the shield through a 47 ohm resistance without any other connections.

Even in this case, the signal propagates mostly in the differential mode. Also, what is clearly visible here is that the wavelength in the shield is longer than in the inner conductor. The propagation velocity of the common mode signal is thus greater than the differential mode signal. This can also be determined from the eigendecomposition of matrix \(\boldsymbol{L} \boldsymbol{C}\). The eigenmodes in fact almost exactly correspond with the differential and common mode currents (another happy almost coincidence), and the corresponding propagation velocities are

\( \begin{align} v_1 & = 1/\sqrt{\lambda_1} \approx 0.690c \\ v_2 & = 1/\sqrt{\lambda_2} \approx 0.995c \end{align} \)

where \(c\) is the speed of light.

How about the initial motivation? What if there is a finite impedance from both conductors to ground? Some references say that it is unsymmetric impedance to ground that causes inbalance in the currents. I'm not sure what is meant exactly, but the following happens when both the inner and the shield are connected to ground through 23 ohm resistances, with no other connections on the right side. On the left side, the shield is grounded and the inner conductor is driven.

So even if the load is symmetric with respect to ground, it may cause common mode currents to flow in the coax. Also, since in such a situation the coax is not terminated at its characteristic impedance, the length of the cable will also affect the behavior.