11.3: Co-axial Cables

Cylindrical co-ordinates are appropriate for the problem of a co-axial cable, Figure (11.2.3). The relevant Maxwell’s equations become

$$\operatorname{curl}(\vec{\text{E}})=-\mu_{0} \frac{\partial \vec{\text{H}}}{\partial \text{t}}, \nonumber$$

and

$$\operatorname{curl}(\vec{\text{H}})=\epsilon \frac{\partial \vec{\text{E}}}{\partial \text{t}}, \nonumber$$

where $\epsilon$ is a real number for a lossless line. Look for solutions of these equations in which, by analogy with a strip-line curved around on itself, the electric field has only a radial component, E_r , that is independent of angle, and the magnetic field has only an angularly independent component H_θ:

$$\begin{align} &\frac{\partial \text{E}_{\text{r}}}{\partial \text{z}}=-\mu_{0} \frac{\partial \text{H}_{\theta}}{\partial \text{t}}, \label{11.12}\\ &\frac{\partial \text{H}_{\theta}}{\partial \text{z}}=-\epsilon \frac{\partial \text{E}_{\text{r}}}{\partial \text{t}}. \end{align}$$

In addition, take E_z = 0 because the tangential components of the electric field must be zero at the perfectly conducting walls of the co-axial cable. But if E_z = 0 it follows from Maxwell’s equations that

$$\operatorname{curl}(\vec{\text{H}})_{z}=0=\frac{1}{\text{r}} \frac{\partial}{\partial \text{r}}\left(\text{rH}_{\theta}\right). \nonumber$$

This implies that

$$\text{H}_{\theta}=\frac{\text{a}(\text{z}, \text{t})}{\text{r}}, \label{11.13}$$

where a(z,t) is a function of time and of position along the cable. Similarly, from div($\vec E$) = 0 one has

$$\frac{1}{\text{r}} \frac{\partial}{\partial \text{r}}\left(\text{rE}_{\text{r}}\right)=0, \nonumber$$

and this is satisfied by

$$\text{E}_{\text{r}}=\frac{\text{b}(\text{z}, \text{t})}{\text{r}}. \label{11.14}$$

By combining the Maxwell Equations ($\ref{11.12}$) the electric and magnetic fields, Equations ($\ref{11.13}$) and ($\ref{11.14}$), must satisfy

$$\begin{align} &\frac{\partial^{2} \text{E}_{\text{r}}}{\partial \text{z}^{2}}=-\mu_{0} \frac{\partial^{2} \text{H}_{\theta}}{\partial \text{z} \partial \text{t}}=\epsilon \mu_{0} \frac{\partial^{2} \text{E}_{\text{r}}}{\partial \text{t}^{2}}, \label{11.15}\\ &\frac{\partial^{2} \text{H}_{\theta}}{\partial \text{z}^{2}}=-\epsilon \frac{\partial^{2} \text{E}_{\text{r}}}{\partial \text{z} \partial \text{t}}=\epsilon \mu_{0} \frac{\partial^{2} \text{H}_{\theta}}{\partial \text{t}^{2}}. \end{align}$$

These have the same form as the strip-line equations (11.2.3). It follows from these equations,and from the requirements ($\ref{11.13}$) and ($\ref{11.14}$), that the general solution for the electric field can be written

$$\text{E}_{\text{r}}(\text{z}, \text{t})=\frac{\text{F}(\text{z}-\text{vt})}{\text{r}}+\frac{\text{G}(\text{z}+\text{vt})}{\text{r}}, \label{11.16}$$

where F(u) and G(u) are arbitrary functions of their arguments, and where

$$\text{v}=\frac{1}{\sqrt{\epsilon \mu_{0}}}. \nonumber$$

The corresponding general solution for the magnetic field is

$$\text{H}_{\theta}(\text{z}, \text{t})=\epsilon \text{v}\left(\frac{\text{F}(\text{z}-\text{vt})}{\text{r}}-\frac{\text{G}(\text{z}+\text{vt})}{\text{r}}\right). \label{11.17}$$

The above electric and magnetic fields satisfy the wave equations ($\ref{11.15}$), they satisfy Equations ($\ref{11.12}$), and they have the form required by Equations ($\ref{11.13}$ and $\ref{11.14}$).

Instead of the electric field strength, the state of the electric field in the cable can be specified by the potential difference between the inner and outer conductors:

$$\text{V}=\int_{\text{R}_{1}}^{\text{R}_{2}} \text{E}_{\text{r}} \text{dr}=\text{F}(\text{z}-\text{vt}) \int_{\text{R}_{1}}^{\text{R}_{2}} \frac{\text{d} \text{r}}{\text{r}}=\text{F}(\text{z}-\text{vt}) \ln \left(\frac{\text{R}_{2}}{\text{R}_{1}}\right) \nonumber$$

for a forward propagating wave. Note that the inner conductor is positive with respect to the outer conductor. The corresponding current on the inner conductor is given by

$$\text{I}=\text{J}_{\text{z}}\left(2 \pi \text{R}_{1}\right)=\text{H}_{\theta}\left(\text{R}_{1}\right)\left(2 \pi \text{R}_{1}\right)=\epsilon \text{v}\left(2 \pi \text{R}_{1}\right) \frac{\text{F}(\text{z}-\text{vt})}{\text{R}_{1}}, \nonumber$$

so that

$$\text{I}=2 \pi \epsilon \text{v} \text{F}(\text{z}-\text{vt}). \nonumber$$

The current flows flows towards +z for the current on the inner conductor; the current flows towards minus z on the outer conductor. That is, on the outer conductor

$$\text{I}=-2 \pi \text{R}_{2} \text{H}_{\theta}\left(\text{R}_{2}\right)=-2 \pi \epsilon \text{v} \text{F}(\text{z}-\text{vt}). \nonumber$$

so that the net current flow through a section of the cable is zero. The characteristic impedance of the cable is given by

$$\text{Z}_{0}=\frac{\text{V}}{\text{I}}=\frac{1}{2 \pi \epsilon \text{v}} \ln \left(\frac{\text{R}_{2}}{\text{R}_{1}}\right)\nonumber$$

or

$$\text{Z}_{0}=\frac{1}{2 \pi} \sqrt{\frac{\mu_{0}}{\epsilon}} \ln \left(\frac{\text{R}_{2}}{\text{R}_{1}}\right). \label{11.18}$$

The potential difference, V, is proportional to the electric field, E_r , and the current, I, is proportional to the magnetic field, H_θ, therefore from Equations ($\ref{11.15}$) the voltage and current satisfy the wave equations

$$\begin{align} &\frac{\partial^{2} \text{V}}{\partial \text{z}^{2}}=\frac{1}{\text{v}^{2}} \frac{\partial^{2} \text{V}}{\partial \text{t}^{2}}, \label{11.19}\\ &\frac{\partial^{2} I}{\partial z^{2}}=\frac{1}{v^{2}} \frac{\partial^{2} I}{\partial t^{2}}, \end{align}$$

where v² = 1/($\epsilon$µ₀). For a forward propagating pulse having the form

$$V(z, t)=F(z-v t) \nonumber$$

the corresponding current pulse is described by

$$\text{I}(\text{z}, \text{t})=\frac{1}{\text{Z}_{0}} \text{F}(\text{z}-\text{vt})=\frac{\text{V}(\text{z}, \text{t})}{\text{Z}_{0}}, \label{11.20}$$

where the characteristic impedance for a co-axial cable is given by Equation (11.18). For a backward propagating potential pulse of the form

$$\text{V}(\text{z}, \text{t})=\text{G}(\text{z}+\text{vt}) \nonumber$$

the corresponding current pulse is described by

$$\text{I}(\text{z}, \text{t})=-\frac{1}{\text{Z}_{0}} \text{V}(\text{z}, \text{t})=-\frac{\text{G}(\text{z}+\text{vt})}{\text{Z}_{0}}. \label{11.21}$$

In the above equations F(z-vt) and G(z+vt) are arbitrary functions of their arguments.