The transmission line
In the last chapter we studied what happened to the lumped elements of circuits when they were operated at very high frequencies, and we were led to see that a resonant circuit could be replaced by a cavity with the fields resonating inside. Another interesting technical problem is the connection of one object to another, so that electromagnetic energy can be transmitted between them. In low-frequency circuits the connection is made with wires, but this method doesn’t work very well at high frequencies because the circuits would radiate energy into all the space around them, and it is hard to control where the energy will go. The fields spread out around the wires; the currents and voltages are not “guided” very well by the wires. In this chapter we want to look into the ways that objects can be interconnected at high frequencies. At least, that’s one way of presenting our subject. Another way is to say that we have been discussing the behavior of waves in free space. Now it is time to see what happens when oscillating fields are confined in one or more dimensions. We will discover the interesting new phenomenon when the fields are confined in only two dimensions and allowed to go free in the third dimension, they propagate in waves. These are “guided waves”—the subject of this chapter. We begin by working out the general theory of the transmission line. The ordinary power transmission line that runs from tower to tower over the countryside radiates away some of its power, but the power frequencies ($50$–$60$ cycles/sec) are so low that this loss is not serious. The radiation could be stopped by surrounding the line with a metal pipe, but this method would not be practical for power lines because the voltages and currents used would require a very large, expensive, and heavy pipe. So simple “open lines” are used. For somewhat higher frequencies—say a few kilocycles—radiation can already be serious. However, it can be reduced by using “twisted-pair” transmission lines, as is done for short-run telephone connections. At higher frequencies, however, the radiation soon becomes intolerable, either because of power losses or because the energy appears in other circuits where it isn’t wanted. For frequencies from a few kilocycles to some hundreds of megacycles, electromagnetic signals and power are usually transmitted via coaxial lines consisting of a wire inside a cylindrical “outer conductor” or “shield.” Although the following treatment will apply to a transmission line of two parallel conductors of any shape, we will carry it out referring to a coaxial line.  We take the simplest coaxial line that has a central conductor, which we suppose is a thin hollow cylinder, and an outer conductor which is another thin cylinder on the same axis as the inner conductor, as in Fig. 24–1. We begin by figuring out approximately how the line behaves at relatively low frequencies. We have already described some of the low-frequency behavior when we said earlier that two such conductors had a certain amount of inductance per unit length or a certain capacity per unit length. We can, in fact, describe the low-frequency behavior of any transmission line by giving its inductance per unit length, $L_0$ and its capacity per unit length, $C_0$. Then we can analyze the line as the limiting case of the $L$-$C$ filter as discussed in Section 22–6. We can make a filter which imitates the line by taking small series elements $L_0\,\Delta x$ and small shunt capacities $C_0\,\Delta x$, where $\Delta x$ is an element of length of the line. Using our results for the infinite filter, we see that there would be a propagation of electric signals along the line. Rather than following that approach, however, we would now rather look at the line from the point of view of a differential equation. Suppose that we see what happens at two neighboring points along the transmission line, say at the distances $x$ and $x+\Delta x$ from the beginning of the line. Let’s call the voltage difference between the two conductors $V(x)$, and the current along the “hot” conductor $I(x)$ (see Fig. 24–2). If the current in the line is varying, the inductance will give us a voltage drop across the small section of line from $x$ to $x+\Delta x$ in the amount \begin{equation*} \Delta V=V(x+\Delta x)-V(x)=-L_0\,\Delta x\,\ddt{I}{t}. \end{equation*}  Or, taking the limit as $\Delta x\to0$, we get \begin{equation} \label{Eq:II:24:1} \ddp{V}{x}=-L_0\,\ddp{I}{t}. \end{equation}  The changing current gives a gradient of the voltage. Referring again to the figure, if the voltage at $x$ is changing, there must be some charge supplied to the capacity in that region. If we take the small piece of line between $x$ and $x+\Delta x$, the charge on it is $q=C_0\,\Delta x V$. The time rate-of-change of this charge is $C_0\,\Delta x\,dV/dt$, but the charge changes only if the current $I(x)$ into the element is different from the current $I(x+\Delta x)$ out. Calling the difference $\Delta I$, we have \begin{equation*} \Delta I=-C_0\,\Delta x\,\ddt{V}{t}. \end{equation*}  Taking the limit as $\Delta x\to0$, we get \begin{equation} \label{Eq:II:24:2} \ddp{I}{x}=-C_0\,\ddp{V}{t}. \end{equation}  So the conservation of charge implies that the gradient of the current is proportional to the time rate-of-change of the voltage. Equations (24.1) and (24.2) are then the basic equations of a transmission line. If we wish, we could modify them to include the effects of resistance in the conductors or of leakage of charge through the insulation between the conductors, but for our present discussion we will just stay with the simple example. The two transmission line equations can be combined by differentiating one with respect to $t$ and the other with respect to $x$ and eliminating either $V$ or $I$. Then we have either \begin{equation} \label{Eq:II:24:3} \frac{\partial^2V}{\partial x^2} =C_0L_0\,\frac{\partial^2V}{\partial t^2} \end{equation}  or \begin{equation} \label{Eq:II:24:4} \frac{\partial^2I}{\partial x^2} =C_0L_0\,\frac{\partial^2I}{\partial t^2} \end{equation}  Once more we recognize the wave equation in $x$. For a uniform transmission line, the voltage (and current) propagates along the line as a wave. The voltage along the line must be of the form $V(x,t)=f(x-vt)$ or $V(x,t)=g(x+vt)$, or a sum of both. Now what is the velocity $v$? We know that the coefficient of the $\partial^2/\partial t^2$ term is just $1/v^2$, so \begin{equation} \label{Eq:II:24:5} v=\frac{1}{\sqrt{L_0C_0}}. \end{equation}  We will leave it for you to show that the voltage for each wave in a line is proportional to the current of that wave and that the constant of proportionality is just the characteristic impedance $z_0$. Calling $V_+$ and $I_+$ the voltage and current for a wave going in the plus $x$-direction, you should get \begin{equation} \label{Eq:II:24:6} V_+=z_0I_+. \end{equation}  Similarly, for the wave going toward minus $x$ the relation is \begin{equation*} V_-=z_0I_-. \end{equation*}  The characteristic impedance—as we found out from our filter equations—is given by \begin{equation} \label{Eq:II:24:7} z_0=\sqrt{\frac{L_0}{C_0}}, \end{equation}  and is, therefore, a pure resistance. To find the propagation speed $v$ and the characteristic impedance $z_0$ of a transmission line, we have to know the inductance and capacity per unit length. We can calculate them easily for a coaxial cable, so we will see how that goes. For the inductance we follow the ideas of Section 17–8, and set $\tfrac{1}{2}LI^2$ equal to the magnetic energy which we get by integrating $\epsO c^2B^2/2$ over the volume. Suppose that the central conductor carries the current $I$; then we know that $B=I/2\pi\epsO c^2r$, where $r$ is the distance from the axis. Taking as a volume element a cylindrical shell of thickness $dr$ and of length $l$, we have for the magnetic energy \begin{equation*} U=\frac{\epsO c^2}{2}\int_a^b\biggl( \frac{I}{2\pi\epsO c^2r} \biggr)^2l\,2\pi r\,dr, \end{equation*}  where $a$ and $b$ are the radii of the inner and outer conductors, respectively. Carrying out the integral, we get \begin{equation} \label{Eq:II:24:8} U=\frac{I^2l}{4\pi\epsO c^2}\ln\frac{b}{a}. \end{equation}  Setting the energy equal to $\tfrac{1}{2}LI^2$, we find \begin{equation} \label{Eq:II:24:9} L=\frac{l}{2\pi\epsO c^2}\ln\frac{b}{a}. \end{equation}  It is, as it should be, proportional to the length $l$ of the line, so the inductance per unit length $L_0$ is \begin{equation} \label{Eq:II:24:10} L_0=\frac{\ln(b/a)}{2\pi\epsO c^2}. \end{equation}  We have worked out the charge on a cylindrical condenser (see Section 12–2). Now, dividing the charge by the potential difference, we get \begin{equation*} C=\frac{2\pi\epsO l}{\ln(b/a)}. \end{equation*}  The capacity per unit length $C_0$ is $C/l$. Combining this result with Eq. (24.10), we see that the product $L_0C_0$ is just equal to $1/c^2$, so $v=1/\sqrt{L_0C_0}$ is equal to $c$. The wave travels down the line with the speed of light. We point out that this result depends on our assumptions: (a) that there are no dielectrics or magnetic materials in the space between the conductors, and (b) that the currents are all on the surfaces of the conductors (as they would be for perfect conductors). We will see later that for good conductors at high frequencies, all currents distribute themselves on the surfaces as they would for a perfect conductor, so this assumption is then valid. Now it is interesting that so long as assumptions (a) and (b) are correct, the product $L_0C_0$ is equal to $1/c^2$ for any parallel pair of conductors—even, say, for a hexagonal inner conductor anywhere inside an elliptical outer conductor. So long as the cross section is constant and the space between has no material, waves are propagated at the velocity of light. No such general statement can be made about the characteristic impedance. For the coaxial line, it is \begin{equation} \label{Eq:II:24:11} z_0=\frac{\ln(b/a)}{2\pi\epsO c}. \end{equation}  The factor $1/\epsO c$ has the dimensions of a resistance and is equal to $120\pi$ ohms. The geometric factor $\ln(b/a)$ depends only logarithmically on the dimensions, so for the coaxial line—and most lines—the characteristic impedance has typical values of from $50$ ohms or so to a few hundred ohms. 