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%TCIDATA{Created=Wed Sep 25 16:04:28 1996}
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\markright{{\sf The Center-Fed Linear Antenna}}
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\section{9.4\quad The Center-Fed Linear Antenna.}

\subsection{9.4.1}

\quad \thinspace \textbf{a.} This is a radiating system of considerable
practical importance, of course. It can be used here as an example to drive
home two points:

\quad (i) If the currents are known, the \textsl{exact} fields in the
radiation zone, hence the \textsl{exact }power radiated, are given by the
vector potential (1.6), page 9.1.3.

\quad (ii) The distribution of currents in not really known without solving
a complicated boundary value problem, because the very process of emission
affects the motion of the charges that are responsible for the radiation.
(This problem of \textsl{radiation reaction} is of fundamental interest and
is taken up by Jackson in Ch.17).

\smallskip\ 

\textbf{b}. If we assume that the only component of $\mathbf{J}$ is 
\[
J_{z}(\mathbf{x}^{\prime })=I\,\sin (\frac{kd}{2}-k\left| z^{\prime }\right|
)\delta (x^{\prime })\delta (y^{\prime }), 
\]
we find from 9.1.3d that the only component of $\mathbf{A}$ is 
\[
A_{z}=\frac{Ie^{ikr}}{cr}\int_{-d/2}^{d/2}\sin (\frac{kd}{2}-k\left|
z^{\prime }\right| )\,e^{-ikz^{\prime }\cos \theta }dz^{\prime } 
\]
\begin{equation}
=\frac{2Ie^{ikr}}{ckr}\frac{\cos \left( \dfrac{kd}{2}\cos \theta \right)
-\cos \left( \dfrac{kd}{2}\right) }{\sin ^{2}\theta }  \label{1}
\end{equation}
where $\theta $ is the polar angle from the antenna's direction. The
particular form of the current is physically reasonable (see below) and has
been chosen to make the integral feasible.

\smallskip\ 

\textbf{c. }Using again $\mathbf{B=\nabla \times A,}$ $\mathbf{E=(}i/k)%
\mathbf{\nabla \times B}$ and keeping only terms of order $1/r$, one finds
that the directions of $\mathbf{B}$ and $\mathbf{E}$ are as in the sketch
(same as for an electric dipole radiative field) and that $B=E=ikA.$

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\newpage

\subsection{9.4.2}

\quad \thinspace \textbf{a.} The radiated power, 
\[
\frac{dP}{d\Omega }=\frac{c}{8\pi }\func{Re}\left[ r^{2}(\mathbf{n\cdot
E\times B}^{*}\right] =\frac{c}{8\pi }r^{2}\left| E\right| ^{2} 
\]
is 
\[
\frac{dP}{d\Omega }=\frac{I^{2}}{2\pi c}\left( \frac{\cos \left( \dfrac{kd}{2%
}\cos \theta \right) -\cos \left( \dfrac{kd}{2}\right) }{\sin \theta }%
\right) ^{2}. 
\]

\textbf{b. }In the limit $kd\ll 1,$ this reduces to 
\[
\frac{I^{2}}{8\pi c}\left( \frac{kd}{2}\right) ^{4}\sin ^{2}\theta . 
\]
The total radiated power in this limit is 
\[
P=\frac{I^{2}}{3c}\left( \frac{kd}{2}\right) ^{4}. 
\]
This agrees with the result 9.2.3e of the dipole approximation when we
identify $I_{0}$ with $Ikd/2.$

\smallskip\ 

\textbf{c. }The next term in the expansion is down by a factor $(kd)^4,$
consistent with the fact that the system has no electric quadrupole or
magnetic dipole moment (or, for that matter, any magnetic moment).

\smallskip\ 

\textbf{d. }The cases of particular interest are those of the half-wave
antenna, $kd=\pi ,$ and the full-wave antenna, $kd=2\pi .$ The radiation
patterns in these two cases are shown by Jackson on page 766. They are still
quite similar to the dipole pattern ``8'', but more localized in the
equatorial plane, like a skinny ``8''$.$ (Note that the antenna points
horizontally in these figures). The distribution of radiated power is
sharper, but the total power is lesser for the full-wave antenna.

\newpage

\subsection{9.4.3}

\quad \thinspace \textbf{a.} The question of the actual current distribution
in the antenna will now be discussed briefly.

\smallskip\ 

\textbf{b. }In the Lorentz gauge, $\phi =-(i/k)\mathbf{\nabla \cdot A}$ and $%
\mathbf{E=(}i/k)(\mathbf{\nabla }\left( \mathbf{\nabla \cdot A}\right) +k^2%
\mathbf{A).}$ When $\mathbf{A}$ has only a z component, $\mathbf{\nabla
\cdot A}=\partial A_z/\partial z$ and 
\begin{equation}
E_z=\frac ik\left( \frac{\partial ^2}{\partial z^2}+k^2\right) A_z.
\label{2}
\end{equation}

\smallskip\ 

\textbf{c. }In the limit where the radius of the antenna, $a,$ is much
smaller than the length $d,$ it is true that only $A_z\neq 0$ and further $%
E_z$ must vanish at $\rho =a.$ Hence, by (\ref{2}) 
\begin{equation}
\left( \frac{\partial ^2}{\partial z^2}+k^2\right) A_z(\rho =a,z)=0.
\label{3}
\end{equation}

\smallskip\ 

\textbf{d.} Now we must compute $A_{z}$ in the near region using the exact
formula (1.2) of page 9.1.2 with $\mathbf{x=(}a\cos \varphi ,\,a\sin \varphi
,\,z)$ and $\mathbf{x}^{\prime }=\mathbf{(}a\cos \varphi ^{\prime },\,a\sin
\varphi ^{\prime },\,z^{\prime })$ (the currents are on the surface of the
antenna, because typically the skin depth $\delta $ is $\ll a).$ Then the
surface current density is $I(z)/2\pi a$ and 
\begin{equation}
A_{z}=\frac{1}{2\pi ac}\int_{-d/2}^{d/2}\int_{0}^{2\pi }\frac{e^{ikR}}{R}%
I(z^{\prime })dz^{\prime }ad\varphi ^{\prime }  \label{4}
\end{equation}
with 
\[
R=\sqrt{(z-z^{\prime })^{2}+a^{2}(\cos \varphi -\cos \varphi ^{\prime
})^{2}+a^{2}(\sin \varphi -\sin \varphi ^{\prime })^{2}} 
\]

\smallskip\ 

\textbf{e. }Simple trigonometry shows that 
\[
R=\sqrt{(z-z^{\prime })^2+4a^2\sin ^2\left( \frac{\varphi -\varphi ^{\prime }%
}2\right) }\,. 
\]
This leads to Jackson's (9.60), with $\beta =(r/2)(\varphi -\varphi ^{\prime
}).$

\smallskip\ 

\textbf{f.} Since the solution of (\ref{3}) is $A_z=a_1\cos kz+a_2\sin kz,$ (%
\ref{4}) can be regarded as an integral equation for the unknown current $%
I(z),$ with the constants $a_1$ and $a_2$ being determined by the boundary
conditions at the free ends of the antenna and at the points where external
leads are attached. Unfortunately the equation is very hard to solve (see
Jackson).

\end{document}
