Radiation from a linear centre-fed antenna

As an illustration of the use of a multipole expansion for a source whose dimensions are comparable to a wavelength, consider the radiation from a linear centre-fed antenna. We assume that the antenna lies along the $z$-axis, and extends from $z=-d/2$ to $z=d/2$. The current flowing along the antenna vanishes at the end points, and is an even function of $z$. Thus, we can write

$$I(z, t) = I(\vert z\vert)\, {\rm e}^{-{\rm i}\,\omega\,t},$$ (1293)

where $I(d/2)=0$. Since the current flows radially, ${\bfm r}\wedge{\bfm j} = 0$. Furthermore, there is no intrinsic magnetization. Thus, according to Eq. (7.91), all of the magnetic multipole coefficients $a_M(l,m)$ vanish. In order to calculate the electric multipole coefficients $a_E(l,m)$, we need expressions for the charge and current densities. In spherical polar coordinates the current density ${\bfm j}$ can be written in the form

$${\bfm j}({\bfm r}) = \hat{\bfm r}\,\frac{I(r)}{2\pi r^2} \,\left[\delta(\cos\theta-1)-\delta(\cos\theta+1)\right],$$ (1294)

for $r<d/2$, where the delta functions cause the current to flow only upwards and downwards along the $z$-axis. From the continuity equation (7.78), the charge density is given by

$$\rho({\bfm r}) = \frac{1}{{\rm i}\,\omega} \frac{d I(r)}{dr}... ...{\delta(\cos\theta-1) -\delta(\cos\theta+1)}{2\pi r^2}\right],$$ (1295)

for $r<d/2$.

These expressions for ${\bfm j}$ and $\rho$ can be substituted into Eq. (7.90) to give

$$a_E(l,m) = \frac{\mu_0 c\,k^2}{2\pi\sqrt{l(l+1)}} \int_0^{d/2} dr\left\{ kr\,j_l(kr)\,I(r) -\frac{1}{k} \frac{dI(r)}{dr} \frac{d[r\,j_l(kr)]}{dr}\right\}$$

$$\mbox{\hspace{1cm}}\int d\Omega \,Y_{lm}^\ast\left[\delta(\cos\theta-1) -\delta(\cos\theta+1)\right].$$ (1296)

The angular integral has the value

$$\int d\Omega \,Y_{lm}^\ast\left[\delta(\cos\theta-1) -\delta... ...] = 2\pi\,\delta_{m,0}\, \left[Y_{l0}(0) - Y_{l0}(\pi)\right],$$ (1297)

showing that only $m=0$ multipoles occur. This is hardly surprising given the cylindrical symmetry of the antenna. The $m=0$ spherical harmonics are even (odd) about $\theta=\pi/2$ for $l$ even (odd). Hence, the only nonvanishing multipoles have $l$ odd. So, the angular integral takes the value

$$\int d\Omega \,Y_{lm}^\ast\left[\delta(\cos\theta-1) -\delta(\cos\theta+1)\right] = \sqrt{4\pi(2l+1)},$$ (1298)

for $l$ odd and $m=0$. After some slight rearrangement, Eq. (7.99) can be written

$$a_E(l,0) = \frac{\mu_0 c\,k}{2\pi} \left[\frac{4\pi (2l+1)} {l(l+1)}\right]^... ...nt_0^{d/2} \left\{ -\frac{d}{dr}\!\left[ r j_l(kr)\,\frac{dI}{dr}\right]\right.$$

$$\mbox{\hspace{2cm}}\left.+r j_l(kr) \left(\frac{d^2 I}{dr^2} + k^2\,I\right)\right\}\,dr.$$ (1299)

In order to evaluate the integral (7.102) we need to specify the current $I(z)$ along the antenna. In the absence of radiation, the sinusoidal time variation at frequency $\omega$ implies a sinusoidal space variation with wavenumber $k=\omega/c$. However, the emission of radiation generally modifies the current distribution. The correct current $I(z)$ can only be found be solving a complicated boundary value problem. For the sake of simplicity, we assume that $I(z)$ is a known function; specifically,

$$I(z) = I\,\sin(kd/2 - k\,\vert z\vert),$$ (1300)

for $z<d/2$, where $I$ is the peak current. With a sinusoidal current the second term in curly brackets in Eq. (7.102) vanishes. The first term is a perfect differential. Consequently, Eqs. (7.102) and (7.103) yield

$$a_E(l, 0) = \frac{\mu_0 c\, I}{\pi d}\left[\frac{4\pi (2l+1)} {l(l+1)}\right]^{1/2} \left(\frac{kd}{2}\right)^2\,j_l(kd/2),$$ (1301)

for $l$ odd.

Let us consider the special cases of a half-wave antenna ($k d=\pi$; i.e., the length of the antenna is half a wavelength) and a full-wave antenna ($kd=2\pi$). For these two values of $kd$ the $l=1$ coefficient is tabulated in Table 3, along with the relative values for $l= 3$, 5. It is clear from the table that the coefficients decrease rapidly in magnitude as $l$ increases, and that higher $l$ coefficients are more important the larger the source dimensions. However, even for a full-wave antenna it is generally adequate to retain only the $l=1$ and $l= 3$ coefficients in order to calculate the angular distribution of the radiation. It is certainly adequate to keep only these two harmonics in order to calculate the total power radiated (which depends on the sum of the squares of the coefficients).

Table 3: The first few electric multipole coefficients for a half-wave and a full-wave antenna

$kd$	$a_E(1,0)$	$a_E(3,0)/a_E(1,0)$	$a_E(5,0)/a_E(1,0)$
$\pi$	$4\sqrt{6\pi}\,(\mu_0 c\,I/4\pi d)$	$4.95\times 10^{-2}$	$1.02\times 10^{-3}$
$2\pi$	$4\pi\sqrt{6\pi}\,(\mu_0 c\,I/4\pi d)$	$0.325$	$3.09\times 10^{-2}$

In the radiation zone the multipole fields (7.54) reduce to

$$c{\bfm B} \simeq \frac{{\rm e}^{\,{\rm i}\,(kr -\omega t)}}{kr} \sum_{l,m} (-{\rm i})^{l+1} \left[ a_E(l,m)\,{\bfm X}_{lm}\right.$$

$$\mbox{\hspace{2cm}}\left.+ a_M(l,m) \,{\bfm n}\wedge{\bfm X}_{lm}\right],$$ (1302)

$${\bfm E} \simeq c{\bfm B}\wedge{\bfm n},$$ (1303)

where use has been made of the asymptotic form (7.16). The time-averaged power radiated per unit solid angle is given by

$$\frac{dP}{d{\mit \Omega}} = \frac{{\rm Re}\,( {\bfm n}\!\cdot\!{\bfm E}\wedge{\bfm B}^{\ast})\,r^2} {2\mu_0},$$ (1304)

or

$$\frac{dP}{d{\mit \Omega}} = \frac{1}{2\,\mu_0 c \, k^2}\left... ...m} +a_M(l,m)\,{\bfm n}\wedge{\bfm X}_{lm}\right]\right\vert^2.$$ (1305)

Retaining only the $l=1$ and $l= 3$ electric multipole coefficients, the angular distribution of the radiation from the antenna is given by

$$\frac{dP}{d{\mit \Omega}} = \frac{\vert a_E(l,0)\vert^2}{4\,... ...{a_E(3,0)}{\sqrt{6}\,a_E(1,0)}\,{\bfm L} Y_{3,0}\right\vert^2,$$ (1306)

where use has been made of Eq. (7.52). The various factors in the absolute square are

$$\vert{\bfm L} Y_{1,0}\vert^2 = \frac{3}{4\pi}\,\sin^2\theta,$$ (1307)

$$\vert{\bfm L} Y_{3,0}\vert^2 = \frac{63}{16\pi}\,\sin^2\theta \,(5\cos^2\theta -1)^2,$$ (1308)

$$({\bfm L} Y_{1,0})^\ast\!\cdot\! ({\bfm L} Y_{3,0}) = \frac{3\sqrt{21}} {8\pi}\,\sin^2\theta\,(5\cos^2\theta-1).$$ (1309)

With these angular factors, Eq. (7.108) becomes

$$\frac{dP}{d{\mit \Omega}} = \lambda\,\frac{3\,\mu_0 c\, I^2}... ...8}}\frac{a_E(3,0)}{a_E(1,0)} \,(5\cos^2\theta-1)\right\vert^2,$$ (1310)

where $\lambda$ equals 1 for a half-wave antenna and $\pi^2/4$ for a full-wave antenna. The coefficient in front of $(5\cos^2\theta -1)$ is $0.0463$ and $0.304$ for the half-wave and full-wave antenna, respectively. It turns out that the radiation pattern from the two-term multipole expansion given above is almost indistinguishable from the exact result for the case of a half-wave antenna. For the case of a full-wave antenna the two-term expansion yields a radiation pattern which differs from the exact result by less than 5%.

The total power radiated by the antenna is

$$P = \frac{1}{2\,\mu_0 c\,k^2} \sum_{l\,{\rm odd}} \vert a_E(l,0)\vert^2,$$ (1311)

where use has been made of Eq. (7.71). It is evident from Table 3 that a two-term multipole expansion gives an accurate expression for the radiated power for both a half-wave and a full-wave antenna. In fact, a one-term multipole expansion gives a fairly accurate result for the case of a half-wave antenna.

It is clear from the above analysis that the multipole expansion converges rapidly when the source dimensions are of order the wavelength of the radiation. It is also clear that if the source dimensions are much less than the wavelength then the multipole expansion is likely to be completely dominated by the term corresponding to the lowest value of $l$.