next up previous
Next: Antenna directivity and effective Up: Radiation and scattering Previous: Radiation and scattering

Basic antenna theory

It possible to solve exactly for the radiation pattern emitted by a linear antenna fed with a sinusoidal current pattern. Assuming that all fields and currents vary in time like ${\rm e}^{-{\rm i}\,\omega t}$, and adopting the Lorentz gauge, it is easily demonstrated that the vector potential obeys the inhomogeneous Helmholtz equation
(\nabla^2+k^2){\bfm A} = - \mu_0\, {\bfm j},
\end{displaymath} (954)

where $k=\omega/c$. The Green's function for this equation, subject to the Sommerfeld radiation condition (which ensures that sources radiate waves instead of absorbing them), is given by Eq. (2.123). Thus, we can invert Eq. (5.1) to obtain
{\bfm A}({\bfm r}) =\frac{\mu_0}{4\pi} \int \frac{{\bfm j}({...
...{\bfm r}'\vert}}{\vert{\bfm r}-{\bfm r}'\vert}\,
d^3{\bfm r}'.
\end{displaymath} (955)

The electric field in the source free region follows from the Ampère-Maxwell equation and ${\bfm B} = \nabla\wedge{\bfm A}$,
{\bfm E} = \frac{\rm i}{k}\nabla\wedge c{\bfm B}.
\end{displaymath} (956)


\vert{\bfm r} - {\bfm r}'\vert = r\,\sqrt{1- 2\,{\bfm n}\!\cdot\!{\bfm r}'/r+
\end{displaymath} (957)

where ${\bfm n} = {\bfm r}/r$. Assuming that $r'\ll r$, this expression can be expanded binomially to give
\vert{\bfm r} -{\bfm r}'\vert = r\left[
1-\frac{{\bfm n}\!\c...
...\frac{2{\bfm n}\!\cdot\!{\bfm r}'}{r}\right)^2
\end{displaymath} (958)

where we have retained all terms up to order $(r'/r)^2$. This expansion occurs in the complex exponential of Eq. (5.2); i.e., it determines the oscillation phase of each element of the antenna. The quadratic terms in the expansion can be neglected provided they can be shown to contribute a phase shift which is significantly less than $2\pi$. Since the maximum possible value of $r'$ is $d/2$, for a linear antenna which extends along the $z$-axis from $z=-d/2$ to $z=d/2$, the phase shift associated with the quadratic terms is insignificant as long as
r \gg \frac{k d^2}{16\pi} = \frac{d^2}{8\lambda},
\end{displaymath} (959)

where $\lambda = 2\pi/k$ is the wavelength of the radiation. This constraint is known as the Fraunhofer limit.

In the Fraunhofer limit we can approximate the phase variation of the complex exponential in Eq. (5.2) by a linear function of $r'$:

\vert{\bfm r} - {\bfm r}'\vert \rightarrow r - {\bfm n}\!\cdot \!{\bfm r}'.
\end{displaymath} (960)

The denominator $\vert{\bfm r}-{\bfm r}'\vert$ in the integrand of Eq. (5.2) can be approximated as $r$ provided that the distance from the antenna is much greater than its length; i.e., provided that
r \gg d.
\end{displaymath} (961)

Thus, Eq. (5.2) reduces to
{\bfm A}({\bfm r}) \simeq \frac{\mu_0}{4\pi}\frac{{\rm e}^{\...
...')\,{\rm e}^{-{\rm i}\,k{\bfm n}\cdot
{\bfm r}'}\,d^3{\bfm r}'
\end{displaymath} (962)

when the constraints (5.6) and (5.8) are satisfied. If the additional constraint
kr\gg 1
\end{displaymath} (963)

is also satisfied, then the electromagnetic fields associated with Eq. (5.9) take the form
$\displaystyle {\bfm B}({\bfm r})$ $\textstyle \simeq$ $\displaystyle {\rm i}\,k\,{\bfm n}\wedge{\bfm A} = {\rm i}\,k\,\frac{\mu_0}{4\p...
...bfm j}({\bfm r}')\,{\rm e}^{-{\rm i}\,k{\bfm n}
\cdot {\bfm r}'}\,d^3{\bfm r}',$ (964)
$\displaystyle {\bfm E}({\bfm r})$ $\textstyle \simeq$ $\displaystyle c{\bfm B}\wedge{\bfm n} = {\rm i}\,ck\,({\bfm n}
\wedge{\bfm A})\wedge{\bfm n}.$ (965)

These are clearly radiation fields, since they are mutually orthogonal, transverse to the radius vector, and satisfy $E=cB\propto r^{-1}$. The three constraints (5.6), (5.8), and (5.10), can be summed up in a single inequality:
d \ll \sqrt{\lambda r} \ll r.
\end{displaymath} (966)

The current density associated with a linear, sinusoidal, centre-fed antenna is

{\bfm j}({\bfm r}) = I \sin (kd/2-k\vert z\vert)\,\delta(x)\,\delta(y)\,\hat{\bfm z}
\end{displaymath} (967)

for $\vert z\vert<d/2$, with ${\bfm j}({\bfm r}) =0$ for $\vert z\vert\geq d/2$. In this case, Eq. (5.9) yields
{\bfm A}({\bfm r}) =\hat{\bfm z}\, \frac{\mu_0 \,I}{4\pi} \f...
...n(kd/2-k\vert z\vert)\,{\rm e}^{-{\rm i}\,k z \cos\theta}\,dz,
\end{displaymath} (968)

where $\cos\theta = {\bfm n}\!\cdot\!\hat{\bfm z}$. The result of this straightforward integration is
{\bfm A}({\bfm r}) = \hat{\bfm z} \,\frac{\mu_0\, I}{4\pi}
...rac{\cos(kd \,\cos\theta/2)
\end{displaymath} (969)

Note from Eqs. (5.11) that the electric field lies in the plane containing the antenna and the radius vector to the observation point. The time-averaged power radiated by the antenna per unit solid angle is
\frac{dP}{d{\mit \Omega}} =\frac{{\rm Re}\,({\bfm n}\!\cdot\...
= \frac{ck^2\,\sin^2\theta\,\vert A\vert^2\,r^2}{2\mu_0}.
\end{displaymath} (970)

\frac{dP}{d{\mit \Omega}} =\frac{\mu_0 c\,I^2}{8\pi^2}
...\cos(kd \,\cos\theta/2)
\end{displaymath} (971)

The angular distribution of power depends on the value of $kd$. In the long wavelength limit $kd\ll 1$ the distribution reduces to

\frac{dP}{d{\mit \Omega}} = \frac{\mu_0 c\,I_0^{~2}}{128\pi^2} \,(kd)^2 \sin^2\theta,
\end{displaymath} (972)

where $I_0 = I\,kd/2$ is the peak current in the antenna. It is easily shown from Eq. (5.13) that the current distribution in the antenna is linear:
I(z) = I_0 (1- 2\vert z\vert/d)
\end{displaymath} (973)

for $\vert z\vert<d/2$. This type of antenna corresponds to a short (compared to the wavelength) oscillating electric dipole, and is generally known as a Hertzian oscillating dipole. The total power radiated is
P = \frac{\mu_0 c\,I_0^{~2} \,(kd)^2}{48\pi}.
\end{displaymath} (974)

In order to maintain the radiation, power must be supplied continuously to the oscillating dipole from some generator. By analogy with the heating power produced in a resistor,
\langle P\rangle_{\rm heat} =\langle I^2\rangle R = \frac{I_0^{~2}\,R}{2},
\end{displaymath} (975)

we can define the factor which multiplies $I_0^{~2}/2$ in Eq. (5.20) as the radiation resistance of the dipole antenna:
R_{\rm rad} = \sqrt{\frac{\mu_0}{\epsilon_0}}\,\frac{(kd)^2}{24\pi}
= 197\left(\frac{d}{\lambda}\right)^2\,\,{\rm ohms}.
\end{displaymath} (976)

Since we have assumed that $\lambda\gg d$, this radiation resistance is necessarily very small. Typically, in devices of this sort the radiated power is swamped by the ohmic losses appearing as heat. Thus, a ``short'' dipole is a very inefficient radiator. Practical antennas have dimensions which are comparable with the wavelength of the emitted radiation.

Probably the most common practical antennas are the half-wave antenna ($k d=\pi$) and the full-wave antenna ($kd=2\pi$). In the former case, Eq. (5.17) reduces to

\frac{d P}{d{\mit\Omega}} =\frac{\mu_0 c\,I^2}{8\pi^2}
\end{displaymath} (977)

In the latter case, Eq. (5.17) yields
\frac{d P}{d{\mit\Omega}} =\frac{\mu_0 c\,I^2}{2\pi^2}
\end{displaymath} (978)

The half-wave antenna radiation pattern is very similar to the characteristic $\sin^2\theta$ pattern of a Hertzian dipole. However, the full-wave antenna radiation pattern is considerably sharper (i.e., it is more concentrated in the transverse directions $\theta=\pm \pi/2$).

The total power radiated by a half-wave antenna is

P = \frac{\mu_0 c\,I^2}{4\pi} \int_0^\pi \frac{\cos^2(\pi \cos\theta/2)}
\end{displaymath} (979)

The integral can be evaluated numerically to give $1.2188$. Thus,
P= 1.2188 \,\frac{\mu_0 c\,I^2}{4\pi}.
\end{displaymath} (980)

Note from Eq. (5.13) that $I$ is equivalent to the peak current flowing in the antenna. Thus, the radiation resistance of a half-wave antenna is given by $P/(I^2/2)$, or
R_{\rm rad} = \frac{0.6094}{\pi} \sqrt{\frac{\mu_0}{\epsilon_0}} = 73 \,\,{\rm ohms}.
\end{displaymath} (981)

This resistance is substantially larger than that for a Hertzian dipole (see Eq. (5.22)). In other words, a half-wave antenna is a far more efficient radiator of electromagnetic radiation than a Hertzian dipole. According to standard transmission line theory, if a transmission line is terminated by a resistor whose resistance matches the characteristic impedance of the line, then all of the power transmitted down the line is dissipated in the resistor. On the other hand, if the resistance does not match the impedance of the line then some of the power is reflected and returned to the generator. We can think of a half-wave antenna, centre-fed by a transmission line, as a 73 ohm resistor terminating the line. The only difference is that the power absorbed from the line is radiated rather than dissipated as heat. Thus, in order to avoid problems with reflected power the impedance of a transmission line feeding a half-wave antenna must be 73 ohms. Not surprisingly, 73 ohm impedance is one of the standard ratings for the co-axial cables used in amateur radio.

next up previous
Next: Antenna directivity and effective Up: Radiation and scattering Previous: Radiation and scattering
Richard Fitzpatrick 2002-05-18