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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
, and adopting the Lorentz gauge,
it is easily demonstrated that the vector potential obeys the inhomogeneous
Helmholtz equation

(954) 
where .
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

(955) 
The electric field in the source free region follows from the AmpèreMaxwell
equation and
,

(956) 
Now

(957) 
where
.
Assuming that , this expression can be expanded binomially to
give

(958) 
where we have retained all terms up to order
. 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 . Since the maximum possible value of is ,
for a linear antenna which extends along the axis from to
, the phase shift associated with the quadratic
terms is insignificant as long as

(959) 
where
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 :

(960) 
The denominator
in the integrand of
Eq. (5.2) can be approximated as provided
that the distance from the antenna is much greater than its length;
i.e., provided that

(961) 
Thus, Eq. (5.2)
reduces to

(962) 
when the constraints (5.6) and (5.8) are satisfied. If the additional
constraint

(963) 
is also satisfied, then the electromagnetic fields associated with Eq. (5.9)
take the form
These are clearly radiation fields, since they are mutually orthogonal,
transverse to the radius vector, and satisfy
.
The three constraints (5.6), (5.8), and (5.10), can be summed up
in a single inequality:

(966) 
The current density associated with a linear, sinusoidal,
centrefed antenna is

(967) 
for , with
for . In this case, Eq. (5.9)
yields

(968) 
where
. The result of this
straightforward integration is

(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 timeaveraged power radiated by the antenna per unit solid angle
is

(970) 
Thus,

(971) 
The angular distribution of power depends on the value of
. In the long wavelength limit the distribution
reduces to

(972) 
where is the peak current in the antenna. It is easily
shown from Eq. (5.13) that the current distribution in the antenna is linear:

(973) 
for . 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

(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,

(975) 
we can define the factor which multiplies in Eq. (5.20)
as the radiation resistance of the dipole antenna:

(976) 
Since we have assumed that , 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 halfwave antenna
() and the fullwave antenna (). In the former case,
Eq. (5.17) reduces to

(977) 
In the latter case, Eq. (5.17) yields

(978) 
The halfwave antenna radiation pattern
is very similar to the characteristic pattern of
a Hertzian dipole. However, the fullwave antenna radiation pattern
is considerably sharper (i.e., it is more concentrated in the
transverse directions
).
The total power radiated by a halfwave antenna is

(979) 
The integral can be evaluated numerically to give . Thus,

(980) 
Note from Eq. (5.13) that is
equivalent to the peak current flowing in the antenna. Thus, the radiation
resistance of a halfwave antenna is given by , or

(981) 
This resistance is substantially larger than that for a
Hertzian dipole (see Eq. (5.22)). In other words, a halfwave 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
halfwave antenna, centrefed 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 halfwave
antenna must be 73 ohms. Not surprisingly, 73 ohm impedance
is one of the standard ratings for the coaxial cables used in amateur
radio.
Next: Antenna directivity and effective
Up: Radiation and scattering
Previous: Radiation and scattering
Richard Fitzpatrick
20020518