Antennae are designed to transmit and receive electromagnetic
waves. As we know from Sec. 12.2, the superposition integrals for the
scalar and vector potentials result in both the radiation and near
fields. If we confine our interest to the fields far from the
antenna, extensive simplifications are achieved.
Many types of antennae are composed of driven conducting elements
that are extremely thin. This often makes it possible to use simple
arguments to approximate the distribution of current over the length
of the conductor. With the current distribution specified at the
outset, the superposition integrals of Sec. 12.3 can then be used to
determine the associated fields.
An element idz' of the current distribution of an antenna
is pictured in Fig. 12.4.1 at the source location r'. If this
element were at the origin of the spherical coordinate system shown,
the associated radiation fields would be as given by (12.2.23) and
(12.2.24). With the distance to the current element r' much less
than r, how do we adapt these expressions so that they represent the
fields when the incremental source is located at r' rather than at
Figure 12.4.1 Incremental current element at
r' is source for radiation field at (r, , ).
The current elements comprising the antenna are typically within
a few wavelengths of the origin. By contrast, the distance r (say,
from a TV transmitting antenna, where the wavelength is on the order
of 1 meter, to a receiver 10 kilometers away) is far larger. For an
observer in the neighborhood of a point (r, , ), there is
little change in sin /r, and hence in the magnitude of the
field, caused by a displacement of the current element from the origin
to r'. However, the phase of the electromagnetic wave
launched by the current element is strongly influenced by changes in
the distance from the element to the observer that are of the order of
a wavelength. This is seen by writing the argument of the exponential
term in terms of the wavelength , jkr = j2 r/.
With the help of Fig. 12.4.1, we see that the distance from the
source to the observer is r - r' ir. Thus, for the
current element located at r' in the neighborhood of the origin,
the radiation fields given by (12.2.23) and (12.2.24) are
Because E and H are vector fields, yet another approximation
is implicit in writing these expressions. In shifting the current
element, there is a slight shift in the coordinate directions at the
observer location. Again, because r is much larger than |r'|,
this slight change in the direction of the field can be ignored.
Thus, radiation fields due to a superposition of current elements can
be found by simply superimposing the fields as though they were
Distributed Current Distribution
A wire antenna, driven by a given current distribution Re [(z) exp (j t)], is shown in Fig. 12.4.2. At the terminals,
the complex amplitude of this current is = Io exp (j
t + o ). It follows from (1) and the superposition
principle that the magnetic radiation field for this antenna is
Figure 12.4.2 Line current distribution as source
of radiation field.
Note that the role played by id for the incremental dipole is now
played by i(z' )dz'. For convenience, we define a field
pattern function o( ) that gives the dependence
of the E and H fields
where l is the length of the antenna and o( ) is
dimensionless. With the aid of o( ), one may write (3)
in the form
By definition, (' ) = Ioexp(jo )
if z' is evaluated at the terminals of the antenna. Thus,
o is neither a function of Io nor of o.
In order to evaluate (5), one needs to know the current dependence on
z' , (z' ). One can show that the current distribution
on a (open-ended) thin wire is made up of a "standing wave" with the
dependence sin (2 s/ ) upon the coordinate s measured
along the wire, from the end of the wire. The proof of this
statement will be presented in Chapter 14, when we shall discuss the
current distribution in a coaxial cable.
Example 12.4.1. Radiation Pattern of Center-Fed Wire Antenna
A wire antenna, fed at its midpoint and on the z axis, is shown
in Fig. 12.4.3. The current distribution is "given" according to
the above remarks.
Figure 12.4.3 Center-fed wire antenna with
standing-wave distribution of current.
In setting up the radiation field superposition integral, (5),
observe that r' ir = z' cos .
Evaluation of the integral7 then gives
7To carry out the integration,
first express the integration over the positive and negative segments
of z' as separate integrals. With the sine functions represented by
the sum of complex exponentials, the integration is reduced to a sum
of integrations of complex exponentials.
The radiation pattern of the wire antenna is proportional to the
absolute value squared of the -dependent factor of o
In viewing the plots of this radiation pattern shown in Fig.
12.4.4, remember that it is the same in any plane of constant .
Thus, a three-dimensional picture of the function ( ,
) is generated by rotating one of these patterns about the z axis.
Figure 12.4.4 Radiation patterns for center-fed
The radiation pattern for a half-wave antenna differs little from
that for the short dipole, shown in Fig. 12.2.7. Because of the
interference between waves generated by segments having different
phases and amplitudes, the pattern for longer wires is more complex.
As the length of the antenna is increased to many wavelengths, the
number of lobes increases.
Desired radiation patterns are often obtained by combining driven
elements into arrays. To illustrate, consider an array of 1 + n
elements, the first at the origin and designated by "0". The others
are designated by i = 1 n and respectively located at
ai. We can find the
radiation pattern for the array by summing over the contributions of
the separate elements. Each of these takes the form of (5),
with r r - ai ir , Io
Ii, o i, and o
( ) ( ).
In the special case where the magnitude (but not the phase) of each
element is the same and the elements are identical, so that
Ii = Io and i = o, this expression can be written as
where the array factor is
Note that the radiation pattern of the array is represented by the
square of the product of o, representing the pattern for a single
element, and the array factor a. If the n + 1 element array is
considered one element in a second array, these same arguments could
be repeated to show that the radiation pattern of the array of arrays
is represented by the square of the product of o, a and
the square of the array factor of the second array.
Figure 12.4.5 Array consisting of two elements
with spacing, a.
Example 12.4.2. Two-Element Arrays
The elements of an array have a spacing a, as shown in Fig.
12.4.5. The array factor follows from evaluation of (12), where
ao = 0 and a1 = aix. The projection of ir
into ix gives (see Fig. 12.4.5)
It follows that
It is convenient to write this expression as a product of a part that
determines the phase and a part that determines the amplitude.
Dipoles in Broadside Array
With the elements short compared to a
wavelength, the individual patterns are those of a dipole. It follows
from (4) that
With the dipoles having a half-wavelength spacing and driven in phase,
The magnitude of the array factor follows from (15).
The radiation pattern for the array follows from (16) and (18).
Figure 12.4.6 geometrically portrays how the single-element pattern
and array pattern multiply to provide the radiation pattern. With
the elements a half-wavelength apart and driven in phase,
electromagnetic waves arrive in phase at points along the y axis
and reinforce. There is no radiation in the x directions,
because a wave
initiated by one element arrives out of phase with the wave being
initiated by that second element. As a result, the waves reinforce
along the y axis, the "broadside" direction, while they cancel
along the x axis.
Figure 12.4.6 Radiation pattern of dipoles in
phase, half-wave spaced, is product of pattern for individual
elements multiplied by the array factor.
Dipoles in End-Fire Array
With quarter-wave spacing and driven 90 degrees out of phase,
the magnitude of the array factor follows from (15) as
The radiation pattern follows from (16) and (21).
Shown graphically in Fig. 12.4.7, the pattern is now in the -x
direction. Waves initiated in the -x direction by the element at
x = a
arrive in phase with those originating from the second element. Thus,
the wave being initiated by that second element in the -x direction
is reinforced. By contrast, the wave initiated in the +x direction
by the element at x = 0 arrives 180 degrees out of phase with the
wave being initiated in the +x direction by the other element.
Thus, radiation in the +x direction cancels, and the array is
Figure 12.4.7 Radiation pattern for dipoles
quarter-wave spaced, 90 degrees out of phase.
Finite Dipoles in End-Fire Array
Finally, consider a pair of
finite length elements, each having a length l, as in Fig. 12.4.3.
The pattern for the individual elements is given by (8). With the
elements spaced as in Fig. 12.4.5, with a = /4 and driven
90 degrees out of phase, the magnitude of the array factor is given
by (21). Thus, the amplitude of the radiation pattern is
For elements of length l = 3 /2 (kl = 3 ), this
pattern is pictured in Fig. 12.4.8.
Figure 12.4.8 Radiation pattern for two center-fed
wire antennas, quarter-wave spaced, 90 degrees out-of-phase, each
having length 3 /2.
The time average power flux density, Sr ( , ),
normalized to the power flux density averaged over the surface of a
sphere, is called the gain of an antenna.
If the direction is not specified, it is implied that G is the
gain in the direction of maximum gain.
The radial power flux density is the Poynting flux, defined by
(11.2.9). Using the time average theorem, (11.5.6), and the fact
that the ratio of E to H for the radiation field is
/, (2), gives
Because the radiation pattern expresses the ( , )
dependence of |E |2 with a multiplicative factor that is in
the numerator and denominator of (24), G can be evaluated using the
radiation pattern for Sr.
Example 12.4.3. Gain of an Electric Dipole
For the electric dipole, it follows from (1) and (2) that the
radiation pattern is proportional to sin2 ( ). The gain in
the direction is then
and the "gain" is 3/2.