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Recall our old buddy the complex Poynting vector for harmonic fields
(J6.132):
|
(13.53) |
The factor of
comes from time averaging the fields. This is the energy
per unit area per unit time that passes a point in space. To find the time
average power per solid angle, we must relate the normal area through which
the energy flux passes to the solid angle:
|
(13.54) |
and project out the appropriate piece of S, i. e. --
. We get (with
)
Re |
(13.55) |
where we must plug in E and H from the expressions above for the
far field.
After a bunch of algebra that I'm sure you will enjoy doing, you will obtain:
|
(13.56) |
The polarization of the radiation is determined by the vector inside the
absolute value signs. By this one means that one can project out each
component of
(and hence the radiation) before evaluating the
square independently, if so desired. Note that the different components
of
need not have the same phase (elliptical polarization, etc.).
If all the components of
(in some coordinate system) have the same
phase, then
necessarily lies along a line and the typical angular
distribution is that of (linearly polarized) dipole radiation:
|
(13.57) |
where
is measured between p and n. When you integrate
over the entire solid angle (as part of your assignment) you obtain the total
power radiated:
|
(13.58) |
The most important feature of this is the
dependence which is, after
all, why the sky is blue (as we shall see, never fear).
Example: A centerfed, linear antenna
In this antenna,
and
|
(13.59) |
From the continuity equation (and a little subtle geometry),
|
(13.60) |
and we find that the linear charge density (participating in the oscillation,
with a presumed neutral background) is independent of
:
|
(13.61) |
where the
sign indicates the upper/lower branch of the antenna and the
means that we are really treating
(which cancels the
related terms in the volume integral below). We can then evaluate the dipole
moment of the entire antenna for this frequency:
|
(13.62) |
The electric and magnetic fields for
in the electric dipole
approximation are now given by the previously derived expressions. The
angular distribution of radiated power is
|
(13.63) |
and the total radiated power is
|
(13.64) |
Remarks. For fixed current the power radiated increases as the square
of the frequency (at least when
, i. e. - long wavelengths
relative to the size of the antenna). The total power radiated by the antenna
appears as a ``loss'' in ``Ohm's Law'' for the antenna. Factoring out
, the remainder must have the units of resistance and is called the
radiation resistance of the antenna:
ohms |
(13.65) |
where we do the latter multiplication to convert the resulting units to ohms.
Note that this resistance is there for harmonic currents even if the
conductivity of the metal is perfect. Note further that by hypothesis this
expression will only be valid for small values of
.
Good golly, this is wonderful. We hopefully really understand electric
dipole radiation at this point. It would be truly sublime if all radiators
were dipole radiators. Physics would be so easy. But (alas) sometimes the
current distribution has no
moment and there is therefore
no dipole term! In that case we must look at the next term or so in the
multipolar expansions
Lest you think that this is a wholly unlikely occurrance, please note that a
humble loop carrying a current that varies harmonically is one such system. So
let us proceed to:
Next: Magnetic Dipole and Electric
Up: Dipole Radiation
Previous: Asymptotic properties in the
Contents
Robert G. Brown
2017-07-11