Practical_Antenna_Handbook_0071639586

164 p a r t I I : F u n d a m e n t a l s
E-field, either. As we did in Chap. 4 (“Transmission Lines and Impedance Matching”),
we can use the principle of superposition to see that a zero voltage between any pair of
points in space or on the earth’s surface can be the result of two separate waves having
equal amplitude and opposite polarities in the region between the two points.
In other words, the boundary conditions at a perfectly conducting surface require us to
conclude that any horizontal E-fields from the original antenna must be matched by
identical fields of equal amplitude and opposite polarity from a second source. But it
turns out (without any proof given here) the only magical entity that can do that for all
locations on the ground plane beneath and near a specific antenna is something we call
an image antenna, looking for all the world like the original antenna viewed in a mirror:
It is located the same distance below the ground plane as the original antenna is above
it, and the horizontal components of its currents, voltages, and fields are equal in amplitude
but opposite in direction (or phase) to those of the original antenna.
Why do we bother with a fictitious image antenna? Because doing so allows us to
predict the effect of a ground plane on the radiation pattern of a nearby antenna simply
by removing the ground plane and replacing it with (fields from) the image antenna! In
other words, we can determine the effect of a nearby ground plane on an antenna’s pattern
by removing the ground plane and treating the antenna and its image as a twoelement
array. Since we’ve already seen how to deal with two-element arrays whose
elements are fed out of phase and in phase, we now can use the same visualizations to
see what happens to far-field patterns from an antenna over perfectly conducting
ground.
Example 5.1 What is the broadside elevation response for a horizontal l/2-dipole of
Fig. 5.6A erected to a height of l/4 above ground?
Solution The dipole is l/4 above ground, so the image antenna must be l/4 below the
surface of the ground. Figure 5.6B shows the signal paths r 1 and r 2 from the two dipoles
to a distant receiver. Because the original dipole is horizontal (with respect to the
ground plane), the image dipole is also horizontal but is considered to be fed 180 degrees
out of phase with it. In your imagination, remove the ground and observe that
the total spacing between the original dipole and the image antenna is twice l/4, or
l/2. Because the antennas are
perfectly out of phase, the
4
4
I
Horizontal dipole
(side view)
combined signal strength at
any far-field receiving point
that is equidistant from them
(which corresponds to all
points on the surface of the
ground) is zero. Further, because
the wave from the image
antenna takes a half-cycle to
travel the half-wavelength
straight up toward the original
dipole, it arrives in perfect
I image
Figure 5.6A A horizontal dipole
and its image antenna.