Simple propagation issues

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Dipoles above a perfectly conducting ground plane.

Vertically polarised antennas.

Consider the situation sketched diagrammatically here.... We have placed a vertically orientated dipole at a height h above a perfectly conducting infinite plane earth. If the earth were not there, the field pattern above the earth would be identical, and would be modelled by the sum of the fields from the source dipole and its image.

                           |+               Source
                           |-


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                           |+               Image
                           |-

The image of the vertical dipole in the ground plane consists of a virtual dipole having the SAME current direction as the source dipole. This is because the electric fields from the source dipole meet the ground plane at right angles, and so the electric field on opposite sides of the plane would be in the same direction if we assumed the plane was not physically there, but the field distribution was entirely created by the source and its image.

We now see that we have a 2-element array antenna, with the source elements fed in phase. The element spacing is 2h and there is axial symmetry about a vertical axis. The general form of the radiation pattern of the array factor, in a vertical plane, for the assumed arbitrary special case of h = three wavelengths, is shown here.

Of course, for the full radiation pattern we must multiply the array pattern by the element pattern, so as there is an element pattern null in the direction "straight up" there will be no radiation straight up.

We see that there are numerous fringes caused by the interference pattern between the antenna and its image; these fringes will increase in number and decrease in width as we increase the height h .

On the surface of the ground plane the fields fall off as 1/distance from the antenna footprint, in the far field. Thus, unlike the horizontal polarisation case, the propagation (power density) loss is an inverse square law, until we consider other factors such as the curvature of the earth, and scattering from objects having vertical conducting surfaces.

Horizontally polarised antennas.

If, however, the antenna above the ground plane is horizontally polarised, so:-

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There are still lobes in the vertical plane, which again narrow and increase in number as we increase the height h. However, the lobe maximum closest to the ground is at an angle to the horizontal, for which angle the differential phase delay between the radiation from the source and the image amounts to one half wavelength. We therefore expect the field strength in this direction again to decrease as 1/distance, and the radiated power density to decrease as 1/(distance^2).

Geometrical factors for both polarisations

There is a further factor in practical situations. We have pointed out that the fields on the ground plane are maximum (vertical polarisation) or null (horizontal polarisation). To receive a signal, the receiving antenna has to be placed at a height H above the ground plane. Let us consider the height H, (which for a vertically polarised quarter wave monopole cannot be less than an eighth of a wavelength). In practice, the ground is obstructed and lossy, as well as being reactive, and often the receive antenna is spaced a number of wavelengths above the ground to get a clear line of sight to the transmitter. The point at the centre of this antenna lies at an angle to the "ground zero" point which is on the ground plane between source and image. As we move the receiving antenna further away, this angle decreases proportionately. However, the fields and power densities only obey the power laws derived above providing the angle from the transmit antenna remains the same. It so happens that we nearly always find ourselves on the edge of a lobe, and the angle decrease moves us further towards the null as the range is increased. The power density and the fields in this scenario therefore fall off faster than the amounts calculated above. An observable "rule of thumb" is that the field strengths fall off, for constant H, as 1/(distance^2) and the power density as 1/(distance^4). For the case of moderate values of h and H, a distance can be found where the power law dependence changes from inverse square to inverse quartic (in received signal power density vs distance). We observe that according to the commonly used Rayleigh distance criterion that we have to be further than about 2[(max dimension of antenna)^2]/(lambda) away from the antenna to be in the far field, so the far field region starts at a range 8(n^2)lambda for radiation of wavelength lambda and antenna height h = n(lambda).