The General Propagation Problem Above 10 GHz

Propagation of radio waves through the atmosphere above 10 GHz involves not only free-space loss but several other important factors. As expressed in Ref. 1, these are as follows:

1. The gaseous contribution of the homogeneous atmosphere due to resonant and nonresonant polarization mechanisms.

2. The contribution of inhomogeneities in the atmosphere.

3. The particulate contributions due to rain, fog, mist, and haze (includes dust, smoke, and salt particles in the air).

Under (1) we are dealing with the propagation of a wave through the atmosphere under the influence of several molecular resonances, such as water vapor (H2O) at 22 and 183 GHz, oxygen (O2) with lines around 60 GHz, and a single oxygen line at 119 GHz. These points and their relative attenuation are shown in Figure 9.1.

Other gases display resonant lines as well, such as N2O, SO2, O3, NO2, and NH3, but because of their low density in the atmosphere, they have negligible effect on propagation.

The major offender is precipitation attenuation [under (2) and (3)]: it can exceed that of all other sources of excess attenuation above 18 GHz. Rainfall and its effect on propagation are the major themes of this chapter.

To better understand these loss mechanisms due to the atmosphere, including rainfall, we very briefly review absorption and scattering. It will be appreciated that when an incident electromagnetic wave passes over an object that has dielectric properties different from the surrounding medium, some energy is absorbed and some is scattered. That which is absorbed heats the absorbing material; that which is scattered is quasi-isotropic and relates to the wavelength of the incident wave. The smaller the scatterer, the more isotropic it is in direction with respect to the wavelength of the incident energy.

We can develop a formula derived from equation (1.9a) to calculate total transmission loss for a given link:

AttenuationdB = 92.45 + 20 log FGHz + 20 log Dkm + a + b + c + d + e

where F = operating frequency in GHz D = link length in kilometers

Attenuation Due Clouds And Fog
Figure 9.1. Specific attenuation due to atmospheric gases: atmospheric pressure, 1013 mb; temperature, 15°C; water vapor, 7.5 g/m3. (From Figure 5, page 16, Rec.ITU-R P.676-6, Ref. 2.)

a = excess attenuation* (dB) due to water vapor b = excess attenuation (dB) due to mist and fog c = excess attenuation (dB) due to oxygen (O2)

*''Excess''in this context means in excess of free-space loss.

d = sum of the absorption losses (dB) due to other gases e = excess attenuation (dB) due to rainfall

Notes and Comments on Equation (9.1)

1. Parameter a varies with relative humidity, temperature, atmospheric pressure, and altitude. The transmission engineer assumes that the water vapor content is linear with these parameters and that the atmosphere is homogeneous (actually horizontally homogeneous but vertically stratified). There is a water vapor absorption band around 22 GHz caused by molecular resonance.

2. Parameters c and d are assumed to vary linearly with atmospheric density, thus directly with atmospheric pressure, and are a function of altitude (e.g., it is assumed that the atmosphere is homogeneous).

3. Parameters b and e vary with the density of the rainfall cell or cloud and the size of the rainfall drops or water particles such as fog or mist. In this case the atmosphere is most certainly not homogeneous. (Droplets smaller than 0.01 cm in diameter are considered mist/fog, while those larger than 0.01 cm are rain.) Ordinary fog produces about 0.1 dB/km of excess attenuation at 35 GHz, rising to 0.6 dB/km at 75 GHz.

In equation (9.1) terms b and d can often be neglected; terms a and c are usually lumped together and called ''excess attenuation due to atmospheric gases'' or simply ''atmospheric attenuation.'' If we were to install a 10-km LOS link at 22 GHz, in calculating transmission loss (at sea level), about 1.6 dB would have to be added for what is called atmospheric attenuation, but which is predominantly water vapor absorption, as shown in Figure 9.1.

It will be noted in Figure 9.1 that there are frequency bands with relatively high levels of attenuation per unit distance. Some of these frequency bands are rather narrow (e.g., H2O at about 22 GHz). Others are wide, such as the O2 absorption band, which covers from about 58 to 62 GHz with skirts down to 50 and up to 70 GHz. At its peak around 60 GHz, the sea-level attenuation is about 15 dB/km. One could ask: ''Of what use are these high-absorption bands?'' Actually, the 5862-GHz band is appropriately assigned for satellite cross-links. These links operate out in space far above the limits of the earth's atmosphere, where the terms a through e may be completely neglected. It is particularly attractive on military cross-links having an inherent protection from earth-based enemy jammers by that significant atmospheric attenuation factor. It is also useful for very short-haul military links such as ship-to-ship secure communication systems. Again, it is the severe atmospheric attenuation that offers some additional security for signal intercept (LPI—low probability of intercept) and against jamming.

On the other hand, Figure 9.1 shows some radio-frequency bands that are relatively open. These openings are often called windows. Three such win-

TABLE 9.1 Windows for Point-to-Point Service
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