## RAIN ATTENUATION

Attenuation due to rain is made up of two components: absorption and scattering. Absorption takes place when the incident signal energy is transformed into mechanical energy, thereby heating the raindrop. Scattering occurs when the incident signal is redirected away from the desired propagation path without energy loss to the raindrop. The relative importance of scattering and absorption is a function of the complex index of refraction of the raindrop and the size of the raindrop relative to the wavelength of the signal (15). The significance of the scattered component generally increases both with the signal frequency and with the size of the raindrop. In general, attenuation due to rain increases with increasing signal frequency for a given rain rate. Typical annual cumulative statistics for frequencies of 12 GHz, 18.7 GHz, 39.6 GHz, and 49.5 GHz on paths at an elevation angle of approximately 38° in Europe are shown in Fig. 4 (some data were extracted from Ref. 16). These curves show a monotonic increase in attenuation with decreasing time percentage, which is typical of temperate climates. There is evidence, however, that in tropical, high-rain-rate regions there is a saturation effect in the attenuation and rainfall rate statistics at low time percentages, leading to a breakpoint in the cumulative statistics (17). This and other features of high-rain-rate regions make the accurate prediction of attenuation due to rain very difficult for those locations.

Rain does not usually consist of a homogeneous collection of raindrops falling at a constant rate. The size, shape, temperature, and fall velocities usually vary in a dynamic manner along the path. To calculate the attenuation of a radio signal passing through rain it is necessary to invoke a drop — size distribution and to integrate the attenuation contributions of the various raindrops along the path through the rain. The characteristics of rain vary so much in space and

Figure 3. Zenith attenuation due to atmospheric gases (from Fig. 3 of Ref. 14). Curve A: standard atmosphere (7.5 g/m3); curve B: dry atmosphere. © ITU. Reproduced with permission. |

Frequency (GHz) |

Step 1. Calculate the effective rain height, hR, for the latitude of the station, ф:

Figure 4. Average annual cumulative statistics of total path attenuation at an elevation angle of about 38° in Europe (some data extracted from Ref. 16). |

5 — 0.075(ф — 23) for

5 for

hR (km) =| 5 for

5 + 0.1(ф + 21) for

0 for

ф> 23° Northern Hemisphere 0° < ф < 23° Northern Hemisphere

0° > ф > -23° Southern Hemisphere

-71° < ф < -21° Southern Hemisphere

ф < -71° Southern Hemisphere

Percentage of a year attenuation is exceeded |

Step 2. For в a 5° compute the slant-path length, Ls, below the rain height from:

Ls = (hR — hg)/ (sinв) km

For в < 5°, the following formula is used:

Ls = [2(hR — hg)]/{[sin2 в + 2(hR — hs)/8500]1/2 + sin в} km

time that it is necessary to resort to empirical methods that employ statistical averaging. The accepted approach is to use a power law representation based on the rainfall rate R (usually a rainfall rate measured at a point on the ground close to the communications link). The specific attenuation у (dB/km) in rain is related to the rainfall rate R (mm/h) by the power law relationship

Step 3. Calculate the horizontal projection, LG, of the slant- path length from:

Lg = Lg cos в km

Step 4. Obtain the rain intensity, R0.0i, exceeded for 0.01% of an average year at the site

Step 5. Calculate the reduction factor, r0.01, for 0.01% of the time for R0.01 ^ 100 mm/h

Figure 5. Schematic presentation of earth-space path (from Fig. 1 in Ref. 13). © ITU. Reproduced with permission. |

Y = kR“ (dB/km) (17)

where k and a are regression coefficients (18) that take account of absorption and scattering.

## Добавить комментарий