Bending Refraction of Radio Waves in Troposphere
The refraction index n = yje^ of the troposphere fluctuates over time and location. In normal conditions the refraction index decreases monotonically versus altitude, because the air density decreases. Because a phenomenon of this kind is a weak function of altitude, it causes slow bending of the ray. Fast changes in the refraction index cause scattering and reflections. Turbulence, where temperature or humidity differs strongly from those of the surroundings, gives rise to scattering. Reflections are caused by horizontal boundaries in the atmosphere due to weather phenomena.
Because n is always close to unity, we often use the socalled refractivity N, which is the difference of the refraction index value from unity in parts per million:
The refractivity for air is obtained from equations
where T is the absolute temperature, p is the barometric pressure (unit mb = hPa), e is the partial pressure of water vapor (mb), and R is the relative humidity. The error of (10.2) is less than 0.5%, if f < 30 GHz, p = 2001,100 mb, T = 240310K, and e < 30 mb. If there is no resonance frequency of oxygen or water vapor molecules in the vicinity, this equation is useful up to 1,000 GHz.
According to an ITUR specification, the average refractivity of the atmosphere versus altitude follows equation
where Na = 315 and bA = 0.136 km 1. These values are calculated using the standardatmosphere model: p = 1,013 mb, its change —12 mb/100m upward, T = 15°C, its change 0.55°C/100m upward, R = 60%. A more accurate model may be obtained by using maps published by ITUR.
Let us now consider a wave that propagates in the troposphere in a direction that makes an angle (f with the horizontal plane, as shown in Figure 10.4. Because the refraction index changes with altitude, the angle ( changes while the wave propagates. According to Snell's law, n cos ( = constant
By derivating this equation with altitude h, we get dn , .
Let us mark the traveled distance in propagation direction as s ; then we get ds 1
We substitute this into (10.6), from which we solve d( 1 dn dn ds n dh cos dh
Rays
Figure 10.4 Refraction (bending) of a wave in the troposphere.
Rays
Figure 10.4 Refraction (bending) of a wave in the troposphere.
cos as n ~ 1, and for terrestrial radio paths in general (f ~ 0°, that is, cos (f~ 1. In an average atmosphere at sea level, the curvature (= — 1/bending radius, i.e., the rate of change of direction with distance) of the ray is df « dn =  10~6Na bA = 43 X 10"6 km"1 (10.9)
as ah that is, the ray bends downward.
Also the Earth's surface bends downward, and its curvature is — 1/R, where the Earth's radius is R = 6,370 km. Therefore the curvature of the ray in reference to the Earth's surface is an 1 1
When analyzing radio paths in the troposphere, we can imagine that the ray is straight, if we use an effective Earth radius KR (see Figure 10.5). In the average atmosphere KR = 8,760 km or K = 1.375. Often we use a value of K = 4/3.
Temporarily the distribution of the refraction index versus altitude may differ considerably from that of the average atmosphere. If an /ah =
—157 X 10 km 1, the ray bends as fast as the Earth's surface (K = «>). If —6 — 1
an /ah < 157 X 10 6 km 1, the ray bends toward the Earth's surface (K< 0). The wave may propagate long distances with successive reflections, as illustrated in Figure 10.6. Propagation with this mechanism is called ducting. Ducting may happen in the socalled inversion layer, where temperature increases rapidly as altitude increases. Such an inversion layer may range in height from a few meters to about 100m and may appear near the ground or at a high altitude.
Figure 10.5 Propagation in the troposphere: (a) refraction; (b) a model of straight propagation above the surface of an extended globe.
10.4 LOS Path
In an LOS path the receiving antenna is above the radio horizon of the transmitting station. Links between two satellites and between an earth station and a satellite are LOS paths, and a path between two terrestrial stations may be such. In the two latter cases the tropospheric effects discussed previously must be taken into account.
According to the ray theory, it is enough that just a ray can propagate unhindered from the transmitting antenna to the receiving antenna. In reality a radio wave requires much more space in order to propagate without extra loss. The free space must be the size of the socalled first Fresnel ellipsoid, shown in Figure 10.7. It is characterized by r 1 + r2  r0 = A/2 (10.11)
where r1 and r2 are distances ofa point on the ellipsoid from the transmitting and receiving points, and r0 is their direct distance. The radius of the first Fresnel ellipsoid is hF = \ (10.12)
We observe that the radius of the ellipsoid in a given path is the smaller the higher the frequency; that is, the ray theory holds better at higher frequencies.
Figure 10.8 shows the extra attenuation due to a knifeedge obstacle in the first Fresnel ellipsoid. According to the ray theory, an obstacle hindering the ray causes an infinite attenuation, and an obstacle just below the ray does not have any effect on the attenuation. However, the result shown in Figure 10.8 is reality and is better explained by Huygens' principle: Every point of the wavefront above the obstacle is a source point of a new spherical wave. This explains diffraction (or bending) of the wave due to an obstacle. Diffraction also helps the wave to propagate behind the obstacle to a space, which is not seen from the original point of transmission. When the knifeedge obstacle is just on the LOS path (h = 0), it causes an extra attenuation of 6 dB. When the obstacle reaches just to the lower boundary of the first Fresnel ellipsoid, the wave arriving at the receiving point may be even stronger than that in a fully obstaclefree case. If there is a hill in the propagation path that cannot be considered a knifeedge, extra attenuation is even higher than in case of a knifeedge obstacle.
When terrestrial radiolink hops are designed, bending of the Earth's surface as well as bending of the ray in the troposphere must be taken into
h/hF
Figure 10.8 Extra attenuation due to the knifeedge diffraction.
h/hF
Figure 10.8 Extra attenuation due to the knifeedge diffraction.
account. The radio horizon is at the distance of ru if the transmitting antenna is at the height of
Example 10.1
Consider a 50km radiolink hop at 10 GHz, when the antenna heights are the same. What are the required antenna heights?
Solution
Taking K = 4/3 we get from (10.13) that the radio horizon of the antennas is in the middle point of the path between the antennas, when hu = 36.8m. By introducing A = 0.03m and r 1 = r2 = 25 km into (10.12) we get the radius of the first Fresnel ellipsoid in the middle point as hp = 19.4m. The antenna heights must be then at least h = hu + hp = 56.2m. In addition, if there are woods in the path, the height of the trees must be taken into account.
At low frequencies, fulfillment of the requirement of leaving the first Fresnel ellipsoid empty is difficult. Often we must compromise. For example, at 100 MHz in the 50km hop of the example, the maximum radius of the first Fresnel ellipsoid is 194m.
A radiolink hop design is aided by using a profile diagram of the terrain, which takes into account the bending of the radio ray in standard conditions. In the diagram, the height scale is much larger than the horizontal scale, and the terrain along the hop is drawn so that perpendicular directions against the Earth's surface are parallel. The middle point of the hop is placed in the middle of the diagram. After drawing the terrain, the antenna heights are selected so that the first Fresnel ellipsoid is fully in free space, if possible. Figure 10.9 presents a profile diagram between Korppoo and Turku in the Finnish archipelago.
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