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The computation of radiowave attenuation in the atmosphere is a series of radio propagation models and methods to estimate the path loss due to attenuation of the signal passing through the radio propagation models and methods to estimate the path loss due to attenuation of the signal passing through the atmosphere by the absorption of its different components. There are many well-known facts on the phenomenon and qualitative treatments in textbooks.[1] A document published by the International Telecommunication Union (ITU) [2] provides some basis for a quantitative assessment of the attenuation. That document describes a simplified model along with semi-empirical formulas based on data fitting. It also recommended an algorithm to compute the attenuation of radiowave propagation in the atmosphere. NASA also published a study on a related subject.[3] Free software from CNES based on ITU-R recommendations is available for download and is available to the public.

The boundary value problem

When a point S communicates with a point T, the orientation of the ray is specified by an elevation angle. In a naïve way, the angle can be given by tracing a straight line from S to T. This specification does not guarantee that the ray will reach T: the variation of refraction index bends the ray trajectory. The elevation angle has to be modified[3] to take into account the bending effect.

For the eikonal equation, this correction can be done by solving a Note: A MATLAB version for the uplink (Telecommunications link) is available from the ITU[2]

When a point S communicates with a point T, the orientation of the ray is specified by an elevation angle. In a naïve way, the angle can be given by tracing a straight line from S to T. This specification does not guarantee that the ray will reach T: the variation of refraction index bends the ray trajectory. The elevation angle has to be modified[3] to take into account the bending effect.

For the eikonal equation, this correction can be done by solving a boundary value problem. As the equation is of second order, the problem is well defined. In spite of the lack of a firm theoretical basis for the ITU method, a trial-error by dichotomy (or binary search) can als

For the eikonal equation, this correction can be done by solving a boundary value problem. As the equation is of second order, the problem is well defined. In spite of the lack of a firm theoretical basis for the ITU method, a trial-error by dichotomy (or binary search) can also be used. The next figure shows the results of numerical simulations.

The curve labeled as bvp is the trajectory found by correcting the elevation angle. The other two are from a fixed step and a variable step (chosen in accordance with the ITU recommendations[6]) solutions without the elevation angle correction. The nominal elevation angle for this case is −0.5-degree. The numerical results obtained at 22.5 GHz were:

Note the way the solution bvp bents over the straight line. A consequence of this property is that the ray can reach locations situated below the horizon of S. This is consistent with observations.[8] The trajectory is a Concave function is a consequence of the fact that the gradient of the refraction index is negative, so the Eikonal equation implies that the second derivative of the trajectory is negative. From the point where the ray is parallel to ground, relative to the chosen coordinates, the ray goes down but relative to ground level, the ray goes up.

Often engineers are interested in finding the limits of a system. In this case, a simple idea is to try some low elevation angle and let the ray reach the desired altitude. This point of view has a problem: if suffice to take the angle for which the ray has a tangent point of lowest altitude. For instance with the case of a source at 5 km altitude, of nominal elevation angle −0.5-degree and the target is at 30 km altitude; the attenuation found by the boundary value method is 11.33 dB. The previous point of view of worst case leads to an elevation angle of −1.87-degree and an attenuation of 170.77 dB. With this kind of attenuation, every system would be unusable! It was found also for this case that with the nominal elevation angle, the distance of the tangent point to ground is 5.84 km; that of the worst case is 2.69 km. The nominal distance from source to target is 6383.84 km; for the worst case, it is 990.36 km.

There are many numerical methods to solve boundary value problems.[9] For the Eikonal

Often engineers are interested in finding the limits of a system. In this case, a simple idea is to try some low elevation angle and let the ray reach the desired altitude. This point of view has a problem: if suffice to take the angle for which the ray has a tangent point of lowest altitude. For instance with the case of a source at 5 km altitude, of nominal elevation angle −0.5-degree and the target is at 30 km altitude; the attenuation found by the boundary value method is 11.33 dB. The previous point of view of worst case leads to an elevation angle of −1.87-degree and an attenuation of 170.77 dB. With this kind of attenuation, every system would be unusable! It was found also for this case that with the nominal elevation angle, the distance of the tangent point to ground is 5.84 km; that of the worst case is 2.69 km. The nominal distance from source to target is 6383.84 km; for the worst case, it is 990.36 km.

There are many numerical methods to solve boundary value problems.[9] For the Eikonal equation, due to the good behavior of the refraction index just a simple Shooting method can be used.