Space Wave Propagation Questions and Answers

Answer: C

Because the impairments grow rapidly with frequency, a margin that would cover a fifty kilometre hop at six gigahertz covers only a few kilometres at thirty eight. Antenna gain in fact rises with frequency for a fixed aperture, which partly offsets it. Far field distance and regulation are not the constraint. Hop length falling as frequency rises is the practical expression of the whole atmospheric impairment picture, and it shapes the architecture of a network.

Answer: D

A profile plots ground height against distance, with the earth curvature for the design k factor included, so the ray and its Fresnel ellipsoid can be drawn on it and clearances read. Digital terrain data has largely replaced manual surveying for the purpose. Signal records, antenna patterns and conductivity maps are separate documents. Drawing the geometry before computing anything is the first step in every fixed link design, and it settles most questions immediately.

Answer: A

Including the curvature for the chosen k factor lets the ray be drawn as a straight line, which is far easier than plotting a curved ray over a flat profile, and changing k simply changes the curvature drawn. Path length, imagery and reflection point are all obtained from the profile rather than being the reason for the curvature. Transferring a complication from the thing being drawn to the reference frame is a general drafting device with wide application.

Answer: B

An obstacle intruding into the first Fresnel zone removes part of the wavefront that would have contributed constructively, and the loss grows rapidly as the intrusion deepens, from six decibels at grazing to twenty or thirty when well obstructed. Spreading is the free space term, absorption in rock is irrelevant since the wave passes around rather than through, and backscatter is negligible. The loss is a diffraction effect and is computed from the geometry rather than from the material.

Answer: C

A rounded hill obstructs the wave over a longer distance along the path, so the diffraction loss exceeds that of a knife edge of the same height, sometimes by ten decibels or more. Real terrain lies between the two idealisations and correction factors are applied. Height alone is insufficient, and frequency dependence remains. Recognising that the idealised case is the optimistic one prevents a path from being declared usable when it is not.

Answer: D

The field arriving at the second obstacle is not a plane wave but the diffracted field of the first, so the losses do not simply add and rigorous treatment becomes involved. Bullington, Epstein-Peterson and Deygout are the standard approximations, each with known biases. Reflection, data accuracy and frequency behaviour are secondary. A problem in which each stage changes the input to the next is inherently harder than one in which contributions are independent.

Answer: A

Engineers use one of the standard approximations, knowing whether it tends to be optimistic or pessimistic, and allow a margin accordingly, verifying by measurement on important links. Abandoning the route is rarely necessary, naive addition of losses is one of the biases being corrected, and calculation is entirely possible if approximate. Using an approximation while knowing the direction of its error is far better practice than either trusting it blindly or refusing to use it.

Answer: B

Leaves and branches contain water and are comparable with the wavelength at microwave frequencies, so a stand of trees can attenuate by several decibels per metre of depth and the loss rises when the foliage is wet. Seasonal variation is substantial in deciduous woodland. Reflection and depolarisation occur too but absorption dominates. A medium whose loss changes with the season is a genuine planning difficulty for a link designed in winter.

Answer: C

Liquid water has a large loss tangent at microwave frequencies, so a wet canopy absorbs considerably more than the same canopy dry, and the effect can amount to several decibels on a short path through trees. Size change and screening are not the mechanism, and reflection is a minor contribution. That the same obstacle behaves differently according to the weather is a reminder that a path survey made in one condition may not describe another.

Answer: D

The first two are standard practice. Rain attenuation becomes significant only above about ten gigahertz, since raindrops are far smaller than the wavelength below that. The three statements between them cover the geometry, the clearance criterion and the dominant impairment, which is most of what a fixed link design consists of, and getting the frequency threshold right determines whether rain appears in the budget at all.