Physics · Ch 13 — Electromagnetic Waves and Communication System

Space wave

13.4.2

Space wave

Space wave propagation covers radio waves that reach the receiving antenna either directly, travelling in a straight line (a line-of-sight path) from the transmitting antenna, or indirectly, after reflecting off the ground, off a satellite, or off the troposphere (the lowest layer of the atmosphere) -- waves reflected specifically by the troposphere are called tropospheric waves. Radio waves of frequency greater than 30 MHz are able to pass through the ionosphere (the atmospheric layer lying roughly between 60 km and 1000 km altitude) with only a small deviation, rather than being reflected back down; this means such high-frequency waves cannot rely on natural sky-wave reflection for long-distance transmission and must instead use a satellite relay if long-range space-wave communication is required. The same limitation applies to TV signals, whose high frequency similarly prevents long-distance transmission by simple ground-based space wave propagation alone.

The maximum distance over which a signal can be usefully received is called its range. Because greater antenna height directly increases the achievable range (as the geometry below makes precise), larger TV coverage requires the transmitting antenna to be mounted as high as possible -- which is exactly why transmitting and receiving antennas are so often placed on top of tall buildings or hills.

The range formula follows from simple right-triangle geometry (Fig. 13.5). Let a transmitting antenna AA' of height $h$ stand at point A on the Earth's surface, with O the centre of the Earth (radius R), and let B be the point on the Earth's surface where the straight-line wave from the antenna top A' just grazes the horizon, at range $d = A'B$ . Triangle OA'B is right-angled at B (a tangent line is perpendicular to the radius at the point of contact), so by Pythagoras:

OA'^{2}=A'B^{2}+OB^{2} \quad\Longrightarrow\quad (R+h)^{2}=d^{2}+R^{2}

Expanding the left side gives $R^{2}+h^{2}+2Rh=d^{2}+R^{2}$ , and since the antenna height $h$ is always very much smaller than the Earth's radius $R$ ( $h\ll R$ ), the $h^{2}$ term can be neglected, leaving

$d\approx\sqrt{2Rh}$ …

Figure Fig.13.5Range of the signal (geometry, not to scale)

What this figure shows. A right-angled-triangle geometry diagram (explicitly noted as not to scale) used to derive the space-wave range formula. A transmitting antenna AA' of height h stands vertically at point A on the Earth's surface; O marks the centre of the Earth, with radius OA = R, so the antenna top A' is at distance OA' = R + h from the centre. B is the point on the Earth's surface where the straight-line (line-of-sight) space wave leaving the antenna top A' just grazes the horizon, with the segment A'B labelled d (the range) drawn tangent to the Earth's surface at B, and OB = R forming the right angle of the triangle OA'B at B (since a tangent line is perpendicular to the radius at the point of tangency). A second, lower elevated point C on the Earth's surface (a receiver at height h') is also shown further along, with its own extra range segment d' out to where the wave's path meets it, illustrating that the total achieva …

Misc Ex.13.7Maximum detection range of a hilltop radar

Worked out. A radar of power 10 kW operating at frequency 20 GHz is located on top of a hill of height 500 m; the problem asks for the maximum distance up to which it can detect an object located on the surface of the Earth, given the radius of the Earth as 6.4x10^6 m. The method applies the space-wave range formula d = sqrt(2Rh) directly with h = 500 m and R = 6.4x10^6 m, giving d = sqrt(2 x 6.4x10^6 x 500) = sqrt(6.4x10^9) = 8x10^4 m = 80 km; the radar's power and operating frequency are given as context but are not needed for this pa …

Misc Ex.13.8Ground area covered by a TV transmitting antenna

Worked out. The height of a TV transmitting antenna is 128 m, and the problem asks how much square area can be covered by the transmitted signal if the receiving antenna is at ground level, given the radius of the Earth as 6400 km. The method first finds the range d = sqrt(2Rh) using h = 128 m and R = 6400 km (converted consistently), obtaining d approximately 40.47 km, and then finds the covered area as that of a circle of radius d, Area = pi*d^2, giving an area of roughly …

Misc Ex.13.9Maximum line-of-sight distance between an elevated transmitter and an elevated receiver

Worked out. A transmitting antenna of height 68 m and a receiving antenna at the top of a 34 m tower are given, along with the radius of the Earth as 6400 km, and the problem asks for the maximum distance between them for satisfactory transmission in line-of-sight mode. The method applies the two-term range formula d_max = sqrt(2Rh_t) + sqrt(2Rh_r), substituting h_t = 68 m and h_r = 34 m together with R = 6.4x10^6 m, adding the two square-root terms (approximately 29.5 km and 20.9 km) to obtain a total maximum distance of about 50.4 km. …

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