Communication Systems: Step-by-Step Worked Examples (1)

easy • 2 min read
Tags Communication Systems

Question

A transmitting antenna of height 80 m is used for line-of-sight communication. (a) Find the maximum distance over which it can communicate with a receiver at ground level. (b) If the receiver is on a 45 m tall tower, find the maximum communication distance. Take radius of Earth R=6.4×106R = 6.4 \times 10^6 m.

Solution — Step by Step

The maximum line-of-sight distance from a transmitting antenna of height hTh_T to a ground receiver:

dT=2RhTd_T = \sqrt{2 R h_T}

If the receiver also has height hRh_R, total maximum distance:

d=2RhT+2RhRd = \sqrt{2 R h_T} + \sqrt{2 R h_R}

dT=2⋅6.4×106⋅80=1.024×109d_T = \sqrt{2 \cdot 6.4 \times 10^6 \cdot 80} = \sqrt{1.024 \times 10^9}

dT≈3.2×104 m=32 kmd_T \approx 3.2 \times 10^4 \text{ m} = 32 \text{ km}

Transmitter contribution: 32 km (from above).

Receiver contribution:

dR=2⋅6.4×106⋅45=5.76×108=24×103 m=24 kmd_R = \sqrt{2 \cdot 6.4 \times 10^6 \cdot 45} = \sqrt{5.76 \times 10^8} = 24 \times 10^3 \text{ m} = 24 \text{ km}

Total:

d=32+24=56 kmd = 32 + 24 = 56 \text{ km}

(a) 32 km, (b) 56 km.

Why This Works

Line-of-sight communication uses VHF/UHF and microwaves, which travel in straight lines. Earth’s curvature limits how far a signal can reach — beyond the horizon, the signal misses the receiver.

Geometrically, if you stand at height hh on a sphere of radius RR, the distance to the horizon is (R+h)2−R2=2Rh+h2≈2Rh\sqrt{(R+h)^2 - R^2} = \sqrt{2Rh + h^2} \approx \sqrt{2Rh} when h≪Rh \ll R. Both transmitter and receiver get their own horizon distance, and these add up.

Quick sanity check: doubling antenna height multiplies range by 2\sqrt{2}, not 2. To double the range, you must quadruple the height. This is why TV transmission towers are kilometres tall.

Alternative Method

Using h2≪2Rhh^2 \ll 2Rh approximation, the formula simplifies. If you keep the full expression (R+h)2−R2\sqrt{(R+h)^2 - R^2} and use exact arithmetic, you get a 0.06% larger answer for h=80h = 80 m — negligible.

Common Mistake

Students forget to add receiver height contribution when both antennas are elevated. They compute only 2RhT\sqrt{2Rh_T} and miss the 2RhR\sqrt{2Rh_R} term — costing easy marks in NEET.

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