Communication Systems: Exam-Pattern Drill (3)

Question

A transmitting antenna at the top of a tower has height $h_T = 32$ m and the receiving antenna height is $h_R = 50$ m. Take Earth’s radius $R = 6.4 \times 10^6$ m. Find (a) the maximum line-of-sight distance for communication, (b) the maximum population covered if population density is $1000$ per km$^2$, and (c) what happens if the transmission switches from line-of-sight to sky-wave.

Solution — Step by Step

Final answers: line-of-sight $d_M \approx 45.5$ km; covered population $\approx 1.29$ million; sky-wave gives much longer range.

Why This Works

The line-of-sight distance formula comes from simple geometry: a tangent from the antenna top to the Earth’s surface. Pythagoras on the right triangle (Earth radius, antenna height, line-of-sight) plus the small-height approximation gives $d \approx \sqrt{2Rh}$.

Two antennas can each see the horizon, so the total communication range is the sum of their individual horizon distances.

Alternative Method

Compute $d^2 = 2 R h$ directly. With $h = 32$ m, $d^2 = 2 \times 6.4 \times 10^6 \times 32 = 4.096 \times 10^8$. So $d \approx 2.024 \times 10^4$ m. Same result via a one-line shortcut.

Common Mistake

Using $d = \sqrt{2Rh}$ without doubling $R$ inside the radical. The formula is $d^2 = 2Rh$, not $Rh$. The factor of $2$ comes from the algebra $(R+h)^2 - R^2 \approx 2Rh$ for $h \ll R$.