Communication Systems: Application Problems (2)

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Question

A TV transmitting antenna is hT=100h_T = 100 m tall. Estimate the maximum line-of-sight range over which a TV signal can be received by an antenna at ground level. Take Earth’s radius R=6.4×106R = 6.4 \times 10^6 m.

Solution — Step by Step

For a transmitting antenna of height hh, the maximum direct (line-of-sight) range to a ground-level receiver:

d=2Rhd = \sqrt{2 R h}

This comes from geometry: the tangent from the antenna top to the spherical Earth.

d=2×6.4×106×100d = \sqrt{2 \times 6.4 \times 10^6 \times 100}

d=1.28×109d = \sqrt{1.28 \times 10^9}

d3.58×104 m35.8 kmd \approx 3.58 \times 10^4 \text{ m} \approx 35.8 \text{ km}

Final answer: d35.8d \approx 35.8 km.

Why This Works

A TV signal travels in a straight line (it’s in the VHF/UHF range, which doesn’t reflect off the ionosphere). The horizon distance from a tower of height hh is set by the tangent line touching Earth. Using the right triangle relation (R+h)2=R2+d2(R + h)^2 = R^2 + d^2 and approximating hRh \ll R gives d22Rhd^2 \approx 2 R h.

Alternative Method

If both transmitting and receiving antennas have heights hTh_T and hRh_R, the range is d=2RhT+2RhRd = \sqrt{2 R h_T} + \sqrt{2 R h_R}. Useful in NEET problems that give both heights.

NEET often asks for the area of coverage. For a tower of height hh, the circular area is πd2=2πRh\pi d^2 = 2\pi R h. Plugging numbers: A4020A \approx 4020 km2^2 for a 100 m tower. Memorise both forms.

Common Mistake

Students sometimes write d=Rhd = \sqrt{R h} (forgetting the factor of 2). The factor comes from the geometry — both the tangent on Earth and the height contribute. Always start from (R+h)2=R2+d2(R+h)^2 = R^2 + d^2 if you forget the formula.

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