Questions on communication systems at Class 12 level are often simple but occasionally you may get questions which are a little bit confusing and time consuming. I give below a couple of practice questions for you:

(1) A television transmitting antenna is mounted at a height of 100 m. For satisfactory line of sight communication at a distance of 40 km, what should be the minimum height of the receiving antenna? (Radius of the earth = 6400 km).

(a) 20 m

(b) 25 m

(c) 30 m

(d) 35 m

(e)** **40 m** **

If *h*_{t }and *h*_{r} represent the heights of the transmitting antenna and the receiving antenna respectively, the maximum separation (*d*) between them for satisfactory line of sight communication is given by

*d = *√(2*R**h*_{t}) + √(2*R**h*_{r}) where *R* is the radius of the earth.

Therefore, we have

40 = √(2×6400×0.1) + √(2×6400×*h*_{r})

[Note that we have expressed all distances in kilometre and hence we will obtain the height of the receiving antenna in km].

Squaring, 1600 = 2×6400 (0.1 + *h*_{r})

Therefore, (0.1 + *h*_{r}) = 1/8 from which

* h*_{r} = 0.025 km = 25 m.

(2) The minimum service area covered by a TV transmitter antenna mounted at a height *h* is (radius of the earth = *R*)

(a) π*R*^{2}*h*

(b) 2π*R*^{2}*h*

(c) π*h*^{2}*R*

(d) π*Rh*

(e)** **2π*Rh*** **

The coverage area will be minimum when the receiving antenna is at the *ground* level. In other words, the height of the receiving antenna is zero.

The distance (*d*) up to which line of sight communication is possible is therefore given by

*d* = √(2*R**h*_{t}) where *h*_{t} is the height of the transmitting antenna.

The minimum coverage area is π*R*^{2}*d =*π[√(2*R**h*_{t})]^{2} = 2π*R**h*_{t} = 2π*Rh*** **

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