Sunday, January 23, 2011

Communication Systems -Two Questions Involving Line of Sight Space Wave Communication

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 ht and hr 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 = √(2Rht) + √(2Rhr) where R is the radius of the earth.

Therefore, we have

40 = √(2×6400×0.1) + √(2×6400×hr)

[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 + hr)

Therefore, (0.1 + hr) = 1/8 from which

hr = 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) πR2h

(b) 2πR2h

(c) πh2R

(d) πRh

(e) 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 = √(2Rht) where ht is the height of the transmitting antenna.

The minimum coverage area is πR2d =π[√(2Rht)]2 = Rht = Rh

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