Two questions from Communication Systems

Questions on topics in Communication Systems are generally simple and not at all time consuming. You should never omit them. Consider the following:
(1) Photo diodes used as light detectors in optical communication systems are operated in the reverse biased manner because
(a) forward biasing will damage the photo diode (b) power requirement is small in reverse biased mode (c) it is more convenient to measure reverse current (d) reverse current alone is affected by illumination (e) the fractional change in reverse current is much greater than the fractional change in forward current.
You know that the forward current is typically of the order of milliamperes while the reverse current is of the order of micro amperes. When light is incident on the photodiode, electrons and holes are produced in equal number, contributing equally to forward current and reverse current. Since the existing reverse current is a small current compared to the forward current, the change in reverse current will be significant compared to the reverse current itself. The change in reverse current is therefore much easier to detect compared to the change in forward current. The correct option therefore is (e).
(2) A TV transmitter tower has a height of 80m. If the range is to be doubled, the height of the tower is to be increased by
(a) 160m (b) 80m (c) 240m (d) 320m (e) 33m
With usual notations, range = √(2Rh). Therefore, to double the range, the height ‘h’ is to be made 4 times (320m). The increment in height is 240m [Option (c)]. In your hurry to answer this simple question, don’t fail to note that it is the increase in height that is asked for.

A Question on TV transmitter Height

Unlike A.M.radio transmitters, TV transmitters can cover a distance of 60km to 65km only. You might have noted that this limitation in the range of a TV transmitter is due to the fact that the carrier used for TV transmission is in the V.H.F and U.H.F. range. They are not reflected by the ionosphere and the curvature of the earth places the above mentioned limit in the range of terrestrial TV transmitters.
You might be remembering the expression for the range ‘d’ of a TV transmitter:
d = √(2Rh) where R is the radius of the earth and h is the height of the TV transmitter tower.
The above expression is the approximation of the expression, d = √(2Rh+h^2), obtained by ignoring h^2 which is small compared to 2Rh.
Now consider the following M.C.Q.:
To cover a population of 3 million, what should be the height of a TV transmitter tower? (Population per square km = 1000)
(a) 25m (b) 45m (c) 55m (d) 65m (e) 75m
If ‘d’ is the required range of the TV transmitter, the coverage area = πd^2 = π×2Rh so that we have, 1000 π×2Rh = 3×10^6.
The radius of the earth is 6400km which you are expected to remember. You may substitute for R in km itself in the above equation to obtain h to be approximately 0.075km = 75m. The correct option therefore is (e).