## Saturday, March 10, 2007

### MCQ on Communication Systems

Questions on communication systems at the higher secondary/plus two level are simple. Here is a question on optical communication systems:
An optical communication system operates at a wave length of 750 nm. The available channel band width for optical communications is only 1% of the optical source frequency. How many TV signals can the system accommodate if each signal requires a band width of 5 MHz?
(a) 8×105 (b) 7.5×105 (c) 6×105 (d) 5×105 (a) 4×105

The optical source frequency, f = c/λ = 3×108/(750×10–9)= 4×1014 Hz.
Total band width available in the system = 1% of 4×1014 Hz = 4×1012 Hz.
Therefore, no. of TV signals that can be accommodated = (4×1012 Hz)/ (5×106 Hz) = 8×105.
Now, consider the following MCQ:
A 9 MHz signal is transmitted from a ground transmitter at a height of 300m. The maximum electron density of the ionosphere is 1.44×1012. A receiver at a distance of 60 km can receive the signal by
(a) space wave only (b) sky wave only (c) space wave and sky wave (d) satellite transponder only (e) sky wave and satellite transponder

The maximum line of sight distance possible is given by d = √(2Rh) where ‘R’ is the radius of the earth (6400 km) and ‘h’ is the transmitter height. Therefore, d = √(2×6400×103×300) = 62×103 m = 62 km.
Reception by space wave is possible since the receiver is at 60 km.
The upper frequency limit for ionospheric reflection (critical frequency) is given by
fc = 9√Nmax = 9√(1.414×1012) = 10.8×106 Hz =10.8 MHz.
The transmitter frequency is 9 MHz only so that the waves can be reflected by the ionosphere.
So, reception by sky wave also is possible and the correct option is (c).
[Note that satellite transponder doesn’t come into the picture since the waves cannot penetrate through the ionosphere].
Here is another MCQ:
A photo detector is made using a semiconductor having a band gap of 1.55 eV. The maximum wave length it can detect is nearly
(a) 500 nm (b) 600 nm (c) 750 nm (d) 800 nm (e) 850 nm
If the wave length is too large, the photon energy will be too small and the incident light will not be able to produce charge carriers in the semiconductor. With a given semiconductor therefore, there is an upper limit for the detectable wave length. Note that the product of the wave length in Angstrom and the energy in electron volt of any photon is 12400 (very nearly). So, the 1.55 eV photon has wave length equal to (12400/1.55) Ǻ = 8000 Ǻ = 800 nm. This is the maximum wave length this semiconductor can detect [Option (d)].