## Friday, February 04, 2011

### Kerala Medical Entrance 2008 (KEAM 2008) Questions on Communication Systems

The following multiple choice questions on communication systems appeared in the Kerala Medical Entrance 2008 (KEAM-2008) examination question paper:

(1) A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then the width of the channel in kHz is

(a)10

(b) 20

(c) 30

(d) 40

(e) 50

Let us assume that the system uses amplitude modulation of the usual double sideband type. (It should have been mentioned in the question)

The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.

[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].

(2) If the critical frequency for sky wave propagation is 12 MHz, then the maximum electron density in the ionosphere is

(a) 1.78×1012/m3

(b) 0.178×1010/m3

(c) 1.12×1012/m3

(d) 0.56×1012/m3

(e) 0.148×1012/m3

The critical frequency fc for reflection by the ionosphere is given by

fc = 9 N1/2 where N is the maximum electron number density.

Therefore, N = fc2/81 = (12×106)2 /81 = 1.78×1012/m3