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 f_c for reflection by the ionosphere is given by

f_c = 9 N^1/2 where N is the maximum electron number density.

Therefore, N = f_c²/81 = (12×10⁶)² /81 = 1.78×10¹²/m³

Kerala Engineering Entrance Questions (MCQ) on Communication Systems

Questions on communication systems are generally simple and interesting and so you can easily score marks by attempting them. The following three multiple choice questions were included in Kerala Engineering Entrance 2005 question paper and will be able to answer them in less than three minutes:

(1) If a radio receiver is tuned to 855 kHz radio wave, the frequency of local oscillator in kHz is

(a) 1510

(b) 455

(c) 1310

(d) 1500

(e) 855

Since the tuned frequency is 855 kHz, the receiver is an AM (amplitude modulation) receiver. Modern AM receivers are super heterodyne receivers employing a higher frequency local oscillator. On mixing the local oscillator output with the incoming amplitude modulated carrier, an amplitude modulated wave at intermediate frequency (IF) of 455 kHz (by convention) is produced. Since the intermediate frequency is 455 kHz, it follows that the frequency of local oscillator is 855 kHz + 455 kHz = 1310 kHz.

(2) If n₁ and n₂ are the refractive indices of the core and the cladding respectively of an optical fibre,

(a) n₁ = n₂

(b) n₁ < n₂

(c) n₂ < n₁

(d) n₂ = 2n₁

(e) n₂ = √(2n₁)

Since the optical fibre confines the light signal within the fibre by total internal reflection, the refractive index of the cladding should be less than that of the core. Therefore, n₂ < n₁ [Option (c)]

(3) A TV tower has a height of 100 m. What is the maximum distance up to which TV transmission can be received? (Radius of the earth, R = 8×10⁶ m)

(a) 34.77 km

(b) 32.7 km

(c) 40 km

(d) 40.7 km

(e) 42.75 km

We have maximum distance, d ≈ √(2Rh) where h is the height of the antenna.

Substituting given values, d ≈ √(2×8×10⁶ ×100) = 40×10³ m = 40 km.

[The mean radius of the earth is nearly 6400 km. The value is given as 8000 km in the problem to make your calculation simple.

You should remember that the height of the transmitting antenna (or receiving antenna) is the height with respect to the ground level. If an antenna is mounted on a mast of height h₁ and the mast is erected on a hill or building of height h₂, the height of the antenna will be h = h₁ + h₂]

Communication Systems: MCQ on Amplitude modulation

Here are some questions (on amplitude modulation) of the type you are most likely to encounter in your entrance exam:

(1) A carrier wave of peak value 16 V is used to transmit a note of frequency 1000 Hz. What should be the peak value of the modulating signal to have a modulation index of 60% using amplitude modulation?

(a) 12 V

(b) 11.2 V

(c) 10.4 V

(d) 9.6 V

(e) 7.2 V

This question demands from you merely a basic knowledge of the modulation process and you are expected to calculate the modulation index (μ) using the equation,

μ = A_m/A_c where A_m and A_c are the amplitudes of the modulating signal and the carrier respectively.

Therefore, 0.6 = A_m/16, from which A_m = 9.6 V. (The frequency of the modulating signal is just a distraction in this question).

[In a practical modulator, the modulating signal amplitude may not be μ times the carrier amplitude. For instance, if the carrier is applied on the base side and the modulating signal is applied on the collector side of a transistor, the carrier amplitude (at the base) will be much less than the modulating signal amplitude, even though the modulation index is less than 100%. Of course, the carrier will appear in the amplified form at the collector. The modulation index (μ) in all situations will be correctly obtained if you consider it as the ratio of the amplitude of variation of the envelope of the amplitude modulated carrier to the amplitude of the unmodulated carrier. But problems with the above type of statement are often seen].

(2) The maximum amplitude of an amplitude modulated wave is 12 V and the minimum amplitude is 4V. The modulation percentage is

(a) 20

(b) 30

(c) 50

(d) 60

(e) 80

The modulation percentage is the modulation index expressed as percentage.

The maximum amplitude of an amplitude modulated (AM) wave is the sum of the amplitude of the unmodulated carrier and the amplitude of variation of the envelope of the AM wave.

The minimum amplitude of an AM wave is the difference between the amplitude (A_c) of the unmodulated carrier and the amplitude (A_m) of variation of the envelope of the AM wave.

Therefore, we have (A_c + A_m) = 12 V

(A_c – A_m) = 4 V

From these equations, A_c = 8 V and A_m = 4 V so that

μ = A_m/A_c = 4/8 = 0.5 = 50%

(3) An unmodulated carrier has a peak value of V volt. When it is amplitude modulated with a sine wave, the maximum peak to peak value (of the modulated carrier) is 3 V volt. The modulation index is

(a) 0.33

(b) 0.5

(c) 0.6

(d) 0.66

(e) 0.8

The maximum peak value (maximum amplitude) of the modulated carrier is 3V/2 volt.

Therefore, (A_c + A_m) = 3V/2

Since the amplitude of the unmodulated carrier (A_c) is V volt, the amplitude (A_m) of the variation of the envelope of the modulated carrier is V/2. Therefore, The modulation index is given by

μ = A_m/A_c = (V/2)/V = 0.5

(4) An amplitude modulated wave has modulation index 0.6. If the power carried by the carrier component in the modulated wave is P_c , what is the power carried by the upper and lower side bands (together)?

(a) 0.6 P_c

(b) 0.54 P_c

(c) 0.36 P_c

(d) 0.46 P_c

(e) 0.18 P_c

The total power (P) carried by an amplitude modulated wave is given by

P = P_c[1+ (μ²/2) where P_c is the carrier power and μ is the modulation index.

Therefore, the power carried by the two side bands together is

P_c ×μ²/2 = P_c×(0.6)²/2 = 0.18 P_c.

[This shows how much power is wasted in the carrier component which does not carry any information. Further, we need only one side band to extract the modulating signal. This points to the benefit of the suppressed carrier single side band transmission].

(5) An AM radio station using double side band transmitted carrier system employs a carrier of frequency 1.2 MHz. If the modulating signal frequency is 2 kHz, the frequencies present in the modulated carrier are

(a) 1.2 MHz and 1.4 MHz

(b) 1 MHz, 1.2 MHz and 1.4 MHz

(c) 1.198 MHz, 1.2 MHz and 1.202 MHz

(d) 1.198 MHz and 1.202 MHz

(e) 1.2 MHz and 1.202 MHz

The double side band transmitted carrier system is the one employed by ordinary broadcast stations. The amplitude modulated wave will contain the carrier frequency (f_c) as well as the upper and lower side frequencies f_c + f_m and f_c – f_m. Therefore, the correct option is (c).