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 h_t and h_r 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 = √(2Rh_t) + √(2Rh_r) where R is the radius of the earth.

Therefore, we have

40 = √(2×6400×0.1) + √(2×6400×h_r)

[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 + h_r)

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

h_r = 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) 2πRh

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 = √(2Rh_t) where h_t is the height of the transmitting antenna.

The minimum coverage area is πR²d =π[√(2Rh_t)]² = 2πRh_t = 2πRh

You can access all posts on communication systems on this site by clicking on the label, ’communication system’ below this post.

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]

Kerala Engineering Entrance 2008 Questions on Communication Systems

Four questions on communication systems appeared in the KEAM (Engineering) 2008 test paper. Generally you will be able to answer all the questions in this section within the stipulated time. It will be a good idea to answer questions from simple sections like this before proceeding to attempt difficult and time consuming ones. Here are the questions with their solutions:

(1) A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively

(a) 2512 kHz and 2508 kHz

(b) 2522 kHz and 2488 kHz

(c) 2502 kHz and 2498 kHz

(d) 2522 kHz and 2498 kHz

(e) 2512 kHz and 2488 kHz

This question should have been stated like this:

A carrier wave wave of frequency 2.51 Mz is amplitude modulated with a signal wave of frequency 12 kHz. The upper and lower side band frequencies are respectively

(a) 2512 kHz and 2508 kHz

(b) 2522 kHz and 2488 kHz

(c) 2502 kHz and 2498 kHz

(d) 2522 kHz and 2498 kHz

(e) 2512 kHz and 2488 kHz

Since there is just one frequency (12 kHz) in the modulating signal, there is just one frequency in the upper side band. The upper side frequency is 2.51 MHz + 12 kHz = 2510 kHz + 12 kHz = 2522 kHz.

Similarly the lower side frequency is 2.51 MHz – 12 kHz = 2510 kHz – 12 kHz = 2498 kHz [Option (d)].

(2) Which of the following statements is wrong?

(a) Ground wave propagation can be sustained at frequencies 500 kHz to 1500 kHz

(b) Satellite communication is useful for the frequencies above 30 MHz

(c) Sky wave propagation is useful in the range of 30 to 40 MHz

(d) Space wave propagation takes place through tropospheric space

(e) The phenomenon involved in sky wave propagation is total internal reflection

The frequency range 500 kHz to 1500 kHz is the medium wave band for AM sound broadcast which you know is based on ground wave propagation. So option (a) is correct. Option (b) is correct since frequencies above 30 MHz will not be reflected by the ionosphere. For the same reason option (c) is wrong.

Since the questions you get in this test are single answer type your answer option is (c).

(3) The principle used in the transmission of signals through an optical fibre is

(a) total internal reflection

(b) reflection

(c) refraction

(d) dispersion

(e) interference

In an optical fibre the refractive index is decreased as the ray proceeds radially outwards from the centre and hence it undergoes total internal reflection[Option (a)].

(4) In satellite communication

1. the frequency used lies between 5 MHz and 10 MHz

2. the uplink and downlink frequencies are different

3. the orbit of geostationary satellite lies in the equatorial plane at an inclination of 0º.

On the above statements

(a) only 2 and 3 are true

(b) all are true

(c) only 2 is true

(d) only 1 and 2 are true

(e) only 1 and 3 are true

Statement 1 is false since 5 MHz and 10 MHz will be reflected by the ionosphere. Statements 2 and 3 are true in the case of geostationary satellite. So the correct option is (a).

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).

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
f_c = 9√N_max = 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)].