Communication Systems Multiple Choice Questions And Answers

Communication Systems

Question 1. From a television tower of height h, the maximum distance up to which TV signals can be received is proportional to

1. h^-1/2
2. h^1/2
3. h²
4. h

Answer: 2. h^1/2

The maximum range for the reception of TV signals is given by

⇒ \(d=\sqrt{2 R_{\mathrm{E}} h}\)

where R_E is the earth’s radius and h is the height of the transmitting antenna.

∴ Thus, d ∝ h^1/2.

Question 2. The area covered by a TV transmitting antenna of a height of 50 m is

1. 120 π km²
2. 1440π km²
3. 640π km²
4. 320π km²

Answer: 3. 640π km²

Given that the height of the transmitting antenna = h = 50 m.

Now, the maximum range for the reception of TV signals is \(d=\sqrt{2 R_{\mathrm{E}} h}\)

∴ area covered = A = nd² = π(2Rh)

= π(2 x 6400 km x 50 m) = 640π km².

Question 3. If the height H of a TV transmitting antenna is doubled, the range d covered initially would become

1. 2d
2. √2d
3. 3d
4. 4d

Answer: 2. √2d

Initial covering range = d = \(\sqrt{2 R_{\mathrm{E}} H}\)

If the height is doubled, the new range is

∴ \(d=\sqrt{2 R_{\mathrm{E}}(2 H)}=\sqrt{2}\left(\sqrt{2 R_{\mathrm{E}} H}\right)=\sqrt{2} d\).

Question 4. In short-wave communications, waves of which of the following frequencies will be reflected by the ionospheric layer having an electron density of 10¹¹ m^-3?

1. 18 MHz
2. 10 MHz
3. 12 MHz
4. 2 MHz

Answer: 4. 2 MHz

The critical frequency is the highest frequency above which the electromagnetic waves penetrate the ionosphere and below which the waves are reflected from the ionosphere. The critical frequency is given by \(f_c=9 \sqrt{N_{\max }}\) Hz, where Nmax is the maximum electron density (inm-3).

Given that N_max = 10¹¹

Hence, \(f_c=9 \sqrt{10^{11}} \mathrm{~Hz}=9 \times 10^5 \sqrt{10} \mathrm{~Hz}=2.8 \times 10^6 \mathrm{~Hz} \approx 2 \mathrm{MHz}\)

Question 5. For satellite communications, which waves are used?

1. Ground waves
2. Space waves
3. Microwaves
4. Sky waves

Answer: 2. Space waves

Space waves travel in a straight line from the transmitting antenna to the receiving antenna. Space waves are used for line-of-sight communications as well as satellite communications.

Question 6. A long-distance communication between two points on the earth is achieved by

1. Satellites
2. Line-of-sight (LoS) transmissions
3. Space waves
4. Sky waves

Answer: 4. Sky waves

Sky waves are used for long-distance communications between two points on the earth.

Question 7. For sky-wave communications of 10-MHz signals, the appropriate number density of electrons in the ionospheric layer will be approximately

1. 1.2 x l0¹² m^-3
2. 10²² m^-3
3. 10⁴ m^-3
4. 10¹⁴ m^-3

Answer: 1. 1.2 x l0¹² m^-3

Given that critical frequency = f_c =10 MHz = 10  x 10⁶ Hz.

We know that \(f_{\mathrm{c}}=9 \sqrt{N_{\max }} \mathrm{Hz}\)

∴ \(10 \times 10^6 \mathrm{~Hz}=9 \sqrt{N_{\max }} \mathrm{Hz}\)

⇒ \(N_{\max }=\frac{\left(10^7\right)^2}{81}=1.2 \times 10^{12}\)

Hence, the number density of electrons must be 1.2 x 10¹² m^-3.

Question 8. The given circuit diagram shows an AM demodulator. For a good demodulation of an AM signal of carrier frequency f_c, the value of RC is given by the relation

1. \(R C=\frac{1}{f_c}\)
2. \(R C \geq \frac{1}{f_c}\)
3. \(R C \leq \frac{1}{f_c}\)
4. \(R C \gg \frac{1}{f_{\mathrm{c}}}\)

Answer: 4. \(R C \gg \frac{1}{f_{\mathrm{c}}}\)

For high-frequency carrier waves, the capacitive reactance \(\left(X_c=\frac{1}{\omega_c C}\right)\) must be small for the carrier to bypass.

Hence, \(R C \gg \frac{1}{f_c}\).

Question 9. If the modulation index is 0.5 and the power of the carrier wave is 2 W, the total power in the modulated wave will be

1. 0.25 W
2. 1.0W
3. 0.5 W
4. 2.25 W

Answer: 4. 2.25 W

If P_c is the power of the carrier wave, the power of the modulated wave will be

⇒ \(P_{\mathrm{m}}=P_{\mathrm{c}}\left(1+\frac{m^2}{2}\right)\) where m the modulation index.

Given that m = 0.5 and P_c = 2 W.

Hence, \(P_{\mathrm{m}}=(2 \mathrm{~W})\left(1+\frac{1}{2} \times \frac{1}{4}\right)=\frac{9}{8} \times 2 \mathrm{~W}=2.25 \mathrm{~W}\)

Question 10. A modulating wave of voltage 5 V and frequency 10 MHz was superimposed on a carrier wave of frequency 20 MHz and voltage 20 V. Then, the modulation index is

1. 2.43
2. 1.25
3. 0.25
4. 64

Answer: 3. 0.25

Given that the amplitude of the modulating voltage is A_m = 5 V and that of the carrier voltage is A_c = 20 V.

Hence, the modulation index is

∴ \(\mu=\frac{A_{\mathrm{m}}}{A_{\mathrm{c}}}=\frac{5 \mathrm{~V}}{20 \mathrm{~V}}=0.25\)

Question 11. If the highest modulating frequency of a wave is 5 kHz, the number of stations that can be accommodated in a bandwidth of 150 kHz will be

1. 10
2. 15
3. 5
4. none of these

Answer: 2. 15

The number of stations to be accommodated is

∴ \(N=\frac{\text { bandwidth }}{\text { bandwidth per station }}=\frac{150 \mathrm{kHz}}{2 \times 5 \mathrm{kHz}}=\frac{150}{10}=15\)

Question 12. A message signal of frequency 100 MHz and peak voltage 100 V is used to execute an amplitude modulation on a carrier wave of frequency of 300 GHz and peak voltage 400 V. The modulation index and the difference between the two sideband frequencies are respectively

1. 4 and 2 x l0⁵ Hz
2. 0.25 and 2 x l0⁷ Hz
3. 4.0 and 1 x 10⁸ Hz
4. 0.25 and 2 x 10⁸ Hz

Answer: 4. 0.25 and 2 x 10⁸ Hz

Modulation index = \(\mu=\frac{V_{\mathrm{m}}}{V_{\mathrm{c}}}=\frac{100 \mathrm{~V}}{400 \mathrm{~V}}=0.25\)

The frequencies of the sidebands are f₁ = f_c + f_m and f₂ = f_c – f_m.

∴ Δf = f₁ – f₂ = 2f_m = 2(100 MHz) = 200 MHz = 2 X 10⁸ Hz.

Question 13. Given below in the left column are different modes of communication using the kinds of waves given in the right column.

table

From the options given below, find the most appropriate match between the entries in the left and the right columns.

1. A-Q B-S C-P D-R
2. A-S B-Q C-R D-P
3. A-R B-P C-S D-Q
4. A-Q B-P C-R D-S

Answer: 1. A-Q B-S C-P D-R

In optical fiber communications with glass fibers, light waves in the infrared region between 1300 nm and 1550nm are used for low attenuation and absorption. So, A → Q.

Radar is a detection system that uses radio waves to detect aircraft, ships, guided missiles, etc. Hence, B → S.

Ultrasound, or ultrasonic, waves (frequencies greater than 20 kHz) are used in sonar, as these waves can penetrate water to long distances (because of their higher frequencies and very short wavelengths). So, C → P.

Mobile phones, or cell phones, operate with radio frequencies, which are a form of electromagnetic energy located in the electromagnetic spectrum around microwaves. Hence, D → R.

Question 14. A signal A cos ωt is transmitted using V₀ sin ω₀t as the carrier wave. The correct amplitude-modulated (AM) signal is

1. \(V_0 \sin \omega_0 t+A \cos \omega t\)
2. \(V_0 \sin \omega_0 t+\frac{A}{2} \sin \left(\omega_0-\omega\right) t+\frac{A}{2} \sin \left(\omega_0+\omega\right) t\)
3. \(\left(V_0+A\right) \cos \omega t \cdot \sin \omega_0 t\)
4. \(V_0 \sin \omega_0(1+0.01 A \sin \omega t) t\)

Answer: 2. \(V_0 \sin \omega_0 t+\frac{A}{2} \sin \left(\omega_0-\omega\right) t+\frac{A}{2} \sin \left(\omega_0+\omega\right) t\)

In amplitude modulation, the amplitude of the carrier wave is varied in accordance with the variation in the signal voltage to be communicated. Thus, the modulated signal is

⇒ \(V(t)=\left(V_0+A \cos \omega t\right) \sin \omega_0 t=V_0 \sin \omega_0 t+\frac{A}{2}\left(2 \cos \omega t \sin \omega_0 t\right)\)

∴ \(V_0 \sin \omega_0 t+\frac{A}{2} \sin \left(\omega_0-\omega\right) t+\frac{A}{2} \sin \left(\omega_0+\omega\right) t\).

Question 15. The wavelength of a carrier wave in a modern optical-fiber communications network is close to

1. 600. nm
2. 100 nm
3. 2400 nm
4. 1500 nm

Answer: 4. 1500 nm

In optical-fiber communications, the carrier waves used are part of the infrared region which has wavelengths close to 1500 nm.

Question 16. A TV transmission tower has a height of 140 m and the receiving antenna is 40 m. What is the maximum distance up to which signals can be broadcast from this tower in the LoS mode? (Given that the radius of the earth = 6400 km.)

1. 65 km
2. 80 m
3. 40 km
4. 48 km

Answer: 1. 65 km

The maximum line-of-sight distance between the transmitting and receiving antennas is given by

⇒ \(d=\sqrt{2 R_{\mathrm{E}} h_{\mathrm{T}}}+\sqrt{2 R_{\mathrm{E}} h_{\mathrm{R}}}\)

where R_E = radius of the earth = 6400 x 10³ m,

h_T = height of the transmitter = 140 m

and h_R = height of the receiver = 40 m.

∴ \(d=\sqrt{2\left(64 \times 10^5 \mathrm{~m}\right)(140 \mathrm{~m})}+\sqrt{2\left(64 \times 10^5 \mathrm{~m}\right)(40 \mathrm{~m})}\)

∴ \((8000 \mathrm{~m})(\sqrt{28}+\sqrt{8})=(8 \mathrm{~km})(5.29+2.83)=64.96 \mathrm{~km} \approx 65 \mathrm{~km}\)

Question 17. In a communication system, only one percent frequency of a signal of wavelength 800 .nm can be used as the bandwidth. How many channels of 6 MHz bandwidth can broadcast this signal?

1. 3.75 x l0⁶
2. 6.25 x 10⁵
3. 3.86 x lO⁶
4. 4.87 x 10⁵

Answer: 3. 3.86 x lO⁶

Signal wavelength = λ = 800 nm = 800 x 10^-9 m.

The corresponding frequency is

⇒ \(f=\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{8 \times 10^{-7} \mathrm{~m}}=3.75 \times 10^{14} \mathrm{~s}^{-1}\)

∴ total bandwidth used for communications = 1% of f

= 3.75 x 10¹² s^-1

∴ the total number of channels is

∴ \(N=\frac{\text { total bandwidth }}{\text { signal bandwidth }}=\frac{3.75 \times 10^{12} \mathrm{~Hz}}{6 \times 10^6 \mathrm{~Hz}}=6.25 \times 10^5\)

Question 18. The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you to get a license, what broadcast frequency will you allot?

1. 2000 kHz
2. 2250 kHz
3. 2900 kHz
4. 2750 kHz

Answer: 1. 2000 kHz

If f_c and f_m respectively are the frequencies of the carrier wave and the modulating signal, the maximum and minimum frequencies of the bandwidth are f_max = f_c+ f_m and f_min = f_c – f_m respectively.

To avoid overlapping, the next broadcast frequencies will be f₁ = f_c ± 2f_m, f₂ = f_c ± 3f_m,…

Hence, for the next broadcast frequency,

f₁ = f_c + 2f_m and f₁‘ = f_c – 2f_m.

Given that f_m = 250 kHz and f_c = 10f_m = 2500 kHz.

f₁ = [2500 + 2(250)] Hz = 3000 kHz

and f₁‘ = [2500- 2(250)] Hz = 2000 kHz.

Question 19. An amplitude-modulated signal is given by V(t) = 10[1 + 0.3 cos (2.2 x 10⁴ )t]sin (5.5 x 10⁵ )t, where t is in seconds. The sideband frequencies are approximately

1. 892.5 kHz and 857.5 kHz
2. 91 kHz and 84 kHz
3. 178.5 kHz and 171.5 kHz
4. 1785 kHz and 1715 kHz

Answer: 2. 91 kHz and 84 kHz

From the given expression of the modulated signal,

ω_m = 2πf_m = 2.2 x 10⁴ rad s^-1

and ω_c = 2πf_c = 5.5 x 10⁵ rad s^-1.

The sideband frequencies are f_c ± f_m.

Thus, the upper sideband is

⇒ \(f_1=f_{\mathrm{c}}+f_{\mathrm{m}}=\frac{10^4}{2 \pi}(55+2.2) \mathrm{Hz}=\frac{5}{\pi}(57.2) \mathrm{kHz}=91.08 \mathrm{kHz}\)

and the lower sideband is

∴ \(f_2=f_{\mathrm{c}}-f_{\mathrm{m}}=\frac{10^4}{2 \pi}(55-2.2) \mathrm{kHz}=\frac{5}{\pi}(52.8) \mathrm{kHz}=84 \mathrm{kHz}\)

These approximately correspond to the option (2).

Question 20. An amplitude-modulated signal is given in the figure. Which of the following best describes the given signal?

1. [1 + 9 sin (2π X 10⁴ )t]sin (2.5π x 10⁵ )t V
2. [9 + sin (2π x 10⁴ )t]sin (2.5π X 10⁵ )t V
3. [9 + sin (4π x 10⁴ )t]sin (5π X 10⁵ )t V
4. [9 + sin (2.5π X 10⁵ )t]sin (2π x 10⁴ )t V

Answer: 2. [9 + sin (2π x 10⁴ )t]sin (2.5π X 10⁵ )t V

From the given diagram of the modulated signals,

maximum amplitude = A_c + A_m = 10 V

and minimum amplitude = A_c – A_m = 8 V.

∴ A_c = 9 V and A_m = 1 V.

∴ the angular frequency of the carrier is

⇒ \(\omega_c=\frac{2 \pi}{T_c}=\frac{2 \pi}{8 \times 10^{-6}} \mathrm{~s}^{-1}=2.5 \pi \times 10^5 \mathrm{~s}^{-1}\)

Similarly \(\omega_{\mathrm{m}}=\frac{2 \pi}{T_{\mathrm{m}}}=\frac{2 \pi}{100 \times 10^{-6}} \mathrm{~s}^{-1}=2 \pi \times 10^4 \mathrm{~s}^{-1}\)

Substituting these values in the equation for the modulated signal, c_m = (A_c + A_m sin ω_mt)sin ω_ct, we get

c_m = [9 + sin (2π x 10⁴)t]sin (2.5K X 10⁵ )t V

Question 21. A 100-V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index?

1. 0.5
2. 0.8
3. 0.4
4. 0.6

Answer: 4. 0.6

V_max = V_c + V_m = 160 V and V_min = V_c – V_m = 40 V.

∴ V_c = 100 V and V_m = 60 V.

∴ modulation index = \(\mu=\frac{V_{\mathrm{m}}}{V_{\mathrm{c}}}=\frac{60 \mathrm{~V}}{100 \mathrm{~V}}=0.6\)

Question 22. To double the covering range of a TV transmission tower, its height should be multiplied by

1. 2
2. 4
3. 2
4. \(\frac{1}{\sqrt{2}}\)

Answer: 2. 4

The covering range for communications is given by \(d=\sqrt{2 R_E h}\), where R_E is the radius of the earth and h is the height of the transmission tower.

When the range is doubled, \(2 d=\sqrt{2 R_E h^{\prime}}\)

∴ \(\frac{h^{\prime}}{h}=4 \Rightarrow h^{\prime}=4 h\)

Question 23. In a line-of-sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, the minimum height of the transmitting antenna should be

1. 51 m
2. 40 m
3. 32 m
4. 20 m

Answer: 3. 32 m

For line-of-sight radio communications,

⇒ \(\sqrt{2 R_{\mathrm{E}} h_{\mathrm{T}}}+\sqrt{2 R_{\mathrm{E}} h_{\mathrm{R}}} \geq d=50 \mathrm{~km}\)

∴ \(\sqrt{2(6400 \mathrm{~km}) h_{\mathrm{T}}}+\sqrt{2(6400 \mathrm{~km})(70 \mathrm{~m})} \geq 50 \mathrm{~km}\)

⇒ \(\sqrt{2(6400 \mathrm{~km}) h_{\mathrm{T}}} \geq 50 \mathrm{~km}-30 \mathrm{~km}=20 \mathrm{~km}\)

∴ \(h_{\mathrm{T}} \geq \frac{(20 \mathrm{~km})^2}{12800 \mathrm{~km}}=31.25 \mathrm{~m} \Rightarrow h_{\mathrm{T}} \geq 31.25 \mathrm{~m}\).

Hence, the minimum height of the transmitter is (h_T)_min = 32 m.

Question 24. The physical sizes of the transmitting and receiving antennas in a communication system are

1. Inversely proportional to the carrier frequency
2. Independent of both the carrier and modulating frequencies
3. Inversely proportional to the modulating frequency
4. Proportional to the carrier frequency

Answer: 1. Inversely proportional to the carrier frequency

In communication systems, the size of the transmitting or receiving antenna is directly proportional to the wavelength (λ).

Thus, \(l \propto \lambda \Rightarrow l \propto \frac{1}{\omega_{\mathrm{c}}}\)

Question 25. In an amplitude-modulator circuit, the carrier wave is given by c(t) = 4 sin 20000πt, while the modulating signal is given by m(t) = 2 sin 2000πt. The values of the modulation index and the lower sideband frequency are respectively

1. 0.2 and 9 kHz
2. 0.5 and 9 kHz
3. 0.5 and 10 kHz
4. 0.4 and 10 kHz

Answer: 2. 0.5 and 9 kHz

Given that A_c = 4 units, A_m = 2 units, ω_c = 20000π and ω_m = 2000π.

∴ f_c = 10000 Hz and f_m = 1000 Hz.

∴ modulation index = \(\mu=\frac{A_{\mathrm{m}}}{A_{\mathrm{c}}}=\frac{2}{4}=0.5\)

Lower sideband = f_c – f_m = 10 kHz – 1 kHz = 9 kHz.