WBCHSE Class 12 Physics Communication System Long Answer Questions

WBCHSE Class 12 Physics Communication System Long Question And Answers

Question 1. Explain briefly the following terms used in communication systems:
Answer:

The following terms used in communication systems:

1. Transducer
2. Repeater
3. Amplification.

1. Transducer:

An energy transformation device. In communication systems, it is used to convert the electrical energy from the modulator into the energy carried by electromagnetic waves. Also, it produces the reverse transformation at the receiving end. These transducers are essential parts of these transmitting and receiving antennas.

2. Repeater:

Space wave communication is a highly efficient mode of long-distance communication. But space waves travel in straight lines and cannot cover a large distance due to the curvature ofthe earth’s surface. So intermediate repeaters or repeater stations are set up to receive the transmitted signal from one side and to send it to other required directions

3. Amplification:

In communication sterns, the generally feeble data signals need adequate amplification before they, after modulating a suitable carrier wave, are transmitted from an antenna

Question 2. In the block diagram of a simple modulator for obtaining an AM signal,  identify the boxes A and B. j Write their functions.

Answer:

A : Linear amplifier; it amplifies the modulated signal and sends it to the transmitting antenna.

B: Transmitting antenna; it transmits the modulated signal in the form of electromagnetic waves.

Question 3. Name the type of waves that are used for line of sight (LOS) communication. What is the range of their frequencies?
Answer:

A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for
satisfactory communication in LOS mode. (Radius of the earth 6.4 × 10⁶ m)

These waves are space waves. The frequency range is from about 30 MHz to 300 MHz and higher.

The maximum distance between the two towers, earth = 6.4 × 10⁶ m )

= \(\sqrt{2 R}(\sqrt{H}+\sqrt{h})\)

= \(=\sqrt{2 \times\left(6.4 \times 10^6\right)} \times(\sqrt{20}+\sqrt{45})\)

= 40000 m ‘

= 40 km

Question 4. Draw a block diagram of a simple modulator for obtaining an amplitude modulated signal A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal to have a modulation index of 75%?
Answer:

 Block diagram of a simple modulator for obtaining amplitude-modulated signal:

A_c = 12 V

μ = 75

= 0.75%

∴ μ = \(\frac{A_m}{A_c}\)

Or, A_m = μ A_c = 0.75 × 12 V

= 9V

∴ The peak voltage of the modulating signal should be 9 V

WBBSE Class 12 Communication System Long Answer Questions

Question 5. A signal of 5 kHz frequency is amplitude modulated carrier wave off-frequency 2 MHz. What are the frequencies ofthe sidebands produced?
Answer:

Upper sideband frequency:

= 5 kHz + 2 MHz =5 × 10^-3 MHz + 2 MHz

= 0.005 + 2 = 2.005 MHz

Lower sideband frequency:

= 2 MHz- 5 kHz = 2 MHz- 0.005 MHz

= 1.995 MHz

Question 6. Why is the baseband signal not transmitted directly? Give any two reasons.
Answer:

The baseband signal is not transmitted directly, because:

1. Its higher wavelength (low frequency) will require the size of the antenna or aerial used for transmission to be very high.
2. For linear antenna (length l) the power radiated is proportional to Q)2> and hence, the effective power radiated by a long wavelength baseband signal would be small.

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Question 7. State the factors which limit its range of propagation
Answer:

Factors that limit the range of propagation:

• At frequencies higher than 40 MHz, the space waves scatter more easily and the communication is limited to line-of-sight propagation.
• Because of the line-of-sight nature of propagation, direct waves get blocked at some point by the curvature of the Earth

Question 8. Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving a reason, the frequency range used in this mode of propagation.
Answer:

• The basic mode of communication used in satellite communication is point-to-point mode.
• Space wave propagation is used in satellite communication
• In space wave propagation, the range of frequency is between 54 MHz to 4.26 GHz since both the ground wave and sky wave propagation fail at such high frequencies

Question 9.

1. What is the line of sight communication?
2. Why is it not possible to use sky waves for the transmission of TV signals? Up to what distance can a signal be transmitted using an antenna ofheight h ?

Answer:

1. A space wave travels in a straight line from the transmitting antenna to the receiving antenna. This is known as line-of-sight communication.

2. TV signals lie in the frequency range of 100 MHz-220 MHz. Waves in this frequency range are not reflected in the ionosphere and hence cannot be transmitted via sky waves.  A signal can be transmitted using an antenna of height h up to a distance of 2Rh

Question 10. Explain any two factors that justify the need for modulating a low-frequency baseband signal.
Answer:

The two factors that justify the need for modulating a low-frequency baseband signal are:

Power radiated by the antenna is given by \(P \propto \frac{l^2}{\lambda^2}\). Hence, there is a need for higher frequency conversation for effective power transmission by the antenna.

To transmit a signal effectively, the size of the antenna should be at least of the size \(\frac{\lambda}{4}\), where λ is the wavelength of the signal to be transmitted. So if λ is small (or frequency is high) signal can be transmitted with reasonable antenna length.

Detailed Explanations of Communication System Principles

Question 11. The frequencies of two sidebands in an AM wave an 640 kHz and 660 kHz respectively. Find the frequencies of carrier and modulating signal. What, is the bandwidth required for amplitude modulation?
Answer:

Let the frequencies of carrier and modulating signal bi denoted by and f respectively

According to the problem

f_c + f_m = 660

f_c – f_m = 640

∴ f_c =  \(\frac{660+640}{2}\) = \(\frac{1300}{2}\)

= 659 kHz

∴ f_m  =  \(\frac{660-640}{2}\) = \(\frac{20}{2}\)

= 10 kHz

Hence, the bandwidth required for amplitude modulation = 660 – 640 = 20 kHz

Question 12.

1. Give three reasons why modulation of a message signal is necessary for long-distance transmission.
2. Show graphically an audio signal, a carrier wave, and an amplitude-modulated wave.

Answer:

Three reasons behind modulating message signals for long-distance transmission are:

1. The audio / audiovisual signals when converted into electromagnetic waves do not have sufficiently high energy to travel up to long distances, because of their lower frequency. Hence these signals must be modulated with high-frequency carrier waves, before beginning

2. For effective transmission by an antenna, its size should at least be of the order \(\frac{\lambda}{4}\), where λ is the wavelength of the signal to be sent For an electromagnetic wave of frequency 20 kHz, we need an antenna of size \(\frac{\lambda}{4}\)  i.e., 3.75 km high, which is practically impossible. Hence, these low-frequency signals are first converted into high frequencies so that transmission can be obtained with reasonable antenna Lengths

3. The power radiated by an antenna is related to its length l and wavelength A of the signal as \(P \propto \frac{l^2}{\lambda^2}\) The relation shows that for die same antenna length, power (P) radiated increases with decreasing wavelength or increasing frequency. Hence, there is a need for higher frequency conversion for effective power transmission by the antenna.

Question 13. A carrier wave of peak voltage 15 V is used to transmit a message signal Find the peak voltage of the modulating signal to have a modulation index of 60%
Answer:

Modulation index is given by μ = \(\frac{A_m}{A_c}\)

Where A_m is the die amplitude of the modulated wave, A_c is the amplitude of the carrier wave.

Here, A_c = 15 V, μ = 60% = 0.6

As,  μ =  \(\frac{A_m}{A_c}\)

Or, A_m= μ A_c

A_m= 0.6 × 15

= 9V

Question 14. What is an antenna? Find the length of a dipole antenna for
Answer:

The antenna is a device used in communication systems for sending electromagnetic waves in all directions from the transmitter and receiving them at the receiver. The wavelength of carrier wave,

λ = \(\frac{\text { velocity }}{\text { frequency }}\)

= \(\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{3 \times 10^8 \mathrm{~s}^{-1}}\)

= 1m

∴ The length of the dipole antenna should be about 1 m

Practice Long Answer Questions on Modulation Techniques

Question 15.  Due to economic reasons, only did upper side bards of an AM wave are transmitted; but at the receiving station there is a facility for generating the carrier.
Answer:

Let the two signals be represented by \(A_1 \cos \left(\omega_c+\omega_m\right) t\) and \(A_c \cos \omega_c t\)

Multiplying the waves, we get,

⇒ \(A_1 A_c \cos \left(\omega_c+\omega_m\right) t \cdot \cos \omega_c t\)

= \(\frac{A_1 A_c}{2} \cos \omega_m t+\frac{A_1 A_c}{2} \cos \left(2 \omega_c+\omega_m\right) t\)

The separation of the relationship indicates that the modulating signal \(\frac{A_c A_1}{2} \cos \omega_m t\)can be easily recovered at the receiving station

Question 16. Write down the Difference between amplitude modulation (AM) and Frequency(FM)
Answer: