Question

NCERT Solutions Class 12 Physics Chapter 15 Communication Systems is one of the best study materials one can find online. Our NCERT Solutions have a complete discussion of all the concepts of physics. One must refer to our NCERT Solutions Class 12 for learning all the difficult concepts with ease.

• Communication Systems – the communication system is a set of all the telecommunication devices present in a network to establish a connection between two devices. It consists of a telecommunication network, transmission system, relay stations, tributary stations, and terminal equipment which are used for the interconnection and interoperation to form an integrated whole system.
• Elements of a Communication System – the main elements are the channel or medium of the communication and the information receiver. It is a system that helps in the exchange of information between two points. The process which helps in the transmission and reception of information is called communication. There are two types of communication systems.
  • Analog system – the transmission of data of various frequencies or amplitude is called the analog system. The broadcast and the telephone transmission are very good examples of analog systems.
  • Digital system – in the digital system the data transmission is in the form of 0s and 1s. The high is represented by 1 and the low is represented by 0.
• Basic Terminology Used in Electronic Communication Systems – the electronic communication system helps in the transmission, process, and receiving the information. The electronic communication system includes a transmitter, a communication channel or medium than a receiver, and noise.
• Bandwidth of Signals – the bandwidth of the signal is the main difference between the upper and lower frequencies of a generated signal. The bandwidth of the signal is equal to the difference between the higher and the lower frequencies.

NCERT Solutions Class 12 Physics is made by the subject matter expert for giving complete clarity to all kinds of concepts. Our expert suggests referring to our solution for a complete understanding of topics.

Answer 1

NCERT Solutions Class 12 Physics Chapter 15 Communication Systems

1. Which of the following frequencies will be suitable for beyond-the horizon communication using sky waves?

(1) 10 kHz

(2) 10 MHz

(3) 1 GHz

(4)  1000 GHz

Answer:

(2) 10 MHz

The signal waves need to travel a large distance for beyond – the – horizon communication.

Due to the antenna size, the 10 kHz signals cannot be radiated efficiently.

The 1 GHz – 1000 GHz (high energy) signal waves penetrate the ionosphere.

The 10 MHz frequencies get reflected easily from the ionosphere. Therefore, for beyond – the – horizon communication signal waves of 10 MHz frequencies are suitable.

2. Frequencies in the UHF range normally propagate by means of :

(1) Ground Waves

(2) Sky Waves

(3) Surface Waves

(4)  Space Waves

Answer:

(4) Space Waves

Due to its high frequency, an ultra-high frequency (UHF) wave cannot travel along the trajectory of the ground also it cannot get reflected by the ionosphere. The ultrahigh-frequency signals are propagated through line – of – sight communication, which is actually space wave propagation.

3. Digital signals

(i) Do not provide a continuous set of values

(ii) Represent value as discrete steps

(iii) Can utilize binary system

(iv) Can utilize decimal as well as binary systems

State which statement(s) are true?

(a) (1), (2) and (3)

(b) (1) and (2) only

(c) All statements are true

(d) (2) and (3) only

Answer:

(a) (1), (2) and (3) 

For transferring message signals the digital signals use the binary (0 and 1) system. Such a system cannot utilise the decimal system. Discontinuous values are represented in digital signals.

4. Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?

Answer:

In line – of – sight communication, between the transmitter and the receiver there is no physical obstruction. So, there is no need for the transmitting and receiving antenna to be at the same height.

Height of the antenna, h = 81 m

Radius of earth, R = 6.4 x 10⁶m

d = √2Rh, for range

The service area of the antenna is given by the relation :

A = πd² = π(2Rh)

= 3.14 x 2 x 6.4 x 10⁶ x 81

= 3255.55 x 10⁶ m² 

= 3255.55 

= 3256 km²

5. 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 in order to have a modulation index of 75%?

Answer:

Given:

Amplitude of carrier wave, A_c = 12 V

Modulation index, m = 75% = 0.75

Amplitude of the modulating wave = A_m

Modulation index is given by the relation :

m = \(\frac{A_{m}}{A_{c}}\)

Therefore, A_m = m.A_c

= 0.75 x 12 V 

= 9 V

6. A modulating signal is a square wave, as shown in the figure.

The carrier wave is given by c(t) = 2sin (8πt) volts.

(1) Sketch the amplitude modulated waveform

(2) What is the modulation index?

Answer:

The amplitude of the modulating signal, A_m = 1v can be easily observed from the given modulating signal.

Carrier wave is given by, c(t) = 2 sin(8nt)

Amplitude of the carrier wave, A_c = 2v

Time period, T_m = 1s

The angular frequency of the modulating signal is given by,

\(\omega _{m} = \frac{2\pi }{T_{m}}\)

= 2π rad s^-1       …(1)

The angular frequency of carrier signal, 

ω_c​ = 8π rad s^-1       …(2)

From eqns.(1) and (2),

we get, ω_c ​= 4ω_m​

The modulating signal having the amplitude modulated waveform is shown in the figure:

(2) Modulation index, \(m = \frac{A_{m}}{A_{c}}​​ = \frac{1}{2} = 0.5\)

7. For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. Determine the modulation index, µ. What would be the value of µ if the minimum amplitude is zero volts?

Answer:

Given,

Maximum Amplitude, A_max = 10 V

Minimum Amplitude, A_min = 2 V

For a wave, modulation index µ, is given by :

\(µ = \frac{A_{max} – A_{min}}{A_{max} + A_{min}}\)

\(=\frac{10 – 2}{10 + 2}​ = \frac{8}{12} = 0.67\) 

If A_min = 0,

Then,

\(µ’ = \frac{A_{max}}{ A_{max}}\)​​ = 10/10 = 1

8. Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

Answer:

Let, ω_c​ be the carrier wave frequency

ω_s​ be the signal wave frequency

Signal received, V = V₁ cos ω_c​ + ω_s​)t

Instantaneous voltage of the carrier wave, V_m = V_c cos ω_c​t

V.V_in = V₁cos(ω_c​ + ω_s​)t. (V_c cos ω_c​t)

= V₁V_c [cos(ω_c​ + ω_s​)t . cos ω_c​t]

The low pass filter allows only the high frequency signals to pass through it. The low frequency signal ω_s​ is obstructed by it.

Thus, at the receiving station, we can record the modulating signal, \(\frac{V_{1}V_{C}}{2}cos\omega _{s}t\) which is the signal frequency.

• The transmitter, transmission channel, and receiver are three basic units of a communication system.
• Low frequencies cannot be transmitted to long distances. Therefore, they are superimposed on a high-frequency carrier signal by a process known as modulation.
• Two important forms of the communication system are Analog and Digital.
• Amplitude modulated waves can be produced by application of the message signal and the carrier wave to a non-linear device, followed by a bandpass filter.