NCERT Solutions for Class 12 Physics Chapter 15 - Communication Systems

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

(a) $10{\text{ kHz}}$

(b) $10{\text{ MHz}}$

(c) $1{\text{ GHz}}$

(d) $1000{\text{ GHz}}$

Ans: (b) 

For beyond-the-horizon communication, the signal waves should travel a large distance. $10{\text{ kHz}}$ signals can’t be radiated well because of the antenna size. The high energy signal waves $(1{\text{ GHz}} - 1000{\text{ GHz}})$ penetrate the ionosphere. $10{\text{ MHz}}$ frequencies get reflected from the ionosphere easily. So, signal waves of such frequencies are suitable for beyond-the-horizon communication.

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

(a) Ground waves.

(b) Sky waves.

(c) Surface waves.

(d) Space waves.

Ans: (d) 

Due to high frequency, an ultra-high frequency wave can neither travel along the trajectory of ground nor can it get reflected by the ionosphere. The signals having ultra-high frequency are propagated through line-of-sight communication, i.e. space wave propagation.

3. Digital signals

(i) Do not provide a continuous set of values,

(ii) Represent values as discrete steps,

(iii) Can utilize binary system, and

(iv) Can utilize decimal as well as binary systems.

Which of the above statements are true?

(a) (i) and (ii) only

(b) (ii) and (iii) only

(c) (i), (ii) and (iii) but not (iv)

(d) All of (i), (ii), (iii) and (iv).

Ans: (c) 

A digital signal uses the binary system for transferring message signals. It cannot use the decimal system (which corresponds to analogue signals). Digital signals represents discontinuous values.

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 $81{\text{ m}}$ tall. How much service area can it cover if the receiving antenna is at the ground level?

Ans:

In line-of-sight communication, there is no physical obstruction between the transmitter and the receiver. It is not necessary for the transmitting antenna and receiving antenna to be at the same height.

Height of the antenna, $h = 81\;{\text{ m}}$

Radius of earth, $R = 6.4 \times {10^5}\;{\text{m}}$

For range,$d = 2Rh$, the service area of the antenna is:

$A = n{d^2}$

$A = n(2Rh)$

$A = 3.14 \times 2 \times 6.4 \times {10^5} \times 81$

$A = 3255.55 \times {10^4}\;{{\text{m}}^2}$

$A = 3255.55\;{\text{ k}}{{\text{m}}^2}$

$A = 3256\;{\text{ k}}{{\text{m}}^2}$

5. A carrier wave of peak voltage $12v$ 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\% $

Ans:

Amplitude of the carrier wave, ${\text{A}} = 12\;{\text{V}}$

Modulation index, $m = 75\%  = 0.75$

Amplitude of the modulating wave $ = {A_n}$

From the relation for modulation index:

$m = \dfrac{{{A_m}}}{{{A_c}}}$

$\therefore {A_m} = m{A_c}$

${A_m} = 0.75 \times 12$

${A_m} = 9V$

6. A modulating signal is a square wave, as shown in Fig. 15.14.

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

(i) Sketch the amplitude modulated waveform

(ii) What is the modulation index?

Ans:

(i) The amplitude of the modulation signal, ${A_n} = 1V$.

The carrier wave is $c(t) = 2\sin (8nt)$.

$\therefore $ Amplitude of the carrier wave, ${A_2} = 2V$

The time period of the modulating signal ${T_n} = 15$

The angular frequency of the modulating signal is:

${\omega _m} = \dfrac{{2\pi }}{{{T_m}}}$

${\omega _m} = 2\pi {\text{rad }}{{\text{s}}^{ - 2}}{\text{        }}...{\text{(i)}}$ 

The angular frequency of the carrier signal is:

${\omega _c} = 8\pi {\text{rad }}{{\text{s}}^{ - 1}}{\text{           }}...{\text{(ii)}}$

From ${\text{(i)}}$ and ${\text{(ii)}}$:

${\omega _c} = 4{\omega _m}$

The amplitude modulated waveform of the modulating signal is as follows:

(ii) Modulation index:

${\text{m}} = \dfrac{{{A_m}}}{{{A_c}}}$

${\text{m}} = \dfrac{1}{2}$

${\text{m}} = 0.5$

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 $V$. What would be the value of $V$ if the minimum amplitude is zero volt?

Ans:

Maximum amplitude, ${A_{\max }} = 10\;V$

Minimum amplitude, ${A_{{\text{min }}}} = 2\;V$

Modulation index $\mu $, is:

$\mu  = \dfrac{{{A_{\max }} - {A_{\min }}}}{{{A_{\max }} + {A_{min}}}}$

$\mu  = \dfrac{{10 - 2}}{{10 + 2}}$

$\mu  = \dfrac{8}{{12}}$

$\mu  = 0.67$

If ${A_{\min }} = 0$,

$\mu  = \dfrac{{{A_{{\text{max }}}}}}{{{A_{{\text{max }}}}}}$

$\mu  = \dfrac{{10}}{{10}}$

$\mu  = 1$

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.

Ans:

Let ${\omega _c}$ and ${\omega _s}$ be the frequencies of the carrier and signal waves.

Signal received at the receiving station, $V = {V_1}\cos \left( {{\omega _c} + {\omega _s}} \right)t$

Instantaneous voltage of the carrier wave, ${V_{in}} = {V_c}\cos {\omega _c}t$

$\therefore V{V_{{\text{in}}}} = {V_1}\cos \left( {{\omega _{\text{c}}} + {\omega _s}} \right){\text{t}} \cdot \left( {{{\text{V}}_c}\cos {\omega _c}{\text{t}}} \right)$

$V{V_{{\text{in}}}} = {V_1}\;{{\text{V}}_c}\left[ {\cos \left( {{\omega _{\text{c}}} + {\omega _{\text{s}}}} \right){\text{t}}{\text{.}}\cos {\omega _c}{\text{t}}} \right]$

$V{V_{{\text{in}}}} = \dfrac{{{V_1}\;{{\text{V}}_{\text{c}}}}}{2}\left[ {2\cos \left( {{\omega _{\text{c}}} + {\omega _{\text{s}}}} \right){\text{t}} \cdot \cos {\omega _{\text{c}}}{\text{t}}} \right]$

$V{V_{{\text{in}}}} = \dfrac{{{V_1}{V_c}}}{2}\left[ {\cos \left\{ {\left( {{\omega _c} + {\omega _s}} \right)t + {\omega _c}t} \right\} + \cos \left\{ {\left( {{\omega _c} + {\omega _s}} \right)t - {\omega _c}t} \right\}} \right]$

$V{V_{{\text{in}}}} = \dfrac{{{V_1}{V_c}}}{2}\left[ {\cos \left\{ {\left( {2{m_c} + {\omega _s}} \right)t + \cos {\omega _s}t} \right\}} \right]$

At the receiving station, the low-pass filter will allow only high frequency signals to pass. It obstructs the low frequency signal. So, at the receiving station, the modulating signal $\dfrac{{{V_1}{V_c}}}{2}\cos {\omega _s}t$ can be recorded, which is the signal frequency.

NCERT Solutions for Class 12 Physics – Free PDF Download

Chapter 15 of NCERT Class 12 Physics is based on Communication Systems. The chapter explores the basic elements of a Communication System that help in communicating information from the source to the receiver. The chapter encompasses topics including the basic terminologies in Communication Systems, the role of the transmission medium, signalling by the propagation of electromagnetic waves, amplitude and frequency modulations, demodulation, etc. NCERT Solutions Class 12 Chapter 15 deals with numerical questions, objective type questions, short answer type questions and long answer type questions.

NCERT Solutions for Class 12 Physics Chapter 15 - Communication System

Chapter 15 – Communication System

NCERT Solutions offers to help students in gaining in-depth knowledge of the chapters. Students not only learn but they also understand the chapters more clearly and efficiently through these NCERT Solutions. Before appearing for your main exams, you need to have an idea about the type of questions that may come, therefore NCERT Solutions helps you practice questions framed by professional teachers.

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This chapter includes a detailed analysis of how the Communication System functions. Block diagrams help explain the basics of Communication System. The communication between the source and receiver is established via the bandwidth of signal and transmission medium. Students will get detailed knowledge about the earth's atmosphere and how do electromagnetic waves propagate in the atmosphere, how to modulate and demodulate amplitude and frequencies for proper communication, etc.

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