NCERT Solutions for Class 12 Physics Chapter 15 – Communication Systems

NCERT Solutions for Class 12 Physics Chapter 15 – Communication Systems – FREE PDF Download

CoolGyan provides All Chapter 15 – Communication Systems Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Physics Chapter 15 – Communication Systems includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Physics Chapter 15 – Communication Systems. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Revision Notes Class 12 Physics
NCERT Solutions Class 12 Physics

Major examples of general communication systems:

E-mail
Internet
Television
Radio
Computer

Concepts involved in NCERT class 12 chapter 15 Communication System are:

1. Introduction
2. Elements Of A Communication System
3. Basic Terminology Used In Electronic Communication Systems
4. Bandwidth Of Signals
5. Bandwidth Of Transmission Medium
6. Propagation Of Electromagnetic Waves
  1. Ground wave
  2. Skywaves
  3. Space wave
7. Modulation And Its Necessity
8. Size of the antenna or aerial
9. Effective power radiated by an antenna Ex 15.7.3 – Mixing up of signals from different transmitters
10. Amplitude Modulation
11. Production Of Amplitude Modulated Wave
12. Detection Of Amplitude Modulated Wave.

Class 12 Physics NCERT Solutions Communication Systems Questions

NCERT Exercises

Question 1.
Which of the following frequencies will be suit-able for beyond-the-horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz

Solution:
(b) : 10 MHz will be suitable frequency for sky waves as lower frequency of 10 kHz will require large radiating antenna and higher frequencies 1 GHz and 1000 GHz will pass through the ionosphere and will not be reflected by it.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves

Solution:
(d) : Frequencies in the UHF range normally propagates by means of space waves. The high frequency space waves are ideal for frequency modulation but do not bend with ground.

Question 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).

Solution:
(c) : Decimal system represents a continuous set of values which cannot be utilized by digital signals.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for the 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?

Solution:
No, for line of sight communication, the two antenna may not be at the same height. Surface area

$A=pi { d }^{ 2 }=pi left( 2hR ight) =cfrac { 22 }{ 7 } imes 2 imes 81 imes 6.4 imes { 10 }^{ 6 }$
$=3258.5 imes { 10 }^{ 6 }sq.metre=3258.5sq.km$

Question 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%?
Solution:
Modulation index, $mu =cfrac { { A }_{ m } }{ { A }_{ c } }$ so, peak voltage
${ A }_{ m }={ mu A }_{ c }=0.75 imes 12=9V$

Question 6.
A modulating signal is a square wave, as shown in Figure

The carrier wave is given by c(t) = 2sin (8πt) volts
(i) Sketch the amplitude modulated wave form
(ii) What is the modulation index?
Solution:
(i) The amplitude modulated wave is shown here’:

(ii) Modulation index, $mu cfrac { { A }_{ m } }{ { A }_{ c } } =cfrac { 1V }{ 2V } =0.5$

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index, p. What would be the value of p if the minimum amplitude is zero volt?
Solution:
We know
Modulation index, $mu =cfrac { { A }_{ m } }{ { A }_{ c } }$
Also, minimum amplitude, ${ A }_{ min }={ A }_{ c }left( 1-mu ight)$
Maximum amplitude, ${ A }_{ max }={ A }_{ c }left( 1+mu ight)$
So, modulation index, $mu =cfrac { { A }_{ max }{ -A }_{ min } }{ { A }_{ max }{ +A }_{ min } }$
or $mu =cfrac { 10-2 }{ 10+2 } =cfrac { 8 }{ 12 } =2/3=0.67$
if A_min = O, then modulation index, $cfrac { { A }_{ mix }-{ A }_{ min } }{ { A }_{ mix }+{ A }_{ min } } =cfrac { 10-0 }{ 10+0 } =cfrac { 10 }{ 10 } =1$

Question 8.
Due to economic reasons, only the upper side band 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.
Solution:
Let, the received signal be cos(ω_c + ω_m)t The carrier signal available at the receiving station is A_c cos ω_ct Multiplying the two signals, we get A₁A_c cos (ω_c + ω_m)t cos ω_ct $cfrac { { A }_{ 1 }{ A }_{ c } }{ 2 } left[ cosleft( 2{ omega }_{ c }+{ omega }_{ m } ight) t+cos{ omega }_{ m }t ight]$ If this signal is passed through a low pass filter, we can recover the modulating signal $cfrac { { A }_{ 1 }{ A }_{ c } }{ 2 } cos{ omega }_{ m }t$

NCERT Solutions for Class 12 Physics Chapter 15 – Communication Systems – Free PDF Download