## Ncert Solutions of Physics Part 2 for Chapter 7 : Communication Systems

### Exercise : Solutions of Questions on Page Number : 530

Q1 :
Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz

10 MHz
Forbeyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz - 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication.
Q2 :
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves.
(b) Sky waves.
(c) Surface waves.
(d) Space waves.

Space waves
Owing to its high frequency, an ultra high frequency (UHF)wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation.
Q3 :
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).

A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilise the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values.
Q4 :  Is it necessary for a transmitting antenna to be at the same height

Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height.
Height of the given antenna, h = 81 m
Radius of earth, R = 6.4 106 m
For range, d = (2Rh)½, the service area of the antenna is given by the relation:
A = Ãâ‚¬d2
= Ãâ‚¬ (2Rh)
= 3.14 2 6.4 106 81
= 3255.55 106 m2
= 3255.55
∝¼ 3256 km2
Q5 :  A carrier wave of peak voltage 12

Amplitude of the carrier wave, Ac= 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = Am
Using the relation for modulation index:
Q6 :
A modulating signal is a square wave, as shown in Fig. 15.14.
The carrier wave is given by
(i) Sketch the amplitude modulated waveform
(ii) What is the modulation index?

It can be observed from the given modulating signal thatthe amplitude of the modulating signal,Am= 1 V
It is given that the carrier wave c(t) = 2 sin (8πt)
Amplitude of the carrier wave, Ac= 2 V
Time period of the modulating signal Tm= 1 s
The angular frequency of the modulating signal is calculated as:
The angular frequency of the carrier signal is calculated as:
From equations (i) and (ii), we get:
The amplitude modulated waveform of the modulating signal is shown in the following figure.
(ii)Modulation index,
Q7 :  For an amplitude modulated wave, the maximum amplitude