Welcome to the Chapter 15 - Communication Systems Physics, Class 12 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 15 - Communication Systems Physics. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
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(b) 10 MHz
For beyond-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.
(d) 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.
(c) 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.
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 × 10 6 m
For range, d = 2Rh,
The service area of the antenna is given by the relation: A = πd 2 = π (2Rh)
= 3.14 × 2 × 6.4 × 10 6 × 81 = 3255.55 × 10 6 m 2 = 3255.55~3256 km 2
Amplitude of the carrier wave, A c = 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = A m
Using the relation for modulation index: m = Am/Ac
Therefore, Am = m Ac
= 0.75 x 12 = 9V
It can be observed from the given modulating signal that the amplitude of the modulating signal, A m = 1 V
It is given that the carrier wave c (t) = 2 sin (8πt)
Therefore, Amplitude of the carrier wave, A c = 2 V
Time period of the modulating signal T m = 1 s
The angular frequency of the modulating signal is calculated as: ωm = 2π/Tm
= 2π rad s-1 .......(i)
The angular frequency of the carrier signal is calculated as: ωc = 8π rad s-1 ....(ii)
From equations (i) and (ii), we get:
= ωc = 4ωm
The amplitude modulated waveform of the modulating signal is shown in the following figure.
( ii ) Modulation index, m = Am/Ac = 1/2 = 0.5
Maximum amplitude, A max = 10 V
Minimum amplitude, A min = 2 V
Modulation index μ, is given by the relation:
Let ω c and ω s be the respective frequencies of the carrier and signal waves.
Signal received at the receiving station, V = V 1 cos (ω c + ω s )t
Instantaneous voltage of the carrier wave, V in = V c cos ω c t
At the receiving station, the low-pass filter allows only high frequency signals to pass through it. It obstructs the low frequency signal ω s .
Thus, at the receiving station, one can record the modulating signal, V1Vs/2 cos ωst which is the signal frequency.
