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|Chapter||15. Communication Systems|
|Category||NCERT Solutions for Class 12|
Class 12 Physics NCERT Solutions are the excellent practice set for students to ace Physics with better understandings and good marks. With the NCERT Solution for Class 12 Physics, students can confidently answer their exam questions and enjoy the best possible marks in the subject. These solutions are prepared keeping in view the latest syllabus and curriculum. In addition, the solutions are designed keeping in mind the requirements of students. Therefore, students can use these solutions for revision or to score high marks in the final examinations. Let’s get started with the solutions of Chapter 15: Communication Systems.
NCERT Solutions for Class 12 Physics Chapter 15
Communication Systems Solutions
Q 15.1) 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
(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.
Q 15.2) Frequencies in the UHF range normally propagate by means of :
(1) Ground Waves
(2) Sky Waves
(3) Surface Waves
(4) Space Waves
(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.
Q 15.3) 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) (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.
Q 15.4) 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
Q 15.5) 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%?
Amplitude of carrier wave, Ac = 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = Am
Modulation index is given by the relation :
Therefore, Am = m.Ac
= 0.75 x 12 V= 9 V
Q 15.6) 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, Am = 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, Ac = 2v
Time period, Tm = 1s
The angular frequency of the modulating signal is given by,
= 2π rad s-1 …(1)
The angular frequency of carrier signal, ωc = 8π rad s-1 …(2)
from eqns.(1) and (2),
we get, ωc = 4ωm
The modulating signal having the amplitude modulated waveform is shown in the figure:
(2) Modulation index, m =
Q 15.7) 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?
Maximum Amplitude, Amax = 10 V
Minimum Amplitude, Amin = 2 V
For a wave, modulation index µ, is given by:
If Amin = 0,
Q 15.8) 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, ωc be the carrier wave frequency
ωs be the signal wave frequency
Signal received, V = V1 cos (ωc + ωs)t
Instantaneous voltage of the carrier wave, Vm = Vc cos ωc t
V.Vin = V1cos (ωc + + ωs)t. (Vc cos ωc t)
The low pass filter allows only the high frequency signals to pass through it. The low frequency signal ωs is obstructed by it.
Thus, at the receiving station, we can record the modulating signal, 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.
That’s it. These were the solutions of NCERT Class 12 Physics Chapter 15 – Communication Systems. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.