Ncert Solution Class 12th Physics Chapter – 15 Communication Systems
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 15 |
| Chapter Name | Communication Systems |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 15 Communication Systems
?Chapter – 15?
✍Communication Systems✍
?Question & Answer?
Q.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
Answer: (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.2: Frequencies in the UHF range normally propagate by means of :
(1) Ground Waves
(2) Sky Waves
(3) Surface Waves
(4) Space Waves
Answer: (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.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) (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.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.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%?
Answer: Given: A carrier wave of peak voltage 12 V and modulation index of the modulating signal 75%.
The modulation index can be given as,
μ= A m A c
Where, Amplitude of modulating wave is A m , Amplitude of carrier wave is A c and the modulating index is m.
By substituting the given values in above equation, we get
75%= A m A c A m = 75 100 ×12 =9 V
Thus, the peak voltage of the modulating signal is 9 V.
Q.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: (i) From the above modulating signal, it can be observed that the amplitude of the modulating signal is,
A m =1 V
The equation of carrier wave is,
c( t )=2sin( 8πt )
Thus,
Amplitude of the carrier wave, A C =2 V
Time period of the modulating signal, T m =1 s
Let, ω m be the angular frequency of the modulating signal
The angular frequency is given as,
ω m = 2π T m
Thus
ω m = 2π 1 =2π rads -1
From above expression of c( t )=2sin( 8πt ), it can be observed that the angular frequency of the carrier signal is,
ω c =8π rads -1
Thus,
ω c =4 ω m
Therefore, the amplitude modulated waveform of modulating signal will be,
(ii) The modulation index is given as,
m= A m A c
Substitute the values in above expression.
m= 1 2 =0.5
Q.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?
Answer: Given,
Maximum amplitude, A max =10 V
Minimum amplitude, A min =2 V
Let μ be the modulation index of the modulated wave.
The modulation index is,
μ= A max − A min A max − A min
Substitute the given values in above expression.
μ= 10−2 10+2 = 8 12 = 2 3 =0.67
If A min =0
Then,
μ= 10−0 10+0 =1
Q.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 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.