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Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
Maximum amplitude, Amax = 10 V
Minimum amplitude, Amin = 2 V
Modulation index μ, is given by the relation:
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
It can be observed from the given modulating signal that the 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,
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
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:
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
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 = πd2
= π (2Rh)
= 3.14 × 2 × 6.4 × 106× 81
= 3255.55 × 106 m2
= 3255.55
∼ 3256 km2
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
Maximum amplitude, Amax = 10 V
Minimum amplitude, Amin = 2 V
Modulation index μ, is given by the relation:
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
It can be observed from the given modulating signal that the 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,
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
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:
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
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 = πd2
= π (2Rh)
= 3.14 × 2 × 6.4 × 106× 81
= 3255.55 × 106 m2
= 3255.55
∼ 3256 km2
Answered on 16/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 10-Communication Systems/COMMUNICATION SYSTEMS/NCERT Solutions/Exercise 15
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