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PHXII15:COMMUNICATION SYSTEMS

356782 A message signal of frequency $ω_{m}$ is superposed on a carrier wave frequency $ω_{c}$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be

1

\frac{ω_{c} - ω_{m}}{2}

2

ω_{m}

3

ω_{c}

4

\frac{ω_{c} + ω_{m}}{2}

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356783 The limitation of amplitude modulation is

1 Clear reception

2 High efficiency

3 Small operating range

4 Good audio quality

PHXII15:COMMUNICATION SYSTEMS

356784 A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $120 V$ and $80 V$ respectively. The amplitude of each side band is

1

10 V

2

5 V

3

20 V

4

15 V

Explanation:

Given: $A_{c} - A_{m} = 80_{V} (1)$
$A_{c} - A_{m} = 120_{V} (2)$
Adding equation (1) and (2), we get $A_{c} = 100 V$
Amplitude of each side bond,
$A = \frac{μ A_{c}}{2} (3)$
where, $\frac{A_{max} - A_{min}}{A_{max} + A_{min}} = μ = \frac{120 - 80}{120 + 80} = \frac{40}{200}$
Putting the value in (3),
$A = (\frac{40}{200}) \times \frac{100}{2} = 10 V$

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356785 Amplitude modulation is used for broadcasting because

1 It is more noise immune

2 It requires less transmitting power

3 It has simple circuit; cheap and simple to manufacture

4 It has high fidelity (faithful reproduction)

PHXII15:COMMUNICATION SYSTEMS

356782 A message signal of frequency $ω_{m}$ is superposed on a carrier wave frequency $ω_{c}$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be

1

\frac{ω_{c} - ω_{m}}{2}

2

ω_{m}

3

ω_{c}

4

\frac{ω_{c} + ω_{m}}{2}

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356783 The limitation of amplitude modulation is

1 Clear reception

2 High efficiency

3 Small operating range

4 Good audio quality

PHXII15:COMMUNICATION SYSTEMS

356784 A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $120 V$ and $80 V$ respectively. The amplitude of each side band is

1

10 V

2

5 V

3

20 V

4

15 V

Explanation:

Given: $A_{c} - A_{m} = 80_{V} (1)$
$A_{c} - A_{m} = 120_{V} (2)$
Adding equation (1) and (2), we get $A_{c} = 100 V$
Amplitude of each side bond,
$A = \frac{μ A_{c}}{2} (3)$
where, $\frac{A_{max} - A_{min}}{A_{max} + A_{min}} = μ = \frac{120 - 80}{120 + 80} = \frac{40}{200}$
Putting the value in (3),
$A = (\frac{40}{200}) \times \frac{100}{2} = 10 V$

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356785 Amplitude modulation is used for broadcasting because

1 It is more noise immune

2 It requires less transmitting power

3 It has simple circuit; cheap and simple to manufacture

4 It has high fidelity (faithful reproduction)

PHXII15:COMMUNICATION SYSTEMS

356782 A message signal of frequency $ω_{m}$ is superposed on a carrier wave frequency $ω_{c}$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be

1

\frac{ω_{c} - ω_{m}}{2}

2

ω_{m}

3

ω_{c}

4

\frac{ω_{c} + ω_{m}}{2}

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356783 The limitation of amplitude modulation is

1 Clear reception

2 High efficiency

3 Small operating range

4 Good audio quality

PHXII15:COMMUNICATION SYSTEMS

356784 A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $120 V$ and $80 V$ respectively. The amplitude of each side band is

1

10 V

2

5 V

3

20 V

4

15 V

Explanation:

Given: $A_{c} - A_{m} = 80_{V} (1)$
$A_{c} - A_{m} = 120_{V} (2)$
Adding equation (1) and (2), we get $A_{c} = 100 V$
Amplitude of each side bond,
$A = \frac{μ A_{c}}{2} (3)$
where, $\frac{A_{max} - A_{min}}{A_{max} + A_{min}} = μ = \frac{120 - 80}{120 + 80} = \frac{40}{200}$
Putting the value in (3),
$A = (\frac{40}{200}) \times \frac{100}{2} = 10 V$

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356785 Amplitude modulation is used for broadcasting because

1 It is more noise immune

2 It requires less transmitting power

3 It has simple circuit; cheap and simple to manufacture

4 It has high fidelity (faithful reproduction)

PHXII15:COMMUNICATION SYSTEMS

356782 A message signal of frequency $ω_{m}$ is superposed on a carrier wave frequency $ω_{c}$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be

1

\frac{ω_{c} - ω_{m}}{2}

2

ω_{m}

3

ω_{c}

4

\frac{ω_{c} + ω_{m}}{2}

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356783 The limitation of amplitude modulation is

1 Clear reception

2 High efficiency

3 Small operating range

4 Good audio quality

PHXII15:COMMUNICATION SYSTEMS

356784 A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $120 V$ and $80 V$ respectively. The amplitude of each side band is

1

10 V

2

5 V

3

20 V

4

15 V

Explanation:

Given: $A_{c} - A_{m} = 80_{V} (1)$
$A_{c} - A_{m} = 120_{V} (2)$
Adding equation (1) and (2), we get $A_{c} = 100 V$
Amplitude of each side bond,
$A = \frac{μ A_{c}}{2} (3)$
where, $\frac{A_{max} - A_{min}}{A_{max} + A_{min}} = μ = \frac{120 - 80}{120 + 80} = \frac{40}{200}$
Putting the value in (3),
$A = (\frac{40}{200}) \times \frac{100}{2} = 10 V$

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356785 Amplitude modulation is used for broadcasting because

1 It is more noise immune

2 It requires less transmitting power

3 It has simple circuit; cheap and simple to manufacture

4 It has high fidelity (faithful reproduction)

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