QBManager

PHXII15:COMMUNICATION SYSTEMS

356777 Modulation is used to

1 Reduce the bandwidth

2 To separate the transmission of different users

3 To ensure that intelligence may be transmitted to long distances

4 To allow the uses of practical antenna

PHXII15:COMMUNICATION SYSTEMS

356778 An AM wave is given by
$V = 1500 [1 + 0.5 \sin 12560 t] \sin (5.26 \times 10^{5} t) .$
The modulating frequency is

1

2.0 k H z

2

1.0 k H z

3

12.5 k H z

4

50 k H z

Explanation:

$C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t$
$f_{m} = \frac{ω_{m}}{2 π} = \frac{12560}{2 π} = 2 k h z$

PHXII15:COMMUNICATION SYSTEMS

356779 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to

1 Poor bandwidth selection of amplifiers

2 Poor selection of modulation index (selected

0 < m < 1

)

3 Loss of energy in transmission

4 Poor selection of carrier frequency

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356780 If a number of sine waves with modulation indices $n_{1}, n_{2}, n_{3} \dots \dots . .$ . Modulate a carrier wave, then total modulation index $(n)$ of the wave is

1

n_{1} + n_{2} \dots . + 2 (n_{1} + n_{2} \dots .)

2

\sqrt{n_{1} - n_{2} + n_{3} \dots \dots}

3

\sqrt{n_{1}^{2} + n_{2}^{2} + n_{3}^{2} \dots}

4 None of these

PHXII15:COMMUNICATION SYSTEMS

356781 An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 \times 10^{8} π t$ volts, the minimum and maximum values of modulated carrier wave are

1

10 V, 20 V

2

4 V, 8 V

3

16 V, 4 V

4

8 V, 20 V

PHXII15:COMMUNICATION SYSTEMS

356777 Modulation is used to

1 Reduce the bandwidth

2 To separate the transmission of different users

3 To ensure that intelligence may be transmitted to long distances

4 To allow the uses of practical antenna

PHXII15:COMMUNICATION SYSTEMS

356778 An AM wave is given by
$V = 1500 [1 + 0.5 \sin 12560 t] \sin (5.26 \times 10^{5} t) .$
The modulating frequency is

1

2.0 k H z

2

1.0 k H z

3

12.5 k H z

4

50 k H z

Explanation:

$C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t$
$f_{m} = \frac{ω_{m}}{2 π} = \frac{12560}{2 π} = 2 k h z$

PHXII15:COMMUNICATION SYSTEMS

356779 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to

1 Poor bandwidth selection of amplifiers

2 Poor selection of modulation index (selected

0 < m < 1

)

3 Loss of energy in transmission

4 Poor selection of carrier frequency

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356780 If a number of sine waves with modulation indices $n_{1}, n_{2}, n_{3} \dots \dots . .$ . Modulate a carrier wave, then total modulation index $(n)$ of the wave is

1

n_{1} + n_{2} \dots . + 2 (n_{1} + n_{2} \dots .)

2

\sqrt{n_{1} - n_{2} + n_{3} \dots \dots}

3

\sqrt{n_{1}^{2} + n_{2}^{2} + n_{3}^{2} \dots}

4 None of these

PHXII15:COMMUNICATION SYSTEMS

356781 An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 \times 10^{8} π t$ volts, the minimum and maximum values of modulated carrier wave are

1

10 V, 20 V

2

4 V, 8 V

3

16 V, 4 V

4

8 V, 20 V

PHXII15:COMMUNICATION SYSTEMS

356777 Modulation is used to

1 Reduce the bandwidth

2 To separate the transmission of different users

3 To ensure that intelligence may be transmitted to long distances

4 To allow the uses of practical antenna

PHXII15:COMMUNICATION SYSTEMS

356778 An AM wave is given by
$V = 1500 [1 + 0.5 \sin 12560 t] \sin (5.26 \times 10^{5} t) .$
The modulating frequency is

1

2.0 k H z

2

1.0 k H z

3

12.5 k H z

4

50 k H z

Explanation:

$C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t$
$f_{m} = \frac{ω_{m}}{2 π} = \frac{12560}{2 π} = 2 k h z$

PHXII15:COMMUNICATION SYSTEMS

356779 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to

1 Poor bandwidth selection of amplifiers

2 Poor selection of modulation index (selected

0 < m < 1

)

3 Loss of energy in transmission

4 Poor selection of carrier frequency

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356780 If a number of sine waves with modulation indices $n_{1}, n_{2}, n_{3} \dots \dots . .$ . Modulate a carrier wave, then total modulation index $(n)$ of the wave is

1

n_{1} + n_{2} \dots . + 2 (n_{1} + n_{2} \dots .)

2

\sqrt{n_{1} - n_{2} + n_{3} \dots \dots}

3

\sqrt{n_{1}^{2} + n_{2}^{2} + n_{3}^{2} \dots}

4 None of these

PHXII15:COMMUNICATION SYSTEMS

356781 An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 \times 10^{8} π t$ volts, the minimum and maximum values of modulated carrier wave are

1

10 V, 20 V

2

4 V, 8 V

3

16 V, 4 V

4

8 V, 20 V

PHXII15:COMMUNICATION SYSTEMS

356777 Modulation is used to

1 Reduce the bandwidth

2 To separate the transmission of different users

3 To ensure that intelligence may be transmitted to long distances

4 To allow the uses of practical antenna

PHXII15:COMMUNICATION SYSTEMS

356778 An AM wave is given by
$V = 1500 [1 + 0.5 \sin 12560 t] \sin (5.26 \times 10^{5} t) .$
The modulating frequency is

1

2.0 k H z

2

1.0 k H z

3

12.5 k H z

4

50 k H z

Explanation:

$C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t$
$f_{m} = \frac{ω_{m}}{2 π} = \frac{12560}{2 π} = 2 k h z$

PHXII15:COMMUNICATION SYSTEMS

356779 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to

1 Poor bandwidth selection of amplifiers

2 Poor selection of modulation index (selected

0 < m < 1

)

3 Loss of energy in transmission

4 Poor selection of carrier frequency

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356780 If a number of sine waves with modulation indices $n_{1}, n_{2}, n_{3} \dots \dots . .$ . Modulate a carrier wave, then total modulation index $(n)$ of the wave is

1

n_{1} + n_{2} \dots . + 2 (n_{1} + n_{2} \dots .)

2

\sqrt{n_{1} - n_{2} + n_{3} \dots \dots}

3

\sqrt{n_{1}^{2} + n_{2}^{2} + n_{3}^{2} \dots}

4 None of these

PHXII15:COMMUNICATION SYSTEMS

356781 An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 \times 10^{8} π t$ volts, the minimum and maximum values of modulated carrier wave are

1

10 V, 20 V

2

4 V, 8 V

3

16 V, 4 V

4

8 V, 20 V

PHXII15:COMMUNICATION SYSTEMS

356777 Modulation is used to

1 Reduce the bandwidth

2 To separate the transmission of different users

3 To ensure that intelligence may be transmitted to long distances

4 To allow the uses of practical antenna

PHXII15:COMMUNICATION SYSTEMS

356778 An AM wave is given by
$V = 1500 [1 + 0.5 \sin 12560 t] \sin (5.26 \times 10^{5} t) .$
The modulating frequency is

1

2.0 k H z

2

1.0 k H z

3

12.5 k H z

4

50 k H z

Explanation:

$C_{m} (t) = A_{c} (1 + μ \sin ω_{m} t) \sin ω_{c} t$
$f_{m} = \frac{ω_{m}}{2 π} = \frac{12560}{2 π} = 2 k h z$

PHXII15:COMMUNICATION SYSTEMS

356779 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to

1 Poor bandwidth selection of amplifiers

2 Poor selection of modulation index (selected

0 < m < 1

)

3 Loss of energy in transmission

4 Poor selection of carrier frequency

NCERT Exemplar

PHXII15:COMMUNICATION SYSTEMS

356780 If a number of sine waves with modulation indices $n_{1}, n_{2}, n_{3} \dots \dots . .$ . Modulate a carrier wave, then total modulation index $(n)$ of the wave is

1

n_{1} + n_{2} \dots . + 2 (n_{1} + n_{2} \dots .)

2

\sqrt{n_{1} - n_{2} + n_{3} \dots \dots}

3

\sqrt{n_{1}^{2} + n_{2}^{2} + n_{3}^{2} \dots}

4 None of these

PHXII15:COMMUNICATION SYSTEMS

356781 An AM wave is expressed as $e = 10 (1 + 0.6 \cos 2000 π t) \cos 2 \times 10^{8} π t$ volts, the minimum and maximum values of modulated carrier wave are

1

10 V, 20 V

2

4 V, 8 V

3

16 V, 4 V

4

8 V, 20 V