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PHXII15:COMMUNICATION SYSTEMS

356790 Depth of modulation in terms of $E_{max}$ and $E_{min}$ is

1

m_{a} = \frac{E_{max} + E_{min}}{E_{min}}

2

m_{a} = \frac{E_{max} - E_{min}}{E_{max}}

3

m_{a} = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}

4

m_{a} = \frac{E_{max} + E_{min}}{E_{max} - E_{min}}

PHXII15:COMMUNICATION SYSTEMS

356791 An oscillator is producing FM waves of $A M$ wire is $24 mV$ and the minimum peak to peak voltage is $8 mV$ . The modulation factor is

1

20 %

2

10 %

3

50 %

4

25 %

Explanation:

Here, $V_{max} = \frac{24}{2} = 12 m V$ and
$V_{min} = \frac{8}{2} = 2 m V$
Now, $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
$= \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50 %$
$h_{R}$ are doubled then s bemomes $\sqrt{2}$ times.

PHXII15:COMMUNICATION SYSTEMS

356792 The antenna current of an AM transmitter is $8 A$ when only the carrier is sent, but it increases to $8.93 A$ when the carrier is modulated by a single wave. The percentage modulation is

1

60.1 %

2

70.1 %

3

80.1 %

4

50.1 %

Explanation:

$I_{f} = I_{C} \sqrt{1 + \frac{m^{2}}{2}}$
$8.93 = 8 \sqrt{1 + \frac{m^{2}}{2}}$
$(1.11)^{2} = \frac{2 + m^{2}}{2}$
$\Rightarrow m = 0.7$
i.e., $70 %$

PHXII15:COMMUNICATION SYSTEMS

356793 If $f_{c}$ and $f_{m}$ are the frequencies of carrier wave and signal, then the bandwidth is

1

f_{m}

2

2 f_{m}

3

f_{c}

4

2 f_{c}

PHXII15:COMMUNICATION SYSTEMS

356794 A modulating signal is a square wave, as shown in the figure.
supporting img

If the carrier wave is given as $C (t) = 2 \sin (8 π t) volts,$ the modulation index is

1

\frac{1}{3}

2 1

3

\frac{1}{2}

4

\frac{1}{4}

Explanation:

From the given diagram
maximum of $m (t) = A_{m} = 1 V$
Also, maximum of $c (t) = A_{c} = 2 V$
$∴$ Modulation index $μ = \frac{A_{m}}{A_{c}} = \frac{1}{2}$
So option (3) is correct.

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356790 Depth of modulation in terms of $E_{max}$ and $E_{min}$ is

1

m_{a} = \frac{E_{max} + E_{min}}{E_{min}}

2

m_{a} = \frac{E_{max} - E_{min}}{E_{max}}

3

m_{a} = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}

4

m_{a} = \frac{E_{max} + E_{min}}{E_{max} - E_{min}}

PHXII15:COMMUNICATION SYSTEMS

356791 An oscillator is producing FM waves of $A M$ wire is $24 mV$ and the minimum peak to peak voltage is $8 mV$ . The modulation factor is

1

20 %

2

10 %

3

50 %

4

25 %

Explanation:

Here, $V_{max} = \frac{24}{2} = 12 m V$ and
$V_{min} = \frac{8}{2} = 2 m V$
Now, $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
$= \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50 %$
$h_{R}$ are doubled then s bemomes $\sqrt{2}$ times.

PHXII15:COMMUNICATION SYSTEMS

356792 The antenna current of an AM transmitter is $8 A$ when only the carrier is sent, but it increases to $8.93 A$ when the carrier is modulated by a single wave. The percentage modulation is

1

60.1 %

2

70.1 %

3

80.1 %

4

50.1 %

Explanation:

$I_{f} = I_{C} \sqrt{1 + \frac{m^{2}}{2}}$
$8.93 = 8 \sqrt{1 + \frac{m^{2}}{2}}$
$(1.11)^{2} = \frac{2 + m^{2}}{2}$
$\Rightarrow m = 0.7$
i.e., $70 %$

PHXII15:COMMUNICATION SYSTEMS

356793 If $f_{c}$ and $f_{m}$ are the frequencies of carrier wave and signal, then the bandwidth is

1

f_{m}

2

2 f_{m}

3

f_{c}

4

2 f_{c}

PHXII15:COMMUNICATION SYSTEMS

356794 A modulating signal is a square wave, as shown in the figure.
supporting img

If the carrier wave is given as $C (t) = 2 \sin (8 π t) volts,$ the modulation index is

1

\frac{1}{3}

2 1

3

\frac{1}{2}

4

\frac{1}{4}

Explanation:

From the given diagram
maximum of $m (t) = A_{m} = 1 V$
Also, maximum of $c (t) = A_{c} = 2 V$
$∴$ Modulation index $μ = \frac{A_{m}}{A_{c}} = \frac{1}{2}$
So option (3) is correct.

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356790 Depth of modulation in terms of $E_{max}$ and $E_{min}$ is

1

m_{a} = \frac{E_{max} + E_{min}}{E_{min}}

2

m_{a} = \frac{E_{max} - E_{min}}{E_{max}}

3

m_{a} = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}

4

m_{a} = \frac{E_{max} + E_{min}}{E_{max} - E_{min}}

PHXII15:COMMUNICATION SYSTEMS

356791 An oscillator is producing FM waves of $A M$ wire is $24 mV$ and the minimum peak to peak voltage is $8 mV$ . The modulation factor is

1

20 %

2

10 %

3

50 %

4

25 %

Explanation:

Here, $V_{max} = \frac{24}{2} = 12 m V$ and
$V_{min} = \frac{8}{2} = 2 m V$
Now, $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
$= \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50 %$
$h_{R}$ are doubled then s bemomes $\sqrt{2}$ times.

PHXII15:COMMUNICATION SYSTEMS

356792 The antenna current of an AM transmitter is $8 A$ when only the carrier is sent, but it increases to $8.93 A$ when the carrier is modulated by a single wave. The percentage modulation is

1

60.1 %

2

70.1 %

3

80.1 %

4

50.1 %

Explanation:

$I_{f} = I_{C} \sqrt{1 + \frac{m^{2}}{2}}$
$8.93 = 8 \sqrt{1 + \frac{m^{2}}{2}}$
$(1.11)^{2} = \frac{2 + m^{2}}{2}$
$\Rightarrow m = 0.7$
i.e., $70 %$

PHXII15:COMMUNICATION SYSTEMS

356793 If $f_{c}$ and $f_{m}$ are the frequencies of carrier wave and signal, then the bandwidth is

1

f_{m}

2

2 f_{m}

3

f_{c}

4

2 f_{c}

PHXII15:COMMUNICATION SYSTEMS

356794 A modulating signal is a square wave, as shown in the figure.
supporting img

If the carrier wave is given as $C (t) = 2 \sin (8 π t) volts,$ the modulation index is

1

\frac{1}{3}

2 1

3

\frac{1}{2}

4

\frac{1}{4}

Explanation:

From the given diagram
maximum of $m (t) = A_{m} = 1 V$
Also, maximum of $c (t) = A_{c} = 2 V$
$∴$ Modulation index $μ = \frac{A_{m}}{A_{c}} = \frac{1}{2}$
So option (3) is correct.

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356790 Depth of modulation in terms of $E_{max}$ and $E_{min}$ is

1

m_{a} = \frac{E_{max} + E_{min}}{E_{min}}

2

m_{a} = \frac{E_{max} - E_{min}}{E_{max}}

3

m_{a} = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}

4

m_{a} = \frac{E_{max} + E_{min}}{E_{max} - E_{min}}

PHXII15:COMMUNICATION SYSTEMS

356791 An oscillator is producing FM waves of $A M$ wire is $24 mV$ and the minimum peak to peak voltage is $8 mV$ . The modulation factor is

1

20 %

2

10 %

3

50 %

4

25 %

Explanation:

Here, $V_{max} = \frac{24}{2} = 12 m V$ and
$V_{min} = \frac{8}{2} = 2 m V$
Now, $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
$= \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50 %$
$h_{R}$ are doubled then s bemomes $\sqrt{2}$ times.

PHXII15:COMMUNICATION SYSTEMS

356792 The antenna current of an AM transmitter is $8 A$ when only the carrier is sent, but it increases to $8.93 A$ when the carrier is modulated by a single wave. The percentage modulation is

1

60.1 %

2

70.1 %

3

80.1 %

4

50.1 %

Explanation:

$I_{f} = I_{C} \sqrt{1 + \frac{m^{2}}{2}}$
$8.93 = 8 \sqrt{1 + \frac{m^{2}}{2}}$
$(1.11)^{2} = \frac{2 + m^{2}}{2}$
$\Rightarrow m = 0.7$
i.e., $70 %$

PHXII15:COMMUNICATION SYSTEMS

356793 If $f_{c}$ and $f_{m}$ are the frequencies of carrier wave and signal, then the bandwidth is

1

f_{m}

2

2 f_{m}

3

f_{c}

4

2 f_{c}

PHXII15:COMMUNICATION SYSTEMS

356794 A modulating signal is a square wave, as shown in the figure.
supporting img

If the carrier wave is given as $C (t) = 2 \sin (8 π t) volts,$ the modulation index is

1

\frac{1}{3}

2 1

3

\frac{1}{2}

4

\frac{1}{4}

Explanation:

From the given diagram
maximum of $m (t) = A_{m} = 1 V$
Also, maximum of $c (t) = A_{c} = 2 V$
$∴$ Modulation index $μ = \frac{A_{m}}{A_{c}} = \frac{1}{2}$
So option (3) is correct.

JEE - 2023

PHXII15:COMMUNICATION SYSTEMS

356790 Depth of modulation in terms of $E_{max}$ and $E_{min}$ is

1

m_{a} = \frac{E_{max} + E_{min}}{E_{min}}

2

m_{a} = \frac{E_{max} - E_{min}}{E_{max}}

3

m_{a} = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}

4

m_{a} = \frac{E_{max} + E_{min}}{E_{max} - E_{min}}

PHXII15:COMMUNICATION SYSTEMS

356791 An oscillator is producing FM waves of $A M$ wire is $24 mV$ and the minimum peak to peak voltage is $8 mV$ . The modulation factor is

1

20 %

2

10 %

3

50 %

4

25 %

Explanation:

Here, $V_{max} = \frac{24}{2} = 12 m V$ and
$V_{min} = \frac{8}{2} = 2 m V$
Now, $m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$
$= \frac{12 - 4}{12 + 4} = \frac{8}{16} = \frac{1}{2} = 0.5 = 50 %$
$h_{R}$ are doubled then s bemomes $\sqrt{2}$ times.

PHXII15:COMMUNICATION SYSTEMS

356792 The antenna current of an AM transmitter is $8 A$ when only the carrier is sent, but it increases to $8.93 A$ when the carrier is modulated by a single wave. The percentage modulation is

1

60.1 %

2

70.1 %

3

80.1 %

4

50.1 %

Explanation:

$I_{f} = I_{C} \sqrt{1 + \frac{m^{2}}{2}}$
$8.93 = 8 \sqrt{1 + \frac{m^{2}}{2}}$
$(1.11)^{2} = \frac{2 + m^{2}}{2}$
$\Rightarrow m = 0.7$
i.e., $70 %$

PHXII15:COMMUNICATION SYSTEMS

356793 If $f_{c}$ and $f_{m}$ are the frequencies of carrier wave and signal, then the bandwidth is

1

f_{m}

2

2 f_{m}

3

f_{c}

4

2 f_{c}

PHXII15:COMMUNICATION SYSTEMS

356794 A modulating signal is a square wave, as shown in the figure.
supporting img

If the carrier wave is given as $C (t) = 2 \sin (8 π t) volts,$ the modulation index is

1

\frac{1}{3}

2 1

3

\frac{1}{2}

4

\frac{1}{4}

Explanation:

From the given diagram
maximum of $m (t) = A_{m} = 1 V$
Also, maximum of $c (t) = A_{c} = 2 V$
$∴$ Modulation index $μ = \frac{A_{m}}{A_{c}} = \frac{1}{2}$
So option (3) is correct.

JEE - 2023