A message signal of $12 k H z$ and peak voltage $20 V$ is used to modulate a carrier wave of frequency $12 M H z$ and peak voltage $30 V$ . Then the modulation index is

0.32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.67

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

67

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.67$
Given - $E_{m} = 20 V; E_{c} = 30 V$
Therefore, modulation index ,
$m_{a} = \frac{E_{m}}{E_{c}} = \frac{20}{30} = 0.67$

Suggest Corrections

Similar questions

Q. A message signal of

12 k H z

and peak voltage

20 V

is used to modulate a carrier wave of frequency

12 M H z

and peak voltage

30 V

. Then the modulation index is

Q. A message signal of peak voltage

20 V

is used to modulate a carrier wave of peak voltage

30 V

. Determine the modulation index.

A message of frequency 5 kHz and peak voltage of 2 V is used to modulate a carrier wave of frequency 2 MHz and a peak voltage of 10 V. The modulation index is

Q. A carrier wave of peak voltage

12 V

is used to transmit a message signal. The peak voltage of the modulating signal in order to have a modulation index of

75

% is

Q. A carrier wave of peak voltage

14 V

is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of

80

% will be:

Join BYJU'S Learning Program

A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Then the modulation index is

The correct option is C 0.67Given - Em=20V;Ec=30VTherefore, modulation index , ma=EmEc=2030=0.67

A message signal of $12 k H z$ and peak voltage $20 V$ is used to modulate a carrier wave of frequency $12 M H z$ and peak voltage $30 V$ . Then the modulation index is

The correct option is C $0.67$
Given - $E_{m} = 20 V; E_{c} = 30 V$
Therefore, modulation index ,
$m_{a} = \frac{E_{m}}{E_{c}} = \frac{20}{30} = 0.67$