JEE Main On 8 April 2017 Question 3
Question: A signal of frequency 20 kHz and peak voltage of 5 volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 volts. Choose the correct statement. [JEE Online 08-04-2017]
Options:
A) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz
B) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz
C) Modulation index = 0.2, side frequency bands are at 1200 kHz and 1180 kHz
D) Modulation index = 5, side frequency bands are at 21/2 kHz and 18.8 kHz
Show Answer
Answer:
Correct Answer: C
Solution:
- Modulation $ idex=m=\frac{V _{m}}{V _0} $ $ =\frac{1}{5}=0.2 $ Frequency $ =12\times 10^{3}kHz $ F = 12.00 kHz $ {F_1} $ = 1200 - 20 = 1180 kHz $ {F_2} $ = 1200 + 20 = 1220 kHz
