A $100V$ carrier wave is made to vary between $160V$ and $40V$ by a modulation signal. What is the modulation index?
Answer
299.4k+ views
Hint: In order to solve this question you have to remember the definition of modulation index and the formula for finding the modulation index. Recall all the concepts related to the modulation, types of modulation, characteristics of the modulation.
Formula used:
$\mu = \dfrac{{{A_m}}}{{{A_c}}}$
Where, ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal
And, $\mu $ is the modulation index
Complete step by step solution:
Here, in the question, we have given the amplitude of the carrier signal,
${A_c} = 100V$
Also given the maximum and minimum amplitude of the modulated wave,
${A_{\max }} = 160V$ and ${A_{\min }} = 40V$
We also know that,
$ \Rightarrow {A_{\max }} = {A_c} + {A_m}$
$ \Rightarrow {A_{\min }} = {A_c} - {A_m}$
Where ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal.
On putting the value of ${A_{\max }}$ in the above equation for the maximum amplitude of the modulated wave, we get,
$ \Rightarrow 160V = {A_c} + {A_m}$
On putting the value of ${A_c}$ in the above equation,
$ \Rightarrow 160V = 100 + {A_m}$
On further solving, we get the value of the amplitude of the modulation signal,
${A_m} = 60V$
Now we have the formula for calculating the modulation index, given by,
modulation index, $\mu = \dfrac{{{A_m}}}{{{A_c}}}$
Where ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal
On putting the values of ${A_m}$ and the ${A_c}$ in the above formula, we get
$\mu = \dfrac{{60V}}{{100V}}$
On further solving we get the value of the modulation index,
$\mu = 0.6$
Thus the value of the modulation index is $0.6$.
Additional Information: There is one more formula for finding the modulation index using the maximum and minimum amplitude of the modulated wave. The formula is given by, $\mu = \dfrac{{{A_{\max }} - {A_{\min }}}}{{{A_{\max }} + {A_{\min }}}}$.
Note: The modulation index is also known as modulation depth. If the modulating index is multiplied by 100 then it is converted into the percentage of modulation. If the value of the modulating index is 1, then it is the example of a perfect modulating, and its percentage of modulation is 100%.
Formula used:
$\mu = \dfrac{{{A_m}}}{{{A_c}}}$
Where, ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal
And, $\mu $ is the modulation index
Complete step by step solution:
Here, in the question, we have given the amplitude of the carrier signal,
${A_c} = 100V$
Also given the maximum and minimum amplitude of the modulated wave,
${A_{\max }} = 160V$ and ${A_{\min }} = 40V$
We also know that,
$ \Rightarrow {A_{\max }} = {A_c} + {A_m}$
$ \Rightarrow {A_{\min }} = {A_c} - {A_m}$
Where ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal.
On putting the value of ${A_{\max }}$ in the above equation for the maximum amplitude of the modulated wave, we get,
$ \Rightarrow 160V = {A_c} + {A_m}$
On putting the value of ${A_c}$ in the above equation,
$ \Rightarrow 160V = 100 + {A_m}$
On further solving, we get the value of the amplitude of the modulation signal,
${A_m} = 60V$
Now we have the formula for calculating the modulation index, given by,
modulation index, $\mu = \dfrac{{{A_m}}}{{{A_c}}}$
Where ${A_m}$ is the amplitude of the modulation signal
${A_c}$ is the amplitude of the carrier signal
On putting the values of ${A_m}$ and the ${A_c}$ in the above formula, we get
$\mu = \dfrac{{60V}}{{100V}}$
On further solving we get the value of the modulation index,
$\mu = 0.6$
Thus the value of the modulation index is $0.6$.
Additional Information: There is one more formula for finding the modulation index using the maximum and minimum amplitude of the modulated wave. The formula is given by, $\mu = \dfrac{{{A_{\max }} - {A_{\min }}}}{{{A_{\max }} + {A_{\min }}}}$.
Note: The modulation index is also known as modulation depth. If the modulating index is multiplied by 100 then it is converted into the percentage of modulation. If the value of the modulating index is 1, then it is the example of a perfect modulating, and its percentage of modulation is 100%.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Effective Nuclear Charge for JEE

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

