Frequency Modulation Vs Amplitude Modulation

1.0Introduction to Modulation

In communication systems, modulation is the process of varying a carrier signal’s property according to the message signal. Without modulation, signals cannot be transmitted efficiently over long distances.

Two major types of modulation commonly studied for JEE physics and engineering are Amplitude Modulation (AM)and Frequency Modulation (FM). Understanding their differences is crucial for solving problems in electromagnetic waves and communication systems.

2.0What is Amplitude Modulation (AM)?

Amplitude Modulation (AM) is a technique where the amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the message signal, while the frequency and phase remain constant.

Key Features of AM:

• Carrier frequency: $f_c$
• Message signal frequency: $f_m$
• Instantaneous amplitude of AM wave: $A_c+A_m\cos \left(2\pi f_{m}t\right)$

Applications: AM is widely used in AM radio broadcasting, long-distance communication, and aviation communication.

Formula for AM signal:

$s(t)=[A_c+A_m\cos (2\pi f_{m}t)]\cos (2\pi f_{c}t)$

Where:

• $A_c$ = carrier amplitude
• $A_m$ = message signal amplitude
• $f_c$ = carrier frequency
• $f_m$ = message signal frequency

3.0What is Frequency Modulation (FM)?

Frequency Modulation (FM) is a technique where the frequency of the carrier wave is varied according to the instantaneous amplitude of the message signal, while amplitude remains constant.

Key Features of FM:

• Frequency deviation is proportional to the message signal amplitude.
• More resistant to noise compared to AM.
• Requires larger bandwidth than AM.

Formula for FM signal:

$s(t)=A_c\cos \left[2\pi f_{c}t+2\pi k_f{\int }_0^{t}m(\tau )d\tau \right]$

Where:

• $A_c$ = carrier amplitude
• $f_c$ = carrier frequency
• m(t) = message signal
• $k_f$ = frequency sensitivity

4.0Comparison: Frequency Modulation vs Amplitude Modulation

This comparison table helps JEE students quickly grasp key differences for theory and numerical problems.

5.0Mathematical Representation of AM and FM

Amplitude Modulation:

Where is the modulation index.

$$\begin{gathered} s_{AM}(t)=A_c[1+m\cos (2\pi f_{m}t)]\cos (2\pi f_{c}t) \\ \text{Where}m=\frac{A_m}{A_c}\text{is the modulation index.} \end{gathered}$$

Frequency Modulation:

$s_{FM}(t)=A_c\cos \left[2\pi f_{c}t+\beta \sin (2\pi f_{m}t)\right]$

$\text{Where}\beta =\frac{\Delta f}{f_m}\text{is the modulation index (frequency deviation / message frequency).}$

Observation for JEE: Modulation index determines signal spectrum, bandwidth, and deviation in transmitted signals.

6.0Advantages and Disadvantages of AM and FM

AM Advantages:

• Simpler transmitter and receiver design.
• Suitable for long-range transmission with less power.

AM Disadvantages:

• Susceptible to noise and interference.
• Lower sound quality in audio transmission.

FM Advantages:

• High noise immunity; better sound quality.
• Frequency deviation reduces interference.

FM Disadvantages:

• Requires larger bandwidth.
• More complex circuitry for transmitter and receiver.

7.0Real-Life Examples of AM and FM

• AM Example: Radio stations like AM 590 kHz.
• FM Example: FM 101.2 MHz for high-fidelity music broadcasting.
• Other Examples:
• • AM: Long-distance aviation and maritime communication.
  • FM: Television audio signals, mobile radio communication.

8.0Applications of AM and FM in Communication

Amplitude Modulation:

• AM radio broadcasting.
• Airplane and marine communication.
• Shortwave communication.

Frequency Modulation:

• FM radio broadcasting.
• TV audio transmission.
• Two-way radio systems and telemetry.

Understanding these applications helps JEE students link theory with practical scenarios.

9.0Solved Numerical Examples

Problem 1: An AM wave has a carrier amplitude of 10 V and a modulation index of 0.5. Find the maximum and minimum amplitude of the wave.

Solution:

$$\begin{gathered} A_{\max }=A_c(1+m)=10(1+0.5)=15\text{V} \\ {A}_{\min }=A_c(1-m)=10(1-0.5)=5\text{V} \end{gathered}$$

Problem 2: A message signal of 3 kHz modulates a carrier of 100 MHz in FM with a frequency deviation of 5 kHz. Calculate the modulation index.

Solution:

$\beta =\frac{\Delta f}{f_m}=\frac{5\text{kHz}}{3\text{kHz}}\approx 1.67$

Problem 3: Compare the bandwidth required for AM and FM signals if the message frequency is 5 kHz and the FM frequency deviation is 75 kHz.

Solution:

$$\begin{gathered} \text{AM bandwidth: }2f_m=2\times 5\text{kHz}=10\text{kHz} \\ \text{FM bandwidth (Carson’s Rule): }2(\Delta f+f_m)=2(75+5)=160\text{kHz} \end{gathered}$$

This shows FM requires significantly larger bandwidth than AM.