Class 12 Physics Notes Chapter 7 (Communication Systems) – Physics Part-II Book

Alright class, let's delve into Chapter 7: Communication Systems. While this chapter might have been rationalized from the latest NCERT syllabus for board exams, its concepts remain fundamental and are often included in the syllabi for various government examinations. So, pay close attention.

Chapter 7: Communication Systems - Detailed Notes

1. Introduction
Communication is the act of transmitting information from one point (source) to another point (destination). A Communication System is the setup used to achieve this transmission. Examples include radio, television, telephony, and the internet.

2. Elements of a Communication System
Every communication system essentially has three main parts:

• Transmitter: Processes the original information signal to make it suitable for transmission through a specific channel.
  • Input: Information signal (e.g., audio from a microphone, video from a camera, data from a computer).
  • Processes:
    • Transducer: Converts the original signal (sound, light, etc.) into an electrical signal.
    • Modulator: Superimposes the electrical message signal onto a high-frequency carrier wave.
    • Amplifier: Boosts the power of the modulated signal for long-distance transmission.
    • Antenna (Transmitting): Radiates the electrical signal as electromagnetic waves.
• Channel: The physical medium that connects the transmitter and the receiver.
  • Examples: Free space (for radio waves), coaxial cables, optical fibers, waveguides.
  • Challenges:
    • Attenuation: Loss of signal strength as it propagates through the medium.
    • Noise: Unwanted signals (atmospheric, man-made, internal electronic noise) that get added to the signal, degrading its quality.
• Receiver: Extracts the desired information signal from the received (often weak and noisy) signal.
  • Processes:
    • Antenna (Receiving): Intercepts the transmitted electromagnetic waves and converts them back to electrical signals.
    • Amplifier: Boosts the weak received signal.
    • Demodulator (Detector): Recovers the original message signal from the carrier wave.
    • Output Transducer: Converts the electrical message signal back into its original form (e.g., sound from a loudspeaker, image on a screen).

Block Diagram:
[Information Source] -> [Transmitter] -> [Channel (Noise added)] -> [Receiver] -> [User of Information]

3. Basic Terminology

• Signal: Information converted into an electrical form suitable for transmission. Can be analog (continuous variation) or digital (discrete steps).
• Transducer: A device that converts one form of energy into another (e.g., microphone, loudspeaker).
• Noise: Unwanted disturbances that interfere with the signal during transmission or reception.
• Attenuation: The loss of strength of a signal while propagating through a medium. Measured in decibels (dB).
• Amplification: The process of increasing the amplitude (and hence strength or power) of a signal using an electronic circuit (amplifier).
• Range: The maximum distance between the source and destination up to which the signal can be received with sufficient strength.
• Bandwidth: The range of frequencies over which a communication system, channel, or signal operates.
  • Signal Bandwidth: Frequency range occupied by the message signal (e.g., Speech: ~300 Hz to 3100 Hz, Bandwidth ≈ 2800 Hz; Music: Higher bandwidth; TV Video: ~4.2 MHz).
  • Channel Bandwidth: Frequency range a transmission medium can pass without significant attenuation. The channel bandwidth must be greater than or equal to the signal bandwidth.
• Modulation: The process of superimposing a low-frequency message signal onto a high-frequency carrier wave.
• Demodulation (Detection): The process of retrieving the original message signal from the modulated carrier wave at the receiver end.
• Repeater: A combination of a receiver and a transmitter used to amplify and retransmit the signal, extending the range of communication (compensates for attenuation). Often used in microwave links and optical fiber communication.

4. Bandwidth of Signals and Transmission Media

• Speech: ~2800 Hz (typically allocated 4 kHz in telephony)

• Music: ~20 kHz

• Video (TV): ~4.2 MHz (allocated 6 MHz including audio for broadcast TV)

• Digital Data: Depends on the data rate.

• Transmission Media Bandwidth (Approximate):

  • Coaxial Cable: ~750 MHz
  • Free Space (Radio Waves): Varies widely depending on the frequency band used.
  • Optical Fiber: > 100 GHz (very large bandwidth)

5. Propagation of Electromagnetic Waves
How radio waves travel from the transmitting antenna to the receiving antenna:

• Ground Wave Propagation:
  • Waves travel along the surface of the Earth (following curvature).
  • Effective for low frequencies (LF, MF range: few kHz up to ~2 MHz). Example: AM radio broadcast (medium wave).
  • Attenuation increases rapidly with frequency and distance due to absorption by the ground.
• Sky Wave Propagation:
  • Uses reflection of radio waves (in the SW range: ~3 MHz to 30 MHz) from the ionosphere (a layer of ionized gas in the upper atmosphere).
  • Allows long-distance communication beyond the horizon. Example: Shortwave radio broadcasts.
  • Depends on ionospheric conditions (which vary with time of day, season, solar activity).
  • Critical Frequency (fc): The maximum frequency of a wave sent vertically upwards that gets reflected back by the ionosphere.
  • Skip Distance: The shortest distance from the transmitter at which a sky wave of a fixed frequency is received after reflection.
• Space Wave Propagation:
  • Waves travel in a straight line from transmitter to receiver (Line-of-Sight, LOS).
  • Used for high frequencies (VHF, UHF, Microwaves: > 40 MHz).
  • Examples: FM radio, TV broadcast, microwave links, satellite communication.
  • Range is limited by the curvature of the Earth.
  • The maximum LOS distance 'd_m' between two antennas of heights h_T (transmitter) and h_R (receiver) above the Earth's surface (Radius R) is given by:
    d_m = sqrt(2 * R * h_T) + sqrt(2 * R * h_R)
  • For TV broadcast (transmitter to receiver on ground, h_R ≈ 0), the range is d = sqrt(2 * R * h_T).

6. Modulation: Need and Types

• Need for Modulation:
  1. Practical Antenna Size: Antenna length should ideally be comparable to the wavelength (λ) of the signal (e.g., λ/4). Audio frequencies (kHz range) have very large wavelengths (km range), requiring impractically large antennas. Modulating onto a high-frequency carrier (MHz range) reduces λ (cm or m range), allowing reasonable antenna sizes.
  2. Effective Power Radiation: Power radiated by an antenna is proportional to (l/λ)². For effective radiation, l/λ needs to be significant, which is achieved with smaller λ (higher frequency).
  3. Avoiding Mixing of Signals: If multiple stations transmit baseband signals (original low-frequency signals) simultaneously without modulation, they will mix up and cannot be distinguished. Modulation allows different stations to use different carrier frequencies, enabling frequency allocation and separation.
• Types of Modulation:
  • Continuous Wave (CW) Modulation: Carrier wave is sinusoidal.
    • Amplitude Modulation (AM): Amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the message signal. Frequency and phase remain constant.
    • Frequency Modulation (FM): Frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the message signal. Amplitude and phase remain constant.
    • Phase Modulation (PM): Phase of the carrier wave is varied in accordance with the instantaneous amplitude of the message signal. Amplitude and frequency remain constant.
  • Pulse Modulation: Carrier wave is a series of pulses. (PAM, PWM, PPM, PCM - Basis for digital communication).

7. Amplitude Modulation (AM)

• Let message signal be m(t) = A_m * sin(ω_m * t)
• Let carrier wave be c(t) = A_c * sin(ω_c * t) (where ω_c >> ω_m)
• The AM wave is given by: c_m(t) = (A_c + A_m * sin(ω_m * t)) * sin(ω_c * t)
c_m(t) = A_c * (1 + (A_m / A_c) * sin(ω_m * t)) * sin(ω_c * t)
• Modulation Index (μ or m_a): Ratio of the amplitude of the message signal to the amplitude of the carrier wave.
  μ = A_m / A_c
  • For avoiding distortion, μ ≤ 1. If μ > 1, overmodulation occurs.
• Frequency Spectrum: An AM wave contains three frequencies:
  • Carrier frequency: f_c (ω_c / 2π)
  • Upper Sideband (USB): f_c + f_m ( (ω_c + ω_m) / 2π )
  • Lower Sideband (LSB): f_c - f_m ( (ω_c - ω_m) / 2π )
• Bandwidth: The difference between the highest and lowest frequencies in the AM signal.
  Bandwidth (AM) = (f_c + f_m) - (f_c - f_m) = 2 * f_m
  (Twice the maximum frequency of the message signal).
• Production of AM: Using a non-linear device (like a diode or transistor) followed by a bandpass filter tuned to f_c.
• Detection of AM: Using a rectifier (diode) followed by an envelope detector (RC low-pass filter) to recover the message signal envelope.

8. Frequency Modulation (FM)

• Frequency of the carrier wave varies according to the message signal amplitude.
• Amplitude of the modulated wave remains constant.
• Advantages over AM: Better noise immunity (less affected by amplitude variations caused by noise), higher fidelity.
• Disadvantages: Requires much larger bandwidth (compared to AM for the same message signal), more complex transmitter and receiver circuits.

Key Formulas Summary:

• Modulation Index (AM): μ = A_m / A_c (Should be ≤ 1)
• Bandwidth (AM): BW = 2 * f_m (where f_m is the highest frequency in the message signal)
• Line-of-Sight (LOS) Range: d_m = sqrt(2 * R * h_T) + sqrt(2 * R * h_R)
• Range of TV Tower: d = sqrt(2 * R * h_T) (R = Radius of Earth ≈ 6400 km)

Multiple Choice Questions (MCQs)

1. The process of superimposing a low-frequency message signal onto a high-frequency carrier wave is called:
   a) Attenuation
   b) Demodulation
   c) Modulation
   d) Amplification

2. Which part of the communication system is responsible for converting the original information (like sound) into an electrical signal?
   a) Amplifier
   b) Transducer
   c) Modulator
   d) Channel

3. Sky wave propagation is used for communication in which frequency range?
   a) VHF and UHF (> 40 MHz)
   b) LF and MF (< 2 MHz)
   c) SW (~3 MHz to 30 MHz)
   d) Microwaves (> 1 GHz)

4. A TV transmitting antenna has a height of 125 m. What is the maximum distance up to which the TV transmission can be received? (Radius of Earth = 6.4 x 10^6 m)
   a) 40 km
   b) 20 km
   c) 80 km
   d) 45 km

5. In Amplitude Modulation (AM), the bandwidth required is:
   a) Equal to the signal frequency (f_m)
   b) Twice the signal frequency (2 * f_m)
   c) Equal to the carrier frequency (f_c)
   d) Half the signal frequency (f_m / 2)

6. For effective transmission using an antenna, the antenna size should be comparable to the:
   a) Frequency of the signal
   b) Amplitude of the signal
   c) Wavelength of the signal
   d) Power of the signal

7. Which type of communication is primarily limited by the line-of-sight (LOS) distance?
   a) Ground wave propagation
   b) Sky wave propagation
   c) Space wave propagation
   d) AM radio broadcast

8. The modulation index (μ) in AM should ideally be:
   a) μ > 1
   b) μ = 0
   c) μ ≤ 1
   d) μ >> 1

9. Which modulation technique offers better immunity to noise?
   a) Amplitude Modulation (AM)
   b) Frequency Modulation (FM)
   c) Phase Modulation (PM)
   d) Pulse Amplitude Modulation (PAM)

10. Unwanted signals that tend to disturb the reception of desired signals are called:
    a) Modulation
    b) Carrier waves
    c) Noise
    d) Attenuation

Answers to MCQs:

1. (c) Modulation
2. (b) Transducer
3. (c) SW (~3 MHz to 30 MHz)
4. (d) 45 km [Calculation: d = sqrt(2 * R * h_T) = sqrt(2 * 6.4 * 10^6 m * 125 m) = sqrt(1600 * 10^6) m = 40 * 10^3 m = 40 km. Wait, let me recheck the calculation. sqrt(2 * 6400 * 10^3 * 125) = sqrt(160 * 10^7) = sqrt(1600 * 10^6) = 40 * 10^3 m = 40 km. Let me check the options again. Ah, perhaps there's a slight variation in R used or approximation. Let's recalculate carefully: d = sqrt(2 * 6.4e6 * 125) = sqrt(1600e6) = 40e3 m = 40 km. Option (a) is 40 km. Option (d) is 45 km. Let me re-read the standard calculation. Sometimes R is taken as 6371 km. Let's use R=6400 km = 6.4 x 10^6 m. d = sqrt(2 * 6.4 * 10^6 * 125) = sqrt(1600 * 10^6) = 40 * 10^3 m = 40 km. The closest option is (a) 40 km. Let me assume option (d) 45 km is a typo or derived using a slightly different R or formula interpretation in some contexts. Standard result is 40 km. Let's stick to the calculated value. Correct Answer should be (a) 40 km based on standard calculation. Self-correction: I will mark (a) as the correct answer based on direct calculation.
5. (b) Twice the signal frequency (2 * f_m)
6. (c) Wavelength of the signal
7. (c) Space wave propagation
8. (c) μ ≤ 1
9. (b) Frequency Modulation (FM)
10. (c) Noise

Revised Answer Key:

1. c
2. b
3. c
4. a (40 km)
5. b
6. c
7. c
8. c
9. b
10. c

Make sure you understand the concepts behind each point and MCQ. Revise these notes thoroughly for your exam preparation. Good luck!