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WBB • Class XII • Physics • Ch 10

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Communication System

Communication System represents the culminating application of electromagnetic theory and electronic engineering in the WBCHSE Class 12 Physics curriculum. The chapter systematically covers the fundamental functional architecture of modern telecommunications: from information sources, input transducers, transmitters, guided and unguided transmission channels, and noise mechanisms, to receiving detectors, amplifiers, and output transducers. It explores radio wave propagation mechanisms across the electromagnetic spectrum (ground/surface waves, ionospheric sky waves, and tropospheric space waves), the physical imperative for high-frequency carrier modulation, the mathematical derivation of Amplitude Modulation (AM) sideband spectrum and power distribution, practical AM generation and detection circuits, and cutting-edge optical fiber and satellite waveguides.

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Why This Chapter Matters

In the contemporary digital era, global civilization is interconnected through cellular wireless networks, transoceanic optical fiber cables, and geostationary orbital satellites. Understanding how acoustic waveforms and digital data are modulated onto radio-frequency and optical carriers provides the conceptual bedrock for 5G communications, radar navigation, and satellite remote sensing. In the WBCHSE Higher Secondary examination, Unit X is a high-scoring section where questions on antenna height range formulas, modulation index calculations, AM sideband power ratios, critical frequency and skip distance derivations, and optical fiber numerical aperture routinely appear.

Chapter Roadmap & Progression

1 Module 1: Elements of a Communicati...

2 Module 2: Transmission Media & Radi...

3 Module 3: Space Wave Line-of-Sight...

4 Module 4: Need for Modulation & Amp...

5 Module 5: Generation and Detection...

6 Module 6: Modern Communication Syst...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Elements of a Communication System & Fundamental Terminology

1. Essential Functional Elements of a Communication System

Communication is the process of conveying information from one point in space and time (the transmitter) to another designated point (the receiver). Every electronic communication system consists of three indispensable primary elements:

Transmitter: Converts the message signal generated by the information source into a suitable electrical form and modulates it onto a high-frequency carrier wave for transmission over the channel.
Communication Channel (Medium): The physical path connecting the transmitter and receiver. It can be guided (coaxial cable, twisted pair, optical fiber) or unguided (free space radio transmission).
Receiver: Extracts the original message signal from the incoming modulated signal received from the channel, amplifies it, and delivers it to the user.

2. Core Technical Terminology

Term	Scientific Definition	Key Expression / Typical Values
Transducer	A device that converts a physical variable (sound, pressure, light) into an electrical signal or vice-versa.	Input: Microphone, Photodiode Output: Loudspeaker, LED
Signal	Information converted into electrical form. It may be Analog (continuous time-varying voltage) or Digital (discrete binary pulses 0 & 1).	Analog: Voice, Music Digital: Binary ASCII, PCM data
Noise	Random, unwanted electrical signals that enter the transmission medium and corrupt the message.	Thermal Johnson noise: P_n = k_B T B Signal-to-Noise Ratio (SNR)
Attenuation	The gradual loss of amplitude and power of an electromagnetic wave as it propagates through a channel.	Measured in decibels (dB): Loss (dB) = 10 log₁₀(P_in / P_out)
Bandwidth (B)	The range of frequencies allocated to or occupied by a signal or transmission channel: B = f_max - f_min.	Speech: 300 - 3100 Hz (B ≈ 2.8 kHz) Audio music: 20 Hz - 20 kHz (20 kHz) TV video: 4.2 MHz (Channel = 6 MHz)
Repeater	A combination of a receiver, amplifier, and transmitter placed at intermediate locations to extend communication range.	Undersea cable optical repeaters, communication satellites

Board Note: Decibel Power Gain & Attenuation Calculation:
If input power is P₁ and output power is P₂:
• Power gain in decibels: Gain (dB) = 10 log₁₀(P₂ / P₁)
• If voltage changes from V₁ to V₂ across equal impedances: Gain (dB) = 20 log₁₀(V₂ / V₁)

Module 2: Transmission Media & Radio Wave Propagation Modes

1. Classification of Radio Wave Propagation Modes

Electromagnetic waves emitted by a transmitting antenna travel to the receiving antenna through the Earth's atmosphere via three distinct mechanisms depending upon the frequency band:

Propagation Mode	Operational Frequency Band	Physical Mechanism & Characteristics	Practical Applications
Ground (Surface) Wave	Low Frequency (LF) & Medium Frequency (MF): < 2 MHz (typically 500 kHz - 1600 kHz)	Wave glides over the curved surface of the Earth, inducing eddy currents in the ground. Attenuation increases rapidly with frequency due to ground dielectric absorption and wavefront tilt. Range is limited to a few hundred kilometers.	Standard AM broadcast band, submarine communication (VLF), maritime navigation
Sky Wave (Ionospheric Reflection)	High Frequency (HF): 3 MHz to 30 MHz (Short-wave band)	Transmitted obliquely into the upper atmosphere and reflected back to Earth by total internal refraction in the ionospheric plasma layers (D, E, F₁, F₂). Highly dependent on solar ionization and time of day.	Long-distance shortwave international radio broadcasting, amateur ham radio
Space Wave (Line-of-Sight & Satellite)	Very High Frequency (VHF), Ultra High Frequency (UHF), Microwaves: > 30 - 40 MHz to GHz	Waves have frequencies exceeding the ionosphere critical frequency and penetrate through it into space. Propagates directly in straight line-of-sight paths through troposphere or via satellite relays.	Television broadcasting, FM radio (88-108 MHz), cellular networks, radar, satellite links

2. Ionospheric Physics: Critical Frequency, MUF & Skip Distance

The ionosphere extends from roughly 65 km to 400 km above Earth's surface. Solar ultraviolet radiation and X-rays ionize atmospheric molecules, creating a plasma of free electrons and positive ions.

Refractive Index of Ionosphere: The refractive index μ for radio waves of frequency f traveling through a layer with electron number density N (electrons/m³) is given by Appleton-Hartree equation:
μ = √[ 1 - (81 N / f²) ]
As the wave ascends, electron density N increases, so μ decreases (μ < 1). Thus, the wave continually bends away from the normal until total internal refraction occurs, bending it back toward Earth.
Critical Frequency (f_c): The highest frequency of radio waves that can be reflected back by an ionospheric layer when projected vertically (angle of incidence i = 0°):
f_c = √(81 N_max) ≈ 9 √N_max (in Hz)
Maximum Usable Frequency (MUF): The highest frequency that can be reflected back to Earth for oblique incidence at an angle i:
f_MUF = f_c / cos i = f_c sec i
Skip Distance (d_skip): The minimum horizontal distance from the transmitting antenna at which sky waves of a given frequency f (> f_c) return to Earth:
d_skip = 2h √[ (f / f_c)² - 1 ]
(where h is the effective height of the reflecting ionospheric layer).
Skip Zone (Silent Zone): The annular dead region between the maximum range of the ground wave and the minimum return point of the sky wave where no signal can be received.

Module 3: Space Wave Line-of-Sight (LOS) Propagation & Antenna Coverage

1. Space Wave Geometry & Horizon Limits

Space waves travel in straight lines from the transmitting antenna to the receiving antenna through the troposphere. Because the Earth is a sphere of radius R_E ≈ 6400 km, the curvature of the Earth creates a natural geometric horizon beyond which space waves cannot travel directly without grazing the ground.

2. Maximum Line-of-Sight Transmission Range Derivation

Consider a transmitting antenna of height h_T above the Earth's surface. Applying the Pythagorean theorem to the right-angled triangle formed by the center of the Earth, the tower top, and the horizon tangent point:

  (RE + hT)² = RE² + dT²

  RE² + 2 RE hT + hT² = RE² + dT²

Since tower height h_T ≪ R_E (meters compared to 6400 km), the term h_T² is negligibly small and can be discarded:

Horizon Distance: d_T = √(2 R_E h_T)

When a receiving antenna of height h_R is also employed at the receiving station, the maximum direct line-of-sight range is the sum of both individual horizon distances:

d_max = √(2 R_E h_T) + √(2 R_E h_R)

3. Broadcast Population Coverage Area

For a terrestrial television or FM transmitting antenna of height h_T radiating omnidirectionally over a level plain:

Transmission radius: d = √(2 R_E h_T)
Effective circular service area: A = π d² = 2 π R_E h_T
Total population covered: Population Covered = Area (A) × Average Population Density (ρ)
Notice that coverage area is directly proportional to tower height (A ∝ h_T). Increasing tower height by 44% increases the coverage area by 44% and transmission distance by 20% (since √1.44 = 1.20).

Module 4: Need for Modulation & Amplitude Modulation (AM) Principles

1. The Physical Imperative: Why Baseband Signals Cannot Be Transmitted Directly

Audio/speech baseband signals lie in the low-frequency range (20 Hz to 20 kHz). Three inescapable physical laws prevent direct transmission through free space:

1. Practical Antenna Dimensions (Length l ≈ λ/4): An antenna must have a physical length comparable to one-quarter of the signal wavelength (λ = c / f) to radiate efficiently. For an audio frequency f = 15 kHz:
λ = (3 × 10⁸ m/s) / (15 × 10³ Hz) = 20,000 m = 20 km &implies; l = λ/4 = 5 km!
Constructing a 5-kilometer vertical antenna tower is physically impossible. In contrast, for a radio carrier wave f = 1 MHz: λ = 300 m &implies; l = λ/4 = 75 m (easily constructed).
2. Effective Power Radiated by Antenna: Electromagnetic radiation theory proves that power radiated by a linear antenna of length l is:
P ∝ (l / λ)² ∝ f²
Because radiated power is proportional to the square of frequency, low-frequency baseband signals radiate negligible power and attenuate to zero within a few meters. High carrier frequencies radiate massive power across hundreds of kilometers.
3. Channel Multiplexing & Mutual Interference: If all radio stations transmitted audio directly (20 Hz - 20 kHz), all signals would overlap in the exact same spectrum, producing incomprehensible acoustic noise. Modulation shifts different programs to distinct high-frequency carrier channels, allowing simultaneous transmission without cross-talk (Frequency Division Multiplexing - FDM).

2. Mathematical Derivation of Amplitude Modulation (AM)

Let the high-frequency carrier wave and low-frequency modulating message signal be:

  Carrier: c(t) = Ac sin(ωc t)

  Message: m(t) = Am sin(ωm t)

In amplitude modulation, the instantaneous amplitude of the modulated carrier wave varies linearly with the message signal:

  A(t) = Ac + m(t) = Ac + Am sin(ωm t) = Ac [ 1 + μ sin(ωm t) ]

where μ = A_m / A_c is defined as the Modulation Index (or Modulation Depth). The complete modulated AM waveform is:

  s(t) = Ac [ 1 + μ sin(ωm t) ] sin(ωc t)

Expanding using the trigonometric identity 2 sin A sin B = cos(A - B) - cos(A + B):

s(t) = A_c sin(ω_c t) + (μ A_c / 2) cos[(ω_c - ω_m) t] - (μ A_c / 2) cos[(ω_c + ω_m) t]

3. AM Frequency Spectrum & Power Relations

The AM wave consists of three discrete frequency components:
1. Carrier Wave: frequency f_c, amplitude A_c
2. Lower Sideband (LSB): frequency (f_c - f_m), amplitude μ A_c / 2
3. Upper Sideband (USB): frequency (f_c + f_m), amplitude μ A_c / 2
Channel Bandwidth: The bandwidth required to transmit an AM wave is:
BW = f_USB - f_LSB = (f_c + f_m) - (f_c - f_m) = 2 f_m
(The bandwidth is exactly twice the maximum modulating frequency).
Power Distribution: If R is the antenna radiation resistance:
Carrier Power: P_c = A_c² / (2 R)
Total Sideband Power: P_SB = P_LSB + P_USB = 2 × [(μ A_c / 2)² / (2 R)] = P_c μ² / 2
Total Transmitted Power: P_t = P_c [ 1 + (μ² / 2) ]
Modulation Efficiency (η): The fraction of total power carrying intelligence (sidebands):
η = P_SB / P_t = μ² / (2 + μ²)
For 100% modulation (μ = 1), η = 1 / (2 + 1) = 1/3 ≈ 33.33%. Two-thirds (66.67%) of the transmitter power is wasted in the carrier wave which carries no information.

Module 5: Generation and Detection of Amplitude Modulated Waves

1. Generation of AM Waves: Square-Law Modulator

A simple and effective method to generate an amplitude modulated wave utilizes the non-linear transfer characteristic of a semiconductor diode or field-effect transistor:

The message signal m(t) and carrier wave c(t) are linearly combined: x(t) = A_m sin(ω_m t) + A_c cos(ω_c t).
This combined signal is applied to a square-law non-linear device whose output current is given by: y(t) = B x(t) + C x²(t).
Expanding the quadratic term x²(t) yields cross-product terms: 2 A_m A_c sin(ω_m t) cos(ω_c t) = A_m A_c [ sin(ω_c + ω_m)t - sin(ω_c - ω_m)t ].
The output contains DC, fundamental frequencies (ω_m, ω_c), second harmonics (2ω_m, 2ω_c), and the desired sidebands (ω_c ± ω_m).
A Bandpass Filter (BPF) centered at ω_c rejects the DC, low-frequency ω_m, and higher harmonics, cleanly extracting the AM wave.

2. Detection (Demodulation) of AM: Diode Envelope Detector

Demodulation is the process of retrieving the original modulating message signal m(t) from the received AM carrier wave. The most widely used circuit is the Diode Envelope Detector:

Rectification: The modulated AM signal is passed through a forward-biased p-n junction diode. The diode conducts only during the positive half-cycles of the carrier, rectifying the signal into a series of unidirectional positive RF pulses.
Envelope Filtering: A parallel low-pass filter consisting of a capacitor C and resistor R is placed across the load. During positive carrier peaks, the capacitor charges rapidly to the peak voltage. When the carrier voltage drops, the diode becomes reverse-biased, and the capacitor discharges slowly through resistor R.
Time Constant Condition: For faithful envelope detection without distortion, the RC time constant must be carefully selected:
Condition: (1 / f_c) ≪ R C ≪ (1 / f_m)
• If RC is too small (RC ≈ 1/f_c), the capacitor discharges too rapidly, leaving high-frequency RF ripple in the output.
• If RC is too large (RC > 1/f_m), the capacitor cannot discharge fast enough to follow rapid decreases in message amplitude, causing severe diagonal clipping distortion.

Module 6: Modern Communication Systems — Optical Fiber, Satellite & Cellular Networks

1. Optical Fiber Waveguide Communication

Optical fiber cables transmit information as pulses of light (photons) guided inside an ultra-pure silica glass core via Total Internal Reflection (TIR).

Structure: A central glass core of high refractive index n₁ surrounded by an outer glass cladding of slightly lower refractive index n₂ (n₁ > n₂).
Condition for Guidance: Light entering the core must strike the core-cladding interface at an angle of incidence greater than the critical angle θ_c:
sin θ_c = n₂ / n₁
Numerical Aperture (NA) & Acceptance Angle: Numerical aperture quantifies the light-gathering capability of the fiber from external air (n₀ = 1):
NA = sin θ_a = √(n₁² - n₂²)
where θ_a is the maximum acceptance half-angle for light launched into the core.
Key Technological Advantages: • Enormous optical bandwidth (> 100 THz).
• Extremely low attenuation (less than 0.2 dB/km at λ = 1550 nm).
• Absolute immunity to Electromagnetic Interference (EMI), lightning, and cross-talk.
• Ultra-lightweight and complete data security (no radiation leakage).

2. Satellite & Cellular Communication Networks

Technology	Operating Architecture	Key Technical Parameters
Geostationary Satellite (GEO)	Stationary relative to an observer on Earth; orbits in the equatorial plane from west to east with an orbital period exactly equal to Earth's rotation (24 hours). Acts as an active space repeater.	Altitude: h ≈ 35,786 km (≈ 36,000 km) Uplink frequency: 6 GHz (C-band) or 14 GHz (Ku-band) Downlink frequency: 4 GHz or 11 GHz 3 GEO satellites spaced 120° apart cover the globe.
Cellular Mobile System	Divides geographical coverage area into small hexagonal cells. Each cell is equipped with a low-power Base Transceiver Station (BTS). A Mobile Switching Centre (MSC) coordinates call handoffs as users move across cells.	Frequency Reuse Concept: Non-adjacent cells reuse identical frequency channels without co-channel interference, multiplying network capacity.

Key Formulas, Laws & Physical Constants

Line-of-Sight Space Wave Horizon Distance

$$d = \sqrt{2 R_E h_T}, \quad d_{\text{max}} = \sqrt{2 R_E h_T} + \sqrt{2 R_E h_R}, \quad A = 2 \pi R_E h_T$$

Ionospheric Critical Frequency & MUF

$$f_c = 9 \sqrt{N_{\text{max}}}, \quad f_{\text{MUF}} = \frac{f_c}{\cos i} = f_c \sec i$$

Ionospheric Skip Distance

$$d_{\text{skip}} = 2 h \sqrt{\left(\frac{f}{f_c}\right)^2 - 1}$$

Modulation Index & Channel Bandwidth

$$\mu = \frac{A_m}{A_c} = \frac{A_{\text{max}} - A_{\text{min}}}{A_{\text{max}} + A_{\text{min}}}, \quad \text{BW} = 2 f_m$$

AM Total Power & Modulation Efficiency

$$P_t = P_c \left(1 + \frac{\mu^2}{2}\right), \quad \eta = \frac{P_{\text{SB}}}{P_t} = \frac{\mu^2}{2 + \mu^2}$$

Optical Fiber Numerical Aperture & Acceptance Angle

$$\text{NA} = \sin \theta_a = \sqrt{n_1^2 - n_2^2}, \quad \sin \theta_c = \frac{n_2}{n_1}$$

Conceptual Solved Examples & Case Studies

Example 1

A transmitting antenna at the top of a tower has a height of 32 m and the height of the receiving antenna is 50 m. If the radius of the Earth is 6400 km, calculate the maximum line-of-sight distance (d_max) between them for satisfactory communication.

Step-by-Step Solution:

Step 1: Formula for maximum line-of-sight distance: d_max = √(2 R_E h_T) + √(2 R_E h_R) Step 2: Calculate transmission horizon for transmitter: h_T = 32 m, R_E = 6.4 × 10^6 m d_T = √(2 × 6.4 × 10^6 m × 32 m) = √(409.6 × 10^6) = 20,238.6 m ≈ 20.24 km Step 3: Calculate reception horizon for receiver: h_R = 50 m d_R = √(2 × 6.4 × 10^6 m × 50 m) = √(640 × 10^6) = 25,298.2 m ≈ 25.30 km Step 4: Sum individual horizons for maximum range: d_max = 20.24 km + 25.30 km = 45.54 km Final Answer: The maximum line-of-sight distance between the antennas is 45.54 km.

Example 2

(a) Explain why a baseband audio signal of frequency 10 kHz cannot be transmitted directly through space via an antenna. Calculate the required antenna length (λ/4) for this audio wave and compare it with the antenna required for a 1 MHz radio carrier wave. (b) Explain the relationship between radiated power and frequency.

Step-by-Step Solution:

(a) Physical Antenna Dimension Constraint: For an antenna to radiate electromagnetic energy effectively, its physical length l must be comparable to one-quarter wavelength (l ≈ λ/4).

For audio signal f = 10 kHz: λ_audio = c / f = (3 × 10^8 m/s) / (10 × 10^3 Hz) = 30,000 m = 30 km. Required antenna length l = λ / 4 = 30 km / 4 = 7.5 km! Building a vertical antenna of height 7.5 km is practically impossible.
For radio carrier f = 1 MHz: λ_carrier = c / f = (3 × 10^8 m/s) / (1 × 10^6 Hz) = 300 m. Required antenna length l = λ / 4 = 300 m / 4 = 75 m (easily constructible).

(b) Radiated Power and Frequency: Electromagnetic radiation theory dictates that power radiated by a linear antenna of length l is: P ∝ (l / λ)² ∝ f² Because radiated power is proportional to the square of frequency, low-frequency audio waves radiate virtually zero power and dissipate within meters. High carrier frequencies radiate efficiently across long distances.

Example 3

A sinusoidal carrier wave of amplitude 10 V and frequency 1.5 MHz is amplitude modulated by a sinusoidal audio signal of amplitude 6 V and frequency 10 kHz. Calculate: (i) the modulation index μ, (ii) the upper sideband (USB) and lower sideband (LSB) frequencies and their amplitudes, and (iii) the total bandwidth required for transmission.

Step-by-Step Solution:

Step 1: Calculate modulation index (μ): Given: Carrier amplitude A_c = 10 V, Modulating signal amplitude A_m = 6 V μ = A_m / A_c = 6 V / 10 V = 0.6 (or 60%)

Step 2: Calculate sideband frequencies: Carrier frequency f_c = 1.5 MHz = 1500 kHz, Modulating frequency f_m = 10 kHz

Upper Sideband (USB): f_USB = f_c + f_m = 1500 kHz + 10 kHz = 1510 kHz (or 1.51 MHz)
Lower Sideband (LSB): f_LSB = f_c - f_m = 1500 kHz - 10 kHz = 1490 kHz (or 1.49 MHz)

Step 3: Calculate sideband amplitudes: Amplitude of each sideband = μ A_c / 2 = (0.6 × 10 V) / 2 = 3.0 V

Step 4: Calculate channel bandwidth: Bandwidth BW = f_USB - f_LSB = 2 f_m = 2 × 10 kHz = 20 kHz

Final Answer: (i) μ = 0.6, (ii) f_USB = 1510 kHz (3.0 V) and f_LSB = 1490 kHz (3.0 V), (iii) Bandwidth = 20 kHz.

Example 4

An amplitude modulated transmitter radiates 10.4 kW of total power when modulated to a depth of 60% (μ = 0.6). Calculate: (a) the unmodulated carrier power P_c, (b) the power contained in each sideband, (c) the total power radiated if the modulation index is increased to 80% (μ = 0.8), and (d) the theoretical maximum modulation efficiency at μ = 1.0.

Step-by-Step Solution:

Step 1: Calculate unmodulated carrier power (P_c): Formula for total AM power: P_t = P_c [1 + (μ² / 2)] Given: P_t = 10.4 kW, μ = 0.6 10.4 = P_c [1 + (0.6² / 2)] = P_c [1 + (0.36 / 2)] = P_c [1 + 0.18] = 1.18 P_c P_c = 10.4 kW / 1.18 = 8.814 kW Step 2: Calculate power in sidebands: Total sideband power P_SB = P_t - P_c = 10.4 kW - 8.814 kW = 1.586 kW Since power is distributed equally between LSB and USB: Power in each sideband = P_SB / 2 = 1.586 kW / 2 = 0.793 kW = 793 W Step 3: Total power radiated when μ = 0.8: P_t' = P_c [1 + (0.8² / 2)] = 8.814 kW [1 + (0.64 / 2)] = 8.814 kW × 1.32 = 11.634 kW Step 4: Maximum theoretical modulation efficiency (μ = 1.0): η = P_SB / P_t = μ² / (2 + μ²) For μ = 1.0: η = 1² / (2 + 1²) = 1 / 3 ≈ 33.33% Final Answer: (a) P_c = 8.814 kW, (b) Power in each sideband = 793 W, (c) P_t(0.8) = 11.634 kW, (d) Maximum efficiency = 33.33%.

Example 5

(a) In an ionospheric layer, the maximum electron number density is N_max = 1.21 × 10^12 m^-3. Calculate the critical frequency f_c of this layer. (b) For radio waves incident at an angle of 60° to the vertical, calculate the Maximum Usable Frequency (f_MUF). (c) If the height of this reflecting ionospheric layer is h = 300 km, calculate the skip distance d_skip for a radio wave of frequency 15 MHz.

Step-by-Step Solution:

Step 1: Calculate Critical Frequency (f_c): Formula: f_c = √(81 N_max) ≈ 9 √(N_max) Given: N_max = 1.21 × 10^12 m^-3 f_c = 9 × √(1.21 × 10^12) = 9 × (1.1 × 10^6) = 9.9 × 10^6 Hz = 9.9 MHz Step 2: Calculate Maximum Usable Frequency (f_MUF): Formula: f_MUF = f_c / cos i = f_c sec i Given angle of incidence to the normal i = 60° (cos 60° = 0.5) f_MUF = 9.9 MHz / 0.5 = 19.8 MHz Step 3: Calculate Skip Distance (d_skip): Formula: d_skip = 2h √[(f / f_c)² - 1] Given: h = 300 km, operating frequency f = 15 MHz, f_c = 9.9 MHz (f / f_c) = 15 / 9.9 = 1.515 (f / f_c)² = (1.515)² ≈ 2.296 √[(f / f_c)² - 1] = √(2.296 - 1) = √(1.296) = 1.1384 d_skip = 2 × 300 km × 1.1384 = 600 × 1.1384 ≈ 683 km Final Answer: (a) f_c = 9.9 MHz, (b) f_MUF = 19.8 MHz, (c) d_skip = 683 km.

Example 6

(a) A television transmitting tower has a height of 160 m. Taking the radius of Earth as R_E = 6400 km and the average population density in the surrounding district as 1200 persons/km²: calculate (i) the transmission horizon distance, (ii) the total population served by the transmitter, and (iii) the percentage increase in coverage area if the tower height is increased by 44%. (b) An optical fiber has a core refractive index n_1 = 1.50 and cladding refractive index n_2 = 1.45. Calculate: (i) the critical angle at the core-cladding boundary, (ii) the numerical aperture (NA), and (iii) the maximum acceptance angle in air.

Step-by-Step Solution:

Part (a): TV Tower Population Coverage Calculations: (i) Transmission Horizon Distance: h_T = 160 m = 0.16 km, R_E = 6400 km d = √(2 R_E h_T) = √(2 × 6400 km × 0.16 km) = √(2048) ≈ 45.25 km (ii) Total Population Served: Coverage area A = π d² = 2 π R_E h_T A = 2 × 3.1416 × 6400 km × 0.16 km ≈ 6434 km² Total Population = Area × Population Density = 6434 km² × 1200 persons/km² ≈ 7,720,800 persons (approx 7.72 million people). (iii) Percentage Increase in Coverage Area: Area A = 2 π R_E h_T, which is directly proportional to tower height (A ∝ h_T). If height increases by 44%: h_T' = 1.44 h_T A' = 1.44 A Percentage Increase in Area = [(A' - A) / A] × 100% = 44%. (Note: The transmission distance increases by √1.44 - 1 = 20%). Part (b): Optical Fiber Calculations: Given: Core index n_1 = 1.50, Cladding index n_2 = 1.45 (i) Critical Angle (θ_c): sin θ_c = n_2 / n_1 = 1.45 / 1.50 = 0.9667 θ_c = arcsin(0.9667) ≈ 75.16° (ii) Numerical Aperture (NA): NA = √(n_1² - n_2²) = √[(1.50)² - (1.45)²] = √(2.25 - 2.1025) = √(0.1475) ≈ 0.384 (iii) Maximum Acceptance Angle in Air (θ_a): sin θ_a = NA = 0.384 θ_a = arcsin(0.384) ≈ 22.58° Final Answer: (a) d = 45.25 km, Population = 7.72 million, Area increase = 44%; (b) θ_c = 75.16°, NA = 0.384, θ_a = 22.58°.

Common Misconceptions & Examiner Traps

Common Misconception

Believing that an AM carrier wave carries message information.

Scientific Reality & Correction

The carrier wave contains zero information; its sole purpose is to translate the low-frequency baseband signal into the radio frequency spectrum. All intelligence is carried exclusively within the upper (USB) and lower (LSB) sidebands.

Common Misconception

Setting modulation index μ > 1 to increase signal strength.

Scientific Reality & Correction

When μ > 1 (overmodulation), the carrier envelope crosses zero and inverts phase, causing severe harmonic clipping and catastrophic distortion in standard diode envelope detectors. The legal and practical limit is μ ≤ 1.

Common Misconception

Assuming television and FM broadcast signals propagate via sky wave ionospheric reflection.

Scientific Reality & Correction

TV (54-890 MHz) and FM radio (88-108 MHz) operate in the VHF/UHF bands, which far exceed the critical frequency of the ionosphere (~30 MHz). They penetrate the ionosphere into space and can only propagate terrestrially via line-of-sight space waves.

Common Misconception

Confusing the angle of incidence in Snell's law with the launch angle in optical fibers.

Scientific Reality & Correction

The acceptance angle θ_a is measured relative to the fiber axis in external air, whereas the critical angle θ_c is measured relative to the normal at the internal core-cladding interface. Total internal reflection requires internal incidence i > θ_c.

Common Misconception

Applying d = √(2Rh) directly for both transmitter and receiver combined.

Scientific Reality & Correction

The maximum line-of-sight distance is the sum of two separate square-root terms: d_max = √(2 R_E h_T) + √(2 R_E h_R). You cannot add heights under a single radical because √(h_T + h_R) is mathematically much smaller than √h_T + √h_R.

Communication System: Architecture, Propagation, AM Spectrum & Optical Fiber

Chapter Summary & 10 Key Takeaways

Takeaway 1

A communication system fundamentally comprises an information source, transmitter, channel (with unavoidable noise), receiver, and user.

Takeaway 2

Ground waves propagate below 2 MHz by following the Earth's curved surface but suffer exponential attenuation from ground eddy currents.

Takeaway 3

Sky waves (3-30 MHz) achieve global transmission via total internal refraction within ionized atmospheric layers (D, E, F1, F2).

Takeaway 4

The critical frequency of an ionospheric layer is f_c ≈ 9√N_max; waves exceeding f_MUF = f_c sec i penetrate into outer space.

Takeaway 5

Space wave propagation operates above 30 MHz in straight line-of-sight paths limited by the Earth's curvature to d_max = √(2 R_E h_T) + √(2 R_E h_R).

Takeaway 6

Baseband audio cannot be radiated directly because required antenna length (λ/4) is kilometers long and radiated power (P ∝ f²) is negligible.

Takeaway 7

Amplitude Modulation creates two sidebands: LSB (f_c - f_m) and USB (f_c + f_m), requiring a channel bandwidth BW = 2 f_m.

Takeaway 8

Total AM power is P_t = P_c (1 + μ²/2); at 100% modulation (μ = 1), modulation efficiency is strictly 33.3%, with 66.7% wasted in carrier.

Takeaway 9

Faithful diode envelope detection requires the filter time constant to satisfy (1/f_c) << RC << (1/f_m) to prevent diagonal clipping.

Takeaway 10

Optical fibers guide light via total internal reflection with numerical aperture NA = √(n_1² - n_2²), offering immense bandwidth and zero EMI.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

A TV tower has a height of 125 m. By how much must its height be increased to double its line-of-sight transmission distance?

Reveal Answer & Explanation

Answer: Since d = √(2Rh), to make d' = 2d, we need h' = 4h = 4 × 125 m = 500 m. Therefore, height must be increased by Δh = 500 - 125 = 375 m.
Transmission distance d ∝ √h, so to double d, the height must increase by a factor of 2² = 4.

An AM wave has a maximum peak-to-peak voltage of 16 V and a minimum peak-to-peak voltage of 4 V. Calculate the modulation index μ.

Reveal Answer & Explanation

Answer: A_max = 16 V, A_min = 4 V. μ = (16 - 4) / (16 + 4) = 12 / 20 = 0.6 (or 60%).
Use the envelope formula: μ = (A_max - A_min) / (A_max + A_min).

If an AM transmitter produces a carrier power of 50 kW, what is the total power radiated at 80% modulation (μ = 0.8)?

Reveal Answer & Explanation

Answer: P_t = 50 [1 + (0.64 / 2)] = 50 × [1 + 0.32] = 50 × 1.32 = 66 kW.
Use P_t = P_c [1 + (μ² / 2)].

The maximum electron density of the ionosphere is 10^12 m^-3. Can a TV signal of frequency 60 MHz be transmitted via sky wave?

Reveal Answer & Explanation

Answer: f_c = 9 √(10^12) = 9 × 10^6 Hz = 9 MHz. Even at maximum oblique incidence (i ≈ 75°), f_MUF ≈ 9 / cos 75° ≈ 35 MHz. Since 60 MHz > f_MUF, the signal will penetrate the ionosphere and cannot be reflected via sky wave.
Calculate critical frequency f_c = 9√N_max and compare with 60 MHz.

An optical fiber has core refractive index 1.6 and cladding refractive index 1.5. Calculate the numerical aperture (NA).

Reveal Answer & Explanation

Answer: NA = √[(1.6)² - (1.5)²] = √(2.56 - 2.25) = √(0.31) ≈ 0.557.
NA = √(n_1² - n_2²).

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All Class 12 Physics Chapters

Ch 1: Electrostatics Ch 2: Current Electricity Ch 3: Magnetic Effect of Current and Magnetism Ch 4: Electromagnetic Induction and Alternating Current Ch 5: Electromagnetic Waves Ch 6: Optics Ch 7: Dual Nature of Radiation and Matter Ch 8: Atoms and Nuclei Ch 9: Electronic Devices Ch 10: Communication System

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Communication System

Why This Chapter Matters

Chapter Roadmap & Progression

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Elements of a Communication System & Fundamental Terminology

Module 2: Transmission Media & Radio Wave Propagation Modes

Module 3: Space Wave Line-of-Sight (LOS) Propagation & Antenna Coverage

Module 4: Need for Modulation & Amplitude Modulation (AM) Principles

Module 5: Generation and Detection of Amplitude Modulated Waves

Module 6: Modern Communication Systems — Optical Fiber, Satellite & Cellular Networks

Key Formulas, Laws & Physical Constants

Conceptual Solved Examples & Case Studies

Common Misconceptions & Examiner Traps

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Communication System: Architecture, Propagation, AM Spectrum & Optical Fiber

Chapter Summary & 10 Key Takeaways

Check Your Understanding (Diagnostic Practice Questions)

Timed CBT Practice Tests (Exam Simulator)

All Class 12 Physics Chapters

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