Communication Engineering (1333201) - Summer 2025 Solution

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Question 1(a) [3 marks]

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Define AM, FM and PM.

Answer:

Mnemonic: “AFaP” - “Amplitude, Frequency and Phase” are the three parameters changed during modulation.

Question 1(b) [4 marks]

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Explain block diagram of communication system.

Answer:

graph LR
    A[Information Source] --> B[Transmitter]
    B --> C[Channel]
    C --> D[Receiver]
    D --> E[Destination]
    F[Noise Source] --> C

Components of Communication System:

• Information Source: Produces message to be communicated
• Transmitter: Converts message to signals suitable for transmission
• Channel: Medium through which signals travel
• Receiver: Extracts original message from received signal
• Destination: Person/device for whom message is intended
• Noise Source: Unwanted signals that interfere with transmitted signal

Mnemonic: “I Transmit Communication Reliably Despite Noise”

Question 1(c) [7 marks]

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Explain Amplitude modulation with waveform and derive voltage equation for modulated signal also Sketch the frequency spectrum of the DSBFC AM.

Answer:

Amplitude Modulation is the process where the amplitude of a high-frequency carrier wave varies according to the instantaneous value of the modulating signal.

Waveform and Equation:

graph TD
    subgraph Amplitude Modulation
    A[Message Signal m(t) = Am cos(ωm·t)]
    B[Carrier Signal c(t) = Ac cos(ωc·t)]
    C[AM Signal s(t) = Ac[1 + μ·cos(ωm·t)]cos(ωc·t)]
    end

Derivation of AM equation:

• Carrier signal: c(t) = Ac cos(ωc·t)
• Modulating signal: m(t) = Am cos(ωm·t)
• Modulation Index: μ = Am/Ac
• AM signal: s(t) = Ac[1 + μ·cos(ωm·t)]cos(ωc·t)
• Expanding: s(t) = Ac·cos(ωc·t) + μ·Ac/2·cos[(ωc+ωm)t] + μ·Ac/2·cos[(ωc-ωm)t]

DSBFC AM Frequency Spectrum:

    |
    |        Carrier
    |          |
    |          |
    |          |
    |    LSB   |   USB
    |     |    |    |
    |_____|____|____|_____
         fc-fm fc  fc+fm

Key Points:

• LSB (Lower Sideband): Located at fc-fm
• USB (Upper Sideband): Located at fc+fm
• Bandwidth: 2fm (twice the highest modulating frequency)

Mnemonic: “CARrying Two SideBands” - DSBFC AM carries both sidebands.

Question 1(c OR) [7 marks]

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Derive the equation for total power in AM, calculate percentage of power savings in DSBFC And SSBSC.

Answer:

Total Power in AM:

For AM signal s(t) = Ac[1 + μ·cos(ωm·t)]cos(ωc·t)

graph TD
    subgraph AM Power Distribution
    A[Carrier Power: Pc = Ac²/2]
    B[Total Sideband Power: PUSB + PLSB = Pc·μ²/2]
    C[Total Power: Pt = Pc(1 + μ²/2)]
    end

Power Calculation:

• Carrier Power: Pc = Ac²/2
• Power in each sideband: PUSB = PLSB = Pc·μ²/4
• Total Sideband Power: PUSB + PLSB = Pc·μ²/2
• Total Power: Pt = Pc + PUSB + PLSB = Pc(1 + μ²/2)

Power Savings:

For μ = 1, SSBSC saves approximately 85% power compared to DSBFC.

Mnemonic: “SSB Saves Power By Cutting Carrier”

Question 2(a) [3 marks]

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Compare AM and FM.

Answer:

Mnemonic: “AM Needs Power, FM Fights Noise”

Question 2(b) [4 marks]

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Draw and explain block diagram for envelope detector.

Answer:

graph LR
    A[AM Signal Input] --> B[Diode]
    B --> C[RC Circuit]
    C --> D[Output Signal]

Components of Envelope Detector:

• Diode: Rectifies the AM signal (allows current flow in one direction)
• RC Circuit: R and C values chosen such that:
  • RC » 1/fc (to filter carrier frequency)
  • RC « 1/fm (to follow the envelope)

Working:

1. Diode conducts during positive half-cycles of carrier
2. Capacitor charges to peak value
3. When input falls, capacitor discharges through resistor
4. Output follows envelope of AM signal

Mnemonic: “Detect, Rect, and Connect” - Detection through Rectification and RC connection.

Question 2(c) [7 marks]

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Draw block diagram of FM radio receiver and explain working of each block.

Answer:

graph LR
    A[Antenna] --> B[RF Amplifier]
    B --> C[Mixer]
    E[Local Oscillator] --> C
    C --> D[IF Amplifier]
    D --> F[Limiter]
    F --> G[FM Detector]
    G --> H[Audio Amplifier]
    H --> I[Speaker]

Working of Each Block:

• Antenna: Receives FM broadcast signals (88-108 MHz)
• RF Amplifier: Amplifies weak RF signals, provides selectivity
• Mixer & Local Oscillator: Converts RF to fixed IF (10.7 MHz) using heterodyning
• IF Amplifier: Provides most of receiver’s gain and selectivity
• Limiter: Removes amplitude variations from FM signal
• FM Detector: Converts frequency variations to audio (uses ratio detector/PLL)
• Audio Amplifier: Amplifies recovered audio signal
• Speaker: Converts electrical signals to sound

Mnemonic: “Really Mighty Instruments Limit Frequency And Make Sound”

Question 2(a OR) [3 marks]

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Define Sensitivity, Selectivity, Fidelity for radio receiver.

Answer:

Mnemonic: “SSF” - “Select Signals Faithfully”

Question 2(b OR) [4 marks]

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Explain ratio detector for FM.

Answer:

graph TD
    A[FM Input] --> B[Secondary Winding]
    B --> C[Diode D1]
    B --> D[Diode D2]
    C --> E[Capacitor C1]
    D --> F[Capacitor C2]
    E --> G[Output]
    F --> G
    E --> H[Stabilizing Capacitor C3]
    F --> H

Working of Ratio Detector:

• Uses balanced circuit with two diodes in series
• Large stabilizing capacitor keeps sum of voltages constant
• Output voltage is proportional to frequency deviation
• Inherently insensitive to amplitude variations (no limiter needed)
• Less susceptible to impulse noise than discriminator

Mnemonic: “RADS” - “Ratio And Diodes Stabilize”

Question 2(c OR) [7 marks]

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Draw block diagram of AM radio receiver and explain working of each block.

Answer:

graph LR
    A[Antenna] --> B[RF Amplifier]
    B --> C[Mixer]
    E[Local Oscillator] --> C
    C --> D[IF Amplifier]
    D --> F[Detector]
    F --> G[AGC]
    G --> B
    G --> D
    F --> H[Audio Amplifier]
    H --> I[Speaker]

Working of Each Block:

• Antenna: Intercepts AM broadcast signals (535-1605 kHz)
• RF Amplifier: Amplifies weak RF signals with good SNR
• Mixer & Local Oscillator: Converts RF to fixed IF (455 kHz)
• IF Amplifier: Provides most gain and selectivity at 455 kHz
• Detector: Extracts audio from AM signal (envelope detector)
• AGC (Automatic Gain Control): Maintains constant output level
• Audio Amplifier: Boosts detected audio to drive speaker
• Speaker: Converts electrical signals to sound waves

Mnemonic: “ARMIDAS” - “Amplify, Mix, IF, Detect, Audio, Speak”

Question 3(a) [3 marks]

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Describe the Nyquist criteria.

Answer:

Nyquist Criteria: To accurately reconstruct a signal from its samples, the sampling frequency (fs) must be at least twice the highest frequency (fmax) present in the signal.

Consequence if violated: Aliasing occurs - higher frequencies appear as lower frequencies in sampled signal.

Mnemonic: “Sample Double to Dodge Aliasing”

Question 3(b) [4 marks]

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Explain Sample and hold Circuit with Waveform.

Answer:

graph LR
    A[Analog Input] --> B[Electronic Switch]
    C[Clock] --> B
    B --> D[Capacitor]
    D --> E[Buffer]
    E --> F[Output]

Sample and Hold Circuit Operation:

• Electronic Switch: Closes briefly during sampling
• Capacitor: Stores sampled voltage
• Buffer Amplifier: Provides high input impedance and low output impedance

Waveform:

Analog Input: ~~~
Clock:        ‾|_|‾|_|‾|_|‾|_|‾
Output:       ‾‾|____|‾‾‾|____|‾‾

Applications:

• Analog-to-Digital Conversion
• Data Acquisition Systems
• Pulse Amplitude Modulation

Mnemonic: “SCAB” - “Switch, Capacitor And Buffer”

Question 3(c) [7 marks]

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Define quantization explain uniform and non-uniform quantization in details.

Answer:

Quantization: Process of mapping a large set of input values to a smaller set of discrete output values.

graph LR
    A[Continuous Amplitude] --> B[Discrete Amplitude]
    B --> C[Digital Code]

Uniform Quantization vs Non-uniform Quantization:

Quantization Error:

• Difference between original and quantized signal
• Maximum error = ±Q/2 (where Q is quantization step size)
• Appears as quantization noise in reconstructed signal

Mnemonic: “UNIQ” - “UNIform has equal steps, non-uniform Quiets noise”

Question 3(a OR) [3 marks]

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Explain aliasing error and how to overcome it.

Answer:

Aliasing Error: Distortion that occurs when a signal is sampled at a rate lower than twice its highest frequency component.

graph TD
    A[Aliasing Error]
    A --> B[Original high frequencies appear as false low frequencies]
    A --> C[Causes distortion that cannot be removed after sampling]

How to Overcome Aliasing:

• Use anti-aliasing filter (low-pass) before sampling
• Increase sampling rate above Nyquist rate (fs > 2fmax)
• Bandlimit the input signal before sampling

Mnemonic: “ALIAS” - “Avoid Low sampling by Increasing And Screening”

Question 3(b OR) [4 marks]

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Draw following signal in time domain and frequency domain: 1) Sawtooth signal 2) Pulse signal

Answer:

Sawtooth Signal:

Time Domain:

    /|  /|  /|  /|
   / | / | / | / |
  /  |/  |/  |/  |
     T   2T  3T

Frequency Domain:

    |
    |
    |\
    | \
    |  \
    |   \
    |____\____________
    0  f0 2f0 3f0 4f0

Pulse Signal:

Time Domain:

    |‾|     |‾|     |‾|
    | |     | |     | |
____|_|_____|_|_____|_|____
    T       2T      3T

Frequency Domain:

    |
    |    sinc function
    |\       /\
    | \     /  \
    |  \___/    \____
    |
    |___________________
    0   f0    2f0    3f0

Mnemonic: “STPF” - “SawTooth slopes down, Pulse has sinc Function”

Question 3(c OR) [7 marks]

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Compare PAM, PWM and PPM with waveform.

Answer:

Waveforms:

Message:    /\/\/\

PAM:        ‖  ‖   ‖ ‖  ‖   ‖
            ‖  ‖   ‖ ‖  ‖   ‖

PWM:        ‖‖‖ ‖‖  ‖ ‖‖‖ ‖‖  ‖
                    
PPM:        ‖  ‖   ‖ ‖  ‖   ‖
            |--|---||-|--|---||

Mnemonic: “APP” - “Amplitude, Pulse-width, Position”

Question 4(a) [3 marks]

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Explain Space wave propagation.

Answer:

Space Wave Propagation: Mode where radio waves travel through lower atmosphere (troposphere) directly or via ground reflection.

graph LR
    A[Transmitter] --> B[Direct Wave]
    A --> C[Ground Reflected Wave]
    B --> D[Receiver]
    C --> D

Characteristics:

• Frequency range: VHF, UHF (30 MHz - 3 GHz)
• Limited to line-of-sight distance
• Range = 4.12(√h₁ + √h₂) km (where h₁, h₂ = heights in meters)
• Affected by terrain, buildings, and atmospheric conditions

Mnemonic: “SLOT” - “Straight Line Over Terrain”

Question 4(b) [4 marks]

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Explain working of Differential PCM (DPCM) transmitter.

Answer:

graph LR
    A[Input Signal] --> B[Quantizer]
    B --> C[Encoder]
    C --> D[Output DPCM]
    C --> E[Inverse Quantizer]
    E --> F[Predictor]
    F -- Predicted value --> G{Subtractor}
    A --> G
    G -- Difference --> B

Working of DPCM Transmitter:

• Predictor: Estimates current sample based on previous samples
• Subtractor: Computes difference between actual and predicted value
• Quantizer: Converts difference signal to discrete levels
• Encoder: Converts quantized values to binary code
• Feedback Loop: Reconstructs signal as receiver would see it

Advantage: Only difference signal is transmitted, which requires fewer bits

Mnemonic: “SPEQIF” - “Subtract, Predict, Encode, Quantize In Feedback”

Question 4(c) [7 marks]

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Explain delta modulator in details also explain slop overload noise and granular noise.

Answer:

Delta Modulation (DM): Simplest form of differential PCM where the difference signal is encoded with just 1 bit.

graph LR
    A[Input Signal] --> B{Comparator}
    B --> C[1-bit Quantizer]
    C --> D[Output DM] 
    C --> E[Integrator]
    E -- Predicted value --> B

Working Principle:

• Compares input signal with integrated version of previous output
• If input > integrated value: transmit 1
• If input < integrated value: transmit 0
• Step size (δ) is fixed

Noise in Delta Modulation:

Mnemonic: “DOGS” - “Delta modulation has Overload and Granular noiseS”

Question 4(a OR) [3 marks]

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Explain Ground wave propagation.

Answer:

Ground Wave Propagation: Radio wave propagation that follows the curvature of the Earth.

graph LR
    A[Transmitter] --> B[Ground Wave]
    B --> C[Receiver]
    D[Earth Surface] --- B

Characteristics:

• Frequency range: LF, MF (30 kHz - 3 MHz)
• Propagates along Earth’s surface (vertically polarized)
• Range depends on transmitter power, ground conductivity, frequency
• Signal strength decreases with distance and frequency
• Used for AM broadcasting, marine communication

Mnemonic: “GEL” - “Ground waves follow Earth at Low frequencies”

Question 4(b OR) [4 marks]

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Explain ADM transmitter.

Answer:

Adaptive Delta Modulation (ADM): Improved version of DM where step size varies according to signal characteristics.

graph LR
    A[Input Signal] --> B{Comparator}
    B --> C[1-bit Quantizer]
    C --> D[Output ADM]
    C --> E[Step Size Controller]
    E --> F[Integrator]
    F -- Predicted value --> B

Working of ADM Transmitter:

• Basic Operation: Similar to standard DM
• Step Size Control: Analyzes recent output bits
• Adaptation Logic:
  • If consecutive bits are same: Increase step size
  • If consecutive bits alternate: Decrease step size

Advantages over DM:

• Reduces both slope overload and granular noise
• Better signal tracking
• Improved SNR

Mnemonic: “ASIC” - “Adapt Step-size, Improve Coding”

Question 4(c OR) [7 marks]

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Explain Block diagram of basic PCM-TDM system.

Answer:

PCM-TDM System: Combines Pulse Code Modulation with Time Division Multiplexing to transmit multiple digital signals over single channel.

graph LR
    subgraph "Transmitter"
    A1[Analog Input 1] --> B1[Sample & Hold]
    A2[Analog Input 2] --> B2[Sample & Hold]
    A3[Analog Input n] --> B3[Sample & Hold]
    B1 --> C[Multiplexer]
    B2 --> C
    B3 --> C
    C --> D[Quantizer]
    D --> E[Encoder]
    E --> F[Frame Formatter]
    end
    
    F --> G[Transmission Channel]
    
    subgraph "Receiver"
    G --> H[Frame Synchronizer]
    H --> I[Decoder]
    I --> J[Demultiplexer]
    J --> K1[LPF 1]
    J --> K2[LPF 2]
    J --> K3[LPF n]
    K1 --> L1[Output 1]
    K2 --> L2[Output 2]
    K3 --> L3[Output n]
    end

Working of PCM-TDM System:

• Transmitter:

  • Multiple analog signals sampled simultaneously
  • Samples time-multiplexed into single stream
  • Stream quantized and encoded into PCM format
  • Framing bits added for synchronization

• Receiver:

  • Frame sync detected for alignment
  • PCM stream decoded to recover samples
  • Demultiplexer separates individual channel samples
  • Low-pass filters reconstruct original analog signals

Mnemonic: “SAMPLE-CODE-MUX” - Sampling, Coding, and Multiplexing

Question 5(a) [3 marks]

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Define radiation pattern, Directivity and Gain for antenna.

Answer:

Relationship: Gain = Directivity × Efficiency

Mnemonic: “RDG” - “Radiation Directs with Gain”

Question 5(b) [4 marks]

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Explain Microstrip Antenna with sketch.

Answer:

Microstrip (Patch) Antenna: Low-profile antenna consisting of a metal patch on a substrate with ground plane.

graph TD
    subgraph "Microstrip Antenna Structure"
    A[Radiating Patch]
    B[Dielectric Substrate]
    C[Ground Plane]
    D[Feed Point]
    
    A --- B
    B --- C
    D --- A
    end

Key Features:

• Patch: Typically rectangular or circular (λ/2 in length)
• Substrate: Low-loss dielectric material (εr = 2.2 to 12)
• Feeding Methods: Microstrip line, coaxial probe, aperture coupling
• Radiation: Primarily from fringing fields at patch edges

Applications: Mobile devices, GPS, RFID, satellite communications

Mnemonic: “PSDG” - “Patch on Substrate with Dielectric over Ground”

Question 5(c) [7 marks]

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Explain PCM transmitter and receiver in details.

Answer:

PCM (Pulse Code Modulation) Transmitter:

graph LR
    A[Analog Input] --> B[Anti-aliasing Filter]
    B --> C[Sample & Hold]
    C --> D[Quantizer]
    D --> E[Encoder]
    E --> F[Parallel to Serial]
    F --> G[Line Coder]
    G --> H[PCM Output]

PCM Receiver:

graph LR
    A[PCM Input] --> B[Regenerative Repeater]
    B --> C[Line Decoder]
    C --> D[Serial to Parallel]
    D --> E[Decoder]
    E --> F[Reconstruction Filter]
    F --> G[Analog Output]

Working Details:

Mnemonic: “SAFE PCR” - “Sample, Amplify, Filter, Encode, Pulse Code Receiver”

Question 5(a OR) [3 marks]

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Explain dipole antenna with sketch.

Answer:

Dipole Antenna: Simplest and most widely used antenna consisting of two conducting elements.

graph LR
    A[Feed Point] --- B[λ/4 Wire]
    A --- C[λ/4 Wire]
    D[Total Length = λ/2] -.-> B
    D -.-> C

Key Characteristics:

• Length: Typically λ/2 (half-wavelength dipole)
• Radiation Pattern: Figure-8 pattern perpendicular to antenna axis
• Impedance: ~73 Ω for half-wave dipole
• Polarization: Same as the orientation of the antenna

Applications: Radio broadcasting, TV reception, amateur radio

Mnemonic: “HALF” - “Half-wavelength Antenna Leads Field”

Question 5(b OR) [4 marks]

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Explain parabolic reflector antenna With Sketch.

Answer:

Parabolic Reflector Antenna: High-gain antenna using parabolic dish to focus electromagnetic waves.

graph LR
    A[Feed Horn] --> B[Parabolic Reflector]
    B --> C[Focused Beam]
    D[Focal Point] -.-> A

Working Principle:

• Feed: Located at focal point of parabola
• Reflector: Parabolic surface reflects waves in parallel direction
• Reflection Property: All paths from focal point to reflector to parallel line are equal

Applications:

• Satellite communications
• Radio astronomy
• Radar systems
• Microwave links

Mnemonic: “PROF” - “Parabola Reflects On Focus”

Question 5(c OR) [7 marks]

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Compare PCM, DM, ADM and DPCM.

Answer:

Key Differences:

• PCM: Encodes absolute amplitude values
• DM: Encodes only 1-bit difference with fixed step
• ADM: Improves DM by adapting step size
• DPCM: Encodes multi-bit difference signal

Mnemonic: “PAID” - “PCM, ADM, Integrate in DPCM”