Principles of Electronic Communication (4331104) - Winter 2023 Solution

Question 1(a) [3 marks]

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Classify Noise signal and explain thermal noise.

Answer:

Noise signals can be classified as:

Thermal Noise:

• Definition: Random motion of electrons in a conductor due to temperature
• Characteristics: White noise with uniform power across frequency spectrum
• Formula: N = kTB (k=Boltzmann constant, T=Temperature, B=Bandwidth)

Mnemonic: “Temperature Excites Random Movements” (TERM)

Question 1(b) [4 marks]

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Comparison between Pre-emphasis and De-emphasis technique.

Answer:

Diagram/Circuit:

graph LR
    A[Input] --> B[Pre-emphasis Circuit]
    B --> C[Modulator]
    C --> D[Transmission]
    D --> E[Demodulator]
    E --> F[De-emphasis Circuit]
    F --> G[Output]
    style B fill:#f96,stroke:#333
    style F fill:#69f,stroke:#333

Mnemonic: “Pump Up Before Transmit, Pull Down After Receive” (PUBTAR)

Question 1(c) [7 marks]

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Derive mathematical expression of AM signal and with help of it explain frequency spectrum of AM signal.

Answer:

Mathematical Expression Derivation:

1. Let the carrier signal be: c(t) = Ac cos(2πfct)

2. Let the modulating signal be: m(t) = Am cos(2πfmt)

3. AM signal: s(t) = Ac[1 + μ·m(t)/Am]cos(2πfct) where μ = modulation index

4. Substituting m(t): s(t) = Ac[1 + μ·cos(2πfmt)]cos(2πfct)

5. Using trigonometric identity cos(A)·cos(B) = ½cos(A+B) + ½cos(A-B): s(t) = Ac·cos(2πfct) + (μAc/2)·cos(2π(fc+fm)t) + (μAc/2)·cos(2π(fc-fm)t)

Frequency Spectrum:

Diagram:

Mnemonic: “Carrier Standing Between Twins” (CSBT)

Question 1(c) OR [7 marks]

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Explain block diagram of Communication System.

Answer:

Block Diagram of Communication System:

graph LR
    A[Input Transducer] --> B[Transmitter]
    B --> C[Channel/Medium]
    C --> D[Receiver]
    D --> E[Output Transducer]
    F[Noise Source] --> C
    style F fill:#f66,stroke:#333

Components and Functions:

Mnemonic: “Input Transmits Through Channel, Receives Output” (ITCRO)

Question 2(a) [3 marks]

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Discuss power distribution among sidebands and carrier in amplitude modulation.

Answer:

Power Distribution in AM Signal:

Diagram:

Mnemonic: “Carrier Takes Two-Thirds” (CTTT)

Question 2(b) [4 marks]

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Why pre-emphases and de-emphases are used? Briefly describe how the signals are modified at transmitter side and receiver side.

Answer:

Purpose of Pre-emphasis and De-emphasis:

Signal Modification Process:

graph LR
    A[Audio Input] --> B[Pre-emphasis at Transmitter]
    B --> C["Boosted High Frequencies<br>(Above 2kHz)"]
    C --> D[FM Modulation]
    D --> E[Transmission]
    E --> F[FM Demodulation at Receiver]
    F --> G[De-emphasis]
    G --> H["Restored Original<br>Frequency Response"]
    style B fill:#f96,stroke:#333
    style G fill:#69f,stroke:#333

Mnemonic: “Boost High, Cut High, Keep Original” (BHCKO)

Question 2(c) [7 marks]

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Explain FM generation techniques. Explain Phase locked loop FM modulator in detail.

Answer:

FM Generation Techniques:

PLL FM Modulator:

graph LR
    A[Modulating Signal] --> B[VCO]
    B --> C[Phase Detector]
    D[Reference Oscillator] --> C
    C --> E[Loop Filter]
    E --> B
    B --> F[FM Output]
    style B fill:#f96,stroke:#333
    style C fill:#69f,stroke:#333

Working Principle:

1. Phase Detector compares VCO output with reference oscillator
2. Loop Filter removes high-frequency components
3. VCO (Voltage Controlled Oscillator) frequency changes with modulating signal
4. Modulating signal directly controls the VCO
5. PLL ensures high stability and linearity

Mnemonic: “Phase Detector Compares, Filter Smooths, VCO Varies” (PDCFV)

Question 2(a) OR [3 marks]

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State advantages and disadvantage of SSB over DSB.

Answer:

Advantages and Disadvantages of SSB over DSB:

Mnemonic: “Power and Bandwidth Saved, But Complex Circuits Needed” (PBSCN)

Question 2(b) OR [4 marks]

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Sketch the frequency spectrum of DSBSC and SSB amplitude modulated wave and pilot carrier.

Answer:

DSBSC Frequency Spectrum:

SSB (Upper Sideband) with Pilot Carrier:

Comparison Table:

Mnemonic: “Two Sides, One Side, or One Side Plus Pilot” (TSOSP)

Question 2(c) OR [7 marks]

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Write a short-note on: Pulse modulation.

Answer:

Pulse Modulation Techniques:

Pulse modulation is a process where continuous analog signal is sampled and converted into pulses.

Waveform Comparison:

Mnemonic: “Amplitude, Width, Position, Code - All Pulse Types” (AWPC)

Question 3(a) [3 marks]

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What is AGC? Draw and explain input-output characteristic curve of simple AGC circuit.

Answer:

Automatic Gain Control (AGC):

• Definition: Circuit that automatically adjusts gain to maintain constant output level
• Purpose: Compensates for varying signal strength in receivers
• Types: Simple AGC, Delayed AGC, Amplified AGC

Input-Output Characteristic Curve:

Working: As input increases, gain decreases to keep output nearly constant after threshold

Mnemonic: “Strong Signals Get Less Gain” (SSLG)

Question 3(b) [4 marks]

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Write a short-note on balanced ratio detector for FM demodulation.

Answer:

Balanced Ratio Detector:

Circuit Diagram:

Working Principle:

• Transformer creates phase-shifted signals for the diodes
• Diodes charge capacitors with different polarities
• As frequency deviates, voltage ratio changes proportionally
• Output is proportional to frequency deviation

Mnemonic: “Balanced Diodes Transform Frequency To Voltage” (BDTFV)

Question 3(c) [7 marks]

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Explain working of various types of FM demodulator circuits.

Answer:

Types of FM Demodulator Circuits:

PLL FM Demodulator Circuit:

graph LR
    A[FM Input] --> B[Phase Detector]
    C[VCO] --> B
    B --> D[Loop Filter]
    D --> C
    D --> E[Demodulated Output]
    style B fill:#f96,stroke:#333
    style C fill:#69f,stroke:#333

Working Principle:

1. Phase detector compares incoming FM with VCO output
2. Error voltage is filtered to remove high frequencies
3. VCO is forced to track input frequency
4. Filter output is proportional to frequency deviation
5. This output is the demodulated FM signal

Mnemonic: “Frequency Variations Drive Phase Errors” (FVDPE)

Question 3(a) OR [3 marks]

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Explain characteristics of a Radio receiver.

Answer:

Characteristics of a Radio Receiver:

Diagram:

graph TD
    A[Selectivity] --> B[Ideal Receiver Characteristics]
    C[Sensitivity] --> B
    D[Fidelity] --> B
    E[Image Rejection] --> B
    style B fill:#f96,stroke:#333

Mnemonic: “Select Signals Faithfully, Ignore Mirrors” (SSFIM)

Question 3(b) OR [4 marks]

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Explain types of distortions occur in AM detector circuit.

Answer:

Types of Distortions in AM Detector Circuit:

Diagram:

Mnemonic: “Diagonal Negative Harmonics Frequency - Distortion Types” (DNHF)

Question 3(c) OR [7 marks]

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Draw the block diagram of a Superheterodyne AM receiver and explain it.

Answer:

Superheterodyne AM Receiver:

graph LR
    A[Antenna] --> B[RF Amplifier]
    B --> C[Mixer]
    D[Local Oscillator] --> C
    C --> E[IF Amplifier]
    E --> F[Detector]
    F --> G[AF Amplifier]
    G --> H[Speaker]
    I[AGC] --> B
    I --> E
    F --> I
    style C fill:#f96,stroke:#333
    style E fill:#69f,stroke:#333

Function of Each Block:

Key Advantages:

• Fixed IF frequency allows optimized amplification
• Better selectivity and sensitivity
• Easier tuning

Mnemonic: “Radio Mixing Local Intermediate Detected Audio Signals” (RMLIDAS)

Question 4(a) [3 marks]

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Explain quantization process used in analog to digital conversion.

Answer:

Quantization Process:

Diagram:

Types of Quantization:

• Uniform: Equal step sizes
• Non-uniform: Varying step sizes
• Adaptive: Adjusts based on signal

Mnemonic: “Sample Levels Assign Binary” (SLAB)

Question 4(b) [4 marks]

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Give the comparison of Sampling techniques.

Answer:

Comparison of Sampling Techniques:

Diagram:

Mnemonic: “Ideal Natural Flat - Sampling Types” (INF)

Question 4(c) [7 marks]

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Draw and explain block diagram of a PCM transmitter and receiver.

Answer:

PCM Transmitter Block Diagram:

graph LR
    A[Input Signal] --> B[Low-pass Filter]
    B --> C[Sample & Hold]
    C --> D[Quantizer]
    D --> E[Encoder]
    E --> F[Multiplexer]
    F --> G[Line Coder]
    G --> H[Channel]
    style D fill:#f96,stroke:#333
    style E fill:#69f,stroke:#333

PCM Receiver Block Diagram:

graph LR
    A[Channel] --> B[Line Decoder]
    B --> C[Demultiplexer]
    C --> D[Decoder]
    D --> E[Reconstruction Filter]
    E --> F[Output Signal]
    style C fill:#f96,stroke:#333
    style D fill:#69f,stroke:#333

Working of PCM System:

Mnemonic: “Filter, Sample, Quantize, Encode, Multiplex, Transmit” (FSQEMT)

Question 4(a) OR [3 marks]

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State and explain Nyquist theorem.

Answer:

Nyquist Theorem:

• Statement: To perfectly reconstruct a bandlimited signal, the sampling frequency must be at least twice the highest frequency component in the signal.

Diagram:

Consequences:

• Undersampling: Aliasing occurs
• Critical sampling: No margin for error
• Oversampling: Better reconstruction but more data

Mnemonic: “Double Maximum Frequency Stops Aliasing” (DMFSA)

Question 4(b) OR [4 marks]

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Compare DM, ADM and DPCM.

Answer:

Comparison of DM, ADM and DPCM:

Diagram:

graph TD
    A[Analog Signal] --> B[DM: Fixed steps]
    A --> C[ADM: Variable steps]
    A --> D[DPCM: Multi-bit coding]
    style B fill:#f69,stroke:#333
    style C fill:#6f9,stroke:#333
    style D fill:#69f,stroke:#333

Mnemonic: “Single-bit, Adaptive-bit, Multi-bit Difference” (SAMD)

Question 4(c) OR [7 marks]

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Explain working of Differential PCM (DPCM) transmitter and receiver.

Answer:

DPCM Transmitter:

graph LR
    A[Input] --> B[Sampler]
    B --> C[Subtractor]
    C --> D[Quantizer]
    D --> E[Encoder]
    E --> F[Transmission Channel]
    E --> G[Decoder]
    G --> H[Predictor]
    H --> C
    style C fill:#f96,stroke:#333
    style H fill:#69f,stroke:#333

DPCM Receiver:

graph LR
    A[Received Signal] --> B[Decoder]
    B --> C[Adder]
    C --> D[Predictor]
    D --> C
    C --> E[Reconstructed Output]
    style C fill:#f96,stroke:#333
    style D fill:#69f,stroke:#333

Working Principle:

Key Advantages:

• Reduced bit rate: Encodes differences which are smaller
• Better quality: Uses signal correlation
• Compatibility: Similar to PCM framework

Mnemonic: “Predict Subtract Quantize Difference” (PSQD)

Question 5(a) [3 marks]

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Describe TDMA frame.

Answer:

TDMA (Time Division Multiple Access) Frame:

Diagram:

TDMA Frame Structure:

• Each user transmits in assigned time slot
• Full frame repeats cyclically
• Frame length depends on number of users

Mnemonic: “Slots In Time Divide Access” (SITDA)

Question 5(b) [4 marks]

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Draw and explain 4 level digital multiplexing hierarchies.

Answer:

4-Level Digital Multiplexing Hierarchy:

graph LR
    A[Level 1: Primary - 24/30 Channels] --> B[Level 2: Secondary - 96/120 Channels]
    B --> C[Level 3: Tertiary - 672/480 Channels]
    C --> D[Level 4: Quaternary - 4032/1920 Channels]
    style A fill:#f96,stroke:#333
    style B fill:#6f9,stroke:#333
    style C fill:#69f,stroke:#333
    style D fill:#96f,stroke:#333

Hierarchy Details:

Mnemonic: “Primary, Secondary, Tertiary, Quaternary Levels” (PSTQ)

Question 5(c) [7 marks]

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Draw and explain block diagram of PCM-TDM system.

Answer:

PCM-TDM System Block Diagram:

graph LR
    subgraph "Transmitter"
    A1[Input 1] --> B1[LPF]
    B1 --> C1[Sampler]
    A2[Input 2] --> B2[LPF]
    B2 --> C2[Sampler]
    A3[Input 3] --> B3[LPF]
    B3 --> C3[Sampler]
    C1 --> D[TDM Multiplexer]
    C2 --> D
    C3 --> D
    D --> E[Quantizer]
    E --> F[Encoder]
    F --> G[Line Coder]
    end
    
    G --> H[Transmission Channel]
    
    subgraph "Receiver"
    H --> I[Line Decoder]
    I --> J[Decoder]
    J --> K[TDM Demultiplexer]
    K --> L1[LPF]
    K --> L2[LPF]
    K --> L3[LPF]
    L1 --> M1[Output 1]
    L2 --> M2[Output 2]
    L3 --> M3[Output 3]
    end
    
    style D fill:#f96,stroke:#333
    style K fill:#69f,stroke:#333

Working of PCM-TDM System:

Key Features:

• Multiple analog channels share a single digital transmission link
• Each channel is sampled sequentially
• Samples are interlaced in time
• Frame synchronization ensures proper demultiplexing

Mnemonic: “Many Analog Channels Share Digital Link” (MACSDL)

Question 5(a) OR [3 marks]

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List advantages and disadvantages of digital communication.

Answer:

Advantages and Disadvantages of Digital Communication:

Mnemonic: “Noise-resistant, Error-correcting, Multiplex-friendly But Bandwidth-hungry” (NEMBB)

Question 5(b) OR [4 marks]

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List Channel Coding Techniques, explain any one of them with example.

Answer:

Channel Coding Techniques:

Block Coding Example: Hamming Code (7,4)

This code takes 4 data bits and adds 3 parity bits to create a 7-bit codeword.

Error Detection:

• If a single bit error occurs, recalculating parity bits identifies error position
• Example: 1011011 → 1111011 (Error at position 2)

Mnemonic: “Parity Bits Protect Data Bits” (PBPDB)

Question 5(c) OR [7 marks]

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Discuss basic time domain digital multiplexing. State advantages & disadvantages of TDM system.

Answer:

Basic Time Domain Digital Multiplexing:

Time Division Multiplexing (TDM) is a technique that allows multiple digital signals to share a common transmission medium by allocating unique time slots to each signal.

TDM System Diagram:

graph LR
    A1[Source 1] --> C[Multiplexer]
    A2[Source 2] --> C
    A3[Source 3] --> C
    C --> D[Transmission Medium]
    D --> E[Demultiplexer]
    E --> F1[Destination 1]
    E --> F2[Destination 2]
    E --> F3[Destination 3]
    
    style C fill:#f96,stroke:#333
    style E fill:#69f,stroke:#333

Advantages of TDM System:

Disadvantages of TDM System:

Mnemonic: “Time Slots Shared But Sync Required” (TSSBSR)