Digital Communication (4341102) - Winter 2023 Solution

Question 1(a) [3 marks]

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Discuss the various communication channels characteristics.

Answer:

Mnemonic: “BBRN” - “Better Bandwidth Requires Nice planning”

Question 1(b) [4 marks]

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Give the difference between even and odd signal.

Answer:

graph TD
    A[Signal x(t)] --> B{Test symmetry}
    B -->|x(-t) = x(t)| C[Even Signal]
    B -->|x(-t) = -x(t)| D[Odd Signal]
    C --> E[Mirror symmetry]
    D --> F[Origin symmetry]

Mnemonic: “EVEN signals are Equal when flipped, ODD signals are Opposite when flipped”

Question 1(c) [7 marks]

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Define repeater. Explain how repeater works with help of necessary circuit and waveforms.

Answer:

Repeater: A device that receives, amplifies, and retransmits a signal to extend the transmission distance without degradation.

Working Principle: Repeaters regenerate digital signals to overcome attenuation and noise accumulation in transmission lines.

Circuit Diagram:

Waveform:

• Signal reception: Detects incoming weak/distorted signals
• Amplification: Strengthens the signal power
• Regeneration: Reconstructs original digital waveform
• Retransmission: Sends restored signal to next segment

Mnemonic: “RARE” - “Receive, Amplify, Regenerate, Emit”

Question 1(c) OR [7 marks]

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Draw block diagram of digital communication system and explain in detail.

Answer:

flowchart LR
    A[Information Source] --> B[Source Encoder]
    B --> C[Channel Encoder]
    C --> D[Digital Modulator]
    D --> E[Channel]
    E --> F[Digital Demodulator]
    F --> G[Channel Decoder]
    G --> H[Source Decoder]
    H --> I[Information Sink]

Mnemonic: “Send Clear Data Messages, Carefully Decode Secure Information”

Question 2(a) [3 marks]

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Define Unit step function, Unit impulse function, Unit ramp function.

Answer:

Mnemonic: “SIR” - “Step Instantly, Impulse Rapidly, Ramp Gradually”

Question 2(b) [4 marks]

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Define Continues time and discrete time signals and explain with example.

Answer:

Diagram:

• Continuous-time: Defined for all time t ∈ R (infinite values)
• Discrete-time: Defined only at specific instants n ∈ Z (countable values)

Mnemonic: “CADD” - “Continuous Always, Discrete Dots”

Question 2(c) [7 marks]

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Explain the block diagram of ASK modulator and de-modulator with waveform.

Answer:

ASK (Amplitude Shift Keying): A digital modulation technique where binary data is represented by varying the amplitude of a carrier wave.

ASK Modulator:

flowchart LR
    A[Digital Input] --> B[Product Modulator]
    C[Carrier Generator] --> B
    B --> D[Bandpass Filter]
    D --> E[ASK Output]

ASK Demodulator:

flowchart LR
    A[ASK Input] --> B[Envelope Detector]
    B --> C[Low Pass Filter]
    C --> D[Comparator]
    D --> E[Digital Output]

Waveforms:

• Modulator: Varies carrier amplitude based on digital input
• Demodulator: Extracts envelope and compares to threshold

Mnemonic: “APE” - “Amplify when Positive, Eliminate when zero”

Question 2(a) OR [3 marks]

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Explain Singularity function.

Answer:

Singularity Function: Mathematical functions that have discontinuities or undefined values at specific points.

Relationships:

• δ(t) = d/dt[u(t)]
• u(t) = ∫δ(t)dt
• r(t) = ∫u(t)dt

Mnemonic: “SIR” - “Singularities Include Rapid changes”

Question 2(b) OR [4 marks]

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Give the difference between bit rate and baud rate.

Answer:

graph TD
    A[Transmission Rate] --> B[Bit Rate]
    A --> C[Baud Rate]
    B -->|"bits/second"| D[Information Transfer Rate]
    C -->|"symbols/second"| E[Modulation Rate]
    F[Modulation Technique] --> G[Bits per Symbol]
    G --> H["Bit Rate = Baud Rate × Bits per Symbol"]

Mnemonic: “BBSR” - “Bits for Binary Speed, Bauds for Symbol Rate”

Question 2(c) OR [7 marks]

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Explain the Principle of 8-PSK signal. Also draw constellation diagram and waveforms of its.

Answer:

8-PSK (Phase Shift Keying): A digital modulation technique where data is encoded by shifting the phase of a carrier signal to 8 different positions.

Principle:

• Each symbol represents 3 bits (log₂8 = 3)
• Phase shifts in multiples of 45° (360°÷8)
• Maintains constant amplitude

Constellation Diagram:

Waveform:

• Bandwidth efficiency: 3 bits per symbol
• Constant amplitude: Better power efficiency
• Error probability: Higher than BPSK/QPSK but lower than 16-PSK

Mnemonic: “8 Points Shifted in K-circle” (8-PSK)

Question 3(a) [3 marks]

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Explain the block diagram of FSK modulator.

Answer:

FSK (Frequency Shift Keying): A digital modulation technique where binary data is represented by varying the frequency of a carrier wave.

flowchart LR
    A[Binary Input] --> B{Switch}
    C[Oscillator f1] --> B
    D[Oscillator f2] --> B
    B --> E[Bandpass Filter]
    E --> F[FSK Output]

Mnemonic: “FISO” - “Frequency Input Selects Oscillator”

Question 3(b) [4 marks]

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Draw the ASK and FSK modulation waveform for the sequence of 1010110011.

Answer:

Explanation:

• ASK: High amplitude for bit ‘1’, low amplitude for bit ‘0’
• FSK: Higher frequency f₁ for bit ‘1’, lower frequency f₂ for bit ‘0’

Mnemonic: “ASK changes Amplitude, FSK changes Frequency”

Question 3(c) [7 marks]

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Explain PSK signal generation and detection with help of its functional diagram.

Answer:

PSK (Phase Shift Keying): A digital modulation technique where data is encoded by changing the phase of a carrier signal.

PSK Modulator:

flowchart LR
    A[Binary Input] --> B[Bipolar Converter]
    B --> C[Product Modulator]
    D[Carrier Generator] --> C
    C --> E[PSK Output]

PSK Demodulator:

flowchart LR
    A[PSK Input] --> B[Product Demodulator]
    C[Carrier Recovery] --> B
    B --> D[Low Pass Filter]
    D --> E[Decision Device]
    E --> F[Binary Output]

Waveforms:

• Generation: Binary 1 → 0° phase, Binary 0 → 180° phase
• Detection: Coherent demodulation with carrier recovery
• Advantages: Better noise immunity than ASK

Mnemonic: “PSK Phases Shift with Knowledge of carrier”

Question 3(a) OR [3 marks]

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Compare Bits PER Symbol for digital modulation techniques-ASK, FSK, PSK, QPSK, 8-PSK and 16-QAM.

Answer:

graph TD
    A[Modulation Techniques]
    A --> B[ASK/FSK/BPSK
1 bit/symbol]
    A --> C[QPSK
2 bits/symbol]
    A --> D[8-PSK
3 bits/symbol]
    A --> E[16-QAM
4 bits/symbol]

Mnemonic: “As Frequency/Phase States Quadruple, Bandwidth Efficiency Doubles”

Question 3(b) OR [4 marks]

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Draw and explain the constellation diagram of 16-QAM.

Answer:

16-QAM (Quadrature Amplitude Modulation): A modulation technique that combines amplitude and phase modulation, where each symbol represents 4 bits.

Constellation Diagram:

Explanation:

• 16 distinct states: Each point represents a unique 4-bit combination
• Carries 4 bits per symbol: log₂16 = 4
• Modulation parameters: Both amplitude and phase are varied
• Symbol mapping: Gray coding used to minimize bit errors

Mnemonic: “16 Quadrants Arranged in Matrix”

Question 3(c) OR [7 marks]

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Explain the Principle of MSK signal. Also draw constellation diagram and waveforms of its.

Answer:

MSK (Minimum Shift Keying): A continuous phase FSK modulation with a modulation index of 0.5, ensuring smooth phase transitions.

Principle:

• Special case of CPFSK (Continuous Phase FSK)
• Frequency separation exactly equals half the bit rate
• Maintains continuous phase, avoiding abrupt transitions
• Modulation index h = 0.5

Constellation Diagram:

Waveforms:

Key Features:

• Constant envelope: Better power efficiency
• Spectral efficiency: Narrower bandwidth than BFSK
• Continuous phase: Smoother transitions, reduced spectral spreading
• OQPSK relation: Can be viewed as offset QPSK with sinusoidal pulse shaping

Mnemonic: “MSK Makes Smooth K-transitions”

Question 4(a) [3 marks]

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Describe the procedure to troubleshoot the FDD multiplexing circuit.

Answer:

flowchart TD
    A[Start] --> B[Check Input Signals]
    B --> C{Signals OK?}
    C -->|Yes| D[Test Filters]
    C -->|No| E[Fix Input Source]
    D --> F{Filters OK?}
    F -->|Yes| G[Test Modulators]
    F -->|No| H[Replace/Adjust Filters]

Mnemonic: “SFMPI” - “Signal, Filter, Modulator, Power, Isolation”

Question 4(b) [4 marks]

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Compare E1 carrier with T1 carrier.

Answer:

Mnemonic: “ET-DR” - “European Thirty, Double Rate”

Question 4(c) [7 marks]

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Explain CDMA technique in detail.

Answer:

CDMA (Code Division Multiple Access): A multiple access technique where multiple users share the same frequency band simultaneously by using unique spreading codes.

flowchart LR
    A[User Data] --> B[Spreading]
    C[Unique Code] --> B
    B --> D[Transmission]
    D --> E[Despreading]
    F[Same Code] --> E
    E --> G[User Data Recovery]

Working Principle:

• Each bit is multiplied by a high-rate spreading code (chips)
• Resulting signal occupies much wider bandwidth
• Receiver uses same code to recover original data
• Other signals appear as random noise, rejected by correlation

Mnemonic: “CUPS” - “Codes Uniquely Provide Separation”

Question 4(a) OR [3 marks]

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Write a short not on classification of multiplexing techniques.

Answer:

Multiplexing Techniques: Methods to combine multiple signals for transmission over a single medium.

graph TD
    A[Multiplexing Techniques] --> B[Frequency Division]
    A --> C[Time Division]
    A --> D[Code Division]
    A --> E[Wavelength Division]
    A --> F[Space Division]

Mnemonic: “FTCWS” - “Five Techniques Create Wide Systems”

Question 4(b) OR [4 marks]

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Draw and explain block diagram of Time Division Multiplexing technique (TDM).

Answer:

Time Division Multiplexing (TDM): A technique where multiple signals share the same channel by allocating different time slots to each signal.

flowchart LR
    A1[Input 1] --> B1[Sampler 1]
    A2[Input 2] --> B2[Sampler 2]
    A3[Input 3] --> B3[Sampler 3]
    A4[Input 4] --> B4[Sampler 4]
    B1 --> C[Commutator]
    B2 --> C
    B3 --> C
    B4 --> C
    C --> D[TDM Channel]
    D --> E[Decommutator]
    E --> F1[Filter 1] --> G1[Output 1]
    E --> F2[Filter 2] --> G2[Output 2]
    E --> F3[Filter 3] --> G3[Output 3]
    E --> F4[Filter 4] --> G4[Output 4]

Mnemonic: “SCTDF” - “Sample, Combine, Transmit, Distribute, Filter”

Question 4(c) OR [7 marks]

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Explain TDMA technique in detail.

Answer:

TDMA (Time Division Multiple Access): A channel access method where multiple users share the same frequency channel by dividing it into different time slots.

flowchart TD
    A[TDMA Frame] --> B[Slot 1
User 1]
    A --> C[Slot 2
User 2]
    A --> D[Slot 3
User 3]
    A --> E[Slot 4
User 4]
    A --> F[Slot 5
User 5]
    A --> G[Slot 6
User 6]

Implementation Details:

• Each user transmits in rapid bursts within assigned slot
• Non-continuous transmission allows handsets to measure signal strengths of nearby cells
• Used in GSM (8 slots per frame), DECT, satellite systems
• Easily adapts to varying data rates by assigning multiple slots

Mnemonic: “TDMA Takes Distinct Moments for Access”

Question 5(a) [3 marks]

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Define probability and write it Significance of in communication.

Answer:

Probability: A measure of the likelihood of an event occurring, expressed as a number between 0 and 1.

Mnemonic: “PRONIS” - “PRObability Numerically Indicates Signal quality”

Question 5(b) [4 marks]

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Explain Huffman code with suitable example.

Answer:

Huffman Code: A variable-length prefix coding algorithm that assigns shorter codes to more frequent symbols.

Example: Consider symbols A, B, C, D with probabilities 0.4, 0.3, 0.2, 0.1

Huffman Coding Process:

graph TD
    A[A:0.4, B:0.3, C:0.2, D:0.1] --> B[A:0.4, B:0.3, CD:0.3]
    B --> C[A:0.4, BCD:0.6]
    C --> D[ABCD:1.0]
    D --> E["A(0) | BCD(1)"]
    E --> F["A(0) | B(10) | CD(11)"]
    F --> G["A(0) | B(10) | C(110) | D(111)"]

Average Code Length = 0.4×1 + 0.3×2 + 0.2×3 + 0.1×3 = 1.9 bits/symbol

Mnemonic: “HEMP” - “Huffman Encodes More Probable symbols with shorter codes”

Question 5(c) [7 marks]

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Explain concept and key features of Internet of Things (IoT).

Answer:

Internet of Things (IoT): A network of physical objects embedded with sensors, software, and connectivity that enables them to collect and exchange data.

graph TD
    A[IoT Ecosystem] --> B[Smart Devices]
    A --> C[Connectivity]
    A --> D[Data Analytics]
    A --> E[User Interface]
    A --> F[Security]
    B --> G[Sensors & Actuators]
    C --> H[Protocols & Standards]
    D --> I[Cloud Computing]
    E --> J[Apps & Services]
    F --> K[Authentication & Encryption]

Applications:

• Smart homes (thermostats, security systems)
• Healthcare (wearable devices, remote monitoring)
• Industrial automation (predictive maintenance)
• Smart cities (traffic management, waste management)
• Agriculture (precision farming, livestock monitoring)

Mnemonic: “CSIA” - “Connect, Sense, Interpret, Automate”

Question 5(a) OR [3 marks]

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Define Channel Capacity in terms of SNR and its importance in communication.

Answer:

Channel Capacity: Maximum rate at which information can be transmitted over a communication channel with arbitrarily small error probability.

Shannon’s Channel Capacity Formula: C = B × log₂(1 + SNR)

Where:

• C = Channel capacity in bits per second
• B = Bandwidth in Hertz
• SNR = Signal-to-Noise Ratio

Mnemonic: “CBLSN” - “Capacity equals Bandwidth times Log of Signal-to-Noise ratio”

Question 5(b) OR [4 marks]

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Explain Shanon Fano code with suitable example.

Answer:

Shannon-Fano Coding: A technique that assigns variable-length codes to symbols based on their probabilities by recursively dividing the set of symbols into two subsets with approximately equal probabilities.

Example: Consider symbols A, B, C, D with probabilities 0.4, 0.3, 0.2, 0.1

Shannon-Fano Procedure:

1. Sort symbols by probability: A(0.4), B(0.3), C(0.2), D(0.1)
2. Divide into groups with nearly equal probability:
   • Group 1: A(0.4) - assigned ‘0’
   • Group 2: B(0.3), C(0.2), D(0.1) = 0.6 - assigned ‘1’
3. Recursively divide Group 2:
   • Group 2.1: B(0.3) - assigned ‘10’
   • Group 2.2: C(0.2), D(0.1) = 0.3 - assigned ‘11’
4. Divide Group 2.2:
   • C(0.2) - assigned ‘110’
   • D(0.1) - assigned ‘111’

Average Code Length = 0.4×1 + 0.3×2 + 0.2×3 + 0.1×3 = 1.9 bits/symbol

Mnemonic: “SFDS” - “Shannon Fano Divides Symbolsets”

Question 5(c) OR [7 marks]

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Draw and explain block diagram of Digital telephone exchange.

Answer:

Digital Telephone Exchange: A system that connects telephone calls by converting analog voice signals to digital form and switching them through digital circuits.

flowchart LR
    A[Subscribers] --> B[Digital Line Units
(DLU)]
    B --> C[Line/Trunk Group
(LTG)]
    C --> D[Switching Network
(SN)]
    D --> E[Central Processor
(CP)]
    E --> D
    D --> C
    C --> B
    B --> A
    F[Operation & Maintenance
Center] --> E

Key Features:

• Time Division Switching: Connects different time slots
• Space Division Switching: Connects different physical paths
• Stored Program Control: Software-based call processing
• Common Channel Signaling: Separate signaling channel (SS7)
• Non-blocking Architecture: All calls can be connected simultaneously

Mnemonic: “DLSCO” - “Digital Lines Switch Calls Orderly”