Digital Communication (4341102) - Summer 2023 Solution

Question 1(a) [3 marks]

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Define signal and give its classification.

Answer: A signal is a physical quantity that varies with time, space, or any other independent variable and contains information.

Classification of Signals:

Mnemonic: “CADEN” (Continuous/Discrete, Analog/Digital, Deterministic/Random, Even/Odd, Energy/Power)

Question 1(b) [4 marks]

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Explain continuous and discrete time signals.

Answer:

Diagram:

graph LR
    A[Signals] --> B[Continuous-time]
    A --> C[Discrete-time]
    B --> D[Defined for all t]
    C --> E[Defined at specific instants nT]
    D --> F["Example: sin(t)"]
    E --> G["Example: sin(nT)"]

Mnemonic: “CAD” - Continuous signals are Analog and Defined for all time; Discrete signals are digital and defined at specific points.

Question 1(c) [7 marks]

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Explain Unit Impulse and Unit Step function.

Answer:

Diagram:

Properties:

• Sampling property: ∫f(t)δ(t-t₀)dt = f(t₀)
• Unit step is integral of impulse: u(t) = ∫δ(τ)dτ from -∞ to t
• Impulse is derivative of unit step: δ(t) = du(t)/dt

Mnemonic: “SHARP-FLAT” - Impulse is Sharp and momentary; Step is Flat and persistent.

Question 1(c) OR [7 marks]

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Explain block diagram of digital communication system.

Answer:

Block Diagram of Digital Communication System:

flowchart LR
    A[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[Destination]

Explanation:

Mnemonic: “SECDCSD” - “Seven Engineers Can Design Communication Systems Diligently”

Question 2(a) [3 marks]

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A signal has a bit rate of 8000 bit/second and a baud rate of 1000 baud. How many data elements are carried by each signal element?

Answer:

Number of data elements (bits) per signal element: = Bit rate ÷ Baud rate = 8000 bits/second ÷ 1000 baud = 8 bits/signal element

Table:

Mnemonic: “Bits Divided By Bauds” (BDBB)

Question 2(b) [4 marks]

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Explain Energy and power signals.

Answer:

Diagram:

graph TD
    A[Signals] --> B[Energy Signals]
    A --> C[Power Signals]
    B --> D[Finite Energy]
    B --> E[Zero Average Power]
    C --> F[Infinite Energy]
    C --> G[Finite Average Power]
    D --> H[Example: Pulse]
    G --> I[Example: Sine Wave]

Mnemonic: “FEZIL” - Finite Energy is Zero in Long-term; Power signals are Infinite in Length

Question 2(c) [7 marks]

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

Answer:

FSK Modulator and Demodulator:

flowchart TD
    subgraph Modulator
        A[Digital Input] --> B[Voltage Controlled Oscillator]
        B --> C[FSK Output]
    end
    subgraph Demodulator
        D[FSK Input] --> E[Bandpass Filter 1\nFrequency f1]
        D --> F[Bandpass Filter 2\nFrequency f2]
        E --> G[Envelope Detector 1]
        F --> H[Envelope Detector 2]
        G --> I[Comparator]
        H --> I
        I --> J[Digital Output]
    end

Waveforms:

Key Principles:

• Bit 0: Transmitted as frequency f₁
• Bit 1: Transmitted as frequency f₂
• Demodulation: Uses bandpass filters to separate frequencies
• Detection: Envelope detectors recover the digital signal

Mnemonic: “FIST” - Frequency Is Shifted for Transmission

Question 2(a) OR [3 marks]

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A signal carries 4 bit/signal elements. If 1000 signal elements sent per second. Find the bit rate.

Answer:

Bit rate = Number of bits per signal element × Signal elements per second Bit rate = 4 bits/signal element × 1000 signal elements/second Bit rate = 4000 bits/second

Table:

Mnemonic: “BBS” - Bit rate equals Bits per symbol times Symbol rate

Question 2(b) OR [4 marks]

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Explain Even and Odd signals.

Answer:

Diagram:

Properties:

• Any signal can be expressed as sum of even and odd components
• Even component: x₁(t) = [x(t) + x(-t)]/2
• Odd component: x₂(t) = [x(t) - x(-t)]/2

Mnemonic: “SAME-FLIP” - Even signals are the SAME when flipped; Odd signals FLIP their sign.

Question 2(c) OR [7 marks]

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Explain the block diagram of QPSK modulator and de-modulator with constellation diagram.

Answer:

QPSK Modulator and Demodulator:

flowchart TD
    subgraph Modulator
        A[Binary Input] --> B[Serial to Parallel\nConverter]
        B --> C[Even Bits]
        B --> D[Odd Bits]
        C --> E[Multiplier]
        D --> F[Multiplier]
        G["cos(2πft)"] --> E
        H["sin(2πft)"] --> F
        E --> I[Summer]
        F --> I
        I --> J[QPSK Output]
    end

    subgraph Demodulator
        K[QPSK Input] --> L[Multiplier 1]
        K --> M[Multiplier 2]
        N["cos(2πft)"] --> L
        O["sin(2πft)"] --> M
        L --> P[Integrator 1]
        M --> Q[Integrator 2]
        P --> R[Decision Device 1]
        Q --> S[Decision Device 2]
        R --> T[Parallel to Serial\nConverter]
        S --> T
        T --> U[Binary Output]
    end

Constellation Diagram:

Key Characteristics:

• Input: 2 bits determine each symbol
• Phases: 4 phases (0°, 90°, 180°, 270°)
• Bits to phases:
  • 00: 45°
  • 01: 135°
  • 11: 225°
  • 10: 315°
• Bandwidth efficiency: 2 bits per symbol

Mnemonic: “QUADrature” - 4 phases for 4 possible 2-bit combinations

Question 3(a) [3 marks]

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Explain the working of ASK modulator with block diagram and output waveforms.

Answer:

ASK Modulator Block Diagram:

flowchart LR
    A[Digital Input] --> B[Multiplier]
    C["Carrier Generator\nsin(2πft)"] --> B
    B --> D[ASK Output]

Waveforms:

Working Principle:

• Digital 1: Carrier signal is transmitted
• Digital 0: No signal (or low amplitude) is transmitted
• Output amplitude varies with input digital signal

Mnemonic: “ASKY” - Amplitude Switches the Carrier? Yes!

Question 3(b) [4 marks]

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Draw the constellation diagram of 8-PSK and 16-QAM.

Answer:

8-PSK Constellation Diagram:

16-QAM Constellation Diagram:

Key Differences:

• 8-PSK: 8 symbols, equal amplitude, phases at 45° intervals
• 16-QAM: 16 symbols, varying amplitudes and phases

Mnemonic: “P-Phase Q-Quantity” - PSK varies Phase only; QAM varies both amplitude (Quantity) and phase

Question 3(c) [7 marks]

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Draw the ASK and FSK modulation waveform for the sequence of 1100101101.

Answer:

Modulation Waveforms:

Key Characteristics:

• ASK: Carrier present for bit 1, absent for bit 0
• FSK: Higher frequency (f₂) for bit 1, lower frequency (f₁) for bit 0

Table of Modulation Methods:

Mnemonic: “AFRO” - Amplitude For 1, Remove for 0 (ASK); Frequency Rises for 1, Off-peak for 0 (FSK)

Question 3(a) OR [3 marks]

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Explain the working of PSK modulator with block diagram and output waveforms.

Answer:

PSK Modulator Block Diagram:

flowchart LR
    A[Digital Input] --> B[Polar Converter\n0→-1, 1→+1]
    B --> C[Multiplier]
    D["Carrier Generator\nsin(2πft)"] --> C
    C --> E[PSK Output]

Waveforms:

Working Principle:

• Digital 1: Carrier signal with 0° phase
• Digital 0: Carrier signal with 180° phase (inverted)
• Amplitude remains constant, only phase changes

Mnemonic: “PSKIT” - Phase Shift Keeps Information True

Question 3(b) OR [4 marks]

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Draw the MSK modulation waveform for the sequence of 1101001101.

Answer:

MSK Modulation Waveform:

Characteristics of MSK:

• Continuous phase transitions (no phase jumps)
• Frequency shifts between f₁ and f₂
• Minimum frequency separation: Δf = 1/(2T)
• Smoother transitions than FSK

Table:

Mnemonic: “MINIMUM SMOOTH” - MSK uses Minimum frequency separation with Smooth transitions

Question 3(c) OR [7 marks]

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Draw BPSK and QPSK modulation waveform for 1100101011.

Answer:

BPSK and QPSK Modulation Waveforms:

Key Differences:

• BPSK: 1 bit per symbol, 2 phases (0° and 180°)
• QPSK: 2 bits per symbol, 4 phases (45°, 135°, 225°, 315°)
• QPSK Pairs: 00, 01, 10, 11 map to different phases

Table:

Mnemonic: “ONE-TWO” - ONE bit for BPSK, TWO bits for QPSK

Question 4(a) [3 marks]

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Encode the data using Huffman code for below probability sequence. P = { 0.4, 0.2, 0.2, 0.1, 0.1}

Answer:

Huffman Coding Process:

Huffman Tree:

Mnemonic: “Higher Probability Means Shorter Code”

Question 4(b) [4 marks]

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Define Probability and Entropy.

Answer:

Key Characteristics:

• Probability Range: 0 ≤ P(A) ≤ 1
• Entropy Units: Bits (using log₂)
• Maximum Entropy: When all events are equally likely
• Minimum Entropy: When outcome is certain (probability = 1)

Mnemonic: “PURE” - Probability Underpins Randomness Estimation

Question 4(c) [7 marks]

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Explain CDMA technique in detail.

Answer:

CDMA (Code Division Multiple Access):

flowchart LR
    A[User Data] --> B[Spreading\nwith Unique Code]
    B --> C[Modulation]
    C --> D[Transmission]
    D --> E[Reception]
    E --> F[Demodulation]
    F --> G[Despreading with\nMatching Code]
    G --> H[Original User Data]

Table of CDMA Characteristics:

Key Advantages:

• Capacity: Higher than FDMA/TDMA in many scenarios
• Security: Inherent encryption through spreading codes
• Multipath Rejection: Rake receivers can combine multipath components
• Soft Handoff: Mobile can communicate with multiple base stations

Mnemonic: “CODES” - Capacity Optimized with Direct-sequence Encoding Schemes

Question 4(a) OR [3 marks]

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Encode the data using Shanon Fano code for below probability sequence. P = { 0.5, 0.25, 0.125, 0.125}

Answer:

Shannon-Fano Coding Process:

Shannon-Fano Tree:

Mnemonic: “Split For Optimum” - Shannon-Fano splits groups for optimum coding

Question 4(b) OR [4 marks]

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Define Information and Channel Capacity.

Answer:

Key Points:

• Information Units: Bits (using log₂)
• Channel Capacity Units: Bits per second
• Factors Affecting Capacity:
  • Bandwidth (B)
  • Signal-to-Noise Ratio (S/N)

Mnemonic: “INCHES” - Information Numerically Calculated, Hopping through Efficient Shannon limit

Question 4(c) OR [7 marks]

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Explain TDMA technique in detail.

Answer:

TDMA (Time Division Multiple Access):

flowchart LR
    A[User 1] --> B[Time Slot 1]
    C[User 2] --> D[Time Slot 2]
    E[User 3] --> F[Time Slot 3]
    G[User 4] --> H[Time Slot 4]
    B --> I[Multiplexer]
    D --> I
    F --> I
    H --> I
    I --> J[Transmission Channel]
    J --> K[Demultiplexer]
    K --> L[Time Slot 1]
    K --> M[Time Slot 2]
    K --> N[Time Slot 3]
    K --> O[Time Slot 4]
    L --> P[User 1]
    M --> Q[User 2]
    N --> R[User 3]
    O --> S[User 4]

Table of TDMA Characteristics:

TDMA Frame Structure:

Mnemonic: “TIME” - Transmission In Measured Epochs

Question 5(a) [3 marks]

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Explain T1 carrier system.

Answer:

T1 Carrier System:

T1 Frame Structure:

Mnemonic: “T1-24-8-8” - T1 has 24 channels, 8 bits, 8kHz

Question 5(b) [4 marks]

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Explain Time Division Multiplexing technique (TDM) in detail.

Answer:

Time Division Multiplexing (TDM):

flowchart LR
    A[Signal 1] --> E[Multiplexer]
    B[Signal 2] --> E
    C[Signal 3] --> E
    D[Signal 4] --> E
    E --> F[Transmission Channel]
    F --> G[Demultiplexer]
    G --> H[Signal 1]
    G --> I[Signal 2]
    G --> J[Signal 3]
    G --> K[Signal 4]

Table of TDM Characteristics:

TDM Frame Structure:

Mnemonic: “TWIST” - Time Windows Interleaving Signals Together

Question 5(c) [7 marks]

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Explain security components of information security in detail.

Answer:

Information Security Components:

graph TD
    A[Information Security] --> B[Confidentiality]
    A --> C[Integrity]
    A --> D[Availability]
    B --> E[Encryption]
    B --> F[Access Control]
    C --> G[Digital Signatures]
    C --> H[Hashing]
    D --> I[Redundancy]
    D --> J[Backup Systems]

Table of Security Components:

Common Security Threats:

• Malware: Viruses, worms, trojans, ransomware
• Social Engineering: Phishing, pretexting
• Man-in-the-Middle Attacks: Intercepting communications
• Denial-of-Service: Preventing legitimate access

Mnemonic: “CIA” - Confidentiality, Integrity, Availability

Question 5(a) OR [3 marks]

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Explain E1 carrier system.

Answer:

E1 Carrier System:

E1 Frame Structure:

Special Time Slots:

• TS0: Frame alignment signal
• TS16: Signaling channel

Mnemonic: “E1-32-8-8” - E1 has 32 channels, 8 bits, 8kHz

Question 5(b) OR [4 marks]

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Explain Frequency Division Multiplexing technique (FDM) in detail.

Answer:

Frequency Division Multiplexing (FDM):

flowchart LR
    A[Signal 1] --> B[Modulator 1\nf1]
    C[Signal 2] --> D[Modulator 2\nf2]
    E[Signal 3] --> F[Modulator 3\nf3]
    G[Signal 4] --> H[Modulator 4\nf4]
    B --> I[Combiner/Mixer]
    D --> I
    F --> I
    H --> I
    I --> J[Transmission Channel]
    J --> K[Filters/Separators]
    K --> L[Demodulator 1\nf1]
    K --> M[Demodulator 2\nf2]
    K --> N[Demodulator 3\nf3]
    K --> O[Demodulator 4\nf4]
    L --> P[Signal 1]
    M --> Q[Signal 2]
    N --> R[Signal 3]
    O --> S[Signal 4]

Table of FDM Characteristics:

FDM Spectrum:

Mnemonic: “FROG” - FRequencies Organized with Gaps

Question 5(c) OR [7 marks]

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

Answer:

Internet of Things (IoT) Concept:

graph TD
    A[Internet of Things] --> B[Connected Devices]
    A --> C[Data Collection]
    A --> D[Data Analytics]
    A --> E[Automation]
    B --> F[Sensors]
    B --> G[Actuators]
    C --> H[Cloud Storage]
    D --> I[AI/Machine Learning]
    E --> J[Smart Applications]

Table of IoT Key Features:

IoT Architecture Layers:

IoT Applications:

• Smart homes and buildings
• Healthcare monitoring
• Industrial automation
• Smart cities
• Agriculture monitoring
• Supply chain management

sMnemonic: “CASED” - Connected, Automated, Sensing, Expressing, Data-driven