Digital Communication (4341102) - Summer 2024 Solution

Question 1(a) [3 marks]

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Define Continuous time Signal and Discrete time Signal with Wave form.

Answer:

graph LR; A[t] --> B[x(t)]; style B fill:#fff,stroke:#333,stroke-width:2px

graph LR; A[n] --> B[x[n]]; style B fill:#fff,stroke:#333,stroke-width:2px

Diagram:

Mnemonic: “Continuous Curves, Discrete Dots”

Question 1(b) [4 marks]

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Explain Energy and power signal.

Answer:

Diagram:

Mnemonic: “Energy Expires, Power Persists”

Question 1(c) [7 marks]

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Explain block diagram of digital communication system.

Answer:

graph 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]

Mnemonic: “Send Signals Carefully, Digital Messages Communicate Data Safely”

Question 1(c) OR [7 marks]

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Explain Unit Step function and Unit impulse function.

Answer:

Diagrams:

Mnemonic: “Step Stays steady after zero, Impulse Instantly appears then vanishes”

Question 2(a) [3 marks]

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A signal carries 8 bit/signal elements. If 1000 signal elements sent per second. Find the bit rate.

Answer:

Mnemonic: “Bits per signal × Signals per second = Bits per second”

Question 2(b) [4 marks]

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Explain Even and Odd signal.

Answer:

Diagram:

Mnemonic: “Even reflects Exactly, Odd reflects Oppositely”

Question 2(c) [7 marks]

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

Answer:

ASK Modulator:

graph LR
    A[Digital Input] --> B[Product Modulator]
    C[Carrier Generator] --> B
    B --> D[ASK Output]

ASK Demodulator:

graph LR
    A[ASK Signal] --> B[Envelope Detector]
    B --> C[Comparator]
    C --> D[Digital Output]

Waveforms:

Mnemonic: “ASK Adjusts Signal’s Knockout amplitude”

Question 2(a) OR [3 marks]

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A signal has a bit rate of 4000 bit/second and a baud rate of 1000 baud. How many data elements are carried by each signal element?

Answer:

Mnemonic: “Bits divided by Bauds equals Bits per signal”

Question 2(b) OR [4 marks]

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Explain Periodic and aperiodic signal.

Answer:

Diagram:

Mnemonic: “Periodic Perfectly repeats, Aperiodic Alters always”

Question 2(c) OR [7 marks]

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

Answer:

PSK Modulator:

graph LR
    A[Digital Input] --> B[Phase Shifter]
    C[Carrier Generator] --> B
    B --> D[PSK Output]

PSK Demodulator:

graph LR
    A[PSK Signal] --> B[Product Detector]
    C[Carrier Recovery] --> B
    B --> D[Low Pass Filter]
    D --> E[Decision Device]
    E --> F[Digital Output]

Waveforms:

Mnemonic: “PSK Phases Shift with Knowledge”

Question 3(a) [3 marks]

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Explain the working of FSK modulator with block diagram and output Waveform.

Answer:

FSK Modulator Block Diagram:

graph LR
    A[Digital Input] --> B[Voltage Controlled Oscillator]
    B --> C[FSK Output]

FSK Waveforms:

• Principle: Digital bit ‘1’ sends carrier with frequency f1, bit ‘0’ sends carrier with frequency f2
• Working: Voltage controlled oscillator changes frequency based on input bit value

Mnemonic: “Frequency Shifts for Knowledge transmission”

Question 3(b) [4 marks]

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Draw the PSK modulation waveform for the sequence of 1010110110.

Answer:

Table for PSK modulation:

Mnemonic: “One-Zero, Phase-Shifts, Keep-Signal Modulated”

Question 3(c) [7 marks]

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Draw the ASK and FSK modulation waveform for the sequence of 1100110101.

Answer:

Digital Input Sequence: 1100110101

Table for comparison:

Mnemonic: “Amplitude Shows Knowledge, Frequency Shifts Knowledge”

Question 3(a) OR [3 marks]

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Explain the working of MSK modulator with block diagram and output Waveform.

Answer:

MSK Modulator Block Diagram:

graph LR
    A[Digital Input] --> B[Serial to Parallel]
    B -->|I-Channel| C[I-Channel Modulator]
    B -->|Q-Channel| D[Q-Channel Modulator]
    E[Carrier Generator] --> C
    E -->|90° Phase Shift| D
    C --> F[Adder]
    D --> F
    F --> G[MSK Output]

MSK Features:

• Continuous phase FSK with frequency deviation exactly half bit rate
• Phase changes occur smoothly (no abrupt phase changes)
• Better spectral efficiency than FSK

Mnemonic: “Minimum Shift Keeps spectrum narrow”

Question 3(b) [4 marks]

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

Answer:

8-PSK Constellation:

16-QAM Constellation:

Mnemonic: “PSK Points on Single circle, QAM Quadrature Amplitude Matrix”

Question 3(c) OR [7 marks]

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Draw BPSK and QPSK modulation waveform for 1010101011.

Answer:

BPSK Modulation:

QPSK Modulation (grouping bits in pairs):

Mnemonic: “Binary Phase Shifts Keys, Quadrature Phase Shifts Keys”

Question 4(a) [3 marks]

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Encode the data using Shanon Fano code for below probability sequence. P = {0.30, 0.25, 0.20, 0.12, 0.08, 0.05}

Answer:

Process:

1. Sort symbols by decreasing probability
2. Split into two groups with nearly equal probability (0.30+0.25=0.55, 0.20+0.12+0.08+0.05=0.45)
3. Assign 0 to first group, 1 to second group
4. Recursively continue for each subgroup

Mnemonic: “Separate, Fano divides, Code efficiently”

Question 4(b) [4 marks]

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Explain Hamming code.

Answer:

Example Hamming(7,4):

Mnemonic: “Hamming Helps Handle Bit Errors”

Question 4(c) [7 marks]

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Compare TDMA and FDMA.

Answer:

Diagram:

Mnemonic: “Time Divides Multiple Access, Frequency Divides Multiple Access”

Question 4(a) OR [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:

Process:

1. Start with sorted probabilities
2. Combine lowest two probabilities (0.1+0.1=0.2)
3. Rearrange and repeat until only two nodes remain
4. Assign bits by traversing tree

Tree Construction:

Mnemonic: “Huffman Helps encode High-frequency data”

Question 4(b) OR [4 marks]

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Explain parity code.

Answer:

Examples:

Mnemonic: “Parity Provides Primitive Error detection”

Question 4(c) OR [7 marks]

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Explain FDMA Technique in detail.

Answer:

FDMA (Frequency Division Multiple Access):

graph TD
    A[Available Bandwidth] --> B[Frequency Division]
    B --> C[User 1 Channel]
    B --> D[User 2 Channel]
    B --> E[User 3 Channel]
    B --> F[User N Channel]

Frequency Allocation:

Mnemonic: “Fixed Division for Multiple Access”

Question 5(a) [3 marks]

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Explain E1 Career system.

Answer:

Diagram:

Mnemonic: “E1 Enables 30 + 2 time slots”

Question 5(b) [4 marks]

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Explain TDMA Access technique.

Answer:

TDMA Frame Structure:

Mnemonic: “Time Divides Multiple Access”

Question 5(c) [7 marks]

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Explain IoT − Concept, Features, Advantages and Disadvantages.

Answer:

IoT Concept:

graph TD
    A[Physical Objects] -->|Sensors| B[Internet Connectivity]
    B --> C[Data Collection]
    C --> D[Data Analysis]
    D --> E[Automated Actions]
    E --> A

Application Areas:

• Smart homes, cities
• Healthcare monitoring
• Industrial automation
• Agriculture
• Transportation

Mnemonic: “Internet of Things: Connected, Automated, Smarter Decisions”

Question 5(a) OR [4 marks]

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Explain T1 Career TDM system.

Answer:

Diagram:

Mnemonic: “T1 Transmits 24 channels”

Question 5(b) OR [3 marks]

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Compare TDM and FDM.

Answer:

Diagram:

Mnemonic: “Time Divides Multiplexing, Frequency Divides Multiplexing”

Question 5(c) OR [7 marks]

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Explain security components of information security.

Answer:

The CIA Triad of Information Security:

graph TD
    A[Information Security] --> B[Confidentiality]
    A --> C[Integrity]
    A --> D[Availability]
    B --> E[Encryption, Access Controls]
    C --> F[Hashing, Digital Signatures]
    D --> G[Redundancy, Fault-tolerance]

Security Threats:

• Malware (viruses, worms, trojans)
• Social engineering
• Denial of Service (DoS)
• Man-in-the-middle attacks
• Insider threats

Mnemonic: “CIA Protects All Network Data”