Digital Communication (4341102) - Summer 2025 Solution

Question 1(a) [3 marks]

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Draw block diagram of digital communication system.

Answer:

graph 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]
    
    J[Noise Source] --> E

Key Components:

• Information Source: Generates message signal
• Source Encoder: Converts analog to digital
• Channel Encoder: Adds error correction codes
• Digital Modulator: Converts digital bits to analog signal

Mnemonic: “Source Channel Modulator travels through Channel to Demodulator Channel Sink”

Question 1(b) [4 marks]

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Write the function of transmitter and receiver of digital communication system.

Answer:

Key Functions:

• Signal Processing: Encoding, modulation, filtering
• Error Control: Detection and correction of transmission errors
• Signal Recovery: Demodulation and decoding at receiver

Mnemonic: “Transmitter Encodes Modulates, Receiver Demodulates Decodes”

Question 1(c) [7 marks]

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Define and explain with example: Continues time and discrete time signals, Real and complex signals and even and odd signals.

Answer:

Even and Odd Signals:

graph TD
    A["Signal x[n]"] --> B{"Check x[-n]"}
    B --> C["x[n] = x[-n]<br/>EVEN SIGNAL"]
    B --> D["x[n] = -x[-n]<br/>ODD SIGNAL"]
    
    C --> E["Example: cos[n]"]
    D --> F["Example: sin[n]"]

Properties:

• Even Signal: Symmetric about y-axis, x(t) = x(-t)
• Odd Signal: Anti-symmetric about origin, x(t) = -x(-t)
• Complex Signal: z(t) = x(t) + jy(t)
• Discrete Signal: Sampled version of continuous signal

Mnemonic: “Continuous Everywhere, Discrete Specific, Real Simple, Complex Combined”

Question 1(c OR) [7 marks]

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Define and explain with example: Unit step function, Unit impulse function, Unit ramp function

Answer:

Applications:

• Unit Step: Switch operations, system response analysis
• Unit Impulse: System impulse response, convolution
• Unit Ramp: System ramp response, integration

Properties:

• Step: Derivative of ramp, integral of impulse
• Impulse: Derivative of step function
• Ramp: Integral of step function

Mnemonic: “Step Sudden, Impulse Instant, Ramp Rising”

Question 2(a) [3 marks]

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Define: bit rate, baud rate and bandwidth.

Answer:

Relationship:

• Bit Rate = Baud Rate × log₂(M)
• M = number of signal levels
• Bandwidth ∝ Baud Rate

Key Points:

• Higher bit rate: More data transmission
• Baud rate: Symbol transmission rate
• Bandwidth: Frequency spectrum occupied

Mnemonic: “Bits Baud Bandwidth - Data Symbol Frequency”

Question 2(b) [4 marks]

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Explain Energy and power signal.

Answer:

Classification:

graph TD
    A[Signal] --> B{Energy Finite?}
    B -->|Yes| C[Energy Signal<br/>P = 0]
    B -->|No| D{Power Finite?}
    D -->|Yes| E[Power Signal<br/>E = ∞]
    D -->|No| F[Neither Energy<br/>nor Power]

Examples:

• Energy Signal: Exponentially decaying signal e^(-t)u(t)
• Power Signal: Sinusoidal signal sin(ωt)
• Neither: Ramp signal t·u(t)

Properties:

• Energy and power signals are mutually exclusive
• Periodic signals are generally power signals
• Non-periodic finite duration signals are energy signals

Mnemonic: “Energy Ends, Power Persists”

Question 2(c) [7 marks]

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Give the comparison between ASK, FSK and PSK modulation techniques and draw their waveforms.

Answer:

Applications:

• ASK: Optical communication, simple radio systems
• FSK: Telephone modems, radio systems
• PSK: Satellite communication, wireless systems

Advantages:

• ASK: Simple implementation, low cost
• FSK: Good noise performance, constant envelope
• PSK: Best noise performance, bandwidth efficient

Mnemonic: “ASK Amplitude, FSK Frequency, PSK Phase”

Question 2(a OR) [3 marks]

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A bit rate of signal generator from 8-bit generator is 1600 bps. Calculate the baud rate of signal.

Answer:

Given:

• Bit rate = 1600 bps
• Number of bits per symbol = 8 bits

Formula: Baud Rate = Bit Rate / Number of bits per symbol

Calculation: Baud Rate = 1600 bps / 8 bits Baud Rate = 200 Baud

Result: The baud rate of the signal is 200 Baud.

Explanation:

• Each symbol carries 8 bits of information
• 1600 bits per second ÷ 8 bits per symbol = 200 symbols per second
• Therefore, baud rate = 200 Baud

Mnemonic: “Bit Rate divided by Bits per Symbol gives Baud”

Question 2(b OR) [4 marks]

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Find whether the signals are even or odd: 1. x(t) = e^(-5t) 2. x(t) = sin 2t 3. x(t) = cos 5t

Answer:

Test Procedure:

1. Even Signal Test: Check if x(t) = x(-t)
2. Odd Signal Test: Check if x(t) = -x(-t)

Properties Used:

• Exponential: e^(-at) is neither even nor odd (a > 0)
• Sine Function: sin(-x) = -sin(x) → Odd function
• Cosine Function: cos(-x) = cos(x) → Even function

Results:

• Signal 1: Neither even nor odd
• Signal 2: Odd signal
• Signal 3: Even signal

Mnemonic: “Cosine Even, Sine Odd, Exponential Neither”

Question 2(c OR) [7 marks]

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Explain the Principle of QPSK signal. Draw its modulator and demodulator diagram. Also draw constellation diagram and waveforms of its.

Answer:

QPSK Principle: QPSK (Quadrature Phase Shift Keying) uses four different phase states to represent 2 bits per symbol.

QPSK Modulator:

graph LR
    A[Data Stream] --> B[Serial to Parallel]
    B --> C[I Channel]
    B --> D[Q Channel]
    C --> E[Mixer 1]
    D --> F[Mixer 2]
    G["Carrier cos(ωt)"] --> E
    H["Carrier sin(ωt)"] --> F
    E --> I[Adder]
    F --> I
    I --> J[QPSK Output]

Constellation Diagram:

QPSK Demodulator:

graph LR
    A[QPSK Input] --> B[Mixer 1]
    A --> C[Mixer 2]
    D["cos(ωt)"] --> B
    E["sin(ωt)"] --> C
    B --> F[LPF]
    C --> G[LPF]
    F --> H[Decision Device]
    G --> I[Decision Device]
    H --> J[Parallel to Serial]
    I --> J
    J --> K[Data Output]

Advantages:

• Bandwidth Efficient: 2 bits per symbol
• Good Noise Performance: Constant envelope
• Widely Used: Standard in digital communication

Applications:

• Satellite communication
• Digital TV broadcasting
• Wireless communication systems

Mnemonic: “QPSK - Quadrature Phase, 2 bits, 4 phases”

Question 3(a) [3 marks]

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Draw the block diagram of FSK modulator

Answer:

graph LR
    A[Digital Data] --> B[Switch]
    C[Oscillator 1<br/>f1] --> B
    D[Oscillator 2<br/>f2] --> B
    B --> E[FSK Output]
    
    F[Data = 1] -.-> C
    G[Data = 0] -.-> D

Components:

• Digital Data Input: Binary data stream (0s and 1s)
• Two Oscillators: f₁ for bit ‘1’, f₂ for bit ‘0’
• Electronic Switch: Selects frequency based on input bit
• FSK Output: Frequency modulated signal

Operation:

• Bit ‘1’: Switch connects oscillator 1 (higher frequency)
• Bit ‘0’: Switch connects oscillator 2 (lower frequency)
• Output: Continuous frequency shifting based on data

Mnemonic: “FSK - Frequency Switch based on data Keys”

Question 3(b) [4 marks]

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Draw and explain block diagram of PSK modulator.

Answer:

graph LR
    A[Digital Data] --> B[Balanced Modulator]
    C["Carrier Oscillator<br/>cos(ωt)"] --> B
    B --> D[PSK Output]
    
    E[Data = 1] -.-> F[0° phase]
    G[Data = 0] -.-> H[180° phase]

Components and Function:

Operation:

• Data ‘1’: Output = +cos(ωt) (0° phase)
• Data ‘0’: Output = -cos(ωt) (180° phase)
• Phase Shift: 180° difference between ‘1’ and ‘0’

Mathematical Expression:

• PSK Signal: s(t) = A·d(t)·cos(ωt)
• Where d(t) = +1 for ‘1’, -1 for ‘0’

Advantages:

• Constant Envelope: Better noise immunity
• Bandwidth Efficient: Occupies same bandwidth as ASK
• Simple Detection: Coherent detection required

Mnemonic: “PSK - Phase Shift using balanced modulator Key”

Question 3(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 Data] --> B[Multiplier]
    C["Carrier cos(ωt)"] --> B
    B --> D[ASK Output]

ASK Demodulator:

graph LR
    A[ASK Input] --> B[Multiplier] 
    C[Local Carrier] --> B
    B --> D[Low Pass Filter]
    D --> E[Decision Device]
    E --> F[Digital Output]
    G[Threshold] --> E

Waveforms:

Modulation Process:

Demodulation Process:

1. Multiplication: ASK signal × Local carrier
2. Low Pass Filtering: Remove high frequency components
3. Decision: Compare with threshold to recover data

Applications:

• Optical Communication: LED/Laser on-off keying
• Simple Radio Systems: AM radio modification
• Short Range Communication: IR remote controls

Advantages/Disadvantages:

Mnemonic: “ASK - Amplitude Switch, multiply and filter Key”

Question 3(a OR) [3 marks]

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Write Principle and draw the constellation diagram of MSK.

Answer:

MSK Principle: MSK (Minimum Shift Keying) is a form of continuous-phase FSK where the frequency deviation is exactly half the bit rate.

Key Properties:

• Continuous Phase: No phase discontinuities
• Minimum Frequency Separation: Δf = Rb/2
• Constant Envelope: Good for nonlinear amplifiers

Constellation Diagram:

Mathematical Representation:

• Bit ‘1’: f₁ = fc + Rb/4
• Bit ‘0’: f₂ = fc - Rb/4
• Frequency Deviation: Δf = Rb/2

Characteristics:

• Spectral Efficiency: Better than conventional FSK
• Continuous Phase: Reduces out-of-band radiation
• Orthogonal Detection: Can be detected as OQPSK

Mnemonic: “MSK - Minimum Shift, Continuous phase Key”

Question 3(b OR) [4 marks]

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Draw and explain the constellation diagram of 16-QAM

Answer:

16-QAM Constellation:

16-QAM Mapping Table:

Key Features:

• 16 Symbol Points: 4 bits per symbol
• Gray Coding: Adjacent symbols differ by 1 bit
• Variable Amplitude: Different power levels
• High Data Rate: 4 times QPSK data rate

Signal Representation: s(t) = I(t)·cos(ωt) - Q(t)·sin(ωt)

Applications:

• Digital Cable TV: High data rate transmission
• Microwave Links: Point-to-point communication
• WiFi Systems: 802.11 standards

Advantages:

• High Spectral Efficiency: 4 bits per symbol
• Good BER Performance: With proper coding
• Flexible Implementation: Software defined radio

Trade-offs:

• Higher Complexity: More complex than QPSK
• Power Variation: Requires linear amplifiers
• Noise Sensitivity: Higher than constant envelope schemes

Mnemonic: “16-QAM - 16 points, 4 bits, Quadrature Amplitude Modulation”

Question 3(c OR) [7 marks]

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

Answer:

Bits per Symbol Comparison:

Detailed Analysis:

graph LR
    A[Digital Modulation] --> B[M-ary Modulation]
    B --> C["Bits per Symbol = log₂(M)"]
    C --> D[Higher M = More bits per symbol]
    D --> E[Higher Data Rate]
    E --> F[But Higher Complexity]

Bandwidth Efficiency:

Power Requirements:

Trade-offs:

• Higher M: More bits per symbol but higher complexity
• Bandwidth vs Power: Trade-off between spectral and power efficiency
• Implementation: Higher order modulation needs better hardware

Applications:

• Low Rate: ASK, FSK, PSK for simple systems
• Medium Rate: QPSK for balanced performance
• High Rate: 8-PSK, 16-QAM for high-speed systems

Formula: Bits per Symbol = log₂(M), where M = number of symbols

Mnemonic: “More symbols, More bits, More complexity”

Question 4(a) [3 marks]

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Define probability and write it Significance of in communication

Answer:

Definition of Probability: Probability is the measure of likelihood that an event will occur, expressed as a number between 0 and 1.

P(Event) = Number of favorable outcomes / Total number of possible outcomes

Significance in Communication:

Key Applications:

• BER Calculation: P(error) = Q(√(2Eb/N0))
• Channel Capacity: Shannon’s theorem uses probability
• Information Theory: Entropy based on probability
• System Design: Performance prediction

Mathematical Tools:

• Gaussian Distribution: For noise analysis
• Rayleigh Distribution: For fading channels
• Poisson Distribution: For arrival processes

Mnemonic: “Probability Predicts Performance in communication systems”

Question 4(b) [4 marks]

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Explain Huffman code with suitable example

Answer:

Huffman Coding Principle: Variable length coding where frequently occurring symbols get shorter codes.

Algorithm:

1. List symbols with probabilities
2. Combine two lowest probability symbols
3. Repeat until single symbol remains
4. Assign codes: left = 0, right = 1

Example:

Huffman Tree Construction:

graph TD
    A1[1.0] --> B1[A: 0.4]
    A1 --> C1[0.6]
    C1 --> D1[B: 0.3] 
    C1 --> E1[0.3]
    E1 --> F1[C: 0.2]
    E1 --> G1[D: 0.1]

Code Assignment:

• A: 0 (1 bit)
• B: 10 (2 bits)
• C: 110 (3 bits)
• D: 111 (3 bits)

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

Advantages:

• Optimal: Minimum average code length
• Prefix Property: No code is prefix of another
• Efficient: Reduces transmission bandwidth

Mnemonic: “Huffman - Frequent symbols get Shorter codes”

Question 4(c) [7 marks]

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

Answer:

IoT Concept: Internet of Things is the network of physical devices embedded with sensors, software, and connectivity to collect and exchange data.

IoT Architecture:

graph TD
    A[Physical Devices] --> B[Connectivity Layer]
    B --> C[Data Processing]
    C --> D[Application Layer]
    D --> E[Business Layer]
    
    F[Sensors] --> A
    G[Actuators] --> A
    H[WiFi/Bluetooth] --> B
    I[Cellular/LoRa] --> B
    J[Cloud Computing] --> C
    K[Edge Computing] --> C

Key Features:

IoT Protocol Stack:

Applications:

• Smart Home: Automated lighting, security
• Industrial IoT: Manufacturing automation
• Healthcare: Remote patient monitoring
• Smart Cities: Traffic management, utilities

Challenges:

• Security: Device vulnerabilities, data privacy
• Scalability: Billions of devices
• Interoperability: Different standards
• Power Consumption: Battery-operated devices

Benefits:

• Automation: Reduced human intervention
• Efficiency: Optimized resource usage
• Real-time Monitoring: Instant data access
• Cost Reduction: Predictive maintenance

Technologies:

• Communication: WiFi, Bluetooth, Cellular, LoRa
• Processing: Edge computing, cloud computing
• Analytics: Big data, machine learning
• Security: Encryption, authentication

Mnemonic: “IoT - Internet of Things, Smart Connected Devices everywhere”

Question 4(a OR) [3 marks]

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Define error correction code and list common error correcting code.

Answer:

Error Correction Code Definition: Error correction codes are techniques that add redundant bits to data to detect and correct transmission errors automatically.

Common Error Correcting Codes:

Applications:

• Memory Systems: ECC RAM
• Storage Devices: Hard drives, CDs
• Communication: Satellite, cellular
• Broadcasting: Digital TV, radio

Mnemonic: “Error Correction Codes - Hamming Reed BCH Convolutional Turbo LDPC”

Question 4(b OR) [4 marks]

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Explain Shanon Fano code with suitable example

Answer:

Shannon-Fano Coding Algorithm: Top-down approach that divides symbols into two groups with approximately equal probabilities.

Algorithm Steps:

1. Arrange symbols in decreasing probability order
2. Divide into two groups with nearly equal total probability
3. Assign ‘0’ to first group, ‘1’ to second group
4. Repeat for each subgroup

Example:

Construction Tree:

graph TD
    A1[A,B,C,D: 1.0] --> B1[A,B: 0.7]
    A1 --> C1[C,D: 0.3]
    B1 --> D1[A: 0.4]
    B1 --> E1[B: 0.3]
    C1 --> F1[C: 0.2]
    C1 --> G1[D: 0.1]

Code Assignment:

• Group 1 (A,B): Code starts with ‘0’
• Group 2 (C,D): Code starts with ‘1’
• A: 00, B: 01, C: 10, D: 11

Comparison with Huffman:

• Shannon-Fano: Top-down approach
• Huffman: Bottom-up approach
• Huffman: Always optimal
• Shannon-Fano: May not be optimal

Average Code Length: L = 0.4×2 + 0.3×2 + 0.2×2 + 0.1×2 = 2.0 bits/symbol

Mnemonic: “Shannon-Fano - Split groups, assign codes Top-down”

Question 4(c OR) [7 marks]

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Explain different standard formats of audio signal.

Answer:

Audio Signal Formats:

Audio Parameters:

graph TD
    A[Audio Signal] --> B[Sampling Rate]
    A --> C[Bit Depth]
    A --> D[Channels]
    A --> E[Compression]
    
    B --> F[44.1 kHz CD Quality]
    C --> G[16-bit Standard]
    D --> H[Mono/Stereo]
    E --> I[Lossy/Lossless]

Sampling Standards:

Compression Types:

• Lossless: Original quality preserved (FLAC, ALAC)
• Lossy: Some quality lost for smaller size (MP3, AAC)
• Uncompressed: No compression (WAV, AIFF)

Applications:

• Broadcasting: AAC for digital radio
• Streaming: MP3, AAC for internet
• Professional: WAV, FLAC for studios
• Mobile: AAC for smartphones

File Size Comparison (3-minute song):

• WAV: 30 MB
• FLAC: 20 MB
• MP3: 3 MB
• AAC: 2.5 MB

Quality vs Size Trade-off:

• Highest Quality: WAV (uncompressed)
• Best Balance: AAC (lossy compressed)
• Smallest Size: Low bitrate MP3
• Lossless Compressed: FLAC

Mnemonic: “WAV MP3 AAC FLAC - Wave, Layer3, Advanced, Free Lossless”

Question 5(a) [3 marks]

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Explain E1 carrier multiplexing hierarchy.

Answer:

E1 Carrier System: European digital transmission standard for multiplexing voice channels.

E1 Hierarchy:

E1 Frame Structure:

Multiplexing Process:

• Level 1: 30 voice channels + 2 control → E1
• Level 2: 4 E1 streams → E2
• Level 3: 4 E2 streams → E3
• Level 4: 4 E3 streams → E4

Applications:

• ISDN: Primary rate interface
• Cellular: Base station connectivity
• Enterprise: Private branch exchange (PBX)
• Internet: Digital subscriber line (DSL)

Mnemonic: “E1 - 30 voices, 2.048 Mbps, European standard”

Question 5(b) [4 marks]

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Compare FDMA with TDMA.

Answer:

FDMA vs TDMA Comparison:

FDMA System:

graph TD
    A[Total Bandwidth] --> B[User 1: f1]
    A --> C[User 2: f2]
    A --> D[User 3: f3]
    A --> E[User N: fn]
    
    F[Guard Band] --> B
    F --> C
    F --> D

TDMA System:

gantt
    title TDMA Time Slots
    dateFormat X
    axisFormat %s
    
    section Frame
    User 1 :done, u1, 0, 1
    User 2 :done, u2, 1, 2
    User 3 :done, u3, 2, 3
    User 4 :done, u4, 3, 4

Advantages/Disadvantages:

Applications:

• FDMA: AMPS (1G), satellite communication
• TDMA: GSM (2G), satellite communication

Mnemonic: “FDMA Frequency, TDMA Time - different domains for multiple access”

Question 5(c) [7 marks]

#

Explain CDMA technique in detail.

Answer:

CDMA Principle: Code Division Multiple Access allows multiple users to share the same frequency and time by using unique spreading codes.

CDMA System Architecture:

graph LR
    A[User Data] --> B[Spreading Code]
    B --> C[Modulator]
    C --> D[Channel]
    D --> E[Correlator]
    E --> F[Despreading]
    F --> G[Data Recovery]
    
    H[Pseudo-random Code] --> B
    I[Same PN Code] --> F

Spreading Process:

CDMA Properties:

Key Features:

CDMA Advantages:

• Capacity: Higher user capacity than FDMA/TDMA
• Security: Encrypted by spreading code
• Soft Handoff: No call dropping during handoff
• Anti-jamming: Spread spectrum immunity
• No Frequency Planning: Same frequency reuse

CDMA Disadvantages:

• Near-Far Problem: Requires power control
• Complexity: More complex than FDMA/TDMA
• Self Interference: Users interfere with each other
• Breathing Effect: Coverage varies with loading

Mathematical Analysis:

• Processing Gain: G = Rc/Rb = 10log₁₀(Rc/Rb) dB
• Capacity: M ≈ 1 + G/(Eb/I₀)
• BER: Depends on number of active users

Power Control:

• Open Loop: Based on received signal strength
• Closed Loop: Base station commands mobile
• Requirement: ±1 dB accuracy needed

Applications:

• IS-95 (cdmaOne): 2G CDMA standard
• WCDMA: 3G UMTS system
• GPS: Satellite navigation
• WiFi: Spread spectrum option

PN Code Properties:

• Autocorrelation: High for synchronized, low for unsynchronized
• Cross-correlation: Low between different codes
• Balance: Equal number of 1s and 0s
• Run Length: Distribution of consecutive bits

Mnemonic: “CDMA - Code Division, same frequency/time, unique codes for Multiple Access”

Question 5(a OR) [3 marks]

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

Answer:

TDM Block Diagram:

graph LR
    A[Input 1] --> E[Multiplexer]
    B[Input 2] --> E
    C[Input 3] --> E
    D[Input N] --> E
    
    E --> F[TDM Signal]
    F --> G[Channel]
    G --> H[Demultiplexer]
    
    H --> I[Output 1]
    H --> J[Output 2] 
    H --> K[Output 3]
    H --> L[Output N]
    
    M[Clock/Sync] --> E
    N[Clock/Sync] --> H

TDM Frame Structure:

Components:

• Multiplexer: Samples inputs sequentially
• Clock/Synchronization: Controls switching timing
• Channel: Transmission medium
• Demultiplexer: Separates multiplexed signal

Operation:

• Each input channel gets dedicated time slot
• Sampling rate must satisfy Nyquist criterion
• Frame synchronization required at receiver

Mnemonic: “TDM - Time Division, sequential sampling, Multiplexing”

Question 5(b OR) [4 marks]

#

Write a short not on classification of multiplexing techniques.

Answer:

Classification of Multiplexing Techniques:

graph TD
    A[Multiplexing] --> B[Analog Multiplexing]
    A --> C[Digital Multiplexing]
    
    B --> D[FDM - Frequency Division]
    B --> E[WDM - Wavelength Division]
    
    C --> F[TDM - Time Division]
    C --> G[CDM - Code Division]
    C --> H[SDM - Space Division]
    
    F --> I[Synchronous TDM]
    F --> J[Asynchronous TDM]

Detailed Classification:

Synchronous vs Asynchronous TDM:

Selection Criteria:

• Nature of Signal: Analog → FDM, Digital → TDM
• Bandwidth: Limited → TDM, Abundant → FDM
• Synchronization: Critical → Synchronous, Flexible → Asynchronous
• Application: Voice → TDM, Data → Statistical TDM

Modern Techniques:

• OFDM: Orthogonal Frequency Division Multiplexing
• MIMO: Multiple Input Multiple Output
• Carrier Aggregation: Multiple frequency bands

Mnemonic: “FDM TDM CDM WDM SDM - Frequency Time Code Wave Space Division Multiplexing”

Question 5(c OR) [7 marks]

#

Describe the procedure to troubleshoot the code division multiplexing circuit

Answer:

CDMA Troubleshooting Procedure:

1. System Overview Check:

graph TD
    A[CDMA System] --> B[Transmitter Section]
    A --> C[Channel Section]
    A --> D[Receiver Section]
    
    B --> E[Data Input OK?]
    B --> F[PN Code Generation OK?]
    B --> G[Spreading Function OK?]
    
    C --> H[Path Loss Measurement]
    C --> I[Interference Check]
    
    D --> J[Correlation OK?]
    D --> K[Despreading OK?]
    D --> L[Data Recovery OK?]

2. Step-by-Step Troubleshooting:

3. Common Problems and Solutions:

4. Test Equipment Required:

5. Measurement Procedures:

Processing Gain Verification:

Gp = 10 log₁₀(Rc/Rb) dB
Where: Rc = chip rate, Rb = bit rate

BER vs Eb/N0 Measurement:

BER = Q(√(2Eb/N0))
Measure at various power levels

Near-Far Effect Check:

• Measure power levels of different users
• Verify power control operation
• Check dynamic range requirements

6. Performance Optimization:

7. Documentation:

• Record all measurements
• Document problem symptoms
• Note solutions applied
• Create troubleshooting log

Systematic Approach:

1. Isolate: Identify faulty section
2. Measure: Use appropriate test equipment
3. Analyze: Compare with specifications
4. Correct: Apply appropriate solution
5. Verify: Confirm problem resolution

Safety Considerations:

• Power levels within safe limits
• Proper grounding procedures
• RF exposure guidelines
• Equipment calibration status

Mnemonic: “CDMA Troubleshoot - Check Data, PN code, Spreading, Channel, Correlation, Recovery”