Digital Communication (4341102) - Winter 2024 Solution

Question 1(a) [3 marks]

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

Answer:

Table: Comparison of Signal Types

Diagram:

graph LR
    subgraph Continuous
        A[Continuous Time Signal] --> B[x(t)]
        B --> C[Defined for all t]
    end
    subgraph Discrete
        D[Discrete Time Signal] --> E[x(n)]
        E --> F[Defined for integer n]
    end

• Amplitude continuity: In continuous signals, amplitude can take any value, while discrete signals have specific amplitude values
• Mathematical notation: Continuous signals use x(t), discrete signals use x[n] or x(n)

Mnemonic: “CoSiDi” - Continuous Signals flow like rivers, Discrete signals are like stepping stones

Question 1(b) [4 marks]

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Explain periodic and aperiodic signal.

Answer:

Table: Periodic vs. Aperiodic Signals

Diagram:

graph TD
    subgraph Periodic
        A[x(t) = x(t+T)] --> B[Repeats exactly]
        B --> C[Fundamental period T]
    end
    subgraph Aperiodic
        D[x(t) ≠ x(t+T)] --> E[Never repeats exactly]
        E --> F[No fundamental period]
    end

• Spectral property: Periodic signals have discrete frequency components, aperiodic have continuous spectrum
• Fourier analysis: Periodic signals use Fourier series, aperiodic use Fourier transform

Mnemonic: “PART” - Periodic signals Always Repeat in Time

Question 1(c) [7 marks]

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Explain block diagram of digital communication system.

Answer:

Diagram: 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]

Table: Functions of Digital Communication System Blocks

• Advantage: Noise immunity, secure transmission, multiplexing capability, integration with digital systems
• Key processes: Sampling, quantization, coding, modulation/demodulation

Mnemonic: “SECMCDS” - Source Encodes, Channel codes, Modulates, Channel, Demodulates, Sink receives

Question 1(c) OR [7 marks]

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Explain singularity functions.

Answer:

Table: Common Singularity Functions

Diagram:

• Integration relationship: Each function is the integral of the previous one
• Mathematical toolkit: Used to analyze complex systems by breaking into simpler components

Mnemonic: “SIPR” - Step Impulse Parable Ramp - functions ordered by increasing order of integration

Question 2(a) [3 marks]

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A signal carries 10 bit/signal elements. If 100 signal elements sent per second. Find the bit rate.

Answer:

Solution:

Bit Rate = Number of bits per signal element × Number of signal elements per second
Bit Rate = 10 bits/signal element × 100 signal elements/second
Bit Rate = 1000 bits/second = 1 kbps

Diagram:

graph LR
    A[Signal Elements: 100/s] --> B[Each Element: 10 bits]
    B --> C[Bit Rate = 1000 bits/s]

• Bit rate: Number of bits transmitted per second (bps)
• Signal element: Physical manifestation of one or more bits

Mnemonic: “BEE” - Bit rate equals Elements times bits per Element

Question 2(b) [4 marks]

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Explain Even and Odd signal.

Answer:

Table: Even vs. Odd Signals

Diagram:

• Decomposition: Any signal can be decomposed as sum of even and odd components
• Even part: f_e(t) = [f(t) + f(-t)]/2
• Odd part: f_o(t) = [f(t) - f(-t)]/2

Mnemonic: “ESOM” - Even Signals have mirror symmetry, Odd signals flip when Mirrored

Question 2(c) [7 marks]

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

Answer:

ASK Modulator Diagram:

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

ASK Demodulator Diagram:

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

Waveform:

Table: ASK Modulation and Demodulation Process

• Binary ASK: Carrier present for ‘1’, absent for ‘0’
• Bandwidth: Minimum BW = bit rate, typically twice bit rate used

Mnemonic: “AMPS” - ASK Modulates carrier Power (amplitude) with digital Signal

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:

Solution:

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

Diagram:

graph LR
    A[Bit Rate: 4000 bps] --> C[Divide]
    B[Baud Rate: 1000 baud] --> C
    C --> D[4 bits/signal element]

• Bit rate: Data transmission speed in bits per second
• Baud rate: Number of signal elements (symbols) per second

Mnemonic: “BBR” - Bits per symbol equals Bit rate divided by Baud Rate

Question 2(b) OR [4 marks]

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Discuss the various communication channels characteristics.

Answer:

Table: Communication Channel Characteristics

Diagram:

graph TD
    A[Channel Characteristics] --> B[Bandwidth]
    A --> C[Noise]
    A --> D[Attenuation]
    A --> E[Distortion]
    A --> F[Channel Capacity]
    C --> G[SNR]
    B --> H[Data Rate]
    F --> H

• SNR (Signal-to-Noise Ratio): Ratio of signal power to noise power
• Channel capacity: C = B·log₂(1+SNR), where B is bandwidth

Mnemonic: “BAND-C” - Bandwidth, Attenuation, Noise, Distortion define Capacity

Question 2(c) OR [7 marks]

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Compare ASK, FSK and PSK.

Answer:

Table: Comparison of Digital Modulation Techniques

Diagram:

• Bit error rate: PSK < FSK < ASK (PSK is best)
• Complexity order: ASK < FSK < PSK (ASK is simplest)

Mnemonic: “AFP” - Amplitude, Frequency, Phase are modified in ASK, FSK, PSK respectively

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:

flowchart LR
    A[Digital Input] --> B[Switch Controller]
    B --> C[Oscillator 1 - f1]
    B --> D[Oscillator 2 - f2]
    C --> E[Output]
    D --> E

Waveform:

Table: FSK Modulation Process

• Binary FSK: Uses two frequencies f₁ and f₂ separated by frequency deviation
• Advantage: Better noise immunity than ASK

Mnemonic: “FAST” - Frequency Alternates between Separate Tones

Question 3(b) [4 marks]

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Draw the PSK modulation waveform for the sequence of 1010110110.

Answer:

BPSK Modulation for 1010110110:

Table: BPSK Mapping

Diagram:

graph TD
    A[Bit Stream 1010110110] --> B[Phase Mapping]
    B --> C[1=0° Phase]
    B --> D[0=180° Phase]
    C --> E[Modulated Signal]
    D --> E

• Phase shift: 180° transition at each bit change
• Constant amplitude: Unlike ASK, amplitude remains constant

Mnemonic: “POPI” - Phase Opposites for bit Pairs represent Information

Question 3(c) [7 marks]

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

Answer:

Input Bit Sequence: 1100110101

ASK Modulation:

FSK Modulation:

Table: Comparison for the Sequence 1100110101

• ASK modulation: Simple on-off keying where carrier is present for ‘1’ and absent for ‘0’
• FSK modulation: Frequency shifts between two distinct values based on bit value

Mnemonic: “AFOP” - ASK switches carrier On-Off while FSK shifts between Pairs of frequencies

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:

flowchart LR
    A[Digital Input] --> B[Serial to Parallel]
    B --> C[Even Bits]
    B --> D[Odd Bits]
    C --> E[Cos Modulator]
    D --> F[Sin Modulator]
    G[90° Phase Shifter] --> F
    H[Carrier Generator] --> E
    H --> G
    E --> I[Combiner]
    F --> I
    I --> J[MSK Output]

Waveform:

Table: MSK Modulation Process

• Phase continuity: Key advantage - reduces bandwidth compared to FSK
• Constant envelope: Resistant to non-linear amplification

Mnemonic: “MCPS” - MSK ensures Continuous Phase Shifts

Question 3(b) OR [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:

Table: Comparison of Constellation Diagrams

• 8-PSK: Points equally spaced around circle with constant amplitude
• 16-QAM: Points arranged in square grid with different amplitudes and phases

Mnemonic: “CEPA” - Constellation points in PSK have Equal amplitudes but different Phases, QAM has both Amplitude and phase variations

Question 3(c) OR [7 marks]

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Draw BPSK and QPSK modulation waveform for 1010101011.

Answer:

Input Bit Sequence: 1010101011

BPSK Modulation:

QPSK Modulation (Grouping bits: 10,10,10,10,11):

Table: BPSK vs. QPSK for 1010101011

• BPSK: Each bit causes a potential 180° phase shift
• QPSK: Processes two bits at once, uses four phase states

Mnemonic: “BQSE” - BPSK takes 1 bit while QPSK takes 2 bits, doubling Spectral Efficiency

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:

Table: Shannon-Fano Coding Process

Diagram:

graph TD
    A[Symbols] --> B[A:0.30, B:0.25, C:0.20, D:0.12, E:0.08, F:0.05]
    B --> C[A:0.30, B:0.25]
    B --> D[C:0.20, D:0.12, E:0.08, F:0.05]
    C --> E[A:0.30]
    C --> F[B:0.25]
    D --> G[C:0.20, D:0.12]
    D --> H[E:0.08, F:0.05]
    G --> I[C:0.20]
    G --> J[D:0.12]
    H --> K[E:0.08]
    H --> L[F:0.05]
    E --> M[Code: 0]
    F --> N[Code: 10]
    I --> O[Code: 110]
    J --> P[Code: 1110]
    K --> Q[Code: 11110]
    L --> R[Code: 11111]

• Shannon-Fano algorithm: Recursively divide symbols into two groups with nearly equal probabilities
• Code efficiency: Not always optimal but generally good compression

Mnemonic: “SPDF” - Split Probabilities and assign Digits based on Frequency

Question 4(b) [4 marks]

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Explain Hamming code.

Answer:

Table: Hamming Code Properties

Diagram:

graph TD
    A[Hamming Code] --> B[Parity Bits]
    A --> C[Data Bits]
    B --> D[Position 1]
    B --> E[Position 2]
    B --> F[Position 4]
    B --> G[Position 8]
    A --> H[Example: Hamming(7,4)]
    H --> I[4 data bits + 3 parity bits]

• Encoding: Calculate parity bits to ensure specific bit positions have even/odd parity
• Decoding: Calculate syndrome to determine error position

Mnemonic: “PSEC” - Parity bits in Power of 2 positions Systematically Enable error Correction

Question 4(c) [7 marks]

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Compare TDMA and FDMA.

Answer:

Table: Comparison of TDMA and FDMA

Diagram:

graph TD
    subgraph TDMA
        A[Time Slots] --> A1[User 1]
        A --> A2[User 2]
        A --> A3[User 3]
        A --> A4[Guard Time]
    end
    subgraph FDMA
        B[Frequency Bands] --> B1[User 1]
        B --> B2[User 2]
        B --> B3[User 3]
        B --> B4[Guard Bands]
    end

• System flexibility: TDMA can dynamically allocate slots, FDMA is fixed allocation
• Implementation: TDMA requires digital technology, FDMA works with analog/digital

Mnemonic: “TIME-FREQ” - TDMA splits Intervals of tiME, FDMA splits Ranges of frEQuency

Question 4(a) OR [3 marks]

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Encode the data using Huffman code for below probability sequence. P = { 0.4, 0.19, 0.16, 0.15, 0.1}

Answer:

Table: Huffman Coding Process

Diagram:

graph TD
    Z[Root: 1.0] --> A[A: 0.4]
    Z --> Y[0.6]
    Y --> B[B: 0.19]
    Y --> X[0.41]
    X --> C[C: 0.16]
    X --> W[0.25]
    W --> D[D: 0.15]
    W --> E[E: 0.1]
    A -- 0 --> AA[Code: 0]
    B -- 1 --> BB[Code: 10]
    C -- 0 --> CC[Code: 110]
    D -- 0 --> DD[Code: 1110]
    E -- 1 --> EE[Code: 1111]

• Huffman algorithm: Build a binary tree from bottom up, starting with least probable symbols
• Optimality: Produces minimal average code length

Mnemonic: “HUMP” - Huffman creates shorter codes for Higher Probabilities

Question 4(b) OR [4 marks]

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

Answer:

Shannon’s Channel Capacity Formula:

C = B × log₂(1 + SNR)

Where:

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

Table: Channel Capacity Characteristics

Diagram:

graph LR
    A[Channel Capacity] --> B[Bandwidth B]
    A --> C[Signal-to-Noise Ratio]
    B --> D[C = B × log₂(1 + SNR)]
    C --> D
    D --> E[Theoretical Maximum]
    E --> F[Error-free Communication]

• Shannon-Hartley theorem: Establishes theoretical maximum data transfer rate
• Error probability: Can be made arbitrarily small if data rate < channel capacity

Mnemonic: “SNRB” - Shannon capacity depends on Noise ratio and Bandwidth

Question 4(c) OR [7 marks]

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Explain FDMA Technique in detail.

Answer:

FDMA (Frequency Division Multiple Access)

Table: FDMA System Characteristics

Diagram:

graph TD
    A[Available Spectrum] --> B[Guard Band]
    A --> C[User 1 Channel]
    A --> D[Guard Band]
    A --> E[User 2 Channel]
    A --> F[Guard Band]
    A --> G[User 3 Channel]
    A --> H[Guard Band]
    A --> I[User 4 Channel]

FDMA Implementation:

• Implementation: Relatively simple using bandpass filters
• Advantages: No synchronization required, continuous transmission
• Disadvantages: Spectrum inefficiency, limited flexibility

Mnemonic: “FDMA-CIGS” - Frequency Division creates Multiple Access through Channels with Individual Guard band Separation

Question 5(a) [3 marks]

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Explain TDMA Access technique.

Answer:

TDMA (Time Division Multiple Access)

Table: TDMA Key Characteristics

Diagram:

graph LR
    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]

• Digital implementation: Requires ADC/DAC for analog signals
• Burst transmission: Users transmit only during assigned slots

Mnemonic: “TIME” - Time slots Individually Managed for Each user

Question 5(b) [4 marks]

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Explain E1 Career system.

Answer:

E1 Carrier System

Table: E1 Carrier System Specifications

Diagram:

graph TD
    A[E1 Frame - 2.048 Mbps] --> B[TS0: Framing]
    A --> C[TS1-15: Voice Channels]
    A --> D[TS16: Signaling]
    A --> E[TS17-31: Voice Channels]
    B --> F[Frame Alignment Signal]
    D --> G[Channel Associated Signaling]

• Multiplexing technique: TDM (Time Division Multiplexing)
• PCM encoding: 8-bit samples at 8 kHz sampling rate

Mnemonic: “E132” - E1 has 32 time slots with 2.048 Mbps

Question 5(c) [7 marks]

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Explain block diagram of Digital telephone exchange, elements of hardware sub systems.

Answer:

Digital Telephone Exchange Block Diagram

flowchart TD
    A[Digital Telephone Exchange] --> B[DLU: Digital Line Unit]
    A --> C[LTG: Line/Trunk Group]
    A --> D[SN: Switching Network]
    A --> E[CP: Central Processor]
    B --> F[Interface to Subscribers]
    C --> G[Interface to Trunks]
    D --> H[Digital Switching]
    E --> I[System Control]

Table: Hardware Subsystems of Digital Telephone Exchange

Hardware Elements Details:

• DLU: Converts analog voice to 64 kbps PCM, handles line signaling
• LTG: Manages E1/T1 trunks, implements protocols like SS7
• SN: Typically time-division switching fabric, non-blocking architecture
• CP: Call processing, billing, maintenance, administrative functions

Mnemonic: “DLSC” - DLU connects subscribers, LTG connects trunks, SN switches calls, CP controls everything

Question 5(a) OR [3 marks]

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Compare TDM and FDM.

Answer:

Table: Comparison of TDM and FDM

Diagram:

• Bandwidth utilization: TDM more efficient for digital, FDM better for analog
• System complexity: TDM requires precise timing, FDM needs precise filters

Mnemonic: “TFDS” - Time and Frequency Division Systems divide different domains

Question 5(b) OR [4 marks]

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Discuss T1 Multiplexing hierarchy.

Answer:

Table: T1 Multiplexing Hierarchy

Diagram:

graph TD
    A[Individual Voice Channels - DS0 64 kbps] --> B[T1/DS1 - 1.544 Mbps]
    B --> C[T2/DS2 - 6.312 Mbps]
    C --> D[T3/DS3 - 44.736 Mbps]
    D --> E[T4/DS4 - 274.176 Mbps]

T1 Frame Structure:

• T1 frame format: 193 bits (24 channels × 8 bits + 1 framing bit)
• Frame duration: 125 μs (8000 frames per second)

Mnemonic: “T-QUAD” - T1, T2, T3, T4 form a QUADruple hierarchy of multiplexing levels

Question 5(c) OR [7 marks]

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List Features, Characteristics, Advantages and Disadvantages of IoT.

Answer:

Table: Internet of Things (IoT) Overview

Features of IoT:

graph TD
    A[IoT Features] --> B[Connectivity]
    A --> C[Intelligence]
    A --> D[Sensing]
    A --> E[Automation]
    A --> F[Cloud Integration]
    A --> G[Data Analytics]

Advantages & Disadvantages:

Characteristics Details:

• Interconnectivity: Anything can be connected to global information & communication infrastructure
• Things-related services: IoT provides thing-related services like privacy protection
• Heterogeneity: Devices based on different hardware/software platforms
• Dynamic changes: Device state changes dynamically (connecting/disconnecting, sleeping/waking)
• Enormous scale: Number of devices requiring management exceeds traditional internet connected devices

Mnemonic: “CASED” - Connectivity, Automation, Sensing, Efficiency, Data analytics - key IoT features