Digital Communication (4341102) - Winter 2024 Solution

Question 1(a) [3 marks]

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

Answer:

Table: Comparison of Signal Types

Diagram:

• 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]

Explain periodic and aperiodic signal.

Answer:

Table: Periodic vs. Aperiodic Signals

Diagram:

• 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]

Explain block diagram of digital communication system.

Answer:

Diagram: Digital Communication System

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]

Explain singularity functions.

Answer:

Table: Common Singularity Functions

Diagram:

goat

• 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]

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:

• 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]

Explain Even and Odd signal.

Answer:

Table: Even vs. Odd Signals

Diagram:

goat

• 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]

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

Answer:

ASK Modulator Diagram:

ASK Demodulator Diagram:

Waveform:

goat

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]

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:

• 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]

Discuss the various communication channels characteristics.

Answer:

Table: Communication Channel Characteristics

Diagram:

• 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]

Compare ASK, FSK and PSK.

Answer:

Table: Comparison of Digital Modulation Techniques

Diagram:

goat

• 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]

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

Answer:

FSK Modulator Block Diagram:

Waveform:

goat

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]

Draw the PSK modulation waveform for the sequence of 1010110110.

Answer:

BPSK Modulation for 1010110110:

goat

Table: BPSK Mapping

Diagram:

• 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]

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

Answer:

Input Bit Sequence: 1100110101

ASK Modulation:

goat

FSK Modulation:

goat

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]

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

Answer:

MSK Modulator Block Diagram:

Waveform:

goat

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]

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

Answer:

8-PSK Constellation Diagram:

goat

16-QAM Constellation Diagram:

goat

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]

Draw BPSK and QPSK modulation waveform for 1010101011.

Answer:

Input Bit Sequence: 1010101011

BPSK Modulation:

goat

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

goat

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]

Encode the data using Shanon Fano code for below probability sequence. P =

Answer:

Table: Shannon-Fano Coding Process

Diagram:

• 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]

Explain Hamming code.

Answer:

Table: Hamming Code Properties

Diagram:

• 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]

Compare TDMA and FDMA.

Answer:

Table: Comparison of TDMA and FDMA

Diagram:

• 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]

Encode the data using Huffman code for below probability sequence. P =

Answer:

Table: Huffman Coding Process

Diagram:

• 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]

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:

• 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]

Explain FDMA Technique in detail.

Answer:

FDMA (Frequency Division Multiple Access)

Table: FDMA System Characteristics

Diagram:

FDMA Implementation:

goat

• 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]

Explain TDMA Access technique.

Answer:

TDMA (Time Division Multiple Access)

Table: TDMA Key Characteristics

Diagram:

• 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]

Explain E1 Career system.

Answer:

E1 Carrier System

Table: E1 Carrier System Specifications

Diagram:

• 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]

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

Answer:

Digital Telephone Exchange Block Diagram

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]

Compare TDM and FDM.

Answer:

Table: Comparison of TDM and FDM

Diagram:

goat

• 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]

Discuss T1 Multiplexing hierarchy.

Answer:

Table: T1 Multiplexing Hierarchy

Diagram:

T1 Frame Structure:

goat

• 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]

List Features, Characteristics, Advantages and Disadvantages of IoT.

Answer:

Table: Internet of Things (IoT) Overview

Features of IoT:

Advantages & Disadvantages:

goat

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