Communication Engineering (1333201) - Summer 2024 Solution

Question 1(a) [3 marks]

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Define modulation and explain its need.

Answer: Modulation is the process of varying one or more properties of a high-frequency carrier signal with a modulating signal containing information.

Table: Need for Modulation

Mnemonic: “ASPIM” - Antenna size, Signal propagation, Proper multiplexing, Interference reduction, Manage bandwidth

Question 1(b) [4 marks]

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Draw & explain block diagram of Communication system

Answer: A communication system transfers information from source to destination through a channel.

graph LR
    A[Information Source] --> B[Transmitter]
    B --> C[Channel]
    C --> D[Receiver]
    D --> E[Destination]
    F[Noise Source] --> C

Table: Communication System Components

Mnemonic: “I Try Communicating Neatly, Receive Data” (I-T-C-N-R-D)

Question 1(c) [7 marks]

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Derive voltage equation for Amplitude modulation.

Answer: Amplitude modulation varies the amplitude of carrier signal proportionally to the message signal.

Mathematical Derivation:

• Let carrier signal be: c(t) = Ac cos(ωct)
• Message signal: m(t) = Am cos(ωmt)
• AM signal: s(t) = Ac[1 + μ·m(t)/Am]cos(ωct)
• Where μ = modulation index = Am/Ac
• Substituting m(t): s(t) = Ac[1 + μ·cos(ωmt)]cos(ωct)
• Expanding: s(t) = Ac·cos(ωct) + μ·Ac·cos(ωmt)·cos(ωct)
• Using identity (cos A·cos B): s(t) = Ac·cos(ωct) + (μ·Ac/2)[cos(ωc+ωm)t + cos(ωc-ωm)t]

Diagram: AM Signal in Time Domain

%%{init: {"theme": "neutral", "themeVariables": {"primaryColor": "#f6f6f6"}}}%%
xychart-beta
    title "AM Signal"
    x-axis "Time" 0 --> 12
    y-axis "Amplitude" -1.5 --> 1.5
    line [0, 0.2, 0.4, 0.6, 0.8, 1, 0.8, 0.6, 0.4, 0.2, 0, -0.2, -0.4]
    line [0, 0.8, 1.2, 0.8, 0, -0.8, -1.2, -0.8, 0, 0.8, 1.2, 0.8, 0]

Mnemonic: “CAMDS” - Carrier Amplitude Modulated by Data Signal

Question 1(c) OR [7 marks]

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Derive the equation for total power in AM, calculate percentage of power savings in DSB and SSB.

Answer: For an AM signal with modulation index μ, the total power consists of carrier power and sideband power.

Table: Power Distribution in AM

Power Savings Calculation:

• In DSB-SC: 100% carrier suppression = (Pc/PT)×100% = 1/(1+μ²/2)×100%
  • For μ = 1: Saving = 2/3×100% = 66.67%
• In SSB: One sideband + carrier suppression = (Pc+PLSB)/PT×100% = (1+μ²/4)/(1+μ²/2)×100%
  • For μ = 1: Saving = 5/6×100% = 83.33%

Mnemonic: “CAPS” - Carrier And Power in Sidebands

Question 2(a) [3 marks]

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Define Image frequency in a radio receiver and explain it with suitable example.

Answer: Image frequency is an unwanted frequency that can produce the same IF (Intermediate Frequency) as the desired signal in a superheterodyne receiver.

Table: Image Frequency

If both 100 MHz and 120 MHz signals exist, both will produce 10 MHz IF, causing interference.

Mnemonic: “LIDS” - Local oscillator plus/minus IF gives Desired signal and Signal image

Question 2(b) [4 marks]

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Draw and explain block diagram for envelope detector.

Answer: Envelope detector extracts the modulating signal from AM wave by following the envelope.

graph LR
    A[AM Input] --> B[Diode]
    B --> C[RC Circuit]
    C --> D[Envelope Output]

Table: Envelope Detector Components

Mnemonic: “DRCT” - Diode Rectifies, Capacitor Tracks

Question 2(c) [7 marks]

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Draw block diagram of AM radio receiver and explain working of each block.

Answer: AM receiver converts radio signal to audio output.

graph LR
    A[Antenna] --> B[RF Amplifier]
    B --> C[Mixer]
    D[Local Oscillator] --> C
    C --> E[IF Amplifier]
    E --> F[Detector]
    F --> G[AF Amplifier]
    G --> H[Speaker]

Table: AM Receiver Blocks

Mnemonic: “ARMLIDAS” - Antenna Receives, Mixer Links Input and Detector, Audio to Speaker

Question 2(a) OR [3 marks]

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Define any FOUR characteristics of radio receiver.

Answer:

Table: Radio Receiver Characteristics

Mnemonic: “SSFIS” - Super Sensitive Fidelity with Image Suppression

Question 2(b) OR [4 marks]

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Explain Ratio detector circuit for FM detection.

Answer: Ratio detector extracts audio from FM signals while rejecting amplitude variations.

graph LR
    A[FM Input] --> B[Transformer]
    B --> C[Diode Circuit]
    C --> D[Stabilizing Capacitor]
    D --> E[Audio Output]

Table: Ratio Detector Components

Mnemonic: “RADS” - Ratio detector Avoids Disturbance from Strength variations

Question 2(c) OR [7 marks]

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Draw and explain block diagram of super heterodyne receiver.

Answer: Superheterodyne receiver converts all incoming RF to fixed IF for better amplification.

graph LR
    A[Antenna] --> B[RF Amplifier]
    B --> C[Mixer]
    D[Local Oscillator] --> C
    C --> E[IF Amplifier]
    E --> F[Detector]
    F --> G[AGC]
    G --> B
    G --> E
    F --> H[AF Amplifier]
    H --> I[Speaker]

Table: Superheterodyne Receiver Components

Mnemonic: “ARMLIADS” - Antenna Receives, Mixer Links, Intermediate Amplifies, Detector Separates

Question 3(a) [3 marks]

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Draw the Time and frequency domain representation of the below signals. 1. Analog signal (sine) 2. Digital signal (square).

Answer:

Table: Signal Representations

Diagram: Signal Representations

Mnemonic: “SOFT” - Sine has One Frequency, square has Timeless harmonics

Question 3(b) [4 marks]

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Explain sampling theorem.

Answer: Sampling theorem states the conditions for accurate signal reconstruction from samples.

Table: Sampling Theorem

Diagram: Aliasing Effect

Mnemonic: “SNAP” - Sample at Nyquist And Prevent aliasing

Question 3(c) [7 marks]

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Explain PAM, PPM and PWM.

Answer: These are pulse modulation techniques where a parameter of pulse is varied.

Table: Pulse Modulation Types

Diagram: Pulse Modulation Techniques

Mnemonic: “AAA-PPW” - Amplitude, Position, Width are modulated in PAM, PPM, PWM

Question 3(a) OR [3 marks]

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Define Nyquist rate and explain.

Answer: Nyquist rate is the minimum sampling frequency required for accurate signal reconstruction.

Table: Nyquist Rate

Mnemonic: “TANS” - Twice As Needed for Sampling

Question 3(b) OR [4 marks]

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Explain quantization process.

Answer: Quantization assigns discrete amplitude levels to sampled values in analog-to-digital conversion.

Table: Quantization Process

Diagram: Quantization Process

Mnemonic: “SLERN” - Sample, Level assign, Error occurs, Resolution determines Noise

Question 3(c) OR [7 marks]

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Explain Ideal, Natural and Flat top sampling.

Answer: These are different practical implementations of sampling process.

Table: Sampling Types Comparison

Diagram: Sampling Types

Mnemonic: “INF” - Ideal is theoretical, Natural is practical, Flat-top holds values

Question 4(a) [3 marks]

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List the advantages and disadvantages of PCM.

Answer:

Table: PCM Advantages and Disadvantages

Mnemonic: “NICHE” vs “BCQPS” - Noise immunity, Integration, Complex circuitry, Higher bandwidth, Error correction vs Bandwidth, Cost, Quantization, Power, Synchronization

Question 4(b) [4 marks]

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Draw and Explain Block Diagram of Delta Modulation.

Answer: Delta modulation transmits only changes in signal level using 1-bit quantization.

graph LR
    A[Input Signal] --> B[Comparator]
    B --> C[1-bit Quantizer]
    C --> D[Output]
    C --> E[Integrator]
    E --> F[1-sample Delay]
    F --> B

Table: Delta Modulation Components

Mnemonic: “CQID” - Compare, Quantize with 1-bit, Integrate, Delay

Question 4(c) [7 marks]

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Compare PCM, DM and DPCM.

Answer:

Table: Comparison of Digital Modulation Techniques

Mnemonic: “PCM-DM-DPCM: More Bits Better Quality, More Complexity Needed”

Question 4(a) OR [3 marks]

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Explain DPCM.

Answer: Differential Pulse Code Modulation encodes difference between actual and predicted sample.

Table: DPCM Characteristics

Mnemonic: “DPCM: Difference Predicted, Correlation Matters”

Question 4(b) OR [4 marks]

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List the advantages and disadvantages of Delta Modulation.

Answer:

Table: Delta Modulation - Pros and Cons

Mnemonic: “SLSRL” vs “SGLSH” - Simple, Low bit-rate, Single bit, Robust, Low cost vs Slope overload, Granular noise, Limited range, Sampling high, SNR low

Question 4(c) OR [7 marks]

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Explain Block diagram of basic PCM-TDM system.

Answer: PCM-TDM combines multiple digitized signals into a single high-speed channel.

graph LR
    A1[Input 1] --> B1[PCM Encoder 1]
    A2[Input 2] --> B2[PCM Encoder 2]
    A3[Input 3] --> B3[PCM Encoder 3]
    B1 --> C[TDM Multiplexer]
    B2 --> C
    B3 --> C
    C --> D[Transmission Channel]
    D --> E[TDM Demultiplexer]
    E --> F1[PCM Decoder 1]
    E --> F2[PCM Decoder 2]
    E --> F3[PCM Decoder 3]
    F1 --> G1[Output 1]
    F2 --> G2[Output 2]
    F3 --> G3[Output 3]

Table: PCM-TDM System Components

Mnemonic: “PETDSF” - PCM Encodes, TDM combines, Digital transmits, Separation occurs, Frames synchronize

Question 5(a) [3 marks]

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Explain Adaptive Delta modulation.

Answer: Adaptive Delta Modulation adjusts step size based on signal characteristics.

Table: Adaptive Delta Modulation

Diagram: Step Size Adaptation

Mnemonic: “ASSG” - Adaptive Step Size Gives better performance

Question 5(b) [4 marks]

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Define the terms 1. Radiation Pattern 2. Antenna gain.

Answer:

Table: Antenna Terms

Diagram: Radiation Pattern Types

Mnemonic: “RPGD” - Radiation Pattern shows Gain Direction

Question 5(c) [7 marks]

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Explain Base station antenna and Mobile station antenna.

Answer: Different antenna designs serve different purposes in wireless communication systems.

Table: Comparison of Base Station and Mobile Station Antennas

Diagram: Antenna Types

Mnemonic: “BHPSTBD” - Base stations Have Power, Size, Tower mounting, Beamforming, Diversity

Question 5(a) OR [3 marks]

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Write down range of frequencies for HF, VHF and UHF.

Answer:

Table: Frequency Bands

Mnemonic: “3-30-300-3000” - Each band starts at 3 times a power of 10 MHz

Question 5(b) OR [4 marks]

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Define the terms 1. Antenna Directivity 2. Polarization.

Answer:

Table: Antenna Properties

Diagram: Polarization Types

Mnemonic: “DIVE POLE” - DIrectivity shows Vector Excellence, POLarization shows Electric field

Question 5(c) OR [7 marks]

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Explain Ground wave propagation and Space wave propagation in detail.

Answer: These are two primary modes of radio wave propagation in the lower atmosphere.

Table: Wave Propagation Comparison

Diagram: Ground Wave vs Space Wave Propagation

Ground Wave Propagation:

• Travels along earth’s surface
• Signal strength decreases with distance
• Better propagation over sea than land
• Affected by ground conductivity and dielectric constant
• Used for AM broadcasting, maritime communication

Space Wave Propagation:

• Consists of direct wave and ground-reflected wave
• Range extended by atmospheric refraction
• Range formula: d = √(2Rh) where R is earth’s radius, h is antenna height
• Affected by diffraction over obstacles
• Used for line-of-sight communications like TV, FM, microwave links

Mnemonic: “GAFFS” - Ground Adheres to earth, Follows surface, Frequencies low, Short wavelengths