Principles of Electronic Communication (4331104) - Summer 2023 Solution

Question 1(a) [3 marks]

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Draw & explain block diagram of Communication system.

Answer:

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

• Input: Message signal originating from source
• Transmitter: Converts message to suitable form for transmission
• Channel: Medium through which signal travels
• Receiver: Extracts original message from received signal
• Output: Delivered message to destination
• Noise Source: Unwanted signals that interfere with communication

Mnemonic: “I Transmit Clearly Receiving Original Messages”

Question 1(b) [4 marks]

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Explain need of modulation. State advantages of modulation.

Answer:

Need for modulation:

graph TD
    A[Practical Antenna Size] --> B[Modulation]
    C[Multiplexing] --> B
    D[Reducing Noise & Interference] --> B
    E[Signal Transmission Distance] --> B

Advantages of modulation:

• Reduced antenna size: Practical antenna length = λ/4, higher frequency means smaller antenna
• Multiplexing possible: Multiple signals transmitted simultaneously through same channel
• Increased range: Modulated signals travel farther than baseband signals
• Noise reduction: Better SNR achieved through modulation techniques

Mnemonic: “Antennas Need Modulation For Reaching Anywhere with Noise Immunity”

Question 1(c) [7 marks]

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Define modulation. Explain Amplitude modulation with waveform and derive voltage equation for modulated signal.

Answer:

Modulation: Process of varying a carrier signal’s parameter (amplitude, frequency, phase) proportionally to the message signal.

Amplitude Modulation Waveform:

Derivation of AM voltage equation:

1. Carrier signal: vc(t) = Vc sin(ωct)
2. Message signal: vm(t) = Vm sin(ωmt)
3. Modulated signal: vAM(t) = [Vc + Vm sin(ωmt)] sin(ωct)
4. Modulation index: μ = Vm/Vc
5. Final AM equation: vAM(t) = Vc[1 + μ sin(ωmt)] sin(ωct)

Mnemonic: “Amplitude Modulation Makes Carrier Value Change”

Question 1(c) OR [7 marks]

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Define noise. Give classification of noise and explain cause of any three internal noise.

Answer:

Noise: Unwanted signals that interfere with communication signals, causing distortion or errors.

Classification of Noise:

Causes of internal noise:

• Thermal noise:

  • Caused by random motion of electrons in conductors
  • Present in all electronic components
  • Directly proportional to temperature and bandwidth

• Shot noise:

  • Occurs due to random arrival of carriers at junctions
  • Found in active devices like diodes and transistors
  • Proportional to DC current flowing through device

• Flicker noise:

  • Results from surface defects and impurities in semiconductors
  • Inversely proportional to frequency (1/f noise)
  • Significant at low frequencies

Mnemonic: “This Shooting Flicker Is Noisy Everywhere”

Question 2(a) [3 marks]

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Define (1) Modulation index for AM (2) Noise Figure (3) Digital Modulation

Answer:

1. Modulation index for AM: Ratio of amplitude of modulating signal to amplitude of carrier signal.

   • μ = Vm/Vc
   • Must be 0 ≤ μ ≤ 1 to avoid distortion

2. Noise Figure: Ratio of input SNR to output SNR of a device.

   • NF = (SNR)input/(SNR)output
   • Indicates noise added by the system
   • Always ≥ 1, expressed in dB

3. Digital Modulation: Technique that represents digital data as variations in carrier signal parameters.

   • Examples: ASK, FSK, PSK, QAM
   • Used for digital data transmission

Mnemonic: “Modulation Measures, Noise Numbers, Digital Data”

Question 2(b) [4 marks]

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Derive equation for total power transmitted for amplitude modulated signal considering carrier power and modulation index.

Answer:

Derivation of total power in AM:

1. AM wave equation: vAM(t) = Vc[1 + μ sin(ωmt)] sin(ωct)

2. For power calculation, consider RMS values:

   • Carrier power (Pc) = Vc²/2R
   • Power in each sideband (PSB) = (μ²Vc²)/(4R)

3. Total power equation:

   • PT = Pc + PUSB + PLSB
   • PT = Pc + 2PSB (since upper and lower sidebands have equal power)
   • PT = Vc²/2R + 2(μ²Vc²)/(4R)
   • PT = (Vc²/2R)[1 + (μ²/2)]

4. Final equation: PT = Pc(1 + μ²/2)

Mnemonic: “Power Total = Power Carrier (1 + μ²/2)”

Question 2(c) [7 marks]

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Explain basic principle of double sideband suppressed carrier amplitude modulation. Derive its voltage equation & draw only balanced modulator circuit using diode.

Answer:

Double Sideband Suppressed Carrier (DSBSC) Principle:

• Carrier is suppressed, only sidebands transmitted
• Contains all information in sidebands
• More power efficient than AM
• Requires complex receiver for demodulation

Voltage equation derivation:

1. AM signal: vAM(t) = Vc[1 + μ sin(ωmt)]sin(ωct)
2. Removing carrier component: vDSBSC(t) = Vc × μ sin(ωmt)sin(ωct)
3. Using trigonometric identity: sin(A)sin(B) = 0.5[cos(A-B) - cos(A+B)]
4. Final equation: vDSBSC(t) = (Vcμ/2)[cos(ωc-ωm)t - cos(ωc+ωm)t]

Balanced Modulator Circuit using Diodes:

Mnemonic: “Delete Carrier, Save Bandwidth, Combine Signals”

Question 2(a) OR [3 marks]

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Define only, w.r.t. radio receiver (1) Sensitivity (2) Selectivity (3) fidelity

Answer:

1. Sensitivity: Ability of a receiver to detect and amplify weak signals.

   • Measured in microvolts (μV)
   • Lower value indicates better sensitivity
   • Typically 1-10 μV for commercial receivers

2. Selectivity: Ability to distinguish between desired signal and adjacent interfering signals.

   • Measured as bandwidth at -3dB points
   • Narrower bandwidth means better selectivity
   • Prevents adjacent channel interference

3. Fidelity: Accuracy with which receiver reproduces original message.

   • Measures quality of reproduction
   • Affected by distortion and noise
   • Higher fidelity means better sound quality

Mnemonic: “Sensitive Selection Faithfully”

Question 2(b) OR [4 marks]

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An AM signal has carrier power of 1 KW with 200 watt in each sideband. Find out modulation index.

Answer:

Given:

• Carrier power (Pc) = 1 KW = 1000 W
• Power in each sideband (PSB) = 200 W

To find: Modulation index (μ)

Solution:

1. Total sideband power: PTSB = 2 × PSB = 2 × 200 = 400 W
2. Using formula: PTSB = Pc × μ²/2
3. 400 = 1000 × μ²/2
4. μ² = (400 × 2)/1000 = 800/1000 = 0.8
5. μ = √0.8 = 0.894 = 0.9 (approx)

Mnemonic: “Sideband Power Reveals Modulation μ”

Question 2(c) OR [7 marks]

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Compare Amplitude modulation with Frequency Modulation considering minimum seven parameters/aspect.

Answer:

Mnemonic: “Bandwidth, Efficiency, Noise, Quality - AM Fails Many Quality Tests”

Question 3(a) [3 marks]

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Draw and label sine wave of 1 KHZ in time domain and frequency domain. State advantage of frequency domain analysis of signal.

Answer:

Time Domain Representation:

Frequency Domain Representation:

Advantages of frequency domain analysis:

• Signal composition: Easily identifies frequency components
• Filter design: Simplified filter response analysis
• Bandwidth determination: Direct visualization of spectrum width
• Noise analysis: Better separation of signal from noise

Mnemonic: “Frequency Shows Components Hidden in Time”

Question 3(b) [4 marks]

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State following frequency (1) IF frequency for AM radio (2) IF frequency for FM radio (3) Frequency Band used in FM radio (4) Frequency Band of Human speech.

Answer:

Mnemonic: “AM455, FM10.7, Band88-108, Speech300-3.4”

Question 3(c) [7 marks]

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Explain Single side band (SSB) modulation with waveform and its advantages. Show how SSB transmission required only 1/6th of power with respect to double sideband full carrier amplitude modulation.

Answer:

Single Side Band (SSB) Modulation:

• Transmits only one sideband (USB or LSB)
• Carrier and other sideband suppressed
• Conserves bandwidth and power

SSB Waveform:

Advantages of SSB:

• Bandwidth efficiency: Uses half bandwidth of AM
• Power efficiency: No power wasted on carrier
• Less fading: Improved performance in long-distance
• Better SNR: More power concentrated in information

Power Comparison:

1. In AM: PT = Pc(1 + μ²/2)
2. For μ = 1, PT = Pc(1 + 0.5) = 1.5Pc
3. AM power distribution: Carrier (Pc) = 67%, Sidebands = 33%
4. SSB uses only one sideband with no carrier
5. SSB power = 16.5% of total AM power = 1/6 approx.

Mnemonic: “Single Side Saves Bandwidth And Power”

Question 3(a) OR [3 marks]

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State following. (1) Bandwidth of modulated signal if modulating frequency is 5 KHZ. (2) Image frequency if selected station frequency is 1000 KhZ in radio (3) Sampling frequency if baseband signal frequency is 10 KHz.

Answer:

1. Bandwidth of AM with modulating frequency 5 kHz:

   • BW = 2 × fm = 2 × 5 kHz = 10 kHz

2. Image frequency for 1000 kHz station with 455 kHz IF:

   • For high-side injection: fimage = fstation + 2 × fIF
   • fimage = 1000 + 2 × 455 = 1000 + 910 = 1910 kHz

3. Sampling frequency for 10 kHz baseband:

   • fs > 2 × fmax (Nyquist rate)
   • fs > 2 × 10 kHz = 20 kHz
   • Sampling frequency should be > 20 kHz

Mnemonic: “Bandwidth Doubles, Image Adds Twice-IF, Sampling Needs Twice-Frequency”

Question 3(b) OR [4 marks]

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Draw following signal stating its mathematical equation. (1) Sine wave (2) Unit step signal (3) Ramp signal (4) Impulse signal.

Answer:

1. Sine Wave:

• Equation: f(t) = A sin(ωt + φ)

2. Unit Step Signal:

• Equation: u(t) = 1 for t ≥ 0, 0 for t < 0

3. Ramp Signal:

• Equation: r(t) = t for t ≥ 0, 0 for t < 0

4. Impulse Signal:

• Equation: δ(t) = ∞ for t = 0, 0 for t ≠ 0

Mnemonic: “Sine Oscillates, Step Jumps, Ramp Climbs, Impulse Spikes”

Question 3(c) OR [7 marks]

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Draw and explain Pre emphasis and De emphasis circuit with its need & characteristic graph. Also compare FM receiver with AM receiver in detail.

Answer:

Pre-emphasis Circuit:

De-emphasis Circuit:

Characteristic Graph:

Need for Pre/De-emphasis:

• Noise reduction: Higher frequencies more susceptible to noise
• Improves SNR: Boosts high frequencies at transmitter, attenuates at receiver
• Time constant: Typically 75μs in FM broadcasting

Comparison between FM and AM Receiver:

Mnemonic: “Pre Boosts Highs, De Cuts Them; FM Filters Noise Better Than AM”

Question 4(a) [3 marks]

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

Answer:

Image Frequency: Unwanted signal frequency that produces the same IF as the desired signal when mixed with local oscillator signal.

Explanation:

• For high-side injection: fimage = fsignal + 2 × fIF
• For low-side injection: fimage = fsignal - 2 × fIF

Example:

• Desired signal: 1000 kHz
• IF: 455 kHz
• Local oscillator frequency (high-side): fLO = 1000 + 455 = 1455 kHz
• Image frequency: fimage = fLO + 455 = 1455 + 455 = 1910 kHz
• Both 1000 kHz and 1910 kHz will produce 455 kHz IF when mixed with 1455 kHz

Mnemonic: “Image In radio Is Interfering 2IF away”

Question 4(b) [4 marks]

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Draw and explain envelope detector circuit for demodulation of Amplitude modulated signal.

Answer:

Envelope Detector Circuit:

Working Principle:

• Diode: Rectifies AM signal, removing negative half-cycles
• RC Circuit: Acts as low-pass filter
• Time Constant: RC must satisfy: 1/fm » RC » 1/fc
• Output: Envelope of AM signal, which is the original message

Envelope Detection Process:

1. Diode conducts during positive half-cycles
2. Capacitor charges to peak value
3. During negative half-cycles, capacitor discharges through resistor
4. Output follows envelope of AM signal

Mnemonic: “Diode Rectifies, RC Smooths Envelope”

Question 4(c) [7 marks]

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

Answer:

AM Radio Receiver (Superheterodyne) Block Diagram:

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

Functions of each block:

• RF Amplifier:

  • Selects desired station signal using tuned circuit
  • Provides initial amplification
  • Improves sensitivity and selectivity
  • Reduces image frequency interference

• Local Oscillator:

  • Generates frequency higher than incoming by IF value
  • Typically fLO = fRF + 455 kHz
  • Tuned simultaneously with RF amplifier

• Mixer:

  • Combines RF signal with local oscillator
  • Produces sum and difference frequencies
  • Outputs intermediate frequency (IF)

• IF Amplifier:

  • Fixed-frequency amplifier (455 kHz)
  • Provides majority of receiver gain
  • Determines selectivity of receiver

• Detector:

  • Extracts original audio from IF signal
  • Usually envelope detector with diode
  • Removes RF component, recovers audio

• AF Amplifier:

  • Amplifies recovered audio signal
  • Includes volume control
  • Drives speaker to audible levels

• Speaker:

  • Converts electrical signals to sound waves

Mnemonic: “Radio Mixing Intermediate Detected Audio For Speaker”

Question 4(a) OR [3 marks]

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State and explain Nyquist Criteria for sampling of signal.

Answer:

Nyquist Criteria: To reconstruct a bandlimited signal without distortion, sampling frequency must be at least twice the highest frequency component in the signal.

Mathematical statement:

• fs ≥ 2fmax
• fs = sampling frequency
• fmax = maximum frequency in signal

Explanation:

• Ensures no aliasing (frequency overlap) occurs
• Minimum sampling rate called Nyquist rate
• Sampling below Nyquist rate causes irreversible distortion
• In practice, fs > 2.2fmax used to allow for filtering

Example:

• For audio with fmax = 20 kHz
• Nyquist rate = 2 × 20 kHz = 40 kHz
• CD sampling rate = 44.1 kHz (>40 kHz)

Mnemonic: “Sample at least Twice as Fast as Highest Frequency”

Question 4(b) OR [4 marks]

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Explain slope overload and granular noise for a delta modulation.

Answer:

Delta Modulation Issues:

graph TD
    A[Delta Modulation Problems] --> B[Slope Overload]
    A --> C[Granular Noise]
    B --> D[Step size too small]
    C --> E[Step size too large]

Slope Overload:

• Occurs when input signal changes faster than DM can track
• Step size too small for rapidly changing signals
• DM output cannot “catch up” with input
• Creates distortion at sharp transitions
• Solution: Increase step size or sampling rate

Granular Noise:

• Occurs during relatively constant signal portions
• Step size too large for slowly changing signals
• Output oscillates around input value
• Creates “roughness” in reconstructed signal
• Solution: Decrease step size

Adaptive Delta Modulation (ADM): Dynamically adjusts step size to minimize both problems.

Mnemonic: “Slopes Need Bigger Steps, Flats Need Smaller Steps”

Question 4(c) OR [7 marks]

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Draw and explain PCM transmitter and receiver in detail.

Answer:

PCM Transmitter:

graph LR
    A[Input Signal] --> B[Anti-aliasing Filter]
    B --> C[Sample & Hold]
    C --> D[Quantizer]
    D --> E[Encoder]
    E --> F[Digital Output]

PCM Receiver:

graph LR
    A[Digital Input] --> B[Decoder]
    B --> C[D/A Converter]
    C --> D[Reconstruction Filter]
    D --> E[Output Signal]

Transmitter Components:

• Anti-aliasing filter: Limits input bandwidth to prevent aliasing
• Sample & Hold: Captures instantaneous values at regular intervals
• Quantizer: Approximates samples to predefined discrete levels
• Encoder: Converts quantized values to binary code

Receiver Components:

• Decoder: Converts binary code back to quantized values
• D/A Converter: Transforms discrete values to continuous voltage
• Reconstruction filter: Removes sampling frequency components, smooths output

PCM Parameters:

• Resolution: Determined by bits per sample (n)
• Quantization levels: L = 2^n
• Bit rate: R = n × fs (bits per second)
• SNR: Improves by ~6dB per bit added

Mnemonic: “Sample, Quantize, Encode; Decode, Convert, Reconstruct”

Question 5(a) [3 marks]

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Define Bit, Bit rate and Baud rate with suitable example.

Answer:

• Bit: Smallest unit of digital information, representing either 0 or 1.

  • Example: 10110 contains 5 bits

• Bit Rate: Number of bits transmitted per second.

  • Unit: bps (bits per second)
  • Example: 9600 bps means 9600 bits transmitted in one second

• Baud Rate: Number of signal changes (symbols) per second.

  • Unit: Baud
  • Example: In QPSK, each symbol represents 2 bits, so 9600 bps = 4800 Baud

Relationship:

• Bit Rate = Baud Rate × number of bits per symbol
• For binary signaling (1 bit/symbol): Bit Rate = Baud Rate
• For multilevel coding: Bit Rate > Baud Rate

Mnemonic: “Bits Build Data, Baud Brings Symbols”

Question 5(b) [4 marks]

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Define multiplexing. State its types. Explain Frequency division multiplexing with suitable diagram.

Answer:

Multiplexing: Technique that allows multiple signals to share a common transmission medium.

Types of Multiplexing:

• Frequency Division Multiplexing (FDM)
• Time Division Multiplexing (TDM)
• Code Division Multiplexing (CDM)
• Wavelength Division Multiplexing (WDM)

Frequency Division Multiplexing:

FDM Working Principle:

• Each signal modulated to different carrier frequency
• Bandwidth allocated to each channel with guard bands
• All channels transmitted simultaneously
• Receiver uses filters to separate channels
• Used in radio/TV broadcasting, cable systems

Mnemonic: “Frequency Divides Multiple Signals Simultaneously”

Question 5(c) [7 marks]

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Draw and explain basic PCM-TDM diagram with diagram.

Answer:

PCM-TDM System Block Diagram:

graph LR
    %% Transmitter
    A1[Source 1] --> B1[LPF 1]
    A2[Source 2] --> B2[LPF 2]
    A3[Source 3] --> B3[LPF 3]
    B1 --> C[Commutator/MUX]
    B2 --> C
    B3 --> C
    C --> D[Sampler]
    D --> E[Quantizer] 
    E --> F[Encoder]
    F --> G[TDM Output]
    
    %% Receiver
    G --> H[Decoder]
    H --> I[DEMUX]
    I --> J1[LPF 1]
    I --> J2[LPF 2]
    I --> J3[LPF 3]
    J1 --> K1[Output 1]
    J2 --> K2[Output 2]
    J3 --> K3[Output 3]

PCM-TDM System Operation:

Transmitter Side:

• Input Sources: Multiple analog signals
• Low-Pass Filters: Limit bandwidth of input signals
• Commutator/MUX: Sequentially samples each input
• Sampler: Converts continuous signals to discrete samples
• Quantizer: Approximates samples to nearest discrete levels
• Encoder: Converts quantized values to binary code
• TDM Output: Transmits frames containing samples from all channels

Receiver Side:

• Decoder: Converts binary code back to quantized values
• DEMUX: Distributes samples to appropriate channel paths
• Low-Pass Filters: Reconstruct original signals, remove sampling components
• Outputs: Recovered original signals

TDM Frame Format:

Mnemonic: “Pulse Code TDM: Sample, Quantize, Encode, Multiplex”

Question 5(a) OR [3 marks]

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State types of TDM and explain any one of them.

Answer:

Types of TDM:

• Synchronous TDM
• Asynchronous TDM (Statistical TDM)
• Intelligent TDM

Synchronous TDM:

• Fixed time slots allocated to each channel
• Time slots transmitted in fixed sequence
• Time slots remain empty if channel has no data
• Simpler implementation but less efficient
• Example: T1 carrier system (24 channels × 8 bits × 8000 samples/sec = 1.544 Mbps)

Frame Structure:

Mnemonic: “Synchronous Slots Stay Steady”

Question 5(b) OR [4 marks]

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Explain TDM. Also State its advantages and disadvantages.

Answer:

Time Division Multiplexing (TDM): Technique where multiple signals share same transmission medium by allocating different time slots to each signal.

Working Principle:

• Each signal sampled at regular intervals
• Samples interleaved in time domain
• Complete frame contains one sample from each channel
• Receiver separates samples to reconstruct original signals

Advantages of TDM:

• Single medium: Efficiently uses one transmission path
• Digital compatibility: Naturally suits digital systems
• Crosstalk elimination: No interference between channels
• Flexible capacity: Easy to add/remove channels
• Cost-effective: Reduces hardware requirements

Disadvantages of TDM:

• Synchronization critical: Timing errors cause major problems
• Complex equipment: Requires precise timing circuits
• Bandwidth limitation: High bit rate needed for many channels
• Inefficiency: Wastes capacity when channels inactive (in synchronous TDM)
• Buffer delays: Can cause latency issues

Mnemonic: “Time Divided Multiple signals Save costs But Need Precise timing”

Question 5(c) OR [7 marks]

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State desirable properties of line coding. Draw waveform in time relation for unipolar RZ, Polar NRZ, and Manchester line coding for a 8 bit digital data 01001110.

Answer:

Desirable Properties of Line Coding:

• DC component: Should be minimal or absent
• Self-synchronization: Should provide timing information
• Error detection: Should allow detection of transmission errors
• Bandwidth efficiency: Should require minimum bandwidth
• Noise immunity: Should be resistant to noise and interference
• Cost & complexity: Should be simple to implement

Line Coding Waveforms for 01001110:

Key characteristics:

• Unipolar RZ: Returns to zero in middle of bit, only positive voltages
• Polar NRZ: No return to zero, uses positive and negative voltages
• Manchester: Mid-bit transition, rising edge = 0, falling edge = 1

Mnemonic: “Unipolar Rises then Zeros, Polar Never Returns, Manchester Always Transitions”