2024 question paper

Principles of Communication Engineering

24 questions

1. Q1a. What is the purpose of pre-emphasis in FM transmission? (i) Reduce low-frequency noise components (ii) Improve signal-to-noise ratio (iii) Suppress higher-frequency (iv) Increase bandwidth20242m

   Module 3: Probability & Random Processes

   What is the purpose of pre-emphasis in FM transmission?
   (i) Reduce low-frequency noise components
   (ii) Improve signal-to-noise ratio
   (iii) Suppress higher-frequency
   (iv) Increase bandwidth

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   Worked Solution

   Answer

   Correct Option: (ii) Improve signal-to-noise ratio (high-frequency boost)

   Explanation

   Pre-emphasis in FM is high-pass filtering (75 µs, cutoff ~2.1 kHz) at transmitter that boosts high-frequency components of $m(t)$ before modulation.

   • FM noise PSD at detector output ∝ $f^2$ (parabolic) → high frequencies noisier.
   • Boosting highs before TX → de-emphasis (low-pass) at receiver restores flat response while suppressing high-freq noise.

   Result: SNR improvement ~10 dB. Complementary to de-emphasis at receiver.

   Final Answer

   Improve SNR → Option (ii) (high-frequency noise reduction). If option (i) “reduce low-frequency noise” is opposite.

2. Q1b. The Viterbi algorithm is used for: (i) Encoding cyclic codes (ii) Decoding convolutional codes (iii) Noise filtering (iv) Error detection20242m

   Module 6: Information Coding and Decoding

   The Viterbi algorithm is used for:
   (i) Encoding cyclic codes
   (ii) Decoding convolutional codes
   (iii) Noise filtering
   (iv) Error detection

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   Worked Solution

   Answer

   Correct Option: (ii) Decoding convolutional codes

   Explanation

   Viterbi algorithm is maximum-likelihood trellis decoder for convolutional codes (and trellis-coded modulation). Uses dynamic programming over trellis states to find most likely path.

   Not for cyclic code encoding (LFSR), filtering, or pure error detection.

   Final Answer

   Decoding convolutional codes → Option (ii).

3. Q1c. The minimum sampling frequency required to avoid aliasing for a signal with a maximum frequency of 8 kHz is: (i) 8 kHz (ii) 12 kHz (iii) 16 kHz (iv) 20 kHz20242m

   Module 4: Pulse Modulation-I

   The minimum sampling frequency required to avoid aliasing for a signal with a maximum frequency of 8 kHz is:
   (i) 8 kHz
   (ii) 12 kHz
   (iii) 16 kHz
   (iv) 20 kHz

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   Answer

   Correct Option: (iii) $16$ kHz

   Explanation

   Nyquist Sampling Theorem: $f_s \ge 2f_m$ to avoid aliasing.

   Given $f_m = 8$ kHz → $f_s^{min}=2\times8=16$ kHz.

   Sampling below 16 kHz causes aliasing (foldover), reconstruction fails.

   Final Answer

   $16$ kHz → Option (iii).

4. Q1d. Which of the following is an advantage of PCM over analog modulation techniques? (i) Requires less bandwidth (ii) Higher noise immunity (iii) Simpler receiver design (iv) Lower power requirements20242m

   Module 4: Pulse Modulation-I

   Which of the following is an advantage of PCM over analog modulation techniques?
   (i) Requires less bandwidth
   (ii) Higher noise immunity
   (iii) Simpler receiver design
   (iv) Lower power requirements

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   Worked Solution

   Answer

   Correct Option: (ii) Higher noise immunity

   Explanation

   PCM quantizes PAM to binary, regenerative repeaters restore levels via threshold, noise does not accumulate. Analog PAM/AM accumulates noise.

   PCM requires more bandwidth ($n f_s$) and complex codec, but SNR greatly higher ($\approx6n$ dB). Not less bandwidth, not simpler.

   Final Answer

   Higher noise immunity → Option (ii).

5. Q1e. If a noise signal has a mean of 0 and a standard deviation of 1, what is its variance? (i) 0 (ii) 1 (iii) 2 (iv) 420242m

   Module 3: Probability & Random Processes

   If a noise signal has a mean of 0 and a standard deviation of 1, what is its variance?
   (i) 0
   (ii) 1
   (iii) 2
   (iv) 4

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   Answer

   Correct Option: (ii) $1$

   Explanation

   Variance $\sigma^2 = E[(X-\mu)^2]$. Standard deviation $\sigma =\sqrt{\text{variance}}$.

   Given $\sigma =1$ → variance $\sigma^2 =1^2=1$.

   Mean 0 irrelevant.

   Final Answer

   $1$ → Option (ii).

6. Q1f. Which of the following techniques is used to reduce noise in PCM? (i) Companding (ii) Differential PCM (iii) Delta Modulation (iv) FDM20242m

   Module 5: Pulse Modulation-II

   Which of the following techniques is used to reduce noise in PCM?
   (i) Companding
   (ii) Differential PCM
   (iii) Delta Modulation
   (iv) FDM

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   Worked Solution

   Answer

   Correct Option: (i) Companding

   Explanation

   Companding ($\mu$-law/A-law compress-expand) non-uniform quantization: small amplitudes fine steps, large coarse. Reduces quantization noise for small signals, improves SQNR vs uniform PCM. Differential PCM and Delta Modulation also reduce but companding is standard for PCM noise reduction.

   Final Answer

   Companding → Option (i).

7. Q1g. What is the main advantage of SSB over DSB? (i) Requires less power (ii) Better noise immunity (iii) Requires less bandwidth (iv) Both (i) & (ii)20242m

   Module 1: Analog Modulation

   What is the main advantage of SSB over DSB?
   (i) Requires less power
   (ii) Better noise immunity
   (iii) Requires less bandwidth
   (iv) Both (i) & (ii)

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   Worked Solution

   Answer

   Correct Option: (iii) Requires less bandwidth

   Explanation

   SSB vs DSB-SC (both suppressed carrier): DSB $2f_m$, SSB $f_m$ → 50% saving. Power: SSB saves half sideband power but main saving is bandwidth. Both save carrier power vs AM. Better noise immunity not primary; SSB still needs coherent detection.

   Final Answer

   Less bandwidth → Option (iii).

8. Q1h. A QPSK signal consists of how many distinct phase states? (i) 2 (ii) 4 (iii) 8 (iv) 1620242m

   Module 2: Digital Modulation

   A QPSK signal consists of how many distinct phase states?
   (i) 2
   (ii) 4
   (iii) 8
   (iv) 16

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   Worked Solution

   Answer

   Correct Option: (ii) $4$

   Explanation

   QPSK (Quadrature PSK): $M=4$, phases $0°,90°,180°,270°$ or $45°,135°,225°,315°$, 2 bits/symbol ($00,01,10,11$).

   Final Answer

   $4$ phase states → Option (ii).

9. Q1i. A frequency-modulated (FM) signal with a frequency deviation of 75 kHz and a modulating frequency of 5 kHz has a modulation index of: (i) 15 (ii) 50 (iii) 75 (iv) 10020242m

   Module 1: Analog Modulation

   A frequency-modulated (FM) signal with a frequency deviation of 75 kHz and a modulating frequency of 5 kHz has a modulation index of:
   (i) 15
   (ii) 50
   (iii) 75
   (iv) 100

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   Worked Solution

   Answer

   Correct Option: (i) $15$

   Explanation

   FM index $\beta=\Delta f/f_m=75\text{kHz}/5\text{kHz}=15$.

   Final Answer

   $15$ → Option (i).

10. Q1j. What is the primary factor that determines the Bit Error Rate (BER) in a communication system? (i) Power spectral density (ii) Bandwidth efficiency (iii) Signal-to-noise ratio (SNR) (iv) Carrier frequency20242m

    Module 2: Digital Modulation

    What is the primary factor that determines the Bit Error Rate (BER) in a communication system?
    (i) Power spectral density
    (ii) Bandwidth efficiency
    (iii) Signal-to-noise ratio (SNR)
    (iv) Carrier frequency

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    Worked Solution

    Answer

    Correct Option: Depends on phrasing — typically $E_b/N_0$ (SNR) determines BER

    Explanation

    BER error probability $P_e=Q(\sqrt{2E_b/N_0})$ for BPSK, depends on $E_b/N_0$ (energy per bit to noise density), which relates to PSD $N_0/2$ and bandwidth efficiency. Higher $E_b/N_0$ → lower BER.

    If options: power spectral density, bandwidth efficiency, SNR — SNR/$E_b/N_0$ is primary.

    Final Answer

    $E_b/N_0$ (SNR) → power spectral density related.

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12. Q2a. Explain the working principle of Phase Shift Keying (PSK) and its advantages over Amplitude Shift Keying (ASK).20247m

    Module 2: Digital Modulation

    Explain the working principle of Phase Shift Keying (PSK) and its advantages over Amplitude Shift Keying (ASK).

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    Worked Solution

    Answer

    PSK Phase Shift Keying encodes bits in carrier phase, amplitude constant.

    Principle: $s(t)=A\cos(\omega_c t+\theta(t))$, BPSK $\theta=0°$ for 0, $180°$ for 1. Generation: bits → NRZ ±1 → multiply carrier. Coherent detection correlates with reference.

    Advantages over ASK:

    	ASK	PSK
    Amplitude	Varies → fading sensitive	Constant → robust
    Power	Wasteful	Constant envelope efficient
    BER	$Q(\sqrt{E_b/N_0})$	$Q(\sqrt{2E_b/N_0})$ 3dB better
    Noise	Amplitude noise errors	Phase noise only

    Final: PSK 3dB better, constant power, preferred for noisy links.

13. Q2b. Discuss the impact of white noise in amplitude modulation systems and derive the signal-to-noise ratio (SNR).20247m

    Module 3: Probability & Random Processes

    Discuss the impact of white noise in amplitude modulation systems and derive the signal-to-noise ratio (SNR).

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    Answer

    White noise $n(t)$ PSD $N_0/2$ flat, $R(\tau)=N_0/2\delta(\tau)$.

    In AM, received $y=A_c[1+\mu m(t)]\cos\omega_c t + n(t)$. After envelope detection, noise in-phase $n_I$ and quadrature $n_Q$.

    Input SNR $\gamma_i=(A_c^2/2)/(N_0B)$, $B=2f_m$.

    Output SNR $\gamma_o = \frac{\mu^2P_m}{1+\mu^2P_m}\gamma_i$, figure $\gamma<1$ (AM wastes carrier). DSB-SC $\gamma=1$.

    Derivation: noise power $N_0B$, signal $A_c^2/2+\overline{m^2}/2$, LPF extracts baseband. Threshold ~10dB.

14. Q3a. Derive the bit error rate (BER) expressions for PSK and FSK under additive white Gaussian noise (AWGN).20247m

    Module 2: Digital Modulation

    Derive the bit error rate (BER) expressions for PSK and FSK under additive white Gaussian noise (AWGN).

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    Worked Solution

    Answer

    AWGN BER: BPSK coherent $P_e=Q(\sqrt{2E_b/N_0})$, distance $2\sqrt{E_b}$. BFSK orthogonal $P_e=Q(\sqrt{E_b/N_0})$, non-coherent $0.5e^{-E_b/2N_0}$. BPSK 3dB better than FSK. Derivation threshold 0 decision.

15. Q3b. If an AM transmitter operates at 1000 kHz carrier frequency with a total power of 10 kW and modulation index of 0.8, determine the power in the carrier and sidebands.20247m

    Module 1: Analog Modulation

    If an AM transmitter operates at 1000 kHz carrier frequency with a total power of 10 kW and modulation index of 0.8, determine the power in the carrier and sidebands.

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    Worked Solution

    Answer

    Given $P_t=10$kW $\mu=0.8$: $P_t=P_c(1+\mu^2/2)$ → $P_c=10/1.32=7.58$kW, $P_{SB}=2.42$kW total, each 1.21kW. Efficiency $\mu^2/(2+\mu^2)=24.2\%$.

16. Q4a. Explain the concept of probability density function (PDF) and cumulative distribution function (CDF) in random processes.20247m

    Module 3: Probability & Random Processes

    Explain the concept of probability density function (PDF) and cumulative distribution function (CDF) in random processes.

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    Answer

    PDF $f_X(x)$: $P(x<X\le x+dx)=f_X(x)dx$, $\int f=1$, $F_X(x)=\int_{-\infty}^x f$. CDF $F_X(x)=P(X\le x)$: monotonic 0→1, $f=dF/dx$. Gaussian $f=(1/\sqrt{2\pi}\sigma)e^{-(x-\mu)^2/2\sigma^2}$, $F=\Phi((x-\mu)/\sigma)$. Use for BER via $Q$.

17. Q4b. Compare Pulse Amplitude Modulation (PAM), Pulse Position Modulation (PPM) and Pulse Code Modulation (PCM) with their applications.20247m

    Module 4: Pulse Modulation-I

    Compare Pulse Amplitude Modulation (PAM), Pulse Position Modulation (PPM) and Pulse Code Modulation (PCM) with their applications.

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    Answer

    PAM: Sample amplitude ∝ $m(t)$, generation via FET switch, BW $f_s$, noise-sensitive. PPM: Pulse position varies, constant amplitude, better than PAM, needs sync. PCM: PAM→quantize→binary (8-bit), BW $n f_s$ large, high immunity, TDM. Applications: PAM backbone, PPM telemetry, PCM telephony 64kbps, CD. Table BW $f_s$, $f_s$, $n f_s$; noise low, medium, high.

18. Q5b. Explain cyclic redundancy check (CRC) and its application in digital communication.20247m

    Module 6: Information Coding and Decoding

    Explain cyclic redundancy check (CRC) and its application in digital communication.

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    Answer

    CRC: Polynomial division $r(x)=m(x)x^r \bmod g(x)$, transmit $m||r$. Generation LFSR taps $g(x)=x^3+x+1$ CRC-3. Detection divide $y(x)/g(x)$ remainder 0 OK else error. Detects burst ≤$r$ bits. Uses Ethernet, HDLC, disk. Overhead $r$ bits per $k$ data.

19. Q6a. A carrier signal c(t) = 10cos(2π x 10^6t) is amplitude modulated by a message signal m(t)= 5 cos(2πx10^3t). Determine the modulation index and sketch the frequency spectrum.20247m

    Module 1: Analog Modulation

    A carrier signal c(t) = 10cos(2π x 10^6t) is amplitude modulated by a message signal m(t)= 5 cos(2πx10^3t). Determine the modulation index and sketch the frequency spectrum.

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    Answer

    Given $c=10\cos2\pi10^6t$, $m=5\cos2\pi10^3t$, $A_m=5$, $A_c=10$, $\mu=A_m/A_c=0.5$ (50%). Envelope $A_c[1+0.5\cos\omega_m t]$ max15 min5. $P_c=50$W, $P_{SB}=6.25$W total.

20. Q6b. Explain the concept of Differential Pulse Code Modulation (DPCM) and its advantages over PCM.20247m

    Module 5: Pulse Modulation-II

    Explain the concept of Differential Pulse Code Modulation (DPCM) and its advantages over PCM.

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    Answer

    DPCM: Quantize difference $d[n]=x[n]-\hat{x}[n]$, predictor $\hat{x}=a_1\tilde{x}[n-1]$. Block $x$→subtract predictor→Quantizer→Encoder→Channel→Decoder→predictor loop. Variance of $d$ << variance of $x$ → fewer bits, prediction gain $G_p=\sigma_x^2/\sigma_d^2$. Better SNR for correlated speech/video vs PCM.

21. Q7a. Compare Amplitude Modulation (AM) and Frequency Modulation (FM) in terms of bandwidth, power efficiency, and noise immunity.20247m

    Module 1: Analog Modulation

    Compare Amplitude Modulation (AM) and Frequency Modulation (FM) in terms of bandwidth, power efficiency, and noise immunity.

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    Answer

    AM vs FM: AM BW $2f_m$ narrow, power 33% max carrier wasteful, noise poor envelope. FM BW $2(\Delta f+f_m)$ wide, constant envelope efficient, capture effect, high noise immunity ~10dB better, complex VCO/discriminator. FM for VHF high fidelity, AM MW simple.

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23. Q7b. If a Delta Modulation system has step size of 0.1V and a sampling rate of 10 kHz, determine the maximum input signal frequency without slope overload distortion.20247m

    Module 5: Pulse Modulation-II

    If a Delta Modulation system has step size of 0.1V and a sampling rate of 10 kHz, determine the maximum input signal frequency without slope overload distortion.

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    Answer

    Slope overload condition: $|dm/dt|_{max}\le\Delta f_s$. For $m=A_m\cos\omega_m t$, $|dm/dt|_{max}=A_m\omega_m$. So $f_m\le\Delta f_s/(2\pi A_m)$. Given $\Delta=0.1$V $f_s=10$kHz $A_m=1$V → $f_{max}=159$Hz. General $f_{max}=\Delta f_s/(2\pi A_m)$.

24. Q8. A (7,4) Hamming code is used for error correction. If the received codeword is 1011010, determine the original data and correct the error.202414m

    Module 6: Information Coding and Decoding

    A (7,4) Hamming code is used for error correction. If the received codeword is 1011010, determine the original data and correct the error.

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    Answer

    Hamming (7,4): Parity bits at 1,2,4. Received 1011010, even parity checks: $p_1=1\oplus1\oplus0\oplus0=0$, $p_2=0\oplus1\oplus1\oplus0=0$, $p_4=1\oplus0\oplus1\oplus0=0$ syndrome 000 → no error. Data bits 3,5,6,7 =1010. If odd parity syndrome 111 → position 7 error. Method: syndrome binary = error position, flip.

25. Q9a. Write short note on: Huffman Codes20247m

    Module 6: Information Coding and Decoding

    Write short note on: Huffman Codes

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    Answer

    Huffman: Variable-length prefix optimal. Merge two smallest probabilities iteratively, assign 0/1. Example 0.4,0.3,0.2,0.1→ codes 0,10,110,111 average 1.9 bits near entropy 1.84. ZIP, JPEG. Prefix-free uniquely decodable.

26. Q9d. Write short note on: Superheterodyne receiver20247m

    Module 1: Analog Modulation

    Write short note on: Superheterodyne receiver

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    Answer

    Superheterodyne: Converts RF to fixed IF $f_{IF}=|f_{LO}-f_{RF}|$ via mixer+LO. IF amp 455kHz AM, 10.7MHz FM, sharp IF filter gives selectivity, sensitivity. RF amp→Mixer←LO→IF amp→Detector→AF. Image $f_{im}=f_{RF}+2f_{IF}$ needs preselection.