2025 question paper

Principles of Communication Engineering

26 questions

1. Q1a. In amplitude modulation, the total bandwidth required is (i) fm (ii) 2fm (iii) fc + fm (iv) fc - fm20252m

   Module 1: Analog Modulation

   In amplitude modulation, the total bandwidth required is
   (i) fm
   (ii) 2fm
   (iii) fc + fm
   (iv) fc - fm

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   Answer

   Correct Option: (ii) $2f_m$

   Explanation

   Amplitude Modulation: $s(t)=A_c[1+\mu m(t)]\cos\omega_c t$ with baseband $m(t)$ bandwidth $f_m$.

   Spectrum: carrier at $f_c$ plus upper sideband $f_c+f_m$ and lower $f_c-f_m$.

   Total transmission bandwidth:
   $$B_T = (f_c+f_m)-(f_c-f_m)=2f_m$$

   If $f_m=5$ kHz, $B=10$ kHz.

   Other options: $f_m$ would be SSB, $f_c\pm f_m$ are sideband frequencies, not bandwidth.

   Final Answer

   AM bandwidth $=2f_m$ → Option (ii).

2. Q1b. In FM, the parameter varied in proportion to the message signal is (i) Amplitude (ii) Frequency (iii) Phase only (iv) Amplitude and phase20252m

   Module 1: Analog Modulation

   In FM, the parameter varied in proportion to the message signal is
   (i) Amplitude
   (ii) Frequency
   (iii) Phase only
   (iv) Amplitude and phase

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   Answer

   Correct Option: (ii) Frequency

   Explanation

   Frequency Modulation: Instantaneous frequency $f_i(t)=f_c+k_f m(t)$ varied proportional to message $m(t)$. Amplitude $A_c$ constant, phase $\theta(t)=2\pi\int f_i dt$. PM varies phase, FM varies frequency (integral of PM).

   Final Answer

   Frequency varied → Option (ii).

3. Q1c. SSB modulation saves (i) Power (ii) Bandwidth (iii) Both power and bandwidth (iv) Only modulation index20252m

   Module 1: Analog Modulation

   SSB modulation saves
   (i) Power
   (ii) Bandwidth
   (iii) Both power and bandwidth
   (iv) Only modulation index

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   Answer

   Correct Option: (iii) Both power and bandwidth

   Explanation

   DSB-SC transmits both sidebands ($2f_m$) with suppressed carrier → saves power vs AM but still $2f_m$ bandwidth.

   SSB suppresses one sideband + carrier → saves both $50\%$ bandwidth ($f_m$) and carrier power.

   Final Answer

   Both power and bandwidth → Option (iii).

4. Q1d. VSB modulation is mainly used in (i) AM radio (ii) FM radio (iii) TV video transmission (iv) Satellite uplink20252m

   Module 1: Analog Modulation

   VSB modulation is mainly used in
   (i) AM radio
   (ii) FM radio
   (iii) TV video transmission
   (iv) Satellite uplink

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   Answer

   Correct Option: (iii) TV video transmission

   Explanation

   VSB compromises: one sideband fully + vestige of other. Bandwidth $f_m+f_v$ ($f_v\approx0.75$ MHz). SSB filter too sharp for video $4.2$ MHz, DSB wastes $8.4$ MHz. VSB saves ~50% vs DSB with practical filter.

   TV video uses VSB (AM-VSB). AM radio uses AM, FM radio uses FM, satellite uses FM.

   Final Answer

   TV video → Option (iii).

5. Q1e. BER stands for (i) Binary error range (ii) Bit energy ratio (iii) Bit error rate (iv) Bandwidth efficiency rate20252m

   Module 2: Digital Modulation

   BER stands for
   (i) Binary error range
   (ii) Bit energy ratio
   (iii) Bit error rate
   (iv) Bandwidth efficiency rate

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   Answer

   Correct Option: (iii) Bit error rate

   Explanation

   BER = Bit Error Rate = bits in error / total bits transmitted, measure of link quality vs $E_b/N_0$.

   Not binary error range, bit energy ratio ($E_b$), or bandwidth efficiency ($\eta$).

   Final Answer

   Bit error rate → Option (iii).

6. Q1f. The power spectral density of white noise is (i) Zero (ii) Constant (iii) Exponential (iv) Parabolic20252m

   Module 3: Probability & Random Processes

   The power spectral density of white noise is
   (i) Zero
   (ii) Constant
   (iii) Exponential
   (iv) Parabolic

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   Answer

   Correct Option: (ii) Constant

   Explanation

   White noise $n(t)$ PSD $S_n(f)=N_0/2$ flat over $-\infty$ to $\infty$, autocorrelation $R(\tau)=N_0/2\delta(\tau)$. Zero mean Gaussian.

   Not zero, exponential, parabolic.

   Final Answer

   Constant → Option (ii).

7. Q1g. Cyclic codes are represented using (i) Frequency response (ii) Generator polynomial (iii) Trellis diagram (iv) Fourier series20252m

   Module 6: Information Coding and Decoding

   Cyclic codes are represented using
   (i) Frequency response
   (ii) Generator polynomial
   (iii) Trellis diagram
   (iv) Fourier series

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   Worked Solution

   Answer

   Correct Option: (ii) Generator polynomial

   Explanation

   Cyclic codes $(n,k)$ are ideals in $\mathbb{F}_2[x]/(x^n-1)$, generated by $g(x)$ dividing $x^n-1$. Codewords $c(x)=m(x)g(x)$. Generator $g(x)$ degree $n-k$, parity check $h(x)=(x^n-1)/g(x)$. Example Hamming (7,4) $g(x)=x^3+x+1$.

   Not frequency response, trellis, Fourier.

   Final Answer

   Generator polynomial → Option (ii).

8. Q1h. Error detection in block coding is based on (i) Modulation index (ii) Parity checking and syndrome (iii) Quantization noise (iv) Clock synchronization20252m

   Module 6: Information Coding and Decoding

   Error detection in block coding is based on
   (i) Modulation index
   (ii) Parity checking and syndrome
   (iii) Quantization noise
   (iv) Clock synchronization

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   Worked Solution

   Answer

   Correct Option: (ii) Parity checking and syndrome

   Explanation

   Block coding $(n,k)$: $r=n-k$ parity bits, generator $G$, parity $H$, syndrome $s=yH^T$. Error detection via syndrome non-zero. Not modulation index, quantization noise, clock sync.

   Final Answer

   Parity/syndrome → Option (ii).

9. Q1i. The minimum Hamming distance required for single-bit error correction is (i) 1 (ii) 2 (iii) 3 (iv) 420252m

   Module 6: Information Coding and Decoding

   The minimum Hamming distance required for single-bit error correction is
   (i) 1
   (ii) 2
   (iii) 3
   (iv) 4

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   Answer

   Correct Option: (iii) $3$

   Explanation

   Single-bit correction: $t=1$ → Hamming distance $d_{min}\ge2t+1=3$. Detect $t$ errors needs $d\ge t+1$.

   Final Answer

   $3$ → Option (iii).

10. Q1j. Which modulation uses the position of a pulse to convey information? (i) PAM (ii) PCM (iii) PPM (iv) PWM20252m

    Module 4: Pulse Modulation-I

    Which modulation uses the position of a pulse to convey information?
    (i) PAM
    (ii) PCM
    (iii) PPM
    (iv) PWM

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    Worked Solution

    Answer

    Correct Option: (iii) PPM

    Explanation

    PAM varies amplitude, PWM varies width, PPM varies position of pulse relative to reference. PPM constant amplitude → better noise immunity vs PAM.

    Final Answer

    PPM → Option (iii).

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12. Q2a. Explain in detail the frequency-domain representation of signals.20257m

    Module 1: Analog Modulation

    Explain in detail the frequency-domain representation of signals.

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    Worked Solution

    Answer

    Frequency-domain representation shows signal spectral content via Fourier.

    Transform

    $X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt$.

    • Periodic: Fourier Series → line spectrum at $nf_0$.
    • Aperiodic: Continuous $X(f)$.
    • AM: $s(t)=A_c[1+\mu m(t)]\cos\omega_c t$ → impulses at $\pm f_c$ plus sidebands $M(f\pm f_c)$.
    • FM: Bessel sidebands $J_n(\beta)$.

    Importance

    Reveals bandwidth, filter design, modulation analysis.

    Final: Time → frequency via Fourier series/transform.

13. Q2b. Discuss Single Sideband (SSB) and Vestigial Sideband (VSB) modulation in detail.20257m

    Module 1: Analog Modulation

    Discuss Single Sideband (SSB) and Vestigial Sideband (VSB) modulation in detail.

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    Worked Solution

    Answer

    SSB vs VSB

    SSB: Suppress carrier + one sideband. BW = $f_m$. Generation via sharp filter or phase shift (Hilbert). Coherent detection. Saves 50% BW and power. Need sharp filter, carrier reinserted.

    VSB: One sideband fully + vestige of other. BW = $f_m+f_v$ (TV: $f_v\approx0.25$ MHz). VSB filter gradual. Used for TV video (SSB filter too sharp for 4.2 MHz video).

    	SSB	VSB
    BW	$f_m$	$f_m+f_v$
    Filter	Very sharp	Gradual
    Complexity	High	Moderate
    Use	Telephony	TV

    Final: SSB max saving, VSB practical for wideband.

14. Q3a. Explain Angle Modulation with representation of FM and PM signals.20257m

    Module 1: Analog Modulation

    Explain Angle Modulation with representation of FM and PM signals.

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    Answer

    Angle Modulation: $s(t)=A_c\cos(\omega_c t+\phi(t))$. FM: $\phi(t)=k_f\int m$, $s_{FM}=A_c\cos(\omega_c t+k_f\int m)$, $\beta=\Delta f/f_m$. PM: $\phi=k_p m$, $s_{PM}=A_c\cos(\omega_c t+k_p m)$. Bessel sidebands, Carson $B=2(\Delta f+f_m)$.

15. Q3b. Describe the generation and detection of FM.20257m

    Module 1: Analog Modulation

    Describe the generation and detection of FM.

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    Answer

    FM Generation: Direct VCO (varactor) or Armstrong (NBFM then multiply). Detection: Slope discriminator (FM→AM+envelope), PLL tracking, zero-crossing. Block limiter→discriminator→LPF→$m(t)$.

16. Q4a. Explain the fundamentals of a digital communication system with a neat block diagram.20257m

    Module 2: Digital Modulation

    Explain the fundamentals of a digital communication system with a neat block diagram.

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    Answer

    Digital Communication System: Source→Source Encoder (compression, Huffman)→Channel Encoder (error control, Hamming/conv, adds parity)→Modulator (ASK/PSK/FSK/QAM, digital→analog)→Channel (AWGN)→Demodulator→Channel Decoder→Source Decoder→Destination. Metrics $R_b$, BER, $E_b/N_0$. Source removes redundancy, channel adds controlled redundancy.

17. Q4b. Compare analog and digital communication systems.20257m

    Module 2: Digital Modulation

    Compare analog and digital communication systems.

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    Answer

    Analog vs Digital: Analog continuous, accumulates noise, low BW, simple, FDM, hard encryption. Digital discrete samples+quantized, regenerative threshold, high immunity, higher BW (PCM $n f_s$), complex codec, TDM efficient, easy encryption. Digital dominates modern due to robustness despite BW cost.

18. Q5a. Describe the basic nomenclature used in digital communication.20257m

    Module 2: Digital Modulation

    Describe the basic nomenclature used in digital communication.

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    Answer

    Nomenclature: Source/destination, message $m(t)$, bits/symbols, $R_b$ bps, baud $R_s=R_b/\log_2M$, bandwidth $B$ Hz, channel AWGN $N_0$, modulation ASK/PSK/QAM, BER $P_e$, SNR $E_b/N_0$, constellation I-Q, encoding source vs channel.

19. Q5b. Define power spectral density and bandwidth efficiency. Explain their importance in digital modulation.20257m

    Module 2: Digital Modulation

    Define power spectral density and bandwidth efficiency. Explain their importance in digital modulation.

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    Answer

    PSD $S(f)$: Power per Hz $P=\int S(f)df$, via autocorrelation $R(\tau)$. White $N_0/2$ flat, BPSK sinc. $\eta=R_b/B$ bps/Hz, Nyquist $\le2\log_2M$, Shannon $\le\log_2(1+SNR)$. PSD guides interference/filter, $\eta$ measures efficiency. Trade higher $\eta$ needs more power.

20. Q6a. Explain the basics of probability and random processes used in communication systems.20257m

    Module 3: Probability & Random Processes

    Explain the basics of probability and random processes used in communication systems.

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    Answer

    **Probability $P(A)\in[0,1]$, $P(A|B)=P(A\cap B)/P(B)$. RV $X$, PDF $f_X$, CDF $F_X$, process $X(t)$ ensemble, mean $\mu(t)=E[X]$, $R(t_1,t_2)=E[X_1X_2]$. Stationary: stats independent shift, WSS mean const $R(\tau)$. Ergodic time=ensemble. AWGN zero-mean Gaussian $N_0/2$. Use for BER via $Q$.

21. Q6b. What are pre-emphasis and de-emphasis networks? Explain their need in communication systems.20257m

    Module 3: Probability & Random Processes

    What are pre-emphasis and de-emphasis networks? Explain their need in communication systems.

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    Answer

    Pre-emphasis: High-pass at transmitter boosting highs ($75\mu$s) before FM, high-freq noise dominates. De-emphasis low-pass at receiver restores flat and suppresses high-freq noise. Improves SNR ~10dB. Block $m$→HPE→FM→chan→FM demod→LPF→$m$.

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23. Q7a. Explain the generation and detection of Pulse Amplitude Modulation (PAM).20257m

    Module 4: Pulse Modulation-I

    Explain the generation and detection of Pulse Amplitude Modulation (PAM).

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    Answer

    PAM: Sample via FET switch (natural) or S/H (flat-top). Detection via LPF $f_s/2$ with $\text{sinc}$ equalizer. Spectrum replicated at $nf_s$ weighted $\text{sinc}$. Applications intermediate for PCM, TDM.

24. Q7b. Compare PAM, PPM, and PCM in terms of bandwidth, noise immunity, complexity, and applications.20257m

    Module 4: Pulse Modulation-I

    Compare PAM, PPM, and PCM in terms of bandwidth, noise immunity, complexity, and applications.

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    Answer

    Comparison PAM/PPM/PCM: PAM amplitude BW $f_s$ low noise, PPM position BW $f_s$ medium, PCM binary $n f_s$ high immunity. PAM TDM backbone, PPM optical/control, PCM telephony 64kbps CD. PCM most robust but BW heavy.

25. Q8a. Explain Differential Pulse Code Modulation (DPCM) with a neat block diagram.20257m

    Module 5: Pulse Modulation-II

    Explain Differential Pulse Code Modulation (DPCM) with a neat block diagram.

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    Answer

    DPCM block: $x[n]$→subtract predictor→Quantizer→Encoder→Channel→Decoder→predictor loop. $e[n]=x[n]-\hat{x}[n]$, $\tilde{x}=\hat{x}+e_q$. Exploits correlation, fewer bits. See Q45.

26. Q8b. Describe Adaptive Delta Modulation (ADM) and explain how it reduces distortion.20257m

    Module 5: Pulse Modulation-II

    Describe Adaptive Delta Modulation (ADM) and explain how it reduces distortion.

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    Answer

    ADM: Adaptive step $\Delta[n+1]=\Delta[n]\cdot K$ if consecutive bits same (slope overload) increase, if alternating decrease. Handles steep and flat. SNR +10dB vs DM, wider dynamic range.

27. Q9a. Explain the need for error detection and correction in digital communication systems.20257m

    Module 6: Information Coding and Decoding

    Explain the need for error detection and correction in digital communication systems.

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    Answer

    Need for error control: Channel noise causes bit errors → unreliable. Techniques: Detection (parity, CRC) + ARQ, Correction (FEC Hamming, convolutional, Turbo) with $t=\lfloor(d_{min}-1)/2\rfloor$. Ensures reliability.

28. Q9b. Describe Shannon-Fano and Huffman coding techniques. Compare their coding efficiencies.20257m

    Module 6: Information Coding and Decoding

    Describe Shannon-Fano and Huffman coding techniques. Compare their coding efficiencies.

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    Answer

    Shannon-Fano top-down split near equal sum, not always optimal. Huffman bottom-up merging smallest, optimal. Huffman minimal expected length $L=\sum p_i l_i$, efficiency $\eta=H/L$ higher. Both prefix-free.