🇮🇳 UPSC ESE E&T · subject

UPSC ESE E&T Communication Systems Syllabus

The Communication Systems syllabus for UPSC ESE E&T in full: 3 chapters and 12 topics, every one listed below, plus 50 Communication Systems flashcards with answers.

3Chapters
12Topics
50Flashcards

Communication Systems syllabus: all 3 chapters

Expand any chapter to see its topics and sub-topics.

  1. Analog Communication

    4 topics
    • Amplitude Modulation
    • Frequency Modulation
    • Phase Modulation
    • Transmitters and Receivers
  2. Digital Communication

    4 topics
    • Pulse Code Modulation
    • Digital Modulation Techniques
    • Error Control Coding
    • Data Compression
  3. Information Theory and Coding

    4 topics
    • Entropy
    • Channel Capacity
    • Source Coding
    • Error Detection and Correction

UPSC ESE E&T Communication Systems flashcards — 50 questions with answers

  1. What is amplitude modulation (AM)?

    A modulation scheme in which the amplitude (envelope) of a high-frequency carrier is varied in proportion to the instantaneous amplitude of the message signal, while the carrier frequency and phase remain constant.

  2. Define modulation index (m) for an AM wave and its standard expression.

    The modulation index m = Am/Ac, the ratio of message amplitude to carrier amplitude. Equivalently m = (Vmax - Vmin)/(Vmax + Vmin). For distortionless AM, 0 < m ≤ 1; m > 1 causes overmodulation and envelope distortion.

  3. What fraction of total transmitted power lies in the sidebands of a standard AM (DSB-FC) wave at 100% modulation?

    At m = 1, total power Pt = Pc(1 + m²/2) = 1.5Pc, so sidebands carry Pc/2 out of 1.5Pc = 1/3 (about 33.3%); the carrier carries 2/3 (66.7%) and conveys no information.

  4. What is the transmission bandwidth of a standard AM (DSB) signal?

    BW = 2·fm, twice the highest message frequency, because both upper and lower sidebands are transmitted.

  5. What is the power and bandwidth advantage of SSB-SC over standard AM?

    Single-sideband suppressed carrier transmits only one sideband and no carrier, so it needs half the bandwidth (BW = fm instead of 2fm) and uses far less power, since all transmitted power conveys information.

  6. In DSB-SC, why is coherent (synchronous) detection required instead of envelope detection?

    Because the carrier is suppressed, the envelope no longer follows the message and reverses phase at zero crossings; recovering the message requires multiplying by a locally generated carrier of identical frequency and phase, then low-pass filtering.

  7. Define frequency modulation (FM).

    An angle modulation scheme in which the instantaneous frequency of the carrier is varied linearly with the instantaneous amplitude of the message signal, while the carrier amplitude stays constant.

  8. Write the expression for frequency deviation and modulation index of an FM signal.

    Frequency deviation Δf = kf·Am (kf is the frequency sensitivity). FM modulation index β = Δf/fm, the ratio of peak frequency deviation to the modulating frequency.

  9. State Carson's rule for the bandwidth of an FM signal.

    BW ≈ 2(Δf + fm) = 2fm(β + 1), where Δf is peak frequency deviation, fm is the highest message frequency, and β is the modulation index.

  10. Distinguish narrowband FM (NBFM) from wideband FM (WBFM).

    NBFM has β << 1 (typically β < 1), occupies a bandwidth ≈ 2fm similar to AM, and has essentially one significant sideband pair. WBFM has β >> 1, occupies a much larger bandwidth (per Carson's rule), and gives better noise immunity.

  11. Why does FM provide better noise immunity than AM?

    Because information is carried in frequency, not amplitude, amplitude disturbances (noise) can be removed by a limiter before detection. FM also benefits from wider bandwidth and capture effect, improving SNR.

  12. What is the FM capture effect?

    The tendency of an FM receiver to suppress the weaker of two signals at the same frequency; the stronger signal 'captures' the demodulator, so a relatively small signal-strength advantage results in the weaker signal being largely rejected.

  13. What are pre-emphasis and de-emphasis in FM, and why are they used?

    Pre-emphasis boosts high-frequency components of the message before transmission; de-emphasis attenuates them by the inverse amount at the receiver. This improves the high-frequency SNR because FM noise power rises with frequency (parabolic noise spectrum).

  14. Define phase modulation (PM).

    An angle modulation scheme in which the instantaneous phase of the carrier is varied linearly with the instantaneous amplitude of the message signal, while the carrier amplitude remains constant.

  15. State the relationship between FM and PM.

    PM and FM are closely related forms of angle modulation: FM of a message m(t) is equivalent to PM of its integral, and PM of m(t) is equivalent to FM of its derivative. One can be generated from the other using an integrator or differentiator.

  16. What is the modulation index of a PM signal?

    β_PM = kp·Am, where kp is the phase sensitivity (rad/volt) and Am is the peak message amplitude. Unlike FM, the PM index is independent of the modulating frequency fm.

  17. How does the modulation index of FM versus PM vary with modulating frequency?

    In FM, β = Δf/fm decreases as fm increases (inversely proportional). In PM, β = kp·Am is independent of fm. This difference distinguishes the two schemes experimentally.

  18. What is the difference between direct and indirect (Armstrong) methods of FM generation?

    Direct method varies a VCO/reactance modulator with the message (good deviation but poor carrier stability). Indirect (Armstrong) method generates NBFM via phase modulation of a crystal-stable carrier, then uses frequency multipliers to obtain WBFM (excellent stability, more complex).

  19. What is the function of a transmitter in a communication system?

    To convert the information (baseband) signal into a form suitable for transmission over a channel—typically by modulation onto a carrier, amplification, and coupling to the antenna/medium so the signal can travel efficiently with adequate power.

  20. What is a superheterodyne receiver and its key advantage?

    A receiver that converts every incoming RF signal to a fixed intermediate frequency (IF) using a local oscillator and mixer before detection. Its advantage is that amplification, selectivity, and filtering occur at a constant IF, giving uniform gain and selectivity across the tuning band.

Show the other 30 Communication Systems flashcards
  1. In a superheterodyne receiver, what is the image frequency and how is it rejected?

    The image is the unwanted frequency located 2·IF away from the desired signal (fimage = fsignal + 2·IF for high-side injection) that also produces the IF. It is rejected by RF-stage selectivity (preselector); a higher IF improves image rejection.

  2. Define sensitivity and selectivity of a radio receiver.

    Sensitivity is the receiver's ability to amplify and detect weak signals (minimum input for a usable output). Selectivity is its ability to reject adjacent/unwanted channels and select only the desired signal.

  3. What is the standard intermediate frequency (IF) for commercial AM broadcast receivers?

    455 kHz is the standard IF for AM broadcast receivers (FM broadcast receivers commonly use 10.7 MHz).

  4. What is pulse code modulation (PCM)?

    A digital representation of an analog signal produced by sampling the signal at a sufficient rate, quantizing each sample to the nearest of a finite set of levels, and encoding each quantized level into a binary codeword.

  5. State the Nyquist sampling theorem.

    A band-limited signal with maximum frequency fm can be completely reconstructed from its samples if the sampling rate fs ≥ 2fm. The minimum rate 2fm is the Nyquist rate; sampling below it causes aliasing.

  6. What is quantization error (quantization noise) in PCM?

    The difference between the actual analog sample value and its nearest quantized level. For a uniform quantizer with step size Δ, the quantization noise power is Δ²/12, and the error is bounded by ±Δ/2.

  7. Give the SNR (in dB) of a uniform PCM system in terms of the number of bits n.

    SNR_q ≈ 6.02n + 1.76 dB for a full-load sinusoid. Each additional bit improves the quantization SNR by about 6 dB.

  8. What is companding in PCM and why is it used?

    Companding (compressing then expanding) applies nonlinear quantization—finer steps for small signals, coarser for large—to keep SNR roughly constant over the signal range. The A-law (Europe) and μ-law (North America/Japan) are the two standard companding laws.

  9. Compare PCM, DPCM, and Delta Modulation.

    PCM encodes each absolute quantized sample (n bits/sample). DPCM encodes the difference between a sample and its prediction, reducing bit rate by exploiting correlation. Delta modulation (DM) is 1-bit DPCM that transmits only the sign of the difference; it is simple but suffers slope overload and granular noise.

  10. What are slope overload distortion and granular noise in delta modulation?

    Slope overload occurs when the signal changes faster than the step size can track (step too small). Granular (hunting) noise occurs when the signal is nearly constant and the output oscillates by ±Δ (step too large). Adaptive DM adjusts step size to mitigate both.

  11. Name the three basic binary digital modulation (keying) techniques.

    Amplitude-shift keying (ASK), frequency-shift keying (FSK), and phase-shift keying (PSK)—the carrier's amplitude, frequency, or phase respectively is switched according to the binary data.

  12. Why is BPSK more power-efficient (lower BER for a given Eb/N0) than ASK and FSK?

    BPSK uses antipodal signals (phases 180° apart), giving the maximum possible distance between the two symbols for a given energy, which minimizes the probability of error. ASK/coherent-FSK use orthogonal/on-off signals with smaller separation, so they need ~3-4 dB more Eb/N0 for the same BER.

  13. What is QPSK and its bandwidth advantage over BPSK?

    Quadrature phase-shift keying transmits 2 bits per symbol using four carrier phases (90° apart). For the same bit rate it needs half the bandwidth of BPSK (or doubles data rate in the same bandwidth) while keeping the same BER as BPSK.

  14. How many bits per symbol does M-ary modulation carry, and what is the trade-off?

    An M-ary scheme carries log2(M) bits per symbol, improving bandwidth/spectral efficiency. The trade-off is that symbols are packed closer together, so a higher Eb/N0 is required to maintain the same error rate (worse noise immunity).

  15. What is the difference between coherent and non-coherent detection?

    Coherent detection requires the receiver to know/recover the exact carrier phase (e.g., PSK, coherent FSK) and gives better BER. Non-coherent detection (e.g., ASK envelope, non-coherent FSK, DPSK) needs no phase reference, is simpler, but performs worse for a given Eb/N0.

  16. What is the purpose of error control coding?

    To add controlled redundancy to transmitted data so the receiver can detect and/or correct errors caused by channel noise, thereby improving reliability without retransmission (FEC) or by enabling retransmission requests (ARQ).

  17. Distinguish between FEC and ARQ error control strategies.

    Forward Error Correction (FEC) adds redundancy that lets the receiver correct errors without a return channel—good for one-way/real-time links. Automatic Repeat reQuest (ARQ) only detects errors and asks the transmitter to resend the data—needs a feedback channel but is simpler.

  18. Distinguish block codes from convolutional codes.

    Block codes (e.g., Hamming, BCH, Reed-Solomon) encode fixed k-bit message blocks into n-bit codewords independently (memoryless). Convolutional codes encode a continuous bit stream using memory (shift registers); each output depends on current and previous input bits, decoded with the Viterbi algorithm.

  19. Define Hamming distance and minimum distance of a code.

    Hamming distance between two codewords is the number of bit positions in which they differ. The minimum distance dmin of a code is the smallest Hamming distance between any two distinct codewords; it determines the code's error-detecting and -correcting power.

  20. How do dmin, error detection, and error correction capability relate?

    A code with minimum distance dmin can detect up to (dmin − 1) errors and correct up to t = ⌊(dmin − 1)/2⌋ errors. Detection and correction capabilities together require dmin ≥ t + e + 1 for correcting t and detecting e errors.

  21. What is a (7,4) Hamming code and how many errors can it correct?

    A linear block code with n = 7 total bits, k = 4 data bits, and 3 parity bits, having minimum distance dmin = 3. It can detect up to 2 errors and correct any single-bit error.

  22. What is data compression and the difference between lossless and lossy compression?

    Data compression reduces the number of bits needed to represent information by removing redundancy. Lossless compression (e.g., Huffman, LZW, run-length) allows exact reconstruction; lossy compression (e.g., JPEG, MP3) discards perceptually less important data for higher compression at the cost of fidelity.

  23. What is Huffman coding and its key property?

    A lossless, variable-length prefix coding algorithm that assigns shorter codewords to more probable symbols and longer ones to rarer symbols, built from a binary tree merging lowest-probability symbols. It produces an optimal prefix code with minimum average codeword length for known symbol probabilities.

  24. Define the entropy H(X) of a discrete source and its formula.

    Entropy is the average information content per symbol: H(X) = −Σ p(xi) log2 p(xi) bits/symbol. It measures the uncertainty of the source and sets the lower bound on the average bits needed to represent it losslessly.

  25. For a source with M equally likely symbols, what is its entropy, and when is entropy maximum?

    H = log2(M) bits/symbol when all M symbols are equiprobable. Entropy is maximum precisely when all symbols are equally likely; any nonuniformity reduces entropy below this maximum.

  26. Define information content of a message of probability p and its units.

    Self-information I = log2(1/p) = −log2 p, measured in bits (when log base 2). A less probable event carries more information; a certain event (p = 1) carries zero information.

  27. State the Shannon-Hartley channel capacity theorem.

    For an AWGN channel, C = B·log2(1 + S/N) bits/second, where B is the bandwidth (Hz) and S/N is the signal-to-noise power ratio. It gives the maximum error-free data rate achievable over the channel.

  28. According to Shannon's channel capacity, what is the trade-off between bandwidth and SNR?

    Capacity can be maintained by trading bandwidth against SNR—e.g., wider bandwidth allows the same capacity at lower SNR. However, as B → ∞, capacity approaches a finite limit C_max = 1.44·(S/N0) because noise power also grows with bandwidth (Shannon limit Eb/N0 ≈ −1.6 dB).

  29. State Shannon's source coding theorem and define coding efficiency.

    The source coding (noiseless coding) theorem states that the minimum average codeword length per symbol cannot be less than the source entropy H(X), i.e., L̄ ≥ H(X). Coding efficiency η = H(X)/L̄ (often ×100%), and redundancy = 1 − η.

  30. How do parity check, checksum, and CRC compare as error-detection methods?

    Single parity adds one bit and detects only odd numbers of errors (weak). Checksum sums data words and detects many errors but can miss reordering/compensating errors. Cyclic Redundancy Check (CRC) uses polynomial division by a generator polynomial, giving strong detection of burst and random errors, and is the most reliable of the three.

UPSC ESE E&T Communication Systems FAQ

What is in the UPSC ESE E&T Communication Systems syllabus?

UPSC ESE E&T Communication Systems is split into 3 chapters: Analog Communication, Digital Communication and Information Theory and Coding. Together they contain 12 topics.

How many chapters are in UPSC ESE E&T Communication Systems?

3 chapters, containing 12 topics. The largest chapter is Analog Communication, with 4 topics.

Are there flashcards for UPSC ESE E&T Communication Systems?

Yes — a 50-card Communication Systems deck, printed in full in the flashcards section of this page and available in the Examius app with spaced repetition. The free plan covers one exam of your choice, including its flashcards; Examius Pro, a one-time purchase, unlocks every other exam and deck.