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AEE Telecom and Electronics Core · Chapter 8

Analog and Digital Communication

What to remember

  • AM: total power Pt = Pc (1 + m²/2); bandwidth = 2 fm. FM: Carson bandwidth = 2 (Δf + fm); modulation index β = Δf / fm.
  • PCM: sampling at fs ≥ 2 fm (Nyquist), then quantizing and encoding. Bit rate = fs × n bits. Shannon capacity C = B log2 (1 + S/N).
  • Digital modulation: ASK changes amplitude, FSK changes frequency, PSK changes phase; M-ary schemes carry log2 M bits per symbol.

Amplitude modulation (AM)

A carrier Ac cos ωc t is modulated by a message signal. For a single tone, s(t) = Ac [1 + m cos ωm t] cos ωc t, where m = Am / Ac is the modulation index. Keep m ≤ 1 to avoid over-modulation and distortion in an envelope detector.

  • Bandwidth = 2 fm (upper and lower sidebands).
  • Total power Pt = Pc (1 + m²/2). Each sideband carries Pc m²/4.
  • Efficiency = m² / (2 + m²). At m = 1 it is 33.3%.
  • DSB-SC: carrier suppressed; bandwidth 2 fm; needs coherent detection (Costas loop).
  • SSB: one sideband only; bandwidth fm; saves power and bandwidth. Made by the filter or phase-shift method.
  • VSB: a vestige of one sideband is kept; used for the picture signal in TV.
  • Detectors: envelope detector (diode, RC) for AM; product detector for DSB-SC and SSB.

Frequency and phase modulation

In FM the carrier frequency changes with the message amplitude; in PM the phase changes. Frequency deviation Δf = kf × Am.

  • Modulation index β = Δf / fm.
  • Carson's rule: BW ≈ 2 (Δf + fm) = 2 fm (β + 1).
  • Narrow-band FM has β much below 1; BW ≈ 2 fm. Wide-band FM has β above 1.
  • Broadcast FM: maximum deviation 75 kHz; audio up to 15 kHz; channel spacing 200 kHz.
  • FM is more noise-resistant than AM (it trades bandwidth for noise performance). Pre-emphasis at the transmitter and de-emphasis at the receiver improve high-frequency SNR. The capture effect lets the stronger of two signals win.
  • FM generators: varactor (direct) or Armstrong (indirect). Detectors: slope detector, balanced discriminator, ratio detector, PLL.

Receivers and noise

  • Superheterodyne receiver: RF amplifier, mixer, local oscillator, IF amplifier, detector, audio amplifier. Selectivity comes mainly from the IF stage.
  • Typical IF: 455 kHz for AM; 10.7 MHz for FM.
  • Local oscillator frequency fLO = fs + IF. Image frequency = fs + 2 IF.
  • Thermal noise power: N = kTB, with k = 1.38 × 10⁻²³ J/K.
  • Noise figure F = (S/N)in / (S/N)out; it is always at least 1 (0 dB). Friis formula for cascaded stages: F = F1 + (F2 − 1)/G1 + …. The first stage should have low noise and high gain.

Pulse and digital communication

Sampling theorem: a signal band-limited to fm is fully recovered if fs ≥ 2 fm. Below that, aliasing occurs. Voice is limited to about 300 Hz to 3400 Hz, sampled at 8 kHz.

PCM steps: sampling, quantizing, encoding.

  • Number of levels L = 2ⁿ for n bits. Step size Δ = Vpp / L.
  • Quantization SNR for a full-scale sine ≈ (6n + 1.76) dB. Each extra bit adds about 6 dB.
  • Bit rate = fs × n. Telephone PCM: 8000 × 8 = 64 kbps.
  • Companding (A-law is used in India and Europe; μ-law in North America) gives better SNR for weak signals.
  • Delta modulation (DM) sends one bit per sample and uses a simple staircase; it suffers slope overload and granular noise.

Line codes: NRZ, RZ, Manchester (a transition in every bit, no DC), AMI (bipolar, no DC).

Multiplexing: FDM divides bandwidth; TDM divides time. Multiple access: FDMA, TDMA and CDMA.

Digital modulation and information theory

SchemeVariesNotes
ASK (OOK)AmplitudeSimple; most affected by noise
FSKFrequencyGood noise immunity; modems, paging
BPSKPhase (0 or 180°)1 bit per symbol; better noise performance than ASK/FSK
QPSKFour phases2 bits per symbol; half the bandwidth of BPSK for the same bit rate
M-QAMAmplitude and phaselog2 M bits per symbol; 16-QAM = 4 bits; needs higher SNR
  • Bit rate = symbol rate × log2 M.
  • ISI (inter-symbol interference) is limited by the Nyquist criterion; the minimum bandwidth is Rb/2 (ideal); raised-cosine filters are used in practice.
  • Entropy: H = − Σ pi log2 pi bits per symbol; for N equally likely symbols H = log2 N.
  • Shannon capacity: C = B log2 (1 + S/N). Capacity rises with bandwidth and SNR.
  • Error control: minimum Hamming distance dmin detects up to dmin − 1 errors and corrects up to (dmin − 1)/2 (rounded down). Parity detects single-bit errors; CRC detects burst errors; Hamming codes correct single-bit errors.

AM and FM compared

PointAMFM
Information carried inAmplitude of the carrierFrequency of the carrier
Bandwidth2 fm (narrow)2 (Δf + fm) (wide)
Noise immunityPoor; noise adds to amplitudeGood; amplitude limiter removes noise
Transmitter power efficiencyLow (carrier wastes power)Higher; constant amplitude allows Class C amplifiers
Band used in broadcastingMF (medium wave) and HFVHF (88 to 108 MHz)
Receiver complexitySimpleMore complex

Why digital? A digital signal can be regenerated at every repeater, so noise does not build up. It supports error control, encryption, multiplexing and storage. The price is a larger bandwidth and the need for synchronization.

Synchronization: a digital receiver needs bit timing and, for coherent detection, carrier recovery. A phase-locked loop (PLL) is the standard building block; it has a phase detector, a low-pass filter and a voltage-controlled oscillator (VCO).

Bandwidth of digital signals: for a binary baseband signal at bit rate Rb the main lobe is about Rb wide for NRZ. BPSK and ASK need about 2 Rb; QPSK needs about Rb; 16-QAM needs about Rb/2 (all for rectangular pulses, main lobe, simple view).

Probability of error: coherent BPSK has the lowest error rate for the same Eb/N0 among the simple binary schemes. Higher-order schemes (16-QAM, 64-QAM) give more bits per hertz but need more signal power.

Worked examples

  • 1. AM power: Pc = 100 W, m = 0.5. Pt = 100 (1 + 0.125) = 112.5 W.
  • 2. FM: Δf = 75 kHz, fm = 15 kHz. β = 5; BW = 2 (75 + 15) = 180 kHz.
  • 3. PCM: 4 kHz signal, 8 bits: fs = 8 kHz; rate = 64 kbps; SNR ≈ 6 × 8 + 1.76 = 49.76 dB.
  • 4. Image frequency: fs = 1000 kHz, IF 455 kHz. fLO = 1455 kHz; image = 1910 kHz.
  • 5. Shannon: B = 3 kHz, S/N = 1000 (30 dB). C = 3000 × log2 1001 ≈ 29.9 kbps.
  • 6. 16-QAM: 1 Mbaud gives 4 Mbps.
  • 7. Thermal noise: T = 290 K, B = 1 MHz. N = 1.38×10⁻²³ × 290 × 10⁶ ≈ 4 × 10⁻¹⁵ W, about −114 dBm.

Exam traps

  • AM bandwidth is 2 fm; SSB bandwidth is fm.
  • Carson's rule uses the deviation plus the message frequency, not the carrier frequency.
  • Image frequency is fs + 2 IF, not fs + IF.
  • Nyquist rate is 2 fm; sampling at fm is not enough.
  • AM power efficiency at m = 1 is 33.3%, not 50%.
  • QPSK carries 2 bits per symbol, not 4.
  • Hamming distance detects dmin − 1 errors but corrects only (dmin − 1)/2.
  • FM improves noise performance by using more bandwidth; AM has no such trade.
  • India uses A-law companding; the USA uses μ-law.
  • DM uses one bit per sample; PCM uses n bits.

One-liners

  • 1. Modulation index of AM must not exceed 1.
  • 2. VSB is used for TV picture transmission.
  • 3. Superhet IF for AM is 455 kHz.
  • 4. Broadcast FM channel spacing is 200 kHz.
  • 5. Pre-emphasis boosts high audio frequencies at the transmitter.
  • 6. Voice is sampled at 8 kHz in telephony.
  • 7. Quantization SNR rises about 6 dB per bit.
  • 8. Manchester code has a transition in every bit period.
  • 9. TDM shares a channel by time slots.
  • 10. BPSK carries 1 bit per symbol.
  • 11. Entropy of N equal symbols is log2 N bits.
  • 12. Shannon capacity grows with bandwidth and SNR.

Practice questions

  1. The bandwidth of a conventional AM signal with message bandwidth fm is

    1. fm/2
    2. 2 fm
    3. 4 fm
    4. fm
    Answer

    B. 2 fm

    Two sidebands each of width fm give 2 fm.

  2. The minimum sampling rate for a signal band-limited to fm is

    1. fm
    2. fm/2
    3. 4 fm
    4. 2 fm
    Answer

    D. 2 fm

    Nyquist sampling theorem: fs ≥ 2 fm.

  3. The standard intermediate frequency of an AM broadcast superheterodyne receiver is

    1. 38 MHz
    2. 455 kHz
    3. 10.7 MHz
    4. 1 MHz
    Answer

    B. 455 kHz

    AM receivers use 455 kHz; FM receivers use 10.7 MHz.

  4. Which modulation scheme carries 2 bits per symbol?

    1. QPSK
    2. ASK (binary)
    3. BFSK
    4. BPSK
    Answer

    A. QPSK

    QPSK uses four phases, so log2 4 = 2 bits per symbol.

  5. Carson's rule gives the approximate bandwidth of an FM signal as

    1. 2 Δf only
    2. fm only
    3. Δf − fm
    4. 2 (Δf + fm)
    Answer

    D. 2 (Δf + fm)

    Carson's rule: BW = 2 (Δf + fm).

  6. Which modulation is used for the picture signal of analog TV?

    1. Pulse code modulation
    2. Frequency modulation
    3. Vestigial sideband
    4. Double sideband suppressed carrier
    Answer

    C. Vestigial sideband

    VSB saves bandwidth while keeping simple detection.

  7. In PCM, the correct order of processing is

    1. encoding, sampling, quantizing
    2. quantizing, sampling, encoding
    3. sampling, quantizing, encoding
    4. sampling, encoding, quantizing
    Answer

    C. sampling, quantizing, encoding

    The signal is sampled, its samples are quantized, then coded into bits.

  8. The quantity (1 + m²/2) in AM relates the total power to

    1. sideband bandwidth
    2. carrier power
    3. message frequency
    4. noise power
    Answer

    B. carrier power

    Pt = Pc (1 + m²/2).

  9. The image frequency in a superheterodyne receiver with signal fs and intermediate frequency IF is

    1. fs − IF
    2. 2 fs
    3. fs + IF
    4. fs + 2 IF
    Answer

    D. fs + 2 IF

    Image = fLO + IF = fs + 2 IF.

  10. Shannon's channel capacity formula is

    1. C = B log2 (1 + S/N)
    2. C = B (1 + S/N)
    3. C = B / log2 (1 + S/N)
    4. C = 2B log10 (S/N)
    Answer

    A. C = B log2 (1 + S/N)

    The capacity of a noisy channel is B log2(1 + S/N).

  11. The entropy of a source with four equally likely symbols is

    1. 0.25 bit
    2. 2 bits
    3. 1 bit
    4. 4 bits
    Answer

    B. 2 bits

    H = log2 4 = 2 bits/symbol.

  12. FM broadcast in the 88 to 108 MHz band uses a maximum frequency deviation of

    1. 15 kHz
    2. 200 kHz
    3. 5 kHz
    4. 75 kHz
    Answer

    D. 75 kHz

    Standard broadcast FM allows ±75 kHz deviation.

  13. What is the total power of an AM wave with carrier 100 W and m = 0.5?

    1. 112.5 W
    2. 150 W
    3. 106.25 W
    4. 125 W
    Answer

    A. 112.5 W

    Pt = 100 (1 + 0.25/2) = 112.5 W.

  14. For m = 1 the power efficiency of AM is

    1. 66.7%
    2. 50%
    3. 33.3%
    4. 100%
    Answer

    C. 33.3%

    m²/(2+m²) = 1/3.

  15. A frequency deviation of 75 kHz and message frequency 15 kHz give a modulation index of

    1. 90
    2. 0.2
    3. 1125
    4. 5
    Answer

    D. 5

    β = Δf/fm = 75/15 = 5.

  16. The Carson bandwidth for Δf = 75 kHz and fm = 15 kHz is

    1. 180 kHz
    2. 30 kHz
    3. 150 kHz
    4. 90 kHz
    Answer

    A. 180 kHz

    2 (75 + 15) = 180 kHz.

  17. The bit rate of a PCM system that samples at 8 kHz with 8 bits per sample is

    1. 8 kbps
    2. 16 kbps
    3. 64 kbps
    4. 128 kbps
    Answer

    C. 64 kbps

    8000 × 8 = 64000 bps.

  18. An 8-bit uniform quantizer has a full-scale sine SNR of about

    1. 24 dB
    2. 49.8 dB
    3. 60 dB
    4. 38.4 dB
    Answer

    B. 49.8 dB

    6n + 1.76 = 48 + 1.76 = 49.76 dB.

  19. A receiver with IF 455 kHz is tuned to 1000 kHz (LO above signal). The image frequency is

    1. 1455 kHz
    2. 545 kHz
    3. 1910 kHz
    4. 1455 MHz
    Answer

    C. 1910 kHz

    Image = 1000 + 2 × 455 = 1910 kHz.

  20. A 16-QAM signal at 1 Mbaud has a bit rate of

    1. 4 Mbps
    2. 1 Mbps
    3. 16 Mbps
    4. 2 Mbps
    Answer

    A. 4 Mbps

    Rb = 1 × log2 16 = 4 Mbps.

  21. A channel with bandwidth 3 kHz and S/N of 1000 has a capacity closest to

    1. 300 kbps
    2. 9 kbps
    3. 3 kbps
    4. 30 kbps
    Answer

    D. 30 kbps

    3000 × log2(1001) ≈ 3000 × 9.97 ≈ 29.9 kbps.

  22. A code with minimum Hamming distance 5 can correct up to how many errors?

    1. 1
    2. 2
    3. 3
    4. 4
    Answer

    B. 2

    t = floor((5−1)/2) = 2 errors corrected.

  23. Number of quantization levels for a 6-bit quantizer is

    1. 128
    2. 36
    3. 32
    4. 64
    Answer

    D. 64

    2⁶ = 64.

  24. A system sends 1 bit per sample using a staircase approximation. This is

    1. delta modulation
    2. PSK
    3. QAM
    4. PCM
    Answer

    A. delta modulation

    Delta modulation sends one bit per sample.

  25. Which line code has a transition in every bit period and no DC component?

    1. Manchester
    2. NRZ unipolar
    3. RZ unipolar
    4. Return to bias only
    Answer

    A. Manchester

    Manchester coding guarantees a mid-bit transition and zero average level.

  26. A message of 5 kHz modulates a carrier in SSB. The bandwidth needed is

    1. 15 kHz
    2. 5 kHz
    3. 10 kHz
    4. 2.5 kHz
    Answer

    B. 5 kHz

    SSB needs one sideband, equal to fm.

  27. For a given bit rate, QPSK needs about what fraction of the bandwidth of BPSK?

    1. One quarter
    2. Twice
    3. One half
    4. Four times
    Answer

    C. One half

    QPSK sends 2 bits per symbol, halving the symbol rate.

  28. In a cascade, the first amplifier stage should have

    1. high noise figure and high gain
    2. low noise figure and high gain
    3. high noise figure and low gain
    4. low gain and low impedance
    Answer

    B. low noise figure and high gain

    By the Friis formula the first stage dominates the noise figure.

  29. For an AM signal with carrier amplitude 10 V and a message amplitude 4 V, the modulation index is

    1. 2.5
    2. 0.04
    3. 4
    4. 0.4
    Answer

    D. 0.4

    m = Am/Ac = 4/10 = 0.4.

  30. Thermal noise power in a 1 MHz bandwidth at 290 K is about

    1. 4 × 10⁻⁹ W
    2. 4 × 10⁻²¹ W
    3. 4 × 10⁻¹⁵ W
    4. 1.4 × 10⁻¹⁵ W
    Answer

    C. 4 × 10⁻¹⁵ W

    kTB = 1.38×10⁻²³ × 290 × 10⁶ ≈ 4×10⁻¹⁵ W.

  31. Which block of a phase-locked loop produces the output frequency?

    1. Mixer with fixed LO
    2. Voltage-controlled oscillator
    3. Phase detector
    4. Low-pass filter
    Answer

    B. Voltage-controlled oscillator

    The VCO output frequency follows the control voltage.

  32. Consider: 1. In an envelope detector m greater than 1 causes distortion. 2. DSB-SC can be detected by a simple envelope detector.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Over-modulation distorts the envelope; DSB-SC needs coherent detection.

  33. Consider: 1. FM trades extra bandwidth for better noise performance. 2. Pre-emphasis is applied at the receiver.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Pre-emphasis is applied at the transmitter; de-emphasis at the receiver.

  34. Consider: 1. Aliasing occurs when sampling below the Nyquist rate. 2. Companding improves SNR for weak signals.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both are standard PCM facts.

  35. Consider: 1. Hamming code can correct a single-bit error. 2. A single parity bit can detect any number of errors.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    A single parity bit detects only odd numbers of bit errors.

  36. Consider: 1. Entropy of a source is maximum when all symbols are equally likely. 2. Entropy increases when one symbol has probability near 1.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Entropy is highest for equal probabilities and falls when one symbol dominates.

  37. Consider: 1. TDM divides the channel into time slots. 2. FDM allots each user a separate frequency band.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both define the two multiplexing methods.

  38. Match the scheme to its feature: 1. FSK 2. BPSK 3. 16-QAM.

    1. Frequency change; 2 bits per symbol; 1 bit per symbol
    2. Phase change; frequency change; 2 bits per symbol
    3. Amplitude change; 4 bits per symbol; 1 bit per symbol
    4. Frequency change; 1 bit per symbol; 4 bits per symbol
    Answer

    D. Frequency change; 1 bit per symbol; 4 bits per symbol

    FSK varies frequency; BPSK sends 1 bit/symbol; 16-QAM sends 4.

  39. Match the receiver type to the standard IF: 1. AM broadcast 2. FM broadcast.

    1. 455 kHz; 455 kHz
    2. 455 kHz; 10.7 MHz
    3. 10.7 MHz; 455 kHz
    4. 10.7 MHz; 10.7 MHz
    Answer

    B. 455 kHz; 10.7 MHz

    AM uses 455 kHz, FM uses 10.7 MHz.

  40. Match the companding law to its region: 1. A-law 2. μ-law.

    1. North America; India and Europe
    2. India only; Japan only
    3. India and Europe; North America
    4. Europe only; India only
    Answer

    C. India and Europe; North America

    A-law is used in the Indian and European standard; μ-law in North America.

  41. Which of the following reduces inter-symbol interference in a band-limited channel?

    1. Increasing carrier power only
    2. Raised-cosine pulse shaping
    3. Using a lower sampling rate
    4. Removing the filter
    Answer

    B. Raised-cosine pulse shaping

    Raised-cosine pulses satisfy the Nyquist criterion for zero ISI.

  42. An FM receiver can reject amplitude noise because of its

    1. IF filter only
    2. audio amplifier
    3. local oscillator
    4. amplitude limiter
    Answer

    D. amplitude limiter

    The limiter removes amplitude variations before detection.

  43. For a bit rate of 2 Mbps with ideal Nyquist signalling, the minimum channel bandwidth is

    1. 4 MHz
    2. 2 MHz
    3. 0.5 MHz
    4. 1 MHz
    Answer

    D. 1 MHz

    Minimum bandwidth = Rb/2 = 1 MHz.

  44. For a PCM system using 10 bits per sample at 8 kHz, the bit rate is

    1. 64 kbps
    2. 80 kbps
    3. 8 kbps
    4. 800 kbps
    Answer

    B. 80 kbps

    8000 × 10 = 80,000 bps.

  45. A signal of bandwidth 4 kHz is sampled at the Nyquist rate. The sampling rate is

    1. 4 kHz
    2. 2 kHz
    3. 16 kHz
    4. 8 kHz
    Answer

    D. 8 kHz

    fs = 2 fm = 8 kHz.

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