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1. A communication system broadly contains transmitter, channel, and:
Explanation: The standard three-part model: Transmitter converts information to a signal; Channel carries it; Receiver reconstructs the original message.
2. A signal whose amplitude takes only two discrete levels (0 or 1) is called:
Explanation: Digital signals are two-level (binary) representations. Analog signals vary continuously. Modern communication increasingly uses digital because it is more immune to noise.
3. Which of the following is an example of an analog signal?
Explanation: Microphone output is a continuously varying voltage proportional to sound pressure — a classic analog signal. Binary data, Morse code, and ASCII are discrete/digital representations.
4. Modulation is the process of:
Explanation: Modulation varies one or more properties (amplitude, frequency, phase) of a high-frequency carrier in accordance with the message signal, making transmission efficient.
5. Why is modulation required for long-distance communication?
Explanation: At audio frequencies (20 Hz – 20 kHz), antenna length λ/2 would be tens of kilometres. Modulation shifts the signal to high carrier frequencies where λ/2 is metres. It also allows frequency-division multiplexing.
6. The bandwidth of a signal is:
Explanation: Bandwidth = f max − f min . A voice signal (300 Hz – 3400 Hz) has a bandwidth of 3100 Hz. Larger bandwidth means more information can be sent per second.
7. The device that converts a message signal into a form suitable for transmission is called:
Explanation: The transmitter processes the information source signal (modulation, amplification, etc.) and feeds it to the antenna for radiation.
8. Noise in a communication channel is:
Explanation: Noise is any random electrical disturbance not part of the message. It can arise from thermal agitation (thermal noise), atmospheric effects, or interference.
9. For efficient radiation, the length of a transmitting antenna should be approximately:
Explanation: A half-wave dipole antenna (length = λ/2) is the standard resonant antenna. At 1 MHz (λ = 300 m), the antenna is 150 m. At 100 MHz (FM), it is 1.5 m — practical.
10. In amplitude modulation (AM), which parameter of the carrier wave changes?
Explanation: AM: carrier amplitude ∝ instantaneous message amplitude. Carrier frequency and phase remain constant.
11. In frequency modulation (FM), which parameter of the carrier wave changes?
Explanation: FM: carrier frequency deviates around its centre frequency in proportion to the message amplitude. The carrier amplitude stays constant, making FM less susceptible to amplitude noise.
12. Ground wave propagation is most effective for frequencies:
Explanation: Ground waves travel along Earth's surface. Attenuation increases sharply with frequency, so this mode is practical only for LF and MF (AM broadcast 530–1710 kHz).
13. Space wave (line-of-sight) propagation is used for:
Explanation: Above ~30 MHz the ionosphere cannot refract waves back to Earth. VHF/UHF/SHF travel in straight lines (LOS). FM (88–108 MHz) and TV (VHF 54–216 MHz, UHF 470–890 MHz) use space waves.
14. Sky-wave propagation relies on:
Explanation: The ionosphere (layers D, E, F₁, F₂ at 60–400 km altitude) contains free electrons that refract HF waves (3–30 MHz) back to Earth, enabling long-distance short-wave radio communication.
15. AM broadcast radio occupies the frequency band:
Explanation: AM medium-wave (MW) broadcast: 530–1710 kHz. FM broadcast: 88–108 MHz. Short-wave (HF): 3–30 MHz. Microwave: 300 MHz – 300 GHz.
16. FM broadcast radio uses the frequency band:
Explanation: FM radio: 88–108 MHz (VHF). This high frequency means only space-wave (LOS) propagation is possible, limiting range to ~50–80 km but providing better sound quality than AM.
17. The process of retrieving the original message signal from a modulated carrier at the receiver is called:
Explanation: Demodulation (detection) is the inverse of modulation. An AM demodulator uses a rectifier and envelope detector; an FM demodulator uses a frequency discriminator or PLL.
18. Frequency Division Multiplexing (FDM) allows multiple signals to share one channel by:
Explanation: FDM assigns each message signal its own frequency band (sub-carrier). Radio stations use FDM — each occupies a specific frequency. Contrast with TDM where signals share time slots.
19. Pulse Code Modulation (PCM) involves:
Explanation: PCM steps: (1) Sampling at f s ≥ 2B (Nyquist); (2) Quantization into 2 n levels; (3) Encoding each sample as n-bit binary word. This is the basis of all digital telephony and audio (CD: 44.1 kHz, 16-bit).
20. Nyquist's sampling theorem states that a band-limited signal of bandwidth B must be sampled at a rate of at least:
Explanation: Nyquist rate f s ≥ 2f max = 2B. Sampling below this rate causes aliasing — different signals become indistinguishable. Voice telephony (B = 4 kHz) is sampled at 8000 samples/s.
21. The modulation index μ of an AM signal is defined as:
Explanation: μ = A m /A c where A m is message amplitude and A c is carrier amplitude. For undistorted AM, μ ≤ 1 (i.e., 0–100%).
22. When μ > 1 in AM, the signal is said to be:
Explanation: Over-modulation (μ > 1) causes the envelope to go negative — the carrier is cut off for part of the cycle, producing severe distortion at the receiver.
23. The bandwidth of an AM signal carrying a single-tone message of frequency f_m is:
Explanation: AM produces two sidebands: upper sideband at f c + f m and lower sideband at f c − f m . Total BW = (f c +f m ) − (f c −f m ) = 2f m .
24. A carrier of 500 kHz is amplitude-modulated by a 5 kHz audio signal. The upper sideband frequency is:
Explanation: USB = f c + f m = 500 + 5 = 505 kHz . LSB = 500 − 5 = 495 kHz. Total BW = 10 kHz.
25. In AM, the total power P_t is related to carrier power P_c by:
Explanation: P t = P c (1 + μ²/2). For μ = 1 (100% mod.), P t = 1.5 P c . The two sidebands together carry μ²P c /2 = 0.5P c . Only sideband power carries information; the carrier power is 'wasted.'
26. If carrier power is 10 kW and modulation index is 0.6, the total transmitted power is:
Explanation: P t = P c (1 + μ²/2) = 10(1 + 0.36/2) = 10(1 + 0.18) = 10 × 1.18 = 11.8 kW .
27. In FM, the frequency deviation Δf is proportional to:
Explanation: In FM, Δf = k f × A m . The rate of frequency variation follows the message frequency. So amplitude controls deviation and message frequency controls rate of change.
28. The modulation index β in FM is defined as:
Explanation: FM modulation index β = Δf / f m . For wideband FM (β >> 1), bandwidth ≈ 2(Δf + f m ) (Carson's rule: BW ≈ 2(β+1)f m ). Commercial FM: Δf = 75 kHz, f m,max = 15 kHz → β = 5.
29. Which property makes FM superior to AM for music broadcasting?
Explanation: Lightning, ignition, and other noise sources add amplitude spikes. Since FM information is in frequency (not amplitude), a limiter at the receiver removes amplitude variations — noise suppression is far better than AM.
30. The maximum distance for space-wave LOS transmission between a transmitting tower of height h_T and a receiving tower of height h_R is:
Explanation: Line-of-sight range: d T = √(2Rh T ) to horizon from transmitter; d R = √(2Rh R ) to horizon from receiver. Total d = d T + d R . R = 6400 km.
31. A TV tower has height 400 m. The maximum LOS range (R = 6400 km) is approximately:
Explanation: d = √(2 × 6.4×10⁶ × 400) = √(5.12×10⁹) ≈ 71,554 m ≈ 71.5 km . To cover a city of 100 km radius, a taller tower or relay stations are needed.
32. The ionospheric layer most important for long-distance short-wave communication at night is:
Explanation: At night the D and E layers largely disappear (recombination). The F layer (200–400 km) merges to F₂ and persists, enabling long-distance short-wave skip. HF waves can bounce multiple times between ionosphere and ground.
33. The minimum antenna height required to radiate a signal of wavelength λ efficiently is:
Explanation: A quarter-wave monopole above a ground plane (or half-wave dipole) is the minimum for efficient radiation. The ground plane provides the other quarter-wave image. At 1 MHz (AM), λ/4 = 75 m; at 100 MHz (FM), λ/4 = 0.75 m.
34. Microwaves used in satellite communication occupy frequencies in:
Explanation: Satellite uplink/downlink uses C-band (4–8 GHz), Ku-band (12–18 GHz), and Ka-band (26–40 GHz) — all in SHF. These frequencies penetrate the ionosphere and support high-bandwidth LOS links.
35. A geostationary satellite orbits at an altitude of approximately:
Explanation: Geostationary orbit: ~35,786 km above equator. Period = 24 hours (synchronous with Earth rotation). The satellite appears stationary from the ground — ideal for fixed dish antennas and broadcast applications.
36. The communication satellite acts as a:
Explanation: An uplink signal is received by the transponder, frequency-converted (e.g., 6 GHz → 4 GHz), amplified, and retransmitted as the downlink. One satellite can cover ~40% of Earth's surface.
37. Signal transmission in optical fibre is based on:
Explanation: Core has higher refractive index than cladding. Rays entering within the acceptance cone undergo TIR at every core-cladding interface and propagate along the fibre with very low loss.
38. Which frequency range is used in optical fibre communication?
Explanation: Optical fibres operate at 850 nm, 1310 nm, or 1550 nm (minimum loss window). These correspond to ~3×10¹⁴ Hz. The enormous carrier frequency allows terabit/s bandwidths — far beyond microwaves.
39. The concept of 'cells' in mobile telephony allows:
Explanation: Cellular concept: divide service area into small cells, each with a low-power base station. Non-adjacent cells can reuse the same frequencies without interference. A set of N cells using all available frequencies is a cluster; capacity = (total spectrum) × (cells per cluster).
40. Signal-to-Noise Ratio (SNR) is defined as:
Explanation: SNR = P signal / P noise (or in dB: 10 log₁₀(P s /P n )). Higher SNR means cleaner reception. FM receivers typically achieve better SNR than AM for the same transmitter power.
41. In an AM signal, the modulation index μ = 0.8 and the carrier power is 100 W. The total sideband power is:
Explanation: Sideband power P sb = μ²P c /2 = (0.64 × 100)/2 = 32 W . Total power = 100 + 32 = 132 W. Only sideband power (32 W out of 132 W) carries useful information.
42. A 10 kHz audio signal AM-modulates a 1 MHz carrier. The channel bandwidth occupied is:
Explanation: BW = 2f m = 2 × 10 kHz = 20 kHz . The signal spans from 990 kHz to 1010 kHz. AM broadcast stations are spaced 10 kHz apart (BW per station = 10 kHz, single-sideband equivalent), but full AM needs 20 kHz.
43. In which modulation technique does the carrier amplitude remain constant?
Explanation: In FM, only the instantaneous frequency varies; amplitude is constant. This makes FM resistant to amplitude-type noise (rain fade, ignition interference) compared to AM.
44. For sky-wave propagation, the critical frequency f_c is the maximum frequency that can be reflected by the ionosphere at normal incidence. It is related to electron density N_max by:
Explanation: f c = 9√N max Hz where N max is peak electron density in electrons/m³. If N max = 10¹² /m³, f c = 9×10⁶ = 9 MHz. Above f c at normal incidence, the wave passes through into space.
45. The Maximum Usable Frequency (MUF) for sky-wave propagation is related to critical frequency f_c by MUF = f_c / cos θ where θ is the angle of incidence. If f_c = 10 MHz and θ = 60°, MUF is:
Explanation: MUF = f c / cos θ = 10 / cos 60° = 10 / 0.5 = 20 MHz . The MUF is always higher than f c . In practice, the operating frequency is chosen as ~85% of MUF (the Optimum Working Frequency, OWF).
46. A radio station broadcasts at 900 kHz. The wavelength and the minimum antenna height required are:
Explanation: λ = c/f = 3×10⁸ / 9×10⁵ = 333.3 m. Minimum antenna height (λ/4 monopole) = 333/4 ≈ 83.3 m . This explains why AM radio towers are enormous structures.
47. For a geostationary satellite, the time delay (round-trip propagation delay) for a signal to travel to the satellite and back is approximately:
Explanation: Round-trip path: 2 × 36,000 km = 72,000 km = 7.2×10⁷ m. Time = d/c = 7.2×10⁷ / 3×10⁸ = 0.24 s. This 240 ms delay is noticeable in phone calls via satellite (total two-way: ~480 ms).
48. An advantage of optical fibre over copper cables is:
Explanation: Optical fibres: (1) Carrier frequency ~10¹⁴ Hz → terabit/s capacity; (2) EMI immune (non-conductive glass); (3) Low loss (~0.2 dB/km at 1550 nm vs ~10 dB/km for coax); (4) Secure (hard to tap without disrupting signal).
49. A voice signal with maximum frequency 4 kHz is to be digitised using PCM with 8-bit encoding. The minimum bit rate required is:
Explanation: Nyquist sampling rate = 2 × 4000 = 8000 samples/s. Each sample is encoded in 8 bits. Bit rate = 8000 × 8 = 64,000 bps = 64 kbps . This is the standard G.711 telephony rate.
50. Time Division Multiplexing (TDM) allows multiple signals to share a channel by:
Explanation: TDM interleaves time slots from different signals. A T1 line carries 24 voice channels (each 64 kbps) by TDM → 1.544 Mbps total. Contrast with FDM which separates in frequency domain.
51. Commercial FM radio has a channel bandwidth of 200 kHz. The maximum frequency deviation Δf allowed is 75 kHz. Using Carson's rule, the maximum audio frequency that can be transmitted is:
Explanation: Carson's rule: BW = 2(Δf + f m ). So 200 = 2(75 + f m ) → 100 = 75 + f m → f m = 15 kHz . This is why FM has better audio quality — it covers the full audible range (20 Hz – 15 kHz vs. 4.5 kHz for AM voice).
52. In a superheterodyne receiver, the purpose of the local oscillator and mixer is to:
Explanation: The mixer beats the incoming signal (f RF ) with the local oscillator (f LO ) to produce IF = f RF − f LO (typically 455 kHz for AM, 10.7 MHz for FM). The fixed IF allows sharp, optimised filtering regardless of the tuned station.
53. Skip distance in sky-wave propagation is:
Explanation: Skip distance is the minimum range of sky-wave coverage. Between the end of ground-wave range and the start of sky-wave skip distance is a 'dead zone' with no reception. It depends on ionospheric height, frequency, and launch angle.
54. The ratio of frequencies used for uplink and downlink in C-band satellite communication is typically:
Explanation: C-band: uplink = 5.925–6.425 GHz, downlink = 3.7–4.2 GHz. Higher uplink frequency (from powerful ground station) compensates for more path loss; the satellite's lower power amplifier benefits from lower downlink frequency.
55. If μ₁ = 0.3 and μ₂ = 0.4 are modulation indices of two sinusoidal tones modulating the same AM carrier, the total modulation index is:
Explanation: For multiple tones, μ total = √(μ₁² + μ₂²) = √(0.09 + 0.16) = √0.25 = 0.5 . This must be ≤ 1 to avoid over-modulation. Total sideband power = μ total ² × P c / 2.
56. What is the main difference between 2G and 3G mobile communication?
Explanation: 1G: analog (AMPS). 2G: digital voice (GSM/CDMA, ~9.6 kbps data). 3G: UMTS/WCDMA, 384 kbps – 14.4 Mbps (HSPA). 4G LTE: up to 150 Mbps+ using OFDMA. 5G: sub-6 GHz + mmWave, Gbps peak rates, low latency.
57. The numerical aperture (NA) of an optical fibre is related to core and cladding refractive indices n₁ and n₂ by:
Explanation: NA = √(n₁² − n₂²) = n₁ sin θ max , where θ max is the half-angle of the acceptance cone. Higher NA → larger acceptance cone → easier coupling but more modal dispersion. Typical single-mode fibre NA ≈ 0.1–0.15.
58. Shannon's channel capacity formula C = B log₂(1 + S/N) states that capacity:
Explanation: Shannon's theorem gives the theoretical maximum bit rate. Doubling SNR increases capacity by B bits/s; doubling bandwidth doubles capacity (roughly). This is the fundamental limit — no coding scheme can exceed it.
59. A channel has bandwidth 3 kHz and SNR = 63. By Shannon's formula, the maximum channel capacity is:
Explanation: C = B log₂(1 + S/N) = 3000 × log₂(64) = 3000 × 6 = 18,000 bps = 18 kbps . Note log₂(64) = 6 since 2⁶ = 64. Telephone voice channels achieve close to this with modern codecs.
60. Population coverage P (number of people served) by a TV tower of height h placed in an area of population density η is:
Explanation: Service area = πd² = π × 2Rh (circular area with radius = LOS distance). Population = η × π × 2Rh. Doubling tower height doubles coverage area and population served (linear relationship with h).
61. The channel in a communication system could be:
Explanation: The communication channel is any medium through which signals travel: free space (radio), copper (telephone), optical fibre (internet backbone), or any combination.
62. Quantization in PCM introduces a type of error called:
Explanation: Quantization rounds each sample to the nearest discrete level. The difference between the actual sample and its quantized value is the quantization error (or noise). For n-bit quantization, maximum quantization noise = Δ/2 where Δ = full scale / 2ⁿ.
63. Phase modulation (PM) varies the _______ of the carrier in proportion to the message.
Explanation: PM: φ(t) = φ₀ + k p m(t). PM and FM are closely related — differentiating a PM signal gives FM. Most modern cellular systems use phase-based modulation (QPSK, QAM).
64. Walkie-talkies typically use which frequency band?
Explanation: Walkie-talkies use VHF (136–174 MHz) or UHF (400–512 MHz). These are short-range LOS devices. Police, fire, and amateur radios also use these bands. UHF penetrates buildings better than VHF.
65. VSAT (Very Small Aperture Terminal) uses satellite links for:
Explanation: VSAT provides two-way satellite internet with dish sizes of 0.75–1.2 m. Used for rural broadband, maritime communication, ATM networks, and disaster relief. Operates in Ku or Ka band.
66. In an AM broadcast station, the information is carried in the:
Explanation: Standard AM (DSB-FC): both upper and lower sidebands plus the carrier are transmitted. The carrier helps the envelope detector at the cheap receiver to work simply. SSB (Single Sideband) transmission is more efficient (half the bandwidth, no carrier power) but needs synchronous detection.
67. Which communication system is NOT affected by line-of-sight (LOS) limitations?
Explanation: Sky-wave HF radio bounces off the ionosphere and can reach the other side of the Earth without LOS. FM, TV, and microwave relay are all LOS limited (range ∝ √h).
68. In CDMA (Code Division Multiple Access), multiple users share the same frequency band by:
Explanation: CDMA spreads each user's signal across the entire spectrum using a unique pseudo-random code. The receiver uses the same code in a correlator to extract only the desired user's signal. Used in 3G (WCDMA, CDMA2000).
69. Modal dispersion in optical fibres causes:
Explanation: In multimode fibre, different modes (ray angles) travel at slightly different speeds → pulses broaden → intersymbol interference (ISI) limits bit rate. Single-mode fibre (core diameter ≈ 8–10 μm) eliminates modal dispersion; only chromatic dispersion remains.
70. In a communication system, repeaters are used to:
Explanation: Signals attenuate as they travel. Repeaters restore the signal level. In digital links, regenerative repeaters re-time and re-shape bits, preventing noise accumulation (unlike analog amplifiers which also amplify noise).
71. Pre-emphasis in FM broadcasting means:
Explanation: Pre-emphasis boosts high-frequency audio content (above 2.1 kHz) before FM modulation. The receiver applies de-emphasis (inverse filter) that also reduces high-frequency noise. Net result: better SNR at high frequencies where FM noise is greatest.
72. A 100% AM signal (μ = 1) transmits total power of 15 kW. The carrier power is:
Explanation: P t = P c (1 + μ²/2) → 15 = P c (1 + 0.5) = 1.5P c → P c = 15/1.5 = 10 kW . Each sideband carries μ²P c /4 = 2.5 kW, total sideband = 5 kW. Only 1/3 of total power carries information.
73. For LOS transmission from a tower of height 100 m on Earth (R = 6400 km), the coverage area in km² is:
Explanation: d = √(2Rh) = √(2 × 6400 × 0.1) = √1280 ≈ 35.8 km. Area = πd² = π × 1280 ≈ 1280π km² ≈ 4021 km². Note: h must be in km → h = 0.1 km.
74. OFDM (Orthogonal Frequency Division Multiplexing) is used in 4G LTE and Wi-Fi because it:
Explanation: OFDM uses hundreds/thousands of orthogonal narrow-band sub-carriers. Each sub-carrier is narrow enough to experience flat fading. A cyclic prefix eliminates inter-symbol interference from multipath. 4G LTE uses 1200 sub-carriers at 15 kHz spacing in a 20 MHz channel.
75. LEO (Low Earth Orbit) satellites at 500–2000 km altitude are preferred for some applications over GEO satellites because they have:
Explanation: LEO: altitude 500–2000 km → propagation delay 5–15 ms → ideal for latency-sensitive applications (Starlink broadband, Iridium voice). Disadvantage: each satellite passes overhead in minutes → need large constellations (Starlink: ~5000 satellites) for continuous coverage.
76. Multiplexing gain in communication systems is:
Explanation: Multiplexing (FDM, TDM, CDMA, OFDM) allows a single expensive channel (satellite transponder, optical fibre, radio spectrum) to carry many conversations simultaneously, drastically reducing per-user cost.
77. Wavelength Division Multiplexing (WDM) in optical fibre allows:
Explanation: DWDM (Dense WDM) carries 40–160 channels at different wavelengths (around 1550 nm, channel spacing 0.8 nm). Each channel can carry 100 Gbps → single fibre capacity > 10 Tbps. The Internet backbone relies on DWDM.
78. Error correction coding in digital communication adds:
Explanation: Channel coding (Hamming, Reed-Solomon, Turbo, LDPC, Polar codes) adds structured redundancy. The receiver uses this redundancy to identify and correct bit errors. Rate = k/n (k info bits in n total bits). Used in CDs, DVDs, deep-space communication, 4G/5G.
79. Handoff (handover) in mobile cellular systems means:
Explanation: As a mobile user crosses cell boundaries, the network automatically transfers the radio link to the new cell's base station without dropping the call. Hard handoff (GSM): break-before-make. Soft handoff (CDMA): simultaneous connection to both cells → diversity gain.
80. Spread spectrum communication technique (as used in CDMA) provides which key advantage over narrow-band systems?
Explanation: Spreading a signal over a wide bandwidth lowers power spectral density (hard to detect/jam). The receiver's correlator provides processing gain = BW_spread / BW_message (e.g., 1000× = 30 dB). Also provides frequency diversity against narrow-band fading. Originally developed for military communication.
81. An AM signal is represented as v(t) = A_c(1 + μ cos ω_m t)cos ω_c t. The three frequency components present are:
Explanation: Expanding: v(t) = A c cos ω c t + (μA c /2)cos(ω c +ω m )t + (μA c /2)cos(ω c −ω m )t. Three components: carrier (ω c ), upper sideband (ω c +ω m ), lower sideband (ω c −ω m ). The message frequency ω m itself does NOT appear.
82. For an AM wave with μ = 0.6, the ratio of sideband power to total power is:
Explanation: P sb = μ²P c /2 = 0.18P c . P t = P c (1 + 0.18) = 1.18P c . Ratio = 0.18/1.18 = 15.25% ≈ 15.3% . Even at 100% modulation, sideband fraction = 0.5/1.5 = 33.3%. AM is very power-inefficient compared to SSB.
83. By Bessel function analysis, an FM signal with β = 2.4 theoretically has how many significant sidebands on each side?
Explanation: For FM modulation index β, the number of significant sideband pairs is approximately β + 1 to β + 2 from Bessel function analysis. For β = 2.4, significant sidebands extend to n = 4 (J₁ through J₄ are significant). Bandwidth = 2 × 4 × f m (exact) vs. Carson's rule 2(Δf + f m ) = 2(2.4+1)f m = 6.8f m ≈ 2×4f m .
84. The Friis transmission equation gives received power P_r = P_t G_t G_r (λ/4πd)². For a satellite link at 6 GHz (λ = 5 cm), distance d = 36,000 km, with G_t = G_r = 10⁴ (40 dBi), and P_t = 100 W, the received power is:
Explanation: P r = 100 × 10⁴ × 10⁴ × (0.05/(4π × 3.6×10⁷))² = 10¹⁰ × (0.05/4.52×10⁸)² = 10¹⁰ × (1.107×10⁻¹⁰)² = 10¹⁰ × 1.22×10⁻²⁰ ≈ 1.22×10⁻¹⁰ W ≈ 0.12 nW. Even with high-gain antennas and 100 W transmitter, received power is sub-nanowatt — explaining why satellite dishes need LNBs.
85. In QAM-64 modulation, each symbol carries how many bits?
Explanation: QAM-M carries log₂(M) bits per symbol. QAM-64 (64 = 2⁶): 6 bits/symbol . 4G LTE uses QAM-64/256 to achieve high spectral efficiency. QAM-256 carries 8 bits/symbol but requires better SNR.
86. The bandwidth efficiency (bits/s/Hz) of QAM-16 with half the bandwidth used for filtering guard band is:
Explanation: QAM-16: 4 bits/symbol. Nyquist channel capacity = 2B log₂(M) = 2B × 4. With 50% overhead for guard bands, effective BW = 2 × symbol rate → efficiency = 4/2 = 2 bps/Hz . Practical OFDM systems achieve 3–6 bps/Hz.
87. For a geostationary satellite to remain fixed over the equator, its orbital velocity must match Earth's angular velocity ω_E = 7.27 × 10⁻⁵ rad/s. The required orbital radius (G = 6.67×10⁻¹¹, M_E = 6×10²⁴ kg) is approximately:
Explanation: For geostationary orbit: GM/r² = ω²r → r³ = GM/ω² = (6.67×10⁻¹¹ × 6×10²⁴)/(7.27×10⁻⁵)² = 4×10¹⁴/5.285×10⁻⁹ = 7.57×10²² → r = (7.57×10²²)^(1/3) ≈ 42,241 km from Earth's centre, i.e., altitude ≈ 35,786 km. Matches the known value.
88. In a step-index optical fibre, core n₁ = 1.50 and cladding n₂ = 1.47. The critical angle θ_c and numerical aperture NA are:
Explanation: sin θ c = n₂/n₁ = 1.47/1.50 = 0.98 → θ c = sin⁻¹(0.98) ≈ 78.5°. NA = √(n₁²−n₂²) = √(2.25−2.1609) = √0.0891 ≈ 0.298 . Acceptance angle in air = sin⁻¹(NA) ≈ 17.3°.
89. The figure of merit of a communication system is often measured by the Eb/N0 (energy per bit to noise spectral density ratio) because:
Explanation: E b /N 0 = (C/R) / (N 0 ) where C is SNR and R is bit rate. Since bandwidth and SNR both scale with data rate, E b /N 0 normalises these out — it's the fundamental figure comparing different modulations at different rates. BPSK requires E b /N 0 ≈ 9.6 dB for 10⁻⁵ BER; QAM-64 needs ≈ 18 dB.
90. In OFDM (used in 4G/5G), the cyclic prefix serves to:
Explanation: The cyclic prefix (CP) is a copy of the last N_cp samples of each OFDM symbol prepended at the start. In time domain, multipath delays up to CP length appear as a circular convolution, which is easily equalized per-subcarrier after FFT. CP must be ≥ maximum multipath delay spread. LTE uses CP of 4.7 μs (normal) or 16.7 μs (extended).
91. The Nyquist formula for noiseless channel capacity is C = 2B log₂(M) where M is the number of signal levels. A noiseless channel of 3 kHz bandwidth using 16 signal levels has capacity:
Explanation: C = 2 × 3000 × log₂(16) = 6000 × 4 = 24,000 bps = 24 kbps . This is the Nyquist (noiseless) limit. In practice, Shannon's theorem (which includes SNR) further limits capacity. For 3 kHz with SNR = 255: C = 3000 × log₂(256) = 24 kbps — same numerically by coincidence here.
92. In Double Sideband Suppressed Carrier (DSB-SC), the transmitted signal is v(t) = A_m cos ω_m t × A_c cos ω_c t. Expanding this gives frequency components at:
Explanation: Using product-to-sum: cos A cos B = ½[cos(A−B) + cos(A+B)]. So DSB-SC: v(t) = (A_mA_c/2)[cos(ω_c−ω_m)t + cos(ω_c+ω_m)t]. No carrier term at ω_c. Advantage: 100% of power in sidebands. Disadvantage: requires coherent (synchronous) demodulation — complex receiver.
93. The Optimum Working Frequency (OWF) for HF sky-wave communication is chosen as approximately:
Explanation: OWF ≈ 0.85 × MUF. Operating exactly at MUF risks signal loss if the ionosphere weakens slightly. OWF provides an 15% safety margin. In practice, operators choose an OWF that remains below MUF 90% of the time for reliable communication.
94. Delta modulation (DM) is a simplified PCM where each sample is encoded as:
Explanation: Delta modulation: transmit 1 if signal went up, 0 if it went down relative to prediction. Very simple (1 bit/sample) but suffers from slope overload (can't follow fast changes) and granular noise (oscillates around flat segments). Adaptive DM (ADM) varies step size to address these problems.
95. The EIRP (Effective Isotropic Radiated Power) of a satellite transmitter is defined as:
Explanation: EIRP = P t × G t (linear) or EIRP(dBW) = P t (dBW) + G t (dBi). It is the power an isotropic radiator would need to produce the same flux density in the direction of maximum gain. Higher EIRP = stronger signal at receiver. Satellite EIRP typically 40–60 dBW.
96. In 5G NR, millimeter-wave (mmWave) bands (24–100 GHz) offer high capacity but suffer from:
Explanation: mmWave: very high bandwidth (GHz of spectrum available) → Gbps data rates. However: (1) poor building penetration (concrete blocks ~40 dB); (2) atmospheric absorption (O₂ at 60 GHz: ~15 dB/km); (3) rain fade; (4) short range (~200 m). Requires massive MIMO beamforming and dense deployment of small cells.
97. In BPSK modulation, two symbols are represented by phase shifts of 0° and 180°. The minimum bandwidth required to transmit at symbol rate f_s is:
Explanation: By Nyquist, the minimum (theoretical) bandwidth for a passband signal with symbol rate f_s is f_s Hz (double-sided) or f_s/2 each side around the carrier. In practice, raised-cosine filtering with roll-off factor α gives BW = f_s(1+α). BPSK at 1 Msymbols/s needs minimum 1 MHz bandwidth.
98. In a communication receiver, the noise figure F is defined as:
Explanation: Noise Figure F = SNR in /SNR out (linear), or NF = 10 log₁₀(F) dB. F ≥ 1 always (a real device always adds noise). An ideal noiseless amplifier has F = 1 (0 dB). LNA (Low Noise Amplifier) at the front-end of a receiver should have NF
99. An erbium-doped fibre amplifier (EDFA) is used in optical communication because it:
Explanation: EDFA: Er³⁺ ions in the fibre are pumped (980 nm or 1480 nm laser) to an excited state. Signal photons at 1530–1565 nm trigger stimulated emission → direct optical gain (~20–40 dB). Can amplify entire WDM comb simultaneously → key enabler of terabit/s fibre networks. Gain bandwidth ~35 nm.
100. FM capture effect means that an FM receiver:
Explanation: FM capture: if one signal is 3–6 dB stronger than a co-channel interferer, the FM limiter-discriminator captures the stronger signal and essentially ignores the weaker one. This makes FM immune to co-channel interference above a threshold — unlike AM where both signals add linearly. Basis for FM cellular re-use planning.
101. The three basic elements of a communication system in order are:
Explanation: Standard model: Information source → Transmitter → Channel (+ noise) → Receiver → Destination. This model (Shannon-Weaver, 1948) underpins all communication theory.
102. The carrier wave in communication is typically at a _____ frequency than the message signal.
Explanation: Carrier frequency >> message frequency. Example: AM radio 1000 kHz carrier vs. 1 kHz audio; FM radio 100 MHz carrier vs. 15 kHz audio. The high carrier allows practical antenna sizes and avoids signal mixing.
103. Which propagation mode is used by standard AM (medium-wave) radio broadcast stations?
Explanation: AM (530–1710 kHz) propagates as ground wave during the day (range ~500 km). At night, the D layer disappears and sky-wave propagation takes over, extending range to thousands of kilometres — that's why distant AM stations can be received at night.
104. In an AM transmitter, the modulation index must be kept ≤ 1 to avoid:
Explanation: μ > 1 means the carrier is turned off during part of the cycle → the envelope no longer faithfully follows the message → severe nonlinear distortion at the demodulator output.
105. The frequency band used for radar and microwave ovens (2.45 GHz) is:
Explanation: SHF: 3–30 GHz. Microwave ovens use 2.45 GHz (same ISM band as Wi-Fi 2.4 GHz). Radar uses various bands: X-band (10 GHz for weather), Ku-band (police radar), Ka-band (speed guns). All are SHF/EHF microwaves.
106. What is the main advantage of LEO satellite constellations (like Starlink) over single GEO satellites?
Explanation: GEO: 36,000 km altitude, ~250 ms delay, cannot see poles (>75° latitude). LEO: 500–1200 km, ~5–15 ms delay, polar coverage, but needs 500–12,000+ satellites for global coverage. Starlink (SpaceX) has ~5,000 operational satellites at ~550 km.
107. The primary advantage of digital communication over analog is:
Explanation: Digital signals can be regenerated — noisy 0s and 1s are detected and retransmitted as clean pulses. Analog amplifiers amplify noise along with the signal. Digital also enables error correction, encryption, and compression.
108. Single-mode optical fibre has a very small core diameter (8–10 μm) in order to:
Explanation: When core diameter ≈ wavelength (V-number 100 GHz·km). Used for all long-haul (>1 km) links.
109. A broadcasting station uses a 500 W carrier modulated 80% by audio. The efficiency η (fraction of total power in sidebands) is:
Explanation: P sb = μ²P c /2 = 0.64×500/2 = 160 W. P t = 500+160 = 660 W. η = 160/660 = 24.24% ≈ 24.2% . This poor efficiency motivates SSB transmission where 100% of power is in the sideband.
110. A digital telephone channel has bandwidth 4 kHz and SNR = 255. Using Shannon's formula the maximum bit rate is:
Explanation: C = B log₂(1+SNR) = 4000 × log₂(256) = 4000 × 8 = 32,000 bps = 32 kbps . Note log₂(256) = 8. In practice, voice is encoded at 64 kbps (PCM, G.711) within this channel — possible because PCM doesn't approach the Shannon limit; advanced codecs (G.729) achieve toll-quality at 8 kbps.
111. A shortwave (HF) signal at 15 MHz is sent from a transmitter. The critical frequency is 8 MHz. Since 15 > 8 MHz, the wave:
Explanation: At normal incidence, f = 15 MHz > f c = 8 MHz → wave passes through to space. But at an oblique angle θ, effective vertical frequency = 15 cos θ. For reflection: 15 cos θ ≤ 8 → cos θ ≤ 0.533 → θ ≥ 57.8°. So the signal must be directed at least 57.8° from vertical (i.e., within 32.2° of horizontal) for ionospheric return.
112. The bit error rate (BER) of BPSK in AWGN is given by Q(√(2Eb/N0)). If Eb/N0 = 9 dB (≈ 7.94), BER is approximately:
Explanation: At E b /N 0 = 9 dB: Q(√(2×7.94)) = Q(√15.88) = Q(3.985) ≈ Q(4) ≈ 3.17×10⁻⁵ ≈ 10⁻⁵ . This is a standard design target for voice (10⁻³) is acceptable; data (10⁻⁶) needed before FEC. After turbo/LDPC coding, 10⁻⁵ uncoded BER becomes effectively zero.
113. In a cellular network with a 7-cell reuse pattern and total available spectrum of 25 MHz, each cell gets spectrum of approximately:
Explanation: With N=7 reuse pattern, each cell uses Total/N = 25/7 ≈ 3.57 MHz . A smaller reuse factor N increases per-cell spectrum (more capacity) but requires better signal separation. Modern networks use fractional frequency reuse and interference coordination for N=1 (each cell uses all spectrum) with sophisticated interference management.
114. In an optical fibre system, chromatic dispersion causes pulse broadening because:
Explanation: Chromatic (material) dispersion: n = n(λ) (dispersion relation of glass). A source with spectral width Δλ causes pulse broadening Δτ = D × L × Δλ where D is dispersion coefficient (ps/nm/km). At 1310 nm, D ≈ 0 for standard SMF (zero-dispersion wavelength). At 1550 nm, D ≈ 17 ps/nm/km — requires dispersion-compensating fibre for long links.
115. The term 'channel capacity' as defined by Shannon represents:
Explanation: Shannon's channel capacity C = B log₂(1 + S/N) is the theoretical maximum rate at which information can be transmitted with arbitrarily low error probability. Actual systems approach but never exceed C. It is achieved only with infinitely complex coding in the limit. This theorem underpins all modern coding theory.
116. A half-wave dipole antenna transmitting at 150 MHz has a physical length of approximately:
Explanation: λ = c/f = 3×10⁸ / 1.5×10⁸ = 2 m. Half-wave dipole length = λ/2 = 1 m . At 150 MHz (VHF), a 1 m dipole is physically practical. This is why VHF/UHF antennas are manageable in size while AM radio towers must be tens of metres tall.
117. A message signal m(t) = 2 cos(2π×5000t) modulates a carrier A_c cos(2π×10⁶t) in AM with μ = 0.5. The amplitude of each sideband component is:
Explanation: AM: v(t) = A c [1 + μcos ω m t]cos ω c t = A c cos ω c t + (μA c /2)cos(ω c +ω m )t + (μA c /2)cos(ω c −ω m )t. Each sideband amplitude = μA c /2 = 0.5A c /2 = A c /4 . Sideband power = (A c /4)²/2 = A c ²/32.
118. The telephone network originally used 4 kHz bandwidth channels (300–3400 Hz voice). With PCM at 8-bit, 8000 samples/s, the bit rate per channel is 64 kbps. A T1 line carrying 24 such channels plus framing has a total bit rate of:
Explanation: 24 channels × 64 kbps = 1536 kbps. Plus framing bit: 1 bit/frame × 8000 frames/s = 8 kbps. Total = 1536 + 8 = 1544 kbps = 1.544 Mbps . This is the North American T1 standard. Europe uses E1: 32 channels (30 voice + 2 signalling) = 2.048 Mbps.
119. Which statement about ground wave propagation is correct?
Explanation: Ground waves follow Earth's curvature but are attenuated by ground absorption (energy dissipated as heat in soil/sea) and 1/r geometric spreading. Seawater (high conductivity) supports longer ground-wave range than dry land. Ground wave range at 1 MHz ≈ 500–1000 km over sea, ~100–300 km over land.
120. In a superheterodyne AM receiver, the intermediate frequency (IF) is typically 455 kHz. If the receiver is tuned to 1000 kHz, the local oscillator frequency is:
Explanation: f LO = f RF + f IF = 1000 + 455 = 1455 kHz (high-side injection, standard for AM receivers). The mixer produces |f LO − f RF | = 455 kHz IF regardless of which station is tuned. The image frequency = f LO + f IF = 1910 kHz must be rejected by the RF preselector filter.