Question 1

Propagation delay in geosynchronous satellite links What is the approximate maximum signal propagation time (end-to-end round-trip delay) in a GEO satellite transmission, accounting for up-and-down paths via the satellite?

Accepted Answer

Introduction / Context: GEO satellite links exhibit noticeable latency because signals must travel about 36,000 km up to the satellite and back down to Earth—often twice (uplink and downlink in each direction). Understanding this delay is vital for voice, data, and interactive applications over satellite links. Given Data / Assumptions: Speed of electromagnetic waves ~3 x 10^8 m/s in free space. One-way GEO hop (ground → satellite → ground) ≈ 2 x 35,786 km of vacuum path plus slant geometry. Round-trip (two GEO hops in a conversation) is roughly twice the one-way hop delay. Concept / Approach: A typical one-way end-to-end GEO path is on the order of 250–280 ms depending on geometry. The round-trip time (RTT) is then approximately double, around 500–560 ms. Industry shorthand often quotes ~540 ms as a representative maximum end-to-end delay for GEO links. Step-by-Step Solution: Compute one-way hop: ~36,000 km up + ~36,000 km down ≈ 72,000 km.Time ≈ 72,000,000 m / (3 x 10^8 m/s) ≈ 0.24 s → ~240–270 ms allowing for slant paths.Round-trip ≈ 2 x 270 ms ≈ 540 ms. Verification / Alternative check: Network measurements over GEO confirm about half-second RTT under typical conditions. Why Other Options Are Wrong: 270 ms (b) is closer to a one-way delay; 135 ms (c) is too small; 1080 ms (d) is roughly double RTT; 50 ms (e) reflects terrestrial or LEO links, not GEO. Common Pitfalls: Confusing one-way and round-trip delays; ignoring processing delays, which add modestly to propagation time. Final Answer: 540 milliseconds

Question 2

Multiple access in satellite transponders Which access methods allow many Earth stations to share a single satellite transponder simultaneously without mutual interference?

Accepted Answer

Introduction / Context: Satellite transponders are expensive shared resources. Multiple access techniques allow multiple users to efficiently share bandwidth and power while keeping interference manageable through separation in frequency, time, or code domain. Given Data / Assumptions: One bent-pipe transponder with fixed bandwidth and gain/back-off limits. Users require simultaneous or quasi-simultaneous access. Orthogonality or separation is needed to avoid unacceptable interference. Concept / Approach: FDMA allocates distinct frequency slots; TDMA assigns non-overlapping time slots; CDMA overlays users with distinct spreading codes that are ideally orthogonal. Each approach permits concurrent usage under proper planning, synchronization, and power control. Step-by-Step Solution: List classic multiple access families: FDMA, TDMA, CDMA.Explain orthogonality: frequency separation, time slotting, or code separation.Conclude that all three are valid means to share a transponder. Verification / Alternative check: Commercial and military satellite networks have historically deployed each of these methods, sometimes hybridized (e.g., MF-TDMA). Why Other Options Are Wrong: Choosing only one technique ignores the other established methods; “None of the above” is contradicted by decades of practice. Common Pitfalls: Assuming a single method is universal; in reality, the selection depends on service type, equipment, and link budgets. Final Answer: All of the above

Question 3

Envelope detection of amplitude-modulated (AM) signals For a simple diode (envelope) detector to recover the message correctly from an AM waveform, which spectral components must be present?

Accepted Answer

Introduction / Context: An envelope detector tracks the amplitude of a passband signal to recover the baseband message. For the envelope to mirror the original modulation, the transmitted AM spectrum must include sufficient carrier amplitude and at least one sideband component. Given Data / Assumptions: Simple diode detector followed by RC smoothing (no synchronous carrier recovery). Conventional full-carrier AM (DSB-LC) or single-sideband with carrier (SSB+C) can be envelope-detected if the carrier is strong enough. Suppressed-carrier systems require coherent detection instead. Concept / Approach: In DSB-LC, the transmitted waveform’s envelope equals the message (scaled) provided the modulation index is within limits and carrier is large relative to sidebands. Even SSB+C can be envelope detected because the residual carrier restores a detectable envelope. However, if the carrier is suppressed (as in SSB-SC or DSB-SC), an envelope detector fails because there is no stable amplitude reference. Step-by-Step Solution: State envelope detection requirement: presence of carrier and at least one sideband.Show that a full-amplitude carrier guarantees a faithful envelope proportional to the message.Conclude that “one sideband plus full carrier” is sufficient and necessary for simple envelope detection. Verification / Alternative check: AM broadcast uses DSB-LC specifically to enable cheap envelope detectors in receivers. Why Other Options Are Wrong: (b) While both sidebands are used in DSB-LC, strictly one sideband is enough if the carrier is full—thus (a) is the minimum sufficient condition; (c) with suppressed carrier cannot be envelope detected; (d) a small fraction of carrier may fail under fading; (e) SSB-SC needs synchronous detection. Common Pitfalls: Assuming both sidebands are mandatory; forgetting that the carrier amplitude must be adequate to prevent envelope distortion. Final Answer: One sideband and a full-amplitude carrier

Question 4

Satellite communication link budget: identify the incorrect statement about free-space path loss and antenna apertures Which of the following statements about link loss (free-space path loss plus antenna effects) in a satellite link is incorrect? Co…

Accepted Answer

Introduction / Context: Satellite link budgets balance transmitted power, antenna gains, and propagation losses to predict received power. Understanding how free-space path loss varies with frequency, distance, and antenna aperture is fundamental to designing reliable satellite links. Given Data / Assumptions: Line-of-sight space link (free-space propagation). Standard relationships: FSPL = (4 * pi * R * f / c)^2 in linear terms. Antenna effective aperture A_e relates to gain G via A_e = (G * lambda^2) / (4 * pi). Concept / Approach: In free space, path loss grows with frequency squared and distance squared. However, when physical antenna apertures are fixed, their gain increases with f^2 (since G ∝ A_phys * (4 * pi / lambda^2)), which can compensate the frequency dependence if apertures are constant. Confusion often arises between holding gain constant versus holding physical size constant. Step-by-Step Solution: FSPL (linear) = (4 * pi * R * f / c)^2 ⇒ increases as f^2 and R^2.Received power Pr ∝ Pt * Gt * Gr / FSPL.For fixed frequency, increasing A_e,t and A_e,r increases G and raises Pr, effectively reducing link loss.For fixed gains and f, Pr falls as 1/R^2; thus link loss grows as R^2.Therefore, any statement claiming link loss is inversely proportional to f^2 for fixed apertures is incorrect. Verification / Alternative check: Using the Friis equation in linear form, Pr = Pt * Gt * Gr * (lambda / (4 * pi * R))^2. Since lambda = c/f, higher f reduces (lambda)^2 term, lowering Pr if gains are held constant. Why Other Options Are Wrong: Option B: Correct: Pr grows with A_e,t * A_e,r at fixed f, so effective link loss falls. Option C: Correct: Higher gain reduces required Pt for the same Pr. Option D: Correct: FSPL grows as R^2. Option E: Incorrect as a general rule; downlink loss is not inherently negligible and depends on EIRP, G/T, and frequency bands. Common Pitfalls: Mixing up fixed-aperture versus fixed-gain scenarios; forgetting that higher frequency also allows higher gain for a given physical size. Final Answer: For fixed antenna sizes (physical apertures), free-space path loss increases with frequency approximately as f^2, not decreases.

Question 5

Telephone echo on a satellite voice channel: practical suppression method In a geostationary satellite voice link where long round-trip delay causes audible echo to the telephone user, which technique is commonly used in practice to remove or greatl…

Accepted Answer

Introduction / Context: Geostationary links have a one-way delay of roughly 240 ms, leading to a round-trip delay near 480 ms. This delay makes hybrid reflections and impedance mismatches audible as echo in telephony. Echo control is therefore essential for user comfort. Given Data / Assumptions: Long GEO satellite propagation delay. Two-wire/four-wire hybrids in telephony can reflect part of the signal. Goal: suppress audible echo at the talker’s ear. Concept / Approach: Echo suppressors and echo cancellers detect the presence of far-end speech and attenuate or subtract the echo path. Modern systems use adaptive echo cancellers that model the echo path and remove the delayed replica of the talker’s own speech. Step-by-Step Solution: Identify the echo source: hybrid reflections and impedance mismatch.Recognize long GEO delays exacerbate echo audibility.Apply an echo suppressor/canceller which monitors speech activity and attenuates or cancels the echo path adaptively.Result: echo power is reduced to inaudible levels. Verification / Alternative check: Telecom standards and satellite telephony deployments universally specify echo control for long-delay circuits; field experience confirms their effectiveness. Why Other Options Are Wrong: Vocoder/Multiplexer: Compression or multiplexing does not inherently cancel echo. “Digital techniques”: Too vague; echo requires specific cancellation algorithms. Higher IF bandwidth: Unrelated to echo delay/reflection. Common Pitfalls: Confusing noise reduction with echo cancellation; they solve different problems. Final Answer: An echo suppressor (or echo canceller)

Question 6

Altitude of a geostationary satellite relative to Earth’s surface The distance of a geostationary (synchronous, equatorial) satellite from Earth’s surface is approximately how many kilometres?

Accepted Answer

Introduction / Context: Geostationary satellites orbit with a period equal to Earth’s rotational period and remain fixed relative to a point on the equator. Their altitude is set by orbital mechanics. Given Data / Assumptions: Geostationary orbit radius from Earth’s center ≈ 42,164 km. Mean Earth radius ≈ 6,378 km. Concept / Approach: Altitude above Earth’s surface = orbital radius − Earth’s mean radius ≈ 42,164 − 6,378 ≈ 35,786 km. Rounded values in practice are often quoted as ~35,800–35,900 km. Step-by-Step Solution: R_orbit ≈ 42,164 km.R_Earth ≈ 6,378 km.Altitude ≈ 42,164 − 6,378 ≈ 35,786 km.Closest listed choice ≈ 35,900 km. Verification / Alternative check: Standard references list geostationary altitude at about 35,786 km; engineering texts commonly round to ~35,800–35,900 km. Why Other Options Are Wrong: 300 km and 10,000 km correspond to low-Earth and medium-Earth regimes; 5 km is nonsensical; 22,300 km is miles (≈ 22,300 miles ≈ 35,900 km), not kilometres. Common Pitfalls: Confusing miles and kilometres; ensure consistent units. Final Answer: 35,900

Question 7

Order of received power from a GEO communications satellite at the earth station A typical carrier power received at the input of a properly pointed earth-station receiver from a geostationary communications satellite is usually of the order of a fe…

Accepted Answer

Introduction / Context: Despite large Effective Isotropic Radiated Power (EIRP) at the satellite and high-gain earth-station antennas, the enormous free-space path loss over ~36,000 km reduces received carrier power to very small levels. Given Data / Assumptions: GEO range ~ 3.6 * 10^7 m. Ku/Ka-band links with high gains but large FSPL. Receiver input sees narrowband carrier powers on the order of 10^−12 to 10^−9 W, depending on system. Concept / Approach: Friis transmission formula indicates Pr = Pt * Gt * Gr * (lambda / (4 * pi * R))^2. Even with large Gt and Gr, R is so large that Pr is in the pico- to nanowatt range. “A few picowatts” is a common order-of-magnitude description for many GEO downlinks. Step-by-Step Solution: Estimate FSPL at, say, 12 GHz and R ≈ 36,000 km.Compute lambda = c/f ≈ 0.025 m; (lambda/(4piR))^2 ~ very small.Combine with typical EIRP and G/T to infer Pr ~ 10^−12…10^−9 W. Verification / Alternative check: Link budget examples in textbooks yield received carrier powers typically in the picowatt to nanowatt range for GEO links. Why Other Options Are Wrong: Milliwatts/Watts/Kilowatts are many orders of magnitude too large for post-FSPL carrier at LNB input under normal operation. Microjoules are energy units, not power. Common Pitfalls: Confusing EIRP with received power; ignoring FSPL magnitude at GEO distances. Final Answer: Picowatts

Question 8

Why do satellite solar arrays degrade in orbit? The gradual loss of power output from a communications satellite’s solar cell arrays over mission life is mainly caused by:

Accepted Answer

Introduction / Context: Solar array degradation is a key factor in end-of-life (EOL) power budgeting for satellites. Understanding the physics of degradation helps in selecting cell types and shielding strategies. Given Data / Assumptions: GEO environment includes trapped electrons and protons, and solar radiation. Solar cells typically GaAs/Ge or multi-junction designs with coverglass. Concept / Approach: Radiation damages the semiconductor lattice, creating defects that act as recombination centers. This reduces minority carrier lifetime, open-circuit voltage, and fill factor, thus decreasing power output. Coverglass thickness is chosen to mitigate electron/proton damage while balancing mass. Step-by-Step Solution: Identify dominant GEO hazards: electrons and protons from radiation belts; solar UV.Recognize displacement damage in cell materials reduces efficiency.Model degradation as percent loss per year; include in EOL power sizing. Verification / Alternative check: Flight data consistently shows radiation-induced degradation; meteoric dust is negligible in GEO; outgassing/oxidation minimal in vacuum at GEO. Why Other Options Are Wrong: Meteoric dust: Insignificant deposition at GEO. Intrinsic resistivity increase/leakage: Not the primary mechanism; radiation damage dominates. Oxidation at GEO: Negligible atmospheric oxygen. Common Pitfalls: Attributing degradation primarily to thermal cycling; it contributes but radiation is the main driver. Final Answer: Bombardment by energetic charged particles (e.g., electrons) in the space environment

Question 9

Typical lifetime of a geostationary communications satellite What is a representative design lifetime for a modern geostationary communications satellite bus (from launch to end-of-life), considering fuel for station-keeping and expected component d…

Accepted Answer

Introduction / Context: Satellite lifetime planning trades off launch mass, propellant for station-keeping, power subsystem margins, and radiation effects to ensure reliable service over the intended mission duration. Given Data / Assumptions: Traditional GEO comsats historically designed for ~10–15 years. Fuel usage for north-south and east-west station-keeping is a limiting factor. Solar array and battery degradation also limit usable lifetime. Concept / Approach: Classical textbook and exam figures often cite 10 years as the nominal design life, with many modern platforms achieving 12–15+ years. For standardized MCQ context, 10 years is the canonical choice. Step-by-Step Solution: Identify major life limiters: propellant, solar array degradation, battery cycle life.Select the typical value used in foundational questions: ~10 years.Note that actual missions may exceed nominal life depending on fuel margins. Verification / Alternative check: Many legacy GEO platforms exceeded 10 years, but exam conventions commonly use the 10-year benchmark. Why Other Options Are Wrong: 5 years is short for GEO comsats; 20 or 50 years exceed typical propellant and component limits for conventional buses. Common Pitfalls: Confusing modern extended-life buses with standard textbook values used for concept checks. Final Answer: 10 years

Question 10

Meaning of Shannon’s law in communications Shannon’s fundamental theorem relates which pair of quantities for a channel with a given bandwidth and noise level?

Accepted Answer

Introduction / Context: Shannon’s channel capacity theorem sets an upper bound on error-free data rate over a noisy channel given its bandwidth and signal-to-noise ratio, forming the foundation of modern digital communication system design. Given Data / Assumptions: Additive white Gaussian noise (AWGN) channel model. Bandwidth B and SNR are known. Concept / Approach: The theorem states C = B * log2(1 + S/N), where C is capacity in bits/s. It ties achievable information rate to both bandwidth and SNR, independent of modulation specifics, assuming ideal coding with arbitrarily low error probability. Step-by-Step Solution: Identify capacity formula: C = B * log2(1 + S/N).Thus capacity depends on both bandwidth and SNR.Select the option matching this relationship. Verification / Alternative check: As S/N → ∞, C grows approximately as B * log2(S/N); as S/N → 0, capacity trends to zero. Why Other Options Are Wrong: Other options cite unrelated pairs (antenna gain, losses) or omit bandwidth; Shannon’s law fundamentally involves capacity, SNR, and bandwidth. Common Pitfalls: Confusing capacity with throughput; practical systems approach but never reach Shannon capacity due to coding and implementation constraints. Final Answer: Information-carrying capacity to signal-to-noise ratio (and bandwidth)

Question 11

Primary selection criterion for a satellite system from a service planner’s perspective When selecting a satellite system for a service deployment, the first determining factor (before detailed link budgeting) is typically its:

Accepted Answer

Introduction / Context: Before fine technical details, a provider must ensure the satellite can physically illuminate the required geographic market. This is dictated by the satellite’s beam coverage (footprint). Given Data / Assumptions: Service targets a defined region (country/continent/offshore zone). Different satellites offer different beams and footprints. Concept / Approach: Coverage area is a gating criterion: if the satellite cannot cover the target region with sufficient EIRP and G/T, no amount of link optimization will make it suitable. Only after confirming footprint do planners compare EIRP, transponder bandwidth, polarization, and spectrum. Step-by-Step Solution: Identify the business region.Select satellites whose footprints cover that region with adequate levels.Then perform link budgets using EIRP, G/T, and bandwidth to pick among candidates. Verification / Alternative check: Operator coverage maps are consulted first; only covered areas proceed to technical and commercial evaluation. Why Other Options Are Wrong: Raw EIRP, antenna size, gain, or IF are secondary unless the region is covered; they cannot compensate for absent footprint. Common Pitfalls: Over-focusing on EIRP numbers without checking whether the beam actually covers the service area of interest. Final Answer: Coverage area (footprint over the intended service region)

Question 12

Effective noise temperature of Earth as seen from space What is the approximate average noise temperature of the Earth when observed from space in the microwave region (clear-sky, broad view)?

Accepted Answer

Introduction / Context: Effective noise temperature describes the equivalent blackbody temperature that would produce the observed radiometric emission. For Earth as viewed from space, this value is close to its effective radiating temperature used in climate models. Given Data / Assumptions: Earth’s effective radiating temperature ≈ 255 K. Microwave observations integrate land, ocean, and cloud cover. Concept / Approach: Radiative balance and radiometry establish that the Earth emits like a ~255 K blackbody on average. Satellite system noise calculations often use a similar figure for Earth-viewed background temperature. Step-by-Step Solution: Recall T_effective ≈ 255 K from planetary energy balance.Choose the closest option: 254 K. Verification / Alternative check: Remote sensing references routinely cite 250–260 K as the typical range for effective Earth radiative temperature. Why Other Options Are Wrong: 303 K corresponds to warm surface temperatures; 100 K and 500 K are far from Earth’s mean radiative properties. Common Pitfalls: Confusing surface temperature with effective radiative temperature; they are not identical. Final Answer: 254 K

Question 13

Typical microwave sky noise temperature for satellite links In clear-sky conditions and away from strong galactic sources, what is a representative noise temperature of the sky as seen by an earth-station antenna in the microwave band?

Accepted Answer

Introduction / Context: The sky noise temperature is an important contributor to the G/T (gain-over-temperature) figure of an earth station. Clear-sky microwave observations typically see relatively low background temperatures. Given Data / Assumptions: Clear sky, low elevation sidelobes minimized. Microwave band away from atmospheric absorption peaks. Concept / Approach: At many satellite communication bands (e.g., C/Ku under clear sky), the antenna sees background dominated by cosmic microwave background, residual atmospheric emission, and spillover to warm ground. Well-designed antennas achieve sky noise temperatures ~20–50 K; 30 K is a common representative value. Step-by-Step Solution: Consider clear sky at suitable elevation angles.Account for atmospheric emission and cosmic background.Representative T_sky ≈ 30 K fits standard link budget assumptions. Verification / Alternative check: Link budget examples and G/T calculations often adopt T_sky in the tens of kelvin under clear conditions. Why Other Options Are Wrong: 0 K is unattainable; 273 K corresponds to ambient thermal ground; 100 K is high for clear sky; 600 K is unrealistic. Common Pitfalls: Ignoring spillover and atmospheric absorption peaks, which can raise system noise significantly if not controlled. Final Answer: 30 K

Question 14

Indian space communications context The 1150 kg geosynchronous satellite INSAT-1A parked about 36,000 km above India significantly enhanced which national capabilities?

Accepted Answer

Introduction / Context: INSAT-1A was an early Indian National Satellite placed in a geosynchronous orbit (~36,000 km altitude) to support nationwide services. Understanding what geostationary platforms primarily deliver helps distinguish civil communication/meteorology roles from defense intelligence payloads. Given Data / Assumptions: Mass about 1150 kg. Orbit: geosynchronous over India (geostationary slot). Payload mix typical of INSAT generation: communication transponders and meteorology instruments. Concept / Approach: Geostationary satellites remain fixed relative to an observer on Earth, making them ideal for continuous broadcast, telephony trunking, data relay, and weather imaging over a broad footprint. Intelligence gathering is generally associated with specialized reconnaissance satellites in very different orbits (e.g., LEO sun-synchronous) and is not a core INSAT objective. Step-by-Step Solution: Identify civil services commonly provided: national telecommunication links, television distribution, and meteorological imaging/telemetry.Map these to the options: domestic communications → option (b); meteorology → option (c).Exclude defense-oriented intelligence collection because it is not a standard INSAT mission.Therefore the combined correct choice is both (b) and (c). Verification / Alternative check: Historical profiles of INSAT satellites list transponders (for TV, telephony, data) and a VHRR/meteorology package for continuous weather monitoring over the subcontinent. Why Other Options Are Wrong: Intelligence gathering capacity: not a stated objective for INSAT-1A. Only communications or only meteorology: incomplete; INSAT combined both. None of these: contradicted by documented mission roles. Common Pitfalls: Confusing geostationary civil platforms with military ISR satellites. Assuming any geostationary satellite supports all space services; payload defines capability. Final Answer: Both (b) and (c)

Question 15

Signal formats over satellites For satellite transmission, which device converts an analog baseband signal into a digital bit stream (and vice versa) before uplinking?

Accepted Answer

Introduction / Context: Modern satellite links often carry digitally encoded voice, video, and data. The first step for an analog source (e.g., voice) is conversion to a digital representation. Distinguishing among front-end devices—codec, modem, and transponder—prevents conceptual mix-ups. Given Data / Assumptions: Analog source (voice/video) must be digitized for PCM or related schemes. Satellite payload may be a bent-pipe (nonregenerative) transponder. Ground segment includes baseband processing and RF equipment. Concept / Approach: A codec performs analog-to-digital and digital-to-analog conversion along with source coding (e.g., PCM, ADPCM). A modem maps bits to waveforms suitable for the RF channel (e.g., QPSK) but does not perform the actual A/D conversion of the program material. A transponder simply frequency-translates and amplifies the passband; it does not digitize content. A compandor only adjusts dynamic range, typically within analog systems. Step-by-Step Solution: Need: analog → digital conversion of baseband.Device that does A/D and D/A: the codec.Modem and transponder roles are downstream (mapping bits to RF and repeating). Verification / Alternative check: Any PCM chain starts with a codec sampling at the Nyquist rate (or higher) with defined quantization bits per sample before channel coding and modulation. Why Other Options Are Wrong: Transponder: passband repeater; no source A/D. Modem: bit-to-waveform mapping; not baseband A/D. Compandor: dynamic range processing only. Frequency synthesizer: generates LO frequencies, not A/D. Common Pitfalls: Calling any box that “handles digital” a modem; the codec is the A/D front end. Assuming the satellite digitizes content; most GEO payloads are bent-pipe. Final Answer: Codec (coder–decoder / A-to-D and D-to-A)

Question 16

Frequency allocations overview The lowest commonly used frequency band for satellite communications in the classical fixed-satellite service is approximately how many gigahertz (GHz)?

Accepted Answer

Introduction / Context: Satellite links are assigned specific microwave bands. Historically, the fixed-satellite service used C-band (around 4/6 GHz) and later Ku-band (11/14 GHz) and Ka-band (20/30 GHz). The question asks for the lowest frequency broadly associated with satellite communications in this classical context. Given Data / Assumptions: Classical FSS allocations (C-band upward) are intended. Options provided: 0.8, 3, 18, and 30 GHz. We select the closest representative threshold value. Concept / Approach: Traditional satellite textbooks treat the microwave window for FSS starting near a few gigahertz. While some systems use L-band around 1–2 GHz, the earliest mainstream commercial satcom is widely referenced from roughly 3–4 GHz upward. Among the options, 3 GHz best represents the lower end of “typical” satellite microwave operations. Step-by-Step Solution: Compare given values to canonical bands: C (4/6 GHz), Ku (12/14 GHz), Ka (20/30 GHz).Identify the lowest plausible marker among options that aligns with classical FSS: 3 GHz. Verification / Alternative check: A quick band ladder shows 3–4 GHz as the start of many reference tables for C-band downlinks, making 3 GHz the best rounding choice presented here. Why Other Options Are Wrong: 0.8 GHz: below typical commercial FSS microwave allocations discussed in standard intros. 18 GHz and 30 GHz: these map to Ku/Ka upper portions, not the lowest. Common Pitfalls: Confusing mobile-satellite L-band with the legacy FSS entry point. Assuming the exact downlink number; the question wants a representative lower bound. Final Answer: 3 GHz

Question 17

Error performance on satellite circuits A typical uncoded bit error rate (BER) figure for satellite links is often cited as approximately one bit error in how many transmitted bits?

Accepted Answer

Introduction / Context: Bit error rate (BER) benchmarks help designers specify link budgets, coding, and modulation. Satellite links without heavy forward-error correction historically achieve BERs around 10^-5 to 10^-7, while modern coded systems may reach 10^-9 or better. Given Data / Assumptions: “Typical” satellite circuit with reasonable carrier-to-noise and modest coding. Comparative options span poor to excellent BER values. Concept / Approach: For legacy or minimally coded links, BER ≈ 10^-6 to 10^-7 is a common planning value. Therefore, “one error in 10^7 bits” is a representative figure frequently quoted in introductory texts and exams. Step-by-Step Solution: Translate phrasing: “one error in N bits” → BER ≈ 1/N.Select N consistent with typical uncoded or lightly coded satellite links: N ≈ 10^7. Verification / Alternative check: Link budget tools often target post-demodulation BER ≤ 10^-6 for voice/data acceptability before higher-layer error detection/correction. Why Other Options Are Wrong: 10^3 or 10^2: unacceptably high BER for most services. 10^10: extremely stringent and usually requires strong coding and high C/N. 10^5: achievable but not the “typical” textbook figure requested here. Common Pitfalls: Confusing pre-FEC and post-FEC BER targets. Assuming a single universal BER; practical values depend on service class. Final Answer: 10^7 bits

Question 18

Why GEO launches favor the equatorial plane Geosynchronous satellites are placed in the Earth’s equatorial plane primarily because only this plane yields what key property?

Accepted Answer

Introduction / Context: A satellite is geosynchronous if its orbital period equals Earth’s rotation. It is geostationary only if the orbit is both geosynchronous and equatorial with zero eccentricity, making the satellite appear fixed in the sky. Given Data / Assumptions: Target: continuous coverage of a fixed Earth footprint. Orbit altitude: ~35,786 km (geosynchronous radius). Eccentricity ≈ 0 and inclination ≈ 0 for truly “stationary.” Concept / Approach: The equatorial plane with zero inclination ensures no north–south apparent motion. Zero eccentricity ensures no east–west oscillation in longitude. Together these conditions produce a geostationary satellite fixed above one longitude, ideal for broadcast and point-to-point services requiring stationary ground antennas. Step-by-Step Solution: Recognize distinction: geosynchronous (period match) vs geostationary (period + equatorial + circular).Only equatorial, circular orbits provide a “stationary” sub-satellite point.Therefore the unique property provided by the equatorial plane is stationarity. Verification / Alternative check: Inclined geosynchronous orbits produce figure-eight ground tracks (analemmas), requiring tracking antennas—not stationary. Why Other Options Are Wrong: 24-hour period occurs for any inclination if the semi-major axis is correct. Global communication is not possible with a single GEO; multiple satellites are needed. Zero-gravity environment is unrelated; microgravity exists in all orbits. Common Pitfalls: Using “geosynchronous” and “geostationary” interchangeably. Ignoring the role of eccentricity in apparent satellite drift. Final Answer: A stationary (geostationary) satellite relative to Earth’s surface

Question 19

PCM rate for digitizing a colour TV baseband A colour TV video signal has 4.6 MHz bandwidth and is digitized using 10 bits per sample. Using basic PCM (Nyquist) sampling, what is the required bit rate in kilobits per second (kb/s)?

Accepted Answer

Introduction / Context: Pulse Code Modulation (PCM) digitizes an analog baseband by sampling at or above twice its highest frequency component and quantizing each sample to a fixed number of bits. Determining the raw bitrate helps appreciate why video compression is necessary in practice. Given Data / Assumptions: Video baseband bandwidth B = 4.6 MHz. Quantization = 10 bits per sample. Use Nyquist sampling: f_s = 2 * B for a low-pass baseband. No compaction, coding overhead, or compression considered. Concept / Approach: For baseband PCM, the minimum sampling rate is f_s = 2 * B. The raw bit rate R is then R = f_s * bits_per_sample. Unit conversion from bits per second to kilobits per second is straightforward (divide by 10^3). Step-by-Step Solution: Compute sampling rate: f_s = 2 * 4.6 * 10^6 = 9.2 * 10^6 samples/s.Multiply by quantization: R = 9.2 * 10^6 * 10 = 92 * 10^6 bits/s.Convert to kb/s: 92 * 10^6 / 10^3 = 92,000 kb/s. Verification / Alternative check: Order-of-magnitude sanity check: several tens of megabits per second are expected for raw uncompressed video; 92 Mb/s is consistent. Why Other Options Are Wrong: 920 or 230 kb/s: far too low for raw 4.6 MHz video at 10 bits/sample. 23,000 or 46,000 kb/s: underestimates because they imply sampling below 2B or fewer bits per sample. Common Pitfalls: Using f_s = B instead of 2B for baseband signals. Forgetting to multiply by bits per sample. Ignoring overheads; here the question asks for basic PCM only. Final Answer: 92,000 kb/s

Question 20

Velocity of a geosynchronous satellite Assuming Earth radius = 6400 km and geosynchronous satellite altitude ≈ 36,000 km, estimate the orbital velocity in km/h.

Accepted Answer

Introduction / Context: Geosynchronous orbits have a period equal to Earth’s rotation (~24 h). At the geostationary radius, a satellite’s tangential velocity follows directly from distance traveled in one orbital period. Given Data / Assumptions: Earth radius R_E ≈ 6400 km. GEO altitude h ≈ 36,000 km. Orbital radius R = R_E + h ≈ 42,400 km. Period T ≈ 24 h (using 24 for simplicity; sidereal day is ~23.934 h). Concept / Approach: The distance traveled per orbit is the circumference: 2 * pi * R. The average orbital speed v is distance / time. A quick calculation provides the expected ~11,000 km/h velocity. Step-by-Step Solution: Compute orbital radius: R = 6400 + 36,000 = 42,400 km.Compute circumference: C = 2 * pi * 42,400 ≈ 2 * 3.1416 * 42,400 ≈ 266,000 km (approx.).Compute speed: v = C / T ≈ 266,000 km / 24 h ≈ 11,083 km/h ≈ 11,100 km/h. Verification / Alternative check: Using the sidereal day T = 23.934 h yields ~11,120 km/h, confirming the order of magnitude and the option rounding. Why Other Options Are Wrong: 28,000 or 36,000 km/h: too high for GEO; such speeds correspond to lower orbits. 15,000 km/h: still high compared to the computed value. 7,900 km/h: closer to LEO orbital speeds in m/s (≈7.9 km/s), not km/h. Common Pitfalls: Mixing units: km/s vs km/h. Forgetting to add Earth’s radius to altitude for orbital radius. Final Answer: 11,100 km/h

Satellite Communication Questions