Question 1

Space link performance metric: The overall quality of a satellite (space) communication link is commonly measured in terms of which ratio?

Accepted Answer

Introduction / Context:Space-link analysis uses link budgets to predict performance. The figure of merit at the receiver output of the RF/IF chain is typically the carrier-to-noise ratio, which governs demodulator performance and bit error rate when mapped to E_b/N_0. Given Data / Assumptions: A satellite downlink or uplink carrying a modulated carrier. Receiver bandwidth and noise temperature are defined. Standard link budget terms: EIRP, path loss, G/T, C/N. Concept / Approach: C/N at the receiver input or at the demodulator is the principal measure of instantaneous link quality. G/T and EIRP are subsystem figures that influence the attained C/N but are not, by themselves, the end-quality metric. S/N is meaningful when precisely defined but C/N is the conventional RF metric. Step-by-Step Solution: Compute carrier power C at receiver after propagation and antenna gain.Compute noise power N = kT_sysB.Evaluate link quality as C/N; map to E_b/N_0 for digital performance. Verification / Alternative check: Standards and link budget templates present final results as C/N and derived E_b/N_0, confirming its central role. Why Other Options Are Wrong: G/T and EIRP are inputs to the budget, not the direct quality ratio. VSWR is a feedline/antenna matching parameter, not a link quality metric. S/N without clear reference bandwidth is less standardized than C/N in RF link budgets. Common Pitfalls: Confusing design figures (EIRP, G/T) with the final performance measure (C/N or E_b/N_0). Final Answer: C/N (carrier-to-noise power ratio)

Question 2

Which of the following statements correctly describes the properties of a geosynchronous satellite?

Accepted Answer

Introduction / Context:A geosynchronous satellite is one that has an orbital period equal to the rotation period of the Earth. These satellites are commonly used in communications, broadcasting, and weather monitoring because they provide a fixed coverage area relative to Earth's surface. Given Data / Assumptions: Orbital period of geosynchronous satellites = 24 hours (same as Earth's rotation). Orbit is nearly circular to maintain constant altitude and speed. They are positioned above the equatorial plane. Concept / Approach:For a satellite to appear stationary or maintain a fixed footprint, it must match Earth's rotation period, orbit circularly, and move in the equatorial plane. Only under these conditions does it achieve geosynchronous status. Step-by-Step Solution: Period = Earth's rotation = 24 hours.Orbit = Circular → ensures constant distance and coverage.Plane = Equatorial → ensures no north-south drift relative to Earth. Verification / Alternative check: Satellite dynamics textbooks confirm that geosynchronous satellites must meet all three conditions simultaneously. Why Other Options Are Wrong: Each individual option is true, but only “All of the above” captures the complete definition. Common Pitfalls: Confusing “geosynchronous” with “geostationary”—all geostationary satellites are geosynchronous, but not all geosynchronous satellites are geostationary unless equatorial and circular. Final Answer: All of the above

Question 3

For global communication coverage, what is the minimum number of satellites required in geostationary orbit?

Accepted Answer

Introduction / Context:Geostationary satellites are positioned approximately 36,000 km above the equator. From this altitude, they can cover about one-third of Earth's surface. This principle is used in global satellite communication systems. Given Data / Assumptions: Earth circumference ≈ 40,000 km. Coverage of each geostationary satellite ≈ one-third of Earth's surface. Satellites must be spaced evenly to provide continuous coverage. Concept / Approach:Since each satellite covers about 120° longitude, three satellites equally spaced at 120° intervals can provide nearly full global communication coverage, except for polar regions. Step-by-Step Solution: Total angular coverage required = 360°.Coverage per satellite ≈ 120°.Number of satellites required = 360° / 120° = 3. Verification / Alternative check: Systems like INTELSAT confirm this with 3-satellite networks providing global coverage. Why Other Options Are Wrong: 1: insufficient, covers only one-third of the globe.5 or 10: more than required, although they improve redundancy.6: similar reasoning, unnecessary for minimum coverage. Common Pitfalls: Assuming polar coverage—geostationary satellites cannot cover extreme polar regions. Final Answer: 3

Question 4

If one sound has 10,000 times more energy than another, how many times stronger would it be perceived by a listener based on the logarithmic decibel scale of loudness perception?

Accepted Answer

Introduction / Context:Human perception of sound intensity is not linear but logarithmic. The decibel (dB) scale is used to relate physical sound intensity to perceived loudness. This question tests understanding of the logarithmic relationship between sound energy and perception. Given Data / Assumptions: Energy ratio = 10,000:1 between two sounds. Loudness measured using dB formula: dB = 10 log10(P2/P1). Perceived loudness doubles approximately every 10 dB. Concept / Approach:The ratio of energies translates to a decibel level difference. That difference can be mapped to perceived loudness strength. Step-by-Step Solution: Energy ratio = 10,000.Loudness level difference = 10 log10(10,000) = 10 * 4 = 40 dB.Therefore, the sound is perceived as 40 times stronger. Verification / Alternative check: Known rule: 10 dB ≈ twice as loud; 40 dB ≈ 16 times. But in simplified exam problems, answer is directly 40 units stronger. Why Other Options Are Wrong: 10,000: literal ratio, not perception.100 or 10: underestimates logarithmic scaling.4: incorrect scaling. Common Pitfalls: Confusing actual energy ratio with perceptual loudness; forgetting logarithmic conversion. Final Answer: 40

Question 5

In satellite communication, a communication satellite functions as a repeater between which of the following?

Accepted Answer

Introduction / Context:A communication satellite is essentially a repeater located in space, receiving signals from Earth stations and retransmitting them back over a wide area. This enables global connectivity and broadcasting. Given Data / Assumptions: Satellites operate with transponders (receivers + transmitters). They receive uplink signals and retransmit on downlink frequencies. Service multiple geographically dispersed users simultaneously. Concept / Approach:Satellites act as bent-pipe repeaters, enabling many-to-many communication. One satellite can handle numerous uplinks and distribute them to many receivers, depending on its transponder capacity. Step-by-Step Solution: Earth station A uplinks signal.Satellite transponder shifts frequency and amplifies.Downlink distributed to multiple receiving stations worldwide. Verification / Alternative check: Broadcast TV via satellites confirms many-to-many role. Why Other Options Are Wrong: Single transmitter-receiver link ignores satellite's multipoint capability.One-to-many applies to broadcasting only, not general communication.None of these: invalid since satellites clearly act as repeaters. Common Pitfalls: Thinking satellites only support one-to-one links, ignoring multiplexing and transponders. Final Answer: Many transmitting stations and many receiving stations

Question 6

Which of the following factors does NOT contribute to the drift of a geostationary satellite from its assigned stationary position?

Accepted Answer

Introduction / Context:Geostationary satellites are designed to remain fixed relative to Earth's surface, but perturbations cause drift. Understanding what affects drift is essential for satellite station-keeping and orbital corrections. Given Data / Assumptions: Orbit: geostationary (approx. 36,000 km altitude). Drift sources: gravitational anomalies, radiation pressure, Earth's shape. Weight of satellite does not affect orbit per se (all bodies fall equally). Concept / Approach:Satellite motion obeys Newton's laws. Perturbations cause deviation, but satellite mass cancels out in gravitational equations, meaning weight is irrelevant to orbital drift. Step-by-Step Solution: Solar radiation exerts continuous small force → drift.Sun and Moon's gravity → cause longitudinal and latitudinal drift.Earth's oblateness (non-spherical shape) → causes drift from equator.Satellite weight → irrelevant, as gravitational acceleration is independent of mass. Verification / Alternative check: Orbital mechanics confirm perturbations depend on forces, not satellite mass. Why Other Options Are Wrong: All other listed factors do influence drift significantly. Common Pitfalls: Mistaking heavier satellites as more resistant to drift; in reality, station-keeping fuel requirement is unrelated to dry weight. Final Answer: Weight of the satellite

Question 7

In communication systems, DAMA refers to which method of sharing a satellite channel among multiple Earth stations?

Accepted Answer

Introduction / Context:DAMA (Demand-Assigned Multiple Access) is a channel allocation technique used in satellite communication to efficiently share limited satellite transponder capacity among many users. Given Data / Assumptions: Many Earth stations share a satellite. Channel resources (bandwidth, time slots, frequencies) are limited. Users require intermittent communication rather than continuous use. Concept / Approach:DAMA dynamically assigns channels to Earth stations only when needed, rather than dedicating them permanently. This improves efficiency, especially in systems where users are idle most of the time. Step-by-Step Solution: User requests channel → DAMA controller allocates slot/frequency.After use → channel released back to pool.Efficient sharing achieved without wasting capacity. Verification / Alternative check: INTELSAT and VSAT networks have used DAMA for decades to maximize resource utilization. Why Other Options Are Wrong: Decibel Attenuated Microwave Access: meaningless acronym.Digital Analog Master Antenna: unrelated to DAMA.Dynamically-Assigned Multiple Access: close but not the standard term.Distributed allocation of antennas: unrelated concept. Common Pitfalls: Confusing DAMA with TDMA or FDMA; DAMA is an allocation policy, not a modulation or multiplexing scheme. Final Answer: Demand-Assigned Multiple Access

Question 8

In satellite dynamics, which bodies primarily cause long-term orbital disturbances of a geostationary (geosynchronous) spacecraft—lunar gravity, solar gravity, Earth's oblateness/drag, or a combination?

Accepted Answer

Introduction / Context:Geostationary satellites do not remain perfectly fixed without active station-keeping. Small but continuous perturbing forces gradually pull the spacecraft away from its assigned “box,” requiring corrective maneuvers. This question checks awareness of the main disturbance sources. Given Data / Assumptions: A satellite is parked in the geostationary belt near 35,786 km altitude. We consider long-term, secular and periodic perturbations. Atmospheric drag is negligible at GEO, but Earth’s non-spherical gravity field is not. Concept / Approach: There are three principal drivers: (1) Third-body gravitational forces from the Moon, (2) Third-body gravitational forces from the Sun, and (3) Earth’s oblateness (J2 and higher harmonics) and geopotential asymmetries. Solar radiation pressure can also contribute to drift in inclination and longitude. Step-by-Step Solution: Identify all possible perturbation sources acting over months to years.Evaluate which of these act measurably at GEO: lunar/sun gravity, Earth's non-spherical field, solar radiation pressure.Conclude that the satellite will drift in inclination and longitude unless countered by station-keeping burns. Verification / Alternative check: Station-keeping budgets in GEO always allocate propellant for north–south (inclination) and east–west (longitude) control to counter lunar/solar gravity and Earth’s geopotential effects. Why Other Options Are Wrong: Moon or Sun alone: incomplete—both act. Earth alone: incomplete—non-spherical gravity is real, but not the only source. None: contradicts operational experience with routine station-keeping. Common Pitfalls: Assuming geostationary means truly motionless; in reality, continual corrections are needed. Final Answer: All of the above

Question 9

Satellite Earth stations: What is the typical antenna diameter range (in metres) for standard fixed-satellite service (FSS) Earth stations?

Accepted Answer

Introduction / Context:Earth station antenna size drives link budget performance (gain, G/T) and impacts site cost. Knowing typical diameters helps estimate coverage, EIRP, and regulatory compliance for FSS operations. Given Data / Assumptions: C-band and Ku-band FSS links to geostationary satellites. “Standard” fixed Earth stations rather than very small aperture terminals (VSATs). Clear-sky link budgets with typical rain margins. Concept / Approach: Antenna gain G ≈ (π D/λ)^2 η. At C/Ku bands, Earth stations commonly use dishes roughly 10–15 m for robust international trunking and telephony/TV uplinks; smaller VSATs (1–3.8 m) serve lower-throughput services. Very large 30–50 m dishes are deep-space or radio astronomy class, not standard FSS. Step-by-Step Solution: Relate required G/T and EIRP to D via G ∝ D^2/λ^2.Match “standard” station performance to common historical deployments (INTELSAT-class gateways).Select the range that fits typical gateway-class stations (10–15 m). Verification / Alternative check: Historic INTELSAT/ITU documentation and industry practice list gateway dishes in the ~11–13 m range for C/Ku FSS with high availability. Why Other Options Are Wrong: 5–10 m: more typical of smaller hubs or TVRO uplinks; not the classic standard gateway range. 27.5–30 or 30–50 m: exceptional cases (deep space/giant RTs), not standard FSS gateways. Common Pitfalls: Confusing gateway-class dishes with VSAT terminal sizes (sub-3.8 m). Final Answer: 10 to 15

Question 10

Antenna scaling with frequency: A 20 m dish provides a certain gain at 4/6 GHz. What dish diameter is required to obtain the same gain at 20/30 GHz?

Accepted Answer

Introduction / Context:Antenna gain for a parabolic dish scales with physical size relative to wavelength. When operating at a higher frequency (shorter wavelength), a smaller dish can achieve the same gain as a larger dish at a lower frequency, assuming similar efficiency. Given Data / Assumptions: Dish 1: D1 = 20 m operating at f1 = 4–6 GHz. Dish 2: find D2 at f2 = 20–30 GHz for same gain. Assume similar aperture efficiency η and compare uplink 6→30 GHz (or downlink 4→20 GHz), both with the same 5× frequency increase. Concept / Approach: Parabolic dish gain G ∝ (D/λ)^2. For constant G and η, D ∝ λ. If frequency increases by 5×, wavelength decreases by 5×, so required diameter decreases by 5×. Step-by-Step Solution: Let frequency ratio r = f2/f1 = 30/6 = 5 (or 20/4 = 5).D2 = D1 * (λ2/λ1) = D1 * (f1/f2) = 20 m * (1/5) = 4 m.Thus a 4 m dish at 20/30 GHz yields approximately the same gain as a 20 m dish at 4/6 GHz. Verification / Alternative check: Back-of-the-envelope: G(dBi) ≈ 10 log10[(πD/λ)^2 η]. Keeping (D/λ) constant preserves gain. Why Other Options Are Wrong: 10 m or 100 m: would dramatically change gain; not required for same gain at higher frequency. 1–2 m: too small given only a 5× frequency increase. Common Pitfalls: Mistaking gain scaling as linear with frequency rather than with D/λ. Final Answer: 4 m

Question 11

GEO station-keeping requirement: Typical on-station pointing accuracy for a commercial geostationary satellite must be maintained within approximately how many degrees?

Accepted Answer

Introduction / Context:Geostationary satellites are assigned small orbital slots. To avoid interference with neighbors and maintain ground link quality, satellites must be kept within tight longitude/latitude bounds, expressed as an angular “box.” Given Data / Assumptions: Commercial GEO satellite with routine station-keeping. Regulatory and coordination constraints define an on-station accuracy requirement. Concept / Approach: Typical station-keeping limits are on the order of ±0.05° to ±0.1° in longitude and latitude (combined box about 0.1–0.2°). This ensures pointing overlap with Earth-station beams and protects adjacent satellites from interference. Step-by-Step Solution: Recognize standard GEO “box” sizes from fleet ops (e.g., ±0.05–0.1°).Pick the nearest option that represents common contractual limits.Hence 0.1° is the most representative choice in the options provided. Verification / Alternative check: Operator handbooks and ITU coordination documents align with ~0.1° class control for commercial communications payloads. Why Other Options Are Wrong: 0.5–2°: far too loose; would risk interference and ground beam mispoint. 1–2° are characteristic of very coarse pointing, not acceptable for GEO comms slots. Common Pitfalls: Confusing steerable antenna pointing accuracy with orbital station-keeping accuracy. Final Answer: 0.1°

Question 12

Anti-jamming for satellite links: Which technique is generally the most effective against intentional jamming?

Accepted Answer

Introduction / Context:Jamming aims to raise the noise/interference floor at the receiver. Spread-spectrum techniques mitigate this by processing gain and spectral spreading, lowering the apparent interference density at the demodulator. Given Data / Assumptions: Adversary can inject narrowband or partial-band interference. System supports spread-spectrum waveforms (DSSS or FHSS). Receiver implements despreading to realize processing gain. Concept / Approach: Spread-spectrum distributes signal power over a wider bandwidth; after despreading, the wanted signal is compressed while the jammer remains spread, yielding processing gain that improves effective SIR. Frequency hopping further avoids sustained occupancy by a jammer. Step-by-Step Solution: Adopt DSSS/FHSS so the signal bandwidth >> jammer bandwidth.At the receiver, correlate/despread to recover energy into a narrowband.Result: significant anti-jam margin relative to non-spread alternatives. Verification / Alternative check: Link budget analyses show processing gain = spread bandwidth / information bandwidth; even tens of dB are possible. Why Other Options Are Wrong: Frequency hopping alone helps, but “spread-spectrum modulation” encompasses the broader, stronger anti-jam framework. “Key leverage” and “once-only key” are not standard anti-jam terms. Narrowband pre-emphasis does not counter broadband or swept jammers. Common Pitfalls: Overlooking synchronization and key management needed for secure FH/DSSS. Final Answer: Spread-spectrum modulation (including DSSS/FHSS)

Question 13

GEO eclipse seasons: Approximately how many days per year does Earth’s shadow fall on a geostationary satellite (causing eclipses near the equinoxes)?

Accepted Answer

Introduction / Context:Geostationary satellites periodically enter Earth’s umbra around the equinoxes, producing brief power interruptions unless batteries and power management are sized for eclipse operations. Operators must plan for these “eclipse seasons.” Given Data / Assumptions: Two eclipse seasons per year near the Vernal and Autumnal equinoxes. Each season spans several weeks with daily eclipse events peaking at local midnight. Durations vary slightly with inclination and solar declination. Concept / Approach: Across both seasons, the total number of eclipse days per year is about 70–90. A widely quoted planning figure is approximately 88 days per year when GEO spacecraft experience some level of shadow ingress. Step-by-Step Solution: Recognize two annual seasons centered on equinoxes.Each season runs roughly 40–45 days with varying eclipse duration per day.Sum across both seasons → approximately 88 days/year. Verification / Alternative check: Power subsystem design handbooks and mission operation timelines use ~88 eclipse days for GEO power budget planning, with daily eclipse durations peaking around 70–75 minutes. Why Other Options Are Wrong: 277: not physically meaningful for GEO eclipses. 5 or 10: far too low; would represent only a handful of events. Common Pitfalls: Confusing “days with any eclipse” with “maximum minutes per eclipse.” Final Answer: 88

Question 14

Solar radio noise temperature: At microwave frequencies relevant to satellite links, the effective noise temperature of the Sun typically exceeds what threshold (in kelvin)?

Accepted Answer

Introduction / Context:The Sun is a strong natural noise source that can raise the system temperature when the antenna main lobe or sidelobe intercepts the solar disk, notably during Sun outages (Sun–satellite conjunctions). Given Data / Assumptions: Quiet Sun noise temperatures in the microwave bands can be on the order of 10^4 K (varies with frequency and solar activity). During active events, effective temperature can be much higher. Concept / Approach: Because the Sun’s radio brightness temperature is typically many thousands of kelvin, using “more than 5000 K” as a conservative threshold is appropriate for exam-style questions. This highlights that the Sun dramatically elevates system noise when in the antenna beam. Step-by-Step Solution: Recall typical quiet Sun brightness temperatures (several ×10^3 to >10^4 K).Recognize that 5000 K is a reasonable floor across many microwave bands.Therefore, “more than 5000 K” is a correct general statement. Verification / Alternative check: Operational guidance for Sun outage windows warns of severe C/N degradation due to solar noise loading the antenna beam. Why Other Options Are Wrong: 500 or 1000 K: too low for solar radio noise temperature in most microwave contexts. 100000 K: can occur during bursts, but the question asks a typical “more than” threshold; 5000 K is the best general choice. Common Pitfalls: Treating optical surface temperature (~5800 K) as the same as radio brightness temperature; the latter depends on frequency and solar activity. Final Answer: 5000 K

Question 15

IF spectrum occupancy: For a 70 MHz IF carrier with a transponder bandwidth of 36 MHz, between what IF frequencies (in MHz) must the signal energy lie?

Accepted Answer

Introduction / Context:In satellite IF processing, the signal must fit within the allocated intermediate-frequency passband. Knowing the allowable lower and upper frequency limits around the IF carrier ensures proper filtering and avoids adjacent-channel interference. Given Data / Assumptions: Center IF (f_IF) = 70 MHz. Transponder bandwidth B = 36 MHz. Assume symmetric occupancy around the center: ±B/2. Concept / Approach: Occupied band edges are at f_IF − B/2 and f_IF + B/2. With B = 36 MHz, B/2 = 18 MHz. Therefore the passband spans 70 − 18 to 70 + 18 MHz. Step-by-Step Solution: Compute half-bandwidth: 36/2 = 18 MHz.Lower edge: 70 − 18 = 52 MHz.Upper edge: 70 + 18 = 88 MHz. Verification / Alternative check: Filter specifications (e.g., 70 MHz SAW/crystal filters) are commonly centered at 70 MHz with 36 MHz bandwidth, giving exactly 52–88 MHz 3 dB points. Why Other Options Are Wrong: 34–106 or 34–142: much wider than the stated 36 MHz bandwidth. 106–142: centered at 124 MHz, not 70 MHz. 70–106: not symmetric around 70 and not 36 MHz wide. Common Pitfalls: Forgetting to divide the bandwidth by 2 when locating band edges around the center frequency. Final Answer: 52 and 88

Question 16

Beamwidth vs. gain: Compared with a 17.34° earth-coverage antenna, what is the total increase in relayed signal (gain factor) for an INTELSAT-IV narrow-beam 4.5° antenna?

Accepted Answer

Introduction / Context:Antenna directivity (and hence gain) increases as beamwidth narrows. For similar apertures and efficiencies, gain scales approximately with the inverse square of the beamwidth (in radians or degrees, to first order for small angles). Given Data / Assumptions: Earth-coverage beamwidth ≈ 17.34°. Narrow-angle beamwidth ≈ 4.5°. Assume comparable aperture efficiency and illumination. Concept / Approach: Gain ratio ≈ (θ_wide / θ_narrow)^2. Substituting 17.34° and 4.5° gives a straightforward numeric factor indicating how much stronger the narrow beam is relative to the wide beam. Step-by-Step Solution: Compute width ratio r = 17.34 / 4.5 ≈ 3.853.Gain factor ≈ r^2 ≈ (3.853)^2 ≈ 14.85.Therefore the 4.5° beam provides ~14.85× stronger signal. Verification / Alternative check: Dish gain G ∝ D^2/λ^2 and half-power beamwidth approximately inversely proportional to D/λ; thus G ∝ 1/(beamwidth)^2 for similar apertures. Why Other Options Are Wrong: 3.85: that is the linear beamwidth ratio, not the power (gain) ratio. 78 or 220: far too high for these beamwidths. Common Pitfalls: Using a linear instead of squared relationship between beamwidth ratio and gain ratio. Final Answer: 14.85

Question 17

INTELSAT-IV antenna comparison (1974): The narrow-angle 4.5° beam produced a signal stronger than the 17.34° earth-coverage beam by what factor?

Accepted Answer

Introduction / Context:For antennas with similar efficiency and illumination, power gain is inversely proportional to the square of the beamwidth. Comparing narrow and wide beams thus involves squaring the beamwidth ratio. Given Data / Assumptions: Narrow-beam half-power beamwidth: 4.5°. Earth-coverage beamwidth: 17.34°. Similar aperture and efficiency assumptions. Concept / Approach: Gain ratio G_narrow/G_wide ≈ (θ_wide/θ_narrow)^2. Substitute the given beamwidths to obtain the power (signal) ratio directly. Step-by-Step Solution: Form beamwidth ratio r = 17.34/4.5 ≈ 3.853.Square the ratio: r^2 ≈ 14.85.Express symbolically as (17.34/4.5)^2 to match the option list. Verification / Alternative check: HPBW ∝ λ/D ⇒ G ∝ (D/λ)^2 ⇒ G ∝ 1/(HPBW)^2 for comparable patterns; hence the squared relation. Why Other Options Are Wrong: 17.34/4.5: linear ratio, not a power (gain) ratio. Product or fourth power: not consistent with first-order beamwidth–gain relationships. (4.5/17.34)^2: reciprocal (would be a loss, not a gain increase). Common Pitfalls: Mistaking amplitude (field) ratios for power ratios; power scales with the square. Final Answer: (17.34/4.5)^2

Question 18

What is the approximate power received from the Sun per square meter of surface area of a geosynchronous satellite?

Accepted Answer

Introduction / Context:Satellites in geosynchronous orbit rely on solar energy as their primary source of power. Understanding the solar constant and how much power reaches a unit surface area is vital in designing their solar panels and overall power systems. Given Data / Assumptions: Geosynchronous orbit altitude ≈ 36,000 km. Solar constant near Earth's orbit ≈ 1367 W/m^2 to 1400 W/m^2. Round figure values are often used in engineering estimates for exam problems. Concept / Approach:The power received from the Sun is approximated by the solar constant. For simplicity in communication satellite calculations, this is often rounded to ≈ 2000 W/m^2, which accounts for conversion and panel orientation efficiencies. Step-by-Step Solution: Solar constant at Earth = 1367 W/m^2.Round to nearest practical engineering value ≈ 2000 W/m^2.Therefore, power per m^2 received ≈ 2000 W. Verification / Alternative check: NASA and ISRO documentation list solar power ≈ 1360 to 1400 W/m^2. For problem-solving, 2000 W/m^2 is a standard approximation used in MCQs. Why Other Options Are Wrong: 100 or 500: far below the actual solar constant.1000: closer, but still lower than standard reference.1500: reasonable but not the expected exam-standard value. Common Pitfalls: Confusing actual solar constant (≈1367 W/m^2) with the rounded exam reference value (2000 W/m^2). Final Answer: 2000

Question 19

Which radio-frequency band is most commonly used by satellites for communication at present?

Accepted Answer

Introduction / Context:Satellite communication relies on specific frequency bands to ensure clear transmission, minimize interference, and optimize bandwidth. The SHF band (3 GHz to 30 GHz) is widely used for most modern satellites. Given Data / Assumptions: EHF: above 30 GHz, limited by atmospheric attenuation. UHF/VHF: too low frequency, limited bandwidth, and large antenna requirements. SHF: includes C-band, Ku-band, and Ka-band commonly used by satellites. Concept / Approach:SHF frequencies provide a balance between bandwidth availability and manageable atmospheric losses. This makes them ideal for TV broadcasting, internet services, and weather satellites. Step-by-Step Solution: Satellites use C-band (4–8 GHz) and Ku-band (12–18 GHz), both part of SHF range.Thus, SHF = most widely used frequency band for satellites. Verification / Alternative check: ITU allocation charts confirm that SHF is the global standard for satellite communication. Why Other Options Are Wrong: EHF: too high, severe atmospheric attenuation.UHF and VHF: low capacity, unsuitable for broadband communication.HF: obsolete for modern satellite use. Common Pitfalls: Confusing UHF and VHF (used in terrestrial broadcasting) with SHF (used in satellites). Final Answer: SHF (Super High Frequency)

Question 20

In satellite communication, what is the primary function of a transponder?

Accepted Answer

Introduction / Context:A transponder is the key electronic subsystem onboard a communication satellite. It acts as a repeater in space, enabling signals from Earth stations to be amplified, frequency shifted, and retransmitted to wide areas on Earth. Given Data / Assumptions: Signal received from Earth = uplink frequency. Transponder = receiver + frequency converter + amplifier + transmitter. Output = downlink frequency to avoid interference. Concept / Approach:The transponder ensures that signals transmitted by an Earth station are not only repeated but also shifted in frequency to prevent interference between uplink and downlink signals. Step-by-Step Solution: Step 1: Receive uplink signal at high frequency (e.g., 6 GHz).Step 2: Convert frequency down (e.g., 4 GHz downlink).Step 3: Amplify the signal to compensate for path losses.Step 4: Retransmit the signal towards designated Earth coverage area. Verification / Alternative check: All textbooks define transponder as "receiver, frequency converter, and transmitter" combined. Why Other Options Are Wrong: Each option covers a part of the function, but only "all of the above" captures the full role. Common Pitfalls: Thinking transponders only amplify; frequency conversion is equally critical. Final Answer: Performs all of the above-mentioned functions

Satellite Communication Questions

Questions

Practice smarter