esa-sceva/satcom-mcqa huggingface dataset:
Topic: Wireless communication systems
Question: Calculate the received power if transmitted power is 1 W, antenna gains are 10 dBi each, and path loss is 100 dB. UsePr(dBm)=Pt(dBm)+Gt+Grâˆ’LP_r (dBm) = P_t(dBm) + G_t + G_r - LPr(dBm)=Pt(dBm)+Gt+Grâˆ’L,Pt(dBm)=10logâ¡10(Pt(W)Ã—1000)P_t(dBm) = 10 \log_{10} (P_t(W) \times 1000)Pt(dBm)=10log10(Pt(W)Ã—1000)
Option 1: -50 dBm
Option 2: -60 dBm
Option 3: -70 dBm
Option 4: -80 dBm
Answer: Option 3
Topic: Uplink
Question: Calculate the required uplink power if path loss is 200 dB, antenna gain is 40 dBi, and minimum received power needed is -100 dBW.
Option 1: 160 dBW
Option 2: 140 dBW
Option 3: 130 dBW
Option 4: 120 dBW
Option 5: 100 dBW
Answer: Option 3
Topic: Non-Terrestrial Networks
Question: Calculate the round-trip delay for a GEO satellite link if the altitude is approximately 36,000 km.(Ignore processing delays)
Option 1: ~240 ms
Option 2: ~120 ms
Option 3: ~160 ms
Answer: Option 1
Topic: SNR
Question: Calculate the SNR in dB if signal power is 1 mW and noise power is 0.1 mW
Option 1: 10 dB
Option 2: 20 dB
Option 3: 30 dB
Answer: Option 1
Topic: Telecommunication Payload
Question: Calculate the symbol rate if a telecommunication payload transmits data at 100 Mbps using QPSK (2 bits/symbol) and FEC rate 3/4. UseSymbolRate=DataRateBitsPerSymbolÃ—FECSymbolRate = \frac{DataRate}{BitsPerSymbol \times FEC}SymbolRate=BitsPerSymbolÃ—FECDataRate
Option 1: 66.7 Msps
Option 2: 75 Msps
Option 3: 100 Msps
Answer: Option 1
Topic: Non-Terrestrial Networks
Question: Calculate the wavelength of a 12 GHz signal used in NTN communications. 
Option 1: 0.025 m
Option 2: 0.0025 m
Option 3: 0.25 m
Answer: Option 1
Topic: Wireless communication systems
Question: Calculate the wavelength of a 2.4 GHz wireless signal. 
Option 1: 0.125 m
Option 2: 0.013 m
Option 3: 1.25 m
Option 4: 0.5 m
Option 5: 0.05 m
Answer: Option 1
Topic: Link Budget
Question: Calculate wavelength of signal with frequency 900 MHz
Option 1: 0.33 m
Option 2: 0.0033 m
Option 3: 3.3 m
Option 4: 33 m
Answer: Option 1
Topic: Gateway station
Question: Channel BW=20 MHz, SNR=10. Find capacity.
Option 1: Capacity=20×10 = 200 Mbps
Option 2: Capacity=20 × log₁₀(10)
Option 3: C = 20 × log₂(1 + 10) = 69.2 Mbps
Option 4: Capacity=20 × SNR²
Formula: C = B log₂(1+SNR)
Answer: Option 3
Topic: DVB-S2
Question: Data rate = 60 Mbps, symbol rate = 30 Msps. Find bits per symbol.
Option 1: bps = 60/30 = 2 bits/symbol
Option 2: bps = 30/60
Option 3: bps = 30 + 60
Option 4: bps = 2 × 30
Formula: M = R/Rs
Answer: Option 1
Topic: BER
Question: Describe the probability Pmi that a cell will be misinserted in a channel with burst errors: {P_{mi}\approx(1/6)\times CLR}
Option 1: the probability Pmi that a cell will be misinserted in a channel with burst errors is  {P_{mi}\approx(1/6)\times CLR} 
Option 2: the probability Pmi that a cell will be misinserted in a channel with burst errors is  {P_{mi}\approx(2\pi r)\times CLR} 
Option 3: the probability Pmi that a cell will be misinserted in a channel with burst errors is  {P_{mi}\approx(1/6)}
Option 4: the probability Pmi that a cell will be misinserted in a channel with burst errors is  {P_{mi}\approx EIRP(\frac{\lambda}{4\pi r})^{2}} 
Formula: {P_{mi}\approx(1/6)\times CLR}
Answer: Option1
Topic: Downlink
Question: Downlink SNR of 20 dB corresponds approximately to a linear ratio of:
Option 1: 10
Option 2: 40
Option 3: 100
Answer: Option3
Topic: Handover procedure
Question: During handover, if delay is 200 ms and packet rate is 1000 packets/sec, how many packets affected?
Option 1: 0
Option 2: 10
Option 3: 20
Answer: Option 3
Topic: Handover procedure
Question: During inter-satellite handover, received SNR drops from 18 dB to 9 dB. What is the loss in signal ratio (linear scale)?
Option 1: 4×
Option 2: 14×
Option 3: 8×
Option 4: 18x
Formula: 9 dB difference → 10^(9/10) ≈ 8× signal ratio loss
Answer: Option 3
Topic: DVB-S2
Question: DVB-S2 bandwidth = 30 MHz, SNR = 10. Find capacity.
Option 1: 30 Mbps
Option 2: 100 Mbps
Option 3: 130 Mbps
Option 4: 80 Mbps
Formula: C = B × log₂(1 + SNR)
Answer: Option 2
Topic: DVB-S2
Question: For 16APSK, bits per symbol are:
Option 1: 2
Option 2: 3
Option 3: 4
Option 4: 5
Option 5: 6
Answer: Option 3
Topic: 3gpp
Question: For a 20 MHz LTE bandwidth, how many resource blocks are standard?
Option 1: 50
Option 2: 75
Option 3: 100
Option 4: 125
Option 5: 150
Answer: Option 3
Topic: IP over satellite
Question: For a 5 Mbps link and RTT of 600 ms, bandwidth delay product is:
Option 1: 1.5 Mb
Option 2: 2.5 Mb
Option 3: 3 Mb
Answer: Option 1
Topic: BER
Question: For a BPSK link with SNR = 10 dB, approximate BER is closest to:
Option 1: 10⁻²
Option 2: 10⁻³
Option 3: 10⁻⁴
Option 4: 10⁻⁵
Formula: BER ≈ 0.5 × erfc(√SNR) → SNR 10 dB → BER ≈ 10⁻³
Answer: Option 2
Topic: BER
Question: For a BPSK link with SNR = 8 dB, approximate BER is:
Option 1: 10⁻²
Option 2: 10⁻³
Option 3: 10⁻⁴
Option 4: 10⁻⁵
Option 5: 10⁻6
Formula: BER ≈ 0.5 × erfc(√SNR) → BER ≈ 10⁻²
Answer: Option 1
Topic: Carrier-to-noise ratio
Question: For a CNR of 20 dB, if noise power increases by 10 dB, new CNR is:
Option 1: 30 dB
Option 2: 20 dB
Option 3: 10 dB
Option 4: 5 dB
Answer: Option 3
Topic: Downlink
Question: For a downlink throughput of 100 Mbps and a modulation efficiency of 4 bits/symbol, approximate symbol rate is:
Option 1: 25 Msps
Option 2: 40 Msps
Option 3: 100 Msps
Option 4: 400 Msps
Option 5: 500 Msps
Answer: Option 1
Topic: Communication Latency
Question: For a LEO satellite at 1,200 km, find round-trip delay.
Option 1: One-way = 1,200×10³ / 3×10⁸ = 0.004 s; Round-trip = 0.008 s
Option 2: Round-trip delay = 2 × (1,200×10³ / 3×10⁸) = 0.008 s
Option 3: Delay = 1 / 1,200
Option 4: Delay = 0.12 s
Formula: t = 2d / c
Answer: Option 2
Topic: IP over satellite
Question: For a satellite link with bandwidth delay product of 20 Mb, optimal TCP window size is about:
Option 1: 2.5 MB
Option 2: 1 MB
Option 3: 4 MB
Answer: Option 1
Topic: Beamforming
Question: For a uniform linear array, adding more antennas typically:
Option 1: Narrows beamwidth
Option 2: Widens beamwidth
Option 3: Reduces gain
Option 4: Increases sidelobes
Answer: Option 1
Topic: Circular Polarization
Question: For cross-polarized circular reception, if transmit power is 40 W, approximate received power is?
Option 1: 40 W
Option 2: 20 W
Option 3: 10 W
Option 4: 30 W
Option 5: 50 W
Formula: Cross-polar loss ≈ 10 dB → Pr = 40 / 10 = 4 W ≈ 10 W
Answer: Option 3
Topic: Downlink
Question: For downlink data rate of 50 Mbps and coding rate 3/4, the uncoded rate is approximately:
Option 1: 37.5 Mbps
Option 2: 50 Mbps
Option 3: 66.7 Mbps
Option 4: 75 Mbps
Answer: Option 3
Topic: 3gpp
Question: For LTE, how many resource blocks fit into 10 MHz bandwidth?
Option 1: 50
Option 2: 75
Option 3: 100
Option 4: 125
Option 5: 150
Answer: Option 1
Topic: Bit error rate
Question: For transmission of 10^7 bits with BER =10^-5, number of errors expected is:
Option 1: 10
Option 2: 100
Option 3: 1,000
Answer: Option 2
Topic: Link availability
Question: For two independent links each with 99% availability, the combined availability using both links is approximately:
Option 1: 98%
Option 2: 99.50%
Option 3: 99.99%
Answer: Option 3
Topic: Gateway station
Question: Gateway receives 5 W, amplifies by 10 dB. New received power?
Option 1: 25 W
Option 2: 30 W
Option 3: 50 W
Option 4: 40 W
Option 5: 60 W
Formula: 10 dB → 10× amplification → 5 × 10 = 50 W
Answer: Option 3
Topic: Carrier-to-noise ratio
Question: Given a CNR of 15 dB, noise power is 1 mW. Signal power is approximately:
Option 1: 15 mW
Option 2: 30 mW
Option 3: 32 mW
Option 4: 45 mW
Option 5: 100 mW
Answer: Option 3
Topic: Gateway station
Question: Given a gateway-to-satellite link bandwidth of 36 MHz and spectral efficiency of 3 bps/Hz, max throughput is:
Option 1: 100 Mbps
Option 2: 108 Mbps
Option 3: 120 Mbps
Answer: Option 2
Topic: Inter satellite Link
Question: Given inter satellite bandwidth of 2 GHz and spectral efficiency 5 bps/Hz, max data rate is:
Option 1: 5 Gbps
Option 2: 7.5 Gbps
Option 3: 8 Gbps
Option 4: 10 Gbps
Answer: Option 4
Topic: Handover procedure
Question: Handover delay = 50 ms, data rate = 5 Mbps. Find data lost.
Option 1: 250,000 bits
Option 2: 250,000 bits
Option 3: 500,000 bits
Option 4: 50,000 bits
Formula: Data loss = R × t
Answer: Option 2
Topic: NTN
Question: How do you calculate slant‐distance ground to satellite?
Option 1: d_{g}(E)=R\cdot sin^{-1}\left( \frac{d(E)cos(E)}{R+H} \right)
Option 2: d_{g}(E)= sin^{-1}\left( \frac{d(E)cos(E)}{R+H} \right)
Option 3: d_{g}(E)=P\cdot sin\left( \frac{d(E)cos(E)}{R+H} \right)
Answer: Option1
Topic: Circular Polarization
Question: How do you calculate the cross-polarization XP using axial ratio?
Option 1: For circular antennas, the cross-polarization XP can be calculated when the axial ratio is known:  XP_{dB}=20*log\left( \frac{AR_{lin}+1}{AR_{lin}-1} \right)
Option 2: For circular antennas, the cross-polarization XP can be calculated when the axial ratio is known:  XP_{dB}=log\left( \frac{AR_{lin}+1}{AR_{lin}-1} \right)
Option 3: For circular antennas, the cross-polarization XP can be calculated when the axial ratio is known:  XP_{dB}=20*\left( \frac{AR_{lin}+1}{AR_{lin}-1} \right)
Option 4: For circular antennas, the cross-polarization XP can be calculated when the axial ratio is known:  XP_{dB}= \left( \frac{AR_{lin}+1}{AR_{lin}-1} \right)
Formula: XP_{dB}=20*log\left( \frac{AR_{lin}+1}{AR_{lin}-1} \right)
Answer: Option1
Topic: Link Budget
Question: How do you calculate the maximum gain in the direction of maximum radiation?
Option 1: {{G}_{max}=(4\pi/\lambda^{2}){A}_{eff}}
Option 2: {{G}_{max}=(4\pi/\omega^{2}){A}_{eff}}
Option 3: {{G}_{max}=(3\pi/\lambda^{3}){A}_{eff}}
Option 4: {{G}_{max}=(4/\lambda^{3}){B}_{eff}}
Formula: {{G}_{max}=(4\pi/\lambda^{2}){A}_{eff}}
Answer: Option1
Topic: Non-Terrestrial Networks
Question: How do you calculate TMDA efficiency in the case of one carrier per link?
Option 1: In the case of a ‘one carrier per link’ technique, P = N(N+1)/2, accounting for all possible directional carrier assignments.
Option 2: In the case of a ‘one carrier per link’ technique, P = N, because each station communicates using a single shared carrier frequency.
Option 3: In the case of a ‘one carrier per link’ technique, P= N(N-1).
Option 4: In the case of a ‘one carrier per link’ technique, P = N-1, because each station communicates using a single shared carrier frequency.
Option 5: In the case of a ‘one carrier per link’ technique, P = N+4, because each station communicates using a single shared carrier frequency.
Formula:  P= N(N-1)
Answer: Option3
Topic: Circular Polarization
Question: How do you convert axial ratio from decibel to linear?
Option 1: AR_{lin}=10^{\left( \frac{AR_{dB}}{20} \right)}
Option 2: AR_{lin}=10^{AR_{dB}}
Option 3: AR_{lin}=10^{\left( \frac{AR_{dB}}{4\pi} \right)}
Option 4: AR_{lin}=log_{\left( \frac{AR_{dB}}{20} \right)}
Option 5: AR_{lin}=(4/\lambda^{3}){B}_{eff}}
Formula: AR_{lin}=10^{\left( \frac{AR_{dB}}{20} \right)}
Answer: Option1
Topic: Link Budget
Question: How do you represent link margin?
Option 1: Link margin is simply the di erence in dB between the actual CNR and
 the required CNR: M= EIRP-FSL-L+\left[ \frac{G_{r}}{TS}\right]_{dB}+228.6-10log(B)-\left[ \frac{C}{N} \right]_{req.dB}
Option 2: Link margin is simply the di erence in dB between the actual CNR and
 the required CNR: M=L+\left[ \frac{G_{r}}{TS}\right]_{dB}+228.6-10log(B)-\left[ \frac{C}{N} \right]_{req.dB}
Option 3: Link margin is simply the di erence in dB between the actual CNR and
 the required CNR: M= EIRP-FSL-L +228.6-10log(B)-\left[ \frac{C}{N} \right]_{req.dB}
Option 4: Link margin is simply the di erence in dB between the actual CNR and
 the required CNR: M= EIRP-FSL-L+\left[ \frac{G_{r}}{TS}\right]_{dB}-\left[ \frac{C}{N} \right]_{req.dB}
Option 5: Link margin is simply the di erence in dB between the actual CNR and
 the required CNR: M= EIRP-228.6-10log(B)-\left[ \frac{C}{N} \right]_{req.dB}
Formula:  M= EIRP-FSL-L+\left[ \frac{G_{r}}{TS}\right]_{dB}+228.6-10log(B)-\left[ \frac{C}{N} \right]_{req.dB}
Answer: Option1
Topic: Link Budget
Question: How do you represent noise power at the output of a network? 
Option 1: N = αN₁ + N₂ , where N1 is the input noise, N2 is the noise added by the network, and α is the power gain.
Option 2: N = N₁ + αN₂, where the input noise is amplified after the network adds its own noise.
Option 3: N = α(N₁ + N₂), implying both input and internal noise are equally amplified.
Option 4: N = N₁ + N₂ / α, suggesting internal noise decreases with gain.
Answer: Option1
Topic: Bit error rate
Question: How do you represent received signal under additive white Gauss noise?
Option 1: It is given by  r=\pm A_{p}+\frac{1}{\sqrt{2\pi\sigma}}e^{-x^{2}/2\sigma^{2}}
Option 2: It is given by  r=\pm B_{p}+\frac{1}{\sqrt{2\pi\sigma}}
Option 3: It is given by  r=\pm A_{p}+\frac{1}{{2\pi\sigma}}e^{-x^{2}/\sigma^{4}}
Option 4: It is given by  r=\pm A_{p}+\frac{1}{\sqrt{2\pi\sigma}}e^{-x^{2}/2\sigma^{2}}
Option 5: It is given by  C=\pm A_{p}+\frac{1}{\sqrt{2\pi\sigma}}e^{-x^{2}/2\Ap^{2}}
Formula: r=\pm A_{p}+\frac{1}{\sqrt{2\pi\sigma}}e^{-x^{2}/2\sigma^{2}}
Answer: Option1
Topic: Link Budget
Question: How do you represent the Friis formula?
Option 1:  Friis formula can be written as C=EIRP(\frac{\lambda}{4\pi r})^{2}\frac{G_{r}}{l}
Option 2:  Friis formula can be written as C=EIRP \frac{G_{r}}{l}
Option 3:  Friis formula can be written as C=(\frac{\lambda}{4\pi r})^{2}\frac{G_{r}}{l}
Option 4:  Friis formula can be written as C=EIRP(\frac{\lambda}{4\pi r})^{2}
Formula: C=EIRP(\frac{\lambda}{4\pi r})^{2}\frac{G_{r}}{l}
Answer: Option1
Topic: Beamforming
Question: How do you represent the interelement time delay in a beam?
Option 1: The interelement time delay is given by  \Delta t= -{\frac{d}{c}}cos\phi_{0}
Option 2: The interelement time delay is given by  \Delta t= -{\frac{d}{c}}cos2\pi
Option 3: The interelement time delay is given by  \Delta t= -{\frac{4d}{r}}cos\phi_{0}
Formula:  \Delta t= -{\frac{d}{c}}cos\phi_{0}
Answer: Option1
Topic: Satellite Bandwidth
Question: How does one find the required bandwidth?
Option 1: Required Bandwidth (BW) = Symbol Rate × (1 + Roll-Off Factor)
Option 2: Symbol Rate (SR) = Data Rate / (Modulation Efficiency × FEC Rate)
Option 3: Required Bandwidth (BW) = Data Rate × (Roll-Off Factor)
Option 4: Required Bandwidth (BW) = Symbol Rate × (1 - FEC factor)
Option 5: Required Bandwidth (BW) = Modulation efficiency × (1 + Roll-Off Factor)
Formula: Required Bandwidth (BW) = Symbol Rate × (1 + Roll-Off Factor)
Answer: Option1
Topic: NTN
Question: How is the doppler shift formula represented?
Option 1: Doppler shift is given by  \Delta f=\frac{v_{r}}{c}f_{0}
Option 2: Doppler shift is given by  \Delta f=\frac{T_{0}}{c}f_{0}
Option 3: Doppler shift is given by  \beta f=\frac{v_{r}}{c}f_{0}
Option 4: Doppler shift is given by  \phi f=\frac{v_{r}}{c}f_{0}
Option 5: Doppler shift is given by  \Delta f=\frac{v_{r}}{c}f_{0}
Formula: \Delta f=\frac{v_{r}}{c}f_{0}
Answer: Option1
Topic: Link Budget
Question: How is the power flux density at the satellite situated at earth station antenna boresight calculated?
Option 1: The power flux density is given by {\Phi}_{max}={P}_{T}{G}_{Tmax}/{4{\pi}R^{2}}
Option 2: The power flux density is given by {\Omega}_{max}={P}_{T}{G}_{Tmax}/{4{\pi}R^{2}}
Option 3: The power flux density is given by {\lambda}_{max}={P}_{T}{G}_{Tmax}/{4{\pi}R^{2}}
Option 4: The power flux density is given by {\Phi}_{max}={R}_{T}{G}_{Tmax}/{4{\pi}R^{3}}
Formula: {\Phi}_{max}={P}_{T}{G}_{Tmax}/{4{\pi}R^{2}}
Answer: Option1
Topic: Signal-to-noise ratio
Question: How is the signal-to-noise ratio calculated?
Option 1: Signal to noise ratio is given by SNR=S/N
Option 2: SNR = N / S, representing the ratio of noise to signal power.
Option 3: SNR = S × N, indicating the product of signal and noise power.
Option 4: SNR = (S + N) / 2, the average of signal and noise power levels.
Option 5: SNR = S – N, representing the difference between signal and noise power.
Answer: Option1
Topic: Link Budget
Question: How is uniform illumination represented mathematically?
Option 1: Uniform illumination is given as {\eta_{i}=1}
Option 2: Uniform illumination is given as {\eta_{i}=1+N}
Option 3: Uniform illumination is given as {\omega_{i}=3}
Option 4: Uniform illumination is given as {\eta_{i}=N-2}
Formula: {\eta_{i}=1}
Answer: Option1
Topic: Wireless communication systems
Question: How will you describe Friis equation in wireless communication?
Option 1: Friis equation is given by [Pr = Pt * Gt * Gr * ({\lambda} / 4 * {\pi} * d)^{2}]
Option 2: Friis equation is given by [Pr = Pt * Lx * Gr * ({\lambda} / 4 * {\pi} * d)^{4}]
Option 3: Friis equation is given by [Pr = Pt * Gt * Gr * ({\lambda} / 2 * {\pi} * d)^{3}]
Formula: [Pr = Pt * Gt * Gr * ({\lambda} / 4 * {\pi} * d)^{2}]
Answer: Option1
Topic: 5G NTN
Question: How will you describe the constant angular velocity of a The satellite with on-board angular momentum?
Option 1: {\Omega}_{x}={T}_{d}/{H}
Option 2: {\Omega}_{x}={T}_{d}/{P}
Option 3: {\Omega}_{x}={F}_{d}/{H}
Formula: {\Omega}_{x}={T}_{d}/{H}
Answer: Option1
Topic: Communication Latency
Question: How will you represent attenuation of free space?
Option 1:  The attenuation of free space is represented by {L}_{FS}={(4\pi R/\lambda)^{2}}
Option 2:  The attenuation of free space is represented by {L}_{FS}={(2\pi R/\lambda)^{2}}
Option 3:  The attenuation of free space is represented by {L}_{FS}={(4\pi R/\lambda)^{4}}
Option 4:  The attenuation of free space is represented by {L}_{FS}={(4\pi P/\lambda)^{2}}
Formula: {L}_{FS}={(4\pi R/\lambda)^{2}}
Answer: Option1
Topic: Gateway station
Question: How will you represent closed loop synchronization mathematically?
Option 1: closed loop synchronization is given as {{D}_{n}(j+1)={D}_{n}(j)-{e}_{n}(j) \ (s)}
Option 2: closed loop synchronization is given as {{R}_{n}(j+1)={R}_{n}(j)-{e}_{n}(j) \ (s)}
Option 3: closed loop synchronization is given as {{E}_{n}(j+1)={E}_{n}(j)-{e}_{n}(j) \ (s)}
Option 4: closed loop synchronization is given as {{G}_{n}(j+1)={G}_{n}(j)-{e}_{n}(j) \ (s)}
Option 5: closed loop synchronization is given as {{D}_{n}(j+3)={D}_{n}(j-1)-{e}_{n}(j) \ (s)}
Formula: {{D}_{n}(j+1)={D}_{n}(j)-{e}_{n}(j) \ (s)}
Answer: Option1
Topic: Circular Polarization
Question: How would you calculate axial ratio from dB to linear? 
Option 1: {AR}_{lin}={10}^{\frac{AR_{dB}}{20}}
Option 2: {AR}_{lin}={30}^{\frac{AR_{dB}}{10}}
Option 3: {AR}_{lin}={5}^{\frac{AR_{dB}}{20}}
Formula: {AR}_{lin}={10}^{\frac{AR_{dB}}{20}}
Answer: Option1
Topic: Telecommunication Payload
Question: How would you calculate the payload length?
Option 1: The payload length is given by: Payload length = Total packet length - Header length.
Option 2: The payload length is given by: Payload length = Total packet length + Header length
Option 3: The payload length is given by: Payload length = Header length - Total packet length
Option 4: The payload length is given by: Payload length = Total packet length only
Formula:  Payload length = Total packet length - Header length
Answer: Option1
Topic: 3gpp
Question: If 3GPP defines throughput target as 1 Gbps but the NTN link achieves 750 Mbps, what is the efficiency percentage?
Option 1: 50%
Option 2: 65%
Option 3: 75%
Option 4: 85%
Formula: Efficiency = (750 / 1000) × 100 = 75%
Answer: Option 3
Topic: Bit error rate
Question: If 50 bits are received in error out of 10⁶, find BER.
Option 1: BER = 50 / 10⁶ = 5×10⁻⁵
Option 2: BER = 10⁶ / 50
Option 3: BER = 10⁻⁶
Option 4: BER = 50 × 10⁶
Formula: BER = N_errors / N_total
Answer: Option 1
Topic: 5G NTN
Question: If a 5G NTN link has a delay of 250 ms, and we reduce propagation distance by half, what will be the new delay (approx.)?
Option 1: 125 ms
Option 2: 250 ms
Option 3: 500 ms
Formula: Delay ∝ distance → halving distance halves propagation delay
Answer: Option 1
Topic: Circular Polarization
Question: If a circularly polarized satellite transmits 20 W and polarization loss = 1.5 dB, find received power ignoring other losses.
Option 1: Pr = 20 − 1.5 = 18.5 W
Option 2: Pr (dB) = 10 log₁₀(20) − 1.5 ≈ 12.5 dBW
Option 3: Pr = 20 × 1.5 = 30 W
Option 4: Pr (dB) = 10 log₁₀(20) − 1.8 ≈ 15.5 dBW
Formula: Pr (dB) = 10 log₁₀(Pt) − Lp
Answer: Option 2
Topic: Downlink
Question: If a downlink channel uses 16 QAM with 256 subcarriers at 15 kHz spacing, the total bandwidth is:
Option 1: 3 MHz
Option 2: 5 MHz
Option 3: 10 MHz
Answer: Option 3
Topic: DVB-S2
Question: If a DVB-S2 transponder bandwidth is 36 MHz and spectral efficiency is 3 bps/Hz, max throughput is:
Option 1: 100 Mbps
Option 2: 108 Mbps
Option 3: 120 Mbps
Option 4: 140 Mbps
Option 5: 1080 Mbps
Answer: Option 2
Topic: Gateway station
Question: If a gateway station transmits at 40 dBm power and path loss is 100 dB, received power at satellite is:
Option 1: -60 dBm
Option 2: -40 dBm
Option 3: -20 dBm
Option 4: 0 dBm
Answer: Option 1
Topic: Gateway station
Question: If a gateway transmits 80 W and the satellite receives 20 W, what is the link loss?
Option 1: Link Loss = 80 - 20 = 50 W
Option 2: Link Loss = 10 × log₁₀(80 / 20) = 6 dB
Option 3: Link Loss = 20 / 80
Option 4: Link Loss = 80 × 20
Formula: L = 10 × log₁₀(Pₜ / Pᵣ)
Answer: Option2
Topic: Satellite bus
Question: If a satellite bus weighs 2000 kg and power system is 30% efficient generating 2 kW, what is the input solar power?
Option 1: 6666.7 W
Option 2: 600 W
Option 3: 2000 W
Answer: Option 1
Topic: Satellite Bandwidth
Question: If a satellite has a transponder bandwidth of 36 MHz and roll-off factor of 0.4, calculate the maximum symbol rate.
Option 1: 25.7 Msps
Option 2: 36.0 Msps
Option 3: 51.0 Msps
Option 4: 20.0 Msps
Option 5: 15.2 Msps
Answer: Option 1
Topic: NTN
Question: If a satellite in an NTN link orbits at 36,000 km and propagation speed is 3×10⁸ m/s, what is the one-way delay?
Option 1: 120 ms
Option 2: 240 ms
Option 3: 360 ms
Option 4: 180 ms
Formula: Delay = Distance / Speed = (36,000×10³) / (3×10⁸) = 0.12 s = 120 ms
Answer: Option 1
Topic: Link Budget
Question: If a satellite link loss decreases by 3 dB, what happens to Pr?
Option 1: Power decreases
Option 2: Power increases by 3 dB
Option 3: No change
Option 4: Power doubles
Formula: ΔPr = −ΔLp
Answer: Option 2
Topic: Communication Latency
Question: If a signal takes 50 ms round trip, one-way latency is:
Option 1: 25 ms
Option 2: 50 ms
Option 3: 100 ms
Option 4: 200 ms
Answer: Option 1
Topic: TCP over satellite
Question: If a TCP window size is 100 KB and RTT is 600 ms, find throughput in Mbps. Throughput = Window Size / RTT.
Option 1: 1.33 Mbps
Option 2: 1.2 Mbps
Option 3: 13.3 Mbps
Option 4: 133 Mbps
Answer: Option 1
Topic: Telecommunication Payload
Question: If a transponder receives at 12 GHz ans downconverts to 2 GHz, what is the local oscillator frequency?
Option 1: 10 GHz
Option 2: 14 GHz
Option 3: 4 GHz
Option 4: 8 GHz
Answer: Option 1
Topic: Circular Polarization
Question: If amplitude ratio of components is 2:1 with 90⁰ phase difference, axial ratio is:
Option 1: 1
Option 2: 3
Option 3: 5
Option 4: Greater than 1
Option 5: Less than 1
Answer: Option 4
Topic: Beamforming
Question: If antenna spacing doubles, the main lobe width:
Option 1: Doubles
Option 2: Halves
Option 3: Quadruples
Answer: Option 2
Topic: DVB-S2
Question: If bandwidth doubles in DVB-S2, what happens to capacity (same SNR)?
Option 1: Capacity halves
Option 2: Capacity doubles
Option 3: No change
Option 4: Capacity increases by 50%
Formula: C ∝ B
Answer: Option 2
Topic: DVB-S2
Question: If BER = 10⁻⁵, how many errors in 10⁶ bits?
Option 1: 5
Option 2: 10
Option 3: 15
Option 4: 20
Formula: Errors = BER × total bits = 10⁻⁵ × 10⁶ = 10
Answer: Option 2
Topic: Link Budget
Question: If calculated link margin = 6 dB and expected fading = 10 dB, what is the required additional margin?
Option 1: Required margin = 6 + 10 = 16 dB
Option 2: Required margin = 10 − 6 = 4 dB
Option 3: Required margin = fading − link margin = 10 − 6 = 4 dB
Formula: Needed = Fade_depth − Link_margin (if positive)
Answer: Option 3
Topic: DVB-S2
Question: If code rate is 3/4 and modulation order is 8PSK (3 bits/symbol), net bits per symbol are:
Option 1: 2.25
Option 2: 3
Option 3: 4
Answer: Option 1
Topic: Communication Latency
Question: If distance doubles, propagation delay:
Option 1: Halves
Option 2: Doubles
Option 3: Stays same
Option 4: Quadruples
Answer: Option 2