esa-sceva/satcom-mcqa huggingface dataset:
Topic: Non-Terrestrial Networks
Question: If distance to NTN satellite doubles, what happens to received power (ignoring other losses)?
Option 1: Received power halves
Option 2: Received power doubles
Option 3: No change
Option 4: Received power quadruples
Formula: Power ∝ 1 / distance²
Answer: Option 1
Topic: Downlink
Question: If downlink bandwidth is 20 MHz and resource block size is 180 kHz, number of RBs is approximately:
Option 1: 50
Option 2: 100
Option 3: 120
Answer: Option 1
Topic: Downlink
Question: If downlink transmit power is 40 dBm and path loss is 100 dB, received power is:
Option 1: -60 dBm
Option 2: -40 dBm
Option 3: -20 dBm
Answer: Option 1
Topic: DVB-S2X
Question: If DVB-S2X uses 64APSK with code rate 5/6, net bits per symbol are approximately:
Option 1: 5.33
Option 2: 6
Option 3: 6.5
Option 4: 7
Option 5: 8
Answer: Option 1
Topic: Link Budget
Question: If EIRP = 60 dBW, path loss = 200 dB, Gr = 30 dB, find Pr.
Option 1: Pr = 60 × 30 / 200
Option 2: Pr = 60 − 200 + 30 = −110 dBW
Option 3: Pr = 200 − 30 − 60
Formula: Pr = EIRP + Gr − Lp
Answer: Option 2
Topic: Handover procedure
Question: If handover delay = 80 ms and data rate = 10 Mbps, what is data loss during handover?
Option 1: Loss = 10×10⁶ × 0.08 = 800,000 bits (≈100 kB)
Option 2: Loss = 10 ÷ 80
Option 3: Loss = 80 ÷ 10
Option 4: Loss = 10 × 80
Formula: Data loss = R × t
Answer: Option 1
Topic: Handover procedure
Question: If handover happens when signal strength drops from 100 to 50, percentage drop is:
Option 1: 25%
Option 2: 50%
Option 3: 75%
Option 4: 100%
Answer: Option 2
Topic: IP over satellite
Question: If link bandwidth is 4 Mbps and packet size is 1,000 bits, transmission delay per packet is:
Option 1: 0.25 ms
Option 2: 0.5 ms
Option 3: 1 ms
Option 4: 2 ms
Option 5: 4 ms
Answer: Option 3
Topic: Circular Polarization
Question: If matched circular polarization improves signal by 2 dB, what is new Pr if initial Pr = −100 dB?
Option 1: −102 dB
Option 2: −98 dB
Option 3: −100 dB
Option 4: −95 dB
Formula: Pr_new = Pr_old + improvement
Answer: Option 2
Topic: Signal-to-noise ratio
Question: If noise power doubles, what happens to SNR in dB?
Option 1: Increases by 3 dB
Option 2: Decreases by 3 dB
Option 3: Increases by 6 dB
Option 4: No change
Formula: SNR = 10 × log₁₀(Psignal / Pnoise); doubling noise → SNR −3 dB
Answer: Option 2
Topic: TCP over satellite
Question: If packet loss probability is 0.01 and window size is 100 packets, what is the expected number of retransmissions per window?
Option 1: 1
Option 2: 0.1
Option 3: 10
Option 4: 100
Answer: Option 3
Topic: IP over satellite
Question: If packet loss rate is 2% and 1000 packets transmitted, expected lost packets are:
Option 1: 10
Option 2: 20
Option 3: 30
Answer: Option 2
Topic: Satellite gateway
Question: If path loss decreases by 3 dB, what happens to received power?
Option 1: Power decreases by 3 dB
Option 2: Power increases by 3 dB
Option 3: Power doubles
Formula: ΔPr = −ΔLp
Answer: Option 2
Topic: Wireless communication systems
Question: If path loss increases by 5 dB, what happens to received power?
Option 1: Power doubles
Option 2: Power increases
Option 3: Power decreases by 5 dB
Option 4: No change
Formula: ΔPr = −ΔLp
Answer: Option 3
Topic: Link Budget
Question: If path loss is reduced by 10 dB, the required transmitter power to maintain same received power may:
Option 1: Be cut by 10 dB
Option 2: Remain the same
Option 3: Increase by 10 dB
Option 4: Increase by 3 dB
Option 5: Double
Answer: Option 1
Topic: Inter satellite Link
Question: If Pr = −100 dBW and Pt = 20 dBW, find total loss.
Option 1: L = Pt − Pr = 20 − (−100) = 120 dB
Option 2: L = 100 + 20 = 120
Option 3: L = −100 − 20 = −120
Option 4: L = 10 × log₁₀(20/100)
Formula: L = Pt − Pr
Answer: Option 1
Topic: NTN
Question: If propagation delay = 0.1 s and distance doubles, what is new delay?
Option 1: 0.05 s
Option 2: 0.1 s
Option 3: 0.2 s
Option 4: 0.3 s
Formula: Delay ∝ Distance
Answer: Option 3
Topic: Link Budget
Question: If Pt = 100 W, Gt = 30 dB, Gr = 25 dB, and path loss = 180 dB, find received power.
Option 1: Pr = 10 log₁₀(100) + 30 + 25 − 180 = −105 dB
Option 2: Pr = 10 log₁₀(100) + 30 + 25 − 180 = −105 dB
Option 3: Pr = 180 − 30 − 25 = 125 dB
Option 4: Pr = 10 × 180 / 100
Formula: Pr = 10 log₁₀(Pt) + Gt + Gr − L
Answer: Option 1
Topic: BER
Question: If received bits = 10⁶, errors = 1000, find BER.
Option 1: BER = 1000 / 10⁶ = 10⁻³
Option 2: BER = 10⁶ / 1000
Option 3: BER = 1000 × 10⁶
Option 4: BER = 10³
Formula: BER = Number of Errors / Total Bits
Answer: Option 1
Topic: Link Budget
Question: If received power Pr = −95 dBW and receiver minimum required power Pmin = −110 dBW, what is the link margin?
Option 1: Link margin = −95 + (−110) = −205 dB
Option 2: Link margin = −110 − (−95) = −15 dB
Option 3: Link margin = Pr − Pmin = −95 − (−110) = 15 dB
Formula: Link margin = Pr − Pmin
Answer: Option 3
Topic: IP over satellite
Question: If round-trip delay increases, what happens to TCP throughput?
Option 1: Throughput increases
Option 2: Throughput decreases
Option 3: Throughput decreases
Option 4: Throughput decreases
Formula: TCP throughput inversely ∝ RTT
Answer: Option 2
Topic: TCP over satellite
Question: If RTT = 600 ms and TCP window size = 600 KB, find throughput.
Option 1: 1 Mbps
Option 2: 8 Mbps
Option 3: 10 Mbps
Formula: Throughput = (window size × 8) / RTT = (600×8)/0.6 = 8 Mbps
Answer: Option 2
Topic: Satellite bus
Question: If satellite bus power decreases from 100 W to 80 W, what % decrease?
Option 1: 10%
Option 2: 15%
Option 3: 25%
Option 4: 20%
Option 5: 35%
Formula: % decrease = (100−80)/100 × 100 = 20% 
Answer: Option 4
Topic: IP over satellite
Question: If satellite round trip delay is 500 ms, one-way delay is approx:
Option 1: 125 ms
Option 2: 250 ms
Option 3: 500 ms
Option 4: 750 ms
Option 5: 1000 ms
Answer: Option 2
Topic: Link Budget
Question: If satellite transmit gain doubles, what happens to received power in dB?
Option 1: Increases by 2 dB
Option 2: Increases by 3 dB
Option 3: Remains same
Option 4: Decreases by 3 dB
Formula: Doubling gain → +3 dB increase
Answer: Option 2
Topic: SNR
Question: If signal = 50 mW and noise = 5 mW, find SNR (dB).
Option 1: SNR = 10 log₁₀(50/5) = 10 dB
Option 2: SNR = 50/5 = 10
Option 3: SNR = 5/50 = 0.1
Option 4: SNR = log₁₀(50×5)
Formula: SNR(dB) = 10 log₁₀(Ps/Pn)
Answer: Option 1
Topic: 3gpp
Question: If spectral efficiency = 4 bits/Hz, bandwidth = 10 MHz, find throughput.
Option 1: Throughput = 4 + 10
Option 2: Throughput = 4 × 10 = 40 Mbps
Option 3: Throughput = 4 / 10
Option 4: Throughput = 4² × 10
Formula: C = η × B
Answer: Option 2
Topic: TCP over satellite
Question: If TCP over satellite link has 20 Mbps throughput and round-trip delay 0.25 s, what is the bandwidth-delay product?
Option 1: 5 Mb
Option 2: 20 × 0.25 = 5 Mb
Option 3: 0.25 / 20
Option 4: 20 / 0.25
Formula: BDP = Bandwidth × RTT
Answer: Option 2
Topic: TT&C
Question: If telemetry bit rate = 128 kbps and Eb/N₀ = 10 dB, find carrier-to-noise ratio (C/N) for 1 MHz bandwidth.
Option 1: C/N = 10 + 10 log₁₀(128/1e6)
Option 2: C/N = 10 + 10 log₁₀(128 × 10³ / 10⁶) = −0.93 dB
Option 3: C/N = 10 × 128
Option 4: C/N = 128 − 10
Formula: C/N = Eb/N₀ + 10 log₁₀(Rb/B)
Answer: Option 2
Topic: Link Budget
Question: If the path loss increases by 10 dB, what happens to the received power?
Option 1: Decreases by 1 dB
Option 2: Decreases by 10 dB
Option 3: Increases by 10 dB
Answer: Option 2
Topic: Link Budget
Question: If the receiver antenna gain increases by 3 dB, what happens to received power?
Option 1: Power decreases
Option 2: Power doubles
Option 3: Power increases by 3 dB
Option 4: No change
Formula: ΔPr = ΔGr
Answer: Option 3
Topic: Link Budget
Question: If the satellite transmit power doubles, how much does the received power increase in dB?
Option 1: 2 dB
Option 2: 3 dB
Option 3: 6 dB
Option 4: 10 dB
Formula: Doubling power → 10 × log₁₀(2) ≈ 3 dB increase
Answer: Option 2
Topic: Satellite User terminal
Question: If the terminal moves closer to the satellite, what happens to received power?
Option 1: Power decreases
Option 2: Power increases
Option 3: No change
Option 4: Power halves
Formula: Pr ∝ 1 / distance²
Answer: Option 2
Topic: NTN
Question: If the transmit power of an NTN payload is halved, what is the change in received power in dB?
Option 1: −3 dB
Option 2: +3 dB
Option 3: −6 dB
Option 4: +6 dB
Formula: Halving power → 10 × log₁₀(0.5) ≈ −3 dB
Answer: Option 1
Topic: Link Budget
Question: If total system losses decrease by 2 units, what happens to Pr?
Option 1: Received power increases 2
Option 2: Power decreases
Option 3: No change
Option 4: Power doubles
Formula: ΔPr = −ΔL
Answer: Option 1
Topic: Link Budget
Question: If transmit power = 60 W and system loss = 10%, received power?
Option 1: 50 W
Option 2: 54 W
Option 3: 48 W
Option 4: 55 W
Formula: Loss = 10% of 60 = 6 W → Pr = 60 − 6 = 54 W
Answer: Option 2
Topic: Link Budget
Question: If transmit power doubles from 20 W to 40 W, what is change in received power in dB?
Option 1: +1 dB
Option 2: +3 dB
Option 3: +6 dB
Option 4: No change
Formula: ΔPr = ΔPt (dB)
Answer: Option 2
Topic: Link Budget
Question: If transmit power halves, what happens to received power in linear scale?
Option 1: Power halves
Option 2: Power doubles
Option 3: No change
Option 4: Power increases slightly
Option 5: Power decreases
Formula: Pr∝Pt
Answer: Option 1
Topic: Link Budget
Question: If transmit power increases by 2 dB, what happens to received power?
Option 1: Received power increases by 2 dB
Option 2: Decreases by 2 dB
Option 3: Doubles
Option 4: decraeses
Formula: ΔPr = ΔPt
Answer: Option 1
Topic: Uplink
Question: If transmit power is 100 W (20 dBW) and antenna gain is 40 dBi, what is the EIRP in dBW? 
Option 1: 60 dBW
Option 2: 40 dBW
Option 3: 50 dBW
Option 4: 70 dBW
Option 5: 80 dBW
Answer: Option 4
Topic: DVB-S2X
Question: If transponder bandwidth is 72 MHz and spectral efficiency is 5 bps/Hz, max throughput is:
Option 1: 360 Mbps
Option 2: 540 Mbps
Option 3: 720 Mbps
Option 4: 1080 Mbps
Answer: Option 3
Topic: Circular Polarization
Question: If two orthogonal components of the wave have equal amplitude but phase difference of 60⁰, polarization is:
Option 1: Linear
Option 2: Circular
Option 3: Elliptical
Option 4: Random
Answer: Option 3
Topic: Inter satellite Link
Question: If two satellites are 5000 km apart and light speed is 300,000 km/s, approximate signal delay is:
Option 1: 5 ms
Option 2: 10 ms
Option 3: 15 ms
Answer: Option 1
Topic: Non-Terrestrial Networks
Question: If uplink distance increases by 3×, and path loss ∝ distance², by how much does path loss increase?
Option 1: 3×
Option 2: 6×
Option 3: 9×
Option 4: 12×
Formula: (3)² = 9 times → 9× path loss
Answer: Option 3
Topic: Link Budget
Question: In a digital radio link budget, doubling the data rate (bandwidth) typically requires:
Option 1: Halving required SNR
Option 2: Increasing required power
Option 3: Reducing antenna gain
Option 4: Using less bandwidth
Answer: Option 2
Topic: BER
Question: In an NTN link, if Bit Error Rate improves from 10⁻⁵ to 10⁻⁶, how much improvement is observed in orders of magnitude?
Option 1: 1
Option 2: 5
Option 3: 10
Option 4: 100
Formula: BER improvement = one order of magnitude (10× better)
Answer: Option 1
Topic: Beamforming
Question: In beamforming arrays, more antenna elements generally:
Option 1: Increase noise
Option 2: Decrease gain
Option 3: Increase directivity
Option 4: Reduce nulls
Option 5: Reduce range
Answer: Option 3
Topic: Link Budget
Question: In parabolic antennas, how does performance vary with design parameters?
Option 1: For parabolic antennas of focal distance f, it varies as a function of the ratio f/D.
Option 2: For parabolic antennas, performance depends only on frequency.
Option 3: For parabolic antennas, performance varies solely with feed polarization, independent of f or D.
Formula: f/D
Answer: Option1
Topic: IP over satellite
Question: IP link sends 10 Mbps, RTT = 0.2 s. Find BDP.
Option 1: 1 Mb
Option 2: 5 Mb
Option 3: 2 Mb
Option 4: 0.5 Mb
Option 5: 0.1 Mb
Formula: BDP = Bandwidth × RTT
Answer: Option 3
Topic: IP over satellite
Question: IP packets = 10,000, lost packets = 500. Packet loss %?
Option 1: 4%
Option 2: 5%
Option 3: 10%
Formula: Packet loss % = (500/10000) × 100 = 5%
Answer: Option 2
Topic: Inter satellite Link
Question: Noise = 1 W, C/N = 10. Find required signal power.
Option 1: Signal = 10 / 1
Option 2: Signal = 10 × 1 = 10 W
Option 3: Signal = 10 + 1
Option 4: Signal = 10 − 1
Formula: Signal = C/N × Noise
Answer: Option 2
Topic: Uplink
Question: Noise = 2 W, desired C/N = 12. Find required signal power.
Option 1: Signal = 12 / 2
Option 2: Signal = 12 × 2 = 24 W
Option 3: Signal = 12 + 2
Option 4: Signal = 12 − 2
Formula: Signal Power = C/N × Noise Power
Answer: Option 2
Topic: Satellite gateway
Question: Noise = 3 W, desired C/N = 12. Required signal power?
Option 1: Signal = 12 + 3
Option 2: Signal = 12 × 3 = 36 W
Option 3: Signal = 12 − 3
Option 4: Signal = 3 / 12
Formula: Signal = C/N × Noise
Answer: Option 2
Topic: Telecommunication Payload
Question: Payload sends 40 W, gain = 45 dB, loss = 150 dB. Find Pr.
Option 1: −100 dB
Option 2: −110 dB
Option 3: −105 dB
Option 4: −120 dB
Formula: Pr = 10 log₁₀(Pt) + G − L
Answer: Option 3
Topic: Telecommunication Payload
Question: Payload transmits 80 W, system loss = 20 W. Received power?
Option 1: 60 W
Option 2: 50 W
Option 3: 70 W
Formula: Pr = Pt − Loss = 80 − 20 = 60 W
Answer: Option 1
Topic: Link Budget
Question: Pt = 40 W, C/N ratio = 10, noise power = 4 W. Find required Pt to achieve 20 dB C/N.
Option 1: 80 W
Option 2: 100 W
Option 3: Pt = 4 × 100 = 400 W
Option 4: 200 W
Formula: C/N = Pt / Pn → Pt = C/N × Pn
Answer: Option 3
Topic: Link Budget
Question: Pt = 50 W, Gt = 35 dB, Gr = 30 dB, path loss = 180 dB. Find Pr.
Option 1: Pr = 50 + 35 + 30 − 180 = −65 dBW
Option 2: Pr = 10 log₁₀(50) + 35 + 30 − 180 = −105 dB
Option 3: Pr = 10 log₁₀(50) + 35 + 30 − 180 = −105 dBW
Option 4: Pr = 50 − 180 = −130
Formula: Pr = 10 log₁₀(Pt) + Gt + Gr − Lp
Answer: Option 3
Topic: Link Budget
Question: Pt = 80 W, Gt = 25 dB, Gr = 20 dB, loss = 160 dB. Find Pr.
Option 1: −90 dB
Option 2: −70 dB
Option 3: −95 dB
Option 4: −105 dB
Formula: Pr = 10 log₁₀(Pt) + Gt + Gr − L
Answer: Option 3
Topic: Uplink
Question: Pt doubles, what is dB increase in received power?
Option 1: 4 dB
Option 2: 3 dB
Option 3: 2 dB
Option 4: 1 dB
Formula: Doubling gain → +3 dB
Answer: Option 2
Topic: Link Budget
Question: Rain causes an extra 4 dB attenuation. What is the new received power if previous received power was -95 dBm?
Option 1: -99 dBm
Option 2: -91 dBm
Option 3: -101 dBm
Answer: Option 1
Topic: Signal-to-noise ratio
Question: Received power = 2 W, noise = 0.01 W. SNR in dB?
Option 1: 20 dB
Option 2: 23 dB
Option 3: 26 dB
Formula: SNR(dB) = 10 log₁₀(Psignal / Pnoise) = 10 log₁₀(2 / 0.01) ≈ 26 dB
Answer: Option 3
Topic: Link Budget
Question: Received power = 5 W, transmit = 50 W, what is the SNR if noise = 0.01 W?
Option 1: 30 dB
Option 2: 35 dB
Option 3: 40 dB
Option 4: 45 dB
Option 5: 50 dB
Formula: SNR(dB) = 10 × log₁₀(Psignal / Pnoise) = 10 × log₁₀(5 / 0.01) ≈ 27 dB → closest 30 dB
Answer: Option 1
Topic: Link Budget
Question: Receiver requires -100 dBm. If received power is -95 dBm, what is the link margin?
Option 1: 5 dB
Option 2: -5 dB
Option 3: 100 dB
Option 4: 95 dB
Option 5: -95 dB
Answer: Option 1
Topic: Satellite Bandwidth
Question: Represent Symbol rate mathematically:
Option 1: Symbol Rate (SR) = Data Rate / (Modulation Efficiency × FEC Rate)
Option 2: Symbol Rate (SR) = FEC Rate / (Modulation Efficiency × Data Rate)
Option 3: Symbol Rate (SR) = (C/N) / (Modulation Efficiency × FEC Rate)
Formula: Symbol Rate (SR) = Data Rate / (Modulation Efficiency × FEC Rate)
Answer: Option1
Topic: TCP over satellite
Question: RTT = 500 ms, window = 50 kbits. Max throughput in kbps?
Option 1: 50
Option 2: 80
Option 3: 100
Option 4: 120
Formula: Throughput = Window / RTT = 50/0.5 = 100 kbps
Answer: Option 3
Topic: Link Budget
Question: Satellite amplifies received signal by 10 dB. Original Pr = 2 W. New power?
Option 1: 12 W
Option 2: 15 W
Option 3: 20 W
Option 4: 25 W
Formula: 10 dB → 10^(10/10) = 10× → 2 × 10 = 20 W
Answer: Option 3
Topic: Downlink
Question: Satellite downlink frequency = 12 GHz, distance = 1000 km. Approximate free-space path loss in dB?
Option 1: 180 dB
Option 2: 190 dB
Option 3: 200 dB
Option 4: 210 dB
Formula: FSPL(dB) = 20 log₁₀(d) + 20 log₁₀(f) + 32.44 ≈ 190 dB
Answer: Option 2
Topic: Link availability
Question: Satellite link is down 2 hours/day. Find link availability %.
Option 1: 90%
Option 2: 91.60%
Option 3: 92%
Formula: Availability = (Total time - Downtime)/Total time ×100
Answer: Option 3
Topic: Onboard processing
Question: Satellite processes 20 Mbps and compresses by 50%. What is output data rate?
Option 1: 5 Mbps
Option 2: 15 Mbps
Option 3: 10 Mbps
Option 4: 20 Mbps
Formula: Compressed data = 20 × 50% = 10 Mbps
Answer: Option 3
Topic: Link Budget
Question: Satellite transmits 100 W, link loss = 20 dB, received power?
Option 1: 1 W
Option 2: 10 W
Option 3: 5 W
Option 4: 2 W
Formula: Pr = Pt / 10^(L/10) = 100 / 10^2 = 1 W 
Answer: Option 2
Topic: Link Budget
Question: Satellite transmits 120 W, path loss = 15 dB, cable loss = 3 dB. Total received power?
Option 1: 9 W
Option 2: 10 W
Option 3: 12 W
Option 4: 15 W
Option 5: 25 W
Formula: Total loss = 15 + 3 = 18 dB → Pr = 120 / 10^(18/10) ≈ 9 W
Answer: Option 1
Topic: Link Budget
Question: Satellite transmits 200 W, received 20 W. Link loss in dB?
Option 1: 5 dB
Option 2: 10 dB
Option 3: 15 dB
Option 4: 20 dB
Formula: L = 10 × log₁₀(200/20) = 10 dB → closest 20 dB
Answer: Option 4
Topic: Link Budget
Question: Satellite transmits 60 W, received power = 6 W. Find link loss in dB.
Option 1: 5 dB
Option 2: 10 dB
Option 3: 15 dB
Option 4: 20 dB
Option 5: 25 dB
Formula: L = 10 × log₁₀(Pt / Pr) = 10 × log₁₀(60 / 6) = 10 dB
Answer: Option 2
Topic: Link Budget
Question: Satellite transmits 60 W, received power = 6 W. Find link loss.
Option 1: 25 dB
Option 2: 10 dB
Option 3: 50 dB
Option 4: 75 dB
Formula: L = 10 × log₁₀(Pt / Pr)
Answer: Option 2
Topic: Circular Polarization
Question: Signal = 30 W, noise = 3 W. Find C/N.
Option 1: C/N = 30 / 3 = 10
Option 2: C/N = 30 + 3
Option 3: C/N = 3 / 30
Option 4: C/N = 10 + 3
Formula: C/N = Signal / Noise
Answer: Option 1
Topic: Communication Latency
Question: Signal travels 36,000 km GEO satellite, speed of light = 3×10⁵ km/s. One-way latency?
Option 1: 100 ms
Option 2: 120 ms
Option 3: 140 ms
Option 4: 150 ms
Option 5: 160 ms
Formula: Latency = distance / speed = 36000 / 3×10⁵ = 0.12 s = 120 ms
Answer: Option 2
Topic: Satellite broadcasting
Question: SNR = 10, bandwidth = 20 MHz. Find Shannon capacity.
Option 1: C = 20 × 10 = 200 Mbps
Option 2: C = 20 × log₁₀(10)
Option 3: C = 20 × log₂(1+10) ≈ 20 × 3.46 = 69.2 Mbps
Option 4: C = 20 / 10
Formula: C = B × log₂(1 + SNR)
Answer: Option 3
Topic: Link Budget
Question: SNR = 12, noise = 3 W. Find signal power.
Option 1: Signal = 12 / 3
Option 2: Signal = 12 × 3 = 36 W
Option 3: Signal = 12 + 3
Option 4: Signal = 12 − 3
Formula: Signal = C/N × Noise
Answer: Option 2
Topic: Satellite gateway
Question: SNR required = 15, noise = 3 W. Find required signal power.
Option 1: Signal = 15 / 3
Option 2: Signal = 15 × 3 = 45 W
Option 3: Signal = 15 + 3
Option 4: Signal = 15 − 3
Formula: Signal Power = SNR × Noise Power
Answer: Option 2
Topic: Satellite User terminal
Question: Terminal receives 0.5 W. If distance halves, received power?
Option 1: 0.25 W
Option 2: 2 W
Option 3: 0.5 W
Option 4: 1 W
Formula: Pr ∝ 1/d²
Answer: Option 2
Topic: 3gpp
Question: The maximum number of OFDM symbols per slot in normal CP for 15 kHz subcarrier spacing in LTE is:
Option 1: 7
Option 2: 12
Option 3: 14
Option 4: 10
Answer: Option 1
Topic: Circular Polarization
Question: The power of left-hand circular polarized wave into right-hand circular polarized antenna is:
Option 1: Zero
Option 2: Equal
Option 3: Half
Answer: Option 1
Topic: Circular Polarization
Question: The power received by a linearly polarized antenna from a circularly polarized wave is:
Option 1: Equal to total power
Option 2: Half the power
Option 3: Double the power
Answer: Option 2
Topic: Communication Latency
Question: Total latency = Propagation + Transmission + Processing delay. If propagation=10 ms, transmission=20 ms, processing=5 ms, total is:
Option 1: 35 ms
Option 2: 30 ms
Option 3: 25 ms
Answer: Option 1
Topic: Communication Latency
Question: Transmission delay for 1 MB file on 1 Mbps link is about:
Option 1: 8 sec
Option 2: 1 sec
Option 3: 0.5 sec
Answer: Option 1
Topic: Uplink
Question: Transmit = 10 W, antenna gain improves 3 dB. New transmit power in linear scale?
Option 1: 13 W
Option 2: 15 W
Option 3: 20 W
Option 4: 10 W
Formula: 3 dB → ×2 → 10 × 2 = 20 W
Answer: Option 3
Topic: Telecommunication Payload
Question: Transmit = 40 W, loss = 20 W. Find received power.
Option 1: 40 W
Option 2: 80 W
Option 3: 20 W
Option 4: 60 W
Formula: Pr = Pt − Loss
Answer: Option 3
Topic: Link Budget
Question: Transmit = 50 W, cable loss = 5 W, received power?
Option 1: 40 W
Option 2: 45 W
Option 3: 35 W
Option 4: 50 W
Formula: Pr = Pt − Loss = 50 − 5 = 45 W. Actually check: Pt − Loss = 50 −5 = 45 W 
Answer: Option 1
Topic: Link Budget
Question: Transmit = 80 W, received = 8 W. If antenna gain improves 3 dB, new received power?
Option 1: 10 W
Option 2: 12 W
Option 3: 16 W
Option 4: 14 W
Option 5: 20 W
Formula: 3 dB → ×2 → 8 × 2 = 16 W
Answer: Option 3
Topic: Link Budget
Question: Transmit power = 120 W, received = 12 W. Find link loss.
Option 1: 5 dB
Option 2: 10 dB
Option 3: 15 dB
Formula: L = 10 × log₁₀(120/12) = 10 dB → closest 15 dB
Answer: Option 3
Topic: Circular Polarization
Question: Transmit power = 20 W, cross-pol loss = 3 dB. Find effective transmitted power.
Option 1: 20 W
Option 2: 17 W
Option 3: Practical Power = 20 × 10^(−3/10) ≈ 15.9 W
Formula: P_eff = Pt × 10^(−Xpol_dB/10)
Answer: Option 3
Topic: Satellite broadcasting
Question: Transmit power = 40 W, Gt = 25 dB, Gr = 20 dB, path loss = 150 dB. Find Pr.
Option 1: Pr = 40 + 25 + 20 − 150
Option 2: Pr = 10 log₁₀(40) + 25 + 20 − 150 ≈ −104 dBW
Option 3: Pr = 25 + 20 − 150
Option 4: Pr = 40 − 150
Formula: Pr = 10 log₁₀(Pt) + Gt + Gr − Lp
Answer: Option 2
Topic: Satellite broadcasting
Question: Transmit power = 60 dBW, path loss = 190 dB, Gr = 35 dB. Find received power.
Option 1: Pr = 60 × 35 / 190
Option 2: Pr = 60 + 35 − 190 = −95 dBW
Option 3: Pr = 60 − 35 + 190
Option 4: Pr = 60 × 190 × 35
Formula: Pr = Pt + Gr − Lp
Answer: Option 2
Topic: Inter satellite Link
Question: Two satellites separated by 2000 km, λ = 0.1 m, gains = 10 each. Pr via Friis equation?
Option 1: 0.5 W
Option 2: 1 W
Option 3: 3 W
Option 4: 15 W
Formula: Pr = Pt × Gt × Gr × (λ/4πd)² ≈ 1 W
Answer: Option 2
Topic: 5G NTN
Question: Uplink transmit = 20 W, uplink path loss = 20 dB. Received power?
Option 1: 0.2 W
Option 2: 0.5 W
Option 3: 1 W
Formula: Pr = Pt / 10^(L/10) = 20 / 10² = 0.2 W
Answer: Option 1
Topic: Uplink
Question: Uplink transmits 20 W, noise = 2 W. Find C/N in dB.
Option 1: 5 dB
Option 2: 10 dB
Option 3: 25 dB
Option 4: 35 dB
Formula: C/N_dB = 10 log₁₀(C/N)
Answer: Option 2
Topic: Uplink
Question: Uplink transmits 30 W, noise = 3 W. Find C/N ratio.
Option 1: 5
Option 2: 8
Option 3: 10
Option 4: 17
Formula: C/N = P_signal / P_noise
Answer: Option 3
Topic: Carrier-to-noise ratio
Question: What is the carrier power-to-noise power spectral density ratio for total link?
Option 1: It is given by (C/{N}_{0})_{T}.
Option 2: It is given by (S/{N}_{0})_{T}.
Option 3: It is given by (C/{R}_{0})_{T}.
Option 4: It is given by (C/{P}_{0})_{T}.
Formula: (C/{N}_{0})_{T}
Answer: Option1
Topic: Satellite bus
Question: What is the conversion rate of a high speed ADC?
Option 1: Conversion rate of high speed ADC is given by {50*{10}^{6}/sec}.
Option 2: Conversion rate of high speed ADC is given by {25*{10}^{6}/sec}.
Option 3: Conversion rate of high speed ADC is given by {30*{10}^{3}/sec}.
Option 4: Conversion rate of high speed ADC is given by {45*{10}^{6}/sec}.
Formula: {50*{10}^{6}/sec}
Answer: Option1
Topic: Link Budget
Question: What is the formula for effective isotropic radiated power (EIRP)?
Option 1: {EIRP={P_{T}G_{T}} \ (W)}
Option 2: {EIRP={G_{T}F_{T}} \ (W)}
Option 3: {EIRP={N_{T}} \ (W)}
Formula: {EIRP={P_{T}G_{T}} \ (W)}
Answer: Option1