Questions stringlengths 12 307 | Option1 stringlengths 1 262 | Option2 stringlengths 1 257 | Option3 stringlengths 1 245 | Option4 stringlengths 1 207 ⌀ | Option5 stringlengths 1 147 ⌀ | Answer stringclasses 8 values | Topic stringclasses 32 values | Mathematical Reasoning Required stringclasses 2 values | Raw Latex Formula stringclasses 127 values |
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If link bandwidth is 4 Mbps and packet size is 1,000 bits, transmission delay per packet is: | 0.25 ms | 0.5 ms | 1 ms | 2 ms | 4 ms | Option 3 | IP over satellite | Yes | null |
If matched circular polarization improves signal by 2 dB, what is new Pr if initial Pr = −100 dB? | −102 dB | −98 dB | −100 dB | −95 dB | null | Option 2 | Circular Polarization | Yes | Pr_new = Pr_old + improvement |
If noise power doubles, what happens to SNR in dB? | Increases by 3 dB | Decreases by 3 dB | Increases by 6 dB | No change | null | Option 2 | Signal-to-noise ratio | Yes | SNR = 10 × log₁₀(Psignal / Pnoise); doubling noise → SNR −3 dB |
If packet loss probability is 0.01 and window size is 100 packets, what is the expected number of retransmissions per window? | 1 | 0.1 | 10 | 100 | null | Option 3 | TCP over satellite | Yes | null |
If packet loss rate is 2% and 1000 packets transmitted, expected lost packets are: | 10 | 20 | 30 | null | null | Option 2 | IP over satellite | Yes | null |
If path loss decreases by 3 dB, what happens to received power? | Power decreases by 3 dB | Power increases by 3 dB | Power doubles | null | null | Option 2 | Satellite gateway | Yes | ΔPr = −ΔLp |
If path loss increases by 5 dB, what happens to received power? | Power doubles | Power increases | Power decreases by 5 dB | No change | null | Option 3 | Wireless communication systems | Yes | ΔPr = −ΔLp |
If path loss is reduced by 10 dB, the required transmitter power to maintain same received power may: | Be cut by 10 dB | Remain the same | Increase by 10 dB | Increase by 3 dB | Double | Option 1 | Link Budget | Yes | null |
If Pr = −100 dBW and Pt = 20 dBW, find total loss. | L = Pt − Pr = 20 − (−100) = 120 dB | L = 100 + 20 = 120 | L = −100 − 20 = −120 | L = 10 × log₁₀(20/100) | null | Option 1 | Inter satellite Link | Yes | L = Pt − Pr |
If propagation delay = 0.1 s and distance doubles, what is new delay? | 0.05 s | 0.1 s | 0.2 s | 0.3 s | null | Option 3 | NTN | Yes | Delay ∝ Distance |
If Pt = 100 W, Gt = 30 dB, Gr = 25 dB, and path loss = 180 dB, find received power. | Pr = 10 log₁₀(100) + 30 + 25 − 180 = −105 dB | Pr = 10 log₁₀(100) + 30 + 25 − 180 = −105 dB | Pr = 180 − 30 − 25 = 125 dB | Pr = 10 × 180 / 100 | null | Option 1 | Link Budget | Yes | Pr = 10 log₁₀(Pt) + Gt + Gr − L |
If received bits = 10⁶, errors = 1000, find BER. | BER = 1000 / 10⁶ = 10⁻³ | BER = 10⁶ / 1000 | BER = 1000 × 10⁶ | BER = 10³ | null | Option 1 | BER | Yes | BER = Number of Errors / Total Bits |
If received power Pr = −95 dBW and receiver minimum required power Pmin = −110 dBW, what is the link margin? | Link margin = −95 + (−110) = −205 dB | Link margin = −110 − (−95) = −15 dB | Link margin = Pr − Pmin = −95 − (−110) = 15 dB | null | null | Option 3 | Link Budget | Yes | Link margin = Pr − Pmin |
If round-trip delay increases, what happens to TCP throughput? | Throughput increases | Throughput decreases | Throughput decreases | Throughput decreases | null | Option 2 | IP over satellite | Yes | TCP throughput inversely ∝ RTT |
If RTT = 600 ms and TCP window size = 600 KB, find throughput. | 1 Mbps | 8 Mbps | 10 Mbps | null | null | Option 2 | TCP over satellite | Yes | Throughput = (window size × 8) / RTT = (600×8)/0.6 = 8 Mbps |
If satellite bus power decreases from 100 W to 80 W, what % decrease? | 10% | 15% | 25% | 20% | 35% | Option 4 | Satellite bus | Yes | % decrease = (100−80)/100 × 100 = 20% |
If satellite round trip delay is 500 ms, one-way delay is approx: | 125 ms | 250 ms | 500 ms | 750 ms | 1000 ms | Option 2 | IP over satellite | Yes | null |
If satellite transmit gain doubles, what happens to received power in dB? | Increases by 2 dB | Increases by 3 dB | Remains same | Decreases by 3 dB | null | Option 2 | Link Budget | Yes | Doubling gain → +3 dB increase |
If signal = 50 mW and noise = 5 mW, find SNR (dB). | SNR = 10 log₁₀(50/5) = 10 dB | SNR = 50/5 = 10 | SNR = 5/50 = 0.1 | SNR = log₁₀(50×5) | null | Option 1 | SNR | Yes | SNR(dB) = 10 log₁₀(Ps/Pn) |
If spectral efficiency = 4 bits/Hz, bandwidth = 10 MHz, find throughput. | Throughput = 4 + 10 | Throughput = 4 × 10 = 40 Mbps | Throughput = 4 / 10 | Throughput = 4² × 10 | null | Option 2 | 3gpp | Yes | C = η × B |
If TCP over satellite link has 20 Mbps throughput and round-trip delay 0.25 s, what is the bandwidth-delay product? | 5 Mb | 20 × 0.25 = 5 Mb | 0.25 / 20 | 20 / 0.25 | null | Option 2 | TCP over satellite | Yes | BDP = Bandwidth × RTT |
If telemetry bit rate = 128 kbps and Eb/N₀ = 10 dB, find carrier-to-noise ratio (C/N) for 1 MHz bandwidth. | C/N = 10 + 10 log₁₀(128/1e6) | C/N = 10 + 10 log₁₀(128 × 10³ / 10⁶) = −0.93 dB | C/N = 10 × 128 | C/N = 128 − 10 | null | Option 2 | TT&C | Yes | C/N = Eb/N₀ + 10 log₁₀(Rb/B) |
If the path loss increases by 10 dB, what happens to the received power? | Decreases by 1 dB | Decreases by 10 dB | Increases by 10 dB | null | null | Option 2 | Link Budget | Yes | null |
If the receiver antenna gain increases by 3 dB, what happens to received power? | Power decreases | Power doubles | Power increases by 3 dB | No change | null | Option 3 | Link Budget | Yes | ΔPr = ΔGr |
If the satellite transmit power doubles, how much does the received power increase in dB? | 2 dB | 3 dB | 6 dB | 10 dB | null | Option 2 | Link Budget | Yes | Doubling power → 10 × log₁₀(2) ≈ 3 dB increase |
If the terminal moves closer to the satellite, what happens to received power? | Power decreases | Power increases | No change | Power halves | null | Option 2 | Satellite User terminal | Yes | Pr ∝ 1 / distance² |
If the transmit power of an NTN payload is halved, what is the change in received power in dB? | −3 dB | +3 dB | −6 dB | +6 dB | null | Option 1 | NTN | Yes | Halving power → 10 × log₁₀(0.5) ≈ −3 dB |
If total system losses decrease by 2 units, what happens to Pr? | Received power increases 2 | Power decreases | No change | Power doubles | null | Option 1 | Link Budget | Yes | ΔPr = −ΔL |
If transmit power = 60 W and system loss = 10%, received power? | 50 W | 54 W | 48 W | 55 W | null | Option 2 | Link Budget | Yes | Loss = 10% of 60 = 6 W → Pr = 60 − 6 = 54 W |
If transmit power doubles from 20 W to 40 W, what is change in received power in dB? | +1 dB | +3 dB | +6 dB | No change | null | Option 2 | Link Budget | Yes | ΔPr = ΔPt (dB) |
If transmit power halves, what happens to received power in linear scale? | Power halves | Power doubles | No change | Power increases slightly | Power decreases | Option 1 | Link Budget | Yes | Pr∝Pt |
If transmit power increases by 2 dB, what happens to received power? | Received power increases by 2 dB | Decreases by 2 dB | Doubles | decraeses | null | Option 1 | Link Budget | Yes | ΔPr = ΔPt |
If transmit power is 100 W (20 dBW) and antenna gain is 40 dBi, what is the EIRP in dBW? | 60 dBW | 40 dBW | 50 dBW | 70 dBW | 80 dBW | Option 4 | Uplink | Yes | null |
If transponder bandwidth is 72 MHz and spectral efficiency is 5 bps/Hz, max throughput is: | 360 Mbps | 540 Mbps | 720 Mbps | 1080 Mbps | null | Option 3 | DVB-S2X | Yes | null |
If two orthogonal components of the wave have equal amplitude but phase difference of 60⁰, polarization is: | Linear | Circular | Elliptical | Random | null | Option 3 | Circular Polarization | Yes | null |
If two satellites are 5000 km apart and light speed is 300,000 km/s, approximate signal delay is: | 5 ms | 10 ms | 15 ms | null | null | Option 1 | Inter satellite Link | Yes | null |
If uplink distance increases by 3×, and path loss ∝ distance², by how much does path loss increase? | 3× | 6× | 9× | 12× | null | Option 3 | Non-Terrestrial Networks | Yes | (3)² = 9 times → 9× path loss |
In a digital radio link budget, doubling the data rate (bandwidth) typically requires: | Halving required SNR | Increasing required power | Reducing antenna gain | Using less bandwidth | null | Option 2 | Link Budget | Yes | null |
In an NTN link, if Bit Error Rate improves from 10⁻⁵ to 10⁻⁶, how much improvement is observed in orders of magnitude? | 1 | 5 | 10 | 100 | null | Option 1 | BER | Yes | BER improvement = one order of magnitude (10× better) |
In beamforming arrays, more antenna elements generally: | Increase noise | Decrease gain | Increase directivity | Reduce nulls | Reduce range | Option 3 | Beamforming | Yes | null |
In parabolic antennas, how does performance vary with design parameters? | For parabolic antennas of focal distance f, it varies as a function of the ratio f/D. | For parabolic antennas, performance depends only on frequency. | For parabolic antennas, performance varies solely with feed polarization, independent of f or D. | null | null | Option1 | Link Budget | Yes | f/D |
IP link sends 10 Mbps, RTT = 0.2 s. Find BDP. | 1 Mb | 5 Mb | 2 Mb | 0.5 Mb | 0.1 Mb | Option 3 | IP over satellite | Yes | BDP = Bandwidth × RTT |
IP packets = 10,000, lost packets = 500. Packet loss %? | 4% | 5% | 10% | null | null | Option 2 | IP over satellite | Yes | Packet loss % = (500/10000) × 100 = 5% |
Noise = 1 W, C/N = 10. Find required signal power. | Signal = 10 / 1 | Signal = 10 × 1 = 10 W | Signal = 10 + 1 | Signal = 10 − 1 | null | Option 2 | Inter satellite Link | Yes | Signal = C/N × Noise |
Noise = 2 W, desired C/N = 12. Find required signal power. | Signal = 12 / 2 | Signal = 12 × 2 = 24 W | Signal = 12 + 2 | Signal = 12 − 2 | null | Option 2 | Uplink | Yes | Signal Power = C/N × Noise Power |
Noise = 3 W, desired C/N = 12. Required signal power? | Signal = 12 + 3 | Signal = 12 × 3 = 36 W | Signal = 12 − 3 | Signal = 3 / 12 | null | Option 2 | Satellite gateway | Yes | Signal = C/N × Noise |
Payload sends 40 W, gain = 45 dB, loss = 150 dB. Find Pr. | −100 dB | −110 dB | −105 dB | −120 dB | null | Option 3 | Telecommunication Payload | Yes | Pr = 10 log₁₀(Pt) + G − L |
Payload transmits 80 W, system loss = 20 W. Received power? | 60 W | 50 W | 70 W | null | null | Option 1 | Telecommunication Payload | Yes | Pr = Pt − Loss = 80 − 20 = 60 W |
Pt = 40 W, C/N ratio = 10, noise power = 4 W. Find required Pt to achieve 20 dB C/N. | 80 W | 100 W | Pt = 4 × 100 = 400 W | 200 W | null | Option 3 | Link Budget | Yes | C/N = Pt / Pn → Pt = C/N × Pn |
Pt = 50 W, Gt = 35 dB, Gr = 30 dB, path loss = 180 dB. Find Pr. | Pr = 50 + 35 + 30 − 180 = −65 dBW | Pr = 10 log₁₀(50) + 35 + 30 − 180 = −105 dB | Pr = 10 log₁₀(50) + 35 + 30 − 180 = −105 dBW | Pr = 50 − 180 = −130 | null | Option 3 | Link Budget | Yes | Pr = 10 log₁₀(Pt) + Gt + Gr − Lp |
Pt = 80 W, Gt = 25 dB, Gr = 20 dB, loss = 160 dB. Find Pr. | −90 dB | −70 dB | −95 dB | −105 dB | null | Option 3 | Link Budget | Yes | Pr = 10 log₁₀(Pt) + Gt + Gr − L |
Pt doubles, what is dB increase in received power? | 4 dB | 3 dB | 2 dB | 1 dB | null | Option 2 | Uplink | Yes | Doubling gain → +3 dB |
Rain causes an extra 4 dB attenuation. What is the new received power if previous received power was -95 dBm? | -99 dBm | -91 dBm | -101 dBm | null | null | Option 1 | Link Budget | Yes | null |
Received power = 2 W, noise = 0.01 W. SNR in dB? | 20 dB | 23 dB | 26 dB | null | null | Option 3 | Signal-to-noise ratio | Yes | SNR(dB) = 10 log₁₀(Psignal / Pnoise) = 10 log₁₀(2 / 0.01) ≈ 26 dB |
Received power = 5 W, transmit = 50 W, what is the SNR if noise = 0.01 W? | 30 dB | 35 dB | 40 dB | 45 dB | 50 dB | Option 1 | Link Budget | Yes | SNR(dB) = 10 × log₁₀(Psignal / Pnoise) = 10 × log₁₀(5 / 0.01) ≈ 27 dB → closest 30 dB |
Receiver requires -100 dBm. If received power is -95 dBm, what is the link margin? | 5 dB | -5 dB | 100 dB | 95 dB | -95 dB | Option 1 | Link Budget | Yes | null |
Represent Symbol rate mathematically: | Symbol Rate (SR) = Data Rate / (Modulation Efficiency × FEC Rate) | Symbol Rate (SR) = FEC Rate / (Modulation Efficiency × Data Rate) | Symbol Rate (SR) = (C/N) / (Modulation Efficiency × FEC Rate) | null | null | Option1 | Satellite Bandwidth | Yes | Symbol Rate (SR) = Data Rate / (Modulation Efficiency × FEC Rate) |
RTT = 500 ms, window = 50 kbits. Max throughput in kbps? | 50 | 80 | 100 | 120 | null | Option 3 | TCP over satellite | Yes | Throughput = Window / RTT = 50/0.5 = 100 kbps |
Satellite amplifies received signal by 10 dB. Original Pr = 2 W. New power? | 12 W | 15 W | 20 W | 25 W | null | Option 3 | Link Budget | Yes | 10 dB → 10^(10/10) = 10× → 2 × 10 = 20 W |
Satellite downlink frequency = 12 GHz, distance = 1000 km. Approximate free-space path loss in dB? | 180 dB | 190 dB | 200 dB | 210 dB | null | Option 2 | Downlink | Yes | FSPL(dB) = 20 log₁₀(d) + 20 log₁₀(f) + 32.44 ≈ 190 dB |
Satellite link is down 2 hours/day. Find link availability %. | 90% | 91.60% | 92% | null | null | Option 3 | Link availability | Yes | Availability = (Total time - Downtime)/Total time ×100 |
Satellite processes 20 Mbps and compresses by 50%. What is output data rate? | 5 Mbps | 15 Mbps | 10 Mbps | 20 Mbps | null | Option 3 | Onboard processing | Yes | Compressed data = 20 × 50% = 10 Mbps |
Satellite transmits 100 W, link loss = 20 dB, received power? | 1 W | 10 W | 5 W | 2 W | null | Option 2 | Link Budget | Yes | Pr = Pt / 10^(L/10) = 100 / 10^2 = 1 W |
Satellite transmits 120 W, path loss = 15 dB, cable loss = 3 dB. Total received power? | 9 W | 10 W | 12 W | 15 W | 25 W | Option 1 | Link Budget | Yes | Total loss = 15 + 3 = 18 dB → Pr = 120 / 10^(18/10) ≈ 9 W |
Satellite transmits 200 W, received 20 W. Link loss in dB? | 5 dB | 10 dB | 15 dB | 20 dB | null | Option 4 | Link Budget | Yes | L = 10 × log₁₀(200/20) = 10 dB → closest 20 dB |
Satellite transmits 60 W, received power = 6 W. Find link loss in dB. | 5 dB | 10 dB | 15 dB | 20 dB | 25 dB | Option 2 | Link Budget | Yes | L = 10 × log₁₀(Pt / Pr) = 10 × log₁₀(60 / 6) = 10 dB |
Satellite transmits 60 W, received power = 6 W. Find link loss. | 25 dB | 10 dB | 50 dB | 75 dB | null | Option 2 | Link Budget | Yes | L = 10 × log₁₀(Pt / Pr) |
Signal = 30 W, noise = 3 W. Find C/N. | C/N = 30 / 3 = 10 | C/N = 30 + 3 | C/N = 3 / 30 | C/N = 10 + 3 | null | Option 1 | Circular Polarization | Yes | C/N = Signal / Noise |
Signal travels 36,000 km GEO satellite, speed of light = 3×10⁵ km/s. One-way latency? | 100 ms | 120 ms | 140 ms | 150 ms | 160 ms | Option 2 | Communication Latency | Yes | Latency = distance / speed = 36000 / 3×10⁵ = 0.12 s = 120 ms |
SNR = 10, bandwidth = 20 MHz. Find Shannon capacity. | C = 20 × 10 = 200 Mbps | C = 20 × log₁₀(10) | C = 20 × log₂(1+10) ≈ 20 × 3.46 = 69.2 Mbps | C = 20 / 10 | null | Option 3 | Satellite broadcasting | Yes | C = B × log₂(1 + SNR) |
SNR = 12, noise = 3 W. Find signal power. | Signal = 12 / 3 | Signal = 12 × 3 = 36 W | Signal = 12 + 3 | Signal = 12 − 3 | null | Option 2 | Link Budget | Yes | Signal = C/N × Noise |
SNR required = 15, noise = 3 W. Find required signal power. | Signal = 15 / 3 | Signal = 15 × 3 = 45 W | Signal = 15 + 3 | Signal = 15 − 3 | null | Option 2 | Satellite gateway | Yes | Signal Power = SNR × Noise Power |
Terminal receives 0.5 W. If distance halves, received power? | 0.25 W | 2 W | 0.5 W | 1 W | null | Option 2 | Satellite User terminal | Yes | Pr ∝ 1/d² |
The maximum number of OFDM symbols per slot in normal CP for 15 kHz subcarrier spacing in LTE is: | 7 | 12 | 14 | 10 | null | Option 1 | 3gpp | Yes | null |
The power of left-hand circular polarized wave into right-hand circular polarized antenna is: | Zero | Equal | Half | null | null | Option 1 | Circular Polarization | Yes | null |
The power received by a linearly polarized antenna from a circularly polarized wave is: | Equal to total power | Half the power | Double the power | null | null | Option 2 | Circular Polarization | Yes | null |
Total latency = Propagation + Transmission + Processing delay. If propagation=10 ms, transmission=20 ms, processing=5 ms, total is: | 35 ms | 30 ms | 25 ms | null | null | Option 1 | Communication Latency | Yes | null |
Transmission delay for 1 MB file on 1 Mbps link is about: | 8 sec | 1 sec | 0.5 sec | null | null | Option 1 | Communication Latency | Yes | null |
Transmit = 10 W, antenna gain improves 3 dB. New transmit power in linear scale? | 13 W | 15 W | 20 W | 10 W | null | Option 3 | Uplink | Yes | 3 dB → ×2 → 10 × 2 = 20 W |
Transmit = 40 W, loss = 20 W. Find received power. | 40 W | 80 W | 20 W | 60 W | null | Option 3 | Telecommunication Payload | Yes | Pr = Pt − Loss |
Transmit = 50 W, cable loss = 5 W, received power? | 40 W | 45 W | 35 W | 50 W | null | Option 1 | Link Budget | Yes | Pr = Pt − Loss = 50 − 5 = 45 W. Actually check: Pt − Loss = 50 −5 = 45 W |
Transmit = 80 W, received = 8 W. If antenna gain improves 3 dB, new received power? | 10 W | 12 W | 16 W | 14 W | 20 W | Option 3 | Link Budget | Yes | 3 dB → ×2 → 8 × 2 = 16 W |
Transmit power = 120 W, received = 12 W. Find link loss. | 5 dB | 10 dB | 15 dB | null | null | Option 3 | Link Budget | Yes | L = 10 × log₁₀(120/12) = 10 dB → closest 15 dB |
Transmit power = 20 W, cross-pol loss = 3 dB. Find effective transmitted power. | 20 W | 17 W | Practical Power = 20 × 10^(−3/10) ≈ 15.9 W | null | null | Option 3 | Circular Polarization | Yes | P_eff = Pt × 10^(−Xpol_dB/10) |
Transmit power = 40 W, Gt = 25 dB, Gr = 20 dB, path loss = 150 dB. Find Pr. | Pr = 40 + 25 + 20 − 150 | Pr = 10 log₁₀(40) + 25 + 20 − 150 ≈ −104 dBW | Pr = 25 + 20 − 150 | Pr = 40 − 150 | null | Option 2 | Satellite broadcasting | Yes | Pr = 10 log₁₀(Pt) + Gt + Gr − Lp |
Transmit power = 60 dBW, path loss = 190 dB, Gr = 35 dB. Find received power. | Pr = 60 × 35 / 190 | Pr = 60 + 35 − 190 = −95 dBW | Pr = 60 − 35 + 190 | Pr = 60 × 190 × 35 | null | Option 2 | Satellite broadcasting | Yes | Pr = Pt + Gr − Lp |
Two satellites separated by 2000 km, λ = 0.1 m, gains = 10 each. Pr via Friis equation? | 0.5 W | 1 W | 3 W | 15 W | null | Option 2 | Inter satellite Link | Yes | Pr = Pt × Gt × Gr × (λ/4πd)² ≈ 1 W |
Uplink transmit = 20 W, uplink path loss = 20 dB. Received power? | 0.2 W | 0.5 W | 1 W | null | null | Option 1 | 5G NTN | Yes | Pr = Pt / 10^(L/10) = 20 / 10² = 0.2 W |
Uplink transmits 20 W, noise = 2 W. Find C/N in dB. | 5 dB | 10 dB | 25 dB | 35 dB | null | Option 2 | Uplink | Yes | C/N_dB = 10 log₁₀(C/N) |
Uplink transmits 30 W, noise = 3 W. Find C/N ratio. | 5 | 8 | 10 | 17 | null | Option 3 | Uplink | Yes | C/N = P_signal / P_noise |
What is the carrier power-to-noise power spectral density ratio for total link? | It is given by (C/{N}_{0})_{T}. | It is given by (S/{N}_{0})_{T}. | It is given by (C/{R}_{0})_{T}. | It is given by (C/{P}_{0})_{T}. | null | Option1 | Carrier-to-noise ratio | Yes | (C/{N}_{0})_{T} |
What is the conversion rate of a high speed ADC? | Conversion rate of high speed ADC is given by {50*{10}^{6}/sec}. | Conversion rate of high speed ADC is given by {25*{10}^{6}/sec}. | Conversion rate of high speed ADC is given by {30*{10}^{3}/sec}. | Conversion rate of high speed ADC is given by {45*{10}^{6}/sec}. | null | Option1 | Satellite bus | Yes | {50*{10}^{6}/sec} |
What is the formula for effective isotropic radiated power (EIRP)? | {EIRP={P_{T}G_{T}} \ (W)} | {EIRP={G_{T}F_{T}} \ (W)} | {EIRP={N_{T}} \ (W)} | null | null | Option1 | Link Budget | Yes | {EIRP={P_{T}G_{T}} \ (W)} |
what is the given formula used for? {p}_{r}=\frac{{p}_{r}{g}_{t}{g}_{r}{c}^{2}}{(4\pi)^{2}{R}^{2}{f}^{2}} | The link between the satellite and Earth station is governed by the optical free-space propagation equation, rather than the basic microwave radio link equation. | It is the link between the satellite and Earth station is governed by the basic microwave
radio link equation. | The link between the satellite and Earth station is governed by the basic microwave radio link equation only for uplink signals, not for the downlink. | null | null | Option2 | Link Budget | Yes | {p}_{r}=\frac{{p}_{r}{g}_{t}{g}_{r}{c}^{2}}{(4\pi)^{2}{R}^{2}{f}^{2}} |
What is the length of a 5G NR slot at 30 kHz subcarrier spacing? | 0.25 ms | 0.5 ms | 1 ms | 2 ms | 4 ms | Option 2 | 3gpp | Yes | null |
What is the max bits per symbol for 256QAM? | 6 | 7 | 8 | 9 | 10 | Option 3 | 3gpp | Yes | null |
Wireless system transmits 25 W, path loss = 100 dB. Find Pr. | −80 Db | −75 dB | −110 dB | −90 dB | −120 dB | Option 2 | Wireless communication systems | Yes | Pr = 10 log₁₀(Pt) − L |
With turbo code rate 2/3 and 8PSK modulation, effective bits per symbol are: | 1.5 | 2 | 2.67 | null | null | Option 3 | DVB-S2 | Yes | null |
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