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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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