Consider a 100 Mbps link between an earth station (sender) and a satellite…
2022
Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3x10^8 m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is_________.
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Correct answer: 7.08
Key idea: the total time is transmission delay plus propagation delay.
Transmission delay = packet size (bits) ÷ link rate.
Propagation delay = distance ÷ propagation speed.
Compute transmission delay:
Packet = 1000 bytes = 8000 bits. Link = 100 Mbps = 100 × 10^6 b/s.
Transmission delay = 8000 / (100 × 10^6) = 8 × 10^-5 s = 0.08 ms.
Compute propagation delay:
Distance = 2100 km = 2,100,000 m. Speed = 3 × 10^8 m/s.
Propagation delay = 2,100,000 / (3 × 10^8) = 0.007 s = 7.00 ms.
Total time = 7.00 ms + 0.08 ms = 7.08 ms.
Rounded to two decimal places: 7.08 ms.
Note: the existing stored answer (7.07 ms) appears to differ by 0.01 ms, likely due to a rounding or unit-conversion discrepancy. The step-by-step calculation above yields 7.08 ms, which is the value consistent with the stated numbers.
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