ITPEC FE Morning April 2020 Question 32
Locations A and B, both on the ground, communicate via a geostationary satellite at an altitude of about 36,000 km. When the distance between each location and the satellite is 37,500 km, and the delay caused by the relay at the satellite is 10 milliseconds, what is the total delay in seconds for the data sent from A to arrive at B? Here, the propagation speed of radio waves is 3×108 m per second.
Explanation
0.26 seconds — the signal travels up to the satellite and down again, plus the 10 ms relay delay.
Use the given 37,500 km per leg, not the 36,000 km altitude: the locations are not directly under the satellite. Radio waves travel at 3 × 108 m per second, which is 300,000 km per second.
- •One leg: 37,500 ÷ 300,000 = 0.125 s
- •Two legs, A to satellite to B: 2 × 0.125 = 0.25 s
- •Add the relay: 0.25 + 0.01 = 0.26 s
Answer (b)
Why not others:
- •(a) 0.13 — counts only one leg, 0.125 + 0.01 = 0.135
- •(c) 0.35 — takes the 10 ms relay delay as 0.1 s, 0.25 + 0.1
- •(d) 0.52 — the round trip from A to B and back, four legs and two relays, 0.5 + 0.02
Key rule: Delay = distance ÷ speed for every leg the signal travels, plus fixed processing delays; convert kilometres and milliseconds before adding.
Networks · AI-written, may contain errors
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.