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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 3 x 108 m/s. The time taken (in milliseconds runoff to 2 decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _____.
Correct answer is '7.08'. Can you explain this answer?
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Consider a 100 Mbps link between an earth station (sender) and a satel...
Given information:
- Link speed between earth station and satellite: 100 Mbps (100 million bits per second)
- Altitude of the satellite: 2100 km
- Signal propagation speed: 3 x 10^8 m/s
- Packet size: 1000 bytes
Calculating the time taken:
To determine the time taken for the receiver to completely receive a packet, we need to consider two factors: the transmission time and the propagation time.
1. Transmission Time:
The transmission time is the time taken to send the packet from the sender to the receiver over the link.
- The link speed is given in bits per second, so we need to convert the packet size from bytes to bits.
- 1 byte = 8 bits, so the packet size in bits is 1000 bytes * 8 = 8000 bits.
- The transmission time can be calculated using the formula: Transmission time = Packet size / Link speed.
- Substituting the values, we get: Transmission time = 8000 bits / 100 Mbps = 0.08 milliseconds.
2. Propagation Time:
The propagation time is the time taken for the signal to travel from the sender to the receiver.
- The altitude of the satellite is given as 2100 km.
- The speed of signal propagation is given as 3 x 10^8 m/s.
- The distance traveled by the signal can be calculated using the formula: Distance = Altitude of satellite.
- Substituting the values, we get: Distance = 2100 km = 2100 * 1000 m = 2,100,000 m.
- The propagation time can be calculated using the formula: Propagation time = Distance / Signal propagation speed.
- Substituting the values, we get: Propagation time = 2,100,000 m / (3 x 10^8 m/s) = 7 milliseconds (rounded to 2 decimal places).
Total Time:
The total time taken is the sum of the transmission time and the propagation time.
- Total time = Transmission time + Propagation time
- Total time = 0.08 milliseconds + 7 milliseconds = 7.08 milliseconds (rounded to 2 decimal places).
Therefore, the correct answer is 7.08 milliseconds.
- Link speed between earth station and satellite: 100 Mbps (100 million bits per second)
- Altitude of the satellite: 2100 km
- Signal propagation speed: 3 x 10^8 m/s
- Packet size: 1000 bytes
Calculating the time taken:
To determine the time taken for the receiver to completely receive a packet, we need to consider two factors: the transmission time and the propagation time.
1. Transmission Time:
The transmission time is the time taken to send the packet from the sender to the receiver over the link.
- The link speed is given in bits per second, so we need to convert the packet size from bytes to bits.
- 1 byte = 8 bits, so the packet size in bits is 1000 bytes * 8 = 8000 bits.
- The transmission time can be calculated using the formula: Transmission time = Packet size / Link speed.
- Substituting the values, we get: Transmission time = 8000 bits / 100 Mbps = 0.08 milliseconds.
2. Propagation Time:
The propagation time is the time taken for the signal to travel from the sender to the receiver.
- The altitude of the satellite is given as 2100 km.
- The speed of signal propagation is given as 3 x 10^8 m/s.
- The distance traveled by the signal can be calculated using the formula: Distance = Altitude of satellite.
- Substituting the values, we get: Distance = 2100 km = 2100 * 1000 m = 2,100,000 m.
- The propagation time can be calculated using the formula: Propagation time = Distance / Signal propagation speed.
- Substituting the values, we get: Propagation time = 2,100,000 m / (3 x 10^8 m/s) = 7 milliseconds (rounded to 2 decimal places).
Total Time:
The total time taken is the sum of the transmission time and the propagation time.
- Total time = Transmission time + Propagation time
- Total time = 0.08 milliseconds + 7 milliseconds = 7.08 milliseconds (rounded to 2 decimal places).
Therefore, the correct answer is 7.08 milliseconds.
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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 3 x 108 m/s. The time taken (in milliseconds runoff to 2 decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _____.Correct answer is '7.08'. Can you explain this answer?
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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 3 x 108 m/s. The time taken (in milliseconds runoff to 2 decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _____.Correct answer is '7.08'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 3 x 108 m/s. The time taken (in milliseconds runoff to 2 decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _____.Correct answer is '7.08'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 3 x 108 m/s. The time taken (in milliseconds runoff to 2 decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _____.Correct answer is '7.08'. Can you explain this answer?.
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