CS601 Assignment No 02 Spring 2020 Solution & Discussion Due Date: 13-06-2020
Question No. 1
After carefully analyzing the following figure that represents encoding scheme based on IEEE 802.3 Manchester approach, you are required to identify the bit pattern associated with the given figure:
Note: Provide your answer in the following table:
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Question No. 2
Suppose a computer in an army headquarters is transferring a file of 10 Megabits to a computer placed in a check post situated at the distance of 2000 KM. Assuming that sender and receiver computers are connected through a point to point link of 1 Gigabit and propagation speed as 2.4 × 108 meters per seconds, you are required to calculate the following:
Note: You need to write all necessary formulas & calculation steps to find the final answers in “Microseconds”.
Tags:
CS601 Data Communication Assignment 2 Solution & Discussion Spring 2020
CS601 Assignment 2 Solution Idea:
After carefully analyzing the following figure that represents encoding scheme based on IEEE 802.3 Manchester approach, you are required to identify the bit pattern associated with the given figure:
Solution:
According to instruction encoding scheme is given below in the table.
1 |
0 |
1 |
1 |
1 |
0 |
1 |
Question No: 2
Suppose a computer in an army headquarters is transferring a file of 10 Megabits to a computer placed in a check post situated at the distance of 2000 KM. Assuming that sender and receiver computers are connected through a point to point link of 1 Gigabit and propagation speed as 2.4 × 108 meters per seconds, you are required to calculate the following:
Solution:
Distance = 2000 KM
Propagation Speed = 2.4 × 108m/s =
= 240000000ms-1
For conversion of meter to Kilometer
Propagation Speed = 240000000/1000
= 240000Kms-1
Propagation time = ?
Propagation time = Distance/propagation time
Propagation time = 2000KM/ 240000Kms-1
Propagation time = 0.008333 sec
Propagation time = 0.008333/1000 milli sec
Propagation time = 8.333 milli sec
Propagation time = 8333.33 µs
Message Size = 10 Megabits
Bandwidth = 1Gigabit/s
Bandwidth = 1000 Megabits/sec
Transmission Time = ?
Transmission time = Message Size/Bandwidth
Transmission Time = 10 Megabits/1000Megabits sec-1
Transmission Time = 0.01 sec
Transmission Time = 10 millisec
Transmission Time = 10000 µs
CS601_Assignment_No_02_Solution_Spring_2020
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