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# CS601 Data Communication Assignment No 02 Fall 2020 Solution / Discussion

CS601 Data Communication Assignment No 02 Fall 2020 Solution / Discussion

Question No. 1

Diagram given below is representing the Differential Manchester encoding scheme. You are required to carefully analyze the diagram and then write the binary pattern represented by the diagram:

Hint: First bit is 0 and its voltage level is negative to positive.

Give your answer in the following table:

Question No. 2

Suppose a telecommunication company’s booster is emitting a signal of 0.45W, which is being passed through 10 devices. It has been observed that 7 of these devices are encountering noise of 3 microwatts whereas other 3 devices are encountering noise of 2 microwatts. Considering this information, you are required to calculate overall Signal to Noise Ratio (SNR) and SNRdB by writing all necessary formulas and calculation steps.

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CS601 Assignment 2 Solution idea Fall 2020

Question No. 1

Diagram given below is representing the Differential Manchester encoding scheme. You are required to carefully analyze the diagram and then write the binary pattern represented by the diagram:

Hint: First bit is 0 and its voltage level is negative to positive.

Give your answer in the following table:

Solution:

 0 1 0 0 1 1 1 0 0

Question No. 2

Suppose a telecommunication company’s booster is emitting a signal of 0.45W, which is being passed through 10 devices. It has been observed that 7 of these devices are encountering noise of 3 microwatts whereas other 3 devices are encountering noise of 2 microwatts. Considering this information, you are required to calculate overall Signal to Noise Ratio (SNR) and SNRdB by writing all necessary formulas and calculation steps.

Solution:

Signal Power = 0.45W

Noise Power of Devices = 7*0.0003

= 0.000021W

Noise Power of Devices = 3*0.000002

= 0.000006W

Total Noise Power = 0.000021 + 0.00006

= 0.000027

SNR – Signal to Noise Ratio = S/W

= 0.45/0.000027

= 16,666.66

SNRdB                                  = 10(logSNR)

= 10(log16,666.66)

= 10(4.22)

= 42.21

CS601_Assignment_No_02_Solution_Fall_2020

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Idea Solution

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