48771: Discrete Communications- Faculty of Engineering and IT Assessment

Solution Code: 1AFJF

Question: IT Assessment

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IT Assessment

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Solution:

Ans Q.No 4 a) Bits required per character : 8 bits , as 27 = 128 b) for 100,000 characters , we will need 7 bits per character, and than 1 bit extra for parity, so total 8 bits *10000 or 8 bits * 23 bit = 186 bits c) Now for a) we will have 9 bits and for part b) we will require 9 bits*23 bits = 207 bits Ans Q 5 Solution As per Nyquest critera Freq required is 240*2 = 480 Hz 20% extra it makes 480+48+48 = 576 Hz To take margin for framing +0.5% or add 29Hz , 576 gets raised to 605 Hz Since we have to transmit 5 signals Minimum Bandwidth required is = 605*5 = 3025 Hz Answer- Q.no6 Solution: given, D+1= 12000-200-20200-20220-2 Ans(a): error is shown in red D+1= 12000-200-20200-20220-2 Ans(b): sequences that are possible: 1100,1000,0100,1010

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