48771: Discrete Communications- Faculty of Engineering and IT Assessment
Solution Code: 1AFJF
Question: IT Assessment
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IT Assessment
Case Scenario/ Task
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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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