Assignment No. 4
Semester: Spring 2015
Data Communication  CS601
Total Marks: 10
Due Date: 05/08/2015
Objective:
To have the understanding of Error Detection And Correction Methods.
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Assignment
Question  1
Consider the following group of bits along with odd parity LRC bits is sent to the receiver, but due to transmission impairments some bits of LRC are lost. You are required to identify the lost LRC bits and complete the given table.
Group of bits LRC bits
10101110 11001101 10101010 11110000 1 ? ? ? 0 ? ? 0
Question  2
Find the binary equivalent of the Polynomial given in following table:
Polynomial Binary Equivalent
X3+X2+1 ?
X4+X+1 ?
X5+X2+1 ?
X6+X+1 ?
X3+X ?
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2nd question solution
1101
10011
100101
1000011
1010
Question  1
Consider the following group of bits along with odd parity LRC bits is sent to the receiver, but due to transmission impairments some bits of LRC are lost. You are required to identify the lost LRC bits and complete the given table.
Group of bits 
LRC bits 

10101110 
11001101 
10101010 
11110000 
1 
0 
0 
0 
0 
1 
1 
0 
Question  2
Find the binary equivalent of the Polynomial given in following table:
Polynomial 
Binary Equivalent 
X^{3}+X^{2}+1 
110+1 
X^{4}+X+1 
1001+1 
X^{5}+X^{2}+1 
1001+1 
X^{6}+X+1 
100001+1 
X^{3}+X 
10+1 

Assignment No 4
CS601
Spring 2015
Question  1
Consider the following group of bits along with odd parity LRC bits is sent to the receiver, but due to transmission impairments some bits of LRC are lost. You are required to identify the lost LRC bits and complete the given table.
Solution
Given group of bits
10101110 11001101 10101010 11110000
Converts this group of bit in to LRC
LRC
10000110
Question  2
Find the binary equivalent of the Polynomial given in following table:
Solution
Polynomial 
Binary Equivalent 
X3+X2+1 
1101 
X4+X+1 
10011 
X5+X2+1 
10011 
X6+X+1 
1000011 
X3+X 
1011 
Second bit from left shouldn't be one?
could you explain this assignment plzzzzzz
1) how to find the polynomial?
2) how we calculate or identified LRC bits lost?
x^5+x^2+1
This equation has x^5 so 1 is written than it don't have x^4 a 0 is written again 0 for x^3 than x^2 is in the equation so a 1 is written than 0 for x and 1 for the constant as a constant is in the equation finally 100101
2. arrange 32 bits in 4x4 matrix than sum each column. for odd parity if answer of a bit is even add 1 else 0. For even parity if answer is odd add 1 else 0
Tariq sb you have pasted 2 different solutions. which one is correct?
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