Textbook:
"Wireless Communications," by Andreas F. Molisch, John Wiley & Sons
chap 14: Problems 1,5,6,10 - end of chapter problems

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Textbook: "Wireless Communications," by Andreas F. Molisch, John Wiley & Sons chap 14: Problems 1,5,6,10  - end of chapter problems

David · Accepted Answer

Textbook:  
"Wireless Communications," by Andreas F. Molisch, John Wiley & Sons  
chap 14: Problems 1,5,6,10  - end of chapter problems
Answer 1:
a) For 3 bit input I the corresponding 7-bit codeword is given by C = iG. 
We first create the generator matrix for a systematic code. 
A generator matrix is constructed by writing the generator polynomial 11101 shifted 
right one bit for each successive row. There are 3 rows corresponding to the choice of 
3 input data bits. 
[1110100] 
[0111010] 
[0011101] 
 This is a valid generator polynomial.However, it does not generate a systematic 
code. 
To adjust the generator matrix for systematic code we need to manipulate the leftmost 
columns so that they form an identity 3×3 matrix. This is done by adding row 1 to row 
0 and then row 2 to row 1. 
Thus, G= 
[1001110] 
[0100111] 
[0011101]  
To encode x
2
 + 1 (101) we multiply it with G. 
Thus codeword = [101]G=[1010011]
b) 
To calculate the syndrome we calculate the check matrix (H
T
) first. The check matrix 
for a systematic code can be found directly from the generator matrix.  
H
T
 =:

Textbook: "Wireless Communications," by Andreas F. Molisch, John Wiley & Sons chap 14: Problems 1,5,6,10 - end of chapter problems

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