To have Y t 0 in the second 1 T thduration we need to have c105 and c2 c 3 025

# To have y t 0 in the second 1 t thduration we need to

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To have Y ( t )=0 in the second 1/ T th duration, we need to have c 1 =0.5 and c 2 = c 3 =0.25. Therefore ) ( 25 . 0 ) ( 25 . 0 ) ( 5 . 0 ) ( 3 2 1 t S t S t S t Y and Y =(0.5, 0.25, 0.25). 15. 15.12.2005 2 nd Midterm Given the parity matrix 111 100 P in a (5,2) systematic code a. Find all codewords b. Complete the syndrome table for single bit errors. Solution a. 01111 10100 P I G and 11011 10100 01111 00000 01111 10100 11 10 01 00 XG C b. 2 5 2 4 2 1 3 2 2 1 1 x c x c x x c x c x c 5 2 3 4 2 2 3 2 1 1 c c s c c s c c c s 000 001 010 100 111 100 00000 00001 00010 00100 01000 10000 _ syndrome bit error For the error positions 10000 and 00100, the syndrome vectors are the same. Therefore the errors for these bits can not be corrected.
151227621 DIGITAL COMMUNICATIONS 12 16. 15.12.2005 2 nd Midterm In the following LRC coded block determine if there are any errors. Mark, if there is any. Give an example error distribution, along with the error you have found, in the case which the errors would go undetected (mark the bits inverted but not possibly detected in such a case.) 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 1 Solution 17. 15.12.2005 2 nd Midterm Given the generator polynomial 1 ) ( 3 p p p g for a (7,4) cyclic code, find the systematic codewords for the information sequences (1001), (0010) and (1011). Solution ) ( ) ( ) ( p p X p p c k n where ) ( ) ( Rem ) ( p g p X p p k n ) ( ) 1 ( ) ( 1 3 3 ) 1 ( p p p p c (for 1001) and ) ( ) ( 2 3 ) 2 ( p p p p c (for 0010) Using the long division, parity error parity error a square error pattern like the inversion of the bits marked would mislead the parity detectors to conclude that there is no error. must be 0
151227621 DIGITAL COMMUNICATIONS 13 So the codeword polynomials are p p p p p p p p p c 2 3 6 2 3 3 ) 1 ( ) 1 ( ) ( and p p p p p p p p c 2 4 2 3 ) 2 ( ) ( . And the codewords are (1001110) and (0010110). Noting that the information sequence (1011) is a linear sum of the others, the corresponding codeword polynomial can easily be found as ) ( ) ( ) 2 ( ) 1 ( p c p c representing the codeword (1011000). In summary, 1011000 0010110 1001110 1011 0010 1001 18. 15.12.2005 2 nd Midterm For a ( n,k ) single bit error correcting block code, how many parity bits would be required if the information sequences were 26 bits. Solution The requirement is that the number of parity bits must be enough to uniquely identify each possible single bit pattern and no error condition. So 1 2 n m where m is the number of parity bits and n is the codeword length. The +1 represents the no error case. Using that, 1 2 26 n n and 31 n is found. If n is selected to be 31 then m =31-26=5 is found. 3 6 p p 3 4 6 p p p 4 p 1 3 p p p p 3 p p p 2 4 p p 2 remainders 4 p 1 3 p p p p p 2 4 p p 2 p
151227621 DIGITAL COMMUNICATIONS 14 19. 19.01.2006 Final Exam Show that all single bit errors can be corrected in the (5,2) systematic code whose parity matrix is given as 101 110 P .

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• Fall '18
• Mr. Bhullar
• Hamming Code, Error detection and correction, Parity bit

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