Only (d, e, and f) This is the solution part (c)

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Q2. A binary linear cyclic block code with a code length of n=14 has the generator polynomial g(x) = 1 + x2 + x a) Determine the number of information and parity check bit bits for each codeword. (2 Marks) b) Draw the diagram of feedback shift register encoder for the Linear Cyclic code. (3 Marks) c) Illustrate the encoding procedure with the message vector 1 1 0 0 1 1 1 1 0 / by using the states of the register(the right most bit is the earliest bit) and obtain the final codeword. (5 Marks) d) Devise a syndrome computation circuit for this code. Let r=[11011001000010] be a received vector. Compute the syndrome of r(t). Display the contents of the syndrome register after each digit of r has been shifted into the syndrome computation circuit. (5 Marks) e) Verify the codeword obtained in partc polynomial division method (5 Marks) f) Compare the syndrome computation performed part d by performing polynomial division method. (5 Marks) with deguee =6 Add (deguee looLLLO vector ta Badh Generatose belynomial, goe) = 1+z2440 and Message vector, na message Message wpictaste belynomial is mine 213 +216 +2 +88127 26 + x + 1 x gox) to get remainder oo x3 + x10 + 209+28+ x 7 x 7+2 + x + 1 + att mu laoill 00.00.000 Now divide mcx cuith 13 2210 ta h x' + x8 6 + x8+ x + x x + xutx x+2 ? 2 CELET 1 rco = The codewoud teolynomial is 8- coa = mcxtu Cx2 = x3 + x + x + x 8 + x7 +1 And the codewand is- C= looli b + O 0