Question

1) Consider a (15,5) linear block code (cyclic) in systematic form. The generator polynomial is given as g(x) = 1 + x + x2 + x5 + x + x10. a. Design and draw the circuit of the feedback shift register encoder and decoder (6 Marks) b. Use the encoder obtained in part a to find the code word for the message (10110). (Assume the right most bit is the earliest bit) (5 Marks) C. Repeat the steps of part b for decoding. (5 Marks) d. Verify the codeword obtained in part b polynomial division method (5 Marks) e. Consider a codeword C = [110110010001100 ). Is this a code word of the above system? Provide suitable justification for your answer. (4 Marks)

Answer 1

Full code word has the length n=15

In this code word there are k=5 message bits

Thus the number of check bits is q=n-k=15-5=10

a)

Flip Flop Flip Flop Flip Flop Flip Flop Flip Flop Flip Flop Flip Flop Flip Flop P2 P3 P1 (LSB) P4 P5 P6 P7 P8 Input (MSB) Message

d)

Given message is [10110]

In order to do this we will translate this to a polynomial: x⁴(1)+x³(0)+x²(1)+x¹(1)+x⁰(0)=x⁴+x²+x=M(x)

x^qM(x)=x¹⁰(x⁴+x²+x)=x¹⁴+x¹²+x¹¹

g(x) x"M(x) 4 X +x 1+x+x2+x"+x8+x20 11 14 12 11 X+X+X 14 129 6 5 4 +X+X+X + x X 11 96 54 х +X +X + x +X 11 9 6 3 x++x+x®+x2+x2 +x 5432 X +X+X+X+X

The addition above is XOR addition and not subtraction.

The remainder is the check bit C(x)= x⁵+x⁴+x³+x²+x

But q=10, so the check bit is C= [ 0 0 0 0 1 1 1 1 1 0 ]

The code vector: [M:C]=[10110:0000111110]