# Solutions_Chapter_02.pdf - Error Control Coding u2014 week...

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Error Control Coding — week 1: Binary field algebra and linear block codes1Solutions to selected problems from Chapter 22.9Denote the equations with the numbers (1) – (4) starting from the top one.Then, summing the equations (3) and (4) (modulo 2) yields(3) + (4)X+Y+Y+Z+Z+W+W=X= 1Similarly, summation of the equations (2) and (3) results in(2) + (3)X+X+Y+Z+Z+W+W=Y= 1From equation (1) we obtain (recall that inGF(2) subtraction is equivalentto addition)(1)W= 1 +X+Y= 1Finally, equation (4) gives(4)Z= 1 +Y+W= 0Thus, the solution is [X Y W Z] = [1 1 1 0].2.10The polynomiala(x) =x5+x3+ 1 is irreducible overGF(2) if it is notdivisible by any polynomialp(x) overGF(2) of degree 1deg(p(x))2 =bdeg(a(x))/2c.Thus, we have to check if there exists a polynomial of the formp(x) =p2x2+p1x+p0overGF(2), which dividesa(x).(a) Asa(0) = 1 we can conclude thatx-0 =xis not a factor ofa(x).(b) Asa(1) = 1 we can conclude thatx-1 =x+ 1 is not a factor ofa(x).If there exists a polynomialp(x) =p2x2+p1x+p0which dividesa(x), we canconclude from (a) thatp0= 1 and from (b) thatp2= 1.Thusp(x) =x2+p1x+ 1 and we have to check the two polynomialsx2+ 1andx2+x+ 1.Clearly, asx2+ 1 = (x+ 1)2and withx+ 1 not being a factor ofa(x), itfollows directly thatx2+ 1 cannot be a factor ofa(x) either.