BCH Codes

In the following questions, wherever required, use the Galois fields F23={0,1,α,α2,α3=1+α,α4=α+α2,α5=1+α+α2,α6=1+α2}and\mathbb{F}_{2^{3}} = \{ 0, 1, \alpha, \alpha^{2}, \alpha^{3} = 1+\alpha, \alpha^{4} = \alpha+\alpha^{2}, \alpha^{5}=1+\alpha+\alpha^{2}, \alpha^{6}= 1+\alpha^{2} \} \, \text{and} F24={0,1,α,α2,α3,α4=1+α,α5=α+α2,α6=α2+α3,α7=1+α+α3,α8=1+α2,α9=α+α3,α10=1+α+α2,α11=α+α2+α3,α12=1+α+α2+α3,α13=1+α2+α3,α14=1+α3}. \begin{align*} \mathbb{F}_{2^{4}} = \{ & 0, 1, \alpha, \alpha^{2}, \alpha^{3}, \alpha^{4} = 1+\alpha, \alpha^{5} = \alpha+\alpha^{2}, \alpha^{6}=\alpha^{2}+\alpha^{3}, \alpha^{7}= 1+\alpha+\alpha^{3}, \\ & \alpha^{8} = 1+\alpha^{2}, \alpha^{9} =\alpha+\alpha^{3}, \alpha^{10}=1+\alpha+\alpha^{2}, \alpha^{11}=\alpha+\alpha^{2}+\alpha^{3},\\ & \alpha^{12}=1+\alpha+\alpha^{2}+\alpha^{3}, \alpha^{13}=1+\alpha^{2}+\alpha^{3}, \alpha^{14} = 1+\alpha^{3} \}. \end{align*} 1. Which one of the following cannot be a binary primitive BCH code of the length?
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2. The generator polynomial g(X)\mathbf{g}(X) of the tt-error correcting binary primitive BCH code is ?
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3. The generator polynomial g(X)\mathbf{g}(X) of the 22-error correcting binary primitive BCH code of length n=15n=15 is
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4. In BCH code, suppose the generator polynomial g(X)\mathbf{g}(X) has 2t+12t+1 number of non-zero coefficients, then g(X)\mathbf{g}(X) is the codeword polynomial of the codeword with the weight equal to dmind_{min}.
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5. Suppose the received polynomial of the (15,5,7)(15, 5, 7) triple-error-correcting BCH code is r(X)=X+X2+X3\mathbf{r}(X) = X+X^{2}+X^{3} , then the corresponding syndrome vector is
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6. Consider the same example which we discussed in the theory of this experiment (Example 3).For the (15,5,7)(15, 5, 7) triple-error-correcting BCH code, suppose that v=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]\mathbf{v} = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] is transmitted, and the vector r=[0,0,0,1,0,1,0,0,0,0,0,0,1,0,0]\mathbf{r} = [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0] is received. Then r(X)=X3+X5+X12\mathbf{r}(X) = X^{3}+X^{5}+X^{12}. Assume that we have performed Berlekamp's iterative algorithm up to the 3rd iteration (i.e, μ=3\mu = 3) and is provided in the Table 6.
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