Virtual Labs

In the following questions, wherever required, use the Galois fields

\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}

\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?

n=127

Explanation

n=63

Explanation

n=1023

Explanation

n=509

Explanation

2. The generator polynomial

\mathbf{g}(X)

of the

t-

error correcting binary primitive BCH code is ?

\mathbf{g}(X) =

LCM

\{\phi_{1}(X), \phi_{2}(X), \ldots , \phi_{2t}(X) \}

Explanation

\mathbf{g}(X) = \text{LCM} \, \, \{\phi_{2}(X), \phi_{4}(X), \phi_{6}(X), \ldots , \phi_{2t}(X) \}

Explanation

\mathbf{g}(X) = \text{LCM} \, \, \{\phi_{1}(X), \phi_{3}(X), \phi_{5}(X), \ldots , \phi_{2t-1}(X) \}

Explanation

d: Both (A) and (C) Explanation

Explanation

3. The generator polynomial

\mathbf{g}(X)

of the

2-

error correcting binary primitive BCH code of length

n=15

1+X+X^{2}+X^{3}+X^{4}

Explanation

1+X+X^{4}+X^{6}+X^{8}

Explanation

1+X^{4}+X^{6}+X^{7}+X^{8}

Explanation

d: None of the above Explanation

Explanation

4. In BCH code, suppose the generator polynomial

\mathbf{g}(X)

has

2t+1

number of non-zero coefficients, then

\mathbf{g}(X)

is the codeword polynomial of the codeword with the weight equal to

d_{min}

a: Yes Explanation

Explanation

b: No Explanation

Explanation

5. Suppose the received polynomial of the

(15, 5, 7)

triple-error-correcting BCH code is

\mathbf{r}(X) = X+X^{2}+X^{3}

, then the corresponding syndrome vector is

\begin{bmatrix} \alpha^{2} & 1 & \alpha^{3} & 1 & \alpha^{6} & \alpha^{5} \end{bmatrix}

Explanation

\begin{bmatrix} 1 & \alpha & \alpha^{10} & \alpha^{4} & \alpha^{10} & \alpha^{5} \end{bmatrix}

Explanation

\begin{bmatrix} \alpha^{11} & \alpha^{7} & \alpha^{11} & \alpha^{14} & 0 & \alpha^{7} \end{bmatrix}

Explanation

\begin{bmatrix} \alpha^{2} & \alpha^{3} & \alpha^{5} & \alpha^{7} & 0 & \alpha^6 \end{bmatrix}

Explanation

6. Consider the same example which we discussed in the theory of this experiment (Example 3).For the

(15, 5, 7)

triple-error-correcting BCH code, suppose that

\mathbf{v} = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

is transmitted, and the vector

\mathbf{r} = [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0]

is received. Then

\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,

\mu = 3

) and is provided in the Table 6.

The value of discrepancy

d_3

is:

0

Explanation

\alpha^{10}

Explanation

1

Explanation

\alpha^{3}

Explanation

BCH Codes