Question

Question: Let C be the (15,7) BCH code of examples 1 and 2. Use the error-correction techniques presented there to correct these received words or to determine that three or more errors have been made.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{110000000000000} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}{\mathbf{x}}^{\mathbf{5}}\mathbf{=}\mathbf{100111000000000} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}{\mathbf{x}}^{\mathbf{7}}\mathbf{=}\mathbf{101010010000000} \\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{}\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{6}}\mathbf{+}{\mathbf{x}}^{\mathbf{7}}\mathbf{+}{\mathbf{x}}^{\mathbf{8}}\mathbf{+}{\mathbf{x}}^{\mathbf{9}}\mathbf{=}\mathbf{100000111100000} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{the}\mathrm{correct}\mathrm{word}\mathrm{is}000000000000000 \\ \left(\mathrm{b}\right)\mathrm{the}\mathrm{correct}\mathrm{word}\mathrm{is}100111001000001 \\ \left(\mathrm{c}\right)\mathrm{the}\mathrm{correct}\mathrm{word}\mathrm{is}101010010000000 \\ \left(\mathrm{d}\right)\mathrm{the}\mathrm{correct}\mathrm{word}\mathrm{is}100000111100000 \end{gathered}$$

Step-by-step solution

Step 1:Given in the question

Let$f\left(x\right)=a_n{x}^n+....+a_r{x}^r+....+a_0$ .....(1)

Here$a_i\in Z_2$

This implies $a_i+a_i=0$ .....(2)

Step 2:Error of 1+x=11000000000000000

Let a (15,7) BCH code with the received word as follows:

$1+\mathrm{x}=11000000000000000$

The objective is to determine the errors.

Use the table at the starting of example 1 and the fact that u + u = 0 for every element u in K.

$$\begin{gathered} r\left(a\right)=1+a \\ =a^4 \\ r\left(a^2\right)=1+a^2 \\ =a^8 \end{gathered}$$

And

$$\begin{gathered} a^3=1+a^3 \\ =a^{14} \end{gathered}$$

The error-locator polynomial is given by the formula:

$$\begin{gathered} D\left(x\right)=x^2+\left(a^4\right)x+\left(a^4+\frac{\left(a^{14}\right)}{\left(a^4\right)}\right) \\ =x^2+a^{4}x+\left(a^8+a^{14-4}\right) \\ =x^2+a^{4}x+\left(a^8+a^{10}\right) \end{gathered}$$

Again, use the table in example 1.

$$\begin{gathered} D\left(x\right)=x^2+\left(a^4\right)x+\left(\left(1+a^2\right)+\left(1+a+a^2\right)\right) \\ =x^2+a^{4}x+\left(\left(1+1\right)+\left(a^2+a^2\right)\right)+a \end{gathered}$$

Use equation (2),

$$\begin{gathered} D\left(x\right)=x^2+\left(a^4\right)x+\left(\left(0\right)+\left(0\right)\right)+a \\ =x^2+a^{4}x+\left(0\right)+a \\ D\left(x\right)=x^2+a^{4}x+a \end{gathered}$$

Now, solve as follows:

$$\begin{gathered} D\left(a^0=1\right)=1^2+a^4\times 1+a \\ =1+a^4+a \\ =1+\left(1+a\right)+a \end{gathered}$$

Use equation (2),

$$\begin{gathered} D\left(a^0=1\right)=\left(1+a\right)+\left(1+a\right) \\ =0+0 \\ =0 \end{gathered}$$

Similarly solved as follows:

$D\left(a\right)=0$

Hence, the correct word is 000000000000000.

Error of 1+x3+x4+x5=100111000000000

Let a (15,7) BCH code with the received word as follows:

$1+{\mathrm{x}}^3+{\mathrm{x}}^4+{\mathrm{x}}^5=100111000000000$

The objective is to determine the errors.

Follow the same procedure, the error-locator polynomial is given by the formula:

$D\left(x\right)=x^2+a^{4}x+a^7$

Here, $a^8\mathrm{and}{a}^{14}$are the roots of the error-locator polynomial.

Hence, the correct word is 100111001000001.

Step 4:Error of 1+x3+x4+x7=101010010000000

Let a (15,7) BCH code with the received word as follows:

$1+{\mathrm{x}}^3+{\mathrm{x}}^4+{\mathrm{x}}^7=101010010000000$

The objective is to determine the errors.

Follow the same procedure, the error-locator polynomial is given by the formula:

$D\left(x\right)=x^2+a^{13}x+a^4$

Here, $a^9\mathrm{and}{a}^{10}$are the roots of the error-locator polynomial.

Hence, the correct word is$101010010000000$

Error of 1+x6+x7+x8+x9=100000111100000

Let a (15,7) BCH code with the received word as follows:

$1+{\mathrm{x}}^6+{\mathrm{x}}^7+{\mathrm{x}}^8+{\mathrm{x}}^9=100000111100000$

The objective is to determine the errors.

Follow the same procedure, the error-locator polynomial is given by the formula:

$D\left(x\right)=x^2+a^{14}x+a^{13}$

Here, is the only root of the error-locator polynomial.

Hence, the correct word is $100000111100000$.