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Question

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\): [Answer all the situations of Hamming Decoding]

Answer

- If the syndrome is \(\mathbf{s} = \begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then the Hamming decoder thinks there are no bit errors in \(\hat{\mathbf{y}}\) (it may be wrong though). In this case, it outputs \(\hat{\mathbf{x}} = \begin{array}{cccc}[ \hat{y}_{3} &\hat{y}_{5} &\hat{y}_{6} &\hat{y}_{7} ]\end{array}^{T}\) since \(y_{3} = x_{1}\), \(y_{5} = x_{2}\), \(y_{6} = x_{3}\) and \(y_{7} = x_{4}\) in \(\mathbf{G}\).

- If the syndrome \(\mathbf{s}\) is not equal to \(\begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then its 3-bit number is converted into a decimal number \(i ∈ [1, 7]\). In this case, the Hamming decoder thinks that the ith bit in \(\hat{\mathbf{y}}\) has been flipped by a bit error (it may be wrong though). The Hamming decoder flips the \(i\)th bit in \(\hat{\mathbf{y}}\) before outputting \(\hat{\mathbf{x}}=\left[\begin{array}{llll}\hat{y}_3 & \hat{y}_5 & \hat{y}_6 & \hat{y}_7\end{array}\right]^T\). If there are multiple bit errors in the received codeword \(\hat{\mathbf{y}}\), the syndrome \(\mathbf{s}\) identifies which bit of \(\hat{\mathbf{y}}\) can be toggled to give the legitimate permutation of \(\mathbf{y}\) that is most similar.

Question

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\): [Answer all the situations of Hamming Decoding]

Answer

?

Question

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\): [Answer all the situations of Hamming Decoding]

Answer

- If the syndrome is \(\mathbf{s} = \begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then the Hamming decoder thinks there are no bit errors in \(\hat{\mathbf{y}}\) (it may be wrong though). In this case, it outputs \(\hat{\mathbf{x}} = \begin{array}{cccc}[ \hat{y}_{3} &\hat{y}_{5} &\hat{y}_{6} &\hat{y}_{7} ]\end{array}^{T}\) since \(y_{3} = x_{1}\), \(y_{5} = x_{2}\), \(y_{6} = x_{3}\) and \(y_{7} = x_{4}\) in \(\mathbf{G}\).

- If the syndrome \(\mathbf{s}\) is not equal to \(\begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then its 3-bit number is converted into a decimal number \(i ∈ [1, 7]\). In this case, the Hamming decoder thinks that the ith bit in \(\hat{\mathbf{y}}\) has been flipped by a bit error (it may be wrong though). The Hamming decoder flips the \(i\)th bit in \(\hat{\mathbf{y}}\) before outputting \(\hat{\mathbf{x}}=\left[\begin{array}{llll}\hat{y}_3 & \hat{y}_5 & \hat{y}_6 & \hat{y}_7\end{array}\right]^T\). If there are multiple bit errors in the received codeword \(\hat{\mathbf{y}}\), the syndrome \(\mathbf{s}\) identifies which bit of \(\hat{\mathbf{y}}\) can be toggled to give the legitimate permutation of \(\mathbf{y}\) that is most similar.

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**Situations of the result of Hamming Decoding**

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\): - If the syndrome is \(\mathbf{s} = \begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then the Hamming decoder thinks there are no bit errors in \(\hat{\mathbf{y}}\) (it may be wrong though). - In this case, it outputs \(\hat{\mathbf{x}} = \begin{array}{cccc}[ \hat{y}_{3} &\hat{y}_{5} &\hat{y}_{6} &\hat{y}_{7} ]\end{array}^{T}\) since \(y_{3} = x_{1}\), \(y_{5} = x_{2}\), \(y_{6} = x_{3}\) and \(y_{7} = x_{4}\) in \(\mathbf{G}\). - If the syndrome \(\mathbf{s}\) is not equal to \(\begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then its 3-bit number is converted into a decimal number \(i ∈ [1, 7]\). - In this case, the Hamming decoder thinks that the ith bit in \(\hat{\mathbf{y}}\) has been flipped by a bit error (it may be wrong though). The Hamming decoder flips the \(i\)th bit in \(\hat{\mathbf{y}}\) before outputting \(\hat{\mathbf{x}}=\left[\begin{array}{llll}\hat{y}_3 & \hat{y}_5 & \hat{y}_6 & \hat{y}_7\end{array}\right]^T\).

Using the resultant \(n − k\) = 3-bit syndrome \(\mathbf{s}\), the Hamming decoder decides if it thinks there are any bit errors in \(\hat{\mathbf{y}}\): - If the syndrome is \(\mathbf{s} = \begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then the Hamming decoder thinks there are no bit errors in \(\hat{\mathbf{y}}\) (it may be wrong though). - In this case, it outputs \(\hat{\mathbf{x}} = \begin{array}{cccc}[ \hat{y}_{3} &\hat{y}_{5} &\hat{y}_{6} &\hat{y}_{7} ]\end{array}^{T}\) since \(y_{3} = x_{1}\), \(y_{5} = x_{2}\), \(y_{6} = x_{3}\) and \(y_{7} = x_{4}\) in \(\mathbf{G}\). - If the syndrome \(\mathbf{s}\) is not equal to \(\begin{array}{ccc}[ 0 &0 &0]\end{array}^{T}\) then its 3-bit number is converted into a decimal number \(i ∈ [1, 7]\). - In this case, the Hamming decoder thinks that the ith bit in \(\hat{\mathbf{y}}\) has been flipped by a bit error (it may be wrong though). The Hamming decoder flips the \(i\)th bit in \(\hat{\mathbf{y}}\) before outputting \(\hat{\mathbf{x}}=\left[\begin{array}{llll}\hat{y}_3 & \hat{y}_5 & \hat{y}_6 & \hat{y}_7\end{array}\right]^T\).

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

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