Chapter 2, Problem 66E

a.

Explanation of Solution

Number of parity bits:

The length of memory word is 10

The number of parity bits for memory word can be calculated using the following formula

$\text{m}\text{+}\text{r}\text{+}\text{1}\le {\text{2}}^{\text{r}}$

Substitute, “10” for “m” in the above formula

$\begin{array}{l}\text{10}\text{+}\text{r}\text{+}\text{1}\le {\text{2}}^{\text{r}}\\ \text{11}\text{+}\text{r}\le {\text{2}}^{\text{r}}\end{array}$

Hence, r must be greater than or equal to 4.

Therefore, the number of parity bits required for 12 length memory word is “4”.

b.

Explanation of Solution

Code word representation using hamming algorithm:

To create the code for memory word with 10-bit length, 4 parity bits are added compulsorily.

The number of bits present in the code word can be calculated using the following formula,

$\text{n}\text{=}\text{m}\text{+}\text{r}$

Substitute, “10” for “m” and “4” for “r” in the above formula,

$\begin{array}{c}\text{n}\text{=}\text{10}\text{+}4\\ \text{=}\text{14}\end{array}$

Thus, the number of bits present in the code word is 14.

The number bits starts from right to left with 1.

The number equal to the power of 2 is known as parity bit.

For example, $2^0=1$, $2^1=2$, $2^2=4$ and $2^3=8$

The positions of the parity bits are 1, 2, 4 and 8.

Now, write the sum of the numbers with powers of 2.

$\begin{array}{l}1=1\\ 2=2\\ 3=1+2\\ 4=4\\ 5=1+4\\ 6=2+4\\ 7=1+2+4\\ 8=8\\ 9=1+8\\ 10=2+8\\ 11=1+2+8\\ 12=4+8\\ 13=1+4+8\\ 14=2+4+8\end{array}$

The data bits are filled with the given code word by leaving the positions related to the parity bits.

$\begin{array}{l}1001100110\\ 1413121110987654321\end{array}$

Parity bit 1 checks parity over 3, 5, 7, 9 and 11.

Now perform the modulo 2 sum on these bits.

$0+1+0+0+1$

Thus, the parity bit of 1 is 0.

Parity bit 2 checks parity over 3, 6, 7, 10, 11 and 14.

Now perform the modulo 2 sum on these bits.

$1+1+0+1+1+1$

Thus, the parity bit of 2 is 0.

Parity bit 3 checks parity over 5, 6, 7, 12, 13 and 14.

Now perform the modulo 2 sum on these bits

$1+1+0+0+0+1$

Thus, the parity bit of 4 is 1.

Parity bit 4 checks parity over 9, 10, 11, 12, 13 and 14.

Now perform the modulo 2 sum on these bits.

$0+1+1+0+0+1$

Thus, the parity bit of 8 is 1.

The data bits are filled as follows

$\begin{array}{l}10011010111000\\ 1413121110987654321\end{array}$

Therefore, the code word for 100100011010 is “10011010111000”.