Chapter 1, Problem 42CRP

a.

Explanation of Solution

Determine the information about the common values:

The values of the bit pattern are $01011$ and $11011$. One of the bits represents the values in excess $-16$ notation and another is the same value in two’s complement notation.

• Consider the two values are in excess $-16$ notation, to find out the value in base ten.
• Convert the $-16$ notation into its equivalent base 10 representation.
• First covert the bit pattern 01011 to base ten representations, multiply every bit of the excess $-16$ representation by power of two from right to left as shown below:

$\begin{array}{c}0\times 2^4\text{+1}\times {\text{2}}^3\text{+0}\times {\text{2}}^2\text{+1}\times {\text{2}}^1\text{+1}\times {\text{2}}^0=0+8+0+2+1\\ =\text{ 11}\end{array}$

• Subtract 16 from the result to find the equivalent value. The excess $-16$ notation is $11-16=-5$. Thus, $01011={\left(-5\right)}_{10}$
• The second bit pattern be $11011$. The conversion of excess $-16$ notation the equivalent base 10 representation is shown below:

$\begin{array}{c}1\times 2^4\text{+1}\times {\text{2}}^3\text{+0}\times {\text{2}}^2\text{+1}\times {\text{2}}^1\text{+1}\times {\text{2}}^0=16+8+0+2+1\\ =\text{ 27}\end{array}$

• The excess $16$ notation is $27-16=11$. Thus, $11011={\left(11\right)}_{10}$
• Thus, $01011={\left(-5\right)}_{10}$ and $11011={\left(11\right)}_{10}$

b.

Explanation of Solution

The two bit pattern is 01011 and 11011. Both the system uses the same bit pattern length.

The relation between the patterns representing a value stored in two’s complement notation with the same value stored in excess notation:

• From the given two bit pattern it can be observed that the all the bits are same expect the high order bit which is 0 in the first bit pattern and 1 in the second pattern...