Chapter 1.7, Problem 1QE

a.

Explanation of Solution

Decode the $01001010$ bit pattern using the floating-point:

First designate the high-order bit of the byte as the sign bit. A $0$ in the sign bit means that the value stored is non-negative, and a $1$ means that the value is negative. The $3$ bits following the sign bit is the exponent field and the remaining $4$  bits are the mantissa field.

An excess notation system using bit patterns of length three is,

Bit Pattern	Value represented
$111$	$3$
$110$	$2$
$101$	$1$
$100$	$0$
$011$	$-1$
$010$	$-2$
$001$	$-3$
$000$	$-4$

Explanation:

• The sign bit is $0$, the exponent part of $01001010$ is $100$ and the mantissa part of $01001010$ is $\mathrm{}$

b.

Explanation of Solution

Decode the $01101101$ bit pattern using the floating-point:

• The sign bit is $0$, the exponent part of $01101101$ is $110$ and the mantissa part of $01101101$ is $.1101$.
• $0$ sign bit indicates that stored value is non-negative.
• The value of the exponent $110$ from Table 1 is $2$.
• As the value of the exponent is $2$, move the radix $2\text{bit}$ to the right of the mantissa $.1101$.

  $\text{The}\text{mantissa}\text{becomes}11.01$

• Convert $11$

c.

Explanation of Solution

Decode the $00111001$ bit pattern using the floating-point:

• The sign bit is $0$, the exponent part of $00111001$ is $011$ and the mantissa part of $00111001$ is $.1001$.
• $0$ sign bit indicates that stored value is non-negative.
• The value of the exponent $011$ from Table 1 is $-1$.
• As the value of the exponent is $-1$, move the radix $-1\text{bit}$ to the left of the mantissa $.1001$.

  $\text{The}\text{mantissa}\text{becomes}.01001$

• Convert $.01001$ into equivalent decimal form as:

$\begin{array}{c}{(\mathrm{}}_{}\end{array}$

d.

Explanation of Solution

Decode the $11011100$ bit pattern using the floating-point:

• The sign bit is $1$, the exponent part of $11011100$ is $101$ and the mantissa part of $11011100$ is $.1100$.
• $1$ sign bit indicates that stored value is negative.
• The value of the exponent $101$ from Table 1 is $1$.
• As the value of the exponent is $1$, move the radix $1\text{bit}$ to the right of the mantissa $.1100$.

  $\text{The}\text{mantissa}\text{becomes}1.100$.

• Convert $1.100$ into equivalent decimal form as:

$\begin{array}{c}{(1}_{}\end{array}$

e.

Explanation of Solution

Decode the $10101011$ bit pattern using the floating-point:

• The sign bit is $1$, the exponent part of $10101011$ is $010$ and the mantissa part of $10101011$ is $.1011$.
• $1$ sign bit indicates that stored value is negative.
• The value of the exponent $010$ from Table 1 is $-2$.
• As the value of the exponent is $-2$, move the radix $2\text{bit}$ to the left of the mantissa $.1011$.

  $\text{The}\text{mantissa}\text{becomes}0.001011$.

• Convert $0.001011$ into equivalent decimal form as:

$\begin{array}{c}{(0}_{}\end{array}$