Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

• From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.
• Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.
• Hence, the floating point representation for given binary value is

• From the above table, it is based on the normalized case.
  • Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is
    • Formula for computing “E”, “E = e – bias”
    • Formula for computing “M”, “M = 1 + f”
    • Formula for computing “V”, “$\text{V}={\text{(}-\text{1)}}^{\text{s}}\times {\text{2}}^{\text{E}}\times \text{M}$”
• From the above table, sign bit “s” is “0”.

$\begin{array}{l}\text{E}=\text{e}-\text{bias}\\ =\text{128}-\text{bias}=>\text{Computing}\text{decimal}\text{value}\text{for}\text{exponent}\text{bit}\text{"10000000"}\text{is}\text{128}\\ =\text{128}-\left({\text{2}}^{\text{8}-\text{1}}-\text{1}\right)=>\text{Formula}\text{determine}\text{bias}\text{is}{\text{2}}^{\text{k}-\text{1}}-\text{1}\end{array}$

B.

Explanation of Solution

Fractional binary representation of “$\frac{\text{22}}{\text{7}}$”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, $\text{n}=\text{11}\text{.00}\text{1 001 000 011 111 101 101 1 }$. Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

$\text{n}=\frac{\text{Y}}{{\text{2}}^{\text{k}}-\text{1}}$

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “$\frac{\text{223}}{\text{71}}$” and “$\frac{\text{22}}{\text{7}}$”.

• When converting “$\frac{\text{223}}{\text{71}}$” to decimal, the value is “3