Chapter 2, Problem 2.83HW

A.

Explanation of Solution

Formula to find the numeric value for infinite string:

The formula to find the numeric value for infinite string using in terms of “Y” and “k” is given below.

From the given question, assume the infinite string of the form,$\text{n}\text{=}\text{0}\text{.yyyyyyyy}\mathrm{........}$ and here “y” is a k-bit sequence.

For a k-bit sequence,

$\begin{array}{l}\text{n}\text{<< k}\text{=}\text{y}\text{.yyyyyyyy}\mathrm{........}\\ \text{ =}\text{Y}\text{+}\text{n}\end{array}$

Hence, to find the numeric value of the string “n” is

$\begin{array}{c}\text{n}\text{<<k}\text{=}\text{Y}\text{+}\text{n}\\ \text{n}\text{<< k - n = Y}\end{array}$

$\begin{array}{c}\text{n *}{\text{2}}^{\text{k}}\text{-}\text{n =}\text{Y}\text{=>}\text{By}\text{x}\text{<<}\text{k}\text{=}\text{x}\text{*}\end{array}$

B.

(a)

Explanation of Solution

Compute the numeric value of the string for “y = 101”:

From the part A, the formula is $\text{n}\text{=}\frac{{\text{B2U}}_{\text{k}}\left(\text{y}\right)}{\left({\text{2}}^{\text{k}}\text{-}\text{1}\right)}$

Here, value of “y” is “101” and value of “k” is number of bits in “y”. So, k = 3.

First compute the value of ${\text{B2U}}_{\text{3}}\text{(y)}$ is given below

$\begin{array}{c}{\text{B2U}}_{\text{3}}\text{(y)}{\text{= B2U}}_{\text{3}}\text{(101)}\\ \text{=}{\text{1×2}}^{\text{2}}\text{+}{\text{0×2}}^{\text{1}}\text{+}{\text{1×2}}^{\text{0}}\\ \text{=}\text{1×4}\text{+}\end{array}$

(b)

Explanation of Solution

Compute the numeric value of the string for “y = 0110”:

From the part A, the formula is $\text{n}\text{=}\frac{{\text{B2U}}_{\text{k}}\left(\text{y}\right)}{\left({\text{2}}^{\text{k}}\text{-}\text{1}\right)}$

Here, value of “y” is “0110” and value of “k” is number of bits in “y”. So, k = 4.

First compute the value of ${\text{B2U}}_{\text{4}}\text{(y)}$ is given below

$\begin{array}{c}{\text{B2U}}_{\text{4}}\text{(y)}{\text{= B2U}}_{\text{3}}\text{(0110)}\\ \text{=}{\text{0×2}}^{\text{3}}\text{+}{\text{1×2}}^{\text{2}}\text{+}{\text{1×2}}^{\text{1}}\text{+}{\text{0×2}}^{\text{0}}\\ \text{=}\text{0}\text{+}\text{1×4}\end{array}$

(c)

Explanation of Solution

Compute the numeric value of the string for “y = 010011”:

From the part A, the formula is $\text{n}\text{=}\frac{{\text{B2U}}_{\text{k}}\left(\text{y}\right)}{\left({\text{2}}^{\text{k}}\text{-}\text{1}\right)}$

Here, value of “y” is “010011” and value of “k” is number of bits in “y”. So, k = 6.

First compute the value of ${\text{B2U}}_{\text{6}}\text{(y)}$ is given below

$\begin{array}{c}{\text{B2U}}_{\text{6}}\text{(y)}{\text{= B2U}}_{\text{3}}\text{(010011)}\\ \text{=}{\text{0×2}}^{\text{5}}\text{+}{\text{1×2}}^{\text{4}}\text{+}{\text{0×2}}^{\text{3}}\text{+}{\text{0×2}}^{\text{2}}\text{+}{\text{1×2}}^{\text{1}}\text{+}{\text{1×2}}^{\text{0}}\\ \text{=}\text{0}\text{+}\text{1×16}\end{array}$