Chapter 84, Problem 45A

To determine

Conversion of Excess-3 code numbers 1010 0110.0011 0101 to a decimal numbers.

Answer to Problem 45A

Decimal numbers is 73.02₁₀.

Explanation of Solution

Given information:

An Excess-3 code numbers 1010 0110.0011 0101.

Calculation:

Excess-3 code is unweighted code.

In excess-3 code decimal numbers are obtained by converting 4 bit binary numbers into decimal numbers and then subtracting 3 in each decimal digit.

Excess-3 code number system uses 2 symbols: The numbers are 0 and 1.

And a decimal number system uses the number 10 as its base i.e. it has 10 symbols; decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Each decimal digit consists of 4 bits excess-3 code.

Excess-3 code numbers are represented as from decimal number

Excess-3 code	0011	0100	0101	0110	0111	1000	1001	1010	1011	1100
Binary code	0000	0001	0010	0011	0100	0101	0110	0111	1000	0101
Decimal	0	1	2	3	4	5	6	7	8	9

Decimal of excess-3 code numbers 1010 0110.0011 0101 is (Starting from right for integer part and starting from left for fractional part)

$Excess-3codenumber=\boxed{1010}\boxed{0110}.\boxed{0011}\boxed{0101}\to fourdecimaldigitsareexist$

$firstdecimaldigit=1\times 2^3+0\times 2^2+1\times 2^1+0\times 2^0$

$firstdecimaldigit=1\times 8+0\times 4+1\times 2+0\times 1$

$firstdecimaldigit=8+0+2+0$

$firstdecimaldigit=10$

$forexcess-3codedecimaldigit=10-3=7$

$\sec onddecimaldigit=0\times 2^3+1\times 2^2+1\times 2^1+0\times 2^0$

$\sec onddecimaldigit=0\times 8+1\times 4+1\times 2+0\times 1$

$\sec onddecimaldigit=0+4+2+0$

$\sec onddecimaldigit=6$

$forexcess-3codedecimaldigit=6-3=3$

$thirddecimaldigit=0\times 2^3+0\times 2^2+1\times 2^1+1\times 2^0$

$\begin{array}{l}thirddecimaldigit=0\times 8+0\times 4+1\times 2+1\times 1\\ thirddecimaldigit=0+0+2+1\\ thirddecimaldigit=3\end{array}$

$\begin{array}{l}forexcess-3codedecimaldigit=3-3=0\\ fourthdecimaldigit=0\times 2^3+1\times 2^2+0\times 2^1+1\times 2^0\\ fourthdecimaldigit=0\times 8+1\times 4+0\times 2+1\times 1\\ fourthdecimaldigit=0+4+0+1\end{array}$

$\begin{array}{l}fourthdecimaldigit=5\\ forexcess-3codedecimaldigit=5-3=2\\ Sodecimalofgivenexcess-3codenumber={73.02}_{10}\end{array}$