Mathematics For Machine Technology
Mathematics For Machine Technology
8th Edition
ISBN: 9781337798310
Author: Peterson, John.
Publisher: Cengage Learning,
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Chapter 86, Problem 46A
To determine

Conversion of Excess-3 code numbers 101 1001 1000.0110 1100 1001 to a decimal numbers.

Expert Solution & Answer
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Answer to Problem 46A

Decimal numbers is 265.39610.

Explanation of Solution

Given information:

An Excess-3 code numbers 101 1001 1000.0110 1100 1001.

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 code0011010001010110011110001001101010111100
    Binary code0000000100100011010001010110011110000101
    Decimal0123456789

Decimal of excess-3 code numbers 101 1001 1000.0110 1100 1001 or 0101 1001 1000.0110 1100 1001 is (Starting from right for integer part and starting from left for fractional part)

Excess3codenumber= 0101 1001 1000 . 0110 1100 1001 sixdecimaldigitsareexist

firstdecimaldigit=0× 2 3 +1× 2 2 +0× 2 1 +1× 2 0

firstdecimaldigit=0×8+1×4+0×2+1×1

firstdecimaldigit=0+4+0+1

firstdecimaldigit=5

forexcess3codedecimaldigit=53=2

seconddecimaldigit=1× 2 3 +0× 2 2 +0× 2 1 +1× 2 0

seconddecimaldigit=1×8+0×4+0×2+1×1

seconddecimaldigit=8+0+0+1

seconddecimaldigit=9

forexcess3codedecimaldigit=93=6

thirddecimaldigit=1× 2 3 +0× 2 2 +0× 2 1 +0× 2 0

thirddecimaldigit=1×8+0×4+0×2+0×1

thirddecimaldigit=8+0+0+0

thirddecimaldigit=8

forexcess3codedecimaldigit=83=5

fourthdecimaldigit=0× 2 3 +1× 2 2 +1× 2 1 +0× 2 0

fourthdecimaldigit=0×8+1×4+1×2+0×1

fourthdecimaldigit=0+4+2+0

fourthdecimaldigit=6

forexcess3codedecimaldigit=63=3

fifthdecimaldigit=1× 2 3 +1× 2 2 +0× 2 1 +0× 2 0

fifthdecimaldigit=1×8+1×4+0×2+0×1

fifthdecimaldigit=8+4+0+0

fifthdecimaldigit=12

forexcess3codedecimaldigit=123=9

sixthdecimaldigit=1× 2 3 +0× 2 2 +0× 2 1 +1× 2 0

sixthdecimaldigit=1×8+0×4+0×2+1×1

sixthdecimaldigit=8+0+0+1

sixthdecimaldigit=9

forexcess3codedecimaldigit=93=6

Sodecimalofgivenexcess3codenumber= 265.396 10

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