Solved examples of binary fractions to decimal number conversion
Problem 1: Convert ( 101.101 )2= ( ? )10
= 1 0 1 . 1 0 1
↑ ↑
MSB LSB
Problem 1: Convert ( 101.101 )2= ( ? )10
= 1 0 1 . 1 0 1
↑ ↑
MSB LSB
= 1 x 22 + 0 x 21 + 1 x 20 . 1 x 2-1 + 0 x 2-2 + 1 x 2-3
= 1 x 4 + 0 x 2 + 1 x 1 . 1 x ( 1 / 2 ) + 0 x ( 1 / 4 ) + 1 x ( 1 / 8 )
= 4 + 0 + 1 . ( 1 / 2 ) + 0 + ( 1 / 8 )
= 5 . 0.5 + 0.125
= 5 . 625
Therefore ( 1 0 1 . 1 0 1 )2 = ( 5.625 )10
Problem 2: Convert ( 0.0001 )2= ( ? )10
= 0 . 0 0 0 1
↑ ↑
MSB LSB
= 0 x 20 . 0 x 2-1 + 0 x 2-2 + 0 x 2-3+ 1 x 2-4
= 0 x 1 . 0 x ( 1 / 2 ) + 0 x ( 1 / 4 ) + 0 x ( 1 / 8 ) + 1 x ( 1 / 16 )
= 0 . 0 + 0 + 0 + ( 1 / 16 )
= 0 . 0.0625
= 0 . 0625
Therefore ( 0 . 0 0 0 1 )2 = ( 0.0625 )10
Problem 3: Convert ( 101010.1111 )2= ( ? )10
= 1 0 1 0 1 0 . 1 1 1 1
↑ ↑
= 1 0 1 0 1 0 . 1 1 1 1
↑ ↑
MSB LSB
= 1 x 25 + 0 x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 0 x 20 . 1 x 2-1 + 1 x 2-2 + 1 x 2-3 + 1 x 2-4 = 1 x 32 + 0 x 16 + 1 x 8 + 0 x 4 + 1 x 2 + 0 x 1 . 1 x ( 1 / 2 ) + 1 x ( 1 / 4 ) + 1 x ( 1 / 8 ) + 1 x ( 1 / 16 )
= 32 + 0 + 8 + 0 + 2 + 0 . ( 1 / 2 ) + ( 1 / 4 ) + ( 1 / 8 ) + ( 1 / 16 )
= 32 + 8 + 2 . ( 0.5 ) + ( 0.25 ) + ( 0.125 ) + ( 0.0625 )
= 42 . 9375
Therefore ( 1 0 1 0 1 0 . 1 1 1 1 )2 = ( 42.9375 )10