written 7.8 years ago by | • modified 2.9 years ago |
Mumbai University > Information Technology> sem 4> computer organization and architecture
Marks: 10M
Year: May16
written 7.8 years ago by | • modified 2.9 years ago |
Mumbai University > Information Technology> sem 4> computer organization and architecture
Marks: 10M
Year: May16
written 7.8 years ago by |
Convert this format is in base2 format For this, first convert 127 in to binary format
28=11100
Convert 0.75 in to binary format
0.75*2=1.50 1
Binary format of 28.75=11100.1
Shifting this binary number
1.110012 *4 Normalized
1.11001 is mantissa
2 **4 is exponent
add exponent 127 +4=131
Move the decimal point left or right so that only a single binary
digit "1" is to the left of the binary decimal point. Compensate by
adjusting the exponent in the opposite direction.
1.11001 times 2*4
Moving the decimal left seven decreases the size of the number; so,
we use an exponent of 6 to compensate and keep the number the same size.
Add 127 to the exponent and convert it to 8-bit binary:
4+ 127 = 131-> 10000011
Since every binary floating-point number (except zero!) is normalized
with "1." at the start, there is no need to store that leftmost "1".
Remove the leading "1." from the mantissa/significand:
1.11001 --> 11001
28.75 is positive - the sign bit is zero: 0
The next eight bits are the exponent: 10000011
The next 23 bits are the mantissa: 11001000000000000000000
Binary result (32 bits): 01000001111001000000000000000000
Convert this format is in base2 format For this, first convert 127 in to binary format
28=11100
Convert 0.75 in to binary format
0.75*2=1.50 1
Binary format of 28.75=11100.1
Shifting this binary number
1.110012 *4 Normalized
1.11001 is mantissa
2 **4 is exponent
add exponent 127 +4=131
Move the decimal point left or right so that only a single binary
digit "1" is to the left of the binary decimal point. Compensate by
adjusting the exponent in the opposite direction.
1.11001 times 2*4
Moving the decimal left seven decreases the size of the number; so,
we use an exponent of 6 to compensate and keep the number the same size.
Add 1023 to the exponent and convert it to 8-bit binary: 4+1023= 1027--> 10000000011
Since every binary floating-point number (except zero!) is normalized
with "1." at the start, there is no need to store that leftmost "1".
Remove the leading "1." from the mantissa/significand:
1.10000000011--> 10000000011
28.75 is positive - the sign bit is zero: 0
The next 11 bits are the exponent: 10000000011
The next 23 bits are the mantissa: 1100100000000000000000000000000000000000000000000000
Binary result (64 bits): 0100000000111100100000000000000000000000000000000000000000000000