written 7.8 years ago by | • modified 2.9 years ago |
Mumbai University > Information Technology> sem 4> computer organization and architecture
Marks: 10M
Year: Dec16
written 7.8 years ago by | • modified 2.9 years ago |
Mumbai University > Information Technology> sem 4> computer organization and architecture
Marks: 10M
Year: Dec16
written 7.8 years ago by |
Convert this format is in base2 format For this, first convert 4 in to binary format
4=100
Convert 0.50 in to binary format
0.50*2=1.0 1
Binary format of 4.50=100.1
Shifting this binary number
1.0012 *2 Normalized
1.001 is mantissa
2 **2 is exponent
add exponent 127 +2=129
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.
1.001 times 2*2
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:
2+ 127 = 129--> 10000001
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:
1.001--> 001
4.50 is positive - the sign bit is zero: 0
The next eight bits are the exponent: 10000001
The next 23 bits are the mantissa: 00100000000000000000000
Binary result (32 bits): 01000000100100000000000000000000
Convert this format is in base2 format For this, first convert 4 in to binary format
4=100
Convert 0.50 in to binary format
0.50*2=1.00 1
Binary format of 4.50=100.1
Shifting this binary number
1.0012 *2 Normalized
1.001 is mantissa
2 **2 is exponent
add exponent 1023+2=1025
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.
1.001 times 2*2
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 11-bit binary:
2+1023= 1025--> 10000000001
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:
1.001--> 001
4.50 is positive - the sign bit is zero: 0
The next 11 bits are the exponent: 10000000001
The next 52 bits are the mantissa:
00100000000000000000000000000000000000000000000000
Binary result (64 bits): 010000000001 00100000000000000000000000000000000000000000000000