How do I work out IEEE 754 64-bit Floating Point Double Precision?
- by yousef gassar
enter code herehello i have done it in 32 but i could dont do it in 62bits please i need help
I am stuck on this question and need help. I don't know how to work it out. This is the question.
Below are two numbers represented in IEEE 754 64-bit Floating Point Double Precision, the bias of the signed exponent is -1023. Any particular real number ‘N’ represented in 64-bit form (i.e. with the following bit fields; 1-bit Sign, 11-bit Exponent, 52-bit Fraction) can be expressed in the form ±1.F2 × 2X by substituting the bit-field values using formula (IV.I):
N = (-1) S × 1.F2 × 2(E – 1023) for 0 < E < 2047.........................….(IV.I)
Where N= the number represented, S=Sign bit-value, E=Exponent=X +1023, F=Fraction or Mantissa are the values in the 1, 11 and 52-bit fields respectively in the IEEE 754 64-bit FP representation.
Using formula (IV.I), express the 64-bit FP representation of each number as:
(i) A binary number of the form:- ±1.F2 × 2X
(ii) A decimal number of the form:- ±0.F10 × 10Y {limit F10 to 10 decimal places}
Sign
0
1
Exponent
1000 0001 001
11
Fraction
1111 0111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
52
Sign
1
1
Exponent
1000 0000 000
11
Fraction
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000
52
I know I have to use the formula for each of the these but how do I work it out? Is it like this?
N = (-1) S × 1.F2 × 2(E – 1023) = 1 x 1.1111 0111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 x 1000 0001 00111 (-1023)?
