833 views
0 votes
0 votes

The following is a scheme for floating point number representation using 16 bits.

Bit Position 15 14 .... 9 8 ...... 0
  s e m
  Sign Exponent Mantissa

 Then the floating point number represented is:

 

                                                        (−1)s(1+m×2−9)2e−31 ,     if the exponent ≠111111

                                                           

                                                             0, otherwise

 

What is the minimum difference between two successive real numbers representable in this system?

 

Any help to solve this question is appreciated.

1 Answer

0 votes
0 votes
let n1=$(-1)^{s}\ast(1+m1\ast 2^{-9})\ast 2^{e-31}$

let n2=$(-1)^{s}\ast(1+m2\ast 2^{-9})\ast 2^{e-31}$

difference=$(-1)^{s}\ast((m2-m1)\ast 2^{-9})\ast 2^{e-31}$     ...........(1)

now for consecutive m2-m1=1
so difference=$2^{e-40}$. to get minimum e=100000,so difference=$2^{-72}$
edited by

Related questions

2 votes
2 votes
1 answer
1
Hitoshi asked Dec 19, 2017
4,826 views
If the decimal number is 3.248 x 104 ,then its equivalent floating number in IEEE 754 standard is ?
1 votes
1 votes
1 answer
3
naveen bhatt asked Dec 10, 2018
1,278 views
Consider a modified 8-bit floating point representation in which 1-bit for sign, 3-bit for exponent and 4-bit for significant. What will be representation for decimal val...