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21 votes
21 votes
A $32$-bit floating-point number is represented by a $7$-bit signed exponent, and a $24$-bit fractional mantissa. The base of the scale factor is $16,$
The range of the exponent is ___________
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4 Answers

Best answer
20 votes
20 votes
  • PS: It is an old question and IEEE format was not there then. Currently we use IEEE format if anything is unspecified.


As given exponent bits are $7$ bits.

So, minimum it could be all $7$ bit are $0's$ and maximum it could all $1's$.

Assuming $2's$ complement representation minimum value $=-2^{6} = -64$ and maximum value $=2^6-1=63.$

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14 votes
14 votes

The given floating point number format is 

Sign bit(1) Exponent(7 bit) Mantissa(24 bit)

A. Since the Number is signed so the range of exponent is

-2n-1 to 2n-1-1 therefore exponent range is -64 to +63.

B. Now scale factor is represents in excess-64 format

Therefore range 

-2n-1  +64 to 2n-1-1 +64

Range will be 0 to127

3 votes
3 votes
a) range of exponent

as in a) part nothing is given about bias  so bias = 27−1−1=63

so exponent range will be 0-63 to 127-63 = - 63 to 64 .

---------------------------

b)
as here given excess 64 so bias number is 64 ,

then range will be -64+64=0 to (63+64)=127
So range is 0 to 127
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1 votes
1 votes
if we consider that the exponent is expressed in 2's complement(in most of the cases it is so because sometimes we get exponent as negative number) form then the range would be -64 to 63  and regarding the bais, it would be 64 and if we take excess 64 then the range would be 0 to 127

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