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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 ___________
asked in Digital Logic by Boss (39.8k points)
edited by | 1.1k views
0
can someone explain this?

4 Answers

+5 votes
Best answer

a) range of exponent

as given exponent bits are 7 bits ....

so min it could be all 7 bit are 0's and maximum it could all 1's ...

minimum number =0 , and maximum number = 127

as in a) part nothing is given about bias .. so bias = $2^{7-1}-1=63$  (nothing is given about single precision also so i used general method )

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

 

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

b) minimum number = 0 , maximum number = 127 ,

as here given excess 64 so bias number is 64 ,

then range will be 0-64=-64 to (127-64=63) 63..

here in question given that base is 16 , it will get simply shifting of mantissa bit ... yes range of value is increased by large base ...(-1)s ( 0.M) 16 E-bias .

correct me if i am wrong

answered by Boss (14.4k points)
selected by
0
@ sonam vyas
for (a) part
wouldn't the range of exponent be -63 to + 63 because in question they have said that 7 bit signed exponent.
And, we usually use bias to prevent using sign bit of exponent to represent the range.
Also, in (b) they have asked us to recalculate the range of exponent if bias was given
so why you have assumed bias in (a) part?
0
Since Signed data is stored in 2'complement form by default ..so range of exponent is -64 to + 63
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Part(b) solution is is confusing and logic is not clear.
0
in part a) bias will be 64 not 63..
+1

reena_kandari why 64?

 

0
Since Signed data is stored in 2'complement form by default ..so range of exponent is -64 to + 63. so we take bias positive of minimum value i.e 64 to make the minimum value to 0.
0
yes we can.. $1000000$  we store exponent value as unsigned integer .

what's the problem in this?
0
@peeyoosh why are you adding sign bit here.  exponent and mantissa  we represent as unsigned integer.

rang of biased exponent is $0<=E<=127$

so the range of actual exponent value is $-64<=E<=63$.
+5 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

 

 

answered by Junior (753 points)
+3

Since the Number is signed so the range of exponent is

-2n-1 to 2n-1-1

It is given that the base is "16" and not "2".

0
But we store the number in binary form so range might be the same like binary for 16 scale factor ? sir plz correct me
@Arjun sir  , then what will be the correct answer ?
0
@Arjun sir plz explain??
0
scale factor 16,plz explain?
0 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
answered by (11 points)
–1 vote
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
answered by Active (3.7k points)
edited by


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