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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 ___________
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can someone explain this?

• 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 $=2^6-1=63.$

answered by Boss (16.6k points)
edited by
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@ 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?
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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.
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in part a) bias will be 64 not 63..
+1

reena_kandari why 64?

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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.
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yes we can.. $1000000$  we store exponent value as unsigned integer .

what's the problem in this?
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@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 $-64<=E<=63$.
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shouldn't it be -166 to 166-1 ??

The base of the scale factor is 16,

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@meghna- If you are saying for range of exponent you are wrong. Base of scale factor means

say your floating point number is

1.M * $B^{e}$ where B is your base and e is your exponent. Here they are asking what range your exponent can have given it cam only be of maximum 7 bits.

So for binary number you follow

1.M * $2^e$
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Here base is mentioned as 16

so range for unbiased exponent will come as -  $-16^{^{6}}$ to $16^{^{6}}-1$ ??

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Little out of context doubt but imp i think ..

say your floating point number is

1.M * B^e where B is your base and e is your exponent. Here they are asking what range your exponent can have given it cam only be of maximum 7 bits

For system other than binary ..for normalization we use same 1.M * B^e??

For other system like base 10 there are digits 0-9...so what should be normalized form thee 1.M only ??

eg 23.54 (base 10)

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@jatin kachane as base 16 given so i think u r right
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How can you store  FFFFFFF in 7 bits ???
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see , its (-1)^S  *  1.M   *  B^E , [b = base and e = exponent} ....

Now in the question given that it is signed exponent , so with 7 bits in signed magnitude u can represent -63 to +63

so the range will be from -63 to +63
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PS: It is an old question and IEEE format was not there then. Currently we use IEEE format if anything is unspecified.

If it is asked after IEEE standard then what will be the Answer?

Is Biasing will get added implicitly because Biasing is Used in IEEE Standard?

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"the base of scale factor is 16" any significance ?

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 (803 points)
+4

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".

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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 ?
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@Arjun sir plz explain??
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scale factor 16,plz explain?
+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 (3.9k points)
edited
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 (29 points)

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