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+12 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 ___________

The range of the exponent is ___________

+8 votes

Best answer

- 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 $=2^6-1=63.$

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?

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?

+1

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

@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 $-64<=E<=63$.

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

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

+1

@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 * $2^e$

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 * $2^e$

0

Here base is mentioned as 16

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

0

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)

0

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

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

+7 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

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

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

Therefore range

-2^{n-1} +64 to 2^{n-1}-1 +64

Range will be 0 to127

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