can someone explain this?

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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 ___________

+5 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?

+6 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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