Question

Given the IEEE-754 Single Precision Floating Point number (stored Excess-127)111111000101011000000000000000002, determine the following: a.The sign of the mantissa b.The magnitude of the exponent (convert to base 10) c.The sign of the exponent

Answer 1

IEEE-754 single precision floating pointing number representation:

Sign (S)	Exponent (E)	Mantissa (M)
1 bit	8 bits	23 bits

Given Floating Point number = 2

Sign	Exponent	Mantissa

Implicit Normal form:

Value = (-1)^S (1.M) x 2^E-127

M = 10101100000000000000000₂

M = 1 x 2^-1 + 0 x 2^-2 + 1 x 2^-3 + 0 x 2^-4 + 1 x 2^-5+ 1 x 2^-6 = 0.5+0+0.125+0+0.03125+0.015625 = 0.671825

Exponent (E) = 11111000₂ = 248₁₀

Decimal Value = (-1)¹ (1.671825) x 2^{248 - 127} =  

(a)

Given Floating Point number = 2

Sign bit is 1. So number is negative and hence sign of mantissa is negative.

(b)

Exponent (E) = 11111000₂ = 1 x 2⁷ + 1 x 2⁶ + 1 x 2⁵+ 1 x 2⁴ + 1 x 2³+ 0 x 2² + 0 x 2¹ + 0 x 2⁰ = 248₁₀

(c)

Decimal Value = (-1)¹ (1.671825) x 2^{248 - 127} =  

The sign of adjusted exponent value is 121. It is positive.