Floating point number

Question

what is the hexadecimal representation of the floating point number (-48.625) in single precision, after rounding off and normalisation

Answer 1

- (48.625)   -->    - (110000.101)

Moving radix point to leading bit 1    --> - (1.10000101)*2⁵

 Since In question it is mentioned ' single precision '

Sign 1 bit,  Exponent 8 bit (in Excess 127), Mantissa 23 bit

1<= E <= 254   and M != 0 , So we can use implicit normalised form 

E = 127 + 5 = (132)₁₀ =(10000100)₂ 

S =(1)₂  (since it's a negative number so sign bit = 1)

M=(10000101000000000000000)₂

S(1)

E(8)

M(23)

1

1

0

0

0

0

1

0

0

1

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

In Hexadecimal form  (C2428000)₁₆ 

Please correct me If anywhere I get wrong.

Comments

perfect

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is there any role of rounding off ????

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why you have choose excess value as 127 why not -127.

I have doubt in, how to choose excess value. Is there any specific method to choose it.

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To understand the relationship between actual value of exponent of 2, stored value in E and bias lets consider the IEEE single precision normalized representation, which is  (-1)^S (1.M) 2^E-127  

The relationship between actual value of 2, stored value in E and bias is

Stored value = Actual value + Bias hence we get the notion 2^E-127

for example consider (1.10000101)*2⁵ 

Here actual value = 5, bias = 127, stored value(value in E) = ?

stored value = 5 + 127 = 132