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Consider the following floating-point format.

Mantissa is a pure fraction in sign-magnitude form.

The decimal number $0.239 \times 2^{13}$ has the following hexadecimal representation (without normalization and rounding off):

  1. $\text{0D 24}$
  2. $\text{0D 4D}$
  3. $\text{4D 0D}$
  4. $\text{4D 3D}$
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3 Answers

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61 votes

Answer is option D in both questions.

$0.239 = (0.00111101)_2$

(a) Stored exponent = actual + biasing                 

$13  +  64  =  77$                            

$(77)_{10} = (1001101)_{2}$      

Answer is:  $\underbrace{0}_{\text{sign}}\;\underbrace{1001101}_{\text{exponent}}\;\underbrace{00111101}_{\text{mantissa}} =\text{0x 4D 3D}$


(b) For normalized representation

$0.00111101 *2^{13} = 1.11101 *2^{10}$

Stored exponent $= 10+64=74$

$(74)_{10}  =(1001010)_{2}$

Answer:  $\underbrace{0}_{\text{sign}}\;\underbrace{1001010}_{\text{exponent}}\;\underbrace{11101000}_{\text{mantissa}}=\text{0x 4A E8}$                .

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  • Option d is right.

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4 votes
1) (.239)base10 can be written in binary as 00111101 in 8 bit

 exp given 13 so 0001101 in seven bit

sign bit 0 positive no.

hence total 0 0001101 00111101

writing it in decimal we have 0D 4D

ans is B

2) in second

biasing by ( 2^n-1)-1

63+13=76

exp= 1001100

sign bit=0

mantisa since implicit normalization is used here

previously m= 00111101

for implicit  m= 11101000

hence we have 0 1001100 11101000

which is equivalent to 4A E8
Answer:

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