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+22 votes

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):

- 0D 24
- 0D 4D
- 4D 0D
- 4D 3D

+24 votes

Best answer

answer = **option D **in both questions.

0.239 = $(0.00111101)_2$

a)Store exponent = actual + biasing

13 + 64 = 77

(77)_{10} = (1001101)_{2}

ans is 1001101 00111101 =$4D 3D$

b) For normalized representation

0.00111101 *2^{13}

1.11101 *2^{10}

Store exponent = 10+64=74

(74)10 = (1001010)_{2}

ans 0 1001010 11101000=$4AE8$

Bias is given. It can be anything till it does the work of converting all negative exponent to positive.

When x bits are for exponent then bias can be 2^(x-1) . need not to be 2^(x-1) - 1

When it is mentioned that 64, it is 64....

I think those people who are confused, their confusion is due to relating it with IEEE standard...where for single precision bias is 127 .. & for double precision bias is 1023 ...

When it is mentioned that 64, it is 64....

I think those people who are confused, their confusion is due to relating it with IEEE standard...where for single precision bias is 127 .. & for double precision bias is 1023 ...

+1 vote

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

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

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