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in Digital Logic by Active (2.5k points)
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i am geting 32A0 ?? plz check

Please Explain ur approach in detail . Solution Given is (b)

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$\text{bias }=64$

$(-1)^{S} (.M)_2 \times 2^{E-\text{bias}}$

$(0.625)_2 = (0.101)_2$

$E-\text{bias} =18$

$E-64 =18$

$E=82 \Rightarrow (1010010)_2$

Floating Point Representation

$\mathbf{ {\color{Red} 0} {\color{Blue} {1010010} } {\color{Green} {10100000}} }$

in Hex : $52A0$
by Boss (21.5k points)
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Without Normalization also means Normalized appraoch ?
@pc..its without normalisation..then why using biasing? and whats the meaning of rounding off here?

@sudsho is the answer 12A0 ?

@pC everything else if fine..just tell me if we use biasing also in without normalisation(not IEEE)..simply without noramlisation like in this question...because exponent is not constant i can write 0.625 as 0.101 or 0.1010 or 0.101000000000000 so on in without normalisation (as we dont have a single representation fr a number here and thats why we use normalisation)
so every time ur exponent will change....or is it like since we are given a format(exponent as bias 64 and all other fields are fixed) in the question to represent we'll go fr it?

@sudsho, I too have some doubts . I posted what I knew .  check this

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