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GO Classes DPP | DIGITAL LOGIC |FLOATING POINT REPRESENTATION

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4 4 votes

In a $16$-bit floating-point format, $1$ bit is for the sign, $6$ bits are for the exponent $($excess-$31$ bias$)$, and $9$ bits are for the fractional mantissa (normalized). What is the decimal value of the bit pattern $0~100001~101000000$?

  1. $1.625 \times 2^2$
     
  2. $1.625 \times 2^{-2}$
     
  3. $6.5$
     
  4. $3.25$

1 Answer

2 2 votes

The bit pattern is $0~100001~101000000 .$

Sign $(S): 0 \Longrightarrow$ Positive.

Exponent $(E): 100001$ in binary is $32+1=33$.

  • The bias is $31$.
     
  • True Exponent $e=E$ - Bias $=33-31=2$.
     

Mantissa $(M): 101 \ldots$ represents $1.101$ in binary

  • Value $=1+\left(1 \times 2^{-1}\right)+\left(0 \times 2^{-2}\right)+\left(1 \times 2^{-3}\right)=1+0.5+0.125=1.625$.

Final Value: $+1.625 \times 2^2=1.625 \times 4=\mathbf{6 . 5}$.

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