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The n-bit fixed-point representation of an unsigned real number $X$ uses $f$ bits for the fraction part. Let $i = n-f$. The range of decimal values for $X$ in this representation is

  1. $2^{-f}$ to $2^{i}$
  2. $2^{-f}$ to $\left ( 2^{i} - 2^{-f} \right )$
  3. 0 to $2^{i}$
  4. 0 to $\left ( 2^{i} - 2^{-f} \right )$
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Should be D........

I approached it this way..


Range is always min to max.


X is an unsigned number therefore it should start from zero. Hence Minimum number in our case is Min = 0.


max will be integer part + fractional part..so  2^i - 1 bits + ( fraction part)


binary in fraction is represented as .1111(in case of maximum) which is converted as 1/2 + 1/4 + 1/8 + .. . .....1/f( f bits as mentioned)


it's sum of GP upto finite terms which is 1/2(1-(1/2)^f)/(1- 1/2) we get 1 - 1/2^f.


now max of integral part + max of fractional part.


2^i - 1 + 1 - 1/2^f
= 2^i + 2^f

Finally we get 0 to 2^i + 2^f



 

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