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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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123 votes

hope this helps!!!!

107 votes
107 votes

Unsigned real number $x.$

Let $n=5, f=2, i = 5-2 = 3.$

1. Minimum value of $x:$

Value on decimal $ = 0.$

2. Maximum value of $x:$

Value on decimal $ = 2^3 - 2^{-2} = 8 - 0.25 = 7.75$ $($Or $2^2+2^1+2^0 + 2^{-1}+2^{-2} = 7.75)$

So (D) is the answer.

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24 votes

A number format representation, where the numbers are unsigned and where we have integer bits $i$ (on the LHS of the decimal point) and $f$ fractional bits (on the RHS of the decimal point) is known as $U(i,f)$ fixed-point format .

Hence, for unsigned number $U(i,f)$, the total bit length is $N = i + f$ .

Ex :-

  • $U(8,8)$ can be written as $00000011.10000010$ .
  • $U(6,2)$ can be written as $000100.10$ .

Now, using the unsigned number range of $N$ - bit word, the range $R$ of unsigned fixed point number can be written as $\frac{0\leq R\leq 2^{N}-1}{2^f}$ OR, $0\leq R\leq 2^{i}- 2^{-f}$ .



Similarly, A number format representation, where the numbers are signed and where we have integer bits $i$ (on the LHS of the decimal point) and $f$ fractional bits (on the RHS of the decimal point) is known as $U(i,f)$ fixed-point format .

Hence, here we have $1$ sign bit, $i$ integer bits and $f$ fractional bits, so total length of $N$- bit word = $1+i+f$ .

Ex :-

  • $A(7,8)$ can be written as $1 0000001.00000000$ = $-127$ , etc. 

 Now, using the signed number range ( 2's complement ) of $N$ - bit word, the range $R$ of signed fixed point number can be written as $\frac{-2^{N-1} \leq R \leq +(2^{N-1}-1)}{2^f}$ OR $-2^{i}\leq R\leq 2^{i}-2^{-f}$ .

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14 votes

answer asks the lower bound of decimal number.

now of any number lower bound is 0

we are left with 2 options

c says 2 but with i bit we can represent number upto 2i-1

so it will be 2i-2-f

so D is correct answer here

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