The n-bit fixed-point representation of an unsigned real number \(X\)uses…
2017
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
- A.
\(2^{-f} to \ 2^i\) - B.
\(2^{-f} to \ \left ( 2^{i} - 2^{-f} \right )\) - C.
0 to
\(2^i\) - D.
\(0 \ to \ \left ( 2^{i} - 2^{-f} \right )\)
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Correct answer: D
Answer: 0 to (2^i - 2^{-f}).
Interpretation: An n-bit unsigned fixed-point number uses f bits for the fractional part and i = n - f bits for the integer part.
Minimum value: When all bits are zero the value is 0.
Maximum value: When all bits are one, the integer part equals 2^i - 1 and the fractional part equals 1 - 2^{-f}. Summing gives (2^i - 1) + (1 - 2^{-f}) = 2^i - 2^{-f}.
Therefore the representable range is 0 to (2^i - 2^{-f}).
Example: For n = 8 and f = 4, i = 4 and the maximum is 2^4 - 2^{-4} = 16 - 1/16 = 15.9375, so the range is 0 to 15.9375.
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