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 isa)2-fb)2-fto ( 2i- 2-f)c)0 to 2-id)0 to 2i- 2-f)Correct answer is option 'D'. Can you explain this answer?

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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

a)
2^-f
b)
2^-f to ( 2ⁱ - 2 ^-f)
c)
0 to 2^-i
d)
0 to 2ⁱ - 2 ^-f)

Correct answer is option 'D'. Can you explain this answer?

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The n-bit fixed-point representation of an unsigned real number X uses...

Since given number is in unsigned bit representation, its decimal value starts with 0.
We have i bit in integral part so maximum value will be 2ⁱ
Thus integral value will be from 0 to 2ⁱ - 1
We know fraction part of binary representation are calculated as (1/0)*2^-f
Thus with f bit maximum number possible = sum of GP series with a = 1/2 and r = 1/2

Thus fmax = {1/2(1 – (1/2)^f}/(1 – 1/2) = 1 – 2^-f

Thus maximum fractional value possible = 2ⁱ – 1 + (1 – 2^-f)

= 2ⁱ - 2^-f

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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 isa)2-fb)2-fto ( 2i- 2-f)c)0 to 2-id)0 to 2i- 2-f)Correct answer is option 'D'. Can you explain this answer?

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