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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 to 2i
- b)2-f to (2i - 2-f)
- c)0 to 2i
- 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...
Diagram:
i represents an integral part of the and f represents the fractional part of the number.
Since, the n number is in unsigned representation, it's decimal value starts with 0. So Minimum value will be zero.
Range of unsigned representation is 0 to 2i - 1.
So, the mum value with i bits goes to 2i - 1.
Fraction of value is in the form of 2(-i). So, when we take the value of i = 1, 2, 3 … n this range of fractional value goes like, 2-1, 2-2, 2-3, …
So, it makes a GP series, with f bit maximum number possible is sum of GP series.
Consider a = ½, r = ½
Maximum value with f bits possible
i represents an integral part of the and f represents the fractional part of the number.
Since, the n number is in unsigned representation, it's decimal value starts with 0. So Minimum value will be zero.
Range of unsigned representation is 0 to 2i - 1.
So, the mum value with i bits goes to 2i - 1.
Fraction of value is in the form of 2(-i). So, when we take the value of i = 1, 2, 3 … n this range of fractional value goes like, 2-1, 2-2, 2-3, …
So, it makes a GP series, with f bit maximum number possible is sum of GP series.
Consider a = ½, r = ½
Maximum value with f bits possible
So, maximum fractional value possible
= maximum value with i bits + maximum value with f bits
= 2i - 1 + 1 - 2-f
= 2i - 2-f
So, require range will be 0 to 2i - 2-f.
= maximum value with i bits + maximum value with f bits
= 2i - 1 + 1 - 2-f
= 2i - 2-f
So, require range will be 0 to 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 is.a)2-fto 2ib)2-fto (2i- 2-f)c)0 to 2id)0 to (2i- 2-f)Correct answer is option 'D'. Can you explain this answer?
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