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+37 votes

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

- $2^{-f}$ to $2^{i}$
- $2^{-f}$ to $\left ( 2^{i} - 2^{-f} \right )$
- 0 to $2^{i}$
- 0 to $\left ( 2^{i} - 2^{-f} \right )$

+2

Important keyword in the question is the number representation is **fixed point, **unlike** **the mindset a student may have of floating point.

i ---> represent integer part of a number

f --> represent fractional part

Say X = 21.75

i is 21 and f is 75

since unsigned number range will start from 0 (zero)

to (2^i -1)+ (1- (2^-f))

+65 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.

+5

good job!!!..the question after reading .. i thought i know nothing... then after reading all other explanation made me felt .. OHH MY goshh... was this even in our syllabus too?...

thenn........,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, after going through ur explaination .. i was like ,.. wow. how simple was that..

thanks a lot..^_^.

thenn........,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, after going through ur explaination .. i was like ,.. wow. how simple was that..

thanks a lot..^_^.

+17 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 $-2^{i}\leq R\leq 2^{i}-2^{-f}$ .

+10 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^{i } but with i bit we can represent number upto 2^{i}-1

so it will be 2^{i}-2^{-f}

so D is correct answer here

+7 votes

Here we know that for any decimal number fixed point lowest bound is 0 alwyas .now we are left with option c and D

now for option c it is wrong. why?

question says:

the range of DECIMAL number

for n bit number we can represent decimal number upto 2^{n}-1... but here it is 2^{n} which can not be

so we can represent upto2^{i}-2^{-f}

so D is correct answer here

+5 votes

**Answer:** **(D)**

**Explanation:** 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 2^{i}

Thus integral value will be from 0 to 2^{i} – 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)

http://www.geeksforgeeks.org/gate-gate-cs-2017-set-1-question-32/

0 votes

i bits of MSB. So the min number that can be represented in this representation i=0000....,

f=00000... so min=0. and the max. number that can be represented i=111111...., f=111111....

0 votes

Unsigned real number X. Lowest number representable should be 0, nothing stops it from being 0, and restrict it to 0.00043 or something. Hence, * A and B eliminated*.

Max number representable $2^{i}$ just looks wrong. What about the bits in f, they'd add up, too. So, * C eliminated*.

Option D seems reasonable.

Solved it under 30 seconds.

A proper solution would be to take small values of n, i and f; first put all 0's (for lower limit) then put all 1's (for upper limit). Then do option elimination.

**Option D**

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