+26 votes
3.4k views

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 $\left ( 2^{i} - 2^{-f} \right )$
asked
edited | 3.4k views
+2
I think answer is D
0

to know more about the fixed representation refer https://en.wikipedia.org/wiki/Q_(number_format)

0
Min val = 0.

Max val = $2^{i}-1$+$\frac{2^{f}-1}{2^{f}}$ = $2^{i}$  $-$ $2^{-f}$.

## 6 Answers

+36 votes
Best answer

so D is the ans.

answered by Boss (7.2k points)
selected
+2
good :)

but how do u confirm it is a unsigned real no?
+3
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..^_^.
+24 votes

hope this helps!!!!

answered by Veteran (27.6k points)
+13 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 $-2^{i}\leq R\leq 2^{i}-2^{-f}$ .

answered by Veteran (51.6k points)
edited by
0
cannot understand how U(-1,16) is 15?

how -1 bit possible?
0
@Srestha

The number will have $16$ bits and range being $0$ to $2^{-1} - 2^{-17}$ and then take any $16$ bit number from the given range. Finally divide by $2^{17}$ .
0

@Kapil

According to ur assumption , U(i,f) where i is number of bits in integer part.

So, can number of bits i=-1?

0
@Srestha

Nothing is there on the LHS . As range is from $0$ to $0.49991234$ something .

Also, this negative notation is only used when we deal with numbers from 0 to 0.something .
+9 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

answered by Veteran (21.4k points)
edited by
+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 2n-1... but here it is 2n which can not be

so we can represent upto2i-2-f

so D is correct answer here

answered by Veteran (21.4k points)
edited
0

DOUBTS :
Question doesnt mention the number being decimal its fixed point representation of unsigned real number !!
real number can be negative but as its unsigned we need to rule out negative

0
question does mention the range of decimal
+3 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 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

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

answered by (157 points)
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