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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 $2^{i}$

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

(C) 0 to $2^{i}$

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

edited | 1.8k views

so D is the ans.

answered by Boss (5.8k points) 4 23 53
selected by
good :)

but how do u confirm it is a unsigned real no?
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..^_^.

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 (50.4k points) 22 90 410
edited by
cannot understand how U(-1,16) is 15?

how -1 bit possible?
@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 $2^{17}$ .

@Kapil

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

So, can number of bits i=-1?

@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 .

hope this helps!!!!

answered by Veteran (17.3k points) 7 12 40

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 (20.4k points) 12 77 174
edited by

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 (20.4k points) 12 77 174
edited

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

question does mention the range of decimal