6.9k views

Given the following binary number in $32$-bit (single precision) $IEEE-754$ format :

$\large 00111110011011010000000000000000$

The decimal value closest to this floating-point number is :

1. $1.45*10^1$
2. $1.45*10^{-1}$
3. $2.27*10^{-1}$
4. $2.27*10^1$

edited | 6.9k views
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Nothing is visible.
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binary no. given is more than 32 bit....seems there is some toping error...plz correct it.
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@Shaik Masthan can you explain how mantisa explore here ???
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$\text{Exponent - 01111100 = 124}$

$\text{True exponent = 124 - 127 = -3}$

$\text{So in normalized form it will be}$

$\text{= 1.1101101 *$2^{-3}$}$

$\text{= 0.237}$

$\text{= 2.37 *$10^{-1} \approx 2.27 * 2^{-1}$}$

In $32$ bit (single precision) IEEE-754 format, binary number is represented as

$S$($1$ bit) $E$($8$ bit) $M$($23$ bits) with implicit normalization and exponent is represented with Excess-127 code.

Here, Sign $bit =$ $0$ $\Rightarrow$ Number is positive.

Exponent bits $= 01111100 = 124$ $\Rightarrow$ $E = 124 - 127 = -3$ (Excess-127)

Mantissa bits $= 11011010000000000000000$ $\Rightarrow$ Number = $1.1101101$  (Implicit normalization).

$\therefore$ Number $= 1.1101101 * 2$$^{-3} = 0.0011101101 = 0.227 = 2.27 x 10$$^{-1}$

$\therefore$ Answer should be C.

by Active (4.6k points)
edited
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0.0011101101101 = 0.227 how? I am getting  .116 :(
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Do it again or you can always use this converter.

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ya my mistake .. the same I did at exam and now at home too . :(
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How come Mantissa bits = 11011011010000000000000000 ??

Bold part is not in question. It is 110 110 1

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Edited. :)
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i am getting 0.231 when converted to decimal ryt?

even in the converter link mentioned there also

Have u chosen the nearest answer?

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Yes, always choose the nearest answer.
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11101101 =237.,how 227 ?

so total 32 bit representation in which 1 bit is for sign, 8 bit for biased exponent, 23 bit for mantissa.

biased exponent part:01111100 =124 in decimal

mantissa part:11011010000000000000000=$\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{128}$

bias=$2^{n-1}-1$   where n=8 so biase=127

value=1.M*$2^{124-127}$

put all values you get closest ans C
by Loyal (6.8k points)

Important: The mistake to avoid here is to avoid the premature conversion of binary numbers to decimal digits.

Use: $-1^{S}*1.M*2^{e-127}$

You'll get $1.1101101 * 2^{-3}$

Now, don't convert 1.1101101 to decimal just yet. Get the whole value, then do the base conversion.

=> $0.0011101101$

=> $\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{128} + \frac{1}{256} + \frac{1}{1024}$

I'll now ignore the last two terms, since they're really small.

=>$0.125+0.0625+0.03125+0.0078125$

=> $0.2265625$

=> $2.26 * 10^{-1}$

Option C

by Active (2.6k points)
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Yup.
After omitting the terms $\frac{1}{256}+\frac{1}{1024}$, the answer gets closer to the option stated in the question.