# Recent questions tagged floating-point-representation

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A 32-bit floating-point number is represented by a 7-bit signed exponent, and a 24-bit fractional mantissa. The base of the scale factor is 16, The range of the exponent is ___________, if the scale factor is represented in excess-64 format.
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In IEEE floationg point representation, the hexadecimal number $0xC0000000$ corresponds to ? $-3.0$ $-1.0$ $-4.0$ $-2.0$
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ans given= -3.75*10-1
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Can someone please validate if the given range is correct, and we can keep all 0's in Mantissa unlike biased exponent.
1 vote
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Why floating point in de-normalized normal form has range between : $\pm1\times2^{-149}$ and $\pm(1 - 2 ^{-23})\times2^{-126}$
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The decimal value of 0.005 in single precision floating point format is __________________
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Question 1 (a) Convert the positive decimal number 17.45 to IEEE 754 single precision representation. Show all of your working out. [15 marks] (b) In IEEE 754 single precision, 1.25 is represented as: 0 01111111 01000000000000000000000 In IEEE 754 single precision 1.26 is ... IEEE 754 representation of 1.26 than 1.25. [10 marks] (c) Why is the exponent biased in IEEE representation? [5 marks]
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Consider the following IEEE-754 single precision format 1 10101010 01010100................0 value represented by above number is _________________________
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If the decimal number is 3.248 x 104 ,then its equivalent floating number in IEEE 754 standard is ?
1 vote
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Source for learning floating point representation
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16 bit Floating Point Representation $(-1)^{sign}*(1.M)*2^{Exp - 63}$ Sign = 1 bit Exponent = 7 bit Mantissa = 8 bit 1) Max positive number 2) Min positive number. 3) Max negative number. 4) Min negative number. 5) What is meant by precision.
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Consider a computer system that stores a floating-point numbers with 16-bit mantissa and an 8-bit exponent, each in two’s complement. find The smallest and largest positive values which can be stored in the system.
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Explain in detail how and what conversion(in binary bit pattern) takes place for following codes: 1) int i=37; float f=*(float *)&i; printf("f=%f",f); [Output:f=0.000000] 2)float f=7.0; short s=*(short *)&f; printf("s=%hd",s); [Output:s=0]
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IEEE 754 32 bit representation for reference:- 1. For the demoralized numbers,why the exponent is -126 and not -127? For 0 we say it is Mantissa=0 and B.E=0, so we get 0*2^0-127=0,but for any number that is having bias exponent as 0 and Mantissa non ... 127 and +128) comes under overflow and underflow condition.I mean if i am 0 or infinity or NAN then can i say that system is overflow/underflow?
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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 : $1.45*10^1$ $1.45*10^{-1}$ $2.27*10^{-1}$ $2.27*10^1$
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Question 1 Explain What is Denormalized Number Give Example Give Representation in IEEE 754 and excess 64 (if any) Question 2 How to Convert $(12.625)_10$ $(12.625)_10 \Leftrightarrow (1100.101)_2$ to IEEE 754 Single Precision (With Normalization) IEEE 754 Single Precision (Without Normalization) Excess-64 (With Normalization) Excess-64 (Without Normalization)
1 vote
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Mantissa is a pure fraction in signed magnitude form .What is the reprentation of decimal number 12.6255*102 12.6255*23 with Normalization and Without Normalization
1 vote
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Are the following statements true? 1. If Biased Exponent > Bias ,then Actual exponent is +ve. 2. If Biased Exponent < Bias ,then Actual exponent is -ve. Please tell reason also,as per me both should be false as Biased is already maximum positive number,adding something to it will cause overflow.So 1st should be -ve and second should be +ve,Please correct if I am wrong
1 vote
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what is the hexadecimal representation of the floating point number (-48.625) in single precision, after rounding off and normalisation
1 vote
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IN FLOATING POINT REPRESENTATION WE KNOW THAT BIAS ADDED IS THE MAXIMUM NUMBER WHICH CAN BE REPRESENTED IN 2S COMPLEMENT. Suppose exponent is represented using 8 bits, so bias added= (2^7)-1=127. But the max negative number representable with 8 bits is -128 and adding ... negative. So how do we get postitive exponent by biassing in this case??. It sounds confusing to me. Plz resolve my confusion.
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How to represnet 1.1 using IEEE 754 Single Precision and Double Precision Number Representation . Please explain in detail
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Its V=(-1)^S...... What will be answer?I am getting 1,is it correct?
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Consider an excess - 50 representation for floating point numbers with $4 BCD$ digit mantissa and $2 BCD$ digit exponent in normalised form. The minimum and maximum positive numbers that can be represented are __________ and _____________ respectively.
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A 32-bit floating-point number is represented by a 7-bit signed exponent, and a 24-bit fractional mantissa. The base of the scale factor is 16, The range of the exponent is ___________
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Consider the following floating-point format. Mantissa is a pure fraction in sign-magnitude form. The normalized representation for the above format is specified as follows. The mantissa has an implicit $1$ preceding the binary (radix) point. Assume that only $0's$ are padded in while shifting a field. The ... of the above number $(0.239 \times 2^{13})$ is: $0A\;20$ $11\;34$ $49\;D0$ $4A\;E8$
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The exponent of a floating-point number is represented in excess-N code so that: The dynamic range is large. The precision is high. The smallest number is represented by all zeros. Overflow is avoided.
The range of representable normalized numbers in the floating point binary fractional representation in a $32$-bit word with $1$-bit sign, $8$-bit excess $128$ biased exponent and $23$-bit mantissa is : $2^{-128}$ to $(1-2^{-23}) \times 2^{127}$ $(1-2^{-23}) \times 2^{-127}$ to $2^{128}$ $(1-2^{-23}) \times 2^{-127}$ to $2^{23}$ $2^{-129}$ to $(1-2^{-23}) \times 2^{127}$
What is the decimal value of the floating-point number $C1D00000$ (hexadecimal notation)? (Assume $32$-bit, single precision floating point $\text{IEEE}$ representation) $28$ $-15$ $-26$ $-28$