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IEEE Floating Point
Revision Guide for Phase
Test
Week 5
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Floating Point
15900000000000000 could be represented as
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15.9 * 1015
1.59 * 1016
A calculator might display 159 E14
159 * 1014Mantissa Exponent
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BinaryThe value of real binary numbers…
Scientific 22 21 20 . 2-1 2-2 2-3
Fractions . ½ ¼ ¾Decimal 4 2 1 . .5 .25 .125
101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625
= 5 ⅝
1 0 1 . 1 0 1
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Binary FractionsThe value of real binary numbers…
Scientific 22 21 20 . 2-1 2-2 2-3
Fractions . ½ ¼ ⅛Decimal 4 2 1 . .5 .25 .125
101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625
= 5 ⅝
1 0 1 . 1 0 1
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Binary FractionsThe value of real binary numbers…
Scientific 22 21 20 . 2-1 2-2 2-3
Fractions . ½ ¼ ⅛Decimal 4 2 1 . .5 .25 .125
101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625
= 5 ⅝
1 0 1 . 1 0 1
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IEEE Single Precision
The number will occupy 32 bits
The first bit represents the sign of the number; 1= negative 0= positive.
The next 8 bits will specify the exponent stored in biased 127 form.
The remaining 23 bits will carry the mantissa normalised to be between 1 and 2.
i.e. 1<= mantissa < 2
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Basic Conversion
Converting a decimal number to a floating point number.
1.Take the integer part of the number and generate the binary equivalent.
2.Take the fractional part and generate a binary fraction
3.Then place the two parts together and normalise.
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IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.1. The Mantissa. The Integer first. 6 / 2 = 3 r 0
3 / 2 = 1 r 1 1 / 2 = 0 r 1
2. Fraction next. .75 * 2 = 1.5
.5 * 2 = 1.0
3. put the two parts together… 110.11Now normalise 1.1011 * 22
= 1102
= 0.112
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= 0.112
IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.1. The Mantissa. The Integer first. 6 / 2 = 3 r 0
3 / 2 = 1 r 1 1 / 2 = 0 r 1
2. Fraction next. .75 * 2 = 1.5
.5 * 2 = 1.0
3. put the two parts together… 110.11Now normalise 1.1011 * 22
= 1102
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IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.1. The Mantissa. The Integer first. 6 / 2 = 3 r 0
3 / 2 = 1 r 1 1 / 2 = 0 r 1
2. Fraction next. .75 * 2 = 1.5
.5 * 2 = 1.0
3. put the two parts together… 110.11Now normalise 1.1011 * 22
= 1102
= 0.112
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IEEE Biased 127 Exponent
To generate a biased 127 exponent
Take the value of the signed exponent and add 127.
Example.
216 then 2127+16 = 2143 and my value for the exponent
would be 143 = 100011112
So it is simply now an unsigned value ....
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Possible Representations of an
Exponent
Binary Sign Magnitude 2's Complement
Biased 127 Exponent.
00000000 0 0 -127 {reserved}
00000001 1 1 -126 00000010 2 2 -125 01111110 126 126 -1 01111111 127 127 0 10000000 -0 -128 1 10000001 -1 -127 2 11111110 -126 -2 127 11111111 -127 -1 128
{reserved}
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Why Biased ?
The smallest exponent 00000000
Only one exponent zero 01111111
The highest exponent is 11111111
To increase the exponent by one simply add 1 to the present pattern.
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Back to the example
Our original example revisited…. 1.1011 * 22
Exponent is 2+127 =129 or 10000001 in binary.
NOTE: Mantissa always ends up with a value of ‘1’ before the Dot. This is a waste of storage therefore it is implied but not actually stored. 1.1000 is stored .1000
6.75 in 32 bit floating point IEEE representation:-
0 10000001 10110000000000000000000
sign(1) exponent(8) mantissa(23)
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Special cases
0 + Infinity and - infinity. Zero is a pattern that only contains ‘0’s
00000000000000000000000000000000 Positive Infinity is the pattern
011111111…. Negative Infinity is the pattern
111111111….
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Truncation and Rounding
Following arithmetic operations on a floating point number we may have increased the number of mantissa bits.
Since we will have a fixed storage (23 places) for the mantissa we require to limit these bits.
The simplest approach is to truncate the result prior to storage
Example0.1101101 stored in 4 bits
stored in 4 bits => 0.1101 ( loss 0.0000101 )
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RoundingIf lost digit is > ½ then add 1 to LSB
Example – in 4 bits
0.1101101 <- 0.1101 + 0.0001 = 0.1110 ( rounded UP)
0.1101011 <- 0.1101 ( rounded DOWN)
NOTE:
Rounding is always preferred to truncation partly because it is intrinsically more accurate , and because we end up with a FAIR error .
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Other Considerations
Truncation always undervalues the result, and can lead to a systematic error situation .
Rounding has one major disadvantage since it requires up to two further arithmetic operations .
Note. When we use floating point care has to be taken when comparing the size of numbers because we are generating binary fractions of a predefined length. There is always going to be the chance of recurring numbers etc like 1/3 in decimal 0.333333333333333333333 etc..
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From Floating Point Binary to Decimal
Example 1 01111011 11100000100000000000000
Sign = 1 therefore this number is a negative number.
Exponent 01111011 = 64+32+16+8+2+1 = 123
subtract the 127 = - 4Mantissa = 1.111000001
1.111000001 * 2- 4 -ve 0.0001111000001 1/16 + 1/32 +1/64+1/128+1/8192
or - 0.1173095703125
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Floating Point Maths
Floating point addition and subtraction.1. Make sure that the two numbers are of the same
magnitude. Their Exponents have to be equal.
2. We then add or subtract the mantissas
3. Starting with the existing exponent re-normalise if needed.
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Example
Example1.1* 23 + 1.1 * 22 Select the smaller number and make the mantissa smaller by
moving the point whilst increasing the exponent until the exponents match.
1.1 * 22 0.11 * 23
Add the mantissas
Re-normalise.
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Example
1.1* 23 001.1 23
+1.1 * 22 000.11 23
010.01 23
Re normalise 010.01 * 23
= 1.001 * 24
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FP math
Floating Point MultiplicationAssume two numbers a x 2m b x 2n
Result (a x 2m ) x (b x 2n) = ( a x b ) x ( 2m+n )
Floating Point DivisionAssume two number a x 2m and b x 2n
Result (a x 2m ) / (b x 2n) = (a/b ) x 2m-n