[SOLVED] IEEE 754 standard

1007ICT / 1807ICT / 7611ICT Computer Systems & Networks
Lecture 5: Decimal real numbers

Last Section: The Processor

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Topics covered:
Central Processing Unit (CPU)
Arithmetic Logic Unit (ALU)
Registers
Data, instructions and instruction sets
Micro programs and the processor datapath
The fetch / decode / execute cycle
Input / Output models
Interrupts
. Revised and updated by,, and Wee Lum 2

Lecture Content
Topics Covered:
Learning objectives
Decimal real numbers
Fixed point binary
Converting fractions
Floating point numbers
Exponential notation
IEEE Floating point formats IEEE Normalisation
IEEE Examples
. Revised and updated by,, and Wee Lum 3

Learning Objectives
In this lecture you will have:
Learnt how to represent decimal real numbers
Learnt how to represent fixed point binary
Learnt how to convert decimal fractions to binary and vice versa
Gainedanunderstandingofbinaryfloating point numbers
Learnt how to represent numbers using exponential notation
Learnt how to represent binary floating point numbers using an IEEE Floating point format
Learnt how to represent binary floating point numbers in normalised form
. Revised and updated by,, and Wee Lum 4

Decimal Real Numbers (Section 4.3)
Real numbers, in particular fractions, are important
When dealing with decimal real numbers, the whole number and fractional parts are separated by a decimal point, e.g. 123.45
The place of each digit gives the power of 10 that the value of the digit represents
Place/Power 2 1 0 -1 -2
10-1 = 10-2 = 10-3 = 10-4 = 10-5 =
1 x 100 = 100
+ 2 x 10 = 20
+ 3x 1 = 3
+ 4 x 0.1 = 0.4
+ 5 x 0.01 = 0.05
1/10 = 1/100 = 1/1000 = 1/10,000 = 1/100,000 =
0.1 0.01 0.001 0.0001 0.00001
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Two models for decimal real numbers
We will discuss two models for representing decimal real numbers
1. Fixed point binary 2. IEEE 754 standard
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Fixed Point Binary
We can represent real numbers in binary by separating the whole and fractional parts of the number by using a binary point . e.g., 01001.11001
For any given word size you need to decide how many bits will be on left side of the . this determines both what the biggest number and what the smallest fraction that can be represented will be.
For example with a 16 bit word you could use 8.8 format so that the biggest unsigned number is 255 and the smallest fraction is 1/256 = 0.00390625
The place of each digit gives the power of 2 that the value of the digit represents
Whole (8 bits)
Fractional (8 bits)
Place/Power
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Fixed Point Binary
Place/Power
Whole (8 bits) Fractional (8 bits)
For example lets look at the binary number
110110.1010
= 1*25 + 1*24 + 0*23 + 1*22 + 1*21 + 0*20 +
= 1*32 + 1*16 + 0*8 + 1*4 + 1*2 + 0*1 + = 54 5/8
Note: 2-1 = 12; 2-2 = 14; 2-3 = 1/8;
2-4 = 1/16 etc
1*2-1 + 0*2-2 + 1*2-3 + 0*2-4
1*12 + 0*14 + 1*1/8 + 0*1/16
2-4 = 2-5 = 2-6 =
1/16 = 1/32 = 1/64 =
0.0625 0.03125 0.015625
2-1 = 2-2 = 2-3 =
1/2 = 0.5 1/4 = 0.25 1/8 = 0.125
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Converting Fractions
You can convert binary to decimal fractions by:
1. Shift the binary point right to the end of the fraction (remember how
many places )
2. Convert the whole binary number to a whole decimal number. 3. Divide the decimal number by 2places
E. g. convert 0.1011012 to decimal:
1. Shift binary point 6 places right: 01011012
2. Convert: 32 + 8 + 4 + 1 = 45
3. Divide 45 by 2places = 45 26 = 45/ 64 = 0.70312510
You can convert from decimal to binary fractions by: 1. Multiply the decimal fraction by 2
2. Save the digit to the left of the point as a result
3. Drop the digit to the left of the point
4. Repeat 1,2,3 until the value 1.0000 is reached . Revised and updated by,, and Wee Lum 9

Example: DecimalBinary Fraction Convert 0.70703125 to binary:
0.70703125 2 0.4140625 2 0.828125 2 0.65625 2 0.3125 2 0.625 2 0.25 2 0.5 2
The resulting binary
results, in order, from left to right: 0.101101012
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= 1.4140625 then save and drop 1 = 0.828125 then save and drop 0 = 1.65625 then save and drop 1 = 1.3125 then save and drop 1 = 0.625 then save and drop 0 = 1.25 then save and drop 1 = 0.5 then save and drop 0 = 1.0 then save 1
fraction is read by appending the partial

Exponent-mantissa format
The main problem with fixed point numbers is that if we want to represent either very small fractions or very large numbers we need to use lots of bits.
One alternative is to use a variable number of bits to represent real numbers but this would make processing extremely complicated if not impossible.
The other alternative is an exponentmantissa format E.g.: represent -0.0000003579 in exponent format
(normal form) sign of exponent sign of mantissa
mantissa -3.579 x 10-7 exponent base
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Exponential Notation
Different scientific notation representations of the same number may be generated by shifting the decimal point left or right and adjusting the exponent value to suit.
To maintain the same value, every place to the right that you move the decimal point, you should decrease the exponent by one.
Similarly, for each place to the left that you move the decimal point, you should increase the exponent by one.
Normalised form
Normalised form has 1 digit (non-zero) before the decimal place (Eg. A.BCDEF x 10x)
433456 = 433456 x 100 = 4.33456 x 105 (normalised form) 0.001102 = 0.001102 x 20 = 1.102 x 2-3 (normalised form)
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IEEE Floating Point Formats
Unlike integer representations, a certain number of bits must be allocated to the exponent and its sign and the remainder to the mantissa and its sign.
There are a number of formats depending upon the particular type of computer/ operating system being used. A format commonly used on Intel and Macintosh PCs is the IEEE Standard 754.
The IEEE standard has a 32- bit, single precision, and 64- bit, double precision, format for binary floating point numbers.
A 32- bit single precision floating point binary number is divided into the following:
1bit -sign of mantissa
8 bit exponent
23 bit mantissa
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IEEE Floating Point Formats
A number of parts that were in the scientific notation, are missing here! This is because this information is somehow implied by the structure of the number.
The mantissa is stored in what is called normalized form. In this form the leading bit is always a 1 and the position of the binary point is always the same so they can be assumed and need not be stored.
The exponent is stored in what is called excess-127 notation (biased exponent with k = 127)
This format eliminates the need for a separate sign bit for the exponent.
The exponent +3 will be stored as 127 + 3 = 130
The exponent -11 will be store as 127 11 = 116
Exponentvaluesof0and255arereservedforspecial circumstances leaving an exponent range of: 2-126 to 2127
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IEEE Floating Point Normalisation
To normalize a binary number for IEEE 754 format:
1. Set the exponent to zero (0)
2. Shift the binary point left or right until the binary number starts with 1. mmmm
3. Adjust the exponent to maintain the original value
4. Discard the least significant bits of the mantissa or add trailing zeroes to keep the number at the specified size (23 digits)
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IEEE Normalisation Examples
1101110.001111001 1. Initialise exponent = 0
2. (Normalise) Shift binary point until = 1.101110001111001 3. Adjust exponent = +6
4. Add trailing 0s = 1.10111000111100100000000
0.0011000111
1. Initialise exponent = 0
2. (Normalise) Shift binary point until = 1.1000111 3. Adjust exponent = -3
4. Add trailing 0s = 1.10001110000000000000000
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IEEE Creation Example
Converting from decimal to IEEE floating point.
275,005.2510 = 1000011001000111101.012
1.(Normalise)Initialiseexponent=0
2. Shift binary point = 1.00001100100011110101
3.(Generateexponent)Adjustexponent=18 1. Excess- 127 = 127 + 18 = 145
2. In binary = 10010001
4.Mantissa(droptheleading1)=00001100100011110101
5. Pad to 23 bits (add 0s)= 00001100100011110101000
6. Sign = 0 (positive)
32-bitIEEEformatis:
01001000100001100100011110101000
sign exponent mantissa
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IEEE unpacking Example
Converting from IEEE floating point format to decimal.
Lets start with a 32- bit number in IEEE format: 1
10101110100111110101100
The first bit (1) tells us that the number is negative.
The next eight (8) bits 10110011 1. Evaluate to: 179
2. Excess- 127: 179 127 = 52
3. Exponent is: 52
The remaining 23 bits are the normalized mantissa:
10101110100111110101100
The leading 1. is implied so mantissa is: 1.10101110100111110101100
1. Initialise places to 0
2. Move binary point 23 places to the right: 110101110100111110101100.0 3. Numerator is 1101011101001111101011002 = 1411063610
4. Denominator is 223 = 8388608 (we shifted 23 times to the right)
5. Numerator/ Denominator = 1.68211889267
5. Decimal real number = -1.68211889267 252
Note: This number is still represented as a power of 2, normally we would convert it to a power of 10
Convert 23 bit mantissa to decimal:
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Importance of type declaration
As you can see, one 32- bit number looks much the same as any other; eg 11011001110101110100111110101100
The above number could be interpreted as an integer, string of characters or any number of other formats. The key, as with integers and characters, is that the computer must know what kind of data it is to decode it successfully.
If interpreted as an:
IEEE floating point number we get: -1.68211889267 2 52
unsigned integer we get: 3654766508
a 2s complement number we get: -640200788
. Revised and updated by,, and Wee Lum 19

Have studied:
Decimal real numbers
Fixed point binary
Converting fractions
Floating point numbers
Exponential notation
IEEE Floating point formats IEEE Normalisation
IEEE Examples
. Revised and updated by,, and Wee Lum 20

Memory organisation
. Revised and updated by,, and Wee Lum 21

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