42510011 0010 1010 1101 0001 0100 1011

S. Rawat

I.I.T. Kanpur

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Floating-point representationIEEE numbers are stored using a kind of scientific notation.

± mantissa * 2exponent

We can represent floating-point numbers with three binary fields:

a sign bit san exponent field eand a fraction field f.

The IEEE 754 standard defines several different precisions.

Single precision numbers include an 8-bit exponent field and a 23-bit fraction, for a total of 32 bits.Double precision numbers have an 11-bit exponent field and a 52-bit fraction, for a total of 64 bits.

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Single Precision and Double PrecisionSingle Precision (Single means One 32 Bit Reg to register)

Double Precision (Double means Two 32 Bit Reg to register)

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Sign

The sign bit is 0 for positive numbers and 1 for negative numbers.

But unlike integers, IEEE values are stored in signed magnitude format.

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MantissaThe field f contains a binary fraction.The actual mantissa of the floating-point value is (1 + f).– In other words, there is an implicit 1 to the left of the

binary point.– For example, if f is 01101…, the mantissa would be

1.01101…There are many ways to write a number in scientific notation, but there is always a unique normalized representation, with exactly one non-zero digit to the left of the point.

0.232 * 103 = 23.2 * 101 = 2.32 * 102 = …A side effect is that we get a little more precision: there are 24 bits in the mantissa, but we only need to store 23 of them.

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Exponent

The exponent field represents the exponent as a biased number.

It contains the actual exponent plus 127 for single precision, or the actual exponent plus 1023 in double precision.This converts all single precision exponents from -127 to 127 into unsigned numbers from 0 to 254, and all double precision exponents from -1023 to 1023 into unsigned numbers from 0 to 2046

Two Examples shown below for single precisionIf the exponent is 4, the e-field will be 4+127 = 131(100000112).If the e-field contains 8’b01011101(9310) the actual exponent is 93-127 = 34

Storing a biased exponent means we can compare IEEE values as if they were signed integers.

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Example IEEE-decimal conversionLet’s find the decimal value of the following IEEE number.

1 01111100 1100000000000000000000

First convert each individual field to decimal.The sign bit s is 1.The e field contains 01111100 = 12410.The mantissa is 0.11000… = 0.7510.

Then just plug these decimal values of s, e and f into our formula.

(1 - 2s) * (1 + f) * 2e-bias

This gives us (1 - 2) * (1 + 0.75) * 2124-127 = (-1.75 * 2-3) = -0.21875.

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Converting a decimal number to IEEE 754

What is the single-precision representation of 347.625?First convert the number to binary:

That is 347.625 = 101011011.1012

Normalize the number by shifting the binary point until there is a single 1 to the left:

101011011.101 x 20 = 1.01011011101 x 28

The bits to the right of the binary point comprise the fractional field f.

The number of times you shifted gives the exponent. The field-e should contain: exponent + 127.

Sign bit: 0 if positive, 1 if negative.

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EXAMPLES

10 0 0 11 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0

10 0 0 11 1 10 1 1 0 1 1 0 1 0 0 1 1 1 1 1 1 1 0 0 0 00 0 0

00 0 1 11 0 00 0 1 0 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 0 11 0 0

01 1 1 01 0 00 1 1 0 1 0 1 1 0 0 1 0 0 1 1 0 1 1 0 0 01 1 0

10 1 1 11 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1

10 1 1 11 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1

23

23.707

Kb=1.38x10-23

-0.03452

+Inf

Nan

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A Review and Features of the ANSI/IEEE Standard Floating Point Representation

42510011 0010 1010 1101 0001 0100 1011

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FP MULTIPLIERGiven two FP numbers n1 and n2, the product of both denoted as n, can be expressed as

2121

2211

2.1

2..12..1

21

21

21

EESS

ESES

pp

pp

nnn

Dadda Tree (24 Bits x 24 Bits)XOR sign Bits

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FP-Mult-Algorithm

1271:&

_;]0:23:1[

_;]0:23:1[

2.1

2..12..1

121

22

11

21

21

21

2121

2211

SSSforrangeallowed

normalifpaspconvert

normalifpaspconvert

pp

pp

nnn

EESS

ESES

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Exponent in FP-Multiplication

12721 EEE

Adding a carry-in to the sum and subtracting 128, which can be done by complementing the most significant bit.

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FP-Multiplier

FP Multiplier Block Diagram

UNPACK

XORAdd

ExponentsDadda_tree

Normalize

Round

Normalize

PACK

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Dadda Tree

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Dadda Tree(contd.)

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Dadda Tree(contd.)

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Dadda Tree(contd.)

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Dadda Tree(contd.)

Do the Reductions 96432

After 2 rows reduction ADD using 47-Bit CLA, Note: you can instantiate CLAs written in previous labs

Result will be of 48 Bits

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FP-Multiplier (contd.)

A 48-bit dadda multiplier is used for the significand multiplication, The 28 most significant bits (MSB) are selected as the tentative result,

where 24 bits are the mantissa result, 1 bit is used for range overflow and 3 extra bits are used for proper rounding.

Next, the result is normalized to the [1, 2) range, Rounded for proper approximation to the closest value and normalized again. The resulting sign, exponent and significand are then packed in single precision format. The normalization consists in a possible single bit right shift and corresponding exponent increment, based on the value of the b1 bit. The rounding scheme used is round to nearest, which can be easily achieved verifying the values of the remaining least significant bits (LSB).

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FP-Multiplier (contd.)

Cases possible:

While feeding in Dadda tree two 24 bits operands

If (both are denormalized)

MULTIPLICATION[47][46]=’00’

Else if (one denormalized and another normal)

MULTIPLICATION[47][46]= ‘00’ or ‘01’

Else if (both normalized)

MULTIPLICATION[47][46]=‘01’

MULTIPLICATION[47][46]=’10’(normalization required)

MULTIPLICATION[47][46]=’11’(normalization required)

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An Example

Lets Multiply 1.5x210 & 1.25x210

Stored Bias will be 10 + 127 = 137,

So 137 + 137 = 274

Stored Exponent of the result must correspond to 20 which is 147.

As in hardware what we are doing is subtracting 127 from the added exponents which were stored in FP registers i.e. 274-127 = 147

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An Example (contd.)

10 0 0 10 0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0

10 0 0 10 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0

n1=1.5x210

n2=1.25x210

Ans=1.875x220

0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1

1

=(223 + 222)10

=(223 + 221)10

Dadda_output=(223 + 222)*(223 + 221)=780000000000H

0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 10 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

7 8

47 46

0 020

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Thanks