Video 1: Intro to Floating point (Unsigned) Fixed-point - - PowerPoint PPT Presentation

Video 1: Intro to Floating point

(Unsigned) Fixed-point representation

The numbers are stored with a fixed number of bits for the integer part and a fixed number of bits for the fractional part. Suppose we have 8 bits to store a real number, where 5 bits store the integer part and 3 bits store the fractional part: 2! 2" 2# 2$ 2%

2!" 2!# 2!$

1 0 1 1 1.0 1 1 !

Smallest number: Largest number:

.¥¥¥÷:¥¥¥..

::i÷÷

(00000 . 001 ) z

(l l l l l

(Unsigned) Fixed-point representation

Suppose we have 64 bits to store a real number, where 32 bits store the integer part and 32 bits store the fractional part: Smallest number: Largest number: "$" … "#"""!. %"%#%$ … %$# # = '

&'! $"

"& 2& + '

&'" $#

%& 2(&

= "!"× 2!"+"!#× 2!#+ ⋯ + "#× 2#+'"× 2$"+'%× 2%+ ⋯ + '!%× 2$!%

,

420,

12-12-2

,

• 32

000,02--0-00%-01=1

2-32=10-9

(l l l

E 109

? ?

¥73

Fixed-point representation

How can we decide where to locate the binary point? More bits on the integer part? More bits on the fractional part? ÷÷:::÷÷÷÷%

.

5 bits → do

,

}

0.0625

#

( a.a.ao.b.b.5888.io

go.es

do

(Unsigned) Fixed-point representation

Range: difference between the largest and smallest numbers possible. More bits for the integer part ⟶ increase range Precision: smallest possible difference between any two numbers More bits for the fractional part ⟶ increase precision Wherever we put the binary point, there is a trade-off between the amount of range and precision. It can be hard to decide how much you need of each! Fix: Let the binary point “float”

!!!"!#. #"#!#$ ! !"!#. #"#!#$#% ! OR

Scientific Notation

In scientific notation, a number can be expressed in the form * = ± , × 10) where , is a coefficient in the range 1 ≤ , < 10 and 2 is the exponent.

1165.7 = 1.1657 × 10! 0.0004728 = 4.728 × 10$&

Note how the decimal point “floats”!

④

• O

Eg

• O

Floating-point numbers

A floating-point number can represent numbers of different order of magnitude (very large and very small) with the same number of fixed digits. In general, in the binary system, a floating number can be expressed as

! = ± $ × 2$

3 is the significand, normally a fractional value in the range [1.0,2.0) 2 is the exponent

=

=

• →

ME [ L

Floating-point numbers

Numerical Form: ! = ±$ × 2+ = ±',. '-'.'/ … '0× 2+

!! ∈ 0,1

Exponent range: * ∈ ,, . Precision: p = 0 + 1

Fractional part of significand (* digits)

(

leading

bit

fractional

0-0=0

000

Video 2: Normalized floating point representation

Converting floating points

Convert (39.6875)"! = 100111.1011 # into floating point representation

• 1. 001111011×25

0.1001111011×26

No Normal alized floating-point numbers

Normalized floating point numbers are expressed as

! = ± 1. '-'.'/ … '0× 2+ = ± 1. 3 × 2+

where " is the fractional part of the significand, # is the exponent and $! ∈ 0,1 .

✓

leading bit

y ✓

M€1403

I

5bits

✓Aoebibzb3b4 → p - 5

\④b,bzb3b4

"t

"↳bit¥7

• Exponent range:
• Precision:
• Smallest positive normalized FP number:
• Largest positive normalized FP number:

Normalized floating-point numbers

% = ± ( × 2&= ± 1. #"#!#$ … #'× 2& = ± 1. - × 2&

OD

• mEE4€

p

1-OO.is#x2=/2I →exponent

1.sissyish

xI

=ftp.z-l#JYfa0Tgeeht

• precision

Normalized floating point number scale

+∞ −∞

p

• htt

I .fx2m

ME [ 40 ]

flow

T

• verflow
• ver

p

"

f

under

.tl#ro?g0.Itia...i

,

ht

l l

• 2

'

2

.

It'll.it)

• gap

?

gap ?

Floating-point numbers: Simple example

A ”toy” number system can be represented as * = ±1. %"%#×2)

for + ∈ [−4,4] and '' ∈ {0,1}.

O

w

• n=2
• M=O

m

I

• .

M

• 2

m

• 3

m - 4

• 1. 00×20=1

1.00×2

i :

fi:% :3.

I

I

m =-3

M

• 4

M

• l

M

• 2

Floating-point numbers: Simple example

A ”toy” number system can be represented as * = ±1. %"%#×2)

for + ∈ [−4,4] and '' ∈ {0,1}.

1.00 ! ×2" = 1 1.01 ! ×2" = 1.25 1.10 ! ×2" = 1.5 1.11 ! ×2" = 1.75 1.00 ! ×2#$ = 0.5 1.01 ! ×2#$ = 0.625 1.10 ! ×2#$ = 0.75 1.11 ! ×2#$ = 0.875 1.00 ! ×2$ = 2 1.01 ! ×2$ = 2.5 1.10 ! ×2$ = 3.0 1.11 ! ×2$ = 3.5 1.00 ! ×2! = 4.0 1.01 ! ×2! = 5.0 1.10 ! ×2! = 6.0 1.11 ! ×2! = 7.0 1.00 ! ×2% = 8.0 1.01 ! ×2% = 10.0 1.10 ! ×2% = 12.0 1.11 ! ×2% = 14.0 1.00 ! ×2& = 16.0 1.01 ! ×2& = 20.0 1.10 ! ×2& = 24.0 1.11 ! ×2& = 28.0 1.00 ! ×2#! = 0.25 1.01 ! ×2#! = 0.3125 1.10 ! ×2#! = 0.375 1.11 ! ×2#! = 0.4375 1.00 ! ×2#% = 0.125 1.01 ! ×2#% = 0.15625 1.10 ! ×2#% = 0.1875 1.11 ! ×2#% = 0.21875 1.00 ! ×2#& = 0.0625 1.01 ! ×2#& = 0.078125 1.10 ! ×2#& = 0.09375 1.11 ! ×2#& = 0.109375

Same steps are performed to obtain the negative numbers. For simplicity, we will show only the positive numbers in this example.

"

}

}

4.0

①

}

0.125

°

foot's

* = ±1. %"%#×2) for + ∈ [−4,4] and '' ∈ {0,1}

• Smallest normalized positive number:
• Largest normalized positive number:

¥

2-4=0.0625

zut

• 2-

0=4 p

4 = ±1. '"'%×2( for + ∈ [−4,4] and '' ∈ {0,1}

Machine epsilon

• Machine epsilon (1%): is defined as the distance (gap) between 1 and the

next larger floating point number.

Em =

1.25

• l

0.25

in general

x # i. f

in,

IEm=2T↳

(1) co

I

• .

• ① I

• .

x I

T.F.co#I-2-nx2o=2-n

Range of integer numbers

4 = ±1. '"'%×2( for + ∈ [−4,4] and '' ∈ {0,1}

Suppose you have this following normalized floating point representation: What is the range of integer numbers that you can represent exactly?

• 48¥

?:O:'o=②.

• c. oooh

(1112=1.10×2

(1001 )z= -

(g) no

÷:÷÷÷÷÷÷¥÷⇒

.