Outline Digital CMOS Design Arithmetic Operators Adders - - PowerPoint PPT Presentation

Pirouz Bazargan Sabet February 2010 Digital Design

Outline

Arithmetic Operators Digital CMOS Design

Adders Comparators Shifters Multipliers Square rooting

Pirouz Bazargan Sabet February 2010 Digital Design

Square root

An real number

• using floating point representation

Find a real number

• such as

( )

y x = +

2

ε

• Calculation cannot be implemented in hardware
• Need iterative operation

Pirouz Bazargan Sabet February 2010 Digital Design

Square root

2

= − y x resolve

• Direct method
• Indirect method

digit-by-digit

Pirouz Bazargan Sabet February 2010 Digital Design

( )

E

M y 2 1 × × − =

( )

2

2 1

E

M y × × − =

E odd ?

[ [

2 , 1 ∈ M

With

( )

E

M y

′

× ′ × − =

2

2 1

[ [

4 , 1 ∈ ′ M

With

( )

E

M y x

′

× ′ × − = = 2 1 Find x such as y x =

( )

E

X x

′

× × − = 2 1

[ [

2 , 1 ∈ X

With

Square root - direct

Pirouz Bazargan Sabet February 2010 Digital Design

Square root - direct

Find x such as y x =

• −

=

× =

n i i i

x X 2

( )

E

M y

′

× ′ × − =

2

2 1

[ [

4 , 1 ∈ ′ M

With

( )

E

X x

′

× × − = 2 1

[ [

2 , 1 ∈ X

With

Iterate on i and evaluate xi

Pirouz Bazargan Sabet February 2010 Digital Design

Square root - direct

Pirouz Bazargan Sabet February 2010 Digital Design

Square root

2

= − y x resolve

• Direct method
• Indirect method

digit-by-digit

Pirouz Bazargan Sabet February 2010 Digital Design

y x

tangent initial guess new estimation

Square root - indirect

( )

x f

( )

= x f

Resolving a non linear equation

Pirouz Bazargan Sabet February 2010 Digital Design

Resolving a non linear equation

( )

= x f Taylor series in the neighborhood of

x

( ) ( ) ( )( ) ( )( )

• +

− ′ ′ + − ′ + =

2

2 1 x x x f x x x f x f x f 1st order :

( ) ( ) ( )( )

x x x f x f x f − ′ + ≈

Square root - indirect

Pirouz Bazargan Sabet February 2010 Digital Design

Resolving a non linear equation

( )

= x f Iterative resolution starting from an initial guess x0

( ) ( ) ( )( )

x x x f x f x f − ′ + ≈

( )

= x f

( ) ( )( )

= − ′ + x x x f x f

( ) ( )

x x f x f x + ′ − = Newton-Raphson method

Square root - indirect

Pirouz Bazargan Sabet February 2010 Digital Design

Resolving y x =

( ) ( ) ( )( )

x x x f x f x f − ′ + ≈ Find a function f such as

( )

= x f for y x =

( )

y x x f − =

2

( ) (

)

( )

2

2 x x x y x x f − + − ≈

( )

= x f

( )

2

2 x x x y x + − =

• +

= 2 1 x x y x

Square root - indirect

Pirouz Bazargan Sabet February 2010 Digital Design

• +

=

+ i i i

x x y x 2 1

1

Each iteration division !!

( )

y x x f − =

2

Square root - indirect

Hard to implement Resolving y x =

Pirouz Bazargan Sabet February 2010 Digital Design

( ) ( ) ( )( )

u u u f u f u f − ′ + ≈ Find a function f such as

( )

= x f for y x =

( )

y u u f − =

−2

( ) (

)

( )

3 2

2 u u u y u u f − − − ≈

− −

( )

= u f

( )

3 2

2 u u y u u + − =

− −

( )

y u u u

2 0 3

2 1 − = y u 1 =

( )

= u f

Square root - indirect

Resolving y x =

Pirouz Bazargan Sabet February 2010 Digital Design

Each iteration multiply !!

( )

y u u u

i i i 2 1

3 2 1 − =

+

y u 1 = y u y y y x ⋅ = = =

Square root - indirect

Resolving y x =

Pirouz Bazargan Sabet February 2010 Digital Design

Resolving y x = y u 1 =

( )

E

M y 2 1 × × − =

( )

2

2 1

E

M y × × − =

E odd ?

[ [

2 , 1 ∈ M

with

( )

E

M y

′

× ′ × − =

2

2 1

[ [

4 , 1 ∈ ′ M

with

( )

E

M y

′

× ′ × − = 2 1

( )

E

M M y

′

× ′ ′ × − = 2 1

Square root - indirect

Pirouz Bazargan Sabet February 2010 Digital Design

Initial guess M u ′ = 1

[ [

4 , 1 ∈ ′ M

with

u

M′

1

4

2

Square root - indirect

1 0.5

Resolving y x = y u 1 =

0.707

] ]

1 , 5 . ∈ u