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CORDIC, Divider, Square Root

EE219A – Spring 2008Special Topics in Circuits and Signal Processing

Lecture 6

Slide 2EE219A – Spring 2008 Lecture 6

Announcements

This week– Thu, Feb 24, 3-5pm, location 3400 BH

● In-class review discussions

Today’s lecture: CORDIC– Iterative division and square rooting


CORDIC

To do the transformation from

and the inverse, we use the CORDIC algorithm.

θjIr eyjyyty ⇒+=)(


CORDIC -COordinate Rotation DIgital Computer

Use rotations to implement a variety of functions.

yzxyz

xyz

xyz

yxz

xyz

xyzyxz

eyxjyxeg xyj

11

22

tan22

costan

sinhsin

tancos

.1

−−

=

=

=

==

=

=+=

+⇔+−


CORDIC (cont’d)How to do it?Start with general rotation by .

Now the trick is to only do rotations by tan which are a power of 2.

φ

[ ][ ]φφ

φφφφφφ

tancos'tancos'sincos'sincos'

xyyyxxxyyyxx

+=−=+=−=

φ


CORDIC (cont’d)

So to rotate to any arbitrary angle, we do a sequence of rotations to get to that value.

672-60.895°

562-51.790°

452-43.576°

342-37.125°

232-214.036°

122-126.565°

01145°

iKTan θθ

Rotation Number


Basic CORDIC Iteration

[ ][ ]iiiiii

iiiiii

dxyKydyxKx

−+

−+

⋅⋅−=

⋅⋅−=

2)(

2)(

1

1

1211)2cos(tan

2

11

±=+

==−

−−

i

ii

d

K

The di is chosen to rotate by ±φ

If we don’t multiply by Ki we get a gain error which is independent of the direction of the rotation.

The error converges to .61 - May not need to compensate it.

We also can accumulate the rotation angle iiii dZZ −−

+ −= 2tan 11


ExampleOK, let’s do an example:

y0

x0

y0

x0

φ

Initial vector(x02+y0

2)0.5

We want to find φ & 20

20 yx +


Example (cont’d)

1) First, check to see if the angle is positive or negative by checking sign of . If positive, rotate by -45º, if negative, rotate by +45º.

0y


Example (cont’d)2) Check the sign of and rotate positive if negative or rotate

negative if is positive by rotation 1 (26º).1y

1y


Example (cont’d)

3) Repeat 2 for each rotation K


The Gain Factor

)642.1()630.1(

)581.1()414.1(

1

43210

3210

210

10

0

==

==

=

KKKKKKKKK

KKKKK

K

( )58.1

7120

20

3

=+

=

yx

Z Shift & Adds of x0,y0

So we did it Rectangular Polar Coordinates

So, start with ; End up with:00 , yx


The Gain Factor (cont’d)How about the reverse?Start with vector on axis:

x0 φ

φ

==

=

=

00

0

,0 zyAx

eAA j

1,01,0+=>−=<

ii

ii

dzdzρ−

+ ⋅−= 21 iii dZZ


CORDIC Algorithm


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

it = 1

it = 2

it = 3

it = 4it = 5it = 6it = 7it = 8

K(0) = 1.000K(1) = 1.414K(2) = 1.581K(3) = 1.630K(4) = 1.642K(5) = 1.646K(6) = 1.646

Φ = 30.00Φ = -15.00Φ = 11.57Φ = -2.47Φ = 4.65Φ = 1.08Φ = -0.71

∆Φ = -45.00∆Φ = 26.57∆Φ = -14.04∆Φ = 7.13∆Φ = -3.58∆Φ = -1.79

Example 1


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

it = 1

it = 2

it = 3

it = 4it = 5it = 6it = 7it = 8

K(0) = 1.000K(1) = 1.414K(2) = 1.581K(3) = 1.630K(4) = 1.642K(5) = 1.646K(6) = 1.646

Φ = 60.00Φ = 15.00Φ = -11.57Φ = 2.47Φ = -4.65Φ = -1.08Φ = 0.71

∆Φ = -45.00∆Φ = -26.57∆Φ = 14.04∆Φ = -7.13∆Φ = 3.58∆Φ = 1.79

Example 2


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

it = 1it = 2it = 3it = 4it = 5it = 6it = 7it = 8

K(0) = 1.000K(1) = 1.414K(2) = 1.581K(3) = 1.630K(4) = 1.642K(5) = 1.646K(6) = 1.646

Φ = 45.00Φ = 0.00Φ = 0.00Φ = 0.00Φ = 0.00Φ = 0.00Φ = 0.00

∆Φ = -45.00∆Φ = 0.00∆Φ = 0.00∆Φ = 0.00∆Φ = 0.00∆Φ = 0.00

Example 3


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

it = 1

it = 2

it = 3

it = 4it = 5it = 6it = 7it = 8

K(0) = 1.000K(1) = 1.414K(2) = 1.581K(3) = 1.630K(4) = 1.642K(5) = 1.646K(6) = 1.646

Φ = 53.13Φ = 8.13Φ = -18.43Φ = -4.40Φ = 2.73Φ = -0.85Φ = 0.94

∆Φ = -45.00∆Φ = -26.57∆Φ = 14.04∆Φ = 7.13∆Φ = -3.58∆Φ = 1.79

Example 4


Calculating SIN & COS

To calculate SIN & COS, start with , rotate by φ.

sin φ

cos φ

θ

(1/1.64)

y0=sin φ

x0=cos φ

0,64.11

00 == yx


Functions


CORDIC DividerTo do a divide, change CORDIC rotations to a linear function calculator.


Generalized CORDIC


Implementation in an FPGA

[R. Andraka 1998]

Three difference equations directly mapped to hardwareThe decision di is driven by the sign of the y or z register– Vectoring: di = -sign(yi)– Rotation: di = sign(zi)

The initial values loaded via muxesOn each clock cycle– the values from the registers are

passed through shifters and add/sub and the values placed in registers

– The shifters are modified on each iteration to cause the desired shift (state machine)

– Elementary angle stored in ROMLast iteration: results read from reg


Iterative Sqrt and Division

Inputs:– a (14 bits), reset (active high)

Outputs:– zs (16 bits), zd (16 bits) Total: 32 bits

~15k gates


Iterative 1/sqrt()

User defined parameters:- data type- wordlength (#bits, binary pt)- quantization- overflow- latency- sample period

wordlength

latency

xs(k+1) = xs(k) / 2· (3–Z· xs

2(k))

User defined parameters– Wordlength (#bits, binary pt)– Quantization, overflow– Latency, sample period

The choice of initial condition– Determines # iterations– and convergence…

xs

Z


Quadratic Convergence

))(3(2)()1( 2kxNkxkx s

ss ⋅−⋅=+

∞→

→k

s Nkx 1)(

))(2()()1( kxNkxkx ddd ⋅−⋅=+

∞→

→k

d Nkx 1)(

))(3(2)()1( 2kykyky s

ss −⋅=+

∞→→

ks ky 1)(

))(3()(21)1( 2 kekeke sss +⋅⋅−=+

1)()( −= kyke ss

))(2()()1( kykyky ddd −⋅=+

∞→→

kd ky 1)(

2)()1( keke dd −=+

1)()( −= kyke dd

Sqrt Div


-4 -2 0 2 4-4

-2

0

2

4

yd(k)

y d(k+1

)

-4 -2 0 2 4-4

-2

0

2

4

ys(k)

y s(k+1

)

√5 √3

Initial Condition

))(3(2)()1( 2kykyky s

ss −⋅=+ ))(2()()1( kykyky ddd −⋅=+

Sqrt Div

∞→→

ks ky 1)(∞→

→kd ky 1)(

3)0(0 << sy

5)0(3 << sy

5)0( >sy

Convergence:Conv. stripes:Divergence:

2)0(0 << dyConvergence:Divergence: o.w.


1/Sqrt() Formula: Convergence

( )21 32 nn

n xNxx ⋅−⋅=+∞→

→n

n Nx 1

EquilibriumsN1,0 ±

Initial condition– Take:

– Find x0 such that:

– Solution:

21)(

−=N

xxV nn

...)4,3,2,1,0(0)()( 1 =∀<−+ nnxVxV nn

})(:{ 00 axVxS <=

“Level set” V(x0)=a is convergence boundLocal convergence (3 equilibriums)

(Eq.1)

(Eq.2)

(Eq.3)


10-1

100

101

-2

-1.5

-1

-0.5

0

0.5

1

1.5

initial condition x0

f3

N=1/64N=1/16N=1/4N=1N=4N=16

Convergence…

0222

5)1(2

300

20

0 <

−−⋅⋅−⋅N

xNxxNx(Eq.2)

Nx M 3

50 =

Roots:

Nx 11 =

( )17121

3,2 ±−⋅

=N

x

Max:


Initial Condition Circuit

>16

>4

>1

>1/4

>1/16

>1/64 x0=8x0=4x0=2x0=1x0=1/2x0=1/4

N

aN

xaN

+<<−11

0(Eq.3)


Top: Convergence, Bottom: Sampled


Convergence in 8 iterations


Mult by 2 (extended range…)



( )21 32 nn

n xNxx ⋅−⋅=+∞→

→n

n Nx 1

Nyx n

n =

( )21 32 nn

n yyy −⋅=+

nn ye −=1

321 2

123

nnn eee ⋅−⋅=+

Error:

∞→→

nny 1



∞→→

nny 1

-4 -2 0 2 4-4

-3

-2

-1

0

1

2

3

4

( )21 32 nn

n yyy −⋅=+

35 30 0 << y

53 0 << y

50 >y

⇒ convergence

⇒ conv. stripes

⇒ divergence


Error

-1.5 -1 -0.5 0 0.5 10

1

2

3

4

5

6

en

e n+1

⇒ divergence

⇒ conv. stripes

⇒ convergence


Results…

0 5 10 15 20 25-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

iteration

solu

tion

1-1+rt(3)+rt(5)-rt(3)++rt(5)--2.15rt(5)-eps


Results…

2 4 6 8 10 12 14 16 18 20-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

iteration

solu

tion

1-1+rt(3)+rt(5)-rt(3)++rt(5)--2.15rt(5)-epsrt(5)+eps


Descending absolute errorSqrt Div

2)1()( −= kk xxE 0)()()( 1 <−= + kkk xExExV k∀

)4()1()1(4

)( 0200

20

00 −+⋅+⋅−⋅= xxxxxxVs )2()1()( 0

2000 −⋅−⋅= xxxxVd

0 1 2 3

-0.2

0

0.2

0.4

0.6

initial condition, x0

sqrt,

V(x

0)

√3

-0.5+√4.25

0 1 2 3

-0.2

0

0.2

0.4

0.6

initial condition, x0

div,

V(x

0)Descending error:


Convergence speed

2 / 12 / 23 / 23 / 3e0: 25%, # iter (sqrt/div)

3 / 24 / 35 / 35 / 4e0: 50%, # iter (sqrt/div)

10%5%1%0.1%Target rel. error (%)

Adaptive algorithm– current result .ic for next iteration

# iterations required for specified accuracy

Documents

CORDIC, Divider, Square Root - University of California ...icslwebs.ee.ucla.edu/dejan/ee219awiki/images/a/a8/Lec-06-Cordic... · CORDIC, Divider, Square Root EE219A – Spring 2008