The CORDIC Examples
The Division by Convergence Examples
June 2007
Computer Arithmetic, Function Evaluation
Slide 1
Rotations
(cos z, sin z)
(1, y)
–1
z
tan y
Key ideas in CORDIC
(1, 0)
start at (1, 0)
rotate by z
get cos z, sin z
start at (1, y)
rotate until y = 0
rotation amount is tan–1y
If we have a computationally efficient way of rotating a vector,
we can evaluate cos, sin, and tan–1 functions
-
Rotation by an arbitrary angle is difficult, so we:
Perform psuedorotations that require simpler operations
Use special angles to synthesize the desired angle z
z = a (1) + a (2) + . . . + a (m)
June 2007
Computer Arithmetic, Function Evaluation
Slide 2
Rotating a Vector (x (i), y (i)) by the Angle a (i)
x(i+1) = x(i) cos a(i) – y(i) sin a(i) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = y(i) cos a(i) + x(i) sin a(i) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
Recall that cos q = 1 / (1 + tan2 q)1/2
E(i+1)
y
E(i+1)
y (i+1)
Rotation
Pseudorotation
R (i+1)
y (i)
E(i)
a (i)
R (i)
O
x (i+1)
June 2007
x (i)
Computer Arithmetic, Function Evaluation
x
Slide 3
Rotating a Vector (x (i), y (i)) by the Angle a (i)
x(i+1) = x(i) cos a(i) – y(i) sin a(i) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = y(i) cos a(i) + x(i) sin a(i) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
Our strategy:
- Eliminate the terms (1 + tan2 a(i))1/2 and
- Choose the angles a(i) so that tan a(i) is a power
of 2
Then:
Need two shift-adds
(one for x(i+1) and the other for y(i+1) )
Slide 4
CORDIC Iterations
–––––––––––––––––––––––––––––––––––––––––
a(i) in degrees tan a(i)
tan a(i)
i
–––––––––––––––––––––––––––––––-–––––––––
0
45.0
1
20
1
26.6
0.5
2-1
2
14.0
0.25
2-2
3
7.1
0.125
2-3
4
3.6
0.062 5
2-4
5
1.8
0.031 25
2-5
6
0.9
0.015 625
2-6
7
0.4
0.007 812 5
2-7
8
0.2
0.003 906 25
2-8
9
0.1
0.001 953 125 2-9
–––––––––––––––––––––––––––––––––––––––––
x(i+1) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
Computer Arithmetic, Function Evaluation
Slide 5
CORDIC Circular Rotation Mode
Given: q = 30O
Find: cos q
Let x(0) = cos0O = 1
x(i+1) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
Start at (1,0) and z = 0o
Rotate by z i
Until Z  q o
To Obtain cos q, sin q
Computer Arithmetic, Function Evaluation
Slide 6
Basic CORDIC Iterations
x(i+1) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
x0
= 1,
y0 = 0,
z0 = 0
Let a0= 45o i.e. Rotate 45o and tana0 = tan45o = 1
Then
x1= (x0 – y0 tan a0) / (1 + tan2 a0)1/2 = 1 (1 + tan2 45o)-1/2 =1 k0
y1= (y0 + x0 tan a0) / (1 + tan2 a0)1/2 = 1 (1 + tan2 45o)-1/2 =1 k0
z1 = z0 – a0 = 45o
where k0 = (1 + tan2 45o)-1/2
45O
Computer Arithmetic, Function Evaluation
Slide 7
CORDIC Iterations
x1= 1k0 y1= 1k0 z1 = 45o
Since 45 > 30 Rotate too much, rotate -26.6O
Let a1= -26.6o i.e. Rotate -26.6o and tana1 = tan-26.6O = -1/2
Then
x2= (x1 – y1 tan a1) / (1 + tan2 a1)1/2
= (1k0 – 1k0 (-1/2)) / (1 + tan2 a1)1/2 =1.5  k0  k1
y2= (y1 + x1 tan a1) / (1 + tan2 a1)1/2
= (1k0 + 1k0 (-1/2)) / (1 + tan2 a1)1/2 = 0.5  k0  k1
z2 = z1 – a1 = 45o-26.6o = 18.4o
+45o-26.6o
Basic CORDIC Iterations
x2= 1.5k0K1 y2= 0.5k0K1 z2 = 18.4o
Since 18.4<30 RotateCCW 14.0 O
Let a2= 14.0o i.e. Rotate +14.0o and tana2 = tan14O = 1/4
Then
x3= (x2 – y2 tan a2) / (1 + tan2 a2)1/2
= (1.5k0K1 – 0.5k0K1 1/4) / (1 + tan2 a2)1/2 =1.375  k0  k1  K2
y3= (y2 + x2 tan a2) / (1 + tan2 a2)1/2
= (0.5k0K1 + 1.5k0K1 1/4) / (1 + tan2 a1)1/2 = 0.875  k0  k1  K2
z2 = z1 – a1 = 18.4o+14o = 32.4o
1
3
2
+45o-26.6o +14O
CORDIC Iterations
x2= 1.375k0K1 y2= 0.875k0K1 z2 = 32.4o
Since 32.4>30 Rotate CW 7.1 O
Let a3= 7.1o i.e. Rotate -7.1o and tana3 = tan-7.1O = -1/8
Then
x3= (x2 – y2 tan a2) / (1 + tan2 a2)1/2
= (1.375k0K1 – 0.875k0K1 (-1/8)) / (1 + tan2 a3)1/2
= 1.484375  k0  k1  K2  K3
y3= (y2 + x2 tan a2) / (1 + tan2 a2)1/2
= (0.875k0K1 + 1.3755k0K1 (-1/8)) / (1 + tan2 a3)1/2
= 0.7030625  k0  k1  K2  K3
z3 = z1 – a1 = 32.4o -7.1o = 25.3O
CORDIC Iterations
k0, k1 , K2 , K3 , and  are always the same
- No matter CW or CCW the vector is rotated ,
tan2 ai are always positive.
We can calculate k0 k1  K2 K3   in advance and
store it for retrieval.
In this example, we set z0 = 0O and we rotate it to obtain z m = 30O
Conversely, we can set z0 = 30O and we rotate it to obtain z m = 0O
CORDIC Linear Vectoring Mode
Given: k
Find: tan-1 k
Let x(0) = 1 y(0) = k
x(i+1) = (x(i) – y(i) tan a(i)) / (1 + tan2 a(i))1/2
y(i+1) = (y(i) + x(i) tan a(i)) / (1 + tan2 a(i))1/2
z(i+1) = z(i) – a(i)
Start at (1,k ) and z = 0 o
Rotate by z i
Until y m 0
To Obtain zm = tan-1 k/1
Computer Arithmetic, Function Evaluation
Slide 12
CORDIC Linear Vectoring Mode
Given: k
Find: tan-1 k
Start at (1,k ) and z = 0 o
Rotate by 45 o, 26.6 o, 14.0 o, …
if yk > 0 Rotate -q
else Rotate + q
And Calculate xk, yk, zk
Until y m 0
To Obtain zm = tan-1k
Computer Arithmetic, Function Evaluation
Slide 13
Formulation for Division
as a Convergence Computation
Idea:
z zx (0 ) x (1)  x ( m 1)
q   (0 ) (1)
d dx x  x ( m 1)
Converges to q
Force to 1
d (i+1) = d (i) x (i)
Set d (0) = d; make d (m) converge to 1
z (i+1) = z (i) x (i)
Set z (0) = z; obtain z/d = q  z (m)
May 2007
Computer Arithmetic, Division
Slide 15
Formulation for Division
as a Convergence Computation
z zx (0 ) x (1)  x ( m 1)
q   (0 ) (1)
d dx x  x ( m 1)
Question 1: How to select the multipliers x (i) ?
Answer:
x (i) = 2 – d (i) if d  [1/2, 1)
Note: d  [1/2, 1)  d =0.1xxxxxxxxxxx…|binary
This choice transforms the recurrence equations into:
d (i+1) = d (i) (2  d (i))
z (i+1) = z (i) (2  d (i))
May 2007
Set d (0) = d; iterate until d (m)  1
Set z (0) = z; obtain z/d = q  z (m)
Computer Arithmetic, Division
Slide 16
Example
of a Convergence Computation for Division
Given:
z 18.0
q 
d 0.7
Z(0) = 18.0 d(0) = 0.7
x (0) = 2 – d (0) = 1.3
z 18*1.3 23.4
q 

d 0.7 *1.3 0.91
May 2007
Computer Arithmetic, Division
Slide 17
Example
of a Convergence Computation for Division
Given:
z 18.0
q 
d 0.7
Z(1) = 23.4 d(1) = 0.91
x (1) = 2 – d (1) = 1.09
z 18  (1.3) 23.4 23.4 1.09 25.506
q 



d 0.7  (1.3) 0.91 0.91 1.09 0.9919
May 2007
Computer Arithmetic, Division
Slide 18
Example
of a Convergence Computation for Division
Z(1) = 25.506 d(1) = 0.9919
x (1) = 2 – d (1) = 1.0081
z 18*1.3*1.09 25.506 25.506*1.0081 25.7125986
q 



d 0.7 *1.3*1.09 0.9919 0.9919*1.0081 0.99993439
 25.7142857
May 2007
Computer Arithmetic, Division
Slide 19