Full Version : Arc-Tangent Routine for PICs
avr >>PIC 8051 ZILOG ARM TI H8 ETC >>Arc-Tangent Routine for PICs


Admin3- 04-18-2006

This routine will take an 8 bit integer that corresponds to the numerator of a fraction whose denominator is 256 and find its arctangent. So the input ranges from 0 to 255 which corresponds to 0 to 255/256 = 0.996 . The output for an arctangent routine that returns a floating point number would be from 0 (atan(0)) to 0.783 (atan(255/256)) radians; or if you prefer, 0 to 44.89 degrees. However, this routine scales the output so that pi/4 radians (or 45 degrees) corresponds to 256. So for the input range of 0 to 255 you get an output of 0 to 255 ( atan(255/256) * 256 / (pi/4) is about 255). It's probably a little more interesting to see an intermediate data point or two:

CODE


  Intger           Float
 x   |  atan(x) |   x   | atan(x)
------+----------+-------+---------
0x4a |  0x5b    |  .289 | .281
0x62 |  0x77    |  .383 | .366
0x6f |  0x84    |  .434 | .409
0xa6 |  0xbb    |  .648 | .575
0xdb |  0xe6    |  .855 | .707


The only thing that's left is combining the fractional division and the swapping of the x and y values if y is greater than x (and then subtracting the result from pi/2 or actually 512 in this case).

CODE

;----------------------------------------------------------
;
;arctan (as adapted from the similar arcsin function)
;
;  The purpose of this routine is to take the arctan of an
;8-bit number that ranges from 0 < x < 255/256. In other
;words, the input, x, is an unsigned fraction whose implicit
;divisor is 256.
;  The output is in a conveniently awkward format of binary
;radians (brads?). The output corresponds to the range of zero
;to pi/4 for the normal arctan function. Specifically, this
;algorithm computes:
;
; arctan(x) = real_arctan(x/256) * 256 / (pi/4)
;  for 0 <= x <= 255
;
;  where, real_arctan returns the real arctan of its argument
;in radians.
;
;  The algorithm is a table look-up algorithm plus first order
;linear interpolation. The psuedo code is:
;
;unsigned char arctan(unsigned char x)
;{
;  unsigned char i;
;
;  i = x >> 4;
;  return(arctan[i] + ((arctan[i+1] - arctan[i]) * (x & 0xf))/16);
;}
;
;
arctan
       SWAPF   x,W
       ANDLW   0xf
       ADDLW   1
       MOVWF   temp                   ;Temporarily store the index
       CALL    arc_tan_table          ;Get a2=atan( (x>>4) + 1)
       MOVWF   result                 ;Store temporarily in result
       DECF    temp,W                 ;Get the saved index
       CALL    arc_tan_table          ;Get a1=atan( (x>>4) )
       SUBWF   result,W               ;W=a2-a1, This is always positive.
       SUBWF   result,F               ;a1 = a1 - (a1-W) = W
       CLRF    temp                   ;Clear the product
       CLRC
       BTFSC   x,0
        ADDWF  temp,F
       RRF     temp,F
       CLRC
       BTFSC   x,1
        ADDWF  temp,F
       RRF     temp,F
       CLRC
       BTFSC   x,2
        ADDWF  temp,F
       RRF     temp,F
       CLRC
       BTFSC   x,3
        ADDWF  temp,F
       RRF     temp,W
       ADDWF   result,F
       RETURN
arc_tan_table
       ADDWF   PCL,F
       RETLW   0
       RETLW   20    ;atan(1/16) = 3.576deg * 256/45
       RETLW   41
       RETLW   60
       RETLW   80
       RETLW   99
       RETLW   117
       RETLW   134
       RETLW   151
       RETLW   167
       RETLW   182
       RETLW   196
       RETLW   210
       RETLW   222
       RETLW   234
       RETLW   245
       RETLW   0      ;atan(32/32) = 45deg * 256/45


The other part of the problem is implementing the division
CODE

FRAC_DIV:
;-------------------
;Fractional division
;
; Given x,y this routine finds:
;  a = 256 * y / x
;

   movlw  8   ;number of bits in the result
   movwf  cnt
   clrf   a   ; the result
   movf   x,w

L1:

   clrc
   rlf    y,f  ;if msb of y is set we know x<y
   rlf    a,f  ;and that the lsb of 'a' should be set
   subwf  y,f  ;But we still need to subtract the
               ;divisor from the dividend just in
               ;case y is less than 256.
   skpnc       ;If y>x, but y<256
    bsf   a,0  ; we still need to set a:0


   btfss  a,0  ;If y<x then we shouldn't have
    addwf y,f  ;done the subtraction

   decfsz cnt,f
    goto  L1

   return


It's easy enough to combine these two routines to obtain a 4-quadrant arctan(y/x) routine. However, you do need to keed in mind that the arctangent routine posted above is only valid over 1/8th of the unit circle. To obtain the other 7/8th's you'll need to apply the appropriate trig identities.

CODE

// pic routines:
extern int arctan(int x);
extern int frac_div(int y, int x);

// Untested c code that implements a 4-quadrant arctan function

int arctan(int x, int y)
{
 int f, swapped;
 int reference_angle;

 swapped = 0;

 if(x < 0) {
   if(y < 0)
     reference_angle = 256 * 2;  // pi
   else
     reference_angle = 256;      // pi/2
 } else {
   if( y < 0)
     reference_angle = 256 * 3;  // 3*pi/2
   else
     reference_angle = 0;
 }

 if (x<=y) {
   f = y;
   y = x;
   x = f;
   swapped = 1;
 }

 f = frac_div(y,x);
 f = arctan(f);
 if (swapped)
   f = 256 - f;

 return f + reference_angle;
}





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