CORDIC算法的matlab实现

大笨象

不知道有木有人研究这个，比较偏向于硬件描述。
今天先发一个sine cose函数的实现。
有兴趣的可以看一下，m文件见附件。
& Q! i  ]# F3 U- Ifunction v=f_cordic(beta,n)
0 I: y. m0 a+ w) f" [+ Gif beta < -pi/2 || beta > pi/2
    if beta < 0
        v=f_cordic(beta + pi,n);
2 f& `0 s" y  g) q0 }+ V% j    else
        v=f_cordic(beta - pi,n);
    end
    v = -v;% flip the sign for second or third quadrant
    return
* h" H0 _4 I3 t4 S8 Nend
6 ^  E; ?6 F7 G, Q; \% Initialization of tables of constants used by CORDIC
3 w! m* U( j5 p( d: r% need a table of arctangents of negative powers of two, in radians:
$ {# `9 T9 U) M8 C' t5 z/ K% angles = atan(2.^-(0:27));
1 N- D7 f& K2 H; Kangles =  [  ...
    0.78539816339745   0.46364760900081   0.24497866312686   0.12435499454676 ...
) w% V3 z! x1 ^! K% R    0.06241880999596   0.03123983343027   0.01562372862048   0.00781234106010 ...
8 l! L' T3 b- N1 Z    0.00390623013197   0.00195312251648   0.00097656218956   0.00048828121119 ...
    0.00024414062015   0.00012207031189   0.00006103515617   0.00003051757812 ...
$ @% h9 @" [  Z$ Z& Q$ e* O    0.00001525878906   0.00000762939453   0.00000381469727   0.00000190734863 ...
    0.00000095367432   0.00000047683716   0.00000023841858   0.00000011920929 ...
    0.00000005960464   0.00000002980232   0.00000001490116   0.00000000745058 ];
+ T0 x) j3 n% e1 `9 N% and a table of products of reciprocal lengths of vectors [1, 2^-j]:
Kvalues = [ ...
    0.70710678118655   0.63245553203368   0.61357199107790   0.60883391251775 ...
    0.60764825625617   0.60735177014130   0.60727764409353   0.60725911229889 ...
  V9 w0 }: f' F2 m. {, ~    0.60725447933256   0.60725332108988   0.60725303152913   0.60725295913894 ...
    0.60725294104140   0.60725293651701   0.60725293538591   0.60725293510314 ...
2 W  B7 ?# U0 e2 J    0.60725293503245   0.60725293501477   0.60725293501035   0.60725293500925 ...
    0.60725293500897   0.60725293500890   0.60725293500889   0.60725293500888 ];
Kn = Kvalues(min(n, length(Kvalues)));

) A2 v* [& x4 [% |$ o/ e' C% Initialize loop variables:
" K0 I4 R* n- l7 F9 L1 D* @' Zv = [1;0]; % start with 2-vector cosine and sine of zero
% W1 p! U/ |/ j+ Y5 Fpoweroftwo = 1;
% Iterations
for j = 1:n;
    if beta < 0
        sigma = -1;
    else
6 q( q/ R$ ?/ r+ n( F        sigma = 1;
0 _( F% X! L. z" z; }7 Z    end
" D5 ^  l9 ^5 N. ]8 f! U    % update the angle from table, or eventually by just dividing by two
; i; ~0 R! a* W- c/ |4 q* r3 v    if j <= length(angles)
        angle = angles(j);
    else
        angle = angle/2;
9 g  u% l' p  k5 P1 c    end
    factor = sigma * poweroftwo;
    R = [1, -factor; factor, 1];
    v = R * v; % 2-by-2 mtrix multiply
4 o6 K' y6 u, r' w7 y; U    beta = beta - sigma * angle; % update the remaining angle
    poweroftwo = poweroftwo / 2;
end
  ]( l1 U- A! f  w2 i- [% g, s% Adjust length of output vector to be [cos(beta), sin(beta)]:
v = v * Kn;
return
4 \% p' G3 E0 v- \- M; G& }9 b/ J
使用20位寄存器，仿真波形如下：
4 }) q2 q5 ]$ S3 W
4 B4 {+ m9 s( B5 `* v* w' J" _1 Z

杨帆

我表示学习了

zhangjichang

这是什么算法？第一次听说

赛才

我表示很给力

大笨象

zhangjichang 发表于 2011-6-8 22:31
& w: n( ~! g' _* ^3 Q这是什么算法？第一次听说

5 e0 h' l: c) Z) Z0 d& b这个，就是那个。感兴趣的话你还是自己搜搜吧。

大笨象

赛才 发表于 2011-6-9 15:01
0 j9 L/ N: b8 X# g- C4 q+ ^( @& c我表示很给力

给力就好。呵呵

魅影骑士

顶一下！

yingzhen

费解。。。。。。。。。。。。。。。。

羅雲琦

有什麽用？求解

jt202010