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。
6 I# X; g6 f4 J& P) ~% {8 }9 v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。
% F% c  L: z+ |/ qfunction v=f_cordic(beta,n)
& d" q# S+ o8 [; b, V* Q% e' J) ^! i% {  _if beta < -pi/2 || beta > pi/2
) Y0 P' P( Q8 k4 F: n- e    if beta < 0
        v=f_cordic(beta + pi,n);
    else
        v=f_cordic(beta - pi,n);
    end
+ q4 ~: n) ~! ]4 `/ I    v = -v;% flip the sign for second or third quadrant
- D' K& O9 h% C1 ^% m    return
end
% Initialization of tables of constants used by CORDIC
8 Q' }5 h* H/ W6 b- c$ t7 l3 c% need a table of arctangents of negative powers of two, in radians:
2 b: q0 M% N  a# `% angles = atan(2.^-(0:27));
angles =  [  ...
8 @3 I' x" H0 Z3 z- g    0.78539816339745   0.46364760900081   0.24497866312686   0.12435499454676 ...
0 ^9 z/ P$ d. \+ {- s) R3 W! {    0.06241880999596   0.03123983343027   0.01562372862048   0.00781234106010 ...
    0.00390623013197   0.00195312251648   0.00097656218956   0.00048828121119 ...
    0.00024414062015   0.00012207031189   0.00006103515617   0.00003051757812 ...
    0.00001525878906   0.00000762939453   0.00000381469727   0.00000190734863 ...
. W: x! v* U# \    0.00000095367432   0.00000047683716   0.00000023841858   0.00000011920929 ...
    0.00000005960464   0.00000002980232   0.00000001490116   0.00000000745058 ];
6 h1 s3 j4 w9 y6 T" o! T" G% 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 ...
    0.60725447933256   0.60725332108988   0.60725303152913   0.60725295913894 ...
    0.60725294104140   0.60725293651701   0.60725293538591   0.60725293510314 ...
    0.60725293503245   0.60725293501477   0.60725293501035   0.60725293500925 ...
    0.60725293500897   0.60725293500890   0.60725293500889   0.60725293500888 ];
Kn = Kvalues(min(n, length(Kvalues)));
$ `; }& N5 h- J& N8 k1 s" K
3 T2 U/ }  `5 g% Initialize loop variables:
v = [1;0]; % start with 2-vector cosine and sine of zero
1 @; u+ t& t  Y- m1 \poweroftwo = 1;
" V0 P0 x: L8 u! {  a6 G% Iterations
for j = 1:n;
    if beta < 0
1 p; C& V6 z7 Q" D" |        sigma = -1;
    else
        sigma = 1;
    end
. F) O  a$ ~% O; ]- Q    % update the angle from table, or eventually by just dividing by two
    if j <= length(angles)
3 y! D. e& _: T; k+ p1 H% M        angle = angles(j);
5 }; D4 m! ]' N$ Q0 M" k    else
* g) i( k" z$ B* @% g9 J        angle = angle/2;
    end
$ L/ q3 L0 r9 ~5 |    factor = sigma * poweroftwo;
    R = [1, -factor; factor, 1];
    v = R * v; % 2-by-2 mtrix multiply
    beta = beta - sigma * angle; % update the remaining angle
    poweroftwo = poweroftwo / 2;
end
1 l/ L2 z6 b8 j9 c$ z3 a. z1 [/ y% Adjust length of output vector to be [cos(beta), sin(beta)]:
  A5 U; }* h3 ^) z3 G; iv = v * Kn;
return
' {$ E  H9 o, a* [" c
使用20位寄存器，仿真波形如下：
3 K3 t( D5 L' Q1 N- x) m& Y3 `

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
5 L8 y$ n; y# r* u( E8 O# A. j5 \: y这是什么算法？第一次听说

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
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