CORDIC算法的matlab实现

大笨象

3 H) p2 G) I' Q1 r% t- x
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。
function v=f_cordic(beta,n)
8 w# V! _3 M& T5 N6 B% {$ ^if beta < -pi/2 || beta > pi/2
    if beta < 0
1 w1 i6 u+ H* m        v=f_cordic(beta + pi,n);
    else
        v=f_cordic(beta - pi,n);
    end
. S! J0 N! y+ a# W" g$ L) Q    v = -v;% flip the sign for second or third quadrant
    return
end
% Initialization of tables of constants used by CORDIC
, J  l* @" h8 g- B% need a table of arctangents of negative powers of two, in radians:
( x' \! P# G) P# @% angles = atan(2.^-(0:27));
) `% V+ n7 m, @$ x3 W  O: H. D" Dangles =  [  ...
    0.78539816339745   0.46364760900081   0.24497866312686   0.12435499454676 ...
    0.06241880999596   0.03123983343027   0.01562372862048   0.00781234106010 ...
" H! U6 Z9 w  f. u    0.00390623013197   0.00195312251648   0.00097656218956   0.00048828121119 ...
    0.00024414062015   0.00012207031189   0.00006103515617   0.00003051757812 ...
    0.00001525878906   0.00000762939453   0.00000381469727   0.00000190734863 ...
7 i9 x- j; F7 D. d( [5 i' u8 \    0.00000095367432   0.00000047683716   0.00000023841858   0.00000011920929 ...
    0.00000005960464   0.00000002980232   0.00000001490116   0.00000000745058 ];
# p$ Q! r' I1 @' w. P% and a table of products of reciprocal lengths of vectors [1, 2^-j]:
- a. z6 |, B' A9 V% q- @Kvalues = [ ...
    0.70710678118655   0.63245553203368   0.61357199107790   0.60883391251775 ...
; S# F0 u8 ^; _% c: e4 Z    0.60764825625617   0.60735177014130   0.60727764409353   0.60725911229889 ...
( W3 B( X9 w# ~  X$ z9 n. C0 g+ h    0.60725447933256   0.60725332108988   0.60725303152913   0.60725295913894 ...
    0.60725294104140   0.60725293651701   0.60725293538591   0.60725293510314 ...
    0.60725293503245   0.60725293501477   0.60725293501035   0.60725293500925 ...
* T1 \/ m; z1 a0 v' ^7 r    0.60725293500897   0.60725293500890   0.60725293500889   0.60725293500888 ];
Kn = Kvalues(min(n, length(Kvalues)));
" F0 q" O, t9 Q" b8 G
% Initialize loop variables:
5 {/ R) Y/ ^  T9 ^v = [1;0]; % start with 2-vector cosine and sine of zero
, D3 c" s8 Y$ g1 epoweroftwo = 1;
% Iterations
for j = 1:n;
    if beta < 0
        sigma = -1;
    else
        sigma = 1;
$ n0 ?1 m/ [" d, l/ b% J    end
$ O( i/ H2 x7 g$ b    % update the angle from table, or eventually by just dividing by two
. R+ T5 E0 Q2 q: v3 W$ }    if j <= length(angles)
  G9 c( G9 g0 J; j  n. o; G) c        angle = angles(j);
5 h& P6 t* |2 c9 R: i' k4 @    else
        angle = angle/2;
    end
    factor = sigma * poweroftwo;
# i& I* y2 p# W    R = [1, -factor; factor, 1];
5 [- A% s  I; a  W  r  z    v = R * v; % 2-by-2 mtrix multiply
    beta = beta - sigma * angle; % update the remaining angle
+ u  w0 D3 k$ z! j    poweroftwo = poweroftwo / 2;
6 J, J5 N# |/ @; E$ y! P7 rend
% Adjust length of output vector to be [cos(beta), sin(beta)]:
; `4 T" `# C$ E+ K: x, Y+ t, Nv = v * Kn;
; e+ t' f3 a4 A& E' ~, r: V" x) Wreturn
6 C- F( O6 g9 |
使用20位寄存器，仿真波形如下：
. C4 P- ~( u+ z  w/ o) @
& f7 B8 H+ m$ y1 _& ]

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
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
5 T: H& \6 s( a5 C; v; ]: v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