本文主要是介绍坐标旋转数字计算法（Coordinate Rotation Digital Computer, CORDIC），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
目录
- 1. CORDIC原理
- 2. 圆周系统
- 3. 双曲系统
介绍CORDIC基本原理及应用。CORDIC为RTL解决三角函数，反三角函数，对数函数，指数函数等函数提供了解决方案。这些函数也可通过查找表法解决，但是CORDIC比查找表法更省面积和资源，耗时也更长。
cordic算法详解
CORDIC算法详解(一)-CORDIC 算法之圆周系统之旋转模式
《CORDIC v6.0LogiCORE IP Product GuideVivado Design SuitePG105 August 6, 2021》
CORDIC算法基本原理
三角函数计算，Cordic 算法入门
1. CORDIC原理
首先明确，CORDIC本质上解决的是一类近似的问题。其基本问题是：直角坐标系 x O y xOy xOy下，任意一点 P 0 = ( x 0 , y 0 ) P_0=(x_0,y_0) P0=(x0,y0)，将 O P 0 ⃗ \vec{OP_0} OP0按照某种仿射变换 [ x s , y s ] T = f ( θ ) ⋅ [ x 0 , y 0 ] [x_s,y_s]^T=f(\theta)·[x_0,y_0] [xs,ys]T=f(θ)⋅[x0,y0]得到 O P s ⃗ = ( x s , y s ) \vec{OP_s}=(x_s,y_s) OPs=(xs,ys)。将该变换通过RTL实现
RTL的限制在于只能使用位操作和加法，无法使用乘除、三角函数的组合逻辑，所以需要一种近似求解的方法。
CORDIC算法可以解决该问题，非常类似于逐次逼近的思想，CORDIC的基本思想是：将放射变换 [ x s , y s ] T = f ( θ ) ⋅ [ x 0 , y 0 ] [x_s,y_s]^T=f(\theta)·[x_0,y_0] [xs,ys]T=f(θ)⋅[x0,y0]看作是 多次子变换 f i ( θ i ) f_i(\theta_i) fi(θi)的叠加，其中 ∣ θ i ∣ |\theta_i| ∣θi∣单调递减。当迭代的次数足够多，就越接近 O P s ⃗ \vec{OP_s} OPs。
即，CORDIC的遵循的定理就是： lim n → + ∞ ∏ i = 0 n [ f i ( θ i ) ] = f ( θ ) \lim\limits_{n\to+\infty} \prod_{i=0}^n[f_i(\theta_i)]=f(\theta) n→+∞lim∏i=0n[fi(θi)]=f(θ)成立
2. 圆周系统
若CORDIC基本问题中 O P 0 ⃗ \vec{OP_0} OP0到 O P s ⃗ \vec{OP_s} OPs模长不变，那么就是圆周系统上的CORDIC问题。变换如下
[ x s y s ] = ( c o s θ − s i n θ s i n θ c o s θ ) [ x 0 y 0 ] (2.1) \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cos\theta & -sin\theta\\ sin\theta & cos\theta \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right] \tag{2.1} [xsys]=(cosθsinθ−sinθcosθ)[x0y0](2.1)
( 2.1 ) (2.1) (2.1)的变换是具备几何意义的，等价于坐标系下的向量旋转，旋转角度就是 θ \theta θ，若为逆时针旋转 θ > 0 \theta>0 θ>0，若为顺时针旋转 θ < 0 \theta<0 θ<0。
上述坐标旋转公式可在圆坐标系下证明，记 α 0 \alpha_0 α0为x轴正半轴逆时针旋转至 O P 0 ⃗ \vec{OP_0} OP0旋转的角度， α s \alpha_s αs为x轴正半轴逆时针旋转至 O P s ⃗ \vec{OP_s} OPs旋转的角度，则有 α s = α 0 + θ \alpha_s=\alpha_0+\theta αs=α0+θ，再使用三角形角度关系证明。
根据CORDIC原理，迭代过程的几何意义就是多次子旋转，如下图所示， O P 0 ⃗ → O P 1 ⃗ → O P 2 ⃗ → O P 3 ⃗ → O P 4 ⃗ ≈ O P s ⃗ \vec{OP_0}→\vec{OP_1}→\vec{OP_2}→\vec{OP_3}→\vec{OP_4}≈\vec{OP_s} OP0→OP1→OP2→OP3→OP4≈OPs。若累计旋转后的角度小于 θ \theta θ，那么下次旋转就逆时针转。若累计旋转的角度大于 θ \theta θ，那么下次旋转就顺时针转
之后就可以迭代求解了，根据 ( 2.1 ) (2.1) (2.1)我们有：
[ x i + 1 y i + 1 ] = c o s ( θ i ) [ 1 − t a n ( θ i ) t a n ( θ i ) 1 ] [ x i y i ] , i = 0 , . . . , n (2.2) \left[\begin{array}{lcr} x_{i+1}\\ y_{i+1} \end{array}\right] =cos(\theta_i)\left[\begin{array}{lcr} 1 & -tan(\theta_i)\\ tan(\theta_i) & 1 \end{array}\right] \left[\begin{array}{lcr} x_i\\y_i \end{array}\right], i=0,...,n \tag{2.2} [xi+1yi+1]=cos(θi)[1tan(θi)−tan(θi)1][xiyi],i=0,...,n(2.2)
其中，将三角函数转化为RTL中位移，则有
θ i = d i ⋅ a r c t a n ( 1 2 i ) , d i = { − 1 , ( θ 0 + . . . + θ i ) > θ 1 , e l s e \theta_i=d_i·arctan(\frac{1}{2^{i}}), \ \ \ \begin{split} d_i= \left \{ \begin{array}{ll} -1, & (\theta_0+...+\theta_i)>\theta\\ 1, & else\\ \end{array} \right. \end{split} θi=di⋅arctan(2i1), di={−1,1,(θ0+...+θi)>θelse
● 根据旋转的几何意义，子旋转的叠加就等价于直接旋转。 ∑ i = 0 n ( θ i ) \sum_{i=0}^n(\theta_i) ∑i=0n(θi)的边界决定着 θ \theta θ的取值范围，即 θ ≈ ∑ i = 0 n ( θ i ) ∈ ( − ∑ i = 0 ∞ [ a r c t a n ( 1 2 i ) ] , ∑ i = 0 ∞ [ a r c t a n ( 1 2 i ) ] ) ≈ ( − 99.8829 ° , 99.8829 ° ) \theta≈\sum_{i=0}^n(\theta_i) \in (-\sum_{i=0}^∞[arctan(\frac{1}{2^{i}})],\sum_{i=0}^∞[arctan(\frac{1}{2^{i}})])≈(-99.8829°,99.8829°) θ≈i=0∑n(θi)∈(−i=0∑∞[arctan(2i1)],i=0∑∞[arctan(2i1)])≈(−99.8829°,99.8829°)
● 若 ∑ j = 0 1 ( θ i ) = θ \sum_{j=0}^{1}(\theta_i)=\theta ∑j=01(θi)=θ，则 d i ≠ 0 d_i≠0 di=0。这是因为 d i = 0 d_i=0 di=0会导致 x i + 1 x_{i+1} xi+1和 y i + 1 y_{i+1} yi+1停止迭代，角度对了但是幅值错误。
于是将 ( 2.2 ) (2.2) (2.2)多次迭代可以得到 ( 2.3 ) (2.3) (2.3)，将 c o s θ i cos\theta_i cosθi连乘放在一起，其余的部分就可以使用RTL进行迭代了。
[ x n y n ] = ∏ i = 0 n − 1 c o s ( θ i ) ⋅ ( ∏ i = 0 n − 1 ( 1 − d i 2 i d i 2 i 1 ) [ x 0 y 0 ] ) = ∏ i = 0 n − 1 c o s ( θ i ) ⋅ [ x ^ n y ^ n ] = ∏ i = 0 n − 1 c o s ( θ i ) ⋅ ( 1 − d n − 1 ⋅ 1 2 n − 1 d n − 1 ⋅ 1 2 n − 1 1 ) ⋅ [ x ^ n − 1 y ^ n − 1 ] (2.3) \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =\prod_{i=0}^{n-1} cos(\theta_i)· \left(\prod_{i=0}^{n-1}\left(\begin{array}{lcr} 1 & -\frac{d_i}{2^{i}}\\ \frac{d_i}{2^{i}} & 1 \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right]\right) =\prod_{i=0}^{n-1} cos(\theta_i)·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \\ =\prod_{i=0}^{n-1} cos(\theta_i)·\left(\begin{array}{lcr} 1 & -d_{n-1}·\frac{1}{2^{n-1}}\\ d_{n-1}·\frac{1}{2^{n-1}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{n-1}\\ \hat{y}_{n-1} \end{array}\right] \tag{2.3} [xnyn]=i=0∏n−1cos(θi)⋅(i=0∏n−1(12idi−2idi1)[x0y0])=i=0∏n−1cos(θi)⋅[x^ny^n]=i=0∏n−1cos(θi)⋅(1dn−1⋅2n−11−dn−1⋅2n−111)⋅[x^n−1y^n−1](2.3)
那么 ∏ i = 0 n − 1 c o s ( θ i ) = ∏ i = 0 n − 1 c o s [ a r c t a n ( 1 2 i ) ] \prod_{i=0}^{n-1} cos(\theta_i)=\prod_{i=0}^{n-1} cos[arctan(\frac{1}{2^{i}})] ∏i=0n−1cos(θi)=∏i=0n−1cos[arctan(2i1)]是多少呢，看下表可知，当n越来越大时 ∏ i = 0 n − 1 c o s ( θ i ) \prod_{i=0}^{n-1} cos(\theta_i) ∏i=0n−1cos(θi)越来越趋近于一个定值 0.607253，所以如果迭代的次数足够多，可以直接用这个定值。
2.1. 迭代算法
综上所述，CORDIC解算过程为：输入为 ( x 0 , y 0 , θ ∈ ( − 0.554905 π , 0.554905 π ) , n ) (x_0,y_0,\theta \in (-0.554905\pi,0.554905\pi),n) (x0,y0,θ∈(−0.554905π,0.554905π),n)，基于迭代公式 ( 2.4 ) ( 2.5 ) ( 2.6 ) ( 2.7 ) (2.4)(2.5)(2.6)(2.7) (2.4)(2.5)(2.6)(2.7)，可得到输出 ( x n , y n , △ α n ) (x_n,y_n,△α_n) (xn,yn,△αn)
[ x ^ i + 1 y ^ i + 1 ] = ( 1 − d i ⋅ 1 2 i d i ⋅ 1 2 i 1 ) ⋅ [ x ^ i y ^ i ] , [ x ^ 0 y ^ 0 ] = [ x 0 x 0 ] (2.4) \left[\begin{array}{lcr} \hat{x}_{i+1}\\ \hat{y}_{i+1} \end{array}\right] =\left(\begin{array}{lcr} 1 & -d_{i}·\frac{1}{2^{i}}\\ d_{i}·\frac{1}{2^{i}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{i}\\ \hat{y}_{i} \end{array}\right], \left[\begin{array}{lcr} \hat{x}_0\\ \hat{y}_0 \end{array}\right]=\left[\begin{array}{lcr} x_0\\ x_0 \end{array}\right] \tag{2.4} [x^i+1y^i+1]=(1di⋅2i1−di⋅2i11)⋅[x^iy^i],[x^0y^0]=[x0x0](2.4)
△ α i + 1 = △ α i + d i ⋅ a r c t a n ( 1 2 i ) , △ α 0 = 0 (2.5) △α_{i+1}=△α_{i}+d_{i}·arctan(\frac{1}{2^{i}}), △α_0=0 \tag{2.5} △αi+1=△αi+di⋅arctan(2i1),△α0=0(2.5)
d i + 1 = { − 1 , △ α i + 1 > θ 1 , e l s e , d 0 = s i g n ( θ ) (2.6) \begin{split} d_{i+1}= \left \{ \begin{array}{ll} -1, & △α_{i+1}>\theta\\ 1, & else\\ \end{array} \right. \end{split},d_0=sign(\theta) \tag{2.6} di+1={−1,1,△αi+1>θelse,d0=sign(θ)(2.6)
[ x n y n ] = 0.607253 ⋅ [ x ^ n y ^ n ] (2.7) \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =0.607253·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \tag{2.7} [xnyn]=0.607253⋅[x^ny^n](2.7)
其中 n n n为迭代次数， △ α i △α_{i} △αi用于记录向量实际旋转角度的累计值，用该值判断应该旋转方向。
预旋转
从上述算法可以看到，只有在 θ ∈ ( − 0.554905 π , 0.554905 π ) \theta \in (-0.554905\pi,0.554905\pi) θ∈(−0.554905π,0.554905π)时才能收敛，那么如何覆盖 [ − π , π ] [-\pi,\pi] [−π,π]区间呢？转两次
即，若 ∣ θ ∣ ≤ π 2 |\theta|≤\frac{\pi}{2} ∣θ∣≤2π直接用上述算法旋转即可，若 ∣ θ ∣ > π 2 |\theta|>\frac{\pi}{2} ∣θ∣>2π先作 ∣ θ ∣ = ∣ π 2 ∣ |\theta|=|\frac{\pi}{2}| ∣θ∣=∣2π∣一次预旋转，再作 ∣ θ − π 2 ∣ |\theta-\frac{\pi}{2}| ∣θ−2π∣旋转。
预旋转如下
[ x p r e y p r e ] = ( c o s ( π 2 ) − s i n [ s i g n ( θ ) ⋅ π 2 ] s i n [ s i g n ( θ ) ⋅ π 2 ] c o s ( π 2 ) ) [ x 0 y 0 ] = ( 0 − s i g n ( θ ) s i g n ( θ ) 0 ) [ x 0 y 0 ] , ∣ θ ∣ > π 2 \left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right] =\left(\begin{array}{lcr} cos(\frac{\pi}{2}) & -sin[sign(\theta)·\frac{\pi}{2}]\\ sin[sign(\theta)·\frac{\pi}{2}] & cos(\frac{\pi}{2}) \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right]=\left(\begin{array}{lcr} 0 & -sign(\theta)\\ sign(\theta) & 0 \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right] ,|\theta|>\frac{\pi}{2} [xpreypre]=(cos(2π)sin[sign(θ)⋅2π]−sin[sign(θ)⋅2π]cos(2π))[x0y0]=(0sign(θ)−sign(θ)0)[x0y0],∣θ∣>2π
进而重新得到圆周系统下CORDIC算法：
CORDIC解算过程为：输入为 ( x 0 , y 0 , θ ∈ ( − π , π ) , n ) (x_0,y_0,\theta \in (-\pi,\pi),n) (x0,y0,θ∈(−π,π),n)，基于迭代公式 ( 2.8 ) ( 2.9 ) ( 2.10 ) ( 2.11 ) ( 2.12 ) (2.8)(2.9)(2.10)(2.11)(2.12) (2.8)(2.9)(2.10)(2.11)(2.12)，可得到输出 ( x n , y n , △ α n ) (x_n,y_n,△α_n) (xn,yn,△αn)
[ x ^ i + 1 y ^ i + 1 ] = ( 1 − d i ⋅ 1 2 i d i ⋅ 1 2 i 1 ) ⋅ [ x ^ i y ^ i ] , [ x ^ 0 y ^ 0 ] = [ x p r e y p r e ] , i = 0 , . . . , n − 1 (2.8) \left[\begin{array}{lcr} \hat{x}_{i+1}\\ \hat{y}_{i+1} \end{array}\right] =\left(\begin{array}{lcr} 1 & -d_{i}·\frac{1}{2^{i}}\\ d_{i}·\frac{1}{2^{i}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{i}\\ \hat{y}_{i} \end{array}\right], \left[\begin{array}{lcr} \hat{x}_0\\ \hat{y}_0 \end{array}\right]=\left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right],i=0,...,n-1 \tag{2.8} [x^i+1y^i+1]=(1di⋅2i1−di⋅2i11)⋅[x^iy^i],[x^0y^0]=[xpreypre],i=0,...,n−1(2.8)
[ x p r e y p r e ] = { [ − s i g n ( θ ) ⋅ y 0 s i g n ( θ ) ⋅ x 0 ] , ∣ θ ∣ > π 2 [ x 0 y 0 ] , e l s e (2.9) \begin{split} \left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right]= \left \{ \begin{array}{ll} \left[\begin{array}{lcr} -sign(\theta)·y_0\\ sign(\theta)·x_0 \end{array}\right], & |\theta|>\frac{\pi}{2} \\ \\ \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right], & else\\ \end{array} \right. \end{split} \tag{2.9} [xpreypre]=⎩ ⎨ ⎧[−sign(θ)⋅y0sign(θ)⋅x0],[x0y0],∣θ∣>2πelse(2.9)
△ α i + 1 = △ α i + d i ⋅ a r c t a n ( 1 2 i ) , △ α 0 = △ α p r e = { s i g n ( θ ) ⋅ π 2 , ∣ θ ∣ > π 2 0 , e l s e (2.10) △α_{i+1}=△α_{i}+d_{i}·arctan(\frac{1}{2^{i}}), \begin{split} △α_0=△α_{pre}= \left \{ \begin{array}{ll} sign(\theta)·\frac{\pi}{2}, & |\theta|>\frac{\pi}{2} \\ \\ 0, & else\\ \end{array} \right. \end{split} \tag{2.10} △αi+1=△αi+di⋅arctan(2i1),△α0=△αpre=⎩ ⎨ ⎧sign(θ)⋅2π,0,∣θ∣>2πelse(2.10)
d i + 1 = { − 1 , △ α i + 1 > θ 1 , e l s e , d 0 = s i g n ( θ ) (2.11) \begin{split} d_{i+1}= \left \{ \begin{array}{ll} -1, & △α_{i+1}>\theta\\ 1, & else\\ \end{array} \right. \end{split},d_0=sign(\theta) \tag{2.11} di+1={−1,1,△αi+1>θelse,d0=sign(θ)(2.11)
[ x n y n ] = 0.607253 ⋅ [ x ^ n y ^ n ] (2.12) \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =0.607253·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \tag{2.12} [xnyn]=0.607253⋅[x^ny^n](2.12)
其中 n n n为迭代次数， △ α i △α_{i} △αi用于记录向量实际旋转角度的累计值，用该值判断应该旋转方向。
经python验证，上述算法正确
旋转模式—— y = c o s ( x ) y=cos(x) y=cos(x) and y = s i n ( x ) y=sin(x) y=sin(x)
预旋转算法中若令 P 0 ( x 0 = 1 , y 0 = 0 ) P_0(x_0=1,y_0=0) P0(x0=1,y0=0)，就可以看作是从 P 0 P_0 P0旋转至某个向量，因此有
[ x s y s ] = ( c o s θ − s i n θ s i n θ c o s θ ) [ 1 0 ] = [ c o s θ d ⋅ s i n θ ] \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cos\theta & -sin\theta\\ sin\theta & cos\theta \end{array}\right) \left[\begin{array}{lcr} 1\\0 \end{array}\right] =\left[\begin{array}{lcr} cos\theta\\d·sin\theta \end{array}\right] [xsys]=(cosθsinθ−sinθcosθ)[10]=[cosθd⋅sinθ]
可见旋转模式下 ( x s , y s ) (x_s,y_s) (xs,ys)恰好为正余弦函数，因此有正余弦函数的CORDIC解算过程为：令输入为 ( x 0 = 1 , y 0 = 0 , θ ∈ ( − π , π ) , n ) (x_0=1,y_0=0,\theta \in (-\pi,\pi),n) (x0=1,y0=0,θ∈(−π,π),n)，基于迭代公式 ( 2.8 ) ( 2.9 ) ( 2.10 ) ( 2.11 ) ( 2.12 ) (2.8)(2.9)(2.10)(2.11)(2.12) (2.8)(2.9)(2.10)(2.11)(2.12)，输出就为 ( x n ≈ c o s ( θ ) , y n ≈ s i n ( θ ) , △ α n ≈ θ ) (x_n≈cos(\theta),y_n≈sin(\theta),△α_n≈\theta) (xn≈cos(θ),yn≈sin(θ),△αn≈θ)
2.2. 向量模式—— y = a r c t a n ( x ) y=arctan(x) y=arctan(x) and l = x 2 + y 2 l=\sqrt{x^2 + y^2} l=x2+y2
预旋转算法中，若要求 P 0 P_0 P0旋转至x轴正半轴，即令 θ = − a r c t a n [ y 0 x 0 ] \theta=-arctan[\frac{y_0}{x_0}] θ=−arctan[x0y0]，因此有
[ x s y s ] = ( c o s θ − s i n θ s i n θ c o s θ ) [ x 0 y 0 ] → [ x s y s ] = [ x 0 2 + y 0 2 0 ] \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cos\theta & -sin\theta\\ sin\theta & cos\theta \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right] → \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left[\begin{array}{lcr} \sqrt{x_0^2+y_0^2} \\0 \end{array}\right] [xsys]=(cosθsinθ−sinθcosθ)[x0y0]→[xsys]=[x02+y020]
既然是旋转至x正半轴，所以 θ = − a r c t a n [ y 0 x 0 ] \theta=-arctan[\frac{y_0}{x_0}] θ=−arctan[x0y0]是已知的。但在RTL设计中，我们不能进行arctan运算，所以要在原算法上进行修改，不能出现 θ \theta θ
因此输入为 ( x 0 , y 0 , n ) (x_0,y_0,n) (x0,y0,n)，基于迭代公式 ( 2.8 ) ( 2.13 ) ( 2.14 ) ( 2.15 ) ( 2.12 ) (2.8)(2.13)(2.14)(2.15)(2.12) (2.8)(2.13)(2.14)(2.15)(2.12)，可得到输出 ( x n ≈ x 0 2 + y 0 2 , y n ≈ 0 , △ α n ≈ θ = − a r c t a n ( y 0 x 0 ) ) (x_n≈\sqrt{x_0^2+y_0^2},y_n≈0,△α_n≈\theta=-arctan(\frac{y_0}{x_0})) (xn≈x02+y02,yn≈0,△αn≈θ=−arctan(x0y0))
[ x p r e y p r e ] = { [ s i g n ( y 0 ) ⋅ y 0 − s i g n ( y 0 ) ⋅ x 0 ] , x 0 < 0 [ x 0 y 0 ] , e l s e (2.13) \begin{split} \left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right]= \left \{ \begin{array}{ll} \left[\begin{array}{lcr} sign(y_0)·y_0\\ -sign(y_0)·x_0 \end{array}\right], & x_0<0 \\ \\ \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right], & else\\ \end{array} \right. \end{split} \tag{2.13} [xpreypre]=⎩ ⎨ ⎧[sign(y0)⋅y0−sign(y0)⋅x0],[x0y0],x0<0else(2.13)
△ α i + 1 = △ α i + d i ⋅ a r c t a n ( 1 2 i ) , △ α 0 = △ α p r e = { − s i g n ( y 0 ) ⋅ π 2 , x 0 < 0 0 , e l s e (2.14) △α_{i+1}=△α_{i}+d_{i}·arctan(\frac{1}{2^{i}}), \begin{split} △α_0=△α_{pre}= \left \{ \begin{array}{ll} -sign(y_0)·\frac{\pi}{2}, & x_0<0 \\ \\ 0, & else\\ \end{array} \right. \end{split} \tag{2.14} △αi+1=△αi+di⋅arctan(2i1),△α0=△αpre=⎩ ⎨ ⎧−sign(y0)⋅2π,0,x0<0else(2.14)
d i + 1 = − s i g n ( y ^ i + 1 ) , d 0 = − s i g n ( y 0 ) (2.15) d_{i+1}=-sign(\hat{y}_{i+1}),d_0=-sign(y_0) \tag{2.15} di+1=−sign(y^i+1),d0=−sign(y0)(2.15)
3. 双曲系统
若CORDIC基本问题中，若遵循双曲旋转变换，那么就是双曲系统上的CORDIC问题。
[ x s y s ] = ( c o s h θ s i n h θ s i n h θ c o s h θ ) [ x 0 y 0 ] (3.1) \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cosh\theta & sinh\theta\\ sinh\theta & cosh\theta \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right] \tag{3.1} [xsys]=(coshθsinhθsinhθcoshθ)[x0y0](3.1)
其中
c o s h x = e x + e − x 2 coshx=\frac{e^x+e^{-x}}{2} coshx=2ex+e−x
s i n h x = e x − e − x 2 sinhx=\frac{e^x-e^{-x}}{2} sinhx=2ex−e−x
t a n h x = s i n h x c o s h x = e x − e − x e x + e − x tanhx=\frac{sinhx}{coshx}=\frac{e^x-e^{-x}}{e^x+e^{-x}} tanhx=coshxsinhx=ex+e−xex−e−x
非常类似于三角函数，其他内容见详解双曲函数(对比三角函数推导)
需要说明的是双曲变换没有明显的几何意义， O P s ⃗ \vec{OP_s} OPs并不是 O P 0 ⃗ \vec{OP_0} OP0旋转 θ \theta θ的结果。实际上，我们并不care到底旋转了多少度，我们只关心子变换 f i ( θ i ) f_i(\theta_i) fi(θi)的表达式。
经过计算实际旋转角度为
θ h y p = a r c t a n [ t a n h ( θ ) ⋅ ( x 0 2 − y 0 2 ) x 0 2 + y 0 2 + 2 t a n h ( θ ) ⋅ x 0 y 0 ] (3.2) \theta_{hyp}=arctan[\frac{tanh(\theta)·(x_0^2-y_0^2)}{x_0^2+y_0^2+2tanh(\theta)·x_0y_0}] \tag{3.2} θhyp=arctan[x02+y02+2tanh(θ)⋅x0y0tanh(θ)⋅(x02−y02)](3.2)
需要说明的是CORDIC算法详解(四)-CORDIC 算法之双曲系统及其数学应用中存在错误，(3.116)和(3.117)公式错误，不满足 y p / x p = t a n ( α ) y_p/x_p=tan(\alpha) yp/xp=tan(α),但(3.119)公式正确。
迭代求解如下，但注意，由于 a r c t a n h ( 1 ) = ∞ arctanh(1)=∞ arctanh(1)=∞所以双曲系统要求 i i i从1开始。
并且为了保证收敛性，每迭代到 i = 3 ⋅ k + 1 , k ∈ + Z i=3·k+1,k\in +\Z i=3⋅k+1,k∈+Z时，该次迭代都要再重复一次，即 i = 1 , 2 , 3 , 4 , 4 , 5 , 6 , 7 , 7 , 8 , 9 , 10 , 10 , . . . i=1,2,3,4,4,5,6,7,7,8,9,10,10,... i=1,2,3,4,4,5,6,7,7,8,9,10,10,...，以此类推。
[ x i + 1 y i + 1 ] = c o s h ( θ i ) [ 1 t a n h ( θ i ) t a n h ( θ i ) 1 ] [ x i y i ] , i = 1 , 2 , 3 , 4 , 4 , 5 , 6 , 7 , 7 , . . . \left[\begin{array}{lcr} x_{i+1}\\ y_{i+1} \end{array}\right] =cosh(\theta_i)\left[\begin{array}{lcr} 1 & tanh(\theta_i)\\ tanh(\theta_i) & 1 \end{array}\right] \left[\begin{array}{lcr} x_i\\y_i \end{array}\right], i=1,2,3,4,4,5,6,7,7,... [xi+1yi+1]=cosh(θi)[1tanh(θi)tanh(θi)1][xiyi],i=1,2,3,4,4,5,6,7,7,...
其中
θ i = d i ⋅ a r c t a n h ( 1 2 i ) , d i = { − 1 , ( θ 1 + . . . + θ i ) > θ 1 , e l s e \theta_i=d_i·arctanh(\frac{1}{2^{i}}), \ \ \ \begin{split} d_i= \left \{ \begin{array}{ll} -1, & (\theta_1+...+\theta_i)>\theta\\ 1, & else\\ \end{array} \right. \end{split} θi=di⋅arctanh(2i1), di={−1,1,(θ1+...+θi)>θelse
θ ≈ ∑ i = 0 n ( θ i ) ∈ ( − ∑ i = 0 ∞ [ a r c t a n h ( 1 2 i ) ] , ∑ i = 1 ∞ [ a r c t a n h ( 1 2 i ) ] ) ≈ ( − 1.0554693737354848 , 1.0554693737354848 ) (3.3) \theta≈\sum_{i=0}^n(\theta_i) \in (-\sum_{i=0}^∞[arctanh(\frac{1}{2^{i}})],\sum_{i=1}^∞[arctanh(\frac{1}{2^{i}})])≈(-1.0554693737354848,1.0554693737354848) \tag{3.3} θ≈i=0∑n(θi)∈(−i=0∑∞[arctanh(2i1)],i=1∑∞[arctanh(2i1)])≈(−1.0554693737354848,1.0554693737354848)(3.3)
将 c o s h ( θ i ) cosh(\theta_i) cosh(θi)连乘放在一起，其余的部分就可以使用RTL进行迭代了。
[ x n y n ] = ∏ i = 1 n c o s h ( θ i ) ⋅ ( ∏ i = 1 n ( 1 d i 2 i d i 2 i 1 ) [ x 1 y 1 ] ) = ∏ i = 0 n − 1 c o s h ( θ i ) ⋅ [ x ^ n y ^ n ] = ∏ i = 1 n c o s h ( θ i ) ⋅ ( 1 d n − 1 ⋅ 1 2 n − 1 d n − 1 ⋅ 1 2 n − 1 1 ) ⋅ [ x ^ n − 1 y ^ n − 1 ] \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =\prod_{i=1}^{n} cosh(\theta_i)· \left(\prod_{i=1}^{n}\left(\begin{array}{lcr} 1 & \frac{d_i}{2^{i}}\\ \frac{d_i}{2^{i}} & 1 \end{array}\right) \left[\begin{array}{lcr} x_1\\y_1 \end{array}\right]\right) =\prod_{i=0}^{n-1} cosh(\theta_i)·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \\ =\prod_{i=1}^{n} cosh(\theta_i)·\left(\begin{array}{lcr} 1 & d_{n-1}·\frac{1}{2^{n-1}}\\ d_{n-1}·\frac{1}{2^{n-1}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{n-1}\\ \hat{y}_{n-1} \end{array}\right] [xnyn]=i=1∏ncosh(θi)⋅(i=1∏n(12idi2idi1)[x1y1])=i=0∏n−1cosh(θi)⋅[x^ny^n]=i=1∏ncosh(θi)⋅(1dn−1⋅2n−11dn−1⋅2n−111)⋅[x^n−1y^n−1]
那么 ∏ i = 1 n c o s h ( θ i ) = ∏ i = 1 n c o s h [ a r c t a n h ( 1 2 i ) ] \prod_{i=1}^{n} cosh(\theta_i)=\prod_{i=1}^{n} cosh[arctanh(\frac{1}{2^{i}})] ∏i=1ncosh(θi)=∏i=1ncosh[arctanh(2i1)]也会越来越趋近于一个定值 1.205136358446461。迭代的次数足够多，可以直接用这个定值。
3.1. 迭代算法
综上所述，CORDIC解算过程为：输入为 ( x 0 , y 0 , θ ∈ ( − 1.0554693737354848 , 1.0554693737354848 ) , n ) (x_0,y_0,\theta \in (-1.0554693737354848,1.0554693737354848),n) (x0,y0,θ∈(−1.0554693737354848,1.0554693737354848),n)，基于迭代公式 ( 3.4 ) ( 3.5 ) ( 3.6 ) ( 3.7 ) (3.4)(3.5)(3.6)(3.7) (3.4)(3.5)(3.6)(3.7)，可得到输出 ( x n , y n , △ α n ) (x_n,y_n,△α_n) (xn,yn,△αn)
[ x ^ i + 1 y ^ i + 1 ] = ( 1 d i ⋅ 1 2 i d i ⋅ 1 2 i 1 ) ⋅ [ x ^ i y ^ i ] , [ x ^ 1 y ^ 1 ] = [ x 0 y 0 ] , i = 1 , 2 , 3 , 4 , 4 , . . . (3.4) \left[\begin{array}{lcr} \hat{x}_{i+1}\\ \hat{y}_{i+1} \end{array}\right] =\left(\begin{array}{lcr} 1 & d_{i}·\frac{1}{2^{i}}\\ d_{i}·\frac{1}{2^{i}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{i}\\ \hat{y}_{i} \end{array}\right], \left[\begin{array}{lcr} \hat{x}_1\\ \hat{y}_1 \end{array}\right]=\left[\begin{array}{lcr} x_0\\ y_0 \end{array}\right], i=1,2,3,4,4,... \tag{3.4} [x^i+1y^i+1]=(1di⋅2i1di⋅2i11)⋅[x^iy^i],[x^1y^1]=[x0y0],i=1,2,3,4,4,...(3.4)
△ α i + 1 = △ α i + d i ⋅ a r c t a n h ( 1 2 i ) , △ α 1 = 0 (3.5) △α_{i+1}=△α_{i}+d_{i}·arctanh(\frac{1}{2^{i}}), △α_1=0 \tag{3.5} △αi+1=△αi+di⋅arctanh(2i1),△α1=0(3.5)
d i + 1 = { − 1 , △ α i + 1 > θ 1 , e l s e , d 0 = s i g n ( θ ) (3.6) \begin{split} d_{i+1}= \left \{ \begin{array}{ll} -1, & △α_{i+1}>\theta\\ 1, & else\\ \end{array} \right. \end{split},d_0=sign(\theta) \tag{3.6} di+1={−1,1,△αi+1>θelse,d0=sign(θ)(3.6)
[ x n y n ] = 1.205136358446461 ⋅ [ x ^ n y ^ n ] (3.7) \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =1.205136358446461·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \tag{3.7} [xnyn]=1.205136358446461⋅[x^ny^n](3.7)
其中 n n n为迭代次数， △ α i △α_{i} △αi用于记录向量实际旋转角度的累计值，用该值判断应该旋转方向。
预旋转
双曲系统也是一样， θ \theta θ收敛范围未覆盖整个 R \R R区间，要使用CORDIC旋转，则要保证每一次CORDIC运算都使 θ \theta θ在式 ( 3.3 ) (3.3) (3.3)的范围内。否则会存在 θ \theta θ与 ∑ i = 0 ∞ [ d i ⋅ a r c t a n h ( 1 2 i ) ] \sum_{i=0}^∞[d_i·arctanh(\frac{1}{2^{i}})] ∑i=0∞[di⋅arctanh(2i1)]的递增误差。假设输入 θ = θ i n ∈ R \theta=\theta_{in} \in \R θ=θin∈R，有两种思路。
● 可以先作预旋转，给定一个 θ 0 \theta_0 θ0满足 ∣ θ 0 − θ i n ∣ < 1.0554693737354848 |\theta_0-\theta_{in}|<1.0554693737354848 ∣θ0−θin∣<1.0554693737354848，先使用 ( 3.1 ) (3.1) (3.1)完成一次 θ = θ 0 \theta=\theta_0 θ=θ0的预旋转，再完成一次 θ = s i g n ( θ i n ) ⋅ ∣ θ 0 − θ i n ∣ \theta=sign(\theta_{in})·|\theta_0-\theta_{in}| θ=sign(θin)⋅∣θ0−θin∣的CORDIC旋转。
但问题是 确定 θ 0 \theta_0 θ0后， c o s h ( θ 0 ) cosh(\theta_0) cosh(θ0)和 s i n h ( θ 0 ) sinh(\theta_0) sinh(θ0) 如何计算？要知道双曲三角函数不具备周期性，而且极限都是∞。所以对于RTL来说只能用查表法计算 c o s h ( θ 0 ) cosh(\theta_0) cosh(θ0)和 s i n h ( θ 0 ) sinh(\theta_0) sinh(θ0)，那为何不直接查表计算 c o s h ( θ ) cosh(\theta) cosh(θ)和 s i n h ( θ ) sinh(\theta) sinh(θ)呢？
● 另一种思路是令 n c = θ 0 / 1.0554693737354848 n_c=\theta_0/1.0554693737354848 nc=θ0/1.0554693737354848，然后执行 n c n_c nc次CORDIC算法，这样耗时太大。
综上所述，双曲系统预旋转CORDIC算法并不适用于RTL实现。所以推荐使用查找表的方法实现相关函数
但本文依旧给出上述的第一种预旋转CORDIC算法：CORDIC解算过程为：输入为 ( x 0 , y 0 , θ ∈ R , n ) (x_0,y_0,\theta \in \R,n) (x0,y0,θ∈R,n)，基于迭代公式 ( 3.8 ) ( 3.9 ) ( 3.10 ) ( 3.11 ) ( 3.12 ) (3.8)(3.9)(3.10)(3.11)(3.12) (3.8)(3.9)(3.10)(3.11)(3.12)，可得到输出 ( x n , y n , △ α n ) (x_n,y_n,△α_n) (xn,yn,△αn)
[ x ^ i + 1 y ^ i + 1 ] = ( 1 d i ⋅ 1 2 i d i ⋅ 1 2 i 1 ) ⋅ [ x ^ i y ^ i ] , [ x ^ 1 y ^ 1 ] = [ x p r e y p r e ] , i = 1 , 2 , 3 , 4 , 4 , . . . (3.8) \left[\begin{array}{lcr} \hat{x}_{i+1}\\ \hat{y}_{i+1} \end{array}\right] =\left(\begin{array}{lcr} 1 & d_{i}·\frac{1}{2^{i}}\\ d_{i}·\frac{1}{2^{i}} & 1 \end{array}\right)·\left[\begin{array}{lcr} \hat{x}_{i}\\ \hat{y}_{i} \end{array}\right], \left[\begin{array}{lcr} \hat{x}_1\\ \hat{y}_1 \end{array}\right]=\left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right], i=1,2,3,4,4,... \tag{3.8} [x^i+1y^i+1]=(1di⋅2i1di⋅2i11)⋅[x^iy^i],[x^1y^1]=[xpreypre],i=1,2,3,4,4,...(3.8)
[ x p r e y p r e ] = { [ c o s h ( k π 3 ) ⋅ x 0 + s i n h [ s i g n ( θ ) ⋅ k π 3 ] ⋅ y 0 s i n h [ s i g n ( θ ) ⋅ k π 3 ] ⋅ x 0 + c o s h ( k π 3 ) ⋅ y 0 ] , k π 3 < ∣ θ ∣ ≤ ( k + 1 ) π 3 , k ∈ + Z [ x 0 y 0 ] , e l s e (3.9) \begin{split} \left[\begin{array}{lcr} x_{pre}\\ y_{pre} \end{array}\right]= \left \{ \begin{array}{ll} \left[\begin{array}{lcr} cosh(\frac{k\pi}{3})·x_0+sinh[sign(\theta)·\frac{k\pi}{3}]·y_0 \\ sinh[sign(\theta)·\frac{k\pi}{3}]·x_0+cosh(\frac{k\pi}{3})·y_0 \end{array}\right], & \frac{k\pi}{3}<|\theta|≤\frac{(k+1)\pi}{3},k \in +\Z \\ \\ \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right], & else \\ \end{array} \right. \end{split} \tag{3.9} [xpreypre]=⎩ ⎨ ⎧[cosh(3kπ)⋅x0+sinh[sign(θ)⋅3kπ]⋅y0sinh[sign(θ)⋅3kπ]⋅x0+cosh(3kπ)⋅y0],[x0y0],3kπ<∣θ∣≤3(k+1)π,k∈+Zelse(3.9)
△ α i + 1 = △ α i + d i ⋅ a r c t a n h ( 1 2 i ) , △ α 1 = △ α p r e = { s i g n ( θ ) ⋅ k π 3 , k π 3 < ∣ θ ∣ ≤ ( k + 1 ) π 3 , k ∈ + Z 0 , e l s e (3.10) △α_{i+1}=△α_{i}+d_{i}·arctanh(\frac{1}{2^{i}}), △α_1=△α_{pre}= \left \{ \begin{array}{ll} sign(\theta)·\frac{k\pi}{3} , & \frac{k\pi}{3}<|\theta|≤\frac{(k+1)\pi}{3},k \in +\Z \\ \\ 0, & else \\ \end{array} \right. \tag{3.10} △αi+1=△αi+di⋅arctanh(2i1),△α1=△αpre=⎩ ⎨ ⎧sign(θ)⋅3kπ,0,3kπ<∣θ∣≤3(k+1)π,k∈+Zelse(3.10)
d i + 1 = { − 1 , △ α i + 1 > θ 1 , e l s e , d 1 = s i g n ( θ ) (3.11) \begin{split} d_{i+1}= \left \{ \begin{array}{ll} -1, & △α_{i+1}>\theta\\ 1, & else\\ \end{array} \right. \end{split},d_1=sign(\theta) \tag{3.11} di+1={−1,1,△αi+1>θelse,d1=sign(θ)(3.11)
[ x n y n ] = 1.205136358446461 ⋅ [ x ^ n y ^ n ] (3.12) \left[\begin{array}{lcr} x_{n}\\ y_{n} \end{array}\right] =1.205136358446461·\left[\begin{array}{lcr} \hat{x}_{n}\\ \hat{y}_{n} \end{array}\right] \tag{3.12} [xnyn]=1.205136358446461⋅[x^ny^n](3.12)
其中 n n n为迭代次数， △ α i △α_{i} △αi用于记录向量实际旋转角度的累计值，用该值判断应该旋转方向。
经python验证，上述算法正确
旋转模式—— y = c o s h ( x ) y=cosh(x) y=cosh(x)、 y = s i n h ( x ) y=sinh(x) y=sinh(x) and y = e x y=e^x y=ex
在迭代算法的基础上，令 P 0 ( x 0 = 1 , y 0 = 0 ) P_0(x_0=1,y_0=0) P0(x0=1,y0=0)，因此有
[ x s y s ] = ( c o s h θ s i n h θ s i n h θ c o s h θ ) [ 1 0 ] = [ c o s h θ s i n h θ ] \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cosh\theta & sinh\theta\\ sinh\theta & cosh\theta \end{array}\right) \left[\begin{array}{lcr} 1\\0 \end{array}\right] =\left[\begin{array}{lcr} cosh\theta\\ sinh\theta \end{array}\right] [xsys]=(coshθsinhθsinhθcoshθ)[10]=[coshθsinhθ]
= > x s + y s = c o s h ( θ ) + s i n h ( θ ) = e θ => x_s + y_s = cosh(\theta) + sinh(\theta) = e^{\theta} =>xs+ys=cosh(θ)+sinh(θ)=eθ
可见旋转模式下 ( x s , y s ) (x_s,y_s) (xs,ys)相加就是指数函数。下面给出不带预旋转的双曲CORDIC算法
因此指数函数的CORDIC解算过程为：输入为 ( x 0 = 1 , y 0 = 0 , θ ∈ ( − 1.0554693737354848 , 1.0554693737354848 ) , n ) (x_0=1,y_0=0,\theta \in (-1.0554693737354848,1.0554693737354848),n) (x0=1,y0=0,θ∈(−1.0554693737354848,1.0554693737354848),n)，基于迭代公式 ( 3.4 ) ( 3.5 ) ( 3.6 ) ( 3.7 ) (3.4)(3.5)(3.6)(3.7) (3.4)(3.5)(3.6)(3.7)，可得到输出 ( x n ≈ c o s h ( θ ) , y n ≈ s i n h ( θ ) , △ α n ≈ θ ) (x_n≈cosh(\theta),y_n≈sinh(\theta),△α_n≈\theta) (xn≈cosh(θ),yn≈sinh(θ),△αn≈θ)，进而得到 x n + y n ≈ e θ x_n+y_n≈e^{\theta} xn+yn≈eθ
3.2. 向量模式—— y = x y=\sqrt{x} y=x and y = ln x y=\ln{x} y=lnx
同样地，若要求 P 0 P_0 P0旋转至x轴正半轴，那么旋转角就是 − a r c t a n ( y 0 x 0 ) -arctan(\frac{y_0}{x_0}) −arctan(x0y0)，此处就需要用到 ( 3.2 ) (3.2) (3.2)式来求解 θ \theta θ
θ h y p = − a r c t a n ( y 0 x 0 ) → θ = − a r c t a n h [ y 0 x 0 ] → 0 < ∣ y 0 x 0 ∣ < t a n h ( 1.0554693737354848 ) ≈ 0.7839236595945456 (3.13) \theta_{hyp}=-arctan(\frac{y_0}{x_0})→ \theta=-arctanh[\frac{y_0}{x_0}] \\ → 0<|\frac{y_0}{x_0}|<tanh(1.0554693737354848)≈0.7839236595945456 \tag{3.13} θhyp=−arctan(x0y0)→θ=−arctanh[x0y0]→0<∣x0y0∣<tanh(1.0554693737354848)≈0.7839236595945456(3.13)
代入 ( 3.1 ) (3.1) (3.1)得到
[ x s y s ] = ( c o s h θ s i n h θ s i n h θ c o s h θ ) [ x 0 y 0 ] → [ x s y s ] = [ x 0 2 − y 0 2 0 ] \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left(\begin{array}{lcr} cosh\theta & sinh\theta\\ sinh\theta & cosh\theta \end{array}\right) \left[\begin{array}{lcr} x_0\\y_0 \end{array}\right] → \left[\begin{array}{lcr} x_s\\ y_s \end{array}\right] =\left[\begin{array}{lcr} \sqrt{x_0^2-y_0^2} \\0 \end{array}\right] [xsys]=(coshθsinhθsinhθcoshθ)[x0y0]→[xsys]=[x02−y020]
上述公式蕴含 ∣ y 0 x 0 ∣ ≤ 1 |\frac{y_0}{x_0}|≤1 ∣x0y0∣≤1，再结合 ( 3.13 ) (3.13) (3.13)，得到 0 < ∣ y 0 x 0 ∣ < 0.7839236595945456 0<|\frac{y_0}{x_0}|<0.7839236595945456 0<∣x0y0∣<0.7839236595945456
因此输入为 ( x 0 , y 0 , θ = − a r c t a n h [ y 0 x 0 ] , n ) (x_0,y_0, \theta=-arctanh[\frac{y_0}{x_0}],n) (x0,y0,θ=−arctanh[x0y0],n)，且满足 0 < ∣ y 0 x 0 ∣ < 0.7839236595945456 0<|\frac{y_0}{x_0}|<0.7839236595945456 0<∣x0y0∣<0.7839236595945456，基于迭代公式 ( 3.8 ) ( 3.9 ) ( 3.10 ) ( 3.11 ) ( 3.12 ) (3.8)(3.9)(3.10)(3.11)(3.12) (3.8)(3.9)(3.10)(3.11)(3.12)，可得到输出 ( x n ≈ x 0 2 − y 0 2 , y n ≈ 0 , △ α n ≈ θ = − a r c t a n h ( y 0 x 0 ) ) (x_n≈\sqrt{x_0^2-y_0^2},y_n≈0,△α_n≈\theta=-arctanh(\frac{y_0}{x_0})) (xn≈x02−y02,yn≈0,△αn≈θ=−arctanh(x0y0))
如果还满足 x 0 = a + 1 , y 0 = a − 1 x_0=a+1, y_0=a-1 x0=a+1,y0=a−1那么就可以得出开方与ln函数。注意此处 a a a的范围，即
0 < ∣ y 0 x 0 ∣ < 0.7839236595945456 → a ∈ ( 0.12112 , 8.25599 ) ， a ≠ 1 (3.14) 0<|\frac{y_0}{x_0}|<0.7839236595945456→ a \in (0.12112,8.25599)， a≠1 \tag{3.14} 0<∣x0y0∣<0.7839236595945456→a∈(0.12112,8.25599)，a=1(3.14)
则CORDIC解算过程为：输入为 ( x 0 = a + 1 , y 0 = a − 1 , n ) (x_0=a+1,y_0=a-1,n) (x0=a+1,y0=a−1,n)， a a a满足 ( 3.14 ) (3.14) (3.14)，并基于迭代公式 ( 3.4 ) ( 3.5 ) ( 2.14 ) ( 3.7 ) (3.4)(3.5)(2.14)(3.7) (3.4)(3.5)(2.14)(3.7)，可得到输出 ( x n ≈ 2 a , y n ≈ 0 , △ α n ≈ θ = − a r c t a n h ( a − 1 a + 1 ) = − 1 2 l n a ) (x_n≈2\sqrt{a},y_n≈0,△α_n≈\theta=-arctanh(\frac{a-1}{a+1})=-\frac{1}{2}lna) (xn≈2a,yn≈0,△αn≈θ=−arctanh(a+1a−1)=−21lna)
这篇关于坐标旋转数字计算法（Coordinate Rotation Digital Computer, CORDIC）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
