基于角速度的欧拉角沃尔什近似输出方法

摘要

本发明公开了一种基于角速度的欧拉角沃尔什近似输出方法，用于解决现有的飞行器机动飞行时欧拉角输出精度差的技术问题。技术方案是通过引入多个参数并将滚转、俯仰、偏航角速度按照沃尔什函数多项式展开，按照依次求解俯仰角、滚转角、偏航角，直接对欧拉角的表达式进行高阶逼近积分，使得欧拉角的求解按照超线性逼近，保证了确定欧拉角的时间更新迭代计算精度，从而提高了惯性设备输出飞行姿态的准确性。

主权项

1.一种基于角速度的欧拉角沃尔什近似输出方法，其特征在于包括以下步骤：步骤1、(a)根据欧拉方程：式中：分别指滚转、俯仰、偏航角；p，q，r分别为滚转、俯仰、偏航角速度；全文参数定义相同；这三个欧拉角的计算按照依次求解俯仰角、滚转角、偏航角的步骤进行；滚转、俯仰、偏航角速度p，q，r的展开式分别为p(t)＝[p₀ p₁ L p_n-1 p_n][ξ₀(t) ξ₁(t)L ξ_n-1(t) ξ_n(t)]^Tq(t)＝[q₀ q₁ L q_n-1 q_n][ξ₀(t) ξ₁(t)L ξ_n-1(t) ξ_n(t)]^Tr(t)＝[r₀ r₁ L r_n-1 r_n][ξ₀(t) ξ₁(t)L ξ_n-1(t) ξ_n(t)]^T其中，${\xi }_k\left(t\right)=\underset{j=0}{\overset{\rho -1}{\Pi }}\mathrm{sgn}\left\{\cos \right[k_j{2}^j\mathrm{\pi t}/\left(\mathrm{NT}\right)\left]\right\}$(0≤t≤NT，k＝0，1，2，L)为沃尔什函数(Walsh Function)；k_j为0或1-k的二进制表示式的二进制数值，ρ为二进制值位数，sgn表示符号函数；T为采样周期，全文符号定义相同；(b)俯仰角的时间更新求解式为：式中：$a_1=1+\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]{\int }_{\mathrm{kT}}^{(k+1)T}\left[\xi \left(t\right){\xi }^T\left(t\right)\right]\mathrm{dt}{H}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$+\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]{\int }_{\mathrm{kT}}^{(k+1)T}\left[\xi \left(t\right){\xi }^T\left(t\right)\right]\mathrm{dt}{H}^T{\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]}^T$$+\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]{\int }_{\mathrm{kT}}^{(k+1)T}\left[\xi \left(t\right){\xi }^T\left(t\right)\right]\mathrm{dt}{H}^T{\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]}^T$$-\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}.$${\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}_{\mathrm{kT}}{H}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$-\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}.$${\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}_{\mathrm{kT}}{H}^T{\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]}^T$$-\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}.$${\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}_{\mathrm{kT}}{H}^T{\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]}^T$$+0.25{\left\{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}\right\}}^2$$+0.25{\left\{\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}\right\}}^2$$-0.25{\left\{\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}\right\}}^2$$a_2=\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}$$-0.5\begin{array}{c}\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]\end{array}{\int }_{\mathrm{kT}}^{(k+1)T}\left[\xi \left(t\right){\xi }^T\left(t\right)\right]{\mathrm{dtH}}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$+0.5\begin{array}{c}\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]\end{array}H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}$$·{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}_{\mathrm{kT}}{H}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$a_3=\left[\begin{array}{ccccc}{r}_0& r_1& L& r_{n-1}& r_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T$$+0.5\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]{\int }_{\mathrm{kT}}^{(k+1)T}\left[\xi \left(t\right){\xi }^T\left(t\right)\right]\mathrm{dt}{H}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$-0.5\left[\begin{array}{ccccc}{q}_0& q_1& L& q_{n-1}& q_n\end{array}\right]H{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}^T{|}_{\mathrm{kT}}^{(k+1)T}$$·{\left[\begin{array}{ccccc}{\xi }_0\left(t\right)& {\xi }_1\left(t\right)& L& {\xi }_n\left(t\right)& {\xi }_{n+1}\left(t\right)\end{array}\right]}_{\mathrm{kT}}{H}^T{\left[\begin{array}{ccccc}{p}_0& p_1& L& p_{n-1}& p_n\end{array}\right]}^T$$H=\left[\begin{array}{ccccc}\frac{1}{2}& & -\frac{2}{n}{I}_{\frac{n}{8}}& & \\ & O& & -\frac{1}{n}{I}_{\frac{n}{4}}& \\ \frac{2}{n}{I}_{\frac{n}{8}}& & 0_{\frac{n}{8}}& & -\frac{1}{2n}{I}_{\frac{n}{2}}\\ & \frac{1}{n}{I}_{\frac{n}{4}}& & 0_{\frac{n}{4}}& \\ & & \frac{1}{2n}{I}_{\frac{n}{2}}& & 0_{\frac{n}{2}}\end{array}\right]$