汽车运动学模型的线性化推导过程

1. 方法

假设有一非线性微分方程
y ˙ = f ( x , u ) \dot y = f(x,u) y˙=f(x,u)
且有工作点满足 y ˙ 0 = f ( x 0 , u 0 ) \dot y_0=f(x_0,u_0) y˙0=f(x0,u0),在该点处对其进行泰勒级数展开,忽略二阶及以上高阶项,有

y ˙ ≈ f ( x 0 , u 0 ) + ∂ f ( x , u ) ∂ x ( x − x 0 ) + ∂ f ( x , u ) ∂ u ( u − u 0 ) \dot y \approx f(x_0,u_0)+\frac{\partial f(x,u)}{\partial x}(x-x_0)+\frac{\partial f(x,u)}{\partial u}(u-u_0) y˙f(x0,u0)+xf(x,u)(xx0)+uf(x,u)(uu0)
由此可得
y ˙ − y ˙ 0 = ∂ f ( x , u ) ∂ x ( x − x 0 ) + ∂ f ( x , u ) ∂ u ( u − u 0 ) \dot y-\dot y_0=\frac{\partial f(x,u)}{\partial x}(x-x_0)+\frac{\partial f(x,u)}{\partial u}(u-u_0) y˙y˙0=xf(x,u)(xx0)+uf(x,u)(uu0)
定义 Δ y ˙ = y ˙ − y ˙ 0 , Δ x = x − x 0 , Δ u = u − u 0 , J x = ∂ f ( x , u ) ∂ x , J u = ∂ f ( x , u ) ∂ u \Delta \dot y=\dot y-\dot y_0,\Delta x=x-x_0,\Delta u=u-u_0,J_x=\frac{\partial f(x,u)}{\partial x},J_u=\frac{\partial f(x,u)}{\partial u} Δy˙=y˙y˙0,Δx=xx0,Δu=uu0Jx=xf(x,u),Ju=uf(x,u),上式可写为
Δ y ˙ = J x Δ x + J u Δ u \Delta \dot y=J_x \Delta x+J_u \Delta u Δy˙=JxΔx+JuΔu
显然 J x , J u J_x,J_u Jx,Ju为常数(定常矩阵),由此可将非线性系统转化为线性定常系统。

2.实例

以车辆二自由度模型为例,对该系统进行建模并将其线性化。
假设车辆速度为v,航偏角为 φ \varphi φ,前轮转角为 δ \delta δ,车辆轴距为l,根据运动学原理可知

v x = v c o s φ = x ˙ v_x=vcos\varphi=\dot{x} vx=vcosφ=x˙

v y = v s i n φ = y ˙ v_y=vsin\varphi=\dot{y} vy=vsinφ=y˙

ω = v t a n δ l = φ ˙ \omega=\frac{vtan\delta}{l}=\dot{\varphi} ω=lvtanδ=φ˙

X ˙ = [ x ˙ y ˙ φ ˙ ] = [ cos ⁡ φ sin ⁡ φ tan ⁡ δ l ] v = [ f 1 f 2 f 3 ] = f ( X , u ) \dot X = \left[ \begin{array}{l} {\dot x}\\ {\dot y}\\ {\dot \varphi } \end{array} \right] = \left[ \begin{array}{l} \cos \varphi \\ \sin \varphi \\ \frac{{\tan \delta }}{l} \end{array} \right]v = \left[ \begin{array}{l} {f_1}\\ {f_2}\\ {f_3} \end{array} \right] = f\left( {X,u} \right) X˙=x˙y˙φ˙=cosφsinφltanδv=f1f2f3=f(X,u)
对上式在点 X r = [ x r , y r , φ r ] T , u r = [ v r , δ r ] T X_r=[x_r,y_r,\varphi_r]^T,u_r=[v_r,\delta_r]^T Xr=[xr,yr,φr]T,ur=[vr,δr]T处进行线性化处理, u = [ v , δ ] T u=[v,\delta]^T u=[v,δ]T为输入变量,由此可得
X ˙ − X ˙ r = [ x ˙ − x ˙ r y ˙ − y ˙ r φ ˙ − φ ˙ r ] = ( ∂ f 1 ∂ x ∂ f 1 ∂ y ∂ f 1 ∂ φ ∂ f 2 ∂ x ∂ f 2 ∂ y ∂ f 2 ∂ φ ∂ f 3 ∂ x ∂ f 3 ∂ y ∂ f 3 ∂ φ ) [ x − x r y − y r φ − φ r ] + ( ∂ f 1 ∂ v ∂ f 1 ∂ δ ∂ f 2 ∂ v ∂ f 2 ∂ δ ∂ f 3 ∂ v ∂ f 3 ∂ δ ) [ v − v r δ − δ r ] \dot X - {\dot X_r} = \left[ \begin{array}{l} \dot x - {{\dot x}_r}\\ \dot y - {{\dot y}_r}\\ \dot \varphi - {{\dot \varphi }_r} \end{array} \right] = \left( {\begin{array}{l} {\frac{{\partial {f_1}}}{{\partial x}}}&{\frac{{\partial {f_1}}}{{\partial y}}}&{\frac{{\partial {f_1}}}{{\partial \varphi }}}\\ {\frac{{\partial {f_2}}}{{\partial x}}}&{\frac{{\partial {f_2}}}{{\partial y}}}&{\frac{{\partial {f_2}}}{{\partial \varphi }}}\\ {\frac{{\partial {f_3}}}{{\partial x}}}&{\frac{{\partial {f_3}}}{{\partial y}}}&{\frac{{\partial {f_3}}}{{\partial \varphi }}} \end{array}} \right)\left[ \begin{array}{l} x - {x_r}\\ y - {y_r}\\ \varphi - {\varphi _r} \end{array} \right] + \left( {\begin{array}{l} {\frac{{\partial {f_1}}}{{\partial v}}}&{\frac{{\partial {f_1}}}{{\partial \delta }}}\\ {\frac{{\partial {f_2}}}{{\partial v}}}&{\frac{{\partial {f_2}}}{{\partial \delta }}}\\ {\frac{{\partial {f_3}}}{{\partial v}}}&{\frac{{\partial {f_3}}}{{\partial \delta }}} \end{array}} \right)\left[ {\begin{array}{l} {v - {v_r}}\\ {\delta - {\delta _r}} \end{array}} \right] X˙X˙r=x˙x˙ry˙y˙rφ˙φ˙r=xf1xf2xf3yf1yf2yf3φf1φf2φf3xxryyrφφr+vf1vf2vf3δf1δf2δf3[vvrδδr]
由此可得
Δ X ˙ = A m Δ X + B m Δ u \Delta \dot X = A_m\Delta X + B_m\Delta u ΔX˙=AmΔX+BmΔu
其中 A m = [ 0 0 − sin ⁡ φ v 0 0 cos ⁡ φ v 0 0 0 ] A _m= \left[ {\begin{array}{|} 0&0&{ - \sin \varphi v}\\ 0&0&{\cos \varphi v}\\ 0&0&0 \end{array}} \right] Am=000000sinφvcosφv0, B m = [ cos ⁡ φ 0 sin ⁡ φ 0 tan ⁡ δ l v l cos ⁡ 2 δ ] B _m= \left[ {\begin{array}{|} {\cos \varphi }&0\\ {\sin \varphi }&0\\ {\frac{{\tan \delta }}{l}}&{\frac{v}{{l{{\cos }^2}\delta }}} \end{array}} \right] Bm=cosφsinφltanδ00lcos2δv,对其进行离散化可得
Δ X ( k + 1 ) = A Δ X ( k ) + B Δ u ( k ) \Delta X(k+1) = A\Delta X(k) + B\Delta u(k) ΔX(k+1)=AΔX(k)+BΔu(k)
其中 A = [ 1 0 − sin ⁡ φ v T 0 1 cos ⁡ φ v T 0 0 1 ] A = \left[ {\begin{array}{|} 1&0&{ - \sin \varphi vT}\\ 0&1&{\cos \varphi vT}\\ 0&0&1 \end{array}} \right] A=100010sinφvTcosφvT1, B = [ cos ⁡ φ T 0 sin ⁡ φ T 0 tan ⁡ δ T l v T l cos ⁡ 2 δ ] , T B = \left[ {\begin{array}{|} {\cos \varphi T }&0\\ {\sin \varphi T}&0\\ {\frac{{\tan \delta T}}{l}}&{\frac{vT}{{l{{\cos }^2}\delta }}} \end{array}} \right],T B=cosφTsinφTltanδT00lcos2δvT,T为采样时间,至此完成了该非线性系统的线性化。

3.说明

1.只有当在静态点附近非常小的变化时,线性化才有意义,否则线性化误差会非常大,即线性化的小偏差思想;
2.当工作点不同时,线性化的结果也不同;
3.线性化后的方程一般为增量形式。

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