四旋翼的非线性模型预测控制(MPC)

1. 四旋翼的动力学模型

这里不加推导地给出四旋翼的动力学模型:
m [ x ¨ y ¨ z ¨ ] = [ ∑ i = 1 4 F i ( cos ⁡ φ sin ⁡ θ cos ⁡ ψ + sin ⁡ φ sin ⁡ ψ ) ∑ i = 1 4 F i ( cos ⁡ φ sin ⁡ θ sin ⁡ ψ − sin ⁡ φ cos ⁡ ψ ) ∑ i = 1 4 F i cos ⁡ φ cos ⁡ θ − m g ] (1) m \left[ \begin{matrix} \ddot x \\ \ddot y \\ \ddot z \end{matrix} \right] = \left[ \begin{matrix} \sum_{i=1}^4 F_i \left( \cos \varphi \sin \theta \cos \psi + \sin \varphi \sin \psi \right) \\ \sum_{i=1}^4 F_i \left( \cos \varphi \sin \theta \sin \psi - \sin \varphi \cos \psi \right) \\ \sum_{i=1}^4 F_i \cos \varphi \cos \theta - mg \end{matrix} \right] \tag{1} m x¨y¨z¨ = i=14Fi(cosφsinθcosψ+sinφsinψ)i=14Fi(cosφsinθsinψsinφcosψ)i=14Ficosφcosθmg (1) [ J x p ˙ J y q ˙ J z r ˙ ] = [ l ( F 2 − F 2 ) − ( J z − J y ) q r l ( F 1 − F 3 ) − ( J x − J z ) p r M D 1 − M D 2 + M D 3 − M D 4 − ( J y − J x ) p q ] (2) \left[ \begin{matrix} J_x \dot p \\ J_y \dot q \\ J_z \dot r \end{matrix} \right] = \left[ \begin{matrix} l \left( F_2 - F_2 \right) - \left( J_z - J_y \right) qr \\ l \left( F_1 - F_3 \right) - \left( J_x - J_z \right) pr \\ M_{D1} - M_{D2} + M_{D3} - M_{D4} - \left( J_y - J_x \right) pq \end{matrix} \right] \tag{2} Jxp˙Jyq˙Jzr˙ = l(F2F2)(JzJy)qrl(F1F3)(JxJz)prMD1MD2+MD3MD4(JyJx)pq (2) [ φ ˙ θ ˙ ψ ˙ ] = [ 1 sin ⁡ φ tan ⁡ θ cos ⁡ φ tan ⁡ θ 0 cos ⁡ φ − sin ⁡ φ 0 sin ⁡ φ cos ⁡ θ cos ⁡ φ cos ⁡ θ ] [ p q r ] \left[ \begin{matrix} \dot \varphi \\ \dot \theta \\ \dot \psi \end{matrix} \right] = \left[ \begin{matrix} 1 & \sin \varphi \tan \theta & \cos \varphi \tan \theta \\ 0 & \cos \varphi & - \sin \varphi \\ 0 & \frac{\sin \varphi}{\cos \theta} & \frac{\cos \varphi}{\cos \theta} \end{matrix} \right] \left[ \begin{matrix} p \\ q \\ r \end{matrix} \right] φ˙θ˙ψ˙ = 100sinφtanθcosφcosθsinφcosφtanθsinφcosθcosφ pqr 其中 φ , θ , ψ \varphi, \theta, \psi φ,θ,ψ均为机体欧拉角, p , q , r p,q,r p,q,r为欧拉角的角速度。(关于更加详细的四旋翼建模过程可以参考笔者博客Backstepping反步法控制四旋翼无人机(一)

2. 模型预测控制(MPC)基本方程

模型预测控制(Model Predictive Control, MPC)本质上是最优控制的一种,尝试在约束条件下达到最优的系统表现。它通过预测模型在未来一段时间内的表现来进行优化控制,通过不断迭代,最终使系统逐渐优化。

一般地,对于一个系统
x ( k + 1 ) = A X ( k ) + B U ( k ) A ∈ R n × n , B ∈ R n × p x \left( k+1 \right) = AX(k)+ BU(k) \\ A \in R^{n \times n}, \quad B \in R^{n \times p} x(k+1)=AXk)+BU(k)ARn×n,BRn×p建立向量
X k = [ x ( k ∣ k ) x ( k + 1 ∣ k ) ⋮ x ( k + N − 1 ∣ k ) ] , U k = [ u ( k ∣ k ) u ( k + 1 ∣ k ) ⋮ u ( k + N − 1 ∣ k ) ] X_k = \left[ \begin{matrix} x(k \vert k) \\ x(k+1 \vert k) \\ \vdots \\ x(k+N-1 \vert k) \end{matrix} \right], \quad U_k = \left[ \begin{matrix} u(k \vert k) \\ u(k+1 \vert k) \\ \vdots \\ u(k+N-1 \vert k) \end{matrix} \right] Xk= x(kk)x(k+1∣k)x(k+N1∣k) ,Uk= u(kk)u(k+1∣k)u(k+N1∣k) 同时假设输出即为 y = x y=x y=x,那么误差即为
E = y − x = 0 E = y-x = 0 E=yx=0另外,性能指标具有如下形式
J = ∑ i = 0 N − 1 [ x ( k + i ∣ k ) T Q x ( k + i ∣ k ) + u ( k + i ∣ k ) t R u ( k + i ∣ k ) ] + x ( k + N ) T F x ( k + N ) J = \sum_{i=0}^{N-1} \left[ x(k+i \vert k)^T Q x(k+i \vert k) + u(k+i \vert k)^t R u(k+i \vert k)\right] + x(k+N)^TFx(k+N) J=i=0N1[x(k+ik)TQx(k+ik)+u(k+ik)tRu(k+ik)]+x(k+N)TFx(k+N)其中 Q , R , F Q,R,F Q,R,F分别为状态控制量静态误差权重。通过调整三者的相对大小,可以实现对三者的不同权重调整,系统表现也会有所不同。

Q ˉ = [ Q Q ⋱ Q F ] , R ˉ = [ R R ⋱ R ] \bar Q = \left[ \begin{matrix} Q & \quad & \quad & \quad & \quad \\ \quad & Q & \quad & \quad & \quad \\ \quad & \quad & \ddots & \quad & \quad \\ \quad & \quad & \quad & Q & \quad \\ \quad & \quad & \quad & \quad & F \end{matrix} \right], \quad \bar R = \left[ \begin{matrix} R & \quad & \quad & \quad \\ \quad & R & \quad & \quad \\ \quad & \quad & \ddots & \quad \\ \quad & \quad & \quad & R \end{matrix} \right] Qˉ= QQQF ,Rˉ= RRR M = [ I A A 2 ⋮ A N ] , C = [ 0 0 ⋯ 0 B 0 ⋯ 0 A B B ⋯ 0 ⋮ ⋮ ⋱ ⋮ A N − 1 B A N − 2 B ⋯ B ] M = \left[ \begin{matrix} I \\ A \\ A^2 \\ \vdots \\ A^N \end{matrix} \right], \quad C = \left[ \begin{matrix} 0 & 0 & \cdots & 0 \\ B & 0 & \cdots & 0 \\ AB & B & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ A^{N-1}B & A^{N-2}B & \cdots & B \end{matrix} \right] M= IAA2AN ,C= 0BABAN1B00BAN2B000B G = M T Q ˉ , E = M T Q ˉ C , H = C T Q ˉ C + R ˉ G = M^T \bar Q, \quad E = M^T \bar Q C,\quad H = C^T \bar Q C + \bar R G=MTQˉ,E=MTQˉC,H=CTQˉC+Rˉ那么性能指标可以写为
J = x k T G x k + 2 x k T E U k + U k T H U k (3) J = x_k^T G x_k + 2 x_k^T E U_k +U_k^THU_k \tag{3} J=xkTGxk+2xkTEUk+UkTHUk(3)该式即为二次规划的一般形式。二次规划的一般形式为
min ⁡ ( z T Q z + c T z ) (4) \min \left( z^T Q z + c^T z \right) \tag{4} min(zTQz+cTz)(4)(3)式中的 U U U即对应(4)式中的 z z z

3. 非线性模型预测控制

不难看出,以上给出的方法仅仅适用于线性系统。当系统为非线性时, A , B A,B A,B矩阵将很难写出,进而 M , C , G , E , H M,C,G,E,H M,C,G,E,H矩阵无法计算。这部分针对非线性系统给出了模型预测控制的算法。

很容易看出,非线性模型预测控制的难点就在于如何获取矩阵 A , B A,B A,B。不妨设系统的期望值为 x r x_r xr,利用泰勒展开将系统的非线性表达式在期望值附近展开,随后利用拉格朗日方程得到矩阵 A , B A,B A,B即可。

具体计算公式如下:
X ˙ = f i ( x i , u j ) \dot X = f_i \left(x_i, u_j \right) X˙=fi(xi,uj) A = [ ∂ f 1 ∂ x 1 ∂ f 1 ∂ x 2 ⋯ ∂ f 1 ∂ x n ∂ f 2 ∂ x 1 ∂ f 2 ∂ x 2 ⋯ ∂ f 2 ∂ x n ⋮ ⋮ ⋱ ⋮ ∂ f n ∂ x 1 ∂ f n ∂ x 2 ⋯ ∂ f n ∂ x n ] x i = x r A = \left[ \begin{matrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \cdots & \frac{\partial f_n}{\partial x_n} \end{matrix} \right] _{x_i = x_r} A= x1f1x1f2x1fnx2f1x2f2x2fnxnf1xnf2xnfn xi=xr B = [ ∂ f 1 ∂ u 1 ∂ f 1 ∂ u 2 ⋯ ∂ f 1 ∂ u p ∂ f 2 ∂ u 1 ∂ f 2 ∂ u 2 ⋯ ∂ f 2 ∂ u p ⋮ ⋮ ⋱ ⋮ ∂ f n ∂ u 1 ∂ f n ∂ u 2 ⋯ ∂ f n ∂ u p ] u i = u r B = \left[ \begin{matrix} \frac{\partial f_1}{\partial u_1} & \frac{\partial f_1}{\partial u_2} & \cdots & \frac{\partial f_1}{\partial u_p} \\ \frac{\partial f_2}{\partial u_1} & \frac{\partial f_2}{\partial u_2} & \cdots & \frac{\partial f_2}{\partial u_p} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial u_1} & \frac{\partial f_n}{\partial u_2} & \cdots & \frac{\partial f_n}{\partial u_p} \end{matrix} \right] _{u_i = u_r} B= u1f1u1f2u1fnu2f1u2f2u2fnupf1upf2upfn ui=ur值得注意的是, A , B A,B A,B中的 x i , u i x_i, u_i xi,ui应当代入期望值 x r , u r x_r, u_r xr,ur

下面以四旋翼为例进行计算。

设状态量为
x 1 = φ x 2 = x ˙ 1 = φ ˙ x 3 = θ x 4 = x ˙ 3 = θ ˙ x 5 = ψ x 6 = x ˙ 5 = ψ ˙ x 7 = x x 8 = x ˙ 7 = x ˙ x 9 = y x 10 = x ˙ 9 = y ˙ x 11 = z x 12 = x ˙ 11 = z ˙ \begin{aligned} x_1 &= \varphi \\ x_2 &= \dot x_1 = \dot \varphi \\ x_3 &= \theta \\ x_4 &= \dot x_3 = \dot \theta \\ x_5 &= \psi \\ x_6 &= \dot x_5 = \dot \psi \\ x_7 &= x \\ x_8 &= \dot x_7 = \dot x \\ x_9 &= y \\ x_{10} &= \dot x_9 = \dot y \\ x_{11} &= z \\ x_{12} &= \dot x_{11} = \dot z \end{aligned} x1x2x3x4x5x6x7x8x9x10x11x12=φ=x˙1=φ˙=θ=x˙3=θ˙=ψ=x˙5=ψ˙=x=x˙7=x˙=y=x˙9=y˙=z=x˙11=z˙对(1)(2)求导有
X ˙ = f i + g U \dot X = f_i + g U X˙=fi+gU A = [ 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 J y − J z J x x 6 0 J y − J z J x x 4 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 J z − J x J y x 6 0 0 0 J z − J x J y x 2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 J x − J y J z x 4 0 J x − J y J z x 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 cos ⁡ x 1 sin ⁡ x 5 − sin ⁡ x 1 sin ⁡ x 3 cos ⁡ x 5 m U 1 0 cos ⁡ x 1 cos ⁡ x 3 cos ⁡ x 5 m U 1 0 sin ⁡ x 1 cos ⁡ x 5 − cos ⁡ x 1 sin ⁡ x 3 sin ⁡ x 5 m U 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 − sin ⁡ x 1 sin ⁡ x 3 sin ⁡ x 5 − cos ⁡ x 1 cos ⁡ x 5 m U 1 0 cos ⁡ x 1 cos ⁡ x 3 sin ⁡ x 5 m U 1 0 cos ⁡ x 1 sin ⁡ x 3 cos ⁡ x 5 + sin ⁡ x 1 sin ⁡ x 5 m U 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 − sin ⁡ x 1 cos ⁡ x 3 m U 1 0 − cos ⁡ x 1 sin ⁡ x 3 m U 1 0 0 0 0 0 0 0 0 0 ] A = \left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{J_y - J_z}{J_x}x_6 & 0 & \frac{J_y - J_z}{J_x}x_4 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{J_z - J_x}{J_y}x_6 & 0 & 0 & 0 & \frac{J_z - J_x}{J_y}x_2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{J_x - J_y}{J_z}x_4 & 0 & \frac{J_x - J_y}{J_z}x_2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \frac{\cos x_1 \sin x_5 - \sin x_1 \sin x_3 \cos x_5}{m}U_1 & 0 & \frac{\cos x_1 \cos x_3 \cos x_5}{m}U_1 & 0 & \frac{\sin x_1 \cos x_5 - \cos x_1 \sin x_3 \sin x_5}{m}U_1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \frac{-\sin x_1 \sin x_3 \sin x_5 - \cos x_1 \cos x_5}{m}U_1 & 0 & \frac{\cos x_1 \cos x_3 \sin x_5}{m}U_1 & 0 & \frac{\cos x_1 \sin x_3 \cos x_5 + \sin x_1 \sin x_5}{m}U_1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \frac{-\sin x_1 \cos x_3}{m}U_1 & 0 & \frac{- \cos x_1 \sin x_3}{m}U_1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right] A= 0000000mcosx1sinx5sinx1sinx3cosx5U10msinx1sinx3sinx5cosx1cosx5U10msinx1cosx3U1100JyJzJxx60JzJxJyx40000000000000mcosx1cosx3cosx5U10mcosx1cosx3sinx5U10mcosx1sinx3U10JxJyJzx6100JzJxJyx20000000000000msinx1cosx5cosx1sinx3sinx5U10mcosx1sinx3cosx5+sinx1sinx5U1000JxJyJzx40JyJzJxx210000000000000000000000000100000000000000000000000001000000000000000000000000010 B = [ 0 0 0 0 0 l J x 0 0 0 0 0 0 0 0 l J y 0 0 0 0 0 0 0 0 1 J z 0 0 0 0 cos ⁡ x 1 sin ⁡ x 3 cos ⁡ x 5 + sin ⁡ x 1 sin ⁡ x 5 m 0 0 0 0 0 0 0 cos ⁡ x 1 sin ⁡ x 3 sin ⁡ x 5 − sin ⁡ x 1 cos ⁡ x 5 m 0 0 0 0 0 0 0 cos ⁡ x 1 cos ⁡ x 3 m 0 0 0 ] B = \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & \frac{l}{J_x} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{l}{J_y} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{J_z} \\ 0 & 0 & 0 & 0 \\ \frac{\cos x_1 \sin x_3 \cos x_5 + \sin x_1 \sin x_5}{m} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{\cos x_1 \sin x_3 \sin x_5 - \sin x_1 \cos x_5}{m} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{\cos x_1 \cos x_3}{m} & 0 & 0 & 0 \end{matrix} \right] B= 0000000mcosx1sinx3cosx5+sinx1sinx50mcosx1sinx3sinx5sinx1cosx50mcosx1cosx30Jxl0000000000000Jyl0000000000000Jz1000000 可见 n = 12 , p = 4 n=12, p=4 n=12,p=4

代入期望值
x 1 r = φ r = 0.02 x 2 r = φ ˙ r = 0 x 3 r = θ r = 0.01 x 4 r = θ ˙ r = 0 x 5 r = ψ r = 0.1 x 6 r = ψ ˙ r = 0 x 7 r = x r = 5 x 8 r = x ˙ r = 0 x 9 r = y r = 3 x 10 r = y ˙ r = 0 x 11 r = z r = − 2 x 12 r = z ˙ r = 2 \begin{aligned} x_{1r} &= \varphi_r = 0.02 \\ x_{2r} &= \dot \varphi_r = 0 \\ x_{3r} &= \theta_r = 0.01 \\ x_{4r} &= \dot \theta_r = 0 \\ x_{5r} &= \psi_r = 0.1 \\ x_{6r} &= \dot \psi_r = 0 \\ x_{7r} &= x_r = 5 \\ x_{8r} &= \dot x_r = 0 \\ x_{9r} &= y_r = 3 \\ x_{10r} &= \dot y_r = 0 \\ x_{11r} &= z_r = -2 \\ x_{12r} &= \dot z_r = 2 \end{aligned} x1rx2rx3rx4rx5rx6rx7rx8rx9rx10rx11rx12r=φr=0.02=φ˙r=0=θr=0.01=θ˙r=0=ψr=0.1=ψ˙r=0=xr=5=x˙r=0=yr=3=y˙r=0=zr=2=z˙r=2以及其他常量
m = 1 g = 9.8 l = 0.3 J x = 2.4 × 1 0 − 2 J y = 1.4 × 1 0 − 2 J z = 1.4 × 1 0 − 2 \begin{aligned} m &= 1 \\ g &= 9.8 \\ l &= 0.3 \\ J_x &= 2.4 \times 10^{-2} \\ J_y &= 1.4 \times 10^{-2} \\ J_z &= 1.4 \times 10^{-2} \\ \end{aligned} mglJxJyJz=1=9.8=0.3=2.4×102=1.4×102=1.4×102共迭代 k = 20 k=20 k=20步,迭代步长 N = 5 N = 5 N=5
则计算得到
A = [ 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0.0996 0 0.9948 0 0.0189 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 − 0.9948 0 0.0998 0 0.0119 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 − 0.02 0 − 0.01 0 0 0 0 0 0 0 0 0 ] A = \left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0.0996 & 0 & 0.9948 & 0 & 0.0189 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ -0.9948 & 0 & 0.0998 & 0 & 0.0119 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -0.02 & 0 & -0.01 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right] A= 00000000.099600.994800.0210000000000000000000.994800.099800.0100100000000000000000.018900.011900000010000000000000000000000000100000000000000000000000001000000000000000000000000010 B = [ 0 0 0 0 0 12.5 0 0 0 0 0 0 0 0 21.4286 0 0 0 0 0 0 0 0 71.4286 0 0 0 0 0.0119 0 0 0 0 0 0 0 0.0119 0 0 0 0 0 0 0 0.9998 0 0 0 ] B = \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 12.5 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 21.4286 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 71.4286 \\ 0 & 0 & 0 & 0 \\ 0.0119 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0.0119 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0.9998 & 0 & 0 & 0 \end{matrix} \right] B= 00000000.011900.011900.9998012.5000000000000021.4286000000000000071.4286000000
得到状态变量如下图所示。
MPC控制结果
注意,图中表示的是6个状态量的误差值。可以看出,6个状态量很快地达到了稳定态,证明了模型预测控制的有效性。

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转载自blog.csdn.net/weixin_58399148/article/details/125252983