﻿ 基于邻域系统的智能车辆最优轨迹规划方法
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 智能系统学报  2019, Vol. 14 Issue (5): 1040-1047  DOI: 10.11992/tis.201805004 0

引用本文

WANG Xing, ZHAO Hailiang, WANG Zhigang. Optimal trajectory planning method of intelligent vehicles based on neighborhood system[J]. CAAI Transactions on Intelligent Systems, 2019, 14(5): 1040-1047. DOI: 10.11992/tis.201805004.

文章历史

Optimal trajectory planning method of intelligent vehicles based on neighborhood system
WANG Xing , ZHAO Hailiang , WANG Zhigang
Department of Information and Computation Science, School of Mathematics, Southwest Jiaotong University, Chengdu 610031, China
Abstract: This study mainly investigates the trajectory planning and control problems in the driving process of smart vehicles. First, based on the theory of neighborhood system, the dynamic control of intelligent vehicles on complex roads is transformed into a simple static control in the neighborhood, and then the optimal trajectory is selected in the neighborhood. The bending resistance index of the curve is defined by the integral of the curvature, and based on this, the optimal trajectory curve evaluation model and algorithm in the neighborhood is given. Finally, a satisfactory trajectory curve established by the interpolation method is taken as an example for simulation. The results show that this method has some advantages in selecting the steadiness and smoothness of the driving trajectory of an intelligent vehicle.
Key words: intelligent vehicle    automatic driving    trajectory planning    obstacle avoidance    neighborhood system    optimal trajectory    satisfaction curve    comprehensive evaluation

1 智能车辆的邻域系统

1)根据控制过程中观测变量的特点选择合适的邻域系统；

2)通过一些优化方法，在当前状态下确定一个满意的可行邻域；

3)在满意的邻域中，根据一些规划方法给予满意的决策，实施新的决策行为；

4)在可行邻域的持续时间内保持相同的行为，直到产生下一个决策；

5)上述过程是一个完整的决策循环。当一个周期完成后，一个新的循环开始依次执行步骤2)～4)。

2 最优轨迹曲线的评判指标 2.1 基于邻域系统的最优轨迹曲线

 $J\left[ {x\left( t \right)} \right] = \int_0^T {F\left( {t,x\left( t \right),x'\left( t \right)} \right){\rm{d}}t}$ (1)

1)用时较短；

2)行走路径较短；

3)乘客感到舒适。

2.2 最优轨迹曲线的评判指标

1)轨迹曲线的首尾位置是固定的，起始位置为当前邻域相对坐标系的原点，末位置为当前标准可行邻域的尾部中点向左平移一个车体长度。

2)轨迹曲线的方程为单值函数，即对于自变量x，其对应的函数值f(x)是唯一的。

3)轨迹曲线是连续的。

 $J = J\left[ {y\left(x \right)} \right]{\rm{ = }}\int_{{x_0}}^{{x_1}} {\sqrt {1{\rm{ + }}{{\left( {y'\left( x \right)} \right)}^2}} } {\rm{d}}x$ (2)

 $K = K\left[ {y\left( x \right)} \right] = \int_{{x_0}}^{{x_1}} {\frac{{\left| {y''\left( x \right)} \right|}}{{\sqrt {{{\left( {1{\rm{ + }}y'\left( x \right)} \right)}^3}} }}} {\rm{d}}x$ (3)

2.3 评判指标的标准化

 $J^* = 1{\rm{ - }}\displaystyle\frac{{\displaystyle\frac{{\int_{{x_0}}^{{x_1}} {\sqrt {1{\rm{ + }}{{\left( {y'\left( x \right)} \right)}^2}} } dx}}{{{\rm{1 + }}\int_{{x_0}}^{{x_1}} {\sqrt {1{\rm{ + }}{{\left( {y'\left( x \right)} \right)}^2}} } dx}} - \displaystyle\frac{{\left( {{x_1} - {x_0}} \right)}}{{{\rm{1 + }}\left( {{x_1} - {x_0}} \right)}}}}{{1 - \displaystyle\frac{{\left( {{x_1} - {x_0}} \right)}}{{{\rm{1 + }}\left( {{x_1} - {x_0}} \right)}}}}$ (4)

 $J_{\min }^*{\rm{ = }}\frac{{\left( {{x_1} - {x_0}} \right)}}{{{\rm{1 + }}\left( {{x_1} - {x_0}} \right)}}$ (5)

 $J^*=1{\rm{ - }}\frac{{\frac{J}{{{\rm{1 + }}J}} - J_{\min }^*}}{{1 - J_{\min }^*}} = \frac{1}{{\left( {{\rm{1 + }}J} \right)\left( {1 - J_{\min }^*} \right)}}$ (6)

 $K^* = 1{\rm{ - }}\displaystyle\frac{{\int_{{x_0}}^{{x_1}} {\displaystyle\frac{{\left| {y''\left( x \right)} \right|}}{{\sqrt {{{\left( {1{\rm{ + }}y'\left( x \right)} \right)}^3}} }}} {\rm{d}}x}}{{1{\rm{ + }}\int_{{x_0}}^{{x_1}} {\displaystyle\frac{{\left| {y''\left( x \right)} \right|}}{{\sqrt {{{\left( {1{\rm{ + }}y'\left( x \right)} \right)}^3}} }}} {\rm{d}}x}}$ (7)

 $K^* = 1{\rm{ - }}\frac{K}{{1 + K}} = \frac{1}{{1 + K}}$ (8)

 $C^* = {w_1} \cdot J^*{\rm{ + }}{w_2} \cdot K^*$ (9)

3 最优轨迹曲线模型及其求解方法 3.1 最优轨迹曲线模型及其满意条件

 $\left\{ \begin{array}{l} \min T \\ \min J\left[ {y\left( x \right)} \right] \\ \min K\left[ {y\left( x \right)} \right] \\ \end{array} \right.$ (10)

 $\left\{ \begin{array}{l} y\left( {{x_0}} \right) \subset \varGamma \left( {{x_0}} \right) \subset Q \\ T \leqslant {T_{\max }} \\ \end{array} \right.$ (11)

 $B = \left\{ {y\left( x \right){\rm{|}}{w_1} \cdot J^*\left[ {y\left( x \right)} \right]{\rm{ + }}{w_2} \cdot K^*\left[ {y\left( x \right)} \right] \geqslant p} \right\}$ (12)

3.2 满意曲线的求解算法

1)确定初始值。

 $\left\{ {\begin{array}{*{20}{l}} {y\left( {{x_1}} \right){\rm{ = }}0,}&{{x_1}{\rm{ = }}0}\\ {y\left( {{x_2}} \right){\rm{ = }}{y_2},}&{{x_1} < {x_2} < {x_3}}\\ {y\left( {{x_3}} \right) = s,}&{{x_3}{\rm{ = }}H - l} \end{array}} \right.$ (13)

 $\left\{ \begin{array}{l} y'\left( {{x_1}} \right){\rm{ = }}0\\ y'\left( {{x_2}} \right) = y_2'\\ y'\left( {{x_3}} \right){\rm{ = }}0 \end{array} \right.$ (14)

 ${k_0} = s/h - l$ (15)
 Download: 图 5 智能车的邻域坐标系 Fig. 5 Neighborhood of intelligent vehicles coordinate system

2)计算轨迹曲线指标。

 ${H_{2n + 1}}\left( x \right){\rm{ = }}\sum\limits_{i = 1}^n {{H_i}\left( x \right)} \cdot {y_i}{\rm{ + }}\sum\limits_{k = 1}^n {{h_k}\left( x \right)} \cdot {y_k}'$ (16)

3)计算导数值。

 ${k_i} = \left( {\alpha + ti} \right){k_0}$ (17)

1)节点坐标为 $\left( {\displaystyle\frac{{H - l}}{2},\frac{s}{2}} \right)$ ，为曲线的中点。

2)节点一阶导数为 $\displaystyle\frac{{3s}}{{2\left( {H{\rm{ - l}}} \right)}}$

3.3 满意曲线的理论分析

 $\zeta = \sup C^*\left[ {y\left( x \right)} \right]$ (18)

 $C_{1}^* \leqslant C_{2}^* \leqslant \cdots \leqslant C_{{i - 1}}^* \leqslant C_{i}^*$ (19)

 $\mathop {\lim}\limits_{i \to \infty } {\rm{ }}C_{i}^* = \zeta$ (20)

4 仿真结果以及同其他文献的对比

 $\begin{array}{l} {y_{\rm{A}}}\left( x \right){\rm{ = }} - 1.21 \cdot {10^{ - 8}} \cdot {x^7} + 2.08 \cdot {10^{ - 6}} \cdot {x^6} - 1.40 \cdot {10^{ - 4}} \cdot {x^5} {\rm{ + }} \\ \qquad\qquad 4.65 \cdot {10^{ - 3}} \cdot {x^4} - 7.63 \cdot {10^{ - 2}} \cdot {x^3} + 5.09 \cdot {10^{ - 1}} \cdot {x^2} \\ \end{array}$
 $\begin{array}{l} {y_{\rm{B}}}\left( x \right){\rm{ = }} 3.13 \cdot {10^{ - 7}} \cdot {x^5} - 3.13 \cdot {10^{ - 5}} \cdot {x^4} + {\rm{ }} \\\quad\qquad 8.75 \cdot {10^{ - 4}} \cdot {x^3} - 2.50 \cdot {10^{ - 3}} \cdot {x^2} \\ \end{array}$
 ${y_{\rm{C}}}\left( x \right){\rm{ = }}\frac{3}{2} \cdot \sin \left( {\frac{{{\text{π}} \cdot x}}{{40}} - \frac{\pi }{2}} \right) + \frac{3}{2}$

 $J\left[ {{y_A}\left( x \right)} \right] = 44.832\text{、}J\left[ {{y_B}\left( x \right)} \right] = 40.304\text{、}J\left[ {{y_C}\left( x \right)} \right] = 40.138$

 $K\left[y_{A}(x)\right]\!=\!333\;18\text{、}\!K\left[y_{B}(x)\right]\!=\!0.402\;2\text{、}\!K\left[y_{C}(x)\right]\!=\!0.993\;1$

 $J^{*}\left[y_{A}(x)\right]=0.903\text{、}J^{*}\left[y_{B}(x)\right]=0.992\text{、}J^{*}\left[y_{C}(x)\right]=0.996$
 $K^{*}\left[y_{A}(x)\right]=0.231\text{、}K^{*}\left[y_{B}(x)\right]=0.713\text{、}K^{*}\left[y_{C}(x)\right]=0.501$

 $C^*\left[y_{A}(x)\right]=0.433\text{、}C^*\left[y_{B}(x)\right]=0.797\text{、}C^{*}\left[y_{C}(x)\right]=0.649$

 $C^{*}\left[y_{A}(x)\right]=0.499\text{、}C^{*}\left[y_{B}(x)\right]=0.824\text{、}C^{*}\left[y_{C}(x)\right]=0.699$

 $C^{*}\left[y_{A}(x)\right]=0.533\text{、}C^{*}\left[y_{B}(x)\right]=0.838\text{、}C^{*}\left[y_{C}(x)\right]=0.723$