【Paper】2017_Distributed control for high-speed trains movements
Y. Zhao and T. Wang, “Distributed control for high-speed trains movements,” 2017 29th Chinese Control And Decision Conference (CCDC), 2017, pp. 7591-7596, doi: 10.1109/CCDC.2017.7978561.
文章目錄
- Abstract
- 1 Introduction
- 2 Preliminaries
- 3 Distributed control for a single high-speed train
- 3.1 The model of a high-speed train
- 3.2 The distributed controller design
- 3.3 Example 1
- 4 Conclusion
Abstract
1 Introduction
2 Preliminaries
3 Distributed control for a single high-speed train
3.1 The model of a high-speed train
假設有 n1n_1n1? 輛車,把這些車分為三類:首車,中間車,尾車。動力學模型分別如下:
{x˙i(t)=vi,i=1,2,?,n1m1v˙1(t)=u1?k(x1?x2?l)?(c0+c1v1+c2v12)m1miv˙i(t)=ui+k(xi?1?xi?l)?k(xi?xi+1?l)?(c0+c1vi+c2vi2)mi,i=2,3,?,n1?1mn1v˙n1(t)=un1+k(xn1?1?xn1?l)?(c0+c1vn1+c2vn12)mn1(1)\left\{\begin{aligned} \dot{x}_i(t) &= v_i, ~~~~ i = 1,2,\cdots, n_1 \\ m_1 \dot{v}_1(t) &= u_1 - k(x_1 - x_2 - l) - (c_0 + c_1 v_1 + c_2 v_1^2) m_1 \\ m_i \dot{v}_i(t) &= u_i + k(x_{i-1} - x_{i} - l) - k(x_{i} - x_{i+1} - l) - (c_0 + c_1 v_i + c_2 v_i^2) m_i, ~~~~ i=2,3,\cdots,n_1-1 \\ m_{n_1} \dot{v}_{n_1}(t) &= u_{n_1} + k(x_{n_1-1} - x_{n_1} - l) - (c_0 + c_1 v_{n_1} + c_2 v_{n_1}^2) m_{n_1} \end{aligned}\right. \tag{1}????????????x˙i?(t)m1?v˙1?(t)mi?v˙i?(t)mn1??v˙n1??(t)?=vi?,????i=1,2,?,n1?=u1??k(x1??x2??l)?(c0?+c1?v1?+c2?v12?)m1?=ui?+k(xi?1??xi??l)?k(xi??xi+1??l)?(c0?+c1?vi?+c2?vi2?)mi?,????i=2,3,?,n1??1=un1??+k(xn1??1??xn1???l)?(c0?+c1?vn1??+c2?vn1?2?)mn1???(1)
符號表示為:
xix_ixi?:表示位置,
mim_imi?:表示質量,
lll:表示原始彈性長度。
當達到平衡狀態后,用以下符號進行表示。
首先速度均達到一致,公式表示為
vˉ1=vˉ2=?=vˉn1=vr\bar{v}_1 = \bar{v}_2 = \cdots = \bar{v}_{n_1} = v_rvˉ1?=vˉ2?=?=vˉn1??=vr?
距離達到均等間隔,公式表示為
xˉi?1?xˉi=l\bar{x}_{i-1} - \bar{x}_i = lxˉi?1??xˉi?=l
xˉi=vˉit\bar{x}_i = \bar{v}_i txˉi?=vˉi?t
平衡狀態時的控制輸入為
uˉi=c0mi+c1mivr+c2mivr2(2)\bar{u}_i = c_0 m_i + c_1 m_i v_r + c_2 m_i v_r^2 \tag{2}uˉi?=c0?mi?+c1?mi?vr?+c2?mi?vr2?(2)
這個公式相當于 Davis formula 和 Newton’s second law (uˉi=Ri/mi\bar{u}_i = R_i/m_iuˉi?=Ri?/mi?)的變形。
接下來定義誤差動力學方程。
令 x^i(t)=xi(t)?xˉi(t)\hat{x}_{i}(t) = {x}_{i}(t) - \bar{x}_{i}(t)x^i?(t)=xi?(t)?xˉi?(t),v^i(t)=vi(t)?vr(t)\hat{v}_{i}(t) = {v}_{i}(t) - {v}_{r}(t)v^i?(t)=vi?(t)?vr?(t),u^i(t)=ui(t)?uˉi(t)\hat{u}_{i}(t) = {u}_{i}(t) - \bar{u}_{i}(t)u^i?(t)=ui?(t)?uˉi?(t)。整合公式(1)和(2)就可以得到如下誤差動力學方程。
{x^˙i=v^i,i=1,2,?,n1miv^˙i=u^i+k∑j=1n1adij(x^j?x^i)?2c2mivrv^i?c1miv^i,i=2,3,?,n1?1(3)\left\{\begin{aligned} \dot{\hat{x}}_i &= \hat{v}_i, ~~~~ i = 1,2,\cdots, n_1 \\ m_i \dot{\hat{v}}_i &= \hat{u}_i + k \sum_{j=1}^{n_1} a_{dij} (\hat{x}_{j} - \hat{x}_{i}) - 2 c_2 m_i v_r \hat{v}_i - c_1 m_i \hat{v}_i, ~~~~ i=2,3,\cdots,n_1-1 \\ \end{aligned}\right. \tag{3}????????x^˙i?mi?v^˙i??=v^i?,????i=1,2,?,n1?=u^i?+kj=1∑n1??adij?(x^j??x^i?)?2c2?mi?vr?v^i??c1?mi?v^i?,????i=2,3,?,n1??1?(3)
符號表示為:
adij>0a_{dij} > 0adij?>0:表示 i,ji,ji,j 是一組,否則為 adij=0a_{dij} = 0adij?=0。
3.2 The distributed controller design
設計控制律:
u^i=ui(t)?uˉi=u1i(t)+u2i(t)+u3i(t)+u4i(t)=?kddi(xi?xˉi)?miδidvi(vi(t)?vr)+miδi∑j=in1avij(vj(t)?vi(t))?mi∑j=1n1adij?xiUij(4)\begin{aligned} \hat{u}_i &= u_i(t) - \bar{u}_i \\ &= u_{1i}(t) + u_{2i}(t) + u_{3i}(t) + u_{4i}(t) \\ &= -k d_{di} (x_i - \bar{x}_i) \\ &- m_i \delta_i d_{vi} (v_i(t) - v_r) \\ &+ m_i \delta_i \sum_{j=i}^{n_1} a_{vij} (v_j(t) - v_i(t)) \\ &- m_i \sum_{j=1}^{n_1} a_{dij} \nabla_{xi} U_{ij} \end{aligned} \tag{4}u^i??=ui?(t)?uˉi?=u1i?(t)+u2i?(t)+u3i?(t)+u4i?(t)=?kddi?(xi??xˉi?)?mi?δi?dvi?(vi?(t)?vr?)+mi?δi?j=i∑n1??avij?(vj?(t)?vi?(t))?mi?j=1∑n1??adij??xi?Uij??(4)
控制律分別由以下四部分組成:
u1i(t)=?kddi(xi?xˉi)u_{1i}(t) = -k d_{di} (x_i - \bar{x}_i)u1i?(t)=?kddi?(xi??xˉi?)
u2i(t)=?miδidvi(vi(t)?vr)u_{2i}(t) = - m_i \delta_i d_{vi} (v_i(t) - v_r)u2i?(t)=?mi?δi?dvi?(vi?(t)?vr?)
u3i(t)=+miδi∑j=in1avij(vj(t)?vi(t))u_{3i}(t) = + m_i \delta_i \sum_{j=i}^{n_1} a_{vij} (v_j(t) - v_i(t))u3i?(t)=+mi?δi?j=i∑n1??avij?(vj?(t)?vi?(t))
u4i(t)=?mi∑j=1n1adij?xiUiju_{4i}(t) = - m_i \sum_{j=1}^{n_1} a_{dij} \nabla_{xi} U_{ij}u4i?(t)=?mi?j=1∑n1??adij??xi?Uij?
3.3 Example 1
4 Conclusion
總結
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