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【Paper】2015_异构无人机群鲁棒一致性协议设计_孙长银

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原文地址：[1]孫長銀,余瑤,張?zhí)m.異構無人機群魯棒一致性協(xié)議設計[J].中國科學:技術科學,2015,45(06):573-582.

2015_異構無人機群魯棒一致性協(xié)議設計_孫長銀

4 分布式魯棒一致性協(xié)議設計
6 Simulations
- Condition 1
- Condition 2
- Condition 3
- Condition 4

先把無人機的通信拓撲圖放出來，這樣在接下來的公式分析時能夠便于理解。

4 分布式魯棒一致性協(xié)議設計

Laplacian Matrix 是
$\left[\begin{matrix} 0 && 0 & 0 & 0 & 0 \\ \\ 0 && 1 & -1 & 0 & 0 \\ 0 && 0 & 1 & -1 & 0 \\ -1 && 0 & 0 & 0 & 0 \\ 0 && 0 & 0 & -1 & 1 \\ \end{matrix}\right]$

$s1=[s1,1Ts2,1T?sM,1T]s_1 = \left[\begin{matrix} s^T_{1,1} \\ s^T_{2,1} \\ \vdots\\ s^T_{M,1} \end{matrix}\right]$

$y1=[y1Ts2T?sMT]y_1 = \left[\begin{matrix} y^T_{1} \\ s^T_{2} \\ \vdots\\ s^T_{M} \end{matrix}\right]$

第1步：
$s˙i,1=∑j=1Maij(y˙i?y˙j)+bi(y˙i?r˙)=∑j=1Maij(gi(Θi)xi,2?gj(Θj)xj,2)+bi(gi(Θi)xi2?r˙)=?∑j=1Maijgj(Θj)xj,2?bir˙+(∑j=1Mai,j+bi)gi(Θi)xi,2=\begin{aligned} \dot{s}_{i,1} &= \sum_{j=1}^{M} a_{ij} (\dot{y}_i - \dot{y}_j) + b_i (\dot{y}_i - \dot{r}) \\ &= \sum_{j=1}^{M} a_{ij} (~g_i(\Theta_i)x_{i,2} - g_j(\Theta_j)x_{j,2}~) + b_i (~g_i(\Theta_i)x_{i_2} - \dot{r}~) \\ &= -\sum_{j=1}^{M} a_{ij} g_j (\Theta_j) x_{j,2} - b_i \dot{r} + (\sum_{j=1}^{M} a_{i,j} + b_i) g_i(\Theta_i)x_{i,2}\\ &= \end{aligned}$

6 Simulations

$M$ 個跟隨者的動態(tài)模型為
$x˙i,1=gi(Θi)xi,2x˙i,2=ui+Φi(xi,1,xi,2,κi)yi=xi,1\begin{aligned} &{\dot{x}_{i,1}} = g_i(\varTheta_i) x_{i,2} \\ &\dot{x}_{i,2} = u_i + \varPhi_i(x_{i,1}, x_{i,2}, \kappa_i) \\ &y_i = x_{i,1} \end{aligned}$

轉換成矩陣寫法：
$[x˙1,1x˙2,1x˙3,1x˙4,1]=[x1,2x2,2x3,2x4,2][x˙1,2x˙2,2x˙3,2x˙4,2]=[u1u2u3u4]+[Φ1(x1,1,x1,2,κ1)Φ2(x2,1,x2,2,κ2)Φ3(x3,1,x3,2,κ3)Φ4(x4,1,x4,2,κ4)]\begin{aligned}& \left[\begin{matrix} \dot{x}_{1,1} \\ \dot{x}_{2,1} \\ \dot{x}_{3,1} \\ \dot{x}_{4,1} \\ \end{matrix}\right]= \left[\begin{matrix} {x}_{1,2} \\ {x}_{2,2} \\ {x}_{3,2} \\ {x}_{4,2} \\ \end{matrix}\right] \\& \left[\begin{matrix} \dot{x}_{1,2} \\ \dot{x}_{2,2} \\ \dot{x}_{3,2} \\ \dot{x}_{4,2} \\ \end{matrix}\right]= \left[\begin{matrix} {u}_{1} \\ {u}_{2} \\ {u}_{3} \\ {u}_{4} \\ \end{matrix}\right] + \left[\begin{matrix} {\varPhi}_{1}(x_{1,1}, x_{1,2}, \kappa_1) \\ {\varPhi}_{2}(x_{2,1}, x_{2,2}, \kappa_2) \\ {\varPhi}_{3}(x_{3,1}, x_{3,2}, \kappa_3) \\ {\varPhi}_{4}(x_{4,1}, x_{4,2}, \kappa_4) \\ \end{matrix}\right] \end{aligned}$

整個一致性協(xié)議為：
$ui=?ρsi,2+fwiwi=?(1+ρs)si,2si,1=∑j=1Maij(yi?yj)+bi(yi?r)si,2=xi,2?1(di+bi)gi?1(Θi)(?ρsi,1+∑j=1Maijgj(Θj)xj,2+bir˙)\begin{aligned} & u_i = -\rho \red{s_{i,2}} + f \blue{w_i} \\ & \blue{w_i} = - (1+\frac{\rho}{s}) \red{s_{i,2}} \\ & \green{s_{i,1}} = \sum_{j=1}^{M} a_{ij} (y_i - y_j) + b_i(y_i - r) \\ & \red{s_{i,2}} = x_{i,2} - \frac{1}{(d_i + b_i)} g_i^{-1}(\varTheta_i) (-\rho \green{s_{i,1}} + \sum_{j=1}^{M} a_{ij} g_j (\varTheta_j) x_{j,2} + b_i \dot{r}) \end{aligned}$

轉換成矩陣寫法：
$[u1u2u3u4]=?ρ[s1,2s2,2s3,2s4,2]+f[w1w2w3w4][w1w2w3w4]=?(1+ρs)[s1,2s2,2s3,2s4,2][s1,1s2,1s3,1s4,1]=L?[y1y2y3y4]+[b1?(y1?r)b2?(y2?r)b3?(y3?r)b4?(y4?r)]=L?[x1,1x2,1x3,1x4,1]+[b1b2b3b4][x1,1?rx2,1?rx3,1?rx4,1?r][s1,2s2,2s3,2s4,2]=[x1,2x2,2x3,2x4,2]?[1d1+b1(?ρs1,1+xj,2+b1r˙)1d2+b2(?ρs2,1+xj,2+b2r˙)1d3+b31d4+b4]\begin{aligned} \left[\begin{matrix} {u}_{1} \\ {u}_{2} \\ {u}_{3} \\ {u}_{4} \\ \end{matrix}\right]&=-\rho \left[\begin{matrix} {s}_{1,2} \\ {s}_{2,2} \\ {s}_{3,2} \\ {s}_{4,2} \\ \end{matrix}\right]+f \left[\begin{matrix} {w}_{1} \\ {w}_{2} \\ {w}_{3} \\ {w}_{4} \\ \end{matrix}\right] \\ \left[\begin{matrix} {w}_{1} \\ {w}_{2} \\ {w}_{3} \\ {w}_{4} \\ \end{matrix}\right]&=-(1+\frac{\rho}{s}) \left[\begin{matrix} {s}_{1,2} \\ {s}_{2,2} \\ {s}_{3,2} \\ {s}_{4,2} \\ \end{matrix}\right] \\ \left[\begin{matrix} {s}_{1,1} \\ {s}_{2,1} \\ {s}_{3,1} \\ {s}_{4,1} \\ \end{matrix}\right]&= L \cdot \left[\begin{matrix} {y}_{1} \\ {y}_{2} \\ {y}_{3} \\ {y}_{4} \\ \end{matrix}\right]+ \left[\begin{matrix} b_1 \cdot ({y}_{1}-r) \\ b_2 \cdot ({y}_{2}-r) \\ b_3 \cdot ({y}_{3}-r) \\ b_4 \cdot ({y}_{4}-r) \\ \end{matrix}\right] \\&= L \cdot \left[\begin{matrix} {x}_{1,1} \\ {x}_{2,1} \\ {x}_{3,1} \\ {x}_{4,1} \\ \end{matrix}\right]+ \left[\begin{matrix} _{1} \\ &_{2} \\ &&_{3} \\ &&&_{4} \\ \end{matrix}\right] \left[\begin{matrix} {x}_{1,1}-r \\ {x}_{2,1}-r \\ {x}_{3,1}-r \\ {x}_{4,1}-r \\ \end{matrix}\right] \\ \left[\begin{matrix} {s}_{1,2} \\ {s}_{2,2} \\ {s}_{3,2} \\ {s}_{4,2} \\ \end{matrix}\right]&= \left[\begin{matrix} {x}_{1,2} \\ {x}_{2,2} \\ {x}_{3,2} \\ {x}_{4,2} \\ \end{matrix}\right]- \left[\begin{matrix} \frac{1}{d_1+b_1}(-\rho s_{1,1} + x_{j,2} + b_1 \dot{r}) \\ \frac{1}{d_2+b_2}(-\rho s_{2,1} + x_{j,2} + b_2 \dot{r}) \\ \frac{1}{d_3+b_3} \\ \frac{1}{d_4+b_4} \\ \end{matrix}\right] \end{aligned}$

仿真部分也給出了干擾的表述方式：
$Φi(xi,1,xi,2,κi)≤∣xi,1∣+∣xi,2∣+rand(5)\begin{aligned} \varPhi_i(x_{i,1}, x_{i,2}, \kappa_i) \le |x_{i,1}| + |x_{i,2}| + \text{rand}(5) \end{aligned}$

把兩個公式合并一下（忽略轉換矩陣的作用）：
$x˙i,1=xi,2x˙i,2=ui+Φi(xi,1,xi,2,κi)=ui+∣xi,1∣+∣xi,2∣+rand(5)=?ρsi,2+fwi+∣xi,1∣+∣xi,2∣+rand(5)=?ρsi,2+f[?(1+ρs)si,2]+∣xi,1∣+∣xi,2∣+rand(5)\begin{aligned} \dot{x}_{i,1} &= x_{i,2} \\ \dot{x}_{i,2} &= u_i + \varPhi_i(x_{i,1}, x_{i,2}, \kappa_i) \\ &= u_i + |x_{i,1}| + |x_{i,2}| + \text{rand}(5) \\ &= -\rho s_{i,2} + f w_i + |x_{i,1}| + |x_{i,2}| + \text{rand}(5) \\ &= -\rho s_{i,2} + f [-(1+\frac{\rho}{s})s_{i,2}] + |x_{i,1}| + |x_{i,2}| + \text{rand}(5) \\ \end{aligned}$

Condition 1

Condition 2

Condition 3

Condition 4

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