#  SCIENTIA SINICA Informationis, Volume 48 , Issue 7 : 947-962(2018) https://doi.org/10.1360/N112017-00282

## Virtual equivalent system theory for adaptive control and simulation verification More info
• ReceivedDec 24, 2017
• AcceptedFeb 14, 2018
• PublishedJul 12, 2018
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Appendix

proof 本引理是引理3的特殊情形, 在引理3中令$\omega(k)=0$, 即得本引理结论.

proof 首先, 我们知道 \begin{equation}\sum _{k=1}^{k=N}\|\tilde{\phi}(k)\|\le \sum _{k=1}^{k=N}\|\tilde{\phi}_1(k)\| + \sum _{k=1}^{k=N}\|\tilde{\phi}_2(k)\|. \tag{44}\end{equation} 下面采用反证法进行证明, 假设 $\|\tilde{\phi}(k)\|$ 无界. 那么必有无穷子列 $\|\tilde{\phi}(p_{k})\|\to~\infty$ 满足 \begin{equation}\sum _{k=1}^{k=N}\|\tilde{\phi}(p_k)\|\le \sum _{k=1}^{k=N}\|\tilde{\phi}_1(p_k)\| + \sum _{k=1}^{k=N}\|\tilde{\phi}_2(p_k)\|. \tag{45}\end{equation} 接下来分别考察(A2)式右面两项的性质. 首先考察第一项, 根据 Stolz 定理 以及 $\|\tilde{\phi}_1(p_k)\|<\infty$, 知道 \begin{equation}\frac{\sum _{k=1}^{k=N}\|\tilde{\phi}_1(p_k)\|}{\sum _{k=1}^{k=N}\|\tilde{\phi}(p_k)\|}\to\frac{\|\tilde{\phi}_1(p_k)\|}{\|\tilde{\phi}(p_k)\|}\to 0. \tag{46}\end{equation} 接下来考察第二项, 注意到 $~\|\tilde{\phi}(k)\|=0,~k\le~0$, 那么, \begin{equation}\sum _{k=1}^{k=N}\|\tilde{\phi}(k-i)\| \le \sum _{k=1}^{k=N}\|\tilde{\phi}(k)\|, i=1,\ldots,s. \tag{47}\end{equation} 因此, \begin{equation}\begin{split} \sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(k)\|+\cdots+\|\tilde{\phi}(k-s)\|) \le(s+1)\sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(k)\|). \end{split} \tag{48}\end{equation} 上式结合本引理的给定条件导致 \begin{equation}\sum _{k=1}^{k=N}\|\tilde{\phi}_2(k)\|=o\left(\sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(k)\|)\right). \tag{49}\end{equation} 进一步, 考虑到收敛序列及其子序列的极限相同, 我们有 \begin{equation}\sum _{k=1}^{k=N}\|\tilde{\phi}_2(p_k)\|=o\left(\sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(p_k)\|)\right). \tag{50}\end{equation} 考虑到 (A2) 和(A3)式, 并利用夹逼原理有 \begin{equation}\frac{\sum _{k=1}^{k=N}\|\tilde{\phi}(p_k)\|}{\sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(p_k)\|)}\to 0. \tag{51}\end{equation} 上式与如下事实相矛盾(根据 Stolz 定理): \begin{equation}\frac{\sum _{k=1}^{k=N}(\|\tilde{\phi}(p_k)\|)}{\sum _{k=1}^{k=N}(\alpha+\|\tilde{\phi}(p_k)\|)}\to 1. \tag{52}\end{equation} 因此原假设不成立. 从而证得结论, 即 $\|\tilde{\phi}(k)\|<\infty$.

proof 用数学归纳法给出证明. 首先, 我们知道, 对于 $k~\le~0~$ 有 \begin{equation}y(k)=y_1(k)+y_2(k)+y_3(k), u(k)=u_1(k)+u_2(k)+u_3(k). \tag{53}\end{equation} 接下来, 假设(A10)式对于$~k,k-1,\ldots,1.~$ 成立. 分别考虑图11$\sim$13, 有 \begin{align}& y(k+1)=\phi^{\rm T}(k-d+1)\hat\theta(t_k)+\omega(k+1)+e_i(k+1), \tag{54} \\ & y_1(k+1)=\phi^{\rm T}_1(k-d+1)\hat\theta(t_k)+\omega(k+1), \tag{55} \\ & y_2(k+1)=\phi^{\rm T}_2(k-d+1)\hat\theta(t_k)+e_i(k+1), \tag{56} \\ & y_3(k+1)=\phi^{\rm T}_3(k-d+1)\hat\theta(t_k). \tag{57} \end{align} 其中 \begin{eqnarray}\phi^{\rm T}_1(k-d+1)&=&[y_1(k),\ldots,y_1(k-n+1),u(k-d+1),\ldots,u_1(k-d-m+1)], \tag{58} \\ \phi^{\rm T}_2(k-d+1)&=&[y_2(k),\ldots,y_2(k-n+1),u_2(k-d+1),\ldots,u_2(k-d-m+1)], \tag{59} \\ \phi^{\rm T}_3(k-d+1)&=&[y_3(k),\ldots,y_3(k-n+1),u_3(k-d+1),\ldots,u_3(k-d-m+1)]. \tag{60} \end{eqnarray} 按照前面的假设, 即, (A10)式对于$~k,k-1,\ldots,1.~$ 成立, 显然有 \begin{equation}\phi^{\rm T}_1(k-d+1)+\phi^{\rm T}_2(k-d+1)+\phi^{\rm T}_3(k-d+1)=\phi^{\rm T}(k-d+1). \tag{61}\end{equation} 因此, 可以得到 \begin{equation}\begin{split} y_1(k+1)+y_2(k+1)+y_3(k+1)=\phi^{\rm T}(k-d+1)\hat\theta(t_k)+\omega(k+1)+e_i(k+1)=y(k+1). \end{split} \tag{62}\end{equation} 下面考察 $u_1(k+1)$, $u_2(k+1)$ 和 $u_3(k+1)$. 在图10中, 有 \begin{equation}u(k+1)=\phi^{\rm T}_{c}(k+1)\theta_c(t_{k+1})+\Delta u'(k+1). \tag{63}\end{equation} 类似地, 在图11$\sim$13中, 有 \begin{eqnarray}u_1(k+1)&=&\phi^{\rm T}_{c1}(k+1)\theta_c(t_{k+1}), \tag{64} \\ u_2(k+1)&=&\phi^{\rm T}_{c2}(k+1)\theta_c(t_{k+1}), \tag{65} \\ u_3(k+1)&=&\phi^{\rm T}_{c3}(k+1)\theta_c(t_{k+1})+\Delta u'(k+1), \tag{66} \end{eqnarray} 其中 \begin{eqnarray}\phi^{\rm T}_c(k+1)&=&[y(k+1),\ldots,y(k+1-s_1),u(k),\ldots,u(k+1-s_2),y_r(k+1),\ldots,y_r(k+1-s_3)], \tag{67} \\ \phi^{\rm T}_{c1}(k+1)&=&[y_1(k+1),\ldots,y_1(k+1-s_1),u_1(k),\ldots,u_1(k+1-s_2),y_r(k+1),\ldots,y_r(k+1-s_3)], \tag{68} \\ \phi^{\rm T}_{c2}(k+1)&=&[y_2(k+1),\ldots,y_2(k+1-s_1),u_2(k),\ldots,u_2(k+1-s_2),0,\ldots,0], \tag{69} \\ \phi^{\rm T}_{c3}(k+1)&=&[y_3(k+1),\ldots,y_3(k+1-s_1),u_3(k),\ldots,u_3(k+1-s_2),0,\ldots,0], \tag{70} \end{eqnarray} 其中, $s_1\geq~1,~s_2\geq~1,~s_3\geq~1$为正整数, 由不同的控制策略决定. 进一步基于 (A10)及(A19)式, 显然有 \begin{equation}\phi^{\rm T}_{c1}(k+1)+\phi^{\rm T}_{c2}(k+1)+\phi^{\rm T}_{c3}(k+1)=\phi^{\rm T}_c(k+1). \tag{71}\end{equation} 因此, \begin{equation}\begin{split} u_1(k+1)+u_2(k+1)+u_3(k+1) =\phi^{\rm T}_c(k+1)\theta_c(t_{k+1})+\Delta u'(k+1) =u(k+1). \end{split} \tag{72}\end{equation} 所以 (A10)式 对所有 $k$皆成立.

proof 利用三角不等式以及显然的事实 $2ab\le~a^2+b^2$, 有 \begin{equation}\begin{split} \frac{1}{n}\sum_{k=1}^{n}\|\widetilde{\phi}(k)\|^2 =\frac{1}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_1(k)+\widetilde{\phi}_2(k)\|^2 \le\frac{2}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_1(k)\|^2+\frac{2}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_2(k)\|^2. \end{split} \tag{73}\end{equation} 下面用反证法得出结论. 假设 $\frac{1}{n}\sum^{n}_{k=1}\|\widetilde{\phi}(k)\|^2$ 无界, 则一定存在无穷子列 $\frac{1}{n}\sum^{n}_{k=1}\|\widetilde{\phi}(p_k)\|^2\to\infty$, 满足 \begin{equation}\begin{split} \frac{1}{n}\sum_{k=1}^{n}\|\widetilde{\phi}(p_k)\|^2 =\frac{1}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_1(p_k)+\widetilde{\phi}_2(p_k)\|^2 \le\frac{2}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_1(p_k)\|^2+\frac{2}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_2(p_k)\|^2. \end{split} \tag{74}\end{equation} 进一步考虑到收敛序列及其子序列的极限相同, 于是得到 \begin{equation}\frac{1}{n}\sum_{k=1}^{n}\|\widetilde{\phi}_2(p_k)\|^2= o\left(\frac{1}{n}\sum^{n}_{k=1}\|\widetilde{\phi}(p_k)\|^2\right). \tag{75}\end{equation} 上式结合(A31)式 以及事实 $\frac{1}{n}\sum^{n}_{k=1}\|\widetilde{\phi}_1(p_k)\|^2<\infty$, 利用夹逼定理, 得出明显的错误结论 $\frac{\sum^{n}_{k=1}\|\widetilde{\phi}(p_k)\|^2}{\sum^{n}_{k=1}\|\widetilde{\phi}(p_k)\|^2}\to~0$, 因此, 之前的假设不成立, 从而得到 $\frac{1}{n}\sum^{n}_{k=1}\|\widetilde{\phi}(k)\|^2<\infty.$

proof 显然 \begin{equation}\begin{split} \frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)+y_2(k)]^2 &=\frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]^2+\frac{1}{n}\sum^{n}_{k=1}[y_2(k)]^2 +\frac{2}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]y_2(k) \\ &\to \frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]^2+\frac{2}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]y_2(k). \end{split} \tag{76}\end{equation} 根据Cauchy不等式, 有 \begin{equation}\begin{split} 0\le\left\{\frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]y_2(k)\right\}^2\le\left\{\frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]^2\right\}.\left\{\frac{1}{n}\sum^{n}_{k=1}[y_2(k)]^2\right\} \to 0.\end{split} \tag{77}\end{equation} 于是, 运用夹逼定理得到 \begin{equation}\left\{\frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]y_2(k)\right \}^2\to 0, \frac{1}{n}\sum^{n}_{k=1}[y_1(k)-y_r(k)]y_2(k)\to 0. \tag{78}\end{equation}

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• Figure 1

Self-tuning control system

• Figure 2

Deterministic VES I

• Figure 3

Deterministic VES II

• Figure 4

Subsystem 1 of deterministic VES II

• Figure 5

Subsystem 2 of deterministic VES II

• Figure 6

Subsystem 3 of deterministic VES II

• Figure 7

VES for one-step-ahead STC (self-tuning control)

• Figure 8

Stochastic self-tuning control system

• Figure 9

Stochastic VES I

• Figure 10

Stochastic VES II

• Figure 11

Subsystem 1 of stochastic VES II

• Figure 12

Subsystem 2 of stochastic VES II

• Figure 13

Subsystem 3 of stochastic VES II

• Figure 14

VES for minimum variance self-tuning control

• Figure 15

Equivalence of output signals of VES before and after decomposition

• Figure 16

Equivalence of control signals of VES before and after decomposition