Adaptive Modified Input and State Estimation for Linear Discrete-Time System with Unknown Inputs

Abstract

This paper is concerned with the problem of simultaneously estimating the state and the input of a linear discrete-time system. In order to improve the accuracy of estimation, a reasonable assumption about the unknown input is proposed. Based on this assumption, a modified input estimation is obtained, and then the optimal state estimation is derived strictly. Moreover, in order to effectively deal with the possible saltations of the unknown input, an adaptive estimation method is also developed. Simulation results illustrate the effectiveness and applicability of the proposed approach.

Appendix

Proof of Theorem 1

In order to prove the invertible of ${\tilde{R}}^*_{k}$, by (36), we only need to prove that $(I_m-C_k{G_{k-1}}K^d_{k-1} M_{k})$ is invertible. To do this, it is only need to show that all the eigenvalue of $(I_m-C_k{G_{k-1}}K^d_{k-1} M_{k})$ are not equal to zero. Let the singular value decomposition of ${\tilde{R}}^{-1/2}_k C_k G_{k-1}$ be

$$\begin{aligned} {\tilde{R}}^{-1/2}_k C_k G_{k-1} = U_k \left[ \begin{array}{c}\varDelta _k\\ 0\end{array} \right] V^\mathrm {T}_k \end{aligned}$$

(74)

where $\varDelta _k$ is a diagonal matrix with order q, $U_k$ and $V_k$ are orthogonal matrices.

Then matrix $C_k G_{k-1}$ and $M_k$ can be expressed as

$$\begin{aligned} C_k G_{k-1}= & {} {\tilde{R}}^{1/2}_k U_k \left[ \begin{array}{c}\varDelta _k\\ 0\end{array} \right] V^\mathrm {T}_k\end{aligned}$$

(75)

$$\begin{aligned} M_k= & {} V_k \left[ \varDelta ^{-1}_k \quad 0 \right] U^\mathrm {T}_k {\tilde{R}}^{-1/2} \end{aligned}$$

(76)

Then it follows that

$$\begin{aligned} C_k G_{k-1} K^d_{k-1} M_k = {\tilde{R}}^{1/2}_k U_k \left[ \begin{array}{cc} \varDelta _k V^\mathrm {T}_k K^d_{k-1} V_k \varDelta ^{-1}_k &{} 0\\ 0 &{} 0\end{array} \right] U^\mathrm {T}_k {\tilde{R}}^{-1/2}_k \end{aligned}$$

(77)

This indicates that matrix $C_k G_{k-1} K^d_{k-1} M_k$ contains all the eigenvalues of $K^d_{k-1}$ as its own eigenvalues, and the rest $(m-q)$ eigenvalues are equal to 0. Denote the eigenvalues of $K^d_{k-1}$ as $\alpha _i$, $(i=1,\ldots ,q)$, then the eigenvalues of $C_k G_{k-1} K_k M_k$ can be represented by $(\alpha _1, \ldots , \alpha _q, 0, \ldots , 0)$, where the number of 0 is $(m-q)$. Therefore, it is easy to show that the eigenvalues of $(I_m - C_k {G_{k-1}}K^d_{k-1} M_{k})$ are $(1-\alpha _1, \ldots , 1-\alpha _q, 1, \ldots , 1)$, where the number of 1 is $(m-q)$. By Lemma 1(ii), we know that $0<\alpha _i<1$, $(i=1,\ldots ,q)$, this means that all the eigenvalue of $(I_m - C_k {G_{k-1}}K^d_{k-1} M_{k})$ is not equal to zero. Hence, matrix $(I_m - C_k {G_{k-1}}K^d_{k-1} M_{k})$ is invertible and therefore ${\tilde{R}}^*_{k}$ is invertible. $\square$