Time-Delay Estimation in State and Output Equations of Nonlinear Systems Using Optimal Computational Approach

Abstract

Many real-world dynamics can be modeled as nonlinear time-delay systems. In order to capture a more realistic model for system dynamics, the exact values of time-delay should be taken into account. For nonlinear time-delay systems, the estimation of delays in both state and output equations is discussed. A cost function is defined based on least-square error between actual and estimated values of the output measurement. The value of time-delays in the nonlinear system are then derived using a gradient-based optimization method. Because of the implicit description of the cost function with respect to the delay value, its gradients cannot be obtained by standard analytical differentiation rules. In this case, the optimal computational methods are utilized to derive two formulas for computing the gradient. An optimization scheme is then formulated to estimate both state and output delays. The effectiveness of the proposed estimation method is finally demonstrated using the simulation results on a benchmark chemical process.

Appendix: Proof of Lemma 4.1 [24, 29]

First, we define

$$\begin{aligned} I := \{ \epsilon \in \mathbb {R};\tau + \epsilon ^k \in \mathfrak {T} \}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} I=[-\tau _k,\bar{\tau }-\tau _k]. \end{aligned}$$

Thus, I is a closed interval of positive values, and $0 \in I$. For each $\epsilon \in I$, we define

$$\begin{aligned} \varphi ^\epsilon (t) := x^\epsilon (t)-x(t), \quad t\le T, \end{aligned}$$

and

$$\begin{aligned} \theta ^{\epsilon ,i}(t)=x^\epsilon (t-\tau _i-\epsilon \delta _{ki})-x(t-\tau _i),\quad t\le T, \quad i=1,\ldots ,m, \end{aligned}$$

If $i \ne k$, then $\delta _{ki}=0$, and $\theta ^{\epsilon ,i}(t)=x^\epsilon (t-\tau _i)-x(t-\tau _i)$, which means

$$\begin{aligned} \theta ^{\epsilon ,i}(t)=\varphi ^\epsilon (t-\tau _i),\quad t\le T,\quad i\ne k, \end{aligned}$$

(46)

and from (3) one achieves

$$\begin{aligned} \varphi ^\epsilon (t)=0,\quad t\le 0. \end{aligned}$$

(47)

Lemma A.1

There exists a positive real number $L_2>0$ such that for all $\epsilon \in I$

$$\begin{aligned} |x^\epsilon (t)|,|\gamma (t)|\le L_2,\quad t\in [-\bar{\tau },T], \end{aligned}$$

(48)

and

$$\begin{aligned} |\varphi ^\epsilon (t)|, \max _{i=1,\ldots ,m} |\theta ^{\epsilon ,i}(t)|\le L_2 |\epsilon |,\quad t \in [0,T]. \end{aligned}$$

(49)

The following result has been proved in [24]. This lemma can be proved by following similar arguments to those used in [38].

Lemma A.2

For almost all $t \in [0,T]$,

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} \frac{\theta ^{\epsilon ,k}(t)-\varphi ^\epsilon (t-\tau _k)}{\epsilon }=-\gamma (t-\tau _k). \end{aligned}$$

(50)

Let $k\in \{1,\ldots ,m\}$ and $\tau \in \mathfrak {T}$ be arbitrary but fixed. The vector $x^\epsilon (t)$ denotes $x(t|\tau +\epsilon ^k)$, and x(t) represents $x(t|\tau )$. For each $\epsilon \in I\backslash \{0\}$ , we define

$$\begin{aligned} \rho (\epsilon )=\int _{0}^{T}|\epsilon ^{-1} \theta ^{\epsilon ,k}(s)-\epsilon ^{-1}\varphi ^\epsilon (s-\tau _k)+\gamma (s-\tau _k)|\hbox {d}s. \end{aligned}$$

(51)

It follows from (47), (48), and (49) that for each $\epsilon \in I\backslash \{0\}$,

$$\begin{aligned}&|\epsilon ^{-1} \theta ^{\epsilon ,k}(s)-\epsilon ^{-1} \varphi ^\epsilon (s-\tau _k)+\gamma (s-\tau _k)| \\&\le |\epsilon ^{-1}|| \theta ^{\epsilon ,k}(s)|+|\epsilon ^{-1}| |\varphi ^\epsilon (s-\tau _k)|+|\gamma (s-\tau _k)| \le 3L_2, \quad s\in [0,T]. \end{aligned}$$

Hence, the right-hand side of the equation (51) is bounded on $\epsilon \in I\backslash \{0\}$. Furthermore, it follows from (50) that $\rho (\epsilon )$ in (51) converges to zero almost everywhere on [0, T] as $\epsilon \rightarrow 0$. Therefore, the Lebesgue dominated convergence theorem [39] leads to,

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} \rho (\epsilon ) =\lim _{\epsilon \rightarrow 0}\int _{0}^{T}|\epsilon ^{-1} \theta ^{\epsilon ,k}(s)-\epsilon ^{-1} \varphi ^\epsilon (s-\tau _k)+\gamma (s-\tau _k)|\hbox {d}s=0. \end{aligned}$$

For a fixed value $\epsilon \in I\backslash \{0\}$, we define

$$\begin{aligned} \bar{f}(s,\alpha ) =f(s,x(s)+\alpha \varphi ^\epsilon (s),\tilde{x}(s)+\alpha \theta ^\epsilon (s)) \quad (s,\alpha )\in [0,T]\times [0,1]. \end{aligned}$$

Applying chain rule yields

$$\begin{aligned} \frac{\partial \bar{f}(s,\alpha )}{\partial \alpha } = \frac{\partial \bar{f}(s,\alpha )}{\partial x}\varphi ^\epsilon (s)+\sum _{i=1}^{m} \frac{\partial \bar{f}(s,\alpha )}{\partial \tilde{x}^i} \theta ^{\epsilon ,i}(s), \end{aligned}$$

(52)

where

$$\begin{aligned} \frac{\partial \bar{f}(s,\alpha )}{\partial x}= & {} \frac{\partial f (s,x(s)+\alpha \varphi ^ \epsilon (s),\tilde{x}(s)+ \alpha \theta ^ \epsilon (s))}{\partial x}, \end{aligned}$$

(53)

$$\begin{aligned} \frac{\partial \bar{f}(s,\alpha )}{\partial x}= & {} \frac{\partial f(s,x(s)+\alpha \varphi ^\epsilon (s),\tilde{x}(s)+\alpha \theta ^ \epsilon (s))}{\partial \tilde{x}^i}. \end{aligned}$$

(54)

Now, the auxiliary system (8) can be rewritten by using (53) and (54) as follows:

$$\begin{aligned} \dot{\psi }^k (s)= & {} \frac{\partial \bar{f}(s,0)}{\partial x}\psi _k (s)+\sum _{i=1}^{m}\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i}\psi _k (s-\tau _i) \\&-\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^k}\gamma ^k (s-\tau _k). \end{aligned}$$

Thus,

$$\begin{aligned} \psi _k (t)= & {} \int _{0}^{t}\frac{\partial \bar{f}(s,0)}{\partial x}\psi _k (s)\hbox {d}s + \sum _{i=1}^{m}\int _{0}^{t}\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i}\psi _k (s-\tau _i)\hbox {d}s \nonumber \\&\quad -\int _{0}^{t}\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^k}\gamma ^k (s-\tau _k)\hbox {d}s. \end{aligned}$$

(55)

The Eq. (52) can be rewritten as:

$$\begin{aligned} \frac{\partial \bar{f}(s,\alpha )}{\partial \alpha }= & {} \triangle _1 (s,\alpha )+\triangle _2 (s,\alpha )+ \frac{\partial \bar{f}(s,0)}{\partial x}\varphi ^ \epsilon (s) \\&+ \sum _{i=1}^{m} \frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i} \varphi ^{\epsilon }(s-\tau _i)+\sum _{i=1}^{m} \frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i} \{\theta ^{\epsilon ,i}(s)-\varphi ^\epsilon (s-\tau _i)\}, \end{aligned}$$

where

$$\begin{aligned} \triangle _1 (s,\alpha )= & {} \left\{ \frac{\partial \bar{f}(s,\alpha )}{\partial x}- \frac{\partial \bar{f}(s,0)}{\partial x}\right\} \varphi ^\epsilon (s), \end{aligned}$$

(56)

$$\begin{aligned} \triangle _2 (s,\alpha )= & {} \left\{ \frac{\partial \bar{f}(s,\alpha )}{\partial \tilde{x}^i}- \frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i}\right\} \theta ^{\epsilon ,i}(s). \end{aligned}$$

(57)

Thus, by using (46), we deduce

$$\begin{aligned} \frac{\partial \bar{f}(s,\alpha )}{\partial \alpha }= & {} \triangle _1 (s,\alpha )+\triangle _2 (s,\alpha )+ \frac{\partial \bar{f}(s,0)}{\partial x}\varphi ^\epsilon (s) \nonumber \\&+\sum _{i=1}^{m} \frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i} \varphi ^{\epsilon }(s-\tau _i)+\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^k} \{\theta ^{\epsilon ,k}(s)-\varphi ^\epsilon (s-\tau _k)\}, \end{aligned}$$

(58)

so

$$\begin{aligned} \varphi ^\epsilon (t)=x^\epsilon (t)-x(t)=\int _{0}^{t}\{\bar{f}(s,1)-\bar{f}(s,0)\}\hbox {d}s. \end{aligned}$$

Therefore, from the fundamental theorem of calculus, one can conclude

$$\begin{aligned} \varphi ^\epsilon (t)=\int _{0}^{t}\{\bar{f}(s,1)-\bar{f}(s,0)\}\hbox {d}s \nonumber \\ =\int _{0}^{t} \int _{0}^{1}\frac{\partial {\bar{f}(s,0)}}{\partial \alpha }d\alpha \hbox {d}s, \end{aligned}$$

(59)

and merging (58) and (59) gives

$$\begin{aligned} \varphi ^\epsilon (t)= & {} \int _{0}^{t} \int _{0}^{1}\{\triangle _1 (s,\alpha )+\triangle _2 (s,\alpha )\} d \alpha \hbox {d}s \nonumber \\&+ \int _{0}^{t}\frac{\partial \bar{f}(s,0)}{\partial x}\varphi ^\epsilon (s)\hbox {d}s+\sum _{i=1}^{m} \int _{0}^{t} \frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i} \varphi ^{\epsilon }(s-\tau _i)\hbox {d}s \nonumber \\&+ \int _{0}^{t}\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^k} \{\theta ^{\epsilon ,k}(s)-\varphi ^\epsilon (s-\tau _k)\}\hbox {d}s \end{aligned}$$

(60)

Clearly, there exists a compact set $\Gamma \subset \mathbb {R}^n$ such that

$$\begin{aligned} x(s) \in \Gamma \quad s\in [-\bar{\tau },T], \end{aligned}$$

where $x(s)=x(s|\tau )$. While f is continuously differentiable (Assumption 3.1), some constants $K_1>0$ and $K_2>0$ exist such that satisfy following equation,

$$\begin{aligned} |\frac{\partial \bar{f}(s,0)}{\partial x}|\le K_1, \qquad |\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i}|\le K_2, \quad s\in [0,T], \end{aligned}$$

Using (60) and (55), it can be deduced that

$$\begin{aligned}&|\epsilon ^{-1} \varphi ^\epsilon (t)-\psi _k (t)|\le K_2\rho (\epsilon )+\int _{0}^{t} K_1|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s \nonumber \\&+ \sum _{i=1}^{m}\int _{0}^{t} K_2|\epsilon ^{-1} \varphi ^\epsilon (s-\tau _i)-\psi _k (s-\tau _i)|\hbox {d}s + |\epsilon |^{-1} \nonumber \\&\int _{0}^{t} \int _{0}^{1} \{|\triangle _1(s,\alpha )|+|\triangle _2(s,\alpha )|\}d\alpha \hbox {d}s, \end{aligned}$$

(61)

we try to simplify the second term on the right-hand side of (61) as

$$\begin{aligned} \sum _{i=1}^{m}\int _{0}^{t} K_2|\epsilon ^{-1} \varphi ^\epsilon (s-\tau _i)-\psi _k (s-\tau _i)|\hbox {d}s= & {} \sum _{i=1}^{m}\int _{-\tau _i}^{t-\tau _i} K_2|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s \\\le & {} \sum _{i=1}^{m}\int _{0}^{t} K_2|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s \\= & {} \int _{0}^{t} k K_2|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s. \end{aligned}$$

Then, (61) changes to

$$\begin{aligned}&|\epsilon ^{-1} \varphi ^\epsilon (t)-\psi _k (t)|\le K_2\rho (\epsilon )+\int _{0}^{t} \bar{K}|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s \nonumber \\&+ |\epsilon |^{-1} \int _{0}^{t} \int _{0}^{1} \{|\triangle _1(s,\alpha )|+|\triangle _2(s,\alpha )|\}d\alpha \hbox {d}s, \end{aligned}$$

(62)

where $\bar{K}=K_1+kK_2$. Since f is continuously differentiable and $x^\epsilon$ is uniformly bounded with respect to $\epsilon$, $\frac{\partial \bar{f}}{\partial x}$ and $\frac{\partial \bar{f}}{\partial \tilde{x}^i}$ are uniformly continuous on $[0,T]\times [0,1]$. Furthermore, by (49), $x(s)+\alpha \varphi ^\epsilon (s)\rightarrow x(s)$ and $\tilde{x}(s)+\alpha \theta ^\epsilon (s)\rightarrow \tilde{x}(s)$ uniformly on $[0,T]\times [0,1]$ as $\epsilon \rightarrow 0$ [40]. Thus, for each $\delta >0$, there exists an ${\varepsilon }>0$ such that for all $\epsilon$ satisfying $|\epsilon |<{\varepsilon }$,

$$\begin{aligned}&|\frac{\partial \bar{f}(s,\alpha )}{\partial x}-\frac{\partial \bar{f}(s,0)}{\partial x}|<\delta , \quad (s,\alpha )\in [0,T]\times [0,1], \\&|\frac{\partial \bar{f}(s,\alpha )}{\partial \tilde{x}^i}-\frac{\partial \bar{f}(s,0)}{\partial \tilde{x}^i}|<\delta , \quad (s,\alpha )\in [0,T]\times [0,1]. \end{aligned}$$

Computing the norm of (56) and (57), and then using the above inequalities together with (49), gives

$$\begin{aligned} |\triangle _1(s,\alpha )| \le \delta L_2 |\epsilon |, \qquad |\triangle _2(s,\alpha )| \le \delta k L_2 |\epsilon |, \end{aligned}$$

(63)

where $|\epsilon |<{\varepsilon }$. Substituting these inequalities into (62) gives,

$$\begin{aligned} |\epsilon ^{-1} \varphi ^\epsilon (t)-\psi _k (t)| \le K_2\rho (\epsilon )+(L_2T+kL_2T)\delta +\int _{0}^{t} \bar{K}|\epsilon ^{-1} \varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s. \end{aligned}$$

We know that $\rho (\epsilon ) \rightarrow 0$ as $\epsilon \rightarrow 0$, so there exists a constant ${\acute{\varepsilon }}>0$ that $\rho (\epsilon )<\delta$ whenever $|\epsilon |<\acute{\varepsilon }$. Hence, for all $\epsilon$ that $|\epsilon |<\min \{\varepsilon ,\acute{\varepsilon }\}$,

$$\begin{aligned} |\epsilon ^{-1} \varphi ^\epsilon (t)-\psi _k (t)| \le K_2\delta +(L_2T+kL_2T)\delta +\int _{0}^{t} \bar{K}|\epsilon ^{-1}\varphi ^\epsilon (s)-\psi _k (s)|\hbox {d}s. \end{aligned}$$

Applying the Bellman–Grönwall Lemma [31] yields

$$\begin{aligned} |\epsilon ^{-1} \varphi ^\epsilon (t)-\psi _k (t)|\le \delta (K_2+L_2T+kL_2T)\exp \{\bar{M}T\}, \end{aligned}$$

where $|\epsilon |<\min \{\varepsilon ,\acute{\varepsilon }\}$. While $\delta$ is an arbitrary value, we can conclude that $\epsilon ^{-1} \varphi ^\epsilon (t) \rightarrow \psi _k (t)$ as $\epsilon \rightarrow 0$, and we have

$$\begin{aligned} \frac{\partial x(t|\tau )}{\partial \tau _k}=\lim _{\epsilon \rightarrow 0}\frac{x^\epsilon (t)-x(t)}{\epsilon }= \lim _{\epsilon \rightarrow 0} \epsilon ^{-1} \varphi ^\epsilon (t)=\psi _k (t) \end{aligned}$$

(64)

$\square$