Asymptotic properties of zero dynamics for nonlinear discretized systems with time delay via Taylor method

Abstract

Most real world plants often operate in continuous-time case and involve time delay. These models are typically described by ordinary differential equations. However, to utilize and analyze these data, the control signals must first be discretized. In this paper, a new discretization method for obtaining an approximate sampled-data model of a nonlinear continuous-time system with time delay is proposed. The presented approach is based on the Taylor method and zero-order hold assumption, which can be used to approximate the system output and its derivatives in such a way as to obtain a local truncation error between the output of the resulting sampled-data model and true continuous-time system output. More importantly, on the basis of this discretized representation, we explicitly derive the mathematical structure of nonlinear discrete-time zero dynamics in the case of time delay. The main contribution is to analyze the stability of sampling zero dynamics for the proposed model with time delay. The ideas presented here generalize the well-known results from the linear system to nonlinear case.

Appendix: Proof of Lemma 2

Consider the $n$th-order integrator

$$\begin{aligned} G(s)=\frac{1}{s^{n}} \end{aligned}$$

(60)

with an input $u(t)$ and an output $y(t)$. In the following, we discuss the sampled-data model obtained from the linear transfer function (60) with a time delay. First, we define the state variables such as $x_{1}(t)=y(t), x_{2}(t)=\dot{y}(t), \ldots , x_{n}(t)=y^{n-1}(t)$.

Applying a higher-order Taylor expansion such as

$$\begin{aligned} x_{j}(k+1)&= y^{j-1}(k+1)=\sum \limits _{i=0}^{\infty }\frac{T^{i}}{i!}y_{k}^{(i+j-1)},\nonumber \\&\quad j=1,2,\ldots ,n \end{aligned}$$

(61)

Thus, the matrix state equation is expressed as

$$\begin{aligned} \left\{ \begin{array}{l} x_{k+1}=\left[ \begin{array}{cccc} 1 &{} T &{} \cdots &{} \frac{T^{n-1}}{(n-1)!} \\ &{} \ddots &{} &{} \vdots \\ &{} O &{} \ddots &{} T \\ &{} &{} &{} 1 \end{array} \right] x_{k}+\left[ \begin{array}{c} \frac{T^{n}}{n!}c_{m}^{n}(\alpha ) \\ \vdots \\ \frac{T^{2}}{2!}c_{m}^{2}(\alpha ) \\ Tc_{m}^{1}(\alpha ) \end{array} \right] \\ {\tilde{u}}_{k}=A_{q}x_{k}+B_{q}{\tilde{u}}_{k} \\ y_{k}=[ \begin{array}{cccc} 1 &{} 0 &{} \cdots &{} 0 \\ \end{array} ]x_{k}=C_{q}x_{k} \end{array}\right. \nonumber \\ \end{aligned}$$

(62)

where $x_{k}=[x_{1}(k), x_{2}(k), \ldots , x_{n}(k)]^{T}$ and ${\tilde{u}}_{k}=u(t-D)$.

The zeros of the sampled-data model above are determined by $y(k+1)=y(k)=0$ in (62) as follows:

$$\begin{aligned} \left\{ \begin{array}{l} 0=\left[ \begin{array}{cccc} T &{} \frac{T^{2}}{2!} &{} \cdots &{} \frac{T^{n-1}}{(n-1)!} \\ \end{array} \right] \overline{x}_{k}+\frac{T^{n}}{n!}c_{m}^{n}(\alpha ){\tilde{u}}_{k} \\ \overline{x}_{k+1}=\left[ \begin{array}{cccc} 1 &{} T &{} \cdots &{} \frac{T^{n-2}}{(n-2)!} \\ &{} O &{} \ddots &{} \vdots \\ &{} &{} &{} 1 \end{array} \right] \overline{x}_{k}+\left[ \begin{array}{c} \frac{T^{n-1}}{(n-1)!}c_{m}^{n-1}(\alpha ) \\ \vdots \\ Tc_{m}^{1}(\alpha ) \end{array} \right] {\tilde{u}}_{k} \end{array}\right. \nonumber \\ \end{aligned}$$

(63)

where $\overline{x}_{k}=[x_{2}(k), \ldots , x_{n}(k)]^{T}$.

Using $z$-transform to (63), it leads to

$$\begin{aligned} \phi _{n}(z)\left[ \begin{array}{c} Z\left[ \overline{x}_{k}\right] \\ Z\left[ u_{k}\right] \\ Z\left[ u_{k-1}\right] \end{array}\right] =0 \end{aligned}$$

(64)

and then we see that the zeros are solution of $|\phi _{n}(z)|=0$, where

$$\begin{aligned} \phi _{n}(z)=\left| \begin{array}{cccccc} -T &{} -\frac{T^{2}}{2!} &{} \cdots &{} -\frac{T^{n-1}}{(n-1)!} &{} -\frac{T^{n}}{n!}c_{m}^{n}(\alpha ) &{} \frac{T^{n}}{n!}c_{m}^{n}(\alpha ) \\ z-1 &{} -T &{} \cdots &{} -\frac{T^{n-2}}{(n-2)!} &{} -\frac{T^{n-1}}{(n-1)!}c_{m}^{n-1}(\alpha ) &{} \frac{T^{n-1}}{(n-1)!}c_{m}^{n-1}(\alpha )\\ &{} \ddots &{} &{} &{} &{} \vdots \\ &{} &{} z-1 &{} -T &{} -\frac{T^2}{2!}c_{m}^{2}(\alpha ) &{} \frac{T^2}{2!}c_{m}^{2}(\alpha ) \\ &{} &{} &{} z-1 &{} -Tc_{m}^{1}(\alpha ) &{} Tc_{m}^{1}(\alpha ) \\ O &{} &{} &{} &{} -1 &{} z \end{array} \right| \end{aligned}$$

(65)

Here, the matrix $S$ is defined as

$$\begin{aligned} S&= \left[ \begin{array}{cc} I &{} \mathbf 0 \\ s^{T} &{} 1 \end{array}\right] ,\nonumber \\ s&= -\frac{n!}{Tc_{m}^{n}(\alpha )} \left[ \begin{array}{ccccc} \frac{1}{T^{n-2}}&\frac{1}{2!T^{n-3}}&\cdots&\frac{1}{(n-1)!}&0 \end{array}\right] ^{T} \end{aligned}$$

(66)

Adding the final $n+1$th column to the $n$th column in (65), Multiplying it by the matrix $S$ from the right hand side yields

$$\begin{aligned} \phi _{n}(z)S&= \left[ \begin{array}{cc} \mathbf O^{T} &{} \frac{T^{n}}{n!}c_{m}^{n}(\alpha )\\ \overline{\varPhi }_{n}(z) &{} *\end{array}\right] \end{aligned}$$

(67)

Hence, we have

$$\begin{aligned} |\phi _{n}(z)|=\frac{(-1)^{n}T^{n}}{n!}|\overline{\varPhi }_{n}(z)| \end{aligned}$$

(68)

because $|S|=1$. Further, recall here that, $|\overline{ \varPhi }_{n}(z)|$ is a monic polynomial with order $n$ from the definition.

When the transfer function (60) is discretized by a ZOH in the case of time delay, the pulse transfer function $H_\mathrm{D}(z)$ is given by

$$\begin{aligned} H_\mathrm{D}(z)=\frac{T^{n}B_{n}(z,D)}{n!(z-1)^{n}z} \end{aligned}$$

(69)

where $B_{n}(z,D)$ is a monic polynomial with the order $n$ [15]. Therefore, we have obtained

$$\begin{aligned} |B_{n}(z,D)|=|\overline{\varPhi }_{n}(z)| \end{aligned}$$

(70)

Also, we consider the sampled-data model for the $n$th-order integrator given by (62), and the state space similarity transformation $z=Tx$, where the nonsingular matrix $T$ is given by

$$\begin{aligned} T=\left[ \begin{array}{c|c} 1 &{} \mathbf{0} \\ \hline T_{21} &{} { I_{n-1}} \end{array}\right] \Longleftrightarrow T^{-1}=\left[ \begin{array}{c|c} 1 &{} \mathbf{0} \\ \hline -T_{21} &{} { I_{n-1}} \end{array}\right] \end{aligned}$$

(71)

where

$$\begin{aligned} T_{21}= \left[ \begin{array}{ccc} -\frac{n}{T}\frac{c_{m}^{n-1}(\alpha )}{c_{m}^{n}(\alpha )}&\cdots&-\frac{n!}{T^{n-1}}\frac{c_{m}^{1}(\alpha )}{c_{m}^{n}(\alpha )} \end{array}\right] ^{T} \end{aligned}$$

(72)

Then, the new state space representation is given by the following matrices

$$\begin{aligned} {\overline{A}}_{q}&= TA _{q}T^{-1}= \left[ \begin{array}{c|c} g_{11} &{} G_{12} \\ \hline G_{21} &{} G_{22} \end{array}\right] \nonumber \\&= \left[ \begin{array}{c|c} 1-A_{12}T_{21} &{} A_{12} \\ \hline -(T_{21}A_{12}+A_{22})T_{21} &{} T_{21}A_{12}+A_{22} \end{array}\right] \end{aligned}$$

(73)

where

$$\begin{aligned} A_{12}&= \left[ \begin{array}{cccc} T&\frac{T^{2}}{2!}&\cdots&\frac{T^{n-1}}{(n-1)!} \end{array}\right] ,\nonumber \\ A_{22}&= \left[ \begin{array}{cccc} 1 &{} T &{} \cdots &{} \frac{T^{n-2}}{(n-2)!} \\ \vdots &{} \ddots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} 0 &{} T \\ 0 &{} 0 &{} \cdots &{} 1 \end{array}\right] \end{aligned}$$

(74)

and

$$\begin{aligned} \overline{B}_{q}= TB_{q}=\left[ \begin{array}{cccc} \dfrac{T^{n}}{n!}c_{m}^{n}(\alpha )&0&\cdots&0 \end{array}\right] ^{T}, \overline{C}_{q}&\!=\!&C_{q}T^{\!-\!1}\!=\!C_{q}\qquad \end{aligned}$$

(75)

These state space matrices give the normal form that appears in (9).

To prove (10), we first note that

$$\begin{aligned} B_{n}(z,D)\left[ \begin{array}{c|c} 0 &{} { I_{n-1}} \\ \hline 1 &{} 0 \end{array}\right] \!=\!\left[ \begin{array}{c|c} \frac{T^{n\!-\!1}}{n!}c_{m}^{n}(\alpha ) &{} A_{12} \\ \hline \!-\!\frac{T^{n\!-\!1}}{n!}c_{m}^{n}(\alpha )T_{21} &{} A_{22}\!-\!z { I_{n-1}} \end{array}\right] \nonumber \\ \end{aligned}$$

(76)

Computing the determinant of the matrices on both side of the equation and using determinant formula about the matrix inversion [17], we have that

$$\begin{aligned} |B_{n}(z,D)|(\!-\!1)^{n\!-\!1}\!=\!\frac{T^{n\!-\!1}}{n!}c_{m}^{n}(\alpha )|A_{22}-z { I_{n\!-\!1}}\!+\!T_{21}A_{12}| \end{aligned}$$

(77)

where, from definition of $G_{22}$ in (73), we finally have that

$$\begin{aligned}&|B_{n}(z,D)|=\frac{T^{n-1}}{n!}c_{m}^{n}(\alpha )(-1)^{n-1}|-z { I_{n-1}} \nonumber \\&\quad +A_{22}+T_{21}A_{12}|=\frac{T^{n-1}}{n!}c_{m}^{n}(\alpha )|z { I_{n-1}}-G_{22}|\nonumber \\ \end{aligned}$$

(78)