2. Problem formulation

The diagram of a typical networked control system is shown in Fig. 1. It is observed that the measurement from the plant is processed by a quantization block before transmitted through a communication channel, which is normally wireless and lossy. The controller unit receives information from the channel and generates a control signal, which is then quantized and transmitted to the actuator through a communication channel with the same behavior.

Figure 1. Typical networked control system

In terms of the plant model, we consider a single-input continuous-time linear system

(1) \begin{equation}\dot{x}(t) =Ax(t) +B\bar{q}\left( u(t)\right) \end{equation}

with a quantized state feedback control law

(2) \begin{equation}u(t) =-Kq\left( x(t) \right). \end{equation}

In (1), $x(t)\in R^{n}$ is the state vector with an arbitrary initial value x(0), $q\left( x(t) \right) $ is the result of quantization of $x(t) $ , $u\in R$ is the control input, $\bar{q}\left( u(t) \right) $ is the result of quantization of $u(t) ,$ A, B are known matrices or vectors of proper dimensions. B is assumed full rank and K is a feedback gain properly designed to ensure all eigenvalues of $A-BK$ have non-positive real parts that is according to Lyapunov stability theory, there exist positive definite symmetric matrices $Q$ and D so that

(3) \begin{equation}(A-BK)^{T}Q+Q(A-BK)=-D. \end{equation}

A static logarithmic quantizer has the form [Reference Fu and Xie12]

(4) \begin{align}V_{s} & = \left\{ \pm v_{si}\,:\,v_{si}=\frac{1}{\rho ^{i}}v_{s0},i=0,1,2\cdots M\right\} \cup \{0\}\nonumber\\[5pt]q_{s}(a) & =\left\{\begin{array}{c@{\quad}c}v_{sM} & a>\dfrac{1}{1-\delta }v_{sM} \\[10pt]v_{si} & \dfrac{1}{1+\delta }v_{si}<a\leq \dfrac{1}{1-\delta }v_{si} \\[11pt]& i=0,1,...,M \\[3pt]0 & 0<a\leq \dfrac{1}{1+\delta }v_{s0} \\[13pt]-q\left( -a\right) & a< 0\end{array}\right.\\[8pt]\delta & =\frac{1-\rho }{1+\rho },0<\rho <1,v_{s0}>0 \nonumber \end{align}

where $2M+1$ is the order of quantizer determined by the characteristics of the communication channel, a is a scalar to be quantized, $v_{s0}$ is an initialization parameter, $V_{sM}$ and $\rho $ are the range and density of the quantizer, respectively. A single parameter $\delta $ is related to $\rho$ and also reflects the quantizer density.

In Fig. 2, the quantization of x(t) is used as an example to demonstrate the mechanism of sensitivity varying quantization. It is assumed that the quantizer order $2M+1$ is given. The sensitivity varying algorithm involves two stages that is “zoom-in” stage and “zoom-out” stage. When the initial input makes the quantizer saturated, $\delta (t)$ is increased at a speed faster than the dynamic of x(t), and such process is called the “zoom-out” stage. When the system state once gets into the non-saturated area, $\delta (t)$ decreases following a certain law, which is called the “zoom-in” stage.

Figure 2. Flowchart of sensitivity varying quantizing mechanism

Therefore, with the dynamic logarithmic quantization scheme adopted in the following sections of this paper, the derivation of $q\left( x(t) \right) $ in (2) follows the quantization algorithm as in (5), which is a significant extension from (4)

(5) \begin{align}V(t) & =\left\{ \pm v_{i}(t) \,:\,v_{i}(t) =\frac{1}{\rho ^{i}( t) }v_{0},i=0,1,\cdots M\right\} \cup \{0\},\nonumber\\[5pt]q(x(t)) & = \left[q(x_{1}(t))\ q(x_{2}(t))\ ...\ q(x_{n}(t))\right] ^{T},\nonumber\\[9pt]q(x_{j}(t)) & =\left\{\begin{array}{c@{\quad}c}v_{M}(t) & x_{j}(t)>\dfrac{1}{1-\delta (t) }v_{M}(t) \\[10pt]v_{i}(t) & \dfrac{1}{1+\delta (t) }v_{i}(t) <x_{j}(t)\leq \dfrac{1}{1-\delta (t) }v_{i}(t) \\[10pt]& i=0,1,...,M \\[6pt]0 & 0<x_{j}(t)\leq \dfrac{1}{1+\delta (t) }v_{0} \\[10pt]-q\left({-}x_{j}(t)\right) & x_{j}(t)<0\end{array}\right.\\[9pt] \delta (t) & =\frac{1-\rho (t)}{1+\rho (t)},0<\rho (t)<1,v_{0}>0,\text{for all }1\leq j\leq n.\nonumber\end{align}

It is also easily derived that

(6) \begin{equation}\rho (t)=\frac{1-\delta (t)}{1+\delta (t)},0<\delta (t)<1 \end{equation}

And the derivation of $\bar{q}\left( u(t) \right) $ in (1) follows the quantization algorithm as in (7),

(7) \begin{align}\bar{V}(t) & =\left\{ \pm \bar{v}_{i}(t) :\bar{v}_{i}(t) =\frac{1}{\bar{\rho}^{i}(t) }\bar{v}_{0},i=0,1,2\cdots \bar{M}\right\} \cup \{0\}\nonumber\\[9pt]\bar{q}(u(t)) & =\left\{\begin{array}{c@{\quad}c}\bar{v}_{\bar{M}}(t) & u(t)>\dfrac{1}{1-\bar{\delta}(t) }\bar{v}_{\bar{M}}(t) \\[13pt]\bar{v}_{i}(t) & \dfrac{1}{1+\bar{\delta}(t) }\bar{v}_{i}(t) <u(t)\leq \dfrac{1}{1-\bar{\delta}(t) }\bar{v}_{i}(t) \\[13pt]& i=0,1,...,\bar{M}\\[8pt]0 & 0<u(t)\leq \dfrac{1}{1+\bar{\delta}(t)}\bar{v}_{0} \\[13pt]-q\left(-u(t)\right) & u(t)<0\end{array}\right.\\[9pt]\bar{\delta}(t) & =\frac{1-\bar{\rho}(t)}{1+\bar{\rho}(t)},0<\bar{\rho}(t)<1,\bar{v}_{0}>0\nonumber\end{align}

In (5) and (7), $2M+1$ and $2\bar{M}+1$ are the orders of quantizers, $v_{0}$ and $\bar{v}_{0}$ are initialization parameters, $V(t) ,\bar{V}(t) $ and $\rho (t) ,\bar{\rho}(t) $ are the range and density of quantizers respectively. $\delta (t),\bar{\delta}(t)$ are associated with $\rho (t) ,\bar{\rho}(t) $ respectively and reflect the quantizer densities.

3. Main results

For the considered continuous-time linear system with quantized state feedback controller (2), the closed-loop behavior is as follows

(8) \begin{align}\dot{x}(t) & = Ax(t) +B\bar{q}\left( u(t)\right) \notag \\& = Ax(t) +B\left( -Kq(x(t) )+\bar{e}(t)\right)\notag \\& = (A-BK)x(t) -BKe(t)+B\bar{e}(t) \end{align}

where $e(t)=q\left( x(t) \right) -x(t) ,$ $\bar{e}(t)=\bar{q}\left( u(t) \right) -u(t) .$

Let $x_{ia}(t) =\frac{1}{1+\delta (t)}v_{i}(t),x_{ib}(t) =\frac{1}{1-\delta (t)}v_{i}(t),i=1,2,\ldots ,M$ . Then, it is observed that $x_{Mb}(t)=\frac{1}{1-\delta (t)}v_{M}(t)$ with $v_{M}(t)=\frac{1}{\rho ^{M}(t) }v_{0}$ is the saturation threshold of the finite level state quantizer. It can be also obtained:

(9) \begin{equation}x_{Mb}(t) =\frac{1}{\rho ^{M}(t) \left( 1-\delta(t) \right) }v_{0}=\frac{(1+\delta (t))^{M}}{(1-\delta (t))^{M+1}}v_{0} \end{equation}

considering (6).

It can be derived that if the state quantizer is not saturated and $\left\vert x_{i}(t)\right\vert >\frac{1}{1+\delta (t) }v_{0}$ ,

\begin{equation*}e_{i}(t)=\delta _{i}(t)x_{i}(t),\text{with }\left\vert \delta_{i}(t)\right\vert \leq \delta (t),\forall i=1,2,...,n.\end{equation*}

Also,

\begin{equation*}e(t)=\Delta (t)x(t)\text{ with }\Delta (t)=diag\{\delta _{1}(t),\delta_{2}(t),...,\delta _{n}(t)\}.\end{equation*}

Then,

\begin{equation*}\left\vert e_{i}(t)\right\vert \leq \delta (t)\left\vert x_{i}(t) \right\vert ,\left\Vert e(\cdot )\right\Vert ^{2}\leq \delta^{2}(\cdot )\sum_{i=1}^{n}x_{i}^{2}\left( \cdot \right) =\delta ^{2}(\cdot)\left\Vert x\left( \cdot \right) \right\Vert ^{2}\end{equation*}

Therefore, $\left\Vert e(\cdot )\right\Vert \leq \delta \left( \cdot \right)\left\Vert x\left( \cdot \right) \right\Vert .$ In a similar way,

(10) \begin{equation}u_{\bar{M}b}(t) =\frac{(1+\bar{\delta}(t))^{\overset{-}{M}+1}}{(1-\bar{\delta}(t))^{\bar{M}}}\bar{v}_{0},\text{with }\left\Vert \bar{e}(\cdot )\right\Vert \leq \bar{\delta}\left( \cdot \right) \left\Vert u\left(\cdot \right) \right\Vert \end{equation}

in case no saturation happens to $\bar{q}(u(t)).$

Consider the scenario where the state quantizer is not saturated, and suppose u(t) also falls into the non-saturation area of corresponding quantizer with an executable quantization result, then reminding ourselves of (3) and (8) gives us $\forall t_{i}<t\leq t_{i+1}$

(11) \begin{align}\frac{d}{dt}x^{T}(t) Qx(t) & = x^{T}(t) \left( A-BK\right) ^{T}Qx(t) +\left(-e^{T}(t) \right) K^{T}B^{T}Qx(t) \notag \\[5pt]& \quad +\bar{e}^{T}(t) B^{T}Qx(t) +x^{T}(t)Q\left( A-BK\right) x(t) +x^{T}(t) BK\left( -e(t) \right) +x^{T}(t) QB\bar{e}(t) \notag \\[5pt]&\leq -\left\Vert x(t) \right\Vert ^{2}D+2\delta (t) \left\Vert QBK\right\Vert \left\Vert x(t) \right\Vert ^{2} +2\bar{\delta}(t) \left\Vert QB\right\Vert \left\Vert x(t) \right\Vert \left\Vert K\right\Vert \left( \left\Vert x(t) \right\Vert\right.\nonumber\\[5pt]& \quad \left.+\delta (t) \left\Vert x(t)\right\Vert \right) \notag \\[5pt]&\leq \left\Vert x(t) \right\Vert ^{2}(-\lambda _{\min }\left(D\right) +2\delta (t) \left\Vert QBK\right\Vert +2\bar{\delta}(t) \left( 1+\delta (t) \right)\left\Vert QB\right\Vert \left\Vert K\right\Vert) \end{align}

where $t_{i}$ is time sequence at which $\delta (t) $ and $\bar{\delta}(t) $ are renewed, $t_{0}$ is the time instant when the system gets into the non-saturated area, $\tau _{i}$ is the sequence of dwell time of $\delta (t) $ and $\bar{\delta}(t) $ , that is $\tau _{i}=t_{i+1}-t_{i}.$

It can be seen from (11) that in order to achieve Lyapunov stability, $\delta (t)$ and $\bar{\delta}(t)$ have to be chosen small enough such that

(12) \begin{equation}2\delta (t) \left\Vert QBK\right\Vert +2\bar{\delta}(t) (1+\delta (t) )\left\Vert QB\right\Vert \left\Vert K\right\Vert <\lambda _{\min }\left( D\right) \end{equation}

For all t, let

(13) \begin{equation}2\delta (t) \left\Vert QBK\right\Vert +2\bar{\delta}(t) \left( 1+\delta (t) \right) \left\Vert QB\right\Vert \left\Vert K\right\Vert =\sigma _{ns}(t) \lambda _{\min }\left(D\right) \end{equation}

then obviously one has to have $0<\sigma _{ns}(t)<1$ .

However, it should be noticed that due to characteristics of the actuation component, an executable $\bar{q}\left( u(t) \right) $ cannot be arbitrarily large. That means, the quantizer for $u(t) $ is saturated, if $\left\vert -Kq(x(t))\right\vert \geq u_{\max }$ with $u_{\max}>0$ being the threshold enforced on u(t) due to physical or practical limitations. Set $\bar{q}(u(t) )=\bar{q}(-Kq\left( x(t) \right) )=sgn(-Kq(x(t)))u_{\max }=-\gamma (t) Kq\left(x(t) \right) $ with $0<\gamma (t) =\frac{u_{\max }}{\left\vert -Kq\left( x(t) \right) \right\vert }<1.$ Then although the state quantizer is not saturated, instead of (8) the corresponding closed-loop behavior is as follows

(14) \begin{align}\dot{x}(t) & = Ax(t) -\gamma (t)BKq\left( x(t) \right) \notag \\[5pt]&= \left( A-BK\right) x(t) +\left( 1-\gamma (t)\right) BKx(t) -\gamma (t) BKe(t) \end{align}

As a result, the differentiation of $x^{T}(t) Qx(t) $ is as follows

(15) \begin{align}\frac{d}{dt}x^{T}(t) Qx(t) & = x^{T}(t) \left( A-BK\right) ^{T}Qx(t) +\left(1-\gamma (t) \right) x^{T}(t) K^{T}B^{T}Qx(t) \notag \\&\quad -\gamma (t) e^{T}(t) K^{T}B^{T}Qx(t)+x^{T}(t) Q\left( A-BK\right) x(t) \notag \\&\quad +\left( 1-\gamma (t) \right) x^{T}(t) QBKx(t) -\gamma (t) x^{T}(t) QBKe(t)\notag \\&\leq -D\left\Vert x(t) \right\Vert ^{2}+2\left( 1-\gamma(t) \right) \left\Vert QBK\right\Vert \left\Vert x(t) \right\Vert ^{2} +2\gamma (t) \delta (t) \left\Vert QBK\right\Vert \left\Vert x(t) \right\Vert ^{2} \notag \\& \quad = \left\Vert x(t) \right\Vert ^{2}(-\lambda _{\min }\left(D\right) +(2\left( 1-\gamma (t) \right) +2\gamma (t) \delta (t) )\left\Vert QBK\right\Vert) \end{align}

It can be seen from (15) that in order to achieve Lyapunov stability, $\delta (t)$ has to chosen small enough such that

(16) \begin{equation}1-\gamma (t) +\gamma (t) \delta (t) <\frac{\lambda _{\min }\left( D\right) }{2\left\Vert QBK\right\Vert }\end{equation}

For all t, let

(17) \begin{equation}\sigma _{s}(t) =\frac{2\left\Vert QBK\right\Vert (1-\gamma(t) +\gamma (t) \delta (t) )}{\lambda_{\min }\left( D\right) }, \end{equation}

then obviously one has to have $0<\sigma _{s}(t)<1$ . Also, when $\left\vert\bar{q}\left( u(t) \right) \right\vert \geq u_{\max }$ , we leave $\bar{\delta}(t)$ static until the quantization of u(t) is not saturated. This will happen in finite time as x(t) enters “zoom-in” stage later on and $\left\vert -Kx(t)\right\vert $ shrinks accordingly.

Proof. Firstly we introduce the “zoom-out” stage of $x(t) $ . During this stage, $\left\Vert x(t)\right\Vert >x_{Mb}(t)\sqrt{\frac{\lambda _{\min}(P_{j})}{\lambda _{\max }(P_{j})}}$ , which corresponds to Case 1 in Theorem 1.

The quantizer of $u(t) $ is saturated if $\left\vert -Kq(x(t) )\right\vert \geq u_{\max }.$ Let $\left\vert -\varphi (t) Kq(x(t))\right\vert =u_{\max },$ $0<\varphi (t)<1.$

\begin{align*}\dot{x}(t) & = Ax(t) +Bu_{\max }sgn(-Kq(x(t) )) \notag \\[3pt]& = Ax(t) +B\left\vert -\varphi (t)Kq(x(t))\right\vert sgn(-Kq(x(t) )) \notag \\[3pt]& = Ax(t) +\varphi (t) B(-Kq(x(t))) \notag \\[3pt]& = \left( A-\varphi (t) (I_{n\times n}+\Delta (t))BK\right)x(t)\end{align*}

Define

\begin{equation*}\phi (t)=e^{\left\Vert A-\varphi (t) (I_{n\times n}+\Delta(t))BK\right\Vert t}.\end{equation*}

It is worth noticing that as the ratio between $u_{\max }$ and $\left\vert-Kq(x(t))\right\vert $ , $\varphi (t)$ is calculated every time $\left\vert-Kq(x(t))\right\vert $ is updated.

So during the “zoom-out” stage of x(t), if $\left\vert -Kq(x(t) )\right\vert \geq u_{\max }$ for some t, the quantization of x(t) has to “zoom-out” faster than the rate of growth of x(t) in order to catch the system state in finite time. That means in such cases, it is required that $\delta (t)$ is increased at a speed fast enough to dominate the rate of growth of $e^{m\left\Vert A-\varphi (t)(I_{n\times n}+\Delta (t))BK\right\Vert t}$ , where $m>1$ . Obviously, the greater m is, the earlier $x_{Mb}$ catches up with x(t). But an extremely large m may deteriorate the system performance by incurring much too fast dynamic.

If the quantization of $u(t) $ is not saturated or some t during the “zoom-out” stage of the state quantizer, one has $\left\vert -Kq(x(t) )\right\vert <u_{\max }.$ Let $\bar{e}(t) =\bar{\delta}^{\prime }(t) u(t) =-\bar{\delta}^{\prime }(t) Kq(x(t) ),$ $\left\vert \bar{\delta}^{\prime }(t) \right\vert <\bar{\delta}(t).$

\begin{align*}\dot{x}(t) & = Ax(t) +B\bar{q}(-Kq(x(t) ))\notag \\[4pt]& = Ax(t) -BKq(x(t) )+B\bar{e}\left( u(t)\right) \notag \\[4pt]& = Ax(t) -BKq(x(t) )-\bar{\delta}^{\prime }(t) BKq(x(t) ) \notag \\[4pt]& = \left( A-\left( 1+\bar{\delta}^{\prime }(t) \right)(I_{n\times n}+\Delta (t))BK\right) x(t)\end{align*}

Define

\begin{equation*}\phi (t)=e^{\left\Vert A-\left( 1+\bar{\delta}^{\prime }(t)\right) (I_{n\times n}+\Delta (t))BK\right\Vert t}.\end{equation*}

So during the “zoom-out” stage of x(t), if $\left\vert -Kq(x(t) )\right\vert \leq u_{\max }$ for some t, the quantization of x(t) has to “zoom-out” faster than the rate of growth of x(t) in order to catch the system state in finite time. That means in such cases, it is required that $\delta (t)$ is increased at a speed fast enough to dominate the rate of growth of $e^{m\left\Vert A-\left( 1+\bar{\delta}^{\prime }(t) \right) (I_{n\times n}+\Delta (t))BK\right\Vert t}$ , where $m>1$ . The greater m is, the earlier $x_{Mb}$ catches up with x(t). But an extremely large m may deteriorate the system performance by incurring much too fast dynamic.

Then there exists a positive integer $i_{0}$ such that

(20) \begin{equation}\left\Vert x(t_{i_{0}})\right\Vert \leq x_{Mb}(t_{i_{0}})\sqrt{\frac{\lambda_{\min }(Q)}{\lambda _{\max }(Q)}}\end{equation}

Define

\begin{equation*}i_{0}:=\min \left\{r\geq 0:\left\Vert x(t_{r})\right\Vert \leq x_{Mb}(t_{r})\sqrt{\frac{\lambda _{\min }(Q)}{\lambda _{\max }(Q)}},\right\}\end{equation*}

Then from (20), one has

\begin{eqnarray*}x^{T}(t_{i_{0}})x(t_{i_{0}}) &\leq &x_{Mb}^{2}(t_{i_{0}})\frac{\lambda_{\min }(Q)}{\lambda _{\max }(Q)} \\x^{T}(t_{i_{0}})x(t_{i_{0}})\lambda _{\max }(Q) &\leq &x_{Mb}^{2}(t_{i_{0}})\lambda _{\min }(Q).\end{eqnarray*}

Considering the fact that $x^{T}(t_{i_{0}})Qx(t_{i_{0}})\leq x^{T}(t_{i_{0}})x(t_{i_{0}})\lambda _{\max }(Q),$ one has

\begin{equation*}x^{T}(t_{i_{0}})Qx(t_{i_{0}})\leq x_{Mb}^{2}(t_{i_{0}})\lambda _{\min }(Q)\end{equation*}

Next, we come to the “zoom-in” stage, which corresponds to Case 2 in Theorem 1. Considering (8) or (14) with (12) we have

(21) \begin{align}x^{T}(t_{i_{0}}+1)Qx(t_{i_{0}}+1) & = x^{T}(t_{i_{0}})Qx(t_{i_{0}})+\int_{t_{i_{0}}}^{t_{i_{0}}+\tau _{i_{0}}}\frac{d}{dt}x^{T}(t)Qx(t)dt \notag \\[3pt]&\leq x^{T}(t_{i_{0}})Qx(t_{i_{0}}) +\int_{t_{i_{0}}}^{t_{i_{0}}+\tau _{i_{0}}}-(1-\sigma (t_{i_{0}}))\lambda_{\min }(D)x^{T}(t_{i_{0}})x(t_{i_{0}})dt \notag \\[3pt]&\leq x_{Mb}^{2}(t_{i_{0}})\lambda _{\min }(Q) -\tau _{i_{0}}(1-\sigma (t_{i_{0}}))\lambda _{\min}(D)x^{T}(t_{i_{0}})x(t_{i_{0}}) \end{align}

where $\sigma (t_{i_{0}})$ is chosen either as $\sigma _{ns}(t_{i_{0}})$ in (13) or $\sigma _{s}(t_{i_{0}})$ in (17) depending on whether the actuation unit is saturated. As $x^{T}(t_{i})x(t_{i})=\eta(t_{i})x_{Mb}^{2}(t_{i})$ with $0\leq \eta (t_{i})< 1,\forall i$ , we have

\begin{align*}x^{T}(t_{i_{0}}+1)Qx(t_{i_{0}}+1) \leq &x_{Mb}^{2}(t_{i_{0}})( \lambda _{\min }(Q) -\tau _{i_{0}}\eta (t_{i_{0}})(1-\sigma (t_{i_{0}}))\lambda _{\min }(D))\end{align*}

It is worth noticing that $\eta(t_{i})$ is the ratio between $x^{T}(t_{i})x(t_{i})$ and $x_{Mb}^{2}$ . Its value should be calculated every time $x(t_{i})$ is updated. Then remind ourselves of (21) to derive

\begin{equation}x^{T}(t_{i_{0}+1})Qx(t_{i_{0}+1})\leq x_{Mb}^{2}(t_{i_{0}+1})\lambda_{\min }(Q). \notag \\\end{equation}

Remark 5. Based on Theorem 1, $\tau_{i}$ can be chosen as any value as long as $0< \tau _{i}< \frac{\lambda _{\min }(Q)}{\eta (t_{i})(1-\sigma (t_{i}))\lambda_{\min }(D)}$ is satisfied. But in real applications, trials may be carried out such that a $\tau_{i}$ which leads to better system performance is chosen.

By repeating the procedure above, we have $x^{T}(t_{i})Qx(t_{i})\leq x_{Mb}^{2}(t_{i})\lambda _{\min }(Q)$ . This also guarantees that the system will not go back to “zoom-out” stage.

Due to (19), we have

\begin{align*}\left\Vert u\left( t_{i_{0}}\right) \right\Vert &\leq \left\Vert k_{1}q\left( x_{1}\left( t_{i_{0}}\right) \right) \right\Vert +\left\Vert k_{2}q\left( x_{2}\left( t_{i_{0}}\right) \right) \right\Vert +\cdots +\left\Vert k_{n}q\left( x_{n}\left( t_{i_{0}}\right) \right)\right\Vert \\[3pt]&\leq \left\Vert k_{1}\right\Vert x_{Mb}\left( t_{i_{0}}\right) +\left\Vert k_{2}\right\Vert x_{Mb}\left( t_{i_{0}}\right) +\cdots +\left\Vert k_{n}\right\Vert x_{Mb}\left( t_{i_{0}}\right) \\[3pt]& = u_{\bar{M}b}\left( t_{i_{0}}\right)\end{align*}

Similarly, we have $\left\Vert u\left( t_{i}\right) \right\Vert \leq u_{\bar{M}b}\left( t_{i}\right) ,\forall i\geq i_{0}.$

It is noticed that $f(0)=-\left( \delta (t_{i})+1\right) <0$ and $f(1)=2\left( p(t_{i})-1\right) <0$ . As f(h(k)) is continuously differentiable, the differential of $f(h(t_{i}))$ is shown as

\begin{equation*}f^{\prime }(h(t_{i}))=2(M+1)p(t_{i})h^{M}(t_{i})+\left( \delta(t_{i})-1\right) .\end{equation*}

Let $f^{\prime }(\bar{h}(t_{i}))=0,$ we have

\begin{align*}\bar{h}(t_{i}) & =\left\vert \sqrt[M]{\frac{1-\delta (t_{i})}{2(M+1)p(t_{i})}}\right\vert .\\[6pt]f(\bar{h}(t_{i})) & = \frac{1-\delta (t_{i})}{M+1}\left\vert \sqrt[M]{\frac{1-\delta (t_{i})}{2(M+1)p(t_{i})}}\right\vert -\left( 1-\delta (t_{i})\right) \left\vert \sqrt[M]{\frac{1-\delta (t_{i})}{2(M+1)p(t_{i})}}\right\vert -\left( \delta (t_{i})+1\right) \\[6pt]& = -\frac{M}{M+1}\left( 1-\delta (t_{i})\right) \left\vert \sqrt[M]{\frac{1-\delta (t_{i})}{2(M+1)p(t_{i})}}\right\vert -\left( \delta (t_{i})+1\right)\\[6pt]& < 0\end{align*}

Then, we have the following properties of $f(h(t_{i}))$ :

1. 1. $f(h(t_{i}))$ is monotonically decreasing between 0 and $\bar{h}(t_{i})$ and monotonically increasing between $\bar{h}(t_{i})$ and $\infty $ .

2. 2. $f(\infty )>0$ .

3. 3. If $\bar{h}(t_{i})>1,f(h(t_{i}))$ has single root between $\bar{h}(t_{i})$ and $\infty .$ If $\bar{h}(t_{i})\leq 1,f(h(t_{i}))$ has single root between 1 and $\infty .$

4. 4. Define the root of $f(h(t_{i}))$ as $h^{\ast}(t_{i}),$ such that $f(h^{\ast }(t_{i}))=0$ and $h^{\ast }(t_{i})>1,$ which implies $\delta (t_{i}+1)<\delta(t_{i}).$

5. 5. As i increases, $\delta (t_{i})$ and $h(t_{i})$ decrease, it can then be derived that $\lim\limits_{i\rightarrow \infty }h(t_{i})=1$ and $\lim\limits_{i\rightarrow \infty }\delta (t_{i})=0.$

As a result of (5), $x_{Mb}(t_{i})=\frac{(1+\delta (t_{i}))^{M}}{(1-\delta(t_{i}))^{M+1}}v_{0}\rightarrow v_{0}$ as $i\rightarrow \infty $ and

(22) \begin{eqnarray}\lim\limits_{i\rightarrow \infty }x^{T}(t_{i})Qx(t_{i}) &\leq &\lim\limits_{i\rightarrow \infty }x_{Mb}^{2}(t_{i})\lambda _{\min }(Q)\notag \\&=&v_{0}^{2}\lambda _{\min }(Q) \end{eqnarray}

Obviously, (22) leads to $\left\Vert x(t_{i})\right\Vert \leq v_{0}$ as $i\rightarrow \infty $ , which further guarantees $\left\Vert x(t)\right\Vert \leq v_{0}$ as $t\rightarrow \infty.$ Then, stability of system is guaranteed as $v_{0}$ is finite and can be chosen arbitrarily small.

4. Numerical results

Example 1. Consider a third-order continuous-time linear system

\begin{equation*}\dot{x}(t) =\left[\begin{array}{c@{\quad}c@{\quad}c}0 & 1 & 0 \\[3pt]0 & -0.2 & 0.5 \\[3pt]0 & 0 & 0\end{array}\right] x(t) +\left[\begin{array}{c}0 \\[3pt]0 \\[3pt]5\end{array}\right] u(t)\end{equation*}

with a quantized feedback control law designed as $K=\left[\begin{array}{c@{\quad}c@{\quad}c}0.0024 & 0.0120 & 0.0800\end{array}\right] .$ As a result, $eig(A-BK)=\left[\begin{array}{c@{\quad}c@{\quad}c}-0.1 & -0.2 & -0.3\end{array}\right] .$

Without loss of generality, set $M=\bar{M}=10,v_{0}=0.1,\bar{v}_{0}=0.005,u_{\max }=0.05$ for $x_{1},x_{2},x_{3}.$ $x\left( 0\right) =\left[\begin{array}{c@{\quad}c@{\quad}c}1.6 & 0.9 & 1.8\end{array}\right] ^{T}.$

The quantization of $x_{1},x_{3}$ encounters saturation at $t=0$ , then during the “zoom-out” stage, choose $m=3,$ which results in $i_{01}=13$ and $i_{03}=1.$ $\delta \left( t_{i0}\right) $ and $\bar{\delta}\left( t_{i0}\right) $ are both chosen as 0.13 which satisfy Eq. (12) or (16). Then during the “zoom-in” stage, it takes the states 17 time intervals to enter the neighborhood of the origin, that is $\left\Vert x(t)\right\Vert \leq v_{0}=0.1.$ And, it is observed the states stay within the neighborhood ever after, that is $\left\Vert x(t)\right\Vert \leq v_{0},\forall t\geq t_{30}.$

The state variables $x_{1},x_{2},x_{3}$ are shown as Figs. 3 and 4 gives a 3D view of system states. Figure 5 illustrates how $\left\Vert x(t)\right\Vert $ converges into a neighborhood of the origin the width of which is $v_{0}$ .

Figure 3. Third-order system states using dynamic logarithmic quantizer

Figure 4. 3D view of system states using dynamic logarithmic quantizer

Figure 5. Convergence of system state into a neighborhood of the origin

The control signals u and $\bar{q}(u)$ are shown in Fig. 6. We could see that the control signal goes through four stages, that is saturated, unsaturated, saturated, then enters unsaturated stage and remains in this stage ever after.

Figure 6. Control signal using dynamic logarithmic quantizer

It is observed that under quantized feedback control with proposed dynamic logarithmic quantizer, the system state converge to a spherical neighborhood of the origin whose radius is $v_{0}$ . Figures 7 and 8 show the system states and control signal of the same system with linear state quantization algorithm as in ref. [Reference Brockett and Liberzon13]. Compared with the dynamic logarithmic quantization algorithm proposed in this paper, in order to maintain the same saturation value and initial value of sensitivity, the order of dynamic linear quantizer must satisfies $M^{\prime }\geq 13$ which is larger than the order of dynamic logarithmic quantizer $M=10.$ It is also noticed that the algorithm in ref. [Reference Brockett and Liberzon13] did not consider the quantization of the control signal, which is equivalent to assuming the quantization of the control signal as ideal. However, with the algorithm proposed in this paper, quantizers are applied to both state and control signals which is closer to situation in the actual NCSs. Although more quantization error is inevitably brought into the system, the overall system performance is not deteriorated in comparison to that in ref. [Reference Brockett and Liberzon13]. This is the benefit brought by suitable controller-quantizer co-design, which is another advantage of the proposed algorithm over that in ref. [Reference Brockett and Liberzon13].

Figure 7. System states using dynamic linear quantizer

Figure 8. Control signal using dynamic linear quantizer

Example 2. Consider the uninterruptible power system in ref. [Reference Wang, Wang and Wang20]

\begin{equation*}\dot{x}(t) =\left[\begin{array}{c@{\quad}c@{\quad}c}0.9226 & 0.6330 & 0 \\[3pt]1 & 0 & 0 \\[3pt]0 & 1 & 0\end{array}\right] x(t) +\left[\begin{array}{c}0.5 \\[3pt]0 \\[3pt]0.2\end{array}\right] u(t)\end{equation*}

and choose a quantized feedback controller gain $K=\left[\begin{array}{c@{\quad}c@{\quad}c}10.0150 & 24.2677 & 9.5755\end{array}\right] .$ As a result, $eig(A-BK)=\left[\begin{array}{c@{\quad}c@{\quad}c}-1 & -2 & -3\end{array}\right] .$ Set $M=\bar{M}=10,v_{0}=0.1,\bar{v}_{0}=0.01,u_{\max }=50$ for $x_{1},x_{2},x_{3}.$ $x\left( 0\right) =\left[\begin{array}{c@{\quad}c@{\quad}c}-0.7 & 0.6 & 0.3\end{array}\right] ^{T}.$

During the “zoom-out” stage, choose $m=1.5,$ $\delta \left( t_{i0}\right)=0.1$ and $\bar{\delta}\left( t_{i0}\right)=0.1.$ Then during the “zoom-in” stage, it takes the states 484 time intervals to enter the neighborhood of the origin, that is $\left\Vert x(t)\right\Vert \leq v_{0}=0.1.$ And it is observed the states stay within the neighborhood ever after, that is $\left\Vert x(t)\right\Vert \leq v_{0},\forall t\geq t_{484}.$

The state variables $x_{1},x_{2},x_{3}$ are shown as Figs. 9 and 10 gives a 3D view of system states.

Figure 9. System states of the uninterruptible power system

Figure 10. 3D view of system states